Two-dimensional Shape Functions

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{| border="1" cellpadding="5" cellspacing="0" class="wikitable" style="margin:auto; background:white;"
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! order ''n'' !! new polynomial terms !! number of terms ''p''
 
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| &nbsp; || <math>1 \,</math>  
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| &nbsp; || <math>1 \,</math> || 1
 
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| Linear || <math>x \qquad y \,</math>
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| Linear || <math>x \qquad y \,</math> || 3
 
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| Quadratic  || <math>x^2 \qquad 2 x y  \qquad y^2\,</math>  
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| Quadratic  || <math>x^2 \qquad 2 x y  \qquad y^2\,</math> || 6
 
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| Cubic || <math>x^3 \qquad 3 x^2 y  \qquad 3 x y^2 \qquad y^3\,</math>  
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| Cubic || <math>x^3 \qquad 3 x^2 y  \qquad 3 x y^2 \qquad y^3\,</math> || 10
 
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| Quartic || <math>x^4 \qquad 4 x^3 y  \qquad 6 x^2 y^2 \qquad 4 x y^3 \qquad y^4\,</math>  
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| Quartic || <math>x^4 \qquad 4 x^3 y  \qquad 6 x^2 y^2 \qquad 4 x y^3 \qquad y^4\,</math> || 15
 
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Revision as of 10:30, 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



For example, in the case of a lineal polynomial:


f(x,y) = α1 + α2x + α3y


can only fit polynomial functions of pth order if they content a polynomial function


for any polynomial function of pth order it is enough to use p-1 integration points.




References

  1. Pascal's triangle
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