# 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

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