von Mises yield criterion when all 3 principal stresses are equal and in tension

Question

Based on the von Mises yield criterion, a material begins to yield at a point when the state of stress at that point is such that the scalar known as the von Mises stress, exceeds the yield strength of the material as determined by a tension test. It is given by the equation:

$$\sigma_{VM}=\sqrt{\frac12[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_1-\sigma_3)^2]}$$

where $\sigma_1$, $\sigma_2$, and $\sigma_3$ are the principal stresses. However, based on this yield criterion, if at a given point the 3 principal stresses are equal, the material at that point will not yield even if the 3 principal stresses take on a very large value, be it in compression or in tension. Intuitively, this should not be happening for the latter case. Is there anything that I have missed out?

Answer 1

As you have found, the von Mises stress only depends on the stress deviator, i.e., if you change the stress state by $\sigma\to \sigma+\lambda \mathbb{1}$ for arbitrary $\lambda$, the von Mises stress stays the same.

Hence, the von Mises criterion is a pure shear yielding criterion. For many materials, you don't expect failure for isotropic compression (i.e., all eigenvalues equal and compressive), except for porous or granular "crushable" media. On the other hand, all materials should fail in tension, and thus you need to supplement the von Mises shear yielding criterion with a tensile cutoff. In the simplest case, you could impose tensile yielding whenever $\sigma_3$ (which I use to denote the most tensile principal stress) reaches a tensile strength $\sigma_\text{t}$.

Also note that many materials, such as concrete and most rocks, are much stronger in compression than in tension, so you could not determine the shear strength in a tensile test.