# Ejemplo 1

$L = 0.5 \ \text{m}$, $A = 6.25 \times 10^{-4} \ \text{m}^{2}$ y $E = 200 \ \text{MPa}$

la función que interpola la geometría es

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

las funciones de forma son

\begin{equation*} \mathbf{N} = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi & \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \end{equation*}

el jacobiano es

\begin{equation*} J = \frac{d x}{d \xi} = \frac{1}{2} x_{2} - \frac{1}{2} x_{1} = \frac{l}{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} & \frac{1}{2} \end{bmatrix} \frac{2}{l} = \begin{bmatrix} -\frac{1}{l} & \frac{1}{l} \end{bmatrix} \end{equation*}

## Un elemento de dos nodos

\begin{equation*} \int_{-1}^{+1} \mathbf{B}^{\mathrm{T}} \mathbf{D} \ \mathbf{B} \ J \ d\xi \ \mathbf{u} = \int_{-1}^{+1} q \ \mathbf{N}^{\mathrm{T}} J \ d\xi + \mathbf{F} \end{equation*}

interpolación de desplazamientos

\begin{equation*} \mathbf{N} = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi & \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \end{equation*}

interpolación de deformaciones

\begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -\frac{1}{l} & \frac{1}{l} \end{bmatrix} = \begin{bmatrix} -\frac{1}{0.5} & \frac{1}{0.5} \end{bmatrix} = \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*}

jacobiano

\begin{equation*} J = \frac{dx}{d\xi} = \frac{l}{2} = \frac{0.5}{2} = \frac{1}{4} \end{equation*}

reemplazando

\begin{equation*} \int_{-1}^{+1} \begin{bmatrix} -2 \\ 2 \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -2 & 2 \end{bmatrix} \frac{1}{4} \ d \xi \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \int_{-1}^{+1} 1000 \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi \\ \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \frac{1}{4} \ d \xi + \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 naturales se transforman a coordenadas globales usando

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

despejando $\xi$

\begin{equation*} \xi = \frac{2}{x_{2} - x_{1}} x - \frac{x_{2} + x_{1}}{x_{2} - x_{1}} \end{equation*}

reemplazando $x_{1}=0$ y $x_{2}=0.5$

\begin{equation*} \xi = \frac{2}{0.5 - 0} x - \frac{0.5 + 0}{0.5 - 0} = 4 x - 1 \end{equation*}

desplazamientos

\begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi & \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \begin{bmatrix} 0 \\ 0.002 \end{bmatrix} = 0.001 + 0.001 \xi= 0.004 x \ [\text{m}] \end{equation*}

deformación normal

\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 normal

\begin{equation*} \sigma = E \ \varepsilon = 0.8 \ [\text{MPa}] \end{equation*}

## Dos elementos de dos nodos

\begin{equation*} \int_{-1}^{+1} \mathbf{B}^{\mathrm{T}} \mathbf{D} \ \mathbf{B} \ J \ d\xi \ \mathbf{u} = \int_{-1}^{+1} q \ \mathbf{N}^{\mathrm{T}} J \ d\xi + \mathbf{F} \end{equation*}

Elemento 1

interpolación de desplazamientos

\begin{equation*} \mathbf{N} = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi & \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \end{equation*}

interpolación de deformaciones

\begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -\frac{1}{l} & \frac{1}{l} \end{bmatrix} = \begin{bmatrix} -\frac{1}{0.25} & \frac{1}{0.25} \end{bmatrix} = \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*}

jacobiano

\begin{equation*} J = \frac{dx}{d\xi} = \frac{l}{2} = \frac{0.25}{2} = \frac{1}{8} \end{equation*}

reemplazando

\begin{equation*} \int_{-1}^{+1} \begin{bmatrix} -4 \\ 4 \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -4 & 4 \end{bmatrix} \frac{1}{8} \ d \xi \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \int_{-1}^{+1} 1000 \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi \\ \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \frac{1}{8} \ d \xi + \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{equation*} \mathbf{N} = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi & \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \end{equation*}

interpolación de deformaciones

\begin{equation*} \mathbf{B} = \frac{d \mathbf{N}}{d x} = \begin{bmatrix} -\frac{1}{l} & \frac{1}{l} \end{bmatrix} = \begin{bmatrix} -\frac{1}{0.25} & \frac{1}{0.25} \end{bmatrix} = \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*}

jacobiano

\begin{equation*} J = \frac{dx}{d\xi} = \frac{l}{2} = \frac{0.25}{2} = \frac{1}{8} \end{equation*}

reemplazando

\begin{equation*} \int_{-1}^{+1} \begin{bmatrix} -4 \\ 4 \end{bmatrix} \begin{bmatrix} 1.25 \times 10^{5} \end{bmatrix} \begin{bmatrix} -4 & 4 \end{bmatrix} \frac{1}{8} \ d \xi \begin{bmatrix} u_{1} \\ u_{2} \end{bmatrix} = \int_{-1}^{+1} 1000 \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi \\ \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \frac{1}{8} \ d \xi + \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 naturales se transforman a coordenadas globales usando

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

despejando $\xi$

\begin{equation*} \xi = \frac{2}{x_{2} - x_{1}} x - \frac{x_{2} + x_{1}}{x_{2} - x_{1}} \end{equation*}

Elemento 1

reemplazando $x_{1}=0$ y $x_{2}=0.25$

\begin{equation*} \xi = \frac{2}{0.25 - 0} x - \frac{0.25 + 0}{0.25 - 0} = 8 x - 1 \end{equation*}

desplazamientos

\begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi & \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \begin{bmatrix} 0 \\ 0.00125 \end{bmatrix} = 0.000625 + 0.000625 \xi = 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 $x_{1}=0.25$ y $x_{2}=0.5$

\begin{equation*} \xi = \frac{2}{0.5 - 0.25} x - \frac{0.5 + 0.25}{0.5 - 0.25} = 8 x - 3 \end{equation*}

desplazamientos

\begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} \xi & \frac{1}{2} + \frac{1}{2} \xi \end{bmatrix} \begin{bmatrix} 0.00125 \\ 0.002 \end{bmatrix} = 0.001625 + 0.000375 \xi = 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*}