Question 3.15: Find the potential in the point (1) using the FEM-method.

There are N_{N}=5 nodes denoted as \#=1,2,3,4,5 and N_{E}=4 equal triangular elements #1-4-5, 1-2-5, 1-2-3, and 1-3-4. The only unknown potential is V _{1} and the other four potential have the numerical values: V _{2}= V _{3}=0, V _{4}= V _{5}=\frac{10+0}{2}=5  V

Using the FEM-method, we first calculate the matrices using the known potentials at the points  # = 2,3,4,5 . Following the procedure outlined in the previous example and (3.77), we find that the global sub-matrices required for the calculation in (3.78) are

\left[ S ^{( e )}\right]=\left[\begin{array}{cccr}S _{1,1}^{(1)}+ S _{4,4}^{(2)} & S _{1,2}^{(1)}+ S _{4,6}^{(2)} & S _{1,3}^{(1)} & S _{4,5}^{(2)} \\S _{2,1}^{(1)}+ S _{6,4}^{(2)} & S _{2,2}^{(1)}+ S _{6,6}^{(2)} & S _{2,3}^{(1)} & S _{6,5}^{(2)} \\S _{3,1}^{(1)} & S _{3,2}^{(1)}& S _{3,3}^{(1)} & 0 \\S _{5,4}^{(2)} & S _{5,6}^{(2)} & 0 & S _{5,5}^{(2)}\end{array}\right]        (3.77)

[ V ]_{u}=-[S]_{u, u}^{-1}[S]_{u, k}[V]_{k} \equiv[F][V]_{k}  ( 3.78 )

[S]_{ u , u }=\left[4 S _{1,1}\right] ;[ S ]_{ u , k }=\left[2 S _{1,2}, 2 S _{1,3}, 2 S _{1,4}, 2 S _{1,5}\right] .

Using the expressions in the previous example or from Appendix 2, the elements of the local \left[ S ^{( e )}\right] -matrices can be calculated. The area of each of the triangular elements is equal to 1 and the coordinates of the nodal points can be obtained from the figure. We finally obtain

S _{1,1}=\varepsilon ; S _{1,2}= S _{1,3}= S _{1,4}= S _{1,5}=-\frac{\varepsilon}{2} .

The elements of the [F]-matrix are calculated from (3.78)

[A]=-\frac{1}{4 \varepsilon}[-\varepsilon,-\varepsilon,-\varepsilon,\varepsilon]=\frac{1}{4}[1,1,1,1]

and since [ V ]_{u}=\left[ V _{1}\right] and [V]_{k}=\left[ V _{2}, V _{3}, V _{4}, V _{5}\right]^{ t }, we finally obtain

V _{1}=\frac{1}{4}\left( V _{2}+ V _{3}+ V _{4}+ V _{5}\right)=\frac{1}{4}(0+0+5+5)=2.5   V

This problem can be also evaluated using the FDM-method. We can write the potential in the point (1) as an average value of the potentials in other four points  ( # = 6,7,8,9)

V _{1}=\frac{1}{4}\left( V _{6}+ V _{7}+ V _{8}+ V _{9}\right)=\frac{1}{4}(0+0+10+0)=2.5 V

This is the same result that we had obtained previously. There are cases where the FEM-method has an advantage over the other techniques. In certain cases, the other methods actually fail!