Question 3.14: Find the voltage distribution within the triangular element ...
Find the voltage distribution within the triangular element if the voltages at the three nodes have the following values.
V1 = 8 @ (0, 0); V2 = 0 @ (4, 0); and V3 = 0 @ (4, 3).
Evaluate the electric energy that is stored in this element. In order to simplify the calculation, the energy should be expressed in terms of the dielectric constant ε. Compare your analytical results with a MATLAB calculation.
The voltage distribution within the element is computed from (3.67)
V^{(e)}(x, y)=\left[\begin{array}{lll} 1 & x & y \end{array}\right]\left[\begin{array}{lll} 1 & x_{1} & y_{1} \\ 1 & x_{2} & y_{2} \\ 1 & x_{3} & y_{3} \end{array}\right]^{-1}\left[\begin{array}{l} V_{1} \\ V_{2} \\ V_{3} \end{array}\right] (3.67)
\begin{aligned} V ^{(e)}(x, y) &=\left[\begin{array}{lll} 1 & x & y \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 4 & 0 \\ 1 & 4 & 3 \end{array}\right]^{-1}\left[\begin{array}{l} 8 \\ 0 \\ 0 \end{array}\right]=\left[\begin{array}{lll} 1 & x & y \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ -1 / 4 & 1 / 4 & 0 \\ 0 & -1 / 3 & 1 / 3 \end{array}\right]\left[\begin{array}{l} 8 \\ 0 \\ 0 \end{array}\right] \\ &=\left[\begin{array}{lll} 1 & x & y \end{array}\right]\left[\begin{array}{c} 8 \\ -2 \\ 0 \end{array}\right]=8-2 x \end{aligned}
It is easy to check that this linear function satisfies the interpolation criteria (3.65) and Laplace’s equation (3.62).
∇ 2 V = 0 (3.62)
V(e) (xj , yj ) = Vj (j = 1,2, 3) (3.65)
In order to calculate the energy, we must first calculate the numerical values for the explicit expressions for the \left[S^{(e)}\right] -matrix elements. The area enclosed within the triangle is A_{e}=6. We explicitly write using (A2.9) and (A2.10) that
S _{1,1}=\frac{\varepsilon}{4 A _{ e }}\left[\left( y _{2}- y _{3}\right)^{2}+\left( x _{2}- x _{3}\right)^{2}\right] (A.2.9)
S _{1,1}^{(e)}=\frac{\varepsilon}{4 \times 6}\left[(0-3)^{2}+(4-4)^{2}\right]=\frac{9 \varepsilon}{24}
S _{1,2}=\frac{\varepsilon}{4 A _{ e }}\left[\left( y _{2}- y _{3}\right)\left( y _{3}- y _{1}\right)+\left( x _{2}- x _{3}\right)\left( x _{3}- x _{1}\right)\right] \text { etc. } (A.2.10)
S _{1,2}^{( e )}= S _{2,1}^{( e )}=\frac{\varepsilon}{4 \times 6}[(0-3)(3-0)+(4-4)(4-4)]=-\frac{9 \varepsilon}{24}
S _{1,3}^{( e )}= S _{3,1}^{( e )} =\frac{\varepsilon}{4 A _{ e }}\left[\left( y _{3}- y _{2}\right)\left( y _{2}- y _{1}\right)+\left( x _{3}- x _{2}\right)\left( x _{2}- x _{1}\right)\right] \\\\=\frac{\varepsilon}{4 \times 6}[(3-0)(0-0)+(4-4)(4-0)]=0
S _{2,2}^{(e)}=\frac{\varepsilon}{4 A _{ e }}\left[\left( y _{3}- y _{1}\right)^{2}+\left( x _{3}- x _{1}\right)^{2}\right]=\frac{\varepsilon}{4 \times 6}\left[(3-0)^{2}+(4-0)^{2}\right]=\frac{25 \varepsilon}{24}
S _{2,3}^{(e)} = S _{3,2}^{( e )}=\frac{\varepsilon}{4 A _{ e }}\left[\left( y _{3}- y _{1}\right)\left( y _{1}- y _{2}\right)+\left( x _{3}- x _{1}\right)\left( x _{1}- x _{2}\right)\right] \\\\ =\frac{\varepsilon}{4 \times 6}\left[(3-0)(0-0)+(4-0)(0-4)]=-\frac{16 \varepsilon}{24}\right.
S _{3,3}^{(e)}=\frac{\varepsilon}{4 A _{ e }}\left[\left( y _{1}- y _{2}\right)^{2}+\left( x _{1}- x _{2}\right)^{2}\right]=\frac{\varepsilon}{4 \times 6}\left[(0-0)^{2}+(0-4)^{2}\right]=\frac{16 \varepsilon}{24}
This is a written as the matrix
S^{(e)}=\varepsilon\left[\begin{array}{ccc} 9 / 24 & -9 / 24 & 0 \\ -9 / 24 & 25 / 24 & -16 / 24 \\ 0 & -16 / 24 & 16 / 24 \end{array}\right]
The energy is computed from (3.74)
W^{(e)}=\frac{1}{2}[V]^{T}\left[S^{(e)}\right][V] (3.74)
\begin{aligned} \frac{ W _{ e }}{\varepsilon} &=\frac{1}{2 \varepsilon}[ V ]^{t}\left[ S ^{(e)}\right][ V ]=\frac{1}{2}\left[\begin{array}{lll} 8 & 0 & 0 \end{array}\right]\left[\begin{array}{ccc} 9 / 24 & -9 / 24 & 0 \\ -9 / 24 & 25 / 24 & -16 / 24 \\ 0 & -16 / 24 & 16 / 24 \end{array}\right]\left[\begin{array}{l} 8 \\ 0 \\ 0 \end{array}\right] \\ &=\frac{1}{2}\left[\begin{array}{lll} 8 & 0 & 0 \end{array}\right]\left[\begin{array}{c} 72 / 24 \\ -72 / 24 \\ 0 \end{array}\right]=12 \end{aligned}
These results are confirmed using MATLAB.