Homework #5 Solutions


Feel free to use http://www.continuummechanics.org/interactivecalcs.html when applicable.

  1. Show that
    \[ \epsilon_{ijk} \, \epsilon_{ijk} = 6 \]
    Use the epsilon-delta identity.

    \[ \begin{eqnarray} \epsilon_{ijk} \, \epsilon_{ijk} & = & \delta_{jj} \delta_{kk} - \delta_{jk} \delta_{jk} \\ \\ \\ & = & 3 * 3 - \delta_{kk} \\ \\ \\ & = & 9 - 3 \\ \\ \\ & = & 6 \end{eqnarray} \]




  2. Given the stress tensor

    \[ \boldsymbol{\sigma} = \left[ \matrix{ 10 & 20 & 30 \\ 20 & 40 & 50 \\ 30 & 50 & 60 } \right] \qquad \]
    One of the two stress tensors below is equivalent to the one above, differing only by a coordinate transformation. The other one represents a different stress state. Which is equivalent and which is different?

    \[ \left[ \matrix{ 4.0341 & 27.291 & 14.519 \\ 27.291 & 76.619 & 46.048 \\ 14.519 & 46.048 & 29.347 } \right] \qquad \qquad \qquad \left[ \matrix{ \;\;\;30.597 & -5.733 & -15.201 \\ \;-5.733 & \;41.305 & \;\;\;18.926 \\ -15.201 & \;18.926 & \;\;\;48.098 } \right] \]


    The way to tackle this question is to compare principal stresses.

    The principal values of the given stress tensor are



    The principal values of the 1st candidate tensor are



    The above image shows that it is this first candidate stress tensor that is equivalent to the given stress tensor. The fact that the principal values are in a different order only means that the software found principal orientations for the two tensors that are 90° apart. This is nothing but a coordinate transformation.

    The principal values of the 2nd candidate stress tensor are: 20, 30, 70. Therefore, this 2nd candidate is not equivalent to the original stress tensor.







  3. With the same stress tensor

    \[ \boldsymbol{\sigma} = \left[ \matrix{ 10 & 20 & 30 \\ 20 & 40 & 50 \\ 30 & 50 & 60 } \right] \]
    Use \(E = 100\) and \(\nu = 0.333\) (metals) and demonstrate that the deviatoric stress is proportional to the deviatoric strain.


    The hydrostatic stress is (10 + 40 + 60) / 3 = 36.67. Subtracting this from the stress tensor gives the deviatoric stress.

    \[ \boldsymbol{\sigma}' = \left[ \matrix{ -26.67 & 20 & 30 \\ \;\;\;20 & 3.33 & 50 \\ \;\;\;30 & 50 & 23.33 } \right] \]
    Hooke's Law is

    \[ \begin{eqnarray} \left[ \matrix{ \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{xy} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{xz} & \epsilon_{yz} & \epsilon_{zz} } \right] & = & {1 \over 100} \left\{ (1 + 0.333) \left[ \matrix{ 10 & 20 & 30 \\ 20 & 40 & 50 \\ 30 & 50 & 60 } \right] - (0.33) \left[ \matrix { 110 & 0 & 0 \\ 0 & 110 & 0 \\ 0 & 0 & 110 } \right] \right\} \\ \\ \\ \\ & = & {1 \over 100} \left\{ \left[ \matrix{ 13.33 & 26.67 & 40 \\ 26.67 & 53.33 & 66.67 \\ 40 & 66.67 & 80 } \right] - \left[ \matrix{ 36.67 & 0 & 0 \\ 0 & 36.67 & 0 \\ 0 & 0 & 36.67 } \right] \right\} \\ \\ \\ \\ & = & \left[ \matrix{ -0.233 & \;\;0.267 & \;0.400 \\ \;\;0.267 & 0.167 & \;0.667 \\ \;\;0.400 & \;\;0.667 & \;0.433 } \right] \end{eqnarray} \]
    And the hydrostatic part of this is \((-0.233 + 0.167 + 0.433)/3 = 0.122\). Subtracting this from the strain tensor gives the deviatoric strain

    \[ \boldsymbol{\epsilon}' = \left[ \matrix{ -0.355 & 0.267 & \;0.400 \\ \;\;0.267 & 0.045 & \;0.667 \\ \;\;0.400 & 0.667 & \;0.311 } \right] \]
    Will (\(2 \, G \, \boldsymbol{\epsilon}'\)) give \(\boldsymbol{\sigma}'\)? To find out, compute \(G\).

    \[ G \; = \; {E \over 2 ( 1 + \nu) } \; = \; {100 \text{ MPa} \over 2 ( 1 + 0.333) } \; = \; 37.5 \text{ MPa} \]
    So \(2 \, G \, \boldsymbol{\epsilon}'\) equals

    \[ 2 \, G \, \boldsymbol{\epsilon}' = \left[ \matrix{ -26.6 & 20.0 & 30.0 \\ \;\;20.0 & 3.33 & 50.0 \\ \;\;30.0 & 50.0 & 23.3 } \right] \]
    This is indeed the deviatoric stress tensor computed above.