# Homework #13 Solutions

1. 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

${\buildrel \nabla \over {\boldsymbol{\sigma}}} = - \text{tr}({\bf D}) \; \boldsymbol{\sigma} + {1 \over J} \, {\bf F} \cdot \dot{\boldsymbol{\sigma}}^\text{PK2} \cdot {\bf F}^T$

2. 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.

$\sigma^\text{Eng} = 2 \, C_{10} \left( \lambda - {1 \over \lambda^2 } \right)$

$W = C_{10} \left( \lambda_1^2 + \lambda_2^2 + \lambda_3^2 - 3 \right)$
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

$\sigma^\text{Eng} = 2 \, C_{10} \left( \lambda_1 - {1 \over \lambda_1^2 } \right)$

4. 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.