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VIBRATION AND CONTROL OF VIBRATION 31.27
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VIBRATION AND CONTROL OF VIBRATION
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where C1 depends on the assumed deflection shape. The maximum potential energy is of the form Umax = C2 (31.80)
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Neglecting damping, we see that the maximum kinetic energy must be equal to the maximum potential energy. Hence, the natural frequency is 2 = C2 C1 (31.81)
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This estimate will always be higher than the true natural frequency. Example 3. Determine the fundamental natural frequency of a uniform cantilever beam of length L supporting a disk of mass M and diametral mass moment of inertia Id, as shown in Fig. 31.23. The modulus of elasticity of the beam material is E, the mass moment of inertia of the cross section is I, and the mass per unit length of the beam is m.
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FIGURE 31.23 Cantilever with end mass.
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Solution. The deflection shape may be assumed to be y(x) = Cx2, which satisfies the geometric boundary conditions of zero deflection and zero slope at x = 0. The maximum kinetic energy of the structure for harmonic vibration is Tmax = 1 m 2 2
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y2(x) dx +
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1 1 M 2y2(L) + Id 2 y 2 (L) 2 2
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The strain energy is given by Umax = 1 EI 2
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y 2(x) dx
Substituting y(x) = Cx2 in the above expressions and equating Tmax = Umax, we find the natural frequency 2 = 20EI mL4 + 5ML3 + 20IdL
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VIBRATION AND CONTROL OF VIBRATION 31.28
LOAD CAPABILITY CONSIDERATIONS
In the absence of the disk, = 4.47 (EI/mL4)1/2. By comparing this to the exact result = 3.52 (EI/mL4)1/2, the error in the approximation is error = 0.95 (EI/mL4)1/2, or 26.9 percent. This error can be reduced by obtaining the strain energy by using a different method. The shear at any section is obtained by integrating the inertial loading from the free end as V( ) = 2 =
m( )y( ) d + M 2y(L)
1 2 mc(L3 3) + M 2cL2 3
and the moment at any point x is M(x) =
V( ) d + Idy (L) =
1 2 mc(3L4 4L3x + x4) + M 2cL2(L x) + 2IdcL 12
The strain energy is then Umax = When the disk is absent, Umax = 4 m2c2 312 9 L 2EI 144 135 1 2
M 2 (x) dx EI
With Tmax and Umax equated, the natural frequency = 3.53 (EI/ML4)1/2 has an error of only 0.28 percent.
31.4 VIBRATION ISOLATION
Often machines and components which exhibit vibrations have to be mounted in locations where vibrations may not be desirable.Then the machine has to be isolated properly so that it does not transmit vibrations.
31.4.1 Transmissibility Active Isolation and Transmissibility. From Eq. (31.38), the force transmissibility, which is the magnitude of the ratio of the force transmitted to the force applied, is given by T=
2 1 + (2 n) 2 2 2 (1 / n) + (2 /
(31.82)
Equation (31.82) is plotted in Fig. 31.12 for different values of . All the curves cross at / n = 2. For / n 2, transmissibility, although below unity, increases with
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VIBRATION AND CONTROL OF VIBRATION 31.29
VIBRATION AND CONTROL OF VIBRATION
an increase in damping, contrary to normal expectations. At higher frequencies, transmissibility goes to zero. Since the force amplitude me 2 in the case of an unbalanced machine is dependent upon the operating speed of the machine, transmissibility can be defined as T= FT me
2 1 + (2 n) 2 2 2 (1 / n) + (2
2 n)
(31.83)
where FT is the amplitude of the transmitted force. Equation (31.83) is plotted in Fig. 31.24. Transmissibility starts from zero at zero operating frequency, and curves for different damping ratios cross at / n = 2. For higher values of operating speed, transmissibility increases indefinitely with frequency. Passive Isolation. When a sensitive instrument is isolated from a vibrating foundation, it is called passive isolation. Consider the system shown in Fig. 31.15, where the base has a motion u = U0 sin t. The equation of motion of the system is given in Eq. (31.46). The ratio of the response and excitation amplitudes is k+i c X = Xf k m 2 + i c (31.84)
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