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.
-
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.
-
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
\[
\begin{eqnarray}
x & = & X \\
y & = & Y + X \left( {C \over T} \right) t
\end{eqnarray}
\]
The deformation gradient is
\[
{\bf F} =
\left[ \matrix{
1 & 0 \\
\\
\left( {C \over T} \right) t & 1 }
\right]
\]
Its inverse is
\[
{\bf F}^{-1} =
\left[ \matrix{
1 & 0 \\
\\
-\left( {C \over T} \right) t & 1 }
\right]
\]
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, \({\bf L} = \dot {\bf F}\).