Funciones de forma triangulares

Elemento de tres nodos:

\begin{equation} \left[ \begin{array}{ccc} & 1 & \\\ x & & y \end{array} \right] \end{equation}

El campo de desplazamientos debido a los seis grados de libertad es:

\begin{eqnarray} u &=& \alpha_{1} + \alpha_{2} x + \alpha_{3} y \\\ v &=& \alpha_{4} + \alpha_{5} x + \alpha_{6} y \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} u \\\ v \end{matrix} \right ] = \left [ \begin{matrix} 1 & x & y & 0 & 0 & 0 \\\ 0 & 0 & 0 & 1 & x & y \end{matrix} \right ] \left [ \begin{matrix} \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \\\ \alpha_{6} \end{matrix} \right ] \end{equation}

En forma matricial reducida:

\begin{equation} u = R \ \alpha \end{equation}

Interpolando en coordenadas rectangulares:

\begin{eqnarray} \alpha_{1} + \alpha_{2} x_{1} + \alpha_{3} y_{1} &=& u_{1} \\\ \alpha_{4} + \alpha_{5} x_{1} + \alpha_{6} y_{1} &=& v_{1} \\\ \alpha_{1} + \alpha_{2} x_{2} + \alpha_{3} y_{2} &=& u_{2} \\\ \alpha_{4} + \alpha_{5} x_{2} + \alpha_{6} y_{2} &=& v_{2} \\\ \alpha_{1} + \alpha_{2} x_{3} + \alpha_{3} y_{3} &=& u_{3} \\\ \alpha_{4} + \alpha_{5} x_{3} + \alpha_{6} y_{3} &=& v_{3} \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} 1 & x_{1} & y_{1} & 0 & 0 & 0 \\\ 0 & 0 & 0 & 1 & x_{1} & y_{1} \\\ 1 & x_{2} & y_{2} & 0 & 0 & 0 \\\ 0 & 0 & 0 & 1 & x_{2} & y_{2} \\\ 1 & x_{3} & y_{3} & 0 & 0 & 0 \\\ 0 & 0 & 0 & 1 & x_{3} & y_{3} \end{matrix} \right ] \left [ \begin{matrix} \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \\\ \alpha_{6} \end{matrix} \right ] = \left [ \begin{matrix} u_{1} \\\ v_{1} \\\ u_{2} \\\ v_{2} \\\ u_{3} \\\ v_{3} \end{matrix} \right ] \end{equation}

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \\\ \alpha_{6} \end{matrix} \right ] = \left [ \begin{matrix} \frac{x_{2} y_{3} - x_{3} y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{1} x_{3} - x_{1} y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} y_{2} - y_{1} x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 \\\ \frac{y_{2} - y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{3} - y_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{1} - y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 \\\ \frac{x_{3} - x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} - x_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{2} - x_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 \\\ 0 & \frac{x_{2} y_{3} - x_{3} y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{1} x_{3} - x_{1} y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} y_{2} - y_{1} x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \\\ 0 & \frac{y_{2} - y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{3} - y_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{1} - y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \\\ 0 & \frac{x_{3} - x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} - x_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{2} - x_{1}} {y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \end{matrix} \right ] \left [ \begin{matrix} u_{1} \\\ v_{1} \\\ u_{2} \\\ v_{2} \\\ u_{3} \\\ v_{3} \end{matrix} \right ] \end{equation}

Reemplazando:

\begin{equation} u = \left [ \begin{matrix} 1 & x & y & 0 & 0 & 0 \\\ 0 & 0 & 0 & 1 & x & y \end{matrix} \right ] \left [ \begin{matrix} \frac{x_{2} y_{3} - x_{3} y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{1} x_{3} - x_{1} y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} y_{2} - y_{1} x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 \\\ \frac{y_{2} - y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{3} - y_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{1} - y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 \\\ \frac{x_{3} - x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} - x_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{2} - x_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 \\\ 0 & \frac{x_{2} y_{3} - x_{3} y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{1} x_{3} - x_{1} y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} y_{2} - y_{1} x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \\\ 0 & \frac{y_{2} - y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{3} - y_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{y_{1} - y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \\\ 0 & \frac{x_{3} - x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} - x_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{2} - x_{1}} {y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \end{matrix} \right ] \left [ \begin{matrix} u_{1} \\\ v_{1} \\\ u_{2} \\\ v_{2} \\\ u_{3} \\\ v_{3} \end{matrix} \right ] \end{equation}

Multiplicando:

\begin{equation} u = \left[ \begin{matrix} \frac{x_{2} y_{3} - x_{3} y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(y_{2} - y_{3}) x}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(x_{3} - x_{2}) y}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} y_{1} - y_{3} x_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(y_{3} - y_{1}) x}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(x_{1} - x_{3}) y}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} y_{2} - y_{1} x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(y_{1} - y_{2}) x}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(x_{2} - x_{1}) y}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 \\\ 0 & \frac{x_{2} y_{3} - x_{3} y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(y_{2} - y_{3}) x}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(x_{3} - x_{2}) y}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} y_{1} - y_{3} x_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(y_{3} - y_{1}) x}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(x_{1} - x_{3}) y}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & 0 & \frac{x_{1} y_{2} - y_{1} x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(y_{1} - y_{2}) x}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} + \frac{(x_{2} - x_{1}) y}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \end{matrix} \right] \left [ \begin{matrix} u_{1} \\\ v_{1} \\\ u_{2} \\\ v_{2} \\\ u_{3} \\\ v_{3} \end{matrix} \right ] \end{equation}

El área de un triángulo con vértices $(x_{1}, y_{1})$, $(x_{2}, y_{2})$ y $(x_{3}, y_{3})$ es:

\begin{equation} A = \frac{1}{2} \left | \begin{matrix} x_{1} & y_{1} & 1 \\\ x_{2} & y_{2} & 1 \\\ x_{3} & y_{3} & 1 \end{matrix} \right | = \frac{1}{2} (y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}) \end{equation}

Reemplazando en el anterior sistema:

\begin{equation} u = \frac{1}{2 A} \left[ \begin{matrix} (x_{2} y_{3} - x_{3} y_{2}) + (y_{2} - y_{3}) x + (x_{3} - x_{2}) y & 0 & (x_{1} y_{1} - y_{3} x_{3}) + (y_{3} - y_{1}) x + (x_{1} - x_{3}) y & 0 & (x_{1} y_{2} - y_{1} x_{2}) + (y_{1} - y_{2}) x + (x_{2} - x_{1}) y & 0 \\\ 0 & (x_{2} y_{3} - x_{3} y_{2}) + (y_{2} - y_{3}) x + (x_{3} - x_{2}) y & 0 & (x_{1} y_{1} - y_{3} x_{3}) + (y_{3} - y_{1}) x + (x_{1} - x_{3}) y & 0 & (x_{1} y_{2} - y_{1} x_{2}) + (y_{1} - y_{2}) x + (x_{2} - x_{1}) y \end{matrix} \right] \left [ \begin{matrix} u_{1} \\\ v_{1} \\\ u_{2} \\\ v_{2} \\\ u_{3} \\\ v_{3} \end{matrix} \right ] \end{equation}

En forma matricial:

\begin{equation} u = \left[ \begin{matrix} N_{1} & 0 & N_{2} & 0 & N_{3} & 0 \\\ 0 & N_{1} & 0 & N_{2} & 0 & N_{3} \end{matrix} \right] \left [ \begin{matrix} u_{1} \\\ v_{1} \\\ u_{2} \\\ v_{2} \\\ u_{3} \\\ v_{3} \end{matrix} \right ] \end{equation}

En forma matricial reducida:

\begin{equation} u = N \ u \end{equation}

Interpolando en coordenadas de área:

\begin{eqnarray} N_{1} + N_{2} + N_{3} &=& 1 \\\ x_{1} N_{1} + x_{2} N_{2} + x_{3} N_{3} &=& x \\\ y_{1} N_{1} + y_{2} N_{2} + y_{3} N_{3} &=& y \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} 1 & 1 & 1 \\\ x_{1} & x_{2} & x_{3} \\\ y_{1} & y_{2} & y_{3} \end{matrix} \right ] \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \end{matrix} \right ] = \left [ \begin{matrix} 1 \\\ x \\\ y \end{matrix} \right ] \end{equation}

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \end{matrix} \right ] = \left [ \begin{matrix} \frac{y_{3} x_{2} - y_{2} x_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & \frac{y_{2} - y_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & \frac{x_{3} - x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \\\ \frac{y_{1} x_{3} - y_{3} x_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & \frac{y_{3} - y_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & \frac{x_{1} - x_{3}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \\\ \frac{y_{2} x_{1} - y_{1} x_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & \frac{y_{1} - y_{2}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} & \frac{x_{2} - x_{1}}{y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}} \end{matrix} \right ] \left [ \begin{matrix} 1 \\\ x \\\ y \end{matrix} \right ] \end{equation}

El área de un triángulo con vértices $(x_{1}, y_{1})$, $(x_{2}, y_{2})$ y $(x_{3}, y_{3})$ es:

\begin{equation} A = \frac{1}{2} \left | \left [ \begin{matrix} x_{3} - x_{1} & y_{3} - y_{1} & 0 \end{matrix} \right ] \times \left [ \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & 0 \end{matrix} \right ] \right | = \frac{1}{2} (y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}) \end{equation}

También puede ser escrito en forma de determinante:

\begin{equation} A = \frac{1}{2} \left | \begin{matrix} x_{1} & y_{1} & 1 \\\ x_{2} & y_{2} & 1 \\\ x_{3} & y_{3} & 1 \end{matrix} \right | = \frac{1}{2} (y_{1} x_{3} - y_{1} x_{2} - x_{1} y_{3} + x_{1} y_{2} + x_{2} y_{3} - x_{3} y_{2}) \end{equation}

Reemplazando en el anterior sistema:

\begin{equation} \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \end{matrix} \right ] = \frac{1}{2 A} \left [ \begin{matrix} y_{3} x_{2} - y_{2} x_{3} & y_{2} - y_{3} & x_{3} - x_{2} \\\ y_{1} x_{3} - y_{3} x_{1} & y_{3} - y_{1} & x_{1} - x_{3} \\\ y_{2} x_{1} - y_{1} x_{2} & y_{1} - y_{2} & x_{2} - x_{1} \end{matrix} \right ] \left [ \begin{matrix} 1 \\\ x \\\ y \end{matrix} \right ] \end{equation}

Multiplicando:

\begin{eqnarray} N_{1} &=& \frac{1}{2 A} [(y_{3} x_{2} - y_{2} x_{3}) + (y_{2} - y_{3}) x + (x_{3} - x_{2}) y] \\\ N_{2} &=& \frac{1}{2 A} [(y_{1} x_{3} - y_{3} x_{1}) + (y_{3} - y_{1}) x + (x_{1} - x_{3}) y] \\\ N_{3} &=& \frac{1}{2 A} [(y_{2} x_{1} - y_{1} x_{2}) + (y_{1} - y_{2}) x + (x_{2} - x_{1}) y] \end{eqnarray}

El área del triángulo $1$ con vértices $(x, y)$, $(x_{2}, y_{2})$ y $(x_{3}, y_{3})$ es:

\begin{equation} A_{1} = \frac{1}{2} \left | \left [ \begin{matrix} x_{3} - x & y_{3} - y & 0 \end{matrix} \right ] \times \left [ \begin{matrix} x_{2} - x & y_{2} - y & 0 \end{matrix} \right ] \right | = \frac{1}{2} [(y_{3} x_{2} - y_{2} x_{3}) + (y_{2} - y_{3}) x + (x_{3} - x_{2}) y] \end{equation}

En forma de determinante:

\begin{equation} A_{1} = \frac{1}{2} \left | \begin{matrix} x & y & 1 \\\ x_{1} & y_{1} & 1 \\\ x_{2} & y_{2} & 1 \end{matrix} \right | = \frac{1}{2} [(y_{3} x_{2} - y_{2} x_{3}) + (y_{2} - y_{3}) x + (x_{3} - x_{2}) y] \end{equation}

El área del triángulo $2$ con vértices $(x, y)$, $(x_{3}, y_{3})$ y $(x_{1}, y_{1})$ es:

\begin{equation} A_{2} = \frac{1}{2} \left | \left [ \begin{matrix} x_{1} - x & y_{1} - y & 0 \end{matrix} \right ] \times \left [ \begin{matrix} x_{3} - x & y_{3} - y & 0 \end{matrix} \right ] \right | = \frac{1}{2} [(y_{1} x_{3} - y_{3} x_{1}) + (y_{3} - y_{1}) x + (x_{1} - x_{3}) y] \end{equation}

En forma de determinante:

\begin{equation} A_{2} = \frac{1}{2} \left | \begin{matrix} x & y & 1 \\\ x_{3} & y_{3} & 1 \\\ x_{1} & y_{1} & 1 \end{matrix} \right | = \frac{1}{2} [(y_{1} x_{3} - y_{3} x_{1}) + (y_{3} - y_{1}) x + (x_{1} - x_{3}) y] \end{equation}

El área del triángulo $3$ con vértices $(x, y)$, $(x_{1}, y_{1})$ y $(x_{2}, y_{2})$ es:

\begin{equation} A_{3} = \frac{1}{2} \left | \left [ \begin{matrix} x_{2} - x & y_{2} - y & 0 \end{matrix} \right ] \times \left [ \begin{matrix} x_{1} - x & y_{1} - y & 0 \end{matrix} \right ] \right | = \frac{1}{2} [(y_{2} x_{1} - y_{1} x_{2}) + (y_{1} - y_{2}) x + (x_{2} - x_{1}) y] \end{equation}

En forma de determinante:

\begin{equation} A_{3} = \frac{1}{2} \left | \begin{matrix} x & y & 1 \\\ x_{2} & y_{2} & 1 \\\ x_{1} & y_{1} & 1 \end{matrix} \right | = \frac{1}{2} [(y_{2} x_{1} - y_{1} x_{2}) + (y_{1} - y_{2}) x + (x_{2} - x_{1}) y] \end{equation}

Reemplazando:

\begin{eqnarray} N_{1} &=& \frac{2 A_{1}}{2 A} = \frac{A_{1}}{A} = L_{1} \\\ N_{2} &=& \frac{2 A_{2}}{2 A} = \frac{A_{2}}{A} = L_{2} \\\ N_{3} &=& \frac{2 A_{3}}{2 A} = \frac{A_{3}}{A} = L_{3} \end{eqnarray}

Interpolando en coordenadas naturales:

\begin{eqnarray} N_{1} + N_{2} + N_{3} &=& 1 \\\ \xi_{1} N_{1} + \xi_{2} N_{2} + \xi_{3} N_{3} &=& \xi \\\ \eta_{1} N_{1} + \eta_{2} N_{2} + \eta_{3} N_{3} &=& \eta \end{eqnarray}

Reemplazando los valores nodales:

\begin{eqnarray} N_{1} + N_{2} + N_{3} &=& 1 \\\ (0) N_{1} + (1) N_{2} + (0) N_{3} &=& \xi \\\ (0) N_{1} + (0) N_{2} + (1) N_{3} &=& \eta \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} 1 & 1 & 1 \\\ 0 & 1 & 0\\\ 0 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \end{matrix} \right ] = \left [ \begin{matrix} 1 \\\ \xi \\\ \eta \end{matrix} \right ] \end{equation}

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \end{matrix} \right ] = \left [ \begin{matrix} 1 & -1 & -1 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} 1 \\\ \xi \\\ \eta \end{matrix} \right ] \end{equation}

Multiplicando:

\begin{eqnarray} N_{1} &=& 1 - \xi - \eta = L_{1} \\\ N_{2} &=& \xi = L_{2} \\\ N_{3} &=& \eta = L_{3} \end{eqnarray}

Elemento de seis nodos:

\begin{equation} \left[ \begin{array}{ccccc} & & 1 & & \\\ & x & & y & \\\ x^{2} & & x y & & y^{2} \end{array} \right] \end{equation}

El campo de desplazamientos debido a los doce grados de libertad es:

\begin{eqnarray} u &=& \alpha_{1} + \alpha_{2} x + \alpha_{3} y + \alpha_{4} x^{2} + \alpha_{5} x y + \alpha_{6} y^{2} \\\ v &=& \alpha_{7} + \alpha_{8} x + \alpha_{9} y + \alpha_{10} x^{2} + \alpha_{11} x y + \alpha_{12} y^{2} \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} u \\\ v \end{matrix} \right ] = \left [ \begin{matrix} 1 & x & y & x^{2} & x y & y^{2} & 0 & 0 & 0 & 0 & 0 & 0 \\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & x & y & x^{2} & x y & y^{2} \end{matrix} \right ] \left [ \begin{matrix} \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \\\ \alpha_{6} \\\ \alpha_{7} \\\ \alpha_{8} \\\ \alpha_{9} \\\ \alpha_{10} \\\ \alpha_{11} \\\ \alpha_{12} \end{matrix} \right ] \end{equation}

Interpolando en coordenadas naturales:

\begin{eqnarray} N_{1} + N_{2} + N_{3} + N_{4} + N_{5} + N_{6} &=& 1 \\\ \xi_{1} N_{1} + \xi_{2} N_{2} + \xi_{3} N_{3} + \xi_{4} N_{4} + \xi_{5} N_{5} + \xi_{6} N_{6}&=& \xi \\\ \eta_{1} N_{1} + \eta_{2} N_{2} + \eta_{3} N_{3} + \eta_{4} N_{4} + \eta_{5} N_{5} + \eta_{6} N_{6} &=& \eta \\\ \xi_{1}^{2} N_{1} + \xi_{2}^{2} N_{2} + \xi_{3}^{2} N_{3} + \xi_{4}^{2} N_{4} + \xi_{5}^{2} N_{5} + \xi_{6}^{2} N_{6} &=& \xi^{2} \\\ \xi_{1} \eta_{1} N_{1} + \xi_{2} \eta_{2} N_{2} + \xi_{3} \eta_{3} N_{3} + \xi_{4} \eta_{4} N_{4} + \xi_{5} \eta_{5} N_{5} + \xi_{6} \eta_{6} N_{6} &=& \xi \eta \\\ \eta_{1}^{2} N_{1} + \eta_{2}^{2} N_{2} + \eta_{3}^{2} N_{3} + \eta_{4}^{2} N_{4} + \eta_{5}^{2} N_{5} + \eta_{6}^{2} N_{6} &=& \eta^{2} \end{eqnarray}

Reemplazando los valores nodales:

\begin{eqnarray} N_{1} + N_{2} + N_{3} + N_{4} + N_{5} + N_{6} &=& 1 \\\ (0) N_{1} + \Big ( \frac{1}{2} \Big ) N_{2} + (1) N_{3} + \Big ( \frac{1}{2} \Big ) N_{4} + (0) N_{5} + (0) N_{6} &=& \xi \\\ (0) N_{1} + (0) N_{2} + (0) N_{3} + \Big ( \frac{1}{2} \Big ) N_{4} + (1) N_{5} + \Big ( \frac{1}{2} \Big ) N_{6} &=& \eta \\\ (0)^{2} N_{1} + \Big ( \frac{1}{2} \Big )^{2} N_{2} + (1)^{2} N_{3} + \Big ( \frac{1}{2} \Big )^{2} N_{4} + (0)^{2} N_{5} + (0)^{2} N_{6} &=& \xi^{2} \\\ (0) (0) N_{1} + \Big ( \frac{1}{2} \Big ) (0) N_{2} + (1) (0) N_{3} + \Big ( \frac{1}{2} \Big ) \Big ( \frac{1}{2} \Big ) N_{4} + (0) (1) N_{5} + (0) \Big ( \frac{1}{2} \Big ) N_{6} &=& \xi \eta \\\ (0)^{2} N_{1} + (0)^{2} N_{2} + (0)^{2} N_{3} + \Big ( \frac{1}{2} \Big )^{2} N_{4} + (1)^{2} N_{5} + \Big ( \frac{1}{2} \Big )^{2} N_{6} &=& \eta^{2} \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 \\\ 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} \\\ 0 & \frac{1}{4} & 1 & \frac{1}{4} & 0 & 0 \\\ 0 & 0 & 0 & \frac{1}{4} & 0 & 0 \\\ 0 & 0 & 0 & \frac{1}{4} & 1 & \frac{1}{4} \end{matrix} \right ] \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \\\ N_{4} \\\ N_{5} \\\ N_{6} \end{matrix} \right ] = \left [ \begin{matrix} 1 \\\ \xi \\\ \eta \\\ \xi^{2} \\\ \xi \eta \\\ \eta^{2} \end{matrix} \right ] \end{equation}

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \\\ N_{4} \\\ N_{5} \\\ N_{6} \end{matrix} \right ] = \left [ \begin{matrix} 1 & -3 & -3 & 2 & 4 & 2 \\\ 0 & 4 & 0 & -4 & -4 & 0 \\\ 0 & -1 & 0 & 2 & 0 & 0 \\\ 0 & 0 & 0 & 0 & 4 & 0 \\\ 0 & 0 & -1 & 0 & 0 & 2 \\\ 0 & 0 & 4 & 0 & -4 & -4 \end{matrix} \right ] \left [ \begin{matrix} 1 \\\ \xi \\\ \eta \\\ \xi^{2} \\\ \xi \eta \\\ \eta^{2} \end{matrix} \right ] \end{equation}

Multiplicando:

\begin{eqnarray} N_{1} &=& 1 - 3 \xi - 3 \eta + 2 \xi^{2} + 4 \xi \eta + 2 \eta^{2} = (1 - \xi - \eta) [2 (1 - \xi - \eta) - 1] = L_{1} (2 L_{1} - 1) \\\ N_{2} &=& 4 \xi - 4 \xi^{2} - 4 \xi \eta = 4 (1 - \xi - \eta) \xi = 4 L_{1} L_{2} \\\ N_{3} &=& -\xi + 2 \xi^{2} = \xi (2 \xi - 1) = L_{2} (2 L_{2} - 1) \\\ N_{4} &=& 4 \xi \eta = 4 L_{2} L_{3} \\\ N_{5} &=& -\eta + 2 \eta^{2} = \eta (2 \eta - 1) = L_{3} (2 L_{3} - 1) \\\ N_{6} &=& 4 \eta - 4 \xi \eta - 4 \eta^{2} = 4 (1 - \xi - \eta) \eta = 4 L_{1} L_{3} \end{eqnarray}

Elemento de diez nodos:

\begin{equation} \left[ \begin{array}{ccccccc} & & & 1 & & & \\\ & & x & & y & & \\\ & x^{2} & & x y & & y^{2} & \\\ x^{3} & & x^{2} y & & x y^{2} & & y^{3} \end{array} \right] \end{equation}

El campo de desplazamientos debido a los veinte grados de libertad es:

\begin{eqnarray} u &=& \alpha_{1} + \alpha_{2} x + \alpha_{3} y + \alpha_{4} x^{2} + \alpha_{5} x y + \alpha_{6} y^{2} + \alpha_{7} x^{3} + \alpha_{8} x^{2} y + \alpha_{9} x y^{2} + \alpha_{10} y^{3} \\\ v &=& \alpha_{11} + \alpha_{12} x + \alpha_{13} y + \alpha_{14} x^{2} + \alpha_{15} x y + \alpha_{16} y^{2} + \alpha_{17} x^{3} + \alpha_{18} x^{2} y + \alpha_{19} x y^{2} + \alpha_{20} y^{3} \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} u \\\ v \end{matrix} \right ] = \left [ \begin{matrix} 1 & x & y & x^{2} & x y & y^{2} & x^{3} & x^{2} y & x y^{2} & y^{3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & x & y & x^{2} & x y & y^{2} & x^{3} & x^{2} y & x y^{2} & y^{3} \end{matrix} \right ] \left [ \begin{matrix} \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \\\ \alpha_{6} \\\ \alpha_{7} \\\ \alpha_{8} \\\ \alpha_{9} \\\ \alpha_{10} \\\ \alpha_{11} \\\ \alpha_{12} \\\ \alpha_{13} \\\ \alpha_{14} \\\ \alpha_{15} \\\ \alpha_{16} \\\ \alpha_{17} \\\ \alpha_{18} \\\ \alpha_{19} \\\ \alpha_{20} \end{matrix} \right ] \end{equation}

Interpolando en coordenadas naturales:

\begin{eqnarray} N_{1} + N_{2} + N_{3} + N_{4} + N_{5} + N_{6} + N_{7} + N_{8} + N_{9} + N_{10} &=& 1 \\\ \xi_{1} N_{1} + \xi_{2} N_{2} + \xi_{3} N_{3} + \xi_{4} N_{4} + \xi_{5} N_{5} + \xi_{6} N_{6} + \xi_{7} N_{7} + \xi_{8} N_{8} + \xi_{9} N_{9} + \xi_{10} N_{10} &=& \xi \\\ \eta_{1} N_{1} + \eta_{2} N_{2} + \eta_{3} N_{3} + \eta_{4} N_{4} + \eta_{5} N_{5} + \eta_{6} N_{6} + \eta_{7} N_{7} + \eta_{8} N_{8} + \eta_{9} N_{9} + \eta_{10} N_{10} &=& \eta \\\ \xi_{1}^{2} N_{1} + \xi_{2}^{2} N_{2} + \xi_{3}^{2} N_{3} + \xi_{4}^{2} N_{4} + \xi_{5}^{2} N_{5} + \xi_{6}^{2} N_{6} + \xi_{7}^{2} N_{7} + \xi_{8}^{2} N_{8} + \xi_{9}^{2} N_{9} + \xi_{10}^{2} N_{10} &=& \xi^{2} \\\ \xi_{1} \eta_{1} N_{1} + \xi_{2} \eta_{2} N_{2} + \xi_{3} \eta_{3} N_{3} + \xi_{4} \eta_{4} N_{4} + \xi_{5} \eta_{5} N_{5} + \xi_{6} \eta_{6} N_{6} + \xi_{7} \eta_{7} N_{7} + \xi_{8} \eta_{8} N_{8} + \xi_{9} \eta_{9} N_{9} + \xi_{10} \eta_{10} N_{10} &=& \xi \eta \\\ \eta_{1}^{2} N_{1} + \eta_{2}^{2} N_{2} + \eta_{3}^{2} N_{3} + \eta_{4}^{2} N_{4} + \eta_{5}^{2} N_{5} + \eta_{6}^{2} N_{6} + \eta_{7}^{2} N_{7} + \eta_{8}^{2} N_{8} + \eta_{9}^{2} N_{9} + \eta_{10}^{2} N_{10} &=& \eta^{2} \\\ \xi_{1}^{3} N_{1} + \xi_{2}^{3} N_{2} + \xi_{3}^{3} N_{3} + \xi_{4}^{3} N_{4} + \xi_{5}^{3} N_{5} + \xi_{6}^{3} N_{6} + \xi_{7}^{3} N_{7} + \xi_{8}^{3} N_{8} + \xi_{9}^{3} N_{9} + \xi_{10}^{3} N_{10} &=& \xi^{3} \\\ \xi_{1}^{2} \eta_{1} N_{1} + \xi_{2}^{2} \eta_{2} N_{2} + \xi_{3}^{2} \eta_{3} N_{3} + \xi_{4}^{2} \eta_{4} N_{4} + \xi_{5}^{2} \eta_{5} N_{5} + \xi_{6}^{2} \eta_{6} N_{6} + \xi_{7}^{2} \eta_{7} N_{7} + \xi_{8}^{2} \eta_{8} N_{8} + \xi_{9}^{2} \eta_{9} N_{9} + \xi_{10}^{2} \eta_{10} N_{10} &=& \xi^{2} \eta \\\ \xi_{1} \eta_{1}^{2} N_{1} + \xi_{2} \eta_{2}^{2} N_{2} + \xi_{3} \eta_{3}^{2} N_{3} + \xi_{4} \eta_{4}^{2} N_{4} + \xi_{5} \eta_{5}^{2} N_{5} + \xi_{6} \eta_{6}^{2} N_{6} + \xi_{7} \eta_{7}^{2} N_{7} + \xi_{8} \eta_{8}^{2} N_{8} + \xi_{9} \eta_{9}^{2} N_{9} + \xi_{10} \eta_{10}^{2} N_{10} &=& \xi \eta^{2} \\\ \eta_{1}^{3} N_{1} + \eta_{2}^{3} N_{2} + \eta_{3}^{3} N_{3} + \eta_{4}^{3} N_{4} + \eta_{5}^{3} N_{5} + \eta_{6}^{3} N_{6} + \eta_{7}^{3} N_{7} + \eta_{8}^{3} N_{8} + \eta_{9}^{3} N_{9} + \eta_{10}^{3} N_{10} &=& \eta^{3} \end{eqnarray}

Reemplazando los valores nodales:

\begin{eqnarray} N_{1} + N_{2} + N_{3} + N_{4} + N_{5} + N_{6} + N_{7} + N_{8} + N_{9} + N_{10} &=& 1 \\\ (0) N_{1} + \Big (\frac{1}{3} \Big) N_{2} + \Big (\frac{2}{3} \Big) N_{3} + (1) N_{4} + \Big (\frac{2}{3} \Big) N_{5} + \Big (\frac{1}{3} \Big) N_{6} + (0) N_{7} + (0) N_{8} + (0) N_{9} + \Big (\frac{1}{3} \Big) N_{10} &=& \xi \\\ (0) N_{1} + (0) N_{2} + (0) N_{3} + (0) N_{4} + \Big (\frac{1}{3} \Big) N_{5} + \Big (\frac{2}{3} \Big) N_{6} + (1) N_{7} + \Big (\frac{2}{3} \Big) N_{8} + \Big (\frac{1}{3} \Big) N_{9} + \Big (\frac{1}{3} \Big) N_{10} &=& \eta \\\ (0)^{2} N_{1} + \Big (\frac{1}{3} \Big)^{2} N_{2} + \Big (\frac{2}{3} \Big)^{2} N_{3} + (1)^{2} N_{4} + \Big (\frac{2}{3} \Big)^{2} N_{5} + \Big (\frac{1}{3} \Big)^{2} N_{6} + (0)^{2} N_{7} + (0)^{2} N_{8} + (0)^{2} N_{9} + \Big (\frac{1}{3} \Big)^{2} N_{10} &=& \xi^{2} \\\ (0) (0) N_{1} + \Big (\frac{1}{3} \Big) (0) N_{2} + \Big (\frac{2}{3} \Big) (0) N_{3} + (1) (0) N_{4} + \Big (\frac{2}{3} \Big) \Big (\frac{1}{3} \Big) N_{5} + \Big (\frac{1}{3} \Big) \Big (\frac{2}{3} \Big) N_{6} + (0) (1) N_{7} + (0) \Big (\frac{2}{3} \Big) N_{8} + (0) \Big (\frac{1}{3} \Big) N_{9} + \Big (\frac{1}{3} \Big) \Big (\frac{1}{3} \Big) N_{10} &=& \xi \eta \\\ (0)^{2} N_{1} + (0)^{2} N_{2} + (0)^{2} N_{3} + (0)^{2} N_{4} + \Big (\frac{1}{3} \Big)^{2} N_{5} + \Big (\frac{2}{3} \Big)^{2} N_{6} + (1)^{2} N_{7} + \Big (\frac{2}{3} \Big)^{2} N_{8} + \Big (\frac{1}{3} \Big)^{2} N_{9} + \Big (\frac{1}{3} \Big)^{2} N_{10} &=& \eta^{2} \\\ (0)^{3} N_{1} + \Big (\frac{1}{3} \Big)^{3} N_{2} + \Big (\frac{2}{3} \Big)^{3} N_{3} + (1)^{3} N_{4} + \Big (\frac{2}{3} \Big)^{3} N_{5} + \Big (\frac{1}{3} \Big)^{3} N_{6} + (0)^{3} N_{7} + (0)^{3} N_{8} + (0)^{3} N_{9} + \Big (\frac{1}{3} \Big)^{3} N_{10} &=& \xi^{3} \\\ (0)^{2} (0) N_{1} + \Big (\frac{1}{3} \Big)^{2} (0) N_{2} + \Big (\frac{2}{3} \Big)^{2} (0) N_{3} + (1)^{2} (0) N_{4} + \Big (\frac{2}{3} \Big)^{2} \Big (\frac{1}{3} \Big) N_{5} + \Big (\frac{1}{3} \Big)^{2} \Big (\frac{2}{3} \Big) N_{6} + (0)^{2} (1) N_{7} + (0)^{2} \Big (\frac{2}{3} \Big) N_{8} + (0)^{2} \Big (\frac{1}{3} \Big) N_{9} + \Big (\frac{1}{3} \Big)^{2} \Big (\frac{1}{3} \Big) N_{10} &=& \xi^{2} \eta \\\ (0) (0)^{2} N_{1} + \Big (\frac{1}{3} \Big) (0)^{2} N_{2} + \Big (\frac{2}{3} \Big) (0)^{2} N_{3} + (1) (0)^{2} N_{4} + \Big (\frac{2}{3} \Big) \Big (\frac{1}{3} \Big)^{2} N_{5} + \Big (\frac{1}{3} \Big) \Big (\frac{2}{3} \Big)^{2} N_{6} + (0) (1)^{2} N_{7} + (0) \Big (\frac{2}{3} \Big)^{2} N_{8} + (0) \Big (\frac{1}{3} \Big)^{2} N_{9} + \Big (\frac{1}{3} \Big) \Big (\frac{1}{3} \Big)^{2} N_{10} &=& \xi \eta^{2} \\\ (0)^{3} N_{1} + (0)^{3} N_{2} + (0)^{3} N_{3} + (0)^{3} N_{4} + \Big (\frac{1}{3} \Big)^{3} N_{5} + \Big (\frac{2}{3} \Big)^{3} N_{6} + (1)^{3} N_{7} + \Big (\frac{2}{3} \Big)^{3} N_{8} + \Big (\frac{1}{3} \Big)^{3} N_{9} + \Big (\frac{1}{3} \Big)^{3} N_{10} &=& \eta^{3} \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 & \frac{1}{3} \\\ 0 & 0 & 0 & 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{2}{3} & \frac{1}{3} & \frac{1}{3} \\\ 0 & \frac{1}{9} & \frac{4}{9} & 1 & \frac{4}{9} & \frac{1}{9} & 0 & 0 & 0 & \frac{1}{9} \\\ 0 & 0 & 0 & 0 & \frac{2}{9} & \frac{2}{9} & 0 & 0 & 0 & \frac{1}{9} \\\ 0 & 0 & 0 & 0 & \frac{1}{9} & \frac{4}{9} & 1 & \frac{4}{9} & \frac{1}{9} & \frac{1}{9} \\\ 0 & \frac{1}{27} & \frac{8}{27} & 1 & \frac{8}{27} & \frac{1}{27} & 0 & 0 & 0 & \frac{1}{27} \\\ 0 & 0 & 0 & 0 & \frac{4}{27} & \frac{2}{27} & 0 & 0 & 0 & \frac{1}{27} \\\ 0 & 0 & 0 & 0 & \frac{2}{27} & \frac{4}{27} & 0 & 0 & 0 & \frac{1}{27} \\\ 0 & 0 & 0 & 0 & \frac{1}{27} & \frac{8}{27} & 1 & \frac{8}{27} & \frac{1}{27} & \frac{1}{27} \end{matrix} \right ] \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \\\ N_{4} \\\ N_{5} \\\ N_{6} \\\ N_{7} \\\ N_{8} \\\ N_{9} \\\ N_{10} \end{matrix} \right ] = \left [ \begin{matrix} 1 \\\ \xi \\\ \eta \\\ \xi^{2} \\\ \xi \eta \\\ \eta^{2} \\\ \xi^{3} \\\ \xi^{2} \eta \\\ \xi \eta^{2} \\\ \eta^{3} \end{matrix} \right ] \end{equation}

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} N_{1} \\\ N_{2} \\\ N_{3} \\\ N_{4} \\\ N_{5} \\\ N_{6} \\\ N_{7} \\\ N_{8} \\\ N_{9} \\\ N_{10} \end{matrix} \right ] = \left [ \begin{matrix} 1 & -\frac{11}{2} & -\frac{11}{2} & 9 & 18 & 9 & -\frac{9}{2} & -\frac{27}{2} & -\frac {27}{2} & -\frac{9}{2} \\\ 0 & 9 & 0 & -\frac{45}{2} & -\frac{45}{2} & 0 & \frac{27}{2} & 27 & \frac{27}{2} & 0 \\\ 0 & -\frac{9}{2} & 0 & 18 & \frac{9}{2} & 0 & -\frac{27}{2} & -\frac{27}{2} & 0 & 0 \\\ 0 & 1 & 0 & -\frac{9}{2} & 0 & 0 & \frac{9}{2} & 0 & 0 & 0 \\\ 0 & 0 & 0 & 0 & -\frac{9}{2} & 0 & 0 & \frac{27}{2} & 0 & 0 \\\ 0 & 0 & 0 & 0 & -\frac{9}{2} & 0 & 0 & 0 & \frac{27}{2} & 0 \\\ 0 & 0 & 1 & 0 & 0 & -\frac{9}{2} & 0 & 0 & 0 & \frac{9}{2} \\\ 0 & 0 & -\frac{9}{2} & 0 & \frac{9}{2} & 18 & 0 & 0 & -\frac{27}{2} & -\frac{27}{2} \\\ 0 & 0 & 9 & 0 & -\frac{45}{2} & -\frac{45}{2} & 0 & \frac{27}{2} & 27 & \frac{27}{2} \\\ 0 & 0 & 0 & 0 & 27 & 0 & 0 & -27 & -27 & 0 \end{matrix} \right ] \left [ \begin{matrix} 1 \\\ \xi \\\ \eta \\\ \xi^{2} \\\ \xi \eta \\\ \eta^{2} \\\ \xi^{3} \\\ \xi^{2} \eta \\\ \xi \eta^{2} \\\ \eta^{3} \end{matrix} \right ] \end{equation}

Multiplicando:

\begin{eqnarray} N_{1} &=& 1 - \frac{11}{2} \xi - \frac{11}{2} \eta + 9 \xi^{2} + 18 \xi y + 9 \eta^{2} - \frac{9}{2} \xi^{3} - \frac{27}{2} \xi^{2} \eta - \frac{27}{2} \xi \eta^{2} - \frac{9}{2} \eta^{3} = \frac{1}{2} (1 - \xi - \eta) [3 (1 - \xi - \eta) - 2] [3 (1 - \xi - \eta) - 1] = \frac{1}{2} L_{1} (3 L_{1} - 2) (3 L_{1} - 1) \\\ N_{2} &=& 9 \xi - \frac{45}{2} \xi^{2} - \frac{45}{2} \xi \eta + \frac{27}{2} \xi^{3} + 27 \xi^{2} \eta + \frac{27}{2} \xi \eta^{2} = \frac{9}{2} (1 - \xi - \eta) \xi [3 (1 - \xi - \eta) - 1] = \frac{9}{2} L_{1} L_{2} (3 L_{1} - 1) \\\ N_{3} &=& -\frac{9}{2} \xi + 18 \xi^{2} + \frac{9}{2} \xi \eta - \frac{27}{2} \xi^{3} - \frac{27}{2} \xi^{2} \eta = \frac{9}{2} (1 - \xi - \eta) \xi (3 \xi - 1) = \frac{9}{2} L_{1} L_{2} (3 L_{2} - 1) \\\ N_{4} &=& \xi - \frac{9}{2} \xi^{2} + \frac{9}{2} \xi^{3} = \frac{1}{2} \xi (3 \xi - 2) (3 \xi - 1) = \frac{1}{2} L_{2} (3 L_{2} - 2) (3 L_{2} - 1) \\\ N_{5} &=& -\frac{9}{2} \xi \eta + \frac{27}{2} \xi^{2} \eta = \frac{9}{2} \xi \eta (3 \xi - 1) = \frac{9}{2} L_{2} L_{3} (3 L_{2} - 1) \\\ N_{6} &=& -\frac{9}{2} \xi \eta + \frac{27}{2} \xi \eta^{2} = \frac{9}{2} \xi \eta (3 \eta - 1) = \frac{9}{2} L_{2} L_{3} (3 L_{3} - 1) \\\ N_{7} &=& \eta - \frac{9}{2} \eta^{2} + \frac{9}{2} \eta^{3} = \frac{1}{2} \eta (3 \eta - 2) (3 \eta - 1) = \frac{1}{2} L_{3} (3 L_{3} - 2) (3 L_{3} - 1) \\\ N_{8} &=& -\frac{9}{2} \eta + \frac{9}{2} \xi \eta + 18 \eta^{2} - \frac{27}{2} \xi \eta^{2} - \frac{27}{2} \eta^{3} = \frac{9}{2} (1 - \xi - \eta) \eta (3 \xi - 1) = \frac{9}{2} L_{1} L_{3} (3 L_{3} - 1) \\\ N_{9} &=& 9 \eta - \frac{45}{2} \xi \eta - \frac{45}{2} \eta^{2} + \frac{27}{2} \xi^{2} \eta + 27 \xi \eta^{2} + \frac{27}{2} \eta^{3} = \frac{9}{2} (1 - \xi - \eta) \eta [3 (1 - \xi - \eta) - 1] = \frac{9}{2} L_{1} L_{3} (3 L_{1} - 1) \\\ N_{10} &=& 27 \xi \eta - 27 \xi^{2} \eta - 27 \xi \eta^{2} = 27 (1 - \xi - \eta) \xi \eta = 27 L_{1} L_{2} L_{3} \end{eqnarray}