# Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 5

## Problem 1: Part 5: Consistency condition - 1

Use the consistency condition and the expressions you have derived in the previous parts to derive an expression for ${\displaystyle {\dot {\gamma }}}$  in terms of ${\displaystyle \partial f/\partial {\boldsymbol {\sigma }}}$ , ${\displaystyle \partial f/\partial \alpha }$ , ${\displaystyle \partial f/\partial T}$ , ${\displaystyle {\boldsymbol {\mathsf {C}}}}$ , and ${\displaystyle {\dot {\boldsymbol {\varepsilon }}}}$ .

From the consistency condition

${\displaystyle {\dot {f}}=f_{\boldsymbol {\sigma }}:{\dot {\boldsymbol {\sigma }}}+f_{\alpha }~{\dot {\alpha }}+f_{T}~{\dot {T}}=0}$

From the previous parts

${\displaystyle {\dot {\boldsymbol {\sigma }}}={\boldsymbol {\mathsf {C}}}:\left({\dot {\boldsymbol {\varepsilon }}}-{\dot {\gamma }}f_{\boldsymbol {\sigma }}\right)~;\qquad {\dot {\alpha }}={\sqrt {\cfrac {2}{3}}}~{\dot {\gamma }}~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}~;\qquad {\dot {T}}={\cfrac {\chi ~{\dot {\gamma }}}{\rho ~C_{p}}}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}~.}$

Plug into the consistency condition to get

${\displaystyle f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:\left({\dot {\boldsymbol {\varepsilon }}}-{\dot {\gamma }}f_{\boldsymbol {\sigma }}\right)+{\sqrt {\cfrac {2}{3}}}~{\dot {\gamma }}~f_{\alpha }~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}+{\cfrac {\chi }{\rho ~C_{p}}}~{\dot {\gamma }}~f_{T}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}=0~.}$

Collect terms containing ${\displaystyle {\dot {\gamma }}}$ :

${\displaystyle {\dot {\gamma }}\left[-f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}+{\sqrt {\cfrac {2}{3}}}~f_{\alpha }~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}+{\cfrac {\chi }{\rho ~C_{p}}}~f_{T}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}\right]+f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}=0}$

Therefore,

${\displaystyle {\dot {\gamma }}={\cfrac {f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}}{f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}-{\sqrt {\cfrac {2}{3}}}~f_{\alpha }~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}-{\cfrac {\chi }{\rho ~C_{p}}}~f_{T}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}}}~.}$