Funciones de forma unidimensionales

Las funciones de forma unidimensionales sirven para aproximar los desplazamientos:

\begin{equation} u = \alpha_{0} + \alpha_{1} x + \cdots + \alpha_{n} x^{n} = \sum_{i = 0}^{n} \alpha_{i} x^{i} \end{equation}

Elemento barra

Los elementos barra soportan esfuerzos debidos a tracción o compresión.

Elemento de dos nodos

Para un elemento de dos nodos y dos grados de libertad:

\begin{equation} u = \alpha_{0} + \alpha_{1} x = \left [ \begin{matrix} 1 & x \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \end{matrix} \right ] \end{equation}

Reemplazando los valores nodales en coordenadas naturales:

\begin{eqnarray} \alpha_{0} + \alpha_{1}(-1) &=& u_{1}\\\ \alpha_{0} + \alpha_{1}(1) &=& u_{2} \end{eqnarray}

Evaluando:

\begin{eqnarray} \alpha_{0} - \alpha_{1} &=& u_{1}\\\ \alpha_{0} + \alpha_{1} &=& u_{2} \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} 1 & -1 \\\ 1 & 1 \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \end{matrix} \right ] = \left [ \begin{matrix} u_{1} \\\ u_{2} \end{matrix} \right ] \end{equation}

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \end{matrix} \right ] = \left [ \begin{matrix} \frac{1}{2} & \frac{1}{2} \\\ -\frac{1}{2} & \frac{1}{2} \end{matrix} \right ] \left [ \begin{matrix} u_{1} \\\ u_{2} \end{matrix} \right ] \end{equation}

Reemplazando:

\begin{equation} u = \left [ \begin{matrix} 1 & x \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \end{matrix} \right ] = \left [ \begin{matrix} 1 & x \end{matrix} \right ] \left [ \begin{matrix} \frac{1}{2} & \frac{1}{2} \\\ -\frac{1}{2} & \frac{1}{2} \end{matrix} \right ] \left [ \begin{matrix} u_{1} \\\ u_{2} \end{matrix} \right ] = \left [ \begin{matrix} \frac{1}{2} - \frac{1}{2} x & \frac{1}{2} + \frac{1}{2} x \end{matrix} \right ] \left [ \begin{matrix} u_{1} \\\ u_{2} \end{matrix} \right ] \end{equation}

Reescribiendo $u$:

\begin{equation} u = \Big ( \frac{1}{2} - \frac{1}{2} x \Big ) u_{1} + \Big ( \frac{1}{2} + \frac{1}{2} x \Big ) u_{2} = N_{1} u_{1} + N_{2} u_{2} \end{equation}

Elemento de tres nodos

Para un elemento de tres nodos y tres grados de libertad:

\begin{equation} u = \alpha_{0} + \alpha_{1} x + \alpha_{2} x^{2} = \left [ \begin{matrix} 1 & x & x^{2} \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \end{matrix} \right ] \end{equation}

Reemplazando los valores nodales en coordenadas naturales:

\begin{eqnarray} \alpha_{0} + \alpha_{1}(-1) + \alpha_{2}(-1)^{2}&=& u_{1}\\\ \alpha_{0} + \alpha_{1}(0) + \alpha_{2}(0)^{2}&=& u_{2}\\\ \alpha_{0} + \alpha_{1}(1) + \alpha_{2}(1)^{2}&=& u_{3} \end{eqnarray}

Evaluando:

\begin{eqnarray} \alpha_{0} - \alpha_{1} + \alpha_{1} &=& u_{1}\\\ \alpha_{0} &=& u_{2}\\\ \alpha_{0} + \alpha_{1} + \alpha_{1}&=& u_{3} \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} 1 & -1 & 1 \\\ 1 & 0 & 0 \\\ 1 & 1 & 1 \\\ \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \end{matrix} \right ] = \left [ \begin{matrix} u_{1} \\\ u_{2} \\\ u_{3} \end{matrix} \right ] \end{equation}

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \end{matrix} \right ] = \left [ \begin{matrix} 0 & 1 & 0 \\\ -\frac{1}{2} & 0 & \frac{1}{2} \\\ \frac{1}{2} & -1 & \frac{1}{2} \end{matrix} \right ] \left [ \begin{matrix} u_{1} \\\ u_{2} \\\ u_{3} \end{matrix} \right ] \end{equation}

Reemplazando:

\begin{equation} u = \left [ \begin{matrix} 1 & x & x^{2} \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \end{matrix} \right ] = \left [ \begin{matrix} 1 & x & x^{2} \end{matrix} \right ] \left [ \begin{matrix} 0 & 1 & 0 \\\ -\frac{1}{2} & 0 & \frac{1}{2} \\\ \frac{1}{2} & -1 & \frac{1}{2} \end{matrix} \right ] \left [ \begin{matrix} u_{1} \\\ u_{2} \\\ u_{3} \end{matrix} \right ] = \left [ \begin{matrix} -\frac{1}{2} x + \frac{1}{2} x^{2} & 1 - x^{2} & \frac{1}{2} x + \frac{1}{2} x^{2} \end{matrix} \right ] \left [ \begin{matrix} u_{1} \\\ u_{2} \\\ u_{3} \end{matrix} \right ] \end{equation}

Reescribiendo $u$:

\begin{equation} u = \Big ( -\frac{1}{2} x + \frac{1}{2} x^{2} \Big ) u_{1} + (1 - x^{2}) u_{2} + \Big ( \frac{1}{2} x + \frac{1}{2} x^{2} \Big ) u_{3} = N_{1} u_{1} + N_{2} u_{2} + N_{3} u_{3} \end{equation}

Elementos barra de mayor grado polinomial

Los elementos de mayor grado pueden obtenerse mediante polinomios de Lagrange:

\begin{equation} \ell = \prod_{i=0, i \neq j}^{k} \frac{x - x_{i}}{x_{j} - x_{i}} \end{equation}

Elemento de dos nodos

Usando la fórmula:

\begin{eqnarray} N_{1} &=& \frac{x - 1}{-1 - 1} = \frac{1}{2} - \frac{1}{2} x \\\ N_{2} &=& \frac{x - (-1)}{1 - (-1)} = \frac{1}{2} + \frac{1}{2} x \end{eqnarray}

Elemento de tres nodos

Usando la fórmula:

\begin{eqnarray} N_{1} &=& \frac{x - 0}{-1 - 0} \frac{x - 1}{-1 - 1} = -\frac{1}{2} x + \frac{1}{2} x^{2} \\\ N_{2} &=& \frac{x - (-1)}{0 - (-1)} \frac{x - 1}{0 - 1} = 1 - x^{2} \\\ N_{3} &=& \frac{x - 0}{1 - 0} \frac{x - 1}{1 - 1} = \frac{1}{2} x + \frac{1}{2} x^{2} \end{eqnarray}