Funciones de forma unidimensionales

Las funciones de forma unidimensionales sirven para aproximar los desplazamientos:

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

Elemento viga Euler-Bernoulli

Los elementos viga soportan esfuerzos debido a flexión.

Elemento de dos nodos

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

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

La deformación angular:

\begin{equation} \theta = \frac{\partial{w}}{\partial{x}} = \alpha_{1} + 2 \alpha_{2} x + 3 \alpha_{3} x^{2} \end{equation}

Reemplazando los valores nodales en coordenadas naturales:

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

Evaluando:

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

En forma matricial:

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

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \end{matrix} \right ] = \left [ \begin{matrix} \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{matrix} \right ] \left [ \begin{matrix} w_{1} \\\ \theta_{1} \\\ w_{2} \\\ \theta_{2} \end{matrix} \right ] \end{equation}

Reemplazando:

\begin{equation} w = \left [ \begin{matrix} 1 & x & x^{2} & x^{3} \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \end{matrix} \right ] = \left [ \begin{matrix} 1 & x & x^{2} & x^{3} \end{matrix} \right ] \left [ \begin{matrix} \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{matrix} \right ] \left [ \begin{matrix} w_{1} \\\ \theta_{1} \\\ w_{2} \\\ \theta_{2} \end{matrix} \right ] = \left [ \begin{matrix} \frac{1}{4} (x+2) (x-1)^{2} & \frac{1}{4} (x+1) (x-1)^{2} & -\frac{1}{4} (x-2) (x+1)^{2} & \frac{1}{4} (x-1) (x+1)^{2} \end{matrix} \right ] \left [ \begin{matrix} w_{1} \\\ \theta_{1} \\\ w_{2} \\\ \theta_{2} \end{matrix} \right ] \end{equation}

Reescribiendo $w$:

\begin{equation} w = \Big [ \frac{1}{4} (x+2) (x-1)^{2} \Big ] w_{1} + \Big [ \frac{1}{4} (x+1) (x-1)^{2} \Big ] \theta_{1} + \Big [ -\frac{1}{4} (x-2) (x+1)^{2} \Big ] w_{2} + \Big [ \frac{1}{4} (x-1) (x+1)^{2} \Big ] \theta_{2} = H_{01} w_{1} + H_{11} \theta_{1} + H_{02} w_{2} + H_{12} \theta_{2} \end{equation}

Elemento de tres nodos

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

\begin{equation} w = \alpha_{0} + \alpha_{1} x + \alpha_{2} x^{2} + \alpha_{3} x^{3} + \alpha_{4} x^{4} + \alpha_{5} x^{5} = \left [ \begin{matrix} 1 & x & x^{2} & x^{3} & x^{4} & x^{5} \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \end{matrix} \right ] \end{equation}

La deformación angular:

\begin{equation} \theta = \frac{\partial{w}}{\partial{x}} = \alpha_{1} + 2 \alpha_{2} x + 3 \alpha_{3} x^{2} + 4 \alpha_{4} x^{3} + 5 \alpha_{5} x^{4} \end{equation}

Reemplazando los valores nodales en coordenadas naturales:

\begin{eqnarray} \alpha_{0} + \alpha_{1} (-1) + \alpha_{2} (-1)^{2} + \alpha_{3} (-1)^{3} + \alpha_{4} (-1)^{4} + \alpha_{5} (-1)^{5} &=& w_{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} &=& w_{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} &=& w_{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{eqnarray}

Evaluando:

\begin{eqnarray} \alpha_{0} - \alpha_{1} + \alpha_{2} - \alpha_{3} + \alpha_{4} - \alpha_{5} &=& w_{1} \\\ \alpha_{1} - 2 \alpha_{2} + 3 \alpha_{3} - 4 \alpha_{4} + 5 \alpha_{5} &=& \theta_{1} \\\ \alpha_{0} &=& w_{2} \\\ \alpha_{1} &=& \theta_{2} \\\ \alpha_{0} + \alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} + \alpha_{5} &=& w_{3} \\\ \alpha_{1} + 2 \alpha_{2} + 3 \alpha_{3} + 4 \alpha_{4} + 5 \alpha_{5} &=& \theta_{3} \end{eqnarray}

En forma matricial:

\begin{equation} \left [ \begin{matrix} 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{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \end{matrix} \right ] = \left [ \begin{matrix} w_{1} \\\ \theta_{1} \\\ w_{2} \\\ \theta_{2} \\\ w_{3} \\\ \theta_{3} \end{matrix} \right ] \end{equation}

Resolviendo el sistema:

\begin{equation} \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \end{matrix} \right ] = \left [ \begin{matrix} 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{matrix} \right ] \left [ \begin{matrix} w_{1} \\\ \theta_{1} \\\ w_{2} \\\ \theta_{2} \\\ w_{3} \\\ \theta_{3} \end{matrix} \right ] \end{equation}

Reemplazando:

\begin{equation} w = \left [ \begin{matrix} 1 & x & x^{2} & x^{3} & x^{4} & x^{5} \end{matrix} \right ] \left [ \begin{matrix} \alpha_{0} \\\ \alpha_{1} \\\ \alpha_{2} \\\ \alpha_{3} \\\ \alpha_{4} \\\ \alpha_{5} \end{matrix} \right ] = \left [ \begin{matrix} 1 & x & x^{2} & x^{3} & x^{4} & x^{5} \end{matrix} \right ] \left [ \begin{matrix} 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{matrix} \right ] \left [ \begin{matrix} w_{1} \\\ \theta_{1} \\\ w_{2} \\\ \theta_{2} \\\ w_{3} \\\ \theta_{3} \end{matrix} \right ] = \left [ \begin{matrix} \frac{1}{4} x^{2} (3x+4) (x-1)^{2} & \frac{1}{4} x^{2} (x+1) (x-1)^{2} & (x-1)^{2} (x+1)^{2} & x (x-1)^{2} (x+1)^{2} & -\frac{1}{4} x^{2} (3x-4) (x+1)^{2} & \frac{1}{4} x^{2} (x-1) (x+1)^{2} \end{matrix} \right ] \left [ \begin{matrix} w_{1} \\\ \theta_{1} \\\ w_{2} \\\ \theta_{2} \\\ w_{3} \\\ \theta_{3} \end{matrix} \right ] \end{equation}

Reescribiendo $u$:

\begin{equation} w = \Big [ \frac{1}{4} x^{2} (3x+4) (x-1)^{2} \Big ] w_{1} + \Big [ \frac{1}{4} x^{2} (x+1) (x-1)^{2} \Big ] \theta_{1} + \Big [ (x-1)^{2} (x+1)^{2} \Big ] w_{2} + \Big [ x (x-1)^{2} (x+1)^{2} \Big ] \theta_{2} + \Big [ -\frac{1}{4} x^{2} (3x-4) (x+1)^{2} \Big ] w_{3} + \Big [ \frac{1}{4} x^{2} (x-1) (x+1)^{2} \Big ] \theta_{3} = H_{01} w_{1} + H_{11} \theta_{1} + H_{02} w_{2} + H_{12} \theta_{2} + H_{03} w_{3} + H_{13} \theta_{3} \end{equation}

Elementos viga Euler-Bernoulli de mayor grado polinomial

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

\begin{eqnarray} H_{0i} &=& [1 - 2 \ \ell_{(x_{i})}^{\prime} (x - x_{i})] [\ell_{(x)}]^{2} \\\ H_{1i} &=& (x - x_{i}) [\ell_{(x)}]^{2} \end{eqnarray}

Elemento de dos nodos

Usando la fórmula para polinomios de Lagrange:

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

Usando la fórmula para polinomios de Hermite:

\begin{eqnarray} H_{01} &=& \Big \\{ 1 - 2 \Big [ -\frac{1}{2} \Big ] [x - (-1)] \Big \\} \Big ( \frac{1}{2} - \frac{1}{2} x \Big )^{2} = \frac{1}{4} (x+2) (x-1)^{2} \\\ H_{11} &=& [x - (-1)] \Big ( \frac{1}{2} - \frac{1}{2} x \Big )^{2} = \frac{1}{4} (x+1) (x-1)^{2} \\\ H_{02} &=& \Big [ 1 - 2 \Big ( \frac{1}{2} \Big ) (x - 1) \Big ] \Big ( \frac{1}{2} + \frac{1}{2} x \Big )^{2} = -\frac{1}{4} (x-2) (x+1)^{2} \\\ H_{12} &=& (x - 1) \Big ( \frac{1}{2} + \frac{1}{2} x \Big )^{2} = \frac{1}{4} (x-1) (x+1)^{2} \end{eqnarray}

Elemento de tres nodos

Usando la fórmula para polinomios de Lagrange:

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

Usando la fórmula para polinomios de Hermite:

\begin{eqnarray} H_{01} &=& \Big \\{ 1 - 2 \Big [ -\frac{1}{2} + (-1) \Big ] [x - (-1)] \Big \\} \Big ( -\frac{1}{2} x + \frac{1}{2} x^{2} \Big )^{2} = \frac{1}{4} x^{2} (3x+4) (x-1)^{2} \\\ H_{11} &=& [x - (-1)] \Big ( -\frac{1}{2} x + \frac{1}{2} x^{2} \Big )^{2} = \frac{1}{4} x^{2} (x+1) (x-1)^{2} \\\ H_{02} &=& \Big \\{ 1 - 2 \Big [ - 2 (0) \Big ] (x - 0) \Big \\} \Big ( 1 - x^{2} \Big )^{2} = (x-1)^{2} (x+1)^{2} \\\ H_{12} &=& (x - 0) \Big ( 1 - x^{2} \Big )^{2} = x (x-1)^{2} (x+1)^{2} \\\ H_{03} &=& \Big \\{ 1 - 2 \Big [ \frac{1}{2} + (1) \Big ] \Big \\} \Big ( \frac{1}{2} x + \frac{1}{2} x^{2} \Big )^{2} = -\frac{1}{4} x^{2} (3x-4) (x+1)^{2} \\\ H_{13} &=& [x - (1)] \Big ( \frac{1}{2} x + \frac{1}{2} x^{2} \Big )^{2} = \frac{1}{4} x^{2} (x-1) (x+1)^{2} \end{eqnarray}