Strain and stress tensor

Question

I have problem by definition of strain and stress. From Gockenbach's book that our reference for FEM, we have $$\epsilon=\frac{\nabla u+ \nabla u^T}{2},$$

that $u$ is vector displacement, and $\nabla u$ is the Jacobian of $u$. So we have $\epsilon$ is symmetric and also $\sigma$, that is

$$2\mu \epsilon+\lambda tr(\epsilon)I$$

My problem is that I see everywhere this statement: if $\epsilon$ is symmetric or if $\sigma$ is symmetric we have... why? I can not see the case that they not be symmetric,

This is more of a physics question (see Tharsis' answer below). I am migrating accordingly. — Willie Wong, May 26 '13 at 15:27
Symmetry of $\epsilon$ comes from its definition, symmetry of $\sigma$ comes from the absence of internal torque. — Learning is a mess, May 26 '13 at 15:53
Thanks, but what are they defenition when they are not symmetric? sorry , I did not see the icon add comment so I have to right it here — , May 26 '13 at 18:48

score 8 · Answer 1 · answered May 26 '13 at 15:17

Indeed, both the strain tensor

$$\epsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right) \tag{1}$$

and the stress tensor

$$\sigma_{ij}=2\mu\epsilon_{ij}+\lambda\epsilon_{kk}\delta_{ij} \tag{2}$$

are symmetric by definition.

However, bear in mind that these definitions are not always valid; $(1)$ assumes that the deformations are infinitesimal and $(2)$ assumes that the solid is elastic (obeys Hooke's law) and isotropic.

score 4 · Answer 2 · answered May 08 '15 at 12:51

4

It is true they are in genereal both symmetric. The symmetry of the stress tensor is not only a matter of definition, it is a general property consecuence of angular momentum conservation. On the other hand, the strain tensor is found to be symmetric as a consecuence of its definition as a measure of $ds^2-ds_0^2$ and it is so not only in its linear aproximation but in the general non linear case

answered May 08 '15 at 12:51

facenian

416

rdt2 correctly observed that the derivation of the symmetry of the stress tensor is not valid when distributed moments are present. – facenian Jul 09 '22 at 23:26

score 4 · Answer 3 · answered Sep 04 '15 at 14:37

@Learning_is_a_mess notes correctly that stress is only symmetric in the absence of internal torque. This is so when the only moments arising are due to forces at a finite distance and so tend to zero as we zoom in on the body. This is not so if, for example, a magnetic material was placed in a strong magnetic field: the internal moment does not tend to zero as we zoom in. The resulting stress is non-symmetric. In such a case there is no point in defining a symmetric strain and you might as well work with the whole non-symmetric deformation gradient.

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Strain and stress tensor

3 Answers3