Homework #13 Solutions
- Start with the relationship between Caucy stress and 2nd Piola-Kirchhoff stress
and take the time derivative to figure out the relationship between
\( \dot{\boldsymbol{\sigma}} \) and \( \dot{\boldsymbol{\sigma}}^\text{PK2} \).
Hint - You will get a Lie derivative along the way.
\[
\boldsymbol{\sigma} = {1 \over J} \, {\bf F} \cdot \boldsymbol{\sigma}^\text{PK2} \cdot {\bf F}^T
\]
Take the time derivative
\[
\dot{\boldsymbol{\sigma}} = - \left( {\dot J \over J^2} \right) \, {\bf F} \cdot \boldsymbol{\sigma}^\text{PK2} \cdot {\bf F}^T +
{1 \over J} \, \dot{\bf F} \cdot \boldsymbol{\sigma}^\text{PK2} \cdot {\bf F}^T +
{1 \over J} \, {\bf F} \cdot \dot {\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T +
{1 \over J} \, {\bf F} \cdot \boldsymbol{\sigma}^\text{PK2} \cdot \dot{\bf F}^T
\]
And substitute in several identities
\[
\text{tr}({\bf D}) = {\dot J \over J} \quad \qquad \dot{\bf F} = {\bf L} \cdot {\bf F} \qquad \quad
\dot{\bf F}^T = {\bf F}^T \cdot {\bf L}^T
\]
to get
\[
\dot{\boldsymbol{\sigma}}
= - \text{tr}({\bf D}) {1 \over J} {\bf F} \cdot {\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T +
{1 \over J} \, {\bf L} \cdot {\bf F} \cdot {\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T +
{1 \over J} \, {\bf F} \cdot \dot{\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T +
{1 \over J} \, {\bf F} \cdot {\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T \cdot {\bf L}^T
\]
This big equation can be shortened by recognizing that many of the terms contain
\({1 \over J} \, {\bf F} \cdot{\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T\),
which is just the Cauchy stress, \(\boldsymbol{\sigma}\).
\[
\dot{\boldsymbol{\sigma}}
= - \text{tr}({\bf D}) \; \boldsymbol{\sigma} +
{\bf L} \cdot \boldsymbol{\sigma} +
\boldsymbol{\sigma} \cdot {\bf L}^T +
{1 \over J} \, {\bf F} \cdot \dot{\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T
\]
This is the relationship between the time derivative of the 2nd Piola-Kirchhoff stress,
\(\dot{\boldsymbol{\sigma}}^\text{PK2}\),
and the Cauchy stress, \(\dot{\boldsymbol{\sigma}}\).
After a little rearrangement...
\[
\dot{\boldsymbol{\sigma}} - {\bf L} \cdot \boldsymbol{\sigma} - \boldsymbol{\sigma} \cdot {\bf L}^T
= - \text{tr}({\bf D}) \; \boldsymbol{\sigma} +
{1 \over J} \, {\bf F} \cdot \dot{\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T
\]
And the left hand side is the Lie derivative, so the equation can be written more compactly as
\[
{\buildrel \nabla \over {\boldsymbol{\sigma}}}
= - \text{tr}({\bf D}) \; \boldsymbol{\sigma} +
{1 \over J} \, {\bf F} \cdot \dot{\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T
\]
-
Start with this equation for strain energy from the
Mooney-Rivlin page
\[
W = C_{10} \left( I_1 - 3 \right)
\]
and show how to get to the following equation for uniaxial tension.
\[
\sigma^\text{Eng} = 2 \, C_{10} \left( \lambda - {1 \over \lambda^2 } \right)
\]
Start with the invariant expanded out.
\[
W = C_{10} \left( \lambda_1^2 + \lambda_2^2 + \lambda_3^2 - 3 \right)
\]
Substitute
\[
\lambda_3 = {1 \over \lambda_1 \lambda_2}
\]
Substitute
\[
W = C_{10} \left( \lambda_1^2 + \lambda_2^2 + {1 \over \lambda_1^2 \lambda_2^2} - 3 \right)
\]
Take the derivative with respect to \(\lambda_1\).
\[
\sigma^\text{Eng} = {\partial W \over \partial \lambda_1} =
2 \, C_{10} \left( \lambda_1 - {1 \over \lambda_1^3 \lambda_2^2} \right)
\]
For uniaxial tension, substitute \(\lambda_2 = 1 / \sqrt{\lambda_1}\) to get
\[
\sigma^\text{Eng} =
2 \, C_{10} \left( \lambda_1 - {1 \over \lambda_1^2 } \right)
\]
-
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.