User:Eml4500.f08.ateam.boggs.t/Lecture 12







The equation for a two-force element,

$$ K^{(e)}\cdot d^{(e)} = f^{(e)} $$

can also be expressed with axial forces and displacements where,

$$

\begin{bmatrix} P_{1}^{(e)}\\ P_{2}^{(e)} \end{bmatrix} = \begin{bmatrix} K & -K\\ -K & K \end{bmatrix} \begin{bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{bmatrix} $$

This can also be expressed as,

$$ K^{(e)} \cdot \begin{bmatrix} 1 & -1\\    -1 & 1 \end{bmatrix} \begin{bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{bmatrix} =\begin{bmatrix} P_{1}^{(e)}\\ P_{2}^{(e)} \end{bmatrix} $$

where,

$$ K^{(e)} = \frac{EA}{L} $$

The relationship between q(e) and d(e) can be expressed in the form of the equation,

$$ q^{(e)} = T^{(e)}\cdot d^{(e)} $$

where q(e) is a 2X1 matrix, T(e) is a 2X4 matrix, and d(e) is a 4X1 matrix.

Consider the displacement vector of local node one, it is denoted by d[1](e) and can be seen at right where,

$$ d_{[1]}^{(e)} = d_{1}^{(e)}\overrightarrow{i} + d_{2}^{(e)}\overrightarrow{j} $$

relating q to d,

$$ q_{1}^{(e)} = d_{[1]}^{(e)}\overrightarrow{\tilde{i}} $$

$$ q_{1}^{(e)} = (d_{1}^{(e)}\overrightarrow{i} + d_{2}^{(e)}\overrightarrow{j})\cdot \overrightarrow{\tilde{i}} $$

$$ q_{1}^{(e)} = d_{1}^{(e)}(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}}) + d_{2}^{(e)}(\overrightarrow{j}\cdot \overrightarrow{\tilde{i}}) $$

Where,

$$ \overrightarrow{i}\cdot \overrightarrow{\tilde{i}} = \cos\theta = l^{(e)} $$

$$ \overrightarrow{j}\cdot \overrightarrow{\tilde{i}} = \sin\theta = m^{(e)} $$

$$ q_{1}^{(e)} = l^{(e)}d_{1}^{(e)} + m^{(e)}d_{2}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \end{bmatrix} $$

For node 2, denoted by d[2](e),

$$ d_{[2]}^{(e)} = d_{3}^{(e)}\overrightarrow{i} + d_{4}^{(e)}\overrightarrow{j} $$

relating q to d,

$$ q_{2}^{(e)} = d_{[2]}^{(e)}\overrightarrow{\tilde{i}} $$

$$ q_{2}^{(e)} = (d_{3}^{(e)}\overrightarrow{i} + d_{4}^{(e)}\overrightarrow{j})\cdot \overrightarrow{\tilde{i}} $$

$$ q_{2}^{(e)} = d_{3}^{(e)}(\overrightarrow{i}\cdot \overrightarrow{\tilde{i}}) + d_{4}^{(e)}(\overrightarrow{j}\cdot \overrightarrow{\tilde{i}}) $$

$$ q_{2}^{(e)} = l^{(e)}d_{3}^{(e)} + m^{(e)}d_{4}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{3}^{(e)}\\ d_{4}^{(e)} \end{bmatrix} $$

Therefore,

$$ \begin{bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)} \end{bmatrix} $$

Here q(e) is a 2X1 matrix, T(e) is a 2X4 matrix, and d(e) is a 4X1 matrix.