User:Eml4500.f08.delta 6.guzman/Lecture Notes 11/19

It is important to recognize what elements constitute the local stiffness matrix for each element in the system, the following diagrams show the local element degrees of freedom of element 1 and element 2

Where $$ \tilde{k}^{(e)}_{6x6}\tilde{d}^{(e)}_{6x6}=\tilde{f}^{(e)}_{6x6} \! $$ and $$\tilde{d}^{(e)}=\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \cdot \cdot \cdot \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} \! $$ and $$ \tilde{f}^{(e)}=\begin{Bmatrix} \tilde{f}_{1}^{(e)}\\ \cdot \cdot \cdot \\ \tilde{f}_{6}^{(e)}

\end{Bmatrix} \! $$

Note: $$ \tilde{f}^{(e)}_{3}=f^{(e)}_{3} \! $$ and $$ \tilde{f}^{(e)}_{6}=f^{(e)}_{6} \! $$ since these are the moments about the $$ \tilde{z} \! $$ axis.

By using the elements of the diagrams above for element 1 and element 2, the local stiffness matrix $$ \tilde{k}\! $$ would be: $$ \begin{bmatrix}\frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{2EI}{L} \\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} $$

Note: All the matrix coefficients would not have the same dimensions but when they are multiplied with rotations or displacements depending on location they will end up with the right units.

Dimensional Analysis
$$ \left [ \tilde {d}_1 \right ] = L = \left [ \tilde {d}_i \right ] \qquad i=1, 2, 4, 5$$

where: $$ \left [ \tilde {d}_1 \right ] $$ is a dimension of length and $$ \left [ \tilde {d}_i \right ] $$ is displacement.

Note: For the dimensional analysis the $$ [ ] \! $$ symbolizes "dimension of" instead of the $$ [ ] \! $$ commonly used for matrices.

$$ \left [ \tilde {d}_3 \right ] = 1 = \left [ \tilde {d}_6 \right ] $$

where: $$ 1 $$ is dimensionless and $$ \left [ \tilde {d}_6 \right ] $$ is the rotational displacement

From the graph it can be observed that $$ \bar{AB} = R * \theta $$ where $$ \theta $$ is measured in radians and $$\bar{AB}$$ is the arc length from point A to point B. Another way to express $$ \theta $$ is $$\frac{\bar{AB}}{R}$$ where both have units of length making $$ \theta $$  dimensionless: $$ \left [ \theta \right ] = \frac {[\bar{AB}]}{[R]} = \frac {L}{L} = 1 $$

Applying the same non-dimensional analysis principle: $$ \sigma = \Epsilon \varepsilon \rightarrow \left [ \sigma \right ] = \left [ \Epsilon \right ] \left [ \varepsilon \right ] = 1 $$