Two-dimensional Shape Functions

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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 \,



Contents

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) =
 \begin{cases}
  1, & i = j \\
  0, & i \ne j
 \end{cases}
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:



 \begin{bmatrix}
  1 & x_1^{(e)} & y_1^{(e)} \\
  1 & x_2^{(e)} & y_2^{(e)} \\
  1 & x_3^{(e)} & y_3^{(e)} 
 \end{bmatrix}
 \begin{bmatrix}
  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)} 
 \end{bmatrix}
= 2 ·A^{(e)}·
 \begin{bmatrix}
  1 & 0 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1 
 \end{bmatrix}


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


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


Three node triangle.jpg


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


a_i^{(e)}=x_j^{(e)}y_k^{(e)}-x_k^{(e)}y_j^{(e)}
b_i^{(e)}=y_j^{(e)}-y_k^{(e)}
c_i^{(e)}=x_k^{(e)}-x_j^{(e)}


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.


ArealCoordinates.jpg


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:


L_1=\frac{A_1}{A}=\frac{\mathbf{CP}}{\mathbf{C}\mathbf{C'}} \qquad L_2=\frac{A_2}{A}=\frac{\mathbf{B}\mathbf{P}}{\mathbf{B}\mathbf{B'}} \qquad L_3=\frac{A_3}{A}=\frac{\mathbf{A}\mathbf{P}}{\mathbf{A}\mathbf{A'}}


Note that:

  • 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:

x_p=L_1 x_1 + L_2 x_2 + L_3 x_3 \,
y_p=L_1 y_1 + L_2 y_2 + L_3 y_3 \,


with xp and yp the coordinates of P.

References

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