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{align} \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}^p_n)~\boldsymbol{\mathit{1}} +

2~\mu~\boldsymbol{\varepsilon}^p_n] \\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n)\\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n + \boldsymbol{\varepsilon}^e_n)\\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}}):\boldsymbol{\varepsilon}^e_n + (\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{align} $$ 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{align} \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{align} $$ 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{align} \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{align} $$ Hence

\begin{align} \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{align} $$ This shows that

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