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

Question

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:

• Expression 1: http://pajarito.materials.cmu.edu/lectures/Polycrystal_plasticity-Aniso3-11Feb20.pdf

• Expression 2: https://matperso.minesparis.psl.eu/Donnees/data14/1479-CrystalX.pdf - http://mms2.mines-paristech.fr/public/mms_paris/animations/Surface/e_Surface-form-monox.pdf

Thank you very much for your help.

Answer 1

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.