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

$f(\mathit{\sigma },\alpha ,T):=\sqrt{\frac{3}{2}}\Vert 𝐬{\Vert }_{}-{\sigma }_y(\alpha ,T)$

where $𝐬$ is the deviatoric part of the stress $\mathit{\sigma }$. Derive an expression for $\partial f/\partial \mathit{\sigma }$ in terms of the normal to the yield surface

$𝐧=\frac{𝐬}{\Vert 𝐬{\Vert }_{}}.$

The von Mises yield function is

$f=\sqrt{\frac{3}{2}}\sqrt{𝐬:𝐬}-{\sigma }_y.$

We can alternatively write the yield function as

$f=\sqrt{𝐬:𝐬}-\sqrt{\frac{2}{3}}{\sigma }_y.$

in which case the following equations take a slightly different form.

Therefore,

$\begin{array}{cc}\frac{\partial f}{\partial \mathit{\sigma }}=f_{\mathit{\sigma }}& =\sqrt{\frac{3}{2}}\frac{\partial }{\partial \mathit{\sigma }}\left(\sqrt{𝐬:𝐬}\right)\\ & =\left(\sqrt{\frac{3}{2}}\right)\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{𝐬:𝐬}}\right)\frac{\partial }{\partial \mathit{\sigma }}(𝐬:𝐬)\\ & =\left(\sqrt{\frac{3}{2}}\right)\left(\frac{1}{2}\right)\left(\frac{1}{\Vert 𝐬{\Vert }_{}}\right)\frac{\partial }{\partial \mathit{\sigma }}(𝐬:𝐬)\end{array}$

The deviatoric part of $\mathit{\sigma }$ is

$𝐬=\mathit{\sigma }-\frac{1}{3}\text{tr}(\mathit{\sigma })1\text{or}{s}_{ij}={\sigma }_{ij}-\frac{1}{3}{\sigma }_{kk}{\delta }_{ij}$

Therefore,

$\begin{array}{cc}𝐬:𝐬& =(\mathit{\sigma }-\frac{1}{3}\text{tr}(\mathit{\sigma })1):(\mathit{\sigma }-\frac{1}{3}\text{tr}(\mathit{\sigma })1)\\ & =\mathit{\sigma }:\mathit{\sigma }-\frac{2}{3}\text{tr}(\mathit{\sigma })\mathit{\sigma }:1+\frac{1}{9}{(\text{tr}(\mathit{\sigma }))}^{2}1:1\end{array}$

Now,

$\mathit{\sigma }:1={\sigma }_{ij}{\delta }_{ij}={\sigma }_{ii}=\text{tr}(\mathit{\sigma })$

and

$1:1={\delta }_{ij}{\delta }_{ij}={\delta }_{ii}=3.$

Therefore,

$𝐬:𝐬=\mathit{\sigma }:\mathit{\sigma }-\frac{2}{3}{(\text{tr}(\mathit{\sigma }))}^2+\frac{1}{3}{(\text{tr}(\mathit{\sigma }))}^2=\mathit{\sigma }:\mathit{\sigma }-\frac{1}{3}{(\text{tr}(\mathit{\sigma }))}^2.$

The derivative with respect to $\mathit{\sigma }$ is

$\frac{\partial }{\partial \mathit{\sigma }}(𝐬:𝐬)=\frac{\partial }{\partial \mathit{\sigma }}(\mathit{\sigma }:\mathit{\sigma })-\frac{1}{3}\frac{\partial }{\partial \mathit{\sigma }}\left[{(\text{tr}(\mathit{\sigma }))}^2\right]=\frac{\partial }{\partial \mathit{\sigma }}(\mathit{\sigma }:\mathit{\sigma })-\frac{2}{3}\text{tr}(\mathit{\sigma })\frac{\partial }{\partial \mathit{\sigma }}[\text{tr}(\mathit{\sigma })]$

Let us use index notation to find the derivatives. In index notation,

$\frac{\partial }{\partial \mathit{\sigma }}(\mathit{\sigma }:\mathit{\sigma })=\frac{\partial }{\partial {\sigma }_{ij}}({\sigma }_{kl}{\sigma }_{kl})=\frac{\partial }{\partial {\sigma }_{ij}}({\sigma }_{11}^2+{\sigma }_{12}^2+{\sigma }_{13}^2+{\sigma }_{21}^2+{\sigma }_{22}^2+{\sigma }_{23}^2+{\sigma }_{31}^2+{\sigma }_{32}^2+{\sigma }_{33}^2)$

Hence, the components of the second-order tensor are

$\begin{array}{cccccccc}\frac{\partial }{\partial {\sigma }_{11}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{11}& & \frac{\partial }{\partial {\sigma }_{12}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{12}& & \frac{\partial }{\partial {\sigma }_{13}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{13}\\ \frac{\partial }{\partial {\sigma }_{21}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{21}& & \frac{\partial }{\partial {\sigma }_{22}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{22}& & \frac{\partial }{\partial {\sigma }_{23}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{23}\\ \frac{\partial }{\partial {\sigma }_{31}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{31}& & \frac{\partial }{\partial {\sigma }_{32}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{32}& & \frac{\partial }{\partial {\sigma }_{33}}({\sigma }_{kl}{\sigma }_{kl})& =2{\sigma }_{33}\end{array}$

Therefore,

$\frac{\partial }{\partial {\sigma }_{ij}}({\sigma }_{kl}{\sigma }_{kl})=2{\sigma }_{ij}\text{or}\frac{\partial }{\partial \mathit{\sigma }}(\mathit{\sigma }:\mathit{\sigma })=2\mathit{\sigma }.$

Similarly,

$\frac{\partial }{\partial \mathit{\sigma }}[\text{tr}(\mathit{\sigma })]=\frac{\partial }{\partial {\sigma }_{ij}}({\sigma }_{kk})=\frac{\partial }{\partial {\sigma }_{ij}}({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})$

Hence, the components of the second-order tensor are

$\begin{array}{cccccccc}\frac{\partial }{\partial {\sigma }_{11}}({\sigma }_{kk})& =1& & \frac{\partial }{\partial {\sigma }_{12}}({\sigma }_{kk})& =0& & \frac{\partial }{\partial {\sigma }_{13}}({\sigma }_{kk})& =0\\ \frac{\partial }{\partial {\sigma }_{21}}({\sigma }_{kk})& =0& & \frac{\partial }{\partial {\sigma }_{22}}({\sigma }_{kk})& =1& & \frac{\partial }{\partial {\sigma }_{23}}({\sigma }_{kk})& =0\\ \frac{\partial }{\partial {\sigma }_{31}}({\sigma }_{kk})& =0& & \frac{\partial }{\partial {\sigma }_{32}}({\sigma }_{kk})& =0& & \frac{\partial }{\partial {\sigma }_{33}}({\sigma }_{kk})& =1\end{array}$

Therefore,

$\frac{\partial }{\partial {\sigma }_{ij}}({\sigma }_{kk})={\delta }_{ij}\text{or}\frac{\partial }{\partial \mathit{\sigma }}[\text{tr}(\mathit{\sigma })]=1.$

Plugging the above results into the expression for the derivative of $𝐬:𝐬$ we get

$\frac{\partial }{\partial \mathit{\sigma }}(𝐬:𝐬)=2\mathit{\sigma }-\frac{2}{3}\text{tr}(\mathit{\sigma })1=2(\mathit{\sigma }-\frac{1}{3}\text{tr}(\mathit{\sigma })1)=2𝐬.$

Hence, we get

$\frac{\partial f}{\partial \mathit{\sigma }}=f_{\mathit{\sigma }}=\left(\sqrt{\frac{3}{2}}\right)\left(\frac{1}{2}\right)\left(\frac{1}{\Vert 𝐬{\Vert }_{}}\right)2𝐬=\sqrt{\frac{3}{2}}\frac{𝐬}{\Vert 𝐬{\Vert }_{}}=\sqrt{\frac{3}{2}}𝐧$

The required expression is

$\frac{\partial f}{\partial \mathit{\sigma }}=f_{\mathit{\sigma }}=\sqrt{\frac{3}{2}}𝐧.$