Two-dimensional Shape Functions
(→Areal Coordinates) |
(→Areal Coordinates) |
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− | with '''x<sub>p</sub>''' and '''y<sub>p</sub>''' the coordinates of '''P''' or any other point inside the triangle '''(x,y)'''. | + | with '''x<sub>p</sub>''' and '''y<sub>p</sub>''' the coordinates of '''P''' or any other point inside the triangle '''(x,y)'''. This is equivalent to the following system of equations: |
Line 169: | Line 169: | ||
1 | 1 | ||
\end{bmatrix} | \end{bmatrix} | ||
+ | </math> | ||
+ | |||
+ | ::<math>L_1= | ||
+ | \frac{ | ||
+ | \begin{vmatrix} | ||
+ | x_1 & x_2 & x_3 \\ | ||
+ | y_1 & y_2 & y_3 \\ | ||
+ | 1 & 1 & 1 | ||
+ | \end{vmatrix} | ||
+ | } | ||
+ | { | ||
+ | \begin{vmatrix} | ||
+ | x_1 & x_2 & x_3 \\ | ||
+ | y_1 & y_2 & y_3 \\ | ||
+ | 1 & 1 & 1 | ||
+ | \end{vmatrix} | ||
+ | } | ||
+ | \qquad | ||
+ | L_2= | ||
+ | \frac{ | ||
+ | \begin{vmatrix} | ||
+ | x_1 & x_2 & x_3 \\ | ||
+ | y_1 & y_2 & y_3 \\ | ||
+ | 1 & 1 & 1 | ||
+ | \end{vmatrix} | ||
+ | } | ||
+ | { | ||
+ | \begin{vmatrix} | ||
+ | x_1 & x_2 & x_3 \\ | ||
+ | y_1 & y_2 & y_3 \\ | ||
+ | 1 & 1 & 1 | ||
+ | \end{vmatrix} | ||
+ | } | ||
+ | \qquad | ||
+ | L_3= | ||
+ | \frac{ | ||
+ | \begin{vmatrix} | ||
+ | x_1 & x_2 & x_3 \\ | ||
+ | y_1 & y_2 & y_3 \\ | ||
+ | 1 & 1 & 1 | ||
+ | \end{vmatrix} | ||
+ | } | ||
+ | { | ||
+ | \begin{vmatrix} | ||
+ | x_1 & x_2 & x_3 \\ | ||
+ | y_1 & y_2 & y_3 \\ | ||
+ | 1 & 1 & 1 | ||
+ | \end{vmatrix} | ||
+ | } | ||
</math> | </math> | ||
Revision as of 16:11, 4 November 2009
Shape functions are selected to fit as exact as possible the Finite Element Solution. If this solution is a combination of polynomial functions of n^{th} order, these functions should include a complete polynomial of equal order.
That is, a complete polynomial of n^{th} order can be written as:
with: the number of terms.
More specifically:
polynomial order n | number of terms p | |
---|---|---|
Constant: | ||
Linear: | ||
Quadratic: |
A quick way to easily obtain the terms of a complete polynomial is by using the Pascal's triangle:
order n | new polynomial terms | number of terms p |
---|---|---|
Linear | ||
Quadratic | ||
Cubic | ||
Quartic |
Contents |
Shape Functions for Triangular Elements
The Three Node Linear Triangle
- The solution for each triangular element can be approached by their corresponding to be expressed using the shape functions:
- If the shape functions are lineal polynomials (three-node triangular element, n=3), and remembering:
- this expression can be written as:
- with the element area and
- And the system of equations is:
- The element area is computed as the half of the determinant of the coordinates matrix:
- Finally, the different parameters can be expressed in terms of the nodal local coordinates as:
- with
Areal Coordinates
In order to generalise the procedure to obtain the shape functions, the areal coordinates is a very useful transformation.
In a triangle, areal or barycentric coordinates are defined as each of the partial subareas obtained by dividing the triangle in three sections.
That is, if we use a inner point P of the triangle of area A as the common vertex of the three subareas A_{1}, A_{2} and A_{3}, then:
Note that:
- A_{1} + A_{2} + A_{3} = A
- L_{1} + L_{2} + L_{3} = 1
- If P is the Centroid or Center of Mass of the triangle, then L_{1} = L_{2} = L_{3} = 1/3
For the Finite Element Method is also interesting to note that:
with x_{p} and y_{p} the coordinates of P or any other point inside the triangle (x,y). This is equivalent to the following system of equations: