A rotating shaft carrying an unbalance disk at its is shown in Fig. Critical speed occurs when the speed of rotation of the shaft
*See, for example, R. T. Hinkle, Kinematics of Machines, 2d ed., Prentice-Hall, Inc., Englewood Cliffs, NJ,1960, p.107.
It will be shown in Case3for vibration isolation thatif an isolator is adequate for the fundamental frequency it would be adequate for the higher harmonics.
Systems One Degree of Freedom-Applications CHAP.3
Disk
( a ) Vibratory system (b) General positionof disk.
FIG. 3-18. Critical speed of rotating shaft.
is equal to the natural frequency of lateral (beam) vibration of the shaft.
Since the shaft has distributed mass and elasticity along its length, the system has more than one degree of freedom. We assume the mass of the shaft is negligible and its lateral stiffness is k.
The top view of a general position of the rotating disk of mass m is shown in Fig. Let G be the mass center of the disk, P the geometric center, and the center of rotation. Assume the damping force, such as air friction opposing the shaft whirl, is proportional to the linear speed of the geometric center P and that the flexibility of the bearings is negligible compared with that of the shaft.
Resolving the forces in the and y directions gives
+ +
kx=mew2coso f cos wt+
cy+
ky=mew2sin = sin wtApplying the impedance method illustrated in Eq. the equations above become
The phase angle in the second equation above indicates that the
3-5 Damped
3-19. Phase relation of rotating shaft for r=
in the generic and y directions are at with one another. It is evident that the amplitudesX and Yare equal.
Since the two harmonic motions and are
the same frequency, and at to each other, their sum is a circle.
Thus, the radius of the circle is equal X or Y. motion of the geometric center of disk in Fig. 3-18describesacircle of radius u about the center of rotation From Eq. we get.
Substituting and = and simplifying we
This is identical to Eq. if m. This is true because the total mass is also the eccentric mass for a rotating disk. Hence the response curves in Fig.3-16 for rotating unbalance also represents for the whirling of rotating shafts.
The phase relation for various operating frequencies shown in Fig.
3-19. It is interesting to note that, when the frequency ratio r the mass G tends to coincide with the center of rotation This can be demonstrated readily. Assume that an unbalance rotor is rotating in a balancing and that a piece of chalk is moved towards the rotor until it barely touches. When the rotational speed is below critical, the chalk mark is found on the side closer to the mass of the rotor.
When the speed is above critical, the-chalk mark is on the side away from the mass center.
Elasticity Bearings and Supports
Rigid bearings were assumed in the above discussion of critical speed.
Figure3-20 showsapulley assembly, in whichthe brackets can be more easily in the vertical direction than in the lateral
Systemswith One Degree of Freedom-Applications CHAP.3
FIG. 3-20. Pulley assembly with flexible bearing supports.
direction. The effect of the elastic bearings is to render the system more flexible and therefore lowering the critical speed. The critical speed can be lowered by 25 percent in some installations. (a)Derive the equation of motion of the system and briefly discuss the effect of the unequal elastic supports on the system performance.
Solution:
A schematic representation of the system is shown in Fig. The elasticity of the bearings and supports is represented by springs mounted in rigid frames. The equivalent spring constants and are due to the stiffness of the shaft, the bearings, and the supports in the and y directions. A general position of the disk is shown in Fig. which may be compared with Fig. Pis the geometric center and G the mass center of the disk. the center of rotation of the system correspond-ing to the static equilibrium position of shaft.
(a) Schematic (b)General position of rotating disk.
.
FIG. 3-21. System with elastic bearing supports.
3-5 Forced Vibration--Harmonic Exatation
FIG. 3-22. Rotation of disk about for various frequencies.
(a) For simplicity, the system is assumed undamped. The equations of motion are
+
= cos+
sinSince the equations indicate that the system has two natural frequencies and therefore two critical speeds. define =
= = = and =mew2. Eq.
the amplitude ratios are
e
Thetwo harmonic motions and are of the same and at to each other. Since their amplitudes are unequal, the sum of their motions is an ellipse. Thus, the geometric center moves in an ellipse about as shown in Fig.3-22. By neglecting the damping in the system, the phase angle can be either zero for below the critical speed or for above the critical speed. Since there are two natural frequencies, we may consider the operations at speeds above and below the critical.
When and both the disk and P rotate in the same direction with the same speed as shown in Fig. The heavy sideof the disk and the position of the shaft key are marked for purpose of
identification. Assume When the disk and rotate
in opposite directions with the same speed as shown in Fig. When is than both natural frequencies, again the disk and P rotate in the same direction with the same speed as shown in Fig.
It is interesting to note that when the excitation is above or below the critical speeds, thereis no reversal in stresses in the shaft; is, while the shaft is revolving, the compression side of the shaft in compression
ofFreedom-Applications 3
and the tension side remainsintension. When the excitation is between the critical speeds, the shaft undergoes two instress per revolu-tion.
of machines and field balancing' are further examples of speed calculations. Since the subject is usually covered in the dynamics of machines, it not be pursued here.
Machines often mounted on springs and dampers as in Fig.
3-23 to minimize the transmission of forces between the machine m and its foundation.
We shall first consider the system in Fig. a harmonic force is applied to and the deflection of the foundation is negligible, the equation of motion is identical to force transmitted to foundation is the sum of the spring force kx and the damping force
Force transmitted =
If the excitation is harmonic, the magnitude and the phase angle of excitation force and the other forces are as illustrated in Fig. 3-24.
The phase angle is generally is of secondary interest. Using the force transmitted is
The ratio of the amplitude of the force transmitted and the am-plitude of the driving force is called the transmissibilityTR.From the equation above, we have
Foundation Foundation (Base) FIG.3-23. Vibration isolation.
SEC. 3-5 Damped Vibration-Harmonic Excitation
FIG. 3-24. Relation of force transmitted and other vectors.
where = and = The equation is plotted in Fig. 3-25.Note that all the curves in the figure cross at r= Hence transmitted force is greater than the driving force below this frequency ratio and less than force when the machine is operated above this frequency ratio.
a speed machine, the amplitude of the exciting force is constant. Hence the force transmitted is proportional to the value of the
Frequency ratio r
FIG.3-25,. Transmissibility versus frequency ratio; in Fig.
96 Systems One of CHAP.3
Frequencyratio
FIG. 3-26. Force ratio versus frequency ratio for excitation; sys-tem shown in Fig. 3-15.
TR. It i s advantageous to operate a constant speed machine atw
For a variable speed machine, the driving force due to an unbal-ance me, is where is the operating frequency. Let us define a constant force = Substituting =mew2intoEq. dividing both sides of the equation by and simplifying, we obtain
where is as defined in Eq. (3-32).Hence the magnitude of the force transmitted can be high in spite of the low transmissibility. The equation is plotted in Fig. 3-26.
The reduction of the force transmitted in buildings is of interest. For example, the mechanical equipment of a tall office building is often located on the roof directly above the penthouse or the boardroom of the company.
The fractional reduction of the force transmitted is Force reduction
-SEC. Damped Vibration-Harmonic
Staticdeflection
FIG. 3-27. Percentage reduction in force transmitted to foundation in isolation, 0.
where and are the amplitudes of the excitation and the transmitted force, respectively. It is observed in Fig.3-25 that low natural frequency and low damping are desirable for vibration isolation. Assume and 1in Eq.(3-32). Thus, T R=l/(r2- and the force reduction becomes
Force reduction = -r2
-Since r2 and the static deflection of a spring = the equation above reduces to
Force reduction=
-The equation is in Fig. 3-27
Example 17
A n air compressor of 450 kg mass (992lb,) operates at a constant speed of 1,750 rpm. The rotating parts are well balanced. T h e reciprocating parts are of 1 0 kg (22lb,). The crank radius is (4 in.). the damper for the mounting introduces a damping factor 0.15, (a)specify the springs for the mounting such that only 2 0 percent of the unbalance force is transmit-ted to the foundation, and (b)determine the amplitude of the transmitted force.
98 Systems with Degree ofFreedom-Applications CHAP. 3
Solution:
= =183.3
SinceTR=0.20=
+
(b) Amplitudeof force transmitted 0.20
=0.20meo2= 6.72 (1,510