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

${\text{(2)}}\qquad {\boldsymbol {\sigma }}_{n+1}^{\text{trial}}={\boldsymbol {\sigma }}_{n}+{\boldsymbol {\mathsf {C}}}:({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n})$ where ${\boldsymbol {\sigma }}_{n},{\boldsymbol {\varepsilon }}_{n}$ are the stress and strain at $t=t_{n}$ and ${\boldsymbol {\sigma }}_{n+1},{\boldsymbol {\varepsilon }}_{n+1}$ 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

${\text{(3)}}\qquad {\boldsymbol {\sigma }}_{n+1}^{\text{trial}}=[\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n+1}]-[\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n}^{p}]~.$ From equation (2)

${\boldsymbol {\sigma }}_{n+1}^{\text{trial}}={\boldsymbol {\sigma }}_{n}+{\boldsymbol {\mathsf {C}}}:({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n})={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}_{n}^{e}+{\boldsymbol {\mathsf {C}}}:({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{e}-{\boldsymbol {\varepsilon }}_{n}^{p})={\boldsymbol {\mathsf {C}}}:({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p})$ Therefore,

${{\boldsymbol {\sigma }}_{n+1}^{\text{trial}}==(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p})=\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})+2~\mu ~{\boldsymbol {\varepsilon }}_{n+1}-\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})-2~\mu ~{\boldsymbol {\varepsilon }}_{n}^{p}~.}$ Hence shown.