# Strain-Displacement relationship symmetrization

In the context of infinitesimal elastic strain theory, one writes the relationship between displacement and strain as

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

My question is; is the full non-simmetrized displacement tensor a meaningful mechanical quantity? what is the physical interpretation of the anti-simmetric part of the displacement gradient, that is:

$$\omega_{ij}=\frac{1}{2}( \frac{\partial u_i}{\partial x_j} - \frac{\partial u_j}{\partial x_i} )$$

In what circumstances does this physical quantity plays a role in the structural analysis of the stresses in a material?

• If you wish to rotate your principal axis around, you will find it much easier to do with the real, physically meaningful tensor formalism. I've never liked the reduced engineering format - I think it is a relic from the dinosaur ages (much like secant and cosecant for that matter). – Jon Custer Oct 21 '14 at 17:36
• can you point a reference for this and elaborate a bit? – diffeomorphism Oct 21 '14 at 17:59
• Most engineering elasticity texts end up with $\sigma_{i} = c_{i,j}\epsilon_{j}$, using $i$ and $j$ = $x,y,z,xy,xz,yz$ (only 6 elements) and throwing in the appropriate factors of 2 or 1/2 on the off-diagonal terms to make it work out right. That is, they assume that $xy = yx$ and shove it into one index. Physics folk would recognize this better as $\sigma_{ij} = c_{ij,kl}\epsilon_{kl}$. This eliminates the need for the seemingly random factors of 2 or 1/2 on the off-diagonal elements, and makes it straightforward to use Euler formalism to rotate the axes to what you'd like them to be. – Jon Custer Oct 21 '14 at 18:53
• you are referring to the Voigt notation: en.wikipedia.org/wiki/Voigt_notation – diffeomorphism Oct 21 '14 at 19:10
• but in this case the anti-symmetric components of displacements are in principle invariant under such relabeling or axis reorientation. The anti-symmetric component of a 3x3 matrix transforms as an axial vector – diffeomorphism Oct 21 '14 at 19:12

## 1 Answer

The anti-symmetric (or skew-symmetric, if you prefer) part of the deformation gradient represents the rigid-body rotation. As a rigid-body motion, it induces no deformation (if inertial effects are ignored) and so induces no stress. It's therefore removed from the mechanics, leaving only the symmetric part to be related to stress.