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

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