Homework #10 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.


Use the following true strain tensor for the first three problems.

\[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ \;\;\; 0.758 & -0.237 & \;\;\;0.012 \\ -0.237 & -0.546 & \;\;\;0.067 \\ \;\;\; 0.012 & \;\;\; 0.067 & -0.212 \\ } \right] \]
  1. This true strain tensor is for an incompressible material. Demonstrate that this is indeed the case.

    The trace of a true strain tensor gives the natural log of the ratio of the intial and final volumes. For an incompressible material, the ratio is 1 and the log of 1 is zero. So the trace of the tensor should equal zero.

    \[ 0.758 - 0.546 - 0.212 = 0 \]
    So the volume has not changed.


  2. Calculate \(\epsilon_{\text{Hyd}}\), \(\gamma_{\text{Max}}\), and \(\gamma_{\text{Sec}}\), and determine whether the strain tensor represents a state closer to uniaxial tension, shear, or equibiaxial tension.

    We need the principal values of the strain tensor.



    The principal strains are 0.800, -0.600, and -0.200. They add to zero, so \(\epsilon_{\text{Hyd}} = 0.0\).

    The max shear is

    \[ \begin{eqnarray} \gamma_{\text{Max}} & = & \epsilon_{\text{Max}} - \epsilon_{\text{Min}} \\ \\ & = & 0.800 - (\text{-} 0.600) \\ \\ & = & 1.400 \end{eqnarray} \]
    The secondary shear is

    \[ \begin{eqnarray} \gamma_{\text{Sec}} & = & \epsilon_{\text{Mid}} - \epsilon_{\text{Hyd}} \\ \\ & = & \text{-} 0.200 - 0.000 \\ \\ & = & \text{-} 0.200 \end{eqnarray} \]
    And the ratio of the shears is

    \[ { \gamma_{\text{Sec}} \over \;\gamma_{\text{Max}}} \;\; = \;\; {-0.200 \over \;\;\; 1.400} \;\; = \;\; -0.143 \]
    So the strain state is close to pure shear, although on the uniaxial tension side of it.


  3. Figure out the corresponding engineering strain tensor.

    We will need \(\Delta L / L_o\), so get there by taking the exponent of the principal true strain values and subtracting 1.

    \[ \begin{eqnarray} \Delta L_1 / L_o & \; = \; & \exp(\;0.800) & - 1 & \; = \; & \;\;1.226 \\ \\ \Delta L_2 / L_o & \; = \; & \exp(-0.600) & - 1 & \; = \; & -0.451 \\ \\ \Delta L_3 / L_o & \; = \; & \exp(-0.200) & - 1 & \; = \; & -0.181 \\ \end{eqnarray} \]
    This gives each principal engineering strain value. So the principal engineering strain tensor is

    \[ {\bf e}^{\text{Eng}}_\text{Princ} = \left[ \matrix{ 1.226 & \;\;0.0 & \;\;0.0 \\ 0.0 & -0.451 & \;\;0.0 \\ 0.0 & \;\;0.0 & -0.181 } \right] \]
    And now the principal strain tensor needs to be rotated back to the proper orientation. This is done by \({\bf e}^\text{Eng} = {\bf Q}^T \cdot {\bf e}^\text{Eng}_\text{Princ} \cdot {\bf Q}\).

    The \({\bf Q}\) matrix is in the screen shot above. It is

    \[ {\bf Q} = \left[ \matrix{ 0.985 & -0.173 & \;\;\;0.000 \\ 0.171 & \;\;\;0.970 & -0.173 \\ 0.030 & \;\;\;0.170 & \;\;\;0.985 } \right] \]
    And the engineering strain tensor is calculated by

    \[ \begin{eqnarray} {\bf e}^\text{Eng} & = & \left[ \matrix{ \;\;\;0.985 & \;\;\;0.171 & \;\;\;0.030 \\ -0.173 & \;\;\;0.970 & \;\;\;0.170 \\ \;\;\;0.000 & -0.173 & \;\;\;0.985 } \right] \left[ \matrix{ 1.226 & \;\;0.0 & \;\;0.0 \\ 0.0 & -0.451 & \;\;0.0 \\ 0.0 & \;\;0.0 & -0.181 } \right] \left[ \matrix{ 0.985 & -0.173 & \;\;\;0.000 \\ 0.171 & \;\;\;0.970 & -0.173 \\ 0.030 & \;\;\;0.170 & \;\;\;0.985 } \right] \\ \\ \\ & = & \left[ \matrix{ \;\;\;1.907 & -0.397 & \;\;\;0.005 \\ -0.397 & -0.273 & \;\;\;0.031 \\ \;\;\;0.005 & \;\;\;0.031 & -0.171 } \right] \end{eqnarray} \]
    So

    \[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ \;\;\; 0.758 & -0.237 & \;\;\;0.012 \\ -0.237 & -0.546 & \;\;\;0.067 \\ \;\;\; 0.012 & \;\;\; 0.067 & -0.212 \\ } \right] \qquad \text{corresponds to} \qquad {\bf e}^\text{Eng} = \left[ \matrix{ \;\;\;1.176 & -0.285 & \;\;\;0.008 \\ -0.285 & -0.393 & \;\;\;0.045 \\ \;\;\;0.008 & \;\;\;0.045 & -0.189 } \right] \]


  4. Figure out the corresponding Green strain tensor.

    We will need \(\Delta L / L_o\), so get there by taking the exponent of the principal true strain values and subtracting 1.

    \[ \begin{eqnarray} \Delta L_1 / L_o & \; = \; & \exp(\;0.800) & - 1 & \; = \; & \;\;1.226 \\ \\ \Delta L_2 / L_o & \; = \; & \exp(-0.600) & - 1 & \; = \; & -0.451 \\ \\ \Delta L_3 / L_o & \; = \; & \exp(-0.200) & - 1 & \; = \; & -0.181 \\ \end{eqnarray} \]
    Each principal Green strain value is \({\Delta L \over L_o} + {1 \over 2} \left( {\Delta L \over L_o} \right)^2\).

    \[ \begin{eqnarray} E_1 & = & \;\;1.226 + {1 \over 2} ( 1.226 )^2 & = & \;\;1.977 \\ \\ E_2 & = & -0.451 + {1 \over 2} ( -0.451 )^2 & = & -0.349 \\ \\ E_3 & = & -0.181 + {1 \over 2} ( -0.181 )^2 & = & -0.165 \\ \end{eqnarray} \]
    So the principal Green strain tensor is

    \[ {\bf E}_{\text{Princ}} = \left[ \matrix{ 1.977 & \;\;0.0 & \;\;0.0 \\ 0.0 & -0.349 & \;\;0.0 \\ 0.0 & \;\;0.0 & -0.165 } \right] \]
    And now the principal strain tensor needs to be rotated back to the proper orientation. This is done by \({\bf E} = {\bf Q}^T \cdot {\bf E}_{\text{Princ}} \cdot {\bf Q}\).

    The \({\bf Q}\) matrix is in the screen shot above. It is

    \[ {\bf Q} = \left[ \matrix{ 0.985 & -0.173 & \;\;\;0.000 \\ 0.171 & \;\;\;0.970 & -0.173 \\ 0.030 & \;\;\;0.170 & \;\;\;0.985 } \right] \]
    And the Green strain tensor is calculated by

    \[ \begin{eqnarray} {\bf E} & = & \left[ \matrix{ \;\;\;0.985 & \;\;\;0.171 & \;\;\;0.030 \\ -0.173 & \;\;\;0.970 & \;\;\;0.170 \\ \;\;\;0.000 & -0.173 & \;\;\;0.985 } \right] \left[ \matrix{ 1.977 & \;\;0.0 & \;\;0.0 \\ 0.0 & -0.349 & \;\;0.0 \\ 0.0 & \;\;0.0 & -0.165 } \right] \left[ \matrix{ 0.985 & -0.173 & \;\;\;0.000 \\ 0.171 & \;\;\;0.970 & -0.173 \\ 0.030 & \;\;\;0.170 & \;\;\;0.985 } \right] \\ \\ \\ & = & \left[ \matrix{ \;\;\;1.907 & -0.397 & \;\;\;0.005 \\ -0.397 & -0.273 & \;\;\;0.031 \\ \;\;\;0.005 & \;\;\;0.031 & -0.171 } \right] \end{eqnarray} \]
    So

    \[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ \;\;\; 0.758 & -0.237 & \;\;\;0.012 \\ -0.237 & -0.546 & \;\;\;0.067 \\ \;\;\; 0.012 & \;\;\; 0.067 & -0.212 \\ } \right] \qquad \text{corresponds to} \qquad {\bf E} = \left[ \matrix{ \;\;\;1.907 & -0.397 & \;\;\;0.005 \\ -0.397 & -0.273 & \;\;\;0.031 \\ \;\;\;0.005 & \;\;\;0.031 & -0.171 } \right] \]


    And recall

    \[ \boldsymbol{\epsilon}_{\text{True}} = \left[ \matrix{ \;\;\; 0.758 & -0.237 & \;\;\;0.012 \\ -0.237 & -0.546 & \;\;\;0.067 \\ \;\;\; 0.012 & \;\;\; 0.067 & -0.212 \\ } \right] \qquad \text{corresponds to} \qquad {\bf e}^\text{Eng} = \left[ \matrix{ \;\;\;1.176 & -0.285 & \;\;\;0.008 \\ -0.285 & -0.393 & \;\;\;0.045 \\ \;\;\;0.008 & \;\;\;0.045 & -0.189 } \right] \]