# Homework #6 Solutions

Feel free to use http://www.continuummechanics.org/interactivecalcs.html when applicable.

1. For $$E = 10 \, \text{MPa}$$, $$\nu = 0.333$$, and tension in the 1-direction such that $$\epsilon_{11} = 0.1$$, $$\epsilon_{22} = -0.0333$$, and $$\epsilon_{33} = -0.0333$$...

Calculate $$C_{1111}$$, $$C_{1122}$$, and $$C_{1133}$$ and use them with the strains to calculate $$\sigma_{11}$$.

It should simply equal $$\sigma_{11} = 1 \, \text{MPa}$$ because this satisfies $$\sigma_{11} = E \epsilon_{11}$$ for this case of uniaxial tension.

Start with the the equation for $$C_{ijkl}$$.

$C_{ijkl} = {E \over (1 + \nu)} \left[ {1 \over 2} ( \delta_{ik} \, \delta_{jl} + \delta_{jk} \, \delta_{il} ) + { \nu \over (1 - 2 \nu)} \delta_{ij} \, \delta_{kl} \right]$
and insert values for $$E$$, $$\nu$$, and the subscripts to get

$\begin{eqnarray} C_{1111} & = & {E \over (1 + \nu)} \left[ {1 \over 2} ( \delta_{11} \, \delta_{11} + \delta_{11} \, \delta_{11} ) + { \nu \over (1 - 2 \nu)} \delta_{11} \, \delta_{11} \right] \\ \\ \\ & = & {E \over (1 + \nu)} \left[ 1 + { \nu \over (1 - 2 \nu)} \right] \\ \\ \\ & = & {E \, ( 1 - \nu) \over (1 + \nu) (1 - 2 \nu) } \\ \\ \\ & = & {10 \, ( 1 - 0.333) \over (1 + 0.333) (1 - 2 * 0.333) } \\ \\ \\ \\ & = & 14.981 \, \text {MPa} \end{eqnarray}$
$\begin{eqnarray} C_{1122} = C_{1133} & = & {E \over (1 + \nu)} \left[ {1 \over 2} ( \delta_{12} \, \delta_{12} + \delta_{12} \, \delta_{12} ) + { \nu \over (1 - 2 \nu)} \delta_{11} \, \delta_{22} \right] \\ \\ \\ & = & {E \, \nu \over (1 + \nu) (1 - 2 \nu) } \\ \\ \\ & = & {10 \, ( 0.333) \over (1 + 0.333) (1 - 2 * 0.333) } \\ \\ \\ \\ & = & 7.491 \, \text {MPa} \end{eqnarray}$
and since all strain components are zero except $$\epsilon_{11}$$, $$\epsilon_{22}$$, and $$\epsilon_{33}$$

$\begin{eqnarray} \sigma_{11} & = & C_{1111} \, \epsilon_{11} + C_{1122} \, \epsilon_{22} + C_{1133} \, \epsilon_{33} \\ \\ \\ & = & ( 14.981 \, \text{MPa} ) ( 0.100 ) + ( 7.491 \, \text{MPa} ) ( -0.0333 ) + ( 7.491 \, \text{MPa} ) ( -0.0333 ) \\ \\ \\ & = & 1.00 \, \text{MPa} \end{eqnarray}$

Use the following deformation gradient for Problems 2-4.

${\bf F} = \left[ \matrix{ \;\;\; 1.5 & \;\;\; 0.3 & -0.2 \\ -0.1 & \;\;\;1.2 & \;\;\; 0.1 \\ \;\;\; 0.3 & -0.2 & \;\;\; 1.1 } \right]$
1. Determine $${\bf R}$$ and $${\bf U}$$ in $${\bf F} = {\bf R} \cdot {\bf U}$$.
Do it the hard way by multiplying all the matrices out yourself (but not by hand! use software).

${\bf F}^T \cdot {\bf F} = \left[ \matrix{ 2.35 & \;\;\;0.27 & \;\;\;0.02 \\ 0.27 & \;\;\;1.57 & -0.16 \\ 0.02 & -0.16 & \;\;\; 1.26 } \right] = {\bf U}^T \cdot {\bf U}$
The principal values of $${\bf U}^T \cdot {\bf U}$$ are

$({\bf U}^T \cdot {\bf U})' = \left[ \matrix{ 2.4351 & 0 & 0 \\ 0 & 1.5673 & 0 \\ 0 & 0 & 1.1776 } \right]$
and the transformation matrix is

${\bf Q} = \left[ \matrix{ \;\;\;0.9530 & 0.3020 & -0.0249 \\ -0.2795 & 0.8441 & -0.4576 \\ -0.1172 & 0.4431 & \;\;\;0.8888 } \right]$

Take the square root of $$({\bf U}^T \cdot {\bf U})'$$ to get $${\bf U}'$$.

${\bf U}' = \left[ \matrix{ 1.5605 & 0 & 0 \\ 0 & 1.2519 & 0 \\ 0 & 0 & 1.0852 } \right]$
And rotate $${\bf U}'$$ back by using

${\bf U} = {\bf Q}^T \cdot {\bf U}' \cdot {\bf Q}$

$\begin{eqnarray} {\bf U} & = & \left[ \matrix{ \;\;\;0.9530 & -0.2795 & -0.1172 \\ \;\;\;0.3020 & \;\;\;0.8441 & \;\;\;0.4431 \\ -0.0249 & -0.4576 & \;\;\;0.8888 } \right] \left[ \matrix{ 1.5605 & 0 & 0 \\ 0 & 1.2519 & 0 \\ 0 & 0 & 1.0852 } \right] \left[ \matrix{ \;\;\;0.9530 & 0.3020 & -0.0249 \\ -0.2795 & 0.8441 & -0.4576 \\ -0.1172 & 0.4431 & \;\;\;0.8888 } \right] \\ \\ \\ & = & \left[ \matrix{ 1.5295 & \;\;\;0.0973 & \;\;\;0.0101 \\ 0.0973 & \;\;\;1.2473 & -0.0679 \\ 0.0101 & -0.0679 & \;\;\;1.1291 } \right] \end{eqnarray}$
The inverse of $${\bf U}$$ is

${\bf U}^{-1} = \left[ \matrix{ 0.6572 & -0.0518 & -0.0090 \\ -0.0517 & 0.8084 & 0.0491 \\ -0.0090 & 0.0491 & 0.8887 } \right]$
and the rotation matrix is

$\begin{eqnarray} {\bf R} & = & {\bf F} \cdot {\bf U}^{-1} & = & \left[ \matrix{ \;\;\; 1.5 & \;\;\; 0.3 & -0.2 \\ -0.1 & \;\;\;1.2 & \;\;\; 0.1 \\ \;\;\; 0.3 & -0.2 & \;\;\; 1.1 } \right] \left[ \matrix{ \;\;\;0.6572 & -0.0518 & -0.0090 \\ -0.0517 & \;\;\;0.8084 & \;\;\;0.0491 \\ -0.0090 & \;\;\;0.0491 & \;\;\;0.8887 } \right] \\ \\ & & & = & \left[ \matrix{ \;\;\;0.9721 & \;\;\;0.1550 & -0.1765 \\ -0.1287 & \;\;\;0.9802 & \;\;\;0.1487 \\ \;\;\;0.1976 & -0.1232 & \;\;\;0.9651 } \right] \end{eqnarray}$

2. How many degrees of rigid body rotation are in this deformation gradient, and what is the axis of rotation?

$\begin{eqnarray} \cos \alpha & = & \frac{1}{2} \left( \text{tr}({\bf R}) - 1 \right) \\ \\ \\ & = & \frac{1}{2} (0.9718 + 0.9803 + 0.9726 - 1) \\ \\ \\ \alpha & = & 15.8^\circ \end{eqnarray}$
The unit vector for the axis of rotation is

$p_1 = { R_{32} - R_{23} \over 2 \sin \alpha } \qquad \qquad p_2 = { R_{13} - R_{31} \over 2 \sin \alpha } \qquad \qquad p_3 = { R_{21} - R_{12} \over 2 \sin \alpha }$
Inserting values gives

$p_1 = { \text{-}0.1232 - 0.1487 \over 2 \sin 15.8^\circ } \qquad \qquad p_2 = { \text{-}0.1765 - 0.1976 \over 2 \sin 15.8^\circ } \qquad \qquad p_3 = { \text{-}0.1287 - 0.1550 \over 2 \sin 15.8^\circ }$
The axis is

${\bf p} = (-0.499, -0.687, -0.521)$

3. Now that you have $${\bf R}$$ and $${\bf U}$$ from #1, calculate $${\bf V}$$ and use http://www.continuummechanics.org/cm/techforms/VRDecomposition.html to check that you got it right.

${\bf V} = {\bf R} \cdot {\bf U} \cdot {\bf R}^T$
$\begin{eqnarray} {\bf V} & = & \left[ \matrix{ \;\;\;0.9718 & \;\;\;0.1549 & -0.1778 \\ -0.1288 & \;\;\;0.9803 & \;\;\;0.1499 \\ \;\;\;0.1975 & -0.1228 & \;\;\;0.9726 } \right] \left[ \matrix{ 1.5298 & \;\;\;0.0975 & \;\;\;0.0100 \\ 0.0975 & \;\;\;1.2473 & -0.0680 \\ 0.0100 & -0.0680 & \;\;\;1.1204 } \right] \left[ \matrix{ \;\;\;0.9718 & -0.1288 & \;\;\;0.1975 \\ \;\;\;0.1549 & \;\;\;0.9803 & -0.1228 \\ -0.1778 & \;\;\;0.1499 & \;\;\;0.9726 } \right] \\ \\ \\ & = & \left[ \matrix{ 1.5397 & \;\;\;0.0709 & \;\;\;0.0649 \\ 0.0709 & \;\;\;1.2042 & -0.0699 \\ 0.0649 & -0.0698 & \;\;\;1.1537 } \right] \end{eqnarray}$