Homework #13 Solutions


  1. Given \(E = 10\text{ MPa}\) and \(\nu = 0.5\), Create a graph with three separate curves (lines, in fact) of max principal stress versus max principal strain for three cases: (i) uniaxial tension, (ii) plane strain tension (shear), (iii) and equibiaxial tension.

    Plot the graphs for strains up to \(\epsilon = 0.10\), (that's 10%). Don't worry about finite strain effects, or which definitions of stress and strain to use in this case. Use the following page for definitions of the three cases: specialstraintopics.html.


    The problem with using Hooke's Law as a function of strain is that \(\nu = 0.5\), so the last term here becomes indeterminate.

    \[ \sigma_{ij} = {E \over (1 + \nu)} \left[ \epsilon_{ij} + { \nu \over (1 - 2 \nu)} \delta_{ij} \epsilon_{kk} \right] \]
    Instead one must use it in the following form

    \[ \epsilon_{ij} = {1 \over E} \left[ (1 + \nu) \sigma_{ij} - \nu \delta_{ij} \sigma_{kk} \right] \]
    Uniaxial tension is the easiest. It is simply

    \[ \sigma = E \, \epsilon \qquad \qquad \text{(uniaxial tension)} \]
    Equibiaxial tension is the 2nd easiest. Set \(\epsilon_{11} = \epsilon\), \(\sigma_{11} = \sigma_{22} = \sigma\), and \(\sigma_{33} = 0\).

    \[ \epsilon = {1 \over E} \left[ (1 + \nu) \sigma - 2 \, \nu \sigma \right] \]
    This reduces to

    \[ \sigma = { E \over 1 - \nu} \epsilon \qquad \qquad \text{(equibiaxial tension)} \]
    Shear is the trickiest by far. Recall that it can be thought of as plane strain tension. Referring to the sketch on this page: specialstraintopics.html, note that \(\epsilon_{22} = 0\) and \(\sigma_{33} = 0\).

    Starting with

    \[ \epsilon_{22} = {1 \over E} \left[ \sigma_{22} - \nu ( \sigma_{11} + \sigma_{33} ) \right] \]
    Set \(\epsilon_{22} = 0\) and \(\sigma_{33} = 0\), and this leads to

    \[ \sigma_{22} = \nu \sigma_{11} \]
    Substitute this into the Hooke's Law equation for \(\epsilon_{11}\) to get

    \[ \epsilon = {1 \over E} \left[ \sigma - \nu^2 \sigma \right] \]
    Solving for stress gives

    \[ \sigma = { E \over 1 - \nu^2} \epsilon \qquad \qquad \text{(shear)} \]



  2. Plot the max principal engineering stress versus max principal engineering strain, up to \(\epsilon_\text{max} = 0.50\), for uniaxial tension, shear, and equibiaxial tension, for a material having (\( C_{10} = 0.4 \) and \( C_{01} = 0.04 \) ).

    Yes, this question is basically the same as #1 above, but with Mooney-Rivlin terms instead.