Coordenadas locales
Ejemplo 1
\( L = 0.5 \ \text{m} \), \( A = 6.25 \times 10^{-4} \ \text{m}^{2} \) y \( E = 200 \ \text{MPa} \)
Un elemento de dos nodos
interpolación de desplazamientos $$ \begin{align*} \mathbf{N} &= \begin{bmatrix} 1 - \frac{1}{L} x & \frac{1}{L} x \end{bmatrix} \\ &= \begin{bmatrix} 1 - \frac{1}{0.5} x & \frac{1}{0.5} x \end{bmatrix} \\ &= \begin{bmatrix} 1 - 2 x & 2 x \end{bmatrix} \end{align*} $$
interpolación de deformaciones $$ \begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -2 & 2 \end{bmatrix} \end{equation*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 1.25 \times 10^{5} \ [\text{N}] \end{equation*} $$
reemplazando $$ \begin{equation*} \int_{0}^{0.5} \begin{bmatrix} -2 \\ 2 \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -2 & 2 \end{bmatrix} dx \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \int_{0}^{0.5} 1000 \begin{bmatrix} 1 - 2 x \\ 2 x \end{bmatrix} dx + \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 2.5 \times 10^{5} & -2.5 \times 10^{5} \\ -2.5 \times 10^{5} & 2.5 \times 10^{5} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} 250 \\ 250 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
reemplazando las condiciones de contorno $$ \begin{equation*} \begin{bmatrix} 2.5 \times 10^{5} & -2.5 \times 10^{5} \\ -2.5 \times 10^{5} & 2.5 \times 10^{5} \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \end{bmatrix} = \begin{bmatrix} 250 \\ 250 \end{bmatrix} + \begin{bmatrix} F_{1} \\ 250 \end{bmatrix} \end{equation*} $$
sumando $$ \begin{equation*} \begin{bmatrix} 2.5 \times 10^{5} & -2.5 \times 10^{5} \\ -2.5 \times 10^{5} & 2.5 \times 10^{5} \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \end{bmatrix} = \begin{bmatrix} F_{1} + 250 \\ 500 \end{bmatrix} \end{equation*} $$
resolviendo $$ \begin{align*} F_{1} &= -750 \ [\text{N}] \\ u_{2} &= 0.002 \ [\text{m}] \end{align*} $$
Desplazamientos, deformaciones y esfuerzos
Las funciones de forma en coordenadas locales se transforman a coordenadas globales usando: $$ \begin{equation*} x = X - h \end{equation*} $$
reemplazando \( h=0 \) $$ \begin{equation*} x = X \end{equation*} $$
desplazamientos $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} 1 - 2 x & 2 x \end{bmatrix} \begin{bmatrix} 0 \\ 0.002 \end{bmatrix} = 0.004x = 0.004 X \ [\text{m}] \end{equation*} $$
deformaciones $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -2 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 0.002 \end{bmatrix} = 0.004 \end{equation*} $$
esfuerzo $$ \begin{equation*} \sigma = E \ \varepsilon = 0.8 \ [\text{MPa}] \end{equation*} $$
Dos elementos de dos nodos
Elemento 1
interpolación de desplazamientos $$ \begin{align*} \mathbf{N} &= \begin{bmatrix} 1 - \frac{1}{L} x & \frac{1}{L} x \end{bmatrix} \\ &= \begin{bmatrix} 1 - \frac{1}{0.25} x & \frac{1}{0.25} x \end{bmatrix} \\ &= \begin{bmatrix} 1 - 4 x & 4 x \end{bmatrix} \end{align*} $$
interpolación de deformaciones $$ \begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -4 & 4 \end{bmatrix} \end{equation*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 1.25 \times 10^{5} \ [\text{N}] \end{equation*} $$
reemplazando $$ \begin{equation*} \int_{0}^{0.25} \begin{bmatrix} -4 \\ 4 \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -4 & 4 \end{bmatrix} dx \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \int_{0}^{0.25} 1000 \begin{bmatrix} 1 - 4 x \\ 4 x \end{bmatrix} dx + \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 5 \times 10^{5} & -5 \times 10^{5} \\ -5 \times 10^{5} & 5 \times 10^{5} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} 125 \\ 125 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
Elemento 2
interpolación de desplazamientos $$ \begin{align*} \mathbf{N} &= \begin{bmatrix} 1 - \frac{1}{L} x & \frac{1}{L} x \end{bmatrix} \\ &= \begin{bmatrix} 1 - \frac{1}{0.25} x & \frac{1}{0.25} x \end{bmatrix} \\ &= \begin{bmatrix} 1 - 4 x & 4 x \end{bmatrix} \end{align*} $$
interpolación de deformaciones $$ \begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -4 & 4 \end{bmatrix} \end{equation*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 1.25 \times 10^{5} \ [\text{N}] \end{equation*} $$
reemplazando $$ \begin{equation*} \int_{0}^{0.25} \begin{bmatrix} -4 \\ 4 \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -4 & 4 \end{bmatrix} dx \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \int_{0}^{0.25} 1000 \begin{bmatrix} 1 - 4 x \\ 4 x \end{bmatrix} dx + \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 5 \times 10^{5} & -5 \times 10^{5} \\ -5 \times 10^{5} & 5 \times 10^{5} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} 125 \\ 125 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
Ensamblaje y solución
ensamblando matriz global $$ \begin{equation*} \begin{bmatrix} 5 \times 10^{5} & -5 \times 10^{5} & 0 \\ -5 \times 10^{5} & 5 \times 10^{5} + 5 \times 10^{5} & -5 \times 10^{5} \\ 0 & -5 \times 10^{5} & 5 \times 10^{5} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} + u_{1} \\ u_{2} \end{bmatrix} = \begin{bmatrix} 125 \\ 125 + 125 \\ 125 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} + F_{1} \\ F_{2} \end{bmatrix} \end{equation*} $$
sumando $$ \begin{equation*} \begin{bmatrix} 5 \times 10^{5} & -5 \times 10^{5} & 0 \\ -5 \times 10^{5} & 10 \times 10^{5} & -5 \times 10^{5} \\ 0 & -5 \times 10^{5} & 5 \times 10^{5} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} 125 \\ 250 \\ 125 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
reemplazando condiciones de contorno $$ \begin{equation*} \begin{bmatrix} 5 \times 10^{5} & -5 \times 10^{5} & 0 \\ -5 \times 10^{5} & 10 \times 10^{5} & -5 \times 10^{5} \\ 0 & -5 \times 10^{5} & 5 \times 10^{5} \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} 125 \\ 250 \\ 125 \end{bmatrix} + \begin{bmatrix} F_{1} \\ 0 \\ 250 \end{bmatrix} \end{equation*} $$
sumando $$ \begin{equation*} \begin{bmatrix} 5 \times 10^{5} & -5 \times 10^{5} & 0 \\ -5 \times 10^{5} & 10 \times 10^{5} & -5 \times 10^{5} \\ 0 & -5 \times 10^{5} & 5 \times 10^{5} \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} F_{1} + 125 \\ 250 \\ 375 \end{bmatrix} \end{equation*} $$
resolviendo $$ \begin{align*} F_{1} &= -750 \ [\text{N}] \\ u_{2} &= 0.00125 \ [\text{m}] \\ u_{3} &= 0.002 \ [\text{m}] \end{align*} $$
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 $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} 1 - 4 x & 4 x \end{bmatrix} \begin{bmatrix} 0 \\ 0.00125 \end{bmatrix} = 0.005 x = 0.005 X \ [\text{m}] \end{equation*} $$
deformaciones $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -4 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 0.00125 \end{bmatrix} = 0.005 \end{equation*} $$
esfuerzo $$ \begin{equation*} \sigma = E \ \varepsilon = 1 \ [\text{MPa}] \end{equation*} $$
Elemento 2
reemplazando \( h=0.25 \) $$ \begin{equation*} x = X - 0.25 \end{equation*} $$
desplazamientos $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} 1 - 4 x & 4 x \end{bmatrix} \begin{bmatrix} 0.00125 \\ 0.002 \end{bmatrix} = 0.00125 + 0.003 x = 0.0005 + 0.003 X \ [\text{m}] \end{equation*} $$
deformaciones $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -4 & 4 \end{bmatrix} \begin{bmatrix} 0.00125 \\ 0.002 \end{bmatrix} = 0.003 \end{equation*} $$
esfuerzo $$ \begin{equation*} \sigma = E \ \varepsilon = 0.6 \ [\text{MPa}] \end{equation*} $$