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{align} \boldsymbol{\sigma}_{n+1} & = \boldsymbol{\mathsf{C}}:\boldsymbol{\varepsilon}^e_{n+1} = \boldsymbol{\mathsf{C}}:(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_

{n+1}) \\ & = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol

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

{\mathsf{I}}): (\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \sqrt{\cfrac{3}{2}}

~\Delta\gamma~\mathbf{n}_n) \\ & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1})~\boldsymbol{\mathit{1}} - \text{tr}(\boldsymbol{\varepsilon}^p_n)~\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}^p_n - \sqrt{\cfrac

{3}{2}}~\Delta\gamma~\mathbf{n}_n \right] \\ & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol

{\varepsilon}^p_n) - \sqrt{\cfrac{3}{2}}~\Delta\gamma~\text{tr}(\mathbf{n}_n)\right]\boldsymbol{\mathit{1}} + 2~\mu~\left[\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \sqrt{\cfrac

{3}{2}}~\Delta\gamma~\mathbf{n}_n \right] \\ & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol

{\varepsilon}^p_n)\right]\boldsymbol{\mathit{1}} + 2~\mu~\left[\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \sqrt{\cfrac

{3}{2}}~\Delta\gamma~\mathbf{n}_n \right] \qquad~(\text{since}~ \text{tr}(\mathbf{n}) = 0) \end{align} $$ 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{align} \text{tr}(\boldsymbol{\sigma}_{n+1}) & = \lambda~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol

{\varepsilon}^p_n)\right]\text{tr}(\boldsymbol{\mathit{1}}) + 2~\mu~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol{\varepsilon}

^p_n) - \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}^p_n)\right] + 2~\mu~\left[\text{tr}(\boldsymbol{\varepsilon}_{n+1}) - \text{tr}(\boldsymbol{\varepsilon}

^p_n)\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}^p_n) \end{align} $$ Therefore,

\begin{align} \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}^p_n)\right]\boldsymbol{\mathit{1}} + 2~\mu~\left[\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n - \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}^p_n)~\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}^p_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}^p_n)~\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}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}_n - \boldsymbol{\varepsilon}^e_n)~\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}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}_n)~\boldsymbol{\mathit{1}} + \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\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}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\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}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}}\right] - 2~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{n}_n \end{align} $$ The stress-strain relation is

\boldsymbol{\sigma}_n = \lambda~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol

{\mathit{1}} + 2~\mu~\boldsymbol{\varepsilon}^e_n $$ Hence,

\begin{align} \mathbf{s}_n & = \lambda~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}} +

2~\mu~\boldsymbol{\varepsilon}^e_n - \frac{1}{3}~\lambda~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\text{tr}(\boldsymbol{\mathit

{1}})~\boldsymbol{\mathit{1}} - \cfrac{2}{3}~\mu~\text{tr}(\boldsymbol{\varepsilon}^e_n)~\boldsymbol{\mathit{1}} \\ & = 2~\mu\left[\boldsymbol{\varepsilon}^e_n - \frac{1}{3}~\text{tr}(\boldsymbol

{\varepsilon}^e_n)~\boldsymbol{\mathit{1}}\right] \end{align} $$ Plugging into expression for $$\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.