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

Problem 1: Part 5: Consistency condition - 1

From the consistency condition

$\dot{f}=f_{\mathit{\sigma }}:\dot{\mathit{\sigma }}+f_{\alpha }\dot{\alpha }+f_T\dot{T}=0$

From the previous parts

$\dot{\mathit{\sigma }}=𝖢:(\dot{\mathit{\epsilon }}-\dot{\gamma }{f}_{\mathit{\sigma }});\dot{\alpha }=\sqrt{\frac{2}{3}}\dot{\gamma }\frac{{\mathit{\epsilon }}^p:f_{\mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}};\dot{T}=\frac{\chi \dot{\gamma }}{\rho C_p}\mathit{\sigma }:f_{\mathit{\sigma }}.$

Plug into the consistency condition to get

$f_{\mathit{\sigma }}:𝖢:(\dot{\mathit{\epsilon }}-\dot{\gamma }{f}_{\mathit{\sigma }})+\sqrt{\frac{2}{3}}\dot{\gamma }{f}_{\alpha }\frac{{\mathit{\epsilon }}^p:f_{\mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}+\frac{\chi }{\rho C_p}\dot{\gamma }{f}_T\mathit{\sigma }:f_{\mathit{\sigma }}=0.$

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

$\dot{\gamma }[-f_{\mathit{\sigma }}:𝖢:f_{\mathit{\sigma }}+\sqrt{\frac{2}{3}}{f}_{\alpha }\frac{{\mathit{\epsilon }}^p:f_{\mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}+\frac{\chi }{\rho C_p}{f}_T\mathit{\sigma }:f_{\mathit{\sigma }}]+f_{\mathit{\sigma }}:𝖢:\dot{\mathit{\epsilon }}=0$

Therefore,

$\dot{\gamma }=\frac{f_{\mathit{\sigma }}:𝖢:\dot{\mathit{\epsilon }}}{f_{\mathit{\sigma }}:𝖢:f_{\mathit{\sigma }}-\sqrt{\frac{2}{3}}{f}_{\alpha }\frac{{\mathit{\epsilon }}^p:f_{\mathit{\sigma }}}{\Vert {\mathit{\epsilon }}^p{\Vert }_{}}-\frac{\chi }{\rho C_p}{f}_T\mathit{\sigma }:f_{\mathit{\sigma }}}.$