Question 10.7: A hydraulic pipe installation, exactly as described in Examp......
A hydraulic pipe installation, exactly as described in Example 8.1, is to be used in a location where the vibration input, specified at the supports, is random, and defined in terms of acceleration power spectral density. If the input level is defined as 2.0 g²/Hz from 10 to 1000 Hz, find
(a) The RMS displacement at the center of the pipe.
(b) The RMS stress at the center of the pipe.
Note: lbf in. units are used, as in the original Example 8.1.
Part (a):
It was shown in Example 8.1 that the vibration behavior of the hydraulic pipe, considered as a simply supported uniform beam excited by acceleration at the supports, as shown in Fig. 10.13 at (a), can be represented by the equivalent system shown in Fig. 10.13 at (b). Here, \ddot{x}_{B} represents the acceleration at the pipe attachments, and y_{c} represents the maximum displacement of the pipe, relative to the attachments, which occurs at the semi-span position. The equivalent mass, stiffness and damper were shown to be given by:
Equivalent mass, \underline{m} = \frac{1}{2}\mu L, referred to the center of the span;
Equivalent stiffness, \underline{k} = \frac{1}{2} EI(\pi^{4} / L^{3}), also referred to the center of the span;
Equivalent damper, \underline{c} = 2\gamma \sqrt{\underline{km}}, \mathrm{where} \gamma = 0.02 in this case.
Other data for the pipe, from Example 8.1, were
D is the outer diameter of pipe = 0.3125 in., d the inner diameter of pipe = 0.2625 in., E the Young’s modulus = 30 \times 10^{6} \mathrm{Ibf/in².}, I the second moment of area of crosssection = \frac{\pi}{64} (D^{4} – d^{4}) = 0.2350 \times 10^{-3} \mathrm{in^{4}.,} \mu the mass of pipe, plus contained fluid, per inch = 0.0214 \times 10^{-3} \mathrm{Ib in^{-1} s^{2}} per in. and L the center to center spacing of pipe supports = 17 in.
Figure 10.13 can be seen to be identical to Fig. 10.12, and therefore Eq. (10.62) can be used directly to find the RMS displacement, \sigma_{y_{c}} at the center of the pipe, i.e.,
where S_{\ddot{x}_{B}} is the base acceleration spectral density in (in./s²)²/Hz, f_{n} the natural frequency in Hz and γ the damping coefficient = 0.02.
The base PSD is 2.0 g²/Hz, so S_{\ddot{x}_{B}} = (386² × 2.0) = 298 000 (in./s²)² / Hz.
The natural frequency, f_{n}, from Example 8.1, is 98.6 Hz. Substituting these numerical values into Eq. (A) gives \sigma_{y_{c}} = 0.0885 in. RMS.
Part (b):
From Eq. (R) in Example 8.1, the relationship between the amplitude of the stress, \left|s_{c}\right|_{MAX}, at the center of the pipe, and the displacement amplitude at the center of the pipe, \left|\mathrm{y}_{c}\right|_{MAX}, is
The same factor applies to the corresponding RMS values, so
\sigma _{s{c}} = \frac{\pi^{2}DE}{2L^{2}} \sigma_{y_{c}} (C)where \sigma_{s_{c}} is the RMS stress and \sigma_{y_{c}} is the RMS displacement. Inserting the numerical values D = 0.3125 in.; E = 30 \times 10^{6} \mathrm{Ibf/in.^{2}}; L = 17 in. and \sigma_{y_{c}} =0.0885 in. gives the RMS stress at the center of the pipe:
\sigma_{s_{c}} = 14170 \mathrm{Ibf/in.²}