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

Problem 1: Part 13: Trial elastic stress

Starting from equation (3) show that

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

where $\mathbf {s}$ is the deviatoric part of ${\boldsymbol {\sigma }}$ and $\mathbf {e}$ is the deviatoric part of ${\boldsymbol {\varepsilon }}$.

From equation (3) we have

${\begin{aligned}{\boldsymbol {\sigma }}_{n+1}^{\text{trial}}&=[\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n+1}]-[\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n}^{p}]\\&=(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p})\\&=(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}+{\boldsymbol {\varepsilon }}_{n}^{e})\\&=(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):{\boldsymbol {\varepsilon }}_{n}^{e}+(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n})\\&={\boldsymbol {\sigma }}_{n}+(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n})\end{aligned}}$

The deviatoric parts of the stress and strain are

$\mathbf {s} _{n+1}^{\text{trial}}={\boldsymbol {\sigma }}_{n+1}^{\text{trial}}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }}_{n+1}^{\text{trial}})~{\boldsymbol {\mathit {1}}}~;~~\mathbf {e} _{n+1}={\boldsymbol {\varepsilon }}_{n+1}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}~;~~\mathbf {e} _{n}={\boldsymbol {\varepsilon }}_{n}-{\frac {1}{3}}~{\text{tr}}(\mathbf {e} _{n})~{\boldsymbol {\mathit {1}}}$

Therefore,

${\begin{aligned}\mathbf {s} _{n+1}^{\text{trial}}&={\boldsymbol {\sigma }}_{n+1}^{\text{trial}}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\sigma }}_{n+1}^{\text{trial}})~{\boldsymbol {\mathit {1}}}\\&={\boldsymbol {\sigma }}_{n}+(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n})-{\frac {1}{3}}~{\text{tr}}{{\boldsymbol {\sigma }}_{n}}~{\boldsymbol {\mathit {1}}}-{\frac {1}{3}}~{\text{tr}}[\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n})]~{\boldsymbol {\mathit {1}}}\end{aligned}}$

Now,

$(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n})=\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n+1}-\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})~{\boldsymbol {\mathit {1}}}-2~\mu ~{\boldsymbol {\varepsilon }}_{n}$

Therefore,

${\begin{aligned}{\text{tr}}[(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n})]&=\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\text{tr}}({\boldsymbol {\mathit {1}}})+2~\mu ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})~{\text{tr}}({\boldsymbol {\mathit {1}}})-2~\mu ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})\\&=3~\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})+2~\mu ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-3\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})-2~\mu ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})\end{aligned}}$

Hence

${\begin{aligned}\mathbf {s} _{n+1}^{\text{trial}}=&~{\boldsymbol {\sigma }}_{n}-{\frac {1}{3}}~{\text{tr}}{{\boldsymbol {\sigma }}_{n}}+\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n+1}-\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})~{\boldsymbol {\mathit {1}}}-2~\mu ~{\boldsymbol {\varepsilon }}_{n}\\&~~-\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}-{\cfrac {2}{3}}~\mu ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}+\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})~{\boldsymbol {\mathit {1}}}+{\cfrac {2}{3}}~\mu ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})~{\boldsymbol {\mathit {1}}}\\=&~\mathbf {s} _{n}+2~\mu ~\left({\boldsymbol {\varepsilon }}_{n+1}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}\right)-2~\mu ~\left({\boldsymbol {\varepsilon }}_{n}-{\frac {1}{3}}~{\text{tr}}({\boldsymbol {\varepsilon }}_{n})~{\boldsymbol {\mathit {1}}}\right)\\=&~\mathbf {s} _{n}+2~\mu ~\mathbf {e} _{n+1}-2~\mu ~\mathbf {e} _{n}\end{aligned}}$

This shows that

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

Nonlinear finite elements/Homework11/Solutions/Problem 1