Two-dimensional Shape Functions

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{| border="1" cellpadding="5" cellspacing="0" class="wikitable" style="margin:auto; background:white;"
 
{| border="1" cellpadding="5" cellspacing="0" class="wikitable" style="margin:auto; background:white;"
 
! polynomial order ''n'' !!  number of terms ''p'' !! <math>f(x,y) \,</math>
 
! polynomial order ''n'' !!  number of terms ''p'' !! <math>f(x,y) \,</math>
−
|-
+
|- align="center"
 
| Constant: <math>0 \,</math> || <math>1 \,</math> || <math>\alpha \,</math>
 
| Constant: <math>0 \,</math> || <math>1 \,</math> || <math>\alpha \,</math>
−
|-
+
|- align="center"
 
| Linear: <math>1 \,</math> || <math>3 \,</math> || <math>\alpha_1+\alpha_2 x + \alpha_3 y \,</math>
 
| Linear: <math>1 \,</math> || <math>3 \,</math> || <math>\alpha_1+\alpha_2 x + \alpha_3 y \,</math>
−
|-
+
|- align="center"
 
| Quadratic: <math>2 \,</math> || <math>6 \,</math> || <math>\alpha_1+\alpha_2 x + \alpha_3 + \alpha_4 x y +\alpha_5 x^2 + \alpha_6 y^2\,</math>
 
| Quadratic: <math>2 \,</math> || <math>6 \,</math> || <math>\alpha_1+\alpha_2 x + \alpha_3 + \alpha_4 x y +\alpha_5 x^2 + \alpha_6 y^2\,</math>
 
|}
 
|}
  
  
 +
A quick way to easily obtain the terms of a complete polynomial is by using the '''Pascal's triangle''':
  
  
 +
{| border="1" cellpadding="5" cellspacing="0" class="wikitable" style="margin:auto; background:white;"
 +
! order ''n'' !! new polynomial terms !! number of terms ''p''
 +
|- align="center"
 +
| &nbsp; || <math>1 \,</math> || <math>1 \,</math>
 +
|- align="center"
 +
| Linear || <math>x \qquad y \,</math> || <math>3 \,</math>
 +
|- align="center"
 +
| Quadratic  || <math>x^2 \qquad 2 x y  \qquad y^2\,</math>  || <math>6 \,</math>
 +
|- align="center"
 +
| Cubic || <math>x^3 \qquad 3 x^2 y  \qquad 3 x y^2 \qquad y^3\,</math>  || <math>10 \,</math>
 +
|- align="center"
 +
| Quartic || <math>x^4 \qquad 4 x^3 y  \qquad 6 x^2 y^2 \qquad 4 x y^3 \qquad y^4\,</math>  || <math>15 \,</math>
 +
|}
 +
 +
== 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 ==
 +
 +
=== The Three Node Linear Triangle ===
 +
 +
:The solution <math> \varphi^{(e)} (x,y)</math> for each triangular element can be approached by their corresponding <math>\hat \varphi^{(e)} (x,y)</math> to be expressed using the shape functions:
 +
 +
::<math> \varphi^{(e)}(x,y) \cong \hat \varphi^{(e)} (x,y) = \sum_{i=1}^n N_i (x,y) \varphi^{(e)}_i </math>
 +
 +
:If the shape functions are lineal polynomials (three-node triangular element, n=3), and remembering:
 +
 +
::<math>
 +
N_i^{(e)}(x_j,y_j) =
 +
\begin{cases}
 +
  1, & i = j \\
 +
  0, & i \ne j
 +
\end{cases}
 +
</math>
 +
 +
:this expression can be written as:
 +
 +
::<math>N_i^{(e)} (x,y) = \frac{1}{2 A^{(e)}}  \left [ a_i^{(e)} + b_i^{(e)} x  + c_i^{(e)} y \right ] \qquad</math> &nbsp;&nbsp;&nbsp;  with <math>A^{(e)} \,</math> the element area and &nbsp; <math>i=1, 2, 3 \,</math>
 +
 +
 +
:And the system of equations is:
 +
 +
 +
::<math>
 +
\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}
 +
</math>
 +
 +
 +
:The element area is computed as the half of the determinant of the coordinates matrix:
 +
 +
 +
::<math>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}
 +
</math>
 +
 +
 +
:[[Image:Three node triangle.jpg]]
 +
 +
 +
:Finally, the different parameters can be expressed in terms of the nodal local coordinates as:
 +
 +
 +
::<math>a_i^{(e)}=x_j^{(e)}y_k^{(e)}-x_k^{(e)}y_j^{(e)}</math>
 +
 +
::<math>b_i^{(e)}=y_j^{(e)}-y_k^{(e)}</math>
 +
 +
::<math>c_i^{(e)}=x_k^{(e)}-x_j^{(e)}</math>
 +
 +
 +
:with &nbsp;<math>i=1,2,3; \quad j=2,3,1; \quad k=3,1,2 \,</math>
 +
 +
 +
=== 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.
 +
 +
 +
::[[Image:ArealCoordinates.jpg|400 px]]
 +
 +
 +
That is, if we use a inner point '''''P''''' of the triangle of area '''A''' as the common vertex of the three subareas '''A<sub>1</sub>''', '''A<sub>2</sub>''' and '''A<sub>3</sub>''', then:
 +
 +
 +
:<math>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'}} </math>
 +
 +
 +
Note that:
 +
* '''A<sub>1</sub>''' + '''A<sub>2</sub>''' + '''A<sub>3</sub>''' = '''A'''
 +
* '''L<sub>1</sub>''' + '''L<sub>2</sub>''' + '''L<sub>3</sub>''' = '''1'''
 +
* If '''''P''''' is the Centroid or Center of Mass of the triangle, then '''L<sub>1</sub>''' = '''L<sub>2</sub>''' = '''L<sub>3</sub>''' = '''1/3'''
 +
 +
 +
For the Finite Element Method is also interesting to note that:
 +
 +
:<math>x_p=L_1 x_1 + L_2 x_2 + L_3 x_3 \,</math>
 +
 +
:<math>y_p=L_1 y_1 + L_2 y_2 + L_3 y_3 \,</math>
 +
 +
 +
with '''x<sub>p</sub>''' and '''y<sub>p</sub>''' the coordinates of '''P''' or any other point inside the triangle '''(x,y)'''. This is equivalent to the following system of equations:
 +
 +
 +
::<math>
 +
\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}
 +
</math>
 +
 +
::<math>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}
 +
</math>
 +
 +
::<math>
 +
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}
 +
</math>
 +
 +
::<math>
 +
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}
 +
</math>
 +
 +
 +
::<math>L_i (x,y) = \frac{1}{2 A}  \left [ a_i + b_i x  + c_i y \right ] = N_i(x,y)</math>
 +
 +
 +
that is exactly the '''shape functions for a triangular element of three nodes'''.
 +
 +
 +
 +
=== Natural Coordinates ===
 +
 +
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It is usual to define the triangle in terms of a normalised geometry (natural coordinates) as is showed in the figure:
 +
 +
 +
:::[[Image:NaturalCoordinates_2.jpg|400 px]]
  
−
For example, in the case of a lineal polynomial:
 
  
 +
Therefore: &nbsp; <math>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 </math>
  
−
:<math>f(x,y)=\alpha_1+\alpha_2 x + \alpha_3 y</math>
 
  
 +
:<math>N_1=1 - \alpha - \beta \qquad N_2=\alpha \qquad N_3=\beta \,</math>
  
−
can only fit polynomial functions of p<sup>th</sup> order if they content a polynomial function
 
  
 +
For isoparametric elements, we have:
  
−
for any polynomial function of pth order it is enough to use p-1 integration points.
 
  
 +
:<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 Jacobian of these shape functions:
  
 +
::<math>\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)}
 +
</math>
  
 
== References ==
 
== References ==
  
 
# [http://en.wikipedia.org/wiki/Pascal%27s_triangle Pascal's triangle]
 
# [http://en.wikipedia.org/wiki/Pascal%27s_triangle Pascal's triangle]
 +
# [http://en.wikipedia.org/wiki/Barycentric_coordinates_%28mathematics%29 Barycentric Coordinates ''(Areal Coordinates)'']
 +
# [http://en.wikipedia.org/wiki/Centroid Centroid]
 +
# [http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Jacobian]
  
  
 
[[Category: Shape Functions]]
 
[[Category: Shape Functions]]

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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