# Funciones de forma para elementos barra

## Elemento de dos nodos

\begin{equation*} u = N_{1} u_{1} + N_{2} u_{2} = \sum_{i = 0}^{1} \alpha_{i} x^{i} = \alpha_{0} + \alpha_{1} x \end{equation*}

en forma matricial

\begin{equation*} u = \alpha_{0} + \alpha_{1} x = \begin{bmatrix} 1 & x \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \end{bmatrix} \end{equation*}

reemplazando $x_{1} = 0$ y $x_{2} = L$

\begin{align*} \alpha_{0} + \alpha_{1} (0) &= u_{1} \\ \alpha_{0} + \alpha_{1} (L) &= u_{2} \end{align*}

simplificando

\begin{align*} \alpha_{0} &= u_{1} \\ \alpha_{0} + L \alpha_{1} &= u_{2} \end{align*}

en forma matricial

\begin{equation*} \begin{bmatrix} 1 & 0 \\ 1 & L \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \end{bmatrix} = \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \end{equation*}

resolviendo el sistema

\begin{equation*} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{L} & \frac{1}{L} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \end{equation*}

reemplazando las incógnitas

\begin{align*} u &= \begin{bmatrix} 1 & x \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \end{bmatrix} \\ &= \begin{bmatrix} 1 & x \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -\frac{1}{L} & \frac{1}{L} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \\ &= \begin{bmatrix} 1 - \frac{1}{L} x & \frac{1}{L} x \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \\ &= \begin{bmatrix} N_{1} & N_{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} \end{align*}

Reescribiendo $u$

\begin{equation*} u = \bigg( 1 - \frac{1}{L} x \bigg) u_{1} + \bigg( \frac{1}{L} x \bigg) u_{2} \end{equation*}

## Elemento de tres nodos

\begin{equation*} u = N_{1} u_{1} + N_{2} u_{2} + N_{3} u_{3} = \sum_{i = 0}^{2} \alpha_{i} x^{i} = \alpha_{0} + \alpha_{1} x + \alpha_{2} x^{2} \end{equation*}

en forma matricial

\begin{equation*} u = \alpha_{0} + \alpha_{1} x + \alpha_{2} x^{2} = \begin{bmatrix} 1 & x & x^{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} \end{equation*}

reemplazando $x_{1} = 0$, $x_{2} = \frac{L}{2}$ y $x_{3} = L$

\begin{align*} \alpha_{0} + \alpha_{1}(0) + \alpha_{2}(0)^{2} &= u_{1} \\ \alpha_{0} + \alpha_{1} \bigg( \frac{L}{2} \bigg) + \alpha_{2} \bigg( \frac{L}{2} \bigg)^{2} &= u_{2} \\ \alpha_{0} + \alpha_{1}(L) + \alpha_{2}(L)^{2} &= u_{3} \end{align*}

simplificando

\begin{align*} \alpha_{0} &= u_{1} \\ \alpha_{0} + \frac{L}{2} \alpha_{1} + \frac{L^{2}}{4} \alpha_{2} &= u_{2}\\ \alpha_{0} + L \alpha_{1} + L^{2} \alpha_{1} &= u_{3} \end{align*}

en forma matricial

\begin{align*} \begin{bmatrix} 1 & 0 & 0 \\ 1 & \frac{L}{2} & \frac{L^{2}}{4} \\ 1 & L & L^{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} = \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{align*}

resolviendo

\begin{equation*} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -\frac{3}{L} & \frac{4}{L} & -\frac{1}{L} \\ \frac{2}{L^{2}} & -\frac{4}{L^{2}} & \frac{2}{L^{2}} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{equation*}

reemplazando las incógnitas

\begin{align*} u &= \begin{bmatrix} 1 & x & x^{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} \\ &= \begin{bmatrix} 1 & x & x^{2} \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ -\frac{3}{L} & \frac{4}{L} & -\frac{1}{L} \\ \frac{2}{L^{2}} & -\frac{4}{L^{2}} & \frac{2}{L^{2}} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \\ &= \begin{bmatrix} 1 - \frac{3}{L} x + \frac{2}{L^{2}} x^{2} & \frac{4}{L} x - \frac{4}{L^{2}} x^{2} & -\frac{1}{L} x + \frac{2}{L^{2}} x^{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \\ &= \begin{bmatrix} N_{1} & N_{2} & N_{3} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{align*}

Reescribiendo $u$

\begin{equation*} u = \bigg( 1 - \frac{3}{L} x + \frac{2}{L^{2}} x^{2} \bigg) u_{1} + \bigg( \frac{4}{L} x - \frac{4}{L^{2}} x^{2} \bigg) u_{2} + \bigg( -\frac{1}{L} x + \frac{2}{L^{2}} x^{2} \bigg) u_{3} \end{equation*}