Two-dimensional Shape Functions

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== Derivatives of the shape functions ==
  
  
 +
For isoparametric elements (those using the same shape functions to interpolate the geometry and the unknowns), we have:
 +
 +
 +
::<math>x=\sum_{i=1}^n N_i(\alpha,\beta) x_i \qquad y=\sum_{i=1}^n N_i(\alpha,\beta) y_i </math>
 +
 +
 +
To obtain the derivatives of the shape functions:
 +
 +
 +
::<math>\frac{\partial N_i}{\partial \alpha} = \frac{\partial N_i}{\partial x} \frac{\partial x}{\partial \alpha} + \frac{\partial N_i}{\partial y} \frac{\partial y}{\partial \alpha}</math>
 +
 +
::<math>\frac{\partial N_i}{\partial \beta} = \frac{\partial N_i}{\partial x} \frac{\partial x}{\partial \beta} + \frac{\partial N_i}{\partial y} \frac{\partial y}{\partial \beta}</math>
 +
 +
 +
::<math>
 +
\begin{Bmatrix}
 +
  \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\
 +
  \displaystyle \frac{\partial N_i}{\partial \beta}
 +
\end{Bmatrix}
 +
=
 +
\underbrace{
 +
\begin{bmatrix}
 +
  \displaystyle \frac{\partial x}{\partial \alpha} & \displaystyle \frac{\partial y}{\partial \alpha} \\ \quad \\
 +
  \displaystyle \frac{\partial x}{\partial \beta} & \displaystyle \frac{\partial y}{\partial \beta}
 +
\end{bmatrix}
 +
}_{\mathbf{J^{(e)}}}
 +
\begin{Bmatrix}
 +
  \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\
 +
  \displaystyle \frac{\partial N_i}{\partial y}
 +
\end{Bmatrix}
 +
=
 +
\mathbf{J^{(e)}}
 +
\begin{Bmatrix}
 +
  \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\
 +
  \displaystyle \frac{\partial N_i}{\partial y}
 +
\end{Bmatrix}
 +
</math>
 +
 +
 +
with <math>\mathbf{J^{(e)}} \,</math> the Jacobian matrix with a determinant <math>\mathbf{|J^{(e)}|} \,</math>
 +
 +
 +
::<math>
 +
\begin{Bmatrix}
 +
  \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\
 +
  \displaystyle \frac{\partial N_i}{\partial y}
 +
\end{Bmatrix}
 +
=
 +
\begin{bmatrix}
 +
  \mathbf{J^{(e)}}
 +
\end{bmatrix}^{-1}
 +
\begin{Bmatrix}
 +
  \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\
 +
  \displaystyle \frac{\partial N_i}{\partial \beta}
 +
\end{Bmatrix}
 +
=
 +
\displaystyle
 +
\frac{1}{ \mathbf{|J^{(e)}|}}
 +
\begin{bmatrix}
 +
  \displaystyle \frac{\partial y}{\partial \beta} & \displaystyle -\frac{\partial y}{\partial \alpha} \\ \quad \\
 +
  \displaystyle -\frac{\partial x}{\partial \beta} & \displaystyle \frac{\partial x}{\partial \alpha}
 +
\end{bmatrix}
 +
\begin{Bmatrix}
 +
  \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\
 +
  \displaystyle \frac{\partial N_i}{\partial \beta}
 +
\end{Bmatrix}
 +
</math>
 +
 +
 +
From here is easy to obtain:
 +
 +
 +
::<math>\partial x \partial y = \mathbf{|J^{(e)}|} \partial \alpha \partial \beta \,</math>
 +
 +
 +
::<math>\frac{\partial x}{\partial \alpha} = \sum_{i=1}^n \frac{\partial N_i}{\partial \alpha} x_i \qquad
 +
\frac{\partial x}{\partial \beta} = \sum_{i=1}^n \frac{\partial N_i}{\partial \beta} x_i</math>
 +
 +
 +
::<math>\frac{\partial y}{\partial \alpha} = \sum_{i=1}^n \frac{\partial N_i}{\partial \alpha} y_i \qquad
 +
\frac{\partial y}{\partial \beta} = \sum_{i=1}^n \frac{\partial N_i}{\partial \beta} y_i</math>
 +
 +
 +
::<math>\mathbf{J^{(e)}} = \sum_{i=1}^n
 +
\begin{bmatrix}
 +
  \displaystyle \frac{\partial N_i}{\partial \alpha} x_i & \displaystyle \frac{\partial N_i}{\partial \alpha} y_i \\ \quad \\
 +
  \displaystyle \frac{\partial N_i}{\partial \beta} x_i & \displaystyle \frac{\partial N_i}{\partial \beta} y_i
 +
\end{bmatrix}
 +
</math>
  
 
== Shape Functions for Triangular Elements ==
 
== Shape Functions for Triangular Elements ==
Line 259: Line 349:
  
  
−
For isoparametric elements (those using the same shape functions to interpolate the geometry and the unknowns), we have:
+
For isoparametric elements, we have:
  
  
−
:<math>x = \sum_{i=1}^n N_i(L_1,L_2,L_3) x_i \qquad y = \sum_{i=1}^n N_i(L_1,L_2,L_3) y_i </math>
+
:<math>x = \sum_{i=1}^n N_i(L_1,L_2,L_3) x_i = N_1(\alpha,\beta) x_1 + N_2(\alpha,\beta) x_2 + N_3(\alpha,\beta) x_3 = (1 - \alpha - \beta) x_1 + \alpha x_2 + \beta x_3</math>
  
 +
:<math>y = \sum_{i=1}^n N_i(L_1,L_2,L_3) y_i = N_1(\alpha,\beta) y_1 + N_2(\alpha,\beta) y_2 + N_3(\alpha,\beta) y_3 = (1 - \alpha - \beta) y_1 + \alpha y_2 + \beta y_3</math>
  
−
To obtain the derivatives of the shape functions:
 
  
 +
To obtain the Jacobian of these shape functions:
  
−
::<math>\frac{\partial N_i}{\partial \alpha} = \frac{\partial N_i}{\partial x} \frac{\partial x}{\partial \alpha} + \frac{\partial N_i}{\partial y} \frac{\partial y}{\partial \alpha}</math>
+
::<math>\mathbf{J^{(e)}} =  
−
 
+
−
::<math>\frac{\partial N_i}{\partial \beta} = \frac{\partial N_i}{\partial x} \frac{\partial x}{\partial \beta} + \frac{\partial N_i}{\partial y} \frac{\partial y}{\partial \beta}</math>
+
−
 
+
−
 
+
−
::<math>
+
−
\begin{Bmatrix}
+
−
  \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\
+
−
  \displaystyle \frac{\partial N_i}{\partial \beta}
+
−
\end{Bmatrix}
+
−
=
+
−
\underbrace{
+
 
  \begin{bmatrix}
 
  \begin{bmatrix}
 
   \displaystyle \frac{\partial x}{\partial \alpha} & \displaystyle \frac{\partial y}{\partial \alpha} \\ \quad \\
 
   \displaystyle \frac{\partial x}{\partial \alpha} & \displaystyle \frac{\partial y}{\partial \alpha} \\ \quad \\
 
   \displaystyle \frac{\partial x}{\partial \beta} & \displaystyle \frac{\partial y}{\partial \beta}
 
   \displaystyle \frac{\partial x}{\partial \beta} & \displaystyle \frac{\partial y}{\partial \beta}
 
  \end{bmatrix}
 
  \end{bmatrix}
−
}_{\mathbf{J^{(e)}}}
 
−
\begin{Bmatrix}
 
−
  \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\
 
−
  \displaystyle \frac{\partial N_i}{\partial y}
 
−
\end{Bmatrix}
 
−
=
 
−
\mathbf{J^{(e)}}
 
−
\begin{Bmatrix}
 
−
  \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\
 
−
  \displaystyle \frac{\partial N_i}{\partial y}
 
−
\end{Bmatrix}
 
−
</math>
 
−
 
−
 
−
with <math>\mathbf{J^{(e)}} \,</math> the Jacobian matrix with a determinant <math>\mathbf{|J^{(e)}|} \,</math>
 
−
 
−
 
−
::<math>
 
−
\begin{Bmatrix}
 
−
  \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\
 
−
  \displaystyle \frac{\partial N_i}{\partial y}
 
−
\end{Bmatrix}
 
 
=
 
=
 
  \begin{bmatrix}
 
  \begin{bmatrix}
−
   \mathbf{J^{(e)}}
+
   - x_1 + x_2 & - y_1 + y_2 \\
−
\end{bmatrix}^{-1}
+
   - x_1 + x_3 & - y_1 + y_3
−
\begin{Bmatrix}
+
−
  \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\
+
−
  \displaystyle \frac{\partial N_i}{\partial \beta}
+
−
\end{Bmatrix}
+
−
=
+
−
\displaystyle
+
−
\frac{1}{ \mathbf{|J^{(e)}|}}
+
−
\begin{bmatrix}
+
−
  \displaystyle \frac{\partial y}{\partial \beta} & \displaystyle -\frac{\partial y}{\partial \alpha} \\ \quad \\
+
−
   \displaystyle -\frac{\partial x}{\partial \beta} & \displaystyle \frac{\partial x}{\partial \alpha}
+
 
  \end{bmatrix}
 
  \end{bmatrix}
−
\begin{Bmatrix}
+
\qquad
−
  \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\
+
\mathbf{|J^{(e)}|} = 2 A^{(e)}
−
  \displaystyle \frac{\partial N_i}{\partial \beta}
+
−
\end{Bmatrix}
+
 
</math>
 
</math>
  

Latest revision as of 18:35, 11 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 \,

Contents

Derivatives of the shape functions

For isoparametric elements (those using the same shape functions to interpolate the geometry and the unknowns), we have:


x=\sum_{i=1}^n N_i(\alpha,\beta) x_i \qquad y=\sum_{i=1}^n N_i(\alpha,\beta) y_i


To obtain the derivatives of the shape functions:


\frac{\partial N_i}{\partial \alpha} = \frac{\partial N_i}{\partial x} \frac{\partial x}{\partial \alpha} + \frac{\partial N_i}{\partial y} \frac{\partial y}{\partial \alpha}
\frac{\partial N_i}{\partial \beta} = \frac{\partial N_i}{\partial x} \frac{\partial x}{\partial \beta} + \frac{\partial N_i}{\partial y} \frac{\partial y}{\partial \beta}


\begin{Bmatrix} \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\ \displaystyle \frac{\partial N_i}{\partial \beta} \end{Bmatrix} = \underbrace{ \begin{bmatrix} \displaystyle \frac{\partial x}{\partial \alpha} & \displaystyle \frac{\partial y}{\partial \alpha} \\ \quad \\ \displaystyle \frac{\partial x}{\partial \beta} & \displaystyle \frac{\partial y}{\partial \beta} \end{bmatrix} }_{\mathbf{J^{(e)}}} \begin{Bmatrix} \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\ \displaystyle \frac{\partial N_i}{\partial y} \end{Bmatrix} = \mathbf{J^{(e)}} \begin{Bmatrix} \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\ \displaystyle \frac{\partial N_i}{\partial y} \end{Bmatrix}


with \mathbf{J^{(e)}} \, the Jacobian matrix with a determinant \mathbf{|J^{(e)}|} \,


\begin{Bmatrix} \displaystyle \frac{\partial N_i}{\partial x} \\ \quad \\ \displaystyle \frac{\partial N_i}{\partial y} \end{Bmatrix} = \begin{bmatrix} \mathbf{J^{(e)}} \end{bmatrix}^{-1} \begin{Bmatrix} \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\ \displaystyle \frac{\partial N_i}{\partial \beta} \end{Bmatrix} = \displaystyle \frac{1}{ \mathbf{|J^{(e)}|}} \begin{bmatrix} \displaystyle \frac{\partial y}{\partial \beta} & \displaystyle -\frac{\partial y}{\partial \alpha} \\ \quad \\ \displaystyle -\frac{\partial x}{\partial \beta} & \displaystyle \frac{\partial x}{\partial \alpha} \end{bmatrix} \begin{Bmatrix} \displaystyle \frac{\partial N_i}{\partial \alpha} \\ \quad \\ \displaystyle \frac{\partial N_i}{\partial \beta} \end{Bmatrix}


From here is easy to obtain:


\partial x \partial y = \mathbf{|J^{(e)}|} \partial \alpha \partial \beta \,


\frac{\partial x}{\partial \alpha} = \sum_{i=1}^n \frac{\partial N_i}{\partial \alpha} x_i \qquad \frac{\partial x}{\partial \beta} = \sum_{i=1}^n \frac{\partial N_i}{\partial \beta} x_i


\frac{\partial y}{\partial \alpha} = \sum_{i=1}^n \frac{\partial N_i}{\partial \alpha} y_i \qquad \frac{\partial y}{\partial \beta} = \sum_{i=1}^n \frac{\partial N_i}{\partial \beta} y_i


\mathbf{J^{(e)}} = \sum_{i=1}^n \begin{bmatrix} \displaystyle \frac{\partial N_i}{\partial \alpha} x_i & \displaystyle \frac{\partial N_i}{\partial \alpha} y_i \\ \quad \\ \displaystyle \frac{\partial N_i}{\partial \beta} x_i & \displaystyle \frac{\partial N_i}{\partial \beta} y_i \end{bmatrix}

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 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 & x_2 & x_3 \\ y & 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} } = \frac{x (y_2-y_3)-y (x_2-x_3)+(x_2 y_3 - x_3 y_2)}{2A} = \frac{(x_2 y_3 - x_3 y_2) + x (y_2-y_3) + y (x_3 - x_2)}{2A}
L_2= \frac{ \begin{vmatrix} x_1 & x & x_3 \\ y_1 & y & 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} } = \frac{x_1 (y-y_3)-y_1 (x-x_3)+(x y_3 - x_3 y)}{2A} = \frac{(x_3 y_1 - x_1 y_3) + x (y_3 - y_1) + y (x_1 - x_3)}{2A}
L_3= \frac{ \begin{vmatrix} x_1 & x_2 & x \\ y_1 & y_2 & y \\ 1 & 1 & 1 \end{vmatrix} } { \begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{vmatrix} } = \frac{x_1 (y_2-y)-y_1 (x_2-x)+(x_2 y - x y_2)}{2A} = \frac{(x_1 y_2 - x_2 y_1) + x (y_1 - y_2) + y (x_2 - x_1)}{2A}


L_i (x,y) = \frac{1}{2 A} \left [ a_i + b_i x + c_i y \right ] = N_i(x,y)


that is exactly the shape functions for a triangular element of three nodes.


Natural Coordinates

It is usual to define the triangle in terms of a normalised geometry (natural coordinates) as is showed in the figure:


NaturalCoordinates 2.jpg


Therefore:   N_2 = \frac{A_2}{A} = \frac{\frac{1 \times \alpha }{2}}{\frac{1 \times 1}{2}} \qquad N_3 = \frac{A_3}{A} = \frac{\frac{1 \times \beta}{2}}{\frac{1 \times 1}{2}} \qquad N_1 = \frac{A_1}{A} = \frac{A - A_2 - A_3}{A} = 1 - N_2 - N_3


N_1=1 - \alpha - \beta \qquad N_2=\alpha \qquad N_3=\beta \,


For isoparametric elements, we have:


x = \sum_{i=1}^n N_i(L_1,L_2,L_3) x_i = N_1(\alpha,\beta) x_1 + N_2(\alpha,\beta) x_2 + N_3(\alpha,\beta) x_3 = (1 - \alpha - \beta) x_1 + \alpha x_2 + \beta x_3
y = \sum_{i=1}^n N_i(L_1,L_2,L_3) y_i = N_1(\alpha,\beta) y_1 + N_2(\alpha,\beta) y_2 + N_3(\alpha,\beta) y_3 = (1 - \alpha - \beta) y_1 + \alpha y_2 + \beta y_3


To obtain the Jacobian of these shape functions:

\mathbf{J^{(e)}} = \begin{bmatrix} \displaystyle \frac{\partial x}{\partial \alpha} & \displaystyle \frac{\partial y}{\partial \alpha} \\ \quad \\ \displaystyle \frac{\partial x}{\partial \beta} & \displaystyle \frac{\partial y}{\partial \beta} \end{bmatrix} = \begin{bmatrix} - x_1 + x_2 & - y_1 + y_2 \\ - x_1 + x_3 & - y_1 + y_3 \end{bmatrix} \qquad \mathbf{|J^{(e)}|} = 2 A^{(e)}

References

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