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Preface This textbook represents the Mechanical Vibrations lecture course given to students in the fourth year at the Department of Engineering Sciences (now F.I.L.S.), English Stream, University Politehnica of Bucharest, since 1993. It grew in time from a course taught in Romanian since 1972 to students in the Production Engineering Department, followed by a special course given between 1985 and 1990 to postgraduate students at the Strength of Materials Chair. Mechanical Vibrations, as a stand alone subject, was first introduced in the curricula of mechanical engineering departments in 1974. To sustain it, we published with Professor Gh. Buzdugan the book Vibration of Mechanical Systems in 1975, at Editura Academiei, followed by two editions of Mechanical Vibrations, in 1979 and 1982, at Editura didactică şi pedagogică. In 1984 we published Vibration Measurement at Martinus Nijhoff Publ., Dordrecht, which was the English updated version of a book published in 1979 at Editura Academiei. As seen from the Table of Contents, this book is application oriented and limited to what can be taught in an one-semester (28 hours) lecture course. It also contains material to support the tutorial that includes the use of finite element computer programs and basic laboratory experiments. The course syllabus changed in time due to the growing use of computers. We wrote simple finite element programs to assist students in solving problems as homework. The course aims to: (a) increase the knowledge of vibration phenomena; (b) further the understanding of the dynamic behaviour of structures and systems; and (c) provide the necessary physical basis for analytical and computational approaches to the development of engineering solutions to vibration problems. As a course taught for non-native speakers, it has been considered useful to reproduce as language patterns some sentences from English texts. Computational methods for large eigenvalue problems, model reduction, estimation of system parameters based on the analysis of frequency response data, transient responses, modal testing and vibration testing are treated in the second volume. No reference is made to the dynamics of rotor-bearing systems and the vibration of discs, impellers and blades which are studied in the Dynamics of Machinery lecture course. April 2006

Mircea Radeş

Prefaţă Lucrarea reprezintă cursul de Vibraţii mecanice predat studenţilor anului IV al Facultăţii de Inginerie în Limbi Străine, Filiera Engleză, la Universitatea Politehnica Bucureşti, începând cu anul 1993. Conţinutul cursului s-a lărgit în timp, pornind de la un curs predat din 1972 studenţilor de la facultatea T. C. M. (în prezent I.M.S.T.), urmat de un curs postuniversitar organizat între 1985 şi 1990 în cadrul Catedrei de Rezistenţa materialelor. Vibraţiile mecanice au fost introduse în planul de învăţământ al facultăţilor cu profil mecanic ca un curs de sine stătător în 1974. Pentru a susţine cursul, am publicat, sub conducerea profesorului Gh. Buzdugan, monografia Vibraţiile sistemelor mecanice la Editura Academiei în 1975, urmată de două ediţii ale manualului Vibraţii mecanice la Editura didactică şi pedagogică în 1979 şi 1982. În 1984 am publicat Vibration Measurement la Martinus Nijhoff Publ., Dordrecht, reprezentând versiunea revizuită în limba engleză a monografiei ce a apărut în 1979 la Editura Academiei. După cum reiese din Tabla de materii, cursul este orientat spre aplicaţii inginereşti, fiind limitat la ceea ce se poate preda în 28 ore. Materialul prezentat conţine exerciţii rezolvate care susţin seminarul, în cadrul căruia se utilizează programe cu elemente finite elaborate de autor şi se prezintă lucrări demonstrative de laborator, fiind utile şi la rezolvarea temelor de casă. Cursul are un loc bine definit în planul de învăţământ, urmărind a) descrierea fenomenelor vibratorii întâlnite în practica inginerească; b) modelarea sistemelor vibratoare şi analiza acestora cu metoda elementelor finite; şi c) înarmarea studenţilor cu baza fizică necesară în modelarea analitică şi numerică a structurilor în vibraţie şi a maşinilor, pentru elaborarea soluţiilor inginereşti ale problemelor de vibraţii. Fiind un curs predat unor studenţi a căror limbă maternă nu este limba engleză, au fost reproduse expresii şi fraze din cărţi scrise de vorbitori nativi ai acestei limbi. În volumul al doilea se vor prezenta metode de calcul pentru probleme de valori proprii de ordin mare, reducerea ordinului modelelor, răspunsul tranzitoriu, estimarea pametrilor sistemelor vibratoare pe baza analizei funcţiilor răspunsului în frecvenţă, analiza modală experimentală şi încercările la vibraţii. Nu se tratează dinamica sistemelor rotor-lagăre şi vibraţiile discurilor şi paletelor, acestea fiind studiate în cadrul cursului de Dinamica maşinilor. Aprilie 2006

Mircea Radeş

Contents

Preface

1

Prefaţă

2

Contents

3

1. Modelling Vibrating Systems

5

1.1 Vibrations vs. Oscillations

5

1.2 Discrete vs. Continuous Systems

6

1.3 Simple Vibrating Systems

7

1.4 Vibratory Motions

8

1.5 Damping

10

2. Simple Linear Systems

11

2.1 Undamped Free Vibrations

11

2.2 Undamped Forced Vibrations

22

2.3 Damped Free Vibrations

35

2.4 Damped Forced Vibrations

42

Exercices

73

3. Simple Non-Linear Systems

79

3.1 Non-Linear Harmonic Response

79

3.2 Cubic Stiffness

81

3.3 Combined Coulomb and Structural Damping

92

3.4 Quadratic Damping

97

3.5 Effect of Pre-Loading

103

4

MECHANICAL VIBRATIONS

4. Two-Degree-of-Freedom Systems

105

4.1 Coupled Translation

106

4.2 Torsional Systems

119

4.3 Flexural Systems

130

4.4 Coupled Translation and Rotation

145

4.5 Coupled Pendulums

151

4.6 Damped Systems

156

Exercices

179

5. Several Degrees of Freedom

183

5.1 Lumped Mass Systems

184

5.2 Plane Trusses

210

5.3 Plane Frames

220

5.4 Grillages

234

5.5 Frequency Response Functions

241

Exercices

247

6. Continuous Systems

259

6.1 Lateral Vibrations of Thin Beams

259

6.2 Longitudinal Vibration of Rods

275

6.3 Torsional Vibration of Rods

278

6.4 Timoshenko Beams

280

References

281

Index

289

1. MODELLING VIBRATING SYSTEMS

Vibrations are dynamic phenomena encountered in everyday life, from the heart beating and walking, trees shaking in gusty winds or boats floating on rough waters, vibration of musical instruments and loudspeaker cones, to bouncing of cars on corrugated roads, swaying of buildings due to wind or earthquakes, vibrations of conveyers and road drills. It is customary to term ‘vibrations’ only the undesired repetitive motions, giving rise to noise or potentially damaging stress levels. The effect of vibrations on humans, buildings and machines are of main concern. Modelling vibration phenomena implies describing the structure and parameters of the vibrating body, the excitation function and the response levels. This introductory chapter focuses on definitions and classifications, to give an overview of the main notions used in vibration analysis.

1.1 Vibrations vs. Oscillations The Oxford Dictionary gives “vibration, n. Vibrating, oscillation; (phys) rapid motion to and fro, esp. of the parts of a fluid or an elastic solid whose equilibrium is disturbed”. It comes out that all matter, gaseous, liquid or solid is capable of executing vibrations and, in fact, so are the elementary particles of which the matter is composed. Generally, oscillations are variations of a state parameter about the value corresponding to a stable equilibrium position (or trajectory). Vibrations are oscillations due to an elastic restoring force. To save confusion, a flexible beam or string vibrates while a pendulum oscillates. For practical engineering purposes it is usual to allocate the term ‘vibration’ predominantly to unwanted periodic motions. In music, the opposite is the case, since all musical instruments use periodic vibrations to make sound. We might say that vibration in engineering is more akin to noise in acoustics: an

6

MECHANICAL VIBRATIONS

annoying, but to a degree, inescapable by-product of the machine, either in terms of external sound or damage within itself. Apart from harmful vibrations, there are installations whose operation is based on vibratory motions, namely: concrete tampers, pile driving vibrators, soil compaction machines, vibrating screens, fatigue testing machines, etc. All bodies possessing mass and elasticity are capable of vibration. A vibrating system has both kinetic energy, stored in the mass by virtue of its velocity, and potential energy, stored in the elastic element as strain energy. A major feature of vibrations is the cyclic transformation of potential energy into kinetic and back again. In a conservative system, when there is no dissipation of energy, the total energy is constant. At the point of maximum displacement amplitude, the instantaneous velocity is zero, the system has only potential energy. At the static equilibrium position, the strain energy is zero and the system has only kinetic energy. The maximum kinetic energy must equal the maximum potential energy. Equating the two energies it is possible to obtain the natural frequency of vibration. This is the basis of Rayleigh’s method. Vibrating systems are subject to damping because energy is removed by dissipation or radiation. Damping is responsible for the decay of free vibrations, for the phase shift between excitation and response, and provides an explanation for the fact that the forced response of a vibratory system does not grow without limit.

1.2 Discrete vs. Continuous Systems The number of independent coordinates needed to specify completely the configuration of a vibrating system at any instant gives the number of degrees of freedom of the system. It follows that, in order to describe the motion of every particle of a system, the number of degrees of freedom has to be infinite. However, for practical purposes, it is useful to use systems of approximate dynamical similarity to the actual system, which have a small number of degrees of freedom. The criteria used to determine how many degrees of freedom to ascribe to any system under analysis are practical in nature. For instance, some of the possible system motions may be so small that they are not of practical interest. Some or most of the motions of particles in the system may be practically similar, allowing such particles to be lumped into a single rigid body. The frequency range of the excitation forces may be so narrow that only one, or at most a few, of the natural frequencies of the system can give rise to resonances. Groups of particles experiencing similar motions may be considered single bodies, thereby reducing the number of degrees of freedom necessary to consider. All these practical considerations lead to the concept of lumped masses which are rigid bodies

1. MODELLING VIBRATING SYSTEMS

7

connected by massless flexible members. The motions predicted by using such approximate lumped-parameter or discrete systems are often close enough to the actual vibrations to satisfy all practical demands and to provide useful design data and allowable vibration limits. In some systems, a second approximation can be made, by taking into account the mass of the elastic members. This is necessary only when the flexible members have distributed masses which are comparable in magnitude with the masses of system components modelled as rigid bodies. Finally, there are many systems of practical interest which have such simple shapes that they can be considered as systems possessing an infinite number of degrees of freedom. Such distributed-parameter or continuous systems may be modelled as strings, beams, plates, membranes, shells and combinations of these. In most engineering applications, geometrically complex structures are replaced by discretized mathematical models. A successful discretization approach is the finite element method. The infinite degree of freedom system is replaced by a finite system exhibiting the same behaviour. The actual structure is divided (hypothetically) into well-defined sub-domains (finite elements) which are so small that the shape of the displacement field can be approximated without too much error, leaving only the amplitude to be found. All individual elements are then assembled together in such a way that their displacements are mating each other at the element nodes or at certain points at their interfaces, the internal stresses are in equilibrium with the applied loads reduced at nodes, and the prescribed boundary conditions are satisfied. Modelling errors include inappropriate element types, incorrect shape functions, improper supports and poor mesh.

1.3 Simple Vibrating Systems A surprisingly large number of practical vibration problems which arise in the machines and structures designed by engineers can be treated with a sufficiently high degree of accuracy by imagining the actual system to consist of a single rigid body, whose motion can be described by a single coordinate. In reality, the simplest imaginable system consists of the body whose motion is of interest and the fixed surrounding medium, relative to which the motion is measured. The problem of treating such a simplified system is fourfold. The first part consists in deciding what part of the system is the rigid body and what part are the flexible members. The second part consists in calculating the values of the dynamic parameters of the rigid body and flexible parts. The third part consists in writing the equations of motion of the equivalent system, Finally, the fourth part consists in solving the equations for the prescribed conditions of

MECHANICAL VIBRATIONS

8

free or forced vibrations. Alternatively, methods using the kinetic and potential energies may be used in the place of the last two stages. The first two parts require judgement and experience which come with practice, that is, with the repeated process of assuming equivalent systems, predicting their motions and checking the predictions against actual measurements on the real systems. Model verification and validation may require updating of system parameters or even of the model structure. The adequacy of the solution depends largerly on the skill with which the basic simplifying assumptions are made. A basic choice is between linear and non-linear models. Damping estimation is another source of error, because damping cannot be calculated like the mass and stiffness properties. The last two steps consist in applying procedures worked out by mathematicians. The real engineering work lies in the first two stages, while the last two stages may be considered as mere applications of recipies. One degree of freedom systems are considered in Chapters 2 and 3. Discrete systems are treated in Chapters 4 and 5. Chapter 6 is devoted to straight beams and bars.

1.4 Vibratory Motions According to the cause producing or sustaining the vibratory motion, one can distinguish: free vibrations, produced by an impact or an initial displacement; forced vibrations, produced by external forces or kinematic excitation; parametric vibrations, due to the change, produced by an external cause, of a system parameter; self-excited vibrations, produced by a mechanism inherent in the system, by conversion of an energy obtained from a uniform energy source associated with the system oscillatory excitation. If the system is distorted from the equilibrium configuration and then released, it will vibrate with free vibrations. If any part of the system is struck by a blow, the system will vibrate freely. Musical instruments like drums are struck and strings are plucked. Free vibrations exist when the forces acting on the system arise solely from motion of the system itself. The frequencies of the free vibrations are fixed functions of the mass, stiffness, and damping properties of the system itself. They are called natural frequencies. For any particular system they have definite constant values. When all particles of a body vibrate in a synchronous harmonic motion, the deflected shape is a natural mode shape. Vibrations which take place under the excitation of external forces are forced vibrations. External forces in any system are forces which have their reactions acting on bodies which are not parts of the system isolated for study. The forcing function can be harmonic, complex periodic, impulse, transient, or random.

1. MODELLING VIBRATING SYSTEMS

9

When a system is excited by a periodic external force which has one frequency equal to or nearly equal to a natural frequency of the system, the ensuing vibratory motion becomes relatively large even for small amplitudes of the disturbing force. The system then is in a state of resonance. An example is the swing pushed at the right intervals. Other examples include vibrations of geared systems at the tooth-meshing frequency, torsional vibrations of multi-cylinder engine shafts at the firing frequency, vibrations of rolling element bearings at the ball passing frequencies, etc. There is an effect arising from the damping which causes the resonance frequency to differ slightly from the natural frequency by an amount which increases with the damping. Fortunately the distinction in practice is very small and can be neglected in most engineering structures, unless very high damping is provided on purpose. Resonance relates to the condition where either a maximum motion is produced by a force of constant magnitude, or a minimum force is required to maintain a prescribed motion level. A resonance is defined by a frequency, a response level and a bandwidth of the frequency response curve. Avoidance of large resonant vibration levels can be accomplished by: a) changing the excitation frequency; b) making stiffness and/or mass modifications to change the natural frequencies; c) increasing or adding damping; and d) adding a dynamic vibration absorber. When the driving frequency is an integer multiple of the natural frequency of the associated linear system, non-linear single-degree-of-freedom systems described by Mathieu equations exhibit parametric instabilities, referred to as parametric resonances. The principal parametric resonance occurs when the excitation frequency is twice the natural frequency. Parametric resonances of fractional order also exist. Multi-degree-of-freedom systems can experience parametric resonance if the driving frequency and two or more natural frequencies satisfy a linear relation with integer coefficients. Parametric resonance is a state of vibration in which energy flows into the system from an external source at resonance, increasing the amplitude of the system’s response. This energy is dependent upon both the natural frequency of the system and the frequency of the parameter variation. During resonant vibrations and self-excited vibrations, the system vibrates at its own natural frequency. But while the former are forced vibrations, whose frequency is equal to a whole-number ratio multiple of the external driving frequency, the latter is independent of the frequency of any external stimulus. In a self-excited vibration, the alternating force that sustains the motion is created or controlled by the motion itself. When the motion stops, the alternating force disappears. Well-known examples include the vibrations of a violin string

MECHANICAL VIBRATIONS

10

being excited by a bow, the ‘chatter’ of cutting tools, of a chalk on the blackboard, of a door that screeches when opened or of a water glass whose rim is rubbed with a wet finger. One can add vortex induced vibrations of industrial smokestacks, galloping and flutter of electric transmission lines, the oil-whirl of rotors in hydrodynamic bearings, vibrations of poppet valves, the wheel shimmy, etc. Parametric vibrations occur in systems with variable stiffness like rotating shafts with non-circular cross-section, pendulums of variable length, geared torsional systems, etc.

1.5 Damping Damping represents the dissipation of energy from a system, generally as a result of energy of motion converted into thermal energy. The loss of energy by radiation, sometimes referred to as geometric damping, is not considered herein. Four of the most common damping mechanisms are: a) Coulomb (sliding friction), in which the force magnitude is independent of velocity, b) viscous, where the force is proportional to velocity, c) velocity-nth power, when the force is proportional to the nth power of velocity across the damper, and d) structural (hysteretic, internal, material), in which the force is proportional to the magnitude of displacement from some quiescent position. Hereditary damping and clearance damping are other possible damping mechanisms. From a microscopic point of view, most damping mechanisms involve frictional forces that oppose the motion (velocity) of some part of a physical system, resulting in heat loss. For example, the Coulomb friction force is caused by two surfaces sliding with respect to one another, and this sliding force is independent of velocity, once the initial static friction (stiction) is overcome. Hysteretic damping may be viewed as a sliding friction mechanism between molecular layers in a material, or between components of a riveted or bolted structure, in which the friction force is proportional to the displacement from the undisturbed position but in phase with the velocity. Viscous damping occurs when molecules of a viscous fluid rub together, causing a resistive force that is proportional to, and opposing the velocity of an object moving through the fluid. Actual oil dampers and shock absorbers provide friction forces proportional to some non-integer power of the relative velocity. The influence of structural and non-linear damping mechanisms on the response of mass-excited single-degree-of-freedom systems is treated in Chapter 3. In the study of discrete vibrating systems only viscous and structural damping is considered.

2. SIMPLE LINEAR SYSTEMS

Any vibrating system has mass and elasticity. The simplest vibrating system consists of a mass attached to a linear spring. When its motion can be described by a single coordinate it has a single degree of freedom. Using this simple model, it is possible to introduce basic concepts such as natural frequency, resonance, beats and antiresonance. During vibration, energy is dissipated by damping. This limits the motion at resonance, decreases the amplitude of free vibration, and introduces phase shifts between excitation and response. Measurement of damping is an important issue because it cannot be calculated like the mass and stiffness properties.

2.1 Undamped Free Vibrations The free vibration of a mass-spring system, that takes place in the absence of any external excitation, is a harmonic motion whose frequency depends solely upon the system parameters, the mass and the stiffness, being independent of the motion initial conditions. It is referred to as a natural frequency because it is an intrinsic (natural) system property. Calculation of natural frequencies is based on values of the stiffness of spring elements and of inertia of mass elements.

2.1.1 The Mass-Spring System The system shown in Fig. 2.1 consists of a linear spring of stiffness k and a weight W having a mass m = W g , where g is the acceleration of gravity. The weight is restricted to move in the vertical direction without rotation. The stiffness k is defined as the change in force per unit change in length of the spring. Figure 2.1, a shows the unstretched spring. When the mass m is suspended from the spring (Fig. 2.1, b), its lower end moves downwards and stops in the static equilibrium position, determined by the spring static deflection δ st . In this position, the gravitational force W = mg acting on the mass downwards is

MECHANICAL VIBRATIONS

12

balanced by the spring force k δ st acting upwards (Fig. 2.1, c), so that the static deflection is mg δ st = . (2.1) k If the mass is disturbed from the rest position, the system free vibrations will take place. In order to write the equation of motion, the origin of vibration displacements is chosen at the static equilibrium position, so that only forces due to displacement from this position need be considered.

Fig. 2.1 Letting all vector quantities in the downward direction be positive, in position x the elastic force acting on the mass is − k x (Fig. 2.1, d). Its motion is described by Newton’s second law

m x&& = −k x , which can be written

m x&& + k x = 0 ,

(2.2)

where a dot above a letter denotes differentiation with respect to time. Equation (2.2) is a homogeneous second order differential equation. Its general solution has the form

x = C1 sin ω n t + C 2 cos ω n t ,

(2.3)

ωn =

(2.4)

where k m

[rad/sec]

is the undamped natural circular frequency of the system. The undamped natural frequency is fn =

1 2π

k . m

[Hz]

(2.5)

2. SIMPLE LINEAR SYSTEMS

13

The arbitrary constants C1 and C 2 are evaluated from the initial conditions of the motion. In the most general case, the system may be started from position x0 with velocity v 0 so that the general solution becomes x=

v0

ωn

sin ωn t + x0 cos ωn t .

(2.6)

Another form of the general solution is x = A sin (ωn t + φ )

(2.7)

where the two arbitrary constants are given by A=

x 20 + (v0 ωn ) 2 ,

φ = tan −1

ωn x0

. (2.8) v0 Equation (2.7) indicates that the free vibration of the spring-mass system is harmonic and occurs at a natural frequency f n . The quantity A represents the displacement amplitude from the static equilibrium position and φ is the phase angle. The circular frequency ω n defines the rate of vibration in terms of radians per unit time, 2π rad being equal to one complete cycle of vibration.

The frequency of vibration is the number of complete cycles of motion in a unit of time, and is the reciprocal of the period

T = 1 f n = 2π ω n .

[sec]

(2.9)

The period of vibration is the time required for the motion to begin repeating itself. The undamped natural frequency may be expressed as a function of the static deflection using equation (2.1) fn =

1 2π

g

δ st

,

[Hz]

(2.10)

where g = 9.81 m s 2 .

2.1.2 Stiffness of Elastic Elements Although it is convenient to model a single-degree-of-freedom system as a mass attached to a single helical spring, in many actual systems the spring can take different forms and can also represent an assemblage of several elastic elements. In Fig. 2.2 the stiffnesses of several elastic elements are calculated as the applied force divided by the displacement of its point of application.

MECHANICAL VIBRATIONS

14

Fig. 2.2 In Fig. 2.3 two general types of spring combinations are shown.

Fig. 2.3 For the series arrangement (Fig. 2.3, a) there is a condition of equal force in each spring. Two linear springs, having stiffnesses k1 and k 2 , will deflect statically when loaded by a weight W by an amount

δ st =

W W + =W k1 k 2

⎛1 1 ⎜⎜ + ⎝ k1 k 2

⎞ ⎟⎟ . ⎠

2. SIMPLE LINEAR SYSTEMS

15

The equivalent spring constant, representing the combined effect of k 1 and k 2 , is kS =

W

=

δ st

1 . 1 1 + k1 k 2

(2.11)

For a system with n springs connected in series, the equivalent stiffness k S is given by 1 1 1 1 . (2.12) = + + ... + kS k 1 k 2 kn

The parallel spring arrangement (Fig. 2.3, b) must satisfy the condition of equal displacement in each spring and the sum of forces in each spring must equal the weight W : W = k 1 δ st + k 2 δ st . Thus, for parallel springs, the equivalent stiffness is kP =

W

δ st

= k1 + k 2 .

(2.13)

In general, a system with n parallel springs has an equivalent stiffness given by k P = k 1 + k 2 + ... + k n .

(2.14)

These rules for compounding spring stiffnesses are exactly the same as those for finding the total capacitance of series or parallel circuits in electrical engineering.

2.1.3 Torsional System Consider the torsional system of Fig. 2.4 consisting of a disc of mass moment of inertia J, kg m 2 , suspended from a bar or wire of torsional stiffness K, N m rad . The system is restricted to undergo angular vibrations around the vertical axis. If the instantaneous angular position of the disc is given by the angle θ , the torque acting on the disc is − Kθ so that Newton’s second law for angular motion is J θ&& = − K θ , which can be written

MECHANICAL VIBRATIONS

16

J θ&& + K θ = 0 ,

(2.15)

where a dot above a letter denotes differentiation with respect to time.

Fig. 2.4 Equation (2.15) has been established by Ch. O. Coulomb in 1784. It has the general solution of the form

θ (t ) = C1 sin ωn t + C2 cos ωn t , where

ωn =

K J

[rad/sec]

(2.16)

is the undamped natural circular frequency of the torsional system. The undamped natural frequency is fn =

1 2π

K . J

[Hz]

(2.17)

From Mechanics of Materials it is known that a uniform shaft of diameter d and length l , from a material with shear modulus of elasticity G, acted upon by a torque M t will twist an angle θ =

πd4 Mt l , where I p = is the polar second GIp 32

moment of area of the shaft cross section. The torsional stiffness is then GIp M K= t = . θ l In fact, there is complete analogy between systems in axial and torsional vibration, with the counterparts of springs and masses being torsional springs and rigid discs possessing polar mass moments of inertia.

2. SIMPLE LINEAR SYSTEMS

17

2.1.4 The Energy Method Assuming that the vibrational motion is harmonic, the frequency can be calculated from an energy consideration. When there is no dissipation of energy, the system is called conservative. At any instant, the energy of a conservative system is the constant sum of potential and kinetic energies

U + T = const .

(2.18)

The maximum potential energy, which occurs in an extreme position, where the mass stands still for a moment, must equal the maximum kinetic energy, which occurs when the mass passes through the static equilibrium position with maximum velocity. The spring force is k x , and the work done on an infinitesimal displacement dx is k x d x . The potential energy in the spring, when stretched over x

a distance x , is U =

1

∫ k x dx = 2 k x

2

. Assuming the vibratory motion of the form

0

1 k A2 . 2 1 The kinetic energy at any instant is T = m v 2 . The velocity is 2 1 v = A ωn cos ωn t , so that the maximum kinetic energy is Tmax = m ωn2 A2 . 2

x = A sin ωn t , the maximum potential energy is U max =

1 1 k A2 = m ωn2 A2 wherefrom the 2 2 k m is obtained, independent of the amplitude A .

Equating U max = Tmax , we obtain natural frequency ωn =

Example 2.1 Determine the natural frequency of the fluid oscillations in a U tube (Fig. 2.5).

Solution. Let the total length of the fluid column be l , the tube cross section be A and the fluid mass density be ρ . Assuming all fluid particles to have the same speed at any instant, the 1 kinetic energy can be written T = ρ A l x& 2 . If the fluid oscillates back and forth, 2 the work done is the same as if the fluid column of length x has been transferred from the left side to the right side of the tube, leaving the remaining fluid undisturbed.

MECHANICAL VIBRATIONS

18

The instantaneous potential energy is U = g ρ A x 2 . Substituting the two energies in the condition that the rate of change of total energy must be zero d (T + U ) = 0 dt and dividing out x& , we obtain the differential equation of motion of the fluid 2g &x& + x =0. l

Fig. 2.5 Therefore the natural frequency

ωn =

2g l

is independent of the kind of fluid used, of the tube shape and its cross-sectional area.

2.1.5 Rayleigh’s Method An application of the energy method to systems with distributed mass and/or elasticity is Rayleigh’s method. It is used to reduce a distributed system into an equivalent spring-mass system and to determine its fundamental natural frequency. The kinetic and potential energies are calculated assuming any reasonable deflection curve that satisfies the geometric boundary conditions. If the true deflection curve of the vibrating system is assumed, the fundamental frequency found by Rayleigh’s method will be the correct frequency. For any other curve, the frequency determined by this method will be higher than the correct frequency. This is explained by the fact that any deviation from the true curve requires additional constraints, a condition that implies greater stiffness and higher

2. SIMPLE LINEAR SYSTEMS

19

frequency. In the following, Rayleigh’s method is applied to beam flexural vibrations. A prismatic beam has a bending rigidity E I (where E is Young’s modulus and I is the second moment of area of the cross section) and a mass per unit length ρ A (where ρ is the mass density and A is the area of the cross section). The lateral deflection is assumed harmonic, with frequency ω 1 , synchronous in all points along the beam

y (x ,t ) = v (x ) cos ω 1 t . The instantaneous potential energy is

U=

∫

M 2 dx 1 = 2 EI 2

2

∫

⎛ ∂2 y ⎞ EI ⎜ 2 ⎟ dx ⎜∂x ⎟ ⎝ ⎠

where the linearized differential equation (5.65) of the beam elastic line M = E I ∂ 2 y ∂ x 2 has been used.

(

)

Its maximum value is

U max

1 = 2

2

∫

⎛ ∂2v ⎞ E I ⎜ 2 ⎟ dx . ⎜∂x ⎟ ⎝ ⎠

The instantaneous kinetic energy is 1 T= 2

∫

2

⎛∂ y⎞ 1 ⎟⎟ dm = ω 12 ρ A ⎜⎜ 2 ⎝ ∂t ⎠

∫ y dx , 2

with the maximum value 1 Tmax = ω 12 2

∫ ρ A v dx . 2

Equating the maximum potential energy to the maximum kinetic energy, we obtain the expression of the fundamental natural frequency

∫ E I (∂ v ∂ x ) d x . = 2 ∫ ρ A v dx 2

ω12

2 2

(2.19)

Example 2.2 Determine the fundamental natural frequency of the uniform cantilever beam shown in Fig. 2.6. Solution. Consider the deflection curve of the form

MECHANICAL VIBRATIONS

20

πx⎞ ⎛ v = v0 ⎜1 − cos ⎟. 2l ⎠ ⎝ It can be seen that this function satisfies the boundary conditions x = 0 , v = 0 , dv dx = 0 , and x = l ,

d 2 v d x 2 = 0 , but not the condition x = l ,

d 3 v d x 3 = 0 (zero shear force), so that it is an approximate admissible function.

Fig. 2.6 π4 E I 2 v0 . The maximum 64 l 3 ρA 2 2 ⎛3 2⎞ ω 1 v0 l ⋅ 0.23 . kinetic energy is Tmax = ρ A ω 21 v02 l ⎜ − ⎟ , or Tmax = 2 ⎝4 π ⎠ Equating the two energies, the fundamental frequency of vibration (in rad/sec) is obtained as 3.6638 E I ω1= . ρA l2 The maximum potential energy is U max =

The true solution (6.16) is ω 1 =

3.515 l2

EI , so that the value based on ρA

Rayleigh’s solution is 4 % higher. If the assumed function is the static deflection curve of the massless cantilever beam with a concentrated load at the end v = v0

2 3 1 ⎡ ⎛ x⎞ ⎛ x⎞ ⎤ ⎢3 ⎜ ⎟ − ⎜ ⎟ ⎥ , 2 ⎣⎢ ⎝ l ⎠ ⎝ l ⎠ ⎦⎥

3 EI 2 1 v0 = k v02 and the maximum 3 2 l 2 1 ⎛ 33ρ Al ⎞ 2 2 1 2 kinetic energy is Tmax = ⎜ ⎟ ω 1 v0 = mred ω 1 v0 . 2 ⎝ 140 ⎠ 2 Equating the two energies, the fundamental frequency given by Rayleigh’s formula is

the maximum potential energy is U max =

(

)

2. SIMPLE LINEAR SYSTEMS

ω1 =

21

3E I l3 = (33 140)ρ Al

3.5675 k = mred l2

EI , ρA

which is only 1.47 % higher than the true solution (6.16). The above equation indicates that, for the assumed deflection curve, the beam with uniformly distributed mass has the same natural frequency as a massless beam with a concentrated mass (33 140 )ρ A l attached at the end. This is called a reduced mass.

Example 2.3 Determine the fundamental natural frequency of the free-free uniform beam shown in Fig. 2.7.

Fig. 2.7 Solution. The assumed deflected shape can be taken of the form v = v 0 sin

πx − a. l

The constant a has to be determined from the conservation of momentum for the free-free beam l

l

l

0

0

0

∫ (velocity)⋅ d(mass ) = ∫ (ω v)(ρ A dx) = ω ρ A ∫ v dx = 0 , which yields a = 2 v 0 π . Using the deflected shape of the form πx 2⎞ ⎛ v = v 0 ⎜ sin − ⎟, l π⎠ ⎝ equation (2.19) yields the fundamental natural frequency

ω1 =

22.6 l2

EI

ρA

.

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22

The true solution (6.21) is ω 1 =

22.4 l2

EI so that the discrepancy is only ρA

0.9 %.

2.2 Undamped Forced Vibrations Undamped forced vibrations are produced by variable forces or imposed displacements. If the mass is subjected to a harmonic force of constant amplitude and variable frequency, when the driving frequency approaches the system natural frequency, the response tends to increase indefinitely. This condition is called resonance and is characterised by violent vibrations. For undamped systems, resonance frequencies are equal to the system natural frequencies and in most cases operation at resonance has to be avoided. For damped systems, the response at resonance has finite magnitude. A swing pushed at the right intervals exhibits resonant oscillations. Operation of soil compactors, concrete tampers, vibration conveyers, road drills and vibrating screens is often based on resonant vibrations. However, the main concern with resonance relates to its adverse effects. While operating at resonance, excessive motion and stress amplitudes are generated, causing structural fatigue and failure, harmful effects or discomfort to humans, and a decrease in product accuracy. The nuisance of a noisy component vibrating at resonance can be an obstacle to the sale of a car or a household appliance. When the harmonic force is applied to the spring, the driving point displacement decreases to zero at the system natural frequency. This condition is called antiresonance. Generally, it is a local property, dependent upon the driving location. It helps obtaining points with very low vibration amplitudes.

2.2.1 Mass Excitation with Arbitrary Force Consider a force F (t ) with an arbitrary general time variation (Fig. 2.8). During the short time interval dτ , the force F (τ ) can be considered constant. The cross-hatched area represents an infinitesimal impulse F (τ ) dτ which produces a velocity variation F (τ ) dτ . m The response of mass m due to the differential impulse, over the entire response history for t > τ , is d x& =

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23

F (τ ) dτ 1 sin ω n (t − τ ) , (2.20) m ωn which can be deduced from (2.6) considering that at t = τ , x0 = 0 and v0 = dx& . dx=

The entire loading history may be imagined to consist of a succession of such infinitesimal impulses, each producing its own differential response of the form (2.20).

Fig. 2.8 For a linear system, the total response can be obtained by summing all the differential responses developed during the loading history, that is, by integrating equation (2.20) as follows 1 x (t ) = mωn

t

∫ F (τ ) sin ωn (t − τ ) dτ .

(2.21)

0

Equation (2.21) is generally known as the Duhamel integral for an undamped system.

2.2.2 Mass Excitation with Harmonic Force The mass-spring system from Fig. 2.9, a is excited by a harmonic force f (t ) = F0 cosω t of constant amplitude F0 and driving frequency ω , applied to the mass. Based on the free body diagram of Fig. 2.9, b, its motion is described by Newton’s second law m x&& = −k x + F0 cos ω t , which can be written

m x&& + k x = F0 cosω t .

(2.22)

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The general solution of the linear non-homogeneous equation (2.22) is the sum of the homogeneous solution (2.3) of the equation with zero right-hand side and a particular solution. The particular solution can be found by assuming that it has the same form as the forcing function x P (t ) = X cos ω t ,

(2.23)

where X is the amplitude of the forced response in steady-state conditions.

Fig. 2.9 On substitution of the particular solution (2.23), equation (2.22) becomes − mω 2 X cosω t + k X cosω t = F0 cosω t which can be divided throughout by cosω t yielding

( k − mω ) X = F 2

X =

or

F0 k − mω

2

=

F0 k 2

1 − mω k

0

=

X st

1 − ( ω ωn

)2

.

(2.24)

In (2.24) F0 (2.25) k is the static deflection of the spring under the (constant) load F0 and ω n = k m is the undamped natural circular frequency (2.4). X st =

Provided that ω ≠ ω n , the general solution of equation (2.22) is

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25

x (t ) = C1 sin ωn t + C2 cos ωn t +

X st

1 − (ω ωn ) 2

cos ω t .

(2.26)

Being the sum of two harmonic waves of different frequencies, the solution (2.26) is not a harmonic motion. Let the initial displacement and velocity be given by the constants x0 and

v 0 . Equation (2.26) yields x (0 ) = C2 +

X st

1 − (ω ωn ) 2

x& (0) = C1ωn = v0 ,

= x0 ,

so that the total response is x (t ) =

⎤ ⎡ X st X st sinωn t + ⎢ x0 − cos ω t . ⎥ cosωnt + 2 ωn 1 − (ω ωn ) ⎥⎦ 1 − (ω ωn ) 2 ⎣⎢ v0

(2.27)

For zero initial conditions, x0 = v 0 = 0 , the response (2.27) becomes x (t ) =

X st

1 − (ω ωn ) 2

(cos ω t − cos ωnt ) .

(2.28)

2.2.3 Beats The difference of cosines in equation (2.28) can be expressed as a product x (t ) = where

ωm =

2 X st

1 − (ω ωn ) 2

ωn + ω 2

sin ωm t sin Δω t ,

and

Δω =

(2.29)

ωn − ω 2

.

In the case when Δω becomes very small, since ω m is relatively large, the product in equation (2.29) represents an amplitude modulated oscillation. The harmonic motion with higher frequency ω m is amplitude modulated by the harmonic motion with lower frequency Δω (Fig. 2.10). The resulting motion, which is a rapid oscillation with slowly varying amplitude, is known as beats. The terminology is derived from acoustics. For instance when two strings for the same note on a pianoforte are slightly out of tune, a listener hears the sound waxing and waning (beating). The beats disappear when the strings are in unison, and there is then only one frequency audible.

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Fig. 2.10 Beats can be heard in an airplane when the two engines have slightly different speeds. It occurs also in electric power stations when a generator is started. Just before the generator is connected to the line, the electric frequency of the generator is slightly different from the line frequency. Thus the hum of the generator and the hum of other generators or transformers are of different pitch, and beats can be heard.

2.2.4 Frequency Response Curves It is of interest to examine more closely the frequency dependence of the steady-state response amplitude X=

1

1 − (ω ω n ) 2

X st .

(2.30)

The absolute value of the coefficient of X st in the right hand side of Eq. (2.30) is referred to as the dynamic magnification factor. Figure 2.11, a is a plot of the amplitude X as a function of the driving frequency ω . For ω ωn < 1 the ordinates are positive, the force and motion are in phase, while for ω ωn > 1 the ordinates are negative, the force and motion are 180 0 out of phase (Fig. 2.11, b). Whereas for ω ωn < 1 the mass is below the static equilibrium position when the force pushes downward, for ω ωn > 1 the mass is above the equilibrium position while the force is pushing downward.

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27

Fig. 2.11 Usually this phase relation is considered of slight interest, therefore the resonance curve is plotted as in Fig. 2.11, c with the modulus of amplitude in the ordinate axis. This is often referred to as a frequency response curve.

2.2.5 Resonance At ω ωn = 1 , when the forcing frequency coincides with the system natural frequency, the amplitude becomes infinitely large (because the system is undamped). This phenomenon is known as “resonance”, and the natural frequency is sometimes also called the “resonance frequency”. At ω = ω n the spring force and the inertia force balance each other and the exciting force increases the amplitude of motion of the undamped system without bound. Damped systems have finite amplitudes at resonance and the phase angle between force and displacement is 90 0 (Fig. 2.28). Consider the case when, starting from rest, the mass-spring system is subjected to a force of instantaneous magnitude F0 cos ωn t , where ω n is the

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28

natural frequency. As ω becomes exactly equal to ω n , the solution (2.27) is no

longer valid. Substitution of F (τ ) = F0 cos ωn τ into equation (2.21) yields t

F x (t ) = 0 mωn F ⎡ x (t ) = 0 ⎢ sin ωn t mωn ⎢ ⎣

∫

∫ cos ω τ sin ω (t − τ )dτ , n

n

0

t

t

2

cos ωn τ dτ − cos ωn t

0

x P (t ) =

∫ 0

F0 mωn

⎤ cos ωn τ sin ωnτ dτ ⎥ , ⎥ ⎦

t sin ωn t . 2

(2.31)

Thus, when excited at resonance, the amplitude of an undamped system increases linearly with time. Because the excitation is a cosine function and the response is a sine function, there is a 90 0 phase angle between them. The same result can be obtained using the limit theorems from calculus.

Fig. 2.12 The total solution for non-zero initial conditions is now of the form

x (t ) =

v0

ωn

sin ωn t + x 0 cos ωn t +

F0 t sin ωn t . 2 m ωn

(2.32)

A plot of x (t ) versus time is given in Figure 2.12 for zero initial conditions. It can be seen that x (t ) grows without bound, but it takes a time for the displacement amplitude to build-up.

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29

2.2.6 Acceleration through Resonance For most practical vibrating systems, the steady amplitude is achieved quickly and the rate at which it is approached is of little interest. However, when a vibrating system is driven through the resonance, i.e. when the forcing frequency is swept with some speed ε = dω dt , there is no time to reach a steady-state condition and the resonance amplitude is finite even for undamped systems. Thus the response to a force of variable frequency may be of major interest when running through a resonance. The response exhibits a resonance-like peak, sometimes followed by a beating-like response. If the sweep is upwards in frequency (Fig. 2.13), the peak frequency is higher that that obtained for steady-state conditions, the peak amplitude is lower and the width of the resonance curve is larger. If the sweep is downwards in frequency, the peak frequency is lower than the steady-state π⎞ ⎛1 resonance frequency. In Fig. 2.13, f (t ) = F0 sin ⎜ ε t 2 + ⎟ and ε = const . 2⎠ ⎝2

Fig. 2.13 The effect of sweep rate is dependent on the system damping, because the lighter the damping, the longer the time to reach the steady-state level of vibration. Figure 2.13 is plotted for zero damping.

2.2.7 Resonance for Constant Displacement Amplitude Resonance relates to the condition where either a maximum motion is produced by a force of constant magnitude, or a minimum force is required to maintain a given motion amplitude.

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30

When the force amplitude F is variable and the displacement amplitude X 0 is kept constant, equation (2.24) can be written

[

]

F = k X 0 1 − (ω ωn ) 2 .

(2.33)

Figure 2.14 is a plot of the force modulus as a function of the driving frequency for X 0 = const. For an undamped system, the force at resonance is zero, because the spring force is balanced by the inertia force.

Fig. 2.14 Resonance is a condition whereby a minimum of excitation is required to produce a maximum of dynamic response.

2.2.8 Excitation with Unbalanced Rotating Masses For many systems, vibrations are produced by driving forces from unbalanced rotating masses. In contrast to the constant-force-amplitude case previously discussed, the rotating-mass-type force has an amplitude proportional to the square of the frequency of vibration. The vibratory force is thus m1eω 2 cosω t , where m1 is the eccentric mass located at an eccentricity e (Fig. 2.15, a). The amplitude of the forced vibrations produced by such a force can be obtained by a substitution of m1eω 2 for F0 in equation (2.24). Then

X=

m 1 eω 2 k − m ω2

=

m1eω 2 k

1 − (ω ωn ) 2

=e

(ω ωn ) 2 . 1 − (ω ωn ) 2

(2.34)

2. SIMPLE LINEAR SYSTEMS

31

It should be pointed out that m is the total vibrating mass and includes the mass m1 .

a

b Fig. 2.15

Figure 2.15, b is a plot of the absolute value of X from equation (2.34) as a function of the circular frequency ω , for e = const . The curve starts from zero, goes to infinity at resonance and decreases to e for high frequencies.

2.2.9 Antiresonance Consider the mass-spring ungrounded system from Fig. 2.16, subjected to a harmonic force applied to the base. The equations of motion can be written

− m x&&2 = k (x 2 − x1 ) = F0 cosω t . The magnitude of the driving point displacement is given by X1 =

F0 k − m ω 2 F0 1 − (ω ωn ) 2 = . k k mω 2 (ω ωn ) 2

For constant force amplitude F0 = const . , its modulus has a minimum zero value at the natural frequency. This is a condition of antiresonance. Generally, it takes place at a frequency at which a maximum of force magnitude produces a minimum of motion. Unlike the resonance, which is a global property of a vibrating system, independent of the driving point, antiresonance is a local property, dependent on the driving location.

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Fig. 2.16 In the absence of damping, the antiresonance frequency of the base excited sprung-mass system is the same as the resonance frequency of the grounded massexcited system. If a second mass is attached at the driving point, the resulting massspring-mass system exhibits both a resonance and an antiresonance in the drivingpoint response.

2.2.10 Transmissibility If the mass-spring system is excited by a prescribed motion x 1 = X 1 cosω t applied to the spring end not connected to the mass, then the motion transmitted to the mass x 2 = X 2 cosω t is defined by the amplitude ratio X2 1 = . X 1 1 − (ω ωn ) 2

(2.35)

The ratio TR = X 2 X 1 is called transmissibility and is plotted in Fig. 2.17 as a function of the frequency ratio ω ωn . For ω ωn > 2 , the transmissibility is less than unity and the sprung mass is said to be isolated from the base motion. Vibration isolation is possible only above resonance, for frequencies ω > 2ω n . The spring between the mass and the vibrating base can be designed to ensure a given degree of isolation, by imposing the value of TR . This shows how much the motion of the isolated mass is reduced with respect to the case when it had been directly mounted on the vibrating base.

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33

Fig. 2.17

2.2.11 Critical Speed of Rotating Shafts Consider the rotor shown in Fig. 2.18, consisting of a single rigid disc symmetrically located on a uniform massless shaft supported by two rigid bearings. The disc centre of mass G is at a radial distance e from its geometric centre C. The centre line of the bearings intersects the plane of the disc at point O. As the shaft starts to rotate about the bearing axis, the disc rotates in its own plane about its geometric centre C. A centrifugal force m rG ω 2 is thus applied to the disc, where ω is the speed of rotation, m is the mass of the disc and rG = OG . This force causes the shaft to deflect in its bearings and the shaft is said to be in a state of unbalance. The shaft reacts with a restoring force k rC acting in C, where k is the stiffness of the shaft at the disc and rC = OC . Neglecting the effect of gravity and damping, the disc is under the action of only these two forces. In order to be in equilibrium, these forces must be collinear, equal in magnitude, and opposite in direction k rC = mω 2 (rC + e ) . Solving for rC , we obtain rC = where ω n = zero speed.

m ω 2e k − mω 2

=

e (ω ω n ) 2

1 − (ω ω n ) 2

.

(2.36)

k m is the natural circular frequency of the rotor lateral vibration at

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34

This expression represents the radius of the orbit along which the point C moves about the bearing axis with an angular velocity ω . Because at the same time the disc rotates in its own plane about C with the same angular velocity, the shaft whirling is called synchronous precession.

Fig. 2.18 The radius of the circular orbit of point G is rG = rC + e =

e

1 − (ω ωn ) 2

.

(2.37)

A plot of rC (solid line) and rG (broken line) as a function of ω is given in Fig. 2.19. At a speed ω 1 < ωn the system rotates with the heavy side G1 outside C1 , whereas for ω 2 > ω n the light side, or the side opposite G 2 , is outside C 2 .

For very high speeds, ω >> ω n , the radius rC becomes equal to the eccentricity and the points O and G coincide; the disc rotates about its centre of gravity. When ω = ω n , the radii rC and rG grow without bound, a state defined as a critical speed. Equations (2.36) and (2.37) indicate that the critical speed of the shaft is equal to the natural frequency of the lateral vibration of the rotor. The sudden change of the relative position of points O, C and G at the critical speed is due to the neglection of damping. In damped systems, the segment CG rotates continuously with respect to OC, when the shaft speed varies, so that the “high point” does not coincide with the “heavy point”. At the critical speed, the angle between the two segments is 90 0 (see Sec. 2.4.11)

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35

Fig. 2.19 Although there is an obvious analogy between the analytical results (2.36) and (2.37) on one hand, and the steady-state response of a linear mass-spring system (2.30) and (2.34) on the other hand, the forced motion of the shaft is not a genuine vibration. The shaft does not experience any alternating stresses while executing this motion. It just bows out in a simple bend. The bend is greatest when the angular speed is equal to the circular frequency of bending vibration that the shaft would have if it did not rotate and were simply executing free undamped flexural vibrations.

2.3 Damped Free Vibrations During vibration, energy is dissipated by friction or other resistances. The motion amplitude in free vibration diminishes with time, while the steady amplitude can be maintained only by external forcing. The dissipation of energy is generally termed damping. It is produced by internal friction in materials, by friction between structural components, by fluid-structure interactions, by radiation or by movement in electric or magnetic fields. The simplest damping mechanism is due to movement in a viscous medium, and the viscous damping force is directly proportional to velocity. It is convenient to replace all damping forces by a single equivalent viscous damping force based on the same value of energy dissipated during a cycle of vibration. Structural or hysteretic damping is described by a damping force in phase with velocity but proportional to the displacement. Experience has indicated that in aircraft structures the damping loss is better represented by the hysteretic damping. More complicated mechanisms, such as hereditary damping, can be used to better describe the behaviour of actual systems.

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2.3.1 Viscous Damping The system shown in Fig. 2.20, a consists of a linear spring of stiffness k, a mass m and a viscous damper or dashpot. The force in the dashpot is directly proportional to velocity and of opposite sign. The proportionality coefficient is referred to as the viscous damping coefficient, c, having units of N (m sec ) .

Fig. 2.20 For free vibrations, the differential equation of motion can be obtained by use of Newton’s second law and the free body diagram from Fig. 2.20, b m x&& = −c x& − k x , which can be written m x&& + c x& + k x = 0 .

(2.38)

Assuming solutions of the form x = e s t , we obtain the characteristic equation k c (2.39) s2 + s + = 0 , m m which has two roots 2

c k ⎛ c ⎞ ± ⎜ ⎟ − . 2m m ⎝ 2m ⎠ The general solution for the damped free vibrations is s 1, 2 = −

(2.40)

x ( t ) = C1 e s1 t + C2 e s 2 t ,

(2.41)

where the integration constants are determined from the initial conditions. As a reference quantity, we define critical damping as corresponding to the value of c for which the radical in (2.40) is zero

2. SIMPLE LINEAR SYSTEMS

37

cc = 2m or

k = ωn , m

cc = 2 k m = 2mωn .

(2.42)

The actual damping of the system can be specified by a dimensionless quantity, which is the ratio of the actual system damping to the critical system damping ζ=

c cc

(2.43)

referred to as the damping ratio (or percent of critical damping) Using this notation, equation (2.40) becomes s 1, 2 = ⎛⎜ − ζ ± ⎝

ζ 2 − 1 ⎞⎟ ω n . ⎠

(2.44)

Three possible cases must be considered for the above equations, depending on whether the roots (2.44) are real, complex, or equal. Case I: Underdamped system, ζ < 1

For ζ < 1 , equation (2.44) can be written s 1, 2 = ⎛⎜ − ζ ± i 1 − ζ 2 ⎞⎟ ω n . ⎝ ⎠

(2.45)

Substitution of (2.45) into (2.41) and conversion to trigonometric form with the aid of Euler’s formula eiβ = cosβ + i sinβ , yields ⎛ i x (t ) = e − ζ ω n t ⎜ C1 e ⎝

1− ζ 2 ω n t

+ C2 e

- i 1− ζ 2 ω n t

⎞ ⎟, ⎠

or x (t ) = A e − ζ ω n t sin ⎛⎜ 1 − ζ 2 ωn t + φ ⎞⎟ . ⎝ ⎠

(2.46)

Equation (2.46) indicates that the motion is oscillatory with diminishing amplitude. The decay in amplitude with time is proportional to e − ζ ωn t , as shown by the dashed curves in Fig. 2.21. The frequency of the damped oscillation

ω d = 1− ζ 2 ω n

(2.47)

is less than the undamped natural frequency ω n and is called the damped natural frequency. As ζ → 1 , ω d approaches zero and the motion is no more oscillatory.

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38

Equation (2.44) can be written

where

s 1, 2 = −σ ± iω d

(2.48)

σ = ζ ωn

(2.49)

is the rate of decay of amplitude (slope of tangent to the exponential curve at t = 0 ).

Fig. 2.21 The following equations are useful ζ=

σ ωd2

+σ

2

,

ωn =

Fig. 2.22

σ ζ

= ωd2 + σ 2 .

(2.50)

2. SIMPLE LINEAR SYSTEMS

39

Case II: Overdamped system, ζ > 1

For ζ > 1 , substitution of (2.44) into (2.41) yields ⎛ −ζ + ⎜

x (t ) = C1 e ⎝

ζ 2 −1 ⎞⎟ ω n t ⎠ +C

⎛ −ζ − ⎜

⎝ 2e

ζ 2 −1 ⎞⎟ ω n t ⎠

.

The motion is no longer oscillatory (Fig. 2.22) and is referred to as aperiodic.

Fig. 2.23 Case III: Critically damped system, ζ = 1

Critical damping represents the transition between the oscillatory and nonoscillatory motions. In this case, the general solution is

x ( t ) = ( C1 + C2 t ) e −ω n t . The motion is similar to that with damping greater than critical (Fig. 2.23) but returns to rest in the shortest time without oscillation. This is used in electrical instruments whose moving parts are critically damped to return quick on the measured value.

2.3.2 Logarithmic Decrement A way to determine the amount of damping in a vibrating system is to measure the rate of decay of oscillations. This is conveniently expressed by the logarithmic decrement, which is defined as the natural logarithm of the ratio of any two successive amplitudes. For viscous damping, this ratio is a constant. Consider the recorded curve of a damped vibration (Fig. 2.24), expressed by equation (2.46).

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40

Fig. 2.24 The decaying sinusoid is tangent to the exponential envelope at points that are slightly to the right of the points of maximum amplitude, where the sine function is equal to 1. However, this difference is negligible, so the ratio of two successive amplitudes can be replaced by the ratio of exponential ordinates at a period distance x1 A e − ζ ωn t = = e ζ ωn Td ( ) − + ζ ω t T n d x2 A e where the period of the damped vibration is Td =

2π

ωn 1− ζ 2

=

2π

ωd

.

The logarithmic decrement is

δ = ln

x1 = ζ ωn Td = x2

2πζ 1− ζ2

.

(2.51)

For ζ 2 , the system has to pass through the resonance, where the amplitude is reduced by damping. In some cases, there are provisions for some light damping and the large amplitudes are limited by stops or by acceleration through resonance. A similar problem can be formulated for a mass-excited grounded system (Fig. 2.26). If the driving force to be isolated is F0 ei ω t (2.73), and the steady-state complex displacement amplitude is X (2.74), then the transmitted force through the spring and damper is also harmonic with an amplitude FT =

(k X )2 + (cω X )2

(2.104)

so that the force transmissibility TR = FT F0

(2.105)

is given by equation (2.102). Note that the force transmitted through the spring and damper is phase shifted with respect to the elastic force and the damping force.

2.4.10 Theory of Seismic Instruments There are two basically different instruments for vibration measurement: a) fixed reference instruments or quasistatic devices, in which the vibratory motion is measured relative to some fixed reference point, and b) seismic instruments, in which the vibratory motion is measured relative to the mass of a mass-springdashpot system attached to the vibrating structure.

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63

The seismic instrument (Fig. 2.40) consists of the casing S, rigidly attached to the vibrating system, the mass-spring-dashpot m-k-c system, and the transducer T, that measures the relative motion between the seismic mass and the casing. It is assumed that the vibrating system, hence the instrument base, experiences a harmonic motion x 1 (t ) = X 1 cos ω t .

(2.106)

Neglecting transient terms, the relative displacement between the mass m and the casing S can be defined by x r (t ) = X r cos (ω t − ϕ ) .

(2.107)

The absolute displacement of the mass m, relative to a fixed reference point, is

x 2 = x1 + x r and the absolute acceleration is

&x&2 = &x&1 + &x&r . The equation of motion of the mass m can be written m (&x&1 + &x&r ) + c x&r + k xr = 0

or

m &x&r + c x&r + k xr = −m &x&1 = m X1 ω 2cos ω t .

(2.108)

Fig. 2.40 Equation (2.108) has a steady-state solution for which Xr = X1

(ω

ωn ) 2

[1 − (ω ω ) ] n

2 2

+ (2ζ ω ωn )

, 2

(2.109)

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64

tan ϕ =

2ζ ω ω n

1 − (ω ωn ) 2

.

Fig. 2.41

Fig. 2.42

(2.110)

2. SIMPLE LINEAR SYSTEMS

65

Figure 2.41 shows the variation of the amplitude ratio (2.109) plotted against ω ωn for two values of the damping ratio. Figure 2.42 shows the variation of the phase shift ϕ plotted against ω ωn . Depending on the frequency range utilized, the instrument indicates displacement, velocity or acceleration. Vibrometer. Within the range III, when ω >> ω n , it can be seen that

X r ≅ X 1 , so that the relative motion X r between the mass and casing, sensed by the transducer, is essentially the same as the displacement X 1 of the structure being measured. Figure 2.42 shows that, within this frequency range, the phase shift is ϕ = π for light damping (ζ → 0 ) , so that the casing and the mass m are vibrating 180 0 out of phase. Relative to an inertial frame (fixed reference point) the mass m remains nearly stationary (becomes a fixed point in space) and the casing motion is measured with respect to it. When T is a displacement transducer, the instrument is a seismic absolute displacement pickup (vibrometer). When T is a velocity transducer, the instrument becomes a velocity pickup. Seismic displacement-measuring instruments should have very low natural frequencies (1 to 5 Hz) which are obtained with low values of k, hence with a soft suspension of the seismic mass, respectively, with relatively large masses m. Accelerometer. Within the range I, for ω ω n , the disc rotates with G inside C, and OC decreases. At very high speeds, point G coincides with point O, the radius rC approaches e, and the shaft rotates about its centre of mass.

2.4.12 Hereditary Damping Viscous damping has been considered in the simplest model, consisting of a dashpot in parallel with a spring (Fig. 2.26). It is said that the Kelvin-Voigt model has directly-coupled damping.

Fig. 2.46 Other simple models incorporate elastically-coupled viscous damping mechanisms. In the three-parameter Maxwell model, the dashpot is introduced in series with another spring (Fig. 2.46, a). The system has two degrees of freedom. The equations of motion can be written

m &x& + k x + c (x& − x&1 ) = f ,

c (x& − x&1 ) = k1 x1 .

(2.115)

For an excitation f = F0 ei ω t , assume solutions of the form x = X e i ω t , x1 = X 1 ei ω t ,

where X and X 1 are complex amplitudes.

(2.116)

2. SIMPLE LINEAR SYSTEMS

69

Equations (2.115) become

( k − mω

2

)

+ i ω c X − i ω c X 1 = F0 ,

− i ω c X + ( k1 + i ω c ) X 1 = 0 ,

and can be written

(1 − β

2

)

+ i 2 ζ β X − i 2 ζ β X 1 = F0 k ,

− i 2 ζ β X + ( N + i 2 ζ β ) X 1 = 0,

(2.117)

(2.118)

where

ω n = k m , β = ω ωn , ζ = c 2 m ωn , N = k1 k .

Fig. 2.47 The complex displacement amplitude of the mass m is

(2.119)

MECHANICAL VIBRATIONS

70

X F0 k

=

1+ i 2ζβ N . 1 − β + i (2 ζ β N ) N + 1 − β 2

(

2

)

(2.120)

The displacement amplification factor X F0 k

=

(1 − β )

2 2

1 + (2 ζ β N ) 2

(

+ (2 ζ β N )2 N + 1 − β 2

)

2

.

(2.121)

is graphically presented in Fig. 2.47 for a stiffness ratio N = 5 and different values of the damping ratio. The corresponding phase angle is presented in Fig. 2.48.

Fig. 2.48 The expression (2.121) can be squared and written under the form 2

2 ⎛ X ⎞ ⎜ ⎟ = ψ 2 = C1 + ζ C2 . ⎜ F0 k ⎟ C3 + ζ 2 C4 ⎝ ⎠

Equation (2.122) can also be written as

(

)

C3 ψ 2 − C1 + C4 ψ 2 − C2 ζ 2 = 0 .

(2.122)

(2.123)

All curves represented by equation (2.123) are passing through the crossing point of the curves of equations C 3 ψ 2 − C1 = 0 ,

These can be expressed as

C4 ψ 2 − C2 = 0 .

2. SIMPLE LINEAR SYSTEMS

ψ=

71

ψ=

C1 C3 ,

C2 C4 ,

or X F0 k

=

1 , 1− β 2

(2.124)

=

1 . N +1− β 2

(2.125)

and X F0 k

Equation (2.124) represents the curve (2.123) of parameter ζ = 0 . Equation (2.125) represents the curve (2.123) of parameter ζ = ∞ . The two curves intersect each other at a point of frequency ratio β = X

(F0 k ) = 2

(N + 2) 2

and of ordinate

N . All frequency response curves are passing through this point.

Figure 2.47 indicates that small variations in damping may result in pronounced variations in resonance frequency. This is totally different from systems with directly coupled viscous damping (Fig. 2.28), where the variation of the resonance frequency with damping is negligible. The resonance frequency grows from β = 1 for ζ = 0 , to β = 1 + N for ζ = ∞ . When the damping increases, the peak of the response first decreases, then increases, thereby indicating that an optimum degree of damping exists ζ opt = N

2 (N + 2 ) ,

for which the resonant response is a minimum, equal to the ordinate of the crossing point of all curves drawn for various damping ratios. For N > 2 , over the damping range of 0.7 < ζ < ζ opt , no resonance appears in the frequency response curve. This behaviour can be explained considering the dynamic response of the three parameter spring-dashpot model from Fig. 2.46, b. Its behaviour is described by two equations

f1 = k x + c ( x& − x&1 ) ,

(2.126)

c ( x& − x&1 ) = k1 x1 .

(2.127)

Since we are not interested in the “hidden” coordinate x1 , containing the internal degree of freedom, we solve equation (2.127) for x1 and substitute the result in equation (2.126) to obtain

MECHANICAL VIBRATIONS

72 t

f 1 = k x + G ( t − τ ) x& (τ ) dτ ,

∫

(2.128)

0

where G (t ) = k1 e

−

k1 c

t

(2.129)

with the underlying assumption that for t = 0 the model is unstrained. In equation (2.128) the damping term depends on the past history of the velocity. For this reason it is called “hereditary damping”. When the force f1 is given as a function of time, the solution of equations (2.126) and (2.127) is f (t ) 1 ⎛⎜ k 1 ⎞⎟ x (t ) = 1 + k + k 1 c ⎜⎝ k + k 1 ⎟⎠

2

t

∫e

−

τ τ1

f 1 (t − τ ) dτ ,

(2.130)

0

where ⎛1 ⎜k ⎝

τ 1 = c⎜ +

1 ⎞⎟ k 1 ⎟⎠

is called the “time constant” of the model. The first term in the right of equation (2.130) describes the instantaneous response, actually noticed for many damped systems. Next consider the forced response to harmonic excitation. Substituting the complex solutions (2.116) into equations (2.126) and (2.127), then eliminating the internal coordinate, we obtain f1 = k x ,

(2.131)

where the complex stiffness k is k = k +i

ω c k1 . k1 + iω c

(2.132)

When equation (2.132) is split into its real and imaginary parts, it may be written in the form obtained for the model with directly-coupled viscous damping k = ke + iω ce ,

(2.133)

where the equivalent stiffness and equivalent coefficient of viscous damping are k e = k + k1

ω 2c 2 , k12 + ω 2 c 2

ce = c

k12 k12 + ω 2 c 2

.

(2.134)

2. SIMPLE LINEAR SYSTEMS

73

The model with hereditary damping is thus reduced to a Kelvin-Voigt model with frequency-dependent parameters. The equivalent spring stiffness k e increases with frequency from k to the asymptotic value k + k1 , whilst the equivalent coefficient of viscous damping ce decreases from c to zero. The energy dissipated per cycle is Wd = π X 2ω ce = π X 2

ω c k12 k12 + ω 2 c 2

(2.135)

which is zero for ω = 0 and ω = ∞ , having a maximum value at ω 0 = k1 c where

ce = c 2 .

Fig. 2.49 This is also reflected in the hysteresis curves (Fig. 2.49). At zero frequency we have a straight line corresponding to a pure spring of stiffness k. For ω = ω 0 there is an ellipse of maximum area. When the frequency tends to infinity, we have a straight line of smaller slope, corresponding to a pure spring of stiffness k + k1 .

Exercises 2.E1 An unknown mass m is hung on a spring of unknown stiffness k. When a mass m 1 = 0.5 kg is added to m, the system natural frequency is lowered

from 50 Hz to 49 Hz . a) Determine the values of m and k. When second spring of stiffness k ′ is added in parallel with the first spring, the natural frequency is increased to 50 Hz . b) Determine the value of k ′ . Answer: m = 12.126 kg , k = 1.19 ⋅ 10 6 N m , k ′ = 56 N m .

MECHANICAL VIBRATIONS

74

2.E2 A mass m = 0.5 kg vibrating in a viscous medium has a period T = 0.15 sec and an initial amplitude a 0 = 10 mm . a) Determine the stiffness k and the viscous damping coefficient c if the amplitude after 12 cycles is a12 = 0.2 mm . ′ when a mass m1 = 0.3 kg is added to m. b) Determine the amplitude a12

′ = 0.45 mm . Answer: k = 877 N m , c = 2.173 N s m , a12 2.E3 A spring-mass system with a natural frequency f = 5 Hz vibrates in a viscous medium having a damping coefficient c = 0.002 N s mm . The logarithmic decrement is δ = 0.31 . a) Determine the mass m and the spring stiffness k. b) Find the new value of the logarithmic decrement if a mass m 1 = 1 kg

is added to the first mass. Answer: m = 0.645 kg , k = 636.4 N m , δ ′ = 0.194 . 2.E4 A spring-mass system with viscous damping is displaced a distance a 0 = 20 mm and released. After 8 cycles of vibration the amplitude decays to a8′ = 4 mm . A mass m 1 = 2 kg is attached to the initial mass producing a static

displacement of 4 mm . If the new system is displaced a distance a 0 = 20 mm and released, after 8 cycles of vibration the amplitude decays to a8′′ = 5 mm . Determine the mass m, the stiffness k and the damping coefficient c. Answer: m = 5.77 kg , k = 4905 N m , c = 10.76 Ns m . 2.E5 A vibrating system of mass m = 5 kg and spring stiffness k = 1 N mm is acted upon by a harmonic force of amplitude F0 = 10 N and frequency 2 Hz . Determine a) the displacement amplitude X; b) the displacement amplitude X ′ when a mass m 1 = 2 kg is added to m; c) for the system with the

added mass, how should k be modified so as the displacement amplitude to become X again. Answer: X = 47.34 mm , X ′ = 95.85 mm , add k ′ = 315.5 N m in parallel, or k ′ = 8546.5 N m in series. 2.E6 A weight m g = 20 N attached to a light spring elongates it 1 mm . Determine a) the amplitude F0 of the force that, acting with a frequency of 20 Hz , produces vibrations with an amplitude X = 3.5 mm . b) Find another excitation

2. SIMPLE LINEAR SYSTEMS

75

frequency at which the force of magnitude F0 produces the same displacement X = 3.5 mm . Answer: F0 = 42.68 N , ω ′ = 61.87 rad sec . 2.E7 A small motor of mass m = 2 kg is found to be transmitting a force of 14 N to its supporting springs when running at a speed of 30 rad sec and has an amplitude of vibration of 3.5 mm . Determine a) the amplitude of the unbalanced force F0 ; b) Find another value of the running speed at which the amplitude of vibration has the same magnitude.

Answer: F0 = 7.7 N , ω ′ = 55.67 rad sec . 2.E8 A fragile instrument of mass m = 1 kg is used on a table that is vibrating because of its proximity to machines running in the area. The table has a harmonic motion of amplitude X 1 = 0.2 mm and frequency f = 15 Hz . To reduce the vibration of the instrument, it is isolated from the table by means of rubber pads of stiffness k = 7,000 N m . Determine a) the amplitude X of the displacement of the instrument: b) the value X ′ of the amplitude, if a pad of stiffness k ′ = 2,000 N m is added in parallel with the first pad.

Answer: X = 0.743 mm , X ′ = 15.148 mm . 2.E9 A mass m = 4 kg is supported between two springs attached to fixed points is excited by a harmonic force of amplitude F0 = 10 N . The upper spring has a stiffness k = 5,000 N m and the lower spring has a stiffness 2k. Determine a) the frequencies at which the displacement amplitude is X = 20 mm , and b) the amplitude F0′ of the force transmitted to the lower support through the spring 2k.

Answer: ω 1 = 60.2 rad sec , ω 2 = 62.25 rad sec , F0′ = 200 N . 2.E10 An undamped vibrating system has an equivalent mass m and an equivalent stiffness k. A harmonic force of amplitude F0 = 1 N and frequency ω = 14 rad sec produces steady-state vibrations of amplitude X. When a mass m1 = 2 kg is added to m, the force of amplitude F0 = 1 N must have a frequency of either ω = 10 rad sec or ω = 12 rad sec in order to produce the same displacement amplitude X. Determine the values of m and k.

MECHANICAL VIBRATIONS

76

Answer: For

ω < ω1 ,

m = 2.083 kg ,

k = 498.1 N m ; for

ω > ω1 ,

m = 5.538 kg , k = 919.6 N m . 2.E11 A machine weighing m g = 12,000 N is mounted at the middle of a

beam of length l = 2 m , cross section second moment of area I = 20 mm 4 and Young’s modulus E = 2.1 ⋅ 10 5 N mm 2 , simply supported at the ends. If the machine has a rotating unbalanced weight m0 g = 2,000 N with eccentricity e = 0.1 mm and speed n = 1500 rpm , determine a) the amplitude X of the forced vibrations of the machine, and b) the magnitude of the dynamic force transmitted to only one of the supports. Answer: X = 50.22 μm , Fdin = 632.77 N . 2.E12 A motor of mass m = 10 kg , supported on springs of total stiffness k = 2,000 N m , generates an unbalanced force of magnitude F0 = 20 N at a frequency of 2 Hz . Determine the stiffness k ′ of a spring and how should it be mounted (in series or in parallel) so as to lower to a half the magnitude of the transmitted force FT 0 . Find the value of another frequency of the excitation force having the same effect.

Answer: k ′ = 731.6 N m in parallel, ω ′ = 10.747 rad sec . 2.E13 A vibrating system consists of a mass m = 0.2 kg and a spring of stiffness k = 0.2 N mm . It is acted upon by a harmonic force of amplitude F0 = 5 N and frequency ω = 35 rad sec . Determine a) the amplitude FT 0 of the force transmitted to the support, and b) the amplitude X of the displacement of the mass m.

Answer: FT 0 = 22.22 N , X = 0.11 m . 2.E14 A spring-mass system with viscous damping is acted upon by a harmonic force of amplitude F0 = 10 N . Varying the frequency of the excitation force, a peak amplitude X 1 = 60 mm is measured at the frequency ω 1 = 14 rad sec .

When a mass m 1 = 2 kg is added to the system, the peak amplitude becomes

X 2 = 80 mm . Determine the mass m, the stiffness k and the damping coefficient c. Answer: m = 2.57 kg , k = 503.72 N m , c = 11.9 N s m .

2. SIMPLE LINEAR SYSTEMS

77

2.E15 A mass m = 5 kg is attached to a spring of stiffness k = 1 N mm and is acted upon by a harmonic force of amplitude F0 = 10 N and frequency f = 2 Hz . Determine: a) the amplitude X 1 of the forced vibrations of the mass m; b) the amplitude X 2 of the forced vibrations when a mass m 1 = 2 kg is added to

m; c) the coefficient of viscous damping c of a dashpot connected in parallel with the spring which reduces the amplitude of vibrations to the initial value. Answer: X 1 = 47.3 mm , X 2 = 92.5 mm , c = 10.3 N s m . 2.E16 Using Rayleigh’s method, estimate the fundamental frequency of the vibrating beams shown in Fig. 2.50 using the given deflection function.

π x⎞ ⎛ v (x ) = v0 ⎜1 − cos ⎟. 2l ⎠ ⎝

a

3 ⎡ x ⎛ x⎞ ⎤ v (x ) = v0 ⎢3 − 4 ⎜ ⎟ ⎥ , ⎝ l ⎠ ⎥⎦ ⎢⎣ l

b

0≤ x≤l 2 v (x ) = v0 sin

c

πx l

.

2π x ⎞ ⎛ v (x ) = v0 ⎜1 − cos ⎟. l ⎠ ⎝

d

Fig. 2.50 Answer: a) ω 1 = 1.612 c) ω 1 = 4.935

EI ml

3

; d) ω 1 =

EI ml

3

22.79 l2

; b) ω 1 = 6.928 EI . ρA

EI

ml (1 + 0.4857 ρ Al m ) 3

;

MECHANICAL VIBRATIONS

78

2.E17 Using Rayleigh’s method, estimate the fundamental frequency for the axial vibrations of the beam shown in Fig. 2.51 using a linear deflection function for the longitudinal displacement. Consider m = 2 ρ Al .

Fig. 2.51 Answer: ω 1 =

0.655 l

E

ρ

.

3. SIMPLE NON-LINEAR SYSTEMS

Methods for parameter estimation are presented, based on the analysis of frequency response curves. Only simple non-linear systems are considered, to show the distortion of frequency response curves due to the system slightly non-linear properties. This is not a full account of the non-linear dynamic behaviour of vibrating structures. A comprehensive treatment requires a separate book.

3.1 Non-Linear Harmonic Response If a non-linear system is excited by a harmonic force, the steady-state response is not harmonic, as for linear systems, but it is periodic, so that it can be expressed as a sum of harmonic components. For systems with local weak nonlinearities, a convenient analysis tool is the Harmonic Balance Method in which the basic assumption is that the response is dominated by the fundamental harmonic component. The steady-state response is considered to be a single harmonic at the excitation frequency, neglecting sub-harmonics or supraharmonics. The forced response is studied only in the neighbourhood of the socalled principal resonance. Similar results are obtained using the Equivalent Linearization Method and the concept of Describing Functions. The non-linear restoring force function is approximated by equivalent spring and damper forces. For linear systems, the restoring force function (minus sign omitted) may be taken of the form

f R ,lin (x , x& , t ) = c x& ( t ) + k x ( t ) , containing the contribution of a viscous damper and a linear spring. For slightly non-linear systems, the restoring force can be expressed as

f R (x , x& , t ) ≅ ceq x& ( t ) + keq x ( t ) ,

MECHANICAL VIBRATIONS

80

where ceq is the equivalent viscous damping coefficient, and k eq is an equivalent stiffness. An equivalent structural damping model may be used as well. Cubic Stiffness For systems without pre-loading or clearance, the elastic force may be represented by a cubic stiffness law

(

)

fe = k x + μ x3 ,

(3.1)

where k is the slope of the stiffness function at the origin and μ is a coefficient of non-linearity, positive for hardening and negative for softening springs. For a harmonic displacement

x ( t ) = a cos ω t ,

(

)

(3.2)

the elastic force (3.1) becomes f e = k a cos ω t + μ a cos ω t . 1 3 Substituting cos3ω t = cos ω t + cos 3ω t , and neglecting the higher 4 4 harmonic term in cos 3ω t , yields 2

3

3 ⎛ ⎞ f e ≅ k a ⎜ cos ω t + μ a 2 cosω t + ....⎟ = keq x , 4 ⎝ ⎠ where the equivalent stiffness is ⎛ 3 ⎞ (3.3) keq = k ⎜1 + μ a 2 ⎟ . ⎝ 4 ⎠ The dynamic response of the non-linear system is obtained substituting this amplitude-dependent stiffness into the equations obtained for the linear system.

Non-Linear Damping Non-linear damping may be studied using the equivalent viscous damping concept. This involves the approximation of a non-linear damping force by an equivalent linear viscous damping force. The criterion for equivalence is that the energy Wd dissipated per cycle of vibration by the non-linear damping element be equal to the energy π ω ceq a 2 dissipated by an equivalent viscous damper experiencing the same harmonic relative displacement. Similar analysis can be carried out based on a concept of equivalent structural damping when Wd = π heq a 2 . Steady-state solutions obtained with the assumption of equivalent viscous damping have been shown to be identical to those obtained by use of the averaging

3. SIMPLE NON-LINEAR SYSTEMS

81

method of Ritz. The same value is obtained using the coefficient of the first term of a Fourier series expansion of the non-linear damping force time history. A generalized non-linear damping force can be described mathematically as being proportional to the nth power of the relative velocity across the damper f d = cn x&

n

sgn ( x& ) ,

(3.4)

where cn is defined as the velocity-nth power damping coefficient. The exponent n and the damping coefficient cn are determined according to the nature of the damping element. A value of n = 0 represents a Coulomb damper where the damping coefficient c0 equals the dry-friction force R . Similarly, values of n = 1 and n = 2 represent viscous and quadratic damping, where cn becomes the damping coefficients c and c2 , respectively. Other values of the exponent n may be selected to represent other nonlinear damping characteristics. For example, the damping developed in a car shockabsorber may be described by a value of the exponent n between 2.0 and 3.0 , depending on the particular system configuration.

3.2 Cubic Stiffness A single-degree-of-freedom system with a non-linear spring and a linear structural damping element is shown in Fig. 3.1. A cubic law with positive coefficient will be taken as a first approximation for describing a hardening spring characteristic.

Fig. 3.1 Structural damping is a linear damping phenomenon for which the damping force is in phase with the relative velocity but is proportional to the relative displacement across the damper.

MECHANICAL VIBRATIONS

82 3.2.1 Harmonic Response

If the mass is acted upon by a harmonic force of constant amplitude F0 and frequency ω , the Duffing-type equation of motion of the vibrating mass may be written as h m &x& + x& + k x + μ x 3 = F0 e iω t , (3.5)

(

ω

)

where h = g k , and g can be an equivalent structural damping factor. A first harmonic approximation of the response is chosen of the form x = a~ e iω t = ( a R + i a I ) e iω t = a e i (ω t +θ ) .

(3.6)

Using the method of harmonic linearization, the higher harmonic terms are neglected, so that it is considered that 3 x3 ≅ a 2 x . (3.7) 4 Substitution of (3.6) and (3.7) into (3.5) yields the real and imaginary components of displacement 2

⎛ g k a2 ⎞ ⎟ , a − ⎜⎜ ⎟ F 0 ⎝ ⎠

⎞ k ⎛ 3 a R = ⎜1 + μ a 2 − η 2 ⎟ a 2 = m ⎠ F0 ⎝ 4 aI = − g

2

k 2 a , F0

(3.8) (3.9)

where

η=

ω , ωn

k . m

ωn =

(3.10)

The displacement magnitude a = a R2 + a I2

(3.11)

is implicitly given by F02

3 4

η 2 = 1 + μ a2 ±

2 2

k a

− g2 .

(3.12)

The phase angle is calculated from

tan θ =

g 3 η −1 − μ a2 4 2

g

= ±

F02 k 2 a2

.

(3.13)

− g2

Elimination of a and η 2 between equations (3.8) and (3.9) yields the locus of the end of the vector a~ in the Argand plane which is a circle of equation

3. SIMPLE NON-LINEAR SYSTEMS

83 2

a R2

2

⎛ ⎛ 1 F0 ⎞ 1 F0 ⎞ ⎟⎟ = ⎜⎜ ⎟⎟ . + ⎜⎜ a I + 2g k ⎠ ⎝ ⎝2g k ⎠

(3.14)

It is identical to equation (2.82) derived for linear systems. Elimination of a between equations (3.9) and (3.11) yields the frequency dependence of the quadrature component a I of response 3 4

η 2 = 1+ μ

F0 (− aI ) ± g gk

F0 1 −1 . g k (− aI )

(3.15)

Similarly, the frequency dependence of the in-phase component a R of the response may be obtained under the form 2

⎛ F ⎞ F 3μ ⎜⎜ 0 ⎟⎟ aR2 + 0 aR 2 k ⎛ F ⎞ ⎝ 2g k ⎠ η 2 = 1 + 3μ ⎜⎜ 0 ⎟⎟ − . 2 2⎡ ⎤ ⎝ 2g k ⎠ ⎛ 2g k ⎞ 2 ⎥ ⎛ F ⎞ ⎟⎟ aR 2 ⎜⎜ 0 ⎟⎟ ⎢ 1 ± 1 − ⎜⎜ ⎥ ⎢ ⎝ 2g k ⎠ ⎢ ⎝ F0 ⎠ ⎦⎥ ⎣

(3.16)

3.2.2 Frequency Response Characteristics

Based on equation (3.12), the magnitude-frequency curves are illustrated in Fig. 3.2 for a fixed value of damping g = const. and several values F0 of the amplitude of harmonic force. The response curves are symmetrically disposed with respect to the “skeleton curve” of equation ⎛ ⎝

3 4

⎞ ⎠

ω 2 = ωn2 ⎜ 1 + μ a 2 ⎟

(3.17)

which passes through the points of maximum amplitude. For linear systems it is a vertical line of abscissa ω = ωn . For systems with hardening stiffness (μ > 0) , the skeleton curve is bent towards higher frequencies. For systems with softening stiffness (μ < 0) , it is bent towards lower frequencies. The locus of the points of vertical tangency of the magnitude-frequency curves is given by equation 3 9 2 4 (3.18) η 2 = 1+ μ a2 ± μ a − g2 4 16 and defines the “stability boundary” XLKY (Fig. 3.2).

MECHANICAL VIBRATIONS

84

Points within the region whose limits are marked by this curve define unstable regimes of vibration. On the response curves they are drawn with broken lines.

Fig. 3.2

Fig. 3.3

The same information is contained in Fig. 3.3 where force-displacement curves are plotted at several frequencies for g = const. They are called constantfrequency lines or isochrones. The locus of the points of horizontal tangency of these curves defines the stability boundary. Point K defines the maximum force amplitude for which vibrations are stable irrespective of the displacement magnitude. Point L defines the maximum displacement amplitude for which vibrations are stable irrespective of the force level.

Fig. 3.4 The phase-frequency curves of Fig. 3.4 are based on equation (3.13).

3. SIMPLE NON-LINEAR SYSTEMS

85

Again, a stability boundary XLKY can be defined, of equation ⎛ 3 ⎞ ⎜⎜ tan θ + ⎟ (3.19) tan θ ⎟⎠ ⎝ which is the locus of the points of vertical tangency of the phase-frequency curves.

η 2 = 1+

g 2

The frequency response curves of the coincident (real) component are represented in Fig. 3.5, based on equation (3.16). In this case, the equation of the stability boundary XKY is g 2 1 9μ 2 + aR (3.20) η2 = 1+ 3μ aR2 4 and is valid for aR < 0 .

Fig. 3.5

Fig. 3.6

The frequency response curves of the quadrature (imaginary) component are illustrated in Fig. 3.6, based on equation (3.15). The stability boundary XKY is given by the equation g2 3 η 2 = 1 + μ aI2 + (3.21) 4 μ aI2 and holds for aR < 0 . The best way to represent the frequency response data is to plot the vector components of the displacement a R and a I , equations (3.8) and (3.9), on an Argand diagram, as in Fig. 3.7. The result is a family of circles of equation (3.14).

MECHANICAL VIBRATIONS

86

Polar plots have the advantage of combining the information about magnitude, phase and forcing frequency on a single diagram. At the same time, the region of interest in the neighbourhood of the principal resonance is enlarged and both types of “jump” phenomena can be easier explained. For non-linear systems it is important to plot response displacement components and not receptances (displacement/force) or other FRFs. Each point represented in the complex plane is defined by two parameters – the excitation frequency, ω , and the amplitude F0 of the input force. Consequently, two sets of response loci have to be drawn, i.e. isochrones - connecting constant-frequency points, and Nyquist plots - connecting points of constant excitation level. Since both the displacement amplitude and the phase angle are sensitive to the force amplitude, the distortion of these curves may be employed to provide an indication of the non-linear behaviour. The phase is more sensitive to non-linearities than the amplitude.

Fig. 3.7 In Fig. 3.7, isochrones are drawn by broken lines. Their equation is obtained by eliminating F0 between equations (3.8) and (3.9). This yields

(

)

3 ⎡ ⎤ a a R = ⎢η 2 − 1 − μ a R2 + a I2 ⎥ I . 4 ⎣ ⎦ g

(3.22)

For μ = 0 , i.e. for linear systems, equation (3.22) describes straight lines diverging from the origin of coordinates. For μ ≠ 0 , equation (3.22) describes curves passing through the origin, more distorted as F0 increases. As F0 grows, the isochrones are so much bent that they become tangent to the response curves.

3. SIMPLE NON-LINEAR SYSTEMS

87

The locus of the tangency points of Nyquist plots with the isochrones defines the stability boundary XLKY. This is a hyperbola of equation aR aI =

2g 3μ

(3.23)

(defined only for aR < 0 , aI < 0 ), which is symmetrical with respect to the bisector aR = aI of the coordinate axes. The stability boundary XLKY intersects the bisector aR = aI at the point L, of frequency ω L = ω n 1 + 2 g , at a distance a L = 4 g 3μ from the origin. For lower frequencies or smaller displacement amplitudes jump phenomena cannot occur. Point K, where the stability boundary is tangent to the polar plot of parameter

F0 = k

ω K = ωn 1 +

32 g 3

(9

3μ

)

and

to

the

isochrone

of

parameter

3 g , indicates the lowest force level and forcing frequency at

which unstable vibrations can occur. It corresponds to a phase angle θ K = −120 0 . Points K and L correspond to those marked on the diagrams of Figs. 3.5 - 3.7.

Fig. 3.8 The effect of non-linear stiffnesses is a shift of frequencies along the circular Nyquist plots, clockwise - for softening springs, and anti-clockwise - for hardening springs. The main resonance frequency is no more at maximum response amplitude, as shown in Fig. 3.8. For equal frequency increments, the maximum spacing between successive points is no more at the principal resonance, so that the Kennedy-Pancu criterion cannot be used for resonance location.

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Introducing a new dimensionless coefficient of non-linearity γ = μ F02 g 3k 2 , it means that for either γ > γ K = 32 9 3 or ω > ω K the stability boundary crosses the Nyquist plots and the isochrones. We may consider systems with weak non-linearities those for which γ < γ K , and systems with strong nonlinearities those for which γ > γ K . According to this classification, weak nonlinearities will produce only a frequency shift along the response curves, while strong non-linearities will produce jump phenomena.

Fig. 3.9 Generally, for single-degree-of-freedom systems with cubic stiffness and structural damping, Nyquist plots are circles as in the case of linear systems, and only the “vertex” shape of isochrones denotes the non-linear behaviour. They are bent anti-clockwise for hardening stiffness, and clockwise, in the case of softening

3. SIMPLE NON-LINEAR SYSTEMS

89

stiffness (Fig. 3.9). Usually all isochrones are curved in the same way and there is not a straight isochrone at the principal resonance. 3.2.3 Jump Phenomena

It is now possible to consider the jump phenomena in some detail. One kind of jump phenomenon occurs when the forcing frequency is changed while keeping constant the amplitude F0 of the excitation force (Fig. 3.10, a). As the frequency is gradually increased from rest, the displacement amplitude increases, the end of the response vector follows the portion BF of the respective polar plot until the stability boundary is reached at point C. Then, there is a “jump” in amplitude and phase from C to D, along the isochrone ωu = const . , followed by a continuous change in amplitude along the arc DO of the polar plot. When the frequency is decreased, the end of the response vector moves along ODE and FB on the Nyquist plot, and jumps from E to F along the isochrone ωl = const . The arcs BF and DO of the polar plot define stable regimes of vibration, the arcs FC and ED define conditionally stable regimes of vibration, while the arc CE defines unstable regimes of vibration. This means that, in the presence of strong non-linearities, large portions of the response curve cannot be experimentally obtained.

a

b Fig. 3.10

Another jump phenomenon may be noticed when the amplitude of the exciting force is varied while the forcing frequency is constant (Fig. 3.10, b). When the force amplitude F0 is gradually increased, the response amplitude increases. The end of the displacement vector moves along the respective isochrone ( arc OVS ) until the stability boundary is reached at point S.

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90

It jumps to the point T, following the response curve F0′′ = const . , and then moves again along the isochrone ( arc TZ ). On decreasing the force amplitude, the end of the response vector moves along the portion ZTU of the isochrone until the stability boundary is reached at point U, when it jumps to the point V, following the response curve F0′ = const. and then moves again along the isochrone from V to O.

a

b Fig. 3.11

The same jump phenomena are represented in Fig. 3.11, a on a displacement-frequency diagram, and in Fig. 3.11, b on a force-displacement diagram, using the same notations for the points. 3.2.4 Parameter Estimation

One method for determining the system dynamic parameters requires at least two response circles (Fig. 3.12), plotted for different amplitudes F0′ and F0′′ = f ⋅ F0′ ( f > 1) of the harmonic excitation force. Let denote by OM ' =

F0′ = a1 , gk

OM ′′ =

F0′′ = a2 , gk

the maximum displacement amplitudes on the two response circles. The above equations yield f =

F0′′ a 2 . = F0′ a 1

(3.24)

3. SIMPLE NON-LINEAR SYSTEMS

91

For linear systems, points M ' and M " have the same frequency ω n = k m . At systems with non-linear stiffness characteristic, the frequencies ω ′r and ω ′r′ of these points are given by 3 4

3 4

ω ′r = ωn 1 + μ a 21 ,

ω ′r′ = ωn 1 + μ a 22 .

(3.25)

Fig. 3.12 Using equations (3.24), from (3.25) we obtain the natural frequency of the corresponding linear system

ω 2n =

f 2ω ′r2 − ω ′r′2

(3.26)

f 2 −1

and the coefficient of non-linearity

μ=

ω ′r′2 − ω ′r2

4

3 a 21 f 2ω ′r2 − ω ′r′2

.

(3.27)

An arc of circle of radius OM ' = a 1 and centre at the origin O crosses the second circle at the points P and Q, of frequencies ω P and ω Q , respectively, given by ⎛ ⎝

3 4

ω 2P ,Q = ω 2n ⎜ 1 + μ a 21 m g

⎞ f 2 −1⎟ . ⎠

(3.28)

The structural damping factor is calculated from g=

1

ω 2Q − ω 2P ω 2n

2 2 2 f 2 − 1 ω Q + ω P ω ′r

.

(3.29)

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The stiffness and mass parameters are then given by

k = F0′ g a 1 ,

m = k ω 2n .

(3.30)

It is recommended to draw several Nyquist plots on the same diagram. The adopted model is valid only if the same values are obtained for the dynamic parameters regardless of the pair of polar plots used.

3.3 Combined Coulomb and Structural Damping Coulomb damping results from the relative motion of two bodies sliding one upon the other in the presence of a normal force N holding them in contact. The friction force R is proportional to N, the proportionality constant being the coefficient of friction. Usually, the difference between the static and dynamic values of the coefficient of friction is neglected, and the normal force N is assumed constant and independent of frequency and (relative displacement) amplitude.

Fig. 3.13 The Coulomb damping force can be written fd = R

x& = R sgn (x& ) x&

(3.31)

so that it is + R or − R depending on whether the relative velocity is positive or negative. The energy dissipated per cycle of vibration by a Coulomb damper in which a force f d acts through a relative displacement

x ( t ) = a sin ω t is given by

π 2

T

Wd =

∫ f d dx = ∫ 0

(3.32)

f d x& dx = 4 a

∫ 0

f d cosω t d (ω t ) = 4 R a

(3.33)

3. SIMPLE NON-LINEAR SYSTEMS

93

and is independent of frequency. There is practical evidence that the dynamic behaviour of bolted or riveted structures and systems incorporating hydraulic actuators can approximately be described by a combined structural and dry-friction model. The energy dissipated per cycle by a hysteretic damper (2.65) is Wd = π h a 2 .

(3.34)

In order to approximate the non-linear Coulomb damping by a linear one, the concept of “equivalent structural damping” may be used. The equivalent structural damping coefficient heq has such a value that the energy dissipated in a harmonic displacement cycle of a given amplitude and frequency is the same as the energy loss of the Coulomb damper, in the same displacement cycle. Whence, equating (3.33) and (3.34) yields

π heq a 2 = 4 R a , wherefrom heq =

4R πa

(3.35)

which is amplitude dependent. 3.3.1 Harmonic Response

The equation of motion for a single-degree-of-freedom system with combined structural and Coulomb damping (Fig. 3.13) and excited by a harmonic force may be written m &x& +

h

ω

x& + R sgn (x& ) + k x = f (t ) .

(3.36)

A first harmonic approximation of the response is chosen of the form (3.32)

x (t ) = a sin ω t

and the force is conveniently expressed as

f (t ) = F0 sin (ω t − θ ) .

(3.37)

Upon substitution of (3.37) and use of the concept of equivalent structural damping, equation (3.36) becomes m &x& +

h + heq

ω

x& + k x = F0 sin (ω t − θ ) .

(3.38)

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94

Substituting the solution (3.32) into equation (3.38) and equating the coefficients of the terms in cos ω t and sin ω t of both sides, the following equations are obtained sin θ = − a g − r ,

(

)

(3.39)

cos θ = a 1 − η 2 , where r=

ak ω 4R , a= , η= , ωn = π F0 F0 ωn

k h , g= . m k

(3.40)

Equations (3.39) give the magnitude of the dimensionless displacement

a=

−gr+

(1 − r )( 1 − η ) (1 −η ) + g

2 2

2

2 2

+ g2

2

(3.41)

and the phase angle r θ = tan -1 2 a . η −1 g+

(3.42)

A solution for a is only possible when r < 1 (F0 > 4 R π ) . Equations (3.41) and (3.42) can be directly obtained from (2.66) and (2.67) if the structural damping factor g is replaced by g + g eq = g + r a . Considering a dimensionless complex displacement amplitude a~ k = a R + i a I = a cosθ + i a sinθ F0

(3.43)

(

(3.44)

we obtain

)

aR = a 2 1 − η 2 ,

r⎞ ⎛ aI = a 2 ⎜ g + ⎟ , a⎠ ⎝

where the magnitude a = a R2 + a I2 is given by equation (3.41).

3.3.2 Nyquist Plots and Isochrones

Based on equations (3.41) and (3.42), the polar diagram of the frequency response for the system of Fig. 3.13 can be drawn in the Argand plane. This Nyquist plot has the pear-shape illustrated in Fig. 3.14. It shows that the

3. SIMPLE NON-LINEAR SYSTEMS

95

elongation, parallel to the imaginary axis, of the otherwise circular diagrams of lightly damped linear systems, is the effect of Coulomb damping. The principal resonance frequency ωn may be determined as the parameter of point M, where the diagram crosses the imaginary negative semi-axis. The displacement amplitude at resonance is 1 (1 − r ) . g

aM = OM =

(3.45)

Fig. 3.14 The lines OB and OC, drawn from the origin at angles of 450 each side of the imaginary negative semi-axis, cross the Nyquist plot at points B and C, of frequencies

ω 1, 2 = ωn

1m

g 1− 2 r

.

(3.46)

The corresponding dimensionless vector lengths are

OB = OC =

1 g

⎞ ⎛ 2 ⎟ ⎜ ⎜ 2 − r⎟ ⎠ ⎝

so that

BC = 2 OB =

(

1 1− 2 r g

)

(3.47)

which may be used for determining the Coulomb friction force. It is recommended to plot several response diagrams for different values of the excitation force amplitude F0 , using as coordinates the coincident and

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96

quadrature components of the actual displacement (Fig. 3.15). If isochrones are drawn on the same diagram, they exhibit a “fir tree” pattern. The only straight line is the isochrone corresponding to phase resonance. Isochrones lying in the real positive half-plane are bent anticlockwise to the right, while those lying in the real negative half-plane are bent clockwise to the left. The pattern of isochrones can be used to recognize the presence of Coulomb damping in a system. Their distortion from straightness is a less ambiguous indication of non-linear behaviour than the deviation from circularity of Nyquist plots.

Fig. 3.15

3.3.3 Parameter Estimation

The method of two polar plots can be applied to systems with non-linear damping too. Consider two Nyquist plots drawn for two distinct values F0′ and F0′′ = f ⋅ F0′ ( f > 1) of the harmonic excitation force (Fig. 3.16). The resonance points M ' and M " , of frequency ωn , are located at the crossing points of the plots with the only isochrone which is a straight line. Denoting α = OM " OM ' , the following equation can be established for the calculation of the Coulomb friction force R=

π F0′ α − f . 4 α −1

(3.48)

If an arc of circle of radius a1 = OM ' and centre in the origin O is drawn, it crosses the second polar plot at points P and Q, of frequencies given by

3. SIMPLE NON-LINEAR SYSTEMS

ω 2P ,Q

97

⎛ ⎞ ⎜ ⎟ 2 − 1 g f ⎟ = ω2 = ωn2 ⎜ 1 m n ⎜ 4R ⎟ 1 − ⎜ π F0′ ⎟⎠ ⎝

⎛ ⎜1 m g (α − 1) ⎜ ⎝

f + 1 ⎞⎟ . f − 1 ⎟⎠

(3.49)

The structural damping factor g can be calculated from

g=

ω 2Q − ω 2P ω 2Q

+ ω 2P

1 α −1

f −1 f +1

(3.50)

where

ω 2P + ω 2Q = 2 ω 2n .

(3.51)

Fig. 3.16 The stiffness is further given by

k=

F0′ g a1

⎛ 4R ⎞ ⎜⎜1 − ⎟⎟ . ⎝ π F0′ ⎠

(3.52)

and the mass by m = k ω n2 .

3.4 Quadratic Damping Quadratic damping is a particular case of velocity-nth power damping which can be associated with the turbulent flow of a fluid through an orifice. It practically occurs at relatively high flow velocities, but can be used as a first

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98

approximation in describing a certain class of non-linear damping phenomena arising in systems including oil dampers and shock absorbers. The damping force f d is proportional to the square of the relative velocity across the damper (minus sign omitted) f d = c2 x& 2 sgn ( x& ) ,

(3.53)

where c2 is defined as the quadratic damping coefficient, being a function of damper geometry and fluid properties. The energy dissipated per cycle of vibration by a quadratic damper in which a force f d acts through a relative displacement x (t ) = a sin ω t is given by

Wd =

∫

f d dx =

T

π 2

0

0

∫ f d x& dx = 4 a ∫

8 f d cosω t d(ω t ) = c2 ω 2 a 3 , (3.54) 3

being dependent upon the frequency and amplitude of vibration. 3.4.1 Harmonic Response

The equation of motion for a single-degree-of-freedom system with directly coupled quadratic damping (Fig. 3.17) and excited by a harmonic force acting upon the mass may be written m &x& + c2 x& 2 sgn (x& ) + k x = f (t ) .

(3.55)

A first harmonic approximation of the response is chosen of the form x (t ) = a sin ω t and the force is conveniently expressed as f (t ) = F0 sin (ω t − θ ) .

Fig. 3.17 Upon substitution in (3.55), the following equations are obtained

3. SIMPLE NON-LINEAR SYSTEMS

sin θ = − a 2

99

8 α2 η 2 , 3π

(

)

cos θ = a 1 − η 2 ,

(3.56)

in which

α2 =

c2 F0 ak ω , a= , η= , ωn = km F0 ωn

k , m

(3.57)

and where α 2 is defined as the quadratic damping parameter. Equations (3.56) give the dimensionless frequency η and the phase shift θ as functions of the dimensionless displacement amplitude a : 2

⎛ 8α 2 ⎞ ⎛ 8α 2 ⎜⎜ ⎟⎟ + ⎜⎜ ⎝ 3π ⎠ ⎝ 3π

1 − a2 2 a

1±

η = 2

⎛ 8α ⎞ 1 + a 2 ⎜⎜ 2 ⎟⎟ ⎝ 3π ⎠

a

θ = tan -1

⎞ ⎟⎟ ⎠

2

2

,

(3.58)

8α 2 2 η 3π . η 2 −1

(3.59)

Considering a complex displacement amplitude

a~ = aR + i aI = a cosθ + i a sinθ we obtain aR = a 2

(

)

k 1 −η 2 , F0

aI = − a 3

k 8 c2 2 η , F0 3 π m

(3.60)

where the magnitude a = a R2 + a I2 is given by 2

a=

(1 − η )

2 2

+

(1 − η )

F0 k

2 4

. 2

2

⎛ 16 c2 ⎞ ⎛ F0 ⎞ 4 ⎟⎟ ⎜ ⎟ η + ⎜⎜ ⎝ 3π m ⎠ ⎝ k ⎠

(3.61)

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100 3.4.2 Nyquist Plots and Isochrones

Based on equations (3.60) and (3.61), the polar diagram of the frequency response for the system of Fig. 3.17 can be plotted in the Argand plane as the geometric locus of the affix of the vector a~ . Figure 3.18 shows three such diagrams. The quadratic damping tends to elongate the polar plots in the direction of the real axis, “flattening” the otherwise circular diagrams of systems with slight viscous damping. The isochrones exhibit a “fireworks” pattern. The only straight line is at phase resonance. Isochrones defined by under-resonance frequencies are bent up-frequency, while those defined by over-resonance frequencies are curved down-frequency.

Fig. 3.18 Analysing in more detail the polar plot in Fig. 3.19, it can be seen that the phase resonance frequency ωn is determined at the point M, where the diagram crosses the imaginary negative semi-axis.

Fig. 3.19 The displacement amplitude at phase resonance is

3. SIMPLE NON-LINEAR SYSTEMS

101 3π = F0 8α 2

F0 k

aM = OM =

3π m . 8 k c2

(3.62)

The lines OB and OC, drawn from the origin at angles of 450 on each side of the imaginary negative semi-axis, cross the Nyquist plot at points B and C, of frequencies

η B ,C =

1+ γ 2 m γ ,

(3.63)

α2

(3.64)

where

γ2 =

2 3π

which may be used for the evaluation of damping.

3.4.3 Parameter Estimation

The method of two polar plots can be used for the estimation of system parameters. Figure 3.20 illustrates two Nyquist plots drawn for two distinct values F0′ and F0′′ = f ⋅ F0′ ( f > 1) of the excitation force amplitude. The points M ' and M " , of frequency η = 1 , are the crossing points of the plots with the only isochrone which is a straight line.

Fig. 3.20 It is useful to define a new dimensionless damping parameter

α=

8α 2 8 c2 F0 = 3π 3π k m

(3.65)

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102

that is a function of the amplitude F0 of the excitation force. For the two polar plots of Fig. 3.20, one can write

α′ =

8 c2 F0′ , 3π k m

α ′′ =

8 c2 F0′′ 3π k m

(3.66)

so that the ratio of the excitation force amplitudes can be denoted

F ′′ α ′′ ⎛ OM " ⎞ f = 0 = =⎜ ⎟ . F0′ α ′ ⎝ OM ' ⎠ 2

Fig. 3.21

(3.67)

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103

If an arc of circle of radius a 1 = OM ' and centre in the origin O is drawn, it will cross the second polar plot at the points P and Q, of dimensionless frequencies η P and η Q , given by

η 2P ,Q =

1m

(

)

α ′ f 2 − 1 + f 2α ′2 . 1+α′

(3.68)

Equations (3.68) yield the damping parameter

α′ =

η 2P

2 −1 . + η 2Q

(3.69)

The quadratic damping factor α 2′ is given by

α 2′ =

3π α′ . 8

(3.70)

The stiffness is further obtained from

k=

F0′ a1

1

α′

.

(3.71)

and the quadratic damping coefficient from

c2 =

α 2′ k 2 . ωn2 F0′

(3.72)

Figure 3.21 summarizes some of the effects of Coulomb and quadratic damping on the Nyquist plots and the isochrones of single-degree-of-freedom systems.

3.5 Effect of Pre-Loading The stiffness function (3.1) is anti-symmetric with respect to the origin. It can be easily shown that, replacing x by x + x0 , where x0 is the deflection produced by pre-load, the stiffness function is represented by three terms

(

)

f e = k0 x + μ 0 x 2 + μ x 3 .

(3.73)

The square term due to pre-load has a softening effect irrespective if whether μ 0 is positive or negative. A positive cubic term can induce a hardening effect at larger displacement amplitudes. In this case, isochrones plotted in the Argand plane are represented by double bend curves.

104

MECHANICAL VIBRATIONS

The analysis is complicated by the fact that the first approximation solution must contain also a constant term, because the system vibrates about a new equilibrium position which is displaced from the stiffness equation origin. It is also very difficult in this case to evaluate the constants k, μ 0 and μ from experimental results. Double-bend behaviour of isochrones as an indication of pre-load has been noticed at machine-tools with slackened guides and at systems with loose joints. Similar patterns of isochrones exhibit polyurethane foam isolator pads, but this is determined by the force-deflection characteristic, which is softening at low force levels, then hardening, at larger displacements.

4. TWO-DEGREE-OF-FREEDOM SYSTEMS The degrees of freedom of a vibrating system are equal to the number of independent coordinates required to describe its motion completely. In this chapter, two-degree-of-freedom systems are analyzed as the simplest case of discrete systems, and as an introduction to systems with a larger number of degrees of freedom. As a preparation for multi-degree-of-freedom systems, the matrix notation is introduced as a compact way of expressing the equations of motion and the dynamic response. In Chapter 2 it is shown that the free vibration of an undamped singledegree-of-freedom system is a harmonic motion at the system natural frequency, referred to as a natural vibration. In contrast, the free vibration of an undamped multi-degree-of-freedom system is periodic, and consists of several simultaneous vibrations at the various natural frequencies. It can be expressed as a sum of harmonic components, each implying a certain natural frequency and displacement configuration, called natural modes of vibration. The motion in a natural mode of vibration is synchronous and harmonic at all system coordinates. The dynamic response of a discrete system can be described by simultaneous ordinary differential equations. For a proper choice of coordinates, known as principal or modal coordinates, the equations can be decoupled and solved independently. The modal coordinates represent linear combinations of the actual displacements. Conversely, the motion can be regarded as a superposition of vibrations in the natural modes of vibration defined by the modal coordinates. A two-degree-of-freedom system has two natural frequencies and, in forced vibrations, for small damping it may have two resonances. The free response to initial excitation and the forced response to external excitation can be expressed in terms of the natural modes of vibration whose shapes are defined by mutually orthogonal vectors with respect to the mass and stiffness matrices. The forced response to harmonic excitation can also be calculated by simply using Cramer’s rule. Resonance occurs when the forcing frequency equals any of the two natural frequencies of the system. Antiresonance can also occur at a frequency equal to the natural frequency of a mass-spring subsystem.

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106

The material contained in this chapter refers primarily to the computation of natural frequencies and mode shapes. The forced vibrations are studied first for undamped systems using both the modal analysis and the direct spectral analysis techniques. Then damped forced vibrations are considered.

4.1 Coupled Translation In the following, two-degree-of-freedom mass-spring systems are considered, in which the lumped masses have unidirectional translational motions.

4.1.1 Equations of Motion Consider the system of Fig. 4.1, a which consists of two masses m1 and m 2 , attached to fixed points by springs k1 and k 3 , and tied together by a “coupling spring” k 2 .

Fig. 4.1 Assuming that masses are guided so as to move only in the horizontal direction, the configuration is entirely determined by their instantaneous displacements x1 and x 2 from the equilibrium positions. The system has two degrees of freedom. Using the free-body diagrams of Fig. 4.1, b and d’Alembert’s principle (dynamic equilibrium of impressed and inertia forces), the equations of motion can be written m 1 &x&1 + k 1 x 1 + k 2 x 1 − x 2 = 0 ,

(

) m 2 &x& 2 + k 3 x 2 − k 2 (x 1 − x 2 ) = 0 .

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

107

On rearranging, these equations become

(

) m 2 &x& 2 − k 2 x 1 + ( k 3 + k 2 ) x 2 = 0.

m 1 &x&1 + k 1 + k 2 x 1 − k 2 x 2 = 0,

(4.1)

This is a set of “coupled” linear differential equations of second order, with constant coefficients. The coupling between the two coordinates is due to the stiffness k 2 . If k 2 = 0 , equations (4.1) become independent, and the system of Fig. 4.1 degenerates into two one-degree-of-freedom systems. Equations (4.1) can be written ⎡m 1 0 ⎤ ⎧ &x&1 ⎫ ⎡k1 + k 2 ⎢ 0 m ⎥ ⎨&x& ⎬ + ⎢ 2 ⎦ ⎩ 2 ⎭ ⎣ − k2 ⎣

− k 2 ⎤ ⎧ x 1 ⎫ ⎧0⎫ ⎨ ⎬ = ⎨ ⎬, k3 + k 2 ⎥⎦ ⎩ x 2 ⎭ ⎩0⎭

(4.2)

or in compact form

[ m ]{ &x& } + [ k ]{ x } = { 0 } , (4.3) [ m ] is the mass matrix, [ k ] is the stiffness matrix and { x } is the column

where vector of displacements. Note that square matrices are denoted by brackets while column vectors are denoted by braces. The mass and stiffness matrices are always symmetrical so that they are equal to their transposes

[ m ] = [ m ]T , [ k ] = [ k ]T .

(4.4)

The mass matrix is diagonal. The coupling is produced by the off-diagonal elements of the stiffness matrix.

4.1.2 Free Vibration. Natural Modes Let examine the conditions under which the two masses have synchronous harmonic motions, i.e. when the system behaves like a single degree of freedom system in natural vibration. We assume solutions of the form x 1 ( t ) = a 1 cos (ω t − ϕ ),

x 2 ( t ) = a 2 cos (ω t − ϕ ),

(4.5)

and examine the conditions under which such motion is possible. The motion we are seeking is one in which the ratio between the two instantaneous displacements remains constant throughout the motion x1 (t ) x 2 (t ) = a1 a 2 = const . (4.6) The shape of the system configuration does not change during the motion, the deflected shape resembles itself at any time.

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108

Substituting solutions (4.5) into the differential equations (4.1), the resulting algebraic equations are

(k

1 + k2

)

− m 1 ω 2 a 1 − k2 a2 = 0 ,

(

(4.7)

)

− k2 a 1 + k3 + k2 − m 2 ω 2 a2 = 0.

The simultaneous homogeneous equations (4.7) admit non-trivial solutions if the determinant of the coefficients a1 and a2 is zero

k 1 + k2 − m 1 ω 2

− k2

− k2

k3 + k 2 − m 2 ω 2

= 0.

(4.8)

This can be written ⎛ k1 + k 2 k3 + k 2 ⎞ 2 k1 k3 + (k1 + k3 ) k 2 ⎟ω + + =0 ⎜ m1 ⎟ m m 1 m2 2 ⎠ ⎝

ω4 − ⎜

(4.9)

which represents a quadratic equation in ω 2 called the characteristic equation, or frequency equation. Its roots ω 12 and ω 22 are real and positive. The quantities ω 1 and ω 2 are called natural frequencies or eigenfrequencies because they depend solely upon the system mass and stiffness parameters. Because the system (4.7) is homogeneous, the amplitudes a 1 and a 2 cannot be determined, only their ratio μ = a 2 a 1 . For ω 1 , the amplitude ratio is ⎛ a2 ⎞ k + k − m2 ω 12 ⎟ = 3 2 ⎜ a1 ⎟ k2 ⎝ ⎠1

μ1 = ⎜

(4.10)

and for ω 2 , the amplitude ratio is ⎛ a2 ⎞ k3 + k 2 − m2 ω 22 ⎜ ⎟ . μ2 = = ⎜ a1 ⎟ k2 ⎝ ⎠2

(4.11)

The ratios (4.10) and (4.11) determine the shape of the system during synchronous motion with frequencies ω 1 and ω 2 , respectively. If one element in each ratio is assigned a certain arbitrary value, then the value of the other element results from the above expressions. This process is called normalisation. The natural frequencies and corresponding amplitude ratios determine the conditions to have synchronous harmonic motions, i.e. the natural modes of vibration. A mode of vibration is defined by two parameters: the natural frequency and the mode shape. The mode shapes can be represented by column vectors

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

⎧1⎫ ⎧a 1 ⎫ ⎬ = a 1 1 ⎨ ⎬ = C1 { u }1 , ⎩a2 ⎭1 ⎩u 1 ⎭ ⎧1⎫ ⎧a ⎫ = ⎨ 1 ⎬ = a 1 ⎨ ⎬ = C2 { u }2 , 2 u ⎩a2 ⎭ ⎩ 2⎭

( )

{ a }1 = ⎨ { a }2

109

( )

(4.12)

2

referred to as modal vectors. In (4.12) the modal vectors are normalised with the first element equal to unity. The normalised vectors are said to represent the shape of a normal mode. The two possible synchronous motions are given by

{ x (t ) }1 = C1 { u }1 cos (ω 1 t − ϕ 1 ), { x (t ) }2 = C2 { u }2 cos (ω 2 t − ϕ 2 )

(4.13)

and the general solution of free vibrations is

{ x (t ) } = { x (t ) }1 + { x (t ) }2 = C1 {u }1 cos (ω 1 t − ϕ 1 ) + C2 { u }2 cos (ω2 t − ϕ 2 ).(4.14) In (4.14) the four integration constants C1 , C 2 , ϕ 1 , ϕ 2 are determined

from the four initial conditions, the displacements and velocities at t = 0 . For arbitrary initial conditions, the free vibration of the two-degree-offreedom system is a periodic motion obtained as the superposition of the two natural modes of vibration, i.e. two harmonic motions with frequencies equal to the natural frequencies of the system. It can be shown that for zero initial velocities and initial displacements resembling a mode shape, the free motion is synchronous and purely harmonic, and takes place at the natural frequency of the respective mode. A system can vibrate in a pure mode of vibration if the initial deflected shape is similar to the mode shape.

Example 4.1 Consider the system of Fig. 4.2 and obtain the natural modes of vibration.

Fig. 4.2 Solution. The equations of motion can be written

MECHANICAL VIBRATIONS

110

− k ⎤ ⎧ x1 ⎫ ⎧0⎫ ⎨ ⎬=⎨ ⎬. k ⎥⎦ ⎩ x 2 ⎭ ⎩0⎭

⎡2m 0 ⎤ ⎧ &x&1 ⎫ ⎡ 3k ⎢ 0 m ⎥ ⎨ &x& ⎬ + ⎢− k ⎣ ⎦ ⎩ 2⎭ ⎣

Assuming solutions of the form

{ x } = { u } cos (ω t − ϕ ) we obtain

(3 k − 2ω m) u − k u − k u + ( k − ω m) u 2

1

2

1

2

2

= 0, = 0,

or, dividing by k,

( 3 − 2α ) u 1 − u 2 = 0, − u 1 + ( 1 − α ) u 2 = 0, where

α = mω 2 k . The condition to have nontrivial solutions is 3 − 2α −1 = 0, −1 1−α

2α 2 − 5α + 2 = 0

with solutions

α1 = 1 2 ,

α2 = 2 .

The natural frequencies are

ω1=

1 2

k , m

ω2 = 2

k . m

Assigning the first element to be unity, the first modal vector is given by

− 1 ⎤ ⎧ u1 ⎫ ⎡ 2 − 1⎤ ⎧1⎫ ⎡ 3 − 2 α1 1 ⎥⎨ ⎬=0 ⎨ ⎬ =⎢ ⎢ −1 1 − α 1 ⎥⎦ ⎩u 2 ⎭1 ⎢− 1 ⎥ 2 ⎣ 2 ⎦⎩ ⎭ ⎣ and the second modal vector is calculated from − 1 ⎤ ⎧ u1 ⎫ ⎡ 3 − 2α 2 ⎡ − 1 − 1⎤ ⎧ 1 ⎫ ⎨ ⎬=0. ⎨ ⎬ =⎢ ⎢ −1 ⎥ 1 − α 2 ⎦ ⎩u 2 ⎭ 2 ⎣ − 1 − 1 ⎥⎦ ⎩− 1⎭ ⎣ The mode shapes are graphically presented in Fig. 4.3. In the first mode, the two masses are moving in the same direction, either both to the right, or both to

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

111

the left, the displacement of the mass m being always twice the displacement of the mass 2 m . In the second mode, the two masses move in opposite directions through the same distance. The midpoint of the spring k does not move, hence it is a nodal point. If this point were clamped, no change in the motion would take place. The second spring is split into two springs of stiffness 2k . The mass 2m is thus connected to ground by two springs of stiffness 2k (in parallel) with an equivalent stiffness 4k , while the mass m is connected to one spring of stiffness 2k , both subsystems having the same natural frequency equal to ω 2 .

Fig. 4.3 Summarising, the first normal mode has a natural frequency ω 1 =

1 2

k m

1⎫ ⎬ , while the second normal mode has a natural ⎩2 ⎭ ⎧1⎫ k and a mode shape { u } 2 = ⎨ ⎬ . If the system is given frequency ω 2 = 2 m ⎩− 1⎭ an initial disturbance of x 1 = 1 and x2 = 2 and then released, the ensuing motion and a mode shape

{ u }1 = ⎧⎨

will be purely harmonic with the frequency ω 1 = 0.207 k m . If the initial displacement is x 1 = 1 and x 2 = −1 , the motion will be harmonic with the frequency ω2 = 1.414 k m .

4.1.3 Orthogonality of Natural Modes Substituting solutions (4.13) in equation (4.3) we obtain

[ k ] { u }1 = ω 21 [ m ]{ u }1 ,

(4.15, a)

MECHANICAL VIBRATIONS

112

[ k ] { u }2 = ω22 [ m ]{ u }2 .

(4.15, b)

Multiplying (4.15, a) on the left by { u }T2 we get

{ u }T2 [ k ] { u }1 = ω 12 { u }T2 [ m ]{ u }1 .

(4.16)

Taking the transpose of (4.15, b) and multiplying on the right by { u }1 we obtain

{ u }T2 [ k ] { u }1 = ω22 { u }T2 [ m ]{ u }1 .

(4.17)

Subtracting (4.16) and (4.17) from each other, for ω 1 ≠ ω 2 , we get

and taking the transpose

{ u }T2 [ m ]{ u }1 = 0 ,

(4.18)

{ u }T1 [ m ]{ u }2 = 0 .

(4.19)

Substituting (4.18) in (4.16) we obtain

and taking the transpose

{ u }T2 [ k ] { u }1 = 0

(4.20)

{u }1T [ k ] {u }2 = 0 .

(4.21)

Equations (4.18)-(4.21) show that the modal vectors are orthogonal with respect to the mass matrix and the stiffness matrix. Note the difference from normal orthogonality of two vectors { a } and { b } , which is written { a }T { b } = 0.

4.1.4 Modal Coordinates Let denote in (4.13)

( ) q2 (t ) = C2 cos (ω 2 t − ϕ 2 )

(4.22)

{ x (t ) } = {u }1 q 1 + {u }2 q 2 .

(4.23)

q1 (t ) = C1 cos ω 1 t − ϕ 1 ,

so that (4.14) becomes

Equation (4.23) can be written ⎧ q1 ⎫ ⎬ = [ u ]{q } ⎩q 2 ⎭

{ x (t ) } = [ { u }1 { u }2 ] ⎨

(4.23, a)

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

113

where

[ u ] = [ { u }1 { u }2 ]

(4.24)

is called the modal matrix. Substituting (4.23) in (4.3), multiplying on the left by { u }Tr (r = 1, 2 ) and considering the orthogonality we obtain M r q&&r + K r qr = 0 , where

M r = { u }Tr [ m ] { u }r , K r = { u }Tr [ k ] { u }r

(4.25) (4.26)

are modal masses and modal stiffnesses, respectively. The values of modal masses and stiffnesses depend on the normalization of modal vectors. When the values of the modal masses are imposed the vectors are said to be mass-normalized. For unit modal masses, the modal stiffnesses are equal to the square of the respective natural frequency. From equations (4.15) we obtain the natural frequency in terms of the modal vector { u }Tr [ k ] { u }r 2 ωr = . r = 1, 2 (4.27) { u }Tr [ m ] { u }r Rayleigh’s quotient is defined as R ( {u } ) =

{ u }T [ k ] { u } . { u }T [ m ] { u }

(4.28)

If the vector {u } coincides with one of the system modal vectors, then the quotient reduces to the associated natural frequency squared. Rayleigh’s quotient has a stationary value in the neighbourhood of a modal vector. If {u } is a trial vector with differs slightly from the first modal vector, then R ( {u } ) is very close to the fundamental natural frequency squared, and always higher. The linear transformation (4.23) uncouples the equations of motion. The coordinates (4.22) for which the equations of motion are independent are called principal coordinates or modal coordinates. Substituting (4.23, a) in (4.3) and multiplying on the left by [ u ] T yields

[ u ] T [ m ] [ u ] { q&& } + [ u ] T [ k ] [ u ]{ q } = { 0 } , or

[ M ] { q&& } + [ K ] { q } = { 0 },

(4.29) (4.29, a)

MECHANICAL VIBRATIONS

114

where the diagonal matrices

[ M ] = [ u ]T [ m ] [ u ]

and [ K ] = [ u ] T [ k ] [ u ]

(4.30)

are the modal mass matrix and the modal stiffness matrix, respectively. The coordinate transformation (4.23, a) simultaneously diagonalizes the mass matrix and the stiffness matrix. After solving separately the decoupled equations (4.29, a), the modal coordinates may be substituted back into (4.23, a) to obtain the physical coordinates in the configuration space. This technique is called modal analysis. The modal analysis uses a linear coordinate transformation based on the modal matrix to uncouple the equations of motion of a vibrating system.

4.1.5 Response to Harmonic Excitation Consider the forced vibrations of the system of Fig. 4.4 under the action of forces f1 (t ) and f 2 (t ) applied on mass m1 , respectively m 2 .

Fig. 4.4 The equations of motion are m 1 &x&1 + (k1 + k 2 ) x 1 − k 2 x 2 = f 1 ,

m 2 &x&2 − k 2 x 1 + (k3 + k 2 ) x 2 = f 2 ,

(4.31)

or, in compact matrix form,

[ m ] { &x& } + [ k ] { x } = { f } , where

(4.31, a)

{ f } is the column vector of impressed forces. 4.1.5.1 Solution by Modal Analysis

[u ]

Substituting (4.23, a) in equation (4.31, a), and multiplying on the left by T

where

we obtain

[ M ] { q&& } + [ K ] { q } = { F } ,

(4.32)

{ F } = [ u ]T { f }

(4.33)

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

115

is the column vector of modal forces. For harmonic excitation

{ f } = { ˆf } cos ω t

(4.34)

the steady-state response is

{ x } = { ˆx } cos ω t ,

{ q } = { qˆ } cos ω t ,

(4.35)

where the a hat above a letter denotes amplitude. Substituting (4.34) and (4.35) in (4.32) we obtain

[− ω

2

[ M ]+ [ K ] ] { qˆ } = [ u ] T { ˆf }= {Fˆ },

(4.36)

so that the amplitudes of modal coordinates are

{ u }T1 { ˆf } = , qˆ 1 = K1 − ω 2 M 1 { u }T1 [ k ] { u }1 − ω 2 { u }T1 [ m ] { u }1

(4.37)

{ u }T2 { ˆf } . { u }T2 [ k ] { u }2 − ω 2 { u }T2 [ m ] { u }2

(4.38)

Fˆ1

qˆ 2 =

Fˆ 2 K2 − ω 2M 2

=

Equation (4.23) for amplitudes becomes

{ ˆx } = { u }1 qˆ1 + { u }2 qˆ2 .

(4.39)

Substituting (4.37) and (4.38) in (4.39) we obtain the vector of amplitudes in physical coordinates

{ ˆx } =

{ u }T2 { ˆf }{ u }2 { u }1T { ˆf }{ u }1 + { u }1T [ k ] { u }1 − ω 2 { u }1T [ m ] { u }1 { u }T2 [ k ] { u }2 − ω 2 { u }T2 [ m ] { u }2 (4.40)

The elements of { ˆx } are of the form ˆx 1 =

ˆx 2 =

{ u }T1 {ˆf } K1 − ω 2 M 1

{u }T1 { ˆf } K1 − ω 2 M 1

u 11 +

u 21 +

{ u }T2 { ˆf }

K2 − ω 2M 2

{u }T2 { ˆf } K2 − ω 2M 2

u 12 ,

(4.41)

u 22 .

(4.42)

.

MECHANICAL VIBRATIONS

116

Example 4.2 Consider the system of Fig. 4.2 acted upon by harmonic forces ˆ f1 = f1 cos ω t and f 2 = ˆf 2 cos ω t and obtain the steady-state response amplitudes. Plot the frequency response curves of receptance when f 2 = 0 . Solution. The equations of motion can be written ⎡2m 0 ⎤ ⎧ &x&1 ⎫ ⎡ 3k ⎢ 0 m⎥ ⎨ &x& ⎬ + ⎢− k ⎣ ⎦ ⎩ 2⎭ ⎣

− k ⎤ ⎧ x1 ⎫ ⎧ f1 ⎫ ⎨ ⎬=⎨ ⎬. k ⎥⎦ ⎩ x 2 ⎭ ⎩ f 2 ⎭

Using the coordinate transformation ⎧1⎫ ⎧ 1⎫ ⎡ 1 1 ⎤ ⎧ q1 ⎫ ⎬ q1 + ⎨ ⎬ q2 = ⎢ ⎥⎨ ⎬ ⎩2⎭ ⎩− 1 ⎭ ⎣ 2 − 1 ⎦ ⎩q 2 ⎭

{ x } = [ u ] {q } = ⎨

and multiplying on the left with the transpose of the modal matrix, we obtain ⎡1 2 ⎤ ⎡2m 0 ⎤ ⎡1 1 ⎤ ⎧ q&&1 ⎫ ⎡1 2 ⎤ ⎡ 3k ⎢1 − 1⎥ ⎢ 0 m ⎥ ⎢2 − 1⎥ ⎨q&& ⎬ + ⎢1 − 1⎥ ⎢− k ⎦⎣ ⎣ ⎦⎣ ⎦⎣ ⎦⎩ 2⎭ ⎣

− k ⎤ ⎡1 1 ⎤ ⎧ q1 ⎫ ⎡1 2 ⎤ ⎧ f1 ⎫ ⎨ ⎬ ⎨ ⎬= k ⎥⎦ ⎢⎣2 − 1⎥⎦ ⎩q2 ⎭ ⎢⎣1 − 1⎥⎦ ⎩ f 2 ⎭

or ⎡6m 0 ⎤ ⎧ q&&1 ⎫ ⎡3k ⎢ 0 3m⎥ ⎨q&& ⎬ + ⎢ 0 ⎣ ⎦ ⎩ 2⎭ ⎣

For harmonic excitation

0 ⎤ ⎧ q1 ⎫ ⎧ f1 + 2 f 2 ⎫ ⎧ F1 ⎫ ⎨ ⎬=⎨ ⎬=⎨ ⎬. 6k ⎥⎦ ⎩q 2 ⎭ ⎩ f1 − f 2 ⎭ ⎩ F2 ⎭

{ f } = { ˆf } cos ω t , the steady-state response is

{ q } = { qˆ } cos ω t . The amplitudes of modal coordinates are qˆ1 =

Fˆ1

2

3k − ω 6m

=

ˆf + 2 ˆf 1 2

(

3k 1 − ω

2

ω 12

)

qˆ2 =

,

ˆf − ˆf 1 2

Fˆ2 = 6k − ω 2 3m

(

6k 1 − ω

2

ω 22

The amplitudes in the configuration space are given by

{ }

{ } ⎧⎨−11⎫⎬

⎧1⎫ b 1 2 c ˆf ⎨ ⎬ b 1 − 1 c ˆf ⎩2⎭+ { ˆx } = 3k 1 − ω 2 ω 12 6k 1 − ω 2

(

)

(

⎩

ω

2 2

)

⎭.

When f 2 = 0 and f1 = f , the vector of amplitudes can be written

⎧ ˆf ⎫ ⎧1⎫ ⎧ ˆf ⎫ ⎧ 1 ⎫ b1 2 c ⎨ ⎬ ⎨ ⎬ b1 − 1c ⎨ ⎬ ⎨ ⎬ ⎩ 0 ⎭ ⎩2 ⎭ + ⎩ 0 ⎭ ⎩− 1⎭ { ˆx } = 3k 1 − ω 2 ω 12 6k 1 − ω 2 ω 22

(

)

(

)

)

.

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

117

and the receptances are

ˆx

α11 = ˆ1 = f

1 ⎛ ω ⎞ 3k ⎜1 − 2 ⎟ ⎜ ω1 ⎟ ⎠ ⎝ 2

+

1

, α 21 =

⎛ ω ⎞ 6 k ⎜1 − 2 ⎟ ⎜ ω2 ⎟ ⎠ ⎝ 2

ˆx2 = ˆf

2 ⎛ ω ⎞ 3k ⎜1 − 2 ⎟ ⎜ ω ⎟ 1 ⎠ ⎝ 2

+

−1 ⎛ ω2 ⎞ 6 k ⎜1 − 2 ⎟ ⎜ ω ⎟ 2 ⎠ ⎝

,

or

ˆx 1 = ˆf

(

1

6m ω − ω 2 1

2

)

+

(

1

3m ω − ω 2 2

2

)

,

ˆx2 = ˆf

(

1

3m ω − ω 2 1

2

)

−

(

1

3m ω − ω 2 2

2

)

.

The frequency response curves are given in Fig. 4.5.

a

b

Fig. 4.5 1 k k and ω 2 = 2 , when the m 2 m excitation frequency equals any of the natural frequencies of the system. When k ω = ωa = , the first mass stands still in space, ˆx 1 = 0 , while the second mass m moves, ˆx 2 ≠ 0 , condition defined as an antiresonance. The antiresonance Resonances occur at ω 1 =

frequency is equal to the natural frequency of the subsystem consisting of the spring k and the mass m. This subsystem is called a dynamic vibration absorber. The energy introduced per cycle in the system by the impressed force goes into this part of the vibrating system, keeping the mass 2m still in space, condition desirable in many practical applications.

MECHANICAL VIBRATIONS

118 4.1.5.2 Solution by Spectral Analysis

Substituting (4.34) and (4.35) in equation (4.31, a), we obtain

([ k ] − ω

2

[ m ] ){ ˆx } = { ˆf },

(4.43)

or

{ ˆx } = ( [ k ] − ω 2 [ m ] )

−1

{ ˆf }.

(4.44)

Equation (4.43) represents a linear set of algebraic equations that can be solved using Cramer’s rule. The inversion in equation (4.44) is never performed.

Example 4.3 Consider the system of Fig. 4.2 acted upon by a harmonic driving force ˆ f1 = f1 cos ω t and obtain the steady-state response amplitudes by direct spectral analysis. Solution. The equations of motion (4.31) are

2 m &x&1 + 3 k x 1 − k x 2 = ˆf 1 cosω t , m &x& 2 − k x 1 + k x 2 = 0.

Substituting solution (4.35) into above equations, we obtain a set of two algebraic equations

(3k − ω

2

(

)

2 m ˆx 1 − k ˆx 2 = ˆf 1 ,

)

− k ˆx 1 + k − ω 2 m ˆx 2 = 0.

Using Cramer’s rule, the amplitudes ˆx 1 and ˆx 2 are

ˆx 1 =

ˆx 2 =

−k 1 ˆf 0 k − ω 2m

3k − ω2 2 m −k

−k k − ω 2m

3k −ω2 2m 1 ˆ f −k 0 3k − ω2 2 m −k

=

−k k − ω 2m

=

( k − ω m) ˆf ( 3 k − ω 2 m)( k − ω m)− k 2

2

(3 k − ω

2

k ˆf 2

)(

2

)

2

2m k −ω m − k

or, in a form close to that obtained by modal analysis

2

,

,

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

ˆx 1 =

⎛ 2 m ⎜⎜ ⎝

⎛ k 2⎞ ⎜ − ω ⎟ ˆf ⎝ m ⎠ , k 2 ⎞ ⎛ 2k 2⎞ − ω ⎟⎟ ⎜ −ω ⎟ 2m ⎠ ⎠⎝ m

119

ˆx 2 =

⎛ 2 m ⎜⎜ ⎝

k ˆ f m . k 2 ⎞ ⎛ 2k 2⎞ − ω ⎟⎟ ⎜ −ω ⎟ 2m ⎠ ⎠⎝ m

The denominator can be recognized as the characteristic determinant, which makes the amplitudes to grow indefinitely when the forcing frequency equals either of the natural frequencies. The system has two resonances.

4.2 Torsional Systems In the following, two-degree-of-freedom disc-shaft systems are considered, in which the rigid discs have angular vibrations with respect to the shaft axis.

4.2.1 Equations of Motion Consider the system of Fig. 4.6, a which consists of two rigid discs of polar mass moment of inertia J 1 and J 2 , kg m 2 , attached to massless shafts of torsional stiffness K1 , K 2 and K 3 , N m rad .

Fig. 4.6 The instantaneous angular position of discs with respect to the equilibrium position is denoted by θ 1 and θ 2 . Using the free-body diagrams of Fig. 4.6, b and

MECHANICAL VIBRATIONS

120

d’Alembert’s principle (dynamic equilibrium of impressed and inertia torques), the equations of motion can be written

(

)

J1 θ&&1 + K1 θ 1 + K 2 θ 1 − θ 2 = 0 ,

(

)

J 2 θ&&2 + K 3 θ 2 − K 2 θ 1 − θ 2 = 0 , or

J1 θ&&1 + (K1 + K 2 ) θ 1 − K 2 θ 2 = 0, J 2 θ&&2 − K 2 θ 1 + (K 3 + K 2 ) θ 2 = 0.

(4.45)

This is a set of coupled differential equations resembling equations (4.1). There is a complete analogy between systems in translational and angular vibration, the counterparts of springs, masses and forces being torsional springs, discs possessing mass moments of inertia and torques. All results established in Section 4.1 apply to torsional systems. In the following, only systems admitting rigid-body motions are considered.

4.2.2 Two-Disc Free-Free Systems The shafting of a motor-driven fan or pump may rotate in its bearings as a rigid body. Many engineering systems can be modelled by a two-disc torsional system not tied rigidly to the ground (Fig. 4.7). The two discs, modelling the rotors of the driving and driven machines, are joined by a torsional spring representing the two shafts and the coupling. Let the discs have polar mass moments of inertia J 1 and J 2 , and the massless shaft have a torsional stiffness K = G I p l . The equations of motion J 1 θ&&1 + K θ 1 − K θ 2 = 0, J 2 θ&&2 − K θ 1 + K θ 2 = 0,

(4.46)

can be written in matrix form ⎡ J1 ⎢0 ⎣

0 ⎤ ⎧⎪θ&&1 ⎫⎪ ⎡ K ⎨ ⎬+ J 2 ⎥⎦ ⎪⎩θ&&2 ⎪⎭ ⎢⎣− K

− K ⎤ ⎧θ 1 ⎫ ⎧0⎫ ⎨ ⎬ = ⎨ ⎬, K ⎥⎦ ⎩θ 2 ⎭ ⎩0⎭

(4.46, a)

and in compact form

[ J ]{θ&& }+ [ K ]{ θ } = { 0 } . The stiffness matrix [ K ] is positive semidefinite. Because the system is ungrounded, the stiffness matrix is singular. The system can rotate freely, having a rigid-body motion in which the potential energy is zero.

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

121

Apparently, this is a two-degree-of-freedom system. However, summing up equations (4.46) we obtain J1 θ&&1 + J 2 θ&&2 = 0 ,

(4.47)

so that the two coordinates θ 1 and θ 2 are not independent. Integrating, we may obtain a constraint equation that can be used to eliminate one coordinate from the problem formulation.

Fig. 4.7 Dividing the first equation (4.46) by J1 , the second equation by J 2 and subtracting the results from each other we obtain ⎛K K ⎞ ⎟⎟ θ 1 − θ 2 = 0 . + ⎝ J1 J 2 ⎠

θ&&1 − θ&&2 + ⎜⎜

(

)

(4.48)

Denoting the twist angle θ 1 − θ 2 = θ , equation (4.48) becomes J1 J 2 && θ + Kθ = 0 J1 + J 2

(4.48, a)

which is the equation of motion of a single-degree-of-freedom system. 4.2.2.1 Normal Modes

Assuming solutions of the form

θ 1 (t ) = a 1 cos (ω t − ϕ ) ,

θ 2 (t ) = a 2 cos (ω t − ϕ ) ,

(4.49)

MECHANICAL VIBRATIONS

122

we obtain the set of algebraic equations

(K − J ω ) a − K a = 0, − K a + (K − J ω ) a = 0. 2

1

1

1

2

2

2

(4.50)

2

Dividing by K and denoting

ω 2 J1 K = α

(4.51)

equations (4.50) become

(1 − α ) a1 − a2 = 0, ⎛ J ⎞ − a1 + ⎜⎜1 − 2 α ⎟⎟ a2 = 0. J1 ⎠ ⎝

(4.52)

The simultaneous homogeneous equations (4.52) admit non-trivial solutions if the determinant of the coefficients a 1 and a 2 is zero 1−α

−1 J =0 1− 2 α J1

−1

or

(4.53)

J2 2 ⎛ J2 ⎞ α − ⎜⎜1 + ⎟⎟ α = 0 . J1 J1 ⎠ ⎝ The solutions are

α1 = 0 ,

α 2 = 1 + J1 J 2 .

(4.53, a)

(4.54)

The first natural frequency is given by ω 21 = 0 and the second by ⎛1

1 ⎞

ω 22 = ⎜⎜ + ⎟⎟ K . ⎝ J1 J 2 ⎠

(4.55)

The root ω 21 = 0 indicates that rigid body displacement is possible. This may be due to a static angular displacement or to a uniform rotating speed. It is not a genuine vibration. The solution of equations (4.46) is of the form

θ 1 (t ) = C1 + C2 t + C3 sin ω 2 t + C4 cos ω 2 t ,

(

)

θ 2 (t ) = C1 + C2 t + 1 − J1ω 22 K (C3 sin ω 2 t + C4 cos ω 2 t )

where C1 ,..,C4 are integration constants. The mode shapes are determined by the ratio μ = a2 a1 = 1 − α .

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

123

For the first mode

μ 1 = (a2 a1 ) 1 = 1 − α1 = 1 ,

(4.56, a)

both discs have the same angular displacement defining the rigid-body rotation in which the shaft is not twisted. For the second mode

μ 2 = (a2 a1 ) 2 = 1 − α 2 = − J1 J 2 ,

(4.56, b)

the discs vibrate in opposition. The shaft has a nodal point which is closer to the larger disc. The mode shapes are graphically presented in Fig. 4.8.

Fig. 4.8 4.2.2.2 Response to Harmonic Excitation

Consider the system of Fig. 4.6 where on the second disc acts a harmonic torque M (t ) = M 0 cos ω t (not shown). The equations of motion are J1 θ&&1 + K θ 1 − K θ 2 = 0 , J 2 θ&&2 − K θ 1 + K θ 2 = M 0 cos ω t .

(4.57)

The steady-state vibration of this system is of the form

θ 1 (t ) = Θ1 cos ω t ,

θ 2 (t ) = Θ 2 cos ω t .

(4.58)

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124

The resulting algebraic equations are

(K − J ω ) Θ − K Θ = 0, − K Θ + (K − J ω ) Θ = M 1

1

2

1

2

2

2

2

(4.59)

0.

a

b Fig. 4.9

Dividing by K and denoting ω 2 J1 K = α we obtain

(1 − α ) Θ 1 −Θ 2 = 0, ⎛ J ⎞ M − Θ 1 + ⎜⎜1 − 2 α ⎟⎟ Θ 2 = 0 . J K 1 ⎝ ⎠

(4.60)

Solving for Θ 1 and Θ 2 using Cramer’s rule −1 M0 J 1 1− 2 α K J1 M0 1 Θ1 = = , 1−α −1 K J2 ⎛ J1 + J 2 ⎞ ⎜ ⎟ α− α J J1 ⎜⎝ J 2 ⎟⎠ −1 1− 2 α J1 0

1−α 0 M0 −1 1 K M0 1−α Θ2 = = . 1−α −1 K J2 ⎛ J1 + J 2 ⎞ ⎜ ⎟ α− α J J1 ⎜⎝ J 2 ⎟⎠ −1 1− 2 α J1

(4.61)

(4.62)

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

125

The amplitudes Θ 1 and Θ 2 are graphically presented in Fig. 4.9 as a function of the disturbing frequency. Both angular amplitudes become infinite when the denominator is zero, i.e. when the frequency of the driving torque equals one of the natural frequencies. There is a sort of “resonance at zero frequency” corresponding to the rigid-body mode and a genuine resonance at

ω2 =

⎛ 1 1 ⎞ ⎜⎜ + ⎟⎟ K . ⎝ J1 J 2 ⎠

The amplitude of the second disc is zero when α = 1 . This antiresonance occurs when ω = K J1 , the natural frequency of the subsystem consisting of the shaft and the first disc, which acts as a dynamic absorber and keeps the second disc still in space. 4.2.2.3 Dynamic Stresses

The amplitude of the angle of twist is

ΔΘ = Θ 2 − Θ 1 =

M0 1 . ⎞ K J 2 ⎛ J1 + J 2 ⎜ − α ⎟⎟ J1 ⎜⎝ J 2 ⎠

(4.63)

Denoting the torque in the shaft by M t = M t 0 cos ω t , its amplitude is M t 0 = K ΔΘ =

M0 . ⎞ J 2 ⎛ J1 + J 2 ⎜ − α ⎟⎟ J1 ⎜⎝ J 2 ⎠

(4.64)

The dynamic shear stresses due to the torsion are τ = τ 0 cos ω t , and the amplitude is (4.65) τ 0 = M t 0 Wp where W p is the polar modulus of the cross section. If the shaft transmits a power N at constant angular speed ω N , then the “static” shear stress is

τ N = M t st W p = N (ω N W p ) .

(4.66)

A fatigue calculation can be carried out based on values of τ 0 and τ N , considering a time history as in Fig. 4.10.

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126

Fig. 4.10

Example 4.4 The torsional system of Fig. 4.6 is acted upon by a harmonic torque of amplitude M 0 = 10 4 N m and frequency ω = 314 rad sec (not shown). Both discs have the polar mass moment of inertia J = 57 kg m 2 . The shaft has a length l = 0.4 m , diameter d = 0.14 m and shear modulus G = 81 GPa . Determine the amplitude of the dynamic shear stresses in the shaft. Solution. The shaft cross-section has I p = π d 4 32 = 0.377 ⋅ 108 mm 4 and W p = π d 3 16 = 0.538 ⋅ 106 mm3 . The torsional stiffness is K = G I p l = 7.63 ⋅ 109

N mm rad . The ratio α = J ω 2 K = 0.735 . The twist angle is ΔΘ = M 0 K (2 − α ) = 0.001036 rad . The amplitude of the dynamic torque is M t 0 = M 0 (2 − α ) = 7910 N m . The amplitude of dynamic shear stresses is

τ 0 = M t 0 W p = 14.7 N mm 2

4.2.3 Geared Systems Consider the geared torsional system of Fig. 4.11, a with a gear of pitch radius r1 on shaft 1 and a gear of pitch radius r 2 on shaft 2. Assume the gears are rigid, of negligible inertia and their teeth remain in contact. The gear ratio is i=

r1 r2

=−

n2 n1

=−

θ2 , θ1

(4.67)

where n 1 and n 2 are the rotational speeds of the two shafts, θ 1 and θ 2 are the corresponding angular displacements. The geared system is conveniently reduced to an equivalent non-geared system (Fig. 4.11, b) in which the gears are omitted. In the reduction process, the

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

127

stiffness of the equivalent shaft is determined from the condition of equal potential energies

(K θ ) 2

actual

(

= Kθ2

)

eq ,

wherefrom K eq

⎛θ = ⎜ actual ⎜ θ eq ⎝

2

⎞ ⎛ ⎟ K actual = ⎜ nactual ⎟ ⎜ neq ⎠ ⎝

2

⎞ ⎟ K actual = i 2 K actual . ⎟ ⎠

(4.68)

Fig. 4.11 The polar mass moment of inertia of the equivalent disc is determined from the condition of equal kinetic energies

( J θ& ) 2

actual

(

= J θ& 2

)

eq ,

wherefrom J eq

⎛ θ& = ⎜ actual ⎜ θ&eq ⎝

2

⎞ ⎛ ⎟ J actual = ⎜ nactual ⎜ neq ⎟ ⎝ ⎠

2

⎞ ⎟ J actual = i 2 J actual . ⎟ ⎠

(4.69)

Choosing shaft 1 as reference, the equivalent parameters of shaft 2 are (Fig. 4.11, b) K 2 eq = i 2 K 2 ,

J 2 eq = i 2 J 2 .

(4.70)

The following rule applies for the equivalent systems when gears have negligible inertia : Remove all gears and multiply all stiffnesses and all inertias by i 2 , where -i is the speed ratio of the geared shaft to the reference shaft.

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128

After determining the mode shapes of the equivalent system, the mode shapes and the torques of the actual system are recovered from the compatibility equations

θ actual θ eq = −i ,

M eq M actual = −i .

(4.71)

4.2.4 Geared-Branched Systems Consider the branched system with gears of negligible inertia and massless shafts shown in Fig. 4.12, a. It can be converted to the model with an one-to-one gear shown in Fig. 4.12, b, by multiplying all the stiffnesses and inertias of the branch 2-3 by the square of the speed ratio i. Note that torques and angular displacements in the reduced branch are different from the actual values, according to equations (4.71).

Fig. 4.12 The equations of motion can be written using the finite element approach. A uniform shaft is considered a two-noded finite element of torsional stiffness K. The points of attachment of the shaft to other parts of the vibrating system are called nodes (not to be confused with the stationary points of mode shapes) and are denoted 1 and 2 in Fig. 4.13. The torques M 1 and M 2 may be related to the rotation angles θ 1 and θ 2 using the equilibrium and the torque/rotation equations M1 = − M 2 = K θ 1

when

θ 2 = 0,

M 1 = − M 2 = − K θ 2 when

θ 1 = 0.

Equations (4.72) may be written in matrix form as

(4.72)

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

− K⎤ ⎧ θ1 ⎫ ⎨ ⎬ K ⎥⎦ ⎩ θ 2 ⎭

⎧ M1 ⎫ ⎡ K ⎨ ⎬=⎢ ⎩ M 2 ⎭ ⎣− K

or in shorthand form stiffness matrix.

129

(4.73)

{ M }= [ k ] { θ }, where [ k ] is referred to as the element e

e

Fig. 4.13 Using equation (4.73), the torque-rotation equation for each shaft in Fig. 4.12, b can be written ⎧ M 1 ⎫ ⎡ K1 − K1 ⎤ ⎧ θ 1 ⎫ ⎧ M 2 ⎫ ⎡ K 2 ⎨ ⎬, ⎨ ⎬= ⎬=⎢ ⎨ K1 ⎥⎦ ⎩ θ 3 ⎭ ⎩ M 3 ⎭ ⎢⎣− K 2 ⎩ M 3 ⎭ ⎣− K1 ⎧ M 3 ⎫ ⎡ K3 ⎨ ⎬=⎢ ⎩ M 4 ⎭ ⎣− K 3

− K3 ⎤ ⎧ θ 3 ⎨ K 3 ⎥⎦ ⎩ θ 4

− K2 ⎤ ⎧ θ 2 ⎨ K 2 ⎥⎦ ⎩ θ 3

⎫ ⎬, ⎭

⎫ ⎬. ⎭

Each of these equations can be expanded as follows

⎧ M1 ⎪M ⎪ 2 ⎨ ⎪ M3 ⎪⎩ M 4

⎫ ⎡ K1 ⎪ ⎢ 0 ⎪ ⎢ ⎬= ⎪ ⎢− K1 ⎪⎭ ⎢⎣ 0

0 − K1 0⎤ ⎧ θ 1 ⎫ 0 0 0⎥⎥ ⎪⎪ θ 2 ⎪⎪ ⎨ ⎬, 0 K1 0⎥ ⎪ θ 3 ⎪ ⎥ 0 0 0⎦ ⎪⎩ θ 4 ⎪⎭ ⎧ M1 ⎪M ⎪ 2 ⎨ ⎪ M3 ⎪⎩ M 4

⎫ ⎡0 ⎪ ⎢0 ⎪ ⎢ ⎬=⎢ ⎪ 0 ⎪⎭ ⎢⎣0

⎧ M1 ⎪M ⎪ 2 ⎨ ⎪ M3 ⎪⎩ M 4 0

0 ⎫ ⎡0 ⎪ ⎢0 K ⎪ ⎢ 2 ⎬= ⎢ ⎪ 0 − K2 ⎪⎭ ⎢⎣0 0

0

0 0 0 K3 0 − K3

0 − K2 K2 0

0⎤ ⎧ θ 1 ⎫ 0⎥⎥ ⎪⎪ θ 2 ⎪⎪ ⎨ ⎬, 0⎥ ⎪ θ 3 ⎪ ⎥ 0⎦ ⎪⎩ θ 4 ⎪⎭

0 ⎤ ⎧θ1 ⎫ 0 ⎥⎥ ⎪⎪ θ 2 ⎪⎪ ⎨ ⎬. − K3 ⎥ ⎪ θ3 ⎪ ⎥ K 3 ⎦ ⎪⎩ θ 4 ⎪⎭

The torques in the overall system are obtained by adding all the torques at each node. This can be obtained by adding the expanded stiffness matrices to give

MECHANICAL VIBRATIONS

130

⎧ M1 ⎪M ⎪ 2 ⎨ ⎪ M3 ⎪⎩ M 4

0 ⎫ ⎡ K1 ⎪ ⎢ 0 K2 ⎪ ⎢ ⎬=⎢ ⎪ − K1 − K 2 ⎪⎭ ⎢⎣ 0 0

− K1 − K2 K1 + K 2 + K 3 − K3

0 ⎤ ⎧θ1 ⎫ 0 ⎥⎥ ⎪⎪ θ 2 ⎪⎪ ⎨ ⎬. − K3 ⎥ ⎪ θ3 ⎪ ⎥ K 3 ⎦ ⎪⎩ θ 4 ⎪⎭

(4.74)

Using the boundary condition θ 4 = 0 , the reduced system stiffness matrix is obtained, so the equations of motion can be written

⎡ J1 ⎢0 ⎢ ⎢⎣ 0

0 J2 0

0⎤ ⎧ θ&&1 ⎪ 0⎥⎥ ⎨ θ&&2 0⎥⎦ ⎪⎩ θ&&3

⎫ ⎡ K1 0 ⎪ ⎢ K2 ⎬+⎢ 0 ⎪ ⎢− K − K 2 ⎭ ⎣ 1

⎤ ⎧ θ1 ⎥ ⎪θ − K2 ⎥⎨ 2 K1 + K 2 + K 3 ⎥⎦ ⎪⎩ θ 3 − K1

⎫ ⎧0⎫ ⎪ ⎪ ⎪ ⎬ = ⎨ 0 ⎬. ⎪ ⎪0⎪ ⎭ ⎩ ⎭

(4.75)

Eliminating the coordinate θ 3 using the third equation − K1 θ 1 − K 2 θ 2 + (K1 + K 2 + K 3 )θ 3 = 0 we obtain the two equations of motion J1θ&&1 + J 2 θ&&2 −

K1 (K 2 + K 3 ) K1 K 2 θ1 − θ 2 = 0, K1 + K 2 + K 3 K1 + K 2 + K 3

K (K + K 3 ) K1 K 2 θ1 + 2 1 θ 2 = 0. K1 + K 2 + K 3 K1 + K 2 + K 3

(4.76)

After solving the eigenvalue problem, in order to draw the actual mode shapes, the angular amplitudes Θ 2 determined for the equivalent system must be transformed back to actual values using equation (4.71). A more straightforward approach is presented in Section 5.1.3.

4.3 Flexural Systems Herein, massless beams and planar frames are considered, with attached rigid masses. For such systems it is easier to calculate flexibilities instead of stiffnesses, the former being measurable quantities.

4.3.1 Flexibility Coefficients Let an elastic beam be subjected to forces f1 and f 2 at points 1 and 2 (Fig. 4.14, a). The deflections in the directions of f1 and f 2 are y 1 and y2 , respectively.

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

131

Consider the unloaded beam (Fig. 4.14, b) and apply a single unit force at point 1 in the direction of f1 . Let the deflections at 1 and 2, in the directions y 1 and y2 be δ 11 and δ 21 . Similarly, let a unit force be applied at point 2 in the direction of f 2 and denote the deflections δ 12 and δ 22 (Fig. 4.14, c). The relationship between the forces f1 and f 2 and the total deflections y 1 and y2 can be expressed by the equations y 1 = δ 11 f1 + δ 12 f 2 , y2 = δ 21 f1 + δ 22 f 2 .

(4.77)

The coefficients δ ij are called flexibility (influence) coefficients. By definition, δ ij is the deflection of coordinate i due to unit load applied to coordinate j.

Fig. 4.14 In matrix notation, equations (4.77) become

or where [ δ

⎧ y 1 ⎫ ⎡δ 11 δ 12 ⎤ ⎧ f1 ⎫ ⎨ ⎬ = ⎢δ ⎥⎨ ⎬ ⎩ y2 ⎭ ⎣ 21 δ 22 ⎦ ⎩ f 2 ⎭

(4.77, a)

{ y } = [ δ ]{ f },

(4.78)

] is known as the flexibility matrix.

The flexibility matrix is symmetrical [ δ ] T = [ δ ] , according to Maxwell’s reciprocal theorem: The deflection at one point in a structure due to a unit load applied at another point equals the deflection at the second point when a unit load is applied at the first (deflections are measured in the same direction as the load).

MECHANICAL VIBRATIONS

132

{ f } = [ k ] { y } the matrix [ k ] known as the stiffness matrix, it comes out that [ k ] = [ δ ] −1 or alternatively Because in an equation of the type

[ δ ] = [ k ] −1 .

is

(4.79)

When rigid-body motions are possible, [ δ ] = [ k ] −1 does not exist. There are no “ground” reactions to counterbalance the unit forces that must be applied to the structure to determine [δ ] and equilibrium is not possible.

4.3.2 Equations of Motion The massless beam of Fig. 4.15, a has constant bending rigidity EI and carries masses m 1 and m 2 at points 1 and 2. If only the two lateral displacements of the two masses are of interest, the motion is completely defined by deflections y 1 and y2 , hence the system has two degrees of freedom.

Fig. 4.15 In free vibrations (Fig. 4.15, b), applying d’Alembert’s principle, the only external forces acting on masses are the inertia forces f1 = −m 1 &y&1 ,

f 2 = − m 2 &y& 2 .

(4.80)

Substituting these forces in equations (4.77) yields the differential equations of motion

δ 11 m 1 &y&1 + δ 12 m 2 &y&2 + y 1 = 0 , δ 21 m 1 &y&1 + δ 22 m 2 &y&2 + y2 = 0.

(4.81)

In matrix form

[ b ] { &y& } + { y } = { 0 }.

(4.82)

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

133

The matrix [ b ] can be written b11 b12 ⎤ ⎡δ 11 m1 δ 12 m2 ⎤ ⎥ = [δ ⎥=⎢ ⎣b21 b22 ⎦ ⎣δ 21 m1 δ 22 m2 ⎦

[ b ] = ⎡⎢

] [ m ],

(4.83)

where [ m ] is the diagonal mass matrix. In equations (4.81) the coupling is due to the off-diagonal elements of the flexibility matrix.

4.3.3 Normal Modes Assuming solutions of the form

{ y } = { a } cos (ω t − ϕ ) ,

(4.84)

equation (4.81) becomes

− ω 2 [ b ]{ a } + { a } = { 0 } or

[ b ]{a } =

1

ω2

{a }.

(4.85)

This is the standard eigenvalue problem, in which λ = 1 ω 2 are the eigenvalues and { a } are the eigenvectors. The eigenvalues are the inverses of the natural frequencies squared, and the eigenvectors are the modal vectors that define the shape of the natural modes of vibration. Equations (4.85) may be written under the form

(b

) a + b ω a = 0, ( b ω − 1) a = 0.

2 11ω − 1 b21ω 2 a1 +

1

22

12 2

2

2

(4.85, a)

2

The condition to have non-trivial solutions gives the frequency equation

(b

11ω

or

2

)(

)

− 1 b22 ω 2 − 1 − b12 b21ω 4 = 0

( b11 b22 − b12 b21 )ω 4 − (b11 + b22 )ω 2 + 1 = 0 .

(4.86)

Equation (4.86) has two real positive roots ω 12 and ω22 , the natural frequencies squared. The mode shapes are defined by the ratio μ = a2 a1 , so that

MECHANICAL VIBRATIONS

134

⎛ a2 ⎞ 1 − b11ω 12 ⎟ = , 2 ⎜ a1 ⎟ b ω 12 1 ⎝ ⎠1

μ1 = ⎜

2 ⎛ a2 ⎞ ⎟ = 1 − b11ω2 . ⎜ a1 ⎟ b12 ω22 ⎝ ⎠2

μ2 =⎜

(4.87)

Example 4.5 Calculate the natural modes of vibration for the beam of Fig. 4.15, a where l1 = l 2 = l 2 , m 1 = m2 = m and E I = const .

Solution. The flexibility coefficients are

δ11 = l3 24 EI , δ12 = δ 21 = 5l 3 48EI , δ 22 = l 3 3EI . The flexibility matrix is

[δ ] =

l3 48 EI

⎡2 5 ⎤ ⎢5 16⎥ . ⎣ ⎦

Equation (4.85) can be written

⎡2 5 ⎤ ⎧ a1 ⎫ ⎢5 16⎥ ⎨a ⎬ = λ ⎣ ⎦ ⎩ 2⎭

⎧ a1 ⎫ ⎨ ⎬ ⎩a 2 ⎭

where

λ = 48 E I ml 3ω 2 .

Fig. 4.16 The frequency equation is 2−λ

5

5

16 − λ

=0

or

λ2 − 18 λ + 7 = 0

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

with solutions

λ1 = 17.602

and

135

λ2 = 0.3976 .

The natural frequencies are

ω 1 = 1.6503 E I ml 3 ,

ω 2 = 10.986 E I ml3 .

The mode shapes are defined by the ratio μ = a2 a1 = λ − 2 5 ⎛ a2 ⎞ λ 1 − 2 ⎟ = = 3.12 , ⎜ a1 ⎟ 5 ⎝ ⎠1

μ1 = ⎜

⎛ a2 ⎞ λ −2 ⎟ = 2 = − 0.32 . ⎜ a1 ⎟ 5 ⎝ ⎠2

μ2 =⎜

They are graphically presented in Fig. 4.16.

4.3.4 Free Vibrations Let denote by

[ u ] = [{ u }1 { u } 2 ] = ⎡⎢

u11 u12 ⎤ ⎥ ⎣u 21 u22 ⎦

(4.88)

the matrix of normalized mode shapes (modal matrix). The actual deflections, denoted in the following by { x } instead of may be expressed in terms of the modal coordinates as in (4.23)

{ y },

2

{ x } = [ u ]{ q } = {u }1 q 1 + {u } 2 q 2 = ∑ Ci cos (ω i t − ϕi ) {u }i .

(4.89)

i =1

The constants Ci and ϕi may be evaluated from the initial conditions of the motion 2

{ x (0) } = ∑ Ci cos ϕi {u }i ,

(4.90)

i =1 2

{ x& (0) } = ∑ ω i Ci sin ϕi { u } i .

(4.91)

i =1

Premultiplying equations (4.90) and (4.91) by

{ u }Tj [ m ] ,

using the

orthogonality relationships (4.18) - (4.19) and the definitions (4.25) and (4.28) of the modal masses

{ u }Tj [ m ] { u }i = 0 ,

i≠ j

M i = { u }Ti [ m ] { u } i , i = 1, 2 (4.92)

MECHANICAL VIBRATIONS

136

we obtain

{ u }Tj [ m ] { x (0) } = C j M j cos ϕ j ,

j = 1, 2

(4.93)

{ u }Tj [ m ]{ x& (0) } = ω j C j M j sin ϕ j .

j = 1, 2

(4.94)

Combining equations (4.93) and (4.94) and renaming the index yields tan ϕi =

{ u }Ti [ m ] { x& (0) } , ωi { u }Ti [ m ] { x (0) }

i = 1, 2

(4.95)

i = 1, 2

(4.96)

and Ci =

{ u }Ti [ m ] { x(0) } { u }Ti [ m ] { x& (0) } M i cos ϕi

=

ωi M i sin ϕi

.

Example 4.6 For the system of Fig. 4.17: a) determine the natural modes of vibration; b) derive the equations of the free vibrations when the mass has an initial vertical velocity v and calculate the trajectory of the mass.

Fig. 4.17 Solution. a) Let y and z be the vertical and horizontal components of the instantaneous displacement of mass m. The flexibility coefficients are

δ yy = 4l 3 3EI , δ yz = δ zy = l 3 2 EI , The equations of motion can be written

δ zz = l 3 3EI .

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

137

4l 3 l3 m &y& + m &z& + y = 0 , 3E I 2E I

l3 l3 m &y& + m &z& + z = 0. 2E I 3E I Looking for solutions of the form

z ( t ) = u2 cos (ω t − ϕ )

y ( t ) = u 1 cos (ω t − ϕ ), we obtain

( 8 − β ) u 1 + 3 u 2 = 0, 3 u 1 + ( 2 − β ) u 2 = 0,

where

β = 6 E I m l 3ω 2 . The frequency equation is

with solutions

8−β

3

3

2−β

=0,

β 1 = 9.2426 ,

β 2 − 10 β + 7 = 0 ,

β 2 = 0.7574 .

The natural frequencies are

ω 1 = 0.8057 E I ml3 ,

ω 2 = 2.8146 E I ml3 .

The mode shapes are given by ⎛ u2 ⎞ u21 β1 − 8 ⎟⎟ = = = 0.4142 , 3 ⎝ u1 ⎠1 u11

μ 1 = ⎜⎜

⎛u ⎞

u

μ 2 = ⎜⎜ 2 ⎟⎟ = 22 = ⎝ u1 ⎠ 2 u12

β 2 −8 3

The modal vectors, normalized with unit first element, are

{ u }1 = ⎧⎨

⎫ ⎬, ⎩0.4142⎭ 1

{ u }2 = ⎧⎨

⎫ ⎬. ⎩− 2.4142⎭ 1

It can be seen that u21 = tan γ 1 = 0.4142 , u11 u22 = tan γ 2 = −2.4142 , u12

γ 1 = 22.50 ,

γ 2 = 112.50 = γ 1 + 900 .

= −2.4142 .

MECHANICAL VIBRATIONS

138

The mass m has unidirectional motion in the natural modes of vibration.

The modal vectors are orthogonal { u }T1 { u }2 = 0 . The motion in the first mode is

along direction 1 at 22.50 measured from the vertical direction. The motion in the second mode is along direction 2 at 122.50 , hence normal to direction 1. This happens because the motion in modal coordinates is along the directions of principal flexibilities. The flexibility matrix

[δ ] =

l 3 ⎡8 3⎤ . 6 E I ⎢⎣3 2⎥⎦

represents a deflection-force relation between the components y and z of the deflection and the components f y and f z of a force applied to the point where the mass is located ⎧ y ⎫ ⎡δ yy ⎨ ⎬=⎢ ⎩ z ⎭ ⎣δ zy

δ yz ⎤ ⎧ f y ⎫ . δ zz ⎥⎦ ⎨⎩ f z ⎬⎭

Consider a reference frame y ∗Oz ∗ rotated through an angle γ with respect to the frame yOz . The transformation of displacements can be written ⎧⎪ y ∗ ⎫⎪ ⎡ cosγ ⎨ ∗ ⎬=⎢ ⎪⎩ z ⎪⎭ ⎣− sinγ

sinγ ⎤ ⎧ y ⎫ ⎨ ⎬ cosγ ⎥⎦ ⎩ z ⎭

and the transformation of forces is defined by ⎧⎪ f y∗ ⎫⎪ ⎡ cosγ ⎨ ∗ ⎬=⎢ ⎪⎩ f z ⎪⎭ ⎣− sinγ

sinγ ⎤ ⎧ f y ⎫ ⎨ ⎬. cosγ ⎥⎦ ⎩ f z ⎭

The new deflection-force relation is

[ ] ⎧⎪⎨⎪ ff

where

∗ y ∗ z

⎫⎪ ⎬ ⎪⎭

⎧⎪ y ∗ ⎫⎪ ∗ ⎨ ∗ ⎬= δ ⎪⎩ z ⎪⎭

⎩

[δ ] = ⎡⎢−cs

s⎤ c − s⎤ [ δ ] ⎡⎢ ⎥ ⎥ c⎦ ⎣s c ⎦

∗

⎣

with c = cosγ and s = sinγ . The flexibility matrix in the rotated reference frame is

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

[δ ] ( ∗

139

⎡ δ yy c 2 + δ zz s 2 + 2δ yz cs =⎢ 2 2 ⎢⎣ δ zz − δ yy sc + δ yz c − s

(

)

(δ zz − δ yy )sc + δ yz ( c 2 − s 2 )⎤⎥ .

)

δ yy s 2 + δ zz c 2 − 2δ yz cs ⎥⎦

It can be seen that there are two angles γ ∗ for which the off-diagonal elements vanish, given by tan 2 γ ∗ =

2 δ yz

=

δ yy − δ zz

2⋅3 = 1 , γ 1∗ = 22.50 , 8−2

γ 2∗ = 112.50 .

The two solutions γ 1∗ and γ 2∗ define the principal directions of flexibility. Substituting these angles into the expression of diagonal elements we obtain the principal flexibilities

δ1, 2 =

δ yy + δ zz 2

⎛ δ yy − δ zz ⎜⎜ 2 ⎝

±

δ1 = 1.5404

2

⎞ 5 ± 3 2 l3 2 ⎟⎟ + δ yz = , 6 EI ⎠

l3 , EI

δ 2 = 0.1262

l3 . EI

Their meaning is straightforward. A force applied along 1 (or 2) produces a deflection only along 1 (or 2). The principal directions of flexibility coincide with the directions of vibration in the natural (principal) modes of vibration. The natural frequencies are given by

ω1 =

1 1 = m δ1 1.5404

ω2 =

1 1 = mδ2 0.1262

EI ml

3

EI ml

3

= 0.805

= 2.815

EI m l3 EI m l3

,

.

b) To determine the free response to a vertical impulse of velocity v , let calculate first the modal masses

( = m (u

) ( ) ) = m (1 + 2.4142 ) = 6.8284 m .

2 2 M 1 = m u11 + u21 = m 1 + 0.4142 2 = 1.1716 m ,

M2

2 12

2 + u22

2

The initial conditions are

{ x (0) } = ⎧⎨

0⎫ ⎬, ⎩0⎭

From (4.93) and(4.96) we obtain

{ x& (0) } = ⎧⎨

v⎫ ⎬. ⎩0⎭

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140

cos ϕ 1 = 0 ,

cos ϕ 2 = 0 ,

mv , ω 1 M1

C2 sin ϕ 2 =

C1 sin ϕ 1 =

mv . ω2 M 2

a

b Fig. 4.18

Fig. 4.19 The vertical component of the instantaneous displacement is

(

)

(

)

y (t ) = C1 cos ω 1 t − ϕ1 u11 + C2 cos ω 2 t − ϕ 2 u12 , y (t ) =

mv mv sin ω 1t + sin ω 2t , ω 1M 1 ω 2M 2

(

)

y (t ) = C 1.0593 sin ω 1 t + 0.0520 sin ω 2 t , where

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

C=v

ml 3 , ω 1 = 0.805 v C , ω EI

141

2

= 2.815 v C .

The horizontal component of the instantaneous displacement is

(

)

(

)

z (t ) = C1 cos ω 1 t − ϕ 1 u 21 + C2 cos ω 2 t − ϕ 2 u22 ,

z (t ) =

mv mv u 21 sin ω 1 t + u 22 sin ω 2 t , ω1 M1 ω2 M 2

(

)

z (t ) = C 0.4387 sin ω 1 t − 0.1256 sin ω 2 t . The trajectory of the mass m is plotted in Fig. 4.18, a for a time duration equal to 2π ω 1 and in Fig. 4.18, b for 4π ω 1 . The two components y and z are plotted in Fig. 4.19 as a function of time. It is seen that the horizontal component is almost harmonic, the second component having a relatively small amplitude.

4.3.5 Response to Harmonic Excitation Consider the steady-state vibrations of the beam of Fig. 4.20, a under the action of the force f (t ) = F0 cos ω t acting on mass m2 .

Fig. 4.20 The equations of motion are

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142

( ) y2 = −δ 21 m 1 &y&1 + ( f − m 2 &y&2 ) δ 22

y 1 = −δ 11 m 1 &y&1 + f − m 2 &y&2 δ 12 , or

δ 11 m 1 &y&1 + δ 12 m 2 &y&2 + y 1 = δ 12 f (t ) ,

(4.97)

δ 21 m 1 &y&1 + δ 22 m 2 &y&2 + y2 = δ 22 f (t ) .

Substituting the steady-state solutions y1 (t ) = Y1 cos ω t ,

y2 (t ) = Y 2 cos ω t ,

into equations (4.97) yields

(1 − ω δ 2

)Y −ω δ m Y + (1 − ω δ

11 m 1

− ω 2δ 21

2

1

12

2

1 1

m 2 Y2 = δ 12 F0 ,

(4.98)

)

22 m 2 Y2 = δ 22 F0 .

The solutions are

Y1 = Y2 =

(

)

δ 12

(

)

F0 ,

)

F0 .

2 1 − δ 11 m 1 + δ 22 m 2 ω 2 + m 1 m 2 δ 11δ 22 − δ12 ω4

(

2 δ 22 − m 1ω 2 δ 11δ 22 − δ12

(

)

(

)

2 1 − δ 11 m 1 + δ 22 m 2 ω 2 + m 1 m 2 δ 11δ 22 − δ12 ω4

(4.99)

The denominator can be recognized as the characteristic polynomial (4.86).

a

b Fig. 4.21

The absolute values of the amplitudes Y1 and Y2 are graphically presented in Fig. 4.21. When the forcing frequency equals either of the natural frequencies, the amplitudes grow indefinitely. The system has two resonances marked by peaks in the frequency response curves.

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

143

Antiresonance occurs for Y2 = 0 when

ω 2 = ωa2 =

(

δ 22

2 m 1 δ 11δ 22 − δ 12

)

.

(4.100)

If the amplitudes of the forced response are known, the amplitudes of the inertia forces can be calculated so that the amplitude of the dynamic forces acting on the beam (Fig. 4.20, b) are

Φ 1 = m1ω 2 Y1 , Φ 2 = m2 ω 2 Y2 + F0 .

(4.101)

The diagram of dynamic bending moments (Fig. 4.20, c) may be then constructed, and the dynamic stresses produced by the harmonic force may be calculated.

Example 4.7 The massless beam of Fig. 4.22, a has diameter d = 40 mm , l = 1 m , E = 210 GPa and m = 50 kg . a) Calculate the natural frequencies; b) Determine the amplitudes of forced vibrations produced by a harmonic force of amplitude F0 = 20 N and frequency 0.179 Hz ; c) Draw the diagram of static bending moments and determine the maximum static stress ; d) Draw the diagram of the dynamic bending moments and calculate the amplitude of the maximum dynamic stress. Solution. The flexibility coefficients are

δ 11 = l 3 6 E I , δ 12 = δ 21 = − l3 8 E I , δ 22 = 5 l 3 24 E I . Denoting

λ=

24 E I , ω 2m l3

the frequency equation is

λ −8

3

6

λ −5

=0,

λ2 − 13λ + 22 = 0 ,

with solutions

λ 1 = 11 ,

λ2 = 2.

The natural frequencies are

ω 1 = 1.477 E I ml 3 = 1.073 rad sec ,

MECHANICAL VIBRATIONS

144

ω 2 = 3.464 E I ml 3 = 2.516 rad sec .

Fig. 4.22 For the given numerical data, the forcing frequency corresponds to λ = 10 . From (4.99) we obtain the vibration amplitudes 3λ

F0 = 0.118 mm , mω 2 F0 22 − 5 λ Y2 = − = −1.105 mm . (λ − 11) (λ − 2) mω 2 Y1 = −

(λ − 11) (λ − 2)

For the static loading shown in Fig. 4.22, b, the diagram of static bending moments is presented in Fig. 4.22, c. The maximum bending moment is 368 Nm and the maximum static stress is σ st = 58.5 N mm 2 . The amplitudes (4.101) of dynamic forces are

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

Φ 1 = 2m ω 2 Y1 = −

6λ

145

(λ − 11) (λ − 2) ⎡

Φ 2 = m ω 2 Y2 + F0 = ⎢1 − ⎣

F0 = 150 N ,

22 − 5 λ

⎤

(λ − 11) (λ − 2) ⎥⎦

F0 = −50 N .

For the dynamic loading shown in Fig. 4.22, d, the diagram of dynamic bending moments is given in Fig. 4.22, e. The maximum bending moment is 87.5 Nm and the maximum dynamic stress is σ d = 14 N mm 2 .

4.4 Coupled Translation and Rotation When the forces and reactions acting on an elastically supported single mass do not coincide with the centre of gravity of the mass, the translational and rotational modes of vibration are coupled. Applications are found at the rigid-body bouncing and pitching of cars on suspensions and car engines on flexible mountings, vibrations of rigid rotors in two bearings, vibrations of long masses at the tip of short cables in Stockbridge dynamic absorbers, vibrations of beams with end discs.

4.4.1 Equations of Motion Consider a rigid bar of mass m and mass moment of inertia about the centre of gravity J supported at its ends on springs of stiffnesses k 1 and k 2 (Fig. 4.23, a). The bar is constrained so that any point can only translate in the vertical direction. Let the bar motion be defined by two coordinates: x − the linear displacement of the centre of mass G, and θ − the angle of rotation about G. Using the free-body diagram of Fig. 4.23, b, the equations of motion can be written m &x& + k 1 (x − l 1θ ) + k 2 ( x + l 2 θ ) = 0 , Jθ&& − k 1 ( x − l 1θ ) l 1 + k 2 ( x + l 2 θ ) l 2 = 0

or

( ) ( ) J θ&& + ( k 2 l 2 − k 1 l 1 ) x + ( k 1 l 12 + k 2 l 22 )θ = 0.

m &x& + k 1 + k 2 x + k 2 l 2 − k 1 l 1 θ = 0,

(4.102)

In matrix form k 2 l 2 − k1 l 1 ⎤ ⎧ x ⎫ ⎧ 0 ⎫ ⎡m 0 ⎤ ⎧ &x& ⎫ ⎡ k1 + k 2 ⎢ 0 J ⎥ ⎨ θ&& ⎬ + ⎢k l − k l k l 2 + k l 2 ⎥ ⎨ θ ⎬ = ⎨ 0 ⎬ . 2 2⎦ ⎩ ⎭ ⎩ ⎭ ⎣ ⎦⎩ ⎭ ⎣ 2 2 1 1 1 1

(4.102, a)

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146

Note in the above equations that if k1l 1 = k 2l 2 the system is uncoupled, and has two independent natural frequencies – one for translation and one for rotation. These are defined by

ωx =

(k1 + k2 )

m and ω θ =

(k l

2 1 1

+ k 2 l 22

)J.

(4.103)

With zero coupling, a force applied to the centre of gravity produces only up-and-down bouncing x, whereas a torque applied to the bar produces only pitching motion θ . The coupling is given by the off-diagonal elements of the stiffness matrix, hence it is referred to as static coupling.

Fig. 4.23 Let the bar motion be defined by other two coordinates, the linear displacements of the bar ends (points of spring connection) x1 and x2 . The transformation of coordinates is defined by ⎧x⎫ 1 ⎡ l 2 l 1 ⎤ ⎧ x1 ⎫ ⎨ ⎬= ⎢ ⎥ ⎨ ⎬. ⎩ θ ⎭ l 1 + l 2 ⎣ − 1 1 ⎦ ⎩ x2 ⎭

(4.104)

Substituting (4.104) into (4.102, a) and multiplying to the left by the transpose of the transformation matrix from (4.104) yields the equations of motion ⎡ ml 22 + J ⎢ ⎣⎢ml 1l 2 − J

0 ⎤ ⎧ x 1 ⎫ ⎧0⎫ ml 1l 2 − J ⎤ ⎧&x&1 ⎫ ⎡k 1 (l 1 + l 2 ) ⎥ ⎨ ⎬+⎢ ⎨ ⎬ = ⎨ ⎬ . (4.105) 2 k 2 (l 1 + l 2 )⎥⎦ ⎩ x2 ⎭ ⎩0⎭ 0 ml1 + J ⎦⎥ ⎩ &x&2 ⎭ ⎣

The coupling is given by the off-diagonal elements of the mass matrix, hence it is referred to as dynamic coupling. When J = m l 1l 2 the coupled translation and rotation can be expressed as a sum of two pitching vibrations, one about the the right end and the other about the left end.

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

147

4.4.2 Normal Modes Denoting

(

)

a = (k1 + k 2 ) m , b = (k 2l 2 − k1l 1 ) m , c = k1l 12 + k 2 l 22 J , equations (4.97) become &x& + a x + bθ = 0 , r 2θ&& + b x + c r 2θ = 0 ,

(4.106)

where r = J m is the radius of gyration. Assuming solutions of the form x (t ) = A x cos (ω t − ϕ ) , θ (t ) = Aθ cos (ω t − ϕ ) ,

(4.107)

we obtain the set of algebraic equations

( a − ω ) A + b Aθ = 0 , b A + ( c − ω ) r Aθ = 0 . 2

x

2

x

(4.108)

2

Equations (4.108) admit non-trivial solutions if the determinant of the coefficients A x and Aθ is zero a −ω2

b

(c − ω ) r 2

b or

(

=0

2

(4.109)

)

ω 4 − (a + c )ω 2 + a c − b 2 r 2 = 0 .

(4.109, a)

The solutions of the frequency equation (4.109, a) are given by

ω 12, 2 = (a + c ) 2 ±

(a − c )2

4 + b2 r 2 .

(4.110)

The first natural frequency ω 1 is always less than ω x or ω θ , whichever is smaller, and the second natural frequency ω 2 is always greater than ω x and ω θ . The mode shapes are obtained substituting the natural frequencies in turn in the amplitude ratio

( ) Aθ ) = − b ( a − ω ) .

μ 1 = ( A x Aθ )1 = − b a − ω 12 , μ 2 = ( Ax

2

2 2

(4.111)

MECHANICAL VIBRATIONS

148

Example 4.8 For the system of Fig. 4.23 determine the natural modes of vibration if l 1 = 3l 4 , l 2 = l 4 , k1 = k 2 = k , J = m l 2 8 .

Solution. Equations (4.108) have the form kl ⎛ 2k ⎞ −ω2 ⎟ Ax + Aθ = 0 , ⎜ 2m ⎝ m ⎠ ⎛ 5k ⎞ l2 kl − ω 2 ⎟⎟ A x + ⎜⎜ Aθ = 0 . 2m ⎝ m ⎠ 8 Denoting

α=

ω 2m k

,

the equations become

( 2 − α ) A x + l Aθ 2

= 0,

l l2 Ax + ( 5 −α ) Aθ = 0 . 2 8

The frequency equation is

α 2 − 7α + 8 = 0 with solutions

α 1 = 1.438 ,

α 2 = 5.561 .

The natural frequencies are

ω 1 = 1.199 k m ,

ω 2 = 2.358 k m .

Fig. 4.24

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

149

The mode shapes (shown in Fig. 4.24) are defined by

⎛ Aθ l 4 ⎞ 2 − α 1 ⎟ = ⎜ = 0.28 , ⎜ Ax ⎟ 2 ⎠1 ⎝

⎛ Aθ l 4 ⎞ 2 −α 2 ⎟ = ⎜ = −1.78 . ⎜ Ax ⎟ 2 ⎠2 ⎝

Negative inverses of the above amplitude ratios define the location of the nodal point with respect to the centre of mass, taking a positive value to the right

⎛ Ax ⎞ ⎛ Ax ⎞ ⎟ = − 1 = −3.56 , d 2 = −⎜ ⎟ = 1 = 0.56 . d 1 = −⎜ ⎟ ⎜ Aθ l 4 ⎟ ⎜ 0.28 ⎠1 ⎝ ⎝ Aθ l 4 ⎠ 2 1.78 The frequencies of the uncoupled pure translation and pure rotation are

ω x = 1.414 k m ,

ω θ = 2.236 k m .

The following inequalities hold in this case

ω1 < ω x < ωθ < ω 2 .

Example 4.9 Calculate the natural modes of vibration for the system of Fig. 4.25, a which consists of a long rigid body of mass m and mass moment of inertia

J = m l 2 8 attached at the end of a cantilever beam of bending rigidity EI.

a

b Fig. 4.25

Solution. Let the rigid body motion be defined by the linear displacement x of the centre of mass G, and by the angle of rotation θ about G.

MECHANICAL VIBRATIONS

150

The relationship between the inertia force F = mω 2 x , the inertia torque

M = Jω 2θ and the linear and angular displacements x and θ can be expressed by the equations x = δ 11 F + δ 12 M ,

θ = δ 21 F + δ 22 M . Two bending moment diagrams are constructed (Fig. 4.25, b), mx − for loading with a unit force at point G in the direction of F , and mθ − for loading with a unit torque at point G in the direction of M . They are used to calculate the flexibility coefficients using Mohr-Maxwell’s method. Let the resulting deflections in the directions x and θ be δ 11 and δ 21 , respectively δ 12 and δ 22 . The flexibility influence coefficients are

δ11 = 13 l 3 12 E I ,

δ12 = δ 21 = l 2 E I ,

δ 22 = l E I .

Substitution in the equations of equilibrium yields

3 2 12 3 βθ = x+ θ , 2 l

β x = 13 x + lθ ,

where

β = 12 E I m l 3 ω 2 .

Fig. 4.26 The frequency equation is

2β 2 − 29β + 3 = 0 with solutions β 1 = 14.396 and β 2 = 0.1042 .

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

151

The natural frequencies are

ω 1 = 0.913 E I ml 3 ,

ω 2 = 10.73 E I ml 3 .

The mode shapes (Fig. 4.26) are given by ⎛ Aθ l ⎞ β 1 − 13 ⎜⎜ ⎟⎟ = = 0.930 , 32 ⎝ Ax ⎠1

β − 13 ⎛ Aθ l ⎞ ⎜⎜ ⎟⎟ = 2 = −8.597 . 32 ⎝ Ax ⎠ 2

4.5 Coupled Pendulums An interesting phenomenon arises in the free vibrations of coupled pendulums, where a continuous transfer of motion occurs from one pendulum to the other due to the weak coupling by an elastic element.

4.5.1 Equations of Motion Consider two simple pendulums (Fig. 4.27, a), each of length l and mass m, swinging in the vertical plane and coupled together by a light spring of stiffness k attached at a distance d from the supporting points and unstrained when the pendulums are in the vertical position.

Fig. 4.27 Using the angular coordinates θ 1 and θ 2 and assuming small amplitudes, the equations for the kinetic and potential energies are T=

(

)

1 2 &2 &2 ml θ 1 + θ 2 , 2

(4.112)

MECHANICAL VIBRATIONS

152

(

)

U = m gl 1 − cosθ 1 + m gl ( 1 − cosθ 2 ) +

or U=

(

)

(

1 1 m g l θ 12 + θ 22 + k d 2 θ 2 − θ 1 2 2

(

1 2 k d θ2 − θ 1 2

)2 .

)2 , (4.113)

Using Lagrange’s equations d ∂T ∂T ∂U − + = 0, d t ∂q& r ∂qr ∂qr

r = 1, 2

(4.114)

we obtain the equations of motion

( ) ml 2θ&&2 + m glθ 2 − k d 2 (θ 1 − θ 2 ) = 0 .

ml 2θ&&1 + m glθ 1 + k d 2 θ 1 − θ 2 = 0 ,

(4.115)

In matrix form, the equations of motion are written ⎡m l 2 ⎢ ⎢⎣ 0

0 ⎤ ⎧ θ&&1 ⎫ ⎡ m g l + kd 2 ⎥ ⎨ ⎬+⎢ m l 2 ⎥⎦ ⎩ θ&&2 ⎭ ⎢⎣ − k d 2

− kd 2 ⎤ ⎧θ1 ⎫ ⎧ 0 ⎫ ⎥ ⎨ ⎬ = ⎨ ⎬ . (4.115, a) m g l + k d 2 ⎥⎦ ⎩ θ 2 ⎭ ⎩ 0 ⎭

The coupling is due to the spring k.

4.5.2 Normal Modes Assuming solutions of the form

θ 1 = a 1 cos (ω t − ϕ ) ,

θ 2 = a2 cos (ω t − ϕ ) ,

and substituting into the differential equations (4.110), we obtain

(mgl − ω m l + k d ) a − k d a + (mgl − ω m l 2

2

2

2

2

1

1

− k d 2 a2 = 0 ,

2

+ k d 2 a2 = 0.

)

The frequency equation is

(mgl − ω m l 2

or

⎛g

ω 4 − 2 ⎜⎜

⎝l

The natural frequencies are

+

2

+ kd2

) − (kd ) 2

2 2

=0

k d 2 ⎞⎟ 2 ⎛⎜ g 2 k d 2 g ⎞⎟ + + ω 2 =0. ⎜ l2 ml 2 ⎟⎠ ml 3 ⎟⎠ ⎝

4. TWO-DEGREE-OF-FREEDOM SYSTEMS

g , l

ω1 =

ω2 =

153

g kd2 +2 2 . l ml

(4.116)

The mode shapes are defined by the amplitude ratios ⎛ a2 ⎞ ⎟ = +1 , ⎜ a1 ⎟ ⎝ ⎠1

μ1 = ⎜

⎛ a2 ⎞ ⎟ = −1 . ⎜ a1 ⎟ ⎝ ⎠2

μ2 =⎜

(4.117)

In the first mode (Fig. 4.27, b), the pendulums are swinging in phase with equal amplitudes. The coupling spring is not strained and the pendulums move as if they were uncoupled. The system natural frequency is equal to that of a single pendulum g l . In the second mode (Fig. 4.27, c), the two pendulums swing against each other with equal amplitudes. Due to the stiffening effect of the coupling spring, the natural frequency is higher than in the first mode.

4.5.3

Free Vibrations

The general solution of free vibrations (4.14) is

θ 1 (t ) = C1 cos (ω 1 t − ϕ1 ) u11 + C2 cos (ω 2 t − ϕ 2 ) u12 , θ 2 (t ) = C1 cos (ω 1 t − ϕ1 ) u21 + C2 cos (ω 2 t − ϕ 2 ) u22 ,

or

θ 1 (t ) = a 1 cos (ω 1 t − ϕ 1 ) + a 2 cos (ω 2 t − ϕ 2 ) , θ 2 (t ) = μ 1 a 1 cos (ω 1 t − ϕ 1 ) + μ 2 a 2 cos (ω 2 t − ϕ 2 ) . Differentiating with respect to time yields

θ&1 (t ) = −ω 1 a 1 sin (ω 1 t − ϕ 1 ) − ω 2 a2 sin (ω 2 t − ϕ 2 ) , θ& 2 (t ) = − μ 1 ω 1 a 1 sin (ω 1 t − ϕ 1 ) − μ 2 ω 2 a 2 sin (ω 2 t − ϕ 2 ) . The integration constants are determined from the initial conditions

θ 1 (0 ) = θ 0 , θ 2 (0 ) = 0 , θ&1 (0 ) = 0 , θ& 2 (0 ) = 0 .

(4.118)

Because μ1 = 1 and μ 2 = −1 , we obtain a 1 cos ϕ 1 + a 2 cos ϕ 2 = θ 0 ,

a 1 cos ϕ 1 − a 2 cos ϕ 2 = 0 ,

ω 1 a 1 sinϕ 1 + ω 2 a 2 sinϕ 2 = 0 , ω 1 a 1 sinϕ 1 − ω 2 a 2 sinϕ 2 = 0 , so that

sin ϕ 1 = sin ϕ 2 = 0 , ϕ 1 = ϕ 2 = 0 ,

MECHANICAL VIBRATIONS

154

2 a 1 cosϕ 1 = 2 a 2 cosϕ 2 = θ 0 ,

a 1 = a2 = θ 0 2 .

The instantaneous values of the angular displacements are

(

θ 1 (t ) =

θ0

θ 2 (t ) =

θ0

where ωm = ω 1 + ω 2

2

2

( cos ω 1 t + cos ω 2 t ) = θ0 cos ω m t ⋅ cos Δω t , ( cos ω 1 t − cos ω 2 t ) = θ0 sin ω m t ⋅ sin Δω t ,

(4.119)

) 2 and Δω = ( ω 2 − ω 1 ) 2 .

When Δω is small with respect to ωm , the products in the above equations represent amplitude modulated oscillations known as beats. This condition is equivalent to

ω22 − ω 12