# Homework #12 Solutions

Reminder - you're going to need these webpages: http://www.continuummechanics.org/techforms/index.html to do this homework.... unless you prefer to use Matlab, Mathematica, etc. Don't bother doing anything by hand anymore. Take advantage of software programs to do all the matrix multiplication and other procedures when you have a chance.

Given the following stress tensor

$\boldsymbol{\sigma} = \left[ \matrix{ 10 & 20 & 30 \\ 20 & 40 & 50 \\ 30 & 50 & 60 } \right]$
1. What is the value of the von Mises stress?

The hydrostatic stress is $$(10 + 40 + 60) / 3 = 36.67$$. Subtract this from the diagonal components to get the deviatoric stress.

$\boldsymbol{\sigma}' = \left[ \matrix{ -26.67 & 20 & 30 \\ \;\;\;20 & 3.333 & 50 \\ \;\;\;30 & 50 & 23.33 } \right]$

$\boldsymbol{\sigma}' : \boldsymbol{\sigma}' \; = \; -26.67^2 + 3.333^2 + 23.33^2 + 2 ( 20^2 + 30^2 + 50^2) \; = \; 8,867$

$\sigma_{VM} \; = \; \sqrt{{3 \over 2} \boldsymbol{\sigma}' : \boldsymbol{\sigma}' } \; = \; 115.3$

2. Propose 2 other stress tensors that will have the same von Mises stress?

$\left[ \matrix{ 110 & 20 & 30 \\ 20 & 140 & 50 \\ 30 & 50 & 160 } \right] \qquad \quad \text{and} \qquad \quad \left[ \matrix{ 115.3 & 0 & \;\;0 \\ 0 & 0 & \;\;0 \\ 0 & 0 & \;\;0 } \right]$

3. Do all stress tensors having the same von Mises stress also have the same principal stresses?

No because the von Mises stress in terms of principal stresses is

$\sigma_\text{VM} = \sqrt{{1\over 2}\left[\left(\sigma_1 - \sigma_2\right)^2 + \left(\sigma_2 - \sigma_3\right)^2 + \left(\sigma_3 - \sigma_1\right)^2 \right] }$
So there is only 1 equation here, but 3 variables, so there is an infinite number of combinations of principal stresses that can give the same von Mises stress.

4. Do all stress tensors having the same principal stresses also have the same von Mises stress?

Yes, because of the equation above.

5. 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

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