# Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 6
## Problem 1: Part 6: Continuum elastic-plastic tangent modulus

The continuum elastic-plastic tangent modulus is defined by the following relation
    ${\dot {\boldsymbol {\sigma }}}={\boldsymbol {\mathsf {C}}}^{\text{ep}}:{\dot {\boldsymbol {\varepsilon }}}~.$
Derive an expression for the elastic plastic tangent modulus using the results you have derived in the previous parts.
The stress rate is given by
    ${\dot {\boldsymbol {\sigma }}}={\boldsymbol {\mathsf {C}}}:\left({\dot {\boldsymbol {\varepsilon }}}-{\dot {\gamma }}f_{\boldsymbol {\sigma }}\right)~.$
From the previous part
    ${\dot {\gamma }}={\cfrac {f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}}{f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}-{\sqrt {\cfrac {2}{3}}}~f_{\alpha }~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _}}-{\cfrac {\chi }{\rho ~C_{p}}}~f_{T}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}}}~.$
Plug in expression for stress rate to get
    ${\begin{aligned}{\dot {\boldsymbol {\sigma }}}&={\boldsymbol {\mathsf {C}}}:\left({\dot {\boldsymbol {\varepsilon }}}-{\cfrac {f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}}{f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}-{\sqrt {\cfrac {2}{3}}}~f_{\alpha }~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _}}-{\cfrac {\chi }{\rho ~C_{p}}}~f_{T}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}}}~f_{\boldsymbol {\sigma }}\right)\\&={\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}-{\cfrac {f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}}{f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}-{\sqrt {\cfrac {2}{3}}}~f_{\alpha }~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _}}-{\cfrac {\chi }{\rho ~C_{p}}}~f_{T}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}}}~{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}\end{aligned}}$
We have to express the above in the form
    ${\dot {\boldsymbol {\sigma }}}={\boldsymbol {\mathsf {C}}}^{\text{ep}}:{\dot {\boldsymbol {\varepsilon }}}\qquad \implies \qquad {\dot {\sigma }}_{ij}=C_{ijkl}^{\text{ep}}~{\dot {\varepsilon }}_{kl}~.$
Since the denominator is a scalar, we don't have to worry about it for this calculation. In that case we can write
    ${\dot {\boldsymbol {\sigma }}}={\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}-\left({\cfrac {f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}}{\text{denom.}}}\right){\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}$
In index notation, we can write
    ${\dot {\sigma }}_{ij}=C_{ijkl}~{\dot {\varepsilon }}_{kl}-{\cfrac {f_{pq}^{\boldsymbol {\sigma }}~C_{pqrs}~{\dot {\varepsilon }}_{rs}}{\text{denom.}}}~C_{ijkl}~f_{kl}^{\boldsymbol {\sigma }}$
Let us manipulate the numerator of the second term above so that we get what we need. Thus
    ${\begin{aligned}f_{pq}^{\boldsymbol {\sigma }}~C_{pqrs}~{\dot {\varepsilon }}_{rs}~C_{ijkl}~f_{kl}^{\boldsymbol {\sigma }}&=\left(C_{pqrs}~f_{pq}^{\boldsymbol {\sigma }}\right)~\left(C_{ijkl}~f_{kl}^{\boldsymbol {\sigma }}\right)~{\dot {\varepsilon }}_{rs}\\&=\left(C_{rspq}~f_{pq}^{\boldsymbol {\sigma }}\right)~\left(C_{ijkl}~f_{kl}^{\boldsymbol {\sigma }}\right)~{\dot {\varepsilon }}_{rs}\qquad {\text{Major symmetry of}}~{\boldsymbol {\mathsf {C}}}\implies C_{pqrs}=C_{rspq}\\&=\left(C_{ijkl}~f_{kl}^{\boldsymbol {\sigma }}\right)~\left(C_{rspq}~f_{pq}^{\boldsymbol {\sigma }}\right)~~{\dot {\varepsilon }}_{rs}\\&\equiv A_{ij}~B_{rs}~{\dot {\varepsilon }}_{rs}\equiv M_{ijrs}~{\dot {\varepsilon }}_{rs}\\&={\boldsymbol {\mathsf {M}}}:{\dot {\boldsymbol {\varepsilon }}}~.\end{aligned}}$
In the above
    $M_{ijrs}=A_{ij}~B_{rs}\qquad \implies \qquad {\boldsymbol {\mathsf {M}}}={\boldsymbol {A}}\otimes {\boldsymbol {B}}$
and
    ${\begin{aligned}A_{ij}&=C_{ijkl}~f_{kl}^{\boldsymbol {\sigma }}\qquad \implies \qquad {\boldsymbol {A}}={\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}\\B_{rs}&=C_{rspq}~f_{pq}^{\boldsymbol {\sigma }}\qquad \implies \qquad {\boldsymbol {B}}={\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}\end{aligned}}$
Therefore,
    ${\boldsymbol {\mathsf {M}}}=({\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }})\otimes ({\boldsymbol {\mathsf {C}}}:f_{{\boldsymbol {\sigma }})}~.$
This gives us
    ${\dot {\boldsymbol {\sigma }}}={\boldsymbol {\mathsf {C}}}:{\dot {\boldsymbol {\varepsilon }}}-\left({\cfrac {({\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }})\otimes ({\boldsymbol {\mathsf {C}}}:f_{{\boldsymbol {\sigma }})}}{\text{denom.}}}\right):{\dot {\boldsymbol {\varepsilon }}}$
or,
    ${\dot {\boldsymbol {\sigma }}}=\left[{\boldsymbol {\mathsf {C}}}-\left({\cfrac {({\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }})\otimes ({\boldsymbol {\mathsf {C}}}:f_{{\boldsymbol {\sigma }})}}{\text{denom.}}}\right)\right]:{\dot {\boldsymbol {\varepsilon }}}~.$
Hence
    ${\boldsymbol {\mathsf {C}}}^{\text{ep}}={\boldsymbol {\mathsf {C}}}-\left({\cfrac {({\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }})\otimes ({\boldsymbol {\mathsf {C}}}:f_{{\boldsymbol {\sigma }})}}{\text{denom.}}}\right)~.$
The continuum elastic-plastic tangent modulus is therefore
    ${{\boldsymbol {\mathsf {C}}}^{\text{ep}}={\boldsymbol {\mathsf {C}}}-\left({\cfrac {({\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }})\otimes ({\boldsymbol {\mathsf {C}}}:f_{{\boldsymbol {\sigma }})}}{f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}-{\sqrt {\cfrac {2}{3}}}~f_{\alpha }~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:f_{\boldsymbol {\sigma }}}{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _}}-{\cfrac {\chi }{\rho ~C_{p}}}~f_{T}~{\boldsymbol {\sigma }}:f_{\boldsymbol {\sigma }}}}\right)~.}$