Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 14

Problem 1: Part 14: Return mapping

Show that

${𝐬}_{n+1}={𝐬}_{n+1}^{\text{trial}}-2\mu \Delta \gamma {𝐧}_n.$

We have

$\begin{array}{cc}{𝝈}_{n+1}& =𝖢:{𝜺}_{n+1}^e=𝖢:({𝜺}_{n+1}-{𝜺}_{n+1}^p)\\ & =(\lambda 1\otimes 1+2\mu 𝖨):({𝜺}_{n+1}-{𝜺}_{n+1}^p)\\ & =(\lambda 1\otimes 1+2\mu 𝖨):({𝜺}_{n+1}-{𝜺}_n^p-\sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n)\\ & =\lambda [\text{tr}({𝜺}_{n+1})1-\text{tr}({𝜺}_n^p)1-\sqrt{\frac{3}{2}}\Delta \gamma \text{tr}({𝐧}_n)1]+2\mu [{𝜺}_{n+1}-{𝜺}_n^p-\sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n]\\ & =\lambda [\text{tr}({𝜺}_{n+1})-\text{tr}({𝜺}_n^p)-\sqrt{\frac{3}{2}}\Delta \gamma \text{tr}({𝐧}_n)]1+2\mu [{𝜺}_{n+1}-{𝜺}_n^p-\sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n]\\ & =\lambda [\text{tr}({𝜺}_{n+1})-\text{tr}({𝜺}_n^p)]1+2\mu [{𝜺}_{n+1}-{𝜺}_n^p-\sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n](\text{since}\text{tr}(𝐧)=0)\end{array}$

Now

${𝐬}_{n+1}={𝝈}_{n+1}-\frac{1}{3}\text{tr}({𝝈}_{n+1})1$

The trace of the stress is given by

$\begin{array}{cc}\text{tr}({𝝈}_{n+1})& =\lambda [\text{tr}({𝜺}_{n+1})-\text{tr}({𝜺}_n^p)]\text{tr}(1)+2\mu [\text{tr}({𝜺}_{n+1})-\text{tr}({𝜺}_n^p)-\sqrt{\frac{3}{2}}\Delta \gamma \text{tr}({𝐧}_n)]\\ & =3\lambda [\text{tr}({𝜺}_{n+1})-\text{tr}({𝜺}_n^p)]+2\mu [\text{tr}({𝜺}_{n+1})-\text{tr}({𝜺}_n^p)](\text{since}\text{tr}(𝐧)=0)\\ & =(3\lambda +2\mu )\text{tr}({𝜺}_{n+1})-(3\lambda +2\mu )\text{tr}({𝜺}_n^p)\end{array}$

Therefore,

$\begin{array}{cc}{𝐬}_{n+1}& ={𝝈}_{n+1}-\frac{1}{3}\text{tr}({𝝈}_{n+1})1\\ & =\lambda [\text{tr}({𝜺}_{n+1})-\text{tr}({𝜺}_n^p)]1+2\mu [{𝜺}_{n+1}-{𝜺}_n^p-\sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n]\\ & -(\lambda +\frac{2}{3}\mu )\text{tr}({𝜺}_{n+1})1+(\lambda +\frac{2}{3}\mu )\text{tr}({𝜺}_n^p)1\\ & =2\mu [{𝜺}_{n+1}-\frac{1}{3}\text{tr}({𝜺}_{n+1})1]-2\mu [{𝜺}_n^p-\frac{1}{3}\text{tr}({𝜺}_n^p)1]-2\mu \sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n\\ & =2\mu {𝐞}_{n+1}-2\mu [{𝜺}_n-{𝜺}_n^e-\frac{1}{3}\text{tr}({𝜺}_n-{𝜺}_n^e)1]-2\mu \sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n\\ & =2\mu {𝐞}_{n+1}-2\mu [{𝜺}_n-{𝜺}_n^e-\frac{1}{3}\text{tr}({𝜺}_n)1+\frac{1}{3}\text{tr}({𝜺}_n^e)1]-2\mu \sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n\\ & =2\mu [{𝐞}_{n+1}-{𝐞}_n]+2\mu [{𝜺}_n^e-\frac{1}{3}\text{tr}({𝜺}_n^e)1]-2\mu \sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n\\ & ={𝐬}_{n+1}^{\text{trial}}-{𝐬}_n+2\mu [{𝜺}_n^e-\frac{1}{3}\text{tr}({𝜺}_n^e)1]-2\mu \sqrt{\frac{3}{2}}\Delta \gamma {𝐧}_n\end{array}$

The stress-strain relation is

${𝝈}_n=\lambda \text{tr}({𝜺}_n^e)1+2\mu {𝜺}_n^e$

Hence,

$\begin{array}{cc}{𝐬}_n& =\lambda \text{tr}({𝜺}_n^e)1+2\mu {𝜺}_n^e-\frac{1}{3}\lambda \text{tr}({𝜺}_n^e)\text{tr}(1)1-\frac{2}{3}\mu \text{tr}({𝜺}_n^e)1\\ & =2\mu [{𝜺}_n^e-\frac{1}{3}\text{tr}({𝜺}_n^e)1]\end{array}$

Plugging into expression for ${𝐬}_{n+1}$, we get

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

Therefore,

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

Remark: If we write the yield function as

$f=\sqrt{𝐬:𝐬}-\sqrt{\frac{2}{3}}{\sigma }_y.$

then the above equation takes the form

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

These are equivalent.

Nonlinear finite elements/Homework11/Solutions/Problem 1