Why is the deviatoric stress the stress that causes distortions?

Question

I'm reading this text on stress.

It is mentioned that the Cauchy stress tensor can be split into a sum of two other tensors: hydrostatic pressure $\pi$ and deviatoric stress. Hydrostatic pressure is defined as the mean of the normal stresses. Deviatoric stress tensor is what we get when we subtract a tensor with the pressure on diagonal from the original Cauchy stress tensor.

$$\begin{align} \ \,\\ \left[{\begin{matrix} s_{11} & s_{12} & s_{13} \\ s_{21} & s_{22} & s_{23} \\ s_{31} & s_{32} & s_{33} \end{matrix}}\right] &=\left[{\begin{matrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{matrix}}\right]-\left[{\begin{matrix} \pi & 0 & 0 \\ 0 & \pi & 0 \\ 0 & 0 & \pi \end{matrix}}\right] \\ &=\left[{\begin{matrix} \sigma_{11}-\pi & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22}-\pi & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33}-\pi \end{matrix}}\right]. \end{align}$$

where

$$\pi=\frac{\sigma_{11}+\sigma_{22}+\sigma_{33}}{3}$$

It is then said:

The second component is the Deviatoric stress and is what actually causes distortion of the body.

My question is: Why do the diagonal stresses in the deviatoric stress tensor cause distortions, considering they are still normal stresses?

I've been taught that normal stresses cause only volumetric changes, and don't cause distortions, that is, don't change the shape of the element. It is the shear stresses, which lay off-diagonal, that cause shape changes to the element.

The shear stresses are present unchanged on the deviator tensor, but the normal stresses are now differences between the normal stresses on the element and their average. But they are still normal stresses, so why are they present in the deviator tensor, if the purpose of the deviator tensor is to include only stresses that cause distortions?

Answer 1

Consider a small volume element of cubic shape. The stress tensor encodes the stresses on the cube faces. (There can be normal stresses and tractions, but you can rotate the cube so that there are only normal stresses. This corresponds to the "principal stress axes" and a diagonal stress tensor, which you can always find, as @user8736288 mentioned in his answer.)

Now the isotropic (or hydrostatic) part of the stress tensor, where all diaginal elements are equal, contributes the same pressure along all three directions of the cube -- the volume element is compressed (or stretched) with the same force in all directions. If the material is isotropic, this will lead to the same amount of compresion in all directions, i.e. the volume element gets squashed, but the shape stays the same. (Note that in anisotropic materials, e.g. wood or slate, an isotropic stress may cause different amounts of compression in various directions and thus change the shape.)

The deviatoric part of the stress corresponds to normal stresses on the surface of the cube, but (since the deviatoric stress tensor is traceless) the sum of all normal stresses is zero -- you have compression in (at least) one direction and tension in (at least) one direction. Hence, again assuming an isotropic material, the volume lement gets compressed in one direction and stretched in another, i.e. the shape changes.

Answer 2

The hydrostatic term will produce expansion or contraction of a volume element, and will preserve the shape of the volume element.The remaing part will have a sum of its diagonal terms equal to zero and hence will produce transformations without any volume change ( to demonstrate this rigoroulsy, one has actually to compute the sum of the diagonal terms of the strain tensor from the stress tensor), but will cause "distortion". Also don't forget that a pure shear stress tensor (for instance in 2D) when rotated at 90° becomes a tensor with two opposite terms on its diagonal (and the zero sum of its diagonal terms is naturally invariant by any frame rotation).