Question 10.8: The system used in Example 6.5 is shown again in Fig. 10.14.......
The system used in Example 6.5 is shown again in Fig. 10.14. The beam is assumed massless, and uniform, with EI = 2 × 10^{6} N m² . Two discrete masses m_{1} = 10 \mathrm{kg and} m_{2} = 8 kg are attached, as shown. The total length of the beam, L = 4 m. The viscous damping coefficient is assumed to be 0.05 for both modes. F_{1} is a random force with constant power spectral density of 100 N²/Hz, from 10 to 120 Hz, and F_{2} is zero. Find the RMS value of the displacement, z_{1}.
From Example 6.5, the equations of motion in orthonormal form (i.e. with the mass matrix defined as a unit diagonal) are
\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ] \left \{ \begin{matrix} \ddot{q}_{1} \\ \ddot{q}_{2} \end{matrix} \right \} + \left [ \begin{matrix} 2\gamma_{1}\omega_{1} & 0 \\ 0 & 2\gamma_{2}\omega_{2} \end{matrix} \right ] \left \{ \begin{matrix} \dot{q}_{1} \\ \dot{q}_{2} \end{matrix} \right \} + \left [ \begin{matrix} \omega^{2}_{1} & 0 \\ 0 & \omega^{2}_{2} \end{matrix} \right ] \left \{ \begin{matrix} q_{1} \\ q_{2} \end{matrix} \right \} = \left \{ \begin{matrix} Q_{1} \\ Q_{2} \end{matrix} \right \} (A)where the undamped natural frequencies, in rad/s, are ω_{1} = 102.02 \mathrm{and} ω_{2} = 621.30, and in this case γ_{1} = γ_{2} = 0.05.
Also from Example 6.5, the relationship between the actual displacements,{z}, and the generalized displacements, {q}, is
and the relationship between the actual external forces, [F], and the generalized forces, [Q], is
\left \{ \begin{matrix} Q_{1} \\ Q_{2} \end{matrix} \right \} = \begin{bmatrix} 0.1071 & 0.3326 \\ -0.2975 & 0.1198 \end{bmatrix} \left \{ \begin{matrix} F_{1} \\ F_{2} \end{matrix} \right \} (C)Since Eq. (A) is expressed in normal modes, with all matrices diagonal, it can now be treated as two uncoupled, single-DOF equations:
\ddot{q}_{1} + 2\gamma_{1}\omega_{1}\dot{q}_{1} +\omega^{2}_{1}q_{1} = Q_{1} (D)\\ \mathrm{and} \\ \ddot{q}_{2} + 2\gamma_{2}\omega_{2}\dot{q}_{2} +\omega^{2}_{2}q_{2} = Q_{2} (E)From Chapter 4, the complex receptance expressions, or FRFs , H_{11}(f) \mathrm{and} H_{22}(f), corresponding to Eqs (D) and (E), respectively, are as follows, noting that the generalized mass is unity in both cases, and that (f) has been omitted throughout for clarity:
H_{11} = \frac{\underline{q}_{1}}{\underline{Q}_{1}} = \frac{(1-\Omega^{2}_{1})-\mathrm{i}(2\gamma _{1}\Omega _{1})}{(2\pi f_{1})^{2}\left[(1-\Omega^{2}_{1})^{2}+(2\gamma _{1}\Omega _{1})^{2}\right] } = R_{11} + \mathrm{i}I_{11} (G) \\ \mathrm{and} \\ H_{22} = \frac{\underline{q}_{2}}{\underline{Q}_{2}} = \frac{(1-\Omega^{2}_{2})-\mathrm{i}(2\gamma _{2}\Omega _{2})}{(2\pi f_{2})^{2}\left[(1-\Omega^{2}_{2})^{2}+(2\gamma _{2}\Omega _{2})^{2}\right] } = R_{22} + \mathrm{i}I_{22} (H)where R_{11} \mathrm{and} I_{11} are the real and imaginary parts of H_{11} \mathrm{and} R_{22} \mathrm{and} I_{22} the real and imaginary parts of H_{22}. Also, Ω_{1} = f/f_{1} \mathrm{and} Ω_{2} = f/f_{2}, where f is the excitation frequency in Hz and f_{1} = ω_{1}/2π = 16.236 Hz is the natural frequency of mode 1 and f_{2} = ω_{2}/2π = 98.883 Hz is the natural frequency of mode 2.
Equations (B) and (C) apply equally well when z_{1}, z_{2}, q_{1}, q_{2}, F_{1}, F_{2}, Q_{1}, Q_{2} are replaced by the complex vectors \underline{z} _{1}, \underline{z} _{2},\underline{q} _{1}, \underline{q} _{2}, \underline{F} _{1}, \underline{F} _{2}, \underline{Q} _{1}, \underline{Q} _{2}. Therefore, from Eq. (B):
From Eqs (G) and (H):
\underline{q}_{1} = (R_{11}+\mathrm{i}I_{11})\underline{Q}_{1} \mathrm{and} \underline{q}_{2} = (R_{22}+\mathrm{i}I_{22})\underline{Q}_{2} (J)and from Eq. (C), since F_{2} = \underline{F}_{2} = 0 in this case:
\underline{Q}_{1} = 0.1071 \underline{F}_{1} \mathrm{and} \underline{Q}_{2} = -0.2975 \underline{F}_{1} (K)From Eqs (I), (J) and (K), the complex receptance between the applied force F_{1} and the displacement z_{1} is
H_{z_{1}F_{1}}(f) = \frac{\underline{z_{1}} }{\underline{F_{1}} } = 0.1071 ^{2} (R_{11}+\mathrm{i}I_{11}) + 0.2975^{2}(R_{22}+\mathrm{i}I_{22}) (L)For random response we require only the modulus, which is
Using a standard spreadsheet program, this was plotted from f = 10 – 120 Hz, and appears as Fig. 10.15.
From Eq. (10.44), we can now find the displacement PSD, S_{z_{1}}\mathrm{, of} z_{1}, given that the force PSD of F_{1} is constant at 100 N²/Hz, from 10–120 Hz. In the present notation, Eq. (10.44) can be written as:
S_{\mathrm{y}} (f) = \left|H(f)\right|^{2} S_{x}(f) (10.44) \\ S_{z_{1}} (f)= \left|H_{z_{1}F_{1}} (f)\right|^{2} S_{F_{1}}(f) (N)where \left|H_{z_{1}F_{1}} (f)\right| is given by Eq. (M), and is plotted as Fig. 10.15, and S_{F_{1}}(f) = 100 N²/Hz from f = 10–120 Hz.
The mean square, and hence the RMS value, of the displacement z_{1}, can now be found from Eq. (10.63):
However, the system is excited only between the frequency limits 10–120 Hz, and the response will be zero outside these limits, so the mean square displacement is given by:
\sigma ^{2}_{z_{1}} = \int_{10}^{120 }{S_{z_{1}} (f)df} (O)This was integrated numerically, using a spreadsheet program, giving
\sigma_{z_{1}} = 0.191 mm RMS
