1.0 FUNDAMENTALS OF VIBRATION1.1 What is Vibration?Mechanical vibration is a form of motion oscillation. it occurs in all forms of machinery and equipment. It is what you feel whenyou put your hand on the hood of a car, the engine of which is running, or on the base of an electric motor when the motor isrunning. Perhaps the simplest illustration of a mechanical vibration is a vertical spring with weight, as shown in Figure 1. In thisposition, the deflection of the spring from its free state is just sufficient to counterbalance the weight W. This deflection is calledthe static deflection of the spring. The position in which the spring is at rest is #1. The spring is then slowly extended to position#2, and released. The subsequent motion of the weight as a function of time, when there is negligible resistance to the motion, iswavy and repetitive as shown in the graph. It exhibits many of the basic characteristics of mechanical vibrations. The maximumdisplacement from the rest or mean position is called the AMPLITUDE of the vibration. The vibratory motion repeats itself atregular intervals (A1, A2, A3). The interval of time within which the motion sequence repeats itself is called a CYCLE or PERIOD.The number of cycles executed in a unit time (for example, during one second or during one minute), is known as theFREQUENCY. In a high-speed oscillation the frequency is high and conversely. When, as in Figure 1, the spring-weight system isnot driven by an outside source, the vibration is a FREE VIBRATION and the frequency is called the NATURAL FREQUENCY of thesystem. In general, vibratory motion may or may not be repetitive and its shape as a function of time may be simple or complex. Typical vibrations, which are repetitive and continuous, are those of the base or housing of an electric motor, household fans,vacuum cleaners and sewing machines, for example. Vibrations of short duration and variable intensity are frequently Initiated bya sudden impact or shock load; for example, rocket equipment upon takeoff, equipment subject to impact and drop tests, apackage falling from a height, or a lading in a freight car. In many machines, the vibration is not part of its regular or intended operation and function, but rather it cannot be avoided.The task of vibration isolation is to control this unwanted vibration so that its adverse effects are kept within acceptable limits.

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1.2 What Causes Vibration? The basic cause is already evident in the simple mass-spring system of Figure 1. it is an UNBALANCED FORCE, or system offorces (in Figure 1 it is the spring force acting on the weight) acting on or through an ELASTIC OR RESILIENT MATERIAL (inFigure 1, this is the spring). The unbalanced force may be due to mass unbalance, such as in an eccentrically mounted rotor, or itmay be due to the variable inertia forces in machinery, which does not move uniformly, e.g. crank-and-connecting-rod motion,linkages, cam-follower systems. In the latter, the speeds and directions of motion of machine parts are continuously changing,e.g. the needle motion in a household sewing machine, bucket motions in earth-moving machinery, etc. Force unbalance canarise also from electric, hydraulic and acoustic sources, e.g. transformer hum, water hammer, a loudspeaker, etc.

1.3 Adverse Effects of Uncontrolled Vibrations The objectionable results of machine vibrations, if left uncontrolled, can be several: High stresses and force levels may be set up as a result of vibrations and in extreme cases may lead to part failure. Suchfailure can be sudden or gradual, as in fatigue. More frequently, there is increased wear of parts and unsatisfactory equipmentperformance. This requires increased maintenance and may also involve downtime of equipment. For example, in a machine toolwith excessive vibrations, parts may be inaccurately machined and subsequently rejected. In other cases, an inadequatelycushioned machine may walk away on its foundation. And finally, noise may become excessive, independent of stress levels,consumer product acceptance maybe jeopardized, and working conditions may become unacceptable. Usually, the objectionableresults are a combination of these circumstances.

1.4 Principles of Vibration and Shock Isolation In discussing vibration isolation, it is useful to identify the three basic elements of all vibrating systems: the equipment(component, machine, motor, instrument or part); the vibration mount or isolator (resilient member); and the base (floor, baseplate, concrete foundation, etc.); the vibration mount is a resilient member (rubber pad, spring or the like), which is interposedbetween the equipment and the base. It is usually quite small. If the equipment is the source of the vibration, the purpose of the vibration mount is to reduce the force transmitted fromequipment to base. The direction of force transmission is from equipment to base. This is probably the most common case. If the base is the source of the vibration, the purpose of the vibration mount is to reduce the vibrating motion transmitted fromthe base to the equipment. The direction of motion transmission is from base to equipment. This case arises, for instance, inprotecting delicate measuring instruments from vibrating floors, etc. In either case, the principle of the cushioning action of the vibration isolator is the same. The isolator is a resilient member. Itacts both as a time delay and a source of temporary energy storage, which evens out the force or motion disturbance on one sideof the vibration mount and transmits or meters out a lesser, controlled disturbance, at the other end of the mount. A good vibration mount, therefore, slows equipment response to a force- or motion disturbance. In engineering terms, thecharacteristic of a good vibration mount is that the natural frequency of the equipment with the mount is substantially lower thanthe frequency of the vibration source (forcing frequency). The design of a suitable vibration mount insures that this is the case.Conversely a poorly designed mount, having an undesirable frequency characteristic, can be worse than no mount at all. In addition to its function as a time delay and source of temporary energy storage, vibration mounts can also function asenergy dissipators or absorbers. This effect is usually produced by the damping characteristics of materials, viscous fluids, slidingfriction, and dashpots,

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although in general these may or may not be part of the isolator. The damping, or energy-dissipating effect of an isolator may benegligible or substantial depending on the application. The main purpose of isolator damping is to reduce or to attenuate thevibrations as rapidly as possible. Damping is particularly important at certain speeds, which cause a condition known asRESONANCE. This occurs when the natural frequency of the equipment with isolator coincides with the frequency of the source ofthe vibration. For example, if an electric motor runs at 3600 RPM, then an isolator-equipment natural frequency of 3600 cyclesper minute corresponds to a condition of resonance. If a machine operates near resonance, or has to pass through a resonantspeed in order to attain operating speed, damping is important in preventing the buildup of vibration to an unsatisfactory level. In summary, then, a good vibration mount functions as a time delay, temporary energy absorber and possibly to some extentas an energy dissipator, or damper. The engineering design of a vibration mount consists in identifying the characteristics of thesource of the vibration, the mechanical characteristics of the equipment and the determination of the mount characteristics, inorder to achieve a specified degree of vibration reduction.

1.5 Principles of Noise Reduction A good vibration mount can be effective in reducing noise as well as In reducing the transmission of forces and motions.(a) What is Noise? Sound is a vibration of air. The air in this case is an elastic member. The vibrations of the air have a frequency and an intensity(loudness). The frequency can be expressed in cycles per second or cycles per minute. The audible frequencies range from about100 cycles/sec. to about 18,000 cycles/sec., although sensitive human ears may have a somewhat larger range. Intensity orloudness, is measured in decibels. A sound intensity of 15 decibels would usually be regarded as quiet, while a decibel level of 60and up is usually regarded as loud and objectionable. Noise may be regarded as objectionable sound. More specifically, the decibel is essentially a comparison of the pressure of the sound to that of a standard or reference sound(.0002 microbars, usually). In order to arrive at a reasonable scale of values, the logarithm of this ratio to the base 10 is usedand multipled by twenty. Typical values of levels of sound intensity and noise intensity are shown in the following Tables 1a and 1b:

TABLE 1a VALUES OF SOUNDAND NOISE INTENSITY

Various industrial operations and related noise levels recorded atdistances of from one to three feet from machine.* *

TABLE 1b VALUES OF SOUNDAND NOISE INTENSITY

**From: Acoustical Enclosures Muffle Plant Noise" byS.Wasserman and A.Oppernheim, Plant Engineering, January1965.

From Mark's Mechanical Engineers' HandBook, Sixth Edition,McGraw Hill Book Co. inc., New York, 1958, Section 12,p.153and "How to Specify Audible Noise" by E.A. Harris and W.E.Levine, Machine Design Nov. 9, 1961, p.168.

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(b) What Causes Noise? A common cause is the impact or vibration of a solid material, which sets air In motion; for example, a hammer striking a nail, ora vibrating equipment panel. In machinery, in particular, there are many commonly found sources of noise. These are usuallyassociated with the operating frequency of machine motions, e.g. the RPM of an electric motor or of gears, the rate of toothengagement in gear teeth, the frequencies associated with reciprocating machinery, etc. It is possible also that vibrations may begenerated in one part of the equipment, but may set up noise and vibration in another part of the equipment, such as doors,panels, chassis, flexible lines, printed-circuit boards etc.

(c) The Adverse Effects of Uncontrolled Noise There are several. First, noise may be an indication of faulty equipment operation, e.g. cracked parts, faulty bearings, excessiverotor unbalance, improper lubrication, loose parts, etc. It is possible also, however, for a machine to function satisfactorilymechanically, but to be rejected by the Customer, if it is too noisy, e.g. noisy household appliances, air conditioners, etc. Second,human efficiency and fatigue may be adversely affected, for example in production lines in a noisy factory, or in the office. Shortof expensive and difficult investigations, a good vibration mount can often be an effective way to reduce noise levels to withinacceptable limits.

(d) How Can Noise be Reduced? There are many ways. One of the most practical and effective may be the use of vibration mounts. As a general rule, a welldesigned vibration isolator will also help reduce noise. In the case of panel flutter, for example, a well designed vibration mountcould reduce or eliminate the noise. This can be achieved by eliminating the flutter of the panel itself, or by preventing itstransmission to ground, or by a combination of the two. The range of audible frequencies is so high that the natural frequenciesof a vibration mount can usually be designed to be well below the noise-producing frequency. In order to reduce noise, try to identify the source of the noise, e.g. transformer hum, panel flutter, gear tooth engagement,rotor unbalance, etc. Next identify the noise frequency. A vibration mount designed in accordance with the guidelines forvibration and shock control can then act as a barrier either in not conducting the sound, or in attenuating the vibration, which isthe source of the noise.

2.0 BASIC DEFINITIONS AND CONCEPTS IN VIBRATION ANALYSIS 2.1 Kinematic Characteristics COORDINATE - A quantity, such as a length or an angle, which helps define the position of a moving part. In Figure 1, x is acoordinate, which defines the position of the weight, W.DISPLACEMENT - A change in position. It is a vector measured relative to a specified position, or frame of reference. The changein x (Figure 1) measured upward, say, from the bottom position, is a displacement. A displacement can be positive or negative,translational or rotational. For example, an upward displacement may be positive; and a downward displacement negative.Similarly, a clockwise rotation may be positive and a counterclockwise rotation negative. Units: inches, feet, or, in the case ofrotations: degrees, radians, etc.VELOCITY - The rate of change of displacement. Units: in/sec, M.P.H., etc. Velocity has a direction. It is a vector. Its magnitude isthe speed. Angular velocity might be measured in radians/sec or deg/sec, clockwise or counterclockwise.ACCELERATION - The rate of change of velocity. Units: in/sec² etc. It is a vector and has magnitude and direction, e.g. 5 in/sec²North. Angular acceleration might be measured in radians/sec² or deg/sec², clockwise or counterclockwise.

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VIBRATORY MOTION - An oscillating motion; for example, that of the weight W, in Figure 1.SIMPLE HARMONIC MOTION - A form of vibratory motion. The motion as a function of the time is of the form X = a sin ωt, wherea, ω are constants. The maximum displacement, a, from the mean position (X = 0) is the amplitude; the frequency (rate at whichmotion repeats itself) is ω/2π cycles/sec, where has the dimensions of reciprocal time, e.g. reciprocal seconds. The motion is alsocalled harmonic or sinusoidal motion.AMPLITUDE - Figure 2 shows a vibrating motion, which repeats itself every T seconds. The maximum values of the displacement,x, from the reference position (x = 0) are called amplitudes. These are (a1, a2. . .). The largest of these is called the peakamplitude.

Figure 2

FREQUENCY - Rate at which motion repeats itself per unit time. If the motion repeats itself every T seconds, the frequency is 1/Tcycles per second.PERIOD, CYCLE The interval of time within which the motion repeats itself. In Figure 2, this is T seconds. The term cycle tends torefer also to the sequence of events within one period.STEADY-STATE MOTION - A periodic motion of a mechanical system, e.g. a continuously vibrating pendulum of constantamplitude.TRANSIENT MOTION - A motion which changes with time in a non-periodic manner; often the motion declines (attenuates) to anegligible value after a finite period of time (e.g. impact effects which decay with time, etc.).PERIODIC AND NON-PERIODIC MOTIONS - A motion, which repeats itself is periodic; a motion, which does not repeat itself, isnon-periodic.HARMONICS - Any motion can be considered as made up of a series of simple harmonic motions of different frequencies andamplitudes. The lowest-frequency component is usually called the fundamental frequency; higher frequency components arecalled harmonics or super-harmonics. Their frequencies are exact multiples of the fundamental frequency. Sometimes,components of the frequencies of which are a fraction of driving frequency are significant (e.g. the "half-frequency" whirl ofrotating shafts, etc.). Such components are called subharmonics.PULSE - Usually a displacement-time or force-time function describing an input into a dynamical system.PULSE SHAPE - The shape of the time-displacement or force-displacement curve of a pulse. Typically, this might be a squarewave, a rectangular pulse, or a half sine-wave pulse. In general, however, the shape can be an arbitrary function of the time.

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SHOCK MOTION - A motion in which there is a sharp, nearly sudden change in velocity, e.g. a hammer blow on a nail, a packagefalling to ground from a height, etc. its mathematical idealization is that of a motion in which the velocity changes suddenly. Themathematical idealization of a sudden velocity change often represents a close approximation to the real dynamic behavior of thesystem.

2.2 Rigid-Body Characteristics MASS - Weight in lbs. divided by the gravitational constant (g = 32.2 ft/sec², or 386 in/sec²).CENTER OF GRAVITY - Point of support at which a body would be in balance.MOMENT OF INERTIA - The moment of inertia of a rigid body about a given axis in the body is the sum of the product of the massof each volume element and the square of its distance from the axis. Units are in-lb-sec², for example. Moments of inertia of thestandard shapes are tabulated in handbooks (see Par. 4.2, Mechanical System Characteristics). If instead of the mass of theelement, the volume is used, the result is also called a moment of inertia. Depending on the application, mass-, volume-, or areamoments of inertia can be used.PRODUCT OF INERTIA - The product of inertia of a rigid body about two intersecting, perpendicular axes in the body is the sum ofthe product of the mass (volumes, areas) of a constituent element and the product of the distances of the element from the twoperpendicular axes. Units same as moment of inertia. Tabulations are available in handbooks.PRINCIPAL AXES OF INERTIA - At any point of a rigid body, mutually perpendicular (orthogonal) axes can be chosen so that theproducts of inertia about these axes vanish. The orthogonal set of axes are called principal axes of inertia. These can often beidentified by the symmetry of the body, because the principal axes coincide with the axes of symmetry. (An axis of symmetry is aline in the body, such that the body can be rotated a fraction of a turn about the line without changing its outline in space).

2.3 Spring and Compliance Characteristics TENSION - When a body is stretched from its free configuration, its particles are said to be in tension (e.g. a stretched bar). Thetensile force per unit area is called the tensile stress (Units: lbs/in²).COMPRESSION - When a body is compressed from its free configuration, its particles are said to be in compression (e.g. a columnin axial loading). The compressive force per unit area is called the compressive stress (Units: lbs/in²).SHEAR - When a body is subject to equal and opposite forces, which are not collinear, the forces tend to shear the body in two.The body is said to be in a state of shear, e.g. a rubber pad under forces parallel to its upper and lower faces. The shear force perunit area is called the shear stress (Units: lbs/in²). Most bodies in a state of stress are in tension, compression and shearsimultaneously, e.g. a beam in bending.SPRING CONSTANT - When a helical spring is stretched or compressed (amount x, say), the applied force, F, is proportional tothe displacement, x (Hooke's law). The constant of proportionality (k) is called the spring constant or gradient: F = kx. Moregenerally, k is the ratio of a force increment to the corresponding displacement increment of the spring. In the usual helicalspring this ratio is a constant (independent of displacement) within the elastic limit of the spring material. However, k may bevariable, e.g. in a rubber pad, since the rubber is nearly incompressible. Units: lbs/in. If the spring deflect in torsion, the units ofk may be in-lb/radian or in-lb/degree, etc.FORCE-DEFLECTION CHARACTERISTIC OF A SPRING - This refers to the shape of the force-deflection curve. The most familiarcase is a straight line through the origin of coordinates (constant k). It is possible, however, for the spring "constant" to vary. If itincreases

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with displacement (e.g. as in a rubber pad), the spring is called a hard spring. If it decreases with displacement (e.g. as in aBelleville spring), the spring is called a soft spring.ENERGY STORAGE - This is the area under the force-deflection curve of the spring. It represents the strain energy stored in thespring (Units: in-lbs, or ft-lbs, etc.).PRELOAD - A spring used in equipment may or may not have a rest (on the shelf) position in which it has its natural, free, orunstretched length. If its rest-position length is not its free length, the spring is in tension or compression. The amount of thistension or compression is called the preload. When measured in force units, it is a preload force; when measured in deflectionfrom free position, it is a preload deflection.ELASTIC MODULUS (E) AND SHEAR MODULUS (G) - These are material properties, which characterize material compliance intension or in compression (E) and in shear (G). They are defined as the ratio of stress to strain, where strain refers to the changein length (or deformation) per unit length. E involves tensile or compressive stress and G involves shear stress. Units: lbs/in². Inmany practical applications, E and G can be regarded as constants, within a limit of material stress known as the proportionallimit. Metals loaded below the proportional limit are examples. Rubber and plastics, however, usually have no well definedproportional limit.

2.4 Damping, Friction and Energy-Dissipation CharacteristicsSTATIC FRICTION, SLIDING FRICTION, COULOMB DAMPING - These are all terms used for the frictional resistance encounteredwhen one body slides relative to another, e.g. a weight dragged on the ground. The frictional force is approximately proportionalto the contact force between the two bodies and opposed to the direction of relative motion. The constant of proportionality, m, isknown as the coefficient of friction. It a 10 lb weight is dragged along a horizontal floor with a coefficient of friction, = 0.2, thefrictional resistance is 0.2 x 10 = 2 lbs. Sometimes a distinction is made between the value of the coefficient of friction whenmotion is just impending (starting friction) and the value during motion (kinetic friction). The coefficient of friction in the lattercase is generally somewhat lower. Table 2 shows typical values of the coefficient of friction for various materials and operatingconditions.VISCOUS DAMPING - If a body moves relative to a second body, viscous damping refers to a resisting force, which is proportionalto the relative velocity between the two bodies and opposes the direction of relative velocity between them. The constant ofproportionality is known as the coefficient of viscous damping, c. Units: lbs per unit velocity, i.e. lbs/(in/sec). Such damping isencountered, for example, in hydraulic dashpots and devices, which meter a liquid through an orifice. The more viscous the fluid,the greater the damping. If c = 0.5 lbs/(in/sec) and the body moves through a viscous fluid at 10 in/sec, the viscous dampingforce is 0.5 x 10 = 5 lbs. Typical example: hydraulic door closers.CRITICAL DAMPING - Value of damping constant just sufficiently high in a mass-spring-damping system so as to preventvibration.DAMPING RATIO - The ratio of the damping constant to the critical damping constant for that system.

2.5 Vibration Characteristics of Mechanical Systems MATHEMATICAL MODEL - An idealized representation of the real mechanical system, simplified so that it can be analyzed. Therepresentation often consists of rigid masses and dashpots. Hopefully, the representation is sufficiently realistic so that theresults of the analysis correspond reasonably closely to the behavior of the physical system from which it was derived.

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LUMPED-AND DISTRIBUTED-PARAMETER SYSTEMS - In a lumped-parameter system, the mass-, elastic- and damping propertiesare separated or lumped into distinct components each of which has only mass or only elasticity or only damping, but not morethan one of these per component. In a distributed-parameter system, a component may have combined mass and elasticity anddamping, distributed continuously through the component. The latter models tend to be more realistic, but more difficult toanalyze.DEGREE OF FREEDOM - This is the number of independent quantities (dimensions), which must be known in order to be able todraw the mechanical system in any one position, the fixed dimensions of the system being known. The simple mass-springsystem of Figure 1 has one degree of freedom; a mechanical differential, for example, has two degrees of freedom; a rigid bodymoving freely in space has 6 degrees of freedom.FORCE AND MOTION EXCITATION - If a force is applied to a dynamical system, it usually is a source of vibration (e.g. centrifugalforce due to an unbalanced rotor). The vibrations are then said to be due to force excitation. If, on the other hand, the foundation(or other part) of a machine is subject to a forced motion (vibration or shock), the resulting machine vibration is said to be due tomotion excitation, e.g. an earthquake actuating a seismograph.FREE VIBRATION - If the weight in Figure 1 is moved out of its equilibrium position, and released, the system will vibrate withoutthe action of any external forces. Such an oscillation is called a free vibration.FORCED VIBRATION - if an external force is applied to the weight in Figure 1, which causes it to vibrate (e.g. a force varyingharmonically with time, say), the resulting motion of the spring-mass system is called a forced vibration. If the base, whichsupports the spring, undergoes a forced motion, which in turn causes the weight to vibrate, the vibration is also forced.RANDOM VIBRATION - Equipment may be caused to vibrate by applied forces or motions, the frequency (or frequencycomponents) of which vary in a random manner with time (e.g. wind gusts on a missile). The resulting vibration is called random.NATURAL FREQUENCY - When mechanical equipment vibrates freely, the resulting number of oscillations per unit time is calledthe frequency (cycles/sec). According to whether the system is free without damping, or free with damping, the frequency iscalled the free-undamped natural frequency or the free-damped natural frequency. The natural frequency is a function of themass distribution and compliance of the system. For a simple mass-spring system (Figure 1), which represents a reasonable

approximation to many real mechanical systems, the natural frequency is equal to radians per second,

where k is the spring constant, lbs/in; W is the weight, lbs; g is the gravitational constant, 386 in/sec²; and xst is the staticdeflection of the spring, in. Thus, flexible systems tend to have low natural frequencies and rigid systems tend to have highnatural frequencies. At the same time, the natural frequency can be changed by altering the compliance and mass distribution ofthe system. The simple expressions for natural frequency just given, yield the natural-frequency curve of the basic vibration chartgiven in Par. 4, Case A. In the chart they are plotted on a logarithmic scale and the frequency is given in cycles per minute,rather than in radians per second. A system may have more than one natural frequency, in which case the lowest of these isoften the most significant. In general, the number of natural frequencies is equal to the degree of freedom of the system.FORCING FREQUENCY - The number of oscillations per unit time of an external force or displacement, applied to a vibratingsystem.

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RESONANCE - Displacement and stress levels tend to build up greatly when the forcing frequency coincides with a naturalfrequency. This condition is known as resonance.

NORMAL MODES - A system having more than one degree of freedom has several distinct or independent types of displacementswith the following characteristics: If set vibrating in one such displacement, the motion will remain of this type. Such types ofdisplacements are called normal modes.

FREQUENCY RATIO - The ratio of two frequencies, usually forcing frequency to natural frequency.

TRANSMISSION RATIO, TRANSMISSIBILITY - There are several transmission ratios. Usually these refer to the ratios of themaximum values of an applied force or a forced motion to the maximum values of the transmitted force or displacement; theratio is expressed as transmitted/applied force or displacement. The direction of transmission can be from equipment to base orvice versa. The transmission ratio is used as a factor of merit in the design of vibration mounts. For well insulated systems, itsvalue is substantially less than unity. [Exception: The value of TDl, eq. (B-1), Par.4, however, is slightly more than unity.]

TRANSIENTS AND STEADY-STATE COMPONENTS - Equipment vibration frequently consists of two parts: a temporary vibration,often set up by startup or impact conditions, called a transient; and a permanent, periodic vibration, called a steady-statecomponent.

LINEAR SYSTEMS - Those in which the response under two different external force systems is the algebraic sum of the responseto each force system separately. Mathematically, the equations of motion are called linear and the resultant motion is obtainableby superposition.

NOISE - When the frequency of vibration is in the range of about 100-18000 cycles/sec, the motion is generally accompanied byaudible sound (assuming equipment is not in a vacuum). Objectionable sound (in contrast to music, for example) is called noise.Frequency in the audible range is sensed in human hearing as pitch, i.e. low-pitch sounds corresponding to low frequencies andconversely. Table 3 illustrates frequencies of typical sound sources.

WAVE PHENOMENA - Elastic and acoustic waves travel through materials at characteristic speeds. Mechanical phenomena(stresses, vibrations, sound, etc.) associated with these waves are called wave phenomena.

TABLE 3 SOUND SOURCES

From: "How to Specify Audible Noise" by E.A. Harris and N.E. Levine, Machine Design, Nov 9, 1961. p. 174.

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3.0 VIBRATION MOUNTS AND THEIR CHARACTERISTICS, INCLUDING RUBBER3.1 Types of Vibration MountsVibration mounts and shock mounts are units or components consisting of a resilient material, usually rubber, mounted in asupporting frame, usually metal, which can be bolted or screwed into the equipment, which is to be protected. Typical mounts and their basic characteristics are given in Table 4. For all of these mounts the product pages of the SDP catalog include: 1. The load recommended as a maximum for each mounting. 2. The minimum forced frequency of vibration of the mechanism that will be absorbed effectively by the mounting whensupporting each of the various minimum loads indicated. 3. (For bumpers and wear pads) maximum constant force that can safely be applied. 4. (For bumpers and wear pads) maximum total force that can safely be withstood occasionally. Also included are curves for each type of mounting, plotted to show the amount of static deflection resulting from various loads. A typical mount is cylindrical mount #10Z2-330-B (Catalog p.706). The load-deflection curves in compression and shear arenearly linear. The maximum recommended loads are 260 lbs in compression and 140 lbs in shear. Curves showing naturalfrequency vs. supported weight are given for load in shear and load in compression. For example, when the mount supportsbetween 90 lbs and 260 lbs in compression, the minimum forced frequency, which will be absorbed effectively by the mount is2500 cycles per minute.

3.2 General Design Considerations In the Selection of Vibration Mounts(a) Direction of LoadIn standard rubber mountings, rubber in shear is more widely used than rubber in compression. The main reason for this is thatrubber in shear will undergo a greater deflection for a given thickness of rubber than will rubber in compression. This means thatlarger deflections result in lower natural frequencies for a given load. However, rubber is used in compression in mountings where a greater load-carrying capacity per unit volume of rubber isrequired than is advisable with rubber in shear; rubber in compression is used also where large deflections are not required.

Although rubber is practically incompressible, it is said to be stressed in compression when it supports a load as shown inFigure 3. Note that the steel plates are being forced together to squeeze, or stress the rubber in compression. Rubber, bonded to metal plates and stressed either in shear or in compression, can be used to eliminate sliding friction,abrasion and wear between parts moving in relation to one another.

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(b) Loads in Two Planes Some mountings are designed to be used with the rubber stressed only in compression or only in shear. Many mountings,however, are designed to absorb vibrations through both compression and shear. In such cases, vibrations arrive at the mountingin two planes. Whether a mounting is designed to be used with the rubber in compression or in shear can be detrmined from load-deflectioncurves and data supplied in this catalog. By bonding or adhering metal plates to both sides of a slab of rubber, it is possible to make what is known as a sandwich. Ifone of these plates is fixed in a vertical position and a load supported on the other plate, the rubber will allow the loaded plate tomove downward, or deflect, a certain distance. Since the relative movement is that of the blades of a pair of shears, it is said thatthe rubber is deflected or stressed in shear and such mountings are called shear sandwiches. These are shown in Figure 4. Thesecond of these illustrates what is known as a club sandwich, incorporating two parallel walls of rubber stressed in shear.Ordinarily the two outer plates are subjected to lateral pressure of limited amount; however, this does not appreciably alter theaction in shear.

(C) Direction of Vibration When mountings are desired for a specific installation, it is important to determine the direction of the vibration created by theoperation of the mechanism as well as whether the vibration is to be absorbed by rubber used in compression or in shear. Thetype of mounting selected should be of such a design that the vibration will cause the rubber to move either in shear or incompression, depending on which of these two the deflection calculation was based. With rubber in compression, natural frequencies will apply to the mounting provided the vibration absorption is accomplishedby movement of the rubber in compression. With rubber in shear, natural frequencies obtained from deflection calculations based on rubber in shear will apply to themounting whether the stress in the rubber is in a horizontal or vertical direction, provided the movement of the rubber is inshear. With rubber in compression and shear, It should be remembered that a mounting can be used to support a mechanism bymeans of rubber in compression, and that at the same time the mounting can absorb vibration by means of rubber in shear. Insuch cases the load carrying capacity of the mountings selected should be based on rubber in compression, whereas the naturalfrequency should be calculated from the deflection based on rubber in shear under the same load - even though the load is notsupported in shear. Vibration insulating the unbalanced rotor of an electric motor is a typical case in which rubber might have tomove both in compression and in shear. The problem is to effectively absorb vibration occurring in all directions in a plane atright angles to the longitudinal axis of the rotor.

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Figure 5 Electric Motor andV-belt Drive. Motor mountstake up load due to rotorunbalance (F1.directionrotates in vertical plane),motor weight (F2), outputtorque (1) and lateral force(F3).

Ordinarily, club-sandwich type mountings might be installed with their longest axes at right angles to the rotor. In this position,the rubber would move in shear for every direction of the vibration. Bolt tension, however, might make it necessary to install the mountings with their longest axes parallel to the rotor. In thiscase, the movement of the rubber in shear would absorb only the vertical component of the vibration. The horizontal componentof the vibration would cause the rubber to move in compression. See Figure 5.

(d) Arrangement of MountingsIn Parallel (Figure 6a)Where a given load is divided among two or more mountings, they are used in parallel. When two identical mountings, or stacks (see below), using rubber in shear, are installed in parallel to support a given load, thedeflection of each mounting or stack is approximately one half of what it would be if the same load were supported on only one.Increasing the number of mountings of the same type and size, so as to equally share the same load, decreases the deflectionequally of each; and for all practical purposes the decrease can be considered as being proportionate (See also Par. 4.2e).In Series (Figure 6b)Two or more mountings may be bolted together. Where the same load passes from one such mounting to the others, themountings are used in series and they are said to comprise a stack. Stacks in turn may be used in series or in parallel. When mounts are used in series, the total deflection under load increases directly in proportion to the number of mountings ofthe same type and size. This principle applies whether the mountings are in shear or in compression. (See also Par. 4. 2f).In Combination (Figure 6c)In some installations mountings are used partly in series and partly in parallel. And in addition, bumpers may be added so as tolimit the excessive movement of the mechanism, which might be caused by some abnormal horizontal force (i.e. this avoids"bottoming"). See also Par. 4.2g).Flexibility of ConnectionsThe effectiveness of properly related mountings may be seriously impaired, or even nullified, by the stiffness of connections(piping, conduit, control rods, etc.) between a resiliently supported mechanism and its foundation. Maximum flexibility in suchconnecting members is therefore essential to best performance of the mountings.

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3.3 Properties and Characteristics of Rubber and Rubber Mountings(a) Rubber-to-Metal Bond Rubber when stressed in shear, is usually bonded to steel. The bond has certain limits beyond which it should not be stressed.For instance, in commercial practice the bond between rubber and metal of a shear mounting is stressed from 25 to 70 lbs persquare inch, depending on the service. These values for bond stress are obtained by dividing the load in pounds by the area, insquare inches, of metal bonded to the rubber, Using the bond area either of the supporting metal or of the floating metal,whichever is the smaller. The minimum bond strength of the catalog mountings is guaranteed as follows: 200 lbs/in² for rubber having hardness up to 50 durometer; 250 lbs/in² for rubber of 50 durometer and higher. Higher minimum bond strengths can be furnished by request where needed.

The bond strength in a well-designed mount is usually so high that in the event of failure the rubber breaks in a region, otherthan at the bond interface.

(b) Hardness of Rubber (Durometer Reading) The Shore Durometer (Type A) is an instrument for measuring the hardness of a rubber surface in terms of the indentationproduced in the surface by a small, flat-tipped, tapering needle subjected to a practically constant indenting force. The arbitraryscale of this durometer runs

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with increasing hardness from 0 to 100. In this catalog, a "30 durometer" rubber refers to a rubber compound which gives areading of 30 on the durometer (Type A). A relationship connecting durometer hardness with the elastic modulus can at best be only approximate. The modulus is anintegrated property of the whole volume of the rubber body, whereas the durometer reading is a property of a small region inand near the rubber surface, namely the local surface hardness. The durometer readings of rubber samples with the same elasticmoodulus may vary by 5 to 6 points at room temperature. The durometer is a useful means of roughly determining the modulus.In the production of rubber mountings, where. identical rubber pieces are being cured, the durometer may serve as a fair controlon the modulus. However, it is to be remembered that the durometer reading is often a function of the size of the rubber pieceand of its surface irregularities. Durometer hardness does not indicate the utility or stamina of a rubber compound.

Durometer Hardness of Some Rubber Compounds

Hardness ASTM Designation3040506070

R-325-BFKR-430-BFKR-530-BFKR-830-BFKR-725-BFK

(c) Presence of Oil, Grease, OzoneOil and grease are natural enemies of rubber and cause deterioration if in constant contact with it. However, the small amountsof oil and grease usually encountered do not cause trouble. It is advisable of course, to install mountings so that oil and grease contact will be avoided, if possible. In cases where oil mightdrip onto a mounting, a thin sheet-metal shield, held in place by the mounting bolt, may be placed over the mounting. In someinstallations, it may be advisable to raise the mounting above a floor to avoid excessive oil contact. An oil-resistant coating that slows up the penetration of oil in contact with rubber to about one fifth its original rate is availableon special order for specific installations. If constant oil contamination is unavoidable, however, special oil- and ozone-resistantmaterials, such as neoprene, may be considered. Paint applied to mountings should be free from manganese, copper salts and softening oils. Lacquers containing light solventsinstead of oxidizing oils are preferred. Precautions should be taken to protect the steel parts of mountings from corrosives. Ozones produces cracking in rubber and destroys it, especially if the rubber is under slight tension. In order to obtain themaximum usefulness from rubber, the presence of ozone should be considered when selecting mountings.

(d) TemperatureThe temperature at which rubber mountings are used is important as extremes affect their service life. Generally, operatingtempertures of rubber mounting do not exceed 140ºF. However, the catalog mountings can be used at temperatures up to 170ºFwithout seriously affecting their service performance. If used at higher temperatures, the load-carrying capacity and the servicelife will decrease.

(e) StabilityThe stability of a resiliently supported mechanism is greatest when the mountings are in a horizontal plane passing through thecenter of gravity of the mechanism and when the mountings are as far from the center of gravity as is practicable. However, theoperation of

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most machines does not require that the mountings be placed at the level of the center of gravity. Machinery to be insulated is often driven by belts. To maintain stability and relative position between the drive and drivenunits, it is advisable to place both on a common base and resiliently support this base. Where this is not practical, the mountingsmust be so arranged so that they will not affect the belt tension.

(f) Physical Properties of Rubber The following tables show some of the physical properties of rubber compounds and related materials.

4.0 DESIGN OF VIBRATION MOUNTS

4.1 Vibration Identification and Specification Typical causes of vibrations are unbalanced rotors, eccentrically running rotors and bearings, reciprocating machinery andmachine parts with non-uniform motions, such as cams and linkages. The total spectrum of possible causes is wide. A typicalsample list is given in Table 7(From W. Tustin: Analysis of Complex Vibrations, Machine Design, June 12, 1969, p.197).

TABLE 5 PHYSICAL PROPERTIES OF FIVE STANDARDSTRUCTURAL RUBBER COMPOUNDS

*The logarithmic decrement given here represents the negative of the power to which 10 must be raised in order to obtain theratio of any two consecutive amplitudes (on the same side of zero deflection) as unexcited vibration dies out. For instance, if the

logarithmic decrement is 0.2. the ratio of one amplitude to the preceding one is =

Successive amplitude ratio.(Ordinarily logarithmic decrement is referred to Naperian log base e, and if such values are required, they would be 2.30 timesthe values given here.)** Table from U.S. Rubber Engineering Guide #850 p. 25

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TABLE 7 HOW TO IDENTIFY CAUSES OF VIBRATION

Vibration Cause Comparative Amplitude* Frequency(Cycles/min) Remarks

Unbalance

Misalignment ofcouplings orbearings, and bentshaft

Defective bearingsof antifriction type

Eccentric journals

Defective ordamaged gear

Mechanicallooseness

Drive belts

Electrical

Aerodynamic orhydraulic forces

Reciprocating forces

Proportional to unbalance;largest in radial direction

Large in axial direction;50% or more of radialvibration

Unsteady

Usually small

Very small

Depends on looseness

Erratic or pulsing

Not characteristic

Not characteristic

Not characteristic

1 x rpm

1 x rpm usual, 2 or 3 xrpm sometimes

Very high-several timesrpm

1 x rpm

Very high-number of gearteeth times rpm

2 x rpm

1, 2, 3, or 4 x rpm of belts

1 x rpm or 1 or 2 xsynchronoas frequency

1 x rpm or number ofblades on fan or impeller xrpm

1, 2, or higher orders xrpm

Most common cause of vibration.

Usually observed as severe axial vibration. Use dialindicators or other positive method for diagnosis. Ifit is a sleeve-bearing machine and there Is nocoupling misalignment, the rotor may beunbalanced.

Bearing responsible is most likely the one nearestthe point of largest high-frequency vibration.

If on gears, largest vibration is in line with gearcenters; It on motor or generator, vibrationdisappears when power is turned off.

The familiar gear "whine," If constant, can often besafely ignored. When pitch begins to change, gearsmay require replacement.

Usually accompanied by unbalance or misalignment.

Strobe light is the best tool to identify a faulty belt.

Cause is electrical if vibration amplitude crops offinstantly when power is turned off.

Rare as a cause of trouble except in cases ofresonance.

Inherent in reciprocating machines; can only bereduced by isolation.

*This is a subjective relationship that depends upon other factors such as machine type and operating frequency.

In machine elements such as gears, the frequencies of vibrations could be the gear speeds, or the frequency of toothengagement, or the frequency with which two given teeth come into contact, or the frequency of a machining error introducedthrough a gear-cutting process, such as hobbing, for example. In adition there are the natural frequencies of the gear-drivesystem. Thus any given situation may require careful study (see, e.g., N.F. Riegler, Mach. Des, July 10, 1969, pp.115-119.)

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In cam-driven systems, the natural frequencies may correspond to the cam speed, or to higher frequencies associated with theharmonics of the acceleration-time function of the follower displacement. A vibration source may be external or internal. The external sources represent applied external forces or motions, such as floormotions, motor drives, impacts, wind gusts and the like. The internal sources include those mentioned above. It is desirable toidentify the frequency, and direction of motion and point of application of all external or internal vibration-generating forces. ifthe amplitudes are available, so much the better. These quantities can be measured, or computed. Vibration-measuring equipment may be portable or permanently installed formonitoring purposes. Portable vibration measuring instruments can be battery-powered and can be used for measuringdisplacement, velocity and acceleration at frequencies ranging from very low up to (at least) the limit of the audible range(3,100,000 cycles/min). Typically such vibration meters involves an accelerometer, the output of which is integratedelectronically to indicate velocity and displacement in addition to acceleration. See, for example, "Vibration testing instrumentselection" by W. Tustin, Machine Design, May 29, 1969, pp.116-124. Vibration and Sound Analyzers are also available. Having identified the sources of vibration, it is then necessary to specify, the degree to which this must be isolated orInsulated. In the following guideline this is determined by the relationship between the natural frequency and the forcedfrequency to be insulated. Thus let the ratio of forced frequency to natural frequency be called the insulation ratio (this ratio is also called the frequencyratio).

Then the following represents typical practice:-TABLE 8

Insulation Ratio Vibration Absorption,Percent Results Attained

10.04.03.02.52.01.51.41.0

98.993.387.581.166.720.0

0(resonance)

excellentexcellentvery good

goodfairpoornone

worse than with nomountings

Where vibration must be almost entirely eliminated, the insulation ratio should be 4 or more, thus requiring relatively large staticdeflections, which usually means more costly mountings. Satisfactory results are usually obtained where the ratio is 2.5 orslightly greater. It is important to note that there are cases where, due to extraordinary resilience of the supporting floor orfoundation, an insulation ratio considerably higher than 2.5 will be required. For instance, a light wooden floor in a frame buildingis considerably more resilient than a concrete floor in a building of steel structure and would require a higher insulation ratio tocompensate for the lack of resilience in the supporting structure. The mathematical determination of the degree of vibrationabsorption is given later in Par. 4. Having defined the vibration environment, it is necessary to specify the mechanicalcharacteristics of the vibrating system.

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4.2 Mechanical System CharacteristicsThe following concerns the characteristics usually needed for vibration analysis:MASS - this is weight divided by the gravitational constant (g = 386 in/sec²). Determine by finding weight of part. Either bydirect weighing, or by computation of product of volume and density.CENTER OF GRAVITY - Can be obtained by computation or experiment. The experiment is suggested by the definition of the C.G.as the point of support at which the body will be in equilibrium. For example, a plane body, or one of constant thickness, can besupported on a peg; when in equilibrium, a vertical line drawn through the peg will pass through the center of gravity. If thisexperiment is repeated with a different peg location relative to the body, the center of gravity will be the point of intersection oftwo lines. Similar experiments, though somewhat more difficult to devise, can be conducted for three-dimensional bodies.MOMENTS OF INERTIA - For standard shapes these are tabulated in the handbooks. A few of the more commonly used shapesare tabulated below. To determine the mass moment of inertia of a body, the cross-sectional area of which is constant, multiplythe area moment of inertia by the product of the length of the body and its density [mass density = (lbs/in3)/g]. Consider, forexample, a rectangular, steel bar, 10" long, 2" wide and 2" high, as shown in Figure 8.

Cross-section Area moment of inertia about axis indicated; if the linear dimensions are in inches, the units of area moment of inertia are (inches)4.

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The area moment of inertia about axis AA is bh3 or (2)(2)3 = 1.333in.4 12 12 The mass moment of inertia of the bar about the midplane, M1M2M3M4 (containing axis AA) = Area Moment x Length of Bar x Mass Density Gravitational ConstantAssuming a value of 0.281 lbs/in3 for the density of steel and a value of 386 in/sec2 for the gravitational constant, the massmoment of inertia, in units of in-lb-sec2, is given by 1.333 x 10 x 0.281 386 or 0.00971 in-lb-sec² The moment of inertia of complicated machine parts can be calculated or determined experimentally. Experimental setupsusually involve a compound-pendulum experiment. The part (rotor, etc.) may be suspended by a knife edge or wire, etc. andpermitted to swing about an axis, which is parallel to the axis about which the mass moment of inertia is desired. See Figure 9.

Let d = distance from center of gravity to point of support (knife edge, or end of wire, etc.) T = period of pendulum vibration in seconds; (measure several and divide by their number). W = weight of part, lbs g = 386 in/sec²

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Then the moment of inertia, l, about an axis through the center of gravity parallel to the swing axis is given by

Similar measurements can be made with the part mounted so as to vibrate as a torsional pendulum (see W.l. Senger, MachineDesign, Nov., Dec., 1944, Jan.-Feb., 1945). Products of inertia are required more seldom than moments of inertia. Their experimental determination is more difficult. Oneway, usable in parts which function as rotors, is to mount the rotor on bearings and permit the rotor to rotate at speed. If therotor unbalance is known, the bearing reactions are directly relatable to the products of inertia (see, for example, Housner andHudson: "Applied Mechanics/Dynamics, Van Nostrand, 1959, p.224, Ex. 7.68). If a moment of inertia about an axis through the center of gravity is known, the moment of inertia about a parallel axis, adistance D from the center of gravity is computed by the Parallel-Axis Theorem: l(displaced axis) = l(about parallel axis through C.G.) + WD²/gwhere W is the weight of the body and g is the gravitational constant. For geared systems, see below under elastic compliance.

ELASTIC COMPLIANCE (Spring Constants)For mechanical springs, data in handbooks, etc. cover this subject fully. See, for example,"Mechanical Springs" by A.M. Wahl, Penton Publishing Co., 1944.

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We give here only the main equations, which occur most frequently:(a) Circular-Wire Helical Spring in Tension or Compression k = d4G 8D3Nwhere k = spring constant lbs/in. d = wire diameter of spring material, in. D = mean coil diameter of spring (O.D.-d), in. N = number of active turns of wire (usually total number less one, or one and one-half turns to allow for end effects). G = shear modulus of spring material, lbs/in².(b) Circular Wire Helical Spring in Torsion k = Ed4 in-lbs/radian 64DN where the symbols are defined as in (a), above, and E = elastic modulus (lbs/in²)Springs also have lateral compliance, which is different from their axial compliance, see AM. Wahl above under "ElasticCompliance."(c) Various Beam Configurations For the most common forms of beams, the deflection formulae are as follows:Spring constant, k, lbs/in (this is the weight, W, divided by the beam deflection at the weight). E = elastic modulus of beam material, lbs/in2 l = area moment of inertia of beam cross-sectional area about neutral axis (axis through center of mass of cross-sectionparallel to the bending moment vector exerted by W) = length of beam

Simply-supported beam;concentrated weight atmiddle.

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(e) Springs in Parallel These combine like electrical resistances in series. This is the case when several springs support a single load, as shown. Thesprings are equivalent to a single spring, the spring constant of which is equal to the sum of the spring constants of theconstituent springs. In the above sketch, the spring constant, k, of the single equivalent spring is given by:

(f) Springs in SeriesThese combine like electrical resistances in parallel. The equivalent single spring is weaker than any of the component springs.The spring constant, k, of the equivalent single spring is given by:

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(g) Spring Combinations, Which are Partly in Parallel and Partly in Series Obtain equivalent spring constants for each set of parallel or series springs separately and then combine. For example, in thesketch shown on the left, the springs k1 and k2 are equivalent to a single spring, the spring constant of which, ke1, is given by:

The three springs, k3, k4, k5 in parallel are equivalent to a single spring, the spring constant, ke2, of which, is given by ke2 = k3 + k4 + k5Now equivalent springs ke1 and ke2 are in series. Hence, the spring constant, k, of the equivalent spring for the entire system, isgiven by

(h) Geared Systems In such systems, the system compliance and inertia is often referred to one shaft, usually the motor shaft, or drive shaft. Thefigure below shows a motor of inertia lM driving a load of inertia lL, through two gear-pinion reductions (Inertias lp1, lg2, lp2,lg3). The torsional compliance of the three shafts 1,2,3 of negligible inertia are k1, k2, k3 respectively. Referred to the motorshaft, the overall compliance, k, of the system and the equivalent moment of inertia, l, of the system (also called the reflectedmoment of inertia) are determined as follows: The equivalent spring constant, k, is given by:

where n21, n31 are the gear ratios of shafts 2, 3 to shaft 1 respectively. The moment of inertia, l, of the system referred to themotor shaft (shaft #1) is given by: l = (lM + lp1) + n21² (lg2 + lp2) + n31² (lg3 + lL)

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The springs are in parallel and combine accordingly, but because of the gear reductions all spring constants are multiplied by thesquare of the gear ratio between the shaft involved and the reference shaft. In computing the equivalent moment of inertia, l, the moment of inertia at an intermediate shaft is multiplied by the square ofthe gear ratio between that shaft and the reference shaft.

(i) Experimental Determination of System Compliance It is often easier to measure the system compliance, instead of computing it. The compliance between any two members of amechanical system can be measured by holding one of them fixed and measuring the force (torque) required to deflect the othermember by a unit distance (angle). When the system is complex, this may be the only way. For example, in calculating damagein freight cars due to shock loading, it is necessary to know the compliance of the lading. For a typical lading, such as canned dogfood packed in fibre cartons and loaded in block pattern, the compliance is more readily determined by experiment than bycomputation.

4.3 Damping In the case of viscous damping, it is usually required to make an estimate of the damping ratio (the ratio of damping to criticaldamping). In most mechanical systems, this ratio ranges from near zero to about 5%. A convenient way to measure the ratio isby noting the decay in amplitude of a damped free vibration of the system. The calculation is given in Case A, Par. 4. In the case of sliding friction (Coulomb damping or static friction), the coefficient of friction is a measure of the resistance tothe sliding motion. Sometimes, it is convenient to work with equivalent viscous damping, rather than with static friction, becauseviscous damping gives rise to mathematically linear equations, while static friction does not. In such cases, it is sometimespossible to compute an equivalent coefficient of viscous damping. In the case of a single-degree-of-freedom system of Case A,Par.4, subject to a sinusoidally varying force acting vertically on the weight W, and assuming for the time being that the dashpotis replaced by sliding friction, It is possible to find the equivalent viscous damping coefficient by equating the work done during aquarter cycle of vibration between the actual system and the equivalent viscously damped system. However, care must be takenthat such equivalence is valid for the purposes of the vibration-mount design. This is not always the case and requires separateinvestigation.

4.4 Vibration Analysis of Basic Vibration-Mount and Shock-Mount Configurations The design of a vibration mount, or shock mount, involves a calculation of the force- or motion transmission from equipment tofoundation or vice versa. In order to facilitate this

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computation, it has been recognized that many mechanical systems can be represented by reasonably simple mathematicalmodels. Once these models have been analyzed, the results can be tabulated and used with minimum effort. Below, we haveselected 6 mathematical models of vibrating machinery and systems, which we believe will be applicable to a large number ofvibration-isolation problems encountered by the engineer. The basic results are summarized for each case. Solved problems aregiven in the subsequent section and illustrate the application of these cases to vibration-mount design.

CASE A: UNBALANCED SINUSOIDAL FORCE ACTING ON EQUIPMENT

p = frequency of applied force or forcing frequency, rad/sec ξ = damping ratio = c/cc, where cc = critical damping coefficient ω = (undamped natural frequency of system, rad/sec)

g = gravitational constant, 386 in/sec² p/ω = frequency ratio (ratio of forced to natural) x = equipment displacement, measured from equilibrium, in. The performance of the vibration moment can be measured by factors of transmissibility. The force transmissibility, TF, is definedas follows:

TF = Max. force transmitted to base Max. unbalanced force acting on equipment

The quantity (1 - TF) is sometimes called the isolation or insulation and is expressed as a percentage. The motion amplification of the equipment can be measured by comparing the max. equipment displacement to the staticdisplacement under the unbalanced force Fmax. This leads to the displacement transmissibility of the equipment, TDE:

TDE = Max. displacement of equipment from static Static displacement of equipment under force Fmax

(A-1)

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The condition p/ω = 1 is known as resonance when ξ = 0. When the damping ratio is less than 0.1 (10%) the max value of TF still occurs very nearly when p/ω =1 and the correspondingvalue of TF is 1/2ξ very nearly. TF is less than 1 only when p/ω, is greater than 1.41.

The following basic vibration chart, Table 9, gives static deflection vs. frequency and % vibration isolation (1 - TF). It is basic toall vibration-mount circulations (see problem section).

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TABLE 9

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The following equations are useful in calculating the natural frequency (ω) and the damping ratio (ξ) of the system: Let xst = static deflection of spring, in.

Ω = natural frequency of system with damping, rad/sec. ΩN = same in cycles/min. ωN = natural frequency of system without damping, cycles/min. ξ = damping ratio of system = c/cc.

an = nth max. amplitude of displacement. x, of equipment (on same side of mean position).Then: xst = W (in.) (A-4) K

The sketch below shows a displacement-time curve of a free, damped vibration with successive amplitudes (a1, a2.. .). Equation(A-6) states that the natural undamped frequency depends only on the static deflection of the system, and this is often readilymeasured. The damping ratio, ξ, can be measured by allowing the damped system to vibrate and measuring the rate of decay ofmaximum amplitudes. Equation (A-9) then shows how the damping ratio can be determined.

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CASE B: FOUNDATION MOVES WITH SINUSOIDAL DISPLACEMENT

W = equipment weight. lbsk = vibration mount spring constant, lbs/inc = system damping coefficientw = undamped natural frequency of system, rad/sec.ξ = damping ratio, c/cuY = Ymaxsinpt = forced motion of baseYmax = max. displacement of base, inp = forced frequency of base section (rad/sec)x = equipment displacement from equilibrium position, in

Here we are interested in reducing the displacement transmitted from the base to the equipment. The transmission factors are: TD = Max. displacement of equipment Max. displacement of base TDl = Max. vibration - mount deflection Max. displacement of base TD is numerically equal to TF [eg. (A-1), base A].

When the damping ratio is less than about 0.1, the maximum force transmitted to the equipment is given very nearly bykTDlYmax.

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CASE C: SHOCK MOTION OF BASE (BASE SUDDENLY BROUGHT TO REST OR BASE ACQUIRES SUDDEN VELOCITY)

W = equipment weight, lbs. k = vibration mount spring constant, lbs/in. c = system damping coefficient. x = equipment displacement from equilibrium, in. y = displacement of base, in. t = time, secs.The results in this section are due to R.D. Mindlin, "Dynamics of Package Cushioning", Bell System Technical Journal, Vol.24, pp.,353-361, July-Oct. 1945. This case is applicable to equipment which drops from a height, or to equipment which acquires a sudden velocity. Let V = sudden velocity change of base, in/sec x = damping ratio = c/cc w = = undamped natural frequency of system, rad/sec

g = gravitational constant, 386 in/sec² dmax = max. isolator deflection, measured from equilibrium position, in dst = static isolator def lection = W/k, in. amax = max. equiment acceleration, in/sec² When 0 ≤ ξ ≤ 0.2, the following is nearly exact:

(C-1)

Figure C-1 illustrates eq. (C-1).When the damping is small, max force transmitted to equipment is very nearly kdmax.

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CASE D: SUDDEN IMPACT ON EQUIPMENT

The results in this section are due to R.E.Newton: Shock and Vibration Handbook,McGraw Hill Book Co., Inc., N.Y.("Theory ofShock Isolation Vol. II, p. 3I-28) and R.D.Mindlin: "Dynamics of Package Cushioning".See Case C.

W = equipment weight, lbsK = vibration mount spring constant, lbs/inC = system damping coefficientx = displacement of equipment from equilibrium, (in)

Sudden impact, or a sharp blow is characterized by a large force (Fo) acting for a short period of time (to) as shown in thesketch. For practical purposes, suddenness is taken to mean that to is small in comparison with the natural period of vibration ofthe system. The impulse, l is defined as the area under the force-time curve, i.e. l = Foto lb-secs. (D-1)The impulse, l, results in a sudden downward velocity V of the equipment, given by

(D-2)

The maximum mount deflection and the maximum equipment acceleration (dmax and amax) can be calculated by substituting Vinto equation (C-1) of Case C.

CASE E: SUDDEN IMPACT (VELOCITY CHANGE OF BASE) WITH EQUIPMENT CONTAINING RESILIENT COMPONENTThe results of this are due to RD. Mindlin (Ref. of Case C). WZ = weight of lightweight component (lbs) z = displacement of WZ from equilibrium (in)

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kzxCzx WxkcyV

= spring constant of component support system (lbs/in) = damping constant of component support system= main equipment weight, (lbs) = main-equipment displacement from equilibrium (in) = vibration mount spring constant (lbs/in) = main equipment damping constant = displacement of base (in) = velocity of base (in/sec)

The sudden impact considered is that in which the base is either moving and suddenly brought to rest, or is at rest and acquires asudden upward velocity. The former, for example, is an approximation of what happens to a package which drops to ground (say)and remains in contact with the ground. A sudden velocity change, V, of the base is shown in the sketch. Since the equipmentand base are assumed always to remain in contact, we refer to this case as one of inelastic impact. It is also assumed that thecomponent weight, Wz, is small in comparison with the main equipment weight, W. In view of this assumption we can neglect the forces exerted on the main equipment by the component support system (butnot the converse). This means that the base-mount-main-equipment system is identical to that of Case C. In the event Case E represents a system falling from a height h to ground, the velocity V is given by: V = 2gh in/sec (E-1)where: g = 386 in/sec² h = height of fall, inches

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The amplification factor, A0, is defined by R.D. Mindlin as follows:

The term "quasistatic" means that the component deflection is calculated for the condition in which the acceleration pulse is veryslow in comparison with the natural period of vibration of the component.The following curves (R.D. Mindlin) show plots of the amplification factor against the frequency ratio ω1/ω2 where: ω1 = = undamped natural frequency of component support system (E-3)

and ω2 = kg/w = undamped natural frequency of main equipment and vibration mount (E-4)

Various values of the damping of the component support system are considered: Figure E-1: 1% main equipment damping. Figure E-2: 5% main equipment damping. Figure E-3: 10% main equipment damping. Sometimes the impact involved may be elastic, i.e. there may be rebound of the package. It is then often a fair approximation,which is on the conservative side, to evaluate the isolator design on the basis of no rebound (inelastic impact) as above.However, if a more exact analysis is required, see R.D. Mindlin (Ref. Case C).

Figure E-1 Amplification Factors For Linear Damped Cushioning With No Rebound. β2 = 0.01 (From R.D. Mindlin:Dynamics of Package Cushioning, p.82)

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Figure E-2 Amplification Factors For Linear Damped Cushioning With No Rebound. β2 = 0.05 (From R.D. Mindlin:Dynamics of Package Cushioning, p.82)

Figure E-3 Amplification Factors For Linear Damped Cushioning With No Rebound. β2 = 0.1 (From R.D. Mindlin:Dynamics of Package Cushioning, p.83)

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CASE F: EQUIPMENT SUBJECT TO A DISTURBING FORCE AND/OR DISTURBING TORQUES - 4-POINT MOUNTING

The results of this case are due to E.H. Hull: "The Use of Rubber in Vibration Isolation," ASME Transactions (J. Applied Mechanics)4,3, Sept. 1937, pp.(A-109)-(A-114).

The figure shows equipment, center of gravity C, mounted on 4 supports, which may represent vibration mounts, and acted uponby a disturbing force, Fy, in the y-direction and/or by torques, Tx, Ty, Tz acting singly or in combination about the x, y, z axes,which are principal axes through the center of gravity, C. The four supports are symmetrically disposed relative to the center of gravity, their location defined by distances bx, by, bzfrom the axes, as shown. The mass moments of inertia through C about the coordinate axes are lx, lz respectively. As a result ofthe external force and torques, the equipment motion is a displacement of C, maximum values of which are denoted (a) by thecoordinates (Cx, Cy, Cz) and (b) by the rotation of the equipment (from equilibrium) about the coordinate axes (θx, θy, θz). Thisdisplacement is generally small relative to the major dimensions of the equipment. Let M = mass of equipment (equipment weight/g, where g = 386 in/sec²). ky = total vertical stiffness of the four supports i.e. 4 times the stiffness of each support, lbs/in ks = total horizontal or shear stiffness of the four supports, i.e. 4 times the horizontal stiffness of each support, lbs/in ω = frequency of sinusoidally applied force and torques (rad/sec)

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Damping is assumed to be negligible.1. Equipment Displacement

in these equations Fy, Tx, Ty, Tz represent peak values of the corresponding applied force/torques.2. Undamped Natural Frequencies

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3. Mount Deflections If, due to any one of the force/torques Fy, Tx, Ty, Tz, the equipment motion is (Cx, Cy, Cz, θx, θy, θz) and if the coordinates ofthe point of support of a vibration mounts are (X, Y, Z) in the equilibrium position, then the deflection (∆X, ∆Y, ∆Z) of the mountfrom equilibrium due to the applied force/torques is given by: ∆X = Cx - θzY + θyZ ∆Y = Cy - θxZ + θzX (F-15) ∆Z = Cz - θyX + θxYprovided the displacements are small. However, if the effects of more than one disturbing force/torque are to be combined, the corresponding displacements of themount must be combined vectorially and cannot be added algebraically, as in eq. (F-15).

General Comments1. It is desirable to make sure that the disturbing forces and torques operate at frequencies sufficiently far removed from thecomputed natural frequencies, so that resonnance conditions are avoided.2. The compliance of the vibration mounts in compression and shear should be such that their combined compliance yieldsnatural frequencies, which are sufficiently lower than the natural frequencies of the system (hopefully at least by a factor of 2.5).3. The displacements (max. deflections) of the mounts can be calculated from eq. (F-15), for any given single disturbing force ortorque. If several force/torques act simultaneously, vector addition of forces in different directions is required, and eq. (F.15)cannot be used.4. The case of a horizontal disturbing force has not been considered, If this is necessary, see C.E. Crede: "Vibration and ShockIsolation," J. Wiley, 1951.5. Other things being equal the best arrangement for the mounts is to arrange them so that their resultant force passes throughthe center of gravity of the equipment and that its line of action is a principal axis. If there is a resultant torque about the centerof gravity its direction should be about a principal axis through the center of gravity. However, if this arrangement is impractical,it need not be adhered to.

CASE G: COMPLEX DRIVING FORCESWhen the disturbing forces are neither sinusoidal nor suddenly applied, the vibration analysis becomes more complicated. It iscorrespondingly more difficult to give general guidelines or methods of analysis. One can, however, consider every force-timevariation as composed of components of different frequencies, each being a multiple of the basic (usually driving) frequency.Mathematically, this is known as expanding an arbitrary function into a Fourier series. Once these components are determined,each one being sinusoidal at a different frequency, any component can be analyzed like a sinusoidal force. This can provide atleast

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some understanding of the vibration phenomenon. Usually the lowest-frequency component predominates and is the mostimportant component to analyze. It is possible, however, that the vibration-mount design will appear unfeasible on the basis ofan analysis of only the fundamental component, whereas the exact analysis would show that a vibration mount can be useful, i.e.sometimes an analysis of components of several frequencies may be required. This, however, may be quite difficult. All one cansay, then, is that having a separation of an arbitrary force-time variation into different-frequency components can provide someinsight. The following represents data on the Fourier series (decomposition into components at different frequencies) of somerepresentative force-time variations, which are neither sinusoidal nor sudden. Each force is assumed to be a periodic function ofthe time.

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To illustrate this approach in a particular case, we consider a commonly found machine motion: the connecting-rod motion, orslider-crank motion, as in internal-combustion engines. This motion can be shown to have the following Fourier analysis:In the sketch: r = crank length, in.

= conn. rod length, in. θ = crank angle, rad or deg. x = piston displacement (piston motion in-line with crank pivot) in. ω = crank speed, assumed constant, rad/sec a = piston acceleration, in/sec²

where A2, A4, A6 are given as follows [J. Hirschhorn: "Kinematics and Dynamics of Plane Mechanisms," McGraw-Hill, 1962 Table5-1, page 115]

/r A2 A4 A6 3.03.54.04.55.0

0.34310.29180.25400.22500.2020

0.01010.00620.00410.00280.0021

0.00030.00010.0001 - -

5.0 DESIGN PROBLEMSThe following are a number of problems intended to familiarize the reader with the basic aspects of the application of vibrationmounts.

Problem No. 1A metal tumbling unit weighing 200 lbs and driven by a 950 RPM motor is to be mounted for at least 81% vibration isolationusing 4 cylindrical mounts in shear. Select the mounts.

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The load per mounting is (1/4) x 200 lbs = 50 lbs. From the basic vibration chart, Table 9, a forcing frequency of 950 RPM and81% isolation lead to a point of intersection corresponding to a static deflection of 0.25". Cylindrical mountA10Z 2-311C, loaded in shear, has a deflection 0.35" at 50 lbs. Since this deflection is in excess of 0.25, theisolation will be greater than the design minimum; from the basic vibration chart, it is seen to be between 85-90%.

Problem No.2Consider the tumbling unit of Problem No. 1 and suppose the motor speed were increased to 2500 RPM. What mountings couldbe used, allowing loading both in shear and in compression? From the basic vibration chart, Table 9, for a forcing frequency of 2500 RPM and 81% isolation, we find a static deflection ofabout 0.037". Hence we must look for mountings with a load rating not less than 50 lbs and with a corresponding deflection ofnot less than 0.037". The following mounts can be considered:

Load in Compression A10Z 2-300B (0.10" deflection)A10Z 2-317B (0.10")A10Z 2-310B (0.12")A10Z 2-314C (0.05")

Load inShear A10Z 2-3308 (0.2" deflection)A10Z 2-311C (0.35")A10Z 3-318C (0.20")

Amongst these, the highest percentage of isolation is afforded by the mount with the largest deflection (A10Z 2-311C), providedsuch a deflection is permissable.

Problem No.3 A small business machine is to be mounted for 81% vibration isolation. The weight is 25 lbs and there are 4 mounting points.What additional information is required for the selection of the mounting? Information which is needed is as follows: frequency of disturbing force; direction and point of application of disturbing force;space limitations, if any; ambient conditions, if unusual; mass and compliance distribution of machine - if complex (i.e. if otherthan case A).NOTE: In the following problems unless otherwise stated it is assumed that the loads are evenly distributed among the mounts.

Problem No.4A device contains 4 symmetrically located special-configuration mounts (Finger-Flex) A10R 4-1500A, each mount deflecting justover 0.13" at 20 lb. load. In order to obtain satisfactory vibration isolation, it is desired to increase the deflection from 0.13" to0.26", the load remaining the same. How can this be done? One way is to stack two (identical) mounts in series, each of the mounts being replaced by such a stack.

Problem No.5A unit which is to be mounted for 81% vibration isolation has a forcing frequency of 950 RPM, weighs 1080 lbs and is to use 6mounts in shear. A mounting with a female tap is required. Select a mount. The load per mount is 1080/6 = 180 lbs. At 950 RPM and 81% isolation, the basic vibration chart gives a static deflection of0.25". Mount A10Z 3-318C loaded in shear has a deflection of about 0.67" at 180 lbs. This being in excess of 0.25" the degree ofisolation is certainly satisfactory. A female tap is provided on the mounting.

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Problem No. 6A 275 lb. motor is mounted with cylindrical mounts A10Z 2-311C loaded in shear, at six points, the forcing frequency being 1100RPM. What is the percentage of vibration isolation attained? The load mount = 275/6 = 45.8 lbs, assuming mounts to be symmetrically located, so that load is evenly distributed. From thedesign information furnished In the catalog, the deflection of the mount at this load value shear is approximately 0.32". From the basic vibration chart, Table 9, the point of intersection of 0.32" static deflection and forcing frequency of 1100 RPMgives an isolation percentage of nearly 90%.

Problem No.7 An air conditioner weighs 250 lbs. and is driven by a motor at 1750 RPM. The unit is mounted in shear on 4 style A10Z 2-317Bcylindrical mounts. Is this design satisfactory? The mounting is not properly installed, because the maximum load rating for this mount, as indicated in the catalog, is 18 lbs inshear and 50 lbs in compression. The load per mount is 250/4 = 62.5 lbs. Possibly the mount should be installed so that it isloaded in compression, but even this is not satisfactory, since the load (62.5 lbs) is significantly in excess of the 50 lbsrecommended limit. Mounts, which have sufficient load capacity, are as follows:

Mount A10Z 2-310B (Compression)A10Z 2-31 1C (Shear - marginal)A10Z 2-330B (Shear)AI0Z 3-318C (Shear)

Static Deflection (0.14") (0.40") (0.22") (0.23")

The choice of mounts depends (amongst other matters) on the degree of isolation desired. With any of the above mounts this willbe in excess of 81%.

Problem No.8 If in the preceeding problem, the air conditioner weighs 350 lbs what is the choice of mounts? The load/mount 350/4 = 87.5 lbs.The following mounts can be considered:

MountA10Z 2-314C (Compression)A10Z 2-31 1C (compression)Al OZ 2-330B (Shear)Al OZ 3-318C (Shear or compression)

Static Deflection (0.08") (0.10") (0.31") (0.38"or 0.08")

At 1750 RPM, 81% vibration isolation corresponds to a static deflection of 0.08".

Problem No.9 A computer weighs 200 lbs. It is to be vibration isolated with 4 mounts. The forcing frequency is 1750 RPM. If the mounts are tobe loaded in compression, what mountings are available and what is the percentage of vibration isolation attained in each case? The load per mount is 200/4 = 50 lbs. Hence we require mounts with a load capability of at least 50 lbs in compression. Foreach mount, the catalog gives curves from which we can find static deflection under a 50-lb load. Going to the basic vibrationchart with this value of static deflection and a forcing frequency for 1750 RPM, the point of intersection defines the percentage ofvibration isolation, which is attained. In this way, the following mounts are amongst the selection which can be made:

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Type of Mount

Catalog Number

Static deflectionat 50 lb compression

% VibrationIsolation

CylindricalCylindricalCylindricalSpecial(Finger-Flex)Special(Finger-Flex)

A10Z 2-317BA10Z 2-300BA10Z 2-310B A10R 4-1506B A10R 4-1506C

0.10" 0.10" 0.12"

0.14"

0.08"

86% 86% 88% Approx.91% Approx. 82%

Problem No. 10A 4-cylinder engine weighing 370 lbs and operating at 2800 RPM is to be isolated for 81% vibration isolation. Discuss theselection of mounts. The lowest frequency to be isolated is 2800 RPM. In general, it is desirable to arrange the mounts so that the resultant of theloads supported by the mounts passes through the center of gravity. If the mounts are symmetrically arranged, and each mountcarries the same load, this usually means that a line through the axis of the mounts passes through the center of gravity. In thiscase we are concerned not only with the translational displacement of the engine as a whole, but also with engine rotation. Inaddition, flexible gas lines and the throttle linkage can vibrate and their vibration isolation may pose an additional problem. At 2800 RPM and 81% vibration isolation, the basic vibration chart gives a static deflection of about 0.032".The load is 370/4 =92.5 lbs per mount. Consider channel mount A10Z 5-101 C3 loaded in shear. This has a deflection of about 0.12" in shear, which can accommodatethe rotation of the engine about the torque-roll axis. The mount deflection in compression would serve to accommodate the shockload in translation.

Problem No. 11 (Case A) An 80 lb fan is to be vibration mounted in shear at four points with at least 93% vibration absorption (isolation), the fan turningat 2000 RPM. Specify the mounts. The main source of vibration is rotor unbalance and the transmission to ground is the vertical component of this force, which issinusoidal. Hence, consider as Case A with a negligble damping.

Solution #1: (Chart) From the basic vibration chart, Table 9, 93% insulation at a forced frequency of 2000 cycles/minute corresponds to a staticdeflection of about 0.14 in. and to a natural frequency of the (undamped) system of about 500 cycles/minute. Considering cylindrical vibration mounts, style A10Z 2-305B loaded in shear, deflects about a 0.15 in. at 20 lbs. and appears tobe suitable for this application.

Solution #2: (Analytical):When the insulation ratio is 93%, the force transmissibility, TF, is 1 - 0.93 = 0.07 or 7%. With zero damping (ξ = 0), eq. (A-1) ofCase A gives:

(A-1)

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when PN > ωN, use - sign in eq. (A-1') when PN< ωN, use + sign in eq. (A-1') For good isolation we would like ωN < PN; hence use minus sign. Solving for (pN/ωN,) from eq. (A-1'), we obtain PN/ωN = 3.91. But PN = 2000 cycles/min. Hence, ωN = 510 cycles/min. From eq. (A-6), of Case A,

Solving for Xst, Xst = 0.14 in. These calculations agree adequately with the values found by use of the chart.

Problem No. 12 Data as in problem 11, but damping is estimated at 10 percent of critical. Does this change the mounting specification? The forcing transmissibility, TF, corresponding to 93% vibration isolation, is 0.07 and the forcing frequency is 2000 cycles perminute. From Figure A-1, using these values and the curve for 0.1 damping ratio, we find the ratio of forcing to natural frequencyabout 5 (this gives a TF of about 0.06). Hence, ωN = 2000/5 = 400 cycles/min. From the basic vibration chart. Table 9, this natural frequency corresponds to a static deflection of about 0.22". The loadremains at 20 lbs. per mount. The mount specified for Problem 11 is too stiff. Mount A10Z 2-310B loaded in shear appears to besatisfactory, with a deflection of about 0.27" at 20 lbs. From Figure A2, we see that the maximum mount deflection measured from static is less than 0.04 times the static deflectionand hence the motion amplification (over static) is negligible, i.e. max. mount deflection is very nearly 0.27". This problem could also have been solved by the computer program, or analytically. in the latter case, eq. (A-1) can be solvedfor ωN at the value ξ = 0.1, PN = 2000. In general when the damping is not in excess of 5% of critical and when the ratio of forcing to natural frequency is not lessthan 2.5, the transmissibility differs very little from that computed without damping and the motion amplification (compared tostatic) is negligible. That is to say, for engineering purposes, we can compute the mount design neglecting damping. From the curves in Figures A1, A2, we can conclude also that while damping will reduce the amplitudes in displacement, theforce transmissibility above resonance in the presence of damping is greater than that without damping.

Problem No. 13 (Case A) A machine weighs between 3.5 lbs and 5 lbs depending on load. The natural system frequency is 12 cycles per second. When theforcing frequency, which produces the vibration, is between

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60-90 cycles/sec. and again when it is between 200-400 cycles/sec. the vibration is objectionable. Design a vibration mount for a3-point suspension with not less than 81% vibration isolation. In the absence of more information, we may assume that Case A is applicaple with zero damping. If we isolate for the lowestobjectionable forced frequency (60 cps), this would take care of all the troublesome regions. The forced frequency is 60 x 60 = 3600 cycles/min. From the basic vibration chart, Table 9, an 81 % insulation ratio at a forcedfrequency of 3600 corresponds to a static deflection of about 0.017 in. The weight supported by each mounting ranges from3.5/3 lbs to 5/3 lbs, or from 1.17 to 1.67 lbs. The natural frequency, if needed, is read off from the chart at about 1450 cycles/min. Hence, the vibration mount specificationis: 0.017" deflection 1.17 lbs to 1.67 lbs supported weight. Cylindrical mount A10Z 2-1765A loaded in shear is a possibility. Considering the special-configuration (Finger-Flex) mountings,mount A10R 4-1500A comes to mind. The deflection of the former at 1.17 lbs is only about 0.008", so that a stack of two is desirable, if space permits. In view of itsconstruction, the spring rate of this mount increases rapidly with deflection and the special configuration unit would be both moreeconomical in the use of space and more effective in taking care of overloads, if this should arise.

Problem No. 14 (Cases A and C)Sensitive radio equipment is to be mounted with a 3-point suspension on a boat. Full restraint is required, because of enginedisturbance, plus impact of waves and bumping of pier. The radio equipment weighs 54 lbs and the engine turns at 2000 RPM. Here we have both steady vibrations at 2000 cycles/min as well as shock loads, caused by wave pounding and the bumpingagainst the dock. We have no precise information on the latter and need to do the best we can. For the steady vibration, consider Case A with zero damping. At 81% insulation and a forcing frequency of 2000 cycles/min, thebasic vibration chart, Table 9, gives a static deflection of about 0.055 in. The load per mounting is 54/3 = 18 lbs. The naturalfrequency obtained from the chart is about 810 cycles/min. A A10R 4-1504B ring-style special-configuration (Finger-Flex) mounting approximately fulfills this condition. In order to limit the effect of shock loads, conical bumpers may be added to limit the horizontal shock load, possibly with theA10Z 7-102C type of bumper. We may, however, also make an arbitrary guess and assume that the pier and waves are equivalent approximately to a 0.5mph sudden, horizontal velocity change of the boat and try to design the vibration mount for this condition. This will in any eventprovide some insight into how much of a sudden velocity change we may expect to cushion by vibration mounting. Thiscorresponds to Case C, horizontal motion and negligible damping (ξ = 0). We would also wish to know how much force the sensitive radio equipment can take without damage. Often such a load isexpressed as a g-load, i.e. how many times its own weight the equipment can survive. For example, a (1/2)g-load means thatthe radio equipment can withstand a max. force of (1/2) (54) = 27 lbs without damage. Suppose we assume that the max. safeload on the radio equipment is 1g or 54 lbs. From eq. (C-1), Case C, we have amax = Vω = 1 (C-1) g g where V = 1 mph 2

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= 8.8 in/sec. Hence, ω = (g) = (386) = 43.8 rad/sec. (1) (V) (1)(8.8) ∴ ωN = ω x 60 = 418 cycles/mm. 2π

This frequency is sufficiently low so that it is difficult to isolate effectively without undesirably large vibration mount deflections.This suggests using a cylindrical mount, say, mounted in compression for the vertical (engine) vibrations and having reasonablylarge compliance in the horizontal (shear) mode to take care of some of the shock, with a conical bumper to limit excessivehorizontal deflections. For example, cylindrical mount A10Z 2-300A has a 0.50"deflection at 20 lbs. compressive load, while in shear, the deflection at16 lbs. is about 0.30", or six times as much. This is an overload, but might still be considered due to the intermittent value of theload. The natural frequency in the shear mode based on the 16 lb load is 188 or about 343 cycles/min, which is 18% lower than the 418 cycles/min figure. 0.30 From eq. (C-1), dmax = amax = 1 dst g and we find dmax = 0.30". (Note that dst is computed as if the weight were supported in shear)

This is too large a maximum deflection. A conical bumber should be used to limit the deflection beyond 0.20", say, alternatively,a stronger and stiffer mount should be considered, for example A10Z 2-300B, which deflects 0.035" at 18 lbs. in shear. Theinsulation ratio in compression is reduced to about 65%; and while the insulation ratio in shear is also reduced, so is thecorresponding maximum deflection. In addition, the conical bumpers should be added. The final choice of mount is a matter ofjudgement. Problem No. 15 (Case A) A single-cylinder gasoline engine belt drives a one-cylinder air compressor. Both units are bolted to a light-gage metal pan,which is welded to the top of an air-receiver tank, which in turn is mounted to a four-wheel steel-tired dolly. The whole unitvibrates and walks all over the floor. The engine weighs 100 lbs and turns at 3000 RPM. The compressor weighs 120 lbs andturns at 1200 RPM. The tank weighs 25 lbs and the dolly weighs 50 lbs. What can be done? Possibly good rubber tires on the dolly would help. If the tank is mounted to the dolly, total weight, W = 100 + 120 + 75 = 295 lbs. The lowest-frequency disturbing force is that due to the air compressor, i.e. 1200 cycles/mm. At 81% vibration isolation (CaseA, ξ = 10), Table 9 gives a static deflection of the isolator of about 0.15". Considering a 4-mount suspension, the load per mountis 74 lbs. Cylindrical mount A10Z 2-310B would be a possiblity, loaded in compression. If the dolly continues to move, since it weighsonly 50 lbs., it might require a little softer material than the 40-durometer rubber, in order to effect more isolation, especiallysince operation is intermittent and weight limit is exceeded by the 30-durometer mounting. Next consider mounting the pan that holds the engine and air-compressor unit. The total wieght here is 100 + 120 = 220 lbs.and with the same static deflection of 0.15, A10Z 2-310A mounting would suffice

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in compression, considering the fact that the chart shows the A10Z 2-310B mounting to deflect 0.12" at 55 lbs. Thelower-durometer mounting (Type A, at 30-durometer) should, therefore, approximate the 0.15" required deflection. Note that thelast letter in the mount identification specifies the durometer hardness of the rubber (A = 30, B = 40, C = 50).

Problem No. 16 (Case G)Isolation of a punch press. This is one of the most difficult and impractical applications for isolation. Shock absorption, at most, isall that can be expected. Unit weighs 1500 lbs, sits on four feet, operates at 50-100 RPM, and is driven by a 5 H.P., 1750 RPMelectric motor, the flywheel turning at 250 RPM. While many vibration problems deal with sinusoidal or nearly sinusoidal forces and some (such as in package cushioning) dealwith essentially sudden velocity changes, here we have a suddenly applied force, which is periodic, but not harmonic. Theforce-time variation is essentially that of the "Repeated Step" in Case G. if we assume that the punching operation of the press occurs, say, during 30º of crank rotation, then the λ in this case(Repeated Step, Case G) is 30/360 = 1/12 = 0.08333. From Case G, we find that the amplitude of the fundamental harmonic is (2/π) sin πλ or 0.164. This is only about 16% of theamplitude of the force pulse, and its frequency is operating frequency (50-100 RPM). Consider, however, the 4th harmonic (200-400 cycles/min). Its amplitude is (2 sin 4πλ)/4π = .1376 or 13.8%. This is not much less than the amplitude of the basic (fundamental) frequency. This shows that in the punch-press type ofdisturbing force, the higher harmonics cannot be neglected. The fundamental frequency (50-100 cycles/mm) is so low that isolation with vibration mounts would lead to excessive staticdeflections. However, it is conceivable that a practical vibration mount would be successful in isolating some of the significanthigher harmonics. Scientific mounting of punch presses, particularly at the feet, Is complicated and at best a trial-and-errormethod, keeping in mind a few rules.

1. 2. 3. 4. 5. 6.

Slow-speed presses should be mounted with mounts of greater deflection than high-speed presses.Deflections used to foot-mount presses may vary from 1/32" to 3/4" depending largely on operating speed andstroke length, with the smaller deflection being the more common. Some authorities suggest that footmountings in compression be no more than 1" thick. This would pretty well limit static deflection to 1/4" - 5/16"in order to provide good life.There may be several static deflections that will work, while other static deflections interspaced in between themwill not work, i.e. 1/16" and 3/16" may work, while 1/8" may not work. This can be understood, at least in part,in terms of the idea that a significant set of higher harmonics may be isolated at one deflection, but not atanother.Even the best mounting system will still transmit a considerable amount of vibration and shock.If the ultimate in isolation is required, the punch press must be attached solidly to a base of large mass and theentire press and base mounted on a single unit.For this application it is reasonable also to consider a 10R series "ISO-PAD," which is available in different sizesdepending upon load.

Problem No. 17 (Case B) A sensitive experiment is to be conducted in the laboratory of a textile plant. The laboratory floor vibrates at an amplitude of0.05" due to the operation of industrial sewing machines and

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other textile machinery. The basic floor-vibration frequency is that of the industrial sewing machines, which operate between therange of 1500-5000 RPM. It is desired to vibration isolate the experiment, which weighs 25 lbs, with a four-point mounting at notless than 81% vibration of displacement. This may be considered as Case B with zero damping. At 81% displacement isolation, the displacement transmissibility, TD, is 0.19. It is calculated using the same equation as forTF, i.e. eq: (A-1) of Case A.

For zero damping,

whence xst = 0.10" (Same result follows from basic vibration chart). The isolator specification, therefore, is .010" static deflection at a load of 25/4 = 6.25 lbs. However, the vibration mount mustalso be capable of withstanding the max. deflection. This is computed from the transmissibility factor TDI:

TDI = max. isolator deflection/max. displacement of base. According to eq. (B-1) with ξ = 0,

Hence, max. isolator deflection will be 1.19 x 0.050" = 0.060". Considering cylindrical vibration mounts, mount A 10 Z 2-316B loaded in shear, has a0.10" deflection at about 6.25 lbs. and atthe max. deflection of 0.10. + 0.060" = 0.160", the force is about 9 lbs., which is still within the limit of safe practice for themount. A soft mounting, such as this one, is often obtained advantageously by mounting the unit in shear.

Problem No. 18Data as in Problem 17, except that system damping is estimated at 10% of critical. Reevaluate the specification of the mount. In problem 17 we found that the displacement transmissibility corresponding to 81% isolation is TD = 0.19; and that the lowestforcing frequency, PN = 1500 cycles/min.

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From Figure A1, which applies to TD as well as to TF, we find that the given value of the transmissibility at 0.1 damping ratioyields a frequency ratio PN/ωN of about 2.7. Hence, ωN = 1500/2.7 = 550 cycles/min. At a natural frequency of 550 cycles/min, the basic vibration chart (Table 9) gives a static deflection of about 0.117". The loadper mount, as in Problem 17, is 6.25 lbs. The ratio of max. mount displacement to max. base displacement at a frequency ratio of 2.7 and at a damping ratio of 0.1 iscomputed from eq. (B-1) and is about 1.2, as in Problem 17. Hence, the isolator specification A10Z 2-316B of Problem 17 remains satisfactory. The problem could also have been solvedwith the aid of the computer program.

Problem No. 19 (Case D)An impact testing machine consists of a simple pendulum of length 4 feet and weight 5 lbs. It is initially horizontal. It is releasedand at the bottom of its swing impacts the test object and in this test comes to rest essentially instantaneously. The equipmentwhich is to be tested weighs 100 lbs and is capable of withstanding accelerations upto 2g. Design a vibration mount, so that theequipment will survive the impact test. The velocity acquired by the pendulum in the 4 foot drop is

= 193 in./sec. striking velocity.The momentum of the pendulum just prior to impact is equal to the mass of the pendulum x velocity of pendulum

Consider this as Case D. The impulse, l of eq. (D-1) then is: l = 2.5 lb-secs. if the pendulum retains a residual velocity VP' just after striking the test object, l would be computed from l = (VP - VP') x massof pendulum. The impact results in an essentially sudden velocity change, V, of the equipment, which, according to eq. (D-2) isgiven by:

This value of V is to be used in eq. (C-1) of Case C, with ξ = 0:

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Using a 4-point mounting, the weight supported per mount is 100/4 = 25 lbs. According to the basic vibration chart (Table 9) at ωN = 764, the static deflection is about 0.062". However, according to eq.(C-1), the isolator should be capable of withstanding a max. deflection, dmax = 2dst or 0.124 inches (from equilibrium). For example, vibration mount A10Z 2-330B loaded in shear, appears to be a possible mount with substantial overload capacity. When dmax exceeds dst, the possibility exists that the mount may be loaded in tension for a short period of time. Thiscondition is undesirable. However, if this condition is momentary (occurring only very infrequently), and there is no other choice,the mount may be adequate, especially if there is substantial damping in the system.

Problem No.20 (Case E) A timing device for a space-vehicle application weighs 1 oz. and is packaged in a foam material, the spring constant of which hasbeen measured at 100 lbs/in, and the damping ratio of which is estimated at 10%. The main package weighs 5 lbs and must becapable of surviving a drop test from a height of 10 feet. The timing device is capable of withstanding a force not in excess of500g's while the main package can take up to 300g's, the damping ratio of the main package system being estimated at 10%.Design a 2-point vibration-mount support for this system. We may consider this as Case E, the timing device being light. The numerical data for Case E, then is: Wz = 1 oz = 0.0625 lbs. kzx = 100 lbs/in. ξxz = 0.10 (timing-device support-system damping ratio) W = 5 lbs. ξ (main-package damping ratio) = 10% = 0.1 The velocity, V. acquired in a drop test of 10 feet is

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In Figure E-3, β2 = main-equipment damping ratio = 0.10 β1 = damping ratio of component - support system = 0.10Select (try) ω1/ω2 = 2Then from Figure E-3, we find Ao = 1.50.

Solving for k, k = 2000 lbs/in.∴Max. component deflection = 1.50 x 0.004339 √2000 in. = 0.291 in.Max. force on component ≈ (kzx) (Max. component deflection) or (100) (0.291) lbs or 29.1 lbs.Hence, g-load on timing device is 29.1/0.0625.This is 466g and hence, acceptable.The main equipment mounting is designed according to Case C with

From eq. (C-1), Case C,

∴dmax/dst = 278.

= 0.0025"∴dmax = (0.0025) (278) = 0.70"

T247

This corresponds to a max. force on the package of 1390 lbs.

The vibration-mount specification, then is: 0.0025" static deflection 2.5 lbs supported weight (= 5/2 lbs.)

This corresponds to a spring rate of 1000 lbs/in. However, the vibration mount must also be able to withstand the max.displacement of 0.70". Consider the rectangular mount A10Z 6-520B. The spring rate in shear is about 800 lbs/in. This mount could be combined witha bumper to take care of the deflections in excess of 3/4", say. Alternatively, a stiffer rectangular mount, possibly with a50-durometer rubber, can be considered. See also note at end of preceding problem.

Problem No.21 (Cases F and A) A 70-lb electric motor, turning at 3600 RPM, delivers 5 H.P. by direct connection to an industrial exhaust fan. The motor must bemounted separately from the fan to shield it against a corrosive environment. The dimensions (Case F) for the four mountlocations are bx = 10", by = 6", bz = 6" and the radii of gyration about the principal axes through the motor center of gravity are5", 8", 5", respectively. Design the vibration mounts. In this case, it is necessary that the mounting be flexible enough to absorb the vibrations, but at the same time rigid enough sothat there is relatively little motion at the coupling between motor and exhaust fan. Consider this as Case F with vertical transmission to ground due to motor unbalance (same as Case A) and torque excitation(say, about x-axis) as in Case F, due to motor torque. The torque about the motor ax is (x-axis) corresponding to 5 H.P. at 3600 RPM is given by

= 87.5 in.-lbs. For the vertical force transmission to ground, we consider Case A. W = 70/4 17.5 lbs per mount.At a forcing frequency of 3600 cycles; min. and with negligible damping (ξ = 0), the basic vibration chart (Table 9)using 81%vibration isolation, gives a static deflection of about 0.017". Next we consider the design of the isolator for this excitation. Suppose we consider (try) vibration mount A10Z 2-314C(cylindrical mount), which approximately matches the required characteristics, assuming mounting so that the vertical load istaken up in compression: Using the notation of Case F, the given compliances of this mount lead to ky (vert. compliance for all 4 mounts) = 4 x 1060 = 4240 lbs/in. ks (shear compliance for all 4 mounts) = 4 x 175 = 700 lbs/in.Next, we consider the effect of the torque of 87.5 in-lbs about the motor axis (X-axis).The numerical values, in the notation of Case F (in addition to those already stated) are: Tx = 87.5 in-lbs. M = 70/386 = 0.181 lbs/g. ω = (2π)(60) = 377 rad/sec.

T248

According to eq. (F-11) and (F-13), the lowest natural frequency, ω4, corresponding to this torque excitation, is given by

Substituting the numerical values into these equations, we obtain B = 21,500,

ω4 = 57.3 rad/sec.

= 546 cycles/min.Hence, frequency ratio (forced/natural) = 3600/546 ≅ 6.5, which seems quite adequate. The second natural frequency, ω5,associated with Tx is about 1910 cycles/min. giving a frequency ratio of about 2.0. According to the guideline in Para 4.1,Vibration Identification and Specification, Section 1, this represents only a fair ratio as it is not so far removed from resonance. Ifthe A10Z 2-314C is not satisfactory, a weaker mounting, possibly with 40 or 30 durometer (A10Z 2-314B or A10Z 2-314A), couldbe considered. in any event we need to check the mount deflections. Due to torque about the x-axis there are motion components Cz (translations along the z-axis) of the center of gravity of themotor and θx (equipment rotation of motor about an axis parallel to the x-axis). According to eq. (F-4, F-5), these are given by Cz = (Txbyks)/∆(in.) θx = Tx(ks - Mω²) / ∆(rad),where ∆ = lx Mω4 - ω² lxks + kybz²M + ksby²M + kyksbz² Substituting the given numerical values into these equations we obtain ∆ = 116.8 x 108 Txbyks = 3.68 x 105

Tx(ks - Mω²) = 2.19 x 106

whence Cz = 3.15 x 105 in., θx = - 1.88 x 10-4 rad

T249

The corresponding deflections of the mounts, the point of support of which has coordinates (X,Y,Z) is calculated from eq. (F-15)as follows:

These deflections are negligible. The deflection which predominates is the vertical deflection due to Fy. The static vertical deflection was computed at 0.017". Going back to Case A.

With 0.017" static deflection, ωN = 1450 from Table 9; also PN = 3600. Hence, TDE = 0.195. Assume, for example, that the max.force is due to a rotor unbalance (eccentricity) of 0.002" and that the rotor weighs 50 lbs., then Fmax = (Rotor mass) (0.002)(angular speed)² = (50/386) x 0.002 x (377)² = 36.8 lbs.

The design, therefore, appears to be satisfactory. Due to the displacements and the ω5 frequency, however, it is desirable that aflexible coupling be used between motor and exhaust fan; if necessary, the same style mounting in a 40 or 30 durometer rubbercould be considered.

T250