Jaekwang,
>From this symmetric gradient tensor, do we have one easy way to calculate
> the following "the second invariant of rate-of-strain tensor"?
>
There is SymmetricTensor::second_invariant [1] which is defined as I2 =
1/2[ (trace sigma)^2 - trace (sigma^2) ]
and SymmetricTensor::first_invariant [1] which just the trace I_1=trace
sigma.
According to Wikipedia [2], your \Pi_\gamma can then be expressed as
\Pi_\gamma^2 = 1/2(I1^2 -2 I2).

##
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Best,
Daniel
[1]
https://www.dealii.org/8.4.0/doxygen/deal.II/classSymmetricTensor.html#a7bbdfc57da6931de6a1757a0fa7ee982
[2] https://en.wikipedia.org/wiki/Invariants_of_tensors
>
> <https://lh3.googleusercontent.com/-k8SKQzE3aEk/WAQQnorR5lI/AAAAAAAAA60/4XJBD25ZRSAa2IYsdK4NCMOuY3QXJydmwCLcB/s1600/Screen%2BShot%2B2016-10-16%2Bat%2B6.42.40%2BPM.png>
>
>
>
> Regards,
>
> Jaekwang Kim
>
>
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