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 $𝝈$ and $𝐞$ is the deviatoric part of $𝜺$.

From equation (3) we have

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

The deviatoric parts of the stress and strain are

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

Therefore,

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

Now,

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

Therefore,

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

Hence

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

This shows that

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

Nonlinear finite elements/Homework11/Solutions/Problem 1