# Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 9

## Problem 1: Part 9: Elastic-plastic tangent modulus

Assume that the elastic response of the material is linear, i.e.,

${\displaystyle {\boldsymbol {\mathsf {C}}}=\lambda {\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu {\boldsymbol {\mathsf {I}}}~.}$

Derive the expression for the elastic-plastic tangent modulus for a von Mises yield condition with Johnson-Cook flow stress for a linear elastic material using the expressions that you have derived in the previous parts.

The elastic-plastic tangent modulus is given by

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

From the previous parts

${\displaystyle f_{\boldsymbol {\sigma }}={\sqrt {\cfrac {3}{2}}}~\mathbf {n} ~;~~f_{\alpha }=-n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~;~~f_{T}=\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~.}$

Therefore,

${\displaystyle {\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}={\sqrt {\cfrac {3}{2}}}~(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):\mathbf {n} ={\sqrt {\cfrac {3}{2}}}~(\lambda ~{\text{tr}}(\mathbf {n} )~{\boldsymbol {\mathit {1}}}+2~\mu ~\mathbf {n} )={\sqrt {\cfrac {3}{2}}}~2~\mu ~\mathbf {n} ~.}$

Some of the results used in the above derivation are shown below.

Recall (from previous homework):

${\displaystyle {\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}:\mathbf {n} =\delta _{ij}~\delta _{kl}~n_{kl}=n_{kk}~\delta _{ij}={\text{tr}}(\mathbf {n} )~{\boldsymbol {\mathit {1}}}}$

and

${\displaystyle {\boldsymbol {\mathsf {I}}}:\mathbf {n} ={\frac {1}{2}}~(\delta _{ik}~\delta _{jl}+\delta _{il}~\delta _{jk})n_{kl}={\frac {1}{2}}~(\delta _{jl}~n_{il}+\delta _{jk}~n_{ki})={\frac {1}{2}}~(n_{ij}+n_{ji})=n_{ij}=\mathbf {n} }$

(we have used the symmetry of the stress tensor above.)

Also,

${\displaystyle {\text{tr}}(\mathbf {n} )={\text{tr}}\left({\cfrac {\mathbf {s} }{\lVert \mathbf {s} \rVert _{}}}\right)={\cfrac {{\text{tr}}(\mathbf {s} )}{\lVert \mathbf {s} \rVert _{}}}}$

Now,

${\displaystyle \mathbf {s} =s_{ij}=\sigma _{ij}-{\frac {1}{3}}~\sigma _{kk}~\delta _{ij}\qquad \implies \qquad {\text{tr}}(\mathbf {s} )=s_{ii}=\sigma _{ii}-{\frac {1}{3}}~\sigma _{kk}~\delta _{ii}=\sigma _{ii}-{\frac {1}{3}}~\sigma _{kk}~3=\sigma _{ii}-\sigma _{kk}=0}$

Therefore,

${\displaystyle {\text{tr}}(\mathbf {n} )=0~.}$

Hence,

${\displaystyle ({\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }})\otimes ({\boldsymbol {\mathsf {C}}}:f_{{\boldsymbol {\sigma }})}={\cfrac {3}{2}}~4~\mu ^{2}~\mathbf {n} \otimes \mathbf {n} =6~\mu ^{2}~\mathbf {n} \otimes \mathbf {n} }$

and

${\displaystyle f_{\boldsymbol {\sigma }}:{\boldsymbol {\mathsf {C}}}:f_{\boldsymbol {\sigma }}={\cfrac {3}{2}}~2~\mu ~\mathbf {n} :\mathbf {n} =3~\mu ~\left({\cfrac {\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}\right):\left({\cfrac {\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}\right)=3~\mu ~{\cfrac {\mathbf {s} :\mathbf {s} }{\mathbf {s} :\mathbf {s} }}=3~\mu ~.}$

Plugging in expression for ${\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}}$  we get

${\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}-\left({\cfrac {6~\mu ^{2}~\mathbf {n} \otimes \mathbf {n} }{3~\mu -n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:\mathbf {n} }{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}-{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi }{\rho ~C_{p}}}~\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~{\boldsymbol {\sigma }}:\mathbf {n} }}\right)~.}$

Now,

${\displaystyle {\boldsymbol {\sigma }}:\mathbf {n} ={\cfrac {(\mathbf {s} +{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\boldsymbol {\mathit {1}}}):\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}={\cfrac {\mathbf {s} :\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}+{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\cfrac {{\boldsymbol {\mathit {1}}}:\mathbf {s} }{\sqrt {\mathbf {s} :\mathbf {s} }}}={\sqrt {\mathbf {s} :\mathbf {s} }}+{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }})~{\cfrac {{\text{tr}}(\mathbf {s} )}{\sqrt {\mathbf {s} :\mathbf {s} }}}={\sqrt {\mathbf {s} :\mathbf {s} }}=\lVert \mathbf {s} \rVert _{}}$

Therefore, the elastic-plastic tangent modulus can be written as

${\displaystyle {\boldsymbol {\mathsf {C}}}^{\text{ep}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~\left[{\boldsymbol {\mathsf {I}}}-{\cfrac {3~\mu ~\mathbf {n} \otimes \mathbf {n} }{3~\mu -n~B~\alpha ^{n-1}\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]~{\cfrac {{\boldsymbol {\varepsilon }}^{p}:\mathbf {n} }{\lVert {\boldsymbol {\varepsilon }}^{p}\rVert _{}}}-{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi }{\rho ~C_{p}}}~\left({\cfrac {1}{T_{m}-T_{0}}}\right)\left[\sigma _{0}+B\alpha ^{n}\right]~\lVert \mathbf {s} \rVert _{}}}\right]~.}$