Ejemplo viga Euler-Bernoulli

Discretizando la viga en un elemento de dos nodos:

\begin{eqnarray} \color{Blue} {\boldsymbol{N}} &=& \left [ \begin{matrix} N_{1} & N_{2} & N_{3} & N_{4} \end{matrix} \right ] = \left [ \begin{matrix} \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{matrix} \right ] \\\ \boldsymbol{B_{f}} &=& \left [ \begin{matrix} \frac{\partial^{2} N_{1}}{\partial x^{2}} & \frac{\partial^{2} N_{2}}{\partial x^{2}} & \frac{\partial^{2} N_{3}}{\partial x^{2}} & \frac{\partial^{2} N_{4}}{\partial x^{2}} \end{matrix} \right ] \end{eqnarray}

Usando la regla de la cadena:

\begin{eqnarray} \frac{\partial^{2} N}{\partial x^{2}} &=& \frac{\partial}{\partial x} \Big ( \frac{\partial N}{\partial \xi} \frac{\partial \xi}{\partial x} \Big ) \\\ &=& \frac{\partial^{2} N}{\partial x \partial \xi} \frac{\partial \xi}{\partial x} + \frac{\partial N}{\partial \xi} \frac{\partial^{2} \xi}{\partial x^{2}} \\\ &=& \frac{\partial}{\partial \xi} \Big ( \frac{\partial N}{\partial x} \Big ) \frac{\partial \xi}{\partial x} + \frac{\partial N}{\partial \xi} \frac{\partial^{2} \xi}{\partial x^{2}} \\\ &=& \frac{\partial}{\partial \xi} \Big ( \frac{\partial N}{\partial \xi} \frac{\partial \xi}{\partial x} \Big ) \frac{\partial \xi}{\partial x} + \frac{\partial N}{\partial \xi} \frac{\partial^{2} \xi}{\partial x^{2}} \\\ &=& \Big ( \frac{\partial^{2} N}{\partial \xi^{2}} \frac{\partial \xi}{\partial x} + \frac{\partial N}{\partial \xi} \frac{\partial^{2} \xi}{\partial \xi \partial x} \Big ) \frac{\partial \xi}{\partial x} + \frac{\partial N}{\partial \xi} \frac{\partial^{2} \xi}{\partial x^{2}} \\\ &=& \frac{\partial^{2} N}{\partial \xi^{2}} \Big ( \frac{\partial \xi}{\partial x} \Big )^{2} + \frac{\partial N}{\partial \xi} \frac{\partial^{2} \xi}{\partial \xi \partial x} \frac{\partial \xi}{\partial x} + \frac{\partial N}{\partial \xi} \frac{\partial^{2} \xi}{\partial x^{2}} \\\ &=& \frac{\partial^{2} N}{\partial \xi^{2}} \Big ( \frac{\partial \xi}{\partial x} \Big )^{2} + 2 \frac{\partial N}{\partial \xi} \frac{\partial^{2} \xi}{\partial x^{2}} \\\ &=& \frac{\partial^{2} N}{\partial \xi^{2}} \Big ( \frac{\partial \xi}{\partial x} \Big )^{2} \end{eqnarray}

Las funciones usadas para interpolar los desplazamientos están en coordenadas locales, para obtener la matriz de flexión debemos calcular el jacobiano y el jacobiano inverso:

\begin{eqnarray} J = \frac{\partial x}{\partial \xi} \\\ J^{-1} = \frac{\partial \xi}{\partial x} \end{eqnarray}

La función que interpola la geometría es:

\begin{equation} x = \Big ( \frac{1}{2} - \frac{1}{2} \xi \Big ) x_{1} + \Big ( \frac{1}{2} + \frac{1}{2} \xi \Big ) x_{2} \end{equation}

Derivando:

\begin{equation} \frac{\partial x}{\partial \xi} = \frac{x_{2} - x_{1}}{2} = \frac{l}{2} \end{equation}

La matriz de flexión es:

\begin{eqnarray} \color{Blue} {B_{f}}&=& \left [ \begin{matrix} \frac{\partial^{2} N_{1}}{\partial \xi^{2}} & \frac{\partial^{2} N_{2}}{\partial \xi^{2}} \frac{\partial x}{\partial \xi} & \frac{\partial^{2} N_{3}}{\partial \xi^{2}} & \frac{\partial^{2} N_{4}}{\partial \xi^{2}} \frac{\partial x}{\partial \xi} \end{matrix} \right ] \Big ( \frac{\partial \xi}{\partial x} \Big )^{2} \\\ &=& \left [ \begin{matrix} \frac{3}{2} \xi & \Big ( -\frac{1}{2} + \frac{3}{2} \xi \Big ) \frac{l}{2} & -\frac{3}{2} \xi & \Big ( \frac{1}{2} + \frac{3}{2} \xi \Big ) \frac{l}{2} \end{matrix} \right ] \Big ( \frac{2}{l} \Big )^{2} \\\ &=& \left [ \begin{matrix} \frac{6}{l^{2}} \xi & -\frac{1}{l} + \frac{3}{l} \xi & -\frac{6}{l^{2}} \xi & \frac{1}{l} + \frac{3}{l} \xi \end{matrix} \right ] \end{eqnarray}

La matriz de rigidez es:

\begin{eqnarray} K &=& \int_{-1}^{+1} \left[ \begin{matrix} \frac{36}{l^{4}} \xi^{2} & \frac{6}{l^{3}} \xi \ (3 \xi - 1) & -\frac{36}{l^{4}} \xi^{2} & \frac{6}{l^{3}} \xi \ (3 \xi + 1) \\\ \frac{6}{l^{3}} \xi \ (3 \xi - 1) & \frac{1}{l^{2}} (3 \xi - 1)^{2} & -\frac{6}{l^{3}} \xi \ (3 \xi - 1) & \frac{1}{l^{2}} (3 \xi - 1) (3 \xi + 1) \\\ -\frac{36}{l^{4}} \xi^{2} & -\frac{6}{l^{3}} \xi \ (3 \xi - 1) & \frac{36}{l^{4}} \xi^{2} & -\frac{6}{l^{3}} \xi \ (3 \xi + 1) \\\ \frac{6}{l^{3}} \xi \ (3 \xi + 1) & \frac{1}{l^{2}} (3 \xi - 1) (3 \xi + 1) & -\frac{6}{l^{3}} \xi \ (3 \xi + 1) & \frac{1}{l^{2}} (3 \xi + 1)^{2} \end{matrix} \right] E I \ \frac{l}{2} \ d\xi \\\ &=& \left[ \begin{matrix} \frac{12 \ E I}{l^{3}} & \frac{6 \ E I}{l^{2}} & -\frac{12 \ E I}{l^{3}} & \frac{6 \ E I}{l^{2}} \\\ \frac{6 \ E I}{l^{2}} & \frac{4 \ E I}{l} & -\frac{6 \ E I}{l^{2}} & \frac{2 \ E I}{l} \\\ -\frac{12 \ E I}{l^{3}} & -\frac{6 \ E I}{l^{2}} & \frac{12 \ E I}{l^{3}} & -\frac{6 \ E I}{l^{2}} \\\ \frac{6 \ E I}{l^{2}} & \frac{2 \ E I}{l} & -\frac{6 \ E I}{l^{2}} & \frac{4 \ E I}{l} \end{matrix} \right] \end{eqnarray}

La matriz de cargas uniformes:

\begin{eqnarray} f &=& \int_{-1}^{+1} \left[ \begin{matrix} \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{matrix} \right] (0) \Big ( \frac{l}{2} \Big ) \ d\xi \\\ &=& \left[ \begin{matrix} 0 \\\ 0 \\\ 0 \\\ 0 \end{matrix} \right] \end{eqnarray}

La matriz de cargas nodales:

\begin{equation} q = \left[ \begin{matrix} A_{z} \\\ M_{A} \\\ B_{z} \\\ M_{B} \end{matrix} \right] \end{equation}

Ensamblando matrices:

\begin{equation} \left[ \begin{matrix} \frac{A E}{l} & -\frac{A E}{l} \\\ -\frac{A E}{l} & \frac{A E}{l} \end{matrix} \right] \left[ \begin{matrix} u_{1} \\\ u_{2} \end{matrix} \right] = \left[ \begin{matrix} 500 l \\\ 500 l \end{matrix} \right] + \left[ \begin{matrix} A_{x} \\\ B_{x} \end{matrix} \right] \end{equation}

Agregando las condiciones de contorno y reemplazando datos: