The octahedral plane is the plane that is equally inclined to the directions of the three principal stresses. For any given stress of state there are eight such planes.

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The normal traction on an octahedral plane is given by

Here $(\sigma _{1},\sigma _{2},\sigma _{3})\,$ are the principal stresses and $(I_{\boldsymbol {\sigma }},II_{\boldsymbol {\sigma }})\,$ are the first two invariants of the stress tensor (${\boldsymbol {\sigma }}\,$).

Let us take the basis as the directions of the principal stresses
${\widehat {\mathbf {n} }}{1}$, ${\widehat {\mathbf {n} }}{2}$, ${\widehat {\mathbf {n} }}{3}$. Then the stress tensor is given by

If ${\widehat {\mathbf {n} }}_{o}$ is the direction of the normal to an octahedral plane, then the components of this normal with respect to the principal basis are $n_{o1}\,$, $n_{o2}\,$, and $n_{o3}\,$. The normal is
oriented in such a manner that it makes equal angles with the principal directions. Therefore, $n_{o1}=n_{o2}=n_{o3}=n_{o}\,$. Since $n_{o1}^{2}+n_{o2}^{2}+n_{o3}^{2}=1\,$, we have $n_{o}=1/{\sqrt {(}}3)$.

The traction vector on an octahedral plane is given by