# Elasticity/Stress example 4

## Example 4

Given:

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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1. The normal traction on an octahedral plane is given by
${\displaystyle \mathbf {t} _{n}^{\text{oct}}={\cfrac {1}{3}}\left(\sigma _{1}+\sigma _{2}+\sigma _{3}\right)={\cfrac {1}{3}}~I_{\boldsymbol {\sigma }}}$ .
1. The projected shear traction on an octahedral plane is given by
${\displaystyle \mathbf {t} _{s}^{\text{oct}}={\cfrac {1}{3}}{\sqrt {\left(\sigma _{1}-\sigma _{2}\right)^{2}+\left(\sigma _{2}-\sigma _{3}\right)^{2}+\left(\sigma _{3}-\sigma _{2}\right)^{2}}}={\cfrac {1}{3}}{\sqrt {2~I_{\boldsymbol {\sigma }}^{2}-6~II_{\boldsymbol {\sigma }}}}.}$

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

### Solution

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

${\displaystyle \left[{\boldsymbol {\sigma }}\right]={\begin{bmatrix}\sigma _{1}&0&0\\0&\sigma _{2}&0\\0&0&\sigma _{3}\end{bmatrix}}}$

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

The traction vector on an octahedral plane is given by

${\displaystyle \mathbf {t} _{o}={\widehat {\mathbf {n} }}_{o}\bullet \left[{\boldsymbol {\sigma }}\right]={n_{o}\sigma _{1},n_{o}\sigma _{2},n_{o}\sigma _{3}}}$

The normal traction is,

${\displaystyle N=\mathbf {t} _{o}\bullet {\widehat {\mathbf {n} }}_{o}=n_{o}^{2}\sigma _{1}+n_{o}^{2}\sigma _{2}+n_{o}^{2}\sigma _{3}}$

Now, ${\displaystyle I_{\sigma }=(\sigma _{1}+\sigma _{2}+\sigma _{3})\,}$ . Therefore,

${\displaystyle N=(1/3)(\sigma _{1}+\sigma _{2}+\sigma _{3})=(1/3)I_{\sigma }\,}$

The projected shear traction is given by

${\displaystyle S={\sqrt {\mathbf {t} _{o}\bullet \mathbf {t} _{o}-N^{2}}}}$

Therefore,

${\displaystyle S={\sqrt {n_{o}^{2}\sigma _{1}^{2}+n_{o}^{2}\sigma _{2}^{2}+n_{o}^{2}\sigma _{3}^{2}-(1/9)(\sigma _{1}+\sigma _{2}+\sigma _{3})^{2}}}}$

Also,

${\displaystyle II_{\sigma }=\sigma _{1}\sigma _{2}+\sigma _{2}\sigma _{3}+\sigma _{3}\sigma _{1}\,}$

If you do the algebra for S, you will get the required relations.