Section A-A Elastic Member k, ca Inertia Member-*^' DAMPED VIBRATION ABSORBER SIMPSON MANUFACTURING CO. LITCHFIELD, MICHIGAN

ENGINEERING RESEARCH INSTITUTE THE UNIVERSITY OF MICHIGAN ANN ARBOR Final Report DESIGN OF DAMPED VIBRATION ABSORBERS FOR AUTOMOTIVE ENGINES Frank L. Schwartz William T. Crothers Project 2526 SIMPSON MANUFACTURING COMPANY LITCHFIELD, MICHIGAN April 1957

The University of Michigan * Engineering Research Institute TABLE OF CONTENTS Page ABSTRACT iii OBJECTIVE iii NOMENCLATURE iv INTRODUCTION 1 TORSIONAL SPRING CONSTANT 1 ANGULAR MOMENT OF INERTIA 2 DAMPING CONSTANT 3 NATURAL FREQUENCY OF A SINGLE-MASS SYSTEM WITHOUT DAMPING 5 NATURAL FREQUENCY OF A TWO-MASS SYSTEM WITHOUT DAMPING 6 MULTIPLE MASS SYSTEMS AND FORCED VIBRATION 6 FORCED VIBRATION OF A SINGLE- AND TWO-MASS SYSTEM WITHOUT DAMPING 7 FORCED VIBRATION OF A THREE-MASS SYSTEM WITH UNDAMPED ABSORBER 8 FORCED VIBRATION AND DAMPED ABSORBERS 10 DESIGN OF DAMPED VIBRATION ABSORBERS 17 I. INITIAL DESIGN 17 Ao Torsiograph method 17 B. Harmonic torque method 22 II. ABSORBER COMPONENTS 23 A. Inertia component 23 B. Elastic component 24 III. DESIGN MODIFICATION 52 CONCLUSIONS 35 RECOMMENDATIONS 35 BIBLIOGRAPHY 36 APPENDIX 357 ii

The University of Michigan * Engineering Research Institute ABSTRACT This report deals with the design of damped vibration absorbers used on automotive engines. Vibration theory, which we have simplified and combined with easily obtained test data, gives a means of rapidly analyzing vibration. This is useful in initiating a new absorber design or in modifying an existing absorber for greater effectiveness. A brief summary of torsional vibration theory and its application to simple systems is contained in the introduction, and a development of their mathematical solutions is contained in the Appendix. The Design Sections include the development of an initial design using a torsiograph vibration curve and the relationship of absorber components to fit the design problem. This is accompanied by an illustrative example. A simple method for the modification of an absorber is also offered for adjustments after preliminary testing or modification of an existing absorber. OBJECTIVE The objective of this report is to develop and simplify a mathematical approach to the problem of absorber design, and to correlate the results of calculated design with those of actual tests. iii

The University of Michigan * Engineering Research Institute NOMENCLATURE Symbol Quantity Dimensions c Damping constant in -lb-sec Cc Critical damping constant 2Jyn in-lb-sec c/cc Damping factor Dimensionless Dx Diameter of mass x in. dx Diameter of shaft x in, G Shear modulus lb/in. g Acceleration of gravity = 386 in/sec2 in/sec2 Jx Moment of inertia of mass x lb-in-sec2 Jarea Polar moment of inertia of area in.4 kx Spring Constant (in rotation) of member x lb-in/rad L Length in. M Moment of a force lb-in. Mr Absorber-engine amplitude ratio (max.) Dimensionless mx Mass (W/g) of member x lb R Radius of gyration in. r Radius in. S Shear displacement in. To Amplitude of torque vector lb-in. Te Torque in engine shaft at Ge lb-in. t Time sec V Velocity (angular) rad/sec Wx Weight of mass x lb w Width in. 9x Maximum rotational displacement or maximum amplitude of mass x rad or degrees gst Maximum static rotational displacement (cD=O) rad or degrees 9ae Maximum vibration amplitude between absorber and engine rad 0x Instantaneous amplitude during harmonic motion of mass x rad Inertia ratio Dimensionless T Period of mass x sec 3014159 Dimensionless cou Circular frequency of the force rad sec mOn Circular frequency (natural) rad sec1 Oxlah Natural circular frequency of absorber rad sec n03a Natural circular frequency of engine (first mode) rad sec-l iv

The University of Michigan * Engineering Research Institute INTRODUCTION The objectives of this investigation are: 1) to develop a mathematical approach to engineering problems involved in the design of damped vibration absorbers and 2) to determine a correlation between calculated and actual performance of absorbers in operation. Mechanical systems, consisting of masses and elastic members, will vibrate 1) if one or more of the masses is displaced and 2) if the displaced mass causes a deformation of an elastic member. Such a vibration can take place if a mass is initially displaced and released or if a pulsating external force is applied to the system in some manner. An automobile engine, with its pistons, connecting rods, crankshaft, flywheel, camshaft, and various accessory moving parts, is a typical example of displaced masses and elastic memberso Such a system presents complex vibration problems. In many cases small vibrations in machinery are not harmfulo When certain conditions prevail during mechanical operation, the vibrations may become excessive and cause serious stress, wear, and noise. Such conditions are most likely to exist when the applied external force is pulsating with the natural frequency of the machine or one of its parts. A situation of this kind is called resonanceo To develop some of these concepts in the light of absorber design a few simple torsional systems and their evaluation will be discussed. In developing equations involving the frequency of vibration of elastic members in torsion, several physical constants are required; the torsional spring constant the angular moment of inertia, and the damping constant. Methods of determining these constants will be outlined. TORSIONAL SPRING CONSTANT The spring constant of a shaft in torsion is defined as the in-lb of torque required to twist the shaft one radian. This may be determined by experiment or it may be calculated from the dimensions and material properties of the shaft. Experimental methodO-Secure one end of the shaft against rotation. To the other end attach a lever arm of known length and support the shaft as near the arm as possible to prevent axial distortion under loado Apply a known load to the lever armo Measure the twist or angular displacement of the shafto Figure 1 is a test procedure suitable for shafts. 1

The University of Michigan * Engineering Research Institute 2 L1 L2' /- M: Wr r r load Fig0 1. Calculation method,,-The following formula for the angular displaceM =. ment of a shaft in torsion may be used to calculate the spring constants M GJarea L By definition, the spring constant GJarea k = LL The polar moment of inertia of the cross-sectional area has the dimensions of inch40 If the length of the shaft L is in inches, and the modulus of rigidity G is in lb/ino2 (11,500,000 for steel), k will have the dimensions of in-lb/ radian0 If the shaft is round, tD4 Jarea = For other shapes Jarea can be found in handbooks. For a complicated shaft tion (ine., bearing journals, webs, etc.) and adding these deflections under a given load. ANGULAR MOMENT OF INERTIA Angular moment of inertia is a measure of the resistance that a body offers to any change in its angular velocityo This may be calmculated for

The University of Michigan * Engineering Research Institute simple, elemental geometric shapes. For more complicated bodies the moment'of inertia may be calculated if the object can be divided into simple geometric shapes, The moment of inertia of any shape, simple or complex, may be determined experimentally. Experimental method.-If a mass is suspended on a long thin vertical rod and allowed to oscillate as a torsional pendulum, the torsional spring constant of the rod will be given by 4 2Jmass k = If a disc of uniform thickness is mounted on the lower or free end of the rod and allowed to oscillate (the other end of the rod being fixed), we have 4iaJD k - TD2 JD for the disc can be calculated from Wr2 D =2g JD is measured for the rod and disc combination. Then if a mass M, having a moment of inertia JM, is placed on or under the disc, concentric with the rod, we have k 4 2(J + JM) 4_ 42J T.2 ~ T 2 D+M Di or JD + JM TD + M2 JD TD from which JM can be calculated, Calculation method.-The angular moment of inertia is defined by the formula " WR2 J = I I g For a complex body, reduce it to a series of elemental geometric partso From the dimensional data, calculate the radius of gyration R about the center of rotation for each of the parts. Determine WR2 for each and add them together to arrive at a total. The moment of inertia for the body will then be this total divided by g. Where the dimensions are in inches and g is 386 in/sec2, J will be in terms of in-lb-sec2. DAMPING CONSTANT The frictional resistances, internal molecular and external, which oppose vibration, are termed damping forces and for most evaluations are -

The University of Michigan * Engineering Research Institute closely approximated by values proportional to the velocity of the masses in vibration. The effects of a damping force proportional to velocity and of a restoring force proportional to displacement are to reduce both the natural frequency of vibration and the amplitudes of successive cycles. Since the amplitudes and velocities of natural torsional vibrations are small, their effect on the frequency of vibration is small. The damping force may be so large as to prevent an oscillation, in which case the damping constant is said to be larger than the critical damping constant. If the damping constant is less than critical, a free vibration, once excited, will continue to vibratewith a decreasing amplitude until it comes to rest. The rate of decay of the vibration will depend on the value of the damping constant. The damping constant is a property of the material in the system and the medium in which the oscillation takes place. It cannot be calculated but may be measured experimentally. With a damping constant c/cc < 1.0, free vibration will occur. A means of measuring c would be to measure the rate of decay of the vibration due to c (Fig. 2). The damped system is displaced and allowed to vibrate. A measure of the change in the amplitude for a number of cycles will determine c in the formula Xn CT 6 = log Xn = * I Fig. 2 The motion is a sine wave with a constant period of T seconds. The inertia of the mass may be calculated or measured. With a measure of the amplitude of two successive cycles Xnl and Xn, the damping c may be calculated For more accurate measurement, a number of cycles may be taken, using *The inertia J used for torsional motion is replaced by mass (m) for translator motion. 4

The University of Michigan * Engineering Research Institute - 1 log Xn T * n - m ge Xn 2J If the decay is rapid, the amplitude measurements should be modified to measure the amplitude at the point of tangency of the exponential envelope drawn on the decay curve. The units of c are in-lb-sec. A second method of obtaining the damping constant is explained in the Design Section of this report. NATURAL FREQUENCY OF A SINGLE-MASS SYSTEM WITHOUT DAMPING The simplest system that supports vibration is that of one mass and one torsional spring as shown in Fig, 3. _ _ _ _ _ _ )L ka ma Fig. 35 The elastic member or shaft is fastened to a rigid support on one end and a mass on the other. The amplitude of movement in rotation about the x axis at any given instant is 0 = 9 cos Tht, which is derived along with an illustrative problem in Part I of the Appendix. The equation states that the instantaneous amplitude 0 is dependent on the initial displacement 9 and varies in harmonic fashion, depending on the natural frequency wna where oa -Ja For clarity, oa may be changed from radians per second to cycles per second by dividing by 2ito Increasing the stiffness of the elastic member or decreasing the moment of inertia will increase the natural frequency. A decrease in stiffness or an increase in moment of inertia will lower the natural frequencyo *The inertia J used for torsional motion is replaced by mass (m) for translator motion, 5

The University of Michigan * Engineering Research Institute NATURAL FREQUENCY OF A TWO-MASS SYSTEM WITHOUT DAMPING A vibration of the same type illustrated in the single-mass system occurs in a free body consisting of two masses connected by an elastic member. Assume in the following figure (Fig. 4) that the system is free to rotate and that the shaft is supported by some means on frictionless bearings. This system results in a single natural frequency, as before, with the masses always moving 180~ out of phase with each other. The natural frequency is,l __; ~~~now dependent upon both masses k1 on = _. + kl J1 J2 ml (see Part II of the Appendix), If m2 is made a Figo 4. rigid support, J2 approaches infinity and the system becomes a single-mass system as shown in Fig. 3. The amplitudes of the masses vary inversely as the moments of inertia, 91 J2 8;2 j. the negative sign indicating the phase difference of 180~o Since the masses are moving opposite to each other at all times and connected to the same shaft twisting it as they move, a fixed point must be established in the shaft where the opposing twisting motions meet. Here no vibration occurs. This is the node point of the system. An illustration of this system may be found in Part II of the Appendix. This example is analogous to an engine if the engine components are reduced to a simple form of two masses and a single shaft, ioeo, let one mass represent the crankshaft and associated parts and the other mass represent the flywheel. MULTIPLE MASS SYSTEMS AND FORCED VIBRATION An absorber without damping added to the front of a crankshaft is equivalent to adding an additional mass and an elastic member to Fig. 4 of the previous discussion. This results in the three-mass system shown in Fig. 5. The method of solution is shown in Part III of the Appendix. The solution is now two frequencies, resulting from two modes of vibration. The first mode has two adjacent masses deflecting in the same direction or phase and one end mass out of phase with the movement of the other two. The second mode has the two end masses in phase and the center mass opposed. The first mode results in a node point in one of the shafts in a position which is de_________________________________ 6 ___________

The University of Michigan * Engineering Research Institute termined by the relationship between the masses and the springs in the system. The. L-I-L-I-|second mode has two nodes, one in each shaft, ki l -......... ||the location again depending upon the rela_ A jkt tionship of the masses and the springs in the ~M~~a -~m~2 n system. ml It should be noted that a system has a number of natural frequencies or modes equal Figo 5, to one less than the number of masses in the system. The combination shown in Fig. 6 has six natural frequencies in torsiono The first mode would have one node (with the larger mass m7 on the right, the node will be to the right of the center of the system)o As the modes increase, the nodes also increase in number up to the sixth mode which has a node in every shafto Considering such a system to represent an engine, there are six natural frequencies which could be driven into resonance and which could cause difficulty. The engine speed determines the external forced frequencies to produce resonance. _n an undamped system, any force kl k2 k3 k k5 alternating at a frequency which coincides with or is a submultiple BE f2 n RU4 "5 m of a natural resonance will destroy the system. In a practical sense, Fig. 6. m7 however, all systems have some form of damping. Damping may be due to hysteresis losses in the metal, bearing friction, or air damping. Each tends to decay or reduce the amplitude of a resonating part unless an external force is applied to replace the energy losses of damping. FORCED VIBRATION OF A SINGLE- AND TWO-MASS SYSTEM WITHOUT DAMPING From a study of the previous examples, it is evident that the effect of an external force must be considered. When a system such as Fig. 5 is driven by a periodic external force, forced vibrations occur having the frequency of the external force and an amplitude depending on the intensity of the external force and the relation of the external force frequency to the natural frequency. The relation of the frequency and amplitude is given by the equation T, --------------------- 7 ------------

The University of Michigan * Engineering Research Institute (The derivation for the single-mass system can be found in Part IV of the Appendix.) An illustration of amplitude variations resulting from change in forced frequency is shown in Fig. 7. Qst is the condition where the deflection ~ is caused by To if To were a steady torque. At L/Dna = 1, the curve shows a discontinuity at resonance, indicating the amplitude will increase until failure takes place or some limiting action is taken to restrict the movement. 1 2 3 4 The solid lines represent ly4 5; t I I ~I *the actual movement of the mass and indicate that below resonance the.i mass is in phase or moving with the 3 | force. Above resonance the mass is 2 V moving against the force and is / < \shown below the line. The dotted |y |\ j line is merely the lower portion of — ) l ""-__- the curve which is actually negative GS|T i but is plotted positively according O I — ~-I --— to common practice. -1 i / ^lYa The two-mass system is /If~~ ~much the same as the single-mass system, resulting in a resonant condition where the amplitudes of the masses approach infinite values 18CP -3" out of phase with each other. _^ ta FORCED VIBRATION OF A THREE-MASS SYSTEM WITH UNDAMPED ABSORBER Fig. 7. This three-mass system has two points of resonance where the forced frequency matches each of the two natural modes. Figure 8 shows the relationship between the absorber and the center mass with the third mass rigidly fixed. Two discontinuities appear in the curves; one at the point where the forced frequency equals the natural frequency of the first mode of the three-mass system, and one at a point where the forced frequency equals the natural frequency of the second mode. Also note in Fig. 8 that there is a point where the value of c/ana = 1 which results in zero amplitude for the center mass. Therefore, when the forced frequency equals the absorber natural frequency (the natural frequency of mass a and shaft a), there i no vibration in the mlm2 system. This is the theory of the dynamic absorber. The values given in the example in the Appendix for the single-mass spring system were chosen to be resonant at the same frequency as the natural resonance frequency of the two-mass system. Thus, when these were joined together to form Fig. 5, ma and ka became the absorber, "tuned" to the resonant condition of the two-mass system. The resonant point where the forced frequency pre8

The University of Michigan * Engineering Research Institute 5.1 Ot0 o t center mass O ma @man ea0... absorber 1st Mode ) 2nd Mode:1 Forced vibration near the 1st mode (co> (na) 4 Forced vibration — [ T ---- - near the 2nd m ode /nd Fig. 8. (W) >a)) ----------------- 9 ----------------

The University of Michigan * Engineering Research Institute viously drove the two-mass system into infinite amplitude now results in zero amplitude of the two masses and a finite amplitude in the absorber. The finite amplitude of the absorber depends only on the spring value of the absorber and the external force. Since the other two masses of the original system are no longer responding to the external frequency, the absorber will "absorb" the total force by a deflection of its spring such that an equal and opposite force counteracts the external force at all times. FORCED VIBRATION AND DAMPED ABSORBERS If a machine has a constant or nearly constant speed, the absorber may be "tuned" to the disturbing frequency and effectively reduce the vibration. However, in an automobile engine with its large range of speeds, difficulty is encountered immediately at two resonant points in the absorber modifie system, whereas in the original system, without absorber, only one resonant point may have been encountered. The solution to the new problem is dampingo With the introduction of damping we must now consider several modifications of the concepts we have described. The most noticeable change with damping present in a system is that the infinite points, where the forced frequency equals the natural frequency, now become finite and can be controlled to a great extent by the amount of damping in the systemo A second change is a shifting of the resonant frequency. This effect is much less apparent and often neglected. Figure 9 shows the effect of an additional mass (dampea ab- I sorber) added to a two-mass system with the end mass a fixed body. The outer pairs of dotted lines represent the amplitude of motion of the center mass ml similar to that shown in the upper sketch of Fig. 8, excepting that the negative amplitudes of Fig. 8 are made positive for simplification. These curves are without damping (c/cc = 0)o The inner pair of dotted lines represent infinite damping (c/cc -+ ). Under these circumstances, the absorber mass ma acts as a part of the center mass ml, and k, On = ja +j, which is somewhat lower in natural frequency than the original system. Two other curves are shown representing absorbers tuned to the natural frequency of the two-mass system but with different damping values, The double-peaked curve has less damping than the single-peaked curve in the center. Depending on the amount of damping, the curve may be varied within the extremes of no damping (outside dotted curves) to infinite damping (center dotted curve). Note that all curves intersect at points P and Q regardless of the damping in the system. For optimum absorber effect, the points P and Q on the curves can be adjusted to equal heights by changing the tuning of the ab---------------------- 1 0

The University of Michigan * Engineering Research Institute //1 H /II// I__ / I I''''' Io I I I I i 1~ I J-1O 011' 3 og d0 H\ ~ ~ ~ ~ ~~~~~~~e H LC) II L,,X^ 3\3 (D" "1, ^ ^ C "0 ^ "^ ^ ^~~~~~~~~~~~~~~~~ 0|0 ^ ^ ~~~~~~~~~~~~~~~~~~~~~~~I ----------— _, --- w~~~~~~~~~~~~~,, _ I,,0i,~~~~\,~ c~ tO...:t C)~~~\ r.I~~~~~~~\ ~lg~~~~~~~\ --— ^-~~~~~~~~~~~~~~~~l'

The University of Michigan * Engineering Research Institute sorber. For a given absorber inertia, the two outside dotted curves (c/cc = 0) are fixed relative to each other. By tuning, these can be shifted relative to the center curve (c/cc + o). An increase in absorber inertia will separate the two outside curves, thus separating P and Q. With proper tuning this has the effect of actually lowering P and Q and of reducing the resulting maximum vibration at those points. To find the optimum tuning point for an absorber, the formula is: %na = +1) where Dna = natural frequency of the absorber LL == natural frequency of the original system mikl = inertia ratio Ja/J1 Control of the amount of damping in the system can limit the height of the curve which is a measure of the amplitude of the motion. Damping which produces a maximum at point P or Q, with P and Q equal (see Fig. 10), will give the best possible absorber for a given inertia.* In the figure, P and Q are shown of equal height and a minimum of vibration is obtained throughout the resonant rangeO With modifications, the methods used to determine absorber design for a fixed torsional spring and mass system can be used to approximate multimass conditions. The more complex systems must be reduced to a resultant or equivalent inertia which will behave in a way similar to the original system' The value of the effective inertia is determined by the inertias and the amplitudes involved. The inertia of each mass and its amplitude squared is calculated and the sum of these is obtained (ZJG2). With the two-mass system of Fig. 4, the effective inertia ZJ(2 would be J101 + J292 and the inertia ratio, modified to Ja/_JG2 when ma is added (Fig. 5)o Considering the absorber design for engines, the inertia ratio is the absorber inertia Ja divided by the equivalent engine inertia ZJ22o** The equivalent engine inertia is a breakdown of the complicated spring-mass system to a system of the type shown in Fig. 11 The relationship of the amplitudes of the various masses is shown in the lower portion of this figure, If the end mass is assigned a vibration amplitude of lo0 unit, the related movements of the other masses may be calculated. The effective inertia is then the sum of each mass times its relative amplitude squared of ZjG2. For ideal representation an equivalent system should have kinetic and potential energies equal to the original system. Engine resonance, shown in Fig. 12a,may be determined by *Discussed fully in Den Hartog, Mechanical Vibrations. **Methods for obtaining these figures are explained in Wilson, Practical Solution of Torsional Vibration Problems.. 12

The University of Michigan * Engineering Research Institute..t O0 0 C~l \| \ \ 3 H 0(0 \ VII II 0 0'O-' V5:]3

The University of Michigan * Engineering Research Institute Ml m2 m3 m4m5 60 Amplitude I -.*20 O e X j I I node 1st Mode Engine Vibration Fig. 11. calculation or by test runs with measurements of engine vibration without an absorber. Several points within the speed range will show vibration amplitudes, the maximum usually being at an engine speed which excites the fundamental mode of the engine and produces a node point on the crankshaft near the flywheel0 In an automobile power train, it is possible for the engine, drive shaft, axles, and wheels to constitute a system which, in vibration, causes the engine to act much like a single mass. The predominant use of fluid-type couplings eliminates this condition to a large extent by providing effective isolation. We therefore will consider only the first mode within the engine itself A mode may be excited at any engine speed where the enginets pulsatin torque coincides with or is a submultiple of the mode. The relationship between engine revolution and the cycle of vibration is known as the order,* For example, the engine might have a vibration frequency of the fifth order at 4000 rpm giving a resonant frequency of 5 x 4000 or 20,000 cpm or 333 cps. This same 4000 rpm resonant point may be excited at other speeds by a different ordero For instance, the 333 cps or 20,000 cpm could be excited at the fourth order at 5000 rpm or at the sixth order at 3333 rpmo These excitation points depend on engine factors such as two- or four-cycle design, firing order, number of cylinders, and unbalance of moving partsu *An order is the number of cycles of vibration per revolution of the shafto 14

The University of Michigan * Engineering Research Institute a i No absorber b With absorber tuned too low c With absorber tuned correctly d With absorber tuned too hih Figo 12o 15

The University of Michigan * Engineering Research Institute The absorber will be effective at any engine speed which produces the same basic vibrating frequency. At other frequencies the absorber is ineffective. Thus, in the example above, an absorber properly tuned at 4000 rpm will work equally well at 5000 rpm and at 3333 rpm. The tuning formula can be adapted to an engine using the effective inertia. absorber inertia effective engine inertia Tuning = - x 100% o 1 + A1 This result will give the percent the absorber must be detuned from engine resonance for optimum results with a given absorber inertia. As an example of proper tuning, an engine with an equivalent inertia of.25 in-lb-sec and with an absorber with an inertia of.05 in-lb-sec would result in an absorber detuning of 1 x lOO% = 83.53% 1+ 0.25 The engine vibration amplitude at resonance equipped with an absorber is an integral part of the absorber tuning. Tuning too low will result in a high point above the original resonance and tuning too high will give a similar high point below resonance. Proper tuning will result in equal amplitudes above and below the original resonant point as explained on page 12 (see Fig. 12). The effect of absorber inertia on the balanced resonant points appears in the following formula. 9 2 ^ = 1+ o Gst I The ratio of G/gst is a factor known as the dynamic magnifier and is the ratio of the resonant amplitude of either of the two resonant peaks to the deflection which would be caused by the pulsating torque if this torque were unvarying (a = 0). A greater absorber inertia results in a smaller 9/gst or more effec- I tive absorber but the rate of increase in effectiveness decreases with an increase in absorber inertiao Another important consideration in absorber inertia is the deflection in the-absorber spring (rubber movement in this case) when at resonanceo The smaller the absorber, the more flexible the rubber must be to maintain a given tuned frequency by the formula Dna = a 1 16

The University of Michigan * Engineering Research Institute The results of lower spring constants means a greater deflection must take place in the rubber to produce a torque to oppose the external vibration torque. DESIGN OF DAMPED VIBRATION ABSORBERS The design of an absorber may be developed from a mathematical approach, a trial-and-error approach, or by methods which may include both. A successful design developed purely from the mathematics of vibration into a finished product approaches the impossible. Assumptions must be made to provide mathematical equations which are solvable. Even with these assumptions the equations become highly complex because of the inherent complexity of engines and the vibrations they represent. Much work has been done in the field of vibration to simplify this mathematical approach by reducing the engine components to simple equivalent systems which may be more easily solved| This generally involves a sacrifice in the accuracy of calculations and the final product must be tested to evaluate its worth. The methods presented under Initial Design are simple mathematical approaches correlated with test data. Design Modification gives some suggestions on methods for modifying an existing design for optimum resultso I. INITIAL DESIGN Ao Torsiograph method —The information necessary to design the absorber is as follows: 1. Equivalent engine inertia (with pulley if possible). 2, Torsiograph of the engine without an absorber (with pulley if possible). This information is used in the following steps: 1o Vibration amplitude of engine with absorbero a. Locate resonance rpm, order, and amplitude from torsiograph curve bo Locate a point 200 rpm or less from resonance on the same curve determining rpm and amplitude. co Solve for dynamic magnifier and Gsto do Step c may be checked by taking another point and solving againo eo Determine absorber inertia, or select absorber inertia and determine vibration amplitude. 17

The University of Michigan * Engineering Research Institute 2. Absorber tuning From these calculations we have these results: 1. oan (absorber frequency) a 2. Ja (absorber inertia) These are results using optimum damping as shown in Fig. 13..32 40 5 0.28 -.2X 410 3 204 6 8 16 18 20.16 10 54 0 I 0 2 4 6 8 10 12 14 16 18 20 Fig. 13. Torsiograph Method Example The basic equation for the dynamic magnifier is used and it is a modification of previous equations to include damping. t- i 1cone/ L where Ge = engine amplitude in degrees double amplitude Gst = engine static amplitude at resonance in degrees double amplitude (W = 0) c = force frequency, cps One = natural frequency, first mode, cps, of the engine c/cc = damping factor of the engine Since engine vibration is usually stated in degrees double amplitude (DoD.A.) and frequency in cps, they are used here in that manner. The fol18

The University of Michigan * Engineering Research Institute lowing is an example of this method of development with steps similar to those on page 17 Information 1. Equivalent engine inertia.32 lb-in-sec2 2e Torsiograph data rpm DoDOAG 44o00 45 4500,85 4600 1.40 4700 1,80 resonance, 4th order 4800 1.30 4900.40 Steps Step 1 a. Resonance 4700 rpm, 1.80 D.D.A. 4th order Step 1 b, Second point 4600 rpm, 1.40 D.D.Ao Step 1 Co Solving for the dynamic magnifier 4700 x 4 n = 60 = 313 cps (natural frequency) e 60 X = 47600 = 313 cps (force frequency) 60 At resonance w = oe and the equation reduces to ge 1 c = st st 2 or --, where e = 1.8~ D.D.A. 9st 2 cc 29e cc c - st = st (first equation) cc 2(1.8) 3.6 At the second point 8e = 104~ D.D.A. 46oo00 x 4 =..60 = 306,5 cps (force frequency) nae = 313 cps (natural frequency) 19

The University of Michigan * Engineering Research Institute GSt A[' (513 ) ]2+ F2-c- (o.2 Qe 2 1 &D J.o4i)2+ (1.958 2 e2 s2 x 1'+3.4c- (second equation) yst 17.64 x 104 + 3.834 (cSubstituting the first equation into the second, Oe2 2 st 17.64 x 10-4 + 3.854 f3 2 with Ge = 1,4~ D.D.A. e Ge2 = 1.96 2 1.96 = est Gs2 17.64 x 104 + 3.834 12.96 gst = 1.96 (17.64 x 10-4 +.296 Gst2) Gst2 = 34.58 x 10'4 +.580 Gst2.420 st2 = 34.58 x 10-4 34. 58 x lc74 Gst = 4- = 82.5 x 10-.420 Gst = 9.07 x 10'2 =.0907~ D.D.A.,e s= i.8 = 19.85 dynamic magnifier at resonance Gst.0907 Step 1 d. Checking at another point 20

The University of Michigan * Engineering Research Institute at resonance c/cc = Gst/3.6 as before (first equation) choosing 4800 rpm 1.3~ D.D.A. 4800 x 4 w>='60 = 320 cps Ge = 103~ D.D.A. st G 022 [2 Cf 2-2 eSt ~ (045)2 + (2.46 c \ c^c 8Q2 ]1 st 20.25 x 10-4 + 4.19 (f) Substituting the first equation into the second, Ge2 1 st2 20.25 x 104 + 4.19.t with Qe = 1.3~ D.D.A. ge2 1.69 6st2 1.69 =- |st 20.25 x 10'4 + 4.19 12.96 st2 = 1.69(20.25 x 10-4 +.323 Gst2) Gst2 = 34.2 x 10C4 +.546 st2.454 2 = 54.2 x 104 454 t 3 44 2 x 10 1-gt = - 2424 - 1 21

The University of Michigan * Engineering Research Institute est = 8 69 x 0-2 = o0869 D.DoA - = e <1.8 = 20,7 dynamic magnifier at resonance 9st.0869 Step 1 e. From these findings we can now approximate the effect of ar absorber on the system. Choosing an absorber with an inertia of.053* in-lb-sec2, the new dynamic magnifier is dynamic magnifier = ge 2 1= 066 with absorber Gst. 32 Ge = 1 + 2 3 61 with effective ~c.+- \.166 st 16engine inertia = 32 Ge max. = 3.61(.09) =.320 DoDoAo The maximum engine vibration with absorber is estimated to be.32~ D.D.A. assuming an approximate static deflection of.090 Do.D.A The vibration or harmonic torque is not known in this case. If it were, the static deflection could be calculated more accurately. The.09 is merely a value between the two static deflections originally calculated in Steps 1 c and 1 do Step 2, Tuning = x 100% or na = (tuning %) an 1 _ 1 8- -1 =.858 x lo00 =% 85.8% 0053 1 +.166 1+.32 85o8% of 313 cps = 269 cps - absorber tuning (ana) Bo Harmonic torque method. —The same problem may be attacked without torsiograph information if use can be made of the following to find Gst of the engine. It is advantageous to return to deflection in radians and to frequency in radians/sec. *This can be any value determined by the vibration reduction needed as shown here, the physical size as shown on page 24., heat problems (page 28), and production costs. 22

The University of Michigan * Engineering Research Institute _ ITnl Z 0 st - ne2 Jg2 where Gst = static deflection in radians (a = 0) ane = natural engine resonance in radians/sec ZJ92 = effective engine inertia without absorber ITnIZG = nth order harmonic torque acting on engine without absorber ITnIZG is also given as HTo (H = harmonic coefficient; To = mean torque), These values must be known to find Gst and may be calculated if information is available to do so. Extensive information on this subject is available in Wilson's two volumes.2 From this point the problem is identical to Step 1 e of the previous methods II. ABSORBER COMPONENTS From the previous calculations the absorber natural frequency (cora) and inertia (Ja) are knowne The damping factor (c/cc) can be determined from Fig. 13o These quantities are interrelated by the formula ka /c\ 2 tWna = Ja a This is the basic single-mass equation k modified to include the effects of damping. In the determination of frequency values, the damping has little effect and is neglected. However, in the calculations of amplitude of vibration, it is of great importance~ Determination of ka in the above formula completes the minimum requirements which must be met by the absorber. It is now necessary to find methods which will enable the components to be designed to these requirements and to perform satisfactorily in service. A. Inertia component. -The inertia value needed in the design has been determined. This value may be calculated or a sample measured by methods explained in the introduction, These methods are accurate and should give good results. Here is an application of the calculated method using the present type of absorber inertia member 23 -

f- The University of Michigan * Engineering Research Institute Calculation Moment of inertia of the absorber using cast iron (see Fig. 14)o WR2 - W(D -D2) 39.2 ID2 (Assume: D2 = 4.68 (depending on pulley -w I _ diameter and gap width) w =.5625 (depending on space, etc ) Fig; 14. Ja =.053 in-lb-sec2 (page 22) Unknown: D Known: WR2 = Jag g = 386 in/sec2 WR2 = 386 Ja WR2 = 386 (.053) WR2 = 20.45 Substituting in equation for WR2 56r 4 [1 -( Q.68)4] 204-5 = 59.2 4 D1 - 479.7 = 1425 D4 = 1904.7 2 D1 = 43.64 D1 = 6.60 in. A factor which may be important for some designs is the inertia effect of the spring component. All elastic members throughout the discussion have been considered as weightless. This will lead to error if the spring has significant inertia of its own when compared to the inertia member itself. Methods of calculation can be found in any vibration text, but in view of the difficulty of determining the spring constant of rubber products used in these designs they do not seem important enough to develop here. Bo Elastic component -The spring constant and the damping constant are much more difficult to determine. Testing the absorber on a frequency machine with a known inertia member will give an effective spring constant from

The University of Michigan * Engineering Research Institute Once this has been established, the rubber in the absorber may be used as a basis for whatever design changes it may be necessary to makeo Properties of Rubber as an Elastic Component The stress-strain properties of rubber are not linear. The shearstrain diagrams for rubber show a sharp increase in shear stress for an increase in shear strain as the rubber continues to be deformed. For small deflections, in the range with which we are dealing, most rubber will approximate linear characteristics. Here are a set of calculations which give an approximate indication of the rubber needed for a specific absorber design (see Fig. 15), A comparison of these calculations with those of an actual absorber under test should give a good indication of modifications which may be made on the design or the rubber composition. No consideration has been made for effects of temperature, rubber compression, or rubber overhang (extension of the rubber beyond the inertiamember width)o a = 269 (2i) = 1690 rad/sec (page 22) A^l ( },I j Ja =.053 in-lb-sec (page 22) 1 _ ____ ____ 11 =.166 (page 22) ka~ C ke ka le a To obtain the absorber sprin, aJ,~ I'~ II' constant ka, Ma -t aJ ~ ~ ~ ~ %a Fig. 15. ka = %a ka = (1690)2 (.053) ka = 1.51 x lO5 J-2iin/ad. Let a = ae max. = maximum relative amplitude value between absorber and shaft where 0ae = $a - e (the difference in the instantaneous values of the absorber and engine amplitudes) Also Mr = 8ae/.st 8st = engine static amplitude without absorber, radians Mr = absorber-engine amplitude ratio 25

The University of Michigan * Engineering Research Institute est =.09~ D.DA. 4st = 09 x -1. =.785 x 10-3 radians 2 18o with =.166 1 = 1 6.02.1l66 From Fig, 13, Mr = 12.0o ae = Mr st = 12.0 (.785 x 10l3) = 9o42 x 10-3 rado Ta = ka 9ae where Ta = maximum vibration torqu Ta = (1.51 x 105)(9o42 x 10-3) = 1424 in-lb We now have the information necessary to find the spring characteristics of the rubber. Using Figs. 16 and 17, we can solve for the shear moduluso \\ Inertia / \ \ nSwmbe / / \\ t-tl7 ~2R = D2 (Fig, 14), in. 0~ shaf / / / /;f \ or / w = Absorber width (Figo 14), \ Pul / / / / ino \ / / / =S = Shear deformation, inr T = Gap width, in. \ /ae = MaxO relative amplitude, radians r = Pulley radius, ino Figo 16o2 26

The University of Michigan * Engineering Research Institute 100.V 0.5 1.0 1.5 Shear Strain, S/T Figo 17v Ta 54 G (shear modulus) train, / (shear wemu = S = shear deformation = Q R Ta = 1424 in-lb T = o105 in, (gap) D2 = 4.68 ino R = D2/2 = 2.34 ino r = R - T = 2034 - o105 = 2.24 in. Gae = 9.42 x 10-3 radians w = o562 ino 1424 1 - 4X(2.24)2( o62)(9o42 x 10-3) S = (9o42 x 10-3)(2.54) =.022 ino G = 2753 b/ino2 with S/T = o022/o105 =.209 This shear modulus is correct only with small deflections because of the nonlinear characteristics of rubbero 27

The University of Michigan * Engineering Research Institute Figure 17 shows some typical curves. Curve (a) has the highest hardness (natural rubber, 60 durometer)o These curves are at constant temperature and repeat stress-strain values obtained at a low loading rateo These must now be compared with the necessary point we have obtained at high-compression and high-frequency valueso To obtain these values for comparison, a rubber with a known curve (such as those on Figo 17) should be employed in an absorber, and its natural frequency measuredo Working through these calculations, all factors are known and a point can be established on the graph (Figo 17) for comparison of the curve with no compression and low loading rateo Several other points may be found by changing the inertia member, diameters, and the effect of gap clearance and the resulting rubber compression, This will then give usable design data (ioeo, shear moduli) from available rubber engineering datao Dynamic Fatigue Life The life of rubber (in. service under absorber conditions) is for the most part determined by the fatigue characteristics of the rubbero Extensive data on natural rubber and its engineering properties including fatigue have been published by the UO S, Rubber Companyo Factors that affect rubber life significantly are: ao Static load bo Amplitude of vibration co Foreign elements (oil, gases) and the influence of light do Temperature In our application there is no static load in shearo The Uo So Rubber Company has conducted tests which conclusively prove that static preloading- is an important factor of fatigue life. Their results seem to indicate that the load, caused by compressing the rubber into the present gap spacing, will give a greater life than a nonloaded type of assembly0 Heat Problems The temperature of the rubber is determined by the outdoor air temperature, the heat of the engine, and the internal heat created within the rubber itself through damping (hysteresis)o The absorber design dictates the presence of damping and the resulting heat is unavoidableo There are several ways to relieve this problemo One is to have a large volume of rubber in which to store the heat generatedo This is an advantage when the absorber is in action periodically as it would be in driving at varying speedso A second way would be to have the heat dissipated from the rubber as guickly as possibleo This can be done by an adequate air flow and by large adjoining metal surfaces on the absorber and on the pulleyo Third, increasing the size of the absorber causes a decrease in the oscillation and the damping needed in the absorber for 28

The University of Michigan * Engineering Research Institute it to perform properly. Fourth, heat dissipation is determined by the conductivity of the rubber, and any alteration in the rubber compound which increases its conductivity will in turn increase its life. To obtain an idea. of the heat generation, we can continue with our design problem. From Fig. 13, with an inertia ratio i =.166 or 1/, = 6.02 the damping factor c/cc =.2, where cc = 2Jaan e, or e c =.2(2Jane) =.2(2)(.053)(313)(2n) c = 41.6 in-lb/rad/sec [unit-absorber damping torque (in-lb-sec), i.e., value of the damping torque in in-lb for each unit velocity (rad/sec) difference between absorber and the mass to which it is attached.] 9ae = 9.42 x 10-3 radians ne = 313 cps or 313(2ic) = 1970 rad/sec With Gae the maximum difference in amplitude between absorber and engine, then the instantaneous difference can be shown as 0 = Gae cos Co et differentiating do/dt = s9aeon inuo t; this is the instantaneous velocity difference on which the damping coefficient (c) acts. Instant. velocity = -sae e sin h et e e Maximum velocity = gae on The effective velocity =.707 maximum =.707 Gaene Veff =.707 (9.42 x 10'3)(1970) = 13.1 rad/sec The effective work from internal friction or hysteresis is c(Veff ) c Veff. = 41.6(13.1) = 545 inrlb/sec. Since 1 Btu = 9336 in-lbo 545 =.05838 Btu/sec. 93356 -' 1 Btu = 252 cal 29

The University of Michigan * Engineering Research Institute or 14.71 cal/sec. We have then a total of 14.71 cal/sec generated in the absorber. This is the same figure regardless of the volume of rubber or any other characteristic as long as this absorber is in action and is properly damped. Improper damping or tuning will reduce this figure, but the result is increased engine vibration. It is imperative that the rubber be able to withstand the temperature build-up that will result from this heat. This, along with conductivity, is one of the major properties an elastic product must possess, and any investigation of elastic products should be conducted with this in mind. Specific recommendations in regard to heat dissipation other than conductivity within the heat generator (rubber) might include increasing the contact surfaces (i.e., absorber surface and pulley surface), and to reduce the temperature of the materials the heat is being dissipated into. If ventilation is a problem, it may be possible to provide air movement by some flanges in the pulley on which the absorber is mounted or an additional stamped piece added. The casting may be modified at the three spokes so that they are cantered to perform as a fan. A second method might be to add a thin stamping or to vane the casting on the pulley which would force large quantities of air through the present drilled holes which lead directly to the inner rubber surface. A modified washer could also fit on the crankshaft in front of the pulley which is bladed to act as a centrifugal fan. Another method, if the production problems could be solved, would be to have a metal cylinder imbedded into the rubber and extending out of the rubber as a fin for the transfer of heat to the air. If this cylinder were stamped from sheet metal and used as the knife edge for assembling purposes, and then perhaps the exposed fin cut and bent similar to a turbine blade, conductivity could be greatly increased at very low cost. A combination of this last suggestion along with casting vanes or scoops on the front of the pulley at the holes would provide air cooling which may be adequate to solve the problem'. With surrounding ambient temperatures in the vicinity of 150~F, and the narrow limit of temperature differential above this because of the maximumtemperature rubber limits, efficient cooling must be realized. Comparison Tests for Rubber Because of the many available rubber compounds and the difficulty of determining their usefulness, some method is needed to indicate their possible behavior as absorber components. Some simple and rapid methods are available to compare characteristics of an unknown compound with a known rubber. Both damping constants and spring constants can be compared by the methods shown in Fig. 18. 50

The University of Michigan * Engineering Research Institute 5; > ~ Heavy Weight and Inertia Member l"x 6"x 8' wood plank Flexible Steel Strap (a) Rubber Test Specimen LeRoy Lettering Pon Clip Board and Data Sheet Rubber Specimen (b) Adjuotment ~Screws Amplitude indicator Adrotao or Lettering Pen Record as above or use a calibrated gauge. Fig. 18. The instrumentation shown in Fig. 18a works very well for the present rubber discs used in absorbers. The discs are left stacked as they were cut, using enough of them to make a stack of about 1-1/2 in. in height. A heavy weight (1000 lb) is used to prevent movement between the discs or adjoining surfaces. A total inertia including the extension board, of 108 in-lb-sec2 with the present rubber used in manufacturing stacked to 1.46 in. (before weighted) gave a period T of 71 x 104 seconds per cycle. Because of the high decay rate, the system has to be forced to obtain a reading of cycles vs time. This is done by applying force (with the hands) in rhythm with the oscillations and recording the time for 50 cycles to occur. This gives a relative k value for the spring constant 1 =2 = cps or -= rad/sec = wn (approximately) T- 5T

The University of Michigan * Engineering Research Institute k 1 I k 71 x 10-4 = o8 k =.846 x 108 in-lb/rad For an indication of the damping, a decay curve can be recorded that will be similar to Fig. 2. The recording paper may be drawn manually as the oscillations decay. With some care the period of oscillation that is recorded may be made uniform. Uniformity, however, is only to establish the position of the exponential envelope for measuring the amplitude of vibration and the period (T) used for the calculation taken from the resonant count made earlier. For greater sensitivity the extension arm of the pen may be increased which in turn will increase the pen movement. Figure 18b shows a method of testing small samples of rubber with the same principles involved as in the previous method. Several items should be kept in mind with either system. 1. Conditions under which comparative samples are tested must be the same a. Inertia of the mass and their weight b. Sample size and configuration c. Temperature (may be varied to test effects) 2. These tests will only indicate trends in performance. 35 The frequencies used here are much lower than those used in an absorber. 4. Endurance life is not known. III. DESIGN MODIFICATION After preliminary resonance tests with a frequency machine, some further modifications may be necessary in the absorber design. Methods suggested under Absorber Components may be used for this purpose. The absorber is then correct as far as calculations will permit. An engine test is now in order. Short period tests will indicate any unsatisfactory tuning or inadequa reduction in amplitude. A simple method of accurately correcting the tuning can be accomplished by the use of torsiograph curves. It is important that the data obtained be accurate and care should be taken in measurement. Two torsiograph 32

The University of Michigan * Engineering Research Institute curves are necessary, one with the absorber as normally used, and another with the absorber fixed or rigid. The curves taken with the absorber rigid could be accomplished by substitution of a mass of equal inertia in the form of some type of disc. Any other equipment which is normally rotating with the engine should also be included. This is important because the absorber tuning depends on the inertias in vibration, and the absence of any vibrating part will alter the curves. Conversely, recording instruments should not add to the inertia for the same reasons. Electronic and magnetic devices are ideal in this respect. They are capable of great accuracy with little or no inertia addition. The data for each condition are plotted on a single sheet of graph paper, forming two curves similar to the solid curves in Fig. 9. The torsiograph curve differs from Fig. 9 in the vertical or ordinate values, which are amplitudes divided by static torque, whereas the torsiograph.is amplitude or double amplitude only. This is of no importance in the application which we are using. The curves will intersect at two points P and Q similar to Fig. 9. Choosing one of the two points, say Q, we can evaluate to find the error in tuning. Using the basic equations for a system as shown in Fig. 15, we have Equation. (with absorber). G,.. KX + (X- Y)2 ( s1t'\KX(X + p.X - 1)2 + [.XY - (X - 1)(X Y)]2 L = absorber-engine inertia ratio cK where c/cc is the absorber K = (2 2 )2 Cc damping factor 2 x = and Y = (a)2 ne -e = dynamic magnifier 9st The Ge used here is then the value of the selected point Q on the data. 9st is an unknown in this application. The damping factor c/cc of the absorber can be any value of these particular points (P and Q), so we can select c/cc = 0 to simplify the equation. The force frequency can be taken from the curve at the point Q and a, from the maximum point of the engine without absorber curve. Since c/ce = 0, then K = 0 and Equation 1 reduces to 33

The University of Michigan * Engineering Research Institute Ge (X - Y) 9st WXY - (X - 1)(X - Y) From page 18 we have Equation 2 (without absorber) suitable for the single-mass system with damping (engine curve, fixed absorber). Ge. _st 1 - ^2 2 Again select C/Cc = 0 and the equation reduces to e~ = 1 vle 8st 1 ((D\)2 Using the convention of Equation 1, / 2 = X = then Qe 1 8st 1 - X Since Equations 1 and 2 represent the same point Q e q - (Equation 1) = (Equation 2) Gst st Then X - Y 1- X XY - (X - 1)(X - Y) XY = known values (Ta measured for resonance on an absorber test machine) = unknown or known which is to be modified. Solving for 1i we have the re-evaluated ratio and thus can retune the absorber by 1/(1 +,), the original formula used for tuning. The care with which the absorber resonance and the graph data are measured will be reflected in the results. Damping in the engine and absorber is not important. Both factors can give considerable trouble if they have to be measured or calculated. Once the tuning is corrected, it is a simple matter to check on the absorber damping. Engine amplitude values with the correctly tuned absorber 34

The University of Michigan * Engineering Research Institute should be near the values calculated on page 22 or recalculated with the new,io The two maximum peaks with values much over those calculated would indicate insufficient damping. A curve which gives only one peak close to that of the original resonance (slightly shifted to a lower frequency) indicates an excess of damping. This effect is explained on page 10. The absorber at this stage is properly designed for maximum effectivenesso There is no indication, however, that its life in service is sufficient to satisfy the needo It remains now to examine the elastic member in the light of fatigue, aging, heat resistance, and stability with changes in temperature which will require a knowledge of the properties of the elastic membero As work progresses on the development of new rubber-type compounds for this use, the evaluation of the compound will necessarily include the above items as well as its creep due to compression lord, cost, and ease of assembly. CONCLUSIONS It is not only possible but advisable to begin the initial design of an absorber by using some simplified mathematical approach0 The simplification of the system and the simplification of equations will necessarily result in errors. Therefore, any absorber design must be carefully tested to determine its degree of success in achieving the desired performance. Two accurate ways to test an absorber are: 1. as a separate unit on a frequency testing machine; and 2, as an integral unit on the type of engine with which it is to be used. Fatigue life may be greatly affected by the design, i.e., preloading, amplitudes, temperature RECOMMENDATIONS The methods above should be used to test the absorber for short-time performance. For an indication of its expected life, controlled runs on an engine which will duplicate anticipated operating conditions are needed, as are temperature and amplitude measurements of the absorber. A thorough investigation should be made of any new material proposed for use as the elastic member with particular regard for conductivity, aging, fatigue, and changes due to temperature, 35

The University of Michigan * Engineering Research Institute To determine the spring constants of possible rubber compounds, a test program should be developed which will correlate simple tests such as those explained in this report with results obtained on a frequency-measuring machine or an engine A similar correlation should be developed for the damping constanto BIBLIOGRAPHY lo Den Hartog, Jo P. Mechanical Vibrations. New York:McGraw-Hill Book Co., 1940o 2o Wilson, Wo Ker. Practical Solution of Torsional Vibration Problems. 2 vols. New York:John Wiley and Sons, 1940o 5. Thompson, W, T. Mechanical Vibrations. Englewood Cliffs, N. J.:PrenticeHall, Inc., 1956. 4o Engineering Properties of Rubber. Fort Wayne, Ind.:U.S. Rubber Co., 1950. 5. Timoshenko, S. Vibration Problems in Engineering....2nd edo New York: Do Van Nostrand Co., Inco, 1937. 6. Hansen, Ho Mo and Chenea, P. F. Mechanics of Vibration. New York:John Wiley and Sons, 1952. 7o Rainville, Earl Do Elementary Differential Equations. New York:MacMillan Co., 1954, ---------------------- 6 —----------

APPENDIX

The University of Michigan * Engineering Research Institute PART I NATURAL FREQUENCY OF A SINGLE-MASS SYSTEM WITHOUT DAMPING ma D = diameter of inertia member (in.) W = weight of inertia member (lb) ma = mass of inertia member in slugs (lb-sec2/in.) = /g d = diameter of shaft (in.) 1 = length of shaft (in.) ka = spring constant of the shaft in torsion (lb-in.) Oa = instantaneous angular displacement of inertia during a cycle of harmonic motion == Xa = angular velocity at time of displacement dt - =0 = angular acceleration at time of displacement dt2 Ga maximum angular displacement Ja = mass moment of inertia of ma There are two forces acting on the mass-the inertia force caused by its rotation and the torsion-spring force caused by the twist of the shaft due to the rotation of the mass. The inertia torque T = Ja Va, which is equivalent to translatory motion where F = ma. Dynamic balance of forces requires that the inertia torque T has an equal but opposite torque acting on it at all times, namely, ka0a. Thus, equating the forces Jaia = - kaa (1) or Ja^a + kava = 0 (la) 38

The University of Michigan * Engineering Research Institute 0a + a o= (2) Ja Solving this differential equation we have the root r r + 0 = O Ja r = ~ i a Ja Therefore Oa = AcOs k t + Bsin t (3) VJa VJa Let on = 4ka/Ja (natural circular frequency) Oa = Acos nt + Bsin int (4) This is the general solution to Equation 1 with A and B constants which depend on initial conditions and determine the amplitude of motion. Selecting the starting position t = 0 when the displacement is maximum, Oa = GaY i.e., the angle of rotation would be at its maximum. Therefore, the parameters are arbitrarily fixed at maximum displacement at the beginning of each cycle. The displacement is Oa = Acos (nt + Bsin cit (Equation 4) Differentiating, we obtain the velocity dt = a = - A an sin aot + B cn cos wnt (5) Differentiating again, we obtain the acceleration d2Oa doaA 2 =dt2 dt a = - A On cos ant - B n sin cnt (6) At t = 0 we have chosen i = 6^, as the start of the cycle. At this point the velocity is zero or ~a = 0 and the cos ant = 1 and sin ant = 0. Substituting in Equation 4, 39

The University of Michigan * Engineering Research Institute Ga = A (4a) Substituting in Equation 5, a = B Cn or B= a (5a) Substituting 4a and 5a back into 4 results in a = 4a cos at + nt + sin at (4b) ~n and with t = 0, Oa = 0 a = Ga cos Dnt The following is an example of this condition. Using the figure above: Inertia member Torsion spring D = 4 in. shaft = ka weight = 19.3 lb length = 4 in. diameter =.434 in. material = cold-rolled steel The equation for motion is Oa = Ga cos oant where on = NkJ. Thus the natural frequency can be found if we find the value of ka and Ja. Mass Wr2 Ja = moment of inertia of ma = - 2g J =19. (2)2.1 lb-in-sec2 386 2 Spring k =..Jp G = shear modulus L Jp = polar moment of area of the shaft cross section Jp = = 32 =.oo548 = 34.8 x o-4 in.4 P = 32 = 32 k = (11.5 x 106)(34.8 x 10-4) k = 1 x 104 Ib-in. 40o

The University of Michigan * Engineering Research Institute Natural frequency l /I x 104 J 0 -radians %(tn =.1 = x 102 seor (Dn = _ x 102 = 50.3 cps 41

The University of Michigan * Engineering Research Institute PART II NATURAL FREQUENCY OF A TWO-MASS SYSTEM WITHOUT DAMPING XD 1 X ml I M2 1 D1 = diameter of large inertia member (in.) D2 = diameter of small inertia member (in.) W1 = weight of large inertia member (lb) W2 = weight of small inertia member (lb) ml = mass of large inertia member in slugs (lb-sec2/in.) m2 = mass of small inertia member in slugs (lb-sec2/in.) d = diameter of shaft (in.) 1 = length of shaft (in.) kl = spring constant of the shaft in torsion (lb-in.) 01 = instantaneous angular displacement of m1 02 = instantaneous angular displacement of m2 As in Part I, there are two forces acting upon each mass. The spring force opposing the inertia force ml must also act on m2 in the same manner to provide the twist in the shaft whenever either mass is displaced. To do this, the masses must always be opposed in their motion and their effect on the spring must be zero (at zero displacement) at the same instant. |10, = - kl(0l - 02) (1) J202 = + k(01 - 02) (2) Jl0l + kljl - ki02 = O (la) |J20 - kl0i + k.02 = O (2a) -------------------— 4 ^ -----------

The University of Michigan * Engineering Research Institute Adding la and 2a, JiSl + J202 = 0 (3a) -2 -1x or.. = -... (3b) J2 dt2 J2 dt2 d2^ = - J1 (5 c) d.0 + C2 =- -J do1 + C1 J2 Assuming t = 0 at the instant the masses are released, the velocities are zero, 01 = d./dt = 0, and,1 = d./dt = 0.' C= = C2 This implies that the masses may be rotating at the same velocity and have no effect on the vibration. With C1 = C2 = O, then C = _ _ dt J2 dt d02 = - J, d (5d) 02 + C3 - i 1 + C4 (3e) J2 with C3 and C4 zero (values of initial displacement) and substituting 3b and 5d in 2a, we have Ji0l + kl:l + ki 01 = 0 (4a) J2 01 + k 01 +EL 01= 0 (4b) Vi J2 01 + (01 ( + - 0 (4c) 1 2 Solving this differential equation, we have r2+ ( + ) = 0 Vi Js2/ -------------------— 4 5 -----------

The University of Michigan * Engineering Research Institute _r = + (-+9 ~i/J r CS t+ Bs=in;+ / t 01 = Acos J t+ BJin J,1 J J1 J J2 Let Cj = A +J1+ T (natural circular frequency); then 01 = Acos cOnt + Bsin Cnt This can be simplified as in Part I to 01 = 1l cos cit The same procedure can be used to find 02 02 = G2 Cos ot The following is an example of this system. W2 = 48.3 lb D2 = 4 in. W1 = 48.3 lb D1 = 8 in. shaft diameter =.718 in. shaft length = 15 in. m2 _ 48.3 lb.125 lb sec2/in. g 386 in/sec2 m = =.125 lb sec2/in. g J2 = 1/8 m2 D| = 1/8 x.125 x (4)2 =.250 lb-in-sec2 J1 = 1/8 ml D2 = 1/8 x 125 x (8 = 1.0 b-in-sec =P. = x1.1 8-) = kl = Jp 1 where 44

The University of Michigan * Engineering Research Institute G = 11.5 x 106 lb/in.2 cold-rolled steel D4 (.18).026 532 52 k =_ (11.5 x 106)(.026) 15 k1 = 2 x 104 _ 1 k ~ k1 2it= 2it J+ J2 1 210 +.2 x 104 lo x 5010 cps 2: V 1.0 +.250 2i -------------------— 4 5 —---------

The University of Michigan * Engineering Research Institute PART III NATURAL FREQUENCY OF A TWO-MASS SYSTEM AND VIBRATION ABSORBER WITHOUT DAMPING E, k2 Jama Jz J1 m2 ml (1) Jaa = - k a ( a (1) Jl~l = ka(0a - 0l) - kl(0l' 0z) (2) J2z2 = kl(Z - 2) (5) Ja^a + kava - kal = 0 (la) Jil - ka0a + kail + kl1l - k02 = 0 (2a) J202 - kl+l + k02 = 0 (3a) 0a + Ja a01 = 0 (ib) ka Ja ka k a k k 01 -- a + + 01 - 3 02 = 0 (2b) Ji J J1 J j, 02 - J 02 + J 02 = 0 (5b) J2 J2 To simplify the solution of the above equations, we can substitute values derived previously (Appendix Parts I and II). ______________________4 6 ------------

The University of Michigan * Engineering Research Institute Wa = 19.3 lb ma = 05 Ja =.1 W1 = 48.3 lb m =.125 J = 1.0 W2 = 48.3 lb m2 =.125 J =.250 ka = 1 x 104 k = 2 x 104 Substituting these values in lb, 2b, and 3b, x 1 x 104 1 104 Oa.1 a +'= 0 (lc) 1 x 104 1 x 104 2 x 104 2 x 104 0i - --- Oa + +1 ~ 5 - = ~ (2C) 1.0 1.0 1.0 1.0 2 x 104 2 x 104 -2 - 0 21 +.2 0 = o (3c).250.250 From previous work (see Appendix Parts I and II), solution to equations of motion of this type are 0 = G cos ount, we can solve Equations lc, 2c, 3c, by substituting this solution and solving for cai, the natural circular frequency. If 0 = G cos att, then differentiating with respect to time gives = - 9(o() sin ont 0 = - 0(co)2 cos t Substituting - ca n cos nt + 105 Ga cos Cant - 105 G cos nt = 0 (d) - O1 2< cos ant - 104 ta cos t + 104 G1 cos cant + 2 x 104 Gi cos nt - 2 x 104 G2 cos (at = 0 (2d) - G2 O2 cos at - 8 x 104 01 cos (ant + 8 x 104 G2 cos nt = 0 (3d) Dividing equations by cos crnt - a (Cn2 + 105 Ga - 105 e1 = 0 (le) n 1 (D2 - 104 ~a + 104 ~1 + 2-x 104 ~i - 2 x 104 2 = 0 (2e)

The University of Michigan * Engineering Research Institute - 2 32 - 8 x 104 1 + 8 x 104 g2 = 0 (3e) (105 - 02)~a + (-105)G1 0 (If) (-104)Ga + (3 x 104 - (o2)l + (-2 x 104)g2 = 0 (2f) n (-8 x 104)0 + (8 x 104 - 2)2 0 (3f) Solving by determinants 105 - a2n - 105 0 5x102 - 2x 104 - 104 3 x 104 - - 2 x 104 = 0 - 8 x 104 8 x 104 - wDn (105 - 4)(3 x 104 - C2)(8 x 104 - (2) - (-8 x 104)(-2 x 104)(105 - 2) - (8 x 104 - u)(-104)(-105) = 0 (4a) 24 x 101 - 104 x 108s + 8 x 104 aU4 - 3 x 109 un + 13 x 104 4n -6 - 16 x 1013 + 16 x 108s o2 - 8 x 1013+ 109 + ( = 0 (4b) n - (21 x 104)o4 + (108 x 108)Wn =0 (4c) Factor out w2 which is a trivial solution. 4 C - (21 x 104) 2 + (108 x 108) = 0 Solving for (o2 by the Quadratic Equation, z 21 x 104 +(o21 x 104_)2 - 4(1) (108 x 108) c=n 9 2 2t = 9 x 104 (on = 300 radians/sec a2n = 12 x 104 0)n = 347 radians/sec n —------------ 8

The University of Michigan * Engineering Research Institute 300 - = 47.7 cps (first mode) 2A 3e ae te552 cps (second mode) 2x These are the two natural frequencies of this particular three-mass system. This method of solution is not practical for more complex systems and one of the references listed should be consulted for other means of solution. Holzer's method is one of the more widely used means of solving the more difficult problems. For our purposes, the three-mass system is all that is necessary for basic understanding, and more specifically, the system which has one mass as fixed wall as shown in Fig. 15. 49

The University of Michigan * Engineering Research Institute PART IV FORCED VIBRATION OF A SINGLE-MASS SYSTEM To cos ot --- = ka ma The case of the single-mass system in Part I with a harmonic force applied to the mass gives: JaJ = - ka o + To cos wot (la) Jao + kaO = To cos oCt (lb) This is a linear nonhomogeneous differential equation whose general solution takes the form Og = c +,p where 0c, the complementary solution, is the solution of the equation Ja,- kls = 0, which is the natural frequency condition. is the particular solution where To cos cot is the external torque. Because of some small amount of damping in all vibrating systems, the Xc factor will not be a sustained oscillation and will drop out after a period of time. Therefore, only the particular solution ( need be considered. j/ k To cos (lt 0 + _ 0 _= (lc) a J Solving r a +ka To cos Wt Ja Ja r = + i ka/J roots of the left side of the equation r' = 0 + iAd roots of the right side 0p = cle~ cos wt + c2e0 sin ct = c1 cos ct + c2 sin ct (2) 50

The University of Michigan * Engineering Research Institute Since cl and c2 must satisfy Equation l3c we can solve for cl and c2 by substituting in lc. p = c cos Cot + ca sin t Differentiating p = - cl) sin ct + caJ cos cot Second differentiation $p = - co2 cos ct - c-cD2 sin ct Substituting - CL2 cos ott - c2D2 sin cut + (cl cos ct + c2 sin cot) - cos ot Ja Ja cos ot (CCD + - Cl) + sin cut (-cU2 + a- 2) cos t Ja Ja Ja - cl + ka To and - c2 + ka c - 0 Ja Ja Ja C2 = 0 C1 + k =k To To To To/ka C1 = Ja(k/J - oZ) = ka Ja2 = 1 -(Ja/ka)c2 With ao =N ka/Ja 2 ka n Ja To/ka CI = 1 - C. c2 na To/ka XP -= 1 m? cos cot cna 51

The University of Michigan * Engineering Research Institute To/ka | 1 _c2 maximum deflection ina Note the maximum deflection becomes infinite when the external force frequency equals the natural frequency independent of the amount of external force. Practically, the damping will limit the deflections, and the size of the external force becomes an important factor in determining the resultant amplitude. 52

The University of Michigan * Engineering Research Institute PART V FORCED VIBRATION OF A THREE-MASS SYSTEM To cos Wt ma2 ml mi Ja0a = ka(0a - 0) (1) J1 = ka(0a 0)- kl(l - 0a) + To cos wot (2) J5f2 = ki(0I - 02) (3) Jaa ka0a - ka = 0 (la) J11 - ka0a + kaoi + kol, - k2 = To COs oDt (2a) J202 - k01 + kl0Z = 0 (3a) ~a "a.a + k a 0 a ka: 0= (lb) 01 - - 0a + - 0i + I 01 - 2 (2b) Ji Jl Ji Ji Ji 02 " 01 + Ja 02 = ~ (5b) Using again the values chosen for Part III, all equations being identical with the exception of To cos out for Equation 2, we can advance to Equations If, 2f, and 3f. (105 - 2)oa + (-105)Q1 = O (If) (-104)Qa + (3 x 104 - w2)1 + (-2 x 104)2 = T~ (2f) Ji 55

The University of Michigan * Engineering Research Institute (-8 x lo4)g + (8 x 104 - 2)2 = 0 (of) Solving by determinants, - (8 x 104 - U2)(To/J1)(-lO5) =a (o10 - W2) (8x04 - 2) (3x1O4 - w2)(-8x1l04 )(-2x104) (105 -w)-(8xl04_D2) (-104)(-105) Note - The denominator of the above expression is simplified in Part III of the Appendix in terms of the natural frequency and reduces to the expression shown below. [Wo2] [1)4 - (21 x l04)W2 + (108 x 108)] - (To/J1)(-10 5)(8 x 104 - wa2) 0a, -. a [W2] [a4 - (21 x 104)& 2 + (108 x 108)] - (To/J) (-105)(8 x 104 - a)2) - (To/Jl)(105) ( - 8 x 104) Qa -- (cL2)(o2 - 9 x 104)(L2 - 12 x 104) = (W2)(o2 _ 9 x 104)((2 - 12 x 104) G1 and Q2 can be determined similarly giving the following equations: - (To/Jl)(105 - w2)(8 x 104 -_ ) 1 = (o?2) (o2 - 9 x 104)(W2 - 12 x 104) - (To/Jl)(105 - w2)(-8 x 104) 2 = )(C 2 9 x o - 12 X 104) The important point in this example is to see that all amplitudes of 9a, ~1, and G2 are dependent on the factors of their denominators, which are all identical. If any of these factors were zero, the amplitudes would go to infinity, the points of resonance. Solving for these points, we can set each factor to zero. o2 = 0 trivial solution -2 9 x 104 = 0 w = 300 radians/sec First mode a2- 12 x 104 = 0 X = 347 radians/sec Second mode These values are the same as those found for the natural resonant frequencies of the system in Part III. A comparison of the numerators gives an indication of the absorber action. @1 has two points where the driving frequency results in zero amplitude: ----------- S~~~~~~~54

The University of Michigan * Engineering Research Institute 105 and 8 x 104. The 105 is the resonant point of ma-ka, causing Q1 and 92 to be zero, and 8 x 104 is the resonant point of m2-kj, which causes G1 and Ga to be zero. Both end masses have absorber action at their respective resonant frequencies. Figure 8 is a set of curves similar to those which would be obtained in this example. The methods of solution we have used are not suitable for systems of more than three masses. Holzer's method is a simple form usable for multiplemass systems and the foregoing examples are meant only to show the mechanics involved in solving for such solutions. 55

The University of Michigan * Engineering Research Institute PART VI Damping in a vibrating system is most easily represented by viscous damping. For comparison with undamped systems and the differential equations used, here is a general equation for vibration with damping. JO + c + k0 = To cos wt The viscous damping factor, introduced as co, indicates that damping forces are a function of velocity. We will not pursue the solutions to damped vibration problems with the new factor co because they become far too complex for what little is gained. However, some of the texts listed in the bibliography deal with this. For our purposes, the differences in damped and undamped vibrations may best be explained by a few vector diagrams of a single-mass system, shown on the following page. The torque vector shown represents the force causing the vibration. The length of the vector is To and it is a constant magnitude rotating Xc radians per second. At a given time t, the vector will have rotated ot radians. The force applied to a mass causing torsional vibration is that portion of To projected down to the horizontal line, namely To cos ot. Thus, the torque component will vary from a positive value of To through zero to a negative value and again to a positive value during one revolution (2s radians). With the torque rotating as described, the effect on a system both with and without damping is shown below resonance, at resonance, and above resonance. The dashed lines on the diagram simply indicate that the vectors are in the same position and are separated only for the sake of clarity. All vectors rotate at c rad/sec such as the torque vector shown. Since there is no motion relative to one another, they can be pictured as vectors at some time t. Note that in all cases the vector sum of the spring force kg, the inertia force Jcu2G, the damping force ccDG (present only in cases with damping), and the force To must all add up to zero (form a closed loop). Below Resonance Below resonance the external force To is mainly opposed by the spring force k~. With no damping, the movement of the mass G is in phase with the force T0. With damping, a portion of the torque must oppose the damping force and the amplitude is always some degree removed from the applied torque. 56

The University of Michigan * Engineering Research Institute VECTOR REPRESENTATION OF A SINGLE-MASS SYSTEM Torque To Tocos ct NO DAMPIING Below Resonance To To At Resonance To Vectors are unstable X Above Resonance 57 0 t ------------ 5 —-----------

The University of Michigan * Engineering Research Institute At Resonance At resonance, the torque is 90~ removed from the amplitude and since the spring force must be out of phase and the inertia force in phase with the amplitude, the undamped system cannot be balanced and therefore results in infinite values as shown in Fig. 7 (page 8). In the damped case, the torque is counteracted only by the damping force. Since the damping force is the value of c0Q, the amplitude at resonance must increase to balance the torque. An increase in the damping constant c in the system would serve directly to reduce the amplitude. Above Resonance Above resonance the external torque is mainly opposed by the inertia force and it becomes increasingly dominant as the frequency w increases, causing a decrease in the amplitude. In the undamped case, the motion or amplitude - is 180~ out of phase with the torque (see Fig. 7). The relationship of the vectors shown is only for three specific fbrced frequencies. It should be realized that for every forced frequency above and below resonance the vectors will have varied magnitudes and phase angles with each other. Some relationships, however, are always present. These are: 1. Spring force (kg) 180~ opposed to amplitude (0) 2. Inertia force (Jcw2) in phase with amplitude (0) 3. Damping force 180o opposed to velocity, i.e., 90~ removed from amplitude (o) 58