Deviatoric stress and invariants

The stress tensor can be expressed as the sum of two tensors: 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 calculate the invariants of the deviatoric stress tensor. The invariants of the deviatoric stress are used frequently in failure criteria.

Consider the stress state $\sigma_{ij}$ at a point of 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 the hydrostatic (or volumetric, spherical) stress tensor. The part that tends to distort the body is called the deviatoric stress tensor (or stress deviator). Hence, the stress tensor may be expressed as:

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

where $\delta_{ij}$ is the Kronecker delta and $p$ is the mean stress, given by:

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

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. Strictly speaking, this clean separation of volume change and distortion holds for isotropic materials; in an anisotropic material a hydrostatic stress may also cause distortion.

The deviatoric stress tensor is obtained by subtracting the hydrostatic stress tensor from the stress tensor:

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

In order to calculate the invariants of the deviatoric stress tensor we will follow the same procedure used in the article Principal stresses and stress invariants. Note that the principal directions of the deviatoric stress tensor coincide with the principal directions of the stress tensor. Indeed, from Eq. (1):

$$s_{ij}-s\delta_{ij}=\sigma_{ij}-(p+s)\delta_{ij} \tag{4}$$

so the two eigenvalue problems have the same eigenvectors, and the principal deviatoric stresses are simply the principal stresses reduced by the mean stress:

$$s_{k}=\sigma_{k}-p,\qquad k=1,2,3 \tag{5}$$

The characteristic equation for $s_{ij}$ is:

where $J_{1}$, $J_{2}$ and $J_{3}$ are the first, second and third deviatoric stress invariants, respectively. Since $J_{1}=0$ (see below), the characteristic equation reduces to $s^{3}-J_{2}s-J_{3}=0$. Its roots 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:

$$\begin{array}{rl}J_{1}=&s_{kk}=0\\J_{2}=&\frac{1}{2}s_{ij}s_{ji}\\=&\frac{1}{6}\left[\left(\sigma_{11}-\sigma_{22}\right)^2+\left(\sigma_{22}-\sigma_{33}\right)^2+\left(\sigma_{33}-\sigma_{11}\right)^2\right]\\&+\sigma_{12}^2+\sigma_{23}^2+\sigma_{31}^2\\=&\frac{1}{3}I_{1}^{2}-I_{2}\\J_{3}=&\det(s_{ij})\\=&\frac{1}{3}s_{ij}s_{jk}s_{ki}\\=&\frac{2}{27}I_{1}^{3}-\frac{1}{3}I_{1}I_{2}+I_{3}\end{array} \tag{7}$$

where $I_{1}$, $I_{2}$ and $I_{3}$ are the three invariants of the stress tensor and $\det(s_{ij})$ is the determinant of $s_{ij}$. In the coordinate system of the principal directions, the invariants take the simple form:

$$\begin{array}{l}J_{2}=\frac{1}{6}\left[\left(\sigma_{1}-\sigma_{2}\right)^2+\left(\sigma_{2}-\sigma_{3}\right)^2+\left(\sigma_{3}-\sigma_{1}\right)^2\right]\\J_{3}=s_{1}s_{2}s_{3}\end{array} \tag{8}$$

Note that since $J_{1}=s_{kk}=0$, there is always a coordinate system in which all normal components of $s_{ij}$ vanish, i.e. the deviatoric stress tensor represents a state of pure shear.

Example

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

$$\sigma_{ij}=\left[\begin{array}{ccc}2&-3&4\\-3&-5&1\\4&1&6\end{array}\right] \tag{9}$$

Solution

First, we calculate the mean stress $p$:

$$p=\frac{1}{3}\left(2-5+6\right)=1 \tag{10}$$

From Eq. (3) we calculate the deviatoric stress tensor:

$$s_{ij}=\left[\begin{array}{ccc}1&-3&4\\-3&-6&1\\4&1&5\end{array}\right] \tag{11}$$

For the invariants of the deviatoric stress tensor we use Eqs. (7) and get:

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

As a check, the invariants of the stress tensor are $I_{1}=3$, $I_{2}=-54$ and $I_{3}=-60$, so from the last expressions of Eqs. (7):

$$\begin{array}{l}J_{2}=\frac{1}{3}\cdot 3^{2}-(-54)=3+54=57\\J_{3}=\frac{2}{27}\cdot 3^{3}-\frac{1}{3}\cdot 3\cdot(-54)+(-60)=2+54-60=-4\end{array} \tag{13}$$

in agreement with the values calculated above. Finally, from Eq. (6) the characteristic equation is:

$$s^3-57s+4=0 \tag{14}$$

Its roots are the principal deviatoric stresses (approximate solution):

$$s_{k}=\left(7.514,\ 0.070,\ -7.585\right) \tag{15}$$

whose sum is zero, as expected since $J_{1}=0$. From Eq. (5), the principal stresses of the original tensor are then:

$$\sigma_{k}=s_{k}+p=\left(8.514,\ 1.070,\ -6.585\right) \tag{16}$$