Example - Krylov method

Problem

The given cantilevered beam ($//<![CDATA[ \begin{array}{l}l=0.8\end{array} //]]>$$//<![CDATA[ \begin{array}{l}\mathrm{m}\end{array} //]]>$) has a quadratic cross section ($//<![CDATA[ \begin{array}{l}a=0.01\end{array} //]]>$$//<![CDATA[ \begin{array}{l}\mathrm{m}\end{array} //]]>$) with $//<![CDATA[ \begin{array}{l}E=2.1\times 10^{11}\end{array} //]]>$$//<![CDATA[ \begin{array}{l}\mathrm{N/m^2}\end{array} //]]>$, $//<![CDATA[ \begin{array}{l}\nu=0.3\end{array} //]]>$, $//<![CDATA[ \begin{array}{l}\rho=7850\end{array} //]]>$$//<![CDATA[ \begin{array}{l}\mathrm{kg/m^3}\end{array} //]]>$. It is discretized using the Finite Element Method and excited at the tip with $//<![CDATA[ \begin{array}{l}F=1\end{array} //]]>$$//<![CDATA[ \begin{array}{l}\mathrm{N}\end{array} //]]>$ and of varying circular frequency ($//<![CDATA[ \begin{array}{l}\Omega=\frac{s}{\mathrm{i}}\end{array} //]]>$).

We now include Rayleigh damping ($//<![CDATA[ \begin{array}{l}\mathbf{C}=\alpha\mathbf{M} + \beta\mathbf{K}\end{array} //]]>$) with coefficients $//<![CDATA[ \begin{array}{l}\alpha=1.64\end{array} //]]>$ $//<![CDATA[ \begin{array}{l}\mathrm{1/s}\end{array} //]]>$, $//<![CDATA[ \begin{array}{l}\beta=4.1 e-6\end{array} //]]>$ $//<![CDATA[ \begin{array}{l}\mathrm{s}\end{array} //]]>$.

Solution

Choosing only a few expansion points demonstrates how local the method approximates the full system. Here we choose $//<![CDATA[ \begin{array}{l}s_1=500i \frac{\text{rad}}{\text{s}},\ s_2=6900i\frac{\text{rad}}{\text{s}}\end{array} //]]>$.

Next we choose eight linearly distributed expansion points between $//<![CDATA[ \begin{array}{l}s_1=100i\end{array} //]]>$ and $//<![CDATA[ \begin{array}{l}s_r=7000i\end{array} //]]>$. This reduced model approximates the full model reasonably up to $//<![CDATA[ \begin{array}{l}1 \times 10^4 \frac{\text{rad}}{\text{s}}\end{array} //]]>$.