Homework #9 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. We know that the 1st invariant of a deviatoric strain tensor is automatically zero regardless of the 1st invariant's value for the total strain tensor. But do the 2nd and 3rd invariants change between the total strain and the deviatoric strain tensors?

Compare the invariants of the strain tensor below with those of its deviatoric counterpart to answer the question.

${\bf E} = \left[ \matrix{ \;\;\;0.50 & \;\;\;0.30 & -0.10 \\ \;\;\;0.30 & -0.10 & \;\;\;0.20 \\ -0.10 & \;\;\;0.20 & -0.10 } \right]$
$\begin{eqnarray} I_1 & = & 0.50 + (-0.10) + (-0.10) \\ \\ & = & 0.30 \\ \\ I_2 & = & (0.50)(-0.10) + (-0.10)(-0.10) + (-0.10)(0.50) - (0.30)^2 - (-0.10)^2 - (0.20)^2 \\ \\ & = & -0.23 \\ \\ I_3 & = & \text{det}({\bf E}) \\ \\ & = & -0.017 \end{eqnarray}$
The hydrostatic strain is one-third of $$I_1$$: $$\epsilon_{Hyd} = 0.10$$. Sutracting this from each diagonal component give the deviatoric strain tensor.

${\bf E}' = \left[ \matrix{ \;\;\;0.40 & \;\;\;0.30 & -0.10 \\ \;\;\;0.30 & -0.20 & \;\;\;0.20 \\ -0.10 & \;\;\;0.20 & -0.20 } \right]$
And the invariants of the deviatoric strain tensor are

$\begin{eqnarray} I_1 & = & 0.40 + (-0.20) + (-0.20) \\ \\ & = & 0.00 \\ \\ I_2 & = & (0.40)(-0.20) + (-0.20)(-0.20) + (-0.20)(0.40) - (0.30)^2 - (-0.10)^2 - (0.20)^2 \\ \\ & = & -0.26 \\ \\ I_3 & = & \text{det}({\bf E}') \\ \\ & = & 0.008 \end{eqnarray}$
So the invariants of the (total) strain tensor and the deviatoric strain tensor are indeed different.

2. If an object is sheared, $$(D/T)$$ style, where $$D = C\,t$$ and $$C$$ is a constant, then figure out the velocity gradient, $${\bf L}$$, for this.

The mapping equations for this case are

$\begin{eqnarray} x & = & X \\ y & = & Y + X \left( {C \over T} \right) t \end{eqnarray}$
${\bf F} = \left[ \matrix{ 1 & 0 \\ \\ \left( {C \over T} \right) t & 1 } \right]$
${\bf F}^{-1} = \left[ \matrix{ 1 & 0 \\ \\ -\left( {C \over T} \right) t & 1 } \right]$
And $$\dot {\bf F}$$ is
$\dot {\bf F} = \left[ \matrix{ 0 & 0 \\ \\ \left( {C \over T} \right) & 0 } \right]$
So $${\bf L}$$ is
$\begin{eqnarray} {\bf L} = \dot {\bf F} \cdot {\bf F}^{-1} & = & \left[ \matrix{ 0 & 0 \\ \\ \left( {C \over T} \right) & 0 } \right] \left[ \matrix{ 1 & 0 \\ \\ -\left( {C \over T} \right) t & 1 } \right] \\ \\ \\ & = & \left[ \matrix{ 0 & 0 \\ \\ \left( {C \over T} \right) & 0 } \right] \end{eqnarray}$
So in this rare example, $${\bf L} = \dot {\bf F}$$.