Two-dimensional Shape Functions
Shape functions are selected to fit as exact as possible the Finite Element Solution. If this solution is a combination of polynomial functions of nth order, these functions should include a complete polynomial of equal order.
That is, a complete polynomial of nth order can be written as:
with: the number of terms.
|polynomial order n||number of terms p|
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|
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:
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 A1, A2 and A3, then:
- A1 + A2 + A3 = A
- L1 + L2 + L3 = 1
- If P is the Centroid or Center of Mass of the triangle, then L1 = L2 = L3 = 1/3
For the Finite Element Method is also interesting to note that:
with xp and yp the coordinates of P or any other point inside the triangle (x,y). This is equivalent to the following system of equations:
that is exactly the shape functions for a triangular element of three nodes.
It is usual to define the triangle in terms of a normalised geometry (natural coordinates) as is showed in the figure:
For isoparametric elements (those using the same shape functions to interpolate the geometry and the unknowns), we have:
To obtain the derivatives of the shape functions:
with the Jacobian matrix with a determinant
From here is easy to obtain: