Two-dimensional Shape Functions

(Difference between revisions)
 Revision as of 10:31, 4 November 2009 (view source)JMora (Talk | contribs)← Older edit Revision as of 10:39, 4 November 2009 (view source)JMora (Talk | contribs) Newer edit → Line 44: Line 44: − For example, in the case of a lineal polynomial: + == Shape Functions for Triangular Elements == + === The Three Node Linear Triangle === − :$f(x,y)=\alpha_1+\alpha_2 x + \alpha_3 y$ + :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: + ::$\varphi^{(e)}(x,y) \cong \hat \varphi^{(e)} (x,y) = \sum_{i=1}^n N_i (x,y) \varphi^{(e)}_i$ − can only fit polynomial functions of pth order if they content a polynomial function + :If the shape functions are lineal polynomials (three-node triangular element, n=3), and remembering: + ::$+ N_i^{(e)}(x_j,y_j) = + \begin{cases} + 1, & i = j \\ + 0, & i \ne j + \end{cases} +$ − for any polynomial function of pth order it is enough to use p-1 integration points. + :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: + + + ::$2 ·A^{(e)} = + \begin{vmatrix} + 1 & x_1^{(e)} & y_1^{(e)} \\ + 1 & x_2^{(e)} & y_2^{(e)} \\ + 1 & x_3^{(e)} & y_3^{(e)} + \end{vmatrix} +$ + + + :[[Image: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  $i=1,2,3; \quad j=2,3,1; \quad k=3,1,2 \,$

Revision as of 10:39, 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  $i=1,2,3; \quad j=2,3,1; \quad k=3,1,2 \,$

References

1. Pascal's triangle