Symmetry of the stress 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, pp. 77-78

Section 1.3.2.4. Symmetry of the stress 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

1.3.2.4. Symmetry of the stress tensor

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Let us now consider the equilibrium condition (1.3.2.2) relative to the resultant moment. After projection on the three axes, and using the Cartesian expression (1.1.3.4) of the vectorial products, we obtain $[\textstyle\int\int\limits_{S}\displaystyle \textstyle{1 \over 2}\displaystyle \varepsilon_{ijk} (x_{i} T_{lj} - x_{j} T_{li}) \hbox{d} \sigma_{l} + \textstyle\int\int\int\limits_{V}\displaystyle \left[\textstyle{1\over 2}\displaystyle \varepsilon_{ijk} \rho (x_{i} F_{j} - x_{j} F_{i}) + \Gamma_{k}\right] \hbox{d}\tau = 0. ]$ (including the inertial forces in the volume forces). $[\varepsilon_{ijk}]$ is the permutation tensor. Applying Green's theorem to the first integral and putting the two terms together gives $[\int\int\int\limits_{V}\displaystyle \left\{\textstyle{1\over 2}\displaystyle \varepsilon_{ijk} \left[{\partial \over \partial x_{l}} (x_{i} T_{lj} - x_{j} T_{li}) + \rho (x_{i} F_{j} - x_{j} F_{i})\right] + \Gamma_{k} \right\} \hbox{d} \tau = 0. ]$

In order that this relation applies to any volume V within the solid C, we must have $[\textstyle{1\over 2}\displaystyle \varepsilon_{ijk} \left[{\partial \over \partial x_{l}} (x_{i} T_{lj} - x_{j} T_{li})\right] + \Gamma_{k} = 0 ]$ or $[\textstyle{1\over 2}\displaystyle \varepsilon_{ijk} \left[x_{i} \left({\partial T_{lj}\over \partial x_{l}} + F_{j} \rho \right) - x_{j} \left({\partial T_{li}\over \partial x_{l}} + F_{i} \rho \right) + T_{ij} - T_{ji} \right] + \Gamma_{k} = 0. ]$

Taking into account the continuity condition (1.3.2.5), this equation reduces to $[\textstyle{1\over 2}\displaystyle \varepsilon_{ijk} \rho [T_{ij} - T_{ji}]+ \Gamma_{k} = 0. ]$

A volume couple can occur for instance in the case of a magnetic or an electric field acting on a body that locally possesses magnetic or electric moments. In general, apart from very rare cases, one can ignore these volume couples. One can then deduce that the stress tensor is symmetrical: $[T_{ij} - T_{ji} = 0.]$

This result can be recovered by applying the relation (1.3.2.2) to a small volume in the form of an elementary parallelepiped, thus illustrating the demonstration using Green's theorem but giving insight into the action of the constraints. Consider a rectangular parallelepiped, of sides $[2 \Delta x_{1}]$ , $[2 \Delta x_{2}]$ and $[2 \Delta x_{3}]$ , with centre P at the origin of an orthonormal system whose axes $[Px_{1}]$ , $[Px_{2}]$ and $[Px_{3}]$ are normal to the sides of the parallelepiped (Fig. 1.3.2.4). In order that the resultant moment with respect to a point be zero, it is necessary that the resultant moments with respect to three axes concurrent in this point are zero. Let us write for instance that the resultant moment with respect to the axis $[Px_{3}]$ is zero. We note that the constraints applied to the faces perpendicular to $[Px_{3}]$ do not give rise to a moment and neither do the components $[T_{11}]$ , $[T_{13}]$ , $[T_{22} ]$ and $[T_{23}]$ of the constraints applied to the faces normal to $[Px_{1}]$ and $[Px_{2}]$ (Fig. 1.3.2.4). The components $[T_{12}]$ and $[T_{21}]$ alone have a nonzero moment.

Figure 1.3.2.4 | top | pdf |

Symmetry of the stress tensor: the moments of the couples applied to a parallelepiped compensate each other.

For face 1, the constraint is $[T_{12} + (\partial T_{12}/ \partial x_{1})\Delta x_{1} ]$ if $[T_{12}]$ is the magnitude of the constraint at P. The force applied at face 1 is $[\left[T_{12} + {\partial T_{12}\over \partial x_{1}}\Delta x_{1}\right]4 \Delta x_{2} \Delta x_{3} ]$ and its moment is $[\left[T_{12} + {\partial T_{12}\over \partial x_{1}}\Delta x_{1}\right]4 \Delta x_{2} \Delta x_{3} \Delta x_{1}. ]$

Similarly, the moments of the force on the other faces are $[\displaylines{\hbox{Face 1}':\quad-\left[T_{12} + {\partial T_{12}\over \partial x_{1}}(-\Delta x_{1})\right]4 \Delta x_{2} \Delta x_{3} (-\Delta x_{1})\semi\hfill\cr \hbox{Face 2}\phantom{'}:\quad\phantom{-}\left[T_{21} + {\partial T_{21}\over \partial x_{2}}\Delta x_{2})\right]4 \Delta x_{1} \Delta x_{3} \Delta x_{2};\hfill\cr \hbox{Face 2}':\quad-\left[T_{21} + {\partial T_{21}\over \partial x_{2}}(-\Delta x_{2})\right]4 \Delta x_{1} \Delta x_{3} (-\Delta x_{2}).\hfill} ]$

Noting further that the moments applied to the faces 1 and 1′ are of the same sense, and that those applied to faces 2 and 2′ are of the opposite sense, we can state that the resultant moment is $[[T_{12} - T_{21}]8 \Delta x_{1} \Delta x_{2} \Delta x_{3} = [T_{12} - T_{21}]\Delta \tau, ]$ where $[8 \Delta x_{1} \Delta x_{2} \Delta x_{3} = \Delta \tau ]$ is the volume of the small parallelepiped. The resultant moment per unit volume, taking into account the couples in volume, is therefore $[T_{12} - T_{21} + \Gamma_{3}.]$ It must equal zero and the relation given above is thus recovered.

References

International Tables for Crystallography (2006). Vol. D. ch. 1.3, pp. 77-78