5.3.1 Linear Vector Shape Functions on a Triangular Elements

The following in this subsection refers to the triangular element from (Fig. <4.1>). The field in the element is interpolated similarly as for the three-dimensional case by edge functions. Since there are three edges in the triangular element three interpolation functions are given

$$\vec{N}^e_{12} = l_{12}(\lambda^e_1\vec{\nabla}\lambda^e_2 - \lambda^e_2\vec{\nabla}\lambda^e_1)$$ (5.76)

$$\vec{N}^e_{23} = l_{23}(\lambda^e_2\vec{\nabla}\lambda^e_3 - \lambda^e_3\vec{\nabla}\lambda^e_2)$$ (5.77)

$$\vec{N}^e_{31} = l_{31}(\lambda^e_3\vec{\nabla}\lambda^e_1 - \lambda^e_1\vec{\nabla}\lambda^e_3).$$ (5.78)

The same properties as for the three-dimensional vector element functions can be proved for the two-dimensional ones. Without loss of generality Edge $12$ is used to prove the properties of the edge functions. The divergence of a vector edge function is zero, i.e.

$$\vec{\nabla}\cdot\vec{N}^e_{12} = l_{12}(\vec{\nabla}\lambda^e_1\... ...{\nabla}\lambda^e_2 - \vec{\nabla}\lambda^e_2\cdot\vec{\nabla}\lambda^e_1) = 0.$$ (5.79)

For the rotor of a vector edge function it can be written

$$\vec{\nabla}\times\vec{N}^e_{12} = l_{12}(\vec{\nabla}\lambda^e_1... ...bla}\lambda^e_1) = 2l_{12}\vec{\nabla}\lambda^e_1\times\vec{\nabla}\lambda^e_2.$$ (5.80)

The tangential component of $\vec{N}^e_{12}$ on Edge $23$ can be obtained from

$$\vec{N}^e_{12}\cdot\vec{r}_{23} = (\lambda^e_1\vec{r}_{23})\cdot\vec{\nabla}\lambda^e_2 - \lambda^e_2(\vec{r}_{23}\cdot\vec{\nabla}\lambda^e_1).$$ (5.81)

Since $\lambda^e_1$ vanishes on the Edge $23$ and $\vec{\nabla}\lambda^e_1$ is perpendicular to $\vec{r}_{23}$ , the two terms on the right hand side of (5.81) are zero and $\vec{N}^e_{12}$ has no tangential component on the Edge $23$ :

$$\vec{N}^e_{12}\cdot\frac{\vec{r}_{23}}{l_{23}} = 0.$$

Analogously it can be shown that $\vec{N}^e_{12}$ has no tangential component on Edge $31$ as well. For $\vec{N}^e_{12}$ along the direction of Edge $12$ the following expression is applied

$$\begin{split}\vec{N}^e_{12}\cdot\frac{\vec{r}_{12}}{l_{12}} &... ...{23}}{J}\cdot\vec{e}_z = \lambda^e_1 + \lambda^e_2. \end{split}$$

Since $\lambda^e_3$ is zero and therefore $\lambda^e_1{+}\lambda^e_2$ is one on Edge $12$ , the tangential component of $\vec{N}^e_{12}$ on Edge $12$ is one. The vector function $\vec{H}_{12}$ has a constant tangential component $\lambda^e_1{+}\lambda^e_2$ only along its corresponding Edge $12$ . Along the remaining edges this function does not have a tangential component. Similarly this characteristics applies also to the remaining functions $\vec{H}_{23}$ and $\vec{H}_{31}$ . Hence it follows that in an approach of the kind

$$\vec{f}(r) = c_1\vec{N}^e_{12} + c_2\vec{N}^e_{23} + c_3\vec{N}^e_{31}$$

the arbitrary coefficients $c_1$ , $c_2$ and $c_3$ are to be regarded as values of the projection of $\vec{f}(r)$ in the possible edge directions, respectively. This fact justifies the name edge function or edge element.

It is not simple to imagine, how the edge functions look like. The edge function $\vec{N}^e_{12}$ is visualized in Fig. <5.8>. Fig. <5.9> depicts function $\vec{f}(r)$ for all coefficients set to one.

Figure 5.8: The shape function along Edge $12$ .

Figure 5.9: The sum of the shape functions in the triangle.