Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 12

Problem 1: Part 12: Trial elastic stress
In the radial return algorithm, we define a trial elastic state as
 * $$\text{(2)} \qquad

\boldsymbol{\sigma}_{n+1}^{\text{trial}} = \boldsymbol{\sigma}_n + \boldsymbol{\mathsf{C}}: (\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n) $$ where $$\boldsymbol{\sigma}_n, \boldsymbol{\varepsilon}_n$$ are the stress and strain at $$t = t_n$$ and $$\boldsymbol{\sigma}_{n+1}, \boldsymbol {\varepsilon}_{n+1}$$ are the values at $$t = t_{n+1}$$. Show that, if the elastic response of the material is linear, equation (2) can be written as
 * $$\text{(3)} \qquad

\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] ~. $$

From equation (2)

\boldsymbol{\sigma}_{n+1}^{\text{trial}} = \boldsymbol{\sigma}_n + \boldsymbol{\mathsf{C}}:

(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_n) = \boldsymbol{\mathsf{C}}:\boldsymbol{\varepsilon}^e_n + \boldsymbol{\mathsf{C}}:

(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^e_n - \boldsymbol{\varepsilon}

^p_n) = \boldsymbol{\mathsf{C}}:(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n) $$ Therefore,

{ \boldsymbol{\sigma}_{n+1}^{\text{trial}} = = (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol

{\mathsf{I}}):(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}^p_n) = \lambda~\text{tr}(\boldsymbol{\varepsilon}_{n+1}) + 2~\mu~\boldsymbol{\varepsilon}_{n+1} - \lambda~\text{tr}(\boldsymbol{\varepsilon}^p_n) - 2~\mu~\boldsymbol{\varepsilon}^p_n~. } $$ Hence shown.