# 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.