# 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

\begin{equation*} \int_{x_{1}}^{x_{2}} \mathbf{B}^{\mathrm{T}} \mathbf{D} \ \mathbf{B} \ dx \ \mathbf{u} = \int_{x_{1}}^{x_{2}} q \ \mathbf{N}^{\mathrm{T}} dx + \mathbf{F} \end{equation*}

interpolación de desplazamientos

\begin{align*} \mathbf{N} &= \begin{bmatrix} \frac{x_{2}}{x_{2} - x_{1}} - \frac{1}{x_{2} - x_{1}} x & -\frac{x_{1}}{x_{2} - x_{1}} + \frac{1}{x_{2} - x_{1}} x \end{bmatrix} \\ &= \begin{bmatrix} \frac{0.5}{0.5 - 0} - \frac{1}{0.5 - 0} x & -\frac{0}{0.5 - 0} + \frac{1}{0.5 - 0} 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

desplazamientos

\begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} 1 - 2 x & 2 x \end{bmatrix} \begin{bmatrix} 0 \\ 0.002 \end{bmatrix} = 0.004 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

\begin{equation*} \int_{x_{1}}^{x_{2}} \mathbf{B}^{\mathrm{T}} \mathbf{D} \ \mathbf{B} \ dx \ \mathbf{u} = \int_{x_{1}}^{x_{2}} q \ \mathbf{N}^{T} dx + \mathbf{F} \end{equation*}

Elemento 1

interpolación de desplazamientos

\begin{align*} \mathbf{N} &= \begin{bmatrix} \frac{x_{2}}{x_{2} - x_{1}} - \frac{1}{x_{2} - x_{1}} x & -\frac{x_{1}}{x_{2} - x_{1}} + \frac{1}{x_{2} - x_{1}} x \end{bmatrix} \\ &= \begin{bmatrix} \frac{0.25}{0.25 - 0} - \frac{1}{0.25 - 0} x & -\frac{0}{0.25 - 0} + \frac{1}{0.25 - 0} 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} \frac{x_{2}}{x_{2} - x_{1}} - \frac{1}{x_{2} - x_{1}} x & -\frac{x_{1}}{x_{2} - x_{1}} + \frac{1}{x_{2} - x_{1}} x \end{bmatrix} \\ &= \begin{bmatrix} \frac{0.5}{0.5 - 0.25} - \frac{1}{0.5 - 0.25} x & -\frac{0.25}{0.5 - 0.25} + \frac{1}{0.5 - 0.25} x \end{bmatrix} \\ &= \begin{bmatrix} 2 - 4 x & -1 + 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_{0.25}^{0.5} \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_{0.25}^{0.5} 1000 \begin{bmatrix} 2 - 4 x \\ -1 + 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

Elemento 1

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 \ [\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

desplazamientos

\begin{equation*} u = \mathbf{N} \ \mathbf{u} = \begin{bmatrix} 2 - 4 x & -1 + 4 x \end{bmatrix} \begin{bmatrix} 0.00125 \\ 0.002 \end{bmatrix} = 0.0005 + 0.003 x \ [\text{N}] \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*}