# Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 15

## Problem 1: Part 15: Finding the plastic flow parameter

The discretized form of the Kuhn-Tucker conditions in conjunction with the consistency condition gives us

$f({\boldsymbol {\sigma }}_{n+1},\alpha _{n+1},T_{n+1})=0~.$

Use this condition and the relations you have derived in the previous sections to arrive at a nonlinear equation in $\Delta \gamma$  that can be solved using Newton iterations.

The yield function is

$f={\sqrt {\cfrac {3}{2}}}~{\sqrt {\mathbf {s} :\mathbf {s} }}-\left[\sigma _{0}+B\alpha ^{n}\right]\left[1-\left({\cfrac {T-T_{0}}{T_{m}-T_{0}}}\right)\right]$

Therefore the discretized form of Kuhn-Tucker + consistency is

${\sqrt {\cfrac {3}{2}}}~{\sqrt {\mathbf {s} _{n+1}:\mathbf {s} _{n+1}}}-\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]=0$

or,

$\mathbf {s} _{n+1}:\mathbf {s} _{n+1}={\cfrac {2}{3}}\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}$

Now,

$\mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}$

Therefore,

{\begin{aligned}\mathbf {s} _{n+1}:\mathbf {s} _{n+1}&=\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-4~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+4~\mu ^{2}~{\cfrac {3}{2}}~(\Delta \gamma )^{2}~\mathbf {n} _{n}:\mathbf {n} _{n}\\&=\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-4~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+6~\mu ^{2}~(\Delta \gamma )^{2}\end{aligned}}

Plugging into the discretized yield condition, we have

${\cfrac {2}{3}}\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}=\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-4~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+6~\mu ^{2}~(\Delta \gamma )^{2}$

or,

$\left[\sigma _{0}+B\alpha _{n+1}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n+1}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}={\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+9~\mu ^{2}~(\Delta \gamma )^{2}$

Also

{\begin{aligned}\alpha _{n+1}&=\alpha _{n}+\Delta \gamma ~{\cfrac {{\boldsymbol {\varepsilon }}_{n}^{p}:\mathbf {n} _{n}}{\lVert {\boldsymbol {\varepsilon }}_{n}^{p}\rVert _{}}}\\T_{n+1}&=T_{n}+{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi ~\Delta \gamma }{\rho _{n}~C_{p}}}~\lVert \mathbf {s} _{n}\rVert _{}\end{aligned}}

Therefore,

$\left[\sigma _{0}+B\left\{\alpha _{n}+\Delta \gamma ~{\cfrac {{\boldsymbol {\varepsilon }}_{n}^{p}:\mathbf {n} _{n}}{\lVert {\boldsymbol {\varepsilon }}_{n}^{p}\rVert _{}}}\right\}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n}+{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi ~\Delta \gamma }{\rho _{n}~C_{p}}}~\lVert \mathbf {s} _{n}\rVert _{}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}={\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}+9~\mu ^{2}~(\Delta \gamma )^{2}$

The nonlinear equation in $\Delta \gamma$  is

{\begin{aligned}g(\Delta \gamma )&=0\\&=9~\mu ^{2}~(\Delta \gamma )^{2}-6~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {n} _{n}\\&\qquad -\left[\sigma _{0}+B\left\{\alpha _{n}+\Delta \gamma ~{\cfrac {{\boldsymbol {\varepsilon }}_{n}^{p}:\mathbf {n} _{n}}{\lVert {\boldsymbol {\varepsilon }}_{n}^{p}\rVert _{}}}\right\}^{n}\right]^{2}\left[1-\left({\cfrac {T_{n}+{\sqrt {\cfrac {3}{2}}}~{\cfrac {\chi ~\Delta \gamma }{\rho _{n}~C_{p}}}\lVert \mathbf {s} _{n}\rVert _{}-T_{0}}{T_{m}-T_{0}}}\right)\right]^{2}\\&\qquad +{\cfrac {3}{2}}~\mathbf {s} _{n+1}^{\text{trial}}:\mathbf {s} _{n+1}^{\text{trial}}\end{aligned}}