# Funciones de forma unidimensionales

Las funciones de forma unidimensionales sirven para aproximar los desplazamientos: 

\begin{equation}
u = \alpha_{0} + \alpha_{1} x + \cdots + \alpha_{n} x^{n} = \sum_{i = 0}^{n} \alpha_{i} x^{i}
\end{equation}

## Elemento barra

Los elementos barra soportan esfuerzos debidos a tracción o compresión.

### Elemento de dos nodos

Para un elemento de dos nodos y dos grados de libertad:

\begin{equation}
u = \alpha_{0} + \alpha_{1} x  =
\left [
\begin{matrix}
1 & x
\end{matrix}
\right ]
\left [
\begin{matrix}
\alpha_{0} \\\
\alpha_{1}
\end{matrix}
\right ]
\end{equation}

Reemplazando los valores nodales en coordenadas naturales:

\begin{eqnarray}
\alpha_{0} + \alpha_{1}(-1) &=& u_{1}\\\
\alpha_{0} + \alpha_{1}(1) &=& u_{2}
\end{eqnarray}

Evaluando:

\begin{eqnarray}
\alpha_{0} - \alpha_{1} &=& u_{1}\\\
\alpha_{0} + \alpha_{1} &=& u_{2}
\end{eqnarray}

En forma matricial:

\begin{equation}
\left [
\begin{matrix}
1 & -1 \\\
1 & 1
\end{matrix}
\right ]
\left [
\begin{matrix}
\alpha_{0} \\\
\alpha_{1}
\end{matrix}
\right ] =
\left [
\begin{matrix}
u_{1} \\\
u_{2}
\end{matrix}
\right ]
\end{equation}

Resolviendo el sistema:

\begin{equation}
\left [
\begin{matrix}
\alpha_{0} \\\
\alpha_{1}
\end{matrix}
\right ] =
\left [
\begin{matrix}
\frac{1}{2} & \frac{1}{2} \\\
-\frac{1}{2} & \frac{1}{2}
\end{matrix}
\right ]
\left [
\begin{matrix}
u_{1} \\\
u_{2}
\end{matrix}
\right ]
\end{equation}

Reemplazando:

\begin{equation}
u =
\left [
\begin{matrix}
1 & x
\end{matrix}
\right ]
\left [
\begin{matrix}
\alpha_{0} \\\
\alpha_{1}
\end{matrix}
\right ] =
\left [
\begin{matrix}
1 & x
\end{matrix}
\right ]
\left [
\begin{matrix}
\frac{1}{2} & \frac{1}{2} \\\
-\frac{1}{2} & \frac{1}{2}
\end{matrix}
\right ]
\left [
\begin{matrix}
u_{1} \\\
u_{2}
\end{matrix}
\right ] =
\left [
\begin{matrix}
\frac{1}{2} - \frac{1}{2} x  & \frac{1}{2} + \frac{1}{2} x
\end{matrix}
\right ]
\left [
\begin{matrix}
u_{1} \\\
u_{2}
\end{matrix}
\right ]
\end{equation}

Reescribiendo $u$:

\begin{equation}
u = \Big ( \frac{1}{2} - \frac{1}{2} x \Big ) u_{1} + \Big ( \frac{1}{2} + \frac{1}{2} x \Big ) u_{2} = N_{1} u_{1} + N_{2} u_{2}
\end{equation}

### Elemento de tres nodos

Para un elemento de tres nodos y tres grados de libertad:

\begin{equation}
u = \alpha_{0} + \alpha_{1} x + \alpha_{2} x^{2} =
\left [
\begin{matrix}
1 & x & x^{2}
\end{matrix}
\right ]
\left [
\begin{matrix}
\alpha_{0} \\\
\alpha_{1} \\\
\alpha_{2}
\end{matrix}
\right ]
\end{equation}

Reemplazando los valores nodales en coordenadas naturales:

\begin{eqnarray}
\alpha_{0} + \alpha_{1}(-1) + \alpha_{2}(-1)^{2}&=& u_{1}\\\
\alpha_{0} + \alpha_{1}(0) + \alpha_{2}(0)^{2}&=& u_{2}\\\
\alpha_{0} + \alpha_{1}(1) + \alpha_{2}(1)^{2}&=& u_{3}
\end{eqnarray}

Evaluando:

\begin{eqnarray}
\alpha_{0} - \alpha_{1} + \alpha_{1} &=& u_{1}\\\
\alpha_{0} &=& u_{2}\\\
\alpha_{0} + \alpha_{1} + \alpha_{1}&=& u_{3}
\end{eqnarray}

En forma matricial:

\begin{equation}
\left [
\begin{matrix}
1 & -1 & 1 \\\
1 & 0 & 0 \\\
1 & 1 & 1 \\\
\end{matrix}
\right ]
\left [
\begin{matrix}
\alpha_{0} \\\
\alpha_{1} \\\
\alpha_{2}
\end{matrix}
\right ] =
\left [
\begin{matrix}
u_{1} \\\
u_{2} \\\
u_{3}
\end{matrix}
\right ]
\end{equation}

Resolviendo el sistema:

\begin{equation}
\left [
\begin{matrix}
\alpha_{0} \\\
\alpha_{1} \\\
\alpha_{2}
\end{matrix}
\right ] =
\left [
\begin{matrix}
0 & 1 & 0 \\\
-\frac{1}{2} & 0 & \frac{1}{2} \\\
\frac{1}{2} & -1 & \frac{1}{2}
\end{matrix}
\right ]
\left [
\begin{matrix}
u_{1} \\\
u_{2} \\\
u_{3}
\end{matrix}
\right ]
\end{equation}

Reemplazando:

\begin{equation}
u =
\left [
\begin{matrix}
1 & x & x^{2}
\end{matrix}
\right ]
\left [
\begin{matrix}
\alpha_{0} \\\
\alpha_{1} \\\
\alpha_{2}
\end{matrix}
\right ] =
\left [
\begin{matrix}
1 & x & x^{2}
\end{matrix}
\right ]
\left [
\begin{matrix}
0 & 1 & 0 \\\
-\frac{1}{2} & 0 & \frac{1}{2} \\\
\frac{1}{2} & -1 & \frac{1}{2}
\end{matrix}
\right ]
\left [
\begin{matrix}
u_{1} \\\
u_{2} \\\
u_{3}
\end{matrix}
\right ] =
\left [
\begin{matrix}
-\frac{1}{2} x + \frac{1}{2} x^{2} & 1 - x^{2} & \frac{1}{2} x + \frac{1}{2} x^{2}
\end{matrix}
\right ]
\left [
\begin{matrix}
u_{1} \\\
u_{2} \\\
u_{3}
\end{matrix}
\right ]
\end{equation}

Reescribiendo $u$:

\begin{equation}
u = \Big ( -\frac{1}{2} x + \frac{1}{2} x^{2} \Big ) u_{1} + (1 - x^{2}) u_{2} + \Big ( \frac{1}{2} x + \frac{1}{2} x^{2} \Big ) u_{3} = N_{1} u_{1} + N_{2} u_{2} + N_{3} u_{3}
\end{equation}

## Elementos barra de mayor grado polinomial

Los elementos de mayor grado pueden obtenerse mediante polinomios de Lagrange:

\begin{equation}
\ell = \prod_{i=0, i \neq j}^{k} \frac{x - x_{i}}{x_{j} - x_{i}}
\end{equation}

### Elemento de dos nodos

Usando la fórmula:

\begin{eqnarray}
N_{1} &=& \frac{x - 1}{-1 - 1} = \frac{1}{2} - \frac{1}{2} x \\\
N_{2} &=& \frac{x - (-1)}{1 - (-1)} = \frac{1}{2} + \frac{1}{2} x
\end{eqnarray}

### Elemento de tres nodos

Usando la fórmula:

\begin{eqnarray}
N_{1} &=& \frac{x - 0}{-1 - 0} \frac{x - 1}{-1 - 1} = -\frac{1}{2} x + \frac{1}{2} x^{2} \\\
N_{2} &=& \frac{x - (-1)}{0 - (-1)} \frac{x - 1}{0 - 1} = 1 - x^{2} \\\
N_{3} &=& \frac{x - 0}{1 - 0} \frac{x - 1}{1 - 1} = \frac{1}{2} x + \frac{1}{2} x^{2}
\end{eqnarray}


```

```

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Jupyter Notebook
Funciones de forma/funciones forma barra.ipynb
ClaudioVZ/Teoria-FEM-Python
8a4532f282c38737fb08d1216aa859ecb1e5b209
Artistic-2.0
2021-09-28T00:23:45.000Z
2021-09-28T00:23:45.000Z
Funciones de forma/funciones forma barra.ipynb
ClaudioVZ/Teoria-FEM-Python
8a4532f282c38737fb08d1216aa859ecb1e5b209
Artistic-2.0
Funciones de forma/funciones forma barra.ipynb
ClaudioVZ/Teoria-FEM-Python
8a4532f282c38737fb08d1216aa859ecb1e5b209
Artistic-2.0
2015-12-04T12:42:00.000Z
2019-10-31T21:50:32.000Z
Qwen/Qwen-72B
1. YES
2. YES