Vibration Theory Fundamentals

Vibration Theory Fundamentals 1 1.1 INTRODUCTION Mehanial vibrating systems onsist of elements suh as a spring for storing potential energy, mass and inertia for kineti energy, and damper for dissipating mehanial energy. The vibration proess alternatively onverts energy between its potential and kineti forms. In its general sense the vibration is a periodi motion that repeats itself in all its details after a ertain interval of time, alled the period of vibration. Some energy must be replaed in eah yle of vibration from an external soure to maintain the vibration. In a linear spring the hange in length is proportional to the fore ating along its length, the onstant of proportionality referred to as the spring s flexibility, whih is the reiproal of its stiffness. The ideal spring has no mass, and so the fores ating at its opposite ends are equal and opposite. A mass is a rigid body; so its aeleration, aording to Newton s seond law of motion, is proportional to the fore ating on the mass. In a visous damper the applied fore is proportional to the relative veloity of its onneting points. The damping oeffiient provides the proportionality onstant, and is the harateristi parameter of the damper. The ideal damper also is massless, and so the fore at one end is equal and opposite to the fore at the other end. Besides translational motion, a vibrating system may exeute rotational motion. The elements of a mehanial system exeuting pure rotation of the parts are analogous to the elements exeuting pure translation. In a rotational system inertia stores kineti energy, while stiffness and damping parameters are defined with referene to angular rotation and veloity. The analogy between translational and rotational motion of a vibrating body arries on to the mathematial equations desribing the motion of the system; fore used in linear motion is replaed by torque for rotational motion. A plot of the linear displaement, or angular rotation, against time may be a ompliated urve. The simplest kind of periodi motion is a harmoni motion, with the displaement expressed as a harmoni funtion of time t and angular veloity ω, also referred to as the irular frequeny, and is measured in radians per seond. The maximum value of the displaement, x 0, is alled the amplitude of vibration. The period of vibration, T, measured in seonds, is the reiproal of the frequeny of vibration, f, measured in yles per seond. The relation between ω, f and T are as follows. A full yle of vibration takes plae when ωt has passed through 360, or π radians. Then the sine funtion resumes its previous path. Thus, when ωt = π, the time interval t is equal to the period T, or T = π/ω seonds. Sine f is the reiproal of T, f = ω/π yles per seond. For reiproating mahinery the frequeny is often expressed as yles per minute, so f = 30ω/π. In a harmoni motion displaement is given by 1

Reiproating Mahinery Dynamis x = x 0 sin (ωt). Veloity is determined by differentiation with respet to time, so dx/dt = x 0 ω os (ωt). Thus, veloity is also harmoni, with a maximum value of x 0 ω. Aeleration is given by the seond derivative of x with respet to time t: d x/dt = x 0 ω sin (ωt) (1.1) also harmoni, with a maximum value of x 0 ω. A vibrating system is said to have one degree of freedom if its geometrial position an be expressed at any instant in time by one number only. For example, a piston moving in a ylinder an be speified by giving the distane from the ylinder end, thus lassifying it as a single degree of freedom system. Another example is a rankshaft in rigid bearings; the system is fully desribed by the angle between any one rank and the vertial plane. In general, if it takes n numbers to speify the position of a mehanial system, the system has n degrees of freedom. A disk moving in its plane without restraint has three degrees of freedom: the x and y displaements of its enter of gravity and the angle of rotation about the enter of its gravity. A irular disk rolling down an inlined plane has only one degree of freedom; if it partly slides and rolls, it has two degrees of freedom, one of translation and one of rolling. A rigid body moving freely in spae has six degrees of freedom, three of translations and three of rotation. Thus, it takes six oordinates to express its position, usually denoted as x, y, z, θ, ϕ and ψ. A single degree of freedom system is not neessarily a simple one. For example, a 1- ylinder gas engine with a rigid rankshaft and a rigidly mounted ylinder blok has only one degree of freedom with all its omponents suh as pistons, onneting rods and amshaft. This is beause a single parameter, suh as the rank angle through whih the shaft has turned, determines ompletely the position of all its moving parts. But if the ylinder blok is mounted on flexible springs so that it an freely move in all diretions as in an automobile engine, it has seven degrees of freedom, six in referene to the motion of the blok as a rigid body in free spae and rank angle as the seventh oordinate. 1. VECTOR METHOD OF REPRESENTING RECIPROCATING MOTION The motion of a vibrating partile an be onveniently represented by means of a rotating vetor, x 0 (Fig. 1.1). When time is rekoned from the horizontal position of a vetor rotating at a uniform angular veloity, the horizontal projetion of the vetor is x 0 os (ωt), and the vertial projetion will be x 0 sin (ωt). Either projetion an be taken to represent harmoni motion of a reiproating omponent. This representation gives rise to the name irular frequeny for ω. x 0 t t =0 Fig. 1.1. Horizontal Projetion of a Rotating Vetor

Vibration Theory Fundamentals 3 This method of representing reiproal motion is onvenient. If a point is simultaneously subjeted to two motions of the same frequeny whih differ by the phase angle φ, namely a os (ωt) and b os (ωt φ), the geometri sum of the two vetors gives the total motion (Fig. 1.). Parallelogram a, b rotates in a ounter-lokwise diretion with uniform angular veloity ω, and the horizontal projetions of all the vetors represent displaement as a funtion of time. Line x x represents a speifi instant in time for whih the vetor diagram is drawn. Displaement of the sum is the sum of the ordinates for a and b. y a t a+b 0 x b a+b a b t Fig. 1.. Two Vibrations Added by their Vetors Addition of the two vetors is permissible only if vibrations are of the same frequeny. Motions a sin (ωt) and a sin (ωt) an be represented by two vetors, the first rotating at an angular speed ω, the seond at twie that speed. The relative position of these two vetors is hanging ontinuously, thus a geometri addition of the vetors has no physial impliation. A speial ase of vetor addition often enountered is the addition of sine and osine waves of different amplitudes, a sin (ωt) and b os (ωt). The two vetors are perpendiular (Fig. 1.3), so a sin (ωt) + b os (ωt) = [(a + b ) 1/ ]. sin (ωt + ϕ), where tan ϕ = b/a. For example, onsider the sum of two motions 10 sin (5t) and 0 sin (5t + 1). At time t = 0, the first vetor points vertially up, with no horizontal projetion. The seond vetor is turned 1 radian in a ounterlokwise diretion from the first vetor. The graphial vetor addition shows the sum vetor to be 6.76 units long. 1.3 COMPLEX NUMBERS METHOD The vetor method offers visualization of harmoni motions that is simpler than the onsideration of the sine wave by itself. For numerial alulations, however, it is not well

4 Reiproating Mahinery Dynamis adapted. Complex numbers employing imaginary numbers offers a simpler method. A omplex number an be represented graphially by a point in plane where real numbers are plotted horizontally and imaginary numbers are plotted along the vertial axis. Using j = ( 1) 1/, imaginary numbers are j, j, 3j, et. By joining that point with the origin, the omplex number an be made to represent a vetor (Fig. 1.4). If the angle of the vetor is α with the horizontal axis, and of length a, then the vetor is a(os α + j sin α). Sine harmoni motions are represented by rotating vetors, the variable angle ωt may be substituted for fixed angle α to obtain a rotating vetor. Its horizontal projetion will then be a reiproating motion. The sum of the two motions onsidered earlier by the vetor method an be solved using omplex numbers. The first vetor is 10j and the seond vetor is 0j os (1) + 0 sin (1) = 10.8j + 16.8. The sum of the two vetors is 16.8 0.8j, representing a vetor of length (16.8 + 0.8 ) 1/ = 6.7 units. (a +b ) 1/ b t a Fig. 1.3. Addition of Sine and Cosine Waves 3j j a j 1 3 4 Fig. 1.4. Vetor Representation in a Complex Plane Differentiating a{os (ωt) + j sin (ωt)}, gives jωa{os (ωt) + j sin (ωt)}. Note that by definition, j.j = 1. Thus, differentiation of a omplex number is equivalent to multipliation by jω. In vetor representation differentiation multiplies the length of the vetor by ω and turns it ahead by 90. Thus, multiplying a omplex number by j is equivalent to moving the vetor by 90 without hanging its magnitude. In making extended alulations with omplex numbers the ordinary rules of algebra are followed. Algebrai alulations are performed without muh reourse to their physial meaning, only the final answer is interpreted using the real part for the vetor magnitude and phase angle from both the real and imaginary parts.

Vibration Theory Fundamentals 5 A onept of importane in many appliations is that of work done by a harmonially varying fore on a harmonially varying motion of the same frequeny. Consider a fore P = P o sin (ωt + ϕ) ating upon a body for whih the motion is given by x = x o sin (ωt). The work done by the fore during a small displaement dx is Pdx, whih may be written as Pdt(dx/dt). During one yle of vibration ωt varies from 0 to π/ω. After integration, the work per yle is W = πp o x o sin (ϕ). In order to understand this expression for work a graphial approah is useful. In Fig. 1.5a the ordinates are the displaement x and the in-phase omponent of the fore. Between A and B the fore is positive, say upward, and the body is moving in the same diretion, so the work done is positive. Between B and C, on the other hand, the body moves downward toward the equilibrium point while the fore is still positive, so that negative work is done. The work between A and B anels that between B and C, and over a whole yle the work done is zero. Thus, if a harmoni fore ats on a body subjeted to a harmoni motion of the same frequeny, the omponent of the fore in phase with the displaement does no work. Sine the veloity vetor is 90 ahead of the displaement, it follows that fore does work only with that omponent whih is in phase with the veloity. Displaement X A B x o Po os C D t X A B Po sin x o C D t (a) (b) Fig. 1.5. Work Done on Harmoni Motion, (a) In Phase and (b) 90 Out of Phase 1.4 FREE VIBRATIONS WITHOUT DAMPING Consider a mass m suspended by a spring from a fixed loation, as shown in Fig. 1.6. Stiffness of the spring is defined by k, the fore required to extend it by 1 unit. A damper is also plaed between the mass and the fixed wall. The damper does not transmit fore to the mass when at rest, but as soon as it moves it exerts a damping fore of dx/dt. It is proportional to the veloity and direted opposite to the mass. is the oeffiient of visous damping. An external alternating fore P o sin (ωt) is also ating on the mass. The problem onsists in determining the motion of mass m due to this external fore. The equation of motion an be obtained by applying Newton s seond law of motion, whih states that fore is equal to mass multiplied by aeleration. All fores ating upward will be onsidered positive when ating downward. Spring fore has the magnitude kx, and is negative sine it pulls upward. Damping fore ating on the mass is also negative, dx/dt. The downward fores ating on the mass are:

6 Reiproating Mahinery Dynamis A k m P sin t o x Fig. 1.6. Single Degree of Freedom Vibrating System kx dx + p dt o sin (ωt) (1.) Newton s law gives: md x/dt + dx/dt + kx = P o sin (ωt) (1.3) This equation is known as the differential equation of motion of a single degree of freedom system. Before developing a solution of the general equation, simplified ases will be onsidered first. If there is no external applied fore and no damping, the equation redues to: md x/dt + kx = 0 (1.4) The most general solution of this equation is: x = C 1 sin 1k m t 1 + C k k os ω n m t (1.5) = = T π m π where C 1 and C are arbitrary onstants. Physially, the result as it stands is indefinite, sine the onstants an take any value. This is beause the problem itself was not fully defined. If it is speified that the mass is pulled out of its equilibrium position to x = x o, then released without initial veloity, two onditions are speified: At t = 0: x = x o and dx/dt = 0. The first ondition gives C = x o. For the seond ondition, (1.5) must be differentiated, then get C 1 = 0. Substitution into (1.5) leads to the speifi solution: k x = x o os m t (1.6) This represents an undamped vibration. Denoting the period of vibration by T, then: m T = π (1.7) k Customarily, (k/m) 1/ is denoted by ω n, alled the natural irular frequeny. The natural frequeny f n is given by: f n = (1.8) measured in yles per seond. In the derivation of the differential equation of motion the effet of gravity has been negleted. The amplitude of vibration is measured from the position where the downward

Vibration Theory Fundamentals 7 fore mg is held in equilibrium by an upward fore kδ, δ being the extension of the spring due to gravity. To measure x 1 from the unstressed position of the spring, then x 1 = x + δ. In the differential equation of motion, then, x must be replaed by x 1, and on the right hand side of the equation a fore mg must be added. If it is assumed that the motion is harmoni, the frequeny an be alulated from an energy onsideration. In the enter of the swing the mass has maximum kineti energy, while at either extreme position it has no kineti energy. At the same time the spring is strethed (or ompressed) to the fullest at the extreme positions, and has elasti, or potential, energy stored in it. Between the middle and extreme positions the spring-mass system has both kineti and potential energy, the sum of whih has to be onstant sine there is no external work done on the system. Thus, the kineti energy at the enter position must equal elasti energy at the extreme position. For a linear spring the potential energy when strethed over a distane x is 1/kx. The kineti energy at any instant is 1/mv. If the motion is assumed to be x = x o sin (ωt), then v = x o ω os (ωt). At the extreme position the potential energy is 1/kx o, and the kineti energy at the enter position where the veloity is maximum is 1/mω x o. Equating the two energies, 1/kx o = 1/mω x o, from whih ω = k/m, independent of the amplitude x o. This method is of signifiane when dealing with problems of greater omplexity, where the generalized energy method, known as the method of Rayleigh, leads to a result. 1.5 TORSIONAL PENDULUM Fig. 1.7 shows a disk of moment of inertia I attahed to a shaft of torsional stiffness k, defined as the torque neessary to produe one radian twist at the disk. The twisting motion of the disk is under the influene of an externally applied torque T o sin (ωt). Sine only the torsional displaement ϕ is required to desribe its position from the equilibrium position, this is a single degree of freedom problem. For a rotating body, Newton s law states that torque is equal to moment of inertia times angular aeleration, or: T = Id ϕ/dt (1.9) The three torques ating on the disk are the spring torque, damping torque and external torque. The spring torque kϕ, where ϕ is in radians, is negative, just as the spring fore in the previous ase was kx. The stiffness k will depend on the shaft ross-setion geometry and its material shear modulus. The damping torque is dϕ/dt, aused by the shaft material s internal damping. The damping onstant is the torque on the disk aused by an angular speed of rotation of 1 radian per seond. k I Fig. 1.7. Torsional Pendulum

8 Reiproating Mahinery Dynamis Appliation of Newton s law yields the differential equation of motion: Id ϕ/dt + dϕ/dt + kϕ = T o sin (ωt) (1.10) This equation is similar to the translational ase, and the solution for the undamped free ondition an be readily obtained by substituting I for m, T o for P and ϕ for x in the linear ase. Any system with inertia, elastiity and damping proportional to the veloity, where the displaement an be desribed by a single quantity, an be defined in this manner. Consider two disks of moment of inertia I 1 and I joined by a shaft of torsional stiffness k, (Fig. 1.8). On the first disk torque T o sin (ωt) is ating. Both disks an assume an angular position independent of the other by twisting the shaft. However, if the quantity of interest is angle of twist in the shaft, it is possible to express the motion in terms of this quantity alone. Let ϕ 1 and ϕ be the angular displaements of the disks, then ϕ 1 ϕ is the shaft twist, k(ϕ 1 ϕ ) is the spring torque, d(ϕ 1 ϕ )/dt is the damping torque. Applying Newton s law to the first disk, then the equation of motion is: T o sin (ωt) = I 1 d ϕ 1 /dt + d(ϕ 1 ϕ )/dt + k(ϕ 1 ϕ ) (1.11) T o sin t A k I 1 A I Fig. 1.8. Two Disks Mounted on an Elasti Shaft and to the seond disk, 0 = I d ϕ /dt + d(ϕ ϕ 1 )/dt + k(ϕ ϕ 1 ) (1.1) Rearranging the variables and subtrating the results from the two equations yields: (T o /I 1 ) sin (ωt) = (d ϕ 1 /dt d ϕ /dt ) + (/I 1 + /I ) d(ϕ 1 ϕ )/dt + (k/i 1 + k/i )(ϕ 1 ϕ ) (1.13) Defining the twist angle (ϕ 1 ϕ ) = ψ, then multiplying the equation by the expression I 1 I /(I 1 + I ) gives the equation of motion with a single variable ψ, I 1 I /(I 1 + I )d ψ/d t + dψ/dt + kψ = T 0 I /(I 1 + I ) sin (ωt) (1.14) The solution of this equation gives information on the angle of twist, but angular rotation of individual disks annot be obtained. A variation is shown in Fig. 1.9 where a gear and pinion system is attahed to the shaft. Assume gears G and P to be without any inertia. Also assuming the gear teeth to be rigid, torsional flexibility is limited to the two shafts. Gear ratio is n. The differential equations of motion for this system ould be derived from Newton s law, but the system ould be redued to that of Fig. 1.8 by omitting the gears and replaing I, k and ψ by equivalent parameters. If I is lamped and torque T o is applied to I 1, then I 1 will rotate through an angle ϕ o, so k = T o /ϕ o. Due to the gears the torque in the shaft k is T o /n, and the angle of twist of k is T o /nk. Sine I is fixed, this is the angle of rotation of the pinion P. Angle of the

Vibration Theory Fundamentals 9 gear G is n times smaller, or T o /n k. Adding angle T o /k 1 for shaft k 1 gives angular rotation of I 1. Thus, the equivalent k is obtained from the expression: 1 ϕ 1 1 = o = + (1.5) k T k n k o 1 T o sin t A G I 1 A k 1 n, 1 P k I Fig. 1.9. Geared System Now onsider the inertia. Angular aeleration a in k 1 and G beomes nα in k and P. Hene the torque in k is nαi. This is also the torque at the pinion P. Gear G sees it n times larger, so the torque at A is n αi and the equivalent of I in the gearless system is n I. In general, then, a geared system suh as that of Fig. 1.9 an be redued to an equivalent nongeared system (Fig. 1.8) in the following manner. Divide the system into separate parts eah of whih has the same speed within itself. Figure 1.9 has only two parts, but there may be several. Choose one of these parts as the primary and assign numbers n to eah of the other parts so that n is the speed ratio with respet to the primary. n > 1 for speeds higher than the primary part s speed, while the speed ratio of the primary part is unity. Next, remove all gears and multiply all spring onstants k and inertias I by the fators n. The differential equation of the redued gearless system is then the same as that of the original geared design. In still another variation of Fig. 1.6, instead of the sinusoidal fore ating on the mass, the upper end of the spring is made to move up and down with an amplitude a o, the motion of A being a o sin (ωt). It an be shown that this motion of the top of the spring is equivalent to a fore on the suspended mass. If x is the downward displaement of the mass, spring extension will be x a o sin (ωt), spring fore is k{x a o sin (ωt)} and damper fore is {dx/dt a o ω os (ωt)}. Newton s law then gives: md x/dt + dx/dt + kx = ka o sin (ωt) + a o ω os (ωt) (1.16) As mentioned earlier in setion 1., the sum of a sine and osine waves of the same frequeny is also harmoni, so: md x/dt + dx/dt + kx = [(ka o ) + (a o ω) ] 1/ sin (ωt + ϕ o ) (1.17) Thus, the motion of the top of the spring with amplitude a o is equivalent to a fore of amplitude {(ka o ) + (a o ω) } 1/. Terms ka o and a o ω in this expression are the maxima of the spring and damper fores, while the expression inside the parenthesis provides the peak magnitude of the fore when the mass is lamped.

10 Reiproating Mahinery Dynamis 1.6 FREE VIBRATIONS WITH VISCOUS DAMPING Undamped free vibrations persist indefinitely, whih annot happen; all free vibrations eliminate with time. The term visous damping is assoiated with the expression dx/dt sine it represents the onditions of damping due to the visosity of oil in a dashpot. Other types of damping exist, but their solution annot be determined with as muh ease. Consider the funtion x = e st, where t is the time and s an unknown onstant. Upon differentiation the same funtion multiplied by a onstant results. This funtion when substituted in the equation: md x/dt + dx/dt + kx = 0 (1.18) permits division by e st and leads to an algebrai equation instead of a differential equation. Thus, assuming e st as the solution, then the above equation beomes: (ms + s + k). e st = 0 (1.19) This is a quadrati in s, with two values that will satisfy the equation: s 1, = (1.0) The most general solution of the problem is: s1t st x = C1e + Ce (1.1) where C 1 and C are arbitrary onstants. The physial signifiane of this equation will depend on whether the expressions for s are real or omplex. When (/m) > k/m, both values of s will be real. Also, they are both negative sine the square root is smaller than the first term /m. Thus (1.1) desribes a solution onsisting of the sum of two dereasing exponentials, as shown in Fig. 1.10. The ase of C 1 = 1 and C = is shown as F a dashed line. This figure shows that the motion is not osillatory k ± m H G Ibut rather a reeping bak to the K J original position. This is due to the fat that when (/m) > k/m m the damping is large. For smaller values of pertaining to more pratial ases, (1.0) gives omplex values for s. The damping at whih this transition ours is alled the ritial damping : k = m = mk = mω m n (1.) If the damping is less than ritial damping, the solution onsists of two fators, the first a dereasing exponential and the seond a sine wave. The ombined result is an exponentially dereasing sine wave, lying in the spae between the exponential urve on both sides of the phase angle axis (Fig. 1.11). The smaller the damping onstant, the flatter will be the exponential urve and the more yles it will take for the vibrations to be eliminated. 1 e st 1 e st 0 1 3 t 1 Fig. 1.10. Motion of Single Degree of Freedom System with / > 1

Vibration Theory Fundamentals 11 A 1 B 0 3 4 5 C 1 Fig. 1.11. Free Vibration of a System with / < 1 The rate of this dying down of the vibrations is of interest, and an be alulated by onsidering two onseutive maxima of the urve, A B, B C, and so on. During the time interval between any two suh maxima, the amplitude of vibration diminishes from: m t R S T m t π + q U V W e k to e where q = (1.3) m 4m The latter of these two expressions is equal to the first one multiplied by the onstant fator: mq π. e (1.4) whih fator is smaller than unity. This fator is the same for any two onseutive maxima, and is independent of the amplitude or of time. Thus, the amplitudes derease in a geometri series. The frequeny of vibration diminishes with inreasing damping. If written in a dimensionless form with the help of (1.3), it beomes: q R ω = S U 1 n T VW (1.5) This equation is plotted in Fig. 1.1, where the ordinate q/ω n is the ratio of the damped to the undamped natural frequeny, and the absissa is the ratio of the atual to the ritial damping. The figure is a irle; for ritial damping the natural frequeny is zero. Due to the horizontal tangent of the irle at = 0, the natural frequeny is nearly onstant and equal to ( km / ) for all typially enountered values of damping (/ < 0.). 1.0 q/ n 0.5 0 1.0 / Fig. 1.1. Natural Frequeny as a Funtion of Damping

1 Reiproating Mahinery Dynamis 1.7 FORCED VIBRATIONS WITHOUT DAMPING Another partiular ase of (1.3) is the one where the damping term is made zero: md x/dt + kx = P o sin (ωt) (1.6) Substituting a harmoni solution of the type x = x 0 sin (ωt) into this equation yields: mω x o sin (ωt) + kx o sin (ωt) = P o sin (ωt) (1.7) produing the result: Po / k x = sin (ωt) (1.8) 1 ( ω/ ωn) The expression P o /k has the same physial signifiane of stati defletion of the spring under a onstant load P o. Thus, if P o /k = x st, the solution beomes: x 1 = sin (ωt) (1.9) xst 1 ( ωω / n) Inluding the homogeneous solution ontaining the two integration onstants, the total solution of the problem beomes: xst x = C 1 sin (ω n t) + C os (ω n t) + sin (ωt) (1.30) 1 ( ωω / n) The first two terms represent the undamped free vibrations, the third term is the undamped fored vibration. Examining the impliations of this result, x/x st is a sine wave with an amplitude of 1/{1 (ω/ω n ) }. For values of ω/ω n > 1 the amplitudes are negative, and of ω/ω n < 1 the amplitudes are positive. When the amplitudes are negative, it only means that the amplitudes are positive but with a phase angle lag of 180. It also means that for ω/ω n < 1 fore and motion are in phase, and they are out of phase for ω/ω n > 1. For ω/ω n < 1 the mass is below the equilibrium position when the fore pushes downward; the mass is above the equilibrium position when the fore pushes downward for ω/ω n > 1. At ω/ω n = 1 the fored frequeny oinides with the natural frequeny, so the fore is pushing the mass at the right time in the right diretion. The amplitude then inreases indefinitely. In the ase of a pendulum, it is pushed slightly in the diretion of motion every time it reahes the end of a swing; a small fore is required to produe a large amplitude. This phenomenon is known as resonane, and the natural frequeny is alled the resonant frequeny. Fig. 1.13. Unbalaned Rotating Mahine on a Beam An important appliation of this theory is when the amplitude of the impressed fore varies with ω. Fig. 1.13 shows a beam on two supports arrying a rotating mahine with an unbalaned entrifugal fore of m 1 ω r, where m 1 is the unbalane mass and r is its distane from the enter line of the shaft. This fore an be resolved into a vertial omponent m 1 ω r sin (ωt) and a horizontal omponent m 1 ω r os (ωt). If it is assumed that the beam is rigid for

Vibration Theory Fundamentals 13 displaements in the horizontal diretion, and flexible for displaements in the vertial diretion, then it beomes a single degree of freedom system. Here the motor mass m and beam stiffness k = 48EI/l 3 are ated upon by the frequeny dependent vertial disturbing fore of amplitude m 1 ω r. Considering the motion of the motor and beam system, it is equivalent to motor mass inertia fore of mω a o ating on the beam. The solution for this problem an be found diretly from (1.8) by substituting mω a o for P o. Then: 8 7 C 6 5 o o y/a 4 3 1 A B 0 0 1 3 / n Fig. 1.14. Resonane Diagram for Rotating Mahine on a Beam Example mω ao/ k ( ωω / n ) y o = = a o 1 ( ωω / n ) 1 ( ωω / n ) yo ( ωω / n ) or = (1.31) ao 1 ( ωω / n ) If y o /a o is plotted as a funtion of ω/ω n, the result is shown in Fig. 1.14. At C there is resonane, and so spring extension is infinitely large in the absene of damping. Sine this result is not in agreement with observations, it is neessary to onsider damping. It shows the relative motion of a system in whih the end of the spring is subjeted to an alternating motion of onstant amplitude a o. The plot also shows the absolute motion of a system in whih the mass experienes a fore of variable amplitude mω a o. The ordinates of the three points A, B and C representing (1.31) an be physially explained as follows. At A the frequeny ω is nearly zero, so the top of the spring moves at a slow rate, the mass follows this motion and the spring does not streth, and so y o = 0. At B the motion of the top of the spring is very rapid, so the mass annot follow and stands still in spae. 1.8 FORCED VIBRATIONS WITH VISCOUS DAMPING The omplete equation of motion is: md x/dt + dx/dt + kx = P o sin (ωt) (1.3) The full solution onsists of the omplete solution of the equation with the right hand side set to zero and a partiular solution of the whole equation (1.3), or:

14 Reiproating Mahinery Dynamis m t x = e {C 1 sin (qt) + C os (qt)} partiular solution (1.33) Thus, only the partiular solution needs to be determined. Assuming a solution of the type x = x o sin (ωt ϕ), the partiular solution an be determined by finding x o and ϕ. Newton s law requires that the sum of all the fores be zero at all times, implying that the sum of the vertial and horizontal omponents must also be zero. From these onditions the unknowns x o and ϕ are solved: x o = R S T P / k ω 1 ω o U V W R + ω S U T VW ω n n (1.34) ω tan (ϕ) = ω n (1.35) 1 ( ω / ωn ) The expressions for the amplitude and the phase angles are in terms of ratios of frequeny and damping, where is the ritial damping. P 0 /k may be interpreted as the stati defletion of the spring due to P 0. These relations are plotted in Fig. 1.15. The amplitude diagram ontains a family of urves, one for eah value of damping. All urves lie below the one for zero damping, sine the amplitude of fored vibration is redued by damping. The maxima of the different urves do not our any more at ω/ω n = 1 but at a smaller frequeny. In fat, three different frequenies need to be distinguished, all of whih oinide for = 0: ω n = ( km / ) = undamped natural frequeny U VW k R q = S m T = damped natural frequeny m The frequeny of maximum fored amplitude, or the resonant frequeny. The phase angle diagram is also of interest. For no damping it was seen that the fore and displaement are in phase below resonane and 180 out of phase above resonane. So the phase angle shows a disontinuous jump at the resonane point. For damping values different from zero the other urves in the phase angle diagram are plotted. In general, damping tends to smooth out the sharpness of the undamped diagrams for both amplitude and phase. Energy relations involved in this proess also give a deeper understanding. For very slow motions ϕ = 0, and little work is done over a whole yle, or no mehanial energy is onverted into heat. Starting from the equilibrium position, the external fore moves through a ertain distane before reahing the extreme position, when it does work. But that work is used to store potential or elasti energy into the spring. During the next quarter yle the motion goes against the external fore and the spring gives up its stored energy. At slow speeds, then, the work of the external fore is transferred into elasti energy and nothing is onverted into heat. At resonant frequeny ϕ = 90, and work dissipated per yle is πp o x o. The external fore is equal and opposite to the damping fore in that ase, so the work is dissipated by the damper. The spring fore and the inertia fore are equal and opposite, and also in phase with displaement. Eah of these fores individually perform work during a quarter yle, but store the energy, returning it during the next quarter yle. The work is stored periodially as elasti energy in the spring and as kineti energy in the mass.

Vibration Theory Fundamentals 15 5 o stat X/S 4 3 / = 0 / = 0.15 / = 0.0 / = 0.50 1 / = 1 0 0 1 3 / n (a) 180 / = 0 Phase Angle, degrees 150 10 90 60 / = 0.15 / = 0.5 / = 1.0 / = 0.0 30 0 0 1 3 / n Fig. 1.15. (a) Amplitudes of Fored Vibrations, (b) Phase Angle Between Fore and Displaement as a Funtion of Frequeny for Various Damping Values Returning to the assertion pertaining to the partiular solution, the general solution onsists of the damped free vibration superposed on the fored vibration. After a short time the damped free vibration disappears and the fored vibration alone persists. For this reason the fored vibration may also be referred to as sustained vibration and the free vibration beomes transient vibration. The values of the onstants C 1 and C depend on the onditions at the start and an be alulated from these onditions using the analytial proess desribed earlier. (b)

16 Reiproating Mahinery Dynamis It is possible to desribe the whole proess by physial reasoning. For example, onsider a mass suspended by a spring and ated upon by an external harmoni fore of frequeny eight times as slow as the natural frequeny of the system. The mass is lamped while the external fore is ating. Suddenly the lamp is withdrawn. Assume the damping in the system is suh that the free vibration dereases by 10 perent for eah yle. Assume also that the lamp is released when the fored vibration has its maximum amplitude. From the initial onditions it follows that at the instant of the release the mass has no defletion and no veloity. The presribed fored vibration starts with x = x o and dx/dt = 0. These two onditions an only be satisfied by starting a free vibration with x = x o and dx/dt = 0. Then the ombined motion will start at zero with zero veloity. Fig. 1.16 (a) shows the free vibration, 1.16 (b) shows the fored vibration and 1.16 () the ombined motion. (a) Free Motion (b) Fored Motion () Total Motion Fig. 1.16. Starting Transient Motion of a Fored Vibration Note that the transient vibration disappears quikly and that the maximum amplitude at the start is nearly twie as great as the sustained final amplitude. If the differene between free and fored frequenies is small and damping is also small, the free and the fored vibration

Vibration Theory Fundamentals 17 vetors would retain nearly the same relative motion during one revolution, the inluded angle between them hanging only slightly. The vetors an then be added geometrially, and during one revolution of the two vetors the motion will pratially be a sine wave of frequeny ω 1 ω and amplitude the arithmeti sum of the two amplitudes. After a large number of yles, the relative position of the vetors varies, and the magnitude of the sum vetors also hanges. The resulting motion an then be desribed as an approximate sine wave with frequeny ω 1 and an amplitude varying slowly between the sum and the differene of the amplitudes of the free and fored vibrations. This phenomenon is known as beats. Note that the free vibrations will be eliminated after a while due to energy dissipation from the damping. In order to have sustained beats it is neessary to have two sustained, or fored, vibrations. 1.9 VIBRATION DUE TO ROTATING ECCENTRIC WEIGHT A lassi example of fored vibrations with visous damping is that of a rotating mahine of mass m with a rotating eentri mass m u mounted in a shaft and bearings, Fig. 1.17. The rotating eentri mass m u follows a irular path of radius e with respet to the bearings. Considering only the motion in the vertial diretion, the displaement of mass m u relative to m is x x 1 = e sin (ωt), where x and x 1 are the absolute displaements of m u and m and ω the angular speed of the mahine rotor. The differential equation of motion of the system is: m u d x/dt + m 1 d x/dt + dx 1 /dt + kx 1 = 0 (1.36) Differentiating (x x 1 ) with respet to time and substituting in (1.36), (m u + m 1 )d x/dt + dx 1 /dt + kx 1 = m u eω sint (ωt) (1.37) This equation is of the same form as (1.3): by substituting (m + m u ) for m and m u eω for P o the resulting displaement, veloity and aeleration of the rotating mahine mass m an be readily obtained. Note that the magnitude of eentri mass m u will be insignifiant ompared to the mahine mass m. e m u x m x 1 k Fig. 1.17. Fored Vibrations from Rotating Eentri Weight When the mahine is mounted on an elasti foundation, the motion of the foundation omes into play. As disussed earlier in setion 1.7, displaement fored on an otherwise unfored system is equivalent to a system where a fore is applied on that system. Therefore, a rotating mahine with an eentri mass in the rotor mounted on an elasti foundation behaves as if two distint foring funtions are applied on it, one due to the eentri mass whose frequeny

18 Reiproating Mahinery Dynamis oinides with the angular speed of the rotor, and the other due to motion from the foundation. The frequeny of the foundation motion will depend on the elasti stiffness of the foundation and the mass of the rotating mahine. Damping harateristis of the foundation will ontrol the amplitude of motion of the mahinery from both soures of exitation, sine the foundation will be in the diretion of the load path traveling from the rotor to the foundation. Relative motion between the rotor and mahine housing will be dependent on the material damping of the shaft and the bearings. As an example of the above disussion, onsider a entrifugal ompressor weighing 3,000 lb. mounted on a simply supported elasti beam of negligible weight, of the type shown in Fig. 1.13. The elasti beam sags 9 in. under the weight of the ompressor, and has a damping oeffiient of 8 lb. for a veloity of 1 in. per seond. The elasti beam is moved up and down at resonant speed with an amplitude of 1 in. Assume the enter of the ompressor to oinide with the enter of the beam along the beam s length. The natural frequeny of vibration is (g/ δ st ) 1/ = (386/9) 1/ = 6.6 rad./seond. At resonane the disturbing fore is {(ka o ) + (a o ω) } 1/. Here stiffness of the beam k = 3000/9 = 333 lb./in; a o = 1 in.; damping = 8 lb.-seond/in; and natural frequeny ω = ω n = 6.6 radians/seond. The disturbing fore is then equal to 380 lbs. The amplitude of motion of the ompressor then is disturbing fore divided by the produt of and ω, and is found to be.07 in. 1.10 ROTATING AND RECIPROCATING UNBALANCE An important group of vibration phenomena of pratial importane in reiproating mahines is torsional osillations in the rankshaft and in the shafting of driven mahinery. The effet is aused by a ombination of periodi aelerations of moving parts (piston, onneting rod, rank) and periodi variations in ylinder steam or gas pressure. In a single ylinder vertial reiproating mahine the piston exeutes an alternating motion, and in the proess experienes alternating vertial aelerations. When the piston aelerates downward there must be a downward fore ating on it, and this fore must have a reation pushing upward against the stationary parts of the engine. Thus, the alternating aeleration of the piston is oupled with an alternating fore on the ylinder frame, making itself felt as a vibration in the engine and its supports. In the lateral diretion perpendiular to both the rankshaft and the piston rod, moving parts are also being aelerated, namely the rank pin and part of the onneting rod. The fores that ause these aelerations must have equal and opposite reations on the frame of the mahine. This last effet is the horizontal unbalane. In the rankshaft main axis no inertia fores appear, sine all moving parts remain in planes perpendiular to the rankshaft. These various inertia fores an ause moments. Consider a two ylinder vertial mahine with the ranks set 180 apart. While one piston is aelerated downward the other piston is aelerated upward, and the two inertia fores form a ouple tending to rok the mahine about a lateral axis. Similarly, the horizontal or lateral inertia fores of the two ranks are equal and opposite, forming a ouple tending to rok the mahine about a vertial axis. A roking about the rankshaft axis an our even in a single ylinder mahine. If the piston is aelerated downward by a pull in the onneting rod, this pull exerises a torque about the rankshaft axis. Sine the piston aeleration is alternating, this inertia torque is also alternating. If the whole mahine mounted on weak springs is onsidered as the mehanial system, the external torque is zero, and so any inrease in lokwise angular momentum of the moving parts must be neutralized by an inrease in ounterlokwise angular momentum of the

Vibration Theory Fundamentals 19 stationery parts of the mahine. Taking merely the moving parts of the mahine as the mehanial system, an inrease in the lokwise angular momentum of the moving parts must be aused by a lokwise torque on these parts, whih has a ounterlokwise reation torque on the frame. If the mahine frame is mounted solidly on its foundation, this ounter torque is transmitted to the foundation and may ause trouble. But if the mahine is mounted on soft springs, reation to the foundation annot penetrate through these springs and the ounter torque is absorbed as an inertia torque of the frame and ylinder blok, and the mahine blok must vibrate. In Fig. 1.18 inertia fores are exluded by assuming the mahine to be turning at a slow onstant speed ω, or the moving parts have negligible mass. The pressure fore on the piston is P, and is varying with time, or with rank angle ωt. Besides ating on the piston, the gas pressure also presses upward against the ylinder head. Fore P is transmitted through the piston rod (fore 1) to the ross head. Negleting frition, it is held there in equilibrium by fores and 3. Of the fores ating on the ross head, 3 is a ompression in the onneting rod, and is a reation pressure on the frame to the right, of magnitude P tan (φ). Fore 3 of magnitude P/os (φ) is transmitted through the onneting rod to the rank pin (fore 4). Shifting this fore parallel to itself to O and adding a torque yp/os (φ), the driving torque of the gas pressure is obtained. Fore 5 is taken up by the main bearings at O and an be resolved into a vertial omponent 6 and a horizontal omponent 7. Triangles 1,, 3 and 5, 6, 7 are similar, so the magnitude of 6 is P and that of 7 is P tan (φ). Pressure P P 1 P tan 3 x l D A t r O B y 4 5 O 6 y 7 x Fig. 1.18. Gas Pressure Load on a Single Cylinder Mahine The fores transmitted to the stationery parts of the mahine are P upward on the ylinder head, P tan (φ) to the right on the ross head guide, P downward on the main bearings at O and P tan (φ) to the left on the main bearings at O. The total resultant fore on the frame

0 Reiproating Mahinery Dynamis is zero, but there is a resultant torque Px tan (φ). From Newton s law of ation and reation, this torque must be equal and opposite to the driving torque on the rankshaft, yp/os (φ). This is easily verified, beause y = x sin (φ). Thus, the gas pressure fores do not ause any resultant fores on the mahine, produing only a torque about the main rankshaft axis. In summary, no fores our along the longitudinal rankshaft axis of an mahine, while in the lateral and vertial diretions only inertia fores appear. About the vertial and lateral axes only inertia torques are found, and in the longitudinal diretion both an inertia torque and a ylinder gas pressure torque our. If the mahine is assumed to be built of elastially non-deformable members, the problem is one of stati balane only. The frame and other stationary parts generally fulfill this ondition, but the rankshaft an be twisted signifiantly, making torsional vibrations possible. The subjet may be divided into three ategories: (a) Inertia Balane: This refers to the balaning of the mahine against vertial and lateral fores, and against moments about vertial and lateral axes. (b) Torque Reation: Under this heading the effets of torque due to inertia and gas pressure ating on the stationary parts about the longitudinal axis are analyzed. () Torsional Vibrations of Crankshaft: Consequenes of the longitudinal torque on the moving parts of the reiproating mahine are dealt with. Effet is of partiular importane sine many rankshafts have enountered failures due to the longitudinal torque. 1.11 VIBRATION ISOLATION An unbalaned mahine has to be installed in a struture where the vibrations are undesirable. Prime examples of this situation are gas turbine engines attahed to an airplane wing and automotive engines. The problem is one of mounting the mahine in suh a manner that no vibrations will appear in the struture to whih it is attahed. Probably the most popular solution onsists of mounting the mahine on springs. In Fig. 1.19 the mahine is represented as a mass m with a fore P o sin (ωt) ating on it. In figure a it is solidly attahed to the struture, and in figure b it is mounted on springs with a ombined stiffness k. For simpliity the mahine is assumed to be rigid. Po sin ( t) Po sin ( t) m m k (a) (b) Fig. 1.19. Flexible Spring Mount for Preventing Vibration Transmission to Foundation The problem is one of finding the magnitude of the fore transmitted to the floor by the mahine. Sine only the springs are in ontat with the foundation, the only fore transmitted

Vibration Theory Fundamentals 1 is the spring fore, and that has the amplitude kx, damping being negleted here. If x o is the displaement of the mass and x st = P o /k is the stati displaement, then: x x o st xo kxo spring fore = = = P / k P impressed fore o o transmitted fore = = Transmissibility (1.38) impressed fore The ideal is to have this ratio zero, but pratial onsiderations aim to make it small. If the spring onstant k =, as in figure a, the natural or resonant frequeny is infinite. The operating frequeny is then small ompared to the natural frequeny, ω/ω n is nearly zero and the transmitted fore equals the impressed fore. Physially this is obvious, sine the rigid floor does not permit mass m to move, and all the fore P o is transmitted to the foundation. It is neessary to design the support springs so as to make the natural frequeny of the whole mahine low ompared with the disturbane frequeny. Thus, the springs should be soft. Other interesting observations an be made from (1.9). If ω is smaller than ω n = ( km) /, the springs make matters worse, sine the transmissibility is greater than unity. If the natural frequeny is one-fifth of the disturbing frequeny, the transmissibility is 1/4, a good ratio. So far the support has been onsidered without damping, whih is the ase in steel springs. If ork or rubber padding is used for the purpose, then damping is not negligible and must be aounted for. The system is then represented by Fig. 1.0. Now the displaement urve is not diretly proportional to the amplitude of the transmissibility urve. The fore is made up of spring fore kx o as well as the damping fore ωx 0. As shown earlier the two fores are out of phase by 90. Consequently, their vetor sum is the total transmitted fore. The amplitude x 0 is given by the following expression: Transmitted fore = P o R S T 1 R ω + S U T k VW U V W R + S U T VW ω ω 1 ω ω n n (1.39) Transmissibility is defined as the ratio of transmitted fore to impressed fore. Thus, damping is advantageous only in the region ω/ω n < 1.41, where spring mounting makes matters worse. For all values of ω/ω n where spring mounting helps, the presene of damping makes the transmissibility worse. However, this statement is not as important as it may appear at first sight. For one thing the effet of damping is not great, and an be easily offset by making the springs weaker. Seondly, operation at resonane point ω/ω n = 1 is never the intention, although it may sometimes unfortunately our, and then the presene of damping is highly desirable. Thus, in spite of the ditum, some damping in the springs is generally advantageous.

Reiproating Mahinery Dynamis Po sin ( t) x m k/ k/ Fig. 1.0. A Spring Support with Damping Pratial ases of isolation by means of springs our in many mahines. Their main appliation lies in apparatus that is inherently unbalaned, or as in the ase of most reiproating mahines, inherently has a non-uniform torque. The ase of internal ombustion engines used in automobiles is disussed in detail in the next setion. 1.1 APPLICATION TO AUTOMOBILE ENGINES Before disussing the vibration transmission of the automobile engine, a brief introdution to ylinder-pressure torque and its variation with rank angle, or time, will be useful. This topi will be disussed at length in later hapters. Internal ombustion engines operate in a yle that have four distint regimes. During the intake phase fresh air is suked into the ylinder through the inlet valves. In an engine operating on four-yles this phase ours when the piston is near 0 rank angle, lose to the ylinder head, and lasts till the rank angle reahes approximately 180. At this instant the piston reahes the lower dead enter. At the end of the stroke the inlet valves lose and the ompression yle begins, lasting from about 180 rank angle to 360 rank angle, when the piston is one again at the top dead enter. The firing phase of the yle then begins, with the spark ignition ourring just after the piston has reahed top dead enter. As the flame spreads throughout the ylinder and the rank turning, the gases expand and perform work on the piston by pushing it to the other end of the ylinder. At approximately 180 rank angle the exhaust valves open and the spent gases then exit the ylinder, at the end of whih the piston reahes one again to the top dead enter to ommene the next yle. In a four yle single ylinder engine, at the four dead enter positions during the two revolutions of a firing yle, the torque is zero. Thus, the average torque delivered by the ylinder is only a small fration of the maximum torque that ours during the firing period. The fat that the torque is so irregular onstitutes one of the primary disadvantages of the reiproating engine as ompared with a turbine where the torque urve as a funtion of time is a straight horizontal line. Aording to Newton s law, every ation is aompanied by an opposite reation, and due to the torque generated by the ylinder gas pressure on the rankshaft the engine blok would experiene a ounter rotating torque. With the engine mounted rigidly on the frame, these torque variations have reations on the automobile s frame, and total suh that the natural frequeny of vibration is appreiably lower than n/ times the running speed. Unoupling of dynami motion is aomplished by mounting the engine blok on two journals, fore and aft, supported in bearings attahed to the hassis, enabling the blok to rotate about an axis nearly parallel to the torque axis and passing through the enter of gravity, shown as axis AA in Fig. 1.1. With any other mount arrangement than the one desribed, the engine blok would

Vibration Theory Fundamentals 3 be free to rotate about axis AA. To prevent this, a antilever leaf spring B between the engine blok and the hassis is used, the stiffness of whih is seleted suh as to make the natural frequeny suffiiently low. Besides ylinder pressure torque the engine also experienes horizontal and vertial inertia fores, whih have to be reated at mount points A and B. For this reason both bearings A as well as the hassis end of the leaf spring B are embedded in rubber. In the atual onstrution axis AA is not exatly parallel to the torque axis. This is the right proedure, beause generally the torque axis is not the prinipal axis of inertia, and onsequently does not oinide with the orresponding axis of rotation. Any rigid body has three prinipal axes of inertia. For instane, take an elongated solid retangular steel bar, Fig. 1., and attah to it a weightless shaft passing through its enter of gravity, but not oiniding with one of its prinipal axes of symmetry. Appliation of a sudden torque to the assembly will ause it to aelerate. Multiplying the aeleration with the masses of elements in the bar gives inertia fores, and multiplying the fores by their distane from the axis of rotation form a torque, whih is equal and opposite to the impressed torque. Also these fores when multiplied by their distane from the axis perpendiular to the shaft have a torque about that axis. This will have reation at the two bearings; the right hand bearing experienes a fore out of the plane of the diagram, the left hand bearing fore being pushed into the plane. In the absene of the bearings, under the influene of the torque the bar would not rotate about the torque axis, sine fores are required at the bearings to make it do so. In general, then, a body under the influene of a torque will rotate about an axis not oiniding with the torque axis if that is not a prinipal axis. The axis about whih the automobile engine has to be suspended, therefore, should not be about the torque axis itself but the axis of rotation to whih the torque axis belongs. Only when the torque axis is a prinipal axis do the two oinide. Several other design shemes of spring-supported automobiles are available, most of whih are similar to that of Fig. 1.1. Some have one rubber support at the rear of the engine and two other rubber supports lose together at the same height in the front. These two supports are virtually a ombination of the single bearing A and the restoring spring B of Fig. 1.1. A A A a B l Fig. 1.1. Shemati Arrangement of Automobile Engine Mount

4 Reiproating Mahinery Dynamis Torque Fig. 1.. Rotation of Bar about an Axis Different from its Prinipal Axis of Inertia To illustrate the above disussion, onsider a four ylinder automobile engine weighing 400 lb. supported in the same manner as in Fig. 1.1. The radius of gyration of the engine about axis AA is 6 in., distane a is 18 in. and length l of the antilever is 4 in. The diameter of the rear wheels is 30 in., and in high gear the engine makes three revolutions per revolution of the rear wheels. The requirement is that the engine be in resonane at a speed orresponding to 3.5 miles per hour, or 61.6 in. per seond, in high gear. Wheel irumferene is 30π = 94.5 in. At the ritial speed the wheel makes 61.6/94.5 = 0.65 revolutions per seond, so the engine runs at 3.65 = 1.95 r.p.s. Now, the engine torque urve goes through a full yle for every firing, and sine there are two firings per revolution in a four ylinder, four yle engine, there will be 3.9 firings per seond. For resonane, then, the frequeny of the engine will be f n = 3.9 yles per seond, or, ω n = 4π (3.9) = 600 rad /se = k/i. Here k is defined as the torque required in the antilever per radian twist. The defletion at the end of the antilever for a twist of ϕ radians is 18ϕ in. If k 1 be the linear stiffness of the antilever in lbs/in., the spring fore is 18k 1 ϕ lbs ating on a moment arm of 18 in., so the torque is 18 18k 1 ϕ in.-lbs. Thus, k = 34k 1. Also, sine the moment of inertia I = 400 (6) /386 = 37.3 in.-lb.-se, then ω n = 600 = 34 k 1 /37.3 and k 1 = 37.3 600/34 = 69 lbs per in. Note that in this example if one ylinder was firing inadequately another periodiity is introdued in the torque urve for eah two revolutions of the engine. Sine the disturbane is four times as slow as the one disussed, it will be in resonane with the natural frequeny of the engine at a speed of 4 3.5 = 14 miles per hour. Still another possibility is when the ylinders are not firing at the proper instant in time, when the overall dynami harateristis will hange. 1.13 EXAMPLE PROBLEMS Example Problem 1.1: A fore 5 sin (π 60t) ats on a displaement of (1/10) sin (π 60t π/ 4). What is the work done during the first seond? (Units: inh, pound, seonds). Solution: The fore is π/4 radians (45 ) out of phase with the displaement; so it an be resolved into two omponents, eah of magnitude 5/ lb. in phase and 90 out of phase with displaement. The omponent in phase with the displaement does no work. Work done by the 90 out of phase omponent per yle is: πp o x o = π. (5/ ). (1/10) = 1.11 in-lbs. In the first seond there are 60 yles, so work performed is 60 1.11 = 66.6 in-lbs. Example Problem 1.: In the above example what is the work during the first 1/1000 seond? Solution: In the first 1/1000 seond there will be 60 1/1000 = 0.06 yle, or 0.06 360 = 1.6. For part of the yle, work is determined by integration. z z W = Pdx= P o sin (ωt). x o ω. os (ωt ϕ) dt

Vibration Theory Fundamentals 5 = 5(1/10). os (ωt 45 ) d(ωt) = {(1/4) os (45 ). sin (ωt) + (1/) sin (45 )[ωt/ (1/4) sin (ωt)]} 0 1.6 = (1/4) (0.707) (0.368) + [1.6 0.707/(4 57.3)] 0.707 0.685/8 = + 0.030 in-lb. Example Problem 1.3: In the system shown in Fig. 1.3 the mass weighs lb., spring stiffness is 5 lb./in.; l = 8 in. and a = b = 3 in. Also a damper is attahed at the mid point of the beam, where the spring is fastened. The damper onstant is 0.005 lb./in/se. What is the damper ritial damping? Solution: The undamped natural frequeny is ω n =. The equivalent spring onstant is k. a /l, or 5 (9/64) = 3.5 lb./in. Then: ω n = ( km / ) = ( 3.5 386/ ) = 6.0 radians/se. The ritial damping onstant is: = mω n = (/386) 6.0 = 0.7 lbs/in/se. k a z 1. 6 b 0 l ( km / ) sin ( ωt) m Fig. 1.3 Example Problem 1.4: Find the rate of deay of free vibration in the example problem # 1.3. Solution: The rate of deay is: x/x = δ = π (/ ) = π (0.005/0.7) = 0.116. Example Problem 1.5: A variable length antilever beam onsists of a strip of spring steel 0.0 in. wide and 0.00 in. thik. It arries a weight of 1/4 oz. at its free end. What should be the free length of the strip if it is required to have frequenies from 6 yles/se. to 60 yles/se? Solution: The spring onstant of a antilever beam is 3EI/l 3. The moment of inertia of the ross setion is: I = (1/1)bh 3 =. (.0) 3 /1 = 1.33 10 7 in 4. Bending onstant EI = 30 10 6 1.33 10 7 = 4.0 lbs-in, so spring onstant k = 1/l 3 lbs/ in. At the end, mass m = 1/(4 16 386) = 4.05 10 5 lbs-in 1 -se. Mass per inh length of strip µ = 0. 0.0 0.8/386 =.9 10 5 lbs-in 1 -se. So total mass = (4.05 + 0.9l) 10-5 lbs-in 1 -se.

6 Reiproating Mahinery Dynamis Maximum length frequeny is 6 yles/se, or, ω = (π 6) = 141 (radians/se). Natural frequeny: ω = k/m, or 141 = 1 10 5 /{(4.05 + 0.9l)l 3 } By a method of trial and error, length of the beam is: l = 5.33 in. Example Problem 1.6: A flywheel is represented approximately by a solid steel disk of 4.5 in. diameter and in. thikness. The shaft on whih it is mounted has a torsional stiffness of 9.5 in.-lbs./radian. Find the natural frequeny of vibration of the system. Solution: Weight of the flywheel is: (π/4) (4.5).0 0.8 = 8.9 lbs. and its moment of inertia is: I = 1 (mr ) = (8.9/386)(.5) = 0.0584 lbs-in-seond. The natural frequeny thus is: ω n = = 1.75 radians/seond. or f n = 1.75/π =.03 yles/seond. Example Problem 1.7: A strobosope is a devie for produing intermittent flashes of light with whih rapid vibratory motions an be made to appear to stand still, or to move very slowly. The flashes of light are of extremely short duration. When a vibrating objet illuminated with this light is adjusted to the same frequeny as the vibration, the objet will be seen in a ertain position; then it will be dark, and the objet is invisible while traveling through its yle. When it returns to the first position after one yle another flash of light ours. Thus the objet appears to stand still. Now onsider a point loated 4 in. from the axis of a mahine rotating at 10,000 rpm. If it is required to see the point with 61 ( BkI a / blurring ) = 4 B( ϕ9. 5/ 0 of 6. 059 less 50 ) than 1/3 4 50 in., what ϕ should = be the duration of the light flashes? R R.75.75 Solution: The point in question travels at: 10,000 π 4/60 = 4,189 in/se. Flash time duration multiplied by blur size will give the veloity of the point, whih is 4189 = t(1/3), so: t = 1/(4189 3) = 1/134,000 seond. Example Problem 1.8: A flywheel onsists of a heavy rim of weight W = 475 lbs and mean radius R =.75 in. attahed by four flexible prismati spokes (Fig. 1.4). If the hub is held fixed, find the period of free rotational vibration of the rim about its entral axis O. Neglet the mass of the spokes, and assume spoke bending stiffness B =,50 lbs/in. Solution: Consider the rim to have a small angle of rotation f from its equilibrium position as shown. Eah spoke behaves as a beam built in at the hub and onstrained to move with the rim at the other end. At the outer end of the spoke shear fore Q and bending moment M at, whih may be determined from formulas for beam stiffness: 1B 6 Bϕ 1 50 6 50ϕ Q = = R 3 R. 75 3. 75 =.93D 6.08ϕ M = = 6.08D 395.6ϕ If the rim is assumed to be rigid, the tangent to the elasti line at the outer end of eah spoke will be radial. Then defletion = Rφ, giving: Q = 6Bφ/R = 6.08φ and M = Bφ/R = 197.8φ

Vibration Theory Fundamentals 7 O R R O A M Q Fig. 1.4 Then the total moment on the rim will be: M t = 4QR 4M = 16Bφ/R = 158φ and the rotational spring onstant is: k r = = 158 in.-lbs/radian. M t B φ = 16 Sine the mass moment of inertia of the flywheel R rim is: I = WR /g = (475.75 )/386 = 636.9 in.-lbs/seond, then the period of free natural vibration of the rim is: 3 WR JR 636.9.75 τ = π = π = π = 3.986 seonds. 16gB 16B 16 50 Example Problem 1.9: Using Rayleigh s method for alulating the angular frequeny of the fundamental mode of vibration of the beam with two masses, shown in Fig. 1.5. Assume W 1 =W = W, flexural rigidity of the beam as EI and neglet beam mass. W 1 V y 1 y x y l/ l/ l/ Fig. 1.5

8 Reiproating Mahinery Dynamis Solution: From Rayleigh s method, for n masses on a beam, the formula for angular frequeny is given by the expression: p = g n j = 1 n j = 1 W y j W y j j j It will be assumed that during vibration the beam maintains a shape similar to the stati defletion urve due to the two weights ating in opposite diretions. The orresponding stati defletions are: y 1 = Wl Wl 5Wl + = 48EI 3EI 96EI 1 3 3 3 y = Wl Wl 5Wl + = 3EI 8EI 96EI Substituting these values into the expression for angular frequeny gives: 1 3 3 3 p = 19 /5W 3 EIg l. Example Problem 1.10: Figure 1.6 shows a mehanism used in the design of test mahines to produe vibrations, and is apable of produing sine and osine displaements. Crank r rotates at a onstant angular veloity ω r, and the projetion of point P upon the x or y axes moves with simple harmoni motion. If the radius is 3.75 in., rank speed is 500 rpm and θ r = ω r t = 65, determine the displaement, veloity and aeleration of point Q along the x axis. Solution: Displaement OQ = x = r(os θ r ) = 3.75 [os (65 )] = 1.585 in. y r P r O r Q x Fig. 1.6 Veloity: v = dx/dt = rω r sin (θ r ) = 3.75 (π 500/60) sin (65 ) = 177.95 in/seond.

Vibration Theory Fundamentals 9 Aeleration: a = rω r os (θ r ) = 3.75 (π 500/60) os (65 ) = 4344.87 in./seond. Example Problem 1.11: A variation of the slider and rank mehanism alls for inreasing the size of the rank pin so that it is larger than the shaft to whih it is attahed, and at the same time offsetting the enter of the rank pin from that of the shaft (Fig. 1.7). The enlarged rank pin, sometimes referred to as an eentri, replaes the rank in the original mehanism. Point A is the enter of the eentri and O the enter of the shaft. The motion with the equivalent rank length OA is idential with that of the slider and rank mehanism. Adequate lubriation between the eentri and the rod may pose a problem and limit the amount of transmitted power. If AB is given as L = 34.5 in., rank length R = 4.5 in. and operating speed is 360 rpm, determine the displaement, veloity and aeleration of the slider when rank angle θ = 75. Solution: Expressions for displaement, veloity and aeleration an be readily derived with the aid of figure 1.8. Note that the projetion of rank R along the vertial plane equals the projetion of L. Now, displaement: x = R + L R[os (θ)] L[sin (φ)] Eentri O A L NM O QP dx R = R ω sin ( θ) + sin ( θ) 3 dt L B 1 Fig. 1.7 L NM R = R[1 os (θ)] + L 1 1 L sin ( θ) Simplifiation of the expression under the radial sign an be obtained by using the first two terms in the binomial series, whih will give suffiiently aurate results. Then: x = R [1 os (θ)] + R sin (θ) L = 4.5[1 os (75 )] + 4.5 sin (75 )/( 34.5) = 3.391 in. Expressions for veloity and aeleration of the ram are obtained by differentiating displaement with respet to time. Assume rank angular veloity to be onstant. Veloity: V = O QP

30 Reiproating Mahinery Dynamis R L x L sin ( ) = R sin ( ) Aeleration: Fig. 1.8 = 4.5 (π 360/60) [ sin (75 ) + 4.5 sin (150 )/( 34.5)] = 159.71 in/seond. L NM A = d x R = Rω os ( θ) + os ( θ) dt L A = 4.5 {(π 360/60) [os (75 ) + 4.5 os (150 )/(34.5)]} = 918.81 in/seond. O QP REFERENCES AND BIBLIOGRAPHY Anwar, I., Computerized Time Transient Torsional Analysis of Power Trains, ASME Paper No. 89-DET-74, 1989. Choi, Y. S. and Noah, S. T., Fored Periodi Vibration of Unsymmetri Pieewise Linear Systems, J. of Sound and Vibration, 11(3):117-16,1988. Crede, C. E., Ruzika, J. E., Theory of Vibration Isolation, Shok & Vibration Handbook, MGraw-Hill, New York, 1988. Den Hartog, J. P., Mehanial Vibrations, Dover Publiations, In., New York, 1984. Deutshman, A. D., Mihels, W. J., Wilson, C. E., Mahine Design - Theory & Pratie, MaMillan Publishing Co., New York, 1989. Ehrih, F. F., Rotordynami Response in Nonlinear Anisotropi Mounting Systems, Pro. Of the 4th Intl. Conf. On Rotor Dynamis, IFTOMM, 1-6, Chiago, September 7-9, 1994. Ehrih, F. E., Handbook of Rotordynamis, MGraw-Hill, New York, 199. Eshelman, R. L., Torsional Vibrations in Mahine Systems, Vibration, 3():3, 1987. Gale, N. F., Widener, S. K., Sui, P. C., Zhang. H., Analytial Engine Design Methods: A Review, SAE Paper # 950806, 1995. Goldstein, H., Classial Mehanis, Addison-Wesley Publishing Co. In., Mass, 1987. Hildebrand, F. B., Advaned Calulus for Appliations, Prentie-Hall, New Jersey, 1983. Jeon, H., Tsuda, K., Theoretial Analysis of the Undamped, Coupled Torsional Axial Vibration of Marine Diesel Engine Shaftings, Bulletin of the M. E. S. J., Vol. 4, No. 7, P41, 1989. Law, B., Predition of Crankshaft and Flywheel Dynamis, Institute of Mehanial Engineers, Paper # C38/046, pages 487-498, Perkins Tehnology, 1989. Mabie, H. H., Reinholtz, C. F., Mehanism and Dynamis of Mahinery, John Wiley & Sons, 1987.

Vibration Theory Fundamentals 31 MLahlan, N. W., Ordinary Nonlinear Differential Equations, Oxford University Press, New York, 1956. Meirovith, L., Analytial Methods in Vibration, Mamillan Publishing Co., In., New York, 1984. Rao, S. S., Mehanial Vibration, Addison-Wesley Publishing Co., Reading, Mass., 1990. Rayleigh, The Theory of Sound, Volume # I, Mamillan & Co. Ltd., London, 1894. Roark, R. J., Young, W. C., Formulas for Stress and Strain, 5th ed., MGraw-Hill, New York, 1985. Robeson, G., Cosworth - The Searh for Power, Patrik Stephens Limited, London, 1991. Shroen, B. G., Die Dynamik der Verbrennungskraftmashine, Springer-Verlag, Vienna, 198. Shemeld, D.E., A History of Development in Rotordynamis, A Manufaturer s Perspetive, CP443, NASA, Washington, D. C., 1986. Timoshenko, S., Young, D. H., Weaver, W., Vibration Problems in Engineering, John Wiley & Sons, In., New York, 1984. Wahel, J. C., Tison, J. D., Vibrations in Reiproating Mahinery and Piping Systems, Proeedings of the 3rd Turbomahinery and Piping Systems,Texas A & M University, College Station, TX, 1994. Wehrli, V. C., Uber Kritishe Drezahlen unter Pulsierender Torsion, Ingenieur Arh,33:73-84, 1993.