# Elementos barra

## Elemento de dos nodos

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

reemplazando los valores $\xi = -1$ y $\xi = 1$

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

simplificando

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

en forma matricial

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

resolviendo

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

reemplazando

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

reescribiendo $u$

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

## Elemento de tres nodos

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

reemplazando los valores $\xi = -1$, $\xi = 0$ y $\xi = 1$

\begin{align*} \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{align*}

simplificando

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

en forma matricial

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

resolviendo

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

reemplazando

\begin{equation*} u = \begin{bmatrix} 1 & \xi & \xi^{2} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \end{bmatrix} = \begin{bmatrix} 1 & \xi & \xi^{2} \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} \end{equation*}

reescribiendo $u$

\begin{equation*} u = \bigg( -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) u_{1} + (1 - \xi^{2}) u_{2} + \bigg( \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) 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{\xi - \xi_{i}}{\xi_{j} - \xi_{i}} \end{equation*}

### Elemento de dos nodos

Usando la fórmula

\begin{align*} N_{1} &= \frac{\xi - 1}{-1 - 1} = \frac{1}{2} - \frac{1}{2} \xi \\ N_{2} &= \frac{\xi - (-1)}{1 - (-1)} = \frac{1}{2} + \frac{1}{2} \xi \end{align*}

### Elemento de tres nodos

Usando la fórmula

\begin{align*} N_{1} &= \frac{\xi - 0}{-1 - 0} \frac{\xi - 1}{-1 - 1} = -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \\ N_{2} &= \frac{\xi - (-1)}{0 - (-1)} \frac{\xi - 1}{0 - 1} = 1 - \xi^{2} \\ N_{3} &= \frac{\xi - 0}{1 - 0} \frac{\xi - 1}{1 - 1} = \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{align*}