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.
-
If a stress tensor is
\[
\boldsymbol{\sigma} =
\left[ \matrix{
10 & 20 & 30 \\
20 & 40 & 10 \\
30 & 10 & 50 }
\right]
\]
then calculate the traction vector on a plane with
unit normal \({\bf n} = (0.100, \; 0.700, \; 0.707)\).
\[
\left\{ \matrix{
T_x \\ T_y \\ T_z }
\right\}
=
\left[ \matrix{
10 & 20 & 30 \\
20 & 40 & 10 \\
30 & 10 & 50 }
\right]
\left\{ \matrix{
0.100 \\ 0.700 \\ 0.707 }
\right\}
=
\left\{ \matrix{
36.21 \\ 37.07 \\ 45.35 }
\right\}
\]
So \({\bf T} = 36.21 \, {\bf i} + 37.07 \, {\bf j} + 45.35 \, {\bf k} \, \text{MPa}\).
-
An incompressible rubber object with an initial cross-sectional area
of 100 mm2 is stretched along the x-axis to twice
its original length by a force of 200 N, and rotated 20°
counter clockwise in the x-y plane.
- What is the full 3-D Cauchy stress tensor?
If it's incompressible and its length doubles, then its
cross-sectional area will decrease by half to
50 mm2. 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 z-axis 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 3-D 2nd Piola-Kirchhoff stress tensor?
The only nonzero stress component will be in the xx-slot
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 3-D 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 Piola-Kirchhoff 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
\[
\boldsymbol{\sigma} = {1 \over J} \, {\bf F} \cdot
\boldsymbol{\sigma}^{PK2} \cdot {\bf F}^T
\]