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Having shape functions $N_i(\xi,\eta), i = 1,...,N_n$ and, a normal vector $n = (n_x,n_y,n_z)$, a thickness function $F_\tau (\zeta), \tau = 1,...,N_\tau$ and nodal variables $\mathbf{Q}_u = (Q_u,Q_v,Q_w)$:

If $$ \int_{\Omega} u \delta u dV = \int_{\Omega} F_\tau(\zeta) N_i(\xi,\eta) \mathbf{Q}_{u} F_z(\zeta) N_j(\xi,\eta) \delta \mathbf{Q}_{u} |J| d \xi d\eta d\zeta = <F_\tau F_z> <N_i N_j> \mathbf{Q}_{u} \delta \mathbf{Q}_{u} $$

where $<F_\tau F_z>$ is a $N_\tau \times N_\tau$ matrix and $<N_i N_j>$ is a $N_n \times N_n$ matrix. This later shape function matrix will interpolate $u$, $v$, and $w$ separately.

Which form should the shape function matrix $N_{un}$ have such that $<N_i n N_{un}>$ is $N_n \times N_n$ and interpolates $u$, $v$, and $w$ separetely in

$$ \int_{\Omega} u \mathbf{n} \delta u dV = \int_{\Omega} F_\tau(\zeta) N_i(\xi,\eta) \mathbf{Q}_{u} F_z(\zeta) \mathbf{n} N_{un}(\xi,\eta) \delta \mathbf{Q}_{u} |J| d \xi d\eta d\zeta = <F_\tau F_z> <N_i n N_{un}> \mathbf{Q}_{u} \delta \mathbf{Q}_{u} $$

having in mind that $$ N_{ve} = \begin{bmatrix} N_1 & 0 & 0 & N_2 & ... & N_n & 0 & 0 \\ 0 & N_1 & 0 & 0 & ... & 0 & N_n & 0 \\ 0 & 0 & N_1 & 0 & ... & 0 & 0 & N_n \end{bmatrix} $$

is not a desirable candidate because $<N_i nN_{ve}>$, which interpolates the $u$, $v$, and $w$ conjointly, is $N_n\times 3 N_n$.

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  • $\begingroup$ I have to admit that I don't understand what you're asking. That's because I don't understand what you're trying to do from your first sentence. Can you explain in more detail? $\endgroup$ – Wolfgang Bangerth Apr 26 '18 at 3:16
  • $\begingroup$ I'm trying to find the form of the shape function associated to a normal vector which will discretize a d.o.f. individually. The $N_{ve}$ will discretize all three displacements, but I want to discretize them individually. $\endgroup$ – N Luis Apr 26 '18 at 6:57
  • $\begingroup$ Let's see if I get it. You want to have one basis that spans the normal displacement to a surface and another basis that spans the tangent displacements? $\endgroup$ – nicoguaro Apr 26 '18 at 15:04
  • $\begingroup$ No. I guess the answer is in the question... $\endgroup$ – N Luis Apr 26 '18 at 21:29
  • $\begingroup$ I can't follow your question or answer, sorry. $\endgroup$ – nicoguaro Apr 28 '18 at 15:04
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One can use matrix $N_{ve}$ and program accordingly, i.e., you have to be attentive that in this case, you have the nodal values conjoint and when coding the stiffness matrix these entries should be taken in consideration.

You end up with two different kinds of nodal values:

  • in the case of $<N_i N_j>, i,j = 1,...,3$ you will have $Q_u = \begin{Bmatrix} u_1 & u_2 & u_3 & v_1 & v_2 & v_3 & w_1 & w_2 & w_3 \end{Bmatrix}$,

  • in the case of $<N_i n N_{ve}>, i,ve = 1,...,3$ you will have $Q_u = \begin{Bmatrix} u_1 & v_1 & w_1 & u_2 & v_2 & w_2 & u_3 & v_3 & w_3 \end{Bmatrix}$

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