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

Problem 1: Part 1: Evolution rule for plastic flow
Let $$\alpha$$ be the equivalent plastic strain, defined as

\alpha := \sqrt{\cfrac{2}{3}} \lVert\boldsymbol{\varepsilon}^p\rVert_{} \qquad \text{where} \qquad \lVert\mathbf{a}\rVert = \sqrt{\mathbf{a}:\mathbf{a}} ~. $$ Express the time derivative of $$\alpha$$ in terms of $$\dot{\gamma}$$ and $$\partial f/\partial \boldsymbol{\sigma}$$. This is the evolution law for $$\alpha$$.

The time derivative of $$\alpha$$ is given by

\dot{\alpha} = \sqrt{\cfrac{2}{3}}~\frac{\partial }{\partial t}(\sqrt{\boldsymbol{\varepsilon}

^p:\boldsymbol{\varepsilon}^p}) = \sqrt{\cfrac{2}{3}}~\left(\frac{1}{2}\right)~ \left(\cfrac{1}{\sqrt{\boldsymbol{\varepsilon}^p:\boldsymbol{\varepsilon}^p}}\right) \frac{\partial }{\partial t}(\boldsymbol{\varepsilon}^p:\boldsymbol{\varepsilon}^p) = \sqrt{\cfrac{2}{3}}~\left(\frac{1}{2}\right)~ \left(\cfrac{1}{\lVert\boldsymbol{\varepsilon}^p\rVert_{}}\right) \frac{\partial }{\partial t}(\boldsymbol{\varepsilon}^p:\boldsymbol{\varepsilon}^p) $$ Now,

\begin{align} \frac{\partial }{\partial t}(\boldsymbol{\varepsilon}^p:\boldsymbol{\varepsilon}^p) & =

\frac{\partial }{\partial t}( \varepsilon_{11}^p\varepsilon_{11}^p + \varepsilon_{12}^p\varepsilon_{12}^p + \varepsilon_

{13}^p\varepsilon_{13}^p + \varepsilon_{21}^p\varepsilon_{21}^p + \varepsilon_{22}^p\varepsilon_{22}^p + \varepsilon_

{23}^p\varepsilon_{23}^p + \varepsilon_{31}^p\varepsilon_{31}^p + \varepsilon_{32}^p\varepsilon_{32}^p + \varepsilon_

{33}^p\varepsilon_{33}^p)\\ & = 2\varepsilon_{11}^p\frac{\partial \varepsilon_{11}^p}{\partial t} + 2\varepsilon_{12}^p\frac{\partial \varepsilon_{12}^p}{\partial t} + 2\varepsilon_{13}^p\frac{\partial \varepsilon_{13}^p}{\partial t} + 2\varepsilon_{21}^p\frac{\partial \varepsilon_{21}^p}{\partial t} + 2\varepsilon_{22}^p\frac{\partial \varepsilon_{22}^p}{\partial t} + 2\varepsilon_{23}^p\frac{\partial \varepsilon_{23}^p}{\partial t} + 2\varepsilon_{31}^p\frac{\partial \varepsilon_{31}^p}{\partial t} + 2\varepsilon_{32}^p\frac{\partial \varepsilon_{32}^p}{\partial t} + 2\varepsilon_{33}^p\frac{\partial \varepsilon_{33}^p}{\partial t} \\ & = 2\varepsilon_{11}^p\dot{\varepsilon}_{11}^p + 2\varepsilon_{12}^p\dot{\varepsilon}_

{12}^p + 2\varepsilon_{13}^p\dot{\varepsilon}_{13}^p + 2\varepsilon_{21}^p\dot{\varepsilon}_{21}^p + 2\varepsilon_{22}^p\dot{\varepsilon}_{22}^p +

2\varepsilon_{23}^p\dot{\varepsilon}_{23}^p + 2\varepsilon_{31}^p\dot{\varepsilon}_{31}^p + 2\varepsilon_{32}^p\dot{\varepsilon}_{32}^p +

2\varepsilon_{33}^p\dot{\varepsilon}_{33}^p \\ & = 2 \sum_{i=1}^{3}\sum_{j=1}^{3}\varepsilon_{ij}^p\dot{\varepsilon}_{ij}^p = 2

\boldsymbol{\varepsilon}^p:\dot{\boldsymbol{\varepsilon}}^p \end{align} $$ Therefore,

\dot{\alpha} = \sqrt{\cfrac{2}{3}}~\left(\frac{1}{2}\right)~ \left(\cfrac{1}{\lVert\boldsymbol{\varepsilon}^p\rVert_{}}\right) \left(2~\boldsymbol{\varepsilon}^p:\dot{\boldsymbol{\varepsilon}}^p\right) = \sqrt{\cfrac{2}{3}}~\cfrac{\boldsymbol{\varepsilon}^p:\dot{\boldsymbol{\varepsilon}}^p}

{\lVert\boldsymbol{\varepsilon}^p\rVert_{}} $$ Using

\dot{\boldsymbol{\varepsilon}}^p = \dot{\gamma}\frac{\partial f(\boldsymbol

{\sigma},\alpha,T)}{\partial \boldsymbol{\sigma}} $$ we get

\dot{\alpha} = \sqrt{\cfrac{2}{3}}~\dot{\gamma}~\cfrac{\boldsymbol{\varepsilon}^p:\frac{\partial f}

{\partial \boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} \qquad\text{where}\qquad \boldsymbol{\varepsilon}^p = \int_0^t \dot{\boldsymbol{\varepsilon}}^p~dt = \int_0^t \dot{\gamma}\frac{\partial f}{\partial \boldsymbol{\sigma}}~dt ~. $$