Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 7

Problem 1: Part 7: Flow rule
The $$J_2$$ theory of plasticity also states that the material satisfies the von Mises yield condition

f(\boldsymbol{\sigma},\alpha,T) := \sqrt{\cfrac{3}{2}}\lVert\mathbf{s}\rVert_{} - \sigma_y(\alpha,T) $$ where $$\mathbf{s}$$ is the deviatoric part of the stress $$\boldsymbol {\sigma}$$. Derive an expression for $$\partial f/\partial\boldsymbol{\sigma} $$ in terms of the normal to the yield surface

\mathbf{n} = \cfrac{\mathbf{s}}{\lVert\mathbf{s}\rVert_{}} ~. $$

The von Mises yield function is

f = \sqrt{\cfrac{3}{2}}~\sqrt{\mathbf{s}:\mathbf{s}} - \sigma_y ~. $$

We can alternatively write the yield function as

f = \sqrt{\mathbf{s}:\mathbf{s}} - \sqrt{\cfrac{2}{3}}~\sigma_y ~. $$ in which case the following equations take a slightly different form.

Therefore,

\begin{align} \frac{\partial f}{\partial \boldsymbol{\sigma}} = f_{\boldsymbol{\sigma}} & = \sqrt{\cfrac{3}{2}}~ \frac{\partial }{\partial \boldsymbol{\sigma}}\left(\sqrt{\mathbf {s}:\mathbf{s}}\right) \\ & = \left(\sqrt{\cfrac{3}{2}}\right)\left(\frac{1}{2}\right) \left(\cfrac{1}{\sqrt{\mathbf{s}:\mathbf{s}}}\right) \frac{\partial }{\partial \boldsymbol{\sigma}}\left(\mathbf{s}:\mathbf{s}\right) \\ & = \left(\sqrt{\cfrac{3}{2}}\right)\left(\frac{1}{2}\right) \left(\cfrac{1}{\lVert\mathbf{s}\rVert_{}}\right) \frac{\partial }{\partial \boldsymbol{\sigma}}\left(\mathbf{s}:\mathbf{s}\right) \end{align} $$ The deviatoric part of $$\boldsymbol{\sigma}$$ is

\mathbf{s} = \boldsymbol{\sigma} - \frac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol {\mathit{1}} \qquad~\text{or}~\qquad s_{ij} = \sigma_{ij} - \frac{1}{3}~\sigma_{kk}~\delta_{ij} $$ Therefore,

\begin{align} \mathbf{s}:\mathbf{s} & = \left(\boldsymbol{\sigma} - \frac{1}{3}~\text{tr}(\boldsymbol {\sigma})~\boldsymbol{\mathit{1}}\right): \left(\boldsymbol{\sigma} - \frac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol{\mathit {1}}\right) \\ & = \boldsymbol{\sigma}:\boldsymbol{\sigma} - \cfrac{2}{3}~\text{tr}(\boldsymbol{\sigma}) ~\boldsymbol{\sigma}:\boldsymbol{\mathit{1}} + \cfrac{1}{9}~\left(\text{tr}(\boldsymbol{\sigma})\right)^2~\boldsymbol{\mathit {1}}:\boldsymbol{\mathit{1}} \end{align} $$ Now,

\boldsymbol{\sigma}:\boldsymbol{\mathit{1}} = \sigma_{ij}\delta_{ij} = \sigma_{ii} = \text {tr}(\boldsymbol{\sigma}) $$ and

\boldsymbol{\mathit{1}}:\boldsymbol{\mathit{1}} = \delta_{ij}\delta_{ij} = \delta_{ii} = 3 ~. $$ Therefore,

\mathbf{s}:\mathbf{s} =\boldsymbol{\sigma}:\boldsymbol{\sigma} - \cfrac{2}{3}~\left(\text {tr}(\boldsymbol{\sigma})\right)^2 + \cfrac{1}{3}~\left(\text{tr}(\boldsymbol{\sigma})\right)^2 = \boldsymbol{\sigma}:\boldsymbol{\sigma} - \cfrac{1}{3}~\left(\text{tr}(\boldsymbol {\sigma})\right)^2~. $$ The derivative with respect to $$\boldsymbol{\sigma}$$ is

\frac{\partial }{\partial \boldsymbol{\sigma}}(\mathbf{s}:\mathbf{s}) = \frac{\partial }{\partial \boldsymbol{\sigma}}(\boldsymbol{\sigma}:\boldsymbol{\sigma}) - \cfrac{1}{3}\frac{\partial }{\partial \boldsymbol{\sigma}}\left[\left(\text{tr}(\boldsymbol {\sigma})\right)^2\right] = \frac{\partial }{\partial \boldsymbol{\sigma}}(\boldsymbol{\sigma}:\boldsymbol{\sigma}) - \cfrac{2}{3}~\text{tr}(\boldsymbol{\sigma})~\frac{\partial }{\partial \boldsymbol{\sigma}} \left[\text{tr}(\boldsymbol{\sigma})\right] $$ Let us use index notation to find the derivatives. In index notation,

\frac{\partial }{\partial \boldsymbol{\sigma}}(\boldsymbol{\sigma}:\boldsymbol{\sigma}) = \frac{\partial }{\partial \sigma_{ij}}(\sigma_{kl}\sigma_{kl}) = \frac{\partial }{\partial \sigma_{ij}}\left(\sigma_{11}^2 + \sigma_{12}^2 + \sigma_{13}^2 + \sigma_{21}^2 + \sigma_{22}^2 + \sigma_{23}^2 + \sigma_{31}^2 + \sigma_{32}^2 + \sigma_{33}^2\right) $$ Hence, the components of the second-order tensor are

\begin{align} \frac{\partial }{\partial \sigma_{11}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{11} & & \qquad \frac{\partial }{\partial \sigma_{12}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{12} & & \qquad \frac{\partial }{\partial \sigma_{13}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{13} \\ \frac{\partial }{\partial \sigma_{21}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{21} & & \qquad \frac{\partial }{\partial \sigma_{22}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{22} & & \qquad \frac{\partial }{\partial \sigma_{23}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{23} \\ \frac{\partial }{\partial \sigma_{31}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{31} & & \qquad \frac{\partial }{\partial \sigma_{32}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{32} & & \qquad \frac{\partial }{\partial \sigma_{33}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{33} \end{align} $$ Therefore,

\frac{\partial }{\partial \sigma_{ij}}(\sigma_{kl}\sigma_{kl}) = 2~\sigma_{ij} \qquad \text{or} \qquad \frac{\partial }{\partial \boldsymbol{\sigma}}(\boldsymbol{\sigma}:\boldsymbol{\sigma}) = 2~\boldsymbol{\sigma} ~. $$ Similarly,

\frac{\partial }{\partial \boldsymbol{\sigma}}\left[\text{tr}(\boldsymbol{\sigma})\right] = \frac{\partial }{\partial \sigma_{ij}}(\sigma_{kk}) = \frac{\partial }{\partial \sigma_{ij}}(\sigma_{11} + \sigma_{22} + \sigma_{33}) $$ Hence, the components of the second-order tensor are

\begin{align} \frac{\partial }{\partial \sigma_{11}}(\sigma_{kk}) & = 1 & & \qquad \frac{\partial }{\partial \sigma_{12}}(\sigma_{kk}) & = 0 & & \qquad \frac{\partial }{\partial \sigma_{13}}(\sigma_{kk}) & = 0\\ \frac{\partial }{\partial \sigma_{21}}(\sigma_{kk}) & = 0 & & \qquad \frac{\partial }{\partial \sigma_{22}}(\sigma_{kk}) & = 1 & & \qquad \frac{\partial }{\partial \sigma_{23}}(\sigma_{kk}) & = 0\\ \frac{\partial }{\partial \sigma_{31}}(\sigma_{kk}) & = 0 & & \qquad \frac{\partial }{\partial \sigma_{32}}(\sigma_{kk}) & = 0 & & \qquad \frac{\partial }{\partial \sigma_{33}}(\sigma_{kk}) & = 1 \end{align} $$ Therefore,

\frac{\partial }{\partial \sigma_{ij}}(\sigma_{kk}) = \delta_{ij} \qquad \text{or} \qquad \frac{\partial }{\partial \boldsymbol{\sigma}}\left[\text{tr}(\boldsymbol{\sigma})\right] = \boldsymbol{\mathit{1}} ~. $$ Plugging the above results into the expression for the derivative of $$\mathbf{s}:\mathbf{s}$$ we get

\frac{\partial }{\partial \boldsymbol{\sigma}}(\mathbf{s}:\mathbf{s}) = 2~\boldsymbol{\sigma} - \cfrac{2}{3}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol{\mathit {1}} = 2~\left(\boldsymbol{\sigma} - \cfrac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol {\mathit{1}}\right) = 2~\mathbf{s} ~. $$ Hence, we get

\frac{\partial f}{\partial \boldsymbol{\sigma}} = f_{\boldsymbol{\sigma}} = \left(\sqrt{\cfrac{3}{2}}\right)\left(\frac{1}{2}\right) \left(\cfrac{1}{\lVert\mathbf{s}\rVert_{}}\right)~2~\mathbf{s} = \sqrt{\cfrac{3}{2}}~\cfrac{\mathbf{s}}{\lVert\mathbf{s}\rVert_{}} = \sqrt{\cfrac{3}{2}}~\mathbf{n} $$ The required expression is

\frac{\partial f}{\partial \boldsymbol{\sigma}} = f_{\boldsymbol{\sigma}} = \sqrt{\cfrac{3} {2}}~\mathbf{n} ~. $$