Experiment shows that materials can withstand very large hydrostatic pressures (spherical state of stress) without undergoing plastic deformation. [1] In many problems, particularly in the theory of plasticity, it is desirable to designate the part of the total stress which can be effective in producing plastic deformation. This is known as the stress deviator $\sigma^{\prime}$. The other component is the hydrostatic, component of stress $\sigma_m$.
In the theory of plasticity, it is therefore essential to mathematically separate the total stress state into two components:
- A component that tries to make a change in volume.
- A component that causes a change in shape (distortion), which is responsible for plastic deformation.
The component that causes shape change is known as the deviatoric stress (or the stress deviator), denoted by $\sigma'_{ij}$ (or sometimes by $s_{ij}$). The component that causes volume change is the hydrostatic stress, denoted by $\sigma_m$.
The mean or hydrostatic stress, $\sigma_m$, is the average of the normal stresses or the average of the three principal stresses ($p$ is the mean normal pressure).
$ \sigma_m = \frac{\sigma_1 + \sigma_2 + \sigma_3}{3} =\frac{\sigma_{xx}+\sigma_{yy}+\sigma_{zz}}{3}= -p \tag{1} $The deviatoric stress is what remains of the total stress after the hydrostatic component is subtracted:
$ \boldsymbol{\sigma}^\prime=\boldsymbol{\sigma}-\sigma_m \mathbf{I}=\boldsymbol{\sigma}-\frac{1}{3}\operatorname{tr}(\boldsymbol{\sigma}) \mathbf{I}\tag{2} $or
$ \sigma_{ij}^\prime=\sigma_{ij}-\sigma_m\delta_{ij}\tag{3} $where $\mathbf{I}$ is 3⨉3 unit tensor (matrix), and $\delta_{ij}$ is the Kronecker delta:
$ \delta_{ij}=\left\{\begin{aligned} &1 & \text{if }i=j\\ &0 & \text{if }i\neq j \end{aligned}\right. $$ \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz}\\ & \sigma_{yy} & \sigma_{yz}\\ \text{sym}& & \sigma_{zz} \end{bmatrix}=\begin{bmatrix} \sigma_{m} & 0 & 0\\ & \sigma_{m} & 0\\ \text{sym}& & \sigma_{m} \end{bmatrix}+\begin{bmatrix} \sigma_{xx}^\prime & \sigma_{xy}^\prime & \sigma_{xz}^\prime\\ & \sigma_{yy}^\prime & \sigma_{yz}^\prime\\ \text{sym}& & \sigma_{zz}^\prime \end{bmatrix} $- Notice that the off-diagonal elements of the stress deviator are the same as the corresponding elements of the stress tensor. That is $\sigma_{ij}^\prime=\sigma_{ij}\quad\text{if }i\neq j$
The principal components of the deviatoric stress tensor are given by:
From these definitions, it can be readily shown that the sum of the principal deviatoric stresses is always zero:
$ \sigma'_1 + \sigma'_2 + \sigma'_3 = 0 $$ \operatorname{tr}(\boldsymbol{\sigma}^\prime)=0. $Invariants of the Stress Deviator
When the stress deviator is expressed in an arbitrary coordinate system ($x, y, z$), its principal values can be found as the roots of the following cubic equation:
$ (\sigma')^3 - J_2\sigma' - J_3 = 0 \tag{5} $The coefficients $J_2$ and $J_3$ are the second and third invariants of the stress deviator. They are called "invariants" because their values are independent of the coordinate system used to describe the stress state. These quantities are fundamental in the mathematical theory of plasticity.
The invariants can be calculated from the components of the stress tensor as follows:
First Invariant, $J_1$:
$ J_1=\sigma_{kk}^\prime=\sigma_{xx}^\prime+\sigma_{yy}^\prime+\sigma_{zz}^\prime=\sigma_1^\prime+\sigma_2^\prime+\sigma_3^\prime=0.\tag{6} $Second Invariant, $J_2$:
$ \begin{aligned} J_2 &= \frac{1}{2}\sigma_{ij}^\prime\sigma_{ij}^\prime \qquad\text{(summation over }i\text{ and }j)\\ &=\frac{1}{2}\left[(\sigma_{xx}^\prime)^2+(\sigma_{yy}^\prime)^2+(\sigma_{zz}^\prime)^2+2\sigma_{xy}^2+2\sigma_{xz}^2+2\sigma_{yz}^2\right]\\ &=\frac{1}{2}\left[(\sigma_1^\prime)^2+(\sigma_2^\prime)^2+(\sigma_3^\prime)^2 \right]\\ &=\frac{1}{6}[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2]\\ &=\frac{1}{6}\left[(\sigma_{xx}-\sigma_{yy})^2+(\sigma_{yy}-\sigma_{zz})^2+(\sigma_{xx}-\sigma_{zz})^2\right]+\sigma_{xy}^2+\sigma_{xz}^2+\sigma_{yz}^2 \end{aligned}\tag{7} $Third Invariant, $J_3$:
$ \begin{aligned} J_3 &= \begin{vmatrix} \sigma_{xx}^\prime & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy}^\prime & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz}^\prime \end{vmatrix}\\ &= \frac{1}{3}[(\sigma'_1)^3 + (\sigma'_2)^3 + (\sigma'_3)^3] \\ &= \frac{1}{27}(2\sigma_1 - \sigma_2 - \sigma_3)(2\sigma_2 - \sigma_3 - \sigma_1)(2\sigma_3 - \sigma_1 - \sigma_2)\\ &=\sigma_1^\prime\sigma_2^\prime\sigma_3^\prime. \end{aligned}\tag{8} $We can also show that [2]
$ J_2=-\frac{1}{3}I_1^2+I_2 $and
$ J_3=I_3-\frac{1}{3}I_1I_2+\frac{3}{27}I_1^3. $Strain Deviator
Similar to the way we defined stress deviator, we can define deviatoric strain (or strain deviator). That is, by subtracting the mean strain from the strain tensor $\epsilon_{ij}$, we get strain deviator $\epsilon_{ij}^\prime$:
$ \epsilon_{ij}^\prime=\epsilon_{ij}-\frac{1}{3}\epsilon_{kk}\delta_{ij}\tag{9} $or
$ \boldsymbol{\epsilon}^\prime=\boldsymbol{\epsilon}-\frac{1}{3}\operatorname{tr}(\boldsymbol{\epsilon})\, \mathbf{I}. $If the material is isotropic and obeys Hooke's law, we have
$ \sigma_{ij}^\prime =2G\epsilon_{ij}^\prime,\tag{10} $where $G$ is the shear modulus.
- This is because hydrostatic pressure acts to change the volume of a material but does not cause the shear stresses necessary for dislocation slip, which is the primary mechanism for plastic flow in crystalline structures. Dislocations are a type of one-dimensional defects in crystalline materials.↩
- Spencer, A. J. M. (1980). Continuum mechanics. Dover Publications. (Eq. 3.63) ↩
