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

MECHANICAL VIBRATIONS 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

1. MODELLING VIBRATING SYSTEMS 7

connected by massless flexible members. The motions predicted by using such approximate lumped-parameter or discrete systems are often close enough to the actual vibrations to satisfy all practical demands and to provide useful design data and allowable vibration limits.

In some systems, a second approximation can be made, by taking into account the mass of the elastic members. This is necessary only when the flexible members have distributed masses which are comparable in magnitude with the masses of system components modelled as rigid bodies.

Finally, there are many systems of practical interest which have such simple shapes that they can be considered as systems possessing an infinite number of degrees of freedom. Such distributed-parameter or continuous systems may be modelled as strings, beams, plates, membranes, shells and combinations of these.

In most engineering applications, geometrically complex structures are replaced by discretized mathematical models. A successful discretization approach is the finite element method. The infinite degree of freedom system is replaced by a finite system exhibiting the same behaviour. The actual structure is divided (hypothetically) into well-defined sub-domains (finite elements) which are so small that the shape of the displacement field can be approximated without too much error, leaving only the amplitude to be found. All individual elements are then assembled together in such a way that their displacements are mating each other at the element nodes or at certain points at their interfaces, the internal stresses are in equilibrium with the applied loads reduced at nodes, and the prescribed boundary conditions are satisfied. Modelling errors include inappropriate element types, incorrect shape functions, improper supports and poor mesh.

1.3 Simple Vibrating Systems

A surprisingly large number of practical vibration problems which arise in the machines and structures designed by engineers can be treated with a sufficiently high degree of accuracy by imagining the actual system to consist of a single rigid body, whose motion can be described by a single coordinate.

In reality, the simplest imaginable system consists of the body whose motion is of interest and the fixed surrounding medium, relative to which the motion is measured. The problem of treating such a simplified system is fourfold. The first part consists in deciding what part of the system is the rigid body and what part are the flexible members. The second part consists in calculating the values of the dynamic parameters of the rigid body and flexible parts. The third part consists in writing the equations of motion of the equivalent system, Finally, the fourth part consists in solving the equations for the prescribed conditions of


free or forced vibrations. Alternatively, methods using the kinetic and potential energies may be used in the place of the last two stages.

The first two parts require judgement and experience which come with practice, that is, with the repeated process of assuming equivalent systems, predicting their motions and checking the predictions against actual measurements on the real systems. Model verification and validation may require updating of system parameters or even of the model structure. The adequacy of the solution depends largerly on the skill with which the basic simplifying assumptions are made. A basic choice is between linear and non-linear models. Damping estimation is another source of error, because damping cannot be calculated like the mass and stiffness properties. The last two steps consist in applying procedures worked out by mathematicians. The real engineering work lies in the first two stages, while the last two stages may be considered as mere applications of recipies.

One degree of freedom systems are considered in Chapters 2 and 3. Discrete systems are treated in Chapters 4 and 5. Chapter 6 is devoted to straight beams and bars.

1.4 Vibratory Motions

According to the cause producing or sustaining the vibratory motion, one can distinguish: free vibrations, produced by an impact or an initial displacement; forced vibrations, produced by external forces or kinematic excitation; parametric vibrations, due to the change, produced by an external cause, of a system parameter; self-excited vibrations, produced by a mechanism inherent in the system, by conversion of an energy obtained from a uniform energy source associated with the system oscillatory excitation.

If the system is distorted from the equilibrium configuration and then released, it will vibrate with free vibrations. If any part of the system is struck by a blow, the system will vibrate freely. Musical instruments like drums are struck and strings are plucked. Free vibrations exist when the forces acting on the system arise solely from motion of the system itself. The frequencies of the free vibrations are fixed functions of the mass, stiffness, and damping properties of the system itself. They are called natural frequencies. For any particular system they have definite constant values. When all particles of a body vibrate in a synchronous harmonic motion, the deflected shape is a natural mode shape.

Vibrations which take place under the excitation of external forces are forced vibrations. External forces in any system are forces which have their reactions acting on bodies which are not parts of the system isolated for study. The forcing function can be harmonic, complex periodic, impulse, transient, or random.

1. MODELLING VIBRATING SYSTEMS 9

When a system is excited by a periodic external force which has one frequency equal to or nearly equal to a natural frequency of the system, the ensuing vibratory motion becomes relatively large even for small amplitudes of the disturbing force. The system then is in a state of resonance. An example is the swing pushed at the right intervals. Other examples include vibrations of geared systems at the tooth-meshing frequency, torsional vibrations of multi-cylinder engine shafts at the firing frequency, vibrations of rolling element bearings at the ball passing frequencies, etc.

There is an effect arising from the damping which causes the resonance frequency to differ slightly from the natural frequency by an amount which increases with the damping. Fortunately the distinction in practice is very small and can be neglected in most engineering structures, unless very high damping is provided on purpose.

Resonance relates to the condition where either a maximum motion is produced by a force of constant magnitude, or a minimum force is required to maintain a prescribed motion level. A resonance is defined by a frequency, a response level and a bandwidth of the frequency response curve. Avoidance of large resonant vibration levels can be accomplished by: a) changing the excitation frequency; b) making stiffness and/or mass modifications to change the natural frequencies; c) increasing or adding damping; and d) adding a dynamic vibration absorber.

When the driving frequency is an integer multiple of the natural frequency of the associated linear system, non-linear single-degree-of-freedom systems described by Mathieu equations exhibit parametric instabilities, referred to as parametric resonances.

The principal parametric resonance occurs when the excitation frequency is twice the natural frequency. Parametric resonances of fractional order also exist. Multi-degree-of-freedom systems can experience parametric resonance if the driving frequency and two or more natural frequencies satisfy a linear relation with integer coefficients.

Parametric resonance is a state of vibration in which energy flows into the system from an external source at resonance, increasing the amplitude of the system’s response. This energy is dependent upon both the natural frequency of the system and the frequency of the parameter variation.

During resonant vibrations and self-excited vibrations, the system vibrates at its own natural frequency. But while the former are forced vibrations, whose frequency is equal to a whole-number ratio multiple of the external driving frequency, the latter is independent of the frequency of any external stimulus.

In a self-excited vibration, the alternating force that sustains the motion is created or controlled by the motion itself. When the motion stops, the alternating force disappears. Well-known examples include the vibrations of a violin string


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 gWm = , 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 stδ . In this position, the gravitational force mgW = acting on the mass downwards is


balanced by the spring force stkδ acting upwards (Fig. 2.1, c), so that the static deflection is

kgm

st =δ . (2.1)

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 xk− (Fig. 2.1, d). Its motion is described by Newton’s second law

xx km −=&& ,

which can be written

0km =+ xx&& , (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

tCtC nn ωω cossin 21 +=x , (2.3)

where mkn =ω [rad/sec] (2.4)

is the undamped natural circular frequency of the system.

The undamped natural frequency is

mkfn 2π

1= . [Hz] (2.5)

2. SIMPLE LINEAR SYSTEMS 13

The arbitrary constants 1C and 2C are evaluated from the initial conditions of the motion. In the most general case, the system may be started from position 0x with velocity 0v so that the general solution becomes

txt nnn

ωωω

cossin 00 +=vx . (2.6)

Another form of the general solution is

( )φω += tA nsinx (2.7)

where the two arbitrary constants are given by

( ) 20

20 nA ωvx += ,

0

01 tanvxnωφ −= . (2.8)

Equation (2.7) indicates that the free vibration of the spring-mass system is harmonic and occurs at a natural frequency nf . The quantity A represents the displacement amplitude from the static equilibrium position and φ is the phase angle. The circular frequency nω defines the rate of vibration in terms of radians per unit time, π2 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

nnfT ω2π1 == . [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)

stn

gfδπ2

1= , [Hz] (2.10)

where 2sm9.81=g .

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.


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 1k and 2k , will deflect statically when loaded by a weight W by an amount

⎟⎟⎠

⎞⎜⎜⎝

⎛+=+=

2121

11kk

WkW

kW

stδ .


The equivalent spring constant, representing the combined effect of 1k and

2k , is

21

111

kk

Wkst

S+

==δ

. (2.11)

For a system with n springs connected in series, the equivalent stiffness Sk is given by

nS k...

kkk1111

21+++= . (2.12)

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 :

stst kkW δδ 21 += .

Thus, for parallel springs, the equivalent stiffness is

21 kkWkst

P +==δ

. (2.13)

In general, a system with n parallel springs has an equivalent stiffness given by

nP k...kkk +++= 21 . (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, 2mkg , suspended from a bar or wire of torsional stiffness K,

radmN . 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 θK− so that Newton’s second law for angular motion is

θθ KJ −=&& ,



0=+ θθ KJ && , (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

( ) tCtCt nn ωωθ cossin 21 += ,

where

JKn =ω [rad/sec] (2.16)

is the undamped natural circular frequency of the torsional system.

The undamped natural frequency is

JKfn π2

1= . [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 tM will twist an angle p

tIG

M l=θ , where

32π 4dI p = is the polar second

moment of area of the shaft cross section. The torsional stiffness is then

l

pt IGMK ==

θ.

In fact, there is complete analogy between systems in axial and torsional vibration, with the counterparts of springs and masses being torsional springs and rigid discs possessing polar mass moments of inertia.


2.1.4 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

.constTU =+ (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 xk , and the work done on an infinitesimal displacement xd is xxk d . The potential energy in the spring, when stretched over

a distance x , is 2

021d xkxxkU

x

== ∫ . Assuming the vibratory motion of the form

tAx nωsin= , the maximum potential energy is 2

21 AkUmax = .

The kinetic energy at any instant is 2

21 vmT = . The velocity is

tA nn ωω cos=v , so that the maximum kinetic energy is 22

21 AmT nmax ω= .

Equating maxmax TU = , we obtain 222

21

21 AmAk nω= wherefrom the

natural frequency mkn =ω is obtained, independent of the amplitude A .

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

kinetic energy can be written 2

21 xAT &lρ= . If the fluid oscillates back and forth,

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.


The instantaneous potential energy is 2xAgU ρ= . Substituting the two energies in the condition that the rate of change of total energy must be zero

( ) 0dd

=+UTt

and dividing out x& , we obtain the differential equation of motion of the fluid

02=+ xgx

l&& .

Fig. 2.5

Therefore the natural frequency

lgn 2=ω

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


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

( ) ( ) txt,xy 1cosωv= .

The instantaneous potential energy is

xxyIE

IEdxMU d

21

2

2

2

22

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

== ∫∫

where the linearized differential equation (5.65) of the beam elastic line ( )22 xyIEM ∂∂= has been used.

Its maximum value is

xx

IEUmax d21

2

2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

= ∫v .

The instantaneous kinetic energy is

∫∫ =⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

= xd21

21 22

1

2

yAdmtyT ρω ,

with the maximum value

∫= xv d21 22

1 ATmax ρω .

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

( )∫

∫ ∂∂=

xv

v

d

dω

2

22221

A

xxIE

ρ. (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


⎟⎠⎞

⎜⎝⎛ −=

l2cos10

xπvv .

It can be seen that this function satisfies the boundary conditions 0=x , 0=v , 0dd =xv , and l=x , 0dd 22 =xv , but not the condition l=x ,

0dd 33 =xv (zero shear force), so that it is an approximate admissible function.

Fig. 2.6

The maximum potential energy is 203

4

64π v

l

IEUmax = . The maximum

kinetic energy is ⎟⎠⎞

⎜⎝⎛ −=

πωρ 2

432

021 lvATmax , or 0.23

220

21 ⋅= lvωρ ATmax .

Equating the two energies, the fundamental frequency of vibration (in rad/sec) is obtained as

AIE

ρω 21

3.6638l

= .

The true solution (6.16) is AIE

ρω 21

3.515l

= , so that the value based on

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

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

32

0 321

ll

xxvv ,

the maximum potential energy is 20

203 2

123 vv kIEUmax ==l

and the maximum

kinetic energy is ( ) 201

20

21 2

1140

3321 vv ωωρ

redmax mAT =⎟⎠⎞

⎜⎝⎛=

l .

Equating the two energies, the fundamental frequency given by Rayleigh’s formula is


( ) AIE

mk

AIE

red ρρω 2

3

13.5675

140333

ll

l=== ,

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 ( ) lAρ14033 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

ax−=

l

πsin0vv .

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

( ) ( ) ( )( ) 0d000

===⋅ ∫∫∫lll

dxvv AdxAmassvelocity ρωρω ,

which yields π2 0v=a .

Using the deflected shape of the form

⎟⎠⎞

⎜⎝⎛ −=

π2πsin0

l

xvv ,

equation (2.19) yields the fundamental natural frequency

AIE

ρω 21

22.6l

= .


The true solution (6.21) is AIE

ρω 21

22.4l

= so that the discrepancy is only

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 ( )tF with an arbitrary general time variation (Fig. 2.8).

During the short time interval τd , the force ( )τF can be considered constant. The cross-hatched area represents an infinitesimal impulse ( ) ττ dF which produces a velocity variation

( )m

Fx ττ dd =& .

The response of mass m due to the differential impulse, over the entire response history for τ>t , is


( ) ( )τωω

ττ−= t

mFx n

nsin1dd , (2.20)

which can be deduced from (2.6) considering that at τ=t , 00 =x and x&d=0v .

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

( ) ( ) ( )∫ −=t

nn

tFm

tx0

dsin 1 ττωτω

. (2.21)

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 ( ) tFtf ωcos0= of constant amplitude 0F 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

tFkm ωcos0+−= xx&& ,

which can be written tFkm ωcos0=+ xx&& . (2.22)


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

( ) tXt ωcos=Px , (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

tFtXktXm ωωωω coscoscos 02 =+−

which can be divided throughout by tωcos yielding

( ) 02 FXmk =− ω

or ( ) 22

02

0

11 n

stXkm

kF

mk

FX

ωωωω −=

−=

−= . (2.24)

In (2.24)

kF

X st0= (2.25)

is the static deflection of the spring under the (constant) load 0F and mkn =ω is the undamped natural circular frequency (2.4).

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


( )( )

tXtCtCn

stnn ω

ωωωω cos

1cossin 221

−++=tx . (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 0x and

0v . Equation (2.26) yields

( )( ) 022

10 xXCx

n

st =−

+=ωω

, ( ) 010 v== nCx ω& ,

so that the total response is

( )( ) ( )

tXtXxtn

stn

n

stn ω

ωωω

ωωω

ωcos

1cos

1sin 220

−+

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−+=

n

0vtx . (2.27)

For zero initial conditions, 000 == vx , the response (2.27) becomes

( )( )

( )ttXn

n

st ωωωω

coscos1 2 −

−=tx . (2.28)

2.2.3 Beats

The difference of cosines in equation (2.28) can be expressed as a product

( )( )

ttXm

n

st ωΔωωω

sinsin1

22−

=tx , (2.29)

where

2ωω

ω+

= nm and

2ωω

ωΔ−

= n .

In the case when ωΔ becomes very small, since mω is relatively large, the product in equation (2.29) represents an amplitude modulated oscillation. The harmonic motion with higher frequency mω is amplitude modulated by the harmonic motion with lower frequency ωΔ (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.


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

( ) stn

XX 211

ωω−= . (2.30)

The absolute value of the coefficient of stX 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 1<nωω the ordinates are positive, the force and motion are in phase, while for 1>nωω the ordinates are negative, the force and motion are

0180 out of phase (Fig. 2.11, b). Whereas for 1<nωω the mass is below the static equilibrium position when the force pushes downward, for 1>nωω the mass is above the equilibrium position while the force is pushing downward.


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 1=nωω , 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 nωω = 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 090 (Fig. 2.28).

Consider the case when, starting from rest, the mass-spring system is subjected to a force of instantaneous magnitude tF nωcos0 , where nω is the


natural frequency. As ω becomes exactly equal to nω , the solution (2.27) is no longer valid. Substitution of ( ) τωτ nFF cos0= into equation (2.21) yields

( ) ( )∫ −=t

nnn

tmFtx

0

0 dsin cos ττωτωω

,

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡−= ∫ ∫

t t

nnnnn ttmFtx

0 0

2

n

0 dsin coscosd cossin ττωτωωττωωω

,

( ) ttmF

tx nn

P ωω

sin2

0= . (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 090 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

( ) ttmFtxt n

nnn

nω

ωωω

ωsin

2cossin 0

00 ++=vtx . (2.32)

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


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 tddωε = , 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

resonance frequency. In Fig. 2.13, ( ) ⎟⎠⎞

⎜⎝⎛ +=

221sin 2

0πε tFtf and .const=ε

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.


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

( )[ ]20 1 nXkF ωω−= . (2.33)

Figure 2.14 is a plot of the force modulus as a function of the driving frequency for const.X =0 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 tem 2 ωω cos1 , where 1m 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 2em ω1 for 0F in equation (2.24). Then

( )( )

( ) 2

2

2

21

2

21

11 n

n

ne

kem

mk

emX

ωωωω

ωω

ω

ω

ω

−=

−=

−= . (2.34)


It should be pointed out that m is the total vibrating mass and includes the mass 1m .

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 .conste = 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

( ) tFxkm ωcos0122 =−=− xx&& .

The magnitude of the driving point displacement is given by

( )( ) 2

20

2

20

11

n

nkF

mmk

kFX

ωωωω

ωω −

=−

= .

For constant force amplitude .constF =0 , 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.


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 mass-excited system. If a second mass is attached at the driving point, the resulting mass-spring-mass system exhibits both a resonance and an antiresonance in the driving-point response.

2.2.10 Transmissibility

If the mass-spring system is excited by a prescribed motion tXx ωcos 11 = applied to the spring end not connected to the mass, then the

motion transmitted to the mass tXx ωcos 22 = is defined by the amplitude ratio

( ) 21

2

11

nXX

ωω−= . (2.35)

The ratio 12 XXTR = is called transmissibility and is plotted in Fig. 2.17 as a function of the frequency ratio nωω .

For 2>nωω , 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 nωω 2> . 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.


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 2

Grm ω is thus applied to the disc, where ω is the speed of rotation, m is the mass of the disc and

OGrG = . 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 Crk acting in C, where k is the stiffness of the shaft at the disc and OCrC = .

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

( )ermrk CC += 2ω .

Solving for Cr , we obtain

( )( ) 2

2

2

2

1 n

nC

emk

emrωω

ωωω

ω−

=−

= . (2.36)

where mkn =ω is the natural circular frequency of the rotor lateral vibration at zero speed.


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

( ) 21 nCG

eerrωω−

=+= . (2.37)

A plot of Cr (solid line) and Gr (broken line) as a function of ω is given in Fig. 2.19. At a speed nωω <1 the system rotates with the heavy side 1G outside

1C , whereas for nωω >2 the light side, or the side opposite 2G , is outside 2C . For very high speeds, nωω >> , the radius Cr becomes equal to the eccentricity and the points O and G coincide; the disc rotates about its centre of gravity.

When nωω = , the radii Cr and Gr 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 090 (see Sec. 2.4.11)


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.


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 ( )secmN .

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

xx kxcm −−= &&& ,


0=++ xx kxcm &&& . (2.38)

Assuming solutions of the form tsx e= , we obtain the characteristic equation

02 =++mks

mcs , (2.39)

which has two roots

mk

mc

mcs −⎟

⎠⎞

⎜⎝⎛±−=

2

21, 22. (2.40)

The general solution for the damped free vibrations is

( ) tsts eCeCtx 2121 += , (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


nc

mk

mc ω==2

,

or nc mmkc ω22 == . (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

ccc

=ζ (2.43)

referred to as the damping ratio (or percent of critical damping)

Using this notation, equation (2.40) becomes

ns ω⎟⎠⎞⎜

⎝⎛ −±−= 1ζζ 2

21, . (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, 1ζ <

For 1ζ < , equation (2.44) can be written

ns ω⎟⎠⎞⎜

⎝⎛ −±−= 2

21, ζ1iζ . (2.45)

Substitution of (2.45) into (2.41) and conversion to trigonometric form with the aid of Euler’s formula βββ sinicosei += , yields

( ) ⎟⎠⎞

⎜⎝⎛ += −−− t-tt nnn eCeCetx ωωω 22 ζ1i

2ζ1i

1ζ ,

or

( ) ⎟⎠⎞⎜

⎝⎛ +−= − φωω teAtx n

tn 2ζ ζ1sin . (2.46)

Equation (2.46) indicates that the motion is oscillatory with diminishing amplitude. The decay in amplitude with time is proportional to tnωζe− , as shown by the dashed curves in Fig. 2.21.

The frequency of the damped oscillation

nd ωω 2ζ1−= (2.47)

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


Equation (2.44) can be written

ds ωσ i21, ±−= (2.48)

where nωσ ζ= (2.49)

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

Fig. 2.21

The following equations are useful

22ζ

σω

σ

+=

d

, 22

ζσωσω +== dn . (2.50)

Fig. 2.22


Case II: Overdamped system, 1ζ >

For 1ζ > , substitution of (2.44) into (2.41) yields

( )tt nn

eCeCtxωω ⎟

⎠⎞⎜

⎝⎛ −−−⎟

⎠⎞⎜

⎝⎛ −+−

+=1ζζ

21ζζ

1

22

.

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

Fig. 2.23

Case III: Critically damped system, 1ζ =

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

( ) ( ) tntCCtx ω−+= e21 .

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).


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

( )dn

dn

nT

Tt

te

eA

eAxx ω

ω

ωζ

ζ

ζ

2

1 ==+−

−

where the period of the damped vibration is

dndT

ωω

2π

ζ1

2π2

=−

= .

The logarithmic decrement is

22

1

ζ1

ζπ2ζln−

=== dn Txx ωδ . (2.51)

For 1ζ << , ζ2πδ ≅ .

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

⋅⋅=3

2

2

1

1

00

xx

xx

xx

xx

n( ) δδ nn

n

n eex

x==−1

so that the logarithmic decrement is given by

nxx

n0ln1

=δ . (2.51, a)


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 1ζ << , equation (2.51, a) yields


nxxn ⋅−= πζ2lnln 0

so that the damping ratio ζ is equal to the slope of the line divided by π2 (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) UΔ to the peak potential energy U stored in the system during that cycle

UUη Δ

= . (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 1X and 2X are two consecutive amplitudes of a damped free vibration,

the energy stored in the spring at maximum displacement is 211

21 XkU = ,

222

21 XkU = . The loss of energy divided by the original energy is

δδ 2111 22

1

2

1

2

1

21 ≅−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

− −eXX

UU

UUU

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

δη 2≅ . (2.52, a)

2.4 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 090 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.


2.4.1 Steady State Vibrations with Viscous Damping

Consider the spring-mass-dashpot grounded system subjected to a harmonic force tF ωcos0 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

tFkxcm ωcos0=++ xx &&& . (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

( ) ( )ϕω −= tXtx cos . (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

( ) ( ) ( ) 0coscossincos 02 =+−−−+− tFtXktXctXm ωϕωϕωωϕωω .

The terms in the above equations are projections of force vectors on a (horizontal) line at an angle tω with respect to the force vector (Fig. 2.27).

The force vector 0F is ϕ degrees ahead of the displacement vector X . The spring force Xk is opposite to the displacement, while the inertia force


Xm 2ω is in phase with the displacement. The damping force Xcω is 090 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

ϕω cos02 FXmXk =− , ϕω sin0FXc = . (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

( ) ( ) ( )[ ] ( ) 2 2 2222

0

ζ21 nn

stX

cmk

FXωωωωωω +−

=+−

= , (2.56)

and the phase shift

( )22 1ζ2tan

n

n

mkc

ωωωω

ωωϕ

−=

−= , (2.57)

where mkn =ω and mkc 2ζ = .

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


The amplitude-frequency diagrams are called resonance curves or frequency response curves. Such a curve starts at a value stX , 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 090 at the undamped natural frequency, to approach

0180 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

7070ζ .> , 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

frequency 2ζ21−= nr ωω where the peak value 2ζ1ζ2 −

= stmax

XX of the

steady-state response occurs.


The “phase resonance” is defined to occur at the undamped natural frequency nωω = (when the phase shift is 090 ) when the displacement amplitude

is ζ2st

resX

X = . For small values of damping the two resonances coincide.

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

( ) tXtx ωcos= (2.58)

lagging by an angle ϕ the driving force applied to the mass

( ) ( )ϕω += tFtf cos0 . (2.59)


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

ϕϕ 2

02

0

2

2

2sincos2 =−+

Ff

Xx

Ff

Xx . (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)

( ) ( )tttFXttxfxfWd ωωϕω

πωπ

dsincosdddd

2

00

2

0∫∫∫ +−=== ,

ϕsinπ 0 XFWd = .

Using the second equation (2.55), the above expression becomes 2π XcWd ω= . (2.61)

In order to produce work, the damping force tXxcfd ωω sin=−= & must

be 090 phase shifted with respect to the displacement ( ) tXtx ωcos= .

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


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, nωω = , 090=ϕ , kFX res ζ20= , the ellipse major axis is vertical and the energy dissipated by damping is

20 ππ resnresd XcXFW ω== . (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.


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 ωxh &− (rather than xc&− ). Equation (2.53) becomes

tFkxhm ωω

cos0=++ xx &&& , (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 hkk i+=∗ , since it is decided that a harmonic solution is required

( ) teFhkm ωi0i =++ xx&& . (2.64)

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

2π XhWd = , (2.65)

which is independent of frequency.

Equations (2.56) and (2.57) become

( ) ( )[ ] 222222

0

1 g

X

hmk

FX

n

st

+−=

+−=

ωωω, (2.66)

( )22 1tan

n

gmkh

ωωωϕ

−=

−= , (2.67)

where khg = 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).

When nωω = the resonant amplitude is ζ2st

resX

X = . For small values of

damping, the peak M coincides with the point of phase resonance. The points B and


C, of ordinate ( ) resX22 are referred to as the half-power points. The amplitude

squared is ( ) 221 resX , so that the power dissipated by damping at the corresponding frequencies 1ω and 2ω is half the power dissipated at resonance.

Fig. 2.32

Substituting into equation (2.56) we obtain

( )( ) ( ) 222

2

ζ21

1ζ2

1 21

nn ωωωω +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛,

which yields the equation

( ) ( ) ( ) ( ) 0ζ81ζ212 2224 =−+−− nn ωωωω .

Solving, we obtain the frequencies of half-power points

( ) ( ) 22221 ζ1ζ2ζ21 +±−=,nωω

which for 1ζ << can be approximated by

( ) ζ21221 ±≅,nωω .

Denoting ( )ζ21221 −≅ nωω and ( )ζ2122

2 +≅ nωω , we obtain


21

22

21

22ζ2

ωω

ωω

+

−≅ (2.68)

or

nnnn ωωω

ωωω

ωωω

ω

ωω 1212122

21

22

22ζ2

−≅

+−=

−≅

so that the damping ratio is given by

nωωΔ

2ζ ≅ , (2.69)

where 12 ωωωΔ −= 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 am


(Fig. 2.33). The natural frequencies 1nω and 2nω are determined at peak displacement amplitudes.

From the corresponding equations (2.4) 21nmk ω= , (2.70)

( ) 22nammk ω+= , (2.71)

it is possible to obtain the equivalent mass

( ) 122

21 −

=nn

ammωω

, (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

km

cF

cF

kFX

nres

0

1

0

1

0ζ21

1===

ω

and for the system with the added mass

kmm

cF

cF

kFX a

nres

+=== 0

2

0

2

0ζ21

2 ω.

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

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

( ) teFtf ωi0= , (2.73)

so that the steady-state solution (2.54) becomes

( ) teXtx ωi= , (2.74) where

IR XXeXX ii +== θ (2.75)

is the complex displacement amplitude.

In equation (2.75), X is the modulus, θ is the phase angle, RX is the real


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

θcosXX R = , θsinXX I = , (2.76) 22IR XXX += , RI XX=θtan . (2.77)

For structural damping, the equation of motion (2.63) becomes

teFkxhm ω

ωi

0=++ xx &&& . (2.78)

Substitution of (2.74) in (2.78) yields

( ) 02 i FXkhm =++− ω .

The complex amplitude X is determined as

( ) gX

hmkFX

n

st

i1i 220

+−=

+−=

ωωω, (2.79)

where mkn =ω , khg = , so that

( )( )[ ] st

n

nR X

gX

222

2

1

1

+−

−=

ωω

ωω , ( )[ ] st

n

I Xg

gX2221 +−

−=

ωω (2.80)

( )[ ] st

n

Xg

X2221

1

+−=

ωω,

( )21tan

n

gωω

θ−

−= . (2.81)

Eliminating ω between the expressions of RX and IX yields 2

22

21

21

⎟⎟⎠

⎞⎜⎜⎝

⎛=+⎟⎟

⎠

⎞⎜⎜⎝

⎛+ stRstI X

gXX

gX . (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),


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.


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 0FX=α 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 Iα (Fig. 2.34, d), and by points of inflection (maximum slope or maximum derivative with respect to 2ω ) in θ (Fig. 2.34, b) and Rα (Fig. 2.34, c). The peak in Iα is

sharper than that in α . At resonance, 090−=θ and 0=Rα . 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

( ) ( ) ( )2

22222222 1

1

d

d1

d

d αωωωω

θ

ωωk

gkhs

nnn

=

⎥⎦⎤

⎢⎣⎡ +−

=−= (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 ωΔ , then the arc length sΔ 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

21

22

21

22

ωω

ωω

+

−=g (2.84)

where 1ω and 2ω are the frequencies of the peaks of ( )ωα R or of the ends of the diameter BC, perpendicular to OM, in the Nyquist plot.


The stiffness can be calculated from the value resα at resonance

resres XF

ggk 0111

==α

. (2.85)

Because in the complex plane the velocity is 090 phase shifted ahead the displacement and the acceleration is 090 phase shifted ahead the velocity, the Nyquist plots of mobility and accelerance are rotated 090 and 0180 , respectively, anticlockwise with respect to the polar plot of receptance.

The Nyquist plot of the mobility IR MMFXωM ii 0 +== , which is not a circle, is shown in Fig. 2.35, a and is described by the following equation

( ) 02

2222 =+−+mkgMM

mkgMMM IRR

IR .

a b

Fig. 2.35

The Nyquist plot of the accelerance IRFX ηηωη i02 +=−= is shown in

Fig. 2.35, b. It is a circle of equation

22

222

41

21

21

mgg

mgm IR+

=⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟

⎠⎞

⎜⎝⎛ − ηη .

The half-power points and the point of maximum response amplitude are shown in both figures.


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 0F . Using, instead of α , the general notation

for frequency response functions ( )ωiH , we obtain for viscous damping

( )nn

mcmkF

XHωωωωωω

ωζ2i

1i

1i 2220 +−

=+−

== . (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

( ) ( )( )21 ii1i

ssmH

−−=

ωωω . (2.87)

where d,s ωσ i21 ±−= (2.48) are the roots of the characteristic equation (2.39).

Equation (2.87) can be expressed in terms of partial fractions

( )( ) 2

2

1

1

21 iiii1

sC

sC

ssm

−+

−=

−− ωωωω. (2.88)

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

( )11 i2

121

i2 ii

i1

ss ssCC

sm

== −−

+=− ωω ω

ωω

.

Thus

( ) ( ) ddd

mmss

mCωωσωσ 2i

1ii

11

211 =

−−−+−=

−=

and similarly

d

mCω2i

12 −= ,

so that equation (2.88) can be written, by factoring a constant of i2

1 out of 1C and

2C , under the standard form


( ) ( ) ( )21 ii 2ii2i

sR

sRH

−−

−=

∗

ωωω . (2.89)

where the star denotes the complex conjugate.

In this case, the residues are purely real

d

mRR

ω1

== ∗ . (2.90)

For multi-degree-of-freedom systems they are complex conjugates.


( ) ( ) ( )dd

UUHωωσωωσ

ω++

+−+

=∗

i ii . (2.91)

where

dmUU

ω21i −=−= ∗ . (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

( )d

Uωωσ −+ i

, (2.93)

consider the plot of


( )dωωσ −+ i1 . (2.94)

In the complex plane, expression (2.94) represents a circle (Fig. 2.36, a) with centre ( )021 ,σ and diameter σ1 . At the point M of maximum amplitude, i.e. at the crossing point of the circle with the real axis, the frequency is dω , 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 045± 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 090 and expansion or contraction by an amount dmω21 (Fig. 2.36, b). The obtained circle is called the circle with predominantly positive frequencies. The centre of this circle is at ( )σ2 0 U,− and

its diameter equals 2ζ1ζ

121

−=

kUσ

. The portion drawn with broken line

corresponds to negative frequencies.

a b

Fig. 2.37

Consider now the second term in the summation

( )d

Uωωσ ++

∗

i. (2.95)

The plot of

( )dωωσ ++ i1 . (2.96)

is shown in Fig. 2.37, a.


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 090 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 nωω and ζ .

a b

Fig. 2.38

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

( )d

dRR

Hωσσ

ω2i

i2i2i

+−=

∗

,

( ) ( )⎥⎦⎤

⎢⎣

⎡+

−=ddd

d mmH

ωσωσωω

2i11

i21i . (2.97)

This can be approximated as

( )d

d mRH

ωσσω

2i1

2ii =≅ . (2.98)

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


Therefore, many single-degree-of-freedom models can be represented simply as

( ) ( ) ( )[ ]d

Rs

RHωσωω

ωii2i2

i1 −−

=−

≅ . (2.99)

2.4.9 Transmissibility in Damped Systems

If the mass-spring-dashpot system is base-excited by a prescribed motion teXx ωi

11 = , the motion transmitted to the mass is teXx ωi22 = , where

ϕi22

−= eXX is a complex amplitude. The differential equation of motion is

( ) ( )12122 xxcxxkxm &&&& −−−−=

which may be rearranged as

11222 xkxcxkxcxm +=++ &&&& . (2.100)

The amplitude ratio is

cmkck

XX

ωωω

ii2

1

2

+−+

= . (2.101)

Fig. 2.39


The motion transmissibility is

( )

( )[ ] ( ) 2 2 2

2

1

2

ζ21

ζ21

nn

n

XX

TRωωωω

ωω

+−

+== , (2.102)

The phase shift is given by

( )( ) ( ) 2 2

3

ζ21ζ2tan

nn

n

ωωωωωωϕ

+−= . (2.103)

Equations (2.102) and (2.103) are graphically represented in Fig. 2.39 for various values of the damping ratio ζ .

For 2>nωω , 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 2>nωω , 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 teF ωi

0 (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

( ) ( )22 XcXkFT ω+= (2.104)

so that the force transmissibility

0FFTR T= (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-spring-dashpot system attached to the vibrating structure.


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

( ) tXtx ωcos11 = . (2.106)

Neglecting transient terms, the relative displacement between the mass m and the casing S can be defined by

( ) ( )ϕω −= tXtx rr cos . (2.107)

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

rxxx += 12

and the absolute acceleration is

rxxx &&&&&& += 12 .

The equation of motion of the mass m can be written

( ) 01 =+++ rrr xkxcxxm &&&&&

or tXmxmxkxcxm rrr ωω cos2

11 =−=++ &&&&& . (2.108)

Fig. 2.40

Equation (2.108) has a steady-state solution for which

( )( )[ ] ( ) 2 2 2

2

1 ζ21 nn

nrXX

ωωωω

ωω

+−= , (2.109)


( )21ζ2tan

n

n

ωωωωϕ

−= . (2.110)

Fig. 2.41

Fig. 2.42


Figure 2.41 shows the variation of the amplitude ratio (2.109) plotted against nωω for two values of the damping ratio. Figure 2.42 shows the variation of the phase shift ϕ plotted against nωω .

Depending on the frequency range utilized, the instrument indicates displacement, velocity or acceleration.

Vibrometer. Within the range III, when nωω >> , it can be seen that

1XX r ≅ , so that the relative motion rX between the mass and casing, sensed by the transducer, is essentially the same as the displacement 1X of the structure being measured. Figure 2.42 shows that, within this frequency range, the phase shift is πϕ = for light damping ( )0ζ → , so that the casing and the mass m are

vibrating 0180 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 nωω << , equation (2.109) becomes

( )21 nr XX ωω≅ ,

or

( )212

1 ωω

XXn

r ≅ , (2.111)

where 21ωX 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 nωω ≅ , 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 1XX r is almost constant.


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 70ζ .= has a horizontal response curve down to 4=nωω . 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 nωω is relatively large, and ϕ is

approximately 0180 for all harmonics, thus no phase distortion occurs. For an accelerometer, when the damping ratio is about 70. , a nearly linear relationship exists between phase angle and frequency, ( )( )nωωπϕ 2≅ . 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 df can be assumed to be proportional to the tangential velocity ωCr , so that

ωCd rcf −= , where c is the coefficient of external viscous damping.


The free body diagram of the disc is shown in Fig. 2.44. The elastic restoring force due to the shaft bending Crk acts along OC. The centrifugal force

Grm 2ω , due to the offset eCG = , 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

( ) CC rkerm =+ ϕω cos2 , ωϕω Crcem =sin2 . (2.112)

Fig. 2.44

The circular orbit of point C has a radius

( )( )[ ] ( ) 2 2 2

2

2

2

ζ21

cos

nn

nC e

mkemr

ωωωω

ωωω

ϕω

+−=

−= . (2.113)

At the critical speed, when nωω = , the orbit radius is ζ2erC = .

Fig. 2.45


The angle between the line CG and the line OC is given by

( ) 22 1ζ2tan

n

n

mkc

ωωωω

ωωϕ

−=

−= . (2.114)

This angle ϕ grows with the speed ω . When nωω < (Fig. 2.45, a), the disc rotates with G outside C. With increasing speed, the segment OC increases, and CG rotates with respect to OC. When nωω = , the line CG leads OC by 090 . When nωω > , the disc rotates with G inside C, and OC decreases. At very high speeds, point G coincides with point O, the radius Cr 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

( ) fxxcxkxm =−++ 1&&&& , ( ) 111 xkxxc =− && . (2.115)

For an excitation teFf ωi0= , assume solutions of the form

teXx ωi= , teXx ωi11 = , (2.116)

where X and 1X are complex amplitudes.


Equations (2.115) become

( )( ) ,XckXc

,FXcXcmk0ii

ii

11

012

=++−

=−+−

ωω

ωωω (2.117)

and can be written

( )( ) ,XNX

,kFXX0ζ2i ζ2 i

ζ2 iζ2i1

1

012

=++−

=−+−

ββ

βββ (2.118)

where

mkn =ω , nωωβ = , nmc ω2ζ = , kkN 1= . (2.119)

Fig. 2.47

The complex displacement amplitude of the mass m is


( )( )220 1ζ2i1

ζ2i1βββ

β−++−

+=

NNN

kFX . (2.120)

The displacement amplification factor

( )

( ) ( ) ( )22222

2

0 1ζ21

ζ21

βββ

β

−++−

+=

NN

NkF

X. (2.121)

is graphically presented in Fig. 2.47 for a stiffness ratio 5=N 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

42

3

22

122

0 ζζ

CCCC

kFX

++

==⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ψ . (2.122)

Equation (2.122) can also be written as

( ) 0ζ22

241

23 =−+− CCCC ψψ . (2.123)

All curves represented by equation (2.123) are passing through the crossing point of the curves of equations

012

3 =− CC ψ , 022

4 =− CC ψ .

These can be expressed as


31 CC=ψ , 42 CC=ψ ,

or

20 1

1β−

=kF

X, (2.124)

and

20 1

1β−+

=NkF

X. (2.125)

Equation (2.124) represents the curve (2.123) of parameter 0ζ = . Equation (2.125) represents the curve (2.123) of parameter ∞=ζ . The two curves intersect each other at a point of frequency ratio ( ) 22+= Nβ and of ordinate

( ) NkFX 20 = . 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 1=β for 0ζ = , to N+= 1β for ∞=ζ . When the damping increases, the peak of the response first decreases, then

increases, thereby indicating that an optimum degree of damping exists

( )22ζ += NNopt ,

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 2>N , over the damping range of opt. ζζ70 << , 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

( )11 xxcxkf && −+= , (2.126)

( ) 111 xkxxc =− && . (2.127)

Since we are not interested in the “hidden” coordinate 1x , containing the internal degree of freedom, we solve equation (2.127) for 1x and substitute the result in equation (2.126) to obtain


( ) ( ) τττ dxtGxkft

&∫ −+=0

1 , (2.128)

where

( )t

ck

ektG1

1−

= (2.129)

with the underlying assumption that for 0=t 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 1f is given as a function of time, the solution of equations (2.126) and (2.127) is

( ) ( ) ( ) ττττ

dtfekk

kckk

tftx

t

−⎟⎟⎠

⎞⎜⎜⎝

⎛

++

+= ∫

−

1

0

2

1

1

1

1 1 1 , (2.130)

where

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

11

11kk

cτ

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

xkf =1 , (2.131)

where the complex stiffness k is

ckkc

kkω

ωi

i1

1

++= . (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

ee ckk ωi+= , (2.133)

where the equivalent stiffness and equivalent coefficient of viscous damping are

2221

22

1ck

ckkkeω

ω+

+= , 2221

21

ckk

cceω+

= . (2.134)


The model with hereditary damping is thus reduced to a Kelvin-Voigt model with frequency-dependent parameters. The equivalent spring stiffness ek increases with frequency from k to the asymptotic value 1kk + , whilst the equivalent coefficient of viscous damping ec decreases from c to zero.

The energy dissipated per cycle is

2221

2122

ckkcXcXW ed ω

ωπωπ+

== (2.135)

which is zero for 0=ω and ∞=ω , having a maximum value at ck10 =ω where 2cce = .

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 0ωω = 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 1kk + .

Exercises

2.E1 An unknown mass m is hung on a spring of unknown stiffness k. When a mass kg501 .m = is added to m, the system natural frequency is lowered from Hz50 to Hz49 . 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 Hz50 . b) Determine the value of k ′ .

Answer: kg12612.m = , mN10191 6⋅= .k , mN56=′k .


2.E2 A mass kg50.m = vibrating in a viscous medium has a period sec150.T = and an initial amplitude mm100 =a . a) Determine the stiffness k and

the viscous damping coefficient c if the amplitude after 12 cycles is mm2012 .a = . b) Determine the amplitude 12a′ when a mass kg301 .m = is added to m.

Answer: mN877=k , msN1732.c = , mm45012 .a =′ .

2.E3 A spring-mass system with a natural frequency Hz5=f vibrates in a viscous medium having a damping coefficient mmsN0020.c = . The logarithmic decrement is 310.=δ . a) Determine the mass m and the spring stiffness k. b) Find the new value of the logarithmic decrement if a mass kg11 =m is added to the first mass.

Answer: kg6450.m = , mN4636.k = , 1940.=′δ .

2.E4 A spring-mass system with viscous damping is displaced a distance mm200 =a and released. After 8 cycles of vibration the amplitude decays to

mm48 =′a . A mass kg21 =m is attached to the initial mass producing a static displacement of mm4 . If the new system is displaced a distance mm200 =a and released, after 8 cycles of vibration the amplitude decays to mm58 =′′a . Determine the mass m, the stiffness k and the damping coefficient c.

Answer: kg775.m = , mN4905=k , mNs7610.c = .

2.E5 A vibrating system of mass kg5=m and spring stiffness mmN1=k is acted upon by a harmonic force of amplitude N100 =F and

frequency Hz2 . Determine a) the displacement amplitude X; b) the displacement amplitude X ′ when a mass kg21 =m 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: mm3447.X = , mm8595.X =′ , add mN15.53=′k in parallel, or mN.58546=′k in series.

2.E6 A weight N20=gm attached to a light spring elongates it mm1 . Determine a) the amplitude 0F of the force that, acting with a frequency of Hz20 , produces vibrations with an amplitude mm53.X = . b) Find another excitation


frequency at which the force of magnitude 0F produces the same displacement mm53.X = .

Answer: N68420 .F = , secrad8761.=′ω .

2.E7 A small motor of mass kg2=m is found to be transmitting a force of N14 to its supporting springs when running at a speed of secrad30 and has an amplitude of vibration of mm53. . Determine a) the amplitude of the unbalanced force 0F ; b) Find another value of the running speed at which the amplitude of vibration has the same magnitude.

Answer: N770 .F = , secrad6755.=′ω .

2.E8 A fragile instrument of mass kg1=m 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 mm201 .X = and frequency Hz15=f . To reduce the vibration of the instrument, it is isolated from the table by means of rubber pads of stiffness mN0007,k = . Determine a) the amplitude X of the displacement of the instrument: b) the value X ′ of the amplitude, if a pad of stiffness

mN0002,k =′ is added in parallel with the first pad.

Answer: mm7430.X = , mm14815.X =′ .

2.E9 A mass kg4=m is supported between two springs attached to fixed points is excited by a harmonic force of amplitude N100 =F . The upper spring has a stiffness mN0005,k = and the lower spring has a stiffness 2k. Determine a) the frequencies at which the displacement amplitude is mm20=X , and b) the amplitude 0F ′ of the force transmitted to the lower support through the spring 2k.

Answer: secrad2601 .=ω , secrad25622 .=ω , N2000 =′F .

2.E10 An undamped vibrating system has an equivalent mass m and an equivalent stiffness k. A harmonic force of amplitude N10 =F and frequency

secrad14=ω produces steady-state vibrations of amplitude X. When a mass kg21 =m is added to m, the force of amplitude N10 =F must have a frequency of

either secrad10=ω or secrad12=ω in order to produce the same displacement amplitude X. Determine the values of m and k.


Answer: For 1ωω < , kg0832.m = , mN1498.k = ; for 1ωω > , kg5385.m = , mN6919.k = .

2.E11 A machine weighing N00012,gm = is mounted at the middle of a

beam of length m2=l , cross section second moment of area 4mm20=I and

Young’s modulus 25 mmN1012 ⋅= .E , simply supported at the ends. If the machine has a rotating unbalanced weight N00020 ,gm = with eccentricity

mm10.e = and speed rpm1500=n , 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: μm2250.X = , N77632.Fdin = .

2.E12 A motor of mass kg10=m , supported on springs of total stiffness mN0002,k = , generates an unbalanced force of magnitude N200 =F at a

frequency of Hz2 . 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 0TF . Find the value of another frequency of the excitation force having the same effect.

Answer: mN6731.k =′ in parallel, secrad74710.=′ω .

2.E13 A vibrating system consists of a mass kg20.m = and a spring of stiffness mmN20.k = . It is acted upon by a harmonic force of amplitude

N50 =F and frequency secrad35=ω . Determine a) the amplitude 0TF of the force transmitted to the support, and b) the amplitude X of the displacement of the mass m.

Answer: N22220 .FT = , m110.X = .

2.E14 A spring-mass system with viscous damping is acted upon by a harmonic force of amplitude N100 =F . Varying the frequency of the excitation force, a peak amplitude mm601 =X is measured at the frequency secrad141 =ω . When a mass kg21 =m is added to the system, the peak amplitude becomes

mm802 =X . Determine the mass m, the stiffness k and the damping coefficient c.

Answer: kg572.m = , mN72503.k = , msN911.c = .


2.E15 A mass kg5=m is attached to a spring of stiffness mmN1=k and is acted upon by a harmonic force of amplitude N100 =F and frequency

zf H2= . Determine: a) the amplitude 1X of the forced vibrations of the mass m; b) the amplitude 2X of the forced vibrations when a mass kg21 =m 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: mm3471 .X = , mm5922 .X = , msN310.c = .

2.E16 Using Rayleigh’s method, estimate the fundamental frequency of the vibrating beams shown in Fig. 2.50 using the given deflection function.

a

( ) ⎟⎠⎞

⎜⎝⎛ −=

l2cos10

xπvxv .

b

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

3

0 43ll

xxvxv ,

20 l≤≤ x c

( )l

xπsin0vxv = .

d

( ) ⎟⎠⎞

⎜⎝⎛ −=

l

xπ2cos10vxv .

Fig. 2.50

Answer: a) 31 6121lmIE.=ω ; b)

( )mA.mIE.

ll ρω

4857019286 31

+= ;

c) 31 9354lmIE.=ω ; d)

AIE

ρω 21

22.79l

= .


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 lAm ρ2= .

Fig. 2.51

Answer: ρ

ω E.l

65501 = .

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 non-linearities, 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 supra-harmonics. The forced response is studied only in the neighbourhood of the so-called 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

( ) ( ) ( )txktxct,x,xf lin,R += && ,

containing the contribution of a viscous damper and a linear spring.

For slightly non-linear systems, the restoring force can be expressed as

( ) ( ) ( )txktxct,x,xf eqeqR +≅ && ,


where is the equivalent viscous damping coefficient, and is an equivalent stiffness. An equivalent structural damping model may be used as well.

eqc eqk

Cubic Stiffness

For systems without pre-loading or clearance, the elastic force may be represented by a cubic stiffness law

( )3xxkfe μ+= , (3.1)

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

For a harmonic displacement

( ) tatx ωcos= , (3.2)

the elastic force (3.1) becomes ( )tatakfe ωμω 32 coscos += .

Substituting ttt ωωω 3cos41cos

43cos3 += , and neglecting the higher

harmonic term in tω3cos , yields

xk....tatakf eqe =⎟⎠⎞

⎜⎝⎛ ++≅ ωμω cos

43cos 2 ,

where the equivalent stiffness is

⎟⎠⎞

⎜⎝⎛ += 2

431 akkeq μ . (3.3)

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 dissipated per cycle of vibration by the non-linear damping element be

equal to the energy dissipated by an equivalent viscous damper experiencing the same harmonic relative displacement.

dW2aceqωπ

Similar analysis can be carried out based on a concept of equivalent structural damping when . 2ahW eqd π=

Steady-state solutions obtained with the assumption of equivalent viscous damping have been shown to be identical to those obtained by use of the averaging

3. SIMPLE NON-LINEAR SYSTEMS 81

method of Ritz. The same value is obtained using the coefficient of the first term of a Fourier series expansion of the non-linear damping force time history.

A generalized non-linear damping force can be described mathematically as being proportional to the nth power of the relative velocity across the damper

( )xxcf nnd && sgn= , (3.4)

where is defined as the velocity-nth power damping coefficient. nc

The exponent n and the damping coefficient are determined according to the nature of the damping element. A value of

nc0=n represents a Coulomb

damper where the damping coefficient equals the dry-friction force 0c R . Similarly, values of and 1=n 2=n represent viscous and quadratic damping, where becomes the damping coefficients c and , respectively. nc 2c

Other values of the exponent n may be selected to represent other non-linear damping characteristics. For example, the damping developed in a car shock-absorber may be described by a value of the exponent n between and , depending on the particular system configuration.

02. 03.

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.


3.2.1 Harmonic Response

If the mass is acted upon by a harmonic force of constant amplitude and frequency

0Fω , the Duffing-type equation of motion of the vibrating mass may

be written as

( ) teFxxkxhxm ωμω

i0

3 =+++ &&& , (3.5)

where , and kgh = g can be an equivalent structural damping factor.

A first harmonic approximation of the response is chosen of the form

( ) ( )θωωω +=+== ttIR

t eaeaaea~x iii i . (3.6)

Using the method of harmonic linearization, the higher harmonic terms are neglected, so that it is considered that

xax 23

43

≅ . (3.7)

Substitution of (3.6) and (3.7) into (3.5) yields the real and imaginary components of displacement

2

0

222

0

22

431 ⎟

⎟⎠

⎞⎜⎜⎝

⎛−=⎟

⎠⎞

⎜⎝⎛ −+=

Fakgaa

FkaaR mημ , (3.8)

2

0a

FkgaI −= , (3.9)

where

nωωη = ,

mk

n =ω . (3.10)

The displacement magnitude

22IR aaa += (3.11)

is implicitly given by

222

2022

431 g

akFa −±+= μη . (3.12)

The phase angle is calculated from

222

20

22

431

tan

gak

F

g

a

g

−±

=−−

=μη

θ . (3.13)

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

a 2ηa~


20

202

21

21

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛++

kF

gkF

gaa IR . (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 of response Ia

( ) ( ) 11431 002 −

−±−+=

II akg

Fgakg

Fμη . (3.15)

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

Ra

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−±⎟⎟

⎠

⎞⎜⎜⎝

⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛

−⎟⎟⎠

⎞⎜⎜⎝

⎛+=

22

0

20

022

02

02

2112

2

23

231

R

RR

aF

kgkg

F

akFa

kgF

kgF

μμη . (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 .constg = and several values of the amplitude of harmonic force.

0F

The response curves are symmetrically disposed with respect to the “skeleton curve” of equation

⎟⎠⎞

⎜⎝⎛ += 222

431 an μωω (3.17)

which passes through the points of maximum amplitude. For linear systems it is a vertical line of abscissa nωω = . For systems with hardening stiffness ( 0> )μ , the skeleton curve is bent towards higher frequencies. For systems with softening stiffness ( 0< )μ , it is bent towards lower frequencies.

The locus of the points of vertical tangency of the magnitude-frequency curves is given by equation

24222

169

431 gaa −±+= μμη (3.18)

and defines the “stability boundary” XLKY (Fig. 3.2).


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 .constg = They are called constant-frequency 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).


Again, a stability boundary XLKY can be defined, of equation

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

θθη

tan3tan

212 g (3.19)

which is the locus of the points of vertical tangency of the phase-frequency curves.

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

22

22

491

31 R

Ra

ag μμ

η ++= (3.20)

and is valid for . 0<Ra

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

2

222

431

II a

gaμ

μη ++= (3.21)

and holds for . 0<Ra

The best way to represent the frequency response data is to plot the vector components of the displacement and , 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).

Ra Ia


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 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.

0F

Fig. 3.7

In Fig. 3.7, isochrones are drawn by broken lines. Their equation is obtained by eliminating between equations (3.8) and (3.9). This yields 0F

( )gaaaa I

IRR ⎥⎦⎤

⎢⎣⎡ +−−= 222

431 μη . (3.22)

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

0F 0F


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

μ32gaa IR = (3.23)

(defined only for , 0<Ra 0<Ia ), which is symmetrical with respect to the bisector of the coordinate axes. IR aa =

The stability boundary XLKY intersects the bisector IR aa = at the point L, of frequency gnL 21+= ωω , at a distance μ34gaL = 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 ( )μ3932 30 gkF = and to the isochrone of parameter

gnK 31+=ωω , indicates the lowest force level and forcing frequency at

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

0120−=Kθ

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.


Introducing a new dimensionless coefficient of non-linearity 232

0 kgFμγ = , it means that for either 3932=> Kγγ or Kωω > the stability boundary crosses the Nyquist plots and the isochrones. We may consider systems with weak non-linearities those for which Kγγ < , and systems with strong non-linearities those for which Kγγ > . According to this classification, weak non-linearities 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


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 of the excitation force (Fig. 3.10, a). 0F

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 .constu =ω , 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

.const=lω 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 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.

0F


It jumps to the point T, following the response curve .constF =′′0 , 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 . and then moves again along the isochrone from V to O.

constF =′0

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 and

( of the harmonic excitation force. 0F ′

00 FfF ′⋅=′′ )1>f

Let denote by

10 akg

F'OM =′

= , 20 akg

FMO =′′

=′′ ,

the maximum displacement amplitudes on the two response circles.

The above equations yield

1

2

0

0aa

FFf =′′′

= . (3.24)


For linear systems, points 'M and "M have the same frequency mkn =ω . At systems with non-linear stiffness characteristic, the frequencies

rω′ and rω ′′ of these points are given by

214

31 anr μωω +=′ , 224

31 anr μωω +=′′ . (3.25)

Fig. 3.12

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

12

2222

−

′′−′=

f

f rrn

ωωω (3.26)

and the coefficient of non-linearity

222

22

213

4

rr

rr

fa ωω

ωωμ

′′−′

′−′′= . (3.27)

An arc of circle of radius 1a'OM = and centre at the origin O crosses the

second circle at the points P and Q, of frequencies Pω and Qω , respectively, given by

⎟⎠⎞

⎜⎝⎛ −+= 1

431 22

122 fganQ,P mμωω . (3.28)

The structural damping factor is calculated from

2

2

22

22

2 1

1

r

n

PQ

PQ

fg

ω

ω

ωω

ωω′+

−

−= . (3.29)


The stiffness and mass parameters are then given by

10 agFk ′= , 2nkm ω= . (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

( )xRxxRfd &&

&sgn== (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 acts through a relative displacement df

( ) tatx ωsin= (3.32) is given by

( )∫∫∫ ====2

00

4dcos4ddπ

ωω aRttfaxxfxfW d

T

ddd & (3.33)


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 2ahWd π= . (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 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

eqh

aRaheq 42 =π , wherefrom

aRheq π

4= (3.35)

which is amplitude dependent.


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

( ) ( )tfxkxRxhxm =+++ &&&& sgnω

. (3.36)

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

( ) tatx ωsin=

and the force is conveniently expressed as

( ) ( )θω −= tFtf sin0 . (3.37)

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

( θωω

−=++

+ tFxkxhh

xm eq sin0&&& ) . (3.38)


Substituting the solution (3.32) into equation (3.38) and equating the coefficients of the terms in tωcos and tωsin of both sides, the following equations are obtained

( ,a

,rga21cos

sin

ηθ )θ

−=

−−= (3.39)

where

0

4FRr

π= ,

0Fkaa = ,

nωωη = ,

mk

n =ω , khg = . (3.40)

Equations (3.39) give the magnitude of the dimensionless displacement

( )( )( ) 222

2222

1

11

g

grrga

+−

+−−+−=

η

η (3.41)

and the phase angle

1tan 2

1-

−

+=

ηθ a

rg. (3.42)

A solution for a is only possible when 1<r ( )πRF 40 > .

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 arggg eq +=+ .

Considering a dimensionless complex displacement amplitude

θθ sinicosi0

aaaaF

ka~IR +=+= (3.43)

we obtain

( )22 1 η−= aaR , ⎟⎠⎞

⎜⎝⎛ +=

argaaI

2 , (3.44)

where the magnitude 22IR aaa += 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


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 nω may be determined as the parameter of point M, where the diagram crosses the imaginary negative semi-axis. The displacement amplitude at resonance is

( rg

OMaM −== 11 ) . (3.45)

Fig. 3.14

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

045

rg

n, 21121

−= mωω . (3.46)

The corresponding dimensionless vector lengths are

⎟⎟⎠

⎞⎜⎜⎝

⎛−== r

gOCOB

221

so that

( )rg

OBBC 2112 −== (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 , using as coordinates the coincident and 0F


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


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 and

( of the harmonic excitation force (Fig. 3.16). 0F ′

00 FfF ′⋅=′′ )1>f

The resonance points 'M and "M , of frequency nω , are located at the crossing points of the plots with the only isochrone which is a straight line.

Denoting 'OM"OM=α , the following equation can be established for the calculation of the Coulomb friction force

140

−−′

=ααπ fFR . (3.48)

If an arc of circle of radius 'OMa =1 and centre in the origin O is drawn, it crosses the second polar plot at points P and Q, of frequencies given by


( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛

−+

−=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

′−

−=

111141

11 2

0

222

ffg

FR

fgnnQ,P αω

π

ωω mm . (3.49)

The structural damping factor g can be calculated from

11

11

22

22

+−

−+

−=

ffg

PQ

PQ

αωω

ωω (3.50)

where 222 2 nQP ωωω =+ . (3.51)

Fig. 3.16

The stiffness is further given by

⎟⎟⎠

⎞⎜⎜⎝

⎛′

−′

=01

0 41FR

agFk

π. (3.52)

and the mass by 2nkm ω= .

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


approximation in describing a certain class of non-linear damping phenomena arising in systems including oil dampers and shock absorbers.

The damping force is proportional to the square of the relative velocity across the damper (minus sign omitted)

df

( )xxcfd && sgn22= , (3.53)

where is defined as the quadratic damping coefficient, being a function of damper geometry and fluid properties.

2c

The energy dissipated per cycle of vibration by a quadratic damper in which a force acts through a relative displacement df ( ) tatx ωsin= is given by

( )∫∫∫ ====2

0

322

038dcos4dd

π

ωωω acttfaxxfxfW d

T

ddd & , (3.54)

being dependent upon the frequency and amplitude of vibration.


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

( ) ( )tfxkxxcxm =++ &&&& sgn22 . (3.55)

A first harmonic approximation of the response is chosen of the form ( ) tatx ωsin= and the force is conveniently expressed as ( ) ( )θω −= tFtf sin0 .

Fig. 3.17

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


22

2

38sin ηαπ

θ a−= , ( )21cos ηθ −= a , (3.56)

in which

mkFc 02

2 =α , 0Fkaa = ,

nωωη = ,

mk

n =ω , (3.57)

and where 2α 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 :

222

22

222

22

381

38

3811

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−±

=

πα

πα

πα

η

a

aa

, (3.58)

138

tan 2

22

1

−=

η

ηπα

θa

- . (3.59)

Considering a complex displacement amplitude

θθ sinicosi aaaaa~ IR +=+=

we obtain

( )2

0

2 1 η−=FkaaR , 22

0

3

38 ηπ mc

FkaaI −= , (3.60)

where the magnitude 22IR aaa += is given by

( ) ( ) 42

02

24222

0

31611

2

ηπ

ηη ⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+−+−

=

kF

mc

kF

a . (3.61)


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 nω 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


20

2

083

83

ckmF

kFOMaM

παπ

=== . (3.62)

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

045

γγη m21+=C,B , (3.63) where

22

32 απ

γ = (3.64)

which may be used for the evaluation of damping.


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

and 0F ′ 00 FfF ′⋅=′′ of the excitation force amplitude. The points ( 1>f ) 'M and "M , of frequency 1=η , 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

mkFc 022

38

38

ππαα == (3.65)


that is a function of the amplitude of the excitation force. 0F

For the two polar plots of Fig. 3.20, one can write

02

38 F

mkc ′=′

πα , 0

23

8 Fmk

c ′′=′′π

α (3.66)

so that the ratio of the excitation force amplitudes can be denoted 2

0

0 ⎟⎠⎞

⎜⎝⎛=

′′′

=′′′

='OM"OM

FFf

αα . (3.67)

Fig. 3.21


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

( )α

ααη

′+

′+−′=

111 222

2 ffQ,P

m. (3.68)

Equations (3.68) yield the damping parameter

1222 −

+=′

QP ηηα . (3.69)

The quadratic damping factor 2α′ is given by

απα ′=′8

32 . (3.70)

The stiffness is further obtained from

α′′

=1

1

0aFk . (3.71)

and the quadratic damping coefficient from

02

22

2 Fkc

n ′′

=ωα . (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 0xx + , where is the deflection produced by pre-load, the stiffness function is represented by three terms

0x

( )3200 xxxkfe μμ ++= . (3.73)

The square term due to pre-load has a softening effect irrespective if whether 0μ 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.


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, 0μ 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 single-degree-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.


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 1m and

2m , attached to fixed points by springs 1k and 3k , and tied together by a “coupling spring” 2k .

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 1x and 2x 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

( ) 02121111 =−++ xxkxkxm && , ( ) 02122322 =−−+ xxkxkxm && .

4. TWO-DEGREE-OF-FREEDOM SYSTEMS 107

On rearranging, these equations become

( )( ) .xkkxkxm

,xkxkkxm

0

0

2231222

2212111

=++−

=−++

&&

&& (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 2k . If 02 =k , 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

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+−−+

+⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

00

00

2

1

232

221

2

1

2

1

xx

kkkkkk

xx

mm

&&

&&, (4.2)

or in compact form

[ ]{ } [ ]{ } { }0=+ xkxm && , (4.3)

where [ ]m is the mass matrix, [ ]k is the stiffness matrix and { }x is the column 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

[ ] [ ]Tmm = , [ ] [ ]Tkk = . (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

( ) ( )( ) ( ),tatx

,tatx

ϕω

ϕω

−=

−=

cos

cos

22

11 (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

( ) ( ) .constaatxtx == 2121 (4.6) The shape of the system configuration does not change during the motion, the deflected shape resembles itself at any time.


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

( )( ) .amkkak

,akamkk

0

0

22

22312

2212

121

=−++−

=−−+

ω

ω (4.7)

The simultaneous homogeneous equations (4.7) admit non-trivial solutions if the determinant of the coefficients 1a and 2a is zero

022232

22

121 =−+−

−−+

ωω

mkkkkmkk

. (4.8)

This can be written

( ) 021

231312

2

23

1

214 =++

+⎟⎟⎠

⎞⎜⎜⎝

⎛ ++

+−

mmkkkkk

mkk

mkk ωω (4.9)

which represents a quadratic equation in 2ω called the characteristic equation, or frequency equation. Its roots 2

1ω and 22ω are real and positive. The quantities 1ω

and 2ω 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 1a and 2a cannot be determined, only their ratio 12 aa=μ .

For 1ω , the amplitude ratio is

2

21223

11

21 k

mkkaa ω

μ−+

=⎟⎟⎠

⎞⎜⎜⎝

⎛= (4.10)

and for 2ω , the amplitude ratio is

2

22223

21

22 k

mkkaa ω

μ−+

=⎟⎟⎠

⎞⎜⎜⎝

⎛= . (4.11)

The ratios (4.10) and (4.11) determine the shape of the system during synchronous motion with frequencies 1ω and 2ω , 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


{ } ( ) { }

{ } ( ) { } ,uCu

aaa

a

,uCu

aaa

a

22221

22

12

11111

12

11

1

1

=⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

=

=⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

=

(4.12)

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

( ){ } { } ( )( ){ } { } ( )22222

11111

cos

cos

ϕω

ϕω

−=

−=

tuCtx

,tuCtx (4.13)

and the general solution of free vibrations is

( ){ } ( ){ } ( ){ } { } ( ) { } ( )2222111121 coscos ϕωϕω −+−=+= tuCtuCtxtxtx .(4.14)

In (4.14) the four integration constants 1C , 2C , 1ϕ , 2ϕ are determined from the four initial conditions, the displacements and velocities at 0=t .

For arbitrary initial conditions, the free vibration of the two-degree-of-freedom 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


⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−+

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡003

002

2

1

2

1

xx

kkkk

xx

mm

&&

&&.

Assuming solutions of the form

{ } { } ( )ϕω −= tux cos

we obtain

( )( ) ,umkuk

,ukumk

0

023

22

1

212

=−+−

=−−

ω

ω

or, dividing by k,

( )( ) ,uu

,uu

01

023

21

21

=−+−

=−−

α

α

where

km 2ωα = .

The condition to have nontrivial solutions is

011

123=

−−−−α

α, 0252 2 =+− αα

with solutions

211 =α , 22 =α .

The natural frequencies are

mk

21

1 =ω , mk22 =ω .

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

021

211

12

11123

12

1

1

1 =⎭⎬⎫

⎩⎨⎧

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−

−=

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−−−−

uu

αα

and the second modal vector is calculated from

01

11111

11123

22

1

2

2 =⎭⎬⎫

⎩⎨⎧−⎥

⎦

⎤⎢⎣

⎡−−−−

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−−−−

uu

αα

.

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


the left, the displacement of the mass m being always twice the displacement of the mass m2 . 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 k2 . The mass m2 is thus connected to ground by two springs of stiffness k2 (in parallel) with an equivalent stiffness k4 , while the mass m is connected to one spring of stiffness

k2 , both subsystems having the same natural frequency equal to 2ω .

Fig. 4.3

Summarising, the first normal mode has a natural frequency mk

21

1 =ω

and a mode shape { }⎭⎬⎫

⎩⎨⎧

=21

1u , while the second normal mode has a natural

frequency mk22 =ω and a mode shape { }

⎭⎬⎫

⎩⎨⎧−

=1

12u . If the system is given

an initial disturbance of 11 =x and 22 =x and then released, the ensuing motion

will be purely harmonic with the frequency mk.20701 =ω . If the initial displacement is 11 =x and 12 −=x , the motion will be harmonic with the

frequency mk.41412 =ω .

4.1.3 Orthogonality of Natural Modes

Substituting solutions (4.13) in equation (4.3) we obtain

[ ]{ } [ ]{ }1211 umuk ω= , (4.15, a)


[ ]{ } [ ]{ }2222 umuk ω= . (4.15, b)

Multiplying (4.15, a) on the left by { }T2u we get

{ } [ ]{ } { } [ ]{ }1T2

211

T2 umuuku ω= . (4.16)

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

obtain { } [ ]{ } { } [ ]{ }1T

2221

T2 umuuku ω= . (4.17)

Subtracting (4.16) and (4.17) from each other, for 21 ωω ≠ , we get

{ } [ ]{ } 01T2 =umu , (4.18)

and taking the transpose { } [ ]{ } 02

T1 =umu . (4.19)

Substituting (4.18) in (4.16) we obtain

{ } [ ]{ } 01T2 =uku (4.20)

and taking the transpose { } [ ]{ } 02

T1 =uku . (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 { } { } .ba 0T =

4.1.4 Modal Coordinates

Let denote in (4.13)

( ) ( )( ) ( )2222

1111

cos

cos

ϕω

ϕω

−=

−=

tCtq

,tCtq (4.22)

so that (4.14) becomes

( ){ } { } { } 2211 ququtx += . (4.23)


( ){ } { } { }[ ] [ ]{ }quqq

uutx =⎭⎬⎫

⎩⎨⎧

=2

121 (4.23, a)


where

[ ] { } { }[ ]21 uuu = (4.24)

is called the modal matrix.

Substituting (4.23) in (4.3), multiplying on the left by { }Tru ( )21,r = and

considering the orthogonality we obtain

0=+ rrrr qKqM && , (4.25) where

{ } [ ] { }rrr umuM T= , { } [ ] { }rrr ukuK T= (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

{ } [ ] { }{ } [ ] { }rr

rrr

umuuku

T

T2 =ω . 21,r = (4.27)

Rayleigh’s quotient is defined as

{ }( ) { } [ ] { }{ } [ ] { }umu

ukuuR T

T= . (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 { }( )uR 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 [ ]Tu yields

[ ] [ ] [ ] { } [ ] [ ] [ ]{ } { }0TT =+ qukuqumu && , (4.29) or

[ ] { } [ ] { } { }0=+ qKqM && , (4.29, a)


where the diagonal matrices

[ ] [ ] [ ] [ ]umuM T= and [ ] [ ] [ ] [ ]ukuK T= (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 ( )tf1 and ( )tf2 applied on mass 1m , respectively 2m .

Fig. 4.4

The equations of motion are

( )( ) ,fxkkxkxm

,fxkxkkxm

22231222

12212111

=++−

=−++

&&

&& (4.31)

or, in compact matrix form,

[ ] { } [ ] { } { }fxkxm =+&& , (4.31, a)

where { }f is the column vector of impressed forces.

4.1.5.1 Solution by Modal Analysis

Substituting (4.23, a) in equation (4.31, a), and multiplying on the left by [ ]Tu we obtain

[ ] { } [ ] { } { }FqKqM =+&& , (4.32) where

{ } [ ] { }fuF T= (4.33)


is the column vector of modal forces.

For harmonic excitation

{ } { } tf̂f ωcos= (4.34)

the steady-state response is

{ } { } tx̂x ωcos= , { } { } tq̂q ωcos= , (4.35)

where the a hat above a letter denotes amplitude.

Substituting (4.34) and (4.35) in (4.32) we obtain

[ ] [ ][ ]{ } [ ] { } { }F̂f̂uq̂KM ==+− T2ω , (4.36)

so that the amplitudes of modal coordinates are

{ } { }{ } [ ] { } { } [ ] { }1T

12

1T1

T1

12

1

11

umuuku

f̂u

MK

F̂q̂

ωω −=

−= , (4.37)

{ } { }{ } [ ] { } { } [ ] { }2

T2

22

T2

T2

22

2

22

umuukuf̂u

MK

F̂q̂

ωω −=

−= . (4.38)

Equation (4.23) for amplitudes becomes

{ } { } { } 2211 q̂uq̂ux̂ += . (4.39)

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

{ } { } { }{ }{ } [ ] { } { } [ ] { }

{ } { }{ }{ } [ ] { } { } [ ] { }2

T2

22

T2

2T2

1T1

21

T1

1T1

umuukuuf̂u

umuukuuf̂u

x̂ωω −

+−

= .

(4.40) The elements of { }x̂ are of the form

{ } { } { } { }12

22

2

T2

111

21

T1

1 uMKf̂u

uMK

f̂ux̂

ωω −+

−= , (4.41)

{ } { } { } { }22

22

2

T2

211

21

T1

2 uMKf̂u

uMK

f̂ux̂

ωω −+

−= . (4.42)


Example 4.2 Consider the system of Fig. 4.2 acted upon by harmonic forces

tf̂f ωcos11 = and tf̂f ωcos22 = and obtain the steady-state response amplitudes. Plot the frequency response curves of receptance when 02 =f .

Solution. The equations of motion can be written

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−+

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

2

1

2

1

2

1 30

02ff

xx

kkkk

xx

mm

&&

&&.

Using the coordinate transformation

{ } [ ] { }⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎭⎬⎫

⎩⎨⎧−

+⎭⎬⎫

⎩⎨⎧

==2

121 12

1111

21

qq

qqqux

and multiplying on the left with the transpose of the modal matrix, we obtain

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡−

−⎥⎦

⎤⎢⎣

⎡−

+⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡− 2

1

2

1

2

1

1121

12113

1121

1211

002

1121

ff

qq

kkkk

qq

mm

&&

&&

or

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

−+

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

2

1

21

21

2

1

2

1 26003

3006

FF

ffff

qq

kk

qq

mm

&&

&&.

For harmonic excitation { } { } tf̂f ωcos= , the steady-state response is

{ } { } tq̂q ωcos= . The amplitudes of modal coordinates are

( )2

12

212

11

13

263 ωωω −

+=

−=

k

f̂f̂mk

F̂q̂ , ( )2

22

212

22

1636 ωωω −

−=

−=

k

f̂f̂mk

F̂q̂ .

The amplitudes in the configuration space are given by

{ }{ }

( )

{ }( )2

222

12 16

11

11

13

21

21

ωωωω −⎭⎬⎫

⎩⎨⎧−

−+

−⎭⎬⎫

⎩⎨⎧

=k

f̂

k

f̂x̂

cbcb.

When 02 =f and ff =1 , the vector of amplitudes can be written

{ }( ) ( )2

222

12 16

11

011

13

21

021

ωωωω −

⎭⎬⎫

⎩⎨⎧−⎭

⎬⎫

⎩⎨⎧

−

+−

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

=k

f̂

k

f̂

x̂cbcb


and the receptances are

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

==

22

2

21

2

111

16

1

13

1

ωω

ωω

αkk

f̂x̂

,

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

==

22

2

21

22

21

16

1

13

2

ωω

ωω

αkk

f̂x̂

,

or

( ) ( )222

221

1

3

1

6

1

ωωωω −+

−=

mmf̂x̂

, ( ) ( )22

222

1

2

3

1

3

1

ωωωω −−

−=

mmf̂x̂

.

The frequency response curves are given in Fig. 4.5.

a b

Fig. 4.5

Resonances occur at mk

21

1 =ω and mk22 =ω , when the

excitation frequency equals any of the natural frequencies of the system. When

mk

a ==ωω , the first mass stands still in space, 01 =x̂ , while the second mass

moves, 02 ≠x̂ , condition defined as an antiresonance. The antiresonance 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 m2 still in space, condition desirable in many practical applications.


4.1.5.2 Solution by Spectral Analysis

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

[ ] [ ]( ){ } { }f̂x̂mk =− 2ω , (4.43)

or

{ } [ ] [ ]( ) { }f̂mkx̂12 −

−= ω . (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

tf̂f ωcos11 = and obtain the steady-state response amplitudes by direct spectral analysis.

Solution. The equations of motion (4.31) are

.xkxkxm

,tf̂xkxkxm

0

cos32

212

1211

=+−

=−+

&&

&& ω

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

( )( ) .x̂mkx̂k

,f̂x̂kx̂mk

0

23

22

1

1212

=−+−

=−−

ω

ω

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

( )( )( ) 222

2

2

2

2

12323

01

kmkmk

f̂mk

mkkkmk

f̂mk

k

x̂−−−

−=

−−−−

−−

=ωω

ω

ωω

ω,

( )( ) 2222

2

2

22323

0123

kmkmk

f̂k

mkkkmk

f̂k

mk

x̂−−−

=

−−−−

−−

=ωω

ωω

ω

,

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


⎟⎠⎞

⎜⎝⎛ −⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⎟⎠⎞

⎜⎝⎛ −

=22

2

1 22

2 ωω

ω

mk

mkm

f̂mk

x̂ , ⎟⎠⎞

⎜⎝⎛ −⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=22

2 22

2 ωωmk

mkm

f̂mk

x̂ .

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.


Consider the system of Fig. 4.6, a which consists of two rigid discs of polar mass moment of inertia 1J and 2J , 2mkg , attached to massless shafts of torsional stiffness 1K , 2K and 3K , radmN .

Fig. 4.6

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


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

( )( ) ,KKJ

,KKJ

0

0

2122322

2121111

=−−+

=−++

θθθθ

θθθθ&&

&&

or ( )

( ) .KKKJ

,KKKJ

0

0

2231222

2212111

=++−

=−++

θθθ

θθθ&&

&& (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 1J and 2J , and the massless shaft have a torsional stiffness lpIGK = .

The equations of motion

,KKJ

,KKJ

0

0

2122

2111

=+−

=−+

θθθ

θθθ&&

&& (4.46)

can be written in matrix form

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−+

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡00

00

2

1

2

1

2

1θθ

θθ

KKKK

JJ

&&

&&, (4.46, a)

and in compact form

[ ]{ } [ ]{ } { }0=+ θθ KJ && .

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.


Apparently, this is a two-degree-of-freedom system. However, summing up equations (4.46) we obtain

02211 =+ θθ &&&& JJ , (4.47)

so that the two coordinates 1θ and 2θ 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 1J , the second equation by 2J and subtracting the results from each other we obtain

( ) 02121

21 =−⎟⎟⎠

⎞⎜⎜⎝

⎛++− θθθθ

JK

JK&&&& . (4.48)

Denoting the twist angle θθθ =− 21 , equation (4.48) becomes

021

21 =++

θθ KJJ

JJ && (4.48, a)

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

4.2.2.1 Normal Modes


( ) ( )( ) ( ),tat

,tat

ϕωθ

ϕωθ

−=

−=

cos

cos

22

11 (4.49)


we obtain the set of algebraic equations

( )( ) .aJKaK

,aKaJK

0

0

22

21

212

1

=−+−

=−−

ω

ω (4.50)

Dividing by K and denoting

αω =KJ12 (4.51)

equations (4.50) become

( )

.aJJa

,aa

01

01

21

21

21

=⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

=−−

α

α (4.52)

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

011

11

1

2 =−−

−−

α

α

JJ (4.53)

or 011

22

1

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛+− αα

JJ

JJ . (4.53, a)

The solutions are

01 =α , 212 1 JJ+=α . (4.54)

The first natural frequency is given by 021 =ω and the second by

KJJ ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

21

22

11ω . (4.55)

The root 021 =ω 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

( )( ) ( )( )tCtCKJtCCt

,tCtCtCCt

2423221212

2423211

cossin1

cossin

ωωωθ

ωωθ

+−++=

+++=

where 41 C,..,C are integration constants.

The mode shapes are determined by the ratio αμ −== 112 aa .


For the first mode

( ) 11 11121 =−== αμ aa , (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

( ) 2122122 1 JJaa −=−== αμ , (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 ( ) tMtM ωcos0= (not shown). The equations of motion are

.tMKKJ

,KKJ

ωθθθ

θθθ

cos

0

02122

2111

=+−

=−+

&&

&& (4.57)

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

( ) tt ωΘθ cos11 = , ( ) tt ωΘθ cos22 = . (4.58)


The resulting algebraic equations are

( )( ) .MJKK

,KJK

022

21

212

1 0

=−+−

=−−

ΘωΘ

ΘΘω (4.59)

a b

Fig. 4.9

Dividing by K and denoting αω =KJ12 we obtain

( )

.K

MJJ

,

02

1

21

21

1

01

=⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

=−−

ΘαΘ

ΘΘα

(4.60)

Solving for 1Θ and 2Θ using Cramer’s rule

KM

JJJ

JJ

JJ

KM

JJ

0

2

21

1

2

1

2

0

1

2

11

11

11

11

10

ααα

α

αΘ

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

=

−−

−−

−

−

= , (4.61)

KM

JJJ

JJ

JJ

KM

0

2

21

1

2

1

2

0

21

11

111101

αα

α

α

α

α

Θ

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

−=

−−

−−−−

= . (4.62)


The amplitudes 1Θ and 2Θ 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

KJJ ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

212

11ω .

The amplitude of the second disc is zero when 1=α . This antiresonance occurs when 1JK=ω , 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

KM

JJJ

JJ

0

2

21

1

212

1

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=−=

αΘΘΔΘ . (4.63)

Denoting the torque in the shaft by tMM tt ωcos0

= , its amplitude is

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+==

αΔΘ

2

21

1

2

00

JJJ

JJ

MKMt . (4.64)

The dynamic shear stresses due to the torsion are tωττ cos0= , and the amplitude is

pt WM00 =τ (4.65)

where pW is the polar modulus of the cross section.

If the shaft transmits a power N at constant angular speed Nω , then the “static” shear stress is

( )pNptN WNWMst

ωτ == . (4.66)

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


Fig. 4.10

Example 4.4 The torsional system of Fig. 4.6 is acted upon by a harmonic torque of

amplitude mN1040 =M and frequency secrad314=ω (not shown). Both discs

have the polar mass moment of inertia 2mkg57=J . The shaft has a length m40.=l , diameter m140.d = and shear modulus GPa81=G . Determine the

amplitude of the dynamic shear stresses in the shaft.

Solution. The shaft cross-section has 484 mm10377032 ⋅== .dI p π and 363 mm10538016 ⋅== .dWp π . The torsional stiffness is lpIGK = = 910637 ⋅.

radmmN . The ratio 73502 .KJ == ωα . The twist angle is ( ) =−= αΔΘ 20 KM rad0010360. . The amplitude of the dynamic torque is

( ) mN7910200=−= αMMt . The amplitude of dynamic shear stresses is

20 mmN714

0.WM pt ==τ

4.2.3 Geared Systems

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

1

2

1

2

2

1

θθ

−=−==nn

rr

i , (4.67)

where 1n and 2n are the rotational speeds of the two shafts, 1θ and 2θ 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


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

( ) ( )eqactual KK 22 θθ = ,

wherefrom

actualactualeq

actualactual

eq

actualeq KiK

nnKK 2

22

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

θθ . (4.68)

Fig. 4.11

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

( ) ( )eqactual JJ 22 θθ && = ,

wherefrom

actualactualeq

actualactual

eq

actualeq JiJ

nnJJ 2

22

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

θθ&

&. (4.69)

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

22

2 KiK eq = , 22

2 JiJ eq = . (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

2i , where -i is the speed ratio of the geared shaft to the reference shaft.


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

ieqactual −=θθ , iMM actualeq −= . (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 1M and 2M may be related to the rotation angles 1θ and 2θ using the equilibrium and the torque/rotation equations

.KMM

,KMM

0 when

0 when

1221

2121

=−=−=

==−=

θθ

θθ (4.72)

Equations (4.72) may be written in matrix form as


⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

2

1

2

1θθ

KKKK

MM

(4.73)

or in shorthand form { } [ ] { }θekM = , where [ ]ek is referred to as the element stiffness matrix.

Fig. 4.13

Using equation (4.73), the torque-rotation equation for each shaft in Fig. 4.12, b can be written

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

3

1

11

11

3

1θθ

KKKK

MM

, ⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

3

2

22

22

3

2θθ

KKKK

MM

,

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

4

3

33

33

4

3θθ

KKKK

MM

.

Each of these equations can be expanded as follows

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

4

3

2

1

11

11

4

3

2

1

000000000000

θθθθ

KK

KK

MMMM

,

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

4

3

2

1

22

22

4

3

2

1

000000000000

θθθθ

KKKK

MMMM

,

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

4

3

2

1

33

33

4

3

2

1

0000

00000000

θθθθ

KKKK

MMMM

.

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


⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−++−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

4

3

2

1

33

332121

22

11

4

3

2

1

00

0000

θθθθ

KKKKKKKK

KKKK

MMMM

. (4.74)

Using the boundary condition 04 =θ , the reduced system stiffness matrix is obtained, so the equations of motion can be written

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−−−−

+⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

000

00

0000000

3

2

1

32121

22

11

3

2

1

2

1

θθθ

θθθ

KKKKKKKKK

JJ

&&

&&

&&

. (4.75)

Eliminating the coordinate 3θ using the third equation

( ) 033212211 =+++−− θθθ KKKKK

we obtain the two equations of motion

( )

( ) .KKK

KKKKKK

KKJ

,KKK

KKKKK

KKKJ

0

0

2321

3121

321

2122

2321

211

321

32111

=+++

+++

−

=++

−+++

+

θθθ

θθθ

&&

&&

(4.76)

After solving the eigenvalue problem, in order to draw the actual mode shapes, the angular amplitudes 2Θ 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 1f and 2f at points 1 and 2 (Fig. 4.14, a). The deflections in the directions of 1f and 2f are 1y and 2y , respectively.


Consider the unloaded beam (Fig. 4.14, b) and apply a single unit force at point 1 in the direction of 1f . Let the deflections at 1 and 2, in the directions 1y and 2y be 11δ and 21δ . Similarly, let a unit force be applied at point 2 in the direction of 2f and denote the deflections 12δ and 22δ (Fig. 4.14, c).

The relationship between the forces 1f and 2f and the total deflections 1y and 2y can be expressed by the equations

.ffy

,ffy

2221212

2121111

δδ

δδ

+=

+= (4.77)

The coefficients ijδ are called flexibility (influence) coefficients.

By definition, ijδ is the deflection of coordinate i due to unit load applied to coordinate j.

Fig. 4.14

In matrix notation, equations (4.77) become

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

2

1

2221

1211

2

1

ff

yy

δδδδ

(4.77, a)

or { } [ ]{ }fy δ= , (4.78)

where [ ]δ is known as the flexibility matrix.

The flexibility matrix is symmetrical [ ] [ ]δδ =T , 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).


Because in an equation of the type { } [ ]{ }ykf = the matrix [ ]k is

known as the stiffness matrix, it comes out that [ ] [ ] 1−= δk or alternatively

[ ] [ ] 1−= kδ . (4.79)

When rigid-body motions are possible, [ ] [ ] 1−= kδ 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.


The massless beam of Fig. 4.15, a has constant bending rigidity EI and carries masses 1m and 2m 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

1y and 2y , 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

111 ymf &&−= , 222 ymf &&−= . (4.80)

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

.yymym

,yymym

0

0

222221121

122121111

=++

=++

&&&&

&&&&

δδ

δδ (4.81)

In matrix form

[ ]{ } { } { }0=+ yyb && . (4.82)


The matrix [ ]b can be written

[ ] [ ] [ ]mmmmm

bbbb

b δδδδδ

=⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

222121

212111

2221

1211 , (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


{ } { } ( )ϕω −= tay cos , (4.84)

equation (4.81) becomes

[ ]{ } { } { }02 =+− aabω

or

[ ]{ } { }aab 21ω

= . (4.85)

This is the standard eigenvalue problem, in which 21 ωλ = 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

( )( ) .abab

,abab

01

01

22

2212

21

22

1212

11

=−+

=+−

ωω

ωω (4.85, a)

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

( )( ) 011 42112

222

211 =−−− ωωω bbbb

or ( ) ( ) 012

22114

21122211 =++−− ωω bbbbbb . (4.86)

Equation (4.86) has two real positive roots 21ω and 2

2ω , the natural frequencies squared.

The mode shapes are defined by the ratio 12 aa=μ , so that


2112

2111

11

21

1

ω

ωμ

b

baa −

=⎟⎟⎠

⎞⎜⎜⎝

⎛= , 2

212

2211

21

22

1ωωμ

bb

aa −

=⎟⎟⎠

⎞⎜⎜⎝

⎛= . (4.87)

Example 4.5 Calculate the natural modes of vibration for the beam of Fig. 4.15, a where

221 lll == , mmm == 21 and .constIE =

Solution. The flexibility coefficients are

EI24311 l=δ , EI485 3

2112 l== δδ , EI3322 l=δ .

The flexibility matrix is

[ ] ⎥⎦

⎤⎢⎣

⎡=

16552

48

3

EIlδ .


⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

2

1

2

1

16552

aa

aa

λ

where 2348 ωλ lmIE= .

Fig. 4.16

The frequency equation is

0165

52=

−−

λλ

or 07182 =+− λλ


with solutions 602171 .=λ and 397602 .=λ .


31 65031 lmIE.=ω , 3

2 98610 lmIE.=ω .

The mode shapes are defined by the ratio 5212 −== λμ aa

1235

21

11

21 .

aa

=−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

λμ , 320

522

21

22 .

aa

−=−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

λμ .

They are graphically presented in Fig. 4.16.

4.3.4 Free Vibrations

Let denote by

[ ] { } { }[ ] ⎥⎦

⎤⎢⎣

⎡==

2221

121121 uu

uuuuu (4.88)

the matrix of normalized mode shapes (modal matrix).

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

{ } [ ]{ } { } { } ( ){ }iiii

i utCquququx ϕω −=+== ∑=

cos2

12211 . (4.89)

The constants iC and iϕ may be evaluated from the initial conditions of the motion

( ){ } { }iii

i uCx ϕcos02

1∑=

= , (4.90)

( ){ } { }iii

ii uCx ϕω sin02

1∑=

=& . (4.91)

Premultiplying equations (4.90) and (4.91) by { } [ ]mu jT , using the

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

{ } [ ] { } 0T =ij umu , ji ≠ { } [ ] { }iii umuM T= , 21,i = (4.92)


we obtain

{ } [ ] ( ){ } jjjj MCxmu ϕcos0T = , 21,j = (4.93)

{ } [ ] ( ){ } jjjjj MCxmu ϕω sin0T =& . 21,j = (4.94)

Combining equations (4.93) and (4.94) and renaming the index yields

{ } [ ] ( ){ }{ } [ ] ( ){ }0

0tan T

T

xmu

xmu

ii

ii

ωϕ

&= , 21,i = (4.95)

and

{ } [ ] ( ){ } { } [ ] ( ){ }iii

i

ii

ii M

xmuM

xmuC

ϕωϕ sin0

cos0 TT &

== . 21,i = (4.96)

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

EIyy 34 3l=δ , EIzyyz 23l== δδ , EIzz 33l=δ .



.zzmIE

ymIE

,yzmIE

ymIE

032

023

4

33

33

=++

=++

&&l

&&l

&&l

&&l

Looking for solutions of the form

( ) ( ),tuty ϕω −= cos1 ( ) ( )ϕω −= tutz cos2

we obtain ( )

( ) ,uu

,uu

023

038

21

21

=−+

=+−

β

β

where 236 ωβ lmIE= .


023

38=

−−

ββ

, 07102 =+− ββ ,

with solutions 242691 .=β , 757402 .=β .


31 80570 lmIE.=ω , 3

2 81462 lmIE.=ω .

The mode shapes are given by

414203

81

11

21

11

21 .

uu

uu

=−

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

βμ , 414223

82

12

22

21

22 .

uu

uu

−=−

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

βμ .

The modal vectors, normalized with unit first element, are

{ }⎭⎬⎫

⎩⎨⎧

=414201

1 .u , { }

⎭⎬⎫

⎩⎨⎧−

=414221

2 .u .

It can be seen that

41420tan 111

21 .uu

== γ , 01 522.=γ ,

41422tan 212

22 .uu

−== γ , 01

02 905112 +== γγ . .


The mass m has unidirectional motion in the natural modes of vibration. The modal vectors are orthogonal { } { } 02

T1 =uu . The motion in the first mode is

along direction 1 at 0522. measured from the vertical direction. The motion in the second mode is along direction 2 at 05122. , hence normal to direction 1. This happens because the motion in modal coordinates is along the directions of principal flexibilities.

The flexibility matrix

[ ] ⎥⎦

⎤⎢⎣

⎡=

2338

6

3

IElδ .

represents a deflection-force relation between the components y and z of the deflection and the components yf and zf of a force applied to the point where the mass is located

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

z

y

zzzy

yzyy

ff

zy

δδδδ

.

Consider a reference frame ∗∗Ozy rotated through an angle γ with respect to the frame yOz . The transformation of displacements can be written

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

∗

∗

zy

zy

γγγγ

cossinsincos

and the transformation of forces is defined by

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

∗

∗

z

y

z

yff

ff

γγγγ

cossinsincos

.

The new deflection-force relation is

[ ]⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

∗

∗∗

∗

∗

z

y

ff

zy δ

where

[ ] [ ] ⎥⎦

⎤⎢⎣

⎡ −⎥⎦

⎤⎢⎣

⎡−

=∗

cssc

cssc

δδ

with γcos=c and γsin=s .

The flexibility matrix in the rotated reference frame is


[ ] ( ) ( )( ) ( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

−+−+−

−+−++=∗

cscsscscscsccssc

yzzzyyyzyyzz

yzyyzzyzzzyy

δδδδδδδδδδδδ

δ2

22222

2222

.

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

128322

2tan =−⋅

=−

=∗

zzyy

yz

δδδ

γ , 01 522.=∗γ , 0

2 5112.=∗γ .

The two solutions ∗1γ and ∗

2γ define the principal directions of flexibility. Substituting these angles into the expression of diagonal elements we obtain the principal flexibilities

IEyzzzyyzzyy

,

32

2

21 6235

22l±

=+⎟⎟⎠

⎞⎜⎜⎝

⎛ −±

+= δ

δδδδδ ,

IE.

3

1 54041 l=δ ,

IE.

3

2 12620 l=δ .

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

331

1 80505404111

ll mIE.

mIE

.m===

δω ,

332

2 81521262011

ll mIE.

mIE

.m===

δω .

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

( ) ( ) m..muumM 17161414201 2221

2111 =+=+= ,

( ) ( ) m..muumM 82846414221 2222

2122 =+=+= .

The initial conditions are

( ){ }⎭⎬⎫

⎩⎨⎧

=00

0x , ( ){ }⎭⎬⎫

⎩⎨⎧

=0

0v

x& .

From (4.93) and(4.96) we obtain


0cos 1 =ϕ , 0cos 2 =ϕ ,

1111 sin

MmC

ωϕ v

= , 22

22 sinM

mCω

ϕ v= .

a b

Fig. 4.18

Fig. 4.19

The vertical component of the instantaneous displacement is

( ) ( ) ( ) 1222211111 coscos utCutCty ϕωϕω −+−= ,

( ) tMω

mtMω

mty 222

111

sinsin ωω vv+= ,

( ) ( )t.t.Cty 21 sin05200sin05931 ωω += ,

where


IEmv

3l=C , C. v80501 =ω , C. v81522 =ω .

The horizontal component of the instantaneous displacement is

( ) ( ) ( ) 2222221111 coscos utCutCtz ϕωϕω −+−= ,

( ) tuMω

mtuMω

mtz 22222

12111

sinsin ωω vv+= ,

( ) ( )t.tCtz 21 sin12560sin0.4387 ωω −= .

The trajectory of the mass m is plotted in Fig. 4.18, a for a time duration equal to 12 ωπ and in Fig. 4.18, b for 14 ωπ . 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.


Consider the steady-state vibrations of the beam of Fig. 4.20, a under the action of the force ( ) tFtf ωcos0= acting on mass 2m .

Fig. 4.20



( )( ) 222211212

122211111

δδ

δδ

ymfymy

,ymfymy&&&&

&&&&

−+−=

−+−=

or ( )( ).tfyymym

,tfyymym

22222221121

12122121111

δδδ

δδδ

=++

=++

&&&&

&&&& (4.97)

Substituting the steady-state solutions

( ) ,tYty ωcos11 = ( ) ,tYty ωcos22 =

into equations (4.97) yields

( )( ) .FYmYm

,FYmYm

02222222

11212

01222122

11112

1

1

δδωδω

δδωδω

=−+−

=−− (4.98)

The solutions are

( ) ( )( )

( ) ( ) .Fmmmm

mY

,Fmmmm

Y

04212221121

2222111

2122211

2122

2

04212221121

2222111

121

1

1

ωδδδωδδ

δδδωδ

ωδδδωδδ

δ

−++−

−−=

−++−=

(4.99)

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

a b

Fig. 4.21

The absolute values of the amplitudes 1Y and 2Y 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.


Antiresonance occurs for 02 =Y when

( )21222111

2222

δδδ

δωω

−==

ma . (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

12

11 Ym ωΦ = , 022

22 FYm += ωΦ . (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 mm40=d , m1=l , GPa210=E and kg50=m . a) Calculate the natural frequencies; b) Determine

the amplitudes of forced vibrations produced by a harmonic force of amplitude N200 =F and frequency Hz1790. ; 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

IE6311 l=δ , IE83

2112 l−== δδ , IE245 322 l=δ .

Denoting

3224

lmIE

ωλ = ,

the frequency equation is

056

38=

−−

λλ

, 022132 =+− λλ ,

with solutions

111 =λ , 22 =λ .


secrad07314771 31 .mIE. == lω ,


secrad51624643 32 .mIE. == lω .

Fig. 4.22

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

( ) ( )

( ) ( ) ..m

FY

,.m

FY

mm1051211

522

mm1180211

3

20

2

20

1

−=−−

−−=

=−−

−=

ωλλλ

ωλλλ

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 Nm368

and the maximum static stress is 2mmN558.st =σ .

The amplitudes (4.101) of dynamic forces are


( ) ( ) N150211

62 012

1 =−−

−== FYmλλλωΦ ,

( ) ( ) N50211

5221 0022

2 −=⎥⎦

⎤⎢⎣

⎡−−

−−=+= FFYm

λλλωΦ .

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

Nm587. and the maximum dynamic stress is 2mmN14=dσ .

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.


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 1k and 2k (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

( ) ( )( ) ( ) 0

0

222111

2211

=++−−

=++−+

llll&&

ll&&

θθθ

θθ

xkxkJ

,xkxkxm

or ( ) ( )( ) ( ) .kkxkkJ

,kkxkkxm

0

0222

2111122

112221

=++−+

=−+++

θθ

θ

llll&&

ll&& (4.102)

In matrix form

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+−−+

+⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡00

00

222

2111122

112221

θθx

kkkkkkkkx

Jm

llll

ll&&

&&. (4.102, a)


Note in the above equations that if 2211 ll kk = the system is uncoupled, and has two independent natural frequencies – one for translation and one for rotation. These are defined by

( ) mkkx 21 +=ω and ( ) Jkk 222

211 ll +=θω . (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) 1x and 2x . The transformation of coordinates is defined by

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−+

=⎭⎬⎫

⎩⎨⎧

2

112

21 111

xxx ll

llθ. (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

( )( ) ⎭

⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+

++

⎭⎬⎫

⎩⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−−+

00

00

2

1

212

211

2

12121

2122

xx

kk

xx

JmJmJmJm

ll

ll

&&

&&

lll

lll . (4.105)

The coupling is given by the off-diagonal elements of the mass matrix, hence it is referred to as dynamic coupling. When 21llmJ = the coupled translation and rotation can be expressed as a sum of two pitching vibrations, one about the the right end and the other about the left end.


4.4.2 Normal Modes

Denoting

( ) mkka 21 += , ( ) mkkb 1122 ll −= , ( ) Jkkc 222

211 ll += ,


,rcxbr

,bxax

0

0 22 =++

=++

θθ

θ&&

&& (4.106)

where mJr = is the radius of gyration.


( ) ( ) ( ) ( ) ,tAt,tAtx x ϕωθϕω θ −=−= cos cos (4.107)

we obtain the set of algebraic equations

( )( ) .ArcAb

,AbAa

x

x

0

022

2

=−+

=+−

θ

θ

ω

ω (4.108)

Equations (4.108) admit non-trivial solutions if the determinant of the coefficients xA and θA is zero

( ) 022

2=

−−

rcbbaω

ω (4.109)

or

( ) ( ) 02224 =−++− rbcaca ωω . (4.109, a)

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

( ) ( ) 222221 42 rbcaca, +−±+=ω . (4.110)

The first natural frequency 1ω is always less than xω or θω , whichever is smaller, and the second natural frequency 2ω is always greater than xω and θω .

The mode shapes are obtained substituting the natural frequencies in turn in the amplitude ratio

( ) ( )( ) ( ) .abAA

,abAA

x

x

2222

2111

ωμ

ωμ

θ

θ

−−==

−−== (4.111)


Example 4.8 For the system of Fig. 4.23 determine the natural modes of vibration if 431 ll = , 42 ll = , kkk == 21 , 82lmJ = .

Solution. Equations (4.108) have the form

.AmkA

mk

,Am

kAmk

x

x

08

52

02

2

22

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=+⎟⎠⎞

⎜⎝⎛ −

θ

θ

ω

ω

ll

l

Denoting

km2ωα = ,

the equations become

( )

( ) .AA

,AA

x

x

08

52

02

2

2=−+

=+−

θ

θ

α

α

ll

l


0872 =+− αα

with solutions 43811 .=α , 56152 .=α .


mk.19911 =ω , mk.35822 =ω .

Fig. 4.24


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

2802

24 1

1

.A

A

x=

−=⎟

⎟⎠

⎞⎜⎜⎝

⎛ αθ l , 7812

24 2

2

.A

A

x−=

−=⎟

⎟⎠

⎞⎜⎜⎝

⎛ αθ l .

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

5632801

41

1 ..A

Ad x −=−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

lθ, 560

7811

42

2 ..A

Ad x ==⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

lθ.

The frequencies of the uncoupled pure translation and pure rotation are

mk.x 4141=ω , mk.2362=θω .

The following inequalities hold in this case

21 ωωωω θ <<< x .

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 82lmJ = 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.


The relationship between the inertia force xmF 2ω= , the inertia torque θω2JM = and the linear and angular displacements x and θ can be expressed by

the equations

.MF

,MFx

2221

1211

δδθ

δδ

+=

+=

Two bending moment diagrams are constructed (Fig. 4.25, b), −xm for loading with a unit force at point G in the direction of F , and −θm 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 11δ and 21δ , respectively 12δ and 22δ . The flexibility influence coefficients are

IE1213 311 l=δ , IE2

2112 l== δδ , IEl=22δ .

Substitution in the equations of equilibrium yields

,x

,xx

θθβ

θβ

2312

2313

+=

+=

l

l

where 2312 ωβ lmIE= .

Fig. 4.26


03292 2 =+− ββ

with solutions 396141 .=β and 104202 .=β .



31 9130 lmIE.=ω , 3

2 7310 lmIE.=ω .

The mode shapes (Fig. 4.26) are given by

930023131

1

.AA

x=

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ βθl , 597823132

2

.AA

x−=

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ βθ l .

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.


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 1θ and 2θ and assuming small amplitudes, the equations for the kinetic and potential energies are

( )22

21

2

21 θθ &&l += mT , (4.112)


( ) ( ) ( )212

221 2

1cos1cos1 θθθθ −+−+−= dkgmgmU ll ,

or

( ) ( )212

222

21 2

121 θθθθ −++= dkgmU l . (4.113)

Using Lagrange’s equations

0dd

=∂∂

+∂∂

−∂∂

rrr qU

qT

qT

t &, 21,r = (4.114)

we obtain the equations of motion

( )( ) .dkgmm

,dkgmm

0

0

212

222

212

112

=−−+

=−++

θθθθ

θθθθ

l&&l

l&&l (4.115)

In matrix form, the equations of motion are written

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−−++

⎭⎬⎫

⎩⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡

00

00

2

122

22

2

12

2

θθ

θθ

dkgmdkdkkdgm

mm

l

l&&

&&

l

l . (4.115, a)

The coupling is due to the spring k.

4.5.2 Normal Modes


( )ϕωθ −= ta cos11 , ( )ϕωθ −= ta cos22 ,

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

( )( ) .adkmgmadk

,adkadkmgm

0

0

2222

12

22

1222

=+−+−

=−+−

ll

ll

ω

ω


( ) ( ) 0222222 =−+− dkdkmgm ll ω

or

022 3

2

2

22

2

24 =⎟

⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎟⎠

⎞⎜⎜⎝

⎛+−

llll mgdkg

mdkg ωω .



l

g=1ω , 2

2

2 2ll mdkg

+=ω . (4.116)

The mode shapes are defined by the amplitude ratios

111

21 +=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

aaμ , 1

21

22 −=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

aaμ . (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 lg . 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

( ) ( ) ( ) 12222111111 coscos utCutCt ϕωϕωθ −+−= ,

( ) ( ) ( ) 22222211112 coscos utCutCt ϕωϕωθ −+−= , or

( ) ( ) ( )2221111 coscos ϕωϕωθ −+−= tatat ,

( ) ( ) ( )222211112 coscos ϕωμϕωμθ −+−= tatat .

Differentiating with respect to time yields

( ) ( ) ( )222211111 sinsin ϕωωϕωωθ −−−−= tatat& ,

( ) ( ) ( )22222111112 sinsin ϕωωμϕωωμθ −−−−= tatat& .

The integration constants are determined from the initial conditions

( ) 01 0 θθ = , ( ) 002 =θ , ( ) 001 =θ& , ( ) 002 =θ& . (4.118)

Because 11 =μ and 12 −=μ , we obtain

02211 coscos θϕϕ =+ aa , 0coscos 2211 =− ϕϕ aa ,

0sinsin 222111 =+ ϕωϕω aa , 0sinsin 222111 =− ϕωϕω aa ,

so that 0sinsin 21 == ϕϕ , 021 ==ϕϕ ,


02211 cos2cos2 θϕϕ == aa , 2021 θ== aa .

The instantaneous values of the angular displacements are

( ) ( )

( ) ( ) ,ttttt

,ttttt

m

m

ωΔωθωωθθ

ωΔωθωωθθ

sinsincoscos2

coscoscoscos2

0210

2

0210

1

⋅=−=

⋅=+= (4.119)

where ( ) 221 ωωω +=m and ( ) 212 ωωωΔ −= .

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

21

21

22 ωωω <<− ,

llll

ggm

dkg<<−+ 2

22 or 22 d

mgk l<< ,

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

Fig. 4.28

Equations (4.119) show that 1θ and 2θ are given by sine and cos

functions which are 090 phase shifted. When 01 θθ = , 02 =θ and vice-versa (Fig.4.28).

The phase difference between the two motions is

( ) ( ) ( ) ( ) ϕΔωΔϕϕωωϕωϕωϕ −=−−−=−−−= tttt 212121122


so it varies slowly with time.

At 0=t , 01 θθ = , 02 =θ .

At 2πωΔ =t , 01 =θ , 002 2sin θπ

ωΔωθθ == m if nm =

ωΔω is an integer.

At πωΔ =t , 02 =θ , 001 cos θπωΔ

ωθθ −== m if nm =ωΔ

ω is an integer.

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 1ω and 2ω . 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

0180 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.


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 1c and 2c . 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:

( ) ( )( ) ( ) 0

0

21221222

212212111111

=−−−−

=−+−+++

xxcxxkxm

,xxcxxkxcxkxm&&&&

&&&&&

or

( ) ( ).xkxkxcxcxm

,xkxkkxcxccxm

0

0

2212221222

221212212111

=+−+−

=−++−++

&&&&

&&&& (4.120)

Fig. 4.30

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


In matrix form, these equations can be written

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−++

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−++

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡00

00

2

1

22

221

2

1

22

221

2

1

2

1

xx

kkkkk

xx

ccccc

xx

mm

&

&

&&

&&, (4.121)

or in compact form

[ ]{ } [ ]{ } [ ]{ } { }0=++ xkxcxm &&& , (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

[ ] [ ]Tcc = .

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

[ ] [ ] [ ]kmc βα += , (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

{ } [ ] { }qux = (4.123)

and multiplying on the left by [ ]Tu we obtain

[ ] { } [ ] { } [ ] { } { }0=++ qKqCqM &&& , (4.124)

where

[ ] [ ] [ ] [ ]umuM T= , [ ] [ ] [ ] [ ]ucuC T= , [ ] [ ] [ ] [ ]ukuK T= . (4.125)

The modal matrices [ ]M and [ ]K are diagonal while the matrix [ ]C is diagonal only for proportional damping when


[ ] [ ] [ ]KMC βα += . (4.126)

In this case, the following orthogonality relations can be established

{ } [ ] { } 0T =rs ucu , 21,s,r,sr =≠ (4.127)

For proportional damping, the decoupled modal equations are

0=++ rrrrrr qKqCqM &&& , 21,r = (4.128)

where rM and rK are defined by (4.26) and (4.28), and

{ } [ ] { }rrr ucuC T= , 21,r = (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 rr ωζ2 where rζ is the rth modal damping ratio and rω is the rth natural frequency.


0ζ2 2 =++ rrrrrr qqq ωω &&& , 21,r = (4.130)

which, for 1ζ0 << r have solutions of the form (2.46)

( ) ⎟⎠⎞⎜

⎝⎛ +−= −

rrrt

rr teAtq rr φωω 2ζ ζ1sin . 21,r = (4.131)

The solutions (4.131) can also be obtained directly. Looking for solutions of the form ts

rr ax e= , equations (4.121) become

( ) ( ),akakascascasm

,akakkascasccasm

0

0

2212221222

2

221212212112

1

=+−+−

=−++−++

or

( ) ( )[ ] ( )( ) ( ) .akscsmaksc

,akscakksccsm

0

0

2222

2122

222121212

1

=++++−

=+−++++ (4.132)

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

( ) ( )[ ] ( ) ( ) 022222

222121

21 =+−++++++ ksckscsmkksccsm . (4.133)


For underdamped systems, equation (4.133) has two pairs of complex conjugate roots

1121 i d,s ωσ ±−= , 2243 i d,s ωσ ±−= , (4.134)

where 1dω and 2dω are the damped natural frequencies and 1σ and 2σ 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

rrr ωσ ζ= , 2ζ1 rrrd −=ωω 21,r = (4.135, a)

22ζ

rrd

rr

σω

σ

+= , 22

ζ rrdr

rr σωσω +== . 21,r = (4.135, b)

With these notations, expression (4.131) becomes

( ) ( )rrdt

rr teAtq r φωσ += − sin . 21,r = (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 ( )raa 12 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 21 =m , 21 =k , 11 =c , 12 =m , 12 =k , 502 .c = , in the appropriate units.

Solution. The equations of free vibrations (4.121) are

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−+

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−+

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡00

1113

50505051

1002

2

1

2

1

2

1

xx

xx

....

xx

&

&

&&

&&.

It is seen that

[ ] [ ]k.c 50= .



( ) ( )( ) ( ) .as.sas.

,as.as.s

0150150

01503512

22

1

212

=++++−

=+−++

The characteristic equation is

( ) ( ) ( ) 01501503512 222 =+−++++ s.s.ss.s , or

02255522 234 =++++ ss.s.s , with roots

831i

81

21 ±−=,s , 27i

21

43 ±−=,s .

The imaginary parts are the damped natural frequencies

696008311 .d ==ω , 32291272 .d ==ω .

The real parts are the damping factors (decay rates)

12501 .=σ , 502 .=σ .

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

( ) ( ) 2181831 2221

211 =+=+= σωω d ,

( ) ( ) 22127 2222

222 =+=+= σωω d .

The modal damping ratios are

176021

81ζ1

11 .===

ωσ

, 3530221ζ

2

22 .===

ωσ

.

The amplitude ratio

150

150150

35122

2

1

2

+++

=+

++=

s.ss.

s.s.s

aa

yields

211

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛

aa

, 121

2 −=⎟⎟⎠

⎞⎜⎜⎝

⎛

aa

.

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


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 [ ] [ ]k.c 10= .

Solution. Equations (4.132) become

( ) ( )( ) ( ) .as.sas.

,as.as.s

0110110

01103302

22

1

212

=++++−

=+−++

The characteristic equation 0240025502 234 =++++ s.s.s.s ,

has the roots

40799i

401

21 ±−=,s , 10199i

101

43 ±−=,s

or 0.706665i025021 ±−= .s , , 1.410674i1043 ±−= .s , .

The damped natural frequencies are

7066601 .d =ω , 410712 .d =ω .

The damping factors have the values

02501 .=σ , 102 .=σ . The modal damping ratios are

03530ζ 1 .= , 07070ζ 2 .= . The undamped natural frequencies and the mode shapes are the same as in

Example 4.10.


Consider the vibrations of the two-degree-of-freedom system from Fig. 4.31 under the action of forces ( )tf1 and ( )tf2 .

Fig. 4.31



( ) ( ) ( )( ) ,tfxkxkxcxcxm

,tfxkxkkxcxccxm

22212221222

1221212212111

=+−+−

=−++−++

&&&&

&&&& (4.136)

or in compact matrix form

[ ] { } [ ] { } [ ] { } { }fxkxcxm =++ &&& . (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

[ ] [ ] [ ][ ] ( ){ } ( ){ }sFsXkscsm =++2 (4.137)

or

( )[ ] ( ){ } ( ){ }sFsXsB = (4.138)

where ( )[ ]sB is referred to as the system matrix.

Multiplying on the left by ( )[ ] ( )[ ]sHsB =−1 yields

( )[ ] ( ){ } ( ){ }sXsFsH = (4.139)

where ( )[ ]sH 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

( )[ ]( )

( )[ ] [ ] ( ) 22222

222121

21

21212

122

22222

2

ksckscsmkksccsm

kksccsmkscksckscsm

sH

+−++++++

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+++++

+++

= (4.140)

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

( )[ ] ( )( )( )( )4321det ssssssssAsB −−−−=

where A is a constant and 41 s,..,s 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.


We can also express (4.139) as

( ) ( )( ) ( )

( )( )

( )( ) ⎭

⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡sXsX

sFsF

shshshsh

2

1

2

1

2221

1211 . (4.141)

where ( )sh jl is obtained by exciting the system at point j and measuring the response at point l . For instance ( )sh11 is a driving point transfer function measured by exciting the system with ( )sF1 and measuring the response ( )sX1

( ) ( )( ) ( )( )( )( )4321

222

2

1

111

ssssssssA

kscsmsFsXsh

−−−−

++== . (4.142)

4.6.3.2 Frequency Response Functions

A frequency response function (FRF) is the transfer function evaluated along the ωi (frequency) axis. Substituting ωi=s into (4.141) we obtain

( ) ( )( ) ( )

( )( )

( )( ) ⎭

⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ωω

ωω

ωωωω

2

1

2

1

2221

1211

iiii

XX

FF

hhhh

or

( )[ ] ( ){ } ( ){ }ωωω XFH =i (4.143)

where ( )[ ]ωiH is the frequency response function matrix and ( )ωjX , ( )ωjF 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

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ).FhFhX

,FhFhX

ωωωωω

ωωωωω

2221212

2121111

ii

ii

+=

+= (4.143, a)

The time domain forced response of the system, ( )tx1 and ( )tx 2 , can then

be computed by taking the inverse Fourier transform of ( )ω1X and ( )ω2X .

Example 4.12 Calculate the matrix of frequency response functions for the two-degree-

of-freedom system of Example 4.11 and plot the FRF curves of ( )ωi11h .


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

( )[ ] ( )ωωωω

ωωωωωω

ω400.5i20252

30i2310i110i110i1

i324

2

2

..

....

H+−++−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−+++−

= .

Fig. 4.32

Fig. 4.33


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 090− . 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) { } [ ]{ }qux = and

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

[ ] { } [ ] { } [ ] { } [ ] { } { }FfuqKqCqM ==++ T&&& . (4.144)

For proportional damping, the decoupled modal equations are

rrrrrrr FqKqCqM =++ &&& , 21,r = (4.145) or

rrrrrrrr MFqqq =++ 2ζ2 ωω &&& . 21,r = (4.146)

For harmonic excitation and steady-state response, denote

{ } { } tf̂f ωie= , { } { } tx~x ωie= , (4.147)


{ } { } tF̂F ωie= , { } { } tq~q ωie= , (4.148)

{ } [ ] { } { }rr

r uq~q~ux~ ∑=

==2

1, (4.149)

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

{ } { } { } { }( )rrrr

r

rrr

rr

M

f̂uCMK

f̂uq~

ωωωωωω ζ2ii 22

T

2

T

+−=

+−= . (4.150)

For unit modal masses, 1=rM , substituting (4.150) into (4.149) yields

{ } { } { }{ }∑= +−

=2

122

T

ζ2ir rrr

rr uf̂ux~

ωωωω

or

{ } { }⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+−=

⎭⎬⎫

⎩⎨⎧ ∑

= 2

12

122

T

2

1

ζ2i f̂f̂uu

x~x~

r rrr

rr

ωωωω. (4.151)

Note that the dyadic product { } { }Trr uu is a 2 by 2 square matrix.

The FRF matrix (4.143) can be expressed now in terms of the modal parameters

( )[ ] { } { } { } { }22

222

T22

1122

1

T11

ζ2iζ2ii

ωωωωωωωωω

+−+

+−=

uuuuH . (4.152)

If we denote the jth element of the rth mode shape vector by ( )jru then

the expression for the general FRF function ( )ωiljh in a partial fraction form is

( )( ) ( ) ( ) ( )

2222

2

22

1122

1

11

ζ2iζ2ii

ωωωωωωωωω

+−+

+−= ll

l

uuuuh jj

j . (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.


Fig. 4.35

Example 4.13 Calculate the partial fraction form of the FRF matrix for the two-degree-of-

freedom system of Example 4.12.

Solution. The mass-normalised mode shapes are

{ }T

1 62

61

⎭⎬⎫

⎩⎨⎧

=u , { }T

2 31

31

⎭⎬⎫

⎩⎨⎧

−=u .

The FRF matrix (4.152) becomes

( )[ ]ωωωω

ω20i2

1 111

31

050i504221

61

i 22 ...H

+−

⎥⎦

⎤⎢⎣

⎡−

−

++−

⎥⎦

⎤⎢⎣

⎡

= .

The driving point FRF for the first mass is

( )ωωωω

ω20i2

31050i50

61i 2211 ...h

+−+

+−= .

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).


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 dJ 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 tM ωcos0 , the equations of motion can be written

( )( ) ,cJ

,tMcKJ

ddd

d

0

cos0

=−−

=−++

θθθ

ωθθθθ&&&&

&&&& (4.154)

where θ is the casing rotation and dθ is the rotation of the internal disc. The damping coefficient is

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −+=

1

41

42

2

32

22

hRR

hRbc μπ (4.155)

where μ is the oil viscosity.

Fig. 4.36


tωθθ cos0= , ( )ϕωθθ −= tdd cos0 , and denoting


JK

n =2ω , nωωη = ,

ndJcω2

ζ = , J

Jd=λ , (4.156)

the amplitude of the casing is described by

( ) ( )2222222

22

0

0

1ζ41

ζ4

−++−

+=

ληηηη

ηθMK . (4.157)

For a fixed λ , the plots of 00 MKθ as a function of the dimensionless frequency η are shown in Fig. 4.37 for various values of ζ .

Fig. 4.37

The value 0ζ = is for an undamped system with resonance frequency nω whose vibration amplitude is infinite at 1=η . The value ∞=ζ is for an undamped system with resonance frequency ( ) λω +=+ 1ndJJK , 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

( )λη += 22M . For any other value of damping, the response curve passes through this point. There is an optimum damping

( )( )λλ ++=

2121ζopt (4.158)

for which the peak amplitude is a minimum and is equal to the ordinate ( )λ21+ of point M .


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 1c is eliminated from the system illustrated in Fig. 4.31.

If the damping coefficient 1c is made zero, and the force 2f is cancelled, equations (4.136) may be rewritten as

( ) ( ) ( )( ) ( ) .xxkxxcxm

,tfxxkxkxxcxm

021221222

12121121211

=−−−−

=−++−+

&&&&

&&&& (4.159)

For harmonic excitation and steady-state response tFf ωi

1 e= , tX~x ωi11 e= , tX~x ωi

22 e= , (4.160)


( ) ( )( ) ( ) .X~cmkX~kc

,FX~kcX~cmkk

0ii

ii

222

22122

222122

121

=+−++−

=+−+−+

ωωω

ωωω (4.161)

The vibration amplitude of the primary system 11 X~X = can be expressed

as

DCBA

xX

st ++

= 22

221

ζζ (4.162)

where


2224 θψμ=A , ( )2222 θψμ −=B , ( )222222 14 θμθθψμ −−=C ,

( ) ( )[ ]2222222 1 θψμθθψμ −−−=D , 1kFxst = , 2222 2ζ mc ω= , (4.163)

12 mm=μ , 1ωωθ = , 12 ωωψ = , 111 mk=ω , 222 mk=ω .

Figure 4.38 shows the frequency response curves of stxX1 as function of the frequency ratio θ , for fixed values of μ and ψ , and various values of 2ζ . The effect of the absorber can be seen to reduce the primary system amplitude from infinity down to a small value at 1=θ .

When 0ζ2 = , DBxX st =1 . When ∞=2ζ , CAxX st =1 . 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

2ζ , 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

( )μψ += 11 . (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

μ211 +=stxX . (4.165)

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

( )μ

μθ

++±

=1

2212 . (4.166)

The frequency ratio θ at which the trough between the two peaks occurs is the mean of these two frequency ratios ( ) 2111 ψμ =+ . This is in fact the tuning frequency ratio.


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

1222 2ζ ωψ mc= . The average of these two values is

( )3122 1832 μμω +=mc . (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

( )1221

121 22 ωθμ mck

FXXX =− (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

[ ]{ } [ ]{ } [ ]{ } { }0=++ xkxcxm &&& , (4.169)


where the matrix [ ]c 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

[ ] { } [ ] { } { }0=− xmxm && . (4.170)

Equations (4.169) and (4.170) can be combined to give

[ ] [ ][ ] [ ]

{ }{ }

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }⎭

⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −+

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡000

00

xx

ckm

xx

mm

&&&

& (4.171)

or [ ] { } [ ] { } { }0=+ qBqA & , (4.172)

where the 4 to 4 matrices [ ]A and [ ]B are real

[ ] [ ] [ ][ ] [ ]⎥⎦

⎤⎢⎣

⎡=

mm

A0

0, [ ] [ ] [ ]

[ ] [ ] ⎥⎦⎤

⎢⎣

⎡ −=

ckm

B0

.

Assuming a solution of the form { } { } tsq eΦ= , and { } { } tsux e= , equations (4.172) become

[ ] [ ]( ) { } { }0=+ ΦBAs . (4.173)

There are four eigenvalues rs given by the solutions of

[ ] [ ]( ) 0det =+ BAs , (4.174)

and four eigenvectors { }rΦ satisfying the eigenvalue problem

[ ] { } [ ]{ }rrr AsB ΦΦ −= . ( )41,...,r = (4.175)

Equation (4.175) can be written in the form

[ ] [ ] { } { }rrr sBA ΦΦ =− −1 . ( )41,...,r = (4.176)

where

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]⎥⎦

⎤⎢⎣

⎡−−

=− −−−

cmkmI

BA 111 0

, (4.177)

Equation (4.171) becomes

[ ] [ ]

[ ] [ ] [ ] [ ][ ] [ ][ ] [ ]

{ }{ } { }0

000

11 =⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡−− −− us

uI

Is

cmkmI

. (4.178)


For underdamped systems, equation (4.178) has two pairs of complex conjugate roots

1121 i d,s ωσ ±−= , 2243 i d,s ωσ ±−= , (4.179)

where the imaginary parts, 1dω and 2dω , are the damped natural frequencies and the real parts, 1σ and 2σ , 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

22rrdr σωω += ,

r

rr ω

σ=ζ . 21,r = (4.180)

For overdamped modes of vibration, the real eigenvalues can be written

rrrr s,s τσ m−=+1 , (4.181)

and the relationships (4.180) become

22rrr τσω −= ,

r

rr ω

σ=ζ . (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 kg 1001 =m , kgm 12 = , mN1099 61 ⋅= .k ,

mN1010 62 ⋅= .k , mNs 12521 == cc . Draw the Nyquist plots of complex

receptances.

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

[ ] ⎥⎦

⎤⎢⎣

⎡=

100100

m , [ ] ⎥⎦

⎤⎢⎣

⎡−

−⋅=

111100

105k , [ ] ⎥⎦

⎤⎢⎣

⎡−

−=

125125125250

c .

The eigenvalues with positive imaginary part are

12.6593i78541 ⋅+−= .λ , 12.6853i965582 ⋅+−= .λ .


Fig. 4.39

The corresponding eigenvectors are

{ }⎭⎬⎫

⎩⎨⎧

⋅−=

6982i20711

1 ..u , { }

⎭⎬⎫

⎩⎨⎧

⋅−−=

22235i86581

2 ..u .


The damped natural frequencies are secrad 6593121 .d =ω and secrad 6853122 .d =ω . The absolute values of the eigenvalues are

secrad 69531221

211 .d =+= σωω , secrad 1963182

22

22 .d =+= σωω , and the

modal damping ratios are 01530ζ 111 .== ωσ and 18530ζ 222 .== ωσ .

For the associated undamped system, the undamped natural frequencies are secrad 30001 =ω and secrad 331.66 1110002 ==ω , 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

secrad .0260 , hence Hz .00410 , while the undamped natural frequencies differ with Hz .0395 . Damping coupling approaches the natural frequencies, fact which complicates their experimental determination.

The Nyquist plots of complex receptances jiji fx=α , 21,j,i = , 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 I II III IV

21 mm = kg 0.0259 0.0259 0.0259 0.0259

31 kk = mN 100 100 100 100

2k mN 50 1 50 1

1c mNs 0.3 0.3 3 3

2c mNs 0.2 0.2 2 2

3c mNs 0.1 0.1 1 1

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


Solution. The undamped natural frequencies are given by

2

3201 m

k=ω , ⎟

⎟⎠

⎞⎜⎜⎝

⎛+=

3

2201

202 21

kk

ωω .

The undamped modal vectors are the same in all cases

{ }⎭⎬⎫

⎩⎨⎧

=11

1a { }⎭⎬⎫

⎩⎨⎧−

=1

12a .

The modal mass, stiffness and damping matrices, calculated with the modal matrix built with the above vectors, are respectively

[ ] ⎥⎦

⎤⎢⎣

⎡=

2

1

2002m

mM , [ ] ⎥

⎦

⎤⎢⎣

⎡++

+=

231

31

400

kkkkk

K ,

[ ] ⎥⎦

⎤⎢⎣

⎡++−

−+=

23131

3131

4ccccccccc

C .

For 31 cc ≠ , 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

( )( ) 0

32322

222

2221212

1 =++++−−

−−++++

kksccsmkscksckksccsm

.

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

01ω 02ω 62.13 87.86 62.13 62.75 62.13 87.86 62.13 62.75

1ω 2ω 62.24 87.69 62.16 62.71 64.28 84.92 62.16 62.71

1ζ 2ζ 0.062 0.132 0.055 0.192 0.540 1.409 0.547 1.919

1dω 2dω 62.13 86.93 62.07 61.55 54.09 - 52.01 -

1σ 2σ 3.845 11.594 3.406 12.033 34.734 - 34.044 -


The undamped natural frequencies are different from the absolute value of eigenvalues for the non-proportionally damped system, rr ωω ≠0 . 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, 011 ωω > , while for the second mode, 022 ωω < , and for Case II, 12 dd ωω < .

The second mode is overdamped in Cases III and IV, having damping ratios 1ζ >r . 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 mNs74903 .c ≅ , while the first mode of vibration becomes overdamped for values larger than mNs8913 .c ≅ .

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.


Exercises

4.E1 For the system shown in Fig. 4.1 let kk 21 = , kk =2 , kk 23 = , mm 31 = , mm =2 . Set up the equations of motion and determine the natural

modes of vibration.

Answer: mk.91901 =ω , mk.77612 =ω , 46401 .=μ , 46462 .−=μ .

4.E2 Using the following values: mN1031 =k , mN5002 =k ,

mN102 33 ⋅=k , kg51 =m , kg102 =m , N1001 =f̂ , 02 =f̂ , secrad15=ω ,

determine the amplitudes of the forced vibrations of the two masses for the system of Fig. 4.4.

Answer: m1601 .X = , m3202 .X −= .

4.E3 Determine the amplitudes of the forced vibrations of the system from Fig. 4.4, if mN103=k , kg5=m , N501 =f̂ , N1002 =f̂ , secrad 10=ω .

Answer: m501 .X −= , m802 .X −= .

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 KK =1 , KK 22 = , KK 33 = , and the moments of inertia of the flywheels are

JJ 21 = , JJ =2 .

Answer: JK=1ω , JK.34522 =ω , 501 .=μ , 42 −=μ .

4.E5 Using the following values: radmN1071 =K ,

radmN102 72 ⋅=K , radmN103 7

3 ⋅=K , 21 mkg100=J , 2

2 mkg200=J , secrad300=ω , determine the amplitudes of the torsional vibrations of the system

shown in Fig. 4.6 when a harmonic torque of amplitude N1050 =M and

frequency secrad300=ω is applied on the second disc.

Answer: rad0073501 .=Θ , rad0077202 .=Θ .


4.E6 Calculate the natural frequencies and the mode shapes for the flexural vibrations of the system shown in Fig. 4.15, where mm 31 = , 32 mm = ,

lll == 21 and .constIE =

Answer: 31 5220 lmIE.=ω , 3

2 731 lmIE.=ω , 61 =μ , 512 .−=μ .

4.E7 Determine the normal modes for flexural vibrations of the system shown in Fig. 4.20, where mm 21 = , mm =2 , lll == 21 , 23 ll = and 0=f .


2 4643 lmIE.=ω , 11 −=μ , 22 =μ .

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


2 6883 lmIE.=ω , 04811 .=μ ,

47702 .−=μ .

4.E9 Determine the amplitudes of the forced vibrations of the two masses for the system of Fig. 4.E9 using the following values: 23 sec800 −=lmIE ,

mm230 =IEF l and secrad10=ω .

Fig. 4.E9

Answer: mm51171 .Y = , mm04252 .Y = .


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: mk.70701 =ω , mk.41412 =ω , l66601 1 .=μ , l33301 2 .−=μ .

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: mk.7801 =ω , mk.28112 =ω , l6401 1 .=μ , l3901 2 .−=μ .

4.E12 Calculate the amplitudes of the forced vibrations for the rigid bar of Fig. 4.E12 using the following values: mN103=k , kg5=m , N500 =F ,

m40.=l and secrad10=ω .

Fig. 4.E12

Answer: m01650.ax = , ad.a r0830=ϕ .


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

Receptance magnitude,

Nmm10 3−

Phase angle between response and force,

degrees

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

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

Answer: secrad2631 =ω , 046601 .=ζ , mN7001 =K ,

mNs24801 .C = , kg100151 21

−⋅= .M , secrad4622 =ω , 033902 .=ζ ,

mN18002 =K , mNs26402 .C = , kg10840 22

−⋅= .M .

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 multi-degree-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.


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

5. SEVERAL DEGREES OF FREEDOM 185

mass of the segment at each end. In the figure Am ρ=0 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

Rayleigh’s model it is 022

22

0 ≠⎟⎠⎞

⎜⎝⎛ llm . This is why lumping using Duncan’s

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 has also advantages for stepped beams, when the bending rigidity IE 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 ..9950101 =ωω When the beam is split into four segments (Fig. 5.2, c), the resulting three-degree-of-freedom system has natural frequencies of 980101 .=ωω , 9950202 .=ωω and 9950303 .=ωω .

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 ..70101 =ωω When the beam is split into two segments (Fig. 5.3, c), the


ratio of the first natural frequencies is ..90101 =ωω When the beam is split into three segments (Fig. 5.2, d), the ratio of the first natural frequencies is

..950101 =ωω

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

ll 0mAm == ρ , (5.1)

and the total mass moment of inertia is

IAAImmJ l

ll

l ρρ+=+=

1212

3

0

30 (5.2)

where I is the cross-section second moment of area.

For a cylindrical beam element of diameter d and length l

⎟⎟⎠

⎞⎜⎜⎝

⎛+= 2

20

43

12l

l dmJ . (5.3)

The total mass moment of inertia is composed of two parts. The first part, 122lm , is due to the element mass being distributed along the length of the beam


at the level of the neutral axis. The second part, the “rotatory inertia” Ilρ , 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

121242

23

03

02

0 llll mmmJ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−= .

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

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

12000021000012000021

2

2

0

l

llmme

t (5.4)

and an element mass matrix for rotatory inertia

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

2

2

0

00000000000000

i

imme

r l , (5.5)

where AIi = .

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)

{ } [ ]{ }fy δ= . (5.6)


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

[ ] [ ]{ } { } { }0=+ yym &&δ . (5.7)

The corresponding eigenvalue problem (4.85) is

[ ][ ] [ ] { } { }012 =⎟

⎠⎞

⎜⎝⎛ − aIm

ωδ . (5.8)

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

[ ][ ] [ ] 01det 2 =⎟⎠⎞

⎜⎝⎛ − Im

ωδ (5.9)

whose solutions are the system natural frequencies rω .

The mode shapes are defined by the modal vectors { }ra , satisfying the homogeneous linear set of equations

[ ][ ] [ ] { } { }012 =⎟

⎟⎠

⎞⎜⎜⎝

⎛− r

raIm

ωδ . (5.10)

Example 5.1 Calculate the natural frequencies of lateral vibration for the three-mass

beam of Fig. 5.4, a, where .constIE =

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

( ) ( )222

6bx

IExbFx −−= ll

v .

The flexibility coefficients are

( )222

6 jiij

ji bxIExb

−−= ll

δ .

Because ,bx 431 l== ,bx 222 l== ,bx 4313 l== the flexibility matrix is

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

91171116117119

768

3

IElδ . (5.11)



⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

91171116117119

aaa

aaa

λ

where 23768 ωλ lmIE= .

Fig. 5.4 Fig. 5.5


0287834 23 =−+− λλλ

with roots

5563311 .=λ , 22 =λ , 443603 .=λ .


31 9334 lmIE.=ω , 3

2 59619 lmIE.=ω , 33 60641 lmIE.=ω .

The modal vectors are

{ }⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

1414211

1 .a , { }⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−=

101

2a , { }⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−=

1414211

3 .a .


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

011

10 =++++ −−

nnnn .... αλαλαλα

is an algebraic equation, then the sum of its roots is

0

1

1 αα

λ −=∑=

n

ii .

Considering the frequency equation (5.9), we can write

∑∑∑===

==−=n

i ii

n

iiii

n

i im

12

10

1

12

11ω

δαα

ω,

where 2iiω is the square of the “isolated” natural frequency of the system

containing only the mass im . 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 n... ωωω <<< 21 , all terms in the first sum except the first may be omitted for the approximate determination of the fundamental frequency

∑∑==

=≅n

i ii

n

i i 12

122

1

111ωωω

or

... 1111233

222

211

21

+++≅ωωωω

(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.


For the cantilever beam of Fig. 5.3, a the true value of the fundamental

frequency (6.16) is 310 523 lmIE.=ω . For the model with one segment (Fig.

5.3, b), the lowest natural frequency is 31 442 lmIE.=ω , which is %.730

lower than the true value. When the beam is split into two segments (Fig. 5.3, c),

the first natural frequency is 31 0983 lmIE.=ω , being %12 lower than the true

one. When the beam is split into three segments (Fig. 5.2, d), the first natural

frequency is 31 2863 lmIE.=ω , which is %.127 lower, and when it is split

into four segments, it is %.54 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

IE768

9 3

3311l

== δδ , IE768

16 3

22l

=δ .

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

311

211 9

7681lm

IEm

==δ

ω ,

322

222 16

7681lm

IEm

==δ

ω ,

333

233 9

7681lm

IEm

==δ

ω ,

so that Dunkerley’s formula (5.12) yields

IE

mIE

mIE

mIE

m76834

7689

76816

76891111 3333

233

222

211

21

llll=++=++≅

ωωωω.

The estimated fundamental natural frequency is

31 75274 lmIE.≅ω ,

which is 3.6% lower than the true value calculated at Example 5.1.


5.1.1.5 Rayleigh’s Formula

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

( )

∑∫

=

∂∂= n

iim

dxxIE

1

22221ω

2iv

v, (5.13)

where ( )ii xvv = are static deflections at the mass locations.

If the strain energy is determined from the work done by the corresponding

lumped weights gm i , then ∑=

⋅=n

iimax gmU

121

iv so that Rayleigh’s formula (5.13)

becomes

∑

∑

=

== n

ii

n

ii

m

mg

1

121ω

2i

i

v

v. (5.14)

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 iv

2ωim , which is proportional to the deflection, we can

recalculate the deflection with the modified weights gm1 , 1

22 vvgm ,

1

33 vvgm .

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.



[ ] [ ]⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

111

3

2

1

δδ gmgmgmgm

vvv

wherefrom we obtain

( ) gmIE

gm76827 3

1321111l

=++= δδδv ,

( ) gmIE

gm76838 3

2322122l

=++= δδδv ,

13 vv = .

Substitution in (5.14) yields

33222321 5434831

768290292

9768

273827273827768ω

lll mEI

.mEI

mEI

==++++

=

or

31 93434 lmIE.=ω ,

which is only 0.02% higher than the true value calculated at Example 5.1.

If 40 /mm l= , 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

0

240

186869293434

mIE.

mIE.

ll=⋅=ω ,

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

0

20

2

2 86969m

IE.m

IEtrue

ll==

πω ,

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.


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

3324212ll

IEIEk =⋅

= , (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-of-freedom systems in Section 4.1.1.


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, recm , 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 rotJ . The average inertia per cylinder is

2

21 rmJJ recrot += ,

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 2mkg ⋅ , and with K values in


Nm/radx106 : 863111 .J = , 01122 .J = , 16706543 .JJJJ ==== , 89707 .J = , 06211 .K = , 10162 .K = , 053543 .KKK === , 06746 .K = .

Estimate the fundamental- and second-mode natural frequencies of torsional vibrations. Plot the corresponding mode shapes.

Solution. rad/sec 5521 =ω , rad/sec 11452 =ω . 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.


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 2i . 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 2i .

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 1K and

2K , and disc mass moments of inertia 1J and 2J , 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


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

''rr θθ −= ,

while compatibility of torques requires

r

Mr

M′′

−= or ''MM θθ = .

The transmissibility ratio is defined as

MM'

rr

nn

i′

−=−=′

=−=θθ

1

2 . (5.16)

In the original system, the stiffness matrix of shaft 2 is defined by

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧ ′

222

22

2 θθ '

KKKK

MM

.

Using the transformation equations, based on (5.16),

⎭⎬⎫

⎩⎨⎧ ′

⎥⎦

⎤⎢⎣

⎡−=

⎭⎬⎫

⎩⎨⎧

22 100

MMi

MM

, ⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−=

⎭⎬⎫

⎩⎨⎧

22 100

θθ

θθ i'

,

the stiffness matrix of shaft 2 in the equivalent model is defined as follows

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣

⎡−

−⎥⎦

⎤⎢⎣

⎡−=

⎭⎬⎫

⎩⎨⎧

222

22

2 100

100

θθi

KKKKi

MM

,

⎭⎬⎫

⎩⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

222

222

2 θθ

KKiKiKi

MM

.

The mass matrix is defined likewise

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣

⎡ ′⎥⎦

⎤⎢⎣

⎡−=

⎭⎬⎫

⎩⎨⎧

222 100

00

100

θθ&&

&&iJ

JiMM

,

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ′=

⎭⎬⎫

⎩⎨⎧

22

2

2 00

θθ&&

&&

JJi

MM

.

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


[ ]⎥⎥⎦

⎤

⎢⎢⎣

⎡=

ee

eee

KKiKiKik

2, [ ]

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ′=

e

ee

JJim0

02. (5.17)

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

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

11

111

KKKK

k , [ ] ⎥⎦

⎤⎢⎣

⎡=

JJ

m0

011 .

They can be obtained from the general expressions (5.17) substituting 1−=i . 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 1J , 5J and 6J and three rigid gears of radii r , 2/r and 3/r having mass moments of inertia 2J , 3J and 4J which are connected by three light shafts having torsional stiffnesses 1K , 2K and 3K . Derive the equations of free vibration.

Solution. In the upper branch, the speed ratio is 2−=i . In the lower branch, the speed ratio is 3−=i . 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:

[ ]

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡++−

−

=

33

22

323211

11

030020

329400

KKKK

KKKKKKKK

K

and

[ ]

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡++

=

6

5

432

1

00000000940000

JJ

JJJJ

M .


The global vector of angular displacements is

{ } { }T4321 θθθθΘ = .


[ ]{ } [ ]{ } { }0=+ ΘΘ KM && .

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 1J driven by a low-pressure turbine 5J and a high pressure turbine 6J .

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: 9501 =J , 5422 =J , 74063 .J = , 78654 .JJ == , 55137 .J = ,

1227986 .JJJ === [ ]2mkg ⋅ , 2476=i , 61 1033620 ⋅= .K , 6

2 101358 ⋅= .K ,


63 10221 ⋅= .K , 6

4 104070 ⋅= .K , 65 10631 ⋅= .K , 6

6 10442 ⋅= .K [ ]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

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−+

−−+

++−

−+−−

=

66

6655

44

4433

5352

32

22

2211

11

000000000

000000000000000000000

KKKKKiK

KKKKKiK

iKiKKiKiKKKKKK

KK

K

The global mass matrix is

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

++

=

9

8

7

6

52

42

3

2

1

000000000000000000000000000000000000000000

JJ

JJ

JiJiJJ

J

M


The natural frequencies have the following values: 01 =f , 26142 .f = , 98223 .f = , 34354 .f = , 98415 .f = , 52506 .f = , Hz 17757 .f = .

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

( ) ( ) 011 =−−−+ +− rrrrr xxkxxkxm &&

which, for harmonic motion tax rr ωsin= , can be expressed in terms of amplitudes as

02

12 1

2

1 =+⎟⎟⎠

⎞⎜⎜⎝

⎛−− −+ rrr aa

kma ω .

The solution of this equation is found by substituting

rir ea β=

which leads to the relationship

βω ββcos

221

2=

+=⎟

⎟⎠

⎞⎜⎜⎝

⎛−

−ii eekm ,


which can also be written

( )2

sin4cos12 22 ββω

=−=km . (5.18)

The general solution for ra is

rCrCar ββ sincos 21 += ,

where 1C and 2C are determined from the boundary conditions. At 0=r the amplitude is zero 00 =a , so that 01 =C . At the free end, nr = , the equation of motion is

( ) 01 =−+ −nnn xxkxm &&

which, in terms of amplitudes, becomes

nn akma ⎟⎟⎠

⎞⎜⎜⎝

⎛−=−

2

1 1 ω .

Substituting from the general solution, we obtain the following relationship for the evaluation of β :

( ) ( )[ ] nn βββ sincos1211sin −−=− .

This result can be reduced to the product form

02

sin21cos2 =⎟⎠⎞

⎜⎝⎛ +

ββ n

which is satisfied by

02

sin =β ,

and by 021cos =⎟⎠⎞

⎜⎝⎛ +nβ , or ( )

( )12212

2 +−

=n

r πβ . ( )n,...,r 1=

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

2

sin2 βωmk

= (5.19)

which leads to

( )( )122

12sin2+

−=

nr

mk

rπω . ( )n,...,r 1= (5.20)


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 521 q,...,q,q (Fig. 5.13, b).

The column vector { } { }Tq,...,q,qQ 521 = is called the global vector of nodal

displacements and { } { }Tf,...,f,fF 521 = is the global vector of nodal forces. A displacement or force has a positive value if acting along the positive q direction. At this stage, the boundary condition 05 =q is not imposed.

Table 5.1

Node Element 1 2 1 1 2 2 2 3 3 3 4 4 4 5 5 2 4 6 3 5


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 { } { }Te f,ff 21=

is related to the vector of local nodal displacements { } { }Te q,qq 21= by the

equation { } [ ] { }eee qkf = where the element stiffness matrix is

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

ee

eee

kkkk

k . (5.21)

This can be established from the equilibrium and force-deflection equations: for 02 =q , 121 qkff =−= , and for 01 =q , 221 qkff −=−= .

On the other hand, the global vector of nodal forces { }F is related to the global vector of displacements { }Q by the equation { } [ ] { }QKF = 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 [ ]ek are placed in the appropriate locations in the larger [ ]K matrix and overlapping elements are then summed.

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

{ } [ ]{ } ⎣ ⎦⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−==

4

3

33

3343

3333 2

121

qq

kkkk

qqqkqUT

.

Expanding the stiffness matrix at the system size we obtain

{ } [ ]{ }Qk~Q

qqqqq

kkkk

qqqqq

U T

T

3

5

4

3

2

1

33

33

5

4

3

2

1

3 21

000000000000000000000

21

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=

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


We see that elements of the matrix [ ]3k~ are located in the third and fourth rows and columns of the [ ]K matrix. When adding spring strain energies

{ } [ ]{ }QKQUU T

ee 2

1== ∑ ,

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

e

ek~K .

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

[ ]

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+−−−++−−−−++−

−−++−−

=

6446

454335

636322

525211

11

0000

0000

kkkkkkkkkkkkkkkk

kkkkkkkk

K .

The boundary conditions must now be specified. Node 5 is fixed so that 05 =q 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

[ ]

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

++−−−++−−−++−

−

=

54335

36322

525211

11

00

00

kkkkkkkkkkkkkkkk

kk

K .

Together with the diagonal mass matrix

[ ]

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

4

3

2

1

000000000000

mm

mm

M

it is used to write the equations of free motion

[ ]{ } [ ]{ } { }0=+ QKQM &&


and to solve the corresponding eigenvalue problem

[ ]{ } [ ]{ }ΦωΦ MK 2= ,

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

[ ]

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

++−−−++−−−++−

−

=

54335

36322

525211

11

00

00

cccccccccccccccc

cc

C .

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

[ ]{ } [ ]{ } [ ]{ } { }0=++ QKQCQM &&& .

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: kg 11121 ==== m....mm , N/m 2421111 == kk , N/m 2989102 == kk , N/m 369193 == kk , N/m 455684 == kk , N/m 562575 == kk ,


N/m 180006 =k .

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


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 Hz,dω ,ζ % Mode Hz,dω ,ζ %

1 15.98 0.502 9 68.88 1.378 2 30.86 0.968 10 73.72 1.579 3 43.60 1.364 11 128.87 0.536 4 46.47 0.301 12 136.59 0.506 5 53.35 1.665 13 143.89 0.477 6 53.42 0.670 14 150.87 0.457 7 59.45 1.853 15 157.52 0.437 8 61.62 1.060

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.


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 1x and 2x 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

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

−=

22 21

12

xxx

xxr (5.22)

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


Expressing the physical coordinate in terms of the natural coordinate yields

( ) ( ) 2211 xrNxrNx += , (5.23) where

( ) ( )rrN −= 121

1 and ( ) ( )rrN += 121

2 (5.24)

can be considered as geometric interpolation functions. The graphs of these functions are shown in Figs. 5.19, a, b.

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 1q and 2q as

( ) ( ) ( ) 2211 qrNqrNru += . (5.25)

In matrix notation

⎣ ⎦ { }e

iii qNqNu == ∑

=

2

1, (5.26)

where ⎣ ⎦ ⎣ ⎦21 NNN = and { } { }Te qqq 21= . (5.27)

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


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 ( )t,xuu = , the strains are xu ∂∂=ε and the stresses are εσ E= .

The element strain energy ( )tUe is

xdxuAEdxAEdVU

e

ee

e

ee

e

e

22

221

21 ∫∫∫ ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

=== εεσ . (5.28)

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

rdrdxx

xd e22

12 l=

−= , (5.29)

where 11 +≤≤− r and the length of the element is 12 xxe −=l .

Because { }eqxN

xu

⎥⎦

⎥⎢⎣

⎢∂∂

=∂∂ and

rN

xr

rN

xN

e ∂∂

=∂∂

∂∂

=∂∂

l

2 where 211 −=

∂∂

rN

and 212 +=

∂∂

rN , we can write { } { }e

TTe q

xN

xNq

xu

⎥⎦

⎥⎢⎣

⎢∂∂

⎥⎦

⎥⎢⎣

⎢∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

2

, so that the

element strain energy (5.28) becomes

{ } { }eT

e

eeTee qrd

rN

rNAEqU ∫

+

−

⎥⎦

⎥⎢⎣

⎢∂∂

⎥⎦

⎥⎢⎣

⎢∂∂

=

1

1

221

l. (5.30)

The above equation is of the form

{ } [ ] { }eeTee qkqU

21

= , (5.31)

where the element stiffness matrix [ ]ek is given by

[ ] ∫∫+

−

+

−

⎥⎦⎥

⎢⎣⎢−⎭⎬⎫

⎩⎨⎧−

=⎥⎦

⎥⎢⎣

⎢∂∂

⎥⎦

⎥⎢⎣

⎢∂∂

=

1

1

1

121

21

212122 rdAErd

rN

rNAEk

e

eeT

e

eee

ll

or


[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1 111

e

eee AEkl

. (5.32)

The element kinetic energy ( )tTe is

xdtuAT

e

ee

2

2 ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=ρ . (5.33)

where ρ is the material mass per unit volume and utu &=∂∂ is the velocity at x.

From (5.26) we obtain

⎣ ⎦ { }eqNu && = , (5.34)

where { }eq& is the column vector of nodal velocities.

Substituting (5.34) into equation (5.33) yields

{ } ⎣ ⎦ ⎣ ⎦ { }e

e

Te

Tee qxdNNAqT && ∫= ρ

21 . (5.35)

Equation (5.35) is of the form

{ } [ ] { }eeTee qmqT &&

21

= . (5.36)

where [ ] ⎣ ⎦ ⎣ ⎦∫=

e

Te

e xdNNAm ρ (5.37)

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

[ ] ⎣ ⎦ ⎣ ⎦( )

( )∫∫+

−

+

−⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−−−==

1

1

22

221

11111

82rd

rrrrArdNNAm eeTeee ll ρρ

or

[ ] ⎥⎦

⎤⎢⎣

⎡=

2112

6eee Am lρ . (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


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 12 −j and

j2 , and displacements 12 −jQ and jQ2 .

In Fig. 5.20 we see that 1q equals the sum of projections of 1Q and 2Q onto the x axis. Thus

αα sincos 211 QQq += . (5.39, a) Similarly αα sincos 432 QQq += . (5.39, b)

Fig. 5.20

Equations (5.39) can be written in matrix form as

{ } [ ] { }eee QTq = , (5.40)

where { } { }Te qqq 21= is the element displacement vector in the local

coordinate system, { } { }Te Q,Q,Q,QQ 4321 = is the element displacement vector in the global coordinate system and

[ ] ⎥⎦

⎤⎢⎣

⎡=

αααα

sincos0000sincoseT (5.41)

is a coordinate transformation matrix.

From nodal coordinate data, denoting ( )11 Y,X and ( )22 Y,X the coordinates of nodes 1 and 2, respectively, we obtain


e

XXl

12cos−

=α , e

YYl

12sin−

=α , ( ) ( )212

212 YYXXe −+−=l . (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

{ } [ ] [ ] [ ]{ }eeeTeTee QTkTQU

21

= . (5.43)

The strain energy in global coordinates can be written as

{ } [ ]{ }eeTee QKQU

21

= , (5.44)

where [ ]eK is the element stiffness matrix in global coordinates.

Comparing equations (5.43) and (5.44) we obtain the stiffness matrix in global coordinates as

[ ] [ ] [ ] [ ]eeTee TkTK = . (5.45)

Substituting for [ ]eT from equation (5.41) and for [ ]ek from equation (5.32) we get

[ ]⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

=

22

22

22

22

sscsscsccsccsscsscsccscc

AEKe

eee

l, (5.46)

where αcos=c and αsin=s .

Similarly, the kinetic energy in global coordinates can be written as

{ } [ ]{ }eeTee QMQT &&

21

= . (5.47)

The element consistent mass matrix in global coordinates is

[ ] [ ] [ ] [ ]eeTee TmTM = ,

or, substituting for [ ]eT from equation (5.41) and for [ ]em from equation (5.38)


[ ]⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

22

22

22

22

2222

2222

6sscssc

sccsccsscsscsccscc

AM eee lρ . (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 [ ]eK and [ ]eM 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

{ } [ ] { }QT~Q ee = , (5.49)

where { }eQ is the element displacement vector in global coordinates, { }Q is the

full displacement vector of the truss structure and [ ]eT~ is referred to as a 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)

{ } [ ] [ ] [ ]{ }QT~KT~QU eeTeTe 2

1=

or

{ } [ ]{ }QK~QU eTe 2

1= ,

where the expanded element stiffness matrix

[ ] [ ] [ ] [ ]eeTee T~KT~K~ = (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


[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

0010000000000100000000000010000000000100

4T~

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

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

0010000000000100000000001000000000010000

5T~ .

Fig. 5.21

The corresponding expanded stiffness matrices are of the form

[ ] =4K~

10987654321

10987654321

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

••••••••

••••••••

, [ ] =5K~

10987654321

10987654321

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

••••••••••••••••

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


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

{ } [ ]{ }QKQU T

21

= , (5.51)

can be calculated by simply adding element strain energies

{ } [ ]{ } { } [ ] { }QK~QQK~QUUe

eT

e

eT

ee ∑∑∑ ===

21

21 . (5.52)

Comparing (5.51) and (5.52) we get

[ ] [ ]∑=e

eK~K . (5.53)

The global stiffness matrix is equal to the sum of the expanded element stiffness matrices.

Similarly, the global mass matrix [ ]M is assembled from expanded element mass matrices as

[ ] [ ] [ ] [ ] [ ]∑∑ ==e

eeTe

e

e T~MT~M~M . (5.54)

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

Π−= TL , (5.55)

where T is the kinetic energy

{ } [ ]{ }QMQT T &&21

= (5.56, a)

and Π is the potential energy


{ } [ ]{ } { } { }FQQKQ TT −=21Π . (5.56, b)

In (5.56, b), { }F is the global vector of applied nodal forces.

Hamilton’s principle leads to Lagrange’s equations for undamped systems

{ } { } 0dd

=∂∂

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂

∂QL

Q

Lt &

or { } { } 0

dd

=∂∂

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂

∂QQ

Tt

Π&

. (5.57)

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

[ ] { } [ ]{ } { }FQKQM =+&& . (5.58)

For free vibrations, the force vector is zero. Thus

[ ] { } [ ]{ } { }0=+ QKQM && . (5.59)

Looking for solutions of the form

{ } { } tQ ωφ sin= , (5.60)

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

[ ]{ } [ ] { }rrr MK φωφ 2= , ( )n,...,r 1= , (5.61)

where 2rω are real eigenvalues equal to the natural frequencies squared and { }rφ

are real eigenvectors.

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

Fig. 5.22


Example 5.9 Calculate the first four natural frequencies and mode shapes for the system

shown in Fig. 5.22, a where GPa200=E , 3mkg7850=ρ and 2mm100=A 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

yxvyu

dd

−=−= ϕ , (5.62)


where v is the deflection of the centroidal axis at x and v′=ϕ is the cross section rotation (or slope) at x . Axial strains are

yxv

xu

2x dd

dd 2

−==ε . (5.63)

Normal stresses on the cross section are given by

yxv2xx d

d2EE −== εσ , (5.64)

where E is Young’s modulus of the material.

The bending moment is the resultant of the stress distribution on the cross section

( ) II2x vxvdAyx zz

A

IEIEM ==−= ∫ dd2

σ . (5.65)

where zI is the second moment of area of the section about the neutral axis z.

The shear force is given by

( ) III3 vxv

xMx zz IEIET ===

dd

dd 3

. (5.66)

The transverse load per unit length is

( ) IV4 vxv

xTx zz IEIEp ===

dd

dd 4

. (5.67)

The differential equation of equilibrium is

( )xpIE z =4xv

dd4

. (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 23 −iQ , 13 −iQ and iQ3 , 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


ends. Their properties are the bending rigidity IE , the mass per unit length Aρ 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.


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

{ } { }Te q,q,q,q,q,qq 654321 = (5.69)

and the corresponding vector of element nodal forces can be written

{ } { }Te f,f,f,f,f,ff 654321 = . (5.70)

Forces 6532 f,f,f,f and the corresponding displacements

6532 q,q,q,q describe the element bending (Fig. 5.25, b) while axial forces ,f,f 41 and displacements ,q,q 41 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, 0=p and equation (5.68)

yields 0dd 44 =xv . Integrating four times, we obtain the deflection v described by a third order polynomial

( ) 432

23

1 aaaa +++= xxxxv . (5.71)

In (5.71), the four integration constants 4321 aa,a,a , can be determined from the geometric boundary conditions, involving the transverse displacement and slope at each end:

1xx = , 2q=v , 3dd q=xv , and 2xx = , 5q=v , 6dd q=xv . (5.72)

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

⎣ ⎦ { }eqN=v , (5.73)

where ⎣ ⎦N is a row vector containing the shape functions, which are cubic polynomials, called Hermite polynomials.

Using natural coordinates, with 1−=r at node 1 and 1+=r at node 2, the transverse displacement can be written


( ) ( ) ( ) ( ) ( )2

4231

211 dd

dd

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

rNN

rNN vrvrvrvrrv . (5.74)

Because the coordinates transform by the relationship (5.22)

rxxxx

x22

1221 −+

+= (5.75)

and since 12 xxe −=l is the length of the element, equation (5.29) holds

rdxd e2l

= . (5.76)

Using the chain rule of differentiation

xvv

dd

2dd e

rl

= , (5.77)

equation (5.74) becomes

( ) ( ) ( ) ( ) ( )2

4231

211 dd

2dd

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

xNN

xNN ee vrvrvrvrrv ll (5.78)

or

( ) 64533221 22qNqNqNqN ee ⋅+⋅+⋅+⋅=

llrv . (5.79)

In (5.73) the row vector of shape functions is

⎣ ⎦ ⎥⎦⎥

⎢⎣⎢= 4321 22

N,N,N,NN ee ll . (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

1N 1N ′ 2N 2N ′ 3N 3N ′ 4N 4N ′

1−=r 1 0 0 1 0 0 0 0 1+=r 0 0 0 0 1 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.


( ) ( ) ( ) ( )321 32

4121

41 rrrrrN +−=+−= ,

( ) ( ) ( ) ( )3222 1

4111

41 rrrrrrN +−−=+−= , (5.81)

( ) ( ) ( ) ( )323 32

4121

41 rrrrrN −+=−+= ,

( ) ( ) ( ) ( )3224 1

4111

41 rrrrrrN −−+−=−+−= .

Fig. 5.26

It is easy to check that at node 1, 2q=v and 32dd q

rel=

v , while at node 2,

5q=v and 62dd q

rel=

v .

5.3.4 Stiffness Matrix of a Beam Element

The strain energy eU of a beam element is

xIE

Ue

ee d

dd

2

2

2

2

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

xv . (5.82)


Equation (5.77) yields

re d

d2dd vxv

l= and 2

2

2

2

dd4

dd

re

vxv2 l= .

Substituting (5.73) we obtain

{ }e

eq

rN⎥⎥⎦

⎥

⎢⎢⎣

⎢= 2

2

2

2

dd4

dd

l2xv . (5.83)

The square of the above quantity is calculated as

{ } { }eT

e

TeT

qrN

rNq

⎥⎥⎦

⎥

⎢⎢⎣

⎢

⎥⎥⎦

⎥

⎢⎢⎣

⎢=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛2

2

2

2

4

2222

dd

dd16

dd

dd

dd

l222 xv

xv

xv ,

which can be also written

{ } ⎣ ⎦ ⎣ ⎦ { }er

Tr

e

Te qNNq ′′′′=⎟⎟⎠

⎞⎜⎜⎝

⎛4

22 16dd

l2xv . (5.84)

On substituting (5.76) and (5.84) into (5.82) we get the element strain energy

{ } ⎣ ⎦ ⎣ ⎦ { }er

Tr

e

eTee qrdNNIEqU ∫

+

−

′′′′=1

13

821

l (5.85)

which has the form

{ } [ ] { }eeB

Tee qkqU

21

= . (5.86)

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

[ ] ⎣ ⎦ ⎣ ⎦∫+

−

′′′′=1

13

8 rdNNEIk rT

re

eeB

l (5.87)

or

[ ]( )

( )( )

( )

r

NNNNNNNNNNNNNNNNNNNNNNNNNNNN

EIke

eeB d8

1

12

4342414

432

32313

42322

212

4131212

1

3 ∫+

−⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′

=l

. (5.88)


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

[ ]e

e

eeB

EIk⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

22

22

3

4626612612

2646612612

llll

ll

llll

ll

l. (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, ( )tx,vv = .

The instantaneous kinetic energy of a beam element is

xdt

ATe

ee

2

2 ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=vρ . (5.90)

where ρ is the material mass per unit volume and vv &=∂∂ t is the velocity at x.

From (5.73) we obtain

⎣ ⎦ { }eqN && =v , (5.91)

where { }eq& is the column vector of nodal velocities.

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

{ } ⎣ ⎦ ⎣ ⎦ { }e

e

Te

Tee qxdNNAqT && ∫= ρ

21 . (5.92)

which has the form

{ } [ ] { }eeB

Tee qmqT &&

21

= . (5.93)

where

[ ] ⎣ ⎦ ⎣ ⎦∫+

−

=1

1

d2

rNNAm rT

reee

Blρ (5.94)

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


[ ]e

eeeB

Am⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−−−

=

22

22

422313221561354313422135422156

420llll

ll

llll

ll

lρ . (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

[ ]⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

4

1

4

1

qq

kff e

S (5.96)

where the stiffness matrix (5.32) is

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1 111

e

eeS

EAkl

. (5.97)

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

[ ] ⎥⎦

⎤⎢⎣

⎡=

2112

6eee

SAm lρ . (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

[ ]

e

e

IEIEIEIE

IEIEIEIE

EAEA

IEIEIEIE

IEIEIEIE

EAEA

k

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−

−

−

−

−

=

llll

llll

ll

llll

llll

ll

460260

61206120

0000

260460

61206120

0000

22

2323

22

2323

. (5.99)


The ratio of the bending terms to the stretching terms in (5.99) is of order ( ) 2li , where ‘i’ is the relevant radius of gyration. For slender beams this ratio may be as small as 201 or 501 , 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

[ ]

e

eee Am

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

22

22

42203130221560135400014000703130422013540221560007000140

420

llll

ll

llll

ll

lρ . (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 1q and 2q are related to the global linear displacements 1Q and 2Q by the equations

.QQq

,QQq

αα

αα

cossin

sincos

212

211

+−=

+=. (5.101)

Fig. 5.27

Equations (5.101) can be written in matrix form as


[ ]⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

2

1

2

1

QQ

Rqq

(5.101, a)

where

[ ] ⎥⎦

⎤⎢⎣

⎡−

=cssc

R

(5.102)

is referred to as a rotation matrix and αcos=c and αsin=s .

The angular displacements (rotations) are the same in both coordinate systems

33 Qq = . (5.103)

Adding the similar relationships for node 2

[ ]⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

5

4

5

4

QQ

Rqq

, 66 Qq = ,

we obtain

{ } [ ] { }eee QTq = , (5.104)

where { }eq is the element displacement vector in the local coordinate system,

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

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

1000000000000000010000000000

cssc

cssc

T e (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

[ ] [ ] [ ] [ ]eeTee TkTK = , (5.106)

and

[ ] [ ] [ ] [ ]eeTee TmTM = . (5.107)


5.3.10 Assembly of Stiffness and Mass Matrices

The global stiffness and mass matrices, [ ]K and [ ]M , are assembled from

element matrices [ ]eK and [ ]eM using element connectivity matrices [ ]eT~ that relate the nodal displacements at element level with the nodal displacements at the complete structure level by equations of the form

{ } [ ] { }QT~Q ee = . (5.108)

The global uncondensed stiffness matrix is equal to the sum of the expanded element stiffness matrices

[ ] [ ]∑=e

eK~K . (5.109)

where [ ] [ ] [ ] [ ]eeTee T~KT~K~ = (5.110)

Similarly, the global uncondensed mass matrix [ ]M is assembled from expanded element mass matrices as

[ ] [ ] [ ] [ ] [ ]∑∑ ==e

eeTe

e

e T~MT~M~M . (5.111)

For grounded systems the unreduced stiffness and mass matrices [ ]K and [ ]M 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. 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 GPa207=E , 3mkg7810=ρ , 4mm271=I

and 2mm680.A = for all beams. The frame is mm9606. wide and mm9606. 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.


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 secrad , calculated using the MATLAB 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.

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 GPa210=E ,

3mkg7850=ρ , 47 m100551 −⋅= .I , 24 m10733 −⋅= .A and m50.=l .

Fig. 5.29


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


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 23 −iQ , 13 −iQ and iQ3 , 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

IE , the torsional rigidity tIG , the mass per unit length Aρ and the length l . Only cross sections whose shear centre coincides with the centroid are considered.


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

{ } { }Te q,q,q,q,q,qq 654321 = (5.112)

and the corresponding vector of element nodal forces can be written

{ } { }Te f,f,f,f,f,ff 654321 = . (5.113)

In (5.113) 3 f and 6 f are transverse forces, while 2f and 5 f are couples producing bending (Fig. 5.32, b). The corresponding displacements

63 q,q are translations while 52 q,q are rotations. Their column vectors are related by the flexural stiffness matrix.

Rearranging the matrix (5.89) we obtain

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

6

5

3

2

22

22

3

6

5

3

2

12612664621261266264

qqqq

EI

ffff

e

e

e

ll

llll

ll

llll

l. (5.114)


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

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

6

5

3

2

22

22

6

5

3

2

156225413 22413354 131562213 3224

420qqqq

A

ffff

e

ee

&&

&&

&&

&&

ll

llll

ll

llll

lρ . (5.115)

The axial nodal forces 41 f,f are torques and the nodal displacements

41 q,q 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

( ) ( ) ( ) 4211 qrNqrNr +=θ

which substituted into the strain energy xx

IGU

e

ete d

2

2

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=θ yields, after the

change of coordinates, the stiffness matrix [ ] ⎣ ⎦ ⎣ ⎦∫+

−

′′=1

1

d2

rNNIG

k rT

re

etet

l.

As a consequence of this analogy, the nodal forces are related to the nodal displacements by equation

[ ]⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

4

1

4

1

qq

kff e

t (5.116)

where the stiffness matrix for torsional effects is

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1 111

e

etet

IGk

l. (5.117)

In (5.117), G is the shear modulus of elasticity and etI 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

[ ] ⎥⎦

⎤⎢⎣

⎡=

2112

6eete

tI

mlρ

. (5.118)


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)

[ ]

e

e

ee

aa

aa

EIk

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

=

126012606406200000126012606206400000

22

22

3

ll

llll

ll

llll

l (5.119)

where eeet IEIGa 2l= .

Combining the mass matrices from equations (5.115) and (5.118) we obtain the consistent mass matrix for the grid element in local coordinates

[ ]

e

eee

bb

bb

Am

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

156220541302240133000200

5413015622013302240

00002

42022

22

ll

llll

ll

llll

lρ (5.120)

where eet AIb 70= .

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)

{ } [ ] { }eee QTq = , (5.121)

where { }eq is the element displacement vector (5.112) in the local coordinate system,

{ } { }Te Q,Q,Q,Q,Q,QQ 654321 =


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

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

1000000000000000010000000000

cssc

cssc

T e , (5.122)

where αcos=c and αsin=s , 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

[ ] [ ] [ ] [ ]eeTee TkTK = , (5.123)

and

[ ] [ ] [ ] [ ]eeTee TmTM = . (5.124)

They are used to assemble the unreduced global stiffness and mass matrices, [ ]K and [ ]M , using element connectivity matrices [ ]eT~ 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 mN1000=k each. The

system properties are GPa210=E , GPa81=G , 3mkg7900=ρ , m50.=l and the diameter is mm20=d for all beams.


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).


Example 5.13 The grid shown in Fig. 5.35 is fixed at points 1 and 2, and has m1=l ,

48 m107850 −⋅= .I , 48 m10571 −⋅= .It , 24 m10143 −⋅= .A , 3mkg7900=ρ , GPa210=E and GPa81=G . 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


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)

( )[ ] [ ] [ ] [ ][ ] 12 ii−

++= kcmH ωωω (5.125)

where ω is the driving frequency.

The FRF matrix can be written

( )[ ]

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

ωωω

ωωωωωω

ω

iii

iiiiii

i

21

22221

11211

nnnn

n

n

h...hh............

h...hhh...hh

H (5.126)

where ( )ωiljh 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 § 742 .. ). In complex

notation, if the excitation ( ) tef̂tf ωill = produces the response

( ) ( )φω += tjj ex̂tx i , then the complex receptance can be expressed in terms of its

real and imaginary components

( ) ( ) ( )ωωφφω φ Ij

Rj

jjjjj hh

x̂f̂

x̂f̂e

x̂f̂

xfh ll

lllll isinicosi i +=+=== .


The ratio ll xf is called a direct receptance (driving point receptance), while the ratio jxfl ( )l≠j 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).


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 § 742 .. ). 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


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

Fig. 5.38

For the point receptance 11h , a three-dimensional plot is illustrated in Fig. 5.39, together with the companion coincident, quadrature, and Nyquist plots.

Fig. 5.39


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


Figure 5.41 illustrates the frequency response diagrams of the receptance 153,h 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


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 11h in Fig. 5.42, a and the function 66h 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 mN106=k , kg50741 .mmm === , kg51952 .mmm === ,

kg1863 === mmm .

Fig. 5.E1

Answer: Hz5731 .f = , Hz621242 .f = , Hz781743 .f = , Hz464339 .f = .

5.E2 Calculate the natural frequencies and mode shapes of the system shown in Fig. 5.E2, considering kg1=jm ( )121 ...,,j = , mN100=jk ( )111 ...,,j = , mN10212 =k . Compare the results with those obtained for

mN10012 =k .

Fig. 5.E2

Answer: a) Hz823802 .f = , Hz828503 .f = , Hz591514 .f = , Hz594115 .f = .


b) Hz8238032 .ff == , Hz5915154 .ff == , Hz2508276 .ff == .

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:

2111 .m = , 4412 .m = , 6913 .m = , 9614 .m = , 2525 .m = , 5626 .m = , 8927 .m = , 2438 .m = , 6139 .m = , kg 410 =m , 11001 =k , 12002 =k ,

13003 =k , 14004 =k , 15005 =k , 16006 =k , 17007 =k , 18008 =k , 19009 =k , mN200010 =k , jj k.c 0020= ( )101 ...,,j = .

Fig. 5.E3

Answer: Hz71801 .f = , %.4501 =ζ , Hz34810 .f = , %.24510 =ζ .

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: Hz6521 .f = , %.8101 =ζ , Hz7315615 .f = , %.44015 =ζ .


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: mk.51701 =ω , mk=2ω , mk.41413 =ω , mk.73214 =ω ,

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:

radmMN22342312 .KKK === , radmMN9045 .K = , radmMN24056 .K = , 2

4321 mkg510.JJJJ ==== , 25 mkg35.J = , 2

6 mkg44.J = .


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 the first three eigenmodes of torsional vibrations. Use rpm405=An ,

rpm225=Bn , 29 mkg602=J , 2

10 mkg91=J , 211 mkg1953=J ,

212 mkg4331=J , 2

821 mkg184==== J...JJ , radmMN11689 =K , radmMN513109 .K , = , radmMN1851110 =,K (reduced to Bn ),

radmMN84601211 .K , = , radmMN84782312 ==== K...KK (G. Ziegler, 1977).


Fig. 5.E7

Answer: Hz6822 .f = , Hz60203 .f = , Hz81354 .f = .

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: radmMN21 =K , radmMN612 .K = ,

radmMN13 =K , radmMN44 =K , 21 mkg15=J , 2

2 mkg10=J , 2

3 mkg18=J , 24 mkg6=J . The speed ratio of drive shaft to axle is 4 to 1.

Fig. 5.E8

Answer: Hz26552 .f = , Hz32623 .f = .

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

231 mkg10622 −⋅= .J , 23

32 mkg10655 −⋅== .JJ , 234 mkg10311 −⋅= .J ,

2365 mkg10556 −⋅== .JJ , 23

87 mkg10490 −⋅== .JJ , m915021 .== ll , m01613 .=l , GPa81=G , mm19=d (James et al., 1989).


Fig. 5.E9


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

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

32454023457268404068724523404532

3750

3

IEmlδ .

The natural frequencies are 31 41264 lmIE.=ω ,

32 61117 lmIE.=ω , 3

3 98638 lmIE.=ω , 34 199764 lmIE.=ω .

b) 31 6470 lmIE.=ω , 3

2 4822 lmIE.=ω , 33 655 lmIE.=ω .


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 mm1x6Φ and

m30.=l , GPa772.E = , 3mkg3100=ρ . For all lumped masses kg60.m = .

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 GPa210=E ,

3mkg7800=ρ , m2=l , 2mm2000=A .

Fig. 5.E12


5.E13 Determine the lowest 20 eigenmodes of the continuous four-span beam from Fig. 5.E13. Comment the clustering of natural frequencies. Consider

GPa208=E , 3mkg7850=ρ , m1=l and rectangular cross-section with mm402 == hb .

Fig. 5.E13


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

lEIK 500= , where IE 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 ( )0=Δ . Notice the clustering of natural frequencies. b) Consider a disordered weakly coupled beam, taking the span lengths equal to ( )Δ−50.l and ( )Δ+50.l , where 010.=Δ . 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 m3 , mass per unit

length mkg15 , and bending rigidity 2kNm4 with GPa200=E . 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).


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 mm40=d and length m21.=l is supported by two bearings of stiffnesses mN41 μ=k and mN62 μ=k , as shown in Fig. 5.E17. The two discs have masses kg4001 =m and kg6002 =m , and radii of gyration m301 .r = and m502 .r = , respectively. Determine: a) the natural frequencies of the flexural vibrations, and b) the first four mode shapes, taking GPa210=E and 3mkg7850=ρ .

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 GPa210=E and 3mkg7850=ρ .


Fig. 5.E18

Answer: Hz76361 .f = , Hz431202 .f = .

5.E19 Determine the first two natural frequencies for the folded beam shown in Fig. 5.E19 and plot the mode shapes. a) Consider l== CDAB and

0=BC . Model the two beams by at least three finite elements. b) Consider l== CDAB and 10l=BC and compare the natural frequencies with those

determined at (a).

Fig. 5.E19

Answer: a) Hz711121 .ff == . b) Hz45101 .f = , Hz69102 .f = .

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

2m0060.Avert = , 2m0040.Ahor = , 2m0030.Adiag = , 4m07560.I = . Material

properties: GPa75=E and 3mkg2800=ρ .

Fig. 5.E20


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 50=l m and has a square cross-section of edge

m1250. . All the outer endpoints are simply-supported and the inner endpoints are rigidly jointed together. The material properties are GPa200=E and

3mkg8000=ρ . Use more than two beam elements to discretize each arm.

Fig. 5.E21

Answer: Hz668751 .f = , Hz8485832 .ff == , Hz8856884 .f...f === .

5.E22 Consider the system from Fig. 5.E22 as a horizontal grid and determine the lowest four natural modes of vibration. Take m250.=l , m201 .=l ,

GPa210=E , 30.=ν and 3mkg7850=ρ . All beams have a square cross-

section with the edge mm5=a . 41410 a.It = .


Fig. 5.E22


5.E23 Determine the first six modes of coupled bending-torsion vibrations of the grillage from Fig. 5.E23. Consider mm50=l , GPa210=E , GPa80=G ,

3mkg7800=ρ and circular cross-section with mm5=d .

Fig. 5.E23


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 27 Nm1052 ⋅. , torsional rigidity

27 Nm10921 ⋅. and mass per unit length mkg300 ; b) two horizontal steel shafts

EBD and FCG, rigidly attached to ABC, and having a flexural rigidity 27 Nm105 ⋅ and mass per unit length mkg500 ; c) masses of kg500 at D and E, and masses of kg250 at F and G. You may neglect the moments of inertia of these masses.

Consider GPa210=E , GPa81=G , 3mkg7850=ρ . 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, . The relationship between the bending moment and the curvature (5.65)

becomes ( tx,vv = )


( ) 2xvtx,

∂∂

=2

zIEM , (6.1)

where E is Young’s modulus for the material of the beam and is the second moment of area of the cross-section about the principal axis which lies in the neutral plane.

zI

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

( ) 3xv

xMtx,

∂∂

=∂∂

=3

zIET . (6.2)

The transverse load per unit length is

( ) 4xv

xTtx,

∂∂

=∂∂

=4

zIEp . (6.3)

It consists of two components, the applied external transverse load and the inertial transverse load (mass per unit length times acceleration)

( )t,xq

2tv

∂∂

−2

Aρ . Substituting

( ) ( ) 2tvtx,

∂∂

−=2

At,xqp ρ

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

( )t,xqt

AIE z =∂∂

+∂∂

2

24 vxv4 ρ . (6.4)

6. CONTINUOUS SYSTEMS 261

6.1.2 Natural Frequencies and Mode Shapes

For free vibrations ( ) 0=t,xq . We look for conditions under which is a synchronous harmonic motion

( )tx,v

( ) ( ) ( )φω += txV sintx,v . (6.5)

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

0dd 2

4=− V

IEAV

z

ρω4x. (6.6)

Equation (6.6) is of fourth order so the general solution will contain four constants

( ) xBxBxBxBxV αααα coshsinhcossin 4321 +++= (6.7)

where

zIEAρωα 24 = . (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 these conditions for a uniform beam are:

( )xV

a) Simply supported end:

and 0=V 0dd 22 =xV . (6.9)

b) Clamped end:

and 0=V 0dd =xV . (6.10)

c) Free end:

0dd 22 =xV and 0dd 33 =xV . (6.11)

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 lα whose solutions yield the natural frequencies

( ) 42

ll

AIE z

nn ραω = . ...,,,n 321= (6.12)


Since equation (6.6) is homogeneous, each mode shape function 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

( )xVn

( )xVn 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 and 0=V0dd 22 =xV at and at 0=x l=x . Substituting the conditions at in

equation (6.7) we obtain 0=x

420 BB += and 420 BB +−= .

Thus . 042 == BB

Applying the conditions at l=x , we obtain

ll αα sinhsin0 31 BB += and ll αα sinhsin0 31 BB +−= .

The only non-trivial solution is 03 =B and 0sin =lα , leading to

πα n=l , ...,,,n 321= (6.13)

The first four roots of equation (6.13) are

56612 4259 2836 1423 .,.,.,.n =lα . 4321 ,,,n = .


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

AIEn z

n ρπω

42 ⎟

⎠⎞

⎜⎝⎛=l

. ...,,,n 321= (6.14)

and the modes of vibration are given by

( ) ( )n

txnC n φωπ+= sinsin

ltx,v ,

where C is an arbitrary constant.

The first four modes of vibration, corresponding to 1=n , 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 , , 0=x 0=V 420 BB += .

At , 0=x 0dd =xV . 310 BB += .

At , l=x 0dd 22 =xV .

llll αααα coshsinhcossin0 4321 BBBB ++−−= .

At , l=x 0dd 33 =xV .

llll αααα sinhcoshsincos0 4321 BBBB +++−= .

The above four linear equations have a non-trivial solution if the following determinant vanishes


0

sinhcoshsincoscoshsinhcossin

01011010

=

−−−

llll

llll

αααααααα

which leads to the frequency equation

1coshcos −=⋅ ll αα . (6.15)

The first five roots of equation (6.15) are

13714 99610 8557 6944 8751 .,.,.,.,.n =lα . 543 21 ,,,,n = . (6.16)

For n greater than five the values are given approximately by

πα2

12 −=

nnl . (6.16, a)

Substituting and 24 BB −= 13 BB −= in the last two end conditions yields

( ) ( )( ) ( .BB

,BB

0sinhsincoshcos

0coshcossinhsin

21

21

=−−+ )=+++

llll

llll

αααα

αααα

Equating to zero the determinant of the above equations yields again the frequency equation (6.15). Expressing , and in terms of , we obtain the shape of the nth mode

4B 3B 2B CB =1

( ) ( )[ ]xxkxxCxV nn αααα sinsinhcoscosh −−−= (6.17) where

ll

ll

αααα

sinsinhcoscosh

++

=nk , ...,,,n 321= (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

, ( ) 00 =′′V ( ) 00 =′′′V , ( ) 0=′′ lV , ( ) 0=′′′ lV . (6.19)

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

1coshcos +=⋅ ll αα . (6.20)



13714 99610 8537 7304 .,.,.,.n =lα . 4321 ,,,n = (6.21, a)

For n greater than four the values are given approximately by

πα2

12 −=

nnl . (6.21, b)

Fig. 6.4


( ) ( )[ ]xxkxxCxV nn αααα coscoshsinsinh +−+= (6.22)

where

ll

ll

αααα

coscoshsinsinh

−−

=nk , ...,,,n 321=


Equation (6.6) admits other two solutions which satisfy the end conditions (6.19), namely

( ) ( ) .BxBxV.,constBxV 75 6 +===

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

, ( ) 00 =V ( ) 00 =′V , ( ) 0=lV , ( ) 0=′ lV .

The frequency equation is the same as (6.19) and so are the natural frequencies given by 137149961085377304 .,.,.,.n =lα , for 4321 ,,,n = , and

πα2

12 −=

nnl for n greater than four.


Fig. 6.5


( ) ( )[ ]xxkxxhCxV nn αααα sinsinhcoscos −−−= (6.23)

where

ll

ll

αααα

sinsinhcoscosh

−−

=nk , ...,,,n 321= (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

, ( ) 00 =V ( ) 00 =′V , ( ) 0=lV , ( ) 0=′′ lV .


ll αα tantanh = . (6.25)


35213 21010 0697 9273 .,.,.,.n =lα . 4321 ,,,n = .

For n greater than four the values are given approximately by


πα4

14 +=

nnl . (6.26)


( ) ( )[ ]xxkxxCxV nn αααα sinsinhcoscosh −−−= (6.27)

where

ll

ll

αααα

sinsinhcoscosh

−−

=nk , ...,,,n 321= (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

, where ( )xm ( ) ( )xAxm ρ= is the mass per unit length. Let the bending rigidity be . ( )xIE

Considering the modes of indices r and s, respectively, we can write

, (6.29) ( ) 02 =−′′ rrr Vm"VIE ω

( ) 02 =−′′ sss Vm"VIE ω . (6.30)

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

sV rVl

, (6.31) ( ) ∫∫ =′′ll

0

2

0

dd xmVVxV"VIE srrsr ω

. (6.32) ( ) ∫∫ =′′ll

0

2

0

dd xmVVxV"VIE rssrs ω

Subtracting equation (6.32) from (6.31) gives

( ) ( ) ( )[∫∫ ′′−′′=−ll

00

22 dd xV"VIEV"VIExmVV rssrrssr ωω ] . (6.33)

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


( ) ( ) ( )[ ] ({ }ll

00

22 d srrssrrsrssr VVVVIE'VIEV'VIEVxmVV ′′′−′′′−′′−′′=− ∫ωω ) . (6.34)

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:

( )( ) .'VIEVIE

,'VIEV,VIEV,VV

0 and 00 and 0 0 and 0 0 and 0

=′′=′′=′′=′=′′==′=

(6.35)

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

, ( ) ( ) ( ) 0d0

=∫l

xxVxVxm sr sr ≠ . (6.36)

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 ( )xVr and are orthogonal (with each other) with respect to the weighting function .

( )xVs( )xm

For a uniform beam, equation (6.36) becomes

, ( ) ( ) 0d0

=∫l

xxVxV sr sr ≠ . (6.36, a)

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

, ( ) ( ) ( ) 0d0

=′′′′∫l

xxVxVxIE sr sr ≠ .

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


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 1l

( ) 1111111111 coshsinhcossin xDxCxBxAxV αααα +++=

and for the second segment of length 2l

( ) 2222222222 coshsinhcossin xDxCxBxAxV αααα +++= .

The boundary conditions

( ) ( ) 000 11 =′′=VV , ( ) ( ) 00211 ==VV l , ( ) ( )0211 VV ′=′ l ,

( ) (0211 VV ′′= )′′ l , ( ) ( ) 02222 =′′= ll VV

yield the following five homogeneous algebraic equations

( )( ) .DCA

,DCA

,DCA

,CACA

,CA

0coscoshcoshsin

0coscoshsinhsin

02sinhsin

0coshcos

0sinhsin

2222222

2222222

21111

221111

1111

=+++−

=−++

=−+−

=−−+

=+

llll

llll

ll

ll

ll

αααα

αααα

αα

αα

αα

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

1A 1C 2A 2C 2D

( ) ( ) 0sinh1sinsinsin1sinhsinh =⋅−⋅−⋅−⋅ llllll ααμαμααμαμ ,

where ll1=μ and . 21 lll +=

Table 6.1 gives the values of lnα 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.


Fig. 6.7


Table 6.1

Beam structure Roots of characteristic equation Mode index Nr. of

spans 1 2 3 4 5 1 3.142 6.283 9.425 12.566 15.708 2 3.142 3.927 6.283 7.069 9.425 3 3.142 3.550 4.304 6.283 6.692 4 3.142 3.393 3.927 4.461 6.283 5 3.142 3.299 3.707 4.147 4.555 6 3.142 3.267 3.550 3.927 4.304 7 3.142 3.236 3.456 3.770 4.084 8 3.142 3.205 3.393 3.644 3.927 9 3.142 3.205 3.330 3.550 3.801

10 3.142 3.205 3.299 3.487 3.707 11 3.142 3.173 3.267 3.424 3.613

12 3.142 3.173 3.267 3.393 3.550


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 tF ωsin0 acting at the middle (Fig. 6.8).

Assuming a steady-state solution of the form

( ) ( ) txY ωsin=tx,v (6.37)

where ω is the excitation frequency, equation (6.4) yields


( ) xCxCxCxCxY αααα coshsinhcossin 4321 +++= (6.38)

where

zIEAρωα 24 = . (6.39)

The boundary conditions for half of the beam are

( ) 00 =′′Y , , ( ) 00 =′′′Y ( ) 02 =′ lY , ( ) 22 0FYIE =′′′ l . (6.40)

The amplitude of the forced vibrations at a location 20 l≤≤ x is

( ) ( )

( )G

C

xx

xxIE

FxY

αααααα

αα

αααααα

αα

α

coshcos

2sinh

2cos

2sin

2cosh

2cosh

2cos

sinhsin

2sinh

2cos

2sin

2cosh

2sin

2sinh

4 30

++

+−

−++

−=

llll

ll

llll

ll

(6.41)

The dynamic deflection at the beam midpoint is

2sinh

2cos

2sin

2cosh

2cosh

2cos12

162

330

llll

ll

l

llαααα

αα

α +

+⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

IEF

Y . (6.42)

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

02

tan2

tanh =+ll αα . (6.43)

The first three roots of equation (6.43) are

6398 4985 36522 .,.,.n =lα . 321 ,,n = .

The values

( ) 73041 .s =lα , ( ) 996102 .s =lα , ( ) 278173 .s =lα , (6.44)

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


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

02

cosh2

cos1 =+ll αα , (6.45)

whose first three roots are

8557 6944 87512 .,.,.n =lα 321 ,,n = . or

( ) ( ) 75031 .=lα , ( ) ( ) 38892 .=lα , ( ) ( ) 710153 .=lα .

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 tM ωsin0 acting at the middle (Fig. 6.10).

Fig. 6.10


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

( ) 00 =′′Y , , ( ) 00 =′′′Y ( ) 02 =′ lY , ( ) 22 0MYIE −=′′ l . (6.46)

The amplitude of the forced vibrations at a location 20 l≤≤ x is

( ) ( )

( )G

C

xx

xxIE

MxY

αααααα

αα

αααααα

αα

α

coshcos

2sinh

2cos

2sin

2cosh

2sinh

2sin

sinhsin

2sinh

2cos

2sin

2cosh

2cosh

2cos

4 20

+−

+−

−+−

+=

llll

ll

llll

ll

(6.47)

Equating to zero the denominator of equation (6.47) we obtain the frequency equation

02

tan2

tanh =−ll αα . (6.48)

The first three roots of equation (6.48) are

6398 4985 36522 .,.,.n =lα . 321 ,,n = .

The values

( ) 92731 .a =lα , ( ) 06972 .a =lα , ( ) 21103 .a =lα , (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 tF ωsin0 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)

( ) 42

ll

AIE z

nn ραω = ...,,,n 321=

where lnα is given by (6.44) and (6.49). In can be seen that

11 αα =s , 21 αα =a , 32 αα =s , 42 αα =a , 53 αα =s , 63 αα =a , . . .

This is explained by writing equation (6.20) under the form


1

2tanh1

2tanh1

2tan1

2tan1

2

2

2

2

=−

+⋅

+

−

l

l

l

l

α

α

α

α

which becomes

02

tanh2

tan2

tanh2

tan =⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛ +

llll αααα .

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 ( )t,xuu = due to axial forces. In Fig. 6.11 a free-body 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

( ) 02

2=

∂∂

−−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+tuxdANxd

xNN ρ ,

2

2

tuA

xN

∂∂

=∂∂ ρ , (6.50)

where u is the displacement at x.


The displacement at xdx + will be ( ) xdxuu ∂∂+ . The axial strain is xux ∂∂=ε , the axial stress is xuEx ∂∂=σ , where E is Young’s modulus of

elasticity. The strain can be expressed in terms of the axial force N as

AE

Nxu=

∂∂ . (6.51)

Combining equations (6.50) and (6.51) yields the differential equation of free vibration

2

2

2

2

tu

Exu

∂∂

=∂∂ ρ . (6.52)

The constant ρEc = is the velocity of propagation of longitudinal waves along the rod.

A harmonic synchronous motion is defined by a solution of the form

( ) ( ) ( )φω += txUt,xu sin (6.53)

which substituted in equation (6.52) yields

0dd 2

2

2=+ U

xU β (6.54)

where

Eρωβ 22 = . (6.55)

The general solution of equation (6.52) is

( ) xCxCxU ββ cossin 21 += . (6.56)

The two integration constants and may be determined from the end conditions. At a fixed end the displacement is zero,

1C 2C0=u . At a free end, the stress

and hence the strain is zero, 0=∂∂ xu . The effect of end springs or concentrated masses can also be accounted for.

Free-free rod

For a rod with both ends free 0=∂∂ xu at 0=x and l=x . We obtain and, since must be nonzero in order to have vibration, the frequency

equation (characteristic equation) becomes 01 =C 2C

0sin =lβ . (6.57)

Its solution consists of an infinite set of eigenvalues


πβ nn =l , ...,,,n 321=

The natural frequencies (eigenfrequencies) are thus given by

ρ

πω Ennl

= , ...,,,n 321= (6.58)

and the mode shapes (eigenfunctions) are defined by

( )l

xnCxUn πcos= ,

where C is an undetermined amplitude.

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 , , the dynamic nodal displacements and , , the dynamic nodal forces (positive in the x-direction), as in Section 5.2.1, the end conditions are

1u 2u 1f 2f

( ) 10 uU = , , ( ) 2uU =l ( )AE

fU 10 −=′ , ( )

AEf

U 2=′ l . (6.59)

Using equation (6.56) we obtain

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

2

1

2

1

cossin10

CC

uu

ll ββ, ,

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

2

1

2

1

sincos01

CC

EAff

ll βββ

and, for 0sin ≠lβ ,

, ⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

2

1

2

1

cotancoseccoseccotan

uu

EAff

ll

ll

ββββ

β

so that the dynamic stiffness matrix is


. (6.60) [ ] ⎥⎦

⎤⎢⎣

⎡−

−=

ll

ll

ββββ

βcotancoseccoseccotan

EAk edyn

The Taylor’s series expansions up to three terms, of the coefficients of the dynamic matrix (6.60), are

( ) ....AE

AAAEEA −−−=453

cotan 23

42 ρωρωββ ll

ll ,

( ) ....AE

AAAEEA −−−−=−360

76

cosec23

42 ρωρωββ ll

ll .

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 ( )2ω− times the coefficient of the consistent mass matrix (5.38), both calculated based on static shape functions.

If the frequency dependent stiffness matrix is

[ ] ( ) ⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=

1 87871

451 111 24

AEAAEk e ll

lρω

and the frequency-dependent consistent mass matrix is

[ ] ( ) ⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=

1 87871

452

2 112

622

AEAAme ll

l ρωρ

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

2

2

2

2

tGx ∂∂

=∂∂ θρθ . (6.61)

where θ is the twist at x and G is the shear modulus of elasticity.

In order to calculate the natural modes of vibration, a harmonic solution is looked for, of the form

( ) ( ) ( )φωΘθ += txt,x sin . (6.62)


Upon substitution in equation (6.61) we obtain

0dd 2

2

2=+ ΘγΘ

x (6.63)

where

Gρωγ 22 = . (6.64)

The general solution of equation (6.63) is

( ) xCxCx γγΘ cossin 21 += . (6.65)

The integration constants and may be determined from the end conditions. At a fixed end the rotation angle is zero,

1C 2C0=θ . At a free end,

0=∂∂ xθ (as the angle of twist due to a torque M is tIGxMd d=θ ).

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 0=x , 0=θ , and at l=x , 0=∂∂ xθ . We obtain and the frequency equation

02 =C

0cos =lγ . (6.66)

Equation (6.66) has an infinite set of solutions

πγ2

12 +=

nnl , ...,,,,n 3210=

The natural frequencies are thus given by

ρ

πω Gnn

l212 +

= , ...,,,,n 3210= (6.67)

and the mode shapes are defined by

( )l

xnCxn πΘ2

12sin +=

where C is an undetermined amplitude.

Dynamic stiffness matrix

By analogy with axial vibrations, the dynamic stiffness matrix of a twisted rod segment is


. (6.68) [ ] ⎥⎦

⎤⎢⎣

⎡−

−=

ll

ll

γγγγ

γcotancoseccoseccotan

te

dyn IGk

where γ is given by (6.64) and is the torsional rigidity of the beam cross-section. This matrix relates the nodal dynamic axial torques with the nodal dynamic twist angles.

tIG

Using Taylor series expansions as in Section 6.2, the following frequency-dependent stiffness and mass matrices are obtained

[ ] ( ) ⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=

1 87871

451 111 24

tt

te

IGIIGk ll

lρω ’

[ ] ( ) ⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=

1 87871

452

2 112

622

tt

te

IGIIm ll

l ρωρ .

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

080.rn <l , where AIr = 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 106, 132 Damped systems 156 Damping 10

− coefficient 157 − Coulomb 92

− 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 30

Flexibility coefficients 130, 187


− 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 126, 196 Grillage 234

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 89

Kinetic energy 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

INDEX 291

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 12, 176 Underdamped system 37

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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Transcript of M. Rades - Mechanical Vibrations 1