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.


  1. 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.


    1. 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

      \[ \begin{eqnarray} \boldsymbol{\sigma} & = & \left[ \matrix{ 0.940 & -0.342 & 0 \\ \\ 0.342 & \;\;0.940 & 0 \\ \\ 0 & 0 & 1 } \right] \left[ \matrix{ 4 & \;\; 0 \;\; & 0 \\ \\ 0 & \;\; 0 \;\;& 0 \\ \\ 0 & \;\; 0 \;\;& 0 } \right] \left[ \matrix{ \;\;0.940 & 0.342 & 0 \\ \\ -0.342 & 0.940 & 0 \\ \\ 0 & 0 & 1 } \right] \\ \\ \\ \\ & = & \left[ \matrix{ 3.534 & 1.286 & \;\; 0 \\ \\ 1.286 & 0.468 & \;\; 0 \\ \\ 0 & 0 & \;\; 0 } \right] \end{eqnarray} \]
    2. 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

      \[ \boldsymbol{\sigma}^{\text{PK2}} \; = \; \left[ \matrix{ 1 & 0 & 0 \\ \\ 0 & 0 & 0 \\ \\ 0 & 0 & 0 } \right] \]
    3. 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

      \[ \begin{eqnarray} \boldsymbol{\sigma}'\!_{\text{Eng}} = & = & \left[ \matrix{ 0.940 & -0.342 & 0 \\ \\ 0.342 & \;\;0.940 & 0 \\ \\ 0 & 0 & 1 } \right] \left[ \matrix{ 2 & \;\; 0 \;\; & 0 \\ \\ 0 & \;\; 0 \;\;& 0 \\ \\ 0 & \;\; 0 \;\;& 0 } \right] \left[ \matrix{ \;\;0.940 & 0.342 & 0 \\ \\ -0.342 & 0.940 & 0 \\ \\ 0 & 0 & 1 } \right] \\ \\ \\ \\ & = & \left[ \matrix{ 1.767 & 0.643 & \;\; 0 \\ \\ 0.643 & 0.234 & \;\; 0 \\ \\ 0 & 0 & \;\; 0 } \right] \end{eqnarray} \]
  2. 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

    \[ \boldsymbol{\sigma} = {1 \over J} \, {\bf F} \cdot \boldsymbol{\sigma}^{PK2} \cdot {\bf F}^T \]
  3. 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 \(\sigma = {\bf n} \cdot \boldsymbol{\sigma} \cdot {\bf n}\).

    \[ \begin{eqnarray} \sigma & \; = \; & \matrix{ \left\{ 0.0000 \; -0.1744 \;\; 0.9847 \right\} \\ \\ \\ \\ \\ } \left[ \matrix{ 20 & 5 & 10 \\ 5 & 10 & 15 \\ 10 & 15 & 5 } \right] \left\{ \matrix{ \;\;\;0.0000 \\ -0.1744 \\ \;\;\;0.9847 } \right\} \\ \\ \\ & = & 0.000 \text{ MPa} \qquad \text{(Yippee!)} \end{eqnarray} \]
  4. Given the stress state below, you could search forever and never find an orientation with zero normal stress, because there isn't one. Why?

    \[ \boldsymbol{\sigma} = \left[ \matrix{ 20 & 5 & 10 \\ 5 & 10 & 15 \\ 10 & 15 & 30 } \right] \]
  5. 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.