Some identities involving the stress tensor

Denote the stress tensor by $$$\sigma_{ij}$$$. Assume it is symmetric. A useful quantity, called the "mean stress", is $$$$\sigma_{m} = \mbox{Tr}\sigma/3 = \sigma_{ii}/3 \ .$$$$ Another useful quantity is the traceless part of $$$\sigma$$$, which is called the deviatoric stress, denoted by $$$s_{ij}$$$: $$$$s_{ij} = \sigma_{ij} - \delta_{ij}\sigma_{m} = \sigma_{ij} - \delta_{ij}\sigma_{kk}/3 \ .$$$$ In many calculations it is useful to use the invariants of $$$s$$$, which are $$$$\begin{array}{rcl} J_{2} & = & \frac{1}{2}s_{ij}s_{ij} \ , \\ J_{3} & = & \mbox{det}s \ . \end{array}$$$$ Clearly $$$J_{2} \geq 0$$$, and often the square-root of $$$J_{2}$$$ is written as: $$$$\bar{\sigma} = \sqrt{J_{2}} \ .$$$$ Alternative forms for $$$J_{3}$$$ are $$$$\begin{array}{rcl} J_{3} & = & \mbox{det}s \\ & = & \frac{1}{6}\epsilon_{ijk}\epsilon_{mnp}s_{im}s_{jn}s_{kp} \\ & = & \frac{1}{3}s_{ij}s_{jk}s_{ki} \end{array}$$$$ All scalar functions of the stress tensor can be written in terms of its invariants. For instance, $$$$s_{ij}s_{jk}s_{kl}s_{li} = 2J_{2}^{2}$$$$

In return-map algorithms invoked during plastic deformation, the derivatives of these quantities are used. Some derivatives are as follows: $$$$\begin{array}{rcl} \frac{\partial\sigma_{m}}{\partial \sigma_{ij}} & = & \frac{1}{3}\delta_{ij} \\ \frac{\partial s_{kl}}{\partial \sigma_{ij}}& = & \delta_{ik}\delta_{jl} - \frac{1}{3}\delta_{ij}\delta_{kl} \\ \frac{\partial J_{2}}{\partial \sigma_{ij}} & = & s_{ij} \\ \frac{\partial\bar{\sigma}}{\partial \sigma_{ij}} & = & \frac{s_{ij}}{2\bar{\sigma}} \\ \frac{\partial J_{3}}{\partial \sigma_{ij}} & = & \frac{1}{2}\epsilon_{ikl}\epsilon_{jmn}s_{km}s_{ln} + \frac{1}{3}\delta_{ij}J_{2} \\ & = & s_{ik}s_{kj} - \frac{2}{3}\delta_{ij}J_{2} \\ \delta_{ij}\frac{\partial J_{2}}{\partial \sigma_{ij}} & = & 0 \\ \delta_{ij}\frac{\partial J_{3}}{\partial \sigma_{ij}} & = & 0 \\ s_{ij}\frac{\partial J_{3}}{\partial \sigma_{ij}} & = & 3J_{3} \end{array}$$$$

The "Lode angle", $$$\theta$$$, is defined through $$$$\sin 3\theta = -\frac{3\sqrt{3}J_{3}}{2J_{2}^{3/2}} \ .$$$$ It is bounded: $$$$-\pi/6 \leq \theta \leq \pi/6$$$$ Its derivative is $$$$\begin{array}{rcl} 3\cos 3\theta \frac{\partial \theta}{\partial \sigma_{ij}} & = & -\frac{3\sqrt{3}}{4 J_{2}^{3/2}} \epsilon_{ikl}\epsilon_{jmn}s_{km}s_{ln} - \frac{\sqrt{3}}{2\sqrt{J_{2}}}\delta_{ij} - \frac{3s_{ij}}{2\sqrt{J_{2}}}\sin 3\theta \\ & = & -\frac{3 \sqrt{3}}{2 J_{2}^{3/2}}s_{ik}s_{kj} + \frac{\sqrt{3}}{\sqrt{J_{2}}}\delta_{ij} - \frac{3s_{ij}}{2\sqrt{J_{2}}}\sin 3\theta \end{array}$$$$ The derivative obeys some nice identities: $$$$\begin{array}{rcl} \delta_{ij}\frac{\partial \theta}{\partial \sigma_{ij}} & = & 0 \\ s_{ij}\frac{\partial \theta}{\partial \sigma_{ij}} & = & 0 \\ \frac{\partial \theta}{\partial \sigma_{ij}} \frac{\partial \theta}{\partial \sigma_{ij}} & = & \frac{1}{2J_{2}} \ . \end{array}$$$$ Some trigonometric identities valid for any angle $$$\theta$$$ that are used in manipulations involving the Lode angle are: $$$$\begin{array}{rcl} \sin^{2}(\theta + 2\pi/3) + \sin^{2}\theta + \sin^{2}(\theta - 2\pi/3) & = & 3/2 \ , \\ \sin(\theta + 2\pi/3)\sin(\theta)\sin(\theta - 2\pi/3) & = & -\frac{1}{4}\sin 3\theta \ , \\ \sin^{4}(\theta + 2\pi/3) + \sin^{4}\theta + \sin^{4}(\theta - 2\pi/3) & = & 3/2 \ . \end{array}$$$$

The "principal stresses", $$$\sigma_{I}$$$, $$$\sigma_{II}$$$ and $$$\sigma_{III}$$$, are the eigenvalues of $$$\sigma$$$, and in "principal stress space" the stress tensor is diagonal: $$$\sigma = \mbox{diag}(\sigma_{I}, \sigma_{II}, \sigma_{III})$$$. Obvious identities follow, relating the principal stresses and the invariants: $$$$\begin{array}{rcl} \sigma_{m} & = & \frac{1}{3}(\sigma_{I} + \sigma_{II} + \sigma_{III}) \\ J_{2} & = & \frac{1}{2}\left( (\sigma_{I} - \sigma_{m})^{2} + (\sigma_{II} - \sigma_{m})^{2} + (\sigma_{III} - \sigma_{m})^{2} \right) \\ J_{3} & = & (\sigma_{I} - \sigma_{m})(\sigma_{II} - \sigma_{m}) (\sigma_{III} - \sigma_{m}) \end{array}$$$$ Often the principal stresses are ordered so $$$\sigma_{I}\geq \sigma_{II}\geq \sigma_{III}$$$, and if this is done then $$$$\begin{array}{rcl} \sigma_{I} & = & \frac{2}{\sqrt{3}}\bar{\sigma}\sin\left(\theta + \frac{2\pi}{3}\right) + \sigma_{m} \\ \sigma_{II} & = & \frac{2}{\sqrt{3}}\bar{\sigma}\sin\theta + \sigma_{m} \\ \sigma_{III} & = & \frac{2}{\sqrt{3}}\bar{\sigma}\sin\left(\theta - \frac{2\pi}{3}\right) + \sigma_{m} \ , \end{array}$$$$ where $$$\theta$$$ is the Lode angle introduced above. This means that, in the principal stress space, $$$$s = \frac{2}{\sqrt{3}}\bar{\sigma}\mbox{diag} \left(\sin(\theta + \frac{2\pi}{3}), \sin\theta, \sin(\theta - \frac{2\pi}{3}) \right)$$$$ A useful identity, which is employed in Tensile and Mohr-Coulomb plasticity is: $$$$\begin{array}{rcl} \sigma_{I} - \sigma_{m} & = & \bar{\sigma}\left(\cos\theta - \frac{1}{\sqrt{3}}\sin\theta\right) \end{array}$$$$