Geometrical interpretation of the coefficients of the strain tensor

Authier, A.; Zarembowitch, A.

doi:10.1107/97809553602060000630

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

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International Tables for Crystallography (2006). Vol. D. ch. 1.3, p. 75

Section 1.3.1.3.2. Geometrical interpretation of the coefficients of the strain tensor

A. Authier^a ^* and A. Zarembowitch^b

^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

1.3.1.3.2. Geometrical interpretation of the coefficients of the strain tensor

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Let us consider an orthonormal system of axes with centre P. We remove nothing from the generality of the following by limiting ourselves to a planar problem and assuming that point P′ to which P goes in the deformation lies in the plane $[x_{1}Px_{2} ]$ (Fig. 1.3.1.4). Let us consider two neighbouring points, Q and R, lying on axes $[Px_{1}]$ and $[Px_{2}]$ , respectively ( $[PQ = \hbox{d} x_{1} ]$ , $[PR = \hbox{d} x_{2}]$ ). In the deformation, they go to points Q′ and R′ defined by $[\eqalign{{\bf QQ'}&: \left\{\matrix {\hbox{d}x'_{1} = \hbox{d}x_{1} + \left(\partial u_{1} / \partial x_{1}\right)\hbox{d}x_{1}\hfill\cr \hbox{d}x'_{2} = \left(\partial u_{2} / \partial x_{1}\right)\hbox{d}x_{1}\hfill\cr \hbox{d}x'_{3} = 0\hfill\cr}\right. \cr {\bf RR'}&: \left\{\matrix {\hbox{d}x'_{1} = \left(\partial u_{1} / \partial x_{2}\right)\hbox{d}x_{2}\hfill\cr \hbox{d}x'_{2} = \hbox{d}x_{2} + \left(\partial u_{2} / \partial x_{2}\right)\hbox{d}x_{2}\hfill\cr \hbox{d}x'_{3} = 0.\hfill\cr}\right.} ]$

Figure 1.3.1.4 | top | pdf |

Geometrical interpretation of the components of the strain tensor. $[Ox_{1}]$ , $[Ox_{2}]$ , $[Ox_{3} ]$ : axes before deformation; $[Ox_{1}']$ , $[Ox_{2}']$ , $[Ox_{3}']$ : axes after deformation.

As the coefficients $[\partial u_{i}/\partial x_{j}]$ are small, the lengths of P′Q′ and P′R′ are hardly different from PQ and PR, respectively, and the elongations in the directions $[Px_{1}]$ and $[Px_{2}]$ are $[\eqalign{{P' Q' - PQ \over PQ} &= { \hbox{d}x'_{1} - \hbox{d}x_{1} \over \hbox{d}x_{1}} = {\partial u_{1} \over \partial x_{1}} = S_{1}\cr {P' R' - PR \over PR} &= { \hbox{d}x'_{2} - \hbox{d}x_{2} \over \hbox{d}x_{2}} = {\partial u_{2} \over \partial x_{2}} = S_{2}.\cr} ]$

The components $[S_{1}]$ , $[S_{2}]$ , $[S_{3}]$ of the principal diagonal of the Voigt matrix can then be interpreted as the elongations in the three directions $[Px_{1}]$ , $[Px_{2}]$ and $[Px_{3}]$ . The angles α and β between PQ and P′Q′, and PR and P′R′, respectively, are given in the same way by $[\alpha = \hbox{d}x'_{2}/\hbox{d}x_{1} = \partial u_{2}/\partial x_{1}\semi\ \ \beta =\hbox{d}x'_{1}/dx_{2} = \partial u_{1}/\partial x_{2}. ]$ One sees that the coefficient $[S_{6}]$ of Voigt's matrix is therefore $[S_{6}= {\partial u_{2} \over \partial x_{1}} + {\partial u_{1} \over \partial x_{2}} = \alpha + \beta. ]$ The angle $[\alpha + \beta]$ is equal to the difference between angles $[{\bf PQ}\wedge {\bf PR}]$ before deformation and $[{\bf P'' Q''}\wedge {\bf P' R'}]$ after deformation. The nondiagonal terms of the Voigt matrix therefore represent the shears in the planes parallel to $[Px_{1}]$ , $[Px_{2}]$ and $[Px_{3}]$ , respectively.

To summarize, if one considers a small cube before deformation, it becomes after deformation an arbitrary parallelepiped; the relative elongations of the three sides are given by the diagonal terms of the strain tensor and the variation of the angles by its nondiagonal terms.

The cubic dilatation (1.3.1.5) is $[\Delta (B) - 1 = S_{1}+ S_{2}+ S_{3}]$ (taking into account the fact that the coefficients $[S_{ij}]$ are small).

References

International Tables for Crystallography (2006). Vol. D. ch. 1.3, p. 75