Two-dimensional Shape Functions

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:<math>L_1=\frac{A_1}{A}=\frac{\matbf{CP}}{\matbf{CC'}} \qquad L_2=\frac{A_2}{A}=\frac{\matbf{BP}}{\matbf{BB'}} \qquad L_3=\frac{A_3}{A}=\frac{\matbf{AP}}{\matbf{AA'}} </math>  
:<math>L_1=\frac{A_1}{A}=\frac{\matbf{C}\mathbf{P}}{\matbf{C}\mathbf{C'}} \qquad L_2=\frac{A_2}{A}=\frac{\matbf{B}\mathbf{P}}{\matbf{B}\mathbf{B'}} \qquad L_3=\frac{A_3}{A}=\frac{\matbf{A}\mathbf{P}}{\matbf{A}\mathbf{A'}} </math>  

Revision as of 11:37, 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 nth order, these functions should include a complete polynomial of equal order.

That is, a complete polynomial of nth order can be written as:

f(x,y)=\sum_{i=1}^p \alpha_i x^j y^k \qquad j+k \le n

with:   \qquad p=\frac{(n+1)(n+2)}{2}   the number of terms.

More specifically:

polynomial order n number of terms p f(x,y) \,
Constant: 0 \, 1 \, \alpha \,
Linear: 1 \, 3 \, \alpha_1+\alpha_2 x + \alpha_3 y \,
Quadratic: 2 \, 6 \, \alpha_1+\alpha_2 x + \alpha_3 + \alpha_4 x y +\alpha_5 x^2 + \alpha_6 y^2\,

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
  1 \, 1 \,
Linear x \qquad y \, 3 \,
Quadratic x^2 \qquad 2 x y  \qquad y^2\, 6 \,
Cubic x^3 \qquad 3 x^2 y  \qquad 3 x y^2 \qquad y^3\, 10 \,
Quartic x^4 \qquad 4 x^3 y  \qquad 6 x^2 y^2 \qquad 4 x y^3 \qquad y^4\, 15 \,


Shape Functions for Triangular Elements

The Three Node Linear Triangle

The solution  \varphi^{(e)} (x,y) for each triangular element can be approached by their corresponding \hat \varphi^{(e)} (x,y) to be expressed using the shape functions:
 \varphi^{(e)}(x,y) \cong \hat \varphi^{(e)} (x,y) = \sum_{i=1}^n N_i (x,y) \varphi^{(e)}_i
If the shape functions are lineal polynomials (three-node triangular element, n=3), and remembering:

 N_i^{(e)}(x_j,y_j) =
  1, & i = j \\
  0, & i \ne j
this expression can be written as:
N_i^{(e)} (x,y) = \frac{1}{2 A^{(e)}}  \left [ a_i^{(e)} + b_i^{(e)} x  + c_i^{(e)} y \right ] \qquad     with A^{(e)} \, the element area and   i=1, 2, 3 \,

And the system of equations is:

  1 & x_1^{(e)} & y_1^{(e)} \\
  1 & x_2^{(e)} & y_2^{(e)} \\
  1 & x_3^{(e)} & y_3^{(e)} 
  a_1^{(e)} & a_2^{(e)} & a_3^{(e)} \\
  b_1^{(e)} & b_2^{(e)} & b_3^{(e)} \\
  c_1^{(e)} & c_2^{(e)} & c_3^{(e)} 
= 2 ·A^{(e)}·
  1 & 0 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1 

The element area is computed as the half of the determinant of the coordinates matrix:

2 ·A^{(e)} =
  1 & x_1^{(e)} & y_1^{(e)} \\
  1 & x_2^{(e)} & y_2^{(e)} \\
  1 & x_3^{(e)} & y_3^{(e)} 

Three node triangle.jpg

Finally, the different parameters can be expressed in terms of the nodal local coordinates as:


with  i=1,2,3; \quad j=2,3,1; \quad k=3,1,2 \,

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 A1, A2 and A3, then:

Failed to parse (unknown function\matbf): L_1=\frac{A_1}{A}=\frac{\matbf{C}\mathbf{P}}{\matbf{C}\mathbf{C'}} \qquad L_2=\frac{A_2}{A}=\frac{\matbf{B}\mathbf{P}}{\matbf{B}\mathbf{B'}} \qquad L_3=\frac{A_3}{A}=\frac{\matbf{A}\mathbf{P}}{\matbf{A}\mathbf{A'}}

are also known as area coordinates, because the coordinates of P with respect to triangle ABC are proportional to the (signed) areas of PBC, PCA and PAB. Areal and trilinear coordinates are used for similar purposes in geometry.


  1. Pascal's triangle
  2. Barycentric Coordinates (Areal Coordinates)
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