M. Rades - Mechanical Vibrations 1

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


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


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


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.


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


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


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)


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.


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

⎞ ⎟⎟ . ⎠


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


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.


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.


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


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


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 =

(

)


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

.


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& =


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)


24

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


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.


26

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.


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


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.


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.


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)


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.


32

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.


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


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)


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.


36

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


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.


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)


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).


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 << 1 , δ ≅ 2π ζ . Sometimes the decay after one cycle of vibration is too small so that it is possible to distinguish a smaller amplitude only after n cycles. The ratio

( )

x0 x 0 x1 x 2 xn −1 = ⋅⋅ = eδ xn xn x1 x 2 x3

n

= e nδ

so that the logarithmic decrement is given by 1 n

δ = ln

x0 . xn

(2.51, a)


41

If the peak amplitude of vibration is plotted on a logarithmic scale against the cycle number on an arithmetic scale, the points will fall on a straight line if the damping is of viscous type as assumed in equation (2.38). In practice, the envelopes of peaks and troughs are first drawn (Fig. 2.25). The height between the envelope lines is then measured at each peak and trough. The height is plotted on a logarithmic scale against the number of semicycles and the best line is drawn through the points. The slope of this line is then used to determine the damping ratio.

Fig. 2.25 For ζ << 1 , equation (2.51, a) yields


42

ln x n = ln x 0 − 2 ζ π ⋅ n so that the damping ratio ζ is equal to the slope of the line divided by 2π (or by π , if measurements are made at heights and troughs as in Fig. 2.25).

2.3.3 Loss Factor A convenient measure of damping is obtained by the loss factor which is defined as the ratio of energy lost per cycle (or energy that must be supplied to the system to maintain steady-state conditions) ΔU to the peak potential energy U stored in the system during that cycle η=

ΔU U

.

(2.52)

In general, the loss factor depends on both the amplitude and frequency of the vibration, and can be applied also to nonlinear systems and to systems with frequency-dependent parameters. If X 1 and X 2 are two consecutive amplitudes of a damped free vibration, 1 the energy stored in the spring at maximum displacement is U 1 = k X 12 , 2 1 U 2 = k X 22 . The loss of energy divided by the original energy is 2 ⎛X U1 − U 2 U = 1 − 2 = 1 − ⎜⎜ 2 U1 U1 ⎝ X1

2

⎞ ⎟⎟ = 1 − e − 2δ ≅ 2δ ⎠

where δ is the logarithmic decrement. Hence, for small damping, the loss factor is approximately twice the logarithmic decrement

η ≅ 2δ .

2.4

(2.52, a)

Damped Forced Vibrations

During damped forced vibrations, the response lags the excitation due to the energy dissipation by damping. The response at phase resonance has finite magnitude and is 90 0 phase shifted with respect to the forcing. The amplitude of motion at resonance is related to damping and the width of the resonance curve is directly proportional to the system damping. In the case of harmonic vibrations, the displacement-force diagram is a closed hysteretic loop which for viscous damping is an ellipse whose area is a measure of the energy dissipated by damping.


43

2.4.1 Steady State Vibrations with Viscous Damping Consider the spring-mass-dashpot grounded system subjected to a harmonic force F0 cos ω t applied to the mass (Fig. 2.26, a).

Fig. 2.26 Based on the free body diagram from Fig. 2.26, b the differential equation of motion can be written as

m x&& + c x& + k x = F0 cos ω t .

(2.53)

The complete solution of equation (2.53) consists of the sum of the solution (2.46) of the homogeneous equation (2.38) and a particular solution which corresponds to the type of excitation in the right hand side. Due to the damping, the homogeneous solution soon dies out, leaving only the particular solution which is a harmonic motion having the same frequency as the exciting force and a phase lag due to damping x (t ) = X cos (ω t − ϕ ) .

(2.54)

The displacement amplitude X and the phase shift ϕ between displacement and force are found by substituting the above solution into the equation (2.53). Shifting all the terms to the right hand side, we obtain mω 2 X cos (ω t − ϕ ) + cω X sin (ω t − ϕ ) − k X cos (ω t − ϕ ) + F0 cos ω t = 0 .

The terms in the above equations are projections of force vectors on a (horizontal) line at an angle ω t with respect to the force vector (Fig. 2.27). The force vector F0 is ϕ degrees ahead of the displacement vector X . The spring force k X is opposite to the displacement, while the inertia force


44

mω 2 X is in phase with the displacement. The damping force cω X is 90 0 ahead the spring force. The vectors remain fixed with respect to each other and rotate together with angular velocity ω . The rotating vector (phasor) diagram in Fig. 2.27 is drawn for a forcing frequency lower than the resonance frequency.

Fig. 2.27 Summation of vector projections in the displacement direction and in the normal direction provides the equilibrium equations k X − mω 2 X = F0 cos ϕ ,

cω X = F0 sin ϕ .

(2.55)

A component of the driving force balances the damping force while the other component is necessary to balance the reactive force, i.e. the difference between the elastic force and the inertia force. Solving for X and ϕ yields the amplitude of the forced vibration

X=

F0

( k − mω ) 2

2

+ ( cω )

2

=

X st

[ 1 − (ω ω ) ] n

2 2

+ ( 2 ζ ω ωn

)

, (2.56) 2

and the phase shift tan ϕ = where

2ζ ω ωn cω = , 2 k − mω 1 − (ω ωn ) 2

(2.57)

ω n = k m and ζ = c 2 k m .

These expressions are plotted in Fig. 2.28 for several values of the damping ratio ζ .


45

The amplitude-frequency diagrams are called resonance curves or frequency response curves. Such a curve starts at a value X st , increases to a maximum at the resonance frequency, decreases through the damped natural frequency (2.47) and the undamped natural frequency (2.4), and continues to decrease, approaching zero asymptotically, as the frequency increases.

Fig. 2.28 The phase angle between the force and the displacement varies from zero, at zero frequency, through 90 0 at the undamped natural frequency, to approach 180 0 asymptotically as the frequency increases. When the damping is small, the rate of change of phase shift in passing through a natural frequency is very sharp. For subcritical damping, the frequency response diagram (Fig. 2.28, a) exhibits a resonance peak which is said to occur at the resonance frequency. For ζ > 0.707 , the resonance peak is completely smoothed. Overcritically damped systems do not exhibit resonances. It is important to note that “amplitude resonance” is defined at the peak X st frequency ωr = ωn 1− 2ζ 2 where the peak value X max = of the 2 ζ 1 − ζ2 steady-state response occurs.


46

The “phase resonance” is defined to occur at the undamped natural frequency ω = ω n (when the phase shift is 90 0 ) when the displacement amplitude X is X res = st . For small values of damping the two resonances coincide. 2ζ

Fig. 2.29 The force vector diagram at phase resonance is shown in Fig. 2.29. The spring force balances the inertia force of the mass, and the excitation force overcomes the damping force only. There is a continuous interchange of potential and kinetic energy between the spring and the mass. The only external force that has to be applied to maintain the system vibrating is that needed to supply the energy dissipated by damping. At resonance, the reactive energy (in spring and mass) is zero and the active energy (actually dissipated) is maximum. That is why a minimum of force is required to maintain a given displacement amplitude. On a plot of the dynamic stiffness (force required to produce unit displacement at the driving point) versus frequency, the resonance appears as a trough (as in Fig. 2.14).

2.4.2 Displacement-Force Diagram Consider for convenience the steady-state displacement

x (t ) = X cos ω t

(2.58)

lagging by an angle ϕ the driving force applied to the mass f (t ) = F0 cos (ω t + ϕ ) .

(2.59)


47

The two equations above are the parametric equations for an ellipse. Eliminating the time between equations (2.58) and (2.59) yields x2 X

2

+

2

f

F02

−2

x f cos ϕ = sin 2ϕ . X F0

(2.60)

The displacement-force response forms an elliptical hysteresis loop as shown in Fig. 2.30, which is traversed in the anticlockwise direction.

Fig. 2.30 The area inside this loop is the energy dissipated during a cycle of motion. It is equal to the work done by the force (2.59) acting on the displacement (2.58) 2π ω

Wd =

∫ f dx = ∫ 0

dx f d t = − X F0 dt

2π

∫ cos (ω t + ϕ ) sin ω t d (ω t ) , 0

Wd = π F0 X sin ϕ . Using the second equation (2.55), the above expression becomes

Wd = π c ω X 2 .

(2.61)

In order to produce work, the damping force f d = −c x& = ω X sin ω t must

be 90 0 phase shifted with respect to the displacement x (t ) = X cos ω t .

If the displacement and force are measured with appropriate transducers and the signals are fed to an oscilloscope (displacement as ordinate and force as abscissa) the resulting image is a Lissajous’ figure. At low frequencies the figure is a straight line, its slope depending on the ratio of amplitudes of the two signals (Fig. 2.31, a). As the frequency increases the straight line opens into an ellipse (Fig. 2.31, b) whose major axis increases with frequency. At the undamped natural frequency (Fig. 2.31, c) the ellipse major axis is very large and vertical. As the


48

frequency continues to increase, the major axis continues rotating but decreases in magnitude (Fig. 2.31, d). The width of the ellipse decreases until at frequencies well above resonance the ellipse is again reduced to a line which lies almost parallel to the horizontal axis (Fig. 2.31, e).

Fig. 2.31 At the phase resonance, ω = ω n , ϕ = 90 0 , X res = F0 2 ζ k , the ellipse major axis is vertical and the energy dissipated by damping is 2 Wd = π F0 X res = π c ω n X res .

(2.62)

The energy dissipated per cycle by viscous damping is directly proportional to the excitation frequency (eq. 2.61).

2.4.3 Structural Damping Experiments with aircraft structures and various materials indicate that the energy dissipated per cycle of vibration is independent of frequency and proportional to the square of displacement amplitude. Damping values for engineering structures are relatively low even at high resonant frequencies. Also, if all damping were viscous, then small, high frequency bells would react to a strike with a dull thud, instead of a clear tinkle. This means that the viscous damping, adopted first for its mathematical tractability, should be replaced by a model in which the energy dissipated by damping is independent of frequency. This type of damping is called hysteretic or structural damping.


49

The use of the term “hysteretic” damping is somewhat confusing, since all damping mechanisms involve a hysteresis curve of some sort. Thus the word “structural” is preferred herein to describe this particular mechanism. It implies a resisting force which is in phase with velocity but, unlike the viscous damping, has a magnitude which is not proportional to the velocity but to the displacement. The damping coefficient is inversely proportional to frequency so that the damping force is − h x& ω (rather than − cx& ). Equation (2.53) becomes

m x&& +

h

ω

x& + k x = F0 cos ω t ,

(2.63)

where h is the coefficient of structural damping. The inclusion of ω in the coefficient of x& implies that only solutions with this frequency may be thought. Alternatively, this equation of motion may be written in terms of the complex stiffness k ∗ = k + i h , since it is decided that a harmonic solution is required

m x&& + (k + i h ) x = F0 eiω t .

(2.64)

Since c is replaced by h ω , the dissipation of energy per cycle is

Wd = π h X 2 ,

(2.65)

which is independent of frequency. Equations (2.56) and (2.57) become

X=

F0

( k − mω )

2 2

tan ϕ =

= +h

2

X st

[1 − (ω ω ) ] n

h g , = 2 k − mω 1 − (ω ωn ) 2

2 2

,

+g

(2.66)

2

(2.67)

where g = h k is the structural damping factor.

2.4.4 The Half-Power Points Method The resonance curve of the spring-mass-dashpot system can be used to determine the damping ratio (Fig. 2.32).

X st . For small values of 2ζ damping, the peak M coincides with the point of phase resonance. The points B and When ω = ω n the resonant amplitude is X res =


50

C, of ordinate

(

)

2 2 X res are referred to as the half-power points. The amplitude

2 squared is (1 2) X res , so that the power dissipated by damping at the corresponding frequencies ω 1 and ω 2 is half the power dissipated at resonance.

Fig. 2.32 Substituting into equation (2.56) we obtain 2

1⎛ 1 ⎞ ⎜ ⎟ = 2 ⎜⎝ 2 ζ ⎟⎠ which yields the equation

(ω

1

(1 − (ω ω ) ) n

(

2 2

+ (2 ζ ω ωn )

)

(

, 2

)

ωn ) 4 − 2 1 − 2 ζ 2 (ω ωn ) 2 + 1 − 8 ζ 2 = 0 .

Solving, we obtain the frequencies of half-power points

(ω

(

)

ωn ) 12, 2 = 1 − 2 ζ 2 ± 2 ζ 1 + ζ 2

which for ζ << 1 can be approximated by

(ω

ωn ) 12, 2 ≅ 1 ± 2 ζ .

Denoting ω 12 ≅ ωn2 ( 1 − 2 ζ ) and ω 22 ≅ ωn2 ( 1 + 2 ζ ) , we obtain


2ζ ≅

51

ω 22 − ω 12 ω 22 + ω 12

(2.68)

or

2ζ ≅

ω 22 − ω 12 2 ωn2

=

ω 2 − ω1 ω 2 + ω1 ω 2 − ω1 ≅ ωn 2 ωn ωn

so that the damping ratio is given by ζ≅

Δω , 2 ωn

(2.69)

where Δω = ω 2 − ω1 is the bandwidth of the resonance curve. From the shape of the resonance curve it is difficult to establish if the damping is really of the viscous type. If only a single degree of freedom is considered, and the motion is to be harmonic, it is most convenient to use the concept of “equivalent viscous damping”, in which the viscous damping coefficient has such a value that the energy dissipated in a harmonic displacement cycle of a certain amplitude and frequency is the same as that of the actual damping mechanism in the same displacement cycle. In equation (2.43) the coefficient c is then the coefficient of equivalent viscous damping.

2.4.5 The Added Mass Method In the neighbourhood of an isolated resonance, the behaviour of a general vibrating system resembles the response of a single-degree-of-freedom system. The equivalent mass and equivalent stiffness of the substitute system can be determined by the additional mass method.

Fig. 2.33 Two frequency response curves are experimentally drawn, one for the actual system, and the other for the system with a known additional mass ma


52

(Fig. 2.33). The natural frequencies ω n1 and ω n 2 are determined at peak displacement amplitudes. From the corresponding equations (2.4)

k = m ω n21 ,

(2.70)

k = (m + m a )ω n2 2 ,

(2.71)

it is possible to obtain the equivalent mass

m=

(ω

ma 2 n1

)

ω 2n 2 − 1

,

(2.72)

then, from equation (2.70), the equivalent stiffness k. Note that the resonance response of the system with added mass is larger because for the actual system

X res 1 =

F0 1 F F = 0 = 0 k 2 ζ1 cω n1 c

m k

and for the system with the added mass

X res 2 =

F0 1 F F = 0 = 0 k 2 ζ 2 cω n 2 c

m + ma . k

If the system operating frequency is near ω n 1 , then the forced response of the vibrating system may be decreased by adding a mass ma .

2.4.6 Solution by Complex Algebra For harmonic excitation, the force acting on the mass of the system of Fig. 2.26 can be written f (t ) = F0 e iω t , (2.73) so that the steady-state solution (2.54) becomes

x (t ) = X eiω t ,

(2.74)

X = X eiθ = X R + i X I

(2.75)

where

is the complex displacement amplitude. In equation (2.75), X is the modulus, θ is the phase angle, X R is the real


53

(in-phase) component, and X I is the imaginary (in-quadrature) component

X R = X cos θ ,

X I = X sin θ ,

(2.76)

X = X R2 + X I2 ,

tan θ = X I X R .

(2.77)

For structural damping, the equation of motion (2.63) becomes m x&& +

h

ω

x& + k x = F0 eiω t .

(2.78)

Substitution of (2.74) in (2.78) yields

(− m ω

2

)

+ i h + k X = F0 .

The complex amplitude X is determined as

X=

F0 2

k − mω + i h

=

X st

1 − (ω ωn ) 2 + i g

,

(2.79)

where ω n = k m , g = h k , so that

XR =

1 − (ω ωn ) 2

[ 1 − (ω ω ) ] n

X=

2 2

+ g2

1

[ 1 − (ω ω ) ] n

2 2

+g

2

X st , X I =

−g

[ 1 − (ω ω ) ] n

X st ,

tan θ =

−g

2 2

1 − (ω ωn ) 2

+ g2

X st

.

(2.80)

(2.81)

Eliminating ω between the expressions of X R and X I yields 2

2

⎛ ⎞ ⎛ 1 ⎞ 1 ⎜⎜ X I + X st ⎟⎟ + X R2 = ⎜⎜ X st ⎟⎟ . 2g ⎝ ⎠ ⎝ 2g ⎠

(2.82)

This circle is the locus of the end of the response vector in the complex plane.

2.4.7 Frequency Response Functions Depending on whether the response is a displacement, velocity or acceleration, there are several frequency response functions (FRFs) defined as the complex ratios response/excitation or excitation/response. The following definitions are almost generally accepted and even standardized : displacement / force = receptance, velocity / force = mobility, acceleration / force = accelerance (or inertance),


54

force / displacement = dynamic stiffness, force / velocity = mechanical impedance, force / acceleration = apparent mass.

Fig. 2.34 Because of the harmonic nature of all of the parameters involved, these functions contain basically the same information about the vibrating system, and simple relationships can be established between them.


55

Basically, three different types of plots are used: a) Bodé diagrams of the FRF modulus versus frequency plus the FRF phase angle versus frequency; b) diagrams of the FRF real part versus frequency plus the FRF imaginary part versus frequency; c) Nyquist diagrams of the FRF real part versus the FRF imaginary part. For a system with structural damping, plots of the receptance α = X F0 are presented in Fig. 2.34 for a fixed value of the structural damping factor. Resonance occurs at point M, while the half-power points are denoted B and C. The Nyquist plot (Fig. 2.34, e), which is the locus of the tip of vector α in the complex plane, is a circle. It contains on a single plot the amplitude and phase angle information. In the vicinity of resonance the frequency scale is most expanded so that half of the circle represents the response between the half-power points, irrespective of the level of damping. The effect of lowering the damping is that of increasing the diameter of the circle and of expanding the frequency scale. Resonance is indicated by maxima in α (Fig. 2.34, a) and in α I (Fig. 2.34, d), and by points of inflection (maximum slope or maximum derivative with respect to ω 2 ) in θ (Fig. 2.34, b) and α R (Fig. 2.34, c). The peak in α I is sharper than that in α . At resonance, θ = −90 0 and α R = 0 . In a Nyquist plot (Fig. 2.34, e), resonance occurs at the crossing of the circle with the imaginary axis, where the rate of change of the arc length with respect to frequency attains a maximum. This is based on the observation that the derivative

(

ds

d ω 2 ωn2

)

=−

1 h

(

dθ

d ω 2 ωn2

)

=

k⎡ ⎢⎣

(

1

)

2 1 − ω 2 ωn2

+ g2⎤ ⎥⎦

= k α 2 (2.83)

has a maximum value at resonance. If the system is excited by a harmonic force and the receptance is plotted point by point, at equal frequency increments Δω , then the arc length Δ s between two successive points is a maximum at resonance. This property is the basis of a method for locating natural frequencies developed by Kennedy and Pancu. The structural damping factor can be calculated from

g=

ω 22 − ω 12 ω 22 + ω 12

(2.84)

where ω1 and ω 2 are the frequencies of the peaks of α R (ω ) or of the ends of the diameter BC, perpendicular to OM, in the Nyquist plot.


56

The stiffness can be calculated from the value α res at resonance k=

1 1 1 F0 = . g α res g X res

(2.85)

Because in the complex plane the velocity is 900 phase shifted ahead the displacement and the acceleration is 900 phase shifted ahead the velocity, the Nyquist plots of mobility and accelerance are rotated 90 0 and 180 0 , respectively, anticlockwise with respect to the polar plot of receptance. The Nyquist plot of the mobility M = i ω X F0 = M R + i M I , which is not a circle, is shown in Fig. 2.35, a and is described by the following equation

(M

2 R

+ M I2

)

2

−

M R2 M M + R I =0. 2 gkm g km

a

b Fig. 2.35

The Nyquist plot of the accelerance η = − ω 2 X F0 = η R + iη I is shown in Fig. 2.35, b. It is a circle of equation 2

2

1 ⎞ ⎛ 1 ⎞ 1+ g2 ⎛ ⎜ ⎟ η η . + − = − ⎟ ⎜ R I 2m ⎠ ⎜⎝ 2 g m ⎟⎠ 4 g 2m2 ⎝ The half-power points and the point of maximum response amplitude are shown in both figures.


57

2.4.8 Receptance Polar Plot for Viscous Damping The receptance can be expressed as the ratio of the complex displacement amplitude X to the force amplitude F0 . Using, instead of α , the general notation

for frequency response functions H (iω ) , we obtain for viscous damping

H (iω ) =

1 1m X . = = 2 2 2 F0 k − mω + iω c ωn − ω + i 2ζ ω ωn

(2.86)

Its Nyquist plot is not a circle, which is a drawback for the identification of system parameters. However, it will be shown that it can be decomposed into two circles. Equation (2.86) can be written under the form H (iω ) =

1m . (i ω − s1 )(i ω − s2 )

(2.87)

where s1, 2 = −σ ± iω d (2.48) are the roots of the characteristic equation (2.39). Equation (2.87) can be expressed in terms of partial fractions 1m

(i ω − s1 )(i ω − s2 )

=

C1 C2 . + i ω − s1 i ω − s2

(2.88)

Multiplying both sides of equation (2.88) by (i ω − s1 ) and evaluating the result at iω = s1 we obtain 1m (i ω − s2 )

= C1 + C2 i ω = s1

i ω − s1 i ω − s2

. i ω = s1

Thus C1 =

1m 1m 1m = = s1 − s 2 (− σ + iω d ) − (− σ − iω d ) i 2 ω d

and similarly

C2 = −

1m , i 2ωd

so that equation (2.88) can be written, by factoring a constant of

C 2 , under the standard form

1 out of C1 and 2i


58

R

H (iω ) =

2 i (i ω − s1 )

−

R∗

2 i (i ω − s2 )

.

(2.89)

where the star denotes the complex conjugate. In this case, the residues are purely real

R = R∗ =

1m

ωd

.

(2.90)

For multi-degree-of-freedom systems they are complex conjugates. Equation (2.89) can be written

H (iω ) =

U

+

U∗

σ + i ( ω − ωd ) σ + i ( ω + ωd )

.

(2.91)

where U = −U ∗ = − i

1 . 2 mωd

(2.92)

It is useful to analyse the Nyquist plot obtained from the analytical expression (2.91).

a

b Fig. 2.36

In order to draw the plot of the first term in the summation

U

σ + i ( ω − ωd ) consider the plot of

,

(2.93)


59

1

σ + i ( ω − ωd )

.

(2.94)

In the complex plane, expression (2.94) represents a circle (Fig. 2.36, a) with centre (1 2σ , 0 ) and diameter 1 σ . At the point M of maximum amplitude, i.e. at the crossing point of the circle with the real axis, the frequency is ω d , the damped natural frequency. The decay rate σ equals the frequency spacing measured from M to points B and C whose response vectors make angles of ± 45 0 with the response vector of M. For negative frequencies the circle is drawn with dotted line. Next consider the effect of the imaginary number U in the numerator of expression (2.93). Multiplication of the former plot by this imaginary number results in a clockwise rotation of the diagram by 90 0 and expansion or contraction by an amount 1 2mω d (Fig. 2.36, b). The obtained circle is called the circle with

predominantly positive frequencies. The centre of this circle is at (0, − U 2σ U

) and

1 1 . The portion drawn with broken line σ 2k ζ 1 − ζ 2 corresponds to negative frequencies.

its diameter equals

=

a

b Fig. 2.37

Consider now the second term in the summation U∗ . σ + i ( ω + ωd ) The plot of 1 . σ + i ( ω + ωd ) is shown in Fig. 2.37, a.

(2.95)

(2.96)


60

The expression (2.95) represents also a circle (Fig. 2.37, b), called the circle with predominantly negative frequencies. This circle has the same diameter as that shown in Fig. 2.36, b but is rotated 90 0 anticlockwise from the real axis. The arc of circle corresponding to positive frequencies represents only a very small part of the circle. The remaining part, corresponding to negative frequencies, is drawn with broken line. Combining the diagrams from Fig. 2.36, b and Fig. 2.37, b, the Nyquist plot of Fig. 2.38, a is obtained, which is no more a circle. Several such diagrams are presented in Fig. 2.38, b for different values of ω ωn and ζ .

a

b Fig. 2.38

The value of the FRF (2.89) at the damped natural frequency is

H (iω d ) = H (iωd ) =

1 2i

R 2i

σ

−

R∗ 2 i , σ + i 2 ωd

⎡ 1 ⎤ 1 − ⎢ ⎥. ⎣ mωd σ mωd (σ + i 2 ωd ) ⎦

(2.97)

This can be approximated as

H (iω d ) ≅

R 1 = . i 2 σ i 2 σ mω d

(2.98)

since the second term on the right of equation (2.97) approaches zero as ω d gets large.


61

Therefore, many single-degree-of-freedom models can be represented simply as H (iω ) ≅

R R = . 2 i (ω − s1 ) 2 i [ω − (σ − iωd )]

(2.99)

2.4.9 Transmissibility in Damped Systems If the mass-spring-dashpot system is base-excited by a prescribed motion

x1 = X 1e i ω t , the motion transmitted to the mass is x 2 = X 2 e i ω t , where X 2 = X 2 e − i ϕ is a complex amplitude. The differential equation of motion is m &x&2 = −k (x 2 − x1 ) − c (x& 2 − x&1 ) which may be rearranged as

m &x&2 + c x& 2 + k x 2 = c x&1 + k x1 .

(2.100)

The amplitude ratio is

X2 k + iω c . = X 1 k − mω 2 + i ω c

Fig. 2.39

(2.101)


62

The motion transmissibility is TR =

X2 X1

=

1 + (2ζ ω ωn ) 2

[ 1 − (ω ω ) ] n

2 2

+ (2ζ ω ωn ) 2

,

(2.102)

The phase shift is given by tan ϕ =

2 ζ (ω ωn ) 3

1 − (ω ωn ) 2 + (2ζ ω ωn ) 2

.

(2.103)

Equations (2.102) and (2.103) are graphically represented in Fig. 2.39 for various values of the damping ratio ζ . For ω ωn > 2 , TR is less than unity, as in Fig. 2.17, but as the damping increases, the transmissibility grows, which means a deterioration of the isolation properties. Reducing the damping is not a good solution because, in order to operate in the region ω ωn > 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.


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)


64

tan ϕ =

2ζ ω ω n

1 − (ω ωn ) 2

.

Fig. 2.41

Fig. 2.42

(2.110)


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 , equation (2.109) becomes

X r X 1 ≅ (ω ωn ) 2 ,

or Xr ≅

1

ω n2

(X ω ) , 1

2

(2.111)

where X 1ω 2 is the acceleration of the structure which is being measured. In this case, the instrument measures a quantity directly proportional to the absolute acceleration of the structure and is called an accelerometer. It has a high natural frequency, obtained with a small seismic mass and a hard spring. In the frequency range II, at ω ≅ ω n , the mass exhibits large amplitude vibrations, property used in the design of reed-type frequency-indicating vibrometers and accelerometers used for measuring rolling bearing defects. Amplitude distortions. To reproduce a complex signal without distortion, all harmonic components must be amplified equally along the frequency axis. This can be accomplished if the amplitude ratio X r X 1 is almost constant.


66

Consequently, frequency response ranges are given for each pickup so that the distortions might remain within prescribed limits. Figure 2.43 shows an augmented part of Fig. 2.41, with curves plotted for four damping ratios. The instrument with ζ = 0.7 has a horizontal response curve down to ω ωn = 4 . Amplitude distortions set a lower frequency limit to the displacement-measuring instrument.

Fig. 2.43 Phase distortions. To reproduce a complex signal without a change in its shape, the phase of its harmonic components must be shifted equally along the time axis. This can be accomplished if the phase angle ϕ increases linearly with frequency.

For the vibrometer, the ratio ω ωn is relatively large, and ϕ is approximately 180 0 for all harmonics, thus no phase distortion occurs. For an accelerometer, when the damping ratio is about 0.7 , a nearly linear relationship exists between phase angle and frequency, ϕ ≅ (π 2)(ω ωn ) . This value of damping is also used to minimize the transient response of the instrument.

2.4.11 Rotating Shaft with External Damping Consider the rotor from Fig. 2.18 under the action of a friction force generated by the motion relative to its stationary environment. In synchronous precession, points C and G rotate around the bearing axis O with an angular velocity ω , the same as the shaft speed of rotation about C. The damping force f d can be assumed to be proportional to the tangential velocity rC ω , so that f d = − c rC ω , where c is the coefficient of external viscous damping.


67

The free body diagram of the disc is shown in Fig. 2.44. The elastic restoring force due to the shaft bending k rC acts along OC. The centrifugal force mω 2 rG , due to the offset CG = e , acts along OG. The viscous damping force is perpendicular to OC. Points O, C and G are no more collinear, and the line CG leads the line OC by an angle ϕ .

Dynamic equilibrium of the three forces implies mω 2 ( rC + e cosϕ ) = k rC ,

mω 2 e sinϕ = c rC ω .

(2.112)

Fig. 2.44 The circular orbit of point C has a radius rC =

mω 2 e cosϕ =e k − mω 2

(ω

ωn ) 2

[1 − (ω ω ) ] n

2 2

+ (2ζ ω ωn ) 2

At the critical speed, when ω = ω n , the orbit radius is rC = e 2ζ .

Fig. 2.45

.

(2.113)


68

The angle between the line CG and the line OC is given by 2 ζ ω ωn cω = . (2.114) 2 k − mω 1 − (ω ωn ) 2 This angle ϕ grows with the speed ω . When ω < ω n (Fig. 2.45, a), the disc rotates with G outside C. With increasing speed, the segment OC increases, tan ϕ =

and CG rotates with respect to OC. When ω = ω n , the line CG leads OC by 90 0 . When 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)


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)


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 .


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


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)


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 .


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


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.


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 .


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

;


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 ) ,


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


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.


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


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).


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).


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).


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.


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.


88

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


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.


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)


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)


92

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)


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)


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


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


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


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


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


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)


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


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)


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)


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


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.


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.


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


⎧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


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


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)


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)


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)


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)


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)

.


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

(

)

(

)

)

.


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.


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

,

,


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


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.


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)


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 .


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)


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)


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.


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


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.


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)


− 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


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.


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).


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)


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


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⎥ . ⎣ ⎦


⎡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


with solutions

λ1 = 17.602

and

135

λ2 = 0.3976 .


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)


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 .


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 .


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 .


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


[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⎭


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


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


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.


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.


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


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


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)


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.


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)


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 .


ω 1 = 1.199 k m ,

ω 2 = 2.358 k m .

Fig. 4.24


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.


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 .


151


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)


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.

)


(mgl − ω m l 2

or

⎛g

ω 4 − 2 ⎜⎜

⎝l


+

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 ⎟⎠ ⎝


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 ,


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 << ω 12 ,

g ml g kd2 g g , + 2 2 − << or k << l l l 2d2 ml

so that beats can occur only for small stiffness values k of the coupling spring.

Fig. 4.28 Equations (4.119) show that θ 1 and θ 2 are given by sine and cos functions which are 90 0 phase shifted. When θ 1 = θ 0 , θ 2 = 0 and vice-versa (Fig.4.28). The phase difference between the two motions is

ϕ = (ω 2 t − ϕ 2 ) − (ω 1 t − ϕ 1 ) = (ω 2 − ω 1 ) t − (ϕ 2 − ϕ 1 ) = 2Δω t − Δϕ


155

so it varies slowly with time. At t = 0 , θ1 = θ 0 , θ 2 = 0 . At Δω t = π 2 , θ1 = 0 , θ 2 = θ 0 sin At Δω t = π , θ 2 = 0 , θ 1 = θ 0 cos

ωm π ω = θ 0 if m = n is an integer. Δω 2 Δω

ω ωm π = −θ 0 if m = n is an integer. Δω Δω

The initial conditions (4.118) imply that the left pendulum is pulled out while keeping the other in place, then released (Fig. 4.29, a). For the first few cycles the left pendulum will swing, the right pendulum stands almost still. Then the right pendulum starts oscillating with increasing amplitudes, while the amplitude of oscillation of the left pendulum decreases. After a sufficient time interval, the left pendulum stands still, while the right pendulum swings with the full amplitude (Fig. 4.29, d). Then the phenomenon repeats itself. There is a continuous transfer of energy from one pendulum to the other until the inherent damping (neglected in this analysis) brings the system to rest.

Fig. 4.29 In terms of the component mode shapes, the motion can be regarded as the sum of two harmonic motions at the natural frequencies ω 1 and ω 2 . Let initially the motion in the second mode be such that the pendulums are departed from each other (Fig. 4.29, b). The phase difference between modes grows with the time elapsed. The motion in the second mode is faster than in the first mode, until it is 1800 degrees in advance (Fig. 4.29, c). The motion in the second mode is such that the pendulums are closer to each other. Summing up the motions, it turns out that at a certain time the left pendulum stands still, while the right one swings with maximum amplitude. Then the amplitude wanders to the left pendulum and the whole sequence repeats itself.


156

4.6 Damped Systems In the preceding sections only undamped systems have been considered. Herein the effect of viscous damping is taken into account. Arbitrary damping introduces coupling of coordinates so that only the special case of proportional damping is considered. Low damping values limit the vibration amplitude at resonance. Larger values may make modes to be overdamped so that the corresponding motion is aperiodic.

4.6.1 Proportional Viscous Damping Consider the system of Fig. 4.30, a. Energy dissipation is conveniently modelled by viscous damping represented by the dashpots c1 and c2 . Viscous damping forces are proportional and opposite to the relative velocity of the dashpot ends (Fig. 4.30, b). The equations of motion may be written by summing the forces acting on each mass:

( ) ( m 2 &x& 2 − k 2 ( x 1 − x 2 ) − c2 ( x& 1 − x& 2 ) = 0

)

m 1 &x&1 + k 1 x 1 + c 1 x& 1+ k 2 x 1 − x 2 + c2 x& 1 − x& 2 = 0 ,

or

m 1 &x&1 + ( c 1 + c2 ) x& 1− c2 x& 2 + ( k1 + k 2 ) x 1 − k 2 x 2 = 0 , m 2 &x& 2 − c2 x& 1 + c2 x& 2 − k 2 x 1 + k 2 x 2 = 0 .

(4.120)

Fig. 4.30 The coupling between the two coordinates is due to both the stiffness k 2 and the damping coefficient c2 .


157

In matrix form, these equations can be written ⎡m1 0 ⎤ ⎧ &x&1 ⎫ ⎡c1 + c2 ⎢ 0 m ⎥ ⎨ &x& ⎬ + ⎢ − c 2⎦ ⎩ 2⎭ ⎣ 2 ⎣

− c2 ⎤ ⎧ x& 1 ⎫ ⎡k1 + k 2 ⎨ ⎬+ c2 ⎥⎦ ⎩ x&2 ⎭ ⎢⎣ − k 2

− k 2 ⎤ ⎧ x 1 ⎫ ⎧0⎫ ⎨ ⎬ = ⎨ ⎬, k 2 ⎥⎦ ⎩ x2 ⎭ ⎩0⎭

(4.121)

or in compact form

[ m ]{ &x& } + [ c ]{ x& } + [ k ]{ x } = { 0 }, (4.121, a) where [ c ] is the damping matrix, and { x& } is the column vector of velocities. Note that the damping matrix is symmetrical so that

[ c ] = [ c ]T . The mass matrix is diagonal. The coupling is produced by the off-diagonal elements of the stiffness matrix and the damping matrix. Modal analysis may be used to solve equations (4.121) if the linear transformation based on the modal matrix diagonalizes the damping matrix simultaneously with the mass and the stiffness matrix. This is simply achieved if the damping matrix can be expressed as a linear combination of the mass and stiffness matrices, that is, if

[ c ]= α [ m ] + β [ k ] ,

(4.122)

where α and β are constants. This form of damping is called proportional damping or Rayleigh damping. There are also other conditions when the modal damping matrix becomes diagonal, but they are only special cases which occur seldom.

4.6.2 Damped Free Vibrations If the undamped problem (4.3) is first solved, one obtains the undamped modes of vibration. The modal matrix (4.24) is constructed having the normal modes as columns. Using the transformation of coordinates

{ x } = [ u ] {q }

(4.123)

and multiplying on the left by [ u ] T we obtain

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

(4.124)

where

[ M ] = [ u ]T [ m ] [ u ] , [ C ] = [ u ]T [ c ] [ u ], [ K ] = [ u ]T [ k ] [ u ]. (4.125) The modal matrices [ M ] and [ K ] are diagonal while the matrix [ C ] is diagonal only for proportional damping when


158

[ C ]= α [ M ] + β [ K ] .

(4.126)

In this case, the following orthogonality relations can be established

{ u }Ts [ c ] { u }r = 0 ,

r ≠ s , r , s = 1, 2

(4.127)

For proportional damping, the decoupled modal equations are M r q&&r + Cr q& r + K r qr = 0 ,

r = 1, 2

(4.128)

where M r and K r are defined by (4.26) and (4.28), and Cr = { u }Tr [ c ] { u }r ,

r = 1, 2

(4.129)

are modal damping coefficients. For unit modal masses, the modal stiffnesses are equal to the square of the respective natural frequency and the modal damping coefficients can be expressed as 2 ζ r ωr where ζ r is the rth modal damping ratio and ωr is the rth natural frequency. Equations (4.128) become q&&r + 2ζ r ωr q& r + ω 2r qr = 0 ,

r = 1, 2

(4.130)

which, for 0 < ζ r < 1 have solutions of the form (2.46) qr (t ) = Ar e − ζ r ω r t sin ⎛⎜ 1 − ζ r2 ωr t + φr ⎞⎟ . ⎝ ⎠

r = 1, 2

(4.131)

The solutions (4.131) can also be obtained directly. Looking for solutions of the form xr = ar e s t , equations (4.121) become m 1 s 2 a 1 + (c1 + c2 ) s a 1 − c2 s a2 + (k1 + k 2 ) a 1 − k 2 a2 = 0 , m2 s 2 a2 − c2 s a 1 + c2 s a2 − k 2 a 1 + k 2 a2 = 0 , or

[m s 1

2

]

+ (c1 + c2 ) s + (k1 + k 2 ) a1 − ( c2 s + k 2 ) a2 = 0,

(

)

− ( c2 s + k 2 ) a1 + m 2 s 2 + c2 s + k 2 a2 = 0.

(4.132)

The condition to have non-trivial solutions leads to the characteristic equation

[m s 1

2

+ (c1 + c2 ) s + (k1 + k 2 )

] (m s 2

2

)

+ c2 s + k 2 − ( c2 s + k 2 ) 2 = 0 .

(4.133)


159

For underdamped systems, equation (4.133) has two pairs of complex conjugate roots s 1, 2 = −σ 1 ± i ω d 1 ,

s 3, 4 = −σ 2 ± iω d 2 ,

(4.134)

where ω d 1 and ω d 2 are the damped natural frequencies and σ 1 and σ 2 are the damping factors (exponential decay rates). The above parameters can be related to the absolute values of the eigenvalues (equal to the undamped natural frequencies for systems with proportional damping) and the damping ratios as follows

ω d r = ωr 1 − ζ 2r

σ r = ζ r ωr , ζr =

σr ω 2d r

+ σ r2

,

ωr =

σr ζr

= ω 2d r + σ r2 .

r = 1, 2

(4.135, a)

r = 1, 2

(4.135, b)

With these notations, expression (4.131) becomes

(

)

qr ( t ) = Ar e − σ r t sin ω d r t + φr .

r = 1, 2

(4.131, a)

For relatively high damping values, equation (4.133) may have two real roots and two complex conjugate roots, when one mode of vibration is overdamped, or two pairs of real roots, when both modes are overdamped and the system has no vibratory motion. These cases are not considered herein. Substituting solutions (4.134) into (4.132) we obtain the amplitude ratios (a2 a1 )r which, for proportional damping, define real modes of vibration. The above analysis is valid for distinct eigenfrequencies. The case of equal eigenvalues is treated elsewhere.

Example 4.10 Calculate the modes of vibration for the system of Fig. 4.30 taking for simplification m 1 = 2 , k 1 = 2 , c 1 = 1 , m 2 = 1 , k 2 = 1 , c 2 = 0.5 , in the appropriate units. Solution. The equations of free vibrations (4.121) are ⎡2 0⎤ ⎧ &x&1 ⎫ ⎡ 1.5 − 0.5⎤ ⎧ x& 1 ⎫ ⎡ 3 − 1⎤ ⎧ x 1 ⎫ ⎧ 0 ⎫ ⎢0 1⎥ ⎨ &x& ⎬ + ⎢− 0.5 0.5 ⎥ ⎨ x& ⎬ + ⎢− 1 1 ⎥ ⎨ x ⎬ = ⎨ 0 ⎬ . ⎣ ⎦⎩ 2⎭ ⎣ ⎦⎩ 2⎭ ⎣ ⎦⎩ 2⎭ ⎩ ⎭ It is seen that

[ c ]= 0.5 [ k ] . Equations (4.132) become


160

(2 s

2

)

+ 1.5 s + 3 a 1 − ( 0.5 s + 1 ) a 2 = 0 ,

(

)

− ( 0.5 s + 1) a 1 + s 2 + 0.5 s + 1 a 2 = 0. The characteristic equation is

(2 s

2

+ 1.5 s + 3

)( s

2

)

+ 0.5 s + 1 − ( 0.5 s + 1 )2 = 0 ,

or

2 s 4 + 2.5 s 3 + 5.5 s 2 + 2 s + 2 = 0 , with roots

1 31 s 1, 2 = − ± i , 8 8

1 7 s 3,4 = − ± i . 2 2

The imaginary parts are the damped natural frequencies

ωd 1 = 31 8 = 0.6960 , ωd 2 = 7 2 = 1.3229 . The real parts are the damping factors (decay rates)

σ 1 = 0.125 ,

σ 2 = 0 .5 .

The undamped natural frequencies are equal to the absolute value of the eigenvalue

ω 1 = ωd21 + σ 12 = ω 2 = ωd2 2 + σ 22 =

( 31 8) ( 7 2)

2

+ (1 8) 2 = 1

2

+ (1 2) 2 = 2 .

2,

The modal damping ratios are ζ1 =

σ1 σ2 12 18 = = 0.176 , ζ 2 = = = 0.353 . ω2 ω1 1 2 2

The amplitude ratio a2 a1

=

2 s 2 + 1.5 s + 3 0.5 s + 1 = 2 0.5 s + 1 s + 0.5 s + 1

yields ⎛ a2 ⎞ ⎜ ⎟ =2, ⎜ a1 ⎟ ⎝ ⎠1

⎛ a2 ⎞ ⎟ = −1 . ⎜ ⎜ a1 ⎟ ⎠2 ⎝

Solving the undamped problem in Example 4.1, the same undamped natural frequencies and mode shapes have been obtained.


161

Example 4.11 Calculate the modal parameters for the system of Fig. 4.30 using the same values for masses and stiffnesses but lower damping [ c ]= 0.1 [ k ] . Solution. Equations (4.132) become

(2 s

2

)

+ 0.3 s + 3 a 1 − (0.1 s + 1 ) a 2 = 0 ,

(

)

− ( 0.1 s + 1) a 1 + s 2 + 0.1 s + 1 a 2 = 0.

The characteristic equation 2 s 4 + 0.5 s 3 + 5.02 s 2 + 0.4 s + 2 = 0 , has the roots 1 799 1 199 s 1, 2 = − ±i , s 3, 4 = − ± i 40 40 10 10 or s 1, 2 = −0.025 ± i 0.706665 , s 3, 4 = −0.1 ± i 1.410674 . The damped natural frequencies are

ωd 1 = 0.70666 , ωd 2 = 1.4107 . The damping factors have the values

σ 1 = 0.025 ,

σ 2 = 0 .1 .

The modal damping ratios are ζ 1 = 0.0353 ,

ζ 2 = 0.0707 .

The undamped natural frequencies and the mode shapes are the same as in Example 4.10.

4.6.3 Response to Harmonic Excitation Consider the vibrations of the two-degree-of-freedom system from Fig. 4.31 under the action of forces f1 (t ) and f 2 (t ) .

Fig. 4.31


162

The equations of motion are m 1 &x&1 + (c1 + c2 ) x& 1 − c2 x&2 + (k1 + k 2 ) x 1 − k 2 x2 = f1 (t ) ,

(4.136)

m 2 &x&2 − c2 x& 1 + c2 x&2 − k 2 x 1 + k 2 x2 = f 2 (t ) ,

or in compact matrix form

[ m ] { &x& } + [ c ] { x& } + [ k ] { x } = { f } .

(4.136, a)

4.6.3.1 Transfer Functions

Taking the Laplace transform of (4.136, a) and assuming that all initial conditions are zero, yields

[[ m ] s

2

+ [c ] s + [k ]

] { X (s ) } = { F (s ) }

(4.137)

or

[ B (s ) ] { X (s ) } = { F (s ) }

(4.138)

where [ B (s ) ] is referred to as the system matrix. Multiplying on the left by [ B(s ) ] −1 = [ H (s ) ] yields

[ H (s ) ] { F (s ) } = { X (s ) }

(4.139)

where [ H (s ) ] is referred to as the transfer function matrix. It is the inverse of the system matrix. For the system of Fig. 4.31 this matrix has the form

[H ( s )] =

[m s 1

⎡ m 2 s 2 + c2 s + k 2 ⎢ c2 s + k 2 ⎢⎣ 2

+ (c1 + c2 ) s + k1 + k 2

⎤ ⎥ m 1 s + (c1 + c2 ) s + k1 + k 2 ⎥⎦ c2 s + k 2

2

] [m

2s

2

]

+ c 2 s + k 2 − ( c2 s + k 2

(4.140)

)

2

The denominator of (4.140) is det [B (s )] which is the system characteristic polynomial. This can be expressed as a product

(

)(

)(

)(

det [ B (s ) ] = A s − s 1 s − s 2 s − s 3 s − s 4

)

where A is a constant and s 1 ,.., s 4 are the roots of the characteristic equation (4.133). Since the coefficients of the characteristic equation are real, for low damping levels the roots appear as complex conjugate pairs. They are called the poles of the transfer function.


163

We can also express (4.139) as ⎡ h11 (s ) h12 (s ) ⎤ ⎧ F1 (s ) ⎫ ⎧ X 1 (s ) ⎫ ⎬. ⎢ h (s ) h (s ) ⎥ ⎨ F (s ) ⎬ = ⎨ 22 ⎣ 21 ⎦ ⎩ 2 ⎭ ⎩ X 2 (s ) ⎭

(4.141)

where h l j (s ) is obtained by exciting the system at point j and measuring the

response at point l . For instance h11 (s ) is a driving point transfer function

measured by exciting the system with F1 (s ) and measuring the response X 1 (s ) h11 (s ) =

X 1 (s ) = F1 (s )

(

m2 s 2 + c2 s + k 2

)(

)(

)(

A s − s1 s − s 2 s − s 3 s − s 4

)

.

(4.142)

4.6.3.2 Frequency Response Functions

A frequency response function (FRF) is the transfer function evaluated along the iω (frequency) axis. Substituting s = iω into (4.141) we obtain ⎡ h11 (iω ) h 12 (iω ) ⎤ ⎧ F1 (ω ) ⎫ ⎧ X 1 (ω ) ⎫ ⎬ ⎢ h (iω ) h (iω ) ⎥ ⎨ F (ω ) ⎬ = ⎨ 22 ⎣ 21 ⎦⎩ 2 ⎭ ⎩ X 2 (ω ) ⎭

or

[ H (iω ) ] { F (ω ) } = { X (ω ) }

(4.143)

where [ H (iω ) ] is the frequency response function matrix and X j (ω ) , F j (ω ) are Fourier transforms of the response and excitation, respectively. The response of the system can be defined in the frequency domain by the multiplication and addition of measured frequency response functions and forcing functions X 1 (ω ) = h 11 (iω ) F1 (ω ) + h12 (iω ) F 2 (ω ),

X 2 (ω ) = h 21 (iω ) F1 (ω ) + h 22 (iω ) F 2 (ω ).

(4.143, a)

The time domain forced response of the system, x 1 (t ) and x 2 (t ) , can then

be computed by taking the inverse Fourier transform of X 1 (ω ) and X 2 (ω ) .

Example 4.12 Calculate the matrix of frequency response functions for the two-degreeof-freedom system of Example 4.11 and plot the FRF curves of h11 (iω ) .


164

Solution. For s = iω , the matrix of FRFs (4.140) becomes

[H (iω )] =

⎡1 − ω 2 + i 0.1ω ⎢ ⎢⎣ 1 + i 0.1ω

⎤ 1 + i 0.1ω ⎥ 3 − 2ω 2 + i 0.3ω ⎥⎦

(

2 ω 4 − 5.02 ω 2 + 2 + i − 0.5 ω 3 + 0.4 ω

Fig. 4.32

Fig. 4.33

)

.


165

Fig. 4.34 The plots of the FRF magnitude and phase angle are given in Fig. 4.32. The magnitude plot has two resonance peaks. Because of the larger value of damping in the second mode, the respective resonance peak is lower. At resonances, the phase angle is about − 900 . The plots of the real part and imaginary part are presented in Fig. 4.33. The Nyquist plot is illustrated in Fig. 4.34 with points marked at equal frequency increments. 4.6.3.3 Solution by Modal Analysis

Using the transformation of coordinates (4.123)

{ x } = [ u ]{ q }

and

multiplying on the left by [ u ] T equations (4.136, a) become

[ M ] { q&& } + [ C ] { q& } + [ K ] { q } = [ u ] T { f } = { F }.

(4.144)

For proportional damping, the decoupled modal equations are

M r q&&r + Cr q&r + K r qr = Fr ,

r = 1, 2

(4.145)

or q&&r + 2ζ r ωr q& r + ω 2r qr = Fr M r .

r = 1, 2

(4.146)

For harmonic excitation and steady-state response, denote

{ f } = { ˆf } eiω t , { x } = { ~x }ei ω t ,

(4.147)


166

{ F } = { Fˆ } eiω t , { q } = { q~ }ei ω t ,

(4.148)

2

{ ~x } = [ u ] { q~ } = ∑ q~r { u }r ,

(4.149)

r =1

where a hat above a letter means real amplitude and a tilde above a letter denotes complex amplitude. Substitute (4.148) into equations (4.145) and (4.146) to obtain q~r =

{ u }Tr { ˆf }

K r − ω 2 M r + i ω Cr

=

{ u }Tr { ˆf }

(

M r ω 2r

2

− ω + i 2 ζ r ω ωr

)

.

(4.150)

For unit modal masses, M r = 1 , substituting (4.150) into (4.149) yields 2

{ ~x } = ∑

r =1

{u }Tr { ˆf } {u }r

ω 2r − ω 2 + i 2 ζ rω ωr

or x1 ⎫ 2 { u }r { u }Tr ⎧~ = ⎨~ ⎬ 2 2 ⎩ x2 ⎭ r =1 ω r − ω + i 2 ζ r ω ωr

∑

⎧⎪ ⎨ ⎪⎩

ˆf 1 ˆf

2

⎫⎪ ⎬. ⎪⎭

(4.151)

Note that the dyadic product { u }r { u }Tr is a 2 by 2 square matrix. The FRF matrix (4.143) can be expressed now in terms of the modal parameters

[ H (iω ) ] =

{u }1{u }T1 ω 12 − ω 2 + i 2 ζ1ω ω 1

+

{u }2 {u }T2

ω 22 − ω 2 + i 2 ζ 2ω ω 2

.

(4.152)

( )j

If we denote the jth element of the rth mode shape vector by u r

then

the expression for the general FRF function h j l (iω ) in a partial fraction form is h j l (iω ) =

( u 1 ) j ( u 1 )l

ω 12 − ω 2 + i 2 ζ1ω ω 1

+

( u 2 ) j ( u 2 )l

ω 22 − ω 2 + i 2 ζ 2 ω ω 2

.

(4.153)

This expression shows explicitly the contribution to the response at any frequency of each individual mode of the system. Figure 4.35 shows the polar plot of a two-degree-of-freedom system FRF by plotting each term in the series form expression separately (dotted lines), and then the summation of these terms (solid line) i.e. the complete FRF.


167

Fig. 4.35

Example 4.13 Calculate the partial fraction form of the FRF matrix for the two-degree-offreedom system of Example 4.12. Solution. The mass-normalised mode shapes are

{ u }1 = ⎧⎨

1 ⎩ 6

T

⎧ 1 2 ⎫ ⎬ , {u } 2 = ⎨ 6⎭ ⎩ 3

T

−

1 ⎫ ⎬ . 3⎭

The FRF matrix (4.152) becomes ⎡1 2⎤ 1 ⎡ 1 − 1⎤ ⎢ 2 4⎥ 1⎥⎦ 3 ⎢⎣ − 1 ⎣ ⎦ [ H (iω ) ] = + . 0.5 − ω 2 + i 0.05 ω 2 − ω 2 + i 0 .2 ω 1 6

The driving point FRF for the first mass is h11 (iω ) =

16 13 + . 2 0.5 − ω + i 0.05 ω 2 − ω + i 0.2 ω 2

Figure 4.35 shows the Nyquist plot (solid line) which results from the summation of the plots of each term in the series form expression (broken lines).


168

4.6.4 The Untuned Viscous Damper The untuned viscous vibration damper (Houdaille damper) is used in reciprocating engines to limit the amplitudes of torsional vibrations over a wide operating range. It consists of a rigid disc free to rotate within a cylindrical cavity filled with viscous fluid. In car engines it is located at the end of the crankshaft in the pulley which drives the fan belt. Consider the crankshaft of torsional stiffness K fixed at one end and with the damper attached at the other end in a casing of rotational inertia J (Fig. 4.36). The free rotational mass has a polar mass moment of inertia J d and is acted upon by the damping torque which is assumed to be proportional to the relative motion between the casing and the internal disc. When the damper is excited by a harmonic torque M 0 cosω t , the equations of motion can be written

(

)

J θ&& + K θ + c θ& − θ&d = M 0 cosω t , J θ&& − c θ& − θ& = 0 , d

d

(

d

)

(4.154)

where θ is the casing rotation and θ d is the rotation of the internal disc. The damping coefficient is ⎛ b R 3 R24 − R 14 ⎞ ⎟ (4.155) c = 2π μ ⎜ 2 + ⎜ h2 2h 1 ⎟ ⎝ ⎠ where μ is the oil viscosity.

Fig. 4.36 Assuming solutions of the form

θ = θ 0 cosω t , θ d = θ d 0 cos (ω t − ϕ ) , and denoting


ω n2 =

169

J K c ω ,η= , ζ= , λ= d , J ωn 2 J d ωn J

(4.156)

the amplitude of the casing is described by K θ0 = M0

(

η 2 1 −η 2

)

η 2 + 4 ζ2 2

(

)

+ 4 ζ 2 η 2 + λη 2 − 1

2

.

(4.157)

For a fixed λ , the plots of K θ 0 M 0 as a function of the dimensionless frequency η are shown in Fig. 4.37 for various values of ζ .

Fig. 4.37 The value ζ = 0 is for an undamped system with resonance frequency ωn whose vibration amplitude is infinite at η = 1 . The value ζ = ∞ is for an undamped system with resonance frequency K (J + J d ) = ωn 1 + λ , in which the damper mass and the wheel move together as a single mass. The response curves for these two extreme values of ζ cross each other at the point M of abscissa 2 (2 + λ ) . For any other value of damping, the response curve passes through this point. There is an optimum damping

ηM =

ζ opt =

1

2 (1 + λ )(2 + λ )

(4.158)

for which the peak amplitude is a minimum and is equal to the ordinate 1 + (2 λ ) of point M .


170

Based on equations (4.156) and (4.158) a torsional vibration damper may be designed with a flat response curve, which is effective over a wide frequency range.

4.6.5 The Damped Vibration Absorber A vibration absorber consists of a second mass-spring system added to a primary mass-spring system to protect it from vibrating. The major effect of adding the secondary system is to change from a single degree of freedom system to a two-degree-of-freedom system. The values of the absorber physical parameters are chosen such that the motion of the main system is a minimum. This is accompanied by substantial motion of the added mass which “absorbs” the energy introduced in the system by the force acting on the original mass. If damping in the secondary system is taken into account, the deflection of the primary mass cannot be reduced to zero, but the useful frequency range of the absorber increases, improving its effectiveness. Consider the harmonic response of a two-degree-of-freedom system with undamped primary system and viscously damped absorber. Such a system is obtained if the dashpot c1 is eliminated from the system illustrated in Fig. 4.31. If the damping coefficient c1 is made zero, and the force f 2 is cancelled, equations (4.136) may be rewritten as

( ) ( ) m 2 &x&2 − c2 ( x& 1 − x&2 ) − k 2 ( x 1 − x2 ) = 0 .

m 1 &x&1 + c2 x& 1 − x&2 + k 1 x 1 + k 2 x 1 − x2 = f1 (t ) ,

For harmonic excitation and steady-state response ~ ~ f1 = F eiω t , x 1 = X 1 ei ω t , x2 = X 2 ei ω t , equations (4.159) become

(k + k

)

~ ~ − m 1 ω 2 + i ω c2 X 1 − ( i ω c2 + k 2 ) X 2 = F , ~ ~ − ( i ω c2 + k 2 ) X 1 + k 2 − m 2 ω 2 + i ω c2 X 2 = 0 . 1

2

(

)

(4.159)

(4.160)

(4.161)

~ The vibration amplitude of the primary system X 1 = X 1 can be expressed

as X1 = xst

where

A ζ 22 + B C ζ 22 + D

(4.162)


(

A = 4 μ 2ψ 2θ 2 , B = μ 2 ψ 2 − θ 2

[ (

)(

)

)

2

171

(

, C = 4 μ 2ψ 2θ 2 1 − θ 2 − μ θ 2

D = μ ψ 2 − θ 2 1 − θ 2 − μ 2ψ 2θ 2

]

2

)

2

,

, xst = F k 1 , ζ 2 = c2 2 ω 2 m2 , (4.163)

μ = m2 m 1 , θ = ω ω 1 , ψ = ω 2 ω 1 , ω 1 = k 1 m 1 , ω2 = k 2 m2 . Figure 4.38 shows the frequency response curves of X 1 xst as function of the frequency ratio θ , for fixed values of μ and ψ , and various values of ζ 2 . The effect of the absorber can be seen to reduce the primary system amplitude from infinity down to a small value at θ = 1 . When ζ 2 = 0 , X 1 xst = B D . When ζ 2 = ∞ , X 1 xst = A C . The corresponding response curves cross each other at points R and S. All response curves of the primary mass in the combined system, drawn for various values of ζ 2 , pass through these two points. By changing the frequency ratio ψ , the ordinates of these two points can be increased or decreased. The most favourable case, for the lowest maximum dynamic response throughout the whole frequency range, could be achieved when the following two conditions could be fulfilled: Firstly, arranging for these two points to have equal ordinates. This is obtained when

ψ = 1 (1 + μ ) .

(4.164)

Secondly, arranging the slope of the curve to be zero at each point, that is the two points become the two peaks in the response curve, indicating that maxima have been reached. Unfortunately it is not possible to have equal maxima occurring for the same frequency ratio ψ , but if the slope at one point is zero the slope at the other point is nearly zero. This is obtained when the response of the main system is given by X 1 xst =

1+ 2 μ .

(4.165)

The frequency ratios at which the two peaks occur are given by

θ2 =

1±

2 (2 + μ ) . 1+ μ

(4.166)

The frequency ratio θ at which the trough between the two peaks occurs is the mean of these two frequency ratios frequency ratio.

1 (1 + μ ) = ψ 1 2 . This is in fact the tuning


172

The optimum damping coefficient resulting from the application of these two considerations may be found by differentiating equation (4.162) with respect to θ and equating to zero for each crossover point, giving two values of ζ 2 ψ = c2 2 m2 ω 1 . The average of these two values is c 2 2 m2 ω 1 =

3 μ 8 (1 + μ )3 .

(4.167)

Fig. 4.38 It is also important to keep low the value of the relative motion between the main mass and the absorber X1 − X 2 =

X1 F

(

2 μ k 1 θ c2 2 m 2 ω 1

)

(4.168)

to ensure longer fatigue life for the absorber spring.

4.6.6 Non-Proportional Viscous Damping Consider the system of Fig. 4.30, a with non-proportional viscous damping. The equations of free vibrations (4.121, a) are

[ m ]{ &x& } + [ c ]{ x& } + [ k ]{ x } = { 0 },

(4.169)


where the matrix

[c ]

173

in not proportional with either [ m ] or [ k ] .

For computational purposes, the equation of motion (4.169) is transformed into the first order state space form. Introducing an auxiliary equation

[ m ] { x& } − [ m ] { x& } = { 0 } .

(4.170)

Equations (4.169) and (4.170) can be combined to give ⎡[ m ] ⎢[0 ] ⎣

[ 0 ] ⎤ ⎧{ x& }⎫ ⎡[ 0 ] − [ m ]⎤ ⎧{ x }⎫ ⎧{ 0 }⎫ + = [ m ]⎥⎦ ⎨⎩{ &x& }⎬⎭ ⎢⎣[ k ] [ c ] ⎥⎦ ⎨⎩{ x& }⎬⎭ ⎨⎩{ 0 }⎬⎭

(4.171)

or

[ A ] { q& } + [ B ] { q } = { 0 } , where the 4 to 4 matrices [ A] and [B ] are real [ m ] [ 0 ]⎤ [ 0 ] − [ m ]⎤ [A] = ⎡⎢ [B] = ⎡⎢ , ⎥ ⎥. ⎣ [ 0 ] [ m ]⎦ ⎣[ k ] [ c ] ⎦ { q } = {Φ }e s t ,

Assuming a solution of the form equations (4.172) become

(4.172)

and

{ x } = { u }e s t ,

( s [ A ] + [ B ] ) {Φ } = { 0 }.

(4.173)

There are four eigenvalues s r given by the solutions of det ( s [ A ] + [ B ] ) = 0 , and four eigenvectors {Φ

}r

(4.174)

satisfying the eigenvalue problem

[ B ] {Φ } r = − s r [ A ]{Φ }r .

(r = 1,...,4)

(4.175)

Equation (4.175) can be written in the form

− [ A ] −1[ B ] {Φ } r = s r {Φ

}r . (r = 1,...,4)

(4.176)

where

[0 ] [I ] ⎤ ⎡ − [ A ] −1 [ B ] = ⎢ ⎥, −1 −1 ⎣− [ m ] [ k ] − [ m ] [ c ]⎦

(4.177)

Equation (4.171) becomes ⎛⎡ [0 ] [ I ] ⎤ ⎡[ I ] ⎜⎢ ⎜ − [ m ] −1 [ k ] − [ m ] −1 [ c ]⎥ − s ⎢ [ 0 ] ⎦ ⎣ ⎝⎣

[ 0 ]⎤ ⎞⎟ ⎧ { u } ⎫ = { 0 }. [ I ]⎥⎦ ⎟⎠ ⎨⎩s{ u }⎬⎭

(4.178)


174

For underdamped systems, equation (4.178) has two pairs of complex conjugate roots s 1, 2 = −σ 1 ± iω d 1 ,

s 3, 4 = −σ 2 ± iω d 2 ,

(4.179)

where the imaginary parts, ω d 1 and ω d 2 , are the damped natural frequencies and the real parts, σ 1 and σ 2 , are the damping factors (decay rates). The above parameters can be related to the absolute values of the natural frequencies and the damping ratios as follows

ω r = ω d2 r + σ r2 ,

ζr =

σr . ωr

r = 1, 2

(4.180)

For overdamped modes of vibration, the real eigenvalues can be written s r , s r +1 = − σ r m τ r ,

(4.181)

and the relationships (4.180) become

ω r = σ r2 − τ r2 ,

ζr =

σr . ωr

(4.182)

The motion in an overdamped mode of vibration is aperiodic so that in the frequency response curves there is no related resonance peak or loop in the Nyquist plot even for relatively spaced natural frequencies. For underdamped systems, equation (4.178) yields also two pairs of complex conjugate eigenvectors, whose half upper part gives the mode shapes. The relative phase differences among the degrees of freedom in a particular mode cause the maximum mass displacements to be reached at different times.

Example 4.14 Calculate the modal parameters for the system with closely spaced natural frequencies of Fig. 4.31, where m 1 = 100 kg , m 2 = 1 kg , k 1 = 9.9 ⋅ 106 N m , k 2 = 0.1 ⋅ 106 N m , c 1 = c 2 = 125 Ns m . Draw the Nyquist plots of complex

receptances. Solution. The mass, stiffness and damping matrices are, respectively

100 0⎤ ⎡100 − 1⎤ ⎡ 250 − 125⎤ , [ k ] = 105 ⋅ ⎢ , [c ]= ⎢ ⎥ ⎥ ⎥ . ⎣ 0 1⎦ ⎣ −1 1 ⎦ ⎣− 125 125 ⎦

[ m ] = ⎡⎢

The eigenvalues with positive imaginary part are

λ 1= −4.785 + i ⋅ 312.659 ,

λ 2 = −58.965 + i ⋅ 312.685 .


175

Fig. 4.39 The corresponding eigenvectors are

{ u }1 = ⎧⎨

⎫ ⎬, ⎩1.207 − i ⋅ 2.698⎭

1

{u } 2 = ⎧⎨

⎫ ⎬. ⎩− 8.865 − i ⋅ 35.222⎭

1


176

The damped natural frequencies are

ω d 2 = 312.685 rad sec .

The

absolute

values

ω 1 = ω d2 1+ σ 12 = 312.695 rad sec , ω 2 = ω d2 2 + σ

ω d 1 = 312.659 rad sec of 2 2

the

eigenvalues

and are

= 318.196 rad sec , and the

modal damping ratios are ζ 1 = σ 1 ω 1 = 0.0153 and ζ 2 = σ 2 ω 2 = 0.1853 . For the associated undamped system, the undamped natural frequencies are ω 01 = 300 rad sec and ω 02 = 100 11 = 331.66 rad sec , and they are different from the absolute values of eigenvalues of the system with non-proportional damping. In this case, the difference of damped natural frequencies is 0.026 rad sec , hence 0.0041 Hz , while the undamped natural frequencies differ with 5.039 Hz . Damping coupling approaches the natural frequencies, fact which complicates their experimental determination. The Nyquist plots of complex receptances α i j = x i f j , i , j = 1, 2 , are shown in Fig. 4.39. The excitation frequency is marked along the curves. The unusual shape of the direct receptance plots, with no distinct two loops, is produced by the relative closeness of natural frequencies and the relatively high damping in the second mode of vibration.

Example 4.15 Calculate the modal parameters for the system of Fig. 4.31, with the physical mass , stiffness and damping values given in Table 4.1 for the following four cases: Case I: lightly damped system with relatively separated natural frequencies; Case II: lightly damped system with closely spaced natural frequencies; Case III: highly damped system with relatively separated natural frequencies; Case IV: highly damped system with closely spaced natural frequencies. Table 4.1 Case

m1 = m 2

kg

I 0.0259

II 0.0259

III 0.0259

IV 0.0259

k1 = k 3

N m

100

100

100

100

k2

N m

50

1

50

1

c1

Ns m

0.3

0.3

3

3

c2

Ns m

0.2

0.2

2

2

c3

Ns m

0.1

0.1

1

1

Draw the Nyquist plots of complex receptances for the four systems.


177

Solution. The undamped natural frequencies are given by 2 ω 01 =

k3 m2

⎛ ⎜ ⎝

k2 ⎞ ⎟. k 3 ⎟⎠

2 2 ⎜ = ω 01 ω 02 1+ 2

,

The undamped modal vectors are the same in all cases

{ a }1 = ⎧⎨

1⎫ 1 { a } 2 = ⎧⎨ ⎫⎬ . ⎬ ⎩1 ⎭ ⎩ −1 ⎭ The modal mass, stiffness and damping matrices, calculated with the modal matrix built with the above vectors, are respectively ⎡k 1 + k 3 ⎣ 0

0 ⎤ ⎡ 2m 1 , 2m 2 ⎥⎦ ⎣ 0

[M ] = ⎢

⎡c + c

c1 − c 3

⎤ . c 1 + c 3 + 4c 2 ⎥⎦

[C ] = ⎢c 1 − c 3 ⎣

1

⎤ , k 1 + k 3 + 4k 2 ⎥⎦

0

[K ] = ⎢

3

For c 1 ≠ c 3 , the damping is non-proportional and the modal damping matrix is not diagonal as the mass and stiffness matrices. The eigenvalues are solutions of the fourth-order algebraic equation

(

)

m 1s 2 + c 1 + c 2 s + k 1 + k 2 2

− c 2s − k 2

(

− c 2s − k 2

)

m 2s + c 2 + c 3 s + k 2 + k 3

=0.

For underdamped modes of vibration, the complex eigenvalues can be written under the form (4.179) while for overdamped modes of vibration, the real eigenvalues can be written under the form (4.181). The numerical values of the modal parameters are given in Table 4.2. Table 4.2 Case

I

II

III

IV

ω 01

ω 02

62.13

87.86

62.13

62.75

62.13

87.86

62.13

62.75

ω1

ω2

62.24

87.69

62.16

62.71

64.28

84.92

62.16

62.71

ζ1

ζ2

0.062

0.132

0.055

0.192

0.540

1.409

0.547

1.919

ωd1 σ1

ωd 2 σ2

62.13

86.93

62.07

61.55

54.09

-

52.01

-

3.845

11.594

3.406

12.033

34.734

-

34.044

-


178

The undamped natural frequencies are different from the absolute value of eigenvalues for the non-proportionally damped system, ω 0 r ≠ ω r . In the case of damping coupling, the absolute values of eigenfrequencies, sometimes referred to as resonant frequencies, are closer to each other than the corresponding undamped natural frequencies, while the damped natural frequencies might have reversed order. For the first mode, ω 1 > ω 01 , while for the second mode, ω 2 < ω 02 , and for Case II, ωd 2 < ωd 1 . The second mode is overdamped in Cases III and IV, having damping ratios ζ r > 1 . Indeed, for the set of physical parameters corresponding to systems with relatively spaced undamped natural frequencies, the second mode of vibration becomes overdamped for a value c 3 ≅ 0.749 Ns m , while the first mode of vibration becomes overdamped for values larger than c 3 ≅ 1.89 Ns m .

Fig. 4.40 Polar diagrams of the complex receptances are presented in Fig. 4.40. The two natural frequencies are located on the diagrams. Except for Case I, the polar plots do not exhibit two loops, one for each mode of vibration, so that the number of degrees of freedom is not apparent from the simple inspection of the frequency response plots. Resonance location in such cases requires adequate methods.


179

Exercises 4.E1 For the system shown in Fig. 4.1 let k 1 = 2 k , k 2 = k , k 3 = 2k ,

m 1 = 3 m , m 2 = m . Set up the equations of motion and determine the natural

modes of vibration. Answer: ω 1 = 0.919 k m , ω 2 = 1.776 k m , μ 1 = 0.464 , μ 2 = −6.464 . 4.E2

Using the following values: k 1 = 103 N m , k 2 = 500 N m ,

k 3 = 2 ⋅ 103 N m , m 1 = 5 kg , m 2 = 10 kg , ˆf 1 = 100 N , ˆf 2 = 0 , ω = 15 rad sec ,

determine the amplitudes of the forced vibrations of the two masses for the system of Fig. 4.4. Answer: X 1 = 0.16 m , X 2 = −0.32 m . 4.E3 Determine the amplitudes of the forced vibrations of the system from Fig. 4.4, if k = 103 N m , m = 5 kg , ˆf = 50 N , ˆf = 100 N , ω = 10 rad sec . 1

2

Answer: X 1 = −0.5 m , X 2 = −0.8 m . 4.E4 Find the natural frequencies and associated mode shapes for the torsional vibrations of the system shown in Fig. 4.6. The stiffnesses of the shafts are K 1 = K , K 2 = 2 K , K 3 = 3 K , and the moments of inertia of the flywheels are

J1 = 2J , J 2 = J . Answer: ω 1 = K J , ω 2 = 2.345 K J , μ 1 = 0.5 , μ 2 = −4 . 4.E5

Using

K 2 = 2 ⋅107 N m rad ,

the

following

K 3 = 3 ⋅ 107 N m rad ,

values: J 1 = 100 kg m 2 ,

K 1 = 107 N m rad , J 2 = 200 kg m 2 ,

ω = 300 rad sec , determine the amplitudes of the torsional vibrations of the system shown in Fig. 4.6 when a harmonic torque of amplitude M 0 = 105 N and frequency ω = 300 rad sec is applied on the second disc. Answer: Θ1 = 0.00735 rad , Θ 2 = 0.00772 rad .


180

4.E6 Calculate the natural frequencies and the mode shapes for the flexural vibrations of the system shown in Fig. 4.15, where m 1 = 3 m , m 2 = m 3 ,

l 1 = l 2 = l and E I = const . Answer:

ω 1 = 0.522 E I m l 3 ,

ω 2 =1.73 E I m l3 ,

μ1 = 6 ,

μ 2 = −1.5 . 4.E7 Determine the normal modes for flexural vibrations of the system shown in Fig. 4.20, where m 1 = 2 m , m 2 = m , l 1 = l 2 = l , l 3 = l 2 and f = 0 .

Answer: ω 1 = 1.477 E I m l 3 , ω 2 = 3.464 E I m l 3 , μ 1 = −1 , μ 2 = 2 . 4.E8 Find the natural frequencies and associated mode shapes for the flexural vibrations of the massless beam shown in Fig. 4.E8.

Fig. 4.E8 Answer:

ω 1 = 0.891 E I m l 3 ,

ω 2 = 3.688 E I m l 3 ,

μ 1 = 1.048 ,

μ 2 = −0.477 . 4.E9 Determine the amplitudes of the forced vibrations of the two masses for the system of Fig. 4.E9 using the following values: E I m l 3 = 800 sec −2 ,

F0 l 3 E I = 2 mm and ω = 10 rad sec .

Fig. 4.E9 Answer: Y1 = 117.5 mm , Y2 = 25.04 mm .


181

4.E10 In Fig. 4.E10 use coordinate x at the centre of mass and θ for the rotation of the bar and set up the equations of motion. Determine the natural frequencies and locate the node of the bar for each mode.

Fig. 4.E10 Answer:

ω 1 = 0.707 k m ,

ω 2 = 1.414 k m ,

1 μ 1 = 0.666 l ,

1 μ 2 = −0.333 l . 4.E11 Find the natural frequencies and draw the mode shapes for the coupled translation and rotation vibrations of the rigid bar of Fig. 4.E11.

Fig. 4.E11 Answer: ω 1 = 0.78 k m , ω 2 = 1.281 k m , 1 μ 1 = 0.64 l , 1 μ 2 = −0.39 l . 4.E12 Calculate the amplitudes of the forced vibrations for the rigid bar of

Fig. 4.E12 using the following values: k = 103 N m , m = 5 kg , F0 = 50 N , l = 0.4 m and ω = 10 rad sec .

Fig. 4.E12 Answer: a x = 0.0165 m , aϕ = 0.083 rad .


182

4.E13 The results are given below of a resonance test on a structure. The response was measured at a place some distance away from the point of application of the excitation force and is given as values of the transfer receptance. Estimate for the two modes of vibration recorded the modal mass, stiffness, damping ratio and viscous damping coefficient.

Frequency, Hz 40 41 41.5 41.8 42 42.5 43.5 44 50 60 68 70 72 73 73.5 74 74.5 75 77 Answer: C 1 = 0.248 Ns m ,

Receptance magnitude,

Phase angle between response and force, degrees

10 −3 mm N 10 10.9 10.9 10.4 9.5 8.4 7.1 4.9 3.0 2.8 2.0 6.4 6.9 7.2 6.6 5.5 4.3 3.0 1.8

10 30 50 70 90 110 130 150 152 160 178 190 210 230 250 270 290 310 330

ω 1 = 263 rad sec ,

ζ 1 = 0.0466 ,

M 1 = 1.015 ⋅ 10 −2 kg ,

ω 2 = 462 rad sec ,

K 2 = 1800 N m , C 2 = 0.264 Ns m , M 2 = 0.84 ⋅ 10 −2 kg .

K 1 = 700 N m ,

ζ 2 = 0.0339 ,

5. SEVERAL DEGREES OF FREEDOM

In the previous chapter, systems with two degrees of freedom have been considered as the simplest case of, and an introduction to the general case of multidegree-of-freedom systems. A system has n degrees of freedom if its configuration at any time can be represented by n independent coordinates. Usually, in the configuration space, the coordinates are rectilinear translations and rotations, but velocities and accelerations may be used as well. Systems with a finite number of degrees of freedom are called discrete systems. It is engineering practice to describe the vibration of a continuous system with a finite number of coordinates. Each element of a discrete system is itself a continuous system, but the lowest frequencies of the latter are much greater than those of the idealized discrete system. The simplest approach leads to lumped parameter systems, consisting of lumped masses or discs, springs and dashpots. Their dynamic properties are defined by scalar quantities. Elements can be described by stiffness and damping matrices, relating the end forces to displacements and velocities across the element. Another discretization technique is the finite element method which can be regarded as a Rayleigh-Ritz method. It consists of approximating the solution of a differential eigenvalue problem, having no known closed-form solution, by a finite series of shape functions multiplied by undetermined coefficients. In the finite element method, the shape functions are local low-degree polynomials and the coefficients are the nodal displacements determined to render the Rayleigh quotient for the system stationary. Elemental matrices are defined for each element type which can be assembled in global mass, stiffness and damping matrices. Having these global matrices, one can write the equations of motion. Seeking synchronous harmonic solutions these are transformed into a homogeneous set of algebraic equations, equivalent to an algebraic eigenvalue problem. Solution of the eigenvalue problem yields the natural frequencies and associated mode shapes. Alternatively, natural frequencies can be obtained by minimization of the Rayleigh quotient.


184

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. The motion in a natural mode of vibration is synchronous and harmonic at all system coordinates. Systems having a finite number of degrees of freedom vibrate simultaneously in each of the several natural modes, as many natural modes as there are degrees of freedom. It takes certain combinations of initial conditions or applied forces to cause them to vibrate in only one mode. However, in damped systems, the free vibration is dominated by a few lower modes, sometimes only the lowest one, while the forced vibration can be described by a summation truncated to the modes with resonances within the frequency range of interest, plus some residual terms due to modes with frequencies lower or higher than the operating range. There is no basic difference between the low order discrete systems treated in this chapter and the large order discrete systems treated in a next chapter except that for the latter the solution of the eigenvalue problem requires more efficient computational methods.

5.1 Lumped Mass Systems Systems comprising uni-dimensional members like bars, shafts or beams are modeled by elementary systems consisting of concentrated masses connected by massless elastic elements. The distributed mass of such members is lumped at arbitrary chosen discrete points, regardless of the variation of vibratory amplitude along the element.

5.1.1 Beams with Lumped Masses Beams with lumped masses can have either point masses with only linear transverse displacements, or rigid discs with both translational and rotational degrees of freedom. 5.1.1.1 Linear Displacements Figure 5.1 shows two commonly used ways in which a uniform beam element could be represented. Duncan’s model (Fig. 5.1, b) has all the mass concentrated at the centre of gravity. Rayleigh’s model (Fig. 5.1, c) has half the


185

mass of the segment at each end. In the figure m 0 = ρ A is the mass per unit length, where ρ is the mass density and A is the cross section area.

Fig. 5.1 Comparing the two models it can be noticed that in Duncan’s model the element inertia to rotation with respect to the midpoint is neglected, while in 2

m0 l ⎛ l ⎞ ⎜ ⎟ ≠ 0 . This is why lumping using Duncan’s 2 ⎝2⎠ model generally gives higher natural frequencies, while Rayleigh’s model gives lower natural frequencies. The difference in frequencies calculated with the two models lowers with increasing the number of elements. Rayleigh’s model it is 2

Rayleigh’s model has also advantages for stepped beams, when the bending rigidity E I along the beam changes at the element ends, so that the model with lumped masses has segments with constant cross section. The discrepancy in the values of natural frequencies calculated using Rayleigh’s model can be illustrated by two examples. When the simply supported beam of Fig. 5.2, a is split into two segments (Fig. 5.2, b), the ratio of the natural frequency to the true value (6.14) for the beam with distributed mass is ω 1 ω 10 = 0.995. When the beam is split into four

segments (Fig. 5.2, c), the resulting three-degree-of-freedom system has natural frequencies of ω 1 ω 10 = 0.98 , ω 2 ω 20 = 0.995 and ω 3 ω 30 = 0.995 . For the cantilever beam of Fig. 5.3, a the approximation is lower. For the one segment model (Fig. 5.3, b), the ratio of the natural frequency to the true value (6.16) is ω 1 ω 10 = 0.7. When the beam is split into two segments (Fig. 5.3, c), the


186

ratio of the first natural frequencies is ω 1 ω 10 = 0.9. When the beam is split into three segments (Fig. 5.2, d), the ratio of the first natural frequencies is ω 1 ω 10 = 0.95.

Fig. 5.2

Fig. 5.3

5.1.1.2 Linear Displacements and Rotations

Figure 5.1, d shows an extension of Duncan’s model with the mass and mass moment of inertia of a beam element concentrated at the middle. For a uniform beam element, the total mass is

m = ρ Al = m0 l ,

(5.1)

and the total mass moment of inertia is

J=

m0 l 3 I ρ A l3 + m0 l = + ρlI 12 A 12

(5.2)

where I is the cross-section second moment of area. For a cylindrical beam element of diameter d and length l

J=

m0 l ⎛⎜ 3 d 2 2 ⎞⎟ +l . ⎟ 12 ⎜⎝ 4 ⎠

(5.3)

The total mass moment of inertia is composed of two parts. The first part, 2

ml 12 , is due to the element mass being distributed along the length of the beam


187

at the level of the neutral axis. The second part, the “rotatory inertia” ρ l I , is due to the fact that the beam mass is also distributed away from the beam neutral axis. This part is not activated by translation; it is only activated by rotation. In the extension of Rayleigh’s model (Fig. 5.1, e), one-half of the mass and a negative mass moment of inertia are lumped to the left and right ends. With this distribution, the conservation of both the mass and total mass moment of inertia with respect to the midpoint is satisfied. This can be checked calculating the latter using the Huygens-Steiner theorem of parallel axes ⎛ m 0 l l 2 m 0 l3 ⎞ m 0 l3 ⎟= − . J =2⎜ ⎜ 2 4 12 ⎟ 12 ⎝ ⎠

The above distinction is important in the finite element analysis. Two diagonal lumped-mass matrices are commonly used: an element mass matrix for translational inertia 0 0 0 ⎤ ⎡1 2 ⎢ 0 − l 2 12 0 0 ⎥⎥ e ⎢ mt = m0 l ⎢0 0 12 0 ⎥ ⎢ ⎥ 0 0 − l 2 12⎦ ⎣0

[ ]

(5.4)

and an element mass matrix for rotatory inertia

⎡0 0 ⎢0 i 2 e mr = m0 l ⎢ ⎢0 0 ⎢ ⎣0 0

[ ]

0⎤ 0 ⎥⎥ , 0 0⎥ ⎥ 0 i2 ⎦

0 0

(5.5)

where i = I A . 5.1.1.3 Flexibility Coefficients

As shown in Section 4.3, the equations of motion for flexural systems with lumped masses are written easier using flexibility coefficients instead of stiffnesses. If the linear displacements of the lumped masses are taken as the coordinates that define the system motion, then the vector of mass displacements { y } is related to the vector of forces applied to the lumped masses { f } by the flexibility matrix [δ ] as in equation (4.78)

{y } = [ δ ]{ f }.

(5.6)


188

The equation of motion for free vibrations (4.82) can be written

[δ ] [ m ] { &y& } + { y } = { 0 }.

(5.7)

The corresponding eigenvalue problem (4.85) is ⎛ ⎜ [δ ⎝

][ m ] −

1 2

ω

[I ]⎞⎟ { a } = {0 } . ⎠

(5.8)

The condition to have non-trivial solutions yields the frequency equation ⎛ det ⎜ [ δ ⎝

][ m ] −

1

ω2

[ I ]⎞⎟ = 0 ⎠

(5.9)

whose solutions are the system natural frequencies ωr . The mode shapes are defined by the modal vectors {a }r , satisfying the homogeneous linear set of equations ⎛ ⎜ [δ ⎜ ⎝

][ m ] −

1

⎞

ωr

⎠

[I ]⎟⎟ { a }r = {0} . 2

(5.10)

Example 5.1 Calculate the natural frequencies of lateral vibration for the three-mass beam of Fig. 5.4, a, where E I = const . Solution. Referring to Fig. 5.4, b, the deflection at any point x due to a concentrated load F applied a distance b from the right end, can be determined from the equation v(x)=

(

)

Fbx 2 l − x2 − b2 . 6E I l

The flexibility coefficients are

δi j =

b j xi 6E I l

(l

2

)

− x 2i − b 2j .

Because x 1 = b 3 = l 4 , x 2 = b 2 = l 2 , x 3 = b 1 = 3l 4 , the flexibility matrix is ⎡ 9 11 7 ⎤ l3 ⎢ [δ ] = 11 16 11⎥⎥ . 768 E I ⎢ ⎢⎣ 7 11 9 ⎥⎦

(5.11)


189

Equation (5.8) can be written ⎡ 9 11 7 ⎤ ⎧ a1 ⎫ ⎢11 16 11⎥ ⎪a ⎪ = λ ⎢ ⎥ ⎨ 2⎬ ⎢⎣ 7 11 9 ⎥⎦ ⎪⎩ a3 ⎪⎭

⎧a1 ⎫ ⎪ ⎪ ⎨a 2 ⎬ ⎪a ⎪ ⎩ 3⎭

where

λ = 768 E I ml3ω 2 .

Fig. 5.4

Fig. 5.5


λ 3 − 34λ 2 + 78 λ − 28 = 0 with roots

λ 1 = 31.5563 ,

λ2 = 2,

λ 3 = 0.4436 .


ω 1 = 4.933 E I ml 3 , ω 2 = 19.596 E I ml 3 , ω 3 = 41.606 E I ml3 . The modal vectors are

⎧ 1 ⎫ ⎧1⎫ ⎧ 1 ⎫ ⎪ ⎪ ⎪ ⎪ {a }1 = ⎨1.4142⎬ , {a } 2 = ⎨ 0 ⎬ , {a } 3 = ⎪⎨− 1.4142⎪⎬ . ⎪ 1 ⎪ ⎪− 1⎪ ⎪ 1 ⎪ ⎩ ⎩ ⎭ ⎩ ⎭ ⎭


190 5.1.1.4 Dunkerley’s Formula

For many vibrating systems, the natural frequencies of the second and higher modes are often considerably greater than that of the fundamental mode. This fact allows the estimation of the fundamental frequency with simple formulae. If

α 0 λ n + α 1 λ n −1 + .... + α n −1 λ + α n = 0 is an algebraic equation, then the sum of its roots is

α1

n

∑λi = −α

i =1

.

0

Considering the frequency equation (5.9), we can write n

1

i =1

ω 2i

∑

=−

n α1 n 1 = ∑ δ ii mi = ∑ 2 , α 0 i =1 i =1 ω i i

where ω i2i is the square of the “isolated” natural frequency of the system containing only the mass m i . This frequency is calculated based on the exact deflected curve of the isolated one-mass system, having the same configuration as the analysed system except for the eliminated masses. Because ω 1 < ω 2 < . . . < ω n , all terms in the first sum except the first may be omitted for the approximate determination of the fundamental frequency 1

ω 21

n

≅∑

1

2 i =1 ω i

n

1

i =1

ω 2i i

=∑

or 1

ω 21

≅

1

ω 211

+

1

ω 222

+

1

ω 233

+ . ..

(5.12)

Thus the reciprocal of the fundamental frequency squared can be obtained by adding the squares of the reciprocals of the isolated frequencies. Equation (5.12) is known as Dunkerley’s formula. It was discovered experimentally and published by S. Dunkerley (1895) then justified theoretically by R. V. Southwell (1921). It enables the estimation of the fundamental frequency of a system without solving the associated eigenvalue problem. The formula is valid for grounded systems only, i.e. it cannot be applied to free-free systems. Generally it yields lower estimates of the fundamental frequency.


191

For the cantilever beam of Fig. 5.3, a the true value of the fundamental frequency (6.16) is ω 10 = 3.52 E I ml 3 . For the model with one segment (Fig. 5.3, b), the lowest natural frequency is ω 1 = 2.44 E I ml 3 , which is 30.7% lower than the true value. When the beam is split into two segments (Fig. 5.3, c), the first natural frequency is ω 1 = 3.098 E I ml 3 , being 12% lower than the true one. When the beam is split into three segments (Fig. 5.2, d), the first natural frequency is ω 1 = 3.286 E I ml 3 , which is 7.12% lower, and when it is split

into four segments, it is 4.5% lower than the true frequency.

Example 5.2 Calculate the fundamental frequency of lateral vibration for the three-mass beam of Example 5.1 (Fig. 5.5, a) using Dunkerley’s formula. Solution. From the flexibility matrix (5.11) we obtain

δ 11 = δ 33 =

9l 3 , 768 E I

δ 22 =

16l 3 . 768 E I

For the single-degree-of-freedom systems with isolated masses (Fig. 5.5, b, c, d), the squares of the natural frequencies are respectively 2 ω 11 =

1 m δ 11

ω 222 = ω 323 =

1 m δ 22 1 m δ 33

=

768 E I , 9m l 3

= =

768 E I 16ml 3

,

768 E I , 9m l 3

so that Dunkerley’s formula (5.12) yields 1

ω 21

≅

1

ω 211

+

1

ω 222

+

1

ω 233

=

9ml 3 16ml 3 9m l 3 34ml 3 + + = . 768 E I 768 E I 768 E I 768 E I

The estimated fundamental natural frequency is

ω 1 ≅ 4.7527 E I ml 3 , which is 3.6% lower than the true value calculated at Example 5.1.


192 5.1.1.5 Rayleigh’s Formula

When a beam is represented by a lumped-mass model, consisting of a series of lumped masses m i (i = 1,..., n ) attached to a massless beam at points of abscissae x i , Rayleigh’s formula (2.19) becomes ω12

) dx ∫ ( , = ∑m v E I ∂2v ∂ x2 n

2

(5.13)

2 i i

i =1

( )

where v i = v x i are static deflections at the mass locations. If the strain energy is determined from the work done by the corresponding 1 n lumped weights m i g , then U max = ∑ m i g ⋅ v i so that Rayleigh’s formula (5.13) 2 i =1 becomes n

g ω12 =

∑ m i vi

i =1 n

∑

i =1

.

(5.14)

m i vi2

where g is the acceleration of gravity. As mentioned in Section 2.1.5, if the true deflection curve of the vibrating system is assumed, the fundamental frequency found by Rayleigh’s formula will be the correct frequency. For any other curve, the frequency determined by this method will be higher than the true frequency. If further accuracy is desired, a better approximation to the dynamic deflection curve can be made by using dynamic loads instead of the static weights. Since the dynamic load is m i ω 2 vi , which is proportional to the deflection, we can recalculate the deflection with the modified weights m 1 g , m 2 g

v v2 , m3 g 3 . v1 v1

Example 5.3 Calculate the fundamental frequency of lateral vibration for the three-mass beam of Example 5.1 (Fig. 5.4, a) using Rayleigh’s formula (5.14). Solution. Using flexibility coefficients and applying superposition, the displacement of any mass can be calculated as the sum of the products of flexibility coefficients at the respective location multiplied by the corresponding weights.


193


⎧v1 ⎫ ⎧1⎫ ⎧m g ⎫ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎨v 2 ⎬ = [δ ] ⎨m g ⎬ = m g [δ ] ⎨1⎬ ⎪1⎪ ⎪m g ⎪ ⎪v ⎪ ⎩⎭ ⎩ ⎭ ⎩ 3⎭ wherefrom we obtain

(

)

v 1 = m g δ 11 + δ 1 2 + δ 13 =

(

)

27l 3 mg , 768 E I

v 2 = m g δ 2 1 + δ 2 2 + δ 23 =

38l 3 mg , 768 E I

v 3 = v1. Substitution in (5.14) yields ω12 =

768EI 27 + 38 + 27 768 EI 92 768 EI = = 3 2 2 2 3 2902 31.54348 ml 3 9m l ml 27 + 38 + 27

or

ω 1 = 4.9343 E I ml3 , which is only 0.02% higher than the true value calculated at Example 5.1. If m = m0 l / 4 , then the system from Fig. 5.4, a is the same as the beam from Fig. 5.2, c. Substituting this value in the above formula yields

ω 1 = 4.9343 ⋅ 2

EI 9.8686 E I , = 4 m0 m0l l2

which is lower than the true value obtained for a continuous beam (6.14)

ω true =

π2 l

2

E I 9.8696 E I = , m0 m0 l2

owing to the lumping process in which the total mass was not conserved.

5.1.2 Multi-Story Frames as Shear Buildings A shear building is a structure in which there is no rotation of a horizontal section at the level of the floors. The deflected building resembles a cantilever beam that is deflected by shear forces only, hence the name shear building.


194

Figure 5.6, a shows a three-story shear building modelled as a frame with three degrees of freedom. Only swaying of the frame in its plane is considered, due to the flexural deformation of the vertical members in the plane of the frame. To accomplish such deflection in a building, the following assumptions are made: a) all the floors are rigid and capable of motion only in the horizontal direction, so that joints between floor girders and stanchions are fixed against rotation; b) all columns are inextensible and massless; and c) the mass of the building is concentrated at the floor levels, so that the vibration of the multi-story building is reduced to the vibration of a system with finite degrees of freedom. It is convenient to represent the shear building solely in terms of a single bay. The displacement of a horizontal member is resisted by elastic restoring forces in the stanchions. If the stanchions are treated as beams of uniform cross-section, standard methods of analysing beams with built-in ends, when one end sinks relative to the other, show that the combined stiffness of the stanchions in flexure is k=

12 E ⋅ 2 I 24 E I = 3 , l3 l

(5.15)

where E is the material Young’s modulus, I - the cross-sectional second moment of area of one stanchion and l - the height of the story (length of the stanchion).

Fig. 5.6 The equations of motion can be easily obtained, using d’Alembert’s principle, from the free body diagrams of Fig. 5.6, b by equating to zero the sum of the forces acting on each mass. The procedure is explained for two-degree-offreedom systems in Section 4.1.1.


195

5.1.3 Torsional Systems Torsional vibrations of multicylinder reciprocating machines are studied with a simplified lumped mass torsional system (Fig. 5.7) obtained by lumping the inertias of rotating and reciprocating parts from each cylinder at discrete points on a main equivalent shaft. The parts mounted on the shafting, like the flywheel, the torsional damper, the coupling and the driven rotor or the propeller can also be modelled as rigid discs.

Fig. 5.7 The connecting rod is replaced by an equivalent dynamical system comprised of two concentrated masses, one at the crank end, the other at the piston end. In each cylinder there is a reciprocating mass, mrec , consisting of the piston mass and the reduced mass of the connecting rod, and a rotating mass, consisting of the crank throw, counterweight and the reduced mass of the connecting rod, having an inertia J rot . The average inertia per cylinder is J = J rot +

1 mrec r 2 , 2

where r is the crank throw radius. The torsional stiffness of crankshaft between cylinders, arising from the combined effect of crank pin twisting and crank web bending, can be either approximated numerically or measured experimentally. The same for the stiffness of the crankshaft between the last cylinder and the flywheel, for the coupling and the driven shaft. The whole system can then be reduced to a torsional model with a series of rigid discs connected by massless flexible shafts, as in Fig. 5.7.

Example 5.4 A diesel-motor-generator set has been reduced to the equivalent system shown in Fig. 5.8, with the J values in kg ⋅ m 2 , and with K values in


196

106 x Nm/rad : J 7 = 0.897 ,

J 1 = 11.863 , K 1 = 1.062 ,

J 2 = 2.011 ,

K 2 = 6.101 ,

J 3 = J 4 = J 5 = J 6 = 0.167 ,

K 3 = K 4 = K 5 = 3.05 ,

K 6 = 4.067 .

Estimate the fundamental- and second-mode natural frequencies of torsional vibrations. Plot the corresponding mode shapes. Solution. ω 1 = 552 rad/sec , ω 2 = 1145 rad/sec . The mode shapes are shown in Fig. 5.8.

Fig. 5.8 Geared systems and car engine-driving train systems are modelled likewise. Geared-branched systems can be modelled as in Section 4.2.4. It is convenient to use an equivalent system which has no gears (or has gears that effect no change of speed) and whose natural frequencies are the same as the frequencies of the original system. This is possible because, while the gears effect a change of rotary speed given by the transmission ratio, the vibration is transmitted through the gears without change in frequency, but with proper change of amplitude.


197

In Section 4.2.4, the actual system with branches of different rotary speeds was reduced to an equivalent system with all parts having the same speed, usually equal to that of the reference shaft. Progressing away from that shaft to other branch through a gear of transmission ratio i, the actual J’s and K’s occurring beyond are multiplied by i 2 . If several gears are passed, the subsequent J’s and K’s are multiplied by the squares of the respective gear ratios. For the mating gears, the total inertia is obtained by adding to the moment of inertia of the reference gear the moment(s) of inertia of the (two) reduced gear(s) multiplied by i 2 . The equivalent model so obtained has the same natural frequencies as the actual system, as well as the same position of nodes. However, in the reduced branches, the amplitude of angular displacements is divided by − i , while the torque in the branch is multiplied by − i . A more straightforward approach is given in the following. In the equivalent model, each shaft and disc has the actual speed, only the driven gear is condensed out.

Fig. 5.9 Consider part of a geared-branched system (Fig. 5.9, a) where (the driving) shaft 1 and (the driven) shaft 2 are connected by two gears with inertias J , J ' and of pitch circle radii r and r' , respectively. Apart from shaft stiffnesses K 1 and K 2 , and disc mass moments of inertia J 1 and J 2 , the angular displacements and

torques acting on each disc are shown in the figure. The system motion is described by four torques and four angular displacements. However, only three of each of them are independent, due to the compatibility conditions between the mating gears. Thus it would be useful to


198

consider the torque M ' and the angular displacement θ ' as dependent variables and eliminate them, replacing the given system by the equivalent model shown in Fig. 5.9, b. Shaft 1 is taken as the reference shaft. Compatibility of the angular displacements requires that r θ = − r' θ ' ,

while compatibility of torques requires M M′ =− r r′

M θ =M 'θ' .

or

The transmissibility ratio is defined as i=−

n2 n1

=

r M θ' =− =− . r′ θ M′

(5.16)

In the original system, the stiffness matrix of shaft 2 is defined by ⎧ M ′ ⎫ ⎡ K2 ⎨ ⎬=⎢ ⎩M 2 ⎭ ⎣− K 2

− K 2 ⎤ ⎧θ ' ⎫ ⎨ ⎬. K 2 ⎥⎦ ⎩θ 2 ⎭

Using the transformation equations, based on (5.16), ⎧ M ⎫ ⎡ − i 0⎤ ⎧ M ′ ⎫ ⎨ ⎬=⎢ ⎥ ⎨ ⎬, ⎩ M 2 ⎭ ⎣ 0 1⎦ ⎩M 2 ⎭

⎧θ ' ⎫ ⎡− i 0⎤ ⎧ θ ⎫ ⎨ ⎬=⎢ ⎥ ⎨ ⎬, ⎩θ 2 ⎭ ⎣ 0 1⎦ ⎩θ 2 ⎭

the stiffness matrix of shaft 2 in the equivalent model is defined as follows ⎧ M ⎫ ⎡ − i 0⎤ ⎡ K 2 ⎨ ⎬=⎢ ⎥⎢ ⎩M 2 ⎭ ⎣ 0 1⎦ ⎣− K 2

⎧ M ⎫ ⎡i 2 K 2 ⎨ ⎬=⎢ ⎩M 2 ⎭ ⎣⎢ i K 2

− K 2 ⎤ ⎡ − i 0⎤ ⎧ θ ⎫ ⎨ ⎬, K 2 ⎥⎦ ⎢⎣ 0 1⎥⎦ ⎩θ 2 ⎭

i K2 ⎤ ⎥ K 2 ⎦⎥

⎧θ ⎫ ⎨θ ⎬ . ⎩ 2⎭

The mass matrix is defined likewise ⎧ M ⎫ ⎡− i 0⎤ ⎡ J ′ 0 ⎤ ⎡− i 0⎤ ⎧ θ&& ⎫ ⎨ ⎬=⎢ ⎥⎢ ⎥⎢ ⎥ ⎨ ⎬, ⎩M 2 ⎭ ⎣ 0 1⎦ ⎣ 0 J 2 ⎦ ⎣ 0 1⎦ ⎩θ&&2 ⎭

⎧ M ⎫ ⎡i 2 J ′ 0 ⎤ ⎥ ⎨ ⎬=⎢ J 2 ⎦⎥ ⎩M 2 ⎭ ⎣⎢ 0

⎧⎪ θ&& ⎫⎪ ⎨ && ⎬ . ⎪⎩θ 2 ⎪⎭

In the following, two rotating discs coupled by an elastic shaft segment are considered as a finite element. For an element of the driven shaft, the element matrices are


[ k ] = ⎡⎢ii KK 2

e

⎣⎢

e

e

i Ke ⎤ ⎥ , K e ⎦⎥

199

[ m ] = ⎡⎢i 0J ′

0⎤ ⎥ . J e ⎦⎥

2

e

e

⎣⎢

(5.17)

The element matrices for the shaft 1 in Fig.5.9 are

[ k ]= ⎡⎢−KK

1

1

⎣

1

− K1 ⎤ , K1 ⎥⎦

[ m ] = ⎡⎢ J0

0⎤ . J ⎥⎦

1

1

⎣

They can be obtained from the general expressions (5.17) substituting i = −1 . In general, for elements not adjacent to the gears, one of the end inertias is zero. Another approach is to condense the system matrices based on the constraint equations expressing the compatibility of the angular displacements of mating gears.

Example 5.5 The geared-branched system shown in Fig. 5.10, a consists of three rigid discs having mass moments of inertia J 1 , J 5 and J 6 and three rigid gears of radii r , r / 2 and r / 3 having mass moments of inertia J 2 , J 3 and J 4 which are connected by three light shafts having torsional stiffnesses K 1 , K 2 and K 3 . Derive the equations of free vibration. Solution. In the upper branch, the speed ratio is i = −2 . In the lower branch, the speed ratio is i = −3 . For the equivalent system shown in Fig. 5.10, b the element matrices are calculated from equations (5.17). The global matrices are assembled using the direct stiffness approach, obtaining:

⎡ K1 ⎢− K [K ] = ⎢⎢ 0 1 ⎢ ⎣⎢ 0

− K1 K 1 + 4K 2 + 9K3 2K 2 3K 3

0 2K 2 K2 0

0 ⎤ 3K 3 ⎥⎥ 0 ⎥ ⎥ K 3 ⎦⎥

and

⎡J 1 ⎢0 [M ] = ⎢⎢ 0 ⎢ ⎣⎢ 0

0 J 2 + 4J3 + 9J 4

0

0

0 J5

0

0

0⎤ 0 ⎥⎥ . 0⎥ ⎥ J 6 ⎦⎥


200

The global vector of angular displacements is

{Θ } = {θ 1

θ 2 θ 3 θ 4 }T .

The equations of motion can be written

[ M ]{Θ&& }+ [ K ]{Θ } = { 0 } .

They are used for the estimation of natural frequencies.

Fig. 5.10 The arrangement in Fig. 5.10 is a simplified model of a typical marine propulsion drive, with the propeller J1 driven by a low-pressure turbine J 5 and a high pressure turbine J 6 . The calculated natural frequencies are compared to the excitation frequencies. Usually the main excitation factor is the variation of the torque on propeller due to the variation of the water forces on the blades during their rotation. The frequency of this variation is called the blade frequency. It is the propeller frequency (once per shaft revolution) multiplied by the number of blades.

Example 5.6 The geared-branched system shown in Fig. 5.11 has the following parameters: J 1 = 950 , J 2 = 542 , J 3 = 406.7 , J 4 = J 5 = 6.78 , J 7 = 13.55 ,

J 6 = J 8 = J 9 = 27.12

[ kg ⋅ m ], 2

i = 76 24 , K 1 = 20.336 ⋅ 106 , K 2 = 8.135 ⋅ 106 ,


201

K 3 = 1.22 ⋅ 106 , K 4 = 0.407 ⋅ 106 , K 5 = 1.63 ⋅ 106 , K 6 = 2.44 ⋅ 106

[Nm/rad ] .

Calculate the natural frequencies of torsional vibrations.

Fig. 5.11

Solution. The actual system is reduced to an equivalent model in which the two pinions are condensed out. The element matrices (5.17) are calculated for each segment, then assembled into the global stiffness and mass matrices. The global stiffness matrix is ⎡ K1 ⎢− K 1 ⎢ ⎢ 0 ⎢ [K ] = ⎢ 0 ⎢ 0 ⎢ ⎢ 0 ⎢ 0 ⎣

− K1

0

K1 + K 2

−K2 2

2

0

0

0

0

0

0

−K2

K2 +i K3 +i K5

iK 3

0

iK 5

0

iK 3

K3 +K4

−K4

0

0

0

−K4

K4

0

0

iK 5

0

0

K5 +K6

0

0

0

0

−K6

The global mass matrix is ⎡J 1 ⎢0 ⎢ ⎢0 ⎢ [M ] = ⎢ 0 ⎢0 ⎢ ⎢0 ⎢0 ⎣

0

0 0

J2 2

2

0

0

0

0

0

0

0

J3 +i J 4 +i J5

0

0

0

0

0

J6

0

0

0

0

0

J7

0

0

0

0

0

J8

0

0

0

0

0

0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 0⎥ J 9 ⎥⎦

0 ⎤ 0 ⎥⎥ 0 ⎥ ⎥ 0 ⎥ 0 ⎥ ⎥ − K 6⎥ K 6 ⎥⎦


202

The natural frequencies have the following values: f 1 = 0 , f 2 = 14.26 ,

f 3 = 22.98 , f 4 = 35.34 , f 5 = 41.98 , f 6 = 50.52 , f 7 = 75.17 Hz .

5.1.4 Repeated Structures Vibratory systems are frequently comprised by a number of identical sections which are repeated several times. Examples of repeated structures are given in Fig. 5.12 for an n-story shear building, an n-mass system with translatory motion and an n-disc torsional system. For such systems the method of difference equation is appropriate for the calculation of natural frequencies.

Fig. 5.12 The equation of motion for the rth mass (Fig. 5.12, a) is

m &x&r + k (xr − xr −1 ) − k (xr +1 − xr ) = 0 which, for harmonic motion xr = ar sin ω t , can be expressed in terms of amplitudes as ⎛ ω 2m ⎞ ⎟ ar + ar −1 = 0 . ar +1 − 2 ⎜1 − ⎜ ⎟ 2 k ⎝ ⎠ The solution of this equation is found by substituting

ar = e i β r which leads to the relationship ⎛ ω 2m ⎞ e i β + e −i β ⎜1 − ⎟= = cos β , ⎜ ⎟ 2 2 k ⎝ ⎠


203

which can also be written

ω 2m k

= 2 (1 − cos β ) = 4 sin 2

β 2

.

(5.18)

The general solution for ar is

ar = C 1 cos β r + C2 sin β r , where C 1 and C 2 are determined from the boundary conditions. At r = 0 the amplitude is zero a0 = 0 , so that C 1 = 0 . At the free end, r = n , the equation of motion is

m &x&n + k (xn − xn −1 ) = 0 which, in terms of amplitudes, becomes

⎛ ω 2m ⎞ ⎟ an . an −1 = ⎜1 − ⎜ k ⎟⎠ ⎝ Substituting from the general solution, we obtain the following relationship for the evaluation of β :

sin β (n − 1) = [ 1 − 2 (1 − cos β

)] sin β n .

This result can be reduced to the product form

β 1⎞ ⎛ 2 cos β ⎜ n + ⎟ sin = 0 2⎠ 2 ⎝ which is satisfied by sin

β 2

=0 ,

1⎞ β (2 r − 1)π ⎛ = . cos β ⎜ n + ⎟ = 0 , or 2 2 (2 n + 1) 2⎠ ⎝

and by

(r = 1, . . . , n )

The natural frequencies are then available from equation (5.18) as

ω=2

k β sin m 2

(5.19)

which leads to

ωr = 2

(2 r − 1)π . k sin m 2 (2 n + 1)

(r = 1, . . . , n )

(5.20)


204

Natural frequencies calculated by the method of difference equation are always given by equation (5.19). However, for each repeated structure the quantity β must be determined from the appropriate boundary conditions.

5.1.5 Multi-Mass-Spring-Dashpot Systems In an one-dimensional vibrating system, every mass moves only in one direction. A node is assigned at each lumped mass, each node having only one degree of freedom. A blocked node is assigned at a fixed boundary.

Fig. 5.13 The four-mass model shown in Fig. 5.13, a has four degrees of freedom and five nodes. The nodal displacements are denoted q 1 , q 2 , . . . , q 5 (Fig. 5.13, b).

{ Q } = { q 1 , q 2 , . . . , q 5 }T is called the global vector of nodal displacements and { F } = { f1 , f 2 , . . . , f 5 }T is the global vector of nodal forces. A

The column vector

displacement or force has a positive value if acting along the positive q direction. At this stage, the boundary condition q 5 = 0 is not imposed. Table 5.1 Element 1 2 3 4 5 6

Node 1 1 2 3 4 2 3

2 2 3 4 5 4 5


205

The six springs are numbered according to their index. Each spring has two nodes. The element connectivity information can be conveniently represented as in Table 5.1. In the connectivity table, local node numbers are 1 and 2, while global node numbers are i and j. Such a table establishes the local-global correspondence. Referring to Fig. 5.13, c, the vector of local nodal forces is related to the vector of local nodal displacements equation

{ f }= { f , f e

{ q }= { q , q } e

1

{ f }= [ k ] { q } where the element stiffness matrix is e

e

1

2

T

2

}T

by the

e

[ k ] = ⎡⎢−kk e

e

⎣

e

− ke ⎤ . ke ⎥⎦

(5.21)

This can be established from the equilibrium and force-deflection equations: for q 2 = 0 , f1 = − f 2 = k q 1 , and for q 1 = 0 , f1 = − f 2 = −k q 2 . On the other hand, the global vector of nodal forces { F } is related to the

global vector of displacements { Q } by the equation { F } = [ K ] { Q } where [ K ] is the unreduced global stiffness matrix.

The matrix [ K ] can be obtained by the direct stiffness approach. Using the element connectivity information, the entries of each element matrix placed in the appropriate locations in the larger elements are then summed.

[K]

[ k ] are e

matrix and overlapping

The assembly of the global stiffness matrix can be explained by the summation of element strain energies. The strain energy in, say, spring 3, is U3 =

1 2

{ q } [ k ]{ q }= 12 ⎣q 3 T

3

3

3

⎡ k q4 ⎦ ⎢ 3 ⎣ − k3

− k3 ⎤ ⎧ q3 ⎫ ⎨ ⎬. k3 ⎥⎦ ⎩ q4 ⎭

Expanding the stiffness matrix at the system size we obtain

⎧ q1 ⎫ ⎪q ⎪ 2 1 ⎪⎪ ⎪⎪ U 3 = ⎨ q3 ⎬ 2⎪ ⎪ q ⎪ 4⎪ ⎪⎩ q5 ⎪⎭

[ ]

T

⎡0 ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢⎣ 0

0

0

0

0

0

0

0

k3 0 − k3 0 0

− k3 k3 0

0⎤ ⎧ ⎪ 0 ⎥⎥ ⎪ ⎪ 0⎥ ⎨ ⎥ 0⎥ ⎪ ⎪ 0 ⎥⎦ ⎪⎩

q1 ⎫ q2 ⎪⎪ T ~ ⎪ 1 q3 ⎬ = { Q } k 3 2 q4 ⎪ ⎪ q5 ⎪⎭

~ where k 3 is the expanded stiffness matrix of spring 3.

[ ]{ Q }


206

[ ]

~ We see that elements of the matrix k 3 are located in the third and fourth

rows and columns of the [ K ] matrix. When adding spring strain energies

U = ∑U e =

[ ]

e

1 2

{ Q }T [ K ] { Q },

the elements of k e are placed in the appropriate locations of the global [ K ] matrix, based on spring connectivity. Overlapping elements are simply added so that [ K ] = k~ e .

∑[ ] e

For the system of Fig. 5.13, a, the non-reduced global stiffness matrix is

− k1 ⎡ k1 ⎢−k k +k +k 1 1 2 5 ⎢ [K ] = ⎢ 0 − k2 ⎢ − k5 ⎢ 0 ⎢⎣ 0 0

0

0

− k2 k 2 + k3 + k 6 − k3 − k6

− k5 − k3 k 3 + k 4 + k5 − k4

⎤ 0 ⎥⎥ − k6 ⎥ . ⎥ − k4 ⎥ k 4 + k6 ⎥⎦ 0

The boundary conditions must now be specified. Node 5 is fixed so that q5 = 0 and should be eliminated from the displacement vector. A reduced global stiffness matrix is obtained by eliminating, from the original stiffness matrix, the row and column corresponding to the specified or “support” degree of freedom. For the system of Fig. 5.13, a the reduced global stiffness matrix is

− k1 ⎡ k1 ⎢− k k + k + k [K ] = ⎢⎢ 0 1 1 − k2 5 2 ⎢ 0 k − 5 ⎣⎢

0

⎤ ⎥ −k5 ⎥. ⎥ −k3 ⎥ k 3 + k 4 + k5 ⎦⎥ 0

−k2 k 2 + k3 + k 6 −k3

Together with the diagonal mass matrix

⎡m 1 0 ⎢0 m [M ] = ⎢⎢ 0 02 ⎢ 0 ⎢⎣ 0

0 0

m3 0

0 ⎤ 0 ⎥⎥ 0 ⎥ ⎥ m 4 ⎥⎦

it is used to write the equations of free motion

[ M ]{Q&& }+ [ K ]{ Q } = { 0 }


207

and to solve the corresponding eigenvalue problem

[ K ]{Φ } = ω 2 [ M ]{Φ }, to determine the real modes of vibration of the undamped system. For a system including dashpots, the same procedure is used to assemble the global damping matrix [ C ] . For the model from Fig. 5.14 the same connectivities have been taken for dashpots as for springs, though generally they can be different.

Fig. 5.14 The global damping matrix is

− c1 ⎡ c1 ⎢− c c + c + c [C ] = ⎢⎢ 0 1 1 − c2 5 2 ⎢ 0 c − 5 ⎣⎢

0

−c2 c 2 + c3 + c6 − c3

⎤ ⎥ − c5 ⎥. ⎥ − c3 ⎥ c 3 + c4 + c5 ⎦⎥ 0

The equations of free motion of the damped system can be written

[ M ]{Q&& }+ [ C ]{Q& }+ [ K ]{Q } = { 0 } . In this case the system has complex modes of vibration. The complex eigenvalues yield the damped natural frequencies and the modal damping ratios. The systems with proportional damping have real modal vectors. Vibrations of damped systems are treated as in Section 4.6.2 and are dealt with in a next chapter.

Example 5.7 Calculate the natural frequencies and mode shapes of the system shown in Fig. 5.15. The system parameters are: m 1 = m 2 = .... = m 11 = 1 kg , k 1 = k 11 = 2421 N/m ,

k 2 = k 10 = 2989 N/m , k 3 = k 9 = 3691 N/m , k 4 = k 8 = 4556 N/m , k 5 = k 7 = 5625 N/m ,


208

k 6 = 18000 N/m . Solution. The natural frequencies, in Hz, are 2.74, 2.95, 7.24, 7.80, 11.47, 12.13, 15.00, 15.62, 18.49, 19.32 and 28.57. There are pairs of two close natural frequencies, one for a symmetric and one for an antisymmetric mode. This is typical for symmetric structures.

Fig. 5.15 The first 10 mode shapes are presented in Fig. 5.16.

Fig. 5.16


209

Example 5.8 For the 15-dof system shown of Fig. 5.17 set up the matrices M, K and C. Find the damped natural frequencies and the modal damping ratios.

Fig. 5.17

Solution. Values obtained using the program VIBMKC are listed in Table 5.2. Damping ratio values are multiplied by 100. Table 5.2 Mode

ω d , Hz

ζ ,%

Mode

ωd , Hz

ζ ,%

1 2 3 4 5 6 7 8

15.98 30.86 43.60 46.47 53.35 53.42 59.45 61.62

0.502 0.968 1.364 0.301 1.665 0.670 1.853 1.060

9 10 11 12 13 14 15

68.88 73.72 128.87 136.59 143.89 150.87 157.52

1.378 1.579 0.536 0.506 0.477 0.457 0.437

The five masses located on the right are one order of magnitude smaller than the others. This produces the cluster of five higher natural frequencies. Due to the relatively low damping values, the damped natural frequencies are approximately equal to the undamped natural frequencies.


210

5.2 Plane Trusses A truss structure consists of pin-jointed members. The main simplifying assumption in trusses is that all members are connected together at their ends by frictionless pin joints and thus cannot transmit moments from one another. In practice, the joints of a truss are made by riveting, welding or bolting. However, a simplified model with pin joints is a surprisingly good engineering approximation. Truss elements can only sustain tension or compression, such members being called ties or struts respectively. Rigid jointed structures are dealt with in Section 5.3. In a truss, it is required that all loads and reactions are applied only at the joints and members have constant axial rigidity, so that they are natural finite elements. To account for the spatial orientations of truss members, local and global coordinate systems are used. In the following, the element stiffness and mass matrices are calculated first in the local coordinate system, then in the global coordinate system. The latter can be expanded to the system size, then simply added to obtain the global stiffness and mass matrices to be used in the eigenvalue problem or dynamic response calculations.

5.2.1 Coordinates and Shape Functions for the Truss Element Consider a two-noded pin-jointed element in the own or local coordinate system. Nodes are conveniently numbered 1 and 2, their coordinates in the physical (cartesian) reference system being x 1 and x 2 respectively (Fig. 5.18, a).

Fig. 5.18 We define a natural or intrinsic reference system which permits the specification of a point within the element by a dimensionless number

r=

x + x2 ⎞ 2 ⎛ ⎜x− 1 ⎟ ⎜ 2 ⎟⎠ x 2 − x1 ⎝

so that r = −1 at node 1 and r = +1 at node 2 (Fig. 5.18, b).

(5.22)


211

Expressing the physical coordinate in terms of the natural coordinate yields

x = N1 (r ) x 1 + N 2 (r ) x 2 ,

(5.23)

where

1 ( 1 − r ) and N 2 (r ) = 1 ( 1 + r ) (5.24) 2 2 can be considered as geometric interpolation functions. The graphs of these functions are shown in Figs. 5.19, a, b.

N1 (r ) =

Fig. 5.19 For a two-noded bar element, a linear distribution of displacements can be assumed. The displacement of an arbitrary point within the element can be expressed in terms of the nodal displacements q 1 and q 2 as

u (r ) = N1 (r ) q 1 + N 2 (r ) q 2 .

(5.25)

In matrix notation

{ }

2

u = ∑ N i qi = ⎣N ⎦ q e , i =1

where

⎣N ⎦ = ⎣N1 N 2 ⎦ and

{q }= { q e

{ }

(5.26)

1

q2

}T .

(5.27)

q e is referred to as the element displacement In equation (5.26), vector and ⎣N ⎦ is the row vector of displacement interpolation functions also named shape functions. It is easy to check that u = q 1 at node 1 and u = q 2 at node 2, and that u varies linearly (Fig. 5.19, c).


212

Equations (5.23) and (5.25) show that both the element geometry and the displacement field are interpolated using the same shape functions, which is referred to as the isoparametric formulation.

5.2.2 Element Stiffness and Mass Matrices in Local Coordinates In dynamic analysis, the displacement is a function of both space and time u = u ( x ,t ) , the strains are ε = ∂ u ∂ x and the stresses are σ = Eε . The element strain energy U e (t ) is

Ue =

1 2

∫

1 2

σ ε dV =

e

∫

Ee ε 2 Ae dx =

e

Ee Ae 2

∫ e

2

⎛ ∂u ⎞ ⎜⎜ ⎟⎟ d x . (5.28) ⎝ ∂x ⎠

The transformation from x to r in equation (5.22) yields

x 2 − x1

dx =

2

dr =

le dr , 2

(5.29)

where − 1 ≤ r ≤ +1 and the length of the element is l e = x 2 − x 1 . Because

{ }

∂u ⎢ ∂ N ⎥ e ∂ N1 ∂N ∂N ∂ r 2 ∂N 1 where =⎢ =− = = ⎥ q and ∂x ⎣ ∂x ⎦ ∂r ∂x ∂r ∂ x l e ∂r 2 2

{ }

⎛ ∂u ⎞ 1 ∂ N2 and = + , we can write ⎜⎜ ⎟⎟ = q e 2 ∂r ⎝ ∂x ⎠ element strain energy (5.28) becomes 1 Ue = 2

{q }

e T

2 Ee Ae le

+1

∫

−1

T

{ }

T

⎢∂N ⎥ ⎢∂N ⎥ e ⎢ ∂ x ⎥ ⎢ ∂ x ⎥ q , so that the ⎣ ⎦ ⎣ ⎦

T

⎢∂N ⎥ ⎢∂N ⎥ ⎢ ⎥ ⎢ ⎥dr ⎣ ∂r ⎦ ⎣ ∂r ⎦

{ q }. e

(5.30)

The above equation is of the form Ue =

1 2

{ q } [ k ] { q }, e T

e

e

(5.31)

[ ]

where the element stiffness matrix k e is given by

[k ] e

or

2E A = e e le

+1

∫

−1

T

⎢∂N ⎥ ⎢∂N ⎥ 2E A dr = e e ⎢ ⎥ ⎢ ⎥ le ⎣ ∂r ⎦ ⎣ ∂r ⎦

+1

∫

−1

⎧− 1 2⎫ ⎢ 1 ⎨ ⎬⎢ − ⎩ 1 2 ⎭⎣ 2

1⎥ dr 2 ⎥⎦


213

[ k ] = El A ⎡⎢−11 e

− 1⎤ . 1 ⎥⎦

e e

⎣

e

(5.32)

The element kinetic energy Te (t ) is

Te =

ρ Ae 2

∫ e

2

⎛ ∂u ⎞ ⎜⎜ ⎟⎟ d x . ⎝ ∂t ⎠

(5.33)

where ρ is the material mass per unit volume and ∂u ∂ t = u& is the velocity at x. From (5.26) we obtain

{ }

u& = ⎣N ⎦ q& e ,

where

{ q& } is the column vector of nodal velocities.

(5.34)

e

Substituting (5.34) into equation (5.33) yields Te =

1 2

{ q& }

e T

∫e ⎣N ⎦

ρ Ae

e ⎣N ⎦ d x { q& }.

T

(5.35)

Equation (5.35) is of the form Te = where

1 2

{ q& } [ m ] { q& }. e T

e

[ m ] = ρ A ∫ ⎣N ⎦ e

e

e

T

⎣N ⎦ d x

(5.36) (5.37)

e

is called the element consistent mass matrix. It is calculated using the same procedure and shape functions as the element stiffness matrix. Changing the variable

[m ]= e

or

ρ Ae l e 2

+1

∫

⎣N ⎦ ⎣N ⎦ d r = T

−1

[ m ] = ρ A6 l e

e e

ρ Ae l e 8

⎡2 1 ⎤ ⎢1 2 ⎥ . ⎣ ⎦

+1

∫

−1

⎡(1 − r ) 2 ⎢ 2 ⎢⎣ 1 − r

1− r 2 ⎤ ⎥dr (1 + r ) 2 ⎥⎦

(5.38)

5.2.3 Transformation from Local to Global Coordinates A typical plane truss element is shown in Fig. 5.20 where both the local coordinate system xOy and the global coordinate system XOY are drawn. Nodal


214

displacements are denoted by small letters in the local coordinate system and by capital letters in the global coordinate system. In the global coordinate system, every node has two degrees of freedom (dof’s). A node whose global node index is j has associated with it dof’s 2 j − 1 and 2 j , and displacements Q2 j −1 and Q2 j . In Fig. 5.20 we see that q 1 equals the sum of projections of Q 1 and Q 2 onto the x axis. Thus Similarly

q 1 = Q 1 cos α + Q 2 sin α .

(5.39, a)

q 2 = Q 3 cos α + Q 4 sin α .

(5.39, b)

Fig. 5.20 Equations (5.39) can be written in matrix form as

{ q } = [ T ] { Q }, e

where

{q }= {q e

1

q2

}T

{ } {

e

e

(5.40)

is the element displacement vector in the local

coordinate system, Q e = Q 1 , Q 2 , Q 3 , Q 4

}T

is the element displacement vector

in the global coordinate system and

[T ] = ⎡⎢cos0 α e

⎣

sin α 0

0 cos α

0 ⎤ sin α ⎥⎦

is a coordinate transformation matrix. From nodal coordinate data, denoting

( X 1 ,Y1 )

coordinates of nodes 1 and 2, respectively, we obtain

(5.41)

and

( X 2 ,Y 2 )

the


cos α =

X 2 − X1 le

, sin α =

Y 2 − Y1 le

215

(X 2 − X 1 ) 2 + (Y 2 − Y1 ) 2 .

, le =

(5.42)

The entries of the matrix (5.41) are calculated based on the above equations.

5.2.4 Element Stiffness and Mass Matrices in Global Coordinates Substituting equation (5.40) into the expression (5.31) of element strain energy in local coordinates, we get Ue =

{ Q } [T ] [ k ][T ]{ Q }.

1 2

e T

e

T

e

e

e

(5.43)

The strain energy in global coordinates can be written as Ue =

[ ]

{ Q } [ K ]{ Q },

1 2

e T

e

e

(5.44)

where K e is the element stiffness matrix in global coordinates. Comparing equations (5.43) and (5.44) we obtain the stiffness matrix in global coordinates as

[ K ]= [T ] [ k ][T ] . (5.45) Substituting for [T ] from equation (5.41) and for [ k ] from equation e

e

T

e

e

e

e

(5.32) we get

[K ] e

⎡ c2 cs − c2 − cs⎤ ⎥ ⎢ E A cs s2 − cs − s2 ⎥ , = e e⎢ 2 l e ⎢− c − c s c 2 cs ⎥ ⎥ ⎢ 2 cs s 2 ⎦⎥ ⎣⎢ − c s − s

(5.46)

where c = cos α and s = sin α . Similarly, the kinetic energy in global coordinates can be written as Te =

1 2

{ Q& } [ M ]{ Q& }. e T

e

e

(5.47)

The element consistent mass matrix in global coordinates is

[ M ]= [T ] [ m ][T ] , e

[ ]

e

T

e

e

[ ]

or, substituting for T e from equation (5.41) and for m e from equation (5.38)


216

[M ] e

⎡2 c2 2 c s c2 cs ⎤ ⎥ ⎢ 2 ρ Ae l e ⎢2 c s 2s cs s2 ⎥ = . 6 ⎢ c2 c s 2 c2 2 c s⎥ ⎥ ⎢ s 2 2 c s 2s 2 ⎥⎦ ⎢⎣ c s

(5.48)

This matrix is singular (order 4 and rank 1) because the element is not grounded. The rank deficiency is equal to the three possible rigid body motions.

5.2.5 Assembly of Stiffness and Mass Matrices The global stiffness and mass matrices, [ K ] and [ M ] , are assembled from

[ ]

[ ]

element matrices K e and M e using the element connectivity information. The compatibility of nodal displacements, at element level with the nodal displacements at the whole truss structure level, can be expressed by equations of the form ~ Q e = T e { Q }, (5.49)

{ } [ ]

{ } is the element displacement vector in global coordinates, {Q } is the ~ displacement vector of the truss structure and [T ] is referred to as a

where Q e

e full connectivity or localisation matrix, containing ones at the dof’s of element nodes and zeros elsewhere.

The element strain energy in global coordinates can be written in terms of the global displacement vector, substituting (5.49) into (5.44) Ue =

1 2

{ Q }T [T~ e ] T [ K e ][T~ e ]{ Q }

or

[ ]{ Q },

1 { Q }T K~ e 2 where the expanded element stiffness matrix Ue =

[ K~ ]= [T~ ] [ K ][T~ ] e

e

T

e

e

(5.50)

has the size of the system matrices. To illustrate this, consider the seven-bar truss shown in Fig. 5.21. The connectivity matrix of element 4, where the two nodes are not consecutively numbered, is


[T~ ] 4

⎡0 ⎢0 =⎢ ⎢0 ⎢ ⎣0

217

0 1 0 0 0 0 0 0 0⎤ 0 0 1 0 0 0 0 0 0⎥⎥ 0 0 0 0 0 1 0 0 0⎥ ⎥ 0 0 0 0 0 0 1 0 0⎦

while, for the element 5, where the two nodes are consecutively numbered, it is

[T~ ] 5

⎡0 ⎢0 =⎢ ⎢0 ⎢ ⎣0

0 0 0 0

0 0 0 0

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

0⎤ 0⎥⎥ . 0⎥ ⎥ 0⎦

Fig. 5.21 The corresponding expanded stiffness matrices are of the form

[ K~ ] = 4

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1 2 3 4 5 6 7 8 9 10 ⎤ 1 ⎥ ⎥ 2 ⎥ 3 • • • • ⎥ • • • • ⎥ 4 ⎥ 5 ⎥ ⎥ 6 ⎥ • • • • ⎥ 7 ⎥ 8 • • • • ⎥ ⎥ 9 ⎥ 10 ⎦

,

[ K~ ] = 5

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1 2 3 4 5 6 7 8 9 10 ⎤ 1 ⎥ ⎥ 2 ⎥ 3 ⎥ ⎥ 4 ⎥ 5 • • • • ⎥ • • • • ⎥ 6 ⎥ • • • • ⎥ 7 ⎥ 8 • • • • ⎥ ⎥ 9 ⎥ 10 ⎦

We may imagine that the structure is built by adding elements one by one, with each element being placed in a preassigned location. As elements are added to the structure, contributions are made to the structure load carrying capacity, hence to the structure stiffness matrix. We may add the element stiffness matrices to


218

obtain the stiffness matrix of the entire structure provided that element matrices are of the “structure size” and operate on identical displacement vectors. Simple addition of the expanded element stiffness matrices produces the structure stiffness matrix. The strain energy for the complete truss structure 1 { Q }T [ K ]{ Q }, 2 can be calculated by simply adding element strain energies U=

U=

1 1 ~ ~ ∑U = ∑ 2 { Q } [ K ]{ Q } = 2 { Q } ∑ [ K ] { Q }. T

T

e

e

e

e

e

(5.51)

(5.52)

e

Comparing (5.51) and (5.52) we get [ K ] = K~ e .

∑e [ ]

(5.53)

The global stiffness matrix is equal to the sum of the expanded element stiffness matrices. Similarly, the global mass matrix element mass matrices as [ M ] = M~ e = T~ e

∑e [

[M ]

is assembled from expanded

] ∑ [ ] [ M ][T~ ]. T

e

e

(5.54)

e

The unreduced stiffness and mass matrices [ K ] and [ M ] are used in the case of free-free systems. For grounded systems, they are condensed using the boundary conditions. The effect of lumped masses and springs can be accounted for by adding their values along the main diagonal at the appropriate locations in the respective matrices.

5.2.6 Equations of Motion and Eigenproblem We define the Lagrangean L by L =T −Π ,

(5.55)

where T is the kinetic energy T=

1 2

and Π is the potential energy

{ Q& }T [ M ] { Q& }

(5.56, a)


Π = In (5.56, b),

1 2

219

{ Q }T [ K ] { Q } − { Q } T { F } .

(5.56, b)

{ F } is the global vector of applied nodal forces.

Hamilton’s principle leads to Lagrange’s equations for undamped systems d ⎛⎜ ∂ L dt ⎜ ∂ Q& ⎝

{ }

⎞ ⎟ − ∂L = 0 ⎟ ∂ {Q } ⎠

or

d ⎛⎜ ∂T d t ⎜ ∂ Q& ⎝

{ }

⎞ ⎟ − ∂Π = 0 . ⎟ ∂ {Q } ⎠

(5.57)

Using the rules of differentiation of a scalar with respect to a vector, we obtain the equations of motion

[ M ] {Q&& }+ [ K ] { Q } = { F }.

(5.58)

For free vibrations, the force vector is zero. Thus

[ M ] {Q&& }+ [ K ] { Q } = { 0 }.

(5.59)

Looking for solutions of the form

{ Q } = {φ }sin ω t ,

(5.60)

where {φ } is the vector of nodal amplitudes of vibration, we obtain the generalized eigenvalue problem

[ K ] {φ }r = ωr2 [ M ] {φ }r ,

(r = 1,..., n ) ,

(5.61)

where ωr2 are real eigenvalues equal to the natural frequencies squared and are real eigenvectors.

{φ }r

The program VIBTRUSS (written in MATLAB) outputs the natural frequencies and mode shapes for undamped truss structures.

Fig. 5.22


220

Example 5.9 Calculate the first four natural frequencies and mode shapes for the system shown in Fig. 5.22, a where E = 200 GPa , ρ = 7850 kg m 3 and A = 100 mm 2 for all 20 bars. Solution. The lowest four natural frequencies, in Hz, are 48.8, 168.4, 235.9 and 336.8. The mode shapes are shown in Figs. 5.22, b to e.

5.3 Plane Frames Frames are structures with rigidly connected members called beams. Beams are slender members used to support transverse loading. They are connected by rigid (nodal) joints that have determinate rotations and, apart from forces, transmit bending moments from member to member. In this section, we first present the finite element formulation for beams, then extend it to plane frames. An inclined beam element will be referred to as a frame element.

5.3.1 Static Analysis of a Uniform Beam Beams with cross sections that are symmetric with respect to the plane of loading are considered herein (Fig. 5.23). Transverse shear deformations are neglected.

Fig. 5.23 The axial displacement of any point on the section, at a distance y from the neutral axis, is approximated by dv u = −ϕ y = − y, (5.62) dx


221

where v is the deflection of the centroidal axis at x and ϕ = v′ is the cross section rotation (or slope) at x . Axial strains are du d2 v =− 2 y. dx dx Normal stresses on the cross section are given by d2 v σ x = E ε x = −E 2 y , dx where E is Young’s modulus of the material.

εx =

(5.63)

(5.64)

The bending moment is the resultant of the stress distribution on the cross section M (x ) = − σ x y dA = E I z

∫

A

d2 v = E I z v II . 2 dx

(5.65)

where I z is the second moment of area of the section about the neutral axis z. The shear force is given by dM d3 v = EIz = E I z v III . dx dx 3 The transverse load per unit length is T (x ) =

dT d4 v = EIz = E I z v IV . 4 dx dx The differential equation of equilibrium is p (x ) =

EIz

d4 v = p (x ) . dx4

(5.66)

(5.67)

(5.68)

5.3.2 Finite Element Discretization The plane frame is divided into elements, as shown in Fig. 5.24. Each node has three degrees of freedom, two linear displacements and a rotation. Typically, the degrees of freedom of node i are Q3 i − 2 , Q3 i −1 and Q3 i , defined as the displacement along the X axis, the displacement along the Y axis and the rotation about the Z axis, respectively. Nodes are located by their coordinates in the global reference frame XOY and element connectivity is defined by the indices of the end nodes. Elements are modelled as uniform beams without shear deformations and not loaded between


222

ends. Their properties are the bending rigidity E I , the mass per unit length ρ A and the length l .

Fig. 5.24 In the following, the shape functions are established for the beam element, then the element stiffness and mass matrices are calculated first in the local coordinate system, then in the global coordinate system. The latter are expanded to the structure size, then simply added to get the global uncondensed stiffness and mass matrices. Imposing the boundary conditions, the reduced stiffness and mass matrices are calculated which, together with the damping matrix, are used in the dynamic analysis.

Fig. 5.25 Consider an inclined beam element, as illustrated in Fig. 5.25, a, where the nodal displacements are also shown.


223

In a local physical coordinate system, the x axis, oriented along the beam, is inclined an angle α with respect to the global X axis. Alternatively, an intrinsic (natural) coordinate system can be used. The vector of element nodal displacements is

{ q }= { q , q e

1

2,

q3, q 4, q5, q6

}T

(5.69)

and the corresponding vector of element nodal forces can be written

{ f }= { f e

Forces

f 2, f 3, f 5, f 6

1,

f 2, f 3, f 4, f 5, f 6

and

the

}T .

corresponding

(5.70) displacements

q 2 , q 3 , q 5 , q 6 describe the element bending (Fig. 5.25, b) while axial forces f 1 , f 4 , and displacements q 1 , q 4 , describe the element stretching (Fig. 5.25, c).

Their action is decoupled so that the respective stiffness and mass matrices can be calculated separately.

5.3.3 Static Shape Functions of a Beam Element For a uniform beam not loaded between ends, p = 0 and equation (5.68) yields d 4 v d x 4 = 0 . Integrating four times, we obtain the deflection v described by a third order polynomial v (x ) = a 1 x 3 + a 2 x 2 + a 3 x + a 4 .

(5.71)

In (5.71), the four integration constants a 1 , a 2 , a 3 , a 4 can be determined from the geometric boundary conditions, involving the transverse displacement and slope at each end:

x = x1 , v = q2 , d v d x = q3 , and x = x 2 , v = q5 , d v d x = q6 .

(5.72)

Alternatively, the transverse displacement can be expressed in terms of the nodal displacements as

{ }

v = ⎣N ⎦ q e ,

(5.73)

where ⎣N ⎦ is a row vector containing the shape functions, which are cubic polynomials, called Hermite polynomials. Using natural coordinates, with r = −1 at node 1 and r = +1 at node 2, the transverse displacement can be written


224

⎛dv⎞ ⎛dv⎞ ⎟⎟ + N 3 (r ) v 2 + N 4 (r )⎜⎜ ⎟⎟ . v (r ) = N1 (r ) v 1 + N 2 (r ) ⎜⎜ ⎝ d r ⎠1 ⎝ d r ⎠2

(5.74)

Because the coordinates transform by the relationship (5.22)

x=

x1 + x 2 2

+

x 2 − x1 2

(5.75)

r

and since l e = x 2 − x 1 is the length of the element, equation (5.29) holds dx =

le dr . 2

(5.76)

Using the chain rule of differentiation d v le d v = , dr 2 dx

(5.77)

equation (5.74) becomes v (r ) = N1 (r ) v 1 + N 2 (r )

le 2

⎛dv⎞ ⎜⎜ ⎟⎟ ⎝ d x ⎠2

(5.78)

le l N 2 ⋅ q3 + N 3 ⋅ q5 + e N 4 ⋅ q6 . 2 2

(5.79)

⎛dv⎞ l ⎜⎜ ⎟⎟ + N 3 (r ) v 2 + N 4 (r ) e 2 ⎝ d x ⎠1

or v (r ) = N1 ⋅ q2 +

In (5.73) the row vector of shape functions is ⎥ ⎢ ⎣N ⎦ = ⎢ N1 , 2e N 2 , N 3 , 2e N 4 ⎥ . ⎦ ⎣

l

l

(5.80)

The Hermite shape functions are cubic polynomials which should satisfy the boundary conditions given in Table 5.3 where primes indicate differentiation with respect to r. Table 5.3

r = −1 r = +1

N1

N ′1

N2

N ′2

N3

N ′3

N4

N ′4

1 0

0 0

0 0

1 0

0 1

0 0

0 0

0 1

Imposing the above conditions to cubic polynomials with four arbitrary constants, we obtain the expressions of the beam element shape functions in natural coordinates (5.81), graphically shown in Fig. 5.26.


225

(

)

N 1(r ) =

1 ( 1 − r ) 2 ( 2 + r ) = 1 2 − 3r + r 3 , 4 4

N 2 (r ) =

1 ( 1 − r ) 2 ( 1 + r ) = 1 1 −r − r 2 + r 3 , 4 4

N 3(r ) =

1 ( 1 + r ) 2 ( 2 − r ) = 1 2 + 3r − r 3 , 4 4

(

N 4 (r ) = −

)

(

(5.81)

)

(

)

1 ( 1 + r ) 2 ( 1 − r ) = − 1 1 + r −r 2 − r 3 . 4 4

Fig. 5.26 It is easy to check that at node 1, v = q 2 and v = q 5 and

d v le = q 3 , while at node 2, dr 2

d v le = q6 . dr 2

5.3.4 Stiffness Matrix of a Beam Element The strain energy U e of a beam element is Ue =

E Ie 2

∫ e

2

⎛d 2v⎞ ⎜ ⎟ ⎜ d x2 ⎟ d x . ⎝ ⎠

(5.82)


226

Equation (5.77) yields dv 2 d v = d x le d r

d 2v 4 d 2 v = . d x 2 l 2e d r 2

and

Substituting (5.73) we obtain

{ }

d 2v 4 ⎢ d 2 N ⎥ e = ⎢ ⎥ q . d x 2 l 2e ⎣⎢ d r 2 ⎦⎥

(5.83)

The square of the above quantity is calculated as 2

T

⎛ 2 ⎞ ⎛ d 2v ⎞ ⎟ =⎜d v⎟ ⎜ ⎜ d x2 ⎟ ⎜ d x2 ⎟ ⎠ ⎝ ⎠ ⎝

{ }

⎛ d 2v ⎞ ⎟ = qe ⎜ ⎜ d x2 ⎟ ⎠ ⎝

T

T

{ }

16 ⎢ d 2 N ⎥ ⎢ d 2 N ⎥ e ⎢ ⎥ ⎢ ⎥ q , l 4e ⎣⎢ d r 2 ⎦⎥ ⎢⎣ d r 2 ⎦⎥

which can be also written 2

{ }

⎛ d 2v ⎞ ⎟ = qe ⎜ ⎜ d x2 ⎟ ⎠ ⎝

T

{ }

16 T e ⎣N r′′ ⎦ ⎣N r′′ ⎦ q . l 4e

(5.84)

On substituting (5.76) and (5.84) into (5.82) we get the element strain energy 1 Ue = 2

{q }

e T

8E I e l 3e

+1

∫

T e ⎣N r′′ ⎦ ⎣N r′′ ⎦ d r { q }

(5.85)

−1

which has the form Ue =

1 2

{ q } [ k ] { q }. e T

e B

e

(5.86)

Comparing (5.85) with (5.86) we obtain the element stiffness matrix due to bending

[ ]= k Be

8EI e l 3e

+1

∫

T ⎣N r′′ ⎦ ⎣N r′′ ⎦ d r

(5.87)

−1

or +1

[ k ] = 8lEI ∫ e B

e

3 e

−1

⎡ (N1′′)2 N1′′N 2′′ ⎢ 2 ⎢ N 2′′ N1′′ (N 2′′ ) ⎢ N ′′N ′′ N ′′N ′′ 3 2 ⎢ 3 1 ⎢⎣ N 4′′ N1′′ N 4′′ N 2′′

N1′′N 3′′ N 2′′ N 3′′ (N 3′′ )2 N 4′′ N 3′′

N1′′N 4′′ ⎤ ⎥ N 2′′ N 4′′ ⎥ dr . N 3′′N 4′′ ⎥ ⎥ (N 4′′ )2 ⎥⎦

(5.88)


227

Substituting the shape functions (5.81) and performing the integration yields the stiffness matrix due to bending in local coordinates

[k ] e B

6l − 12 6l ⎤ ⎡ 12 ⎢ 6l 4l 2 − 6l 2l 2 ⎥ EI ⎥ . = 3e ⎢ ⎢ ⎥ l l 12 6 12 6 − − − le ⎢ 2 2⎥ − 6l 4l ⎦ e ⎣ 6l 2l

(5.89)

5.3.5 Consistent Mass Matrix of a Beam Element In dynamic calculations, the beam lateral deflection is a function of both space and time, v = v (x, t ) . The instantaneous kinetic energy of a beam element is Te =

ρ Ae 2

∫ e

2

⎛∂v⎞ ⎜⎜ ⎟⎟ d x . ⎝ ∂t ⎠

(5.90)

where ρ is the material mass per unit volume and ∂ v ∂ t = v& is the velocity at x. From (5.73) we obtain

{ }

v& = ⎣N ⎦ q& e ,

where

(5.91)

{ q& } is the column vector of nodal velocities. e

Substituting (5.76) and (5.91) into equation (5.90) yields Te =

1 2

{ q& }

e T

ρ Ae

∫

T e ⎣N ⎦ ⎣N ⎦ d x { q& }.

(5.92)

e

which has the form Te =

1 2

{ q& } [ m ] { q& }. e T

e B

e

(5.93)

⎣N r ⎦ d r

(5.94)

where

[ ]= mBe

ρ Ae l e 2

+1

∫ ⎣N ⎦ r

T

−1

is the element consistent mass matrix. On substituting the shape functions (5.81) and integrating their products we get the beam element mass matrix in local coordinates


228

[m ] e B

⎡ 156 ⎢ ρ Ae l e ⎢ 22l = 420 ⎢ 54 ⎢ ⎣ − 13l

22l

54

4l

2

13l

13l

156

− 3l 2

− 22l

− 13l ⎤ − 3l 2 ⎥⎥ . − 22l ⎥ ⎥ 4l 2 ⎦ e

(5.95)

This mass matrix is derived by the same approach as the stiffness matrix, so that it is consistent with the stiffness matrix.

5.3.6 Axial Effects The axial nodal forces are related to the nodal displacements by equation ⎧ f1 ⎫ e ⎧ q1 ⎫ (5.96) ⎨ ⎬ = kS ⎨ ⎬ ⎩ f4 ⎭ ⎩ q4 ⎭

[ ]

where the stiffness matrix (5.32) is

[ k ] = EAl e S

⎡ 1 − 1⎤ ⎢− 1 1 ⎥ . ⎣ ⎦

e

e

(5.97)

Similarly, from (5.38) we obtain the element mass matrix due to stretching

[ m ] = ρ A6 l e S

e e

⎡2 1 ⎤ ⎢1 2 ⎥ . ⎣ ⎦

(5.98)

5.3.7 Frame Element Matrices in Local Coordinates For the frame element, combining equations (5.97) and (5.89), and arranging in proper locations we get the element stiffness matrix

[k ] e

⎡ EA ⎢ l ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 =⎢ ⎢− EA ⎢ l ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢⎣

0

0

12 E I l3 6E I l2

6E I l2 4E I l

0

0

−

12 E I l3 6E I l2

6E I l2 2E I l

−

−

EA l 0 0

EA l 0 0

0 12 E I l3 6E I − 2 l

−

0 12 E I l3 6E I − 2 l

⎤ ⎥ 6E I ⎥ ⎥ l2 ⎥ 2E I ⎥ l ⎥ . ⎥ 0 ⎥ ⎥ 6E I ⎥ − 2 ⎥ l 4 E I ⎥⎥ l ⎥⎦ e 0

(5.99)


(i l )

229

The ratio of the bending terms to the stretching terms in (5.99) is of order 2

, where ‘i’ is the relevant radius of gyration. For slender beams this ratio may be as small as 1 20 or 1 50 , so the stiffness matrix may possibly be numerically ill-conditioned. Combining equations (5.98) and (5.95) and arranging in proper locations we obtain the consistent mass matrix for the frame element

[m ] e

0 0 70 0 0 ⎤ ⎡140 ⎢ 0 156 22l 0 54 − 13l ⎥⎥ ⎢ 22l 4l 2 0 13l − 3l 2 ⎥ ρ Ae l e ⎢ 0 = ⎥ . ⎢ 0 0 140 0 0 ⎥ 420 ⎢ 70 ⎢ 0 54 13l 0 156 − 22l ⎥ ⎥ ⎢ 2 0 − 22l 4l 2 ⎦⎥ e ⎣⎢ 0 − 13l − 3l

(5.100)

5.3.8 Coordinate Transformation A frame element is shown in Fig. 5.27 both in the initial and deformed state. For node 1, the local linear displacements q 1 and q 2 are related to the global linear displacements Q 1 and Q 2 by the equations q 1 = Q 1 cos α + Q 2 sin α , q 2 = −Q 1 sin α + Q 2 cos α .

.

Fig. 5.27 Equations (5.101) can be written in matrix form as

(5.101)


230

⎧ q1 ⎫ ⎧ Q1 ⎫ ⎨ q ⎬ = [ R] ⎨ Q ⎬ ⎩ 2⎭ ⎩ 2⎭

(5.101, a)

[ R ] = ⎡⎢

(5.102)

where s⎤ c ⎥⎦

c ⎣− s

is referred to as a rotation matrix and c = cos α and s = sin α . The angular displacements (rotations) are the same in both coordinate systems q 3 = Q3 .

(5.103)

Adding the similar relationships for node 2 ⎧ q4 ⎫ ⎧ Q4 ⎫ ⎨ q ⎬ = [R] ⎨Q ⎬ , ⎩ 5⎭ ⎩ 5⎭ we obtain

{ q } = [ T ] { Q }, e

where

q6 = Q6 ,

e

e

(5.104)

{q } is the element displacement vector in the local coordinate system, e

{Q } is the element displacement vector in the global coordinate system and e

[T ] e

⎡c ⎢− s ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0

s c 0 0 0 0

0 0 0 0 1 0 0 c 0 −s 0 0

0⎤ 0⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 0 1⎥⎦ 0 0 0 s c

(5.105)

is the local-to-global coordinate transformation matrix.

5.3.9 Frame Element Matrices in Global Coordinates Using the same procedure as in § 5.2.4, the stiffness and mass matrices of the frame element in global coordinates are obtained as

[ K ]= [T ] [ k ][T ] ,

(5.106)

[ M ]= [T ] [ m ][T ] .

(5.107)

e

and

e

e

e

T

T

e

e

e

e


231

5.3.10 Assembly of Stiffness and Mass Matrices The global stiffness and mass matrices, [ K ] and [ M ] , are assembled from ~ element matrices K e and M e using element connectivity matrices T e that relate the nodal displacements at element level with the nodal displacements at the complete structure level by equations of the form ~ Q e = T e { Q }. (5.108)

[ ]

[ ]

[ ]

{ } [ ]

The global uncondensed stiffness matrix is equal to the sum of the expanded element stiffness matrices [ K ] = K~ e . (5.109)

∑e [ ]

where

[ K~ ]= [T~ ] [ K ][T~ ] e

e

T

e

e

Similarly, the global uncondensed mass matrix [ M expanded element mass matrices as [ M ] = M~ e = T~ e T M e T~ e .

∑e [

] ∑ [ ] [ ][ ] e

(5.110)

] is assembled from (5.111)

For grounded systems the unreduced stiffness and mass matrices [ K

[ M ] are condensed using the boundary conditions.

] and

The effect of lumped masses and springs can be accounted for by adding their values along the main diagonal at the appropriate locations in the respective matrices. When the external loads include distributed forces, they are replaced by kinematically equivalent nodal forces, calculated by a procedure which is consistent with the derivation of the stiffness matrix and the mass matrix, assuming the validity of the static shape functions. The equations of motion can be regarded as having been derived as soon as the mass matrix, the stiffness matrix and the force vector have been derived.

Example 5.10 Calculate the first 15 natural frequencies and mode shapes for the plane frame shown in Fig. 5.28 where E = 207 GPa , ρ = 7810 kg m 3 , I = 271 mm 4 and A = 80.6 mm 2 for all beams. The frame is 606.9 mm wide and 606.9 mm high and is made up of two vertical columns and two equi-spaced cross beams. Solution. Due to symmetry, only half of the frame can be considered, with appropriate (separate) constraints for symmetric and antisymmetric modes.


232

Each half-plane was modelled with 16 identical planar beam elements, 8 for the column and 4 for each half-cross beam.

Fig. 5.28 The lowest 15 natural frequencies, in rad sec , calculated using the computer program VIBFRAME, have the following values: 107.20, 377.47, 397.25, 475.73, 1099.3, 1316.2, 1504.0, 1911.6, 2061.4, 2447.5, 2695.0, 2903.7, 4171.1, 4618.3 and 4943.6. MATLAB

The mode shapes are shown in Fig. 5.28.

Example 5.11 Calculate the first 15 natural frequencies and mode shapes of the planar vibrations for the frame shown in Fig. 5.29 where E = 210 GPa ,

ρ = 7850 kg m 3 , I = 1.055 ⋅ 10 −7 m 4 , A = 3.73 ⋅ 10 −4 m 2 and l = 0.5 m .

Fig. 5.29


233

Fig. 5.30 Solution. Each segment of length l was modelled by 5 identical beam elements, resulting in 85 elements for the complete frame, hence 86 nodes. With four fixed nodes and three degrees of freedom per node, the condensed system matrices were of order 246. The mode shapes calculated using the program


234

VIBFRAME are presented in Fig. 5.30 together with the rounded values of the

respective natural frequencies.

5.4 Grillages Grids or grillages are planar structural systems subjected to loads applied normally to their plane. They are special cases of tree-dimensional frames in which each joint has only three nodal displacements, a translation and two rotations, describing bending and torsional effects.

Fig. 5.31

5.4.1 Finite Element Discretization The grid is divided into elements, as shown in Fig. 5.31. Each node has three degrees of freedom, two rotations and a linear displacement. Typically, the degrees of freedom of node i are Q3 i − 2 , Q3 i −1 and Q3 i , defined as the rotation about the X axis, the rotation about the Y axis and the displacement along the Z axis, respectively. Nodes are located by their coordinates in the global reference frame XOY and element connectivity is defined by the indices of the end nodes. Elements are modelled as uniform bars with bending and torsional flexibility, without shear deformations and not loaded between ends. Their properties are the flexural rigidity E I , the torsional rigidity G I t , the mass per unit length ρ A and the length l . Only cross sections whose shear centre coincides with the centroid are considered.


235

5.4.2 Element Stiffness and Mass Matrices in Local Coordinates Consider an inclined grid element, as illustrated in Fig. 5.32, a, where the nodal displacements are also shown.

Fig. 5.32 In a local physical coordinate system, the x axis, oriented along the beam, is inclined an angle α with respect to the global X axis. The z axis for the local coordinate system is collinear with the Z axis for the global system. Alternatively, an intrinsic (natural) coordinate system can be used. The vector of element nodal displacements is

{q }= {q , q e

1

2,

q3, q4, q5, q6

}T

(5.112)

and the corresponding vector of element nodal forces can be written

{ f }= { f , f e

In (5.113)

f 3 and

1

2,

f 3, f 4, f 5, f 6

}T .

(5.113)

f 6 are transverse forces, while f 2 and

f 5 are

couples producing bending (Fig. 5.32, b). The corresponding displacements q 3 , q 6 are translations while q 2 , q 5 are rotations. Their column vectors are related by the flexural stiffness matrix. Rearranging the matrix (5.89) we obtain ⎧ ⎪ ⎪ ⎨ ⎪ ⎪⎩

⎡ 4l 2 6l 2l 2 − 6l ⎤ ⎧q 2 ⎫ f2 ⎫ ⎥ ⎪ ⎪ ⎢ f 3 ⎪⎪ EI e ⎢ 6l 12 6l − 12 ⎥ ⎪q 3 ⎪ = ⎨ ⎬. ⎬ f 5 ⎪ l3e ⎢ 2l 2 6l 4l 2 − 6l ⎥ ⎪q 5 ⎪ ⎥ ⎢ f 6 ⎪⎭ ⎢⎣− 6l − 12 − 6l 12 ⎥⎦ e ⎪⎩q 6 ⎪⎭

(5.114)


236

Likewise, the above nodal forces are related to the respective nodal accelerations by the mass matrix. Rearranging the matrix (5.95) we can write ⎧ ⎪ ⎪ ⎨ ⎪ ⎪⎩

f2 ⎫ f 3 ⎪⎪ ρ Ae l e ⎬= f5 ⎪ 420 ⎪ f6 ⎭

⎡ 4l 2 ⎢ ⎢ 22l ⎢ − 3l 2 ⎢ ⎢⎣ 13l

13l ⎤ ⎧q&& 2 ⎫ − 3l 2 ⎥ ⎪ ⎪ 156 − 13l 54 ⎥ ⎪ q&& 3 ⎪ ⎨ ⎬. − 13l 4l 2 − 22l ⎥ ⎪q&& 5 ⎪ ⎥ 54 − 22l 156 ⎥⎦ e ⎪⎩q&& 6 ⎪⎭ 22l

(5.115)

The axial nodal forces f 1 , f 4 are torques and the nodal displacements q 1 , q 4 are twist angles. They describe torsional effects so that their action is decoupled from bending. The respective stiffness and mass matrices can be calculated separately. The derivation of these matrices is essentially identical to the derivation of the stiffness and mass matrices for axial effects in a frame element or in a truss element. The twist angle can be expressed in terms of the shape functions (5.24) as

θ (r ) = N1 (r ) q 1 + N 2 (r ) q 4 G It e

which substituted into the strain energy U e =

2

∫ e

[ ]

2 G It e = le

kte

change of coordinates, the stiffness matrix

2

⎛ ∂θ ⎞ ⎜⎜ ⎟⎟ d x yields, after the ⎝ ∂x ⎠ +1

∫ ⎣N ′ ⎦ r

T

⎣N r′ ⎦ d r .

−1

As a consequence of this analogy, the nodal forces are related to the nodal displacements by equation ⎧ f1 ⎫ e ⎧ q1 ⎫ (5.116) ⎨ ⎬ = kt ⎨ ⎬ ⎩ q4 ⎭ ⎩ f4 ⎭

[ ]

where the stiffness matrix for torsional effects is

[ k ] = GlI e t

te

e

⎡ 1 − 1⎤ ⎢− 1 1 ⎥ . ⎣ ⎦

(5.117)

In (5.117), G is the shear modulus of elasticity and I t e is the torsional constant of the cross section. For axially symmetrical cross sections the latter is the polar second moment of area. Similarly, the element consistent mass matrix due to torsion is

[ m ] = ρ I6 l e t

te e

⎡2 1 ⎤ ⎢1 2 ⎥ . ⎣ ⎦

(5.118)


237

For the grid element, combining the stiffness matrices from equations (5.114) and (5.117), we get the stiffness matrix in local coordinates relating the nodal forces (5.113) and the nodal displacements (5.112)

[k ] e

0 0 0 0 ⎤ −a ⎡ a ⎢ 0 4l 2 6l 2 0 2l − 6l ⎥⎥ ⎢ 6l 12 0 6l − 12 ⎥ EI ⎢ 0 = 3e ⎢ ⎥ 0 0 a 0 0 ⎥ l e ⎢− a ⎢ 0 2l 2 6l 0 4l 2 − 6l ⎥ ⎢ ⎥ ⎢⎣ 0 − 6l − 12 0 − 6l 12 ⎥⎦ e

(5.119)

where a = G I t e l 2e E I e . Combining the mass matrices from equations (5.115) and (5.118) we obtain the consistent mass matrix for the grid element in local coordinates

[m ] e

0 ⎡2b ⎢0 4l 2 ⎢ ρ Ae l e ⎢ 0 22l = ⎢ 0 420 ⎢ b ⎢ 0 − 3l 2 ⎢ 13l ⎣⎢ 0

0 ⎤ 13l ⎥⎥ 156 0 54 ⎥ ⎥ 0 2b 0 0 ⎥ 4l 2 − 22l ⎥ − 13l 0 ⎥ 54 0 − 22l 156 ⎦⎥ e 0 22l

b 0

0 − 3l 2 − 13l

(5.120)

where b = 70 I t e Ae .

5.4.3 Coordinate Transformation It is necessary to transform the matrices (5.119) and (5.120) from the local to the global system of coordinates before their assemblage in the corresponding matrices for the complete grid. As has been indicated, the z direction for local axes coincides with the Z direction for the global axes, so that only the rotational components of displacements should be converted. The transformation of coordinates is defined by equation (5.104)

{ q } = [ T ] { Q }, e

e

e

(5.121)

{ } is the element displacement vector (5.112) in the local coordinate

where q e system,

{ Q }= { Q , Q e

1

2,

Q3, Q4 , Q5, Q6

}T


238

is the element displacement vector in the global coordinate system (Fig. 5.32) and

[T ] e

⎡c ⎢− s ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎣⎢ 0

0 0⎤ 0 0 0⎥⎥ 0 0 0⎥ ⎥, c s 0⎥ − s c 0⎥ ⎥ 0 0 1⎦⎥

s 0 c 0

0

0 1 0 0 0 0 0 0

(5.122)

where c = cos α and s = sin α , is the local-to-global coordinate transformation matrix (5.105). The same transformation matrix (5.122) serves to transform the nodal forces from local to global coordinates.

5.4.4 Element Stiffness and Mass Matrices in Global Coordinates Using the same procedure as in § 5.2.4 and § 5.4.5, we obtain the stiffness and mass matrices of the grid element in global coordinates as

[ K ]= [T ] [ k ][T ] ,

(5.123)

[ M ]= [T ] [ m ][T ] .

(5.124)

e

and

e

e

e

T

T

e

e

e

e

They are used to assemble the unreduced global stiffness and mass ~ matrices, [ K ] and [M ] , using element connectivity matrices T e that relate the nodal displacements at element level with the nodal displacements at the complete structure level by equations of the form (5.108).

[ ]

For grounded systems the unreduced matrices [ K ] and [ M ] are then condensed using the boundary conditions. The effect of lumped masses and springs can be accounted for by adding their values along the main diagonal at the appropriate locations in the respective matrices.

Example 5.12 Calculate the first 9 natural frequencies and mode shapes of the plane grid shown in Fig. 5.33, supported by four springs of stiffness k = 1000 N m each. The system properties are E = 210 GPa , G = 81 GPa , ρ = 7900 kg m 3 , l = 0.5 m and the diameter is d = 20 mm for all beams.


239

Fig. 5.33 Solution. The grid was modelled with 10 elements and 8 nodes, having thus 24 degrees of freedom. The mode shapes calculated using the MATLAB program VIBGRID are presented in Fig. 5.34 together with the rounded values of the respective natural frequencies.

Fig. 5.34 The first three modes of vibration are ‘rigid body’ modes, of rocking and bouncing of the undeformed grid on the suspension springs. Mode 4 is called ‘first bending’ (two transverse nodal lines), mode 5 is ‘first torsion’ (one longitudinal nodal line), mode 6 is ‘second bending’ (three transverse nodal lines), mode 7 is ‘second torsion’, mode 8 is ‘third torsion’ and mode 9 is ‘third bending’ (four transverse nodal lines).


240

Example 5.13 The grid shown in Fig. 5.35 is fixed at points 1 and 2, and has l = 1 m ,

I = 0.785 ⋅ 10 −8 m 4 , I t = 1.57 ⋅ 10 −8 m 4 , A = 3.14 ⋅ 10 −4 m 2 , ρ = 7900 kg m 3 , E = 210 GPa and G = 81 GPa . Calculate the first 9 natural frequencies and mode shapes of vibration.

Fig. 5.35 Solution. The grid was modelled with 14 elements and 8 nodes, having 18 dof’s. The mode shapes calculated using the program VIBGRID are presented in Fig. 5.36 together with the rounded values of the respective natural frequencies.

Fig. 5.36


241

Mode 1 is ‘first bending’, mode 2 is ‘first torsion’, mode 3 is ‘second bending’, mode 4 is ‘second torsion’, mode 5 is ‘third bending’ and mode 6 is a sort of a ‘first bending’ about the longitudinal axis.

5.5 Frequency Response Functions The preceding sections were restricted to the calculation of normal modes of vibration for low order undamped vibrating systems. It is of interest to study the frequency response of damped systems and to illustrate displays of their frequency response functions (FRFs). The analysis of experimentally obtained FRFs is presented in a next chapter devoted to Experimental Modal Analysis and Identification of Vibrating Systems.

5.5.1 Frequency Response Function Matrix For a viscously damped system, the Frequency Response Function (FRF) matrix is defined as (4.143)

[ H (iω )] = [ω 2 [ m ] + iω [c ] + [ k ] ]

−1

(5.125)

where ω is the driving frequency. The FRF matrix can be written ⎡ h 11 (iω ) h 12 (iω ) ⎢ h (iω ) h (iω ) 22 [ H (iω )] = ⎢⎢ 21. . . ... ⎢ ⎢⎣ h n1 (iω ) h n 2 (iω )

. . . h1n (iω ) ⎤ . . . h 2n (iω ) ⎥⎥ ... ... ⎥ ⎥ . . . h nn (iω ) ⎥⎦

(5.126)

where h j l (iω ) is the response measured at the coordinate j produced by a harmonic force applied at coordinate l . There are several frequency response functions, depending on whether the response is a displacement, velocity or an acceleration (see § 2.4.7 ). In complex notation, if the excitation f ( t ) = ˆf eiω t produces the response x j ( t ) = ˆx j e

i (ω t +φ )

l

l

, then the complex receptance can be expressed in terms of its

real and imaginary components h j l (iω ) =

ˆf ˆf ˆf fl = l eiφ = l cosφ + i l sinφ = h Rjl (ω ) + i h Ij l (ω ) . ˆx j ˆx j x j ˆx j


242

The ratio f l xl is called a direct receptance (driving point receptance), while the ratio f l x j ( j ≠ l ) is termed a transfer receptance or a cross receptance depending on whether j is a coordinate at a different location or at the same point as l but in another direction. Note that FRFs are measurable quantities that can be obtained by sinusoidal test techniques.

5.5.2 Frequency Response Function Plots FRF data are complex and thus there are three quantities – the frequency plus the real and imaginary parts (or modulus and phase) of the complex function – to be displayed.

Fig. 5.36 The three most common graphical formats are: a) the Bode plots (a pair of plots of modulus versus frequency and phase versus frequency); b) the Co-Quad plots (real component versus frequency and imaginary component versus frequency); and c) the Nyquist plots (imaginary versus real part, eventually with marks at equal frequency increments).


243

Plotted quantities include motion/force-type functions like receptance, mobility and inertance (accelerance), as well as their inverses, dynamic stiffness, impedance and dynamic mass (see § 2.4.7 ). Scales can be linear, logarithmic and semilogarithmic, with or without gridlines, with one or two cursors, zoom capabilities and overlay facilities.

Fig. 5.37 Figure 5.36 shows Bode plots for the receptance FRF of a three dof damped structure, with the displacement response measured at dofs 1, 2, and 3, due to forcing applied at dof 1. Note the logarithmic vertical scale for the magnitude plots, necessary to reveal the details at the lower levels of the response exhibited at antiresonances. Figure 5.37 shows the respective plots of the real (coincident) and imaginary (quadrature) components of receptance, while the first column in Fig. 5.38 shows the corresponding Nyquist plots. It is seen that for lightly-damped systems with relatively well-separated natural frequencies, the response near resonances can be approximated by circular loops. For comparison, mobility and inertance plots are presented in the second and third columns of Fig. 5.38, for the same FRFs. Note their rotation by 90


244

degrees in the complex plane, due to the phase relationship between displacement, velocity and acceleration.

Fig. 5.38 For the point receptance h11 , a three-dimensional plot is illustrated in Fig. 5.39, together with the companion coincident, quadrature, and Nyquist plots.

Fig. 5.39


245

For lightly damped systems with low modal density, peaks in the FRF plots of the magnitude of either modulus or the quadrature component of receptance, mobility or accelerance indicate resonances. A system with n degrees of freedom, having underdamped modes of vibration, will exhibit n resonances. Visual inspection of FRF plots helps in estimating the system order, by counting the resonance peaks within the frequency range of interest. As shown for two-degree-of-freedom systems (Fig. 4.40), FRF curves for systems with high damping and/or close natural frequencies may not exhibit all resonances as peaks. For systems having moderate damping, best resonance location is achieved on Nyquist plots, at the points where the rate of change of phase angle (and the arc length) with respect to frequency has a local maximum. If marks are made on such a plot at equal frequency increments, resonance is located at maximum spacing between successive points (as in Fig. 4.39). Antiresonances appear as deep troughs in FRF modulus plots of the type motion/force. In a driving point FRF, resonances and antiresonances alternate (Foster’s reactance theorem). Between any two resonance peaks there is a deep antiresonance through. In a transfer FRF, some or all antiresonances are replaced by minima (shallow throughs). The number and location of antiresonances change in FRFs plotted for different response points and directions. Figure 5.40 is an overlay of three receptance curves calculated for three different directions at the same point of a structure. It illustrates the coincidence of resonance peaks and the different number and location of antiresonance troughs. The upper curve, where resonances and antiresonances alternate, is a drive-point receptance.

Fig. 5.40


246

h 3,15

Figure 5.41 illustrates the frequency response diagrams of the receptance calculated for the 15-dof system of Example 5.8, shown in Fig. 5.17.

a

b

c

d Fig. 5.41

Comparing the frequencies of the resonance peaks with the natural frequencies given in Table 5.2, the cluster of five higher frequencies is first seen, despite the low level of the FRF magnitudes, due to the logarithmic scale (Fig. 5.41, a). From the other ten resonances, only nine can be distinguished, the close modes at 53.35 Hz and 53.42 Hz giving a single peak, while the resonances at 59.45 Hz and 73.72 Hz being almost smeared.

a

b Fig. 5.42


247

The appearance of resonance peaks is dependent on the frequency resolution of response data, but for a different selection of the response measurement and excitation points, the function h11 in Fig. 5.42, a and the function h 6 6 in Fig. 5.42, b, some peaks are missing.

Exercises 5.E1 Find the natural frequencies and mode shapes of the system shown in

Fig. 5.E1. Let k = 106 N m , m 1 = m 4 = m 7 = 0.5 kg , m 2 = m 5 = m 9 = 1.5 kg , m 3 = m 6 = m 8 = 1 kg .

Fig. 5.E1 Answer: f 1 = 73.5 Hz , f 2 = 124.62 Hz , f 3 = 174.78 Hz , f 9 = 433.46 Hz . 5.E2 Calculate the natural frequencies and mode shapes of the system shown in Fig. 5.E2, considering m j = 1 kg ( j = 1, ...,12) , k j = 100 N m

( j = 1,...,11) ,

k 12 = 102 N m . Compare the results with those obtained for

k 12 = 100 N m .

Fig. 5.E2 Answer: a) f 2 = 0.8238 Hz , f 3 = 0.8285 Hz , f 4 = 1.5915 Hz , f 5 = 1.5941 Hz .


248

b) f 2 = f 3 = 0.8238 Hz , f 4 = f5 = 1.5915 Hz , f 6 = f 7 = 2.2508 Hz . 5.E3 For the 10-dof system from figure 5.E3 find the damped natural frequencies and the modal damping ratios. Calculate the undamped natural frequencies and plot the corresponding mode shapes. System parameters: m 1 = 1.21 , m 2 = 1.44 , m 3 = 1.69 , m 4 = 1.96 , m 5 = 2.25 , m 6 = 2.56 ,

m 7 = 2.89 ,

m 8 = 3.24 ,

m 9 = 3.61 ,

m 10 = 4 kg ,

k 1 = 1100 ,

k 2 = 1200 ,

k 3 = 1300 , k 4 = 1400 , k 5 = 1500 , k 6 = 1600 , k 7 = 1700 , k 8 = 1800 , k 9 = 1900 , k10 = 2000 N m , c j = 0.002 k j

( j = 1,...,10) .

Fig. 5.E3 Answer: f 1 = 0.718 Hz , ζ 1 = 0.45% , f 10 = 8.34Hz , ζ 10 = 5.24% . 5.E4 For the 15-dof system from figure 5.E4 find the damped natural frequencies and the modal damping ratios.

Fig. 5.E4 Answer: f 1 = 2.65 Hz , ζ 1 = 0.81% , f 15 = 156.73Hz , ζ 15 = 0.44% .


249

5.E5 A ladder-type structure is fixed at both ends as shown in Fig. 5.E5. The mass of each step is m and the lateral stiffness of each section between steps is k. Determine the natural frequencies and plot the mode shapes.

Answer: ω 1 = 0.517 k m , ω 2 = k m , ω 3 =1.414 k m , ω 4 =1.732 k m ,

Fig. 5.E5

Fig. 5.E6

5.E6 A four-cylinder engine with flywheel is used to drive an electrical generator. For the prediction of torsional natural frequencies, the combined system is represented by the equivalent system shown in Fig. 5.E6. Determine the first four natural modes of vibration and plot the mode shapes using the following values: K 12 = K 23 = K 34 = 2.2 MN m rad , K 45 = 0.9 MN m rad , K 56 = 0.24 MN m rad ,

J 1 = J 2 = J 3 = J 4 = 0.51 kg m 2 , J 5 = 5.3 kg m 2 , J 6 = 4.4 kg m 2 . Answer: f 2 = 46.21 Hz , f 3 = 106.11 Hz , f 4 = 291.41 Hz , f 5 = 481.45 Hz . 5.E7 An eight-cylinder diesel engine with flywheel, transmission gear and ship propeller is represented by the torsional system shown in Fig. 5.E7. Determine n A = 405 rpm , the first three eigenmodes of torsional vibrations. Use

nB = 225 rpm ,

J 9 = 602 kg m 2 ,

J 12 = 4331 kg m 2 ,

J 10 = 91 kg m 2 ,

J 1 = J 2 = . . . = J 8 = 184 kg m 2 ,

J 11 = 1953 kg m 2 , K 89 = 116 MN m rad ,

K 9 ,10 = 13.5 MN m rad ,

K 10 ,11 = 185 MN m rad

K 11,12 = 0.846 MN m rad ,

K 12 = K 23 = ... = K 78 = 84 MN m rad

1977).

(reduced

to

nB ),

(G. Ziegler,


250

Fig. 5.E7

Answer: f 2 = 2.68 Hz , f 3 = 20.60 Hz , f 4 = 35.81 Hz . 5.E8 Determine the two lowest (non-zero) natural frequencies of the torsional system shown in Fig. 5.E8, for the following values of the moments of inertia and torsional stiffnesses: K 1 = 2 MN m rad , K 2 = 1.6 MN m rad ,

K 3 = 1 MN m rad ,

K 4 = 4 MN m rad ,

J 1 = 15 kg m 2 ,

J 2 = 10 kg m 2 ,

J 3 = 18 kg m 2 , J 4 = 6 kg m 2 . The speed ratio of drive shaft to axle is 4 to 1.

Fig. 5.E8

Answer: f 2 = 55.26 Hz , f 3 = 62.32 Hz . 5.E9 The gear-and-shaft system for driving the two paddles of a commercial feed mixer is illustrated in Fig. 5.E9. Each bevel gear has 20 teeth (20T), two of the spur gears have 36 teeth (36T) and the common spur gear has 12 teeth (12T). Determine the natural frequencies of torsional vibrations. Use J 1 = 22.6 ⋅ 10 −3 kg m 2 , J 2 = J 3 = 5.65 ⋅10 −3 kg m 2 , J 4 = 11.3 ⋅ 10 −3 kg m 2 ,

J 5 = J 6 = 56.5 ⋅ 10 −3 kg m 2 ,

J 7 = J 8 = 90.4 ⋅ 10 −3 kg m 2 ,

l 3 = 1.016 m , G = 81 GPa , d = 19 mm (James et al., 1989).

l 1 = l 2 = 0.915 m ,


251

Fig. 5.E9

Answer: f 2 = 16.90 Hz , f 3 = 18.41 Hz , f 4 = 37.39 Hz , f 5 = 79.61 Hz . 5.E10 Calculate the natural frequencies and the mode shapes for the flexural vibrations of the massless beams shown in Fig. 5.E10.

Fig. 5.E10

Answer: a) The flexibility matrix is ⎡ 32 ⎢ m l 3 ⎢ 45 [δ ] = 3750 E I ⎢ 40 ⎢ ⎣ 23 The

natural

frequencies

45 40 23 ⎤ 72 68 40 ⎥⎥ . 68 72 45 ⎥ ⎥ 40 45 32 ⎦ are

ω 1 = 4.4126 E I m l3 ,

ω 2 =17.611 E I m l 3 , ω 3 = 38.986 E I m l 3 , ω 4 = 64.1997 E I m l 3 . b) ω 1 = 0.647 E I m l 3 , ω 2 = 2.482 E I m l 3 , ω 3 = 5.65 E I m l 3 .


252

5.E11 Determine the first eight natural modes of vibration of the plane trusses from Fig. 5.E11. The pin-ended bars have cross-section Φ 6 x 1 mm and

l = 0.3 m , E = 72.7 GPa , ρ = 3100 kg m 3 . For all lumped masses m = 0.6 kg .

Fig. 5.E11

Answer: a) 20.51, 83.31, 111.49, 164.82, 215.10, 265.74, 299.27, 361.48 Hz. b) 20.02, 79.07, 117.04, 163.05, 231.06, 272.94, 290.24, 358.46 Hz. 5.E12 For the pin-jointed framework from Fig.5.E12 determine the lowest six natural frequencies and the corresponding mode shapes, using E = 210 GPa ,

ρ = 7800 kg m 3 , l = 2 m , A = 2000 mm 2 .

Fig. 5.E12

Answer: f 1 = 74.61 Hz , f 2 = 142.24 Hz , f 3 = 237.22 Hz , f 6 = 529.15 Hz . 5.E13 Determine the lowest 20 eigenmodes of the continuous four-span beam from Fig. 5.E13. Comment the clustering of natural frequencies. Consider

E = 208 GPa , ρ = 7850 kg m 3 , l = 1 m and rectangular cross-section with b = 2h = 40 mm .

Fig. 5.E13


253

Answer: 46.69, 54.48, 72.98, 94.33 Hz, 187.48, 203.59, 237.82, 274.84 Hz, 427.85, 452.63, 504.23, 557.66 Hz,

829.07, . . . , 1043.98 Hz.

5.E14 A two-span simply supported beam (Fig. 5.E14) has a rotational spring at the middle support. For convenience, its stiffness is defined as K = 500 EI l , where E I is the beam flexural rigidity and l is the total beam length. a) Calculate the mode shapes of the flexural vibrations of the beam with equal span lengths (Δ = 0 ) . Notice the clustering of natural frequencies. b) Consider a disordered weakly coupled beam, taking the span lengths equal to l (0.5 − Δ ) and l (0.5 + Δ ) , where Δ = 0.01 . Explain the mode localization in disordered mechanical structures. c) Extend the analysis to disordered three-span and four-span beams.

Fig. 5.E14 5.E15 A simply supported uniform beam has a length 3 m , mass per unit

length 15 kg m , and bending rigidity 4 kNm 2 with E = 200 GPa . Represent the beam by four finite elements and calculate the first four natural frequencies assuming the mass to be: a) concentrated at the mid point of each element; b) the mass of each element placed half at each end point; c) distributed along the beam (consistent mass matrix). Compare the results against those found by Rayleigh’s method, using a “sine” function for the deflection curve, and by Southwell’s formula. Plot the mode shapes for Case (c).


254

5.E16 Using the finite element method, find estimates for the lowest two natural frequencies for bending vibrations of a uniform cantilevered beam. Model it by 1, 2, 3, 4, 5 and 6 equal-length elements and compare the frequencies so obtained with the exact values. 5.E17 A uniform steel shaft of diameter d = 40 mm and length l = 1.2 m is supported by two bearings of stiffnesses k 1 = 4 N μ m and k 2 = 6 N μ m , as

shown in Fig. 5.E17. The two discs have masses m 1 = 400 kg and m 2 = 600 kg , and radii of gyration r1 = 0.3 m and r 2 = 0.5 m , respectively. Determine: a) the natural frequencies of the flexural vibrations, and b) the first four mode shapes, taking E = 210 GPa and ρ = 7850 kg m 3 .

Fig. 5.E17

Answer:

5.E18 The ends of the solid-steel roll shown in Fig. 5.E18 are simply supported in short bearings. a) Estimate the first natural frequency of flexural vibrations using Rayleigh’s method. b) Check the result against that find using a lumped parameter model. c) Estimate the first two natural frequencies using the program VIBFRAME, taking E = 210 GPa and ρ = 7850 kg m 3 .


255

Fig. 5.E18

Answer: f 1 = 36.76 Hz , f 2 = 120.43 Hz . 5.E19 Determine the first two natural frequencies for the folded beam shown in Fig. 5.E19 and plot the mode shapes. a) Consider AB = CD = l and BC = 0 . Model the two beams by at least three finite elements. b) Consider AB = CD = l and BC = l 10 and compare the natural frequencies with those determined at (a).

Fig. 5.E19

Answer: a) f 1 = f 2 = 11.71 Hz . b) f 1 = 10.45 Hz , f 2 = 10.69 Hz .

5.E20 Find the first eight natural frequencies and mode shapes for the inplane vibrations of the plane frame shown in Fig. 5.E20 using the following values:

A vert = 0.006 m 2 , A hor = 0.004 m 2 , A diag = 0.003 m 2 , I = 0.0756 m 4 . Material properties: E = 75 GPa and ρ = 2800 kg m 3 .

Fig. 5.E20


256

Answer: Using 48 elements and 44 nodes (4 elements for a beam), the natural frequencies are 45.15, 79.07, 227.72, 249.94, 365.62, 444.03, 452.83, 476.83 Hz. 5.E21 Determine the 12 lowest eigenmodes for the double cross shown in Fig. 5.E21. Each arm is of length l = 50 m and has a square cross-section of edge 0.125 m . All the outer endpoints are simply-supported and the inner endpoints are rigidly jointed together. The material properties are E = 200 GPa and

ρ = 8000 kg m3 . Use more than two beam elements to discretize each arm.

Fig. 5.E21

Answer: f 1 = 5.6687 Hz , f 2 = f 3 = 8.8485 Hz , f 4 = ... = f8 = 8.8856 Hz .

5.E22 Consider the system from Fig. 5.E22 as a horizontal grid and determine the lowest four natural modes of vibration. Take l = 0.25 m , l1 = 0.2 m ,

E = 210 GPa , ν = 0.3 and ρ = 7850 kg m 3 . All beams have a square crosssection with the edge a = 5 mm . I t = 0.141 a 4 .


257

Fig. 5.E22

Answer: f 1 = 10.69 Hz , f 2 = 27.31 Hz , f 3 = 44.96 Hz , f 4 = 108.65 Hz .

5.E23 Determine the first six modes of coupled bending-torsion vibrations of the grillage from Fig. 5.E23. Consider l = 50 mm , E = 210 GPa , G = 80 GPa ,

ρ = 7800 kg m3 and circular cross-section with d = 5 mm .

Fig. 5.E23


258

Answer: The first six mode shapes are illustrated below.

5.E24 The grillage shown in Fig. 5.E24 is comprised as follows: a) a

horizontal steel shaft ABC of flexural rigidity 2.5 ⋅ 107 Nm 2 , torsional rigidity 1.92 ⋅ 10 7 Nm 2 and mass per unit length 300 kg m ; b) two horizontal steel shafts

EBD and FCG, rigidly attached to ABC, and having a flexural rigidity 5 ⋅ 107 Nm 2 and mass per unit length 500 kg m ; c) masses of 500 kg at D and E, and masses of 250 kg at F and G. You may neglect the moments of inertia of these masses. Consider E = 210 GPa , G = 81 GPa , ρ = 7850 kg m 3 . Find the first six natural modes of vibration.

Fig. 5.E24

Answer: 2.35, 10.39, 13.39, 36.16, 63.25, 105.22 Hz.

6. CONTINUOUS SYSTEMS

When the mass and elasticity are distributed throughout the vibratory system, a description of its configuration requires an infinite number of coordinates. Such a system may be regarded as having an infinite number of degrees of freedom, hence an infinite number of natural frequencies and mode shapes. In this chapter the vibrations of bars with distributed mass and elasticity are considered, assuming them to be composed of homogeneous, isotropic and elastic materials. The treatment is restricted to only a few particular cases useful as benchmark examples for the problems solved using the computer programs mentioned in the previous chapter.

6.1 Lateral Vibrations of Thin Beams In the analysis which follows we shall adopt the assumptions usual in simple bending theory, that is, the Bernoulli-Euler beam theory: a) the beam is initially straight; b) the depth of the beam is small compared to its radius of curvature at its maximum displacement; c) plane sections remain plane at all phases of an oscillation; d) one principal axis of inertia is perpendicular to the direction of motion in the vibration; e) the beam is free from longitudinal force; f) the deformation due to shearing of one cross section relative to an adjacent one is negligible; g) the beam mass is concentrated at its neutral axis. The last two imply that the effects of transverse shear and rotatory inertia are neglected. 6.1.1 Differential Equation of Motion The beam lateral deflection is a function of both space and time, v = v (x, t ) . The relationship between the bending moment and the curvature (5.65) becomes


260 M (x, t ) = E I z

∂2 v , ∂x2

(6.1)

where E is Young’s modulus for the material of the beam and I z is the second moment of area of the cross-section about the principal axis which lies in the neutral plane. From the equilibrium of an infinitesimal beam element (Fig. 6.1), the relationships between bending moment, shear force and external load are obtained.

Fig. 6.1 The shear force is given by T (x, t ) =

∂M ∂3 v = EIz . ∂x ∂x3

(6.2)

The transverse load per unit length is p (x, t ) =

∂T ∂4 v = EIz . ∂x ∂x4

(6.3)

It consists of two components, the applied external transverse load q (x , t ) and the inertial transverse load (mass per unit length times acceleration) ∂2 v − ρ A 2 . Substituting ∂t

p (x, t ) = q (x , t ) − ρ A

∂2 v ∂ t2

in equation (6.3) we obtain the differential equation of motion

EIz

∂4 v ∂2v + ρ A = q (x ,t ) . ∂ x4 ∂ t2

(6.4)


261

6.1.2 Natural Frequencies and Mode Shapes For free vibrations q ( x ,t ) = 0 . We look for conditions under which v (x, t ) is a synchronous harmonic motion

v (x, t ) = V (x ) sin (ω t + φ ) .

(6.5)

Substituting the solution (6.5) in equation (6.4), we obtain for q = 0

ρA d 4V −ω2 V =0. (6.6) 4 EIz dx Equation (6.6) is of fourth order so the general solution will contain four constants V ( x ) = B 1 sin α x + B 2 cos α x + B 3 sinh α x + B 4 cosh α x where

α 4 = ω2

ρA EIz

.

(6.7)

(6.8)

The four constants are determined from the end conditions. The standard end conditions are: a) simply supported or pinned end, for which the displacement is zero and the bending moment is zero as there is no rotational constraint; b) fixed or clamped end, for which the displacement and slope are zero; c) free end, for which the bending moment and shear force are zero. In terms of the function V (x ) these conditions for a uniform beam are: a) Simply supported end:

V = 0 and d 2 V d x 2 = 0 .

(6.9)

b) Clamped end:

V = 0 and d V d x = 0 .

(6.10)

d 2 V d x 2 = 0 and d 3V d x 3 = 0 .

(6.11)

c) Free end:

Imposing two conditions at each end we obtain a set of four homogeneous equations. A nontrivial solution is obtained by setting the determinant of their coefficients equal to zero. Expansion of the determinant and subsequent reduction gives a transcendental equation in the quantity α l whose solutions yield the natural frequencies

ωn = (α n l ) 2

EIz . ρ A l4

n = 1, 2, 3, . . .

(6.12)


262

Since equation (6.6) is homogeneous, each mode shape function Vn (x ) can describe only relative displacements of the various parts of the beam, and not absolute values. As for discrete systems, the process of assigning absolute values to the function Vn (x ) is known as normalization, and it is one of multiplying it by a suitable constant. The selection of this constant is arbitrary. We may, for example, assign a unit value to the displacement at a certain location. For the graphical display of several mode shapes it is useful to assign a unit value to the maximum displacement in any mode. Another method is to normalize the mode to unit modal mass. The natural frequencies and mode shapes will be determined in the following for some particular end conditions.

Fig. 6.2

Both ends simply supported For a simply supported beam (Fig. 6.2), the end conditions are V = 0 and d V d x 2 = 0 at x = 0 and at x = l . Substituting the conditions at x = 0 in equation (6.7) we obtain 2

0 = B 2 + B 4 and 0 = − B 2 + B 4 . Thus B 2 = B 4 = 0 . Applying the conditions at x = l , we obtain

0 = B 1 sin α l + B 3 sinh α l and 0 = − B1 sin α l + B 3 sinh α l . The only non-trivial solution is B 3 = 0 and sin α l = 0 , leading to

α l = nπ ,

n = 1, 2, 3, . . .

The first four roots of equation (6.13) are

α nl = 3.142, 6.283, 9.425, 12.566 .

n = 1, 2, 3, 4 .

(6.13)


263

Thus for a uniform beam with simply supported ends the natural frequencies are given by 4

⎛ nπ ⎞ E I z . ⎟ ⎝ l ⎠ ρA

ωn2 = ⎜

n = 1, 2, 3, . . .

(6.14)

and the modes of vibration are given by v ( x, t ) = C sin

(

)

nπ x sin ω n t + φ n , l

where C is an arbitrary constant. The first four modes of vibration, corresponding to n = 1 , 2, 3, and 4, and the location of the respective nodal points are given in Fig. 6.2.

Fig. 6.3

Cantilever With the origin at the fixed end, as in Fig. 6.3, and using equations (6.10) and (6.11), the end conditions are the following: At x = 0 , V = 0 ,

0 = B2 + B4 .

At x = 0 , d V d x = 0 .

0 = B1 + B 3 .

At x = l , d 2 V d x 2 = 0 .

0 = − B1 sin α l − B 2 cosα l + B 3 sinh α l + B 4 cosh α l . At x = l , d 3V d x 3 = 0 .

0 = − B1 cos α l + B 2 sin α l + B 3 cosh α l + B 4 sinh α l . The above four linear equations have a non-trivial solution if the following determinant vanishes


264

0

1

0

1

1

0

1

0

− sin α l − cos α l sinh α l cosh α l − cos α l sin α l cosh α l sinh α l

=0

which leads to the frequency equation

cos α l ⋅ cosh α l = −1 .

(6.15)

The first five roots of equation (6.15) are

α n l = 1.875, 4.694, 7.855, 10.996, 14.137 .

n = 1, 2, 3, 4, 5 .

(6.16)

For n greater than five the values are given approximately by

α nl =

2 n −1 π. 2

(6.16, a)

Substituting B 4 = − B 2 and B 3 = − B 1 in the last two end conditions yields

B 1 (sin α l + sinh α l ) + B 2 (cos α l + cosh α l ) = 0, B 1 (cos α l + cosh α l ) − B 2 (sin α l − sinh α l ) = 0.

Equating to zero the determinant of the above equations yields again the frequency equation (6.15). Expressing B 4 , B 3 and B 2 in terms of B 1 = C , we obtain the shape of the nth mode Vn ( x ) = C [ cosh α x − cos α x − kn ( sinh α x − sin α x ) ] where cosh α l + cos α l n = 1, 2, 3, . . . kn = , sinh α l + sin α l

(6.17) (6.18)

The first four modes of vibration and the location of the respective nodal points are given in Fig. 6.3.

Both ends free For a free-free beam (Fig. 6.4), the end conditions are

V ′′ (0) = 0 , V ′′′ (0) = 0 , V ′′ (l ) = 0 , V ′′′ (l ) = 0 .

(6.19)

Following the above procedure and using the appropriate end conditions, it can be shown that the frequency equation is

cos α l ⋅ cosh α l = +1 . The first four roots of equation (6.20) are

(6.20)


265

α n l = 4.730, 7.853, 10.996, 14.137 . n = 1, 2, 3, 4

(6.21, a)

For n greater than four the values are given approximately by

α nl =

2 n −1 π. 2

(6.21, b)

Fig. 6.4 The mode shapes are given by

Vn ( x ) = C [ sinh α x + sin α x − kn ( cosh α x + cos α x ) ] where

kn =

sinh α l − sin α l , cosh α l − cos α l

(6.22)

n = 1, 2, 3, . . .

The first four modes of vibration and the location of the respective nodal points are given in Fig. 6.4. Equation (6.6) admits other two solutions which satisfy the end conditions (6.19), namely V (x ) = B5 = const., V (x ) = B 6 x + B7 . These are the so-called “rigid body” modes of vibration. The former corresponds to the vertical bouncing, the latter to the rocking in the vertical plane, both having zero natural frequency.

Both ends clamped For a beam with clamped ends (Fig. 6.5), the end conditions are

V (0) = 0 , V ′ (0) = 0 , V (l ) = 0 , V ′ (l ) = 0 . The frequency equation is the same as (6.19) and so are the natural frequencies given by α n l = 4.730, 7.853, 10.996, 14.137 , for n = 1, 2, 3, 4 , and 2 n −1 α nl = π for n greater than four. 2


266

Fig. 6.5 The mode shapes are given by

Vn ( x ) = C [ cos hα x − cosα x − kn ( sinh α x − sin α x ) ] where

kn =

cosh α l − cos α l , sinh α l − sin α l

(6.23)

n = 1, 2, 3, . . .

(6.24)


Fig. 6.6

One end clamped and one end pinned When the beam is clamped at one end and simply supported at the other end (Fig. 6.6), the end conditions are

V (0) = 0 , V ′ (0) = 0 , V (l ) = 0 , V ′′ (l ) = 0 . The frequency equation is

tanh α l = tan α l .

(6.25)

The first four roots of equation (6.25) are

α n l = 3.927, 7.069, 10.210, 13.352 .

n = 1, 2, 3, 4 .

For n greater than four the values are given approximately by


267

α nl =

4n +1 π. 4

(6.26)

The mode shapes are given by

Vn ( x ) = C [ cosh α x − cos α x − kn ( sinh α x − sin α x ) ] where

kn =

cosh α l − cos α l , sinh α l − sin α l

n = 1, 2, 3, . . .

(6.27)

(6.28)


6.1.3 Orthogonality of Mode Shape Functions

As for all undamped vibrating systems, the mode shape functions of beams with distributed mass and elasticity are orthogonal. In the general case of beams of variable cross-section, they are orthogonal with respect to a weighting function m (x ) , where m (x ) = ρ A( x ) is the mass per unit length. Let the bending rigidity be E I (x ) . Considering the modes of indices r and s, respectively, we can write

(E I Vr′′)" −ωr2 m Vr = 0 ,

(6.29)

(E I Vs′′)" −ω s2 m Vs = 0 .

(6.30)

Multiplying equations (6.29) and (6.30) by Vs and Vr , respectively, and integrating from 0 to l yields l

l

2 ∫ (E I Vr′′)" Vs dx = ω r

∫ Vr Vs m dx ,

0

0

l

∫ (E I Vs′′)" Vr dx

(6.31)

l

∫ Vs Vr m dx .

= ω 2s

0

(6.32)

0

Subtracting equation (6.32) from (6.31) gives

(

ω 2r

− ω 2s

) ∫ V V m dx = ∫ [ (E I V ′′)" V l

l

s r

0

r

s

− (E I Vs′′)" Vr ] dx .

0

Integrating the right side of (6.33) by parts results in

(6.33)


268

(ω

2 r

− ω 2s

) ∫ V V m dx = { [V (E I V ′′)' −V (E I V ′′)' ] − E I (V ′V ′′ − V ′V ′′ ) } . l

s r

s

r

r

s

s r

r s

l 0

(6.34)

0

The right side of the equation (6.34) vanishes if, at each end of the beam, at least one of the following pairs of boundary conditions is prescribed: V =0 V =0 V′ = 0

and

E I V ′′ = 0

and

and and

V ′ = 0, E I V ′′ = 0 , (E I V ′′)' = 0,

(6.35)

(E I V ′′)' = 0.

Assuming that one of the above pairs of boundary conditions is applied at each end of the beam, equation (6.34) reduces to l

∫ m (x )Vr (x ) Vs (x )dx = 0 ,

r ≠s.

(6.36)

0

Equations of the type (6.36) are known as the orthogonality relations for the natural mode shapes of the beam. It is said that the functions Vr (x ) and Vs ( x ) are orthogonal (with each other) with respect to the weighting function m (x ) . For a uniform beam, equation (6.36) becomes l

∫ Vr (x ) Vs (x )dx = 0 ,

r ≠s.

(6.36, a)

0

Also, for a beam with any combination of the above end conditions (simply-supported, clamped and free) l

∫ E I (x ) Vr′′(x ) Vs′′ (x )dx = 0 ,

r ≠s.

0

Equations (6.36) can be used to determine the free response for given initial conditions. 6.1.4 Multi-Span Beams

In finding the natural frequencies of a beam on multiple supports, the section between each pair of supports is considered as a separate beam with its origin at left support of the section. Equation (6.7) applies to each section. At each end of the beam the usual boundary conditions are applicable, depending on the type of support. At each intermediate support the deflection is


269

zero. Since the beam is continuous, the slope and the bending moment just to the left and to the right of the support are the same.

Example 6.1 Find the frequency equation for a two span beam with the extreme ends simply supported and unequal span lengths. Solution. The equation (6.7) is written for the first segment of length l 1

( )

V1 x 1 = A1 sin α x 1 + B 1 cos α x 1 + C 1 sinh α x 1 + D 1 cosh α x 1 and for the second segment of length l 2

V2 ( x2 ) = A 2 sin α x2 + B 2 cos α x2 + C 2 sinh α x2 + D 2 cosh α x2 . The boundary conditions

V1 (0) = V1′′ (0) = 0 , V1 (l1 ) = V2 (0) = 0 , V1′ (l1 ) = V2′ (0) , V1′′ (l1 ) = V2′′ (0) ,

V2 (l 2 ) = V2′′ (l 2 ) = 0

yield the following five homogeneous algebraic equations A1 sin α l 1 + C 1 sinh α l 1 = 0 , A1 cos α l 1 + C 1 cosh α l 1 − A 2 − C 2 = 0 , − A1 sin α l 1 + C 1 sinh α l 1 − 2 D2 = 0 ,

A 2 sin α l 2 + C 2 sinh α l 2 + D 2 ( cosh α l 2 − cos α l 2 ) = 0 ,

− A 2 sin α l 2 + C 2 cosh α l 2 + D 2 ( cosh α l 2 + cos α l 2 ) = 0.

The above set of linear equations has non-trivial solutions if the determinant of the coefficients of A1 , C 1 , A 2 , C 2 and D 2 is equal to zero. This condition yields the frequency equation

sinh μ α l ⋅ sinh ( 1 − μ )α l ⋅ sin α l − sin μ α l ⋅ sin (1 − μ )α l ⋅ sinh α l = 0 , where μ = l1 l and l = l1 + l 2 . Table 6.1 gives the values of α nl in equation (6.12) for finding the first five natural frequencies of uniform continuous beams on uniformly spaced supports. The mode shapes for the first cluster of natural frequencies plus the first frequency in the next cluster are shown in Fig. 6.7 for beams with two, three, four and five equal spans, respectively.


270

Fig. 6.7


271

Table 6.1 Beam structure Nr. of spans 1 2 3 4 5 6 7 8 9 10 11 12

Roots of characteristic equation Mode index 1 3.142 3.142 3.142 3.142 3.142 3.142 3.142 3.142 3.142 3.142 3.142 3.142

2 6.283 3.927 3.550 3.393 3.299 3.267 3.236 3.205 3.205 3.205 3.173 3.173

3 9.425 6.283 4.304 3.927 3.707 3.550 3.456 3.393 3.330 3.299 3.267 3.267

4 12.566 7.069 6.283 4.461 4.147 3.927 3.770 3.644 3.550 3.487 3.424 3.393

5 15.708 9.425 6.692 6.283 4.555 4.304 4.084 3.927 3.801 3.707 3.613 3.550

6.1.5 Response to Harmonic Excitation

When a uniform beam is subjected to harmonic disturbing forces or couples, they can be accounted for through the boundary conditions, as shown in the following for a free-free beam.

Fig. 6.8 Symmetrical loading Consider a uniform free-free beam subjected to a harmonic force F 0 sin ω t acting at the middle (Fig. 6.8). Assuming a steady-state solution of the form

v (x, t ) = Y (x ) sin ω t where ω is the excitation frequency, equation (6.4) yields

(6.37)


272

Y ( x ) = C 1 sin α x + C 2 cos α x + C 3 sinh α x + C 4 cosh α x where

α 4 = ω2

ρA EIz

.

(6.38)

(6.39)

The boundary conditions for half of the beam are

Y ′′ (0) = 0 , Y ′′′ (0) = 0 , Y ′ (l 2) = 0 , E I Y ′′′ (l 2) = F 0 2 .

(6.40)

The amplitude of the forced vibrations at a location 0 ≤ x ≤ l 2 is Y (x ) =

αl

αl

2 2 (sin α x + sinh α x ) − C 4 E I α 3 cosh αl sin αl + cos αl sinh αl 2 2 2 2 αl αl + cosh cos 2 2 (cos α x + cosh α x )G − αl αl αl αl cosh sin + cos sinh 2 2 2 2 sinh

F0

− sin

(6.41)

The dynamic deflection at the beam midpoint is

αl αl 3 3 1 + cos cosh ⎛ l ⎞ F0 l ⎛ 2 ⎞ 2 2 . Y⎜ ⎟= ⎜ ⎟ ⎝ 2 ⎠ 16 E I ⎝ αl ⎠ cosh αl sin αl + cos αl sinh αl 2 2 2 2

(6.42)

Equating to zero the denominator of equation (6.42) we obtain the resonance frequencies. This yields tanh

αl 2

+ tan

αl 2

= 0.

(6.43)

The first three roots of equation (6.43) are

α n l 2 = 2.365, 5.498, 8.639 .

n = 1, 2, 3 .

The values

(α l )1s = 4.730 , (α l ) 2 s = 10.996 , (α l ) 3s = 17.278 ,

(6.44)

are the roots of equation (6.20) corresponding to the symmetrical modes of vibration.


273

Equating to zero the numerator of equation (6.42) we obtain the antiresonance frequencies. This yields 1 + cos

αl 2

cosh

αl 2

= 0,

(6.45)

whose first three roots are

α n l 2 = 1.875, 4.694, 7.855 or

n = 1, 2, 3 .

(α l ) (1 )= 3.750 , (α l ) ( 2 )= 9.388 , (α l ) ( 3 )= 15.710 .

Fig. 6.9 The driving point receptance curve for the free-free beam from Fig. 6.8 is illustrated in Fig. 6.9 using a logarithmic vertical scale. Anti-symmetrical loading

Consider a uniform free-free beam subjected to a harmonic couple M 0 sin ω t acting at the middle (Fig. 6.10).

Fig. 6.10


274

The lateral deflection is given by (6.38). The boundary conditions for half of the beam are

Y ′′ (0) = 0 , Y ′′′ (0) = 0 , Y ′ (l 2) = 0 , E I Y ′′ (l 2) = − M 0 2 .

(6.46)

The amplitude of the forced vibrations at a location 0 ≤ x ≤ l 2 is Y (x ) =

M0

4EI α

2

C

cos

αl

αl 2

+ cosh

αl

αl 2

αl

αl

(sin α x + sinh α x ) −

− cos sinh 2 2 2 αl αl sin + sinh 2 2 (cos α x + cosh α x )G − αl αl αl αl cosh sin − cos sinh 2 2 2 2

cosh

2

sin

(6.47)

Equating to zero the denominator of equation (6.47) we obtain the frequency equation tanh

αl 2

− tan

αl 2

=0.

(6.48)

The first three roots of equation (6.48) are

α n l 2 = 2.365, 5.498, 8.639 .

n = 1, 2, 3 .

The values

(α l )1a = 3.927 , (α l ) 2 a = 7.069 , (α l ) 3a = 10.21 ,

(6.49)

are the roots of equation (6.20) corresponding to the antisymmetrical modes of vibration. When the beam is acted upon by a harmonic force F 0 sin ω t applied at an arbitrary location, the equivalent loads applied at the beam middle are a force and a couple, and the forced response may be obtained by summation of solutions (6.41) and (6.47). The resonance frequencies are given by (6.12)

ωn = (α n l ) 2

EIz ρ A l4

n = 1, 2, 3, . . .

where α n l is given by (6.44) and (6.49). In can be seen that

α 1s = α 1 , α 1a = α 2 , α 2 s = α 3 , α 2 a = α 4 , α 3s = α 5 , α 3a = α 6 , . . . This is explained by writing equation (6.20) under the form


275

αl 1 + tanh 2 2 =1 2 ⋅ α α l l 1 − tanh 2 1 + tan 2 2 2 1 − tan 2

αl

which becomes

αl αl ⎞ ⎛ αl αl ⎞ ⎛ + tanh − tanh ⎜ tan ⎟ ⎜ tan ⎟ =0. 2 2 ⎠⎝ 2 2 ⎠ ⎝ Equating each factor to zero yields the frequency equations (6.43) and (6.48).

6.2 Longitudinal Vibration of Rods In a thin uniform rod of cross-section area A and mass per unit volume ρ , there are axial displacements u = u (x , t ) due to axial forces. In Fig. 6.11 a freebody diagram of an infinitesimal element of length dx of the rod is shown.

Fig. 6.11 The equation of motion in the x-direction is ⎛ ⎞ ∂N ∂ 2u ⎜⎜ N + d x ⎟⎟ − N − (ρ Ad x ) 2 = 0 , ∂x ∂t ⎝ ⎠

∂N ∂ 2u = ρA 2 , ∂x ∂t

where u is the displacement at x.

(6.50)


276

The displacement at x + d x will be u + (∂u ∂ x ) d x . The axial strain is ε x = ∂ u ∂ x , the axial stress is σ x = E ∂u ∂ x , where E is Young’s modulus of elasticity. The strain can be expressed in terms of the axial force N as ∂u N . = ∂x E A

(6.51)

Combining equations (6.50) and (6.51) yields the differential equation of free vibration ∂ 2 u ρ ∂ 2u . = ∂ x2 E ∂ t 2

(6.52)

The constant c = E ρ is the velocity of propagation of longitudinal waves along the rod. A harmonic synchronous motion is defined by a solution of the form

u (x ,t ) = U (x ) sin (ω t + φ )

(6.53)

which substituted in equation (6.52) yields d 2U + β 2U = 0 d x2

where

β 2 = ω2

ρ E

.

(6.54)

(6.55)

The general solution of equation (6.52) is

U (x ) = C 1 sin β x + C 2 cos β x .

(6.56)

The two integration constants C 1 and C 2 may be determined from the end conditions. At a fixed end the displacement is zero, u = 0 . At a free end, the stress and hence the strain is zero, ∂u ∂ x = 0 . The effect of end springs or concentrated masses can also be accounted for. Free-free rod

For a rod with both ends free ∂u ∂ x = 0 at x = 0 and x = l . We obtain C 1 = 0 and, since C 2 must be nonzero in order to have vibration, the frequency equation (characteristic equation) becomes

sin β l = 0 . Its solution consists of an infinite set of eigenvalues

(6.57)


277

β n l = nπ ,

n = 1, 2, 3, . . .

The natural frequencies (eigenfrequencies) are thus given by

ωn = n

π

E

l

ρ

n = 1, 2, 3, . . .

,

(6.58)

and the mode shapes (eigenfunctions) are defined by U n ( x ) = C cos nπ

where C is an undetermined amplitude.

x , l

The dynamic response of a continuous rod can be regarded as a superposition of motions in the eigenmodes. These are synchronous harmonic motions at the respective natural frequency with all points vibrating in phase (or out-of-phase) with the relative spatial distribution of amplitudes defined by the eigenfunction. In practical applications, the infinite series is truncated to the modes with natural frequencies within the operating frequency range of the structure. Dynamic stiffness matrix

It is of interest to derive the exact dynamic stiffness matrix for an axially vibrating bar segment (a two-noded pin-jointed element) in local coordinates and to compare it with the matrices obtained using static shape functions, (5.32) and (5.38). Denoting u 1 , u 2 , the dynamic nodal displacements and f 1 , f 2 , the dynamic nodal forces (positive in the x-direction), as in Section 5.2.1, the end conditions are

U (0) = u 1 , U (l ) = u 2 , U ′(0) = −

f1 EA

, U ′(l ) =

f2 EA

.

(6.59)

Using equation (6.56) we obtain ⎧u 1 ⎫ ⎡ 0 1 ⎤ ⎧C 1 ⎫ ⎨u ⎬ = ⎢ ⎥ ⎨ ⎬, ⎩ 2 ⎭ ⎣sin β l cos β l ⎦ ⎩C 2 ⎭

⎧ f1⎫ ⎨ f ⎬ = EAβ ⎩ 2⎭

0 ⎤ ⎧C 1 ⎫ ⎡ −1 ⎢cos β l − sin β l ⎥ ⎨C ⎬ , ⎦ ⎩ 2⎭ ⎣

and, for sin β l ≠ 0 , ⎧ f1⎫ ⎨ f ⎬ = EAβ ⎩ 2⎭

⎡ cotan β l − cosec β l ⎤ ⎧ u 1 ⎫ ⎢− cosec β l cotan β l ⎥ ⎨u ⎬ , ⎦ ⎩ 2⎭ ⎣

so that the dynamic stiffness matrix is


278

βl [ k ] = EAβ ⎡⎢−cotan cosec β l e dyn

⎣

− cosec β l ⎤ . cotan β l ⎥⎦

(6.60)

The Taylor’s series expansions up to three terms, of the coefficients of the dynamic matrix (6.60), are EAβ cotan β l =

EA ρ Al l 3 (ρ A)2 −ω2 −ω4 − .... , 3 45 E A l

− EAβ cosec β l = −

7l 3 (ρ A)2 ρ Al EA −ω2 − ω4 − .... . 6 360 E A l

It may be seen that, in each of the above series, the first term is equal to the corresponding stiffness coefficient of the matrix (5.32) and the second term is equal to − ω 2 times the coefficient of the consistent mass matrix (5.38), both calculated based on static shape functions.

(

)

If the frequency dependent stiffness matrix is

[ k ] = ElA ⎡⎢−11

− 1⎤ l ⎡ 1 7 8⎤ + ω 4 (ρ Al ) 2 ⎥ 1⎦ 1 ⎥⎦ 45 E A ⎢⎣7 8

e

⎣

and the frequency-dependent consistent mass matrix is

[ m ] = ρ 6Al ⎡⎢ 12 e

⎣

1⎤ 2l + ω 2 (ρ Al ) 2 ⎥ 2⎦ 45 E A

⎡ 1 7 8⎤ ⎢7 8 1 ⎥⎦ ⎣

it can be seen that the third term in each of the above series comes from the frequency-dependent stiffness and mass coefficients.

6.3 Torsional Vibration of Rods The equation of motion of a rod in torsional vibration is similar to that of longitudinal vibration of rods presented in the preceding section. It can be written ∂ 2θ

=

ρ ∂ 2θ

. ∂ x2 G ∂ t 2 where θ is the twist at x and G is the shear modulus of elasticity.

(6.61)

In order to calculate the natural modes of vibration, a harmonic solution is looked for, of the form

θ (x ,t ) = Θ (x ) sin (ω t + φ ) .

(6.62)


279

Upon substitution in equation (6.61) we obtain d 2Θ + γ 2Θ = 0 2 dx

where

γ 2 = ω2

ρ G

(6.63)

.

(6.64)

The general solution of equation (6.63) is

Θ (x ) = C 1 sin γ x + C 2 cos γ x .

(6.65)

The integration constants C 1 and C 2 may be determined from the end conditions. At a fixed end the rotation angle is zero, θ = 0 . At a free end, ∂θ ∂ x = 0 (as the angle of twist due to a torque M is dθ = M dx G I t ). The effect of end springs, concentrated discs, as well as concentrated forces and torques can also be accounted for. Rod with one end fixed and the other end free At x = 0 , θ = 0 , and at x = l , ∂θ ∂ x = 0 . We obtain C 2 = 0 and the frequency equation

cos γ l = 0 .

(6.66)

Equation (6.66) has an infinite set of solutions 2n + 1 π, 2

γ nl =

n = 0,1, 2, 3, . . .

The natural frequencies are thus given by

ωn =

2n + 1 π 2 l

G

ρ

,

n = 0,1, 2, 3, . . . (6.67)

and the mode shapes are defined by

Θ n (x ) = C sin where C is an undetermined amplitude.

2n + 1 x π 2 l

Dynamic stiffness matrix By analogy with axial vibrations, the dynamic stiffness matrix of a twisted rod segment is


280

γl [k ] = G I γ ⎡⎢−cotan cosec γ l e dyn

t

⎣

− cosec γ l ⎤ . cotan γ l ⎥⎦

(6.68)

where γ is given by (6.64) and G I t is the torsional rigidity of the beam crosssection. This matrix relates the nodal dynamic axial torques with the nodal dynamic twist angles. Using Taylor series expansions as in Section 6.2, the following frequencydependent stiffness and mass matrices are obtained

[ k ] = GlI e

t

⎡ 1 − 1⎤ l 2 4 ⎢− 1 1 ⎥ + ω (ρ I t l ) 45G I ⎣ ⎦ t

[ m ] = ρ 6I l ⎡⎢ 12 e

t

⎣

1⎤ 2l + ω 2 (ρ I t l ) 2 ⎥ 2⎦ 45G I t

⎡ 1 7 8⎤ ’ ⎢7 8 1 ⎥⎦ ⎣ ⎡ 1 7 8⎤ . ⎢7 8 1 ⎥⎦ ⎣

6.4 Timoshenko Beams The results obtained in section 6.1 apply to slender beams, for which rotary inertia and deformation due to shear is neglected. The analysis of beam vibration including these effects, usually known as the Timoshenko beam theory, is presented in a next chapter. These effects lower the natural frequencies compared to the values obtained from equation (6.6). For a uniform beam with both ends simply supported, the values given by the approximate equation (6.14) have an error of less than 5% provided n r l < 0.08 , where r = I A is the radius of gyration and n is the mode number. A feature of Timoshenko beams is the existence of two distinct natural frequencies for each mode shape taken up by the centre line of the beam. For the lower of each pair of frequencies, deformations due to bending and shear are in phase with each other and are summed to give the total lateral deflection. In the higher frequency case, the bending and shear deformations are out-of-phase, with the net transverse deflection equal to their difference. In the Timoshenko theory, the basic assumption is that plane-cross-sections remain plane throughout vibration, but they are not perpendicular to the beam centre line, but rotated by a shear angle which is considered constant in all points (neglecting warping due to shear deformation). This angle is calculated using an effective area obtained multiplying the actual cross-section area by a shear factor which is a function of Poisson’s ratio and cross section geometry.

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Index Absorber, dynamic vibration 117 Accelerance 53, 56 Accelerometer 65 Additional mass 51 Amplitude 13 − complex 53 − distortions 65 − of forced vibration 24, 124 − resonant 45, 49 Antiresonance 31, 117 Aperiodic motion 39 Apparent mass 54 Base excitation 32 Beam element 223 − influence coefficient 187 − lumped masses 184 − multi-span 268 − Rayleigh’s formula 192 Beats 25, 154 Bode diagram 55, 242 Branched system 128, 197 Characteristic equation 108 Complex algebra method 52 Concentrated mass 184 Consistent mass matrix 213, 227 Continuous systems 259 Coordinate transformation 213, 229 Coulomb damping 92 Coupled pendulums 151 − translation 106 − translation and rotation 145 Coupling 146 Critical damping 36 Critical speed 33 Cubic stiffness 81 D’Alembert’s principle Damped systems 156 Damping 10 − coefficient 157 − Coulomb 92

106, 132

− − − − − − − − − − − −

critical 36 energy dissipated by 47 equivalent viscous 51 factor 49 hereditary 68 hysteretic 48 optimum 169 quadratic 97 ratio 37, 158 structural 48 velocity squared 97 viscous 36 Dashpot 36 Decrement, logarithmic 39 Degrees of freedom 6 Diagram 22 − Bode 55, 242 − displacement-force 46 − Nyquist 55, 244 Difference equation 202 Discretisation 221 Distortions 65 Duhamel’s integral 23 Dunkerley’s formula 190 Dynamic absorber 117 − stiffness 54 − stresses 125 Eigenfrequencies 108 Eigenvalues 133 Eigenvectors 133 Energy dissipated by damping 47 − kinetic 17, 213, 227 − method 17 − strain 205, 225, 236 Equivalent viscous damping 51 Excitation, base 31 − harmonic 23, 114 − mass 23 − with unbalanced rotating masses Flexibility coefficients

130, 187

30


290

− matrix 23 Flexural system 130 Forced vibration, damped 42 − − undamped 22 Frame element 228 Free body diagram 106 Free-free beam 21 Free vibration 11, 135, 153 − − damped 35, 157 − − undamped 11, 107 Frequency 12 − damped 37, 159, 174 − equation 108, 264, 266 − fundamental 192 − natural 108 − peak 45 − resonant 45 − response 163 − − curves 26 − − function 163 − − − matrix 163, 241 − undamped 12 Geared system Grillage 234

126, 196

Half-power points 49 Harmonic excitation 23, 161 − motion 13 Houdaille damper 168 Hysteresis loop 47 Hysteretic damping 48 Inertance 53 Influence coefficients 130, 187 Initial conditions 13 Instruments, vibration measuring 62 Isochrones 84 Isolation, vibration 32 Jump phenomena Kinetic energy

89 17, 213, 227

Lagrange’s equations 152, 219 Lateral vibration of beams 259 Logarithmic decrement 39 Longitudinal vibration of rods 275 Loss factor 42 Lumped mass system 184 Magnification factor

26

Mass addition effect 51 Mass-spring system 11 Matrix, damping 157 − flexibility 131 − mass 107, 212, 227, 235 − stiffness 107, 212, 225, 235 Mechanical impedance 54 Mobility 53, 56 Modal analysis 114, 165 − coordinates 112 − damping coefficients 158 − mass 113, 135 − matrix 113 − stiffness 113 − vectors 109 Mode orthogonality 111 − natural 108 − normal 109 − shape 109 Multi-span beam 268 Multi-storey frame 193 Natural coordinates 210 − frequency 108 − − closely spaced 176 − − damped 37, 159, 174 − − fundamental 19 − − undamped 12 Non-linear systems 79 Non-proportional damping 172 Normal coordinates 108 − modes 109 Normalisation 108 Nyquist diagram 55, 244 Orthogonality 111, 267 Oscillations 5 Overdamped modes 174 − system 39 Pendulum 151 Percent of critical damping 37 Period of vibration 13 Phase angle 43 − distortion 66 − shift 14, 62 Plane frame 220 Potential energy 17 Principal coordinates 113 Proportional damping 157

291

INDEX Rayleigh’s method 18, 192 − quotient 113 Receptance 53, 241 Repeated structure 202 Residues 58 Resonance 27 − acceleration through 29 − curve 45 − frequency 45 Rigid body modes 123, 265 Rod, longitudinal vibration of 275 − torsional vibration of 278 Rotating shaft 66 Seismic instruments 62 Shape functions 210, 223 Skeleton curve 83 Spring combinations 14 State space equations 173 Stiffness coefficients 13 − matrix 107, 212, 225, 235 Strain energy 205, 212, 225, 236 Structural damping 48 Transfer function 162 Torsional stiffness 16, 120 − system 15, 119, 195 Transmissibility 32, 61 Truss element 210 Undamped natural frequency Underdamped system 37

12, 176

Vibration absorber 170 − damper 36 − free 8, 11, 135, 153 − forced 8, 22, 42 − parametric 8 Vibrometer 65 Viscous damper, untuned 168 − damping 36 − − non-proportional 172 Whirling shaft 66 Work due to damping

47

M. Rades - Mechanical Vibrations 1

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