Strain and stress tensor 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,
 A: 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.
A: 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
A: @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.
Keywords for searching are couple stress and micropolar.
