# Question about the general Schmid’s law expression to calculate the critically resolved shear stress

I have a question about Schmid’s law for an arbitrary stress state. I found conflicting expressions and I would like to know which one is correct.

• $$\boldsymbol{s}$$: slip direction

• $$\boldsymbol{n}$$: slip plane normal

• $$\boldsymbol{m}$$: Schmid or slip tensor

• $$\boldsymbol{\sigma}$$: Cauchy stress tensor

• $$\tau$$: shear stress

The first expression (1) I found is the following: $$\mathbf{m} = \boldsymbol{s} \otimes \boldsymbol{n}$$ $$\tau = \boldsymbol{m}:\boldsymbol{\sigma}$$

The second expression (2) I found is the following: $$\mathbf{m} = \frac{1}{2}(\boldsymbol{s} \otimes \boldsymbol{n} + \boldsymbol{n} \otimes \boldsymbol{s})$$ $$\tau = \boldsymbol{\sigma}:\boldsymbol{m}$$

It seems like $$\boldsymbol{m}$$ in the first expression would equate $$\boldsymbol{m}$$ in the second one if $$\boldsymbol{m}$$ would be symmetric, but as far as I understand, this is not possible. I do not understand how the second expression is correct. Is the second expression correct? Why?

Sources:

Thank you very much for your help.

• Why not specify the sources, so people can consider the context, underlying definitions, assumptions, and (if warranted) credibility? Equations on their own are near-meaningless without this background. Commented Aug 2, 2023 at 22:03
• Sources added. Thank you for your comment. Commented Aug 2, 2023 at 22:15

The Schmid tensor, as you've mentioned, is not symmetric in general. However, the Cauchy stress $$\boldsymbol{\sigma}$$ is always symmetric. Moreover, for any 2nd order tensors $$\boldsymbol{B}$$ and $$\boldsymbol{A}$$ such that $$\boldsymbol{A} =\boldsymbol{A}^T$$, you can show that $$\boldsymbol{A}:\boldsymbol{B} = \boldsymbol{A}:\text{sym}\boldsymbol{B}$$. Thus, \begin{align} \tau & = \boldsymbol{\sigma}:\left( \boldsymbol{s} \otimes \boldsymbol{n} \right) \\\\ & = \boldsymbol{\sigma}:\text{sym}\left( \boldsymbol{s} \otimes \boldsymbol{n} \right) \end{align}
Some authors pre-emptively define the Schmid tensor to be symmetric. They simply do the above operation ahead of time. I prefer to keep them in their normal form (i.e., $$\boldsymbol{s} \otimes \boldsymbol{n}$$) until a specific computation arises.
• And to be explicit $\frac{1}{2}(\mathbf{s}\otimes\mathbf{n} + \mathbf{n}\otimes\mathbf{s})$ is the symmetric part of the tensor $\mathbf{s}\otimes\mathbf{n}$ Commented Sep 28, 2023 at 19:09