# Elementos viga Euler-Bernoulli

## Elemento de dos nodos

\begin{equation*} v = \alpha_{0} + \alpha_{1} \xi + \alpha_{2} \xi^{2} + \alpha_{3} \xi^{3} = \begin{bmatrix} 1 & \xi & \xi^{2} & \xi^{3} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \end{bmatrix} \end{equation*}

reemplazando los valores $\xi_{1} = -1$ y $\xi_{2} = 1$

\begin{align*} \alpha_{0} + \alpha_{1} (-1) + \alpha_{2} (-1)^{2} + \alpha_{3} (-1)^{3} &= v_{1} \\ \alpha_{1} + 2 \alpha_{2} (-1) + 3 \alpha_{3} (-1)^{2} &= \theta_{1} \\ \alpha_{0} + \alpha_{1} (1) + \alpha_{2} (1)^{2} + \alpha_{3} (1)^{3} &= v_{2} \\ \alpha_{1} + 2 \alpha_{2} (1) + 3 \alpha_{3} (1)^{2} &= \theta_{2} \end{align*}

simplificando

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

en forma matricial

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

resolviendo

\begin{equation*} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{2} & -\frac{1}{4} \\ -\frac{3}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ 0 & -\frac{1}{4} & 0 & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \end{bmatrix} \end{equation*}

reemplazando

\begin{align*} v &= \begin{bmatrix} 1 & \xi & \xi^{2} & \xi^{3} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \end{bmatrix} \\ &= \begin{bmatrix} 1 & \xi & \xi^{2} & \xi^{3} \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{2} & -\frac{1}{4} \\ -\frac{3}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ 0 & -\frac{1}{4} & 0 & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{4} (\xi + 2) (\xi - 1)^{2} & \frac{1}{4} (\xi + 1) (\xi - 1)^{2} & -\frac{1}{4} (\xi - 2) (\xi + 1)^{2} & \frac{1}{4} (\xi - 1) (\xi + 1)^{2} \end{bmatrix} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \end{bmatrix} \end{align*}

Reescribiendo $v$:

\begin{align*} v &= \bigg[ \frac{1}{4} (\xi + 2) (\xi - 1)^{2} \bigg] v_{1} + \bigg[ \frac{1}{4} (\xi + 1) (\xi - 1)^{2} \bigg] \theta_{1} + \bigg[ -\frac{1}{4} (\xi - 2) (\xi + 1)^{2} \bigg] v_{2} + \bigg[ \frac{1}{4} (\xi - 1) (\xi + 1)^{2} \bigg] \theta_{2} \\ &= N_{1} v_{1} + N_{2} \theta_{1} + N_{3} v_{2} + N_{4} \theta_{2} \end{align*}

## Elemento de tres nodos

\begin{equation*} v = \alpha_{0} + \alpha_{1} \xi + \alpha_{2} \xi^{2} + \alpha_{3} \xi^{3} + \alpha_{4} \xi^{4} + \alpha_{5} \xi^{5} = \begin{bmatrix} 1 & \xi & \xi^{2} & \xi^{3} & \xi^{4} & \xi^{5} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \alpha_{4} \\ \alpha_{5} \end{bmatrix} \end{equation*}

reemplazando los valores $\xi_{1} = -1$, $\xi_{2} = 0$ y $\xi_{3} = 1$

\begin{align*} \alpha_{0} + \alpha_{1} (-1) + \alpha_{2} (-1)^{2} + \alpha_{3} (-1)^{3} + \alpha_{4} (-1)^{4} + \alpha_{5} (-1)^{5} &= v_{1} \\ \alpha_{1} + 2 \alpha_{2} (-1) + 3 \alpha_{3} (-1)^{2} + 4 \alpha_{4} (-1)^{3} + 5 \alpha_{5} (-1)^{4} &= \theta_{1} \\ \alpha_{0} + \alpha_{1} (0) + \alpha_{2} (0)^{2} + \alpha_{3} (0)^{3} + \alpha_{4} (0)^{4} + \alpha_{5} (0)^{5} &= v_{2} \\ \alpha_{1} + 2 \alpha_{2} (0) + 3 \alpha_{3} (0)^{2} + 4 \alpha_{4} (0)^{3} + 5 \alpha_{5} (0)^{4} &= \theta_{2} \\ \alpha_{0} + \alpha_{1} (1) + \alpha_{2} (1)^{2} + \alpha_{3} (1)^{3} + \alpha_{4} (1)^{4} + \alpha_{5} (1)^{5} &= v_{3} \\ \alpha_{1} + 2 \alpha_{2} (1) + 3 \alpha_{3} (1)^{2} + 4 \alpha_{4} (1)^{3} + 5 \alpha_{5} (1)^{4} &= \theta_{3} \end{align*}

simplificando

\begin{align*} \alpha_{0} - \alpha_{1} + \alpha_{2} - \alpha_{3} + \alpha_{4} - \alpha_{5} &= v_{1} \\ \alpha_{1} - 2 \alpha_{2} + 3 \alpha_{3} - 4 \alpha_{4} + 5 \alpha_{5} &= \theta_{1} \\ \alpha_{0} &= v_{2} \\ \alpha_{1} &= \theta_{2} \\ \alpha_{0} + \alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} + \alpha_{5} &= v_{3} \\ \alpha_{1} + 2 \alpha_{2} + 3 \alpha_{3} + 4 \alpha_{4} + 5 \alpha_{5} &= \theta_{3} \end{align*}

en forma matricial

\begin{equation*} \begin{bmatrix} 1 & -1 & 1 & -1 & 1 & -1 \\ 0 & 1 & -2 & 3 & -4 & 5 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 & 5 \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \alpha_{4} \\ \alpha_{5} \end{bmatrix} = \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \end{equation*}

resolviendo

\begin{equation*} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \alpha_{4} \\ \alpha_{5} \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & \frac{1}{4} & -2 & 0 & 1 & -\frac{1}{4} \\ -\frac{5}{4} & -\frac{1}{4} & 0 & -2 & \frac{5}{4} & -\frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{4} & 1 & 0 & -\frac{1}{2} & \frac{1}{4} \\ \frac{3}{4} & \frac{1}{4} & 0 & 1 & -\frac{3}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \end{equation*}

reemplazando

\begin{align*} v &= \begin{bmatrix} 1 & \xi & \xi^{2} & \xi^{3} & \xi^{4} & \xi^{5} \end{bmatrix} \begin{bmatrix} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \alpha_{4} \\ \alpha_{5} \end{bmatrix} \\ &= \begin{bmatrix} 1 & \xi & \xi^{2} & \xi^{3} & \xi^{4} & \xi^{5} \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & \frac{1}{4} & -2 & 0 & 1 & -\frac{1}{4} \\ -\frac{5}{4} & -\frac{1}{4} & 0 & -2 & \frac{5}{4} & -\frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{4} & 1 & 0 & -\frac{1}{2} & \frac{1}{4} \\ \frac{3}{4} & \frac{1}{4} & 0 & 1 & -\frac{3}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{4} \xi^{2} (3 \xi + 4) (\xi - 1)^{2} \\ \frac{1}{4} \xi^{2} (\xi + 1) (\xi - 1)^{2} \\ (\xi - 1)^{2} (\xi + 1)^{2} \\ \xi (\xi - 1)^{2} (\xi + 1)^{2} \\ -\frac{1}{4} \xi^{2} (3 \xi - 4) (\xi + 1)^{2} \\ \frac{1}{4} \xi^{2} (\xi - 1) (\xi + 1)^{2} \end{bmatrix}^{\mathrm{T}} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \\ &= \begin{bmatrix} N_{1} \\ N_{2} \\ N_{3} \\ N_{4} \\ N_{5} \\ N_{6} \end{bmatrix}^{\mathrm{T}} \begin{bmatrix} v_{1} \\ \theta_{1} \\ v_{2} \\ \theta_{2} \\ v_{3} \\ \theta_{3} \end{bmatrix} \end{align*}

## Elementos viga Euler-Bernoulli de mayor grado polinomial

Los elementos de mayor grado pueden obtenerse mediante polinomios de Hermite

\begin{align*} H_{0i} &= [1 - 2 \ \ell_{(\xi_{i})}^{\prime} (\xi - \xi_{i})] [\ell_{(\xi)}]^{2} \\ H_{1i} &= (\xi - \xi_{i}) [\ell_{(\xi)}]^{2} \end{align*}

### Elemento de dos nodos

Usando la fórmula para polinomios de Lagrange

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

Usando la fórmula para polinomios de Hermite:

\begin{align*} H_{01} &= \bigg\{ 1 - 2 \bigg[ -\frac{1}{2} \bigg] [\xi - (-1)] \bigg\} \bigg( \frac{1}{2} - \frac{1}{2} \xi \bigg)^{2} = \frac{1}{4} (\xi + 2) (\xi - 1)^{2} \\ H_{11} &= [\xi - (-1)] \bigg( \frac{1}{2} - \frac{1}{2} \xi \bigg)^{2} = \frac{1}{4} (\xi + 1) (\xi - 1)^{2} \\ H_{02} &= \bigg[ 1 - 2 \bigg( \frac{1}{2} \bigg) (\xi - 1) \bigg] \bigg( \frac{1}{2} + \frac{1}{2} \xi \bigg)^{2} = -\frac{1}{4} (\xi - 2) (\xi + 1)^{2} \\ H_{12} &= (\xi - 1) \bigg( \frac{1}{2} + \frac{1}{2} \xi \bigg)^{2} = \frac{1}{4} (\xi - 1) (\xi + 1)^{2} \end{align*}

### Elemento de tres nodos

Usando la fórmula para polinomios de Lagrange

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

Usando la fórmula para polinomios de Hermite

\begin{align*} H_{01} &= \bigg\{ 1 - 2 \bigg[ -\frac{1}{2} + (-1) \bigg] [\xi - (-1)] \bigg\} \bigg( -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg)^{2} = \frac{1}{4} \xi^{2} (3 \xi + 4) (\xi - 1)^{2} \\ H_{11} &= [\xi - (-1)] \bigg( -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg)^{2} = \frac{1}{4} \xi^{2} (\xi + 1) (\xi - 1)^{2} \\ H_{02} &= \{ 1 - 2 [ - 2 (0) ] (\xi - 0) \} \bigg( 1 - \xi^{2} \bigg)^{2} = (\xi - 1)^{2} (\xi + 1)^{2} \\ H_{12} &= (\xi - 0) \bigg( 1 - \xi^{2} \bigg)^{2} = \xi (\xi - 1)^{2} (\xi + 1)^{2} \\ H_{03} &= \bigg\{ 1 - 2 \bigg[ \frac{1}{2} + (1) \bigg] \bigg\} \bigg( \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg)^{2} = -\frac{1}{4} \xi^{2} (3 \xi - 4) (\xi + 1)^{2} \\ H_{13} &= [\xi - (1)] \bigg( \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg)^{2} = \frac{1}{4} \xi^{2} (\xi - 1) (\xi + 1)^{2} \end{align*}