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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?

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The components in the stress tensor depend on the coordinate system being used.

Think about a simple situation like pure shear in two dimensions.

In one coordinate system, the only non-zero entries in the stress tensor are the $\tau_{xy}$ and $\tau_{yx}$ components (which are equal, of course).

If you rotate the coordinate axes by 45 degrees, the $\tau_{xy}$ and $\tau_{yx}$ components become zero and you have equal and opposite values for $\sigma_{xx}$ and $\sigma_{yy}$ components.

But note that the sum of the two direct stress components is zero, so there is no hydrostatic pressure in the stress field.

For a three dimensional stress field, you can always find a set of coordinate directions where all the shear stress terms are zero - namely the "principal stress directions" for the state of stress.

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  • $\begingroup$ Thank you, but I still don't understand how this answers my question. I understand I can find the principal directions, but I can't see how it's related to the deviatoric stress. If I make the rotation in the situation you describe, then the deviatoric tensor is the same as the original stress tensor since we are subtracting zero from the diagonal, right...? $\endgroup$ – S. Rotos Jun 21 at 12:24

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