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.

Show:

1. The normal traction on an octahedral plane is given by

$\mathbf {t} _{n}^{\text{oct}}={\cfrac {1}{3}}\left(\sigma _{1}+\sigma _{2}+\sigma _{3}\right)={\cfrac {1}{3}}~I_{\boldsymbol {\sigma }}$.

1. The projected shear traction on an octahedral plane is given by

$\mathbf {t} _{s}^{\text{oct}}={\cfrac {1}{3}}{\sqrt {\left(\sigma _{1}-\sigma _{2}\right)^{2}+\left(\sigma _{2}-\sigma _{3}\right)^{2}+\left(\sigma _{3}-\sigma _{2}\right)^{2}}}={\cfrac {1}{3}}{\sqrt {2~I_{\boldsymbol {\sigma }}^{2}-6~II_{\boldsymbol {\sigma }}}}.$

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

Solution

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

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

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

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

The normal traction is,

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

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

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

The projected shear traction is given by

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

Therefore,

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

Also,

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

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

Elasticity