Homework #8 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.

We've talked about how a normal component of a Green strain tensor
can never be less than 0.5, no matter how much an object is compressed.
But if you calculate the principal strains of the following Green
strain tensor, you will find that one of them is
less than 0.5. So what's going on here?
\[
{\bf E} =
\left[ \matrix{
0.10 & 0.60 & 0.00 \\
0.60 & 0.00 & 0.00 \\
0.00 & 0.00 & 0.00
} \right]
\]
Start with the principal values. They are 0.652 and 0.552.
This confirms that one of the principal values in this Green strain
tensor is indeed less than 0.50.
Since \({\bf E} = {1 \over 2}({\bf F}^T \cdot {\bf F}  {\bf I})\), then
\(({\bf F}^T \cdot {\bf F} = 2{\bf E} + {\bf I})\). Doing this gives
\[
{\bf F}^T \cdot {\bf F} =
\left[ \matrix{
1.20 & 1.20 & 0.00 \\
1.20 & 1.00 & 0.00 \\
0.00 & 0.00 & 1.00
} \right]
\]
We don't care about rotations, so just assume they're zero. This means that
\({\bf F} = {\bf U}\) for this problem. So we are back to the usual polar
decomposition problem of needing to find \({\bf U}\), given
\[
{\bf F}^T \cdot {\bf F} =
{\bf U}^T \cdot {\bf U} =
\left[ \matrix{
1.20 & 1.20 & 0.00 \\
1.20 & 1.00 & 0.00 \\
0.00 & 0.00 & 1.00
} \right]
\]
Taking the first step of solving for \({\bf U}\) by calculating
the principal values of \({\bf U}^T \cdot {\bf U}\) gives
\[
({\bf U}^T \cdot {\bf U})' =
\left[ \matrix{
2.304 & 0 & 0 \\
0 & 0.104 & 0 \\
0 & 0 & 1.00
} \right]
\]
The next step would be to take the square root of the diagonal values,
but this is impossible because the (2,2) slot is negative.
This is critical because a negative principal value in \({\bf U}_{princ}\),
(actually \({\bf U}^2_{princ}\)) means that the material has turned inside out!

A 2D problem: Take a square with corners at (0,0) and (1,1) and apply
the following coordinate mapping to it.
\[
\begin{eqnarray}
x & = & 0.530 X  0.530 Y + 2 \\
\\
y & = & 0.884 X + 0.884 Y + 1
\end{eqnarray}
\]
Do the following:
 Sketch the undeformed square, and also its deformed shape.
 Calculate principal strains and figure out their orientation.
(It's up to you to pick a strain tensor.)
The deformation gradient is
\[
{\bf F} =
\left[ \matrix{
0.530 & 0.530 \\
0.884 & \;\;\;0.884
} \right]
\]
Calcuate the Green Strain...
\[
{\bf E} =
{1 \over 2} \left( {\bf F}^T \! \cdot {\bf F}  {\bf I} \right) =
\left[ \matrix{
0.031 & 0.250 \\
0.250 & 0.031
} \right]
\]
The principal strains and \({\bf Q}\) matrix are
The principal orientation is
\[
\begin{eqnarray}
\theta & = & \text{Sin}^{1}(Q_{12}) \\
\\
& = & \text{Sin}^{1}(0.7071) \\
\\
& = & 45^\circ
\end{eqnarray}
\]
which means it is 45° from the Xaxis because it is based on
the Green strain, which is "undeformed."
 Sketch an inscribed undeformed square and deformed rectangle
inside the deformed square, and do so at the proper orientation.
(See the red square and rectangle in the figure near the top
of the principal strain page
for an example.)
We need to know how much rigid body rotation is present because this
needs to be added to the 45°: angle above to get the final
deformed orienation. (This is the key insight here.)
So do a polar decomposition...
The rigid body rotation is
\[
\begin{eqnarray}
\theta & = & \text{Sin}^{1}(R_{21}) \\
\\
& = & \text{Sin}^{1}(0.7071) \\
\\
& = & 45^\circ
\end{eqnarray}
\]
So the final orientation of the principal directions are 45° + 45° = 90°,
and then another 90° from that for the second principal strain.
 Is the principal direction for your strain tensor consistent with your
sketch? Why or why not?
It is only when you add in the rigid body rotation to the Green strain
principal directions.
Note  An \({\bf F} = {\bf V}\cdot{\bf R}\) polar decomposition also gives some insight
into this problem. Doing so gives
\[
{\bf V} =
\left[ \matrix{
0.75 & 0.00 & \;\;0 \\
0.00 & 1.25 & \;\;0 \\
0 & 0 & \;\;1
} \right]
\qquad
{\bf R} =
\left[ \matrix{
0.7071 & 0.7071 & \;\;0 \\
0.7071 & \;\;\;0.7071 & \;\;0 \\
0 & 0 & \;\;1
} \right]
\]
The \({\bf V}\) matrix shows that the X and Y directions are the principal directions
because there are no shears present. This is consistent with the above conclusion
that the principal directions are at 45° + 45° = 90° for the 1st principal
direction, and then another 90° from that for the 2nd.

For this strain tensor
\[
{\bf E} =
\left[ \matrix{
0.50 & \;\;\; 0.30 & \;\;\;0.20 \\
0.30 & 0.20 & \;\;\; 0.10 \\
0.20 & \;\;\;0.10 & 0.10
} \right]
\]
 Calculate the 3 strain invariants.
Its invariants are
\[
\begin{eqnarray}
I_1 & = & 0.50 + (0.20) + (0.10) \\
\\
& = & 0.20
\\
\\
I_2 & = & (0.50)(0.20) + (0.20)(0.10) + (0.10)(0.50)  (0.30)^2  (0.20)^2  (0.10)^2 \\
\\
& = & 0.27
\\
\\
I_3 & = & \text{det}({\bf E}) \\
\\
& = & 0.034
\end{eqnarray}
\]
 Calculate the principal strains and their orientation. And use these
to recalculate the invariants and confirm that you get the same answers.
Calculate the principal strains
The principal strain tensor is
\[
{\bf E}_P =
\left[ \matrix{
0.6748 & \;\;\;0.0 & \;\;\;0.0 \\
0.0 & 0.3147 & \;\;\;0.0 \\
0.0 & \;\;\;0.0 & 0.1601
} \right]
\]
And the "new" invariants are
\[
\begin{eqnarray}
I_1 & = & 0.6748 + (0.3147) + (0.1601) \\
\\
& = & 0.20
\\
\\
I_2 & = & (0.6748)(0.3147) + (0.3147)(0.1601) + (0.1601)(0.6748) \\
\\
& = & 0.27
\\
\\
I_3 & = & \text{det}({\bf E}) = (0.6748)(0.3147)(0.1601) \\
\\
& = & 0.034
\end{eqnarray}
\]
So they are indeed the same.