# Continuum Mechanics: Finding Plastic Strain

1. Homework Statement

http://img683.imageshack.us/img683/7060/selection001l.png [Broken]

2. Homework Equations

$\epsilon^{pl} = \epsilon - \epsilon^{el}$

$\epsilon^{pl} = \epsilon - \frac{\bar{\sigma}}{E}$

$r = \frac{\epsilon_w}{\epsilon_t}$

3. The Attempt at a Solution

I'm stuck trying to calculate $\bar{\sigma}$. Can I just assume that $\bar{\sigma} = \sigma$ @ 104 s-1? If so, the axial plastic strain is calculated as follows:

\begin{align} \epsilon_a^{pl} &= \epsilon_a - \frac{\bar{\sigma}}{E} \\ &= (0.10) - \frac{(66.1)}{(200*10^3)} \\ &= 0.09967 \end{align}

and

\begin{align} \epsilon_w^{pl} &= \epsilon_w - \frac{\bar{\sigma}}{E} \\ &= (-0.042) - \frac{(66.1)}{(200*10^3)} \\ &= -0.04233 \end{align}

If this is correct I should be able to related the thickness by $v$, correct?

Also, as far as (b) goes, should I be using $\sigma = k \epsilon^n \dot{\epsilon}^m$?

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$\bar{\sigma}$ might be intended to be taken as Von Mises (Von Mises is the only context where I've personally seen $\bar{\sigma}$). So, for this uniaxial stress, you should (if I'm not mistaken) divide the value that you are using for $\bar{\sigma}$ by $\sqrt{3}$.