Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 13

Problem 1: Part 13: Trial elastic stress

Starting from equation (3) show that

${𝐬}_{n+1}^{\text{trial}}={𝐬}_n+2\mu ({𝐞}_{n+1}-{𝐞}_n)$

where $𝐬$ is the deviatoric part of $\mathit{\sigma }$ and $𝐞$ is the deviatoric part of $\mathit{\epsilon }$.

From equation (3) we have

$\begin{array}{cc}{\mathit{\sigma }}_{n+1}^{\text{trial}}& =[\lambda \text{tr}({\mathit{\epsilon }}_{n+1})1+2\mu {\mathit{\epsilon }}_{n+1}]-[\lambda \text{tr}({\mathit{\epsilon }}_n^p)1+2\mu {\mathit{\epsilon }}_n^p]\\ & =(\lambda 1\otimes 1+2\mu 𝖨):({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n^p)\\ & =(\lambda 1\otimes 1+2\mu 𝖨):({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n+{\mathit{\epsilon }}_n^e)\\ & =(\lambda 1\otimes 1+2\mu 𝖨):{\mathit{\epsilon }}_n^e+(\lambda 1\otimes 1+2\mu 𝖨):({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n)\\ & ={\mathit{\sigma }}_n+(\lambda 1\otimes 1+2\mu 𝖨):({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n)\end{array}$

The deviatoric parts of the stress and strain are

${𝐬}_{n+1}^{\text{trial}}={\mathit{\sigma }}_{n+1}^{\text{trial}}-\frac{1}{3}\text{tr}({\mathit{\sigma }}_{n+1}^{\text{trial}})1;{𝐞}_{n+1}={\mathit{\epsilon }}_{n+1}-\frac{1}{3}\text{tr}({\mathit{\epsilon }}_{n+1})1;{𝐞}_n={\mathit{\epsilon }}_n-\frac{1}{3}\text{tr}({𝐞}_n)1$

Therefore,

$\begin{array}{cc}{𝐬}_{n+1}^{\text{trial}}& ={\mathit{\sigma }}_{n+1}^{\text{trial}}-\frac{1}{3}\text{tr}({\mathit{\sigma }}_{n+1}^{\text{trial}})1\\ & ={\mathit{\sigma }}_n+(\lambda 1\otimes 1+2\mu 𝖨):({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n)-\frac{1}{3}\text{tr}{\mathit{\sigma }}_{n}1-\frac{1}{3}\text{tr}[\lambda 1\otimes 1+2\mu 𝖨):({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n)]1\end{array}$

Now,

$(\lambda 1\otimes 1+2\mu 𝖨):({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n)=\lambda \text{tr}({\mathit{\epsilon }}_{n+1})1+2\mu {\mathit{\epsilon }}_{n+1}-\lambda \text{tr}({\mathit{\epsilon }}_n)1-2\mu {\mathit{\epsilon }}_n$

Therefore,

$\begin{array}{cc}\text{tr}[(\lambda 1\otimes 1+2\mu 𝖨):({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n)]& =\lambda \text{tr}({\mathit{\epsilon }}_{n+1})\text{tr}(1)+2\mu \text{tr}({\mathit{\epsilon }}_{n+1})-\lambda \text{tr}({\mathit{\epsilon }}_n)\text{tr}(1)-2\mu \text{tr}({\mathit{\epsilon }}_n)\\ & =3\lambda \text{tr}({\mathit{\epsilon }}_{n+1})+2\mu \text{tr}({\mathit{\epsilon }}_{n+1})-3\lambda \text{tr}({\mathit{\epsilon }}_n)-2\mu \text{tr}({\mathit{\epsilon }}_n)\end{array}$

Hence

$\begin{array}{cc}{𝐬}_{n+1}^{\text{trial}}=& {\mathit{\sigma }}_n-\frac{1}{3}\text{tr}{\mathit{\sigma }}_n+\lambda \text{tr}({\mathit{\epsilon }}_{n+1})1+2\mu {\mathit{\epsilon }}_{n+1}-\lambda \text{tr}({\mathit{\epsilon }}_n)1-2\mu {\mathit{\epsilon }}_n\\ & -\lambda \text{tr}({\mathit{\epsilon }}_{n+1})1-\frac{2}{3}\mu \text{tr}({\mathit{\epsilon }}_{n+1})1+\lambda \text{tr}({\mathit{\epsilon }}_n)1+\frac{2}{3}\mu \text{tr}({\mathit{\epsilon }}_n)1\\ =& {𝐬}_n+2\mu ({\mathit{\epsilon }}_{n+1}-\frac{1}{3}\text{tr}({\mathit{\epsilon }}_{n+1})1)-2\mu ({\mathit{\epsilon }}_n-\frac{1}{3}\text{tr}({\mathit{\epsilon }}_n)1)\\ =& {𝐬}_n+2\mu {𝐞}_{n+1}-2\mu {𝐞}_n\end{array}$

This shows that

${𝐬}_{n+1}^{\text{trial}}={𝐬}_n+2\mu ({𝐞}_{n+1}-{𝐞}_n)$