# 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
    $\alpha :={\sqrt {\cfrac {2}{3}}}\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _\qquad {\text{where}}\qquad \lVert \mathbf {a} \rVert ={\sqrt {\mathbf {a} :\mathbf {a} }}~.$
Express the time derivative of $\alpha$ in terms of ${\dot {\gamma }}$ and $\partial f/\partial {\boldsymbol {\sigma }}$. This is the evolution law for $\alpha$.
The time derivative of $\alpha$ is given by
    ${\dot {\alpha }}={\sqrt {\cfrac {2}{3}}}~{\frac {\partial }{\partial t}}({\sqrt {{\boldsymbol {\varepsilon }}^{p}:{\boldsymbol {\varepsilon }}^{p}}})={\sqrt {\cfrac {2}{3}}}~\left({\frac {1}{2}}\right)~\left({\cfrac {1}{\sqrt {{\boldsymbol {\varepsilon }}^{p}:{\boldsymbol {\varepsilon }}^{p}}}}\right){\frac {\partial }{\partial t}}({\boldsymbol {\varepsilon }}^{p}:{\boldsymbol {\varepsilon }}^{p})={\sqrt {\cfrac {2}{3}}}~\left({\frac {1}{2}}\right)~\left({\cfrac {1}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _}}\right){\frac {\partial }{\partial t}}({\boldsymbol {\varepsilon }}^{p}:{\boldsymbol {\varepsilon }}^{p})$
Now,
    ${\begin{aligned}{\frac {\partial }{\partial t}}({\boldsymbol {\varepsilon }}^{p}:{\boldsymbol {\varepsilon }}^{p})&={\frac {\partial }{\partial t}}(\varepsilon _{11}^{p}\varepsilon _{11}^{p}+\varepsilon _{12}^{p}\varepsilon _{12}^{p}+\varepsilon _{13}^{p}\varepsilon _{13}^{p}+\varepsilon _{21}^{p}\varepsilon _{21}^{p}+\varepsilon _{22}^{p}\varepsilon _{22}^{p}+\varepsilon _{23}^{p}\varepsilon _{23}^{p}+\varepsilon _{31}^{p}\varepsilon _{31}^{p}+\varepsilon _{32}^{p}\varepsilon _{32}^{p}+\varepsilon _{33}^{p}\varepsilon _{33}^{p})\\&=2\varepsilon _{11}^{p}{\frac {\partial \varepsilon _{11}^{p}}{\partial t}}+2\varepsilon _{12}^{p}{\frac {\partial \varepsilon _{12}^{p}}{\partial t}}+2\varepsilon _{13}^{p}{\frac {\partial \varepsilon _{13}^{p}}{\partial t}}+2\varepsilon _{21}^{p}{\frac {\partial \varepsilon _{21}^{p}}{\partial t}}+2\varepsilon _{22}^{p}{\frac {\partial \varepsilon _{22}^{p}}{\partial t}}+2\varepsilon _{23}^{p}{\frac {\partial \varepsilon _{23}^{p}}{\partial t}}+2\varepsilon _{31}^{p}{\frac {\partial \varepsilon _{31}^{p}}{\partial t}}+2\varepsilon _{32}^{p}{\frac {\partial \varepsilon _{32}^{p}}{\partial t}}+2\varepsilon _{33}^{p}{\frac {\partial \varepsilon _{33}^{p}}{\partial t}}\\&=2\varepsilon _{11}^{p}{\dot {\varepsilon }}_{11}^{p}+2\varepsilon _{12}^{p}{\dot {\varepsilon }}_{12}^{p}+2\varepsilon _{13}^{p}{\dot {\varepsilon }}_{13}^{p}+2\varepsilon _{21}^{p}{\dot {\varepsilon }}_{21}^{p}+2\varepsilon _{22}^{p}{\dot {\varepsilon }}_{22}^{p}+2\varepsilon _{23}^{p}{\dot {\varepsilon }}_{23}^{p}+2\varepsilon _{31}^{p}{\dot {\varepsilon }}_{31}^{p}+2\varepsilon _{32}^{p}{\dot {\varepsilon }}_{32}^{p}+2\varepsilon _{33}^{p}{\dot {\varepsilon }}_{33}^{p}\\&=2\sum _{i=1}^{3}\sum _{j=1}^{3}\varepsilon _{ij}^{p}{\dot {\varepsilon }}_{ij}^{p}=2{\boldsymbol {\varepsilon }}^{p}:{\dot {\boldsymbol {\varepsilon }}}^{p}\end{aligned}}$
Therefore,
    ${\dot {\alpha }}={\sqrt {\cfrac {2}{3}}}~\left({\frac {1}{2}}\right)~\left({\cfrac {1}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _}}\right)\left(2~{\boldsymbol {\varepsilon }}^{p}:{\dot {\boldsymbol {\varepsilon }}}^{p}\right)={\sqrt {\cfrac {2}{3}}}~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:{\dot {\boldsymbol {\varepsilon }}}^{p}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _}}$
Using
    ${\dot {\boldsymbol {\varepsilon }}}^{p}={\dot {\gamma }}{\frac {\partial f({\boldsymbol {\sigma }},\alpha ,T)}{\partial {\boldsymbol {\sigma }}}}$
we get
    ${\dot {\alpha }}={\sqrt {\cfrac {2}{3}}}~{\dot {\gamma }}~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:{\frac {\partial f}{\partial {\boldsymbol {\sigma }}}}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _}}\qquad {\text{where}}\qquad {\boldsymbol {\varepsilon }}^{p}=\int _{0}^{t}{\dot {\boldsymbol {\varepsilon }}}^{p}~dt=\int _{0}^{t}{\dot {\gamma }}{\frac {\partial f}{\partial {\boldsymbol {\sigma }}}}~dt~.$