Ejemplo 2
\( L = 0.5 \ \text{m} \), \( A = 6.25 \times 10^{-4} \ \text{m}^{2} \) y \( E = 200 \ \text{MPa} \)
la función que interpola la geometría es $$ \begin{equation*} x = \bigg( -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) x_{1} + (1 - \xi^{2}) x_{2} + \bigg( \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) x_{3} \end{equation*} $$
las funciones de forma son $$ \begin{equation*} \mathbf{N} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \end{equation*} $$
el jacobiano es $$ \begin{equation*} J = \frac{d x}{d \xi} = \frac{x_{3} - x_{1} + (2 x_{1} - 4 x_{2} + 2 x_{3}) \xi}{2} \end{equation*} $$
deformaciones $$ \begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \frac{d \mathbf{N}}{d \xi} \frac{d \xi}{d x} = \begin{bmatrix} -\frac{1}{2} + \xi & -2 \xi & \frac{1}{2} + \xi \end{bmatrix} \frac{2}{x_{3} - x_{1} + (2 x_{1} - 4 x_{2} + 2 x_{3}) \xi} \end{equation*} $$
Un elemento de tres nodos
funciones de forma $$ \begin{equation*} \mathbf{N} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \end{equation*} $$
las deformaciones se obtienen reemplazando \( x_{1} = 0 \), \( x_{2} = 0.25 \) y \( x_{3} = 0.5 \) $$ \begin{align*} \mathbf{B} &= \begin{bmatrix} -\frac{1}{2} + \xi & -2 \xi & \frac{1}{2} + \xi \end{bmatrix} \frac{2}{x_{3} - x_{1} + (2 x_{1} - 4 x_{2} + 2 x_{3}) \xi} \\ &= \begin{bmatrix} -\frac{1}{2} + \xi & -2 \xi & \frac{1}{2} + \xi \end{bmatrix} \frac{2}{0.5 - 0 + [2 (0) - 4 (0.25) + 2 (0.5)] \xi} \\ &= \begin{bmatrix} -2 + 4 \xi & -8 \xi & 2 + 4 \xi \end{bmatrix} \end{align*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 1.25 \times 10^{5} \ [\text{N}] \end{equation*} $$
el jacobiano se obtiene reemplazando \( x_{1} = 0 \), \( x_{2} = 0.25 \) y \( x_{3} = 0.5 \) $$ \begin{align*} J &= \frac{x_{3} - x_{1} + (2 x_{1} - 4 x_{2} + 2 x_{3}) \xi}{2} \\ &= \frac{0.5 - 0 + [2 (0) - 4 (0.25) + 2 (0.5)] \xi}{2} \\ &= \frac{1}{4} \end{align*} $$
reemplazando $$ \begin{equation*} \int_{-1}^{+1} \begin{bmatrix} -2 + 4 \xi \\ -8 \xi \\ 2 + 4 \xi \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -2 + 4 \xi & -8 \xi & 2 + 4 \xi \end{bmatrix} \frac{1}{4} \ d \xi \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \int_{-1}^{+1} 1000 \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \\ 1 - \xi^{2} \\ \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \frac{1}{4} \ d \xi + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 5.83 \times 10^{5} & -6.67 \times 10^{5} & 8.33 \times 10^{4} \\ -6.67 \times 10^{5} & 1.33 \times 10^{6} & -6.67 \times 10^{5} \\ 8.33 \times 10^{4} & -6.67 \times 10^{5} & 5.83 \times 10^{5} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} 83.33 \\ 333.33 \\ 83.33 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
reemplazando las condiciones de contorno $$ \begin{equation*} \begin{bmatrix} 5.83 \times 10^{5} & -6.67 \times 10^{5} & 8.33 \times 10^{4} \\ -6.67 \times 10^{5} & 1.33 \times 10^{6} & -6.67 \times 10^{5} \\ 8.33 \times 10^{4} & -6.67 \times 10^{5} & 5.83 \times 10^{5} \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} 83.33 \\ 333.33 \\ 83.33 \end{bmatrix} + \begin{bmatrix} F_{1} \\ 0 \\ 250 \end{bmatrix} \end{equation*} $$
sumando $$ \begin{equation*} \begin{bmatrix} 5.83 \times 10^{5} & -6.67 \times 10^{5} & 8.33 \times 10^{4} \\ -6.67 \times 10^{5} & 1.33 \times 10^{6} & -6.67 \times 10^{5} \\ 8.33 \times 10^{4} & -6.67 \times 10^{5} & 5.83 \times 10^{5} \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} F_{1} + 83.33 \\ 333.33 \\ 333.33 \end{bmatrix} \end{equation*} $$
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: $$ \begin{equation*} x = \bigg( -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) x_{1} + (1 - \xi^{2}) x_{2} + \bigg( \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) x_{3} \end{equation*} $$
reemplazando \( x_{1}=0 \), \( x_{2}=0.25 \) y \( x_{3}=0.5 \) $$ \begin{equation*} x = \frac{1}{4} + \frac{1}{4} \xi \end{equation*} $$
despejando \( \xi \) $$ \begin{equation*} \xi = -1 + 4 x \end{equation*} $$
desplazamientos $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \begin{bmatrix} 0 \\ 0.00125 \\ 0.002 \end{bmatrix} = 0.00125 + 0.001 \xi - 0.00025 \xi^{2} = 0.006 x - 0.004 x^{2} \ [\text{m}] \end{equation*} $$
deformación $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -2 + 4 \xi & -8 \xi & 2 + 4 \xi \end{bmatrix} \begin{bmatrix} 0 \\ 0.00125 \\ 0.002 \end{bmatrix} = 0.004 - 0.002 \xi = 0.006 - 0.008 x \end{equation*} $$
esfuerzo $$ \begin{equation*} \sigma = E \ \varepsilon = 1.2 - 1.6 x \ [\text{MPa}] \end{equation*} $$
Dos elementos de tres nodos
Elemento 1
funciones de forma $$ \begin{equation*} \mathbf{N} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \end{equation*} $$
las deformaciones se obtienen reemplazando \( x_{1} = 0 \), \( x_{2} = 0.125 \) y \( x_{3} = 0.25 \) $$ \begin{align*} \mathbf{B} &= \begin{bmatrix} -\frac{1}{2} + \xi & -2 \xi & \frac{1}{2} + \xi \end{bmatrix} \frac{2}{x_{3} - x_{1} + (2 x_{1} - 4 x_{2} + 2 x_{3}) \xi} \\ &= \begin{bmatrix} -\frac{1}{2} + \xi & -2 \xi & \frac{1}{2} + \xi \end{bmatrix} \frac{2}{0.25 - 0 + [2 (0) - 4 (0.125) + 2 (0.25)] \xi} \\ &= \begin{bmatrix} -4 + 8 \xi & -16 \xi & 4 + 8 \xi \end{bmatrix} \end{align*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 1.25 \times 10^{5} \ [\text{N}] \end{equation*} $$
el jacobiano se obtiene reemplazando \( x_{1} = 0 \), \( x_{2} = 0.125 \) y \( x_{3} = 0.25 \) $$ \begin{align*} J &= \frac{x_{3} - x_{1} + (2 x_{1} - 4 x_{2} + 2 x_{3}) \xi}{2} \\ &= \frac{0.25 - 0 + [2 (0) - 4 (0.125) + 2 (0.25)] \xi}{2} \\ &= \frac{1}{8} \end{align*} $$
reemplazando $$ \begin{equation*} \int_{-1}^{+1} \begin{bmatrix} -4 + 8 \xi \\ -16 \xi \\ 4 + 8 \xi \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -4 + 8 \xi & -16 \xi & 4 + 8 \xi \end{bmatrix} \frac{1}{8} \ d \xi \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \int_{-1}^{+1} 1000 \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \\ 1 - \xi^{2} \\ \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \frac{1}{8} \ d \xi + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 1.17 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} \\ -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} \\ 1.67 \times 10^{5} & -1.33 \times 10^{6} & 1.17 \times 10^{6} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} 41.67 \\ 166.67 \\ 41.67 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
Elemento 2
funciones de forma $$ \begin{equation*} \mathbf{N} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \end{equation*} $$
las deformaciones se obtienen reemplazando \( x_{1} = 0.25 \), \( x_{2} = 0.375 \) y \( x_{3} = 0.5 \) $$ \begin{align*} \mathbf{B} &= \begin{bmatrix} -\frac{1}{2} + \xi & -2 \xi & \frac{1}{2} + \xi \end{bmatrix} \frac{2}{x_{3} - x_{1} + (2 x_{1} - 4 x_{2} + 2 x_{3}) \xi} \\ &= \begin{bmatrix} -\frac{1}{2} + \xi & -2 \xi & \frac{1}{2} + \xi \end{bmatrix} \frac{2}{0.5 - 0.25 + [2 (0.25) - 4 (0.375) + 2 (0.5)] \xi} \\ &= \begin{bmatrix} -4 + 8 \xi & -16 \xi & 4 + 8 \xi \end{bmatrix} \end{align*} $$
matriz constitutiva $$ \begin{equation*} \mathbf{D} = E \ A = 1.25 \times 10^{5} \ [\text{N}] \end{equation*} $$
el jacobiano se obtiene reemplazando \( x_{1} = 0.25 \), \( x_{2} = 0.375 \) y \( x_{3} = 0.5 \) $$ \begin{align*} J &= \frac{x_{3} - x_{1} + (2 x_{1} - 4 x_{2} + 2 x_{3}) \xi}{2} \\ &= \frac{0.5 - 0.25 + [2 (0.25) - 4 (0.375) + 2 (0.5)] \xi}{2} \\ &= \frac{1}{8} \end{align*} $$
reemplazando $$ \begin{equation*} \int_{-1}^{+1} \begin{bmatrix} -4 + 8 \xi \\ -16 \xi \\ 4 + 8 \xi \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -4 + 8 \xi & -16 \xi & 4 + 8 \xi \end{bmatrix} \frac{1}{8} \ d \xi \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \int_{-1}^{+1} 1000 \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \\ 1 - \xi^{2} \\ \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \frac{1}{8} \ d \xi + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
integrando $$ \begin{equation*} \begin{bmatrix} 1.17 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} \\ -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} \\ 1.67 \times 10^{5} & -1.33 \times 10^{6} & 1.17 \times 10^{6} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} 41.67 \\ 166.67 \\ 41.67 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
Ensamblaje y solución
ensamblando matriz global $$ \begin{equation*} \begin{bmatrix} 1.17 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} & 0 & 0 \\ -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} & 0 & 0 \\ 1.67 \times 10^{5} & -1.33 \times 10^{6} & 1.17 \times 10^{6} + 1.17 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} \\ 0 & 0 & -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} \\ 0 & 0 & 1.67 \times 10^{5} & -1.33 \times 10^{6} & 1.17 \times 10^{6} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} + u_{1} \\ u_{2} \\ u_{3} \end{bmatrix} = \begin{bmatrix} 41.67 \\ 166.67 \\ 41.67 + 41.67 \\ 166.67 \\ 41.67 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} + F_{1} \\ F_{2} \\ F_{3} \end{bmatrix} \end{equation*} $$
sumando $$ \begin{equation*} \begin{bmatrix} 1.17 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} & 0 & 0 \\ -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} & 0 & 0 \\ 1.67 \times 10^{5} & -1.33 \times 10^{6} & 2.33 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} \\ 0 & 0 & -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} \\ 0 & 0 & 1.67 \times 10^{5} & -1.33 \times 10^{6} & 1.17 \times 10^{6} \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \end{bmatrix} = \begin{bmatrix} 41.67 \\ 166.67 \\ 83.33 \\ 166.67 \\ 41.67 \end{bmatrix} + \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \end{bmatrix} \end{equation*} $$
reemplazando las condiciones de contorno $$ \begin{equation*} \begin{bmatrix} 1.17 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} & 0 & 0 \\ -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} & 0 & 0 \\ 1.67 \times 10^{5} & -1.33 \times 10^{6} & 2.33 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} \\ 0 & 0 & -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} \\ 0 & 0 & 1.67 \times 10^{5} & -1.33 \times 10^{6} & 1.17 \times 10^{6} \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \end{bmatrix} = \begin{bmatrix} 41.67 \\ 166.67 \\ 83.33 \\ 166.67 \\ 41.67 \end{bmatrix} + \begin{bmatrix} F_{1} \\ 0 \\ 0 \\ 0 \\ 250 \end{bmatrix} \end{equation*} $$
sumando $$ \begin{equation*} \begin{bmatrix} 1.17 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} & 0 & 0 \\ -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} & 0 & 0 \\ 1.67 \times 10^{5} & -1.33 \times 10^{6} & 2.33 \times 10^{6} & -1.33 \times 10^{6} & 1.67 \times 10^{5} \\ 0 & 0 & -1.33 \times 10^{6} & 2.67 \times 10^{6} & -1.33 \times 10^{6} \\ 0 & 0 & 1.67 \times 10^{5} & -1.33 \times 10^{6} & 1.17 \times 10^{6} \end{bmatrix} \begin{bmatrix} 0 \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \end{bmatrix} = \begin{bmatrix} F_{1} + 41.67 \\ 166.67 \\ 83.33 \\ 166.67 \\ 291.67 \end{bmatrix} \end{equation*} $$
resolviendo $$ \begin{align*} F_{1} &= -750 \ [\text{N}] \\ u_{2} &= 0.0006875 \ [\text{m}] \\ u_{3} &= 0.00125 \ [\text{m}] \\ u_{4} &= 0.0016875 \ [\text{m}] \\ u_{5} &= 0.002 \ [\text{m}] \\ \end{align*} $$
Desplazamientos, deformaciones y esfuerzos
Las funciones de forma en coordenadas locales se transforman a coordenadas globales usando: $$ \begin{equation*} x = \bigg( -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) x_{1} + (1 - \xi^{2}) x_{2} + \bigg( \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \bigg) x_{3} \end{equation*} $$
Elemento 1
reemplazando \( x_{1}=0 \), \( x_{2}=0.125 \) y \( x_{3}=0.25 \) $$ \begin{equation*} x = \frac{1}{8} + \frac{1}{8} \xi \end{equation*} $$
despejando \( \xi \) $$ \begin{equation*} \xi = -1 + 8 x \end{equation*} $$
desplazamientos $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \begin{bmatrix} 0 \\ 0.0006875 \\ 0.00125 \end{bmatrix} = 0.0006875 + 0.000625 \xi - 0.0000625 \xi^{2} = 0.006 x - 0.004 x^{2} \ [\text{m}] \end{equation*} $$
deformación $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -4 + 8 \xi & -16 \xi & 4 + 8 \xi \end{bmatrix} \begin{bmatrix} 0 \\ 0.0006875 \\ 0.00125 \end{bmatrix} = 0.005 - 0.001 \xi = 0.006 - 0.008 x \end{equation*} $$
esfuerzo $$ \begin{equation*} \sigma = E \ \varepsilon = 1.2 - 1.6 x \ [\text{MPa}] \end{equation*} $$
Elemento 2
reemplazando \( x_{1}=0.25 \), \( x_{2}=0.375 \) y \( x_{3}=0.5 \) $$ \begin{equation*} x = \frac{3}{8} + \frac{1}{8} \xi \end{equation*} $$
despejando \( \xi \) $$ \begin{equation*} \xi = -3 + 8 x \end{equation*} $$
desplazamientos $$ \begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} -\frac{1}{2} \xi + \frac{1}{2} \xi^{2} & 1 - \xi^{2} & \frac{1}{2} \xi + \frac{1}{2} \xi^{2} \end{bmatrix} \begin{bmatrix} 0.00125 \\ 0.0016875 \\ 0.002 \end{bmatrix} = 0.0016875 + 0.000375 \xi - 0.0000625 \xi^{2} = 0.006 x - 0.004 x^{2} \ [\text{m}] \end{equation*} $$
deformación $$ \begin{equation*} \varepsilon = \mathbf{B} \ \mathbf{u} = \begin{bmatrix} -4 + 8 \xi & -16 \xi & 4 + 8 \xi \end{bmatrix} \begin{bmatrix} 0.00125 \\ 0.0016875 \\ 0.002 \end{bmatrix} = 0.003 - 0.001 \xi = 0.006 - 0.008 x \end{equation*} $$
esfuerzo $$ \begin{equation*} \sigma = E \ \varepsilon = 1.2 - 1.6 x \ [\text{MPa}] \end{equation*} $$