Homework #11 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.

An incompressible rubber object with an initial crosssectional area
of 100 mm^{2} is stretched along the xaxis to twice
its original length by a force of 200 N, and rotated 20°
counter clockwise in the xy plane.
 What is the full 3D Cauchy stress tensor?
If it's incompressible and its length doubles, then its
crosssectional area will decrease by half to
50 mm^{2}. So the Cauchy stress component is
\[
\sigma_{xx} \; = \; {F \over A} \; = \; {200 \text{ N} \over 50 \text{ mm}^2} \; = \; 4 \text{ MPa}
\]
The rotation matrix for a 20° rotation around the zaxis is
\[
{\bf R} \; = \;
\left[ \matrix{
\cos 20^\circ & \sin 20^\circ & 0 \\
\\
\sin 20^\circ & \;\;\;\cos 20^\circ & 0 \\
\\
0 & 0 & 1 }
\right]
\; = \;
\left[ \matrix{
0.940 & 0.342 & 0 \\
\\
0.342 & \;\;0.940 & 0 \\
\\
0 & 0 & 1 }
\right]
\]
The final rotated Cauchy stress is
\[
\begin{eqnarray}
\boldsymbol{\sigma}
& = &
\left[ \matrix{
0.940 & 0.342 & 0 \\
\\
0.342 & \;\;0.940 & 0 \\
\\
0 & 0 & 1 }
\right]
\left[ \matrix{
4 & \;\; 0 \;\; & 0 \\
\\
0 & \;\; 0 \;\;& 0 \\
\\
0 & \;\; 0 \;\;& 0 }
\right]
\left[ \matrix{
\;\;0.940 & 0.342 & 0 \\
\\
0.342 & 0.940 & 0 \\
\\
0 & 0 & 1 }
\right]
\\
\\
\\
\\
& = &
\left[ \matrix{
3.534 & 1.286 & \;\; 0 \\
\\
1.286 & 0.468 & \;\; 0 \\
\\
0 & 0 & \;\; 0 }
\right]
\end{eqnarray}
\]
 What is the full 3D 2nd PiolaKirchhoff stress tensor?
The only nonzero stress component will be in the xxslot
for this example. We've seen that for simple tension,
\( \sigma^{\text{PK2}}_{xx} = \sigma_{xx} /
(1 + \epsilon_{\text{Eng}})^2\). This gives
\[
\sigma^{\text{PK2}}_{xx} \; = \; { 4 \text{ MPa} \over
( 1 + 1 )^ 2} \; = \; 1 \text{ MPa}
\]
So the complete stress tensor is
\[
\boldsymbol{\sigma}^{\text{PK2}}
\; = \;
\left[ \matrix{
1 & 0 & 0 \\
\\
0 & 0 & 0 \\
\\
0 & 0 & 0 }
\right]
\]
 Propose a definition for the engineering stress tensor
and use it to calculate a full 3D engineering stress tensor.
One possibility could be \(F / A_o\) in the reference orientation.
This would simply be
\[
\boldsymbol{\sigma}_{\text{Eng}}
\; = \;
\left[ \matrix{
2 & 0 & 0 \\
\\
0 & 0 & 0 \\
\\
0 & 0 & 0 }
\right]
\]
An alternative definition could be in the final rotated orientation.
It would be
\[
\boldsymbol{\sigma}'\!_{\text{Eng}} =
{\bf R} \cdot \boldsymbol{\sigma}_{\text{Eng}}
\cdot {\bf R}^T
\]
This gives
\[
\begin{eqnarray}
\boldsymbol{\sigma}'\!_{\text{Eng}} =
& = &
\left[ \matrix{
0.940 & 0.342 & 0 \\
\\
0.342 & \;\;0.940 & 0 \\
\\
0 & 0 & 1 }
\right]
\left[ \matrix{
2 & \;\; 0 \;\; & 0 \\
\\
0 & \;\; 0 \;\;& 0 \\
\\
0 & \;\; 0 \;\;& 0 }
\right]
\left[ \matrix{
\;\;0.940 & 0.342 & 0 \\
\\
0.342 & 0.940 & 0 \\
\\
0 & 0 & 1 }
\right]
\\
\\
\\
\\
& = &
\left[ \matrix{
1.767 & 0.643 & \;\; 0 \\
\\
0.643 & 0.234 & \;\; 0 \\
\\
0 & 0 & \;\; 0 }
\right]
\end{eqnarray}
\]

Someone gives you a 2nd PiolaKirchhoff stress tensor like the one
below and wants to know the corresponding Cauchy stress tensor.
What prevents you from doing the calculation?
\[
\boldsymbol{\sigma}^{\text{PK2}} =
\left[ \matrix{
10 & 20 & 30 \\
20 & 40 & 50 \\
30 & 50 & 60 }
\right]
\]
You also need to know the deformation gradient
because the two are related by
\[
\boldsymbol{\sigma} = {1 \over J} \, {\bf F} \cdot
\boldsymbol{\sigma}^{PK2} \cdot {\bf F}^T
\]

For the stress tensor here, find at least one unit normal vector
to a cut plane on which the normal
stress (yes normal, not shear) is zero.
\[
\boldsymbol{\sigma} =
\left[ \matrix{
20 & 5 & 10 \\
5 & 10 & 15 \\
10 & 15 & 5 }
\right]
\]
Start by entering the stress tensor in the
tensor transform page
and rotating around to find a direction with zero normal stress.
One such orientation is
So the normal direction is \({\bf n} = (0,\;0.1744,\; 0.9847)\).
This can be checked by doing
\(\sigma = {\bf n} \cdot \boldsymbol{\sigma} \cdot {\bf n}\).
\[
\begin{eqnarray}
\sigma
& \; = \; &
\matrix{
\left\{ 0.0000 \; 0.1744 \;\; 0.9847 \right\} \\
\\
\\
\\
\\
}
\left[ \matrix{
20 & 5 & 10 \\
5 & 10 & 15 \\
10 & 15 & 5 }
\right]
\left\{ \matrix{
\;\;\;0.0000 \\ 0.1744 \\ \;\;\;0.9847 }
\right\}
\\
\\
\\
& = &
0.000 \text{ MPa} \qquad \text{(Yippee!)}
\end{eqnarray}
\]

Given the stress state below, you could search forever and never
find an orientation with zero normal stress, because there isn't one. Why?
\[
\boldsymbol{\sigma} =
\left[ \matrix{
20 & 5 & 10 \\
5 & 10 & 15 \\
10 & 15 & 30 }
\right]
\]
It is because all three principal stresses are positive, so the
normal stress on any face, which is bounded between the largest and
smallest principal stress values, will always be >0.
In the previous problem, it turns out that two of the pincipals were
positive, while the third one was negative.