Continuum Mechanics: Finding Plastic Strain

[lanew]

Homework Statement

http://img683.imageshack.us/img683/7060/selection001l.png

Homework Equations

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

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

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

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}<br /> \epsilon_a^{pl} &amp;= \epsilon_a - \frac{\bar{\sigma}}{E} \\<br /> &amp;= (0.10) - \frac{(66.1)}{(200*10^3)} \\<br /> &amp;= 0.09967<br /> \end{align}

and

\begin{align}<br /> \epsilon_w^{pl} &amp;= \epsilon_w - \frac{\bar{\sigma}}{E} \\<br /> &amp;= (-0.042) - \frac{(66.1)}{(200*10^3)} \\<br /> &amp;= -0.04233<br /> \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?

 

[afreiden]

\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}.

I'm not too familiar with this stuff, but I'm assuming that your table of strains and stresses are all beyond yield, and the higher strain rates are supposed to correlate with stronger material behavior (typos in the table?). Figured I'd take a look since you have no replies, but I'm no plasticity expert.