International
Tables for
Crystallography
Volume D
Physical properties of crystals
Edited by A. Authier

© International Union of Crystallography 2006
International Tables for Crystallography (2006). Vol. D. ch. 1.3, p. 95

Section 1.3.7.3.1. Isotropic media

A. Authiera* and A. Zarembowitchb

a Institut de Minéralogie et de la Physique des Milieux Condensés, Bâtiment 7, 140 rue de Lourmel, 75015 Paris, France, and bLaboratoire de Physique des Milieux Condensés, Université P. et M. Curie, 75252 Paris CEDEX 05, France
Correspondence e-mail:  aauthier@wanadoo.fr

1.3.7.3.1. Isotropic media

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In this case, the strain-energy density becomes [\eqalignno{\Phi &= \textstyle{1\over 2}\displaystyle (\lambda+2\mu)(S_{11})^2 + 2 \mu(S_{12}S_{21} + S_{13}S_{31}) + \textstyle{1\over 3}\displaystyle (l+2m)(S_{11}){}^3&\cr &\quad + 2m S_{11}\left(S_{12}S_{21} + S_{13}S_{31}\right). &(1.3.7.6)\cr} ]Differentiating (1.3.7.6)[link] with respect to the strains, we get [\openup4.5pt\eqalign{{\partial \Phi \over \partial S_{11}} & = (\lambda+2\mu)S_{11} + (l+2m)(S_{11}){}^2 + 2m \left(S_{12}S_{21} + S_{13}S_{31}\right)\cr {\partial \Phi \over \partial S_{12}} & =2 \mu S_{21} + 2m S_{11}S_{21}\cr {\partial \Phi \over \partial S_{13}} & = 2\mu S_{31} + 2m S_{11}S_{31}\cr {\partial \Phi \over \partial S_{21}} & = 2 \mu S_{12} + 2m S_{11}S_{12}\cr {\partial \Phi \over \partial S_{31}} & =2\mu S_{13} + 2m S_{11}S_{13}.\cr} ] All the other [{\partial \Phi / \partial S_{ij}} = 0].

From (1.3.7.5)[link], we derive the stress components: [\openup6pt\eqalign{T_{11} &= \alpha_{1k}{\partial \Phi \over \partial S_{1k}}\semi \ \ T_{12} = \alpha_{2k}{\partial \Phi \over \partial S_{1k}}\semi \ \ T_{13} = \alpha_{3k}{\partial \Phi \over \partial S_{1k}}\semi \cr T_{21} &= \alpha_{1k}{\partial \Phi \over \partial S_{2k}}\semi \ \ T_{22} = \alpha_{2k}{\partial \Phi \over \partial S_{2k}}\semi \ \ T_{23} = \alpha_{3k}{\partial \Phi \over \partial S_{2k}}\semi\cr T_{31} &= \alpha_{1k}{\partial \Phi \over \partial S_{3k}}\semi \ \ T_{32} = \alpha_{2k}{\partial \Phi \over \partial S_{3k}}\semi \ \ T_{33} = \alpha_{3k}{\partial \Phi \over \partial S_{3k}}.\cr} ]Note that this tensor is not symmetric.

For the particular problem discussed here, the three components of the equation of motion are [\eqalign{\rho u_1'' &= {\rm d}T_{11}/{\rm d}X_1,\cr \rho u_2'' &= {\rm d}T_{21}/{\rm d}X_1,\cr \rho u_3'' &= {\rm d}T_{31}/{\rm d}X_1.\cr} ]

If we retain only terms up to the quadratic order in the displacement gradients, we obtain the following equations of motion: [\eqalign{\rho u_{1}'' &= (\lambda + 2\mu){\partial^{2} u_{1} \over \partial X_{1}^{2}} + [3(\lambda + 2\mu) + 2(l + 2m)] {\partial u_{1} \over \partial X_{1}}{\partial^{2} u_{1} \over \partial X_{1}^{2}}\cr &\quad + (\lambda + 2\mu + m) \left[{\partial u_{2} \over \partial X_{1}}{\partial^{2} u_{2} \over \partial X_{1}^{2}} + {\partial u_{3} \over \partial X_{1}}{\partial^{2} u_{3} \over \partial X_{1}^{2}}\right]\cr \rho u_{2}'' &= \mu{\partial^{2} u_{2} \over \partial X_{1}^{2}} + (\lambda + 2\mu + m) \left[{\partial u_{1} \over \partial X_{1}}{\partial^{2} u_{2} \over \partial X_{1}^{2}} + {\partial u_{2} \over \partial X_{1}}{\partial^{2} u_{1} \over \partial X_{1}^{2}}\right]\cr \rho u_{3}'' &= \mu{\partial^{2} u_{3} \over \partial X_{1}^{2}} + (\lambda + 2\mu + m)\left[{\partial u_{1} \over \partial X_{1}}{\partial^{2} u_{3} \over \partial X_{1}^{2}} + {\partial u_{3} \over \partial X_{1}}{\partial^{2} u_{1} \over \partial X_{1}^{2}}\right].\cr} \eqno(1.3.7.7) ]








































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