2D formulation for Electrostatic Problems
From KratosWiki
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(→Source Vector f<sup>(e)</sup>) |
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= A^{(e)} \frac{\rho_S}{3} \left [ 1 \quad 1 \quad 1 \right ]^T | = A^{(e)} \frac{\rho_S}{3} \left [ 1 \quad 1 \quad 1 \right ]^T | ||
</math> | </math> | ||
+ | |||
+ | |||
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+ | :::'''Quadratic case''' ('''n<sub>p</sub>'''=3 integration points): | ||
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+ | ::::<math>p=1 \qquad N=\left [ \frac{1}{2} \quad \frac{1}{2} \quad 0\right ] \qquad W_1=\frac{1}{6}\,</math> | ||
+ | ::::<math>p=2 \qquad N=\left [ 0 \quad \frac{1}{2} \quad \frac{1}{2}\right ] \qquad W_2=\frac{1}{6}\,</math> | ||
+ | ::::<math>p=3 \qquad N=\left [ \frac{1}{2} \quad 0 \quad \frac{1}{2}\right ] \qquad W_3=\frac{1}{6}\,</math> | ||
+ | |||
+ | ::::<math> | ||
+ | \int_{\Omega^{(e)}} \mathbf{N^T} \rho_S \partial \Omega^{(e)} | ||
+ | = 2 A^{(e)} \left { \left [ \frac{1}{2} \quad \frac{1}{2} \quad 0\right ]^T | ||
+ | + [ 0 \quad \frac{1}{2} \quad \frac{1}{2} \right ]^T | ||
+ | + [ \frac{1}{2} \quad 0 \quad \frac{1}{2} \right ]^T | ||
+ | \right } \frac{1}{6} \rho_S | ||
+ | = A^{(e)} \frac{\rho_S}{3} \left [ 1 \quad 1 \quad 1 \right ]^T | ||
+ | </math> | ||
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Revision as of 10:44, 27 November 2009
The 2D Electrostatic Poisson's equation given by the governing PDE and its boundary conditions:
can be written as (see the General formulation for Electrostatic Problems):
with (n is the number of nodes of the element):
2D formulation for Triangular Elements
After applying the numerical integration for triangular elements by using the natural coordinates, we obtain:
Stiffness Matrix K(e)
Source Vector f(e)
- Linear case (np=1 integration point):
- Quadratic case (np=3 integration points):
- Failed to parse (syntax error): \int_{\Omega^{(e)}} \mathbf{N^T} \rho_S \partial \Omega^{(e)} = 2 A^{(e)} \left { \left [ \frac{1}{2} \quad \frac{1}{2} \quad 0\right ]^T + [ 0 \quad \frac{1}{2} \quad \frac{1}{2} \right ]^T + [ \frac{1}{2} \quad 0 \quad \frac{1}{2} \right ]^T \right } \frac{1}{6} \rho_S = A^{(e)} \frac{\rho_S}{3} \left [ 1 \quad 1 \quad 1 \right ]^T