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Continuum Mechanics: Finding Plastic Strain

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data

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


    2. Relevant equations

    [itex]\epsilon^{pl} = \epsilon - \epsilon^{el}[/itex]

    [itex]\epsilon^{pl} = \epsilon - \frac{\bar{\sigma}}{E}[/itex]

    [itex]r = \frac{\epsilon_w}{\epsilon_t}[/itex]


    3. The attempt at a solution

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

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

    and

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

    If this is correct I should be able to related the thickness by [itex]v[/itex], correct?

    Also, as far as (b) goes, should I be using [itex]\sigma = k \epsilon^n \dot{\epsilon}^m[/itex]?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 20, 2011 #2
    [itex]\bar{\sigma}[/itex] might be intended to be taken as Von Mises (Von Mises is the only context where I've personally seen [itex]\bar{\sigma}[/itex]). So, for this uniaxial stress, you should (if I'm not mistaken) divide the value that you are using for [itex]\bar{\sigma}[/itex] by [itex]\sqrt{3}[/itex].

    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.
     
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