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:
- The normal traction on an octahedral plane is given by
.
- 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 }}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca25f7f43c1b553deedb13bcc89fa79749b48573)
Here
are the principal stresses and
are the first two invariants of the stress tensor (
).
Let us take the basis as the directions of the principal stresses
,
,
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c44d39e70c0663e973183805342739acec3ad27f)
If
is the direction of the normal to an octahedral plane, then the components of this normal with respect to the principal basis are
,
, and
. The normal is
oriented in such a manner that it makes equal angles with the principal directions. Therefore,
. Since
, we have
.
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1759d63a98a18c9d27028f1a9efb0cc144b94202)
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec9b256914e078abd4cd6141c0a3db2c6dff59f2)
Now,
. Therefore,
![{\displaystyle N=(1/3)(\sigma _{1}+\sigma _{2}+\sigma _{3})=(1/3)I_{\sigma }\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4843796c88384195cef279f7c1bc65bcd47c9e7)
The projected shear traction is given by
![{\displaystyle S={\sqrt {\mathbf {t} _{o}\bullet \mathbf {t} _{o}-N^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/799ee3d710872277afbb25c5acf367d994fc76d3)
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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34538814b717962dcd6fa63b687dcdb14b90d264)
Also,
![{\displaystyle II_{\sigma }=\sigma _{1}\sigma _{2}+\sigma _{2}\sigma _{3}+\sigma _{3}\sigma _{1}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3686d026c2b3de9e207af4a3f3e2d5f934f16b04)
If you do the algebra for S, you will get the required relations.