Homework #13 Solutions

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

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 MooneyRivlin terms instead.