Hi Simon,
If you compare Holzapfel's definition with that of Wriggers, for instance,
one notices the difference in how they define this 4th-order identity
tensor. This particular term comes from the derivative d*C*/d*C* (i.e.
differentiating a [symmetric] tensor w.r.t itself), and if you do not
account for the symmetries of tensor then you end up with Holzapfel's
result, namely that
d*C*/d*C *-> dC_{ij}/dC_{kl} = \delta{ik}\delta{jl} = I_{ijkl}.
In my version of the book, the text below eq. 6.84 where the reference
deviator tensor is defined points one to eq. 1.160. There you see that this
tensor does not hold the symmetries that a SymmetricTensor does (it doesn't
have the property of minor symmetry) and it also doesn't map all rank-2
tensors onto a symmetric tensor. It also probably makes a difference in how
one defines the transpose of this deviator tensor (which would rely on
d*C*^T/d*C *-> dC_{ji}/dC_{kl} = \delta{jk}\delta{il} = I^{T}_{ijkl},
maybe?). If one considers Wrigger's approach (eq 3.125 and the text below
it; specific reference listed in the step-44 introduction) and explicitly
accounts for the symmetry of the tensor then
d*C*/d*C *-> d(0.5*[C_{ij} + C_{ji}]) /dC_{kl} = 0.5*[\delta{ik}\delta{jl}
+ \delta{jk}\delta{il}] = S_{ijkl}.
Such a tensor definitely maps all rank-2 tensors to symmetric tensors and
helps fulfill the relation that for SymmetricTensors *A*=*A*^{T}.
The differences in the definitions are subtle and your question reminds me
of how many time I've pondered pretty much the same thing. I've also
previously observed the similar differences in results of contractions that
you mentioned in your second email, and have been equally surprised. In the
end, the consequences on the outcome of each operation involving your
definition of the deviator tensor *could* simply depend on which tensor
class you are storing the operators and results in. Certainly the rank-4
tensor *I* is not a SymmetricTensor and therefore *P* = *P*(*I*) would not
be one either. But since the rank-4 tensor *S* is by construction and
definition
<https://dealii.org/current/doxygen/deal.II/symmetric__tensor_8h.html#ab3e890348aa219805e84f7d367e098c3>
a SymmetricTensor, *P* = *P*(*S*) would also be one.
Best,
Jean-Paul
On Wednesday, August 10, 2022 at 12:54:07 AM UTC+2 Wolfgang Bangerth wrote:
> On 8/8/22 15:11, Simon Wiesheier wrote:
> > grafik.png
> > grafik.png
> > The only place I use Dev_P is to compute the above double contractions
> ('P' in
> > the above is what dealii returns as Dev_P)
> > As you can see, the fourth order tangent C_bar consists of outer
> products of
> > second order symmetric tensors:
> > boldsymbol 'I' is the second order unit tensor, C_bar is a symmetric
> second
> > order tensor, and blackboard 'I' is the general fourth order unit tensor,
> > the deltas are just scalars.
> >
> > In my opinion, the double contractions 'P : outer_product(C_bar, C_bar)'
> and
> > 'P : I' produce different results if the fourth order symmetric unit
> tensor
> > 'S' is used to define 'P' .
> > The same holds also for doing the double contractions with P_T.
> > Correct?
>
> Not being a nonlinear mechanics person, I'll have to defer to someone else
> with more knowledge about these matters :-(
>
> Best
> W.
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: bang...@colostate.edu
> www: http://www.math.colostate.edu/~bangerth/
>
>
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