Question 11.1: Figure 11.3 shows a proposed vibration-isolated shelf, to be......
Figure 11.3 shows a proposed vibration-isolated shelf, to be fitted at a location in a high performance aircraft, where the environmental vibration is broad-band random.
The measured PSD levels, in the vertical direction, are all enveloped by the values shown in Table 11.1, which can therefore be regarded as an upper bound of the input PSD level.
The shelf, and the equipment mounted on it, are assumed rigid. It is supported by four vibration isolators, each of which can be considered to consist of a linear spring and viscous damper in parallel.
For vertical behavior only, if the system is assumed balanced, i.e. the center of gravity coincides with the stiffness center, it can be represented schematically by Fig. 11.1. The undamped natural frequency is 20 Hz, and the assembled system has Q = 2.5, equivalent to a viscous damping coefficient, γ = 0.2.
(a) Calculate and plot the maximum vertical acceleration PSD level to be expected on the shelf for the frequency range 10–1000 Hz.
(b) Calculate and compare
(i) the RMS vertical acceleration of the aircraft structure to which the shelf is attached;
(ii) the RMS vertical acceleration of the shelf and supported equipment.
| Table 11.1 | |
| Frequency range (Hz) | Acceleration PSD (g²/Hz) |
| 10–50 | 0.01 |
| 50–1000 | 0.10 |
Part (a):
From Eq. (10.44), the response PSD function of any linear system with a single input is given by multiplying the input PSD function by the square of the frequency response function between input and output, so in this case:
where S_{\ddot{z}} is the response or shelf PSD in g²/Hz; S_{\ddot{x}} the input or base PSD in g²/Hz, given by Table 11.1 and T the transmissibility ratio for acceleration, which from Eqs (4.35) and (11.1) is given by:
\frac{\left|z \right| }{\left|x \right| } = \sqrt{\frac{1 + (2\gamma \Omega )^{2}}{(1-\Omega ^{2})^{2} + (2\gamma \Omega )^{2}}} (4.35) \\ T = \frac{\left|z \right| }{\left|x \right| } = \frac{\left|\dot{z} \right| }{\left|\dot{x} \right| } = \frac{\left|\ddot{z} \right| }{\left|\ddot{x} \right| } (11.1) \\ T = \frac{\left|\ddot{z} \right| }{\left|\ddot{x} \right| } = \frac{\left|z \right| }{\left|x \right| } = \sqrt{\frac{1 + (2\gamma \Omega )^{2}}{(1-\Omega ^{2})^{2} + (2\gamma \Omega )^{2}} } (B)where \Omega = f/f_{n} = f/20 \mathrm{and} \gamma = 0.2.
The transmissibility, T, is plotted in Fig. 11.4(a) for f = 10–1000 Hz.
The shelf PSD, S_{\ddot{z}}, is now given by Eq. (A), as follows.
For the range 10–50 Hz:
For the range 50–1000 Hz:
S_{\ddot{z} } = T^{2} S_{\ddot{x} } = T^{2} \times 0.10The shelf response PSD is plotted, together with the input PSD, for comparison, in Fig. 11.4(b). The large reduction of the shelf response PSD compared with the input PSD at higher frequencies can be seen, as can the increase around the natural frequency of the isolating system at 20 Hz.
Part (b):
From Eq. (10.30), the mean square value (and the variance, if the mean value is zero) of any random waveform is equal to the area under the PSD function.
(i) Applying Eq. (C) to the input PSD, S_{\ddot{x}}, as defined by Table 11.1:
\sigma^{2}_{\ddot{x}} = 0.01 (50 – 10) + 0.10 (1000 – 50) = 95.4 \mathrm{g^{2}}where \sigma^{2}_{\ddot{x}} is the mean square value of the input, in g units.
The RMS value of the input waveform is thus \sigma_{\ddot{x}} = 9.77 g.
(ii) The response PSD, S_{\ddot{z}}, was integrated numerically, giving \sigma^{2}_{\ddot{z}} = 0.970 g², where \sigma^{2}_{\ddot{z}} is the mean square value of the shelf response, in g units.
The RMS value of the shelf response is thus \sigma_{\ddot{z}} = 0.985 g.
The isolators therefore reduce the overall RMS acceleration of the shelf compared with the aircraft structure by nearly a factor of ten in this case.
