Dear Anup,

one option you have is to use a Triangulation<dim, dim+1>.

I would still use two scalar extractors:

const FEValuesExtractors::Scalar u_x
const FEValuesExtractors::Scalar u_y

Tensor<2,dim+1> gradient;

gradient[0] = fe_values_ref[u_x].gradient(k, q_point) 
gradient[1] = fe_values_ref[u_y].gradient(k, q_point) 

But at this point your gradient would be ok. Alternatively, you have to do 
things manually...

Best,

Luca.


> On 13 Dec 2016, at 23:30, Anup Basak <anupbas...@gmail.com> wrote:
> 
> Hello all,
> 
> I am solving for a  3 dimensional displacement vector (dim+1 dimensional)  in 
> a dim=2 dimensional space (for
> a planer problem in elasticity). Hence I consider an FEvaluesExtractor:
> 
> const FEValuesExtractors::Vector u_fe;
> 
> and initialize it as 
> 
> u_fe(0)
> 
> 
> Now I need to compute the gradient of the shape functions and the gradient of 
> the solution vectors. Hence I 
> use the command:
> 
> 
> for (unsigned int q_point = 0; q_point < n_q_points; ++q_point)
> for (unsigned int k = 0; k < dofs_per_cell; ++k)
> fe_values_ref[u_fe].gradient(k, q_point)
> 
> 
> Then I get 2x2 tensors. Could anyone suggest how to get gradient terms 
> corresponding to all three vectors
> and finally get a 3x3 tensors, instead of 2x2. I am expecting to get a 3x3 
> tensor with non-zero entries in the
>  first two columns and all zero entries in the third column.
> 
> 
> Thanks and regards,
> Anup.
> 
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