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VIBRATION AND CONTROL OF VIBRATION 31.11
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VIBRATION AND CONTROL OF VIBRATION
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FIGURE 31.9 Phase-angle frequency response for forced motion.
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Substituting the system response from Eq. (31.27) into Eq. (31.36) gives FT = T sin ( t ) F0 where the nondimensional magnitude of the transmitted force T is given by T= 1 + (2 / n)2 2 (1 2 / n)2 + (2 / n)2
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(31.37)
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(31.38)
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and the phase between FT and F0 is given by = tan 1 2 ( / n)3 1 2/ 2 + 4 2 2/ 2 n n (31.39)
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The transmissibility T is shown in Fig. 31.12 versus forcing frequency. At very low forcing frequencies, the transmissibility is close to unity, showing that the applied force is directly transmitted to the foundation. The transmissibility is very large in the vicinity of the system natural frequency, and for high forcing frequencies the transmitted force decreases considerably. The phase variation between the transmitted force and the applied force is shown in Fig. 31.13. Rotating Imbalance. When machines with rotating imbalances are mounted on elastic supports, they constitute a vibrating system subjected to excitation from the
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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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VIBRATION AND CONTROL OF VIBRATION 31.12
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LOAD CAPABILITY CONSIDERATIONS
FIGURE 31.10 Velocity frequency response.
rotating imbalance. If the natural frequency of the system coincides with the frequency of rotation of the machine imbalance, it will result in severe vibrations of the machine and the support structure. Consider a machine of mass M supported as shown in Fig. 31.14. Let the imbalance be a mass m with an eccentricity e and rotating with a frequency . Consider the motion x of the mass M m, with xm as the motion of the unbalanced mass m relative to the machine mass M. The equation of motion is (M m) + m( + m) + cx + kx = 0 x x x The motion of the unbalanced mass relative to the machine is xm = e sin t Substitution in Eq. (31.40) leads to M + cx + kx = me 2 sin t x (31.42) (31.41) (31.40)
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
VIBRATION AND CONTROL OF VIBRATION 31.13
VIBRATION AND CONTROL OF VIBRATION
FIGURE 31.11 Acceleration frequency response.
This equation is similar to Eq. (31.25), where the force amplitude F0 is replaced by me 2. Hence, the steady-state solution of Eq. (31.42) is similar in form to Eq. (31.27) and is given nondimensionally as ( / n)2 sin ( t ) x M = [(1 2/ 2 )2 + (2 / n)2]1/2 e m n where tan = 2 / n 1 2/ 2 n (31.44) (31.43)
Note that since the excitation is proportional to 2, the response has an 2 term in the numerator and resembles the acceleration response of a system subjected to a force of constant magnitude, given by Eq. (31.35). The complete solution consists of the complementary part of the solution and is
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VIBRATION AND CONTROL OF VIBRATION 31.14
LOAD CAPABILITY CONSIDERATIONS
FIGURE 31.12 Transmissibility plot.
x = exp nt {A exp [( 2 1)1/2 nt] + B exp [ ( 2 1)1/2 nt]} + me( / n)2 sin ( t ) M[(1 2/ 2 )2 + (2 / n)2]1/2 n (31.45)
System Excited at the Foundation. When the system is excited at the foundation, as shown in Fig. 31.15, with a certain displacement u(t) = U0 sin t, the equation of motion can be written as m + c(x u ) + k(x u) = 0 x This equation can be written in the form m + cx + kx = cu0 cos t + ku0 sin t x = F0 sin ( t + ) (31.47) (31.46)
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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