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

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

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

The yield function is

$f=\sqrt{\frac{3}{2}}\sqrt{𝐬:𝐬}-[{\sigma }_0+B{\alpha }^n][1-\left(\frac{T-T_0}{T_m-T_0}\right)]$

Therefore the discretized form of Kuhn-Tucker + consistency is

$\sqrt{\frac{3}{2}}\sqrt{{𝐬}_{n+1}:{𝐬}_{n+1}}-[{\sigma }_0+B{\alpha }_{n+1}^n][1-\left(\frac{T_{n+1}-T_0}{T_m-T_0}\right)]=0$

or,

${𝐬}_{n+1}:{𝐬}_{n+1}=\frac{2}{3}{[{\sigma }_0+B{\alpha }_{n+1}^n]}^2{[1-\left(\frac{T_{n+1}-T_0}{T_m-T_0}\right)]}^2$

Now,

${𝐬}_{n+1}={𝐬}_{n+1}^{\text{trial}}-2\mu \sqrt{\frac{3}{2}}\mathrm{\Delta }\gamma {𝐧}_n$

Therefore,

$\begin{array}{cc}{𝐬}_{n+1}:{𝐬}_{n+1}& ={𝐬}_{n+1}^{\text{trial}}:{𝐬}_{n+1}^{\text{trial}}-4\mu \sqrt{\frac{3}{2}}\mathrm{\Delta }\gamma {𝐬}_{n+1}^{\text{trial}}:{𝐧}_n+4{\mu }^2\frac{3}{2}(\mathrm{\Delta }\gamma {)}^2{𝐧}_n:{𝐧}_n\\ & ={𝐬}_{n+1}^{\text{trial}}:{𝐬}_{n+1}^{\text{trial}}-4\mu \sqrt{\frac{3}{2}}\mathrm{\Delta }\gamma {𝐬}_{n+1}^{\text{trial}}:{𝐧}_n+6{\mu }^2(\mathrm{\Delta }\gamma {)}^2\end{array}$

Plugging into the discretized yield condition, we have

$\frac{2}{3}{[{\sigma }_0+B{\alpha }_{n+1}^n]}^2{[1-\left(\frac{T_{n+1}-T_0}{T_m-T_0}\right)]}^2={𝐬}_{n+1}^{\text{trial}}:{𝐬}_{n+1}^{\text{trial}}-4\mu \sqrt{\frac{3}{2}}\mathrm{\Delta }\gamma {𝐬}_{n+1}^{\text{trial}}:{𝐧}_n+6{\mu }^2(\mathrm{\Delta }\gamma {)}^2$

or,

${[{\sigma }_0+B{\alpha }_{n+1}^n]}^2{[1-\left(\frac{T_{n+1}-T_0}{T_m-T_0}\right)]}^2=\frac{3}{2}{𝐬}_{n+1}^{\text{trial}}:{𝐬}_{n+1}^{\text{trial}}-6\mu \sqrt{\frac{3}{2}}\mathrm{\Delta }\gamma {𝐬}_{n+1}^{\text{trial}}:{𝐧}_n+9{\mu }^2(\mathrm{\Delta }\gamma {)}^2$

Also

$\begin{array}{cc}{\alpha }_{n+1}& ={\alpha }_n+\mathrm{\Delta }\gamma \frac{{\mathit{\epsilon }}_n^p:{𝐧}_n}{\Vert {\mathit{\epsilon }}_n^p{\Vert }_{}}\\ T_{n+1}& =T_n+\sqrt{\frac{3}{2}}\frac{\chi \mathrm{\Delta }\gamma }{{\rho }_n{C}_p}\Vert {𝐬}_n{\Vert }_{}\end{array}$

Therefore,

${[{\sigma }_0+B{\{{\alpha }_n+\mathrm{\Delta }\gamma \frac{{\mathit{\epsilon }}_n^p:{𝐧}_n}{\Vert {\mathit{\epsilon }}_n^p{\Vert }_{}}\}}^n]}^2{[1-\left(\frac{T_n+\sqrt{\frac{3}{2}}\frac{\chi \mathrm{\Delta }\gamma }{{\rho }_n{C}_p}\Vert {𝐬}_n{\Vert }_{}-T_0}{T_m-T_0}\right)]}^2=\frac{3}{2}{𝐬}_{n+1}^{\text{trial}}:{𝐬}_{n+1}^{\text{trial}}-6\mu \sqrt{\frac{3}{2}}\mathrm{\Delta }\gamma {𝐬}_{n+1}^{\text{trial}}:{𝐧}_n+9{\mu }^2(\mathrm{\Delta }\gamma {)}^2$

The nonlinear equation in $\mathrm{\Delta }\gamma$ is

$\begin{array}{cc}g(\mathrm{\Delta }\gamma )& =0\\ & =9{\mu }^2(\mathrm{\Delta }\gamma {)}^2-6\mu \sqrt{\frac{3}{2}}\mathrm{\Delta }\gamma {𝐬}_{n+1}^{\text{trial}}:{𝐧}_n\\ & -{[{\sigma }_0+B{\{{\alpha }_n+\mathrm{\Delta }\gamma \frac{{\mathit{\epsilon }}_n^p:{𝐧}_n}{\Vert {\mathit{\epsilon }}_n^p{\Vert }_{}}\}}^n]}^2{[1-\left(\frac{T_n+\sqrt{\frac{3}{2}}\frac{\chi \mathrm{\Delta }\gamma }{{\rho }_n{C}_p}\Vert {𝐬}_n{\Vert }_{}-T_0}{T_m-T_0}\right)]}^2\\ & +\frac{3}{2}{𝐬}_{n+1}^{\text{trial}}:{𝐬}_{n+1}^{\text{trial}}\end{array}$