Two-dimensional Shape Functions

From KratosWiki
Jump to: navigation, search

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
Personal tools
Categories