Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 1

Problem 1: Part 1: Evolution rule for plastic flow

Let $\alpha$ be the equivalent plastic strain, defined as

$\alpha :=\sqrt{\frac{2}{3}}\Vert {\mathit{\epsilon }}^p{\Vert }_{}\text{where}\Vert 𝐚\Vert =\sqrt{𝐚:𝐚}.$

Express the time derivative of $\alpha$ in terms of $\dot{\gamma }$ and $\partial f/\partial \mathit{\sigma }$. This is the evolution law for $\alpha$.

The time derivative of $\alpha$ is given by

$\dot{\alpha }=\sqrt{\frac{2}{3}}\frac{\partial }{\partial t}(\sqrt{{\mathit{\epsilon }}^p:{\mathit{\epsilon }}^p})=\sqrt{\frac{2}{3}}\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{{\mathit{\epsilon }}^p:{\mathit{\epsilon }}^p}}\right)\frac{\partial }{\partial t}({\mathit{\epsilon }}^p:{\mathit{\epsilon }}^p)=\sqrt{\frac{2}{3}}\left(\frac{1}{2}\right)\left(\frac{1}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}\right)\frac{\partial }{\partial t}({\mathit{\epsilon }}^p:{\mathit{\epsilon }}^p)$

Now,

$\begin{array}{cc}\frac{\partial }{\partial t}({\mathit{\epsilon }}^p:{\mathit{\epsilon }}^p)& =\frac{\partial }{\partial t}({\epsilon }_{11}^p{\epsilon }_{11}^p+{\epsilon }_{12}^p{\epsilon }_{12}^p+{\epsilon }_{13}^p{\epsilon }_{13}^p+{\epsilon }_{21}^p{\epsilon }_{21}^p+{\epsilon }_{22}^p{\epsilon }_{22}^p+{\epsilon }_{23}^p{\epsilon }_{23}^p+{\epsilon }_{31}^p{\epsilon }_{31}^p+{\epsilon }_{32}^p{\epsilon }_{32}^p+{\epsilon }_{33}^p{\epsilon }_{33}^p)\\ & =2{\epsilon }_{11}^p\frac{\partial {\epsilon }_{11}^p}{\partial t}+2{\epsilon }_{12}^p\frac{\partial {\epsilon }_{12}^p}{\partial t}+2{\epsilon }_{13}^p\frac{\partial {\epsilon }_{13}^p}{\partial t}+2{\epsilon }_{21}^p\frac{\partial {\epsilon }_{21}^p}{\partial t}+2{\epsilon }_{22}^p\frac{\partial {\epsilon }_{22}^p}{\partial t}+2{\epsilon }_{23}^p\frac{\partial {\epsilon }_{23}^p}{\partial t}+2{\epsilon }_{31}^p\frac{\partial {\epsilon }_{31}^p}{\partial t}+2{\epsilon }_{32}^p\frac{\partial {\epsilon }_{32}^p}{\partial t}+2{\epsilon }_{33}^p\frac{\partial {\epsilon }_{33}^p}{\partial t}\\ & =2{\epsilon }_{11}^p{\dot{\epsilon }}_{11}^p+2{\epsilon }_{12}^p{\dot{\epsilon }}_{12}^p+2{\epsilon }_{13}^p{\dot{\epsilon }}_{13}^p+2{\epsilon }_{21}^p{\dot{\epsilon }}_{21}^p+2{\epsilon }_{22}^p{\dot{\epsilon }}_{22}^p+2{\epsilon }_{23}^p{\dot{\epsilon }}_{23}^p+2{\epsilon }_{31}^p{\dot{\epsilon }}_{31}^p+2{\epsilon }_{32}^p{\dot{\epsilon }}_{32}^p+2{\epsilon }_{33}^p{\dot{\epsilon }}_{33}^p\\ & =2\sum _{i=1}^3\sum _{j=1}^3{\epsilon }_{ij}^p{\dot{\epsilon }}_{ij}^p=2{\mathit{\epsilon }}^p:{\dot{\mathit{\epsilon }}}^p\end{array}$

Therefore,

$\dot{\alpha }=\sqrt{\frac{2}{3}}\left(\frac{1}{2}\right)\left(\frac{1}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}\right)(2{\mathit{\epsilon }}^p:{\dot{\mathit{\epsilon }}}^p)=\sqrt{\frac{2}{3}}\frac{{\mathit{\epsilon }}^p:{\dot{\mathit{\epsilon }}}^p}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}$

Using

${\dot{\mathit{\epsilon }}}^p=\dot{\gamma }\frac{\partial f(\mathit{\sigma },\alpha ,T)}{\partial \mathit{\sigma }}$

we get

$\dot{\alpha }=\sqrt{\frac{2}{3}}\dot{\gamma }\frac{{\mathit{\epsilon }}^p:\frac{\partial f}{\partial \mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}\text{where}{\mathit{\epsilon }}^p=\underset{0}{\overset{t}{\int }}{\dot{\mathit{\epsilon }}}^{p}dt=\underset{0}{\overset{t}{\int }}\dot{\gamma }\frac{\partial f}{\partial \mathit{\sigma }}dt.$