# Introduction to Elasticity/Stress example 4

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# 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 $\mathbf{t}_n^{\text{oct}} = \cfrac{1}{3}\left(\sigma_1 + \sigma_2 + \sigma_3\right) = \cfrac{1}{3}~I_{\boldsymbol{\sigma}}$.
2. 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.