Hi Jon
The FEValuesExtractors::Vector is for first-order tensors as you say and not
arbitrary size vectors.
You can however create a std::vector of FEValuesExtractors::Scalar extractors.
The following example had displacement (\in R^dim), \gamma (\in R^
n_slip_systems) and yield ( \in R^ n_slip_systems) as nodal variables:
// generate displacement, slip and yield shape function views
const FEValuesExtractors::Vector displacements_fe(0);
std::vector<FEValuesExtractors::Scalar> gamma_fe;
std::vector<FEValuesExtractors::Scalar> yield_fe;
for (unsigned int alpha = 0; alpha < n_slip_systems; alpha++) {
gamma_fe.push_back(FEValuesExtractors::Scalar(deal_II_dimension +
alpha));
yield_fe.push_back(FEValuesExtractors::Scalar(deal_II_dimension +
n_slip_systems + alpha));
}
Regards
Andrew
On 16 Aug 2010, at 6:23 PM, Jonathan Pitt wrote:
> Wolfgang,
>
> Yes, this helps in that it confirms that my present course of action isn't
> foolish, but rather what others would think to do as well!
>
> One question I've run across in this: FEValuesExtractors::Vector creates a
> grouping of scalar elements dim long into a vector, but I need a grouping of
> dim+1 (because my triangulation is 2D, but my FESystem is 3D). Is there a
> way to get this with what is currently implemented?
>
> Thanks,
>
> Jon
>
>
>
>
> On Wed, Aug 11, 2010 at 8:59 AM, Wolfgang Bangerth <[email protected]>
> wrote:
>
> Jon,
>
> > Thanks for the question. I'm trying to create a formulation which retains
> > all three components of displacement, but with no dependence on the \phi
> > coordinate. It is not necessarily true that the u_\phi component will be
> > constant, it just has to be free of dependence on \phi.
>
> I see. So you would have a 2-dimensional domain (x-z) but a 3-component
> problem (x-z-phi displacements)?
>
> In that case, deal.II would give you a 3x2 matrix for the gradients, which you
> will have to extend to a 3x3 matrix. You'll probably want to compute dot
> products of gradients in some simple way, so you should transform the 3x3
> matrix into cartesian coordinates rather than leaving them in cylindrical
> coordinates.
>
> Does that help? I'm not quite sure any more what the exact question was ;-)
>
> Cheers
> W.
>
> -------------------------------------------------------------------------
> Wolfgang Bangerth email: [email protected]
> www: http://www.math.tamu.edu/~bangerth/
>
>
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