Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 12

Problem 1: Part 12: Trial elastic stress

In the radial return algorithm, we define a trial elastic state as

$\text{(2)}{\mathit{\sigma }}_{n+1}^{\text{trial}}={\mathit{\sigma }}_n+𝖢:({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n)$

where ${\mathit{\sigma }}_n,{\mathit{\epsilon }}_n$ are the stress and strain at $t=t_n$ and ${\mathit{\sigma }}_{n+1},{\mathit{\epsilon }}_{n+1}$ are the values at $t=t_{n+1}$. Show that, if the elastic response of the material is linear, equation (2) can be written as

$\text{(3)}{\mathit{\sigma }}_{n+1}^{\text{trial}}=[\lambda \text{tr}({\mathit{\epsilon }}_{n+1})1+2\mu {\mathit{\epsilon }}_{n+1}]-[\lambda \text{tr}({\mathit{\epsilon }}_n^p)1+2\mu {\mathit{\epsilon }}_n^p].$

From equation (2)

${\mathit{\sigma }}_{n+1}^{\text{trial}}={\mathit{\sigma }}_n+𝖢:({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n)=𝖢:{\mathit{\epsilon }}_n^e+𝖢:({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n^e-{\mathit{\epsilon }}_n^p)=𝖢:({\mathit{\epsilon }}_{n+1}-{\mathit{\epsilon }}_n^p)$

Therefore,

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

Hence shown.