# Two-dimensional Shape Functions

(Difference between revisions)
 Revision as of 09:49, 4 November 2009 (view source)JMora (Talk | contribs)← Older edit Latest revision as of 18:35, 11 November 2009 (view source)JMora (Talk | contribs) (→Natural Coordinates) (75 intermediate revisions by one user not shown) Line 7: Line 7: − with $p=frac{(n+1)(n+2)}{2}$ + with:   $\qquad p=\frac{(n+1)(n+2)}{2}$   the number of terms. + More specifically: − can only fit polynomial functions of pth order if they content a polynomial function + {| border="1" cellpadding="5" cellspacing="0" class="wikitable" style="margin:auto; background:white;" + ! polynomial order ''n'' !!  number of terms ''p'' !! $f(x,y) \,$ + |- align="center" + | Constant: $0 \,$ || $1 \,$ || $\alpha \,$ + |- align="center" + | Linear: $1 \,$ || $3 \,$ || $\alpha_1+\alpha_2 x + \alpha_3 y \,$ + |- align="center" + | Quadratic: $2 \,$ || $6 \,$ || $\alpha_1+\alpha_2 x + \alpha_3 + \alpha_4 x y +\alpha_5 x^2 + \alpha_6 y^2\,$ + |} − for any polynomial function of pth order it is enough to use p-1 integration points. + 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" + |   || $1 \,$ || $1 \,$ + |- align="center" + | Linear || $x \qquad y \,$ || $3 \,$ + |- align="center" + | Quadratic  || $x^2 \qquad 2 x y \qquad y^2\,$  || $6 \,$ + |- align="center" + | Cubic || $x^3 \qquad 3 x^2 y \qquad 3 x y^2 \qquad y^3\,$  || $10 \,$ + |- align="center" + | Quartic || $x^4 \qquad 4 x^3 y \qquad 6 x^2 y^2 \qquad 4 x y^3 \qquad y^4\,$  || $15 \,$ + |} + == 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: + + + ::$2 ·A^{(e)} = + \begin{vmatrix} + 1 & x_1^{(e)} & y_1^{(e)} \\ + 1 & x_2^{(e)} & y_2^{(e)} \\ + 1 & x_3^{(e)} & y_3^{(e)} + \end{vmatrix} +$ + + + :[[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  $i=1,2,3; \quad j=2,3,1; \quad k=3,1,2 \,$ + + + === 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 '''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: + + + :::[[Image:NaturalCoordinates_2.jpg|400 px]] + + + 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 == == 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 $x^4 \qquad 4 x^3 y \qquad 6 x^2 y^2 \qquad 4 x y^3 \qquad y^4\,$ $15 \,$

## 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:

$2 ·A^{(e)} = \begin{vmatrix} 1 & x_1^{(e)} & y_1^{(e)} \\ 1 & x_2^{(e)} & y_2^{(e)} \\ 1 & x_3^{(e)} & y_3^{(e)} \end{vmatrix}$

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.

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:

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:

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