# Two-dimensional Shape Functions

(Difference between revisions)
 Revision as of 15:57, 4 November 2009 (view source)JMora (Talk | contribs) (→Areal Coordinates)← Older edit Revision as of 16:11, 4 November 2009 (view source)JMora (Talk | contribs) (→Areal Coordinates)Newer edit → Line 149: Line 149: − with '''xp''' and '''yp''' the coordinates of '''P''' or any other point inside the triangle '''(x,y)'''. Therefore: + 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: Line 169: Line 169: 1 1 \end{bmatrix} \end{bmatrix} + [/itex] + + ::$L_1= + \frac{ + \begin{vmatrix} + x_1 & x_2 & x_3 \\ + y_1 & y_2 & y_3 \\ + 1 & 1 & 1 + \end{vmatrix} + } + { + \begin{vmatrix} + x_1 & x_2 & x_3 \\ + y_1 & y_2 & y_3 \\ + 1 & 1 & 1 + \end{vmatrix} + } + \qquad + L_2= + \frac{ + \begin{vmatrix} + x_1 & x_2 & x_3 \\ + y_1 & y_2 & y_3 \\ + 1 & 1 & 1 + \end{vmatrix} + } + { + \begin{vmatrix} + x_1 & x_2 & x_3 \\ + y_1 & y_2 & y_3 \\ + 1 & 1 & 1 + \end{vmatrix} + } + \qquad + L_3= + \frac{ + \begin{vmatrix} + x_1 & x_2 & x_3 \\ + y_1 & y_2 & y_3 \\ + 1 & 1 & 1 + \end{vmatrix} + } + { + \begin{vmatrix} + x_1 & x_2 & x_3 \\ + y_1 & y_2 & y_3 \\ + 1 & 1 & 1 + \end{vmatrix} + }$ [/itex]

## 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 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: $N_i^{(e)}(x_j,y_j) = \begin{cases} 1, & i = j \\ 0, & i \ne j \end{cases}$
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: $2 ·A^{(e)} = \begin{vmatrix} 1 & x_1^{(e)} & y_1^{(e)} \\ 1 & x_2^{(e)} & y_2^{(e)} \\ 1 & x_3^{(e)} & y_3^{(e)} \end{vmatrix}$ 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. 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 or any other point inside the triangle (x,y). This is equivalent to the following system of equations: $\begin{bmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} L_1 \\ L_2 \\ L_3 \end{bmatrix} = \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}$ $L_1= \frac{ \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{vmatrix} } { \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{vmatrix} } \qquad L_2= \frac{ \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{vmatrix} } { \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{vmatrix} } \qquad L_3= \frac{ \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{vmatrix} } { \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{vmatrix} }$