[deal.II] Re: Easy way to calculate second-deviatoric tensor

[JAEKWANG KIM]

Thank you all guys who replied here!! 

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Re: [deal.II] Re: Easy way to calculate second-deviatoric tensor

[Wolfgang Bangerth]

On 10/17/2016 03:36 AM, Daniel Arndt wrote:


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).


You can also compute it directly: if you have the symmetric gradient eps(u) 
already computed, then

  Pi_gamma = std::sqrt (2*eps*eps);

Best
 W.

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Wolfgang Bangerth  email: bange...@colostate.edu
   www: http://www.math.colostate.edu/~bangerth/

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[deal.II] Re: Easy way to calculate second-deviatoric tensor

[Daniel Arndt]

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).

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

>
> 
>
>
>
> Regards, 
>
> Jaekwang Kim 
>  
>

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