## Deviatoric stress and invariants

The stress tensor can be expressed as the sum of two stress tensors, namely: the hydrostatic stress tensor and the deviatoric stress tensor. In this article we will define the hydrostatic and the deviatoric part of the stress tensor and we will calculate the invariants of the stress deviator tensor. The invariants of the deviatoric stress are used frequently in failure criteria.

Consider a stress tensor $$\sigma_{ij}$$ acting on a body. The stressed body tends to change both its volume and its shape. The part of the stress tensor that tends to change the volume of the body is called mean hydrostatic stress tensor or volumetric stress tensor. The part that tends to distort the body is called stress deviator tensor. Hence, the stress tensor may expressed as:

$\sigma_{ij}=s_{ij}+p\delta_{ij}$
(1)

where $$\delta_{ij}$$ is the Kronecker delta (with $$\delta_{ij}=1$$ if $$i=j$$ and $$\delta_{ij}=0$$ if $$i\neq j$$ ), $$p$$ is the mean stress given by:

$p=\frac{1}{3}\sigma_{kk}=\frac{1}{3}\left(\sigma_{11}+\sigma_{22}+\sigma_{33}\right)=\frac{1}{3}I_{1}$
(2)

where $$I_{1}$$ is the first invariant of the stress tensor (see also: Principal stresses and stress invariants). The product $$p\delta_{ij}$$ is the hydrostatic stress tensor and contains only normal stresses. The deviatoric stress tensor can be obtained by subtracting the hydrostatic stress tensor from the stress tensor:

$\begin{array}{rl}s_{ij}=&\sigma_{ij}-p\delta_{ij}\\=&\left[\begin{array}{ccc}\sigma_{11}-p & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22}-p & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33}-p\end{array}\right]\end{array}$
(3)

In order to calculate the invariants of the stress deviator tensor we will follow the same procedure used in the article Principal stresses and stress invariants. It must be mentioned that the principal directions of the stress deviator tensor coincide with the principal directions of the stress tensor. The characteristic equation for $$s_{ij}$$ is:

$\left| s_{ij}-s\delta_{ij}\right|=s^{3}-J_{1}s^{2}-J_{2}s-J_{3}=0$
(4)

where $$J_{1}$$, $$J_{2}$$ and $$J_{3}$$ are the first, second and third deviatoric stress invariants, respectively. The roots of the polynomial are the three principal deviatoric stresses $$s_{1}$$, $$s_{2}$$ and $$s_{3}$$. $$J_{1}$$, $$J_{2}$$ and $$J_{3}$$ may be calculated by the following expressions:

$\begin{array}{rl}J_{1}=&s_{kk}=0\\J_{2}=&\frac{1}{2}s_{ij}s_{ji}\\=&\frac{1}{6}\left[\left(\sigma_{11}-\sigma_{22}\right)^2+\left(\sigma_{22}-\sigma_{33}\right)^2+\left(\sigma_{33}-\sigma_{11}\right)^2\right]\\&+\sigma_{12}^2+\sigma_{23}^2+\sigma_{31}^2\\=&\frac{1}{3}I_{1}^{2}-I_{2}\\J_{3}=&\det(s_{ij})\\=&\frac{1}{3}s_{ij}s_{jk}s_{ki}\\=&\frac{2}{27}I_{1}^{3}-\frac{1}{3}I_{1}I_{2}+I_{3}\end{array}$
(5)

where $$I_{1}$$, $$I_{2}$$ and $$I_{3}$$ are the three invariants of the stress tensor and $$\det(s_{ij})$$ is the determinant of $$s_{ij}$$. It should be mentioned that since $$J_{1}=s_{kk}=0$$, the stress deviator tensor describes a state of pure shear.

### Exercise

Calculate the stress deviator tensor and its invariants for the following stress tensor:

$\sigma_{ij}=\left[\begin{array}{ccc}2&-3&4\\-3&-5&1\\4&1&6\end{array}\right]$
(6)

Firstly we calculate the mean pressure $$p$$:

$p=\frac{1}{3}\left(2-5+6\right)=1$
(7)

From equation (3) we calculate the stress deviator tensor:

$s_{ij}=\left[\begin{array}{ccc}1&-3&4\\-3&-6&1\\4&1&5\end{array}\right]$
(8)

For the stress deviator tensor invariants we will use equations (5) and we get:

$\begin{array}{rl}J_{1}=&0\\J_{2}=&\frac{1}{2}s_{ij}s_{ji}\\=&\frac{1}{2}\left(s_{11}^2+s_{22}^2+s_{33}^2+2s_{12}^2+2s_{23}^2+2s_{13}^2\right)\\=&\frac{1}{2}\left(1^2+(-6)^2+5^2+2*(-3)^2+2*1^2+2*4^2\right)\\=&57\\J_{3}=&\frac{1}{3}s_{ij}s_{jk}s_{ki}\\=&s_{11}s_{22}s_{33}+2s_{12}s_{23}s_{13}-s_{11}s_{23}^2-s_{22}s_{13}^2-s_{33}s_{12}^2\\=&1*(-6)*5+2*(-3)*1*4\\&-1*1^2-(-6)*4^2-5*(-3)^2\\=&-4\end{array}$
(9)

Finally the characteristic equation is:

$s^3-57s+4=0$
(10)