Ejemplo 3
\( L = 40 \ \text{cm} \), \( A_{1} = 25 \ \text{cm}^{2} \), \( A_{2} = 16 \ \text{cm}^{2} \) y \( E = 2000 \ \text{Kg}/\text{cm}^{2} \)
Un elemento de dos nodos
área $$ \begin{equation*} A(x) = 25 - \frac{9}{40} x \ [\text{cm}^{2}] \end{equation*} $$
desplazamientos $$ \begin{equation*} \mathbf{N} = \begin{bmatrix} 1 - \frac{1}{L} x & \frac{1}{L} x \end{bmatrix} = \begin{bmatrix} 1 - \frac{1}{40} x & \frac{1}{40} x \end{bmatrix} \end{equation*} $$
deformaciones $$ \begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -\frac{1}{40} & \frac{1}{40} \end{bmatrix} \end{equation*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 50000 - 450 x \ [\text{Kg}] \end{equation*} $$
reemplazando $$ \begin{equation*} \int_{0}^{40} \begin{bmatrix} -\frac{1}{40} \\ \frac{1}{40} \end{bmatrix} \begin{bmatrix} 50000 - 450 x \end{bmatrix} \begin{bmatrix} -\frac{1}{40} & \frac{1}{40} \end{bmatrix} dx \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 1025 & -1025 \\ -1025 & 1025 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
reemplazando las condiciones de contorno $$ \begin{equation*} \begin{bmatrix} 1025 & -1025 \\ -1025 & 1025 \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} \\ -300 \end{bmatrix} \end{equation*} $$
resolviendo $$ \begin{align*} F_{1} &= 300 \ [\text{Kg}] \\ u_{2} &= -0.293 \ [\text{cm}] \end{align*} $$
Desplazamientos, deformaciones y esfuerzos
Las funciones de forma en coordenadas locales se transforman a coordenadas globales usando: $$ \begin{equation*} x = X - h \end{equation*} $$
reemplazando $$ \begin{equation*} x = X \end{equation*} $$
desplazamientos $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} 1 - \frac{1}{40} x & \frac{1}{40} x \end{bmatrix} \begin{bmatrix} 0 \\ -0.293 \end{bmatrix} = -0.007325 x = -0.007325 X \ [\text{cm}] \end{equation*} $$
deformación normal $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -\frac{1}{40} & \frac{1}{40} \end{bmatrix} \begin{bmatrix} 0 \\ -0.293 \end{bmatrix} = -0.007325 \end{equation*} $$
esfuerzo normal $$ \begin{equation*} \sigma = E \varepsilon = -14.65 \ [\text{Kg}/\text{cm}^{2}] \end{equation*} $$
Dos elementos de dos nodos
Elemento 1
área $$ \begin{equation*} A(x) = 25 - \frac{9}{40} x \ [\text{cm}^{2}] \end{equation*} $$
desplazamientos $$ \begin{equation*} \mathbf{N} = \begin{bmatrix} 1 - \frac{1}{L} x & \frac{1}{L} x \end{bmatrix} = \begin{bmatrix} 1 - \frac{1}{20} x & \frac{1}{20} x \end{bmatrix} \end{equation*} $$
deformaciones $$ \begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -\frac{1}{20} & \frac{1}{20} \end{bmatrix} \end{equation*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 50000 - 450 x \ [\text{Kg}] \end{equation*} $$
reemplazando $$ \begin{equation*} \int_{0}^{20} \begin{bmatrix} -\frac{1}{20} \\ \frac{1}{20} \end{bmatrix} \begin{bmatrix} 50000 - 450 x \end{bmatrix} \begin{bmatrix} -\frac{1}{20} & \frac{1}{20} \end{bmatrix} dx \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 2275 & -2275 \\ -2275 & 2275 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
Elemento 2
área $$ \begin{equation*} A(x) = \frac{41}{2} - \frac{9}{40} x \ [\text{cm}^{2}] \end{equation*} $$
desplazamientos $$ \begin{equation*} \mathbf{N} = \begin{bmatrix} 1 - \frac{1}{L} x & \frac{1}{L} x \end{bmatrix} = \begin{bmatrix} 1 - \frac{1}{20} x & \frac{1}{20} x \end{bmatrix} \end{equation*} $$
deformaciones $$ \begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -\frac{1}{20} & \frac{1}{20} \end{bmatrix} \end{equation*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 41000 - 450 x \ [\text{Kg}] \end{equation*} $$
reemplazando $$ \begin{equation*} \int_{0}^{20} \begin{bmatrix} -\frac{1}{20} \\ \frac{1}{20} \end{bmatrix} \begin{bmatrix} 41000 - 450 x \end{bmatrix} \begin{bmatrix} -\frac{1}{20} & \frac{1}{20} \end{bmatrix} dx \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 1825 & -1825 \\ -1825 & 1825 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
Ensamblaje y solución
ensamblando matriz global $$ \begin{equation*} \begin{bmatrix} 2275 & -2275 & 0 \\ -2275 & 2275+ 1825 & -1825 \\ 0& -1825 & 1825 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} + u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} \\ F_{2} + F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
sumando $$ \begin{equation*} \begin{bmatrix} 2275 & -2275 & 0 \\ -2275 & 4100 & -1825 \\ 0& -1825 & 1825 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
reemplazando condiciones de contorno $$ \begin{equation*} \begin{bmatrix} 2275 & -2275 & 0 \\ -2275 & 4100 & -1825 \\ 0& -1825 & 1825 \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} F_{1} \\ 0 \\ -300 \end{bmatrix} \end{equation*} $$
resolviendo $$ \begin{align*} F_{1} &= 300 \ [\text{Kg}] \\ u_{2} &= -0.132 \ [\text{m}] \\ u_{3} &= -0.296 \ [\text{m}] \end{align*} $$
Desplazamientos, deformaciones y esfuerzos
Las funciones de forma en coordenadas locales se transforman a coordenadas globales usando: $$ \begin{equation*} x = X - h \end{equation*} $$
Elemento 1
reemplazando \( h=0 \) $$ \begin{equation*} x = X \end{equation*} $$
desplazamientos $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} 1 - \frac{1}{20} x & \frac{1}{20} x \end{bmatrix} \begin{bmatrix} 0 \\ -0.132 \end{bmatrix} = -0.0066 x = -0.0066 X \ [\text{cm}] \end{equation*} $$
deformación normal $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -\frac{1}{20} & \frac{1}{20} \end{bmatrix} \begin{bmatrix} 0 \\ -0.132 \end{bmatrix} = -0.0066 \end{equation*} $$
esfuerzo normal $$ \begin{equation*} \sigma = E \ \varepsilon = -13.2 \ [\text{Kg}/\text{cm}^{2}] \end{equation*} $$
Elemento 2
reemplazando \( h=20 \) $$ \begin{equation*} x = X - 20 \end{equation*} $$
desplazamientos $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} 1 - \frac{1}{20} x & \frac{1}{20} x \end{bmatrix} \begin{bmatrix} -0.132 \\ -0.296 \end{bmatrix} = -0.132 - 0.0082 x = 0.032 - 0.0082 X \ [\text{cm}] \end{equation*} $$
deformación normal $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -\frac{1}{20} & \frac{1}{20} \end{bmatrix} \begin{bmatrix} -0.132 \\ -0.296 \end{bmatrix} = -0.0082 \end{equation*} $$
esfuerzo normal $$ \begin{equation*} \sigma = E \ \varepsilon = -16.4 \ [\text{Kg}/\text{cm}^{2}] \end{equation*} $$