Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 14

Problem 1: Part 14: Return mapping

Show that

$\mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-2~\mu ~\Delta \gamma ~\mathbf {n} _{n}~.$

We have

${\begin{aligned}{\boldsymbol {\sigma }}_{n+1}&={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}_{n+1}^{e}={\boldsymbol {\mathsf {C}}}:({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n+1}^{p})\\&=(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n+1}^{p})\\&=(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\mathsf {I}}}):({\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n})\\&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})~{\boldsymbol {\mathit {1}}}-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})~{\boldsymbol {\mathit {1}}}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~{\text{tr}}(\mathbf {n} _{n})~{\boldsymbol {\mathit {1}}}\right]+2~\mu ~\left[{\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\right]\\&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~{\text{tr}}(\mathbf {n} _{n})\right]{\boldsymbol {\mathit {1}}}+2~\mu ~\left[{\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\right]\\&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]{\boldsymbol {\mathit {1}}}+2~\mu ~\left[{\boldsymbol {\varepsilon }}_{n+1}-{\boldsymbol {\varepsilon }}_{n}^{p}-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}\right]\qquad ~({\text{since}}~{\text{tr}}(\mathbf {n} )=0)\end{aligned}}$

Now

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

The trace of the stress is given by

${\begin{aligned}{\text{tr}}({\boldsymbol {\sigma }}_{n+1})&=\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]{\text{tr}}({\boldsymbol {\mathit {1}}})+2~\mu ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})-{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~{\text{tr}}(\mathbf {n} _{n})\right]\\&=3~\lambda ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]+2~\mu ~\left[{\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\right]\qquad ~({\text{since}}~{\text{tr}}(\mathbf {n} )=0)\\&=(3~\lambda +2~\mu ){\text{tr}}({\boldsymbol {\varepsilon }}_{n+1})-(3~\lambda +2~\mu ){\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{p})\end{aligned}}$

Therefore,

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

The stress-strain relation is

${\boldsymbol {\sigma }}_{n}=\lambda ~{\text{tr}}({\boldsymbol {\varepsilon }}_{n}^{e})~{\boldsymbol {\mathit {1}}}+2~\mu ~{\boldsymbol {\varepsilon }}_{n}^{e}$

Hence,

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

Plugging into expression for $\mathbf {s} _{n+1}$, we get

$\mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-\mathbf {s} _{n}+\mathbf {s} _{n}-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}$

Therefore,

${\mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-2~\mu ~{\sqrt {\cfrac {3}{2}}}~\Delta \gamma ~\mathbf {n} _{n}}$

Remark: If we write the yield function as

$f={\sqrt {\mathbf {s} :\mathbf {s} }}-{\sqrt {\cfrac {2}{3}}}~\sigma _{y}~.$

then the above equation takes the form

${\mathbf {s} _{n+1}=\mathbf {s} _{n+1}^{\text{trial}}-2~\mu ~\Delta \gamma ~\mathbf {n} _{n}}$

These are equivalent.

Nonlinear finite elements/Homework11/Solutions/Problem 1