Yield and Failure

Analysis using elastic properties is only applicable when the stress remains below the yield stress. However yield stress is a single value while the stress state is a tensor. The simplest case is for a material in a uni-axial stress state where the material yields when

$\sigma_{xx} = \sigma_y$

When there is a general (multi-axial) stress state there are a number of possible yield (or failure) criteria.

Principal stresses​

The general stress tensor is

$\sigma = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \\ \end{bmatrix}$

The principal stresses are derived by rotating the stress tensor so that the shear stresses are zero resulting in a set of principal stresses

$\sigma = \begin{bmatrix} \sigma_{I} & 0 & 0 \\ 0 & \sigma_{II} & 0 \\ 0 & 0 & \sigma_{III} \\ \end{bmatrix}$

where

$\sigma_I > \sigma_{II} > \sigma_{III}$

Maximum principal stress – Lamé​

The simplest of these if that the material yields when the maximum principal stress reaches the yield stress.

$\sigma_I = \sigma_y$

This criteria is applicable to brittle materials.

Maximum shear stress – Tresca​

From Mohr's circle, based on the principal stresses $\sigma_I$, $\sigma_{II}$ and $\sigma_{III}$, the largest shear stress is

$\tau_y = max\left(\dfrac{\sigma_I - \sigma_{II}}{2}, \dfrac{\sigma_{II} - \sigma_{III}}{2}, \dfrac{\sigma_I - \sigma_{III}}{2}\right)$

$max\left((\sigma_I - \sigma_{II}), (\sigma_{II} - \sigma_{III}), (\sigma_I - \sigma_{III})\right) = \sigma_y$

Effective stress – von Mises​

Using the principal stresses an effective (distortional) stress can be defined as

$\sigma_e = \sqrt{\dfrac{(\sigma_I - \sigma_{II})^2 + (\sigma_{II} - \sigma_{III})^2 + (\sigma_{III} - \sigma_I)^2}{2}}$

The von Mises yield criterion is then

$\sqrt{\dfrac{(\sigma_I - \sigma_{II})^2 + (\sigma_{II} - \sigma_{III})^2 + (\sigma_{III} - \sigma_I)^2}{2}} = \sigma_y$

As with the Tresca criterion a hydrostatic state of stress $\sigma_I = \sigma_{II} = \sigma_{III}$ will not result in yielding