Homework #6 Solutions
Feel free to use
http://www.continuummechanics.org/interactivecalcs.html when applicable.
Use the following deformation gradient for Problems 1-3.
\[
{\bf F} =
\left[ \matrix{
\;\;\; 1.5 & \;\;\; 0.3 & -0.2 \\
-0.1 & \;\;\;1.2 & \;\;\; 0.1 \\
\;\;\; 0.3 & -0.2 & \;\;\; 1.1
} \right]
\]
-
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).
Then check your result using this page:
http://www.continuummechanics.org/cm/techforms/RUDecomposition.html
\[
{\bf F}^T \cdot {\bf F} =
\left[ \matrix{
2.35 & \;\;\;0.27 & \;\;\;0.02 \\
0.27 & \;\;\;1.57 & -0.16 \\
0.02 & -0.16 & \;\;\; 1.26
} \right]
= {\bf U}^T \cdot {\bf U}
\]
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
\[
\begin{eqnarray}
{\bf R} & = & {\bf F} \cdot {\bf U}^{-1} & = &
\left[ \matrix{
\;\;\; 1.5 & \;\;\; 0.3 & -0.2 \\
-0.1 & \;\;\;1.2 & \;\;\; 0.1 \\
\;\;\; 0.3 & -0.2 & \;\;\; 1.1
} \right]
\left[ \matrix{
\;\;\;0.6572 & -0.0518 & -0.0090 \\
-0.0517 & \;\;\;0.8084 & \;\;\;0.0491 \\
-0.0090 & \;\;\;0.0491 & \;\;\;0.8887
} \right] \\
\\
& & & = &
\left[ \matrix{
\;\;\;0.9721 & \;\;\;0.1550 & -0.1765 \\
-0.1287 & \;\;\;0.9802 & \;\;\;0.1487 \\
\;\;\;0.1976 & -0.1232 & \;\;\;0.9651
} \right]
\end{eqnarray}
\]
-
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
\[
{\bf p} = (-0.499, -0.687, -0.521)
\]
-
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.
\[
{\bf V} = {\bf R} \cdot {\bf U} \cdot {\bf R}^T
\]
\[
\begin{eqnarray}
{\bf V} & = &
\left[ \matrix{
\;\;\;0.9718 & \;\;\;0.1549 & -0.1778 \\
-0.1288 & \;\;\;0.9803 & \;\;\;0.1499 \\
\;\;\;0.1975 & -0.1228 & \;\;\;0.9726
} \right]
\left[ \matrix{
1.5298 & \;\;\;0.0975 & \;\;\;0.0100 \\
0.0975 & \;\;\;1.2473 & -0.0680 \\
0.0100 & -0.0680 & \;\;\;1.1204
} \right]
\left[ \matrix{
\;\;\;0.9718 & -0.1288 & \;\;\;0.1975 \\
\;\;\;0.1549 & \;\;\;0.9803 & -0.1228 \\
-0.1778 & \;\;\;0.1499 & \;\;\;0.9726
} \right] \\
\\
\\
& = &
\left[ \matrix{
1.5397 & \;\;\;0.0709 & \;\;\;0.0649 \\
0.0709 & \;\;\;1.2042 & -0.0699 \\
0.0649 & -0.0698 & \;\;\;1.1537
} \right]
\end{eqnarray}
\]