Basically, I have a hexahedral finite element mesh. I know the coordinates of elements, I used the coordinate transformation into an isoparametric structure and shape(basis) functions to calculate the Jacobian to find the volume.
I also need the surface area of each face separately along with its outward unit normal.
What would be the best possible way to do that? Can the Jacobian be used to do this?
I have tried working with using vector algebra, by finding the cross-product of diagonals of each face to find the direction of the outward unit normal and the area but that doesn't work for an irregular hexahedron( which I think shouldn't be the case). Can this be another general approach?
I can't understand where I am going wrong here. Will be grateful if someone can give me some idea.
Best Answer
The problem is that, in general, the faces of an irregular hexahedron are not flat. There is not one normal, the surface area is curved in space and difficult to calculate exactly (whatever the additional value of "exact" is in a numerical context). So I think you should anyway be satisfied with some sensible approximations. Here comes a proposal.
Let the vertices of the (in general non-planar) face of an irregular hexahedron be given by $\,\vec{r}_A , \vec{r}_B , \vec{r}_C , \vec{r}_D$ , as shown in the picture. Then form the vertices of the gray ( Varignon) quadrilateral as : $$\vec{r}_1 = \frac{1}{2}\left(\vec{r}_A+\vec{r}_C\right) \quad ; \quad \vec{r}_2 = \frac{1}{2}\left(\vec{r}_B+\vec{r}_D\right) \\ \vec{r}_3 = \frac{1}{2}\left(\vec{r}_A+\vec{r}_B\right) \quad ; \quad \vec{r}_4 = \frac{1}{2}\left(\vec{r}_C+\vec{r}_D\right) $$ It can be easily proved that the quadrilateral $\overline{1324}$ is a paralellogram and hence it is planar. The intersection point of the (yellow) diagonals of the Varignon paralellogram is a good place to define the normal, as a vector perpendicular to $\overline{1324}$ . Last but not least the total area of the face is defined by the sum of five planar areas: the area of the paralellogram $\overline{1324}$ and the four areas of the triangles $\,\overline{A31} , \overline{B23} , \overline{D42} , \overline{C14}$ . Hope this helps.