Structural Dynamics for Engineers, 2e

Structural Dynamics for EngineersSecond edition

Structural Dynamicsfor EngineersSecond edition

H.A. Buchholdt and S.E. Moossavi Nejad

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Earthquake Design Practice for Buildings, Second edition.E. Booth. ISBN 978-0-7277-2947-7Designers’ Guide to Eurocode 4: Design of Composite Steel andConcrete Structures, Second edition.R.P. Johnson. ISBN 978-0-7277-4173-8

Finite-element Design of Concrete Structures, Second edition.G.A. Rombach. ISBN 978-0-7277-4189-9Designers’ Guide to EN 1991-1.4 Eurocode 1: Actions on Structures(Wind Actions).N.J. Cook. ISBN 978-0-7277-3152-4Designers’ Guide to Eurocode 8: Design of Structures forEarthquake Resistance.M.N. Fardis. ISBN 978-0-7277-3348-1

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Contents Biographies xi

Preface xiii

01 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Causes and effects of structural vibration 11.1. Introduction 11.2. Vibration of structures: simple harmonic motion 4

1.3. Nature and dynamic effect of man-made andenvironmental forces 6

1.4. Methods of dynamic response analysis 9

1.5. Single-DOF and multi-DOF structures 9Further reading 12

02 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equivalent one degree-of-freedom systems 152.1. Introduction 15

2.2. Modelling structures as 1-DOF systems 152.3. Theoretical modelling by equivalent 1-DOF

mass–spring systems 16

2.4. Equivalent 1-DOF mass–spring systems for linearlyelastic line structures 19

2.5. Equivalent 1-DOF mass–spring systems for linearly

elastic continuous beams 392.6. First natural frequency of sway structures 472.7. Plates 57

2.8. Summary and conclusions 57References 59Further reading 59

03 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Free vibration of one degree-of-freedom systems 613.1. Introduction 613.2. Free un-damped rectilinear vibration 613.3. Free rectilinear vibration with viscous damping 64

3.4. Evaluation of logarithmic decrement of damping fromthe decay function 68

3.5. Free un-damped rotational vibration 70

3.6. Polar moment of inertia of equivalent lumpedmass–spring system of bar element with one free end 72

3.7. Free rotational vibration with viscous damping 76

Further reading 77

04 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Forced harmonic vibration of one degree-of-freedomsystems 794.1. Introduction 79

4.2. Rectilinear response of 1-DOF system with viscousdamping to harmonic excitation 79

4.3. Response at resonance 83

4.4. Forces transmitted to the foundation by unbalancedrotating mass in machines and motors 87

4.5. Response to support motion 92

4.6. Rotational response of 1-DOF systems with viscousdamping to harmonic excitation 98Further reading 101

v

05 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Evaluation of equivalent viscous damping coefficientsby harmonic excitation 1035.1. Introduction 1035.2. Evaluation of damping from amplification of static

response at resonance 103

5.3. Vibration at resonance 1045.4. Evaluation of damping from response functions

obtained by frequency sweeps 106

5.5. Hysteretic damping 1125.6. The effect and behaviour of air and water at

resonance 114Further reading 115

06 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Response of linear and non-linear onedegree-of-freedom systems to random loading:time domain analysis 1176.1. Introduction 1176.2. Step-by-step integration methods 118

6.3. Dynamic response to turbulent wind 1256.4. Dynamic response to earthquakes 1266.5. Dynamic response to impacts caused by falling loads 126

6.6. Response to impulse loading 1336.7. Incremental equations of motion for multi-DOF

systems 133

References 135Further reading 135

07 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Free vibration of multi-degree-of-freedom systems 1377.1. Introduction 1377.2. Eigenvalues and eigenvectors 137

7.3. Determination of free normal mode vibration bysolution of the characteristic equation 138

7.4. Solution of cubic characteristic equations by theNewton approximation method 141

7.5. Solution of cubic characteristic equations by thedirect method 142

7.6. Two eigenvalue and eigenvector theorems 142

7.7. Iterative optimisation of eigenvectors 1467.8. The Rayleigh quotient 1517.9. Condensation of the stiffness matrix in lumped mass

analysis 1517.10. Consistent mass matrices 1547.11. Orthogonality and normalisation of eigenvectors 156

7.12. Structural instability 159References 161Further reading 161

08 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Forced harmonic vibration of multi-degree-of-freedomsystems 1638.1. Introduction 1638.2. Forced vibration of undamped 2-DOF systems 163

8.3. Forced vibration of damped 2-DOF systems 166

vi

8.4. Forced vibration of multi-DOF systems with orthogonaldamping matrices 169

8.5. Tuned mass dampers 173References 175

09 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Damping matrices for multi-degree-of-freedom systems 1779.1. Introduction 1779.2. Incremental equations of motion for multi-DOF

systems 1779.3. Measurement and evaluation of damping in higher

modes 1789.4. Damping matrices 179

9.5. Modelling of structural damping by orthogonaldamping matrices 179Further reading 184

10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . The nature and statistical properties of wind 18510.1. Introduction 18510.2. The nature of wind 18510.3. Mean wind speed and variation of mean velocity with

height 18710.4. Statistical properties of the fluctuating velocity

component of wind 191

10.5. Probability density function and peak factor forfluctuating component of wind 200

10.6. Cumulative distribution function 201

10.7. Pressure coefficients 201Further reading 202

11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dynamic response to turbulent wind:frequency-domain analysis 20311.1. Introduction 20311.2. Aeroelasticity and dynamic response 20311.3. Dynamic response analysis of aeroelastically stable

structures 20411.4. Frequency-domain analysis of 1-DOF systems 20411.5. Relationships between response, drag force and

velocity spectra for 1-DOF systems 20511.6. Extension of the frequency-domain method to

multi-DOF systems 21211.7. Summary of expressions used in the

frequency-domain method for multi-DOF systems 21511.8. Modal force spectra for 2-DOF systems 21611.9. Modal force spectra for 3-DOF systems 217

11.10. Aerodynamic damping of multi-DOF systems 21811.11. Simplified wind response analysis of linear multi-DOF

structures in the frequency domain 225

11.12. Concluding remarks on the frequency-domainmethod 230

11.13. Vortex shedding of bluff bodies 231

11.14. The phenomenon of lock-in 23711.15. Random excitation of tapered cylinders by vortices 240

vii

11.16. Suppression of vortex-induced vibration 24011.17. Dynamic response to the buffeting of wind using

time-integration methods 241References 243

Further reading 243

12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . The nature and properties of earthquakes 24512.1. Introduction 245

12.2. Types and propagation of seismic waves 24512.3. Propagation velocity of seismic waves 24512.4. Recording of earthquakes 24812.5. Magnitude and intensity of earthquakes 248

12.6. Influence of magnitude and surface geology oncharacteristics of earthquakes 249

12.7. Representation of ground motion 252

References 254Further reading 254

13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dynamic response to earthquakes: frequency-domainanalysis 25513.1. Introduction 25513.2. Construction of response spectra 25513.3. Tripartite response spectra 256

13.4. Use of response spectra 25813.5. Response of multi-DOF systems to earthquakes 26013.6. Deterministic response analysis using response spectra 262

13.7. Dynamic response to earthquakes usingtime-domain integration methods 265

13.8. Power spectral density functions for earthquakes 265

13.9. Frequency-domain analysis of single-DOF systemsusing power spectra for translational motion 266

13.10. Influence of the dominant frequency of the groundon the magnitude of structural response 269

13.11. Extension of the frequency-domain method fortranslational motion to multi-DOF structures 270

13.12. Response of 1-DOF structures to rocking motion 274

13.13. Frequency-domain analysis of single-DOF systemsusing power spectra for rocking motion 275

13.14. Assumed power spectral density function for rocking

motion used in examples 27613.15. Extension of the frequency-domain method for

rocking motion to multi-DOF structures 27913.16. Torsional response to seismic motion 282

13.17. Reduction of dynamic response 28613.18. Soil–structure interaction 288

References 291

Further reading 291

14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generation of wind and earthquake histories 29314.1. Introduction 293

14.2. Generation of single wind histories by a Fourierseries 293

viii

14.3. Generation of wind histories by the autoregressivemethod 294

14.4. Generation of spatially correlated wind histories 29714.5. Generation of earthquake histories 299

14.6. Cross-correlation of earthquake histories 30314.7. Design earthquakes 303

References 305

Index 307

ix

Biographies H.A. Buchholdt

Hans Anton Buchholdt is Norwegian and a Professor Emeritus

at the University of Westminster, where he was appointed a

principal lecturer and professor at the Department of Civil

Engineering. During his time there, he studied and developed

methods for generating numerical models of correlated wind

and earthquake histories, and methods for calculating the non-

linear static and dynamic response of pre-tensioned cable roofs

and guyed masts in the time domain. Professor Buchholdt has

published a number of papers on the analysis, testing and

construction of cable roofs and guyed masts, acted as a

consultant to industry and is the author of the book:

Introduction to Cable Roof Structures. He has also been

engaged in cooperative research projects with institutions

abroad, notably the University of Florence and the Norwegian

Building Research Institute.

S.E. Moossavi Nejad

Shodja Edin Moossavi Nejad is an Emeritus Fellow at the

University of Westminster and a chartered structural engineer,

practising structural analysis and testing. He is Director of

Aronson Associates, specialist consultant engineers for

dynamic analysis and testing of structures. During his time at

the University of Westminster, he was employed as course

leader of Civil Engineering courses as well as postgraduate

courses. He also conducted many research projects resulting in

publication of a number of papers on the topic of structural

dynamics and testing, some in association with University of

Patras, Greece.

He developed a dynamic testing and analysis system

(ARONSYS) which is used in house for dynamic analyses of

flexible structures and testing of structures and their

components. His working experience includes dynamic testing

of a number of stadium stands and bridges, including the old

Arsenal Stadium, Aston Villa and many others where damper

systems were designed and installed to overcome undesired

vibrations. He is currently a consultant for dynamic testing of

platforms for equestrian activities for the London 2012

Olympic Games.

xi

Preface This book is intended as an introduction to the dynamics of

civil engineering structures. It has evolved from lectures given

to industrially based MSc students in order to improve their

understanding and implementation of modern design codes,

which increasingly require a greater knowledge and

understanding of vibration caused by either man or the

environment. It is also intended to give practising engineers a

better understanding of the dynamic theories that form the

basis of computer analyses systems.

Experience has shown that it is all too easy to make mistakes

in the input data and still accept the results obtained. It is

hoped that the methods presented will aid the practising

engineer to judge the validity of the dynamic response

calculations obtained using the computer programmes.

Throughout the text, worked examples are provided in order to

illustrate and demonstrate the use of theories presented; it is

hoped that these will prove useful to the reader who will have

the ability to downsize the problems and solve them manually

to obtain general results for comparison with detailed results

from computer programmes.

In this edition, the importance of dynamic testing and its use in

betterment of numerical models are emphasised, as well as the

use of dampers to reduce the amplitudes of vibration (in

particular the use of mass dampers).

Additional information on the movement of surrounding air

and water which vibrates with the structure is also provided.

In order to follow the theoretical work presented, the reader

will need to have some knowledge of differential and integral

calculus, first- and second-order differential equations,

determinants, matrices and matrix formulation of structural

problems. These topics are included in the teaching of

mathematics and the theory of structures in undergraduate

engineering courses. Knowledge of the concept of eigenvalues

and eigenvectors will also be useful, but is not essential.

Chapter 1 provides a number of reasons why the modern

structural engineer needs to have knowledge of vibration.

Many civil engineering structures vibrate predominantly in the

first mode with a simple harmonic motion, and may therefore

be reduced to mass–spring systems with only 1 degree of

freedom (DOF). Also covered in this chapter is the concept

that wind and earthquake histories may be considered to

consist of a summation of harmonic components, and that

xiii

most structures which possess a dominant frequency that falls

within the frequency band of either history will tend to vibrate

at resonance.

Chapter 2 shows how to make an initial estimate of the

dominant first natural frequencies of loaded beam elements,

continuous beams and multi-storey structures by equating the

maximum kinetic energy to the maximum strain energy at

resonance.

The theories of free damped linear and torsional vibration of

1-DOF systems are presented in Chapter 3.

Chapter 4 provides closed-form solutions to the response of

damped 1-DOF systems subjected to rectilinear and torsional

harmonic excitation (caused by the rotation of unbalanced

motors) and to harmonic support excitation.

The evaluation of structural damping is considered in Chapter 5.

Measurements of damping by the two classical methods

described in all dynamic text books – namely the measurement

of logarithmic damping from records of decaying vibrations and

the measurement of damping ratios from amplitude–frequency

curves (the so-called bandwidth method ) – usually leads to

inaccurate results. The two main reasons are that the level of

damping varies with the amplitude of vibration and structural

damping is at best only approximately viscous. The latter

method is also difficult to implement as it is usually difficult to

obtain a set of satisfactory values near the peak of the curve on

either side of resonance. The authors have therefore included a

few methods, not found in most other text books, by which the

accuracy of these methods can be studied and improved upon.

Chapter 6 is devoted to the formulation of step-by-step

methods for calculating time histories of response of 1- and

multi-DOF systems when subjected to impulse loading and

time histories for wind, earthquakes and explosions.

Matrix formulation of the equations of motion of free

vibration and the calculation of natural frequencies and mode

shapes for multi-DOF systems are presented in Chapter 7,

together with methods for reducing the number of degrees of

freedom of structures when this is required. Methods for

determining the natural frequencies and mode shapes of

simplified structures are included.

Chapter 8 presents the classical method of mode superposition

in which the dynamic response of an N-DOF structure is

xiv

sought by transforming the global equations of motion into the

equations of motion for N 1-DOF systems. This

transformation is made possible by the orthogonal properties

of the eigenvectors or mode-shape vectors for the structure and

assumptions made with respect to the properties of the

damping matrix.

The construction of damping matrices is considered in

Chapter 9. It has been pointed out that, in practice, such

matrices need to be assembled only in the case of dynamic

response analysis of non-linear systems such as cable and

cable-stayed structures. These structures may respond in a

number of closely spaced modes for which the method of mode

superposition presented in Chapter 8 is not appropriate. In this

regard, it should be mentioned that the use of inadequate

damping matrices in the case of a cable-stayed bridge model

resulted in calculated amplitudes of strains in the stays that

differed considerably from those measured.

Chapters 10 and 11 deal with the nature and statistical

properties of wind and the response to buffeting and vortex

shedding. Chapter 11, dealing with dynamic response, is mainly

concerned with frequency-domain analysis using power spectra

and the method of mode superposition developed in Chapter 8.

A considerable amount of space is devoted to the use and

importance of cross-spectral density functions, which are not

normally found in detail elsewhere.

Chapters 12 and 13 deal with the nature of, and dynamic

response to, earthquakes. The emphasis is once more on

frequency-domain methods using the method of mode

superposition, response and power spectra. Examples of both

rectilinear and torsional response analyses are given. The new

Eurocodes require that rocking motion caused by earthquakes

should be taken into account in future designs. For this reason,

the authors have constructed a power spectrum for rocking in

order to demonstrate its use in dynamic analysis. The authors

wish to emphasise that such spectra are introduced only to

demonstrate their use when they become agreed and available,

as they do not appear to be covered in the current literature.

The spectrum in this text should not be used for design

purposes.

Chapter 14 presents methods for generating spatially correlated

wind histories and families of correlated earthquake histories;

such histories need to be available in order to use the step-by-

step methods given in Chapters 6, 11 and 13. Earthquake

histories may be generated either with the statistical properties

xv

of recorded earthquakes or the dominant ground frequency of

the site. References to research behind the development of

these methods are provided.

This book is not intended to be an advanced course in

theoretical structural dynamics: for this the authors

recommend Dynamics of Structures by R.W. Clough and

J. Penzien. Some topics have however been developed further

than in most text books, namely the evaluation of damping

values and the use of spectral and cross-spectral density

functions (or power spectra) to predict response to wind and

earthquakes and to generate correlated wind and earthquake

histories required for the analysis of non-linear structures.

As this book is intended for the practising engineer, certain

older techniques (such as the Duhamels integral used in time-

domain analysis) have been omitted. In the authors’

experience, the Newark �-equations or the Wilson �-equations

are easier to understand and equally effective. Also, no

reference has been made to computer methods for solving large

eigenvalue problems; for these the reader should consult

mathematical text books.

This book has two major omissions. The first is that no

reference is made to wave loading such as experienced by dams

and offshore structures. For this the reader is referred to other

publications and, in particular, the original work by C.A.

Brebbia in Dynamic Analysis of Offshore Structures which gives

a very good introduction to the subject. Another omission (in

this case a partial one) is the subject of soil–structure

interaction, which is important as it modifies the dynamic

behaviour of structures. The interaction between the structure

and the ground can be taken into account either by

representing the stiffness and damping properties of soil as

equivalent springs and dampers, respectively, or by modelling

the soil by finite elements. The former is at best an

approximate method, which requires some experience to use.

For the latter a great deal of experience is necessary as this is a

highly specialised field; it is therefore considered to be outwith

the scope of this book. For this reason, only the concept of

numerical modelling of the soil by springs and dampers is

presented. For more detailed information, the reader is referred

to Earthquake Design Practice for Buildings by D.E. Key.

There are a number of other topics which it has not been

possible to include, as the main purpose of this work is to give

the reader an introduction to the vibration of structures and (it

is hoped) to make other more advanced or specialised texts

xvi

easier to follow. Most of the omitted topics can be found in a

handbook on vibration entitled Shock Vibration by

C.M. Harris. Methods for dynamic response analysis of cable

and cable-stayed structures are given in Introduction to Cable

Roof Structures by H.A. Buchholdt.

The authors have learnt a great deal while writing this book

and hope that others will also benefit from this work. We will

be pleased to receive comments and suggestions for a possible

revised edition, and to have our attention drawn to any errors

that must inevitably exist and for which we alone are

responsible.

xvii

Structural Dynamics for Engineers, 2nd edition

ISBN: 978-0-7277-4176-9

ICE Publishing: All rights reserved

http://dx.doi.org/10.1680/sde.41769.001

Chapter 1

Causes and effects of structural vibration

1.1. IntroductionAn understanding of structural vibration and the ability to undertake dynamic analysis are

becoming increasingly important. The reasons for this are obvious. Advances in material and

computational technology have made it possible to design and construct taller masts, buildings

with ever more slender frames and skins that contribute little to the overall stiffness, and roofs

and bridges with increasingly larger spans. In addition, masts, towers and new forms of con-

struction such as offshore structures are being built in more hostile environments than previously.

This, together with increasing vehicle weights and traffic volumes, requires that designers take

vibration of structures into account at the design stage to a much greater extent than they have

done in the past.

Sometimes the trouble caused by vibration is merely the nuisance resulting from sound

transmission or the feeling of insecurity arising from the swaying of tall buildings and light

structures such as certain types of footbridge. Occasionally, however, vibration can lead to

dynamic instability, fatigue cracking or incremental plastic deformations. The first two types

of problem may lead to a reduction in utilisation of a structure. The latter may lead to

costly repairs if discovered in time or, if not, to complete failure with possible loss of human

life.

Unfortunately, there are numerous examples of structural vibrational failures, many of which

have resulted in the loss of life. The disastrous effects of the numerous large earthquakes that

occurred during the twentieth century are obvious examples, but wind and waves have also

taken their toll. Examples are the collapse of the Tacoma Narrows Bridge in the USA, the

Alexander Kjelland platform in the North Sea, the cooling towers at Ferry Bridge in the UK

and numerous large cable-stayed masts. Many failures of such masts have occurred in Arctic

environments, but a number have occurred in countries with more clement climates such as

Britain and Germany. Frequently, the cause of failure has been the development of fatigue

cracks in the attachments of the guys to the tower, caused by the vibration of the guys.

Large-magnitude earthquakes have devastating effects on buildings, as shown by Japan’s 2011

earthquake of magnitude 8.9 on the Richter scale. This was followed by a tsunami, which is

more challenging to design against. If the duration of the earthquake is long, then it produces

rapid fatigue in joints as a result of cyclical movements.

Until quite recently it was assumed that the response of guyed masts to earthquakes need not be

considered. Recent research has shown, however, that an earthquake can be as severe as any

storm if the dominant frequency of the ground coincides with one of the main natural frequencies

of the mast; this can cause local buckling of structural elements, which again can lead to a

complete collapse.

1

Long-term vibration induced by traffic can lead to fatigue in structural elements and should not be

underestimated. A number of old railway bridges have started to develop fatigue cracks in the

gusset plates, which have had to be replaced. Costly repairs and modifications have also had to

be undertaken on relatively new suspension and cable-stayed bridges, because the possibility of

fatigue caused by traffic-induced vibration had not been sufficiently investigated at the design

stages. The suspension bridge across the River Severn near Bristol in the UK is one example

among many.

The effect of traffic is not confined to bridges – it must also be taken into account when the

foundations of buildings situated next to railway lines or roads carrying heavy traffic are being

designed. It is interesting that many ancient buildings such as cathedrals, which have been built

next to main roads, tend to lean towards the road and, in many cases, also show signs of

cracks as a result of centuries of minute amplitude vibrations caused by carts passing on the

cobbled road surfaces.

In factories, rotating machinery can lead to large-amplitude vibrations that can cause fatigue

problems if not considered early enough. Such problems apparently occur more frequently

than has been generally appreciated in the past, and some countries such as Sweden have

produced design guides in an effort to overcome them. Other causes of vibration are currents

in air and water, explosions, impact loading and the rupture of members in tension. Currents

can give rise to vibration as a result of vortex shedding. Explosions such as those used in

demolitions will transmit pressure waves both through the ground and through the air and

can, if insufficient precautions are taken, cause damage to nearby buildings and sensitive

electronic instruments. The dynamic shocks set up by the ruptures of highly tensioned members

such as steel cables in tension can be devastating. A number of mast failures, where one of the

guys or attachments has ruptured because of the development of fatigue cracks, can be

attributed to the magnitude of the bending moments caused by the resulting dynamic shock;

these moments will have been several times greater than those the towers would have experi-

enced if the guys had been removed statically. There are also examples of kilometres of electric

transmission lines with towers collapsing because of the rupture of a single cable, and a

number of hangers in a suspension bridge have snapped because a single hanger was broken

when hit by a lorry.

The sudden release of forces restraining the movements of an element may also lead to struc-

tural failures. After a very heavy snowfall, the steel box space ring containing the pre-tensioned

cable net roof over the Palasport in Milan buckled. The space ring was supported on roller

bearings on the top of inclined columns. A number of explanations for the failure were

suggested; the most likely is that the rollers, which were completely locked at the time the roof

was subjected to resonance testing, suddenly moved under the exceptionally heavy snow load.

The resulting dynamic shock induced bending moments much larger than those for which the

ring had been designed.

From the above it ought to be evident that it is important for engineers not only to develop an

understanding of structural vibration, but also to be able to investigate the effects of dynamic

response at the design stage when a structure can be readily modified (rather than having to

make possible costly alterations later on). This can be achieved by an ‘evaluation of dynamic

behaviour’ procedure. This procedure is complementary to the static design and encompasses

frequency and mode shape analysis in addition to the simple assessments given below.


2

Not every designer needs to be an expert in dynamic analysis, but all ought to have an under-

standing of the ways in which structures are likely to respond to different types of dynamic excita-

tion and of the fact that some types of structure are more dynamically sensitive than others. In

particular, designers should know that all structures possess not one but a number of natural

frequencies, each of which is associated with a particular mode shape of vibration. They should

be aware of the fact that pulsating forces or pulsating force components with the same frequencies

as the structural frequencies will cause the structure to vibrate with amplitudes much greater than

those caused by pulsating forces with frequencies different to the structural frequencies. The

designer therefore needs to be able to calculate the natural frequencies of a structure, identify

and formulate the characteristics of different types of man-made and environmental forces,

and to calculate the total response to these forces in the modes in which the structure will vibrate.

An understanding of the importance of damping and the principles and methods to control and

reduce the amplitudes of vibrations is also required.

The following are some types of structure and structural element that, experience has shown, can

be dynamically sensitive

g tall buildings and tall chimneysg suspension and cable-stayed bridgesg steel-framed railway bridgesg free-standing towers and guyed mastsg cable net roofs and membrane structuresg cable-stayed cantilever roofsg cooling towersg floors with large spans and floors supporting machinesg foundations subjected to vibrationg structures during erection and structural renovationg offshore structuresg electrical transmission lines.

As a general rule, one can use a simple method based on Edin’s Box. Consider a box of size a� b� c

as shown in Figure 1.1. If the building to be investigated is put inside this box and the ratios of a to b

and c remain near 1, then the building is not likely to be dynamically sensitive provided local stiff-

ness deficiency is avoided. However, if one of the sides is considerably larger than the other two, e.g.

a multi-storey building, then the building is dynamic sensitive. Ratios of 10 and above introduce

dynamic sensitivity; in terms of a large-span bridge, a value of b can be as much as 50 times the

value of c, indicating that the building is inherently dynamic sensitive.

Figure 1.1 Simple dynamic sensitivity test

b

ac


3

This list is not intended to be exhaustive, but it indicates the range and variety of civil engineering

structures whose dynamic responses need to be considered before they are constructed. Of the

above, only the membrane and cable and cable-stayed structures are likely to respond in a rela-

tively large number of modes because their dominant frequencies are closely spaced within

their respective frequency spectra.

The relationship between the dominant frequency of a structure and its degree of static struc-

tural stability also deserves attention. Both are functions of stiffness and mass. The criterion

for instability is that the stiffness during any time of a load history becomes zero. If this

happens the dominant frequency will also be zero, and the mode of collapse will be similar to

that of the mode shape of vibration. Frequency analysis is therefore a useful tool for investi-

gating the stability of a structure and the amount of load a structure can support before it

becomes unstable.

Finally, engineers concerned with the design, operation and maintenance of nuclear installations

need to have a thorough understanding of the effects of vibrations caused by any possible source

of excitation, because of the very serious consequences of any failure.

1.2. Vibration of structures: simple harmonic motionThe motion of any point of a structure when vibrating in one of its natural modes closely resem-

bles simple harmonic motion (SHM). An example of simple harmonic motion is the type of

motion obtained when projecting the movement of a point on a flywheel, rotating with a constant

angular velocity, onto a vertical or horizontal axis. The motion of any point of a structure

vibrating in one of its natural modes can therefore be described by

x tð Þ ¼ x0 sin !ntð Þ ð1:1Þ

_xx tð Þ ¼ x0!n cos !ntð Þ ð1:2Þ

€xx tð Þ ¼ �x0!2n sin !ntð Þ ð1:3Þ

where x(t) is the amplitude of motion at time t, _xx tð Þ is the velocity of the motion at time t, €xx tð Þ isthe acceleration of motion at time t, x0 is the maximum amplitude of response and !n is the

natural angular frequency of the structure in rad/s.

Equation 1.1 also represents the motion of a lumped mass suspended by a linear elastic spring

when the mass is displaced from its position of equilibrium and then released to vibrate. It is

therefore possible to model the vibration of a structure in a given mode as an equivalent mass–

spring system where the lumped mass and spring stiffnesses are associated with a given mode

shape. From Newton’s law of motion:

M€xx ¼ �Kx: ð1:4Þ

Substitution of the expressions for x(t) and €xx tð Þ given by Equations 1.1 and 1.2 yields:

M!2n ¼ K : ð1:5Þ

Hence

!n ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðK=MÞ

pð1:6Þ


4

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðK=MÞ

pð1:7Þ

where f is the frequency in cycles per second (Hz).

It should be noted that in Equation 1.7 the stiffness must be in N/m and the mass must be in kg. If

the weight of the vibrating mass is used, since M¼W/g the unit of weight must be N and that of

the gravitational acceleration g must be m/s2. When the weight rather than the mass is used,

Equation 1.7 is expressed:

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiKg=W

p: ð1:8Þ

Single mass–spring systems are referred to as one degree-of-freedom systems, or 1-DOF systems.

They are particularly useful as initial numerical models when first trying to ascertain the possible

dynamic response of a structure, as most civil engineering structures mainly respond in the first

mode. From Equation 1.7 it follows that, to calculate the dominant natural frequency of a

structure, we need only calculate or obtain the equivalent spring stiffness and the magnitude of

the corresponding vibrating mass. The stiffness may be obtained by elastic calculations using

equations derived in linear elastic theory, from a computer program that calculates the deflection

for a specified force or from static testing of a model or a real structure.

Alternatively Equation 1.7 permits the calculation of the equivalent lumped vibrating masses of

structures if the stiffnesses and the frequencies of the structures are known. The frequencies in

such cases must be found by dynamic testing or by a standard eigenvalue computer program.

The determination of a first natural frequency and an equivalent vibrating mass is demonstrated

in the following two examples.

Example 1.1

Determine the frequency of a bridge with a 10 t lorry stationed at mid-span. The bridge itself

may be considered as a simply supported beam of uniform section having a total weight of

200 t. From a static analysis of the bridge, it was found that the deflection at mid-span due

to a force of 1.0 kN applied at mid-span is 1.5 mm.

The stiffness of the bridge at mid-span is given by

K ¼ 1000 N=0:0015 m ¼ 6:67� 105 N=m:

The mass to be included is the sum of the mass of the lorry and the equivalent vibrating mass

of the bridge. In Chapter 2, it is shown that for simply supported beams of uniform section

the equivalent mass is approximately equal to half the total mass. Hence the equivalent

lumped mass is

M ¼ 10 000 kgþ 0:5� 200 000 kg ¼ 110 000 kg

Finally, substitution of the values for M and K into Equation 1.7 yields

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi6:67� 105=110 000ð Þ

q¼ 0:392Hz


5

1.3. Nature and dynamic effect of man-made and environmentalforces

As mentioned above, the significance of the natural frequencies is that if a structure is excited by a

pulsating force with the same frequency as one of the structural frequencies, it will begin to vibrate

with increasing amplitudes. These can be many times greater and therefore more destructive than

the deflection that would have been caused by a static force of the same magnitude as the

maximum pulsating force. When this is the case, the structure is said to be vibrating in resonance.

Thus, a dynamic force or force component

P tð Þ ¼ P0 sin !tð Þ ð1:9Þ

will give rise to large amplitude vibration if !¼!n.

Example 1.2

It has been decided to undertake a preliminary study of the dynamic response characteristics

of the cable-stayed cantilever roof shown in Figure 1.2 by calculating the response of a

1-DOF system representing the vibration of the free end of the roof. Static testing of the

roof has shown that it will deflect 1.0 mm when a force of 1.0 kN is applied at the free end,

and dynamic testing that the frequency of vibration is 1.6 Hz. Calculate the stiffness and

the mass of the equivalent 1-DOF mass–spring system needed for the initial dynamic

investigation.

Figure 1.2 Cable-stayed cantilever roof

The stiffness of the cantilever roof is

K ¼ 1000 N=0:002 m ¼ 5:0� 105 N=m:

Hence, the equivalent lumped mass of the structure is given by

M ¼ K= 2�fð Þ2 ¼ 5 � 0� 105= 2�� 1 � 6ð Þ2 ¼ 4947:32 kg


6

A commonly encountered form of dynamic excitation is that caused by unbalanced rotating

machines and motors. This form of dynamic force can generally be expressed by

P tð Þ ¼ me!2i sin !itð Þ ð1:10Þ

where m is the total unbalanced mass, e is the eccentricity of the mass m, and !i is the speed of the

motor. Even small values of the product me can lead to problems if !¼!n, unless designed

against.

Figure 1.3 shows the recorded histories of wind velocities at different heights along a mast;

Figure 1.4 shows the recorded history of ground acceleration due to an earthquake. Such samples

have been subjected to Fourier analysis, which has shown that both forms of motion can be

Figure 1.3 Records of wind speeds at three levels of a 153m tall guyed mast

12.2 m64.0 m

153.3 m

0 1 2 3 4 5 6 7 8Time: min

Win

d sp

eed:

m/s

30

20

10

0

Figure 1.4 Accelerogram of the NS component of the El Centro earthquake, 18 May 1940

0 5 10 15 20 25Time: s

Acc

eler

atio

n/ac

cele

ratio

n of

gra

vity

0.3

0.2

0.1

0

–0.1

–0.2

–0.3


7

considered to consist of the summation of a large number of sinusoidal waves with varying

frequencies and amplitudes. The velocity of wind at any time may therefore be written as

V tð Þ ¼ ~VV þXNi¼ 1

�i sin !itþ �ið Þ ð1:11Þ

where ~VV is the mean wind speed, �i is the amplitude of fluctuations, !i is the angular frequency

in rad/s, �i is the random phase angle in rad and the subscript i indicates the ith harmonic

component.

The corresponding drag force acting on a structure may be written as

Fd tð Þ ¼ ~FFd þXNi¼ 1

Fdi sin !itþ �ið Þ ð1:12Þ

The fluctuating drag force given by Equation 1.12 can give rise to quite significant amplitudes of

vibration if the frequency of only one of its components is equal to a dominant structural

frequency. The swaying motion of some very tall slender buildings with low first natural frequen-

cies is a direct result of the fact that their dominant frequencies coincide with the frequency

components of wind in the part of the wind frequency spectrum where wind possesses a consider-

able amount of energy. The same phenomenon may occasionally be observed in nature during

periods of strong gusty winds when, for the same reason, a single tree may suddenly vibrate

violently while other trees merely bend in the along-wind direction.

Similarly, the acceleration of the strong motion of an earthquake may be expressed as

€xxg tð Þ ¼XNi¼ 1

€xxi sin �ið Þ ð1:13Þ

where €xxI is the amplitude of acceleration, !i is the angular frequency in rad/s, and �i is the random

phase in rad. The corresponding exciting force acting on a 1-DOF structure of mass is

M€xxg tð Þ ¼XNi¼ 1

M€xxi sin !itþ ’ið Þ: ð1:14Þ

During earthquakes it has been observed that buildings with a first natural frequency equal or

close to the dominant frequency of the ground may vibrate quite violently, while others with

frequencies different from the dominant ground frequency vibrate less. This again underlines

the fact that the larger amplitude vibrations will occur when the dominant frequencies of

structures coincide with one of the frequency components in a random exciting force.

Wind and earthquakes as well as waves (whose effect is not considered in this book) can lead to

vibrations with large amplitudes if one or more of the natural frequencies of a structure are equal

to some of the angular frequencies !i in Equations 1.10, 1.11 or 1.12. Large-amplitude vibration

can be very destructive and, even if the amplitudes are not large, continued vibration may lead to

fatigue failures. The possibility of vibration must therefore be taken into account at the design

stage.


8

1.4. Methods of dynamic response analysisThere are basically two approaches for predicting the dynamic response of structures: time-

domain methods and frequency-domain methods. The first method is used to construct time

histories of such variables as forces, moments and displacements by calculating the response at

the end of a succession of very small time steps. The second method is used to predict the

maximum value of the same quantities by adding the response in each mode in which the structure

vibrates. Time-domain methods can be used to calculate the dynamic response of both linear and

non-linear structures and require that time histories for the dynamic forces be available or can be

generated. Frequency-domain analysis is limited to linear structures, as the natural frequencies of

non-linear structures vary with the amplitude of response. The method has won considerable

popularity in spite of its limitations as it permits the use of power and response spectra, which

to date have been more easily available than time histories. Power spectra for wind are introduced

in Chapter 10 and response and power spectra for earthquakes in Chapter 13. Methods for

generating correlated wind and earthquake histories are presented in Chapter 14.

1.5. Single-DOF and multi-DOF structuresIn general, even the simplest of structures such as simply supported beams and cantilevers are in

reality multi-DOF systems with an infinite number of DOFs. For practical purposes, however,

many simple structures and structural elements may initially (as mentioned above) be analysed

as 1-DOF systems by considering them as simple mass–spring systems with an equivalent

lumped mass and an equivalent elastic spring. Some examples of this form of simplification are

illustrated in Figure 1.5.

When a structure is reduced to a 1-DOF system, it is possible only to calculate the response in one

mode (usually the dominant mode). In order to study the vibration in several modes, a structure

has to be modelled as a multi-DOF mass–spring system. An example is shown in Figure 1.6,

where a three-storey portal frame structure in which the floors are assumed to be rigid is modelled

as a 3-DOF mass–spring system. Figure 1.7 shows how a pin-jointed frame may be modelled as a

Figure 1.5 Equivalent 1-DOF mass–spring systems


9

multi-DOFmass–spring system by lumping the mass of the members at the nodes and considering

the stiffness of the members as weightless springs.

The dynamic response of a large number of structures can, at least initially, be determined

by modelling them as 1-DOF systems. In Chapter 8, it is shown how the dynamic response of

N-DOF structures can be determined by

g transforming them into N 1-DOF mass–spring systems, each with a natural frequency

equal to one of those of the original structure

Figure 1.6 Three-storey portal frame modelled as a 3-DOF mass–spring system

Figure 1.7 Modelling of a pin-jointed frame as a multi-DOF mass–spring system


10

g calculating the response of each of the 1-DOF systemsg transforming the responses of these 1-DOF systems to yield the global response of the

original N-DOF structure.

Thus, not only 1-DOF systems but alsoN-DOF systems require that engineers be fully conversant

with the dynamic response analysis of single-DOF mass–spring systems. Before proceeding with

the response analysis of 1-DOF systems, it is useful to develop some expressions for the lumped

masses and elastic spring stiffnesses for equivalent 1-DOF systems of some simple beam elements,

and to introduce an approximate method for estimating the first natural frequencies of continuous

beams and multi-storey framed structures. This is done in Chapter 2.

1.5.1 Importance of dynamic testingThe writers believe it is important for practising engineers to develop a feeling for how structures

behave dynamically. This can only be achieved by studying the vibration of both models and

real structures. In the case of the latter, recordings of vibration caused by vibrators but also by

environmental forces should be studied. Another helpful way to obtain a feeling for how

structures behave dynamically is by vibrating numerical models on the computer; with today’s

high-speed computers, this be done in the time domain.

To attempt laboratory tests on models of real structures is difficult unless they are made very

large, because the scaling down of the mass of a structure leads to models with large concentrated

masses. This usually leads to difficulties when attempting to scale down the stiffness.

Frequency testing of structural models and structural elements using different types of vibrators,

shaking tables and recording equipment is important, as is the demonstration of different types of

damping. Tests should also include measurements of the behaviour of the surrounding air and/or

water at resonance.

A simple way to introduce the subject of vibration of structures is by demonstrating the vibration

of a small cantilever, the frequency of which can be induced by a load release and varied by the

addition of different concentrated masses at its tip. The effect of additional damping can be shown

by the fixation of an adhesive tape or other simple means.

By vibrating a spring–mass system with the same spring stiffness as the stiffness at the end of

the cantilever and with a lumped mass adjusted to give the same frequency as the cantilever,

the vibration of the cantilever can be modelled as a spring–mass system.

The response to a sinusoidal frequency sweep through resonance can be shown by means of a

small electric motor with an eccentric rotating mass. The buffeting forces of wind, earthquakes

and waves can be considered as the sum of sinusoidal forces of varying frequencies and

magnitudes (see Equations 1.12 and 1.14). If one of the frequency components has a frequency

equal to the natural frequency of the structure it will give rise to large-amplitude vibration, the

magnitude of which may need to be reduced by dampers.

The response of structures with different first frequencies to ground motion can be shown by

attaching three cantilever columns with the same cross-sections but with increasing heights to a

block of wood, and moving the block forward and backwards with increasing frequency. First,

the tallest column will vibrate while the two smaller columns will remain still. The medium


11

column will then vibrate while the tallest and shortest will stay still. Finally, the shortest column

will vibrate with the two taller columns standing still.

Finally, the use of smoke tunnels to demonstrate vortex shedding and turbulence is useful.

1.5.2 The EurocodeThroughout this book a number of solutions and calculation methods are used which are based on

various codes of practice for structural design and analyses. In particular, the requirements for

frequencies and damping which affect the dynamic behaviour of structures are used. At present

time, the main codes of practice are the British Standard publications and the Eurocode. Since

the UK is part of the European Community, British engineers have to follow the Eurocodes.

Since these include the actions of traffic, machinery, wind and earthquakes, it follows that UK

students need some knowledge of the dynamics of structures based on Eurocodes. However,

it does not follow that this book makes particular reference to the Eurocode as a topic, since

problems of vibration are global and are not based on local variations.

Particular sections of the Eurocodes related to dynamics of structures can be found in

g Eurocode 1, Part 1.4 Actions on structures, general actions, wind actionsg Eurocode 1, Part 2 Actions on structures, traffic loads on bridgesg Eurocode 8, Design of structures for earthquake resistance, Part 1 General rules, seismic

actions and rules for buildings.

FURTHER READING

Bolt BA (1978) Earthquakes: A Primer. W.H. Freeman, San Francisco.

Brebbia CA and Walker C (1979) Dynamic Analysis of Offshore Structures. Newnes–

Butterworth, London.

Buchholdt HA (1985) Introduction to Cable Roof Structures. Cambridge University Press,

Cambridge.

Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.

Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.

Key DE (1988) Earthquake Design Practice for Buildings. Thomas Telford, London.

Problem 1.1

A tapering tubular 20 m tall antenna mast supports a disc weighing 10 kN at the top. Analysis

of accelerometer reading shows that the dominant frequency of the mast is 2.3 Hz. A rope

attached to the top of the mast deflects the point of attachment 5 mm horizontally when

the horizontal component of the tension in the rope is 20 kN. Calculate the equivalent elastic

spring stiffness and lumped mass of a mass–spring system which is to be used for studying the

response at the top of the mast to wind.

Problem 1.2

A continuous steel box girder bridge is designed with a central span of 50 m and two outer

spans each of 25 m. The expressions for the mass and spring stiffness of a dynamically equiva-

lent mass–spring system are 0.89 wL/g and 13.7 EI/L3, respectively. Calculate the dominant

frequency of the bridge if L¼ 25 m, w¼ 120 kN/m, E¼ 210 kN/mm2 and I¼ 0.225 m4.


12

Krishna P (1978) Cable Suspended Roofs. McGraw-Hill, New York.

Lawson TW (1990) Wind Effects on Buildings, vols 1 and 2. Applied Science, Barking.

Simue E and Scalan RH (1978) Wind Effects on Structures. Wiley, Chichester.

Warburton GB (1992) Reduction of Vibrations. Wiley, Chichester.

Wolf JH (1985) Dynamic Soil–Structure Interaction. Prentice-Hall, Englewood Cliffs.


13


ISBN: 978-0-7277-4176-9



Chapter 2

Equivalent one degree-of-freedom systems

2.1. IntroductionThe challenges associated with dynamics of structures can be well presented and understood

by modelling a 1-DOF system and showing the parameters associated with the model and their

solutions. This chapter deals with the behaviour of linearly elastic line structures as a 1-DOF

system.

2.2. Modelling structures as 1-DOF systemsThe natural frequency and approximate response of line-like structures such as tall slim buildings,

masts, chimneys, bridges and towers may, as mentioned in Chapter 1, be estimated by assuming

that they mainly respond in the first mode, and by modelling them as single mass–spring systems.

This is made relatively easy in many cases by the fact that the first mode of vibration of these types

of structure has a mode shape very similar to the deflected form caused by the appropriate concen-

trated and/or distributed load. The modelling of such structures requires the evaluation of the

equivalent or generalised mass M, spring stiffness K, damping coefficient C and forcing function

P(t), such that the frequency of the model is the same as that for the structure itself and the

response of the mass is equal in magnitude to the movement of the point of the structure that

is being simulated.

Newton’s law of motion states that force¼mass� acceleration. Thus, if the mass and stiffness are

denoted by M and K, respectively, and the amplitude and acceleration at time t are x and €xx,

respectively, then since force F¼ kx it follows that

Kx ¼ �M€xx: ð2:1Þ

Since the vibration of structures may be assumed to closely resemble that of SHM, the displace-

ment x and acceleration €xx may be written as

x ¼ X sin !tð Þ

€xx ¼ �X!2 sin !tð Þ:

Substitution for x and €xx into Equation 2.1 yields

KX ¼ MX!2 ð2:2Þ

which in turn yields

! ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK=Mð Þ

pð2:3Þ

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK=Mð Þ

p: ð2:4Þ

15

Multiplication of both sides of Equation 2.2 by X/2 yields

12KX

2 ¼ 12MX2!2 ð2:5Þ

which means that the maximum strain energy KX2/2 is equal to the maximum kinetic energy

MX2!2/2. It follows that the spring stiffness and lumped mass of an equivalent mass–spring

system may be found by determining the spring stiffness such that the energy stored in the

spring will be the same as that stored in the structure when both are deflected an amount X. In

addition, the lumped mass will have the same kinetic energy as the structure when both experience

a maximum velocity of X!, given that the motion of the lumped mass–spring system represents

the motion of the structure at the position where the maximum amplitude of vibration is X. A

method based on this approach is presented in the following section.

2.3. Theoretical modelling by equivalent 1-DOF mass–spring systemsIn order to evaluate the expressions for the equivalent lumped mass, spring stiffness, damping

coefficient and generalised dynamic force, consider the cantilever column shown in Figure 2.1

where the flexural rigidity, mass damping coefficient, dynamic force, motion at a distance x

from the base and motion at the top of the column are given by EI(x), m(x), c(x), p(x), y(x, t)

and Y(t), respectively. The height of the un-deformed column is L and the height of the deformed

columnL*.Q is a constant axial force and �(x) a shape function that defines the shape of the mode

Figure 2.1 (a) Cantilever column with flexural rigidity EI(x), mass m(x) and damping coefficient c(x)

subjected to a dynamic load p(x, t) at a distance x above the base; (b) equivalent mass–spring system

with stiffness K, mass M, damping coefficient C and dynamic load P(t)

(a) (b)

L′ L

δx

δy

δL

P(t)

Y(t)

y(t)

p(L, t)

M

C

K

x

Q

m1

m2

m3


16

of vibration and is unity at the point of the structure at which motion is to be modelled by

the mass–spring system. In case of the tower shown in Figure 2.1 the shape function is assumed

to be unity when x¼L, in which case the model will simulate the movement at the top of the

column.

The relationship between y(x, t) and Y(t) may be expressed:

y x; tð Þ ¼ � xð ÞY tð Þ ð2:6aÞ

_yy x; tð Þ ¼ ’ xð Þ _YY tð Þ: ð2:6bÞ

To develop an expression for the equivalent mass it is assumed that the spring is weightless and the

kinetic energy of the mass–spring system is equal to that of the column. We therefore have

1

2M _YY2 tð Þ ¼ 1

2

ðL0m xð Þ ’ xð Þ _YY tð Þ

� �2dxþ 1

2

XNi¼ 1

mi ’ xið Þ _YY tð Þ� �2 ð2:7Þ

and hence

M ¼ðL0m xð Þ ’ xð Þ½ �2 dxþ

XNi¼ 1

mi ’ xið Þ½ �2: ð2:8Þ

The expression for the equivalent elastic spring stiffness is similarly found by equating the strain

energy stored in the spring to that stored in the column, i.e.

1

2KEY

2 tð Þ ¼ 1

2

ðL0M xð Þd�: ð2:9Þ

Because

d2y=dx2 ¼ M xð Þ=EI xð Þ ¼ d�=dx; ð2:10Þ

Equation 2.9 may also be written as

1

2KEY

2 tð Þ ¼ðL0EI xð Þ d2y=dx2

� �2dx ð2:11Þ

or

1

2KEY

2 tð Þ ¼ 1

2

ðL0EI xð Þ d2’=dx2

� �Y tð Þ

� �2dx; ð2:12Þ

hence

KE ¼ðL0EI xð Þ d2’=dx2

� �2dx: ð2:13Þ

In order to take account of the constant axial force Q, it is necessary to define a new stiffness

referred to as the equivalent geometric stiffness KG of the mass–spring system. The expression

for this stiffness is obtained by equating the potential energy of the axial load Q to the strain


17

energy stored in the spring due to this load. Thus,

12KGY

2 tð Þ ¼ Q L� L�ð Þ: ð2:14Þ

To develop an expression for KG it is therefore necessary to obtain an expression for (L� L*).

Consider the element �L; the length of this element may be expressed

�L ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ dy=dxð Þ2� �q

: ð2:15Þ

Expansion of the above square root by the binomial theorem and integration over the vertical

projection of the height L* of the deformed column yield

L ¼ðL�

01þ 1

2dy=dxð Þ2� 1

8dy=dxð Þ4þ � � �

� �dx: ð2:16Þ

If the quadratic and higher-order terms are neglected, then

L� L� ¼ðL�

0

1

2dy=dxð Þ2 dx: ð2:17Þ

The upper limit in Equation 2.17 may be changed from L* to L if it can be assumed that L*� L.

Making this assumption and substituting the above expression for L� L* into Equation 2.14, we

obtain:

1

2KGY tð Þ2¼ 1

2Q

ðL0

d’=dxð ÞY tð Þ½ �2 dx ð2:18Þ

and hence

KG ¼ Q

ðL0

d’=dxð Þ2 dx: ð2:19Þ

The total spring stiffness is therefore given by

K þ KE þ KG ð2:20Þ

or

K ¼ðL0EI xð Þ d2’=dx2

� �2dxþQ

ðL0

d’=dxð Þ2 dx: ð2:21Þ

The total stiffness K increases with increasing axial force and decreases with increasing compres-

sive force.Q is therefore taken as positive if it causes tension and negative if it causes compression.

The critical load Qcrit has been reached when

KE þ KG ¼ 0: ð2:22Þ

The expression for the equivalent damping, indicated as a dashpot in Figure 2.1(b), is found by

equating the virtual work of the damping force in the mass–spring system to the virtual work

of the damping forces in or acting on the column. In Chapter 3, it is explained that the damping


18

forces at a given time t may be expressed as the product of a viscous damping coefficient and the

velocity of the motion of the structure. An expression for C, the damping coefficient for the

equivalent mass–spring system, therefore may be found from

C _YY tð Þ�y tð Þ ¼ðL0c xð Þ ’ xð Þ _YY tð Þ

� �dxþ

XNi¼ 1

ci ’ xið Þ _YY tð Þ� �

’ xið Þ�Y½ � ð2:23Þ

and hence

C ¼ðL0c xð Þ ’ xð Þ½ �2 dxþ

XNi¼ 1

ci ’ xið Þ½ �2: ð2:24Þ

Similarly, the expression for the equivalent dynamic force that should be applied to the mass–

spring system may be found by equating the virtual work of this force to that of the real forces:

P tð Þ�Y tð Þ ¼ðL0p x; tð Þ ’ xð Þ�Y tð Þ½ � dxþ

XNi¼ 1

Pi ’ xið Þ�Y½ �: ð2:25Þ

and hence

P tð Þ ¼ðL0p tð Þ’ xð Þ dxþ

XNi¼ 1

Pi’ xið Þ: ð2:26Þ

The use of Equations 2.8, 2.13, 2.19, 2.21 and 2.26 will yield the equivalent mass, stiffness,

damping and dynamic force for the modelling of a structure as a 1-DOF system, provided the

mode shape of vibration is known. The latter can, as mentioned above, be found by assuming

the mode of vibration to be geometrically similar to the deflected shape caused by a uniform or

concentrated loads or by determining the mode shape by an eigenvalue analysis (Chapter 6).

In practice, the use of Equation 2.24 is very limited as the value for the damping coefficient

C is based on experimental data associated not only with the properties of the material used

and the method of construction, but also with the mode shape of vibration. A damping

coefficient evaluated or assumed for a given mode can therefore be used directly without

Equation 2.24.

In the following sections, expressions for the equivalent mass, stiffness, critical load and natural

frequencies are developed for some simple structures and structural elements which in the first

mode vibrate with mode shapes that are geometrically similar to their statically deformed shapes.

2.4. Equivalent 1-DOF mass–spring systems for linearly elastic linestructures

2.4.1 Cantilevers and columns with uniformly distributed loadAssume the mode shape of vibration of the cantilever shown in Figure 2.2 to be geometrically

similar to the deflected form y(x) caused by the uniformly distributed load wL. The deflected

form may be determined by integration of the expression for the bending moment M(x) at a

distance x from the fixed end, where

M xð Þ ¼ EI d2y=dx2 ¼ 12w L� xð Þ2: ð2:27Þ


19

Integration of Equation 2.27 and imposition of the boundary conditions y(x)¼ dy/dx¼ 0 when

x¼ 0 yields:

y ¼ w=24EIð Þ 6L2x2 � 4Lx3 þ x4� �

ð2:28Þ

yx¼L ¼ wL4=8EI : ð2:29Þ

For the equivalent mass–spring system to model the motion of the free end of the cantilever, the

shape function must be unity at that point. This will be the case when

w ¼ 8EI=L4: ð2:30Þ

Substitution of this expression for w into Equation 2.28 yields the following expressions for the

shape function �(x) and its first and second derivatives:

’ xð Þ ¼ 1=3L4� �

6L2x2 � 4Lx3 þ x4� �

; ð2:31aÞ

’0 xð Þ ¼ 4=3L4� �

3L2x� 3Lx2 þ x3� �

; ð2:31bÞ

’00 xð Þ ¼ 4=L4� �

l2 � 2Lxþ x2� �

: ð2:31cÞ

The weight of the equivalent lumped mass is therefore given by

W ¼ðL0w ’ xð Þ½ �2 dx ¼

ðL0w 1=3L4� �2

6L2x2 � 4Lx3 þ x4� �2

dx ð2:32Þ

and hence

W ¼ 728=2835ð ÞwL: ð2:33Þ

The equivalent elastic spring stiffness is given by

KE ¼ðL0EI ’00 xð Þ� �2

dx ¼ðL0EI 4=L4� �2

L2 � 2Lxþ x2� �2

dx ð2:34Þ

Figure 2.2 Cantilever with uniformly distributed load wL and axial tensile force T

y

T

x

L

EI, wL


20

and hence

KE ¼ 16EI=L3: ð2:35Þ

The equivalent geometrical spring stiffness is given by

KG ¼ðL0T ’0 xð Þ� �2

dx ¼ðL0T 4=3L4� �2

3L2x� 3Lx2 þ x3� �2

dx ð2:36Þ

and hence

KG ¼ 8T=7L: ð2:37Þ

The critical value for the axial force occurs when

K ¼ KE þ KG ¼ 16EI=5L3 þ 8T=7L ¼ 0 ð2:38Þ

or

T ¼ �14EI=5L2: ð2:39Þ

If the geometrical stiffness is neglected, the natural frequency of the cantilever is given by

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKEg=Wð Þ

p¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi16EIg=5L3

728wL=2835

� sð2:40Þ

or

f ¼ 0:5618313ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ

q: ð2:41Þ

When the correct mode shape is used, the natural frequency is given by

f ¼ 0:5602254

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ

q: ð2:42Þ

The error caused by assuming the mode shape to be similar to the deflected form is therefore

0.287%.

2.4.2 Cantilevers and columns with a concentrated load at the free endAssume that the mode shape of the uniformly loaded cantilever subjected to a vertical concen-

trated load P and an axial tensile force T at the free end (as shown in Figure 2.3) is geometrically

similar to the deflected form due to P. The deflected shape y(x) may be determined from the

expression for the bending moment M(x) at a distance x from the fixed end, which is given by

M xð Þ ¼ EId2y=dx2 ¼ P L� xð Þ: ð2:43Þ

Integration of Equation 2.43 twice and imposing the boundary conditions y(x)¼dy/dx¼ 0 when

x¼ 0 yields:

y xð Þ ¼ P=6EIð Þ 3Lx2 � x3� �

ð2:44Þ

yx¼L ¼ PL3=3EI : ð2:45Þ


21

For a mass–spring system to model the motion of the free end of the cantilever, the shape function

must be unity at this point. When this is the case

P ¼ 3EI=L3: ð2:46Þ

Substitution of this value for P into Equation 2.44 yields the following expressions for the shape

function �(x) and its first and second derivatives:

’ xð Þ ¼ 1=2L3� �

3Lx2 � x3� �

; ð2:47aÞ

’0 xð Þ ¼ 3=2L3� �

2Lx� x2� �

; ð2:47bÞ

’00 xð Þ ¼ 3=L3� �

L� xð Þ: ð2:47cÞ


W ¼ PþðL0w ’ xð Þ2� �2

dx ¼ðL0w 1=2L3� �2

3Lx2 � x3� �2

dx ð2:48Þ

and hence

W ¼ Pþ 33=140ð ÞwL: ð2:49Þ



dx ¼ðL0EI 3=L3� �2

L� xð Þ2 dx ð2:50Þ

and hence

KE ¼ 3EI=L3: ð2:51Þ

Figure 2.3 Cantilever beam column with uniformly distributed load, concentrated vertical load P and

axial tensile load T

y

T

Px

L

EI, wL


22



dx ¼ðL0T 3=2L3� �2

2Lx� x2� �2

dx ð2:52Þ

Hence

KG ¼ 6T=5L ð2:53Þ


K ¼ KE þ KG ¼ 3EI=L3 þ 6T=5L ¼ 0 ð2:54Þ

or

T ¼ �5EI=2L3: ð2:55Þ

Comparison of the expressions for the critical load given by Equations 2.55 and 2.39 reveals that

the two assumed mode shapes lead to a difference in the value for T of 12.0%.

If the geometrical stiffness and the concentrated vertical load are neglected, the frequency of the

cantilever with the assumed mode shape is

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKEg

W

� s¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3EIg=L3

33wL=140

� sð2:56Þ

or

f ¼ 0:56779


q: ð2:57Þ

The error in the natural frequency caused by the assumed mode shape when the load is uniformly

distributed is therefore approximately 1.35%.

2.4.3 Simply supported beam with uniformly distributed loadAssume the mode shape of vibration of the simply supported beam shown in Figure 2.4 subjected

to an axial tensile force T to be similar to the deflected form y(x) caused by the distributed load

Figure 2.4 Simply supported beam with uniformly distributed load wL and axial tensile force T

y

TT

x

L

EI, wL


23

wL. The deflected shape y(x) is obtained from the expression for the bending moment at a distance

x from the left-hand support:

M xð Þ ¼ EI d2y=dx2 ¼ � 12wLxþ 1

2wx2: ð2:58Þ

Integration of Equation 2.58 twice and imposing the boundary conditions that y(0)¼ y(L) = 0

yields:

y xð Þ ¼ w=24EIð Þ x4 � 2Lx3 þ L3x� �

ð2:59Þ

yx¼L=2 ¼ 5wL4=384EI : ð2:60Þ

If the mass–spring system is to model the motion at the centre of the beam, then the mode shape at

this point must be equal to unity. When this is the case,

w ¼ 384EI=5L4: ð2:61Þ

Substitution of this value of w into Equation 2.59 yields the following expressions for the shape

function and its first and second derivatives:

’ xð Þ ¼ 16=5L4� �

x4 � 2Lx3 þ Lx3� �

; ð2:62aÞ

’0 xð Þ ¼ 16=5L4� �

4x3 � 6Lx2 þ L3� �

; ð2:62bÞ

’00 xð Þ ¼ 16=5L4� �

12x2 � 12Lx� �

: ð2:62cÞ


W ¼ðL0w ’ xð Þ½ �2 dx ¼

ðL0w 16=5L4� �2

x4 � 2Lx3 þ L3x� �2

dx ð2:63Þ

and hence

W ¼ 3968=7875ð ÞwL: ð2:64Þ



dx ¼ðL0EI 16=5L4� �2

12x2 � 12Lx� �2

dx ð2:65Þ

and hence

KE ¼ 6144EI=125L3: ð2:66Þ



dx ¼ðL0T 16=5L4� �2

4x3 � 6Lx2 þ L3� �2

dx ð2:67Þ

and hence

KG ¼ 4353T=875L: ð2:68Þ


24


K ¼ KE þ KG ¼ 6144EI=125L3 þ 4352T=875L ¼ 0 ð2:69Þ

or

T ¼ �9:8824EI=L2: ð2:70Þ

The natural frequency for the beam, neglecting the axial load, is given by

f ¼ 1

2�


W

� s¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi6144EIg=L3

3968wL=7875

� sð2:71Þ

or

f ¼ 1:571919


q: ð2:72Þ

When the correct mode shape is used, the expression for the natural frequency is

1:5707963


q: ð2:73Þ

The error in the natural frequency resulting from assuming the mode shape to be geometrically

similar to the deflected form caused by the self-weight of the beam is therefore 0.07%.

2.4.4 Simply supported beam with a concentrated load at mid-spanIf a beam supports a concentrated load in addition to its own weight as shown in Figure 2.5, it

may be assumed that the mode shape of vibration is similar to the deflected form caused by P.

The deflected shape y(x) is obtained from the expression for the bending moment M(x) at a

distance x from the left-hand support:

M xð Þ ¼ EI d2y=dx2 ¼ � 12Px: ð2:74Þ

Integration of Equation 2.74 twice and imposing the boundary conditions y¼ 0 when x¼ 0 and

dy/dx¼ 0 when x¼L/2 yields:

y xð Þ ¼ P=48EIð Þ 3L2x� 4x3� �

ð2:75Þ

yx¼L=2 ¼ PL3=48EI : ð2:76Þ

Figure 2.5 Simply supported beam with concentrated load P at mid-span and axial tensile force T

y

T

P

T

x

L /2 L /2

EI wL


25

If the mass–spring system is to model the motion at the centre of the beam then the shape function

at this point must be, as previously, unity. When this is the case,

P ¼ 48EI=L3: ð2:77Þ

Substitution of the above expression for P into Equation 2.75 yields the following expressions for

the shape function and its derivatives:

’ xð Þ ¼ 1=L3� �

3L2x� 4x3� �

; ð2:78aÞ

’0 xð Þ ¼ 1=L3� �

3L2 � 12x2� �

; ð2:78bÞ

’00 xð Þ ¼ 1=L3� �

�24xð Þ: ð2:78cÞ


W ¼ Pþ 2

ðL=20

w ’ xð Þ½ �2 dx ¼ Pþ 2

ðL=20

w 1=L3� �2

3L2x� 4x3� �2

dx ð2:79Þ

and hence

W ¼ Pþ 17=35ð ÞwL: ð2:80Þ


KE ¼ 2

ðL=20

EI ’00 xð Þ� �2

dx ¼ 2

ðL=20

EI �24xð Þ2 dx ð2:81Þ

and hence

KE ¼ 48EI=L3: ð2:82Þ


KG ¼ 2

ðL=20

T ’0 xð Þ� �2

dx ¼ 2

ðL=20

T 1=L3� �2

3L2 � 12x2� �2

dx ð2:83Þ

and hence

KG ¼ 24T=5L: ð2:84Þ


K ¼ KE þ KG ¼ 48EI=L3 þ 24T=5L ¼ 0 ð2:85Þ

or

T ¼ �10EI=L2: ð2:86Þ

Comparison of the expressions for the critical axial load given by Equations 2.70 and 2.86

shows that the two differently assumed mode shapes lead to a difference of 1.17% in the values

for T.


26

The natural frequency for the assumed mode shape, neglecting the concentrated load P and the

axial force T, is given by

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKGg

W

� s¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi46EIg=L3

17wL=35

� sð2:87Þ

or

f ¼ 1:5821597


q: ð2:88Þ

The error caused by the assumed mode shape is therefore in this case equal to 0.723%.

2.4.5 Built-in beam with uniformly distributed loadAssume the mode shape of vibration of the uniformly loaded built-in beam, subjected to an axial

load T and having a constant flexural rigidity EI as shown in Figure 2.6, to be geometrically

similar to the deflected form caused by the load wL. An expression for the deflected form can

be found from the expression for the bending moment at a distance x from the left-hand support,

which is given by:

M xð Þ ¼ EI d2y=dx2 ¼ MA � 12wLxþ 1

2wx2: ð2:89Þ

Integration of Equation 2.89 twice and imposing the boundary conditions y� dy/dx¼ 0 when

x¼ 0, and y¼ 0 when x¼L yields:

y xð Þ ¼ w=24EIð Þ x4 � 2Lx3 þ L2x2� �

ð2:90Þ

yx¼ 1=2 ¼ wL4=384EI : ð2:91Þ

To model the motion at mid-span, the shape function must be unity at this point. When this is the

case,

w ¼ 384EI=L4: ð2:92Þ

Substitution of this value for w into Equation 2.90 yields the following expressions for the shape

function and its first and second derivatives:

’ xð Þ ¼ 16=L4� �

x4 � 2Lx3 þ L2x2� �

; ð2:93aÞ

Figure 2.6 Built-in beam with uniformly distributed load wL and axial load T

y

TT

x

EI , wL

L


27

’0 xð Þ ¼ 32=L4� �

2x3 � 3Lx2 þ L2x� �

; ð2:93bÞ

’00 xð Þ ¼ 32=L4� �

6x2 � 6Lxþ L2� �

: ð2:93cÞ

Thus the weight of the equivalent lumped mass is given by

W ¼ðL0w ’ xð Þ½ �2 dx ¼

ðL0w 16L4� �2

x4 � 2Lx3 þ L2x2� �2

dx ð2:94Þ

and hence

W ¼ 128=315ð ÞwL: ð2:95Þ



dx ¼ðL0EI 32=L4� �2

6x2 � 6Lxþ L2� �2

dx ð2:96Þ

and hence

KG ¼ 1024EI=5L3: ð2:97Þ



dx ¼ðL0T 32=L4� �2

2x3 � 3Lx2 þ L2x� �2

dx ð2:98Þ

and hence

KG ¼ 512T=105L: ð2:99Þ


KE þ KG ¼ 1024EI=5L3 þ 512T=105L ¼ 0 ð2:100Þ

or

T ¼ �42EI=L2: ð2:101Þ

Neglecting the axial force, the natural frequency associated with the assumed mode shape is

given by

f ¼ 1

2�


W

� s¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1024EIg=5L3

128wL=315

� sð2:102Þ

or


q: ð2:103Þ


28

Using the correct mode shape,

f ¼ 3:5608213


q: ð2:104Þ

The error in this case is therefore 0.34%.

2.4.6 Built-in beam with concentrated load at mid-spanConsider the beam shown in Figure 2.7. Because of the concentrated load at mid-span, it is

assumed that the mode shape of vibration is geometrically similar to the deflected form due to

P. The deflected form itself is found from the expression for the bending moment at x between

the left-hand support and P, which is given by

M xð Þ ¼ EI d2y=dx2 ¼ MA � 12Px: ð2:105Þ

Integration of Equation 2.105 twice and imposing the boundary conditions y¼ dy/dx¼ 0 when

x¼ 0 and dy/dx¼ 0 when x¼L/2 yields:

y xð Þ ¼ P=48EIð Þ 3Lx2 � 4x3� �

ð2:106Þ

and hence

yx¼L=2 ¼ PL3=192EI : ð2:107Þ

For the equivalent mass–spring system to represent the motion at mid-span, the shape function at

this point must be unity. This requires that:

P ¼ 192EI=L3: ð2:108Þ

Substitution of this expression for P into Equation 2.106 yields the required shape function and

hence its first and second differentials:

’ xð Þ ¼ 4=L3� �

3LX2 � 4x3� �

; ð2:109aÞ

’0 xð Þ ¼ 24=L3� �

Lx� 2x2� �

; ð2:109bÞ

’00 xð Þ ¼ 24=L3� �

L� 4xð Þ: ð2:109cÞ

Figure 2.7 Built-in beam with concentrated load P at mid-span and axial tension load T

y

T

P

T

x

L /2 L /2

EI wL


29


W ¼ Pþ 2

ðL=20

w ’ xð Þ½ �2 dx ¼ Pþ 2

ðL=20

w 4=L3� �2

3Lx2 � 4x3� �2

dx ð2:110Þ

and hence

W ¼ 13=35ð ÞwL: ð2:111Þ


KE ¼ 2

ðL=20

EI ’00 xð Þ� �2

dx ¼ 2

ðL=20

EI 24=L3� �2

L� 4xð Þ2 dx ð2:112Þ

and hence

KE ¼ 192EI=L3: ð2:113Þ


KG ¼ 2

ðL=20

T ’0 xð Þ� �2

dx ¼ 2

ðL=20

T 24=L3� �2

Lx� 2x2� �2

dx ð2:114Þ

and hence

KG ¼ 24T=5L: ð2:115Þ


K ¼ KE þ KG ¼ 192EI=L3 þ 24T=5L ¼ 0 ð2:116Þ

or

T ¼ �40EI=L2: ð2:117Þ

Comparison of the expressions for the critical axial load given by Equations 2.101 and 2.117

shows that the two differently assumed mode shapes lead to a difference of 5%.

Neglecting the concentrated load P and the axial load T yields the following frequency for a

uniformly loaded built-in beam:

f ¼ 1

2�


W

� s¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi192EIg=L2

13wL=35

� sð2:118Þ

or


q: ð2:119Þ

The error resulting from this form of mode shape is therefore 1.62%.


30

2.4.7 Uniformly loaded beam with one end simply supported and one endbuilt-in

Consider the axially loaded beam shown in Figure 2.8 supporting a uniformly distributed load

wL. In order to model the point of the beam that will vibrate with the greatest amplitude as a

mass–spring system, it is first necessary to determine the mode shape of vibration. This is most

easily done by assuming it to be geometrically similar to the deflected form caused by the

distributed load. The deflected shape y(x) can be found from the bending moment at a distance

x from the hinged end:

M xð Þ ¼ EI d2y=dx2 ¼ �Rxþ 12wx

2: ð2:120Þ

Integration of Equation 2.120 twice and imposition of the boundary conditions y(x) = 0 when

x¼ 0 and x¼L, and dy/dx¼ 0 when x¼L yields:

y xð Þ ¼ w=48EIð Þ L3x� 3Lx3 þ 2x4� �

: ð2:121Þ

The maximum deflection occurs when dy/dx¼ 0. Since dy/dx¼ 0 when x¼L:

8x3 � 9Lx2 þ L3 ¼ x� x1ð Þ x� x2ð Þ x� Lð Þ ¼ 0: ð2:122Þ

Division of the left-hand side of Equation 2.122 by (x� L) and solution of the resulting quadratic

equation with respect to x yields:

x1 ¼ 0:4215351L

x2 ¼ �0:2965351L:

The negative value for x obviously has no practical meaning, thus the maximum displacement is

found by substitution of the value for x1 into Equation 2.121 which yields:

y xð Þmax¼ 0:2599738 wL4=48EI� �

: ð2:123Þ

For the equivalent mass–spring system to model the motion at position x¼x1, the displacement at

this point must be unity. When this is the case,

w=48EI ¼ 3:8465403=L4: ð2:124Þ

Figure 2.8 Beam with one end simply supported and one built-in, supporting a distributed loadwL and

an axial load T

y

TT

x

EI , wL

L


31

Substitution of the above value for w/48EI into Equation 2.121 yields the shape function and

hence its first and second derivatives:

’ xð Þ ¼ 3 � 8465403=L4� �

L3x� 3Lx3 þ 2x4� �

; ð2:125aÞ

’0 xð Þ ¼ 3 � 8465403=L4� �

L3 � 9Lx2 þ 8x3� �

; ð2:125bÞ

’00 xð Þ ¼ 23 � 079242=L4� �

�3Lxþ 4x2� �

: ð2:125cÞ


W ¼ðL0w ’ xð Þ½ �2 dx ¼

ðL0w 3:8465403=L4� �2

L3x� 3Lx3 þ 2x4� �2

dx ð2:126Þ

and hence

W ¼ 0:4462246wL: ð2:127Þ

The equivalent elastic stiffness is given by


dx ¼ðL0EI 23:079242=L4� �2 �3Lxþ 4x4

� �2dx ð2:128Þ

and hence

KE ¼ 106:53028EI=L3: ð2:129Þ

The equivalent geometrical stiffness is given by


dx ¼ðL0T 3:8465403=L4� �2

L3 � 9Lx2 þ 8x3� �2

dx ð2:130Þ

and hence

KG ¼ 5:0728704T=L2: ð2:131Þ

The critical value for the axial load occurs when

K ¼ KE þ KG ¼ 106:53028EI=L3 þ 5:0728704T=L2 ¼ 0 ð2:132Þ

or

T ¼ �21EI=L3: ð2:133Þ

Neglecting the effect of the axial force, the natural frequency of the beam is given by

f ¼ 1

2�


W

� s¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi106:53028EIg=L3

0:4462246wL

� sð2:134Þ

or

f ¼ 2:4591211


q: ð2:135Þ


32

Use of the correct mode shape yields


q: ð2:136Þ

The error in the natural frequency caused by a uniformly distributed load is therefore 0.332%.

In the preceding development, the mass–spring system was modelled to represent the motion of

the beam at the position of maximum static displacement. If the motion of the beam at, say,

mid-span is to be studied, then the shape function at this point must be unity. This is achieved

by determining the value of w that will cause the deflection at mid-span to be equal to 1. From

Equation 2.121, the deflection at mid-span is obtained as

yx¼L=2 ¼ wL4=192EI : ð2:137Þ

When yx¼L/2¼ 1,

w ¼ 192EI=L4: ð2:138Þ

Substitution of this expression for w into Equation 2.121 yields

’ xð Þ ¼ 4=L4� �

L3x� 3Lx3 þ 2x4� �

; ð2:139aÞ

’0 xð Þ ¼ 4=L4� �

L3 � 9Lx2 þ 8x3� �

; ð2:139bÞ

’00 xð Þ ¼ 24=L4� �

�3Lxþ 4x3� �

: ð2:139cÞ


W ¼ðL0w ’ xð Þ½ �2 dx ¼

ðL0w 4=L4� �2

L3x� 3Lx3 þ 2x4� �2

dx ð2:140Þ

and hence

W ¼ 152=315ð ÞwL: ð2:141Þ



dx ¼ðL0EI 24=L4� �

�3Lx3 þ 4x3� �2

dx ð2:142Þ

and hence

KE ¼ 576EI=5L3: ð2:143Þ



dx ¼ðL0T 4=L4� �2

L3 � 9Lx2 þ 8x3� �2

dx ð2:144Þ

and hence

KG ¼ 192T=35L: ð2:145Þ


33

These expressions forW,KE andKG result in the same values for the critical axial load and natural

frequency as given by Equations 2.133 and 2.136, respectively.

2.4.8 Uniformly loaded beam with one end simply supported, one end built-inand a concentrated load at mid-span

Assume the mode shape of the propped cantilever shown in Figure 2.9 to be geometrically similar

to the deflected form caused by the concentrated load at mid-span. The deflected shape is found by

integration of the expression for the bending moment at section x, which is given by

EI d2y=dx2 ¼ �Rxþ P x� 12L

� �ð2:146Þ

and imposing the boundary conditions y¼ 0 when x¼ 0 and x¼L and dy/dx¼ 0 when x¼L.

This yields:

y xð Þ ¼ P=96EIð Þ 3L2xþ 16 x� 12L

� �3�5x3h i

ð2:147Þ

yx¼L=2 ¼ 7PL3=768EI ð2:148Þ

and when yx¼L/2¼ 1

P ¼ 768EI=7L3: ð2:149Þ

Substitution of this expression into Equation 2.148 yields the following expressions for the shape

function �(x) and its first and second derivatives:

’ xð Þ ¼ 8=7L3� �

3L2xþ 16 x� 12L

� �3�5x3h i

; ð2:150aÞ

’0 xð Þ ¼ 8=7L3� �

3L2 þ 48 x� 12L

� �2�15x2h i

; ð2:150bÞ

’00 xð Þ ¼ 8=7L3� �

96 x� 12L

� �� 30x

� �: ð2:150cÞ

The weight of the equivalent lumped mass at the centre of the beam is now given by

W ¼ PþðL=20

w ’ xð Þ½ �2 dxþðLL=2

w ’ xð Þ½ �2 dx ð2:151Þ

Figure 2.9 Uniformly loaded propped cantilever subjected to a load P at mid-span and an axial tensile

force T

y

T

P

T

x

L /2 L /2

EI wL


34

or

W ¼ PþðL=20

w 8=7L3� �2

3L2x� 5x3� �2

dxþðLL=2

w 8=7L3� �2

3L2xþ 16 x� 1

2L

� 3

�5x3

" #2dx:

ð2:152Þ

This yields

W ¼ Pþ 0:2813411wLþ 0:1641398wL ð2:153Þ

and hence

W ¼ Pþ 0:4454809wL: ð2:154Þ

The corresponding equivalent elastic spring stiffness is given by

KE ¼ðL=20

EI ’00 xð Þ� �2

dxþðLL=2

EI ’00 xð Þ� �2

dx ð2:155Þ

or

KE ¼ðL=20

EI 8=7L3� �2 �30xð Þ2 dxþ

ðLL=2

EI 8=7L3� �2

96 x� 1

2L

� � 30x

� �2dx: ð2:156Þ

This yields

KE ¼ 48:979592EI=L3 þ 60:734694EI=L3 ð2:157Þ

and hence

KE ¼ 109:714286EI=L3: ð2:158Þ


KG ¼ðL=20

T ’0 xð Þ� �2

dxþðLL=2

T ’0 xð Þ� �2

dx ð2:159Þ

or

KG ¼ðL=20

T 8=7L3� �2

3L2 � 15x2� �2

dxþðLL=2

T 8=7L3� �2

3L2 � 48 x� L=2ð Þ2�15x2� �2

dx: ð2:160Þ

This yields

KG ¼ 2:8163265T=Lþ 3:2897959T=L ð2:161Þ

and hence

KG ¼ 5:1061224T=L: ð2:162Þ


35

The critical value for the axial load therefore occurs when

KE þ KG ¼ 109:714286EI=L3 þ 5:1061224T=L ¼ 0 ð2:163Þ

which yields

T ¼ �21:486812EI=L2: ð2:164Þ

If the axial force T and the concentrated load P are neglected, the assumed mode shape yields

f ¼ 2:4976822


q: ð2:165Þ

The error in the natural frequency generated by assuming the mode shape to be geometrically

similar to the deflected form caused by a concentrated load at mid-span is therefore 1.905%.

2.4.9 Built-in beam with uniformly distributed load and one end vibratingvertically

Assume the mode shape of vibration to be similar to the deflected form y(x) caused by a vertical

displacement Y of the right-hand support (Figure 2.10). The deflected form due to this displace-

ment is found from the expression of the bending moment at a section x from the left-hand

support:

EI d2y=dx2 ¼ �MA þ VA � 12 x

2: ð2:166Þ

Integration of Equation 2.166 twice and imposing the boundary conditions y(x) = dy/dx¼ 0

when x¼ 0 and x¼L yields:

y xð Þ ¼ Y=L3� �

3Lx2 � 2x3� �

: ð2:167Þ

For a mass–spring system to model the motion of the right-hand support, the shape function and

its first and second derivatives are given by:

’ xð Þ ¼ 1=L3� �

3Lx2 � 2x3� �

; ð2:168aÞ

’0 xð Þ ¼ 6=L3� �

Lx� x2� �

; ð2:168bÞ

’00 xð Þ ¼ 6=L3� �

L� xð Þ: ð2:168cÞ

Figure 2.10 Built-in beam with uniformly distributed load wL, axial force T and one support vibrating

vertically

y

T∆

T

x

EI , wL

L


36

The weight of the equivalent lumped mass system is therefore given by

W ¼ðL0w ’ xð Þ½ �2 dx ¼

ðL0w 1=L3� �2

3Lx2 � 2x3� �2

dx ð2:169Þ

or

W ¼ 13=35ÞwL:ð ð2:170Þ



dx ¼ðL0EI 6=L3� �2

L� 2xð Þ2 dx ð2:171Þ

or

KE ¼ 12EI=L3: ð2:172Þ

The equivalent geometrical stiffness is given by


dx ¼ðL0T 6=L3� �2

Lx� x2� �2

dx ð2:173Þ

or

KG ¼ 6T=5L: ð2:174Þ


K ¼ KE þ KG ¼ 12EI=L3 þ 6T=5L ¼ 0 ð2:175Þ

or

T ¼ �10EI=L2: ð2:176Þ

2.4.10 Beam with uniformly distributed load, one end hinged and the built-inend vibrating vertically

Assume the mode shape of vibration of the propped cantilever shown in Figure 2.11 to be

geometrically similar to the deflected form y(x) caused by a vertical displacement Y of the

built-in support. The deflected shape may be found from the expression for the bending

moment at a distance x from the left-hand support:

M ¼ EI d2y=dx2 ¼ Vx� 12wx

2: ð2:177Þ

Integration of Equation 2.177 twice and imposing the boundary conditions y(x) = 0 when x¼ 0,

y(x)¼Y when x¼L and dy/dx¼ 0 when x¼L yield:

y xð Þ Y=2L3� �

3L2x� x3� �

: ð2:178Þ


37

For an equivalent mass–spring system to model the vertical motion of the built-in end, the shape

function at that point must be unity which requires that Y¼ 1. When this is the case, the shape

function and its first and second derivatives are given by:

’ xð Þ ¼ 1=2L3� �

3L2x� x3� �

; ð2:179aÞ

’0 xð Þ ¼ 3=2L3� �

L2 � x2� �

; ð2:179bÞ

’00 xð Þ ¼ 3=2L3� �

�2xð Þ: ð2:179cÞ


W ¼ðL0w ’ xð Þ½ �2 dx ¼

ðL0w 1=2L3� �2

3L2x� x3� �2

dx ð2:180Þ

or

W ¼ 17=35ð ÞwL: ð2:181Þ

The equivalent elastic stiffness is given by


dx ¼ðL0EI 3=2L3� �2 �2xð Þ2 dx ð2:182Þ

or

KE ¼ 3EI=L3 ð2:183Þ

and the equivalent geometrical stiffness by


dx ¼ðL0T 3=2L3� �2

L2 � x2� �2

dx ð2:184Þ

or

KG ¼ 6T=5L: ð2:185Þ


K ¼ KE þ KG ¼ 3EI=L3 þ 6T=5L ¼ 0 ð2:186Þ

Figure 2.11 Propped cantilever with uniformly distributed load wL, axial force T and the built-in end

vibrating vertically

y

T

∆T

xEI , wL

L


38

or

T ¼ �5EI=2L2: ð2:187Þ

2.5. Equivalent 1-DOF mass–spring systems for linearly elasticcontinuous beams

The method used in the preceding sections to develop expressions for equivalent lumped masses

and spring stiffness for single-span beams using Equations 2.8, 2.13 and 2.19 and assuming the

mode shapes to be geometrically similar to a deflected form can be extended to continuous

beams. This is done by, for example, assuming the mode shape to be similar to the deflected

shape caused by a uniformly distributed load acting alternately downwards and upwards on

succeeding spans, or by assuming the mode shapes to be geometrically similar to the deflected

form caused by a concentrated load at the point where the response is to be studied. Generally,

however, this method of approach is not practical because of the time involved in developing

the required shape functions and the subsequent integrations. When it is necessary to determine

the natural frequencies of multi-span beams, it is better to use one of the many structural analysis

programs available that include the solution of the eigenvalue problem.

At the design stage, however, and in order to check the output from a computer analysis, it is

useful to have available a simple method that enables a quick estimate of the first and perhaps

even the second natural frequencies of continuous beams. Such estimates can be made by

assuming the mode shapes of the individual spans to be similar to the mode shapes of

corresponding simply supported beams (or, if one end or both ends of a continuous beam are

rigidly encased, by beams being simply supported at one end and built-in at the other). Each

span can therefore be modelled by the expressions for the equivalent lumped masses and spring

stiffnesses given by Equations 2.64 and 2.66, or 2.127 and 2.129.

The equivalent lumped mass for a continuous beam is found by equating the maximum kinetic

energy of the equivalent lumped masses of the individual spans. Similarly, the equivalent spring

stiffness is determined by equating the maximum strain energy of the spring to the sum of the

maximum strain energies in the equivalent springs for each span.

Consider a continuous beam with N spans. The maximum kinetic energy of the beam in terms of

the equivalent lumped masses of the individual spans is given by

1

2ME

_YY2 ¼ 1

2

XNi¼ 1

Mi� xið Þ2 _YY2 ð2:188Þ

where ME is the equivalent lumped mass for the continuous beam, Mi is the equivalent lumped

mass for the ith span as given by Equation 2.64 or 2.127, _YY is the maximum velocity of mass

ME and �(xi) is the value of the shape function at position xi of Mi. We therefore have:

ME ¼XNi¼ 1

Mi’ xið Þ2: ð2:189Þ

Similarly, the maximum strain energy of the continuous beam as a function of the equivalent

elastic springs representing the stiffness of each span is given by

1

2KEY

2 ¼ 1

2

XNKEi� xið Þ2Y2 ð2:190Þ


39

where KE is the equivalent elastic spring stiffness of the continuous beam, KEi is the equivalent

spring stiffness of the ith span as given by Equation 2.66 or 2.129 andY is the maximum amplitude

of the equivalent mass M. We therefore have:

KE ¼XNi¼ 1

KEi’ xið Þ2: ð2:191Þ

The natural frequency of a continuous or multi-span beam is therefore given by

f ¼

XNi¼ 1

KEi’ xið Þ2

XNi¼ 1

Mi’ xið Þ2

0BBBB@

1CCCCA

1=2

: ð2:192Þ

The degree of accuracy obtained by use of Equation 2.192 depends very much on the estimates of

the relative values of �(xi). In practice, such estimates can be difficult. The most accurate values

for the frequencies are most easily obtained when the spans are of approximately the same length,

which is often the case in real structures. This statement is demonstrated by the following two

examples.

Example 2.1

The continuous beam ABCDE shown in Figure 2.12 has four equal spans of length L and is

built-in at E. The section of the beam is constant throughout, having a flexural rigidity EI,

and weighs w per unit length. Develop first expressions for the weight of the equivalent

lumped mass and spring stiffness of the beam corresponding to vibration in the first mode,

and hence an expression for its first natural frequency, and expressions for the equivalent

lumped mass, spring stiffness and natural frequency corresponding to vibration in the

second mode shown.

Figure 2.12 Continuous beam with four equal spans and one end built-in at E

2nd mode

1st mode

A B C D E

L L L L


40

In order to develop expressions for the equivalent weight and spring stiffness of the first mode

shown in Figure 2.12, assume that the maximum amplitude of vibration for each of the spans

AB, BC and CD is equal to unity and that the maximum amplitude of span DE is propor-

tional to the maximum amplitude of the other three spans. The amplitude of span DE is

therefore given by

’ xð ÞDEmax¼

5:41403� 10�3wL4=EI

13:0208� 10�3wL4=EI¼ 0:4157985:

The assumed mode shape of vibration for the first mode is therefore

’ xð Þ ¼ 1:0 � 1:0 1:0 � 0:415958f g:

Substitution of these values and the expression for the weights of the equivalent lumped

masses for each of the four spans given by Equations 2.64 and 2.129 into Equation 2.189

yields

M ¼ 3� 3968=7875ð ÞwL� 1:02 þ 0:4462246ð ÞwL� 0:4159582� �

=g

¼ 1:588252wL=g:

Substitution of the same values for �(x) and the expressions for the equivalent elastic spring

stiffness for each of the four spans given by Equations 2.66 and 2.129 into Equation 2.191

yields

KE ¼ 3� 6144EI=125L3� �

� 1:02 þ 106:53628EI=L3� �

� 0:4159582

¼ 165:88902EI=L3:

The first natural frequency is therefore

f1 ¼1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi165:88902EI=L3

1:588252wL=g

� s

or

f1 ¼ 1:6265568ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ

q:

The correct value is

f1 ¼ 1:6392959


q;

implying that the error in this case is 0.78%. This of course is very good but it is rather

fortuitous, as in a real mode shape the ratio of the amplitudes of spans CD and DE would

tend to be less than 1 if it could be assumed that the slopes of the tangents of the two

spans at D were equal. Also, the amplitude of span CD would be less than the amplitudes

of spans AB and BC because of the built-in end at E. In Example 2.2, therefore, it is

shown that mode shapes based on assumptions of the relative displacements of non-

continuous individual spans can lead to considerable errors.


41

The frequency of the beam corresponding to the second mode shape shown in Figure 2.18 can

be determined by assuming the beam to have 8 spans, each of length L/2. The assumed mode

shape vector is therefore given by

’ xð Þ ¼ 1:0 � 1:0 1:0 � 1:0 1:0 � 1:0 1:0 � 0:415958f g:

Substitution of these values and the equivalent lumped mass for each half span into Equation

2.189 yields

ME ¼ 7� 3968=7875ð ÞwL=2� 1:02 þ 0:4462246ð ÞwL=2� 0:4159582� �

=g

¼ 1:8021587wL:

Substitution of the same values for �(x) and the equivalent spring stiffnesses for each half

span into Equation 2.191 yields

KE ¼ 7� 6144� 8EI=125L3� �

� 1:02� �

þ 106:53628� 8EI=L3� �

� 0:4159582� �

=g

¼ 2899:9762EI=L3:

Thus the frequency for this mode is given by

f2 ¼1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2899:9762EIg=L3

1:8021587wL

� s

and hence

f2 ¼ 6:38441089


q:


f2 ¼ 6:4378174


q

and so the error is also small in this case: 0.83%.

Example 2.2

The continuous beam ABCD shown in Figure 2.13 is of uniform section with constant

flexural rigidity EI and self-weight w per unit length. The lengths of spans AB and BC are

equal to L and span BC is of length 2L. Determine the first natural frequency of the beam.

Assume that the first mode shape is geometrically similar to that caused by a uniformly

distributed load acting upwards on spans AB and CD and downwards on span BC.

If it is assumed that the beam is non-continuous at supports B and C, then the following mode

shape vector is obtained for the amplitudes of the central points of each span:

’ xð Þ ¼ 0:125 � 1:0 0:125f g:


42

Figure 2.13 Continuous beam with the length of the central span twice the length of the two

outer ones

A B C D

A B C D

L 2L L

Substitution of the above values for �(x) and the appropriate weight of the equivalent lumped

masses of the individual spans into Equation 2.189 yields

ME ¼ 2� 3968=7875ð ÞwL� 0:1252 þ 3968=7875ð Þw2L� 1:02� �

=g

¼ 1:0234921wL=g:

The equivalent elastic spring stiffness is determined by substitution of the values for �(x) and

the equivalent spring stiffnesses for the three spans into Equation 2.191. This yields

KE ¼ 2� 6144EI=125L3� �

� 0:1252� �

þ 6144EI=125� 8L3� �

� 1:02� �

¼ 7:68EI=L3

and hence

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi7:68EI=L3

1:0234921wL=g

� s

or

f ¼ 0:4359719


q:


f ¼ 0:613798


q;

demonstrating an error in the estimated value of –28.97%. This is obviously not very good,

but is expected as the overall stiffness of the beam is reduced by assuming non-continuity at

the supports.

An alternative estimate may be achieved by assuming that the points of contra-flexure occur

at positions B and C. This yields the mode shape vector

’ xð Þ ¼ 0:5 � 1:0 0:5f g:


43

Use of the above values for �(x) yields

ME ¼ 2� 3968=7875ð ÞwL� 0:52 þ 3968=7875ð Þw2L� 1:02� �

=g

¼ 1:2596825wL=g

KE ¼ 2� 6144EI=125L3� �

� 0:52� �

þ 6144EI=125� 8L3� �

� 1:02� �

¼ 30:72EI=L3:

We therefore have

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi30:72EI=L3

1:2596825wL=g

� s

or

f ¼ 0:7859595


q:

The error is this case is thereforeþ28.05%. This is not very good either, but is again expected as

the assumption made implies that the EI value for the central span is much greater than for the

outer spans. By use of the theorem of three moments it can in fact be shown that the points of

contra-flexure lie to the right of B and to the left of C. This will result in a maximum deflection

of the outer spans relative to the central span somewhere between 0.125 and 0.5. If it is assumed

that these amplitudes are equal to (1.25þ 0.5)/2, we obtain the mode shape vector

’ xð Þ ¼ 0:3125 � 1:0 0:3125f g:

We therefore have

ME ¼ 2� 3968=7875ð ÞwL� 0:31252 þ 3968=7875ð Þw2L� 1:02� �

=g

¼ 1:1061587wL=g

KE ¼ 2� 6144EI=125L3� �

� 0:31252� �

þ 6144EI=125� 8L3� �

� 1:02� �

¼ 15:744EI=L3

and hence

f ¼ 0:600439


q:

The error in this case is therefore only –2.145%, which is acceptable. Obviously, the reasons

for the discrepancies are the geometries of the assumed mode shapes. The above mode shapes

can be compared with the mode shape vector

’ xð Þ ¼ 0:326 � 1:0 0:326f g

whichwas obtained froma computer analysis, andwhichwhen substituted into Equations 2.189

and 2.191 together with the appropriate values for equivalent masses and spring stiffness yields

f ¼ 0:6139793


q:


44

When the correct mode shape is used, the error resulting from considering each beam as

a mass–spring system is only 0.03%. A relatively simple way to decide on a mode shape is

to determine the points of contra-flexure by first constructing the bending moment diagram

and then sketching the corresponding deflected form, knowing that the amplitude of the

central span is 1.0 and that of the outer spans lies between 0.125 and 0.5 m. It is left to the

reader to try this out. When the authors tried it, they obtained the mode shape vector

’ xð Þ ¼ 0:32 � 1:0 0:32f g:

Example 2.3

Determine the first natural frequency of the beam in Example 2.2 by assuming that the mode

shape is geometrically similar to that caused by a point load applied at the centre of span BC.

With the origin at A, the deflected form of section AB can be shown to be

y ¼ 3P=96Pð Þ x3 � L2x� �

ð2:193aÞ

and with the origin at B the deflected shape of section BC is

y ¼ P=96EIð Þ �8x3 þ 9Lx2 þ 6L2x� �

: ð2:193bÞ

The shape functions and their derivatives for spans AB, BC and CD are therefore given by

’AB xð Þ ¼ ’CD xð Þ ¼ 3=7L3� �

x3 � L2x� �

; ð2:194aÞ

’0AB xð Þ ¼ ’0

CD xð Þ ¼ 3=7L3� �

3x2 � L2� �

; ð2:194bÞ

’00AB xð Þ ¼ ’00

CD xð Þ ¼ 18=7L3� �

ðxÞ; ð2:194cÞ

’BC xð Þ ¼ 1=7L3� �

�8x3 þ 9Lx2 þ 6L2x� �

; ð2:194dÞ

’0BC xð Þ ¼ 2=7L3

� ��12x2 þ 9Lxþ 3L2� �

; ð2:194eÞ

’00BC xð Þ ¼ 6=7L3

� ��8xþ 3Lð Þ: ð2:194f Þ

The expression for the weight of the equivalent lumped mass is

WE ¼ Pþ 2

ðL0w ’AB xð Þ½ �2 dxþ 2

ðL0w ’BC xð Þ½ �2 dx: ð2:195Þ

Substitution of the expression for �AB(x) and �BC(x) and integration between limits yield

WE ¼ Pþ 48=1715ð ÞwLþ 1480=1715ð ÞwL ¼ Pþ 1528=1715ð ÞwL: ð2:196Þ

The expression for the equivalent spring stiffness is

KE ¼ 2

ðL0EI ’00

AB xð Þ� �2

dxþ 2

ðL0EI ’00

CD xð Þ� �2

dx: ð2:197Þ


45

Substitution of the expressions for �00AB xð Þ and �00

CD xð Þ and integration between limits yield

KE ¼ 216EI=49L3 þ 456EI=49L3 ¼ 96EI=7L3: ð2:198Þ

If the concentrated load P is neglected, the natural frequency is therefore given by

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi96EI=7L3

1528wL=1715

� s

or

f ¼ 0:6244204


q: ð2:199Þ

This approach therefore yields an error of only 1.73%. The amount of work involved,

however, is considerable.

The equivalent geometrical stiffness due to an axial tensile force T that is applied to either end

of the beam is given by

KG ¼ 2

ðL0T ’0

AB xð Þ� �2

dxþ2

ðL0T ’0

BC xð Þ� �2

dx: ð2:200Þ

Substitution for �0AB xð Þ and �0

BC xð Þ and integration between limits yields

KG ¼ 72T=245Lþ 552=245L ¼ 624T=245L: ð2:201Þ

Example 2.4

Dynamic testing of a continuous beam of the same proportions as shown in Figure 2.13, but

with supports that at either end were partially restrained from moving horizontally, yielded a

first resonance frequency of F¼ 0.753(EIg/wL4). Assuming the mode shape of vibration to be

geometrically similar to the deflected form caused by a concentrated load at the midpoint of

the central span, estimate the additional equivalent spring stiffness or geometrical stiffness

and the axial force caused by these restraints.

The geometrical stiffness may be found from the relationship

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKE þ KG

ME

� s¼ 0:753

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg

wL4

� s:

We therefore have

KG ¼ 0:753� 2�ð Þ2 EIg=wL4� �

� 1528wL=1715gð Þ � 96EI=7L3� �

¼ 6:2295606EI=L3

and from Equation 2.201

T ¼ 245L=624ð Þ � KG ¼ 245L=624ð Þ � 6:2295606EI=L3� �

¼ 2:4459012EI=L2:


46

2.6. First natural frequency of sway structuresThe most common types of sway structure are towers, chimneys and tall multi-storey buildings.

The dominant frequency of the first two can usually be assessed by considering them as cantilever

columns with constant or tapering sections, while the dominant mode and hence the frequency of

the last can be calculated by considering the sway of the columns between each floor level as a

lumped mass. A good approximation to the mode shapes of multi-storey buildings is to assume

them to be geometrically similar to the deflected forms caused by concentrated loads, each applied

horizontally at floor level and in magnitude equal to the weight of the floor. The mode shape

having been determined in this manner, the natural frequency corresponding to this mode

shape can be determined by equating the maximum kinetic energy of the lumped mass system

to the maximum strain energy stored in the columns. The details of the method are most easily

explained through examples.

2.6.1 Multi-storey shear structuresShear structures are structures in which the floors are so stiff compared to the columns that they

can be assumed to be rigid.

Example 2.5

Use an approximate method to determine the shape of the first mode of vibration and the first

natural frequency for the three-storey shear structure shown in Figure 2.14. The shear stiff-

ness k¼ 12EI/L3 is the same for all the columns and the weight of each of the three floors is w

per unit length. If w¼ 20.0 kN/m, calculate the value of the flexural rigidity EI to yield a

dominant frequency of 3.0 Hz. The weight of the columns may be neglected.

Figure 2.14 Three-storey shear structure with EI constant for all columns and the weight of all

floors equal to w/m

4 m

4 m

4 m

10 m 10 m 10 m

The mode shape is determined by applying a force at each floor level proportional to the

weight of each floor. The following horizontal force vector may therefore be used:

P ¼ 3:0 2:0 1:0f g:


47

The displacement due to P at level 1 is

x1 ¼ 3:0þ 2:0þ 1:0ð Þ=4k ¼ 1:5=k;

at level 2 is

x2 ¼ 1:5=kþ 2:0þ 1:0ð Þ=3k ¼ 2:5=k

and at level 3 is

x3 ¼ 2:5=kþ 1:0=2k ¼ 3:0=k:

The assumed mode shape vector is therefore

x ¼ 1:5 2:5 3:0f g:

The maximum kinetic energy is given by

KEmax ¼1

2

X3i¼ 1

mi!2x2i

¼ 12!

2 w=gð Þ 30� 1:52 þ 20� 2:52 þ 10� 3:02 �

¼ 141:25!2w=g:

The maximum strain energy is given by

Umax ¼1

2

X3i¼ 1

ki xi � xi� 1ð Þ2

¼ 12 k 4� 1:52 þ 3� 2:5� 1:5ð Þ2þ2� 3:0� 2:5ð Þ2 �

¼ 6:25k:

Equating the maximum kinetic energy and the maximum strain energy yields

6:25� 12EI=4:03 ¼ 141:25!2w=g

and hence

f ¼ 0:0144966ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wð Þ

p:



p;

implying an error in this case of only 0.94%. The flexural rigidity of the columns is now found

by substitution of the value of 3.0 Hz for f in the above expression for the natural frequency of

the structure,. We therefore have

3:0 ¼ 0:0144966ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEig=20:0Þð

p

and therefore EI ¼ 87 311:447 kN m2.

For low-rise shear structures, an even simpler (but less accurate) method for estimating the

dominant frequency is to lump the masses of all the floors together as an equivalent mass


48

at the first floor level, thus reducing the structure to a 1-DOF system. For the structure in

Figure 2.14, the equivalent mass is

ME ¼ w=gð Þ 10� 12þ 20� 8þ 30� 4ð Þ=4 ¼ 100w=g:

The total shear stiffness of the columns below the first floor level is

KE ¼ 4� 12EI=4:03 ¼ 3EI=4:0

and hence


p:

The error is therefore �4.28%.

An alternative method of estimating the first natural frequency is to consider the structure as

an equivalent 1-DOF system where the lumped mass is at the roof level. The equivalent mass

at this level is given by

ME ¼ w=gð Þ 30� 4:0þ 20� 8:0þ 10� 12:0f g=12 ¼ 400w=12g:

The equivalent spring stiffness KE is determined from

1

KE

¼ 1

k1þ 1

k2þ 1

k3¼ 1

4kþ 1

3kþ 1

2k¼ 13

12k

which yields

KE ¼ 12k

13¼ 12

13� 12EI

4:03¼ 9EI

52

and hence

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9EI � 12g

52� 400w

� s¼ 0:0114683

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wð Þ:

p

The error in this case is therefore�20.15%. The reason for this magnitude of error is that the

implied mode shape assumed in this simplification differs significantly from the real shape.

Example 2.6

Use an approximate method to estimate the equivalent mass, stiffness and dominant

frequency of a ten-storey shear structure, the data for which are given in Table 2.1.

Assume the distance between each floor is 3.0 m.

The first step is to calculate the mode shape vector for the first mode by applying a horizontal

load at each floor level proportional to the weight of the floor. As the weight of all the floors is

the same in this case, 1.0 kN may be applied at each level. The displacements at the various

levels are therefore

level 1: x1 ¼ 10� 1:0=2:5� 10�6 ¼ 4:000� 10�6 m


49

Table 2.1 Data for Example 2.6

Level Mass� 10�3: kg Stiffness� 10�6: N/m

1 1225.0 2500.0

2 1225.0 2500.0

3 1225.0 2500.0

4 1225.0 1024.0

5 1225.0 1024.0

6 1225.0 1024.0

7 1225.0 1024.0

8 1225.0 324.0

9 1225.0 324.0

10 1225.0 324.0

level 2: x2 ¼ 4:000� 10�6 þ 9:0� 1:0=2:500� 10�6 ¼ 7:600� 10�6 m

level 3: x3 ¼ 7:600� 10�6 þ 8:0� 1:0=2:500� 10�6 ¼ 10:800� 10�6 m

level 4: x4 ¼ 10:800� 10�6 þ 7:0� 1:0=1:024� 10�6 ¼ 17:636� 10�6 m

level 5: x5 ¼ 17:636� 10�6 þ 6:0� 1:0=1:024� 10�6 ¼ 23:495� 10�6 m

level 6: x6 ¼ 23:495� 10�6 þ 5:0� 1:0=1:024� 10�6 ¼ 28:378� 10�6 m

level 7: x7 ¼ 28:78� 10�6 þ 4:0� 1:0=1:024� 10�6 ¼ 32:284� 10�6 m

level 8: x8 ¼ 32:284� 10�6 þ 3:0� 1:0=0:324� 10�6 ¼ 41:543� 10�6 m

level 9: x9 ¼ 41:543� 10�6 þ 2:0� 1:0=0:324� 10�6 ¼ 47:716� 10�6 m

level 10: x10 ¼ 47:716� 10�6 þ 1:0� 1:0=0:324� 10�6 ¼ 50:802� 10�6 m

Dividing all the above displacements by x10 yields the mode shape vector

x ¼ 0:0787f 0:1496 0:2126 0:3471 0:4625

0:5586 0:6355 0:8177 0:9392 1:0000g

Substitution of these values together with the given values for masses and shear

stiffnesses yields the following values for the maximum kinetic energy and maximum strain

energy:

KEmax ¼1

2

X10i¼ 1

mi!2i x

2i ¼ 2:25081!2 � 106 Nm

Umax ¼1

2

X10i¼ 1

ki xi � xi� 1ð Þ2 ¼ 50:63286� 106 Nm:


50

Because x10 is assumed to be unity, the equivalent mass, spring stiffness and natural

frequency for studying the motion of the top of the building are:

M ¼ 4:50162� 106 kg;

KE ¼ 101:26572� 106 N=m and

f1 ¼ 0:7549 Hz;

respectively. The value for the first natural frequency obtained from an eigenvalue analysis

is f¼ 0.7419 Hz; the error is therefore 1.75%. Although the method yields a surprisingly

accurate value for the natural frequency, it is not advisable to use the assumed mode shape

when calculating the bending stresses in the columns. The reason for this will become clear

when the elements in the mode shape vector x are compared to those in the vector ~xx

below, which are calculated by an eigenvalue analysis.

x ¼ 0:0787f 0:1496 0:2126 0:3471 0:4625

0:5586 0:6355 0:177 0:9392 1:0000g

~xx ¼ 0:0488f 0:0971 0:1443 0:2559 0:3608

0:4564 0:5401 0:7602 0:9117 1:0000g

Simplification of the structure to a 1-DOF system with the mass concentrated at the first floor

level yields

ME ¼ 1:225� 106 30:0þ 27:0þ 24:0þ 21:0þ 18:0þ 15:0þ 12:0þ 9:0þ 6:0þ 3:0ð Þ=3:0

¼ 67:375� 106 kg

KE ¼ 2500� 106 N=m

and hence

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2500� 106=67:375� 106ð Þ

q¼ 0:9695 Hz:

The error is therefore +30.68%, which is obviously not good enough.

Simplification of the structure to a 1-DOF system with the mass concentrated at roof level

yields

ME ¼ 1:225� 106 30:0þ 27:0þ 24:0þ 21:0þ 18:0þ 15:0þ 12:0þ 9:0þ 6:0þ 3:0ð Þ=30:0

and hence

ME ¼ 6:7375� 106 kg:

The equivalent spring stiffness KE is found from

1

KE

¼ 1

k1þ 1

k2þ 1

k3þ 1

k4þ 1

k5þ 1

k6þ 1

k7þ 1

k8þ 1

k9þ 1

k10


51

2.6.2 Multi-storey structures with flexible floorsEstimates of the dominant frequencies of all buildings that cannot be regarded as shear structures

require that the joint rotations and deformation of the floors, which reduce the overall shear stiff-

ness of a structure, be taken into account. In such cases, it is first necessary to calculate equivalent

shear stiffness by modifying the shear stiffness of the columns at each level by multiplying them by

a reduction factor, which is a function of both the column and the beam rigidities. We then assume

the mode shapes to be similar to the deflected forms caused by concentrated loads applied hori-

zontally at each floor level, whose magnitudes are equal to the weight of the floor at which they are

applied. With some assumptions, the expression for the reduction factor can be shown to be

RF ¼P

EI=Lð ÞbeamsPEI=Lð Þbeams þ 1

2

PEI=Lð Þcolumns

: ð2:202Þ

The error in the above factor depends on the degree of fixity at the foundations, the distribution of

the EI/L values of beams and columns at different floor levels and the size of the frame. For

normally proportioned structures, Equation 2.202 gives reasonable values.

In the following example, the above method is used to determine the first natural frequencies of

two structural models whose frequencies had previously been determined by eigenvalue analysis

using computers and resonance testing.

1

KE

¼ 3

2500� 106þ 4

1024� 106þ 3

324� 106¼ 0:0143655� 10�6

and hence

KE ¼ 69:611176� 106 N=m

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi69:611176� 106=6:7375� 106ð Þ

q¼ 0:5118 Hz:

The error in this case is therefore –31.01%.

Example 2.7

Figure 2.15 shows a steel frame model of a three-storey structure. The values of the EI/L in

Nm are marked against each member. The mass at the first floor level is 0.4228 kg, at the

second level 0.3979 kg and at roof level 0.2985 kg. Calculate the equivalent mass and

spring stiffness at roof level and hence the first natural frequency of the frame.

The sway stiffness k at each level is found by first calculating the sum of the shear stiffnesses

of the columns at each level, assuming zero joint rotations, and then multiplying this stiffness

by a reduction factor calculated in accordance with Equation 2.202. We therefore have

kr ¼XNi¼ 1

12EIiL3i

( )RF


52

Figure 2.15 Steel frame model of a three-storey structure

0.2286 m

0.304 m

0.304 m

4.269

3.202

4.803

4.269

4.803

6.404

4.269

4.803

6.404

4.269

3.202

4.803

2.135 2.135 2.135

2.135 2.135 2.135

3.202 3.202 3.202

3 at 0.4572 m = 1.3716 m

where N is the number of columns between floor levels r and (r� 1). Thus, at level 1:

k1 ¼12 2� 4:803ð Þ þ 2� 6:404ð Þ½ �

0:30482� 3� 3:202

3� 3:202ð Þ þ 4:803þ 6:404½ � ¼ 1336:2219 N=m;

at level 2:

k2 ¼12 2� 3:202ð Þ þ 2� 4:803ð Þ½ �

0:30482� 3� 2:135

3� 2:135ð Þ þ 3:202þ 4:803½ � ¼ 919:1742 N=m;

and at roof level:

k3 ¼12 4� 4:269ð Þ

0:22862� 3� 2:135

3� 2:135ð Þ þ 2� 4:269ð Þ½ � ¼ 1680:7235 N=m:

Assuming the first mode shape to be geometrically similar to the deflected form caused by the

load vector

P ¼ 0:4228 0:3979 0:2985 �

� 9:81 N

yields

x1 ¼ 0:4228þ 0:3979þ 0:2985ð Þ � 9:81=1336:2219 ¼ 8:2167� 10�3 m;

x2 ¼ 8:21671� 10�3 þ 0:3979þ 0:2985ð Þ � 9:81=919:1742 ¼ 15:6491� 10�3 m;

x3 ¼ 15:6491� 10�3 þ 0:2985� 9:81=1680:7235 ¼ 17:3914� 10�3 m:

The mode shape vector if the amplitude at level 3 is taken as unity is therefore:

x ¼ 0:4725 0:8998 1:0000 �


53

and hence the kinetic energy is

KE ¼ 12 0:4228� 0:47252� �

þ 0:3979� 0:89982� �

þ 0:2985� 1:02� ��

!2

¼ 0:3575242!2 N=m:

The corresponding strain energy is

U ¼ 12 1336:2219� 0:47252� �

þ 919:1742 0:8998� 0:4725ð Þ2þ1680:7235 1:0� 0:8998ð Þ2� �

¼ 241:5111 Nm

and

fn ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi241:5111

0:3575242

� s¼ 4:1365 Hz:

The correct value obtained from a computer analysis is 3.630 Hz, indicating an error of

þ13.953% which is not particularly good.

An alternative method of estimating the first natural frequency is to determine the stiffness

and mass of an equivalent 1-DOF system. The equivalent mass at roof level is given by

ME ¼ 0:2985þ 0:3979� 0:6096=0:8382þ 0:4228� 0:3048=0:8382

¼ 0:7435 kg;

the equivalent spring stiffness KE is found from

1

KE

¼ 1

k1þ 1

k2þ 1

k3¼ 1

1336:2219þ 1

919:1742þ 1

1680:7235

¼ 411:30368 N=m

and the frequency by

f ¼ 1

2�


0:7435

� s¼ 3:743 Hz:

The error in this case is only þ2.83%.

Example 2.8

The overall dimensions of the five-storey model shown in Figure 2.16 are 1000� 200�200 mm. Floor levels are 200mm apart. The mass at each level is represented by four steel

cubes of dimensions 40� 40� 40 mm, each cube weighing 0.483 kg. The columns consist

of 5 mm diameter steel rods and the beams of 3 mm diameter rods. The total weight of the

columns is 0.628 kg and that of the beams is 0.200 kg. The second moment of inertia of the

columns is 30.7 mm4 and that of the beams is 3.98 mm4.


54

Figure 2.16 Five-storey flexible steel model with four lumped masses attached at each floor level

200 mm

200 mm

200 mm

200 mm

200 mm200 mm

Determine the dominant frequency by first modelling the structure as a 1-DOF system

with the lumped mass at the top of the model, and then by assuming the mode shape to be

geometrically similar to the deflected form caused by unit point loads applied horizontally

at each floor level. Assume the modulus of elasticity for steel to be 205 kN/mm2. The

measured natural frequency was 2.105 Hz and the calculated frequency using a computer

was 1.95 Hz.

The calculation of the equivalent lumped mass at the top of the model is

ME ¼ 10:488

5

1000

1000þ 800

1000þ 600

1000þ 400

1000þ 200

1000

� ¼ 6:2928 kg;

for which an expression is given by Equation 2.202. We therefore have

RF ¼ 2� 205� 3:98ð Þ=2002� 205� 3:98ð Þ=200þ 0:5� 4� 205� 30:7ð Þ=200 ¼ 0:1147635:


55

The stiffness at each floor level is given by

K ¼ 4� 12EI

L3�RF ¼ 4� 12� 205� 30:7

2003� 0:1147635 ¼ 4:33358� 10�3 kN=mm:

The equivalent spring stiffness at the top of the model can be determined from the equality

1

KE

¼ 1

Kþ 1

Kþ 1

Kþ 1

Kþ 1

K¼ 5

K

and therefore

KE

4:3358� 10�3

5¼ 8:66714� 10�4kN=mm ¼ 866:14 N=m:

The natural frequency of the equivalent mass–spring system is therefore

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi866:714

6:928

� s¼ 1:8678 Hz:

The percentage error compared with the theoretical value obtained from an eigenvalue

analysis is therefore

1:8678� 1:9490

1:9490¼ �4:166%:

The mode shape assumed to be geometrically similar to the deflected form caused by a unit

horizontal point load applied at each floor level is given by

x1 ¼ 5=K

x2 ¼ 5=K þ 4=K

x3 ¼ 5=K þ 4=K þ 3=K

x4 ¼ 5=K þ 4=K þ 3=K þ 2=K

x5 ¼ 5=K þ 4=K þ 3=K þ 2=K þ 1=K

¼ 5=K

¼ 9=K

¼ 12=K

¼ 14=K

¼ 15=K

Division by 5/K yields the mode shape vector

x ¼ 1:0 1:8 2:4 2:8 3:0 �

:

The maximum kinetic energy corresponding to this mode shape is given by

KE ¼ 1

2� 10:488

5!2 1:02 þ 1:82 þ 2:42 þ 2:82 þ 3:02� �

¼ 28:149792!2 Nm:

The corresponding maximum strain energy is

U ¼ 12 � 4333:58� 1:02 þ 1:8� 1:0ð Þ2þ 2:4� 1:8ð Þ2þ

�2:8� 2:4ð Þ2þ 3:0� 2:8ð Þ2

�

¼ 4766:38Nm


56

2.7. PlatesTheoretically, it is possible to determine the first frequency of a plate in the same manner as for

beams by assuming the mode shape to be geometrically similar to that caused by a uniformly and/

or concentrated applied load. However, because both the geometry and the support conditions

can vary considerably from plate to plate, the above approach is not practical except for circular

plates and for rectangular plates simply supported at each corner. For these two cases, it is

possible to calculate a first frequency by assuming the mode shape to be sinusoidal. For plates,

therefore, the reader is referred to handbooks on vibration such as Harris (1988) which give

extensive lists of expressions for the frequencies of plates with different geometries and support

conditions.

2.8. Summary and conclusionsThe first part of this chapter presents a method for determining the dominant frequencies of

beams with uniform and concentrated loading by assuming that the mode shape is geometrically

similar to the deflected form caused by a concentrated and distributed load. The assumed mode

shape is then used to determine the lumped mass and spring stiffness of an equivalent 1-DOF

system using Equations 2.8 and 2.13. It is then shown that the equivalent masses and spring stiff-

ness for beams can be used to estimate the dominant frequencies of continuous beams and how the

accuracies of such calculations are critically dependent on the assumed mode shapes. The method

used is extended to sway structures with both rigid and flexural floors.

In all the examples, the degree of accuracy obtained is given. These examples indicate that the

accuracy in the calculated frequencies depends on the assumed mode shapes. Thus, for the

structure in Example 2.8, it is found that the first mode shape obtained using a finite element

program closely approximates the average of the two mode shapes assumed. When this mode

shape is used, the percentage difference between the theoretically correct frequency and that

obtained by equating the maximum kinetic energy to the maximum strain energy is –1.35%.

The accurate calculation of structural frequencies requires numerical modelling and the use of

a computer. Data preparation takes time, and experience has shown that it is useful to be able

to make an initial estimate of the fundamental frequency at the design stage as well as when

checking the values from any computer analysis.

One question remains: why should a designer wish to change a structure that is designed to with-

stand the static forces stipulated in various design codes? The answer is given in Chapter 1, where

it is pointed out that dynamic forces such as wind, waves and earthquakes may be considered to

consist of a large number of harmonic components with different frequencies and varying levels

of energy. Structures with dominant frequencies that lie within a high-energy frequency band

therefore may respond in resonance. This may have (and has had) catastrophic consequences.

and hence

f ¼ 1

2�


28:149792

� s¼ 2:0711 Hz:

Percentage error ¼ 2:711� 1:9490

1:9490¼ þ6:265%


57

Now we have shown how many structures and structural elements can be reduced to a 1-DOF

mass–spring system, Chapters 3–6 are devoted to the free and forced vibration of such systems.

Problem 2.1

A simply supported beam of length L, flexural rigidity EI and self-weight wL supports three

concentrated loads, each of weight wL, at positionsL/4, L/2 and 3L/4 along the span. Assume

the mode of vibration to be geometrically similar to the deflected shape caused by a uniformly

distributed dead load, and develop an expression for the natural frequency of the equivalent

mass–spring system in terms of w, L, EI and g.

Problem 2.2

In the portal frame structure shown in Figure 2.17, the floor BDF is assumed to be rigid. The

columns AB, CD and EF are uniform with the same flexural rigidity EI. The weight of the

floor is w per metre in length and the weight of the columns is w/10 per metre in length.

Develop an expression for the natural frequency of the structure in terms of EI, w and g.

Figure 2.17 Portal frame structure with rigid floor slab

A C E

B10.0 m 10.0 m

6.0 m

D F

Problem 2.3

A continuous beam ABCD of uniform cross-section, flexural rigidity EI and self-weight w per

unit length has three equal spans, each of length L. The beam is simply supported at B and C

but is built-in at A and D. Use an approximate method to estimate the first and second

natural frequencies of the beam.

Problem 2.4

Develop an expression for the dominant frequency of the five-storey shear structure shown in

Figure 2.17 in terms of the weight of the floor slabs and the flexural rigidity of the columns.

Each floor slab weighs w per unit length of span, and all columns have the same flexural

rigidity EI.


58

REFERENCES


FURTHER READING


Cambridge.


Craig RR Jr (1981) Structural Dynamics. Wiley, Chichester.

Irvine HM (1986) Structural Dynamics for the Practising Engineer. Allen & Unwin, London.

Stroud KA (1970) Engineering Mathematics. Macmillan, London.

Timoshenko SP and Gere JM (1972) Mechanics of Materials. Van Nostrand Reinhold,

New York.

Problem 2.5

Determine the first mode shape of the structural model in Example 2.8 by assuming it to be

geometrically similar to the average of the deflected forms caused by only one horizontally

applied load at the top of the model, and the form caused by equal horizontal concentrated

loads applied at all five levels. Hence, calculate the natural frequency of the model by equating

the maximum kinetic energy to the maximum strain energy occurring for the givenmode shape.

Problem 2.6

What is the expression for the dominant frequency for the structure shown in Figure 2.18, if it

is assumed that the flexural rigidities of all the floors are the same and in turn equal to 5, 10

and 20 times those of the columns? By what percentage will the frequency be reduced in each

case relative to the frequency obtained for the case when all the floors are assumed to be rigid?

Figure 2.18 Five-storey shear structure

3L 3L 3L 3L

L

L

L

L

L


59


ISBN: 978-0-7277-4176-9



Chapter 3

Free vibration of one degree-of-freedomsystems

3.1. IntroductionIn the previous chapter, it was shown how a large number of different types of structures and

structural elements could be modelled as lumped mass–spring systems. The free vibrations of

such systems, both with and without damping, are investigated in this chapter. Expressions for

equivalent structural damping are developed and we examine how it can be measured and how

damping affects the natural frequencies.

3.2. Free un-damped rectilinear vibrationConsider the system shown in Figure 3.1. If the static deflection due to the weight of the lumped

mass is

� ¼ W=K ¼ Mg=K ð3:1Þ

then, from Newton’s law of motion,

M€xx ¼ W � K �þ xð Þ ð3:2Þ

and hence

M€xxþ Kx ¼ 0 ð3:3Þ

or

€xxþ !2nx ¼ 0 ð3:4Þ

where

!2n ¼ K=M: ð3:5Þ

The solution to Equation 3.4 is given by

x ¼ A cos !ntð Þ þ B sin !ntð Þ: ð3:6Þ

From Equation 3.5, it can be seen that the vertical motion of the mass M has a vibratory char-

acter, since both sin(!nt) and cos(!nt) are periodic functions which repeat themselves after an

interval of time T such that

!n T þ tð Þ � !nt ¼ 2�: ð3:7Þ

61

The time T is called the period or periodic time of vibration. From Equations 3.1, 3.5 and 3.7 it

follows that

T ¼ 2�

!n

¼ 2�

ffiffiffiffiffiffiffiffiffiffiffiffiM

K

� �s¼ 2�

ffiffiffiffiffiffiffiffiffiffiffi�

g

� �s: ð3:8Þ

The number of cycles per second is, as seen in Chapter 2, referred to as the frequency of vibration;

one cycle per second is referred to as 1 Hertz (Hz). Then

f ¼ 1

T¼ !n

2�¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiK

M

� �s¼ 1

2�

ffiffiffiffiffiffiffiffiffiffig

�

� �rð3:9Þ

where f is the frequency and !n is the natural angular frequency in rad/s.

3.2.1 Examination of Equation 3.6For structures that vibrate with SHM, the velocity _xx is a maximum when x¼ 0 and the accelera-

tion €xx is a maximumwhen x¼ xmax¼x0. With this information, the values for the constantsA and

B in Equation 3.6 can be determined by choosing convenient starting points for the motion.

t¼ 0 when x¼ 0

Substitution of x¼ 0 and t¼ 0 into Equation 3.6 yields A¼ 0. Hence

x ¼ B sin !ntð Þ: ð3:10Þ

The amplitude x is a maximum when sin(!t)¼ 1.0; therefore B¼ x0 and

x ¼ x0 sin !ntð Þ ð3:11aÞ

Figure 3.1 System of free un-damped rectilinear vibration

M

x

∆


62

_xx ¼ x0!n cos !ntð Þ ð3:11bÞ

€xx ¼ �x0!2n sin !ntð Þ: ð3:11cÞ

t¼ 0 when x¼ x0When x¼ x0 and t¼ 0, A¼x0. Substitution of this value for A into Equation 3.6 and differentia-

tion with respect to t yields

_xx ¼ �x0!n sin !ntð Þ þ B!n cos !ntð Þ: ð3:12Þ

When t¼ 0 and _xx ¼ 0, B¼ 0 and hence

x ¼ x0 cos !ntð Þ ð3:13aÞ

_xx ¼ �x0! sin !ntð Þ ð3:13bÞ

€xx ¼ �x0!2n cos !ntð Þ: ð3:13cÞ

t¼ 0 when x¼�, 0<�< |x0|

When t¼ 0 and x¼�,A¼�. Substitution of this value forA into Equation 3.6 and differentiation

with respect to t yields

_xx ¼ ��!n sin !ntð Þ þ B!n cos !ntð Þ: ð3:14Þ

When t¼ 0 and _xx ¼ _��, B ¼ _��=!n and hence

x ¼ � cos !ntð Þ þ _��

!n

sin !ntð Þ; ð3:15Þ

that is, the amplitude x at time t is a function of the amplitude � and velocity _�� at time t¼ 0.

3.2.2 Equivalent viscous dampingEnergy is dissipated in the form of heat during vibration, and a steady amplitude of vibration

cannot be maintained without its continuous replacement. The heat is generated in a number

of different ways: by dry and fluid friction, by hysteresis effect in individual components (internal

friction), in concrete by the opening and closing of hair cracks and by magnetic, hydrodynamic

and aerodynamic forces. The different forces that contribute to the damping of a structure may

vary with amplitude, velocity, acceleration and stress intensity, and are difficult (if not impossible)

to model mathematically. However, ideal models of damping have been conceived which

represent satisfactory approximations. Of these, the viscous damping model (in which the

damping force is proportional to the velocity) leads to the simplest mathematical treatment

and generally the most satisfactory results, provided damping forces caused by hydrodynamic

and aerodynamic forces (when significant) are taken into account separately.

As a viscous damping force is proportional to the velocity of vibration at any time t, it can be

expressed as

Fd ¼ C _xx ð3:16Þ

whereC is the constant of proportionality or coefficient of viscous damping and _xx is the velocity of

vibration at time t. The coefficient of viscous damping is numerically equal to the damping force

Free vibration of one degree-of-freedom systems

63

when the velocity is unity; the unit of C is therefore Ns/m. Symbolically, viscous damping is

designated by a dashpot as shown in Figure 3.2.

3.3. Free rectilinear vibration with viscous dampingConsider a vibrating damped lumped mass–spring system. If the motion of the mass is resisted by

forces that are proportional to the velocity of the mass then the resisting forces may be assumed to

have viscous characteristics, in which case the damping mechanism is denoted by a dashpot as

shown in Figure 3.2.

From Newton’s law of motion,

M€xx ¼ W � K �þ xð Þ � C _xx ð3:17Þ

or

M€xxþ C _xxþ Kx ¼ 0: ð3:18Þ

Assume that Equation 3.18 is satisfied by a function of the form

x ¼ A est: ð3:19Þ

Substitution of this function into Equation 3.18 yields

Ms2A est þ CsA est þ KA est ¼ 0: ð3:20Þ

Division of each term in Equation 3.20 by MAest yields

s2 þ 2nsþ !2n ¼ 0 ð3:21Þ

where

n ¼ C=2M

Figure 3.2 Lumped mass–spring system with viscous damping

X

∆M

K C


64

and

! ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK=Mð Þ

p

From Equation 3.21 it follows that

s ¼ �n�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin2 � !2

nð Þq

: ð3:22Þ

Thus, when studying free damped vibration further it is necessary to consider the cases

n ¼ !n; n > !n and n < !n

or, alternatively, since n¼C/2M and !n¼p(K/M), the cases when

C ¼ 2ffiffiffiffiffiffiffiffiffiffiffiffiffiKMð Þ

p

C > 2ffiffiffiffiffiffiffiffiffiffiffiffiffiKMð Þ

p

C < 2ffiffiffiffiffiffiffiffiffiffiffiffiffiKMð Þ

p

should be considered.

The term 2p(KM) is referred to as critical damping and is denoted Cc. The ratio C/Cc is called the

damping ratio, noting that �. n¼ �!n, and replacing for Cc results in the expression for C as

C ¼ 2�!nM ð3:23Þ

Substitution of n¼ �!n into Equation 3.22 yields

s ¼ ��!n � !n

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�2 � 1� �q

ð3:24Þ

or

s ¼ ��!n � !in

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q

: ð3:25Þ

From Equations 3.24 and 3.25, it follows that the motion of free damped vibration needs to be

studied for the cases when

� > 1:0

� ¼ 1:0

� < 1:0:

When � > 1.0, Equation 3.25 becomes

s ¼ ��!n � !np

and hence

x ¼ e��!nt C e!npt þD e�!npt� �

: ð3:26Þ


65

Equation 3.26 is not a periodic function and therefore does not represent a periodic motion. If

displaced from its position of equilibrium, the mass will gradually return to its original position.

When �¼ 1.0, Equation 3.25 becomes

s ¼ ��!n

and hence

x ¼ C e��!nt: ð3:27Þ

Equation 3.27 does not contain a periodic function either, and therefore does not represent a

periodic motion. The mass will return to its position of equilibrium if displaced, but more quickly

than in the case for � > 1.0.

When � < 1.0, the concept of damped angular natural frequency is now introduced, defined

!d ¼ !n


ð3:28Þ

and hence

s ¼ ��!n � i!d

x ¼ e��!nt C ei!dt þD e�i!dt� �

: ð3:29Þ

Since

eþi!dt ¼ cos !dtð Þ þ sin !dtð Þ

e�i!dt ¼ cos !dtð Þ � sin !dtð Þ;

it follows that

x ¼ e��!nt A cos !dtð Þ þ B sin !dtð Þ½ �: ð3:30Þ

Equation 3.30 represents a periodic function. Comparison of the period of this function with that

for un-damped free vibration given by Equation 3.6 shows that the period T increases from 2�/!n

to (2�/!n)p(1� �2). When the value of � is small compared to 1.0 the increase is negligible. In

most practical problems it can be assumed with sufficient accuracy that the damping does not

affect the period of vibration, which can be assumed to be equal to 2�/!n.

In order to determine the constants A and B in Equation 3.30, let x¼xmax¼x0 and _xx ¼ 0 when

t¼ 0. This yields:

x ¼ x0 e��!nt cos !dtð Þ þ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1� �2� �q sin !dtð Þ

24

35: ð3:31Þ

From Equation 3.31, it can be seen that for every period T, or for every cycle of vibration, the

amplitude is diminished by the ratio

x0 e��!nT : x0: ð3:32Þ


66

Thus, if xr is the amplitude at the end of the rth oscillation and xs is the amplitude at the end of the

sth oscillation, then

xrxs

¼ x0 e��!nrT

x0 e��!nsT

ð3:33Þ

or

xrxs

¼ e�!n s� rð ÞT ð3:34Þ

and hence

�!nT ¼ 1

s� rlnxrxs

: ð3:35Þ

The product �!nT is referred to as the logarithmic decrement of damping and is usually denoted

by �. We therefore have

� ¼ 1

s� rlnxrxs

: ð3:36Þ

When r¼ 0 and s¼ n,

� ¼ 1

nln

x0xn: ð3:37Þ

As the period for damped vibration is

T ¼ 2�

!n

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q ; ð3:38Þ

it follows that

� ¼ 2��ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q ð3:39Þ

from which we obtain

��ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4�2 þ �2� �q : ð3:40Þ

Since � is usually much smaller than 2�, Equation 3.40 may be simplified and written as

� ¼ �

2�: ð3:41Þ

The motions for systems with � > 1.0 and � < 1.0 are shown in Figures 3.3 and 3.4, respectively.


67

3.4. Evaluation of logarithmic decrement of damping from thedecay function

Decay functions of the type shown in Figure 3.4 can be obtained by the sudden release of a load

from a structure, by vibrating it at resonance and then stopping the vibrator and recording the

ensuing motion of a pen-recorder or by means of a computer (using either an accelerometer or

a displacement transducer).

The expression for the logarithmic decrement of damping given by Equation 3.41 assumes that the

resulting structural damping mechanism has the characteristics of viscous damping, i.e. that the

damping force resisting the motion at any time is proportional to the velocity of vibration.

This assumption can be checked by plotting the values of ln (x0/xn) against n, the number of oscil-

lations, as shown in Figure 3.5. The plotted values will lie along a straight line if the damping is

proportional to the velocity, and along a curve if it is not. In the case of the former, the damping is

independent of the amplitude of response; in the latter case it is dependent upon the amplitude of

response.

Since �¼ ln(x0/xn)/n, it follows that the slope of the straight line is equal to � and that the slope of

any tangent to a curved line is the value of � for the amplitude of vibration corresponding to the

contact point between the tangent and the curve.

Figure 3.3 Motion of a lumped mass–spring system with viscous damping ratio � > 1.0

t

x0

Figure 3.4 Diagram showing the motion of a lumped mass–spring system with viscous damping ratio

� < 1.0

t

x0

x0 e–ξωnt


68

Figure 3.5 Ln(x0/xn) plotted against n for structures with (a) viscous and (b) non-viscous damping

(a) (b)

ln(x

0/x

n)

n n

Example 3.1

A tubular steel antenna-mast supporting a 3.0 m diameter disc is deflected by tensioning a

rope attached to its top and then set in motion by cutting the rope. The first part of the

subsequent motion, recorded by the use of an accelerometer, is shown in Figure 3.6. The

estimated spring stiffness at the top of the mast is 30.81 kN/m. Calculate first the logarithmic

decrement of damping, the damping ratio and the damped and un-damped natural

frequencies of the mast, and then the equivalent mass and damping coefficient for the

generalised mass–spring system.

Figure 3.6 Pen-recorder trace of the motion at the top of a 20 m tall antenna mast (note that

€xx0 ¼ 258 mm=s2; €xx1 ¼ 226 mm=s2; €xx2 ¼ 199 mm=s2; €xx3 ¼ 176 mm=s2)

2 4 6 8 10 12 14

x0 x1 x2 x3

Time: s

Acc

eler

atio

n: m

m/s

2

The logarithmic decrement of damping � is given by

� ¼ 1

nln

x0xn

¼ 1

3ln258

176¼ 0:1274918

and hence

� ¼ 12:75%:


69

3.5. Free un-damped rotational vibrationNewton’s law of motion states that

force ¼ mass� acceleration:

Similarly, d’Alembert’s principle states that

moment of force ¼ polar moment of inertia� angular acceleration:

We therefore have

T ¼ �Ip €��: ð3:42Þ

If it is assumed that the forcing moment T strains a bar element in pure St Venant torsion and

rotates one end of it through an angle �, then

T ¼ Kt� ð3:43Þ

The damping ratio � is given by

� ¼ �

2�¼ 0:1274918

2�¼ 0:0202909

and hence

� ¼ 2:03%:

The damped and un-damped natural frequencies are given by

fd ¼ 3

14� 2¼ 0:25 Hz

fn ¼ fdffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q ¼ 0:25ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1� 0:02029092ð Þp ¼ 0:2501029 Hz:

The damping therefore only has a negligible effect on the natural frequency, and the differ-

ence between the two values is certainly less than any expected accuracy in the measured

value.

The equivalent lumped mass at the top of the mast is

M ¼ K

4�2f 2n¼ 30810

4�2 � 0:25010292¼ 12476:543 kg:

The damping coefficient C for the equivalent mass–spring system is therefore given by

C ¼ 2�!nM ¼ 2� 0:0202909� 0:2501029� 2�ð Þ � 12476:543

¼ 795:654 N s=m:


70

where Kt is the torsional stiffness. Substitution of this expression for the moment of force into

Equation 3.42 yields

Ip €��þ Kt� ¼ 0: ð3:44Þ

If it is further assumed that the motion represented by Equation 3.44 is simple harmonic, i.e. that

� ¼ � sin !tð Þ

€�� ¼ �’!2 sin !tð Þ;

it follows that

�Ip�!2 þ Kt� ¼ 0 ð3:45Þ

and hence that

!n ¼ffiffiffiffiffiffiffiffiffiffiffiffiKt

Ip

� �s: ð3:46Þ

Multiplication of each term in Equation 3.45 by ’=2 yields

12Kt’

2 ¼ 12 Ip’

2!2: ð3:47Þ

Thus, as in the case of rectilinear vibration, the maximum strain energy is equal to the maximum

kinetic energy when energy losses due to damping are neglected.

For a cylindrical element of length L, radius R, mass M and specific density �, the polar moment

of inertia is

Ip ¼ 12 ��LR

4 ¼ 12MR2 ð3:48Þ

Kt ¼�GR4

2L: ð3:49Þ

For a bar element with length L, massM and rectangular cross-sectional area of dimensions a� b,

the polar moment of inertia about the central axis is

Ip ¼ 112M a2 þ b2

� �ð3:50Þ

Kt ¼Gab a2 þ b2

� �12L

: ð3:51Þ

The strain energy stored in the same element when subjected to a forcing moment or torque T

causing a differential end rotation � is

U ¼ T�

2¼ T2L

2GJ¼ GJ�2

2Lð3:52Þ

where J is the second polar moment of area about the central axis. Equation 3.52 implies that

Ip¼ JL�.


71

3.6. Polar moment of inertia of equivalent lumped mass–springsystem of bar element with one free end

The equivalent polar moment of inertia of the mass–spring system shown in Figure 3.7(b) (whose

rotational vibration represents the vibration of the free end of the bar shown in Figure 3.7(a)) can

be determined by equating the rotational kinetic energy of the lumped mass–spring system to that

of the bar. If the angular velocity at the free end of the bar at time t is _�� tð Þ and Ipe is the equivalent

polar moment of inertia of the lumped mass, then

1

2Ipe _��

2tð Þ ¼

ðL0

1

2

Ip

L

_�� tð ÞxL

( )2

dx ð3:53Þ

and hence

Ipe ¼ 13 Ip: ð3:54Þ

Figure 3.7 Equivalent rotational mass–spring system of linear elastic element vibrating about its own

central axis

(c)(b)(a)

Me

Kt ωt

φM

K

Example 3.2

An 8m diameter circular post-tensioned concrete platform is 0.25 m thick and is supported

centrally on a 5.0 m tall circular hollow concrete column with an external diameter of

2.0 m and an internal diameter of 1.5 m. Calculate the natural rotational frequency. The

specific density of concrete is 2400 kg/m3 and the modulus of rigidity is 12.0 kN/mm2.

The rotational stiffness of the hollow column is given by

Kt ¼�G R4 � r4� �2L

¼�� 12:0� 106 1:04 � 0:754

� �2� 5:0

¼ 2:5770877� 106 kN m=rad:


72

The mass of the 8.0 m diameter platform is

M ¼ ��R2t ¼ 2400:0� �� 4:02 � 0:25 ¼ 30 159:289 kg

and the polar moment of inertia of the platform about its central support is therefore

Ip ¼ 12MR2 ¼ 1

2 � 30 159:289� 4:02 ¼ 241 274:31 kg m2:

The additional equivalent polar moment of inertia of the column is

Ipe ¼ 13 Ip ¼ 1

3 � 12 �� R4 � r4

� �L

¼ 13 � 1

2 � 2400:0� �� 1:04 � 0:75� �4�5:0 ¼ 4295:1462 kg m2:

The rotational stiffness of the column is

Kt ¼�G R4 � r4� �2L

¼�� 12:0� 106 1:04 � 0:754

� �2� 5:0

¼ 2:5770877� 106 N m=rad:

The natural rotational angular frequency is therefore

!n

ffiffiffiffiffiffiffiffiffiffiffiffiKt

Ip

� �s¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2:5770877� 106

241 274:31þ 4295:1462

� �s¼ 3:2394958 rad=s

and hence

fn ¼ !n

2�¼ 0:5155817 Hz:

Example 3.3

The square 40 m� 40m platform shown in Figure 3.8 is supported on four 45 m tall hollow

circular concrete columns. The external diameters of the columns are 4.0 m and the internal

diameters 3.0 m. The columns are spaced 20 m centre to centre as shown in the figure, and

may be considered to be rigidly fixed at the base and to the platform. The mass of the

platform is 3.84� 106 kg, and it is so stiff that it may be assumed to be rigid. Calculate the

natural lateral and rotational frequencies of the structure. The specific density of concrete

is 2400 kg/m3, the modulus of elasticity is 30.0 kN/mm2 and the shear modulus of elasticity

12.0 kN/mm2.

Determination of the natural frequency of the lateral motion

The second moments of the cross-sectional areas of the columns are

I ¼ 14� R4 � r4� �

¼ 14� 2:04 � 1:54� �

¼ 8:5902924 m4:


73

Figure 3.8 Elevation and plan view of the 40m� 40m platform in Example 3.3

40 m

40 m

45 m

20 m

4.0 m

The shear stiffness of each column is

K ¼ 12EI

L3¼ 12� 30:0� 106 � 8:5902924

45:03¼ 33:936958� 103 kN=m:

The equivalent mass at the top of each column is

Me ¼ 13=35ð ÞwL ¼ 13=35ð Þ� 2:02 � 1:52� �

� 2400:0� 45:0

¼ 220 539:8 kg:

The total equivalent mass at a height of 45.0 m is therefore

Me ¼ 3:84� 106 þ 4� 0:2205398� 106 ¼ 4:7221592� 106 kg

and hence the lateral natural frequency of the platform is

Fn; lateral ¼1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4� 33:936958� 106

4:7221592� 106

� �s¼ 0:8533 Hz:

Determination of the natural frequency of the rotational motion

The polar moment of inertia of the platform deck is given by

Ip;deck ¼ 112M a2 þ b2

� �¼ 1

12 � 3:84� 106 40:02 þ 40:02� �

¼ 1024:0� 106 kg m2:

The equivalent second polar moment of inertia of each column about its own axis at 45 m

above the supports is

Ipe; column ¼ 13 Ip; column ¼ 1

3 � 12 �� R4 � r4

� �L

¼ 13 � 1

2 � 2400� � 2:04 � 1:54� �

� 45:0


74

and hence

Ipe; column ¼ 618 501:05 kg m2:

The equivalent second polar moment of inertia of each column about the central axis of the

structure 45 m above the supports is determined by using the theorem of parallel axes and

adding the polar moment of inertia of the equivalent mass at the top of each column due

to the shear deformation. We therefore have

Ip;column ¼ Ipe;column þ1

3Mcolumn � h2 þ 13

35Mcolumnh

2

and hence

Ip;column ¼ Ipe þ74

105Mcolumn � h2

Ip; column ¼ 618 501:05þ 74

105� 2400� �� 2:02 � 1:52

� �

¼ 84:310529� 106 kg m2

and therefore

Ip; total ¼ 1024:0� 106� �

þ 4� 84:310529� 106� �

¼ 1361:2421� 106 kg m2:

The equivalent rotational spring stiffness is most conveniently determined by equating the

strain energy stored in the spring to that stored in the columns due to bending and torsion

when both the structure and the spring are rotated through an angle �. We therefore have

1

2Kt; spring’

2 ¼ 41

2� 12EI

L3h’ð Þ2þ 1

2��G R4 � r4� �2L

’2

" #

Kt; spring ¼ 412EIh2

L3þ�G R4 � r4� �2L

" #

Kt; spring ¼ 412� 30:0� 106 � 8:5902924� 200

45:03þ�� 12:0� 106 � 2:04 � 1:54

� �2� 45:0

" #

¼ 45:475523� 106 kN m=rad:

The rotational frequency can now be found by substitution of the values for the equivalent

rotational spring stiffness and total polar moment of inertia into Equation 3.46; we therefore

have

fn; rotational ¼1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi45:475523� 109

1:3612421� 109

� �s¼ 0:9199 Hz:


75

3.7. Free rotational vibration with viscous dampingFrom d’Alembert’s principle, it follows that

Ip €�� ¼ �Kt�� Ct_�� ð3:55Þ

or

Ip €��þ Ct_��þ Kt� ¼ 0 ð3:56Þ

where Ct_�� is a viscous damping force which is proportional to the angular velocity and Ct is a

viscous damping coefficient with units of N/s/rad. The critical damping coefficient Ctc and

damping ratio �t can be shown, in the same way as for rectilinear motion, to be

Ctc ¼ 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiKtIp� �q

ð3:57Þ

�t ¼ Ct=Ctc ð3:58Þ

and hence

Ct ¼ 2�tIp!n: ð3:59Þ

Substitution of Equation 3.59 into Equation 3.56 yields

Ip €��þ 2�tIp!n_�� ¼ Kt�: ð3:60Þ

It can further be shown that

�t ¼�tffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

4�2 þ �2t� �q ð3:61Þ

where �t is the logarithmic decrement of damping, which is determined from decay functions in

exactly the same way as for transverse and lateral motions. �t is usually much smaller than 2�,

hence Equation 3.61 may be written as

�t ¼�t2�

: ð3:62Þ

Problem 3.1

The vibration of an elastic system consisting of a weightW¼ 100 N and a spring with stiffness

K¼ 8.0 N/mm is to be damped with a viscous dashpot so that the ratio of two successive

amplitudes is 1.00 to 0.85. Determine: (a) the natural frequency of the un-damped system;

(b) the required value of the logarithmic decrement of damping; (c) the required damping

ratio; (d ) the corresponding damping coefficient; (e) the resulting damped natural frequency;

and ( f ) the amplitude after the tenth oscillation, if the first amplitude of free vibration is

5.00 mm.


76

FURTHER READING





Paz M (1980) Structural Dynamics. Van Nostrand Reinhold, New York.


Timoshenko SP and Gere JM (1972) Mechanics of Materials. Van Nostrand, New York.

Problem 3.2

The amplitude of vibration of an elastic mass–spring system is observed to decrease by 5.0%

with each successive cycle of the motion. Determine the damping coefficientC of the system if

the spring stiffness of the system K¼ 35.0 kN/m and the mass M¼ 4.5 kg.

Problem 3.3

A structure is modelled as a viscously damped oscillator with a spring constant

K¼ 5900.0 kN/m and un-damped natural frequency !n¼ 25.0 rad/s. Experimentally, it was

found that a force of 0.5 kN produced a relative velocity of 50 mm/s in the damping element.

Determine: (a) the damping ratio �; (b) the damped period Td; (c) the logarithmic decrement

of damping �; and (d ) the ratio between two consecutive amplitudes.

Problem 3.4

Repeat the calculation of the natural lateral and rotational frequencies of the platform in

Example 3.3 with the external column diameters assumed to be 5.0 m and the internal

diameters 4.0 m.


77


ISBN: 978-0-7277-4176-9



Chapter 4

Forced harmonic vibration of onedegree-of-freedom systems

4.1. IntroductionIn Chapters 1–3, it is pointed out that rotating machines tend to generate harmonic pulsating

forces when not properly balanced and that Fourier analyses of random forces such as wind,

waves and earthquakes can be considered as sums of harmonic components (as indeed can

explosions and impulse forces). It is also pointed out that a large number of civil engineering

structures respond mainly in the first mode and that it is possible, in many cases, to reduce

such structures to equivalent 1-DOF mass–spring systems. The different forms of structural

damping can be modelled as viscous damping in the form of a dashpot. This chapter takes a

step forward and considers the response of viscously damped 1-DOF mass–spring systems

when subjected to harmonic forcing functions. A thorough knowledge of how damped equivalent

1-DOF systems respond to harmonic excitation is fundamental to an understanding of how

structures exposed to random dynamic forces are likely to behave.

4.2. Rectilinear response of 1-DOF system with viscous damping toharmonic excitation

Consider the motion of the damped mass–spring system shown in Figure 4.1 when subjected to

the harmonic exciting force

P tð Þ ¼ P0 sin !tð Þ: ð4:1Þ


M€xx ¼ W � K �þ xð Þ � C _xxþ P0 sin !tð Þ ð4:2Þ

and hence

M€xxþ C _xxþ Kx ¼ P0 sin !tð Þ ð4:3Þ

which, on division of each of the elements by M, yields

€xxþ 2�!n _xxþ !2nx ¼ q0 sin !tð Þ ð4:4Þ

where

2�!n ¼ C=M

!2n ¼ K=M

q0 ¼ P0=M

79

If it is assumed that !n> �!n, then from Equation 3.29 the complementary function for Equation

4.4 is of the form

x ¼ e��!nt A cos !dtð Þ þ B sin !dtð Þ½ � ð4:5Þ

where

!d ¼ !n


: ð4:6Þ

The particular integral is found by assuming that

x ¼ C sin !tð Þ þD cos !tð Þ

_xx ¼ D! cos !tð Þ �D! sin !tð Þ

€xx ¼ �C!2 sin !tð Þ �D!2 cos !tð Þ:

Substitution of the expressions for x, _xx and €xx into Equation 4.4 yields

�C!2 sin !tð Þ �D!2 cos !tð Þ þ 2C�!n! cos !tð Þ

�2D�!n! sin !tð Þ þ C!2n sin !tð Þ þD!2

n cos !tð Þ ¼ q0 sin !tð Þ: ð4:7Þ

Equating the coefficients of the cos(!t) and sin(!t) terms in Equation 4.7 yields

�C!2 � 2D�!n!þ C!2n ¼ q0 ð4:8aÞ

�D!2 þ 2C�!n!þD!2n ¼ 0 ð4:8bÞ

Figure 4.1 Forced vibration of damped mass–spring system

P(t)

KC

M∆

x


80

and therefore

C ¼ q0ð1� r2Þ!2nð1� r2Þ2 þ 4�2r2

ð4:9aÞ

D ¼ 2q0�r

!2nð1� r2Þ2 þ 4�2r2

: ð4:9bÞ

The complete solution to Equation 4.4 is therefore

x ¼ e��!n t A cos !dtð Þ þ B sin !dtð Þ½ � þ C sin !tð Þ þD cos !tð Þ½ � ð4:10Þ

where the constants C andD are given by Equations 4.9a and 4.9b. As the value of t increases, the

first term on the right-hand side of Equation 4.10 will gradually decrease until it becomes negli-

gible. The free vibration component represented by the first term is called the transient vibration.

The second term containing the disturbing force represents the forced vibration.

The expression for forced damped vibration given by the second term in Equation 4.10 can be

simplified by considering steady-state response and by using rotating vectors as shown in

Figure 4.2, from which it can be seen that an alternative expression for the dynamic displacement

x is

x ¼ E sin !t� �ð Þ ð4:11Þ

where

tan �ð Þ ¼ D

Cþ 2�r

1� r2: ð4:12Þ

From Figure 4.2,

E ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC2 þD2ð Þ

q¼ q0

!2n

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 4�2r2� �q ð4:13Þ

and, since q0¼P0/M, !2n ¼K/M and xstþ P0/K,

E ¼ xstffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 4�2r2� �q ð4:14Þ

from which it follows that

x ¼ xstffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 4�2r2� �q sin !t� �ð Þ ð4:15Þ

or

x ¼ xstMF sin !t� �ð Þ ð4:16Þ

Forced harmonic vibration of one degree-of-freedom systems

81

where MF is the dynamic magnification factor, defined

MF ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 4�2r2� �q : ð4:17Þ

The value of the frequency ratio r that yields the greatest response at steady-state vibration is

found by differentiating Equation 4.17 with respect to r and equating the result to zero. For

real structures having damping ratios � >p2, the peak response frequency ratio is found to be at

r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q

ð4:18Þ

when the structure is said to be in resonance. The corresponding maximum value of the MF is

MF ¼ 1

2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q : ð4:19Þ

Figure 4.2 Vector diagram for steady-state response

0

D

ωt

π/2 – ωt

α

E

C

x

C s

in(ω

t)D

cos

(ωt)

P 0 s

in(ω

t)

P0


82

In practice, values of the damping ratio � are so small compared to unity that values of �2 may be

neglected. The maximum response can therefore be considered to occur when

r ¼ 1

MF ¼ 1=2�

� ¼ �=2:

The complete solution to Equation 4.4, an expression having been obtained for the particular

integral, is therefore

x ¼ e��!n t A cos !dtð Þ þ B sin !dtð Þ½ � þ xstMF sin !t� �ð Þ: ð4:20Þ

4.3. Response at resonanceAt resonance, i.e. when r¼ 1, the response equation (Equation 4.20) reduces to

x ¼ e��!n t A cos !dtð Þ þ B sin !dtð Þ½ � þ xst2�

cos !tð Þ: ð4:21Þ

If it is assumed that x¼ _xx¼ 0 when t¼ 0, it is found that

A ¼ xst2�

B ¼ xst

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q

and thus that

x ¼ xst2�

e��!n t cos !dð Þ þ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q sin !dtð Þ

0@

1A� cos !tð Þ

8<:

9=;: ð4:22Þ

For the level of damping experienced in most structures, the contribution to the amplitude by the

sine term is negligible. Equation 4.22 may therefore, for practical purposes, be written as

x ¼ sst2�

e��!n t cos !dtð Þ � cos !tð Þ½ ��

: ð4:23Þ

For linear structures vibrated at resonance, !¼!d, Equation 4.23 can therefore be simplified and

the expression of the response ratio at time t, R(t)¼x/xst, can be written as

R tð Þ ¼ 1

2�e��!n t � 1� �

cos !dtð Þ: ð4:24Þ

As t increases, e��!t approaches 0 and Equation 4.24 reduces to

R tð Þ ¼ 1

2�cos !dtð Þ: ð4:25Þ

As can be seen from Equation 4.17, MF is a function of the frequency ratio r¼!/!n and the

damping ratio � and will therefore vary in magnitude with the exciting frequency !. The maximum

value of MF occurs, as given in Equation 4.18, when r¼p(1� �2). The variation of the dynamic

magnification factor MF with the frequency ratio r is shown in Figure 4.3. Bearing in mind that a


83

damping ratio of �¼ 0.1, which is the same as a logarithmic decrement of damping of �¼ 62.4%,

is much greater than that for normal structures, the maximum response can be assumed to occur

when !¼!n. Figure 4.3 also shows the significance of damping when a structure is being vibrated

with an exciting frequency at or near its natural frequency.

The vector diagram shown in Figure 4.2 shows that the force P0 is in phase with the vector OC.

From this and the expression for the response:

x ¼ e��!n t A cos !dtð Þ þ B sin !dtð Þ½ � þ E sin !t� �ð Þ ð4:26Þ

it follows that when r< 1 the response x lags behind the disturbing force P0 sin(!t) by a phase

angle � and reaches its maximum (�/!) seconds after the maximum disturbing force P0 has

occurred. When r> 1, the response leads the disturbing force by a phase angle � and reaches

its maximum (�/!) seconds before the maximum disturbing force occurs.

From Equation 4.12, the expression for � is given by

� ¼ tan�1 2�r

1� r2

� : ð4:27Þ

Figure 4.3 Variation of the dynamic magnification factor MF with the frequency ratio !/!n

ξ = 0.0

ξ = 0.1

ξ = 0.15

ξ = 0.2

ξ = 0.25

ξ = 0.65

0.0 0.5 1.0 1.5 2.0 2.5

Mag

nific

atio

n fa

ctor

5.5

5.0

4.5

4.0

3.5

3.0

2.5

2.0

1.5

1.0

0.5

0.0

ω/ωn


84

When ! < !n;

� < �=2;

when ! ¼ !n,

� ¼ �=2

and when ! > !n;

� > �=2:

In Figure 4.4, values of � are plotted against the ratio !/!n for different levels of damping. The

resulting curves show that in the region of resonance, where !� !n, there is a sharp variation

in the phase angle. In the limit when �¼ 0 the variation follows the broken line 01123.

What happens at resonance can be elucidated further by study of the equation of motion and the

corresponding vector diagrams. From Equation 4.16, the amplitude at any time t is given by

x ¼ x0 sin !t� �ð Þ ð4:28Þ

where

x0 ¼ xstMF

and hence

_xx ¼ x0! cos !t� �ð Þ ð4:29aÞ

€xx ¼ �x0!2 sin !t� �ð Þ: ð4:29bÞ

Figure 4.4 Variation of phase angle � with frequency ratio r¼!/!n for different levels of damping

0 1 2 3

0 1 2 3

ξ = 0.00ξ = 0.125ξ = 0.50

Phas

e an

gle

α

ω/ωn

π

π/2

0


85

Substitution of the expressions for x, _xx and €xx into Equation 4.3 (the equation of motion) yields

�Mx0!2 sin !t� �ð Þ þ Cx0! cos !t� �ð Þ þ Kx0 sin !t� �ð Þ � P0 sin !tð Þ ¼ 0: ð4:30Þ

At resonance, when �¼ �/2, Equation 4.30 reduces to

Mx0!2n cos !ntð Þ þ Cx0!n sin !ntð Þ � Kx0 cos !ntð Þ � P0 sin !ntð Þ ¼ 0: ð4:31Þ

From Equation 4.31, if the coefficients of the sine and cosine terms are equated to zero,

P0 ¼ Cx0!n

kx0 ¼ Mx0!2n:

ð4:32Þ

The work done by the exciting force at resonance is only used to maintain the amplitude of

response by replacing the energy lost through the structural damping mechanism. Equations

4.3 and 4.31 are represented in Figure 4.5.

Figure 4.5 Vector diagrams of dynamic forces acting on a 1-DOF system (a) before and (b) at resonance

(a) (b)

P0P0

Kx0

Kx0

Mx0ω2

Mx0ω2n

Cx0ωn

Cx0ωn

αωt

ωt

Example 4.1

A beam supports at its centre a machine weighing 71.5 kN. The beam is simply supported, has

a span L¼ 3.5 m and a cross-sectional moment of inertia I¼ 5.3444� 107 mm4 and weighs

18.2 kN/m. The motor runs at 300 rev/min and its rotor is out of balance to the extent of

180 N at an eccentricity of 25 cm.What is the amplitude at steady-state response if the equiva-

lent viscous damping ratio is 10% of �critical? Determine also the phase angle � of response

relative to that of the unbalanced force.

The equivalent stiffness of the beam is

K ¼ 48EI=L3 ¼ 48� 200� 5:3444� 107=3:53 � 109

¼ 11:96647 kN=mm:


86

4.4. Forces transmitted to the foundation by unbalanced rotatingmass in machines and motors

Consider a machine of mass M mounted firmly to the foundation as shown in Figure 4.6(a). Let

the machine have an unbalanced rotating mass m at an eccentricity e. At an angular velocity of

! rad/s, this mass will give rise to an unbalanced rotating force

P0 ¼ me!2 ð4:33Þ

with vertical and horizontal components P0 sin(!t) and P0 cos(!t), respectively.

The equivalent weight of the beam at mid-span is given by Equation 2.80. Thus

W ¼ Pþ 17=35ð ÞwL ¼ 71:5þ 17=35ð Þ � 18:2� 3:5 ¼ 102:44 kN:

The natural frequency of the beam is therefore

!n ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKg=Mð Þ

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi11:96647� 1000� 1000� 9:81=102:44� 1000ð Þ

p

¼ 33:851876 rad=s:

The angular forcing frequency of the motor is

! ¼ 300� 2�=60 ¼ 31:415927 rad=s:

Thus the unbalanced force when the motor is running at 300 rev/min is

P0 ¼ me!2 ¼ 180� 0:25� 31:4159272=9:81� 1000 ¼ 4:5273416 kN:

The expression for steady-state response is given by Equation 4.15, which yields the

maximum response when sin(!t� �)¼ 0. We therefore have

xmax ¼P0

KMF ¼ me!2

K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ 2�rÞ2�

�q

where

r ¼ !=!n ¼ 31:415927=33:851176 ¼ 0:9280409

� ¼ 0:1:

Substitution of the values for r and � into the expression for xmax yields

MF ¼ 4:3153508

and

xmax ¼ 4:5273416� 4:3153505=11:96647 ¼ 1:633 mm:

The phase angle � is given by Equation 4.27 as

� ¼ tan�1 2�r

1� r2

� ¼ tan�1 2� 0:1� 0:9280409

1� 0:92804092

� ¼ 53:222�:


87

In order to reduce the effect of the vertical pulsating force P0 sin(!t), the machine may be mounted

on springs with total stiffness K and on dampers with a resultant damping coefficient C as shown

in Figure 4.6(b). The differential equation of motion for the vertical vibration caused by P0 sin(!t)

for this case is

M€xxþ C _xxþ Kx ¼ P0 sin !tð Þ: ð4:34Þ

The solution to Equation 4.34 is given by Equation 4.15, thus

x ¼ P0

K¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð1� r2Þ2 þ 2�rð Þ2� �q sin !t� �ð Þ ð4:35Þ

where

� ¼ tan�1 2�r

1� r2: ð4:36Þ

The forces transmitted to the foundation by the springs and the dampers are Kx and C _xx,

respectively. The total force transmitted to the foundation is therefore

F ¼ Kxþ C _xx ð4:37Þ

or

F ¼ P0MF sin !t� �ð Þ þ P0C!=KMFcos !t� �ð Þ: ð4:38Þ

With reference to the vector diagram shown in Figure 4.7, it can be shown that the expression for

the transmitted force can also be written as

F ¼ P0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 2�rð Þ2� �q

MFsin !t� �� ð Þ ð4:39Þ

Figure 4.6 Motors with rotating unbalance mounted (a) directly to the foundation and (b) on springs

and dampers

P0

m m

e e

P0

ω ω


88

where

� ¼ tan�1 C!=Kð Þ ¼ tan�1 2�rð Þ: ð4:40Þ

The maximum force transmitted to the foundation is therefore

F0 ¼ P0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½1þ ð2�rÞ2 �

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2�rÞ2�

q : ð4:41Þ

The ratio

T ¼ F0

P0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½1þ ð2�rÞ2�

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2�rÞ2�

q ð4:42Þ

is referred to as the transmissibility of the system and T is the transmissibility factor. From

Equation 4.42 it can be shown that when r <ffiffiffi2

p,

T > 1:0;

when r ¼ffiffiffi2

p,

T ¼ 1:0

and when r >ffiffiffi2

p,

T < 1:0:

Figure 4.7 Vector diagram for forces in rotating unbalanced machines and motors

P0 M

F

ωt – α

β

(P 0Cω/K

)MF

P0 √[1 + (2ξr) 2]MF

x


89

Transmissibility curves for different levels of damping are shown in Figure 4.8. From these, it can

be seen that the transmissibility decreases with decreasing damping ratios when the frequency

ratio is greater thanp2, and increases with decreasing damping ratios when the frequency

ratio is less thanp2.

Example 4.2

Amachine which weighs 2.5 kN and is to be operated at frequencies of 1000 and 4000 rev/min

is to be installed in a factory and mounted on isolators with a combined damping ratio of

10% of critical. The machine has a total unbalanced mass of 0.01 kg at an eccentricity of

100 mm. Calculate the force transmitted and the required spring stiffness of the isolators if

the maximum pushing force transmitted to the floor at the operating frequencies is to be

reduced by 75%. Calculate also the maximum pulsating force transmitted to the floor

when the speed of the machine increases from 0 to 4000 rev/min.

The maximum operating pulsating force transmitted occurs when the machine is running at

4000 rev/min. We therefore have

P4000 ¼ me!2 ¼ 0:01� 0:1� 4000� 2�=60ð Þ2¼ 175:45963 N:

The maximum permissible operational force is therefore

F4000 ¼ 0:25� 175:45963 ¼ 43:864908 N:

The required frequency ratio to reduce the pulsating force by 75% can be found by use of

Equation 4.42 which, when the values for T and � are substituted, can be written as

0:25 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½1þ ð2� 0:1� rÞ2�

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2� 0:1� rÞ2�

q

Figure 4.8 Transmissibility versus frequency ratios for vibration isolation

0 1 √2 2 3

Frequency ratio r = ω/ωn

ξ = 0.2

ξ = 0.2

ξ = 0

ξ = 0

ξ = 0.25

ξ = 0.25

ξ = 0.333

ξ = 0.333Tran

smis

sibi

lity

T3

2

1

0


90

which yields

r ¼ 1:6689337:

The natural frequency of the machine mounted on isolators is therefore

!n ¼ !

r¼ 4000� 2�

60� 1:6689337¼ 250:98602 rad=s

and the required spring stiffness is

K ¼ !2nM ¼ 250:986022 � 2500=9:81

¼ 16:053512� 106 N=m

¼ 16:053512 kN=mm:

When the machine runs at 1000 rev/min, the maximum pulsating force is

P1000 ¼ 0:01� 0:1� 1000� 2�=60ð Þ2¼ 10:966227 N:

The frequency ratio when the machine runs at 1000 rev/min is

r ¼ 1000� 2�=60ð Þ=250:98602 ¼ 0:4172334

and the maximum force transmitted to the floor when the machine is running at this speed is

therefore

F1000 ¼P1000


qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2�rÞ2�

q

¼10:966227

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½1þ ð2� 0:1� 0:4172334Þ2�

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� 0:41723342Þ2 þ ð2� 0:1� 0:4172334Þ2�

q

¼ 15:913816 N:

This force is less than the force transmitted when the machine runs at 4000 rev/min; the

calculated spring stiffness is therefore satisfactory.

When r¼ 1, the maximum pulsating force due to the unbalanced mass is

Pn ¼ me!2n ¼ 0:01� 0:1� 250:986022 ¼ 62:993982 N:

The corresponding force transmitted to the floor in this case is given by

Fmax ¼Pn


2�¼

62:993982ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 2� 0:1� 1:0ð Þ2� �q

2� 0:1

¼ 321:20754 N:


91

4.5. Response to support motionIn the previous section, we examined the transmission of harmonic forces, such as those caused

by unbalanced machinery, to the supports. In the following sections, the opposite is considered

as the response of structures to harmonic excitation of the supports themselves is studied. The

shaking of supports or foundations is notably caused by earthquakes; even minor earthquakes

can be particularly devastating if the depth of the soil above the bedrock is such that one of

the dominant frequencies of the ground coincides with the frequency in which the structure is

likely to respond. Other sources of ground motion are traffic and explosions. In the case of the

latter, the travel of shockwaves through the ground is followed by pressure waves through the

air, the effects of both of which have to be considered if there are buildings in the vicinity. In

the study of the response to ground vibration, the response relative to a fixed point, the

absolute response of a structure and the response relative to the support are considered. The

response relative to a fixed point is important when studying the likely effect of ground motion

on, say, sensitive electronic equipment in a building. The absolute response of a structure is

important when assessing the strength of the building itself.

4.5.1 Response relative to a fixed pointConsider the mass–spring system shown in Figure 4.9, where the support subjects the left-hand

end of the spring to the periodic displacement

xg ¼ xg0 sin !tð Þ ð4:43Þ

where xg0 is the maximum amplitude of the support motion and xg and x are absolute

displacements. The equation of motion for this case is given by

M€xxþ C _xx� _xxg� �

þ K x� xg� �

¼ 0 ð4:44Þ

or

M€xxþ C _xxþ Kx ¼ Kxg0 sin !gt� �

þ Cxg0!g cos !gt� �

: ð4:45Þ

With reference to the vector diagram shown in Figure 4.10, it can be seen that Equation 4.45 may

be written as

Mxþ C _xxþ Kx ¼ F0 sin !gt� ��

ð4:46Þ

Figure 4.9 Mass–spring system with left-hand end of spring being subjected to a harmonic

displacement xg

K

C

M

xxg


92

where

F0 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½Cxg0!gÞ2 þ ðKxg0Þ2�

q¼ Kxg0


qð4:47Þ

and

� ¼ tan�1 C!g=K ¼ tan�1 2�t: ð4:48Þ

The solution to Equation 4.46 is given by Equation 4.15; we therefore have

x ¼xg0


ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 2�rð Þ2� �q sin !gt� ��

� �ð4:49Þ

where, from Equation 4.12,

� ¼ tan�1 2�r

1� r2: ð4:50Þ

The ratio

T ¼ x0xg0

¼


ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 2�rð Þ2� �q ð4:51Þ

is the absolute transmissibility or the transmissibility relative to a fixed point, and is a measure of

the extent to which the motion of the support is either magnified or reduced by the structure. It

Figure 4.10 Vector diagram for forces arising from harmonic excitation of support of the mass–spring

system in Figure 4.9

βωgt

F 0 = Kx g0

√[1 + (2ξr)

2 ]

Kxg0

Cxg0

ωg

x


93

should be noted that the right-hand side of Equation 4.51 is identical to the right-hand side of

Equation 4.42. The transmissibility curves shown in Figure 4.8 are therefore also valid for struc-

tures subjected to harmonic excitation of the supports.

Example 4.3

A delicate instrument which weighs 450N is to be spring-mounted to the floor of a test

laboratory, which occasionally vibrates with a frequency of 10 Hz and a maximum amplitude

of 2.5 mm. Determine the stiffness of the isolation springs required to reduce the vertical

amplitude of the instrument to 0.025 mm if the instrument is isolated with dampers with

damping ratios (a) 2.0% and (b) 20.0% of critical.

(a) When �¼ 0.02, the transmissibility from Equation 4.51 is given by

T ¼ 0:025

2:5¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½1þ ð2� 0:02� rÞ2�

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2� 0:02� rÞ2�

q

which yields

r ¼ 10:459366

and hence

!n ¼ !

r¼ 2�� 10

10:459366¼ 6:0072336 rad=s:

The required spring stiffness when �¼ 0.02 is therefore

K ¼ !2M ¼ 6:00723362 � 450=9:81 ¼ 1655:3603 N=m:

(b) When �¼ 0.2, the transmissibility from Equation 4.51 is

T ¼ 0:025

2:5¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½1þ ð2� 0:2� rÞ2�

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2� 0:2� rÞ2�

q

which yields

r ¼ 40:100599

and hence

!n ¼ !

r¼ 2�� 10

40:100599¼ 1:5666451 rad=s:

The required spring stiffness when �¼ 0.2 is therefore

K ¼ !2nM ¼ 1:56664512 � 450=9:81 ¼ 112:58609 N=m:

Comparison of the calculated spring stiffness shows that, as expected, the stiffness reduces as

the value of the damping ratio increases.


94

4.5.2 Response relative to the supportIf the movement of the mass is measured relative to the movement of the spring support, as shown

in Figure 4.11, and the movement of the support is given by Equation 4.43 as in the previous case,

then the equation of motion may be written as

M €xxþ €xxg� �

þ C _xxþ Kx ¼ 0 ð4:52Þ

or

M€xxþ C _xxþ Kx ¼ �Mxg0!2g sin !gt

� �: ð4:53Þ

The negative sign in Equation 4.53 is clearly irrelevant and may be neglected. The solution to

Equation 4.53 is again given by Equation 4.15. The response relative to the support motion is

therefore

x ¼Mxg0!

2g

K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ þ 2�rð Þ2� �q sin !gt� �

� �: ð4:54Þ

Since !n¼K/M, Equation 4.54 may be written as

x ¼xg0r

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 2�rð Þ2� �q sin !gt� �

� �: ð4:55Þ

The transmissibility factor for the response relative to the support is therefore given by

T ¼ x0xg0

¼ r2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 2�rð Þ2� �q : ð4:56Þ

Relative transmissibility curves constructed by substitution of increasing values for the frequency

ratio r and the damping ratio � into Equation 4.56 are shown in Figure 4.12.

Figure 4.11 Mass–spring system with left-hand end of spring being subjected to a harmonic

displacement

K

CM

xxg xg


95

Figure 4.12 Relative transmissibility versus frequency ratio

0 1 2 3Frequency ratio r = ωg/ωn

ξ = 0.5

ξ = 1.0

ξ = 0.4

ξ = 0.15

3

2

1

0

Rela

tive

tran

smis

sibi

lity

Example 4.4

A rigid jointed rectangular steel portal frame has a span of 20 m. Each column is 4.0 m tall

and pinned at the base. The weight of the horizontal beam, which may be assumed to be

rigid, is 4.0 kN/m. The second moment of area and the section modulus of each of the

columns, which may be considered to be weightless, are 3200 cm4 and 286 cm3, respectively.

The damping ratio for the structure may be assumed to be 2.0% critical. Young’s modulus

E¼ 100 kN/mm2. If the frame is subjected to a sinusoidal ground motion xg¼8.0 sin(11.5t) mm, determine: (a) the transmissibility of the motion to the girder; (b) the

maximum shear force in each column; and (c) the maximum bending stress in each column.

The shear stiffness of the frame is

K ¼ 2� 3EI

L3¼ 2� 3� 200� 3200� 104

4:0� 109¼ 0:6 kN=mm:

The natural frequency of the frame is therefore

!n ¼


M

�s¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi600� 103 � 9:81

20� 4000

�s¼ 8:5775871 rad=s

and the frequency ratio is given by

r ¼ !

!n

¼ 11:5

8:5775871¼ 1:3407034:


96

4.5.3 SeismographsMovements of the ground due to earthquakes or other forms of disturbances can be recorded by

the use of seismographs (Figure 4.13). These instruments, which for arbitrarily chosen damping

ratios will measure the relative displacement between the spring-supported mass and the base

of the instrument, can be designed to measure either the displacement or the acceleration of the

base support. From Equation 4.55, the relative response of the mass in the seismograph due to

a base movement of xg is given by

xg ¼ xg0 sin !gt� �

ð4:57Þ

From Equation 4.56, the relative transmissibility, i.e. the motion of the crossbeam relative to

the ground, is

T ¼ x

xg¼ x

8:0

1:34070342ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� 1:34070342Þ2 þ ð2� 0:02� 1:3407034Þ2�

q

¼ 2:2488622

and hence the maximum horizontal motion of the beam is

x ¼ 2:2488622� 8:0 ¼ 17:990898 mm:

The maximum shear force in each column is given by

SF ¼ 12Kx ¼ 1

2 � 0:6� 17:990898 ¼ 5:3972694 kN:

The maximum bending stresses in the columns are therefore

�M ¼ SF�H

Z¼ 5:3972694� 4:0� 103

286� 103¼ 0:0754862 kN=mm2:

Figure 4.13 Seismograph

x

xg


97

x ¼xg0r

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 2�rð Þ2� �q sin !gt� �

� �: ð4:58Þ

From Figure 4.12, the relative transmissibility is approximately equal to 1 when r> 1 and �� 0.5.

When this is the case, the movement of the mass relative to the base is given by

x ¼ xg0 sin !gt� ��

ð4:59Þ

and the seismograph will therefore record the movement of the support.

From Equation 4.54, the relative movement of the mass may be written as

x ¼Mxg0!

2g

K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 2�rð Þ2� �q sin !g � �

� �: ð4:60Þ

Since

€xxg ¼ �xg0!2g sin !gt

� �¼ �€xxg0 sin !gt

� �; ð4:61Þ

it follows that Equation 4.60 may be written as

x ¼M€xxg0K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 2�rð Þ2� �q sin !gt� �

� �: ð4:62Þ

From Figure 4.3, when 0< r< 0.6 and �� 0.65, the MF is approximately equal to 1. When this is

the case, the movement of the mass relative to the base is given by

x ¼M€xxg0K

sin !gt� ��

: ð4:63Þ

The recording of seismographs, designed with the above values for r and �, will therefore be

proportional to the acceleration of the support. In both of the above cases, the range of the

seismograph may be increased by varying the spring stiffness or the size of the mass.

4.6. Rotational response of 1-DOF systems with viscous damping toharmonic excitation

Consider the motion of the damped mass–spring system shown in Figure 4.14 when subjected to

the harmonic exciting moment

T tð Þ ¼ T0 sin !tð Þ ¼ P0e sin !tð Þ: ð4:64Þ

If d’Alembert’s principle is applied,

Ip €�� ¼ �Kt�� Ct_��þ P0e sin !tð Þ ð4:65Þ

and hence

Ip €��þ Kt�þ Ct_�� ¼ P0e sin !tð Þ ð4:66Þ


98

where Ip is the polar moment of inertia of the lumped mass, Kt is the torsional stiffness of the

spring and Ct is the equivalent torsional viscous damping coefficient. From Equation 3.58,

Ct ¼ 2�t!nIp: ð4:67Þ

Following the same steps as when finding the solution to the rectilinear equation of motion, it can

be shown that the steady-state response to torsional excitation is

� ¼ P0e

Kt

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ 2�rð Þ2� �q sin !t� �ð Þ ð4:68Þ

where

� ¼ tan�1 2�tr

1� r2: ð4:69Þ

Figure 4.14 Elevation and plan view of damped mass–spring system subjected to harmonic exciting

moment T(t)¼ P0e sin(!t)

Ct

Ct

xKt

e

ω

P0 sin(ωt)

Example 4.5

The natural frequency of the translational motion of the structure shown in Figure 4.15 is

0.8533 Hz. The corresponding equivalent mass and spring stiffness are 4.722� 106 kg and

135.748� 103 kN/m, respectively. The natural rotational frequency about the vertical axis

is 0.9199 Hz. The corresponding polar moment of inertia and torsional spring stiffness are

1361.2421� 106 kg m2 and 45475.523� 103 kN/rad, respectively. For the purpose of design

it is assumed that the centre of gravity of the equivalent mass of the structure is located

1.0 m above the x axis. Calculate the maximum translational and rotational response to a

horizontal support motion xgt¼ 0.02 sin(6.0t) m if the damping in both the translational

and rotational modes is 2.0% of critical.

The expression for the translational response of 1-DOF systems is given by Equation 4.54.

We therefore have

xmax ¼Mxg0!

2g

K� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

½ð1� r2Þ2 þ ð2�rÞ2�q


99

Figure 4.15 Platform structure with assumed non-symmetric mass distribution subjected to

harmonic support motion

40 m

40 m

45 m

y

x

xg(t)

where

r ¼ 6:0=2�� 0:8533 ¼ 1:1191019

and hence

xmax ¼4:722� 106 � 0:02� 6:02

135:748� 103 � 103� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

½ð1� 1:11910192Þ2 þ ð2� 0:02� 1:1191019Þ2�q

or

xmax ¼ 0:3812 m:

The expression of the rotational response of a 1-DOF system is given by Equation 4.68. We

therefore have

�max ¼M€xxge!

2g

Kt

� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2�rÞ2�

q

where

r ¼ 6:0=2�� 0:9199 ¼ 1:0380799

and hence

�max ¼4:722� 106 � 0:02� 6:02 � 1:0

45 475:523� 103 � 103� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

½ð1� 1:03807992Þ2 þ ð2� 0:02� 1:0380799Þ2�q


100

FURTHER READING


Craig Jr RR (1981) Structural Dynamics. Wiley, Chichester.





or

�max ¼ 9:64985� 10�3 rad ¼ 0:553�:

The maximum rotational displacement of each corner of the platform is therefore

� ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi202 þ 202ð Þ

q� � 9:64985� 10�3 ¼ �0:2729 m:

Problem 4.1

Two parallel simply supported beams support a machine weighing 150.0 kN at their mid-spans.

The beams span 3.4 m, have a total cross-sectional moment of inertia I¼ 5.3444� 107 mm4

and together weigh 18.2 kN/m. The motor runs at 800 rev/min, and its rotor is out of balance

to the extent of 150 N at an eccentricity of 25 cm. What will the amplitude of steady-state

response be if the equivalent viscous damping ratio is 10% of critical? Determine also the

phase angle of response relative to that of the unbalanced force E¼ 200 kN/mm2.

Problem 4.2

Determine the force transmitted by the machine to the supports of the beam whose data are

given in Problem 4.1. Calculate also the force transmitted if the motor runs at a speed equal to

the natural frequency of the system. Start by developing the expression for the appropriate

transmissibility factor and give the values for this factor for the two running speeds in

question.

Problem 4.3

Calculate the response of the top floor of the shear structure shown in Figure 2.14 to the

ground motion x¼ 11.0 sin(18.85 t) mm. The weight of each floor is 20.0 kN/m and the

flexural rigidity of each of the columns is EI¼ 87 311.477 kNm2. Assume the response of

the building to be the same as for an equivalent 1-DOF mass–spring system.


101


ISBN: 978-0-7277-4176-9



Chapter 5

Evaluation of equivalent viscous dampingcoefficients by harmonic excitation

5.1. IntroductionChapter 3 shows how the logarithmic decrement � of viscous damping of a 1-DOF system can

be determined from the decay function of free vibration by use of Equation 3.36 and plotting

ln(x0/xn) against the number of oscillations n. Chapter 3 also shows that the relationship between

the logarithmic decrement of damping and damping ratio is given by

� ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4�2 þ �2� �q ð5:1Þ

which, since � is usually very much smaller than 4�2, may be written

� ¼ �

2�: ð5:2Þ

This chapter discusses various methods by which the viscous and equivalent viscous damping

coefficients can be determined through harmonic excitation of a structure. The following methods

are presented for the evaluation of damping

g amplification of the static response at resonanceg vibration at resonance: (a) balancing of the maximum input and damping forces and (b)

measurement of energy loss per cycleg frequency sweep to obtain response functions using (a) the bandwidth method,

(b) amplitude ratios to obtain values for damping ratios at various points along the

response curve, or (c) equivalent linear viscous response functions to calculate stiffness and

damping ratios.

5.2. Evaluation of damping from amplification of static response atresonance

For weakly damped structures, whose stiffness is known and whose maximum response xn0 at

resonance occurs when the frequency ratio r is approximately equal to 1, the damping ratio � is

most easily obtained by measuring the response amplitude at resonance. From Equation 4.14,

it follows that

� ¼ xst2xn0

¼ P0

2Kxn0: ð5:3Þ

For structures that possess a higher level of damping, Equation 5.1 will lead to an underestimation

of the damping ratio as the maximum amplitude of response will have been reached before the

103

frequency ratio is equal to 1. This underestimation will increase with increasing degrees of

damping, as can be seen from Figure 4.3. However, as the level of damping encountered in

most structures is relatively low, the use of Equation 5.1 will in most cases be satisfactory.

5.3. Vibration at resonance5.3.1 Evaluation of viscous damping by balancing the maximum input and

damping forcesAnother method of determining the damping of a structure is to vibrate it at resonance and then

equate the maximum exciting and damping forces. The equivalent viscous damping coefficient can

then be calculated by use of the first relationship given in Equation 4.32. This yields

C ¼ P0

xn0!n

ð5:4Þ

� ¼ P0

2Mxn0!2n

ð5:5aÞ

or

� ¼ P0

2Kxn0: ð5:5bÞ

Equation 5.5b may be more convenient if the equipment available is sufficiently sensitive to

measure the response near zero frequency, as it permits a value for K to be obtained without

any static testing. The method requires that the instruments used be sufficiently sensitive to

keep the phase angle � at a steady-state response equal to �/2, and is accurate only when the

damping is linearly viscous. In this case, a graph of the exciting or input force plotted against

the displacement will yield an ellipse of area

An ¼ �P0xn0: ð5:6Þ

5.3.2 Evaluation of viscous damping by numerical integrationThe alternative method for calculation of the work done is by numerical integration of the

response through one cycle, as shown in Figure 5.1.

Figure 5.1 One complete cycle for integration

a

x = a0 sin(ωt)


104

The forcing function at resonance where !n¼!e is P0 sin !t, where P0¼me!2 andm and e are the

eccentric mass and distance from the centre, respectively. We then have

work done ¼ð2�0

P0 sin!t dx:

5.3.3 Evaluation of equivalent viscous damping by measurement of energy lossper cycle

If the damping mechanism does not possess a linear viscous characteristic, the plot of the input

force against the amplitude at steady-state response will be a curve similar to the solid line

in Figure 5.2. If the area enclosed by this curve is also denoted An, then the area can also be

calculated by the integration given above and the equivalent maximum force amplitude is,

from Equation 5.6, given by

Pe0 ¼An

�xn0: ð5:7Þ

Substitution of this value for the equivalent maximum input force into Equation 5.4 yields

C ¼ An

�x2n0!n

ð5:8Þ

� ¼ An

2�Mx2n0!2n

ð5:9aÞ

or

� ¼ An

2�Kx2n0: ð5:9bÞ

Figure 5.2 Input force plotted against displacement for linear (broken line) and non-linear (solid line)

viscous damping

Area = An

Linear viscous damping(equivalent area = An)

P(t )

x

xn0

P0

Evaluation of equivalent viscous damping coefficients by harmonic excitation

105

The value for An can then be substituted from Equation 5.6 into Equation 5.9a to obtain

� ¼ 1

2Mxn0!2n

:

However, if Equation 5.6 is substituted in Equation 5.9b, then

� ¼ P0

2Kxn0:

5.4. Evaluation of damping from response functions obtained byfrequency sweeps

Another technique much used for measuring damping is based on frequency sweeps past the point

where resonance occurs by construction of a frequency response curve, where each successive

point is obtained from the steady-state response after an incremental increase in the frequency

of a vibrator. This procedure will lead to curves of the type shown in Figure 4.3, where values

of xmax/xst¼MF are plotted against the frequency ratio r. An examination of the curves in

Figure 4.3 shows that both the magnification factor and the general shape of the curves are

functions of the level of damping in a structure. In particular, it can be noted that the difference

between the two frequencies corresponding to a given magnification factor, referred to as the

bandwidth, is a function of the degree of damping. Three different methods are now presented

for determination of the damping of structures and structural elements from the response

function, the first of which is based on a bandwidth corresponding to a specific magnification

factor or response amplitude.

5.4.1 Bandwidth methodIn evaluation of damping by the bandwidth method, it is convenient to measure the two frequen-

cies and the distance between them at points where the amplitudes of the magnification factor or

amplitudes of response (as shown in Figure 5.2) are equal to 1/p2 of the peak amplitude.

Figure 5.3 Frequency response curve showing the bandwidth at 1/p2 of amplitude at r¼ 1

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

r = ω/ωn

Mag

nific

atio

n fa

ctor

Peak amplitude

Peak amplitude/√2

ξ = (f2 – f1)/2fnf2 f1

12.0

10.0

8.0

6.0

4.0

2.0

0


106

From Equation 4.15, the maximum responses when the frequency ratio r is equal and not equal to

1 are:

xn0 ¼xst2�

ð5:10Þ

x0 ¼xstffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

½ð1� r2Þ2 þ ð2�r2Þ2�q ; ð5:11Þ

respectively. The frequency ratio r at which the amplitudes are equal to 1/p2 of the amplitude

when r¼ 1 can therefore be determined by solving the equality

xstffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2�rÞ2�

q ¼ 1ffiffiffi2

p xst2�

: ð5:12Þ

Squaring both sides of Equation 5.12 and solving for r2 yields

r2 ¼ 1� 2�2 � 2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q

: ð5:13Þ

Since �2 is usually much smaller than 1.0, Equation 5.13 may be written as

r21 ¼ 1� 2�2 þ 2� ð5:14aÞ

r22 ¼ 1� 2�2 � 2�: ð5:14bÞ

Subtraction of Equation 5.14b from Equation 5.14a yields

4� ¼ r21 � r22 ¼ r1 þ r2ð Þ r1 � r2ð Þ ð5:15Þ

or

4�!n ¼ !1 þ !2ð Þ !1 � !2ð Þ: ð5:16Þ

For weakly damped structures it may be assumed that

!n ¼ 12 !1 þ !2ð Þ: ð5:17Þ

Substitution of this expression for !n into Equation 5.16 yields

� ¼ !1 � !2

2!n

¼ �!

2!n

ð5:18aÞ

or

� ¼ f1 � f22fn

¼ �f

2fn: ð5:18bÞ

It should be noted that the size of the frequency step required in a frequency sweep in order to plot

a steady-state response curve accurately, especially between f1 and f2, will depend on both the level


107

of damping and the natural frequency of the structure. When, for example, �¼ 0.01 and

fn¼ 2.0 Hz and the bandwidth is equal to

f1 � f2 ¼ 2� 0:01� 2:0 ¼ 0:04 Hz;

a frequency step as small as (say) 0.004 Hz may be necessary to plot a satisfactory curve. If,

however, the level of damping is higher and/or the natural frequency is greater, a larger step

may be used. Thus, if �¼ 0.05 and fn¼ 20.0 Hz, the bandwidth is therefore

f1 � f2 ¼ 2� 0:05� 20:0 ¼ 2:0 Hz

and a frequency step of (say) 0.2 Hz may be sufficient.

The expression for the damping ratio given by Equation 5.18a or Equation 5.18b assumes that the

response curve shown in Figure 5.3 is obtained by vibrating a structure or structural element with

a pulsating force P(t)¼P0 sin(!t), where P0 is constant. This will not be the case if the vibrator

consists of a motor with a rotating eccentric mass such as is shown in Figure 4.6. When this is

the case, Equation 5.12 must be modified as

me!2

K¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

½ð1� r2Þ2 þ ð2�rÞ2�q ¼ 1ffiffiffi

2p me!2

n

K

1

2�: ð5:19Þ

Simplification and rearrangement of Equation 5.19 yields

1� 8�2� �

r4 � 2� 4�2� �

r2 þ 1 ¼ 0 ð5:20Þ

which, solved with respect to r2, yields

r2 ¼1� 2�2 � 2�


1� 8�2: ð5:21Þ

If it is assumed that

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ �2� �q

¼ 1

then

r21 ¼1� 2�2 þ 2�

1� 8�2ð5:22aÞ

r22 ¼1� 2�2 � 2�

1� 8�2: ð5:22bÞ

Subtraction of Equation 5.22b from Equation 5.22a yields

r21 � r22 ¼4�

1� 8�2ð5:23Þ


108

which, solved with respect to �, gives

� ¼ �1�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 8�f =fnð Þ

p8�f =fn

: ð5:24Þ

If the square root is expanded by the binomial theorem, neglecting the cubic and higher terms and

rejecting the negative sign in front of the square root for obvious reasons, we then have

� ¼ �f

2fnþ 1

8

�f

fn

� �2

: ð5:25Þ

The second term in Equation 5.25 results in increases of 0.25%, 1.25%, 2.5% and 5.0% in 2� when

�f / fn¼ 0.01, 0.05, 0.10 and 0.20, respectively.

Experience has shown that the damping calculated by this method will only be accurate if there is

a pure mode of vibration. The only way to ensure most accurate damping is by the method of

measuring the work done by one cycle of vibration at resonance.

5.4.2 Amplitude ratiosResponse curves similar to those shown in Figures 4.3 and 5.1 can sometimes be difficult to obtain

because of limitations in the equipment available to perform the frequency sweeps. In such cases,

the damping ratios may be determined as follows.

Let the maximum response at resonance due to P(t)¼P0 sin(!nt) be

xn ¼ P0

K

1

2�ð5:26Þ

and that due to P(t)¼P0 sin(!t) be

x ¼ P0

K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2�rÞ2�

q : ð5:27Þ

Elimination of P0/K from Equations 5.26 and 5.27 leads to

xn1x0

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð2�rÞ2�

q

2�ð5:28Þ

which, when solved with respect to �, yields

� ¼ 1� r2

2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�2 � r2ð Þ

p ð5:29Þ

where

� ¼ xnx:


109

For vibrators with eccentric masses, for which P(t)¼me!2 sin(!t), the expression for � can be

shown to be

� ¼ 1� r2

2rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�2r2 � 1ð Þ

p : ð5:30Þ

The percentage error caused by determination of the damping ratio using Equation 5.30

corresponds to the errors resulting from not including the second term in Equation 5.25.

5.4.3 Calculation of stiffness and damping ratios from an equivalent linearviscous response function

The evaluation of damping ratios from decay and response functions will result in the same values

only if the damping is linearly viscous. This is generally not the case, and the methods described so

far and based on such functions will usually yield different values which often differ considerably.

An alternative method for measuring non-linear viscous damping, if the instruments available are

not suitable for measuring the energy lost per cycle, is to undertake a frequency sweep and then

establish an equivalent theoretical linear viscous response function as shown in Figure 5.4.

At a frequency f, let the experimental amplitude be z and the theoretical amplitude x. The variance

of the area between the two curves is therefore given by

�2 ¼XNn� 1

x� zð Þ2�f ð5:31Þ

where z is the experimental amplitude and x is the theoretical response, defined

x ¼ P0

K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð�rÞ2�

q ð5:32Þ

where

� ¼ 2�: ð5:33Þ

Figure 5.4 Experimental and theoretical response functions

Am

plitu

de

Experimental response function

Theoretical response function

Frequency: Hz

f

x

z

zn xn

fn


110

It is now assumed that the best equivalent theoretical linear response curve is the one for which the

variance of the area between the two curves is a minimum with respect to both K and �, i.e. when

the gradient

g ¼@ �2� �

=@K

@ �2� �

=@�

" #¼ 0: ð5:34Þ

This condition can be achieved through the iterative process

Kiþ 1

�iþ 1

� �¼

Ki þ �Ki

�i þ ��i

� �ð5:35Þ

by expanding the gradient vector at the ith iteration through a Taylor series, neglecting cubic and

higher-order terms and assuming that the gradient vector at the (iþ 1)th iteration is zero. This

yields

�K

��

� �

i

¼ �@2 �2� �

=@k2 @2 �2� �

=@K@�

@2 �2� �

=@K@� @2 �2� �

=@�2

" #�1

i

@ �2� �

=@K

@ �2� �

=@�

" #

i

ð5:36Þ

which may also be written as

� K ;�f g ¼ H�1i gis ð5:37Þ

whereHi is the Hessian matrix and gi is the gradient vector of the variance �2 at the ith iterate, or

point (Ki, �i) in optimisation space.

The elements inH and g in Equation 5.37 are found by differentiating the expression for �2 given

by Equation 5.31 with respect to K and �. The differentiations implied in Equation 5.36 yield

@ �2� �@K

¼XNi¼ 1

2 x� zð Þ @x@K

�f ð5:38aÞ

@2 �2� �

@K2¼XNi¼ 1

2@x

@K

@x

@Kþ x� zð Þ

� �@2x

@K2�f ð5:38bÞ

@ �2� �@�

¼XNi¼ 1

2 x� zð Þ @x@�

�f ð5:38cÞ

@2 �2� �

@�2¼XNi¼ 1

2@x

@�

@x

@�þ x� zð Þ

� �@2x

@�2�f ð5:38dÞ

@2 �2� �

@K@�¼XNi¼ 1

2@x

@K

@x

@�þ x� zð Þ

� �@2x

@K@��f ð5:38eÞ

where the expressions for the partial differentials are obtained by differentiation of Equation 5.32

with respect to K and �. We therefore have

@x

@K¼ � P0

K2

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� r2ð Þ2þ �rð Þ2

h ir ð5:39aÞ


111

@2x

@K2¼ 2P0

K3

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð�rÞ2�

q ð5:39bÞ

@x

@�¼ �P0

K

r2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð�rÞ2�

q� 3ð5:39cÞ

@2x

@�2¼ 3P0

K

r4�2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð�rÞ2�

q� 5� P0

K

r2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð�rÞ2�

q� 3ð5:39dÞ

@2x

@K@�¼ P0

K2

r2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r2Þ2 þ ð�rÞ2�

q� 3: ð5:39eÞ

The iterative process described by Equation 5.35 has converged when gTg¼ 0, where the elements

in g are found by substitution of the latest update of K and � in Equations 5.38a and c and 5.39a

and c.

The determination of an equivalent viscous response curve will generally lead to a value for xn0that is slightly different from zn0. Calculations using the value for the damping ratio �¼ �/2 to

model the damping mechanism in the structure tested would therefore lead to a different

maximum amplitude for a given exciting force than that obtained experimentally. To overcome

this problem, it is suggested that the calculation of the damping ratio be modified as

� ¼ �zn=2xn: ð5:40Þ

The justification for this is that, at resonance, Equation 5.32 would yield

zn0 ¼P0

K

1

2�ð5:41Þ

and hence

P0

2�¼ Kzn0: ð5:42Þ

The calculated values for K and � can therefore be verified by plotting P0/2� against zn0 for

different values of P0 and comparing the resulting graph with one obtained from an ordinary

static load test.

The method presented does not require instruments that can read or maintain the phase angle � at

908 at resonance, but it does require a computer that can store and analyse the data from

frequency sweeps.

5.5. Hysteretic dampingThe expression for the viscous damping coefficient given by Equation 3.22 leads to the expression

for the damping force

Fd ¼ C _xx ¼ 2�!nM _xx ð5:43Þ


112

which shows that the damping force is a function not only of the mass and velocity of vibration,

but also of the natural frequency of the structure. This contradicts a great deal of experimental

evidence, which indicates that the damping is often very nearly independent of the mode

frequency. A frequency-independent damping model is the hysteretic damping model, where

the damping force is proportional to the stiffness and displacement of the structure but in

phase with the velocity. Mathematically, the force may be expressed as

Fd ¼ ChK xj j _xx

_xxj j ¼ ChK

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�x

_xx

�2" #vuut _xx: ð5:44Þ

The force displacement diagram for this form of damping for one cycle of vibration at resonance is

shown in Figure 5.5 in which the shaded area representing the energy lost per cycle is

An ¼ 2ChKx2n0: ð5:45Þ

Substitution of this expression for An into Equation 5.9b yields

Ch ¼ ��; ð5:46Þ

which is independent of the mode frequency. The equation of motion for a 1-DOF system with

damping independent of the frequency is therefore given by

M€xxþ ��K

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�x

_xx

�2" #vuut _xxþ Kx ¼ P tð Þ: ð5:47Þ

The solution of Equation 5.47 requires an iterative solution method, as the damping is a function

of both displacement and velocity.

Figure 5.5 Assumed variation in hysteretic damping force with displacement at resonance

P(t)

xn0 xn0

ChKxn0

ChKxn0


113

5.6. The effect and behaviour of air and water at resonanceWhen a structure vibrates, the surrounding air and/or water will tend to oppose the motion.

This opposing force is deemed to be proportional to the square of the velocity of the motion.

At resonance however, the force is proportional to the acceleration of the motion with the air

and/or water moving together with the structure, adding to the total vibrating mass without

increasing the damping. Testing of a small cable roof at the University of Trondheim, Norway

and three large circular cable-roof models at the University of Westminster, UK using a single

vibrator showed that at the (and near the) point of excitation the air moved with the structure.

Further away from the point of excitation, the air gradually reverted to oppose the motion. It

is thought that the reason for this is that it is not possible to vibrate a structure in a pure mode

with only one vibrator. When vibrating a rigid circular plate supported by a spring, it was

found that at resonance the air at all points both above and below moved with the plate and

that the damping remained the same as for the spring loaded with compensatory lumped mass

alone. Similarly, it was found that when vibrating a beam with a hanging plate submerged in

water, at resonance the water moved with the beam. The damping remained the same as

without water but the frequency, because of the additional mass of the moving water, was

reduced.

These results lead to the conclusion that, when using a single vibrator, the measured damping will

always contain an element of air and/or water damping. The magnitude of this damping will

depend on the resonance frequency.

Problem 5.1

The equivalent lumped mass of a shear structure is 100 t, the first natural frequency 3.0 Hz

and the damping 5.0% of critical. Plot the curves for the amplitude response and phase

angle for a frequency sweep from 0.0 to 6.0 Hz when the building is vibrated by a variable

speed motor with a mass of 5.0 kg at 50 cm eccentricity.

Problem 5.2

Having plotted the response function for the structure in Problem 5.1, verify the correctness

of the curve plotted by calculating the damping using (a) the bandwidth method and (b) the

amplitude ratio method. In the latter case, calculate the two values for damping ratios by

selecting a point on the curve corresponding to a frequency of 2.0 Hz and another

corresponding to a frequency of 4.0 Hz. (c) What is the percentage error in the calculated

values if the maximum pulsating force is assumed to be constant and equal to me!n?

Problem 5.3

If the damping is assumed to be linearly viscous, calculate the energy lost per cycle at reso-

nance when the structure in Problem 5.1 is vibrated by a pulsating force P(t)¼ 1000 sin(!t)

N. Calculate also the equivalent linear viscous damping ratio if a plot of the exciting force

against the amplitude of response at resonance during one cycle is a rectangle and the

maximum amplitude of response is 1.5 mm. Hence, calculate the equivalent maximum

damping force at resonance.


114

FURTHER READING





115


ISBN: 978-0-7277-4176-9



Chapter 6

Response of linear and non-linear onedegree-of-freedom systems to randomloading: time domain analysis

6.1. IntroductionThe response of linear 1-DOF systems to harmonic excitation is presented in Chapter 4 in terms of

closed-form solutions. This chapter considers the response of the same type of equivalent

mass–spring systems to random forms of loading, and extends the solution methods to include

non-linear structures. Examples of random types of loading are wind, waves and earthquakes,

and examples of non-linear structures are suspension bridges, cable-stayed footbridges and

canopies, guyed masts and cable and membrane roofs. Even structures that are regarded as

linear may exhibit non-linear characteristics if subjected to strong excitation. Non-linear

structures can be classified as either stiffening or softening. In the case of stiffening structures,

the rate of change of displacements will reduce with increasing deformation. In the case of

stiffening structures, the rate of change of displacements will reduce with increasing deformation.

In the case of softening structures, the reverse will be the case. Figure 6.1 shows typical load

displacement curves for linear and non-linear stiffening and softening 1-DOF structures.

117

Figure 6.1 Load–displacement curves for linear and non-linear 1-DOF structures

Stiffening

Linear

Dis

plac

emen

t

Load

Softening

When the stiffness of a structure varies with the amplitude of response, it follows that the natural

frequencies will also vary. For this reason, the closed-form solutions presented in Chapter 4 are no

longer valid. Figure 6.2 shows examples of frequency response curves for stiffening and softening

1-DOF systems to harmonic excitation.

In general, the damping will also vary with the amplitude of vibration. Generally, however,

structural damping is assumed to remain constant because of lack of information, and because

the values given in codes of practice tend to be conservative. Even after extensive testing it is

only generally possible to produce an approximate numerical model of a structural damping

mechanism based on the damping ratios obtained from vibration in a few of the lower modes.

The response of linear structures to random forms of loading such as wind and earthquakes

may be carried out in the frequency domain using power spectra, as described in Chapters 10

and 12, which enable the use of closed-form solutions. For non-linear structures, this approach

will underestimate the response in the case of softening structures and overestimate it in the

case of stiffening structures. The general approach for predicting the behaviour of non-linear

structures to all types of dynamic loading is to predict the response by a forward integration in

the time domain. Several such methods are presented and discussed in the following sections.

6.2. Step-by-step integration methodsLet the force–time curve shown in Figure 6.3(a) represent the variation of a random force P(t)

acting on a 1-DOF system, and the displacement–time curve in Figure 6.3(b) represent the

resulting dynamic response.

At time t let

K tð Þ ¼ K

P tð Þ ¼ P

x tð Þ ¼ x

_xx tð Þ ¼ _xx

€xx tð Þ ¼ €xx

Figure 6.2 Frequency response curves for non-linear (a) stiffening and (b) softening 1-DOF systems

Frequency

(a)

Am

plitu

de

Frequency

(b)


118

and at time (tþ�t) let

K tþ�tð Þ ¼ K þ�K

P tþ�tð Þ ¼ Pþ�P

x tþ�tð Þ ¼ xþ�x

_xx tþ�tð Þ ¼ _xxþ� _xx

€xx tþ�tð Þ ¼ €xxþ�€xx:

Thus at time t, the equation of motion is

M€xxþ C _xxþ Kx ¼ P ð6:1Þ

and at time (tþ�t) it is

M €xxþ�€xxð Þ þ C þ�Cð Þ _xxþ� _xxð Þ þ K þ�Kð Þ xþ�xð Þ ¼ Pþ�P: ð6:2Þ

Subtraction of Equation 6.1 from Equation 6.2 yields

M�€xxþ C� _xxþ�C _xxþ� _xxð Þ þ K�xþ�K xþ�xð Þ ¼ �P: ð6:3Þ

In practice, it has been found sufficient to let the damping and the stiffness coefficients remain

constant during each time step �t, and update them only at the end of each step. The terms

�C _xxþ� _xxð Þ and �K(xþ�x) may therefore be neglected, and Equation 6.3 reduces to

M�€xxþ C� _xxþ K�x ¼ �P ð6:4Þ

which is referred to as the incremental equation of motion and can be solved only if there exists

a relationship between �€xx, � _xx and �x. A number of such relationships are proposed in the

literature. Here, only the three most commonly used are considered

g the linear change of acceleration methodg the Wilson �-methodg the constant acceleration method.

Figure 6.3 Variation of (a) random force P(t) and (b) response x(t) with time

(a)

P (t) x(t)

tΔt

(b)t

Δt

Response of linear and non-linear 1-DOF systems to random loading

119

6.2.1 Linear change of acceleration methodIn this method, the acceleration during each time step is assumed to vary linearly as shown in

Figure 6.4, from which the slope at time (tþ �) is seen to be constant and can be written as

_€xxtþ �ð Þ ¼ �€xx

�¼ A ð6:5Þ

€xx tþ �ð Þ ¼ A� þ B ð6:6Þ

_xx tþ �ð Þ ¼ 12A�

2 þ B� þ C ð6:7Þ

x tþ �ð Þ ¼ 16A�

3 þ 12B�

2 þ C� þD: ð6:8Þ

The constant A is given by Equation 6.5 and the constants B, C and D may be determined by the

condition that when � ¼ 0,

€xx tþ �ð Þ ¼ €xx tð Þ ¼ €xx

_xx tþ �ð Þ ¼ _xx tð Þ ¼ _xx

x tþ �ð Þ ¼ x tð Þ ¼ x:

We therefore have

B ¼ €xx

C ¼ _xx

D ¼ x

and at time (tþ�t)

_xxþ� _xx ¼ 12�€xx�tþ €xx�tþ _xx ð6:9Þ

Figure 6.4 Assumed change in acceleration during a time step �t in the linear acceleration method

x(t)

x x

Δx

t Δt

τ


120

xþ�x ¼ 16�€xx�t3 þ 1

2€xx�t2 þ _xx�tþ x ð6:10Þ


�€xx ¼ 6

�t2�x� 6

�t_xx� 3€xx: ð6:11Þ

Substitution of this expression for �€xx into Equation 6.9 yields

� _xx ¼ 3

�t�x� 3 _xx� 1

2€xx�t: ð6:12Þ

Substitution of the expressions for� _xx and�€xx given by Equations 6.11 and 6.12 into Equation 6.4

leads to the following formulation of the incremental equation of motion for a 1-DOF

system:

K þ 3

�tC þ 6

�t2M

� ��x ¼ �Pþ C 3 _xxþ�t

2€xx

� �þM

6

�t_xxþ 3€xx

� �ð6:13Þ

or

�x ¼ K�1d �Pd ð6:14Þ

where the dynamic stiffness Kd and the equivalent dynamic load Pd are defined

Kd ¼ K þ 3

�tC þ 6

�t2M ð6:15Þ

�Pd ¼ �Pþ C 3 _xxþ�t

2€xx

� �þM

6

�t_xxþ 3€xx

� �: ð6:16Þ

Once �x has been calculated using Equation 6.14, the values for displacement, velocity and

acceleration to be used at the beginning of the next time step are

x tþ�tð Þ ¼ xþ�x ð6:17Þ

_xx tþ�tð Þ ¼ 3

�t�x� 2 _xx��t

2€xx ð6:18Þ

€xx tþ�tð Þ ¼ 6

�t2�x� 6

�t_xx� 2€xx: ð6:19Þ

The linear acceleration method tends to become unstable if �t> T/2, where T¼ 1/f is the

period of natural vibration. Instability is, however, not usually a problem in the case of 1-DOF

systems where �t needs to be less than, say, T/10 to ensure sufficient accuracy in the predicted

response.

6.2.2 Wilson �-methodIn the Wilson �-method, the acceleration is assumed to vary linearly during a prolonged time step

��t, where � > 1.0 as shown in Figure 6.5. Each new step, however, is started from time (tþ�t)

and not (tþ ��t).


121

From Equations 6.11 and 6.12, it follows that at time (tþ ��t)

�€xx ¼ 6

�2�t2�x� 6

��t_xx� 3€xx ð6:20Þ

� _xx ¼ 3

��t�x� 3 _xx� 1

2��t€xx: ð6:21Þ

Substitution of these expressions for �€xx and � _xx into Equation 6.4 yields

K þ 3

��tC þ 6

�2�t2M

� ��x ¼ �Pþ C 3 _xxþ ��t

2€xx

� �þM

6

��t_xxþ 3€xx

� �ð6:22Þ

or

�x ¼ K�1d �Pd ð6:23Þ

where �x is the incremental displacement at the end of the time step ��t, and

Kd ¼ K þ 3

��tC þ 6

�2�t2M ð6:24Þ

�Pd ¼ �Pþ C 3 _xxþ ��t

2€xx

� �þM

6

��t_xxþ €xx

� �: ð6:25Þ

The acceleration, velocity and displacement at the beginning of the next time step at time (tþ�t)

are found by inspection of Figure 6.5, from which it can be deduced that at time (tþ�t) the

change in acceleration during the time interval �x is

�€xx

�¼ 6

�3�t2�x� 6

�2�t_xx� 3

�€xx: ð6:26Þ

Figure 6.5 Assumed linear acceleration during time step (tþ ��t) in the Wilson �-method

x(t)

x x

ΔxΔx/θ

t Δt

θΔt

τ


122

The acceleration at time (tþ�t) is therefore given by


�3�t2�x� 6

�2�t_xxþ 1� 3

�

� �€xx: ð6:27Þ

The expressions for the velocity and acceleration at time (tþ�t) are then found by substitution of

the expression for �€xx=� given by Equation 6.26 into Equations 6.9 and 6.10, respectively. This

yields


�3�t�þ 1þ 3

�2

� �_xxþ 3�t

21� 1

�

� �€xx ð6:28Þ

x tþ�tð Þ ¼ 1

�2�xþ xþ�t 1� 1

�2

� �_xxþ�t2

64� 3

�

� �€xx: ð6:29Þ

For linear structures, the method is stable when �5 1.37. In general, a value of �¼ 1.4 appears to

be satisfactory. Values of � much in excess of 1.4 result in an increasing overestimation of the

predicted amplitude of response, combined with an increasing phase lag relative to the dynamic

force.

6.2.3 Constant acceleration methodIn this method the acceleration is assumed to remain constant during the time step �t and equal

to €xxþ 12�€xx

� �as shown in Figure 6.6, where it is compared with the assumption of linear

acceleration. From Figure 6.6, the acceleration at time (tþ �) is

€xx tþ �ð Þ ¼ €xxþ 12�€xx ð6:30Þ

_xx tþ �ð Þ ¼ €xx� þ 12�€xx� þ A ð6:31Þ

x tþ �ð Þ ¼ 12€xx�2 þ 1

4�€xx�2 þ A� þ B: ð6:32Þ

Figure 6.6 Assumed acceleration in the constant acceleration method

x(t)

x x

½Δx

½Δx

t Δt

τ


123

When � ¼ 0,

_xx tþ �ð Þ ¼ _xx tð Þ ¼ _xx

x tþ �ð Þ ¼ x tð Þ ¼ x

and hence

A ¼ _xx

B ¼ x:

When � ¼�t,

_xx tþ �ð Þ ¼ _xx tþ�tð Þ ¼ _xxþ� _xx

x tþ �ð Þ ¼ x tþ�tð Þ ¼ xþ�x

and hence

� _xx ¼ �t€xxþ 12�t�€xx ð6:33Þ

�x ¼ �t _xxþ 12�t2€xxþ 1

4�t2�€xx ð6:34Þ


�€xx ¼ 4

�t2�x� 4

�t_xx� 2€xx: ð6:35Þ

Substitution of Equation 6.35 into Equation 6.33 yields

�€xx ¼ 4

�t2�x� 2 _xx: ð6:36Þ

Substitution of the expressions for�€xx and� _xx given by Equations 6.35 and 6.36, respectively, into

Equation 6.4 yields

K þ 2

�tC þ 4

�t2M

� ��x ¼ �Pþ 2C _xxþM

4

�t_xxþ 2€xx

� �ð6:37Þ

or

�x ¼ K�1d �Pd ð6:38Þ

where

Kd ¼ K þ 2

�tC þ 4

�t2M ð6:39Þ

�Pd ¼ �Pþ 2C _xxþM4

�t_xxþ 2€xx

� �: ð6:40Þ


124

Having calculated the incremental displacement �x using Equation 6.38, the displacement,

velocity and acceleration at the end of the time step at time (tþ�t) can be calculated from

x tþ�tð Þ ¼ xþ�x ð6:41Þ


�t�x� _xx ð6:42Þ


�t2�x� 4

�t_xx� €xx: ð6:43Þ

The accuracy with which the linear acceleration, constant acceleration and Wilson �-methods

predict the response for a given load history depends on the size of the time step. This must be

small enough to enable all the significant harmonic components in the load history to be taken

into account, and should not be greater than 0.05 times the period of the 1-DOF system analysed.

Experience indicates that with the same time step, the constant acceleration andWilson �-methods

with �¼ 1.4 yield similar results. The former may therefore be preferable since less computational

effort is required.

6.2.4 The Newmark �-methodWhen studying time-domain methods for the dynamic analysis of linear and non-linear structures,

references to the Newmark �-method will be encountered. Newmark proposed the following

expressions for the velocity and displacement at time (tþ�t)

_xx tþ�tð Þ ¼ _xxþ�t 1� �ð Þ€xxþ�t� €xxþ�€xxð Þ ð6:44Þ

x tþ�tð Þ ¼ xþ�t _xxþ�t2 12 � ��

€xxþ�t2� €xxþ�€xxð Þ ð6:45Þ

where � and � are variable constants. It is usually assumed that �¼ 1/2 while � may be assigned

different values. With �¼ 1/2 and �¼ 1/6 or �¼ 1/4, it can be shown that the Newmark �-method

is identical to the linear acceleration or constant acceleration method, respectively.

6.3. Dynamic response to turbulent windLet the drag force due to wind per unit of velocity of the wind relative to that of the structure be Fd

and the corresponding wind velocities and velocities of the structure at times t and (tþ�t) be V,

Vþ�V, _xx and _xxþ� _xx respectively. If it is assumed that the force due to wind is proportional to

the square of the relative velocity of the wind to that of the structure, then the force exerted on the

structure by the wind is

P ¼ Fd V � _xxð Þ2 ð6:46Þ

at time t and

Pþ�P ¼ Fd V þ�Vð Þ � _xxþ� _xxð Þ½ �2 ð6:47Þ

at time (tþ�t). Subtraction of Equation 6.46 from Equation 6.47, ignoring the terms of second

order of smallness, yields

�P ¼ 2Fd V � _xxð Þ�V � V � _xxð Þ� _xx½ �: ð6:48Þ


125

By using the constant acceleration method, from Equation 6.36 we have

� _xx ¼ 2

�t�x� 2 _xx ð6:49Þ

and hence

�P ¼ 2Fd V � _xxð Þ�V þ 2 v� _xxð Þ _xx� 2

�tV � _xxð Þ�x

� �: ð6:50Þ

Substitution of this expression for �P into Equation 6.37 yields the dynamic equation

K þ 2

�tC þ 4

�t2M þ 4

�tFd V � _xxð Þ

� ��x ¼ 2Fd V � _xxð Þ �V þ 2 _xxð Þ þ 2C _xxþM

4

�t_xxþ 2€xx

� �ð6:51Þ

which reveals that the wind, as well as exciting a structure, also increases its dynamic stiffness.

Equation 6.51 may alternatively be written as

Kd ¼ K þ 2

�tC þ 2Fd V � _xxð Þ½ � þ 4

�t2M: ð6:52Þ

When the dynamic stiffness is expressed in this form it can be seen that the wind not only excites a

structure, but also increases the damping coefficient by the term Fd V � _xxð Þ. For certain types of

structure, such as guyed masts, the aerodynamic damping may be more significant than the

damping caused by friction in joints and hysteresis losses in the structure itself.

6.4. Dynamic response to earthquakesLet the acceleration of the ground at the support of a 1-DOF structure at times t and (tþ�t) be

€xxg and €xxg þ�€xxg, respectively. From Equation 4.53, the inertia force acting on the mass of the

structure is M€xxg tð Þ. If €xxg is the acceleration at time t and €xxg þ�€xxg is the acceleration at time

(tþ�t), the change in dynamic force during the time step �t is therefore

�P ¼ M €xxg þ�€xxg� �

�M€xxg ¼ M�€xxg: ð6:53Þ

Substitution of this expression for �P into Equation 6.37 yields the response equation for

earthquake excitation

K þ 2

�tC þ 4

�t2M

� ��x ¼ M �€xxg þ

4

�t_xxþ 2€xx

� �þ 2C _xx ð6:54Þ

where the displacements x and �x, velocity _xx and acceleration �€xx are relative to the support.

Response analysis in the time domain to determine the effects of turbulent wind and earthquakes

is normally only carried out for non-linear structures. For linear structures, such analysis is

undertaken in the frequency domain.

6.5. Dynamic response to impacts caused by falling loadsConsider the case when a weight of mass m drops from a height H onto a floor having an

equivalent massM and an equivalent spring stiffnessK. If the floor is assumed to respond linearly,

the maximum response x0 can most easily be determined by equating the initial maximum


126

potential energy of the weight to the maximum strain energy stored in the floor, if the loss of

energy at impact is neglected. We therefore have

mg H þ x0ð Þ ¼ 12Kx

20 ð6:55Þ

and hence

x0 ¼ xst �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2st þ 2Hxstð Þ

qð6:56Þ

where xst¼mg/K. The negative sign in front of the square root has no meaning and can be

ignored. Since H is usually much greater than xst the square term in Equation 6.56 may be

neglected, in which case the expression for the response x0 reduces to

x0 ¼ xst þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2Hxstð Þ

p: ð6:57Þ

It is therefore only necessary to undertake dynamic response analysis if the structure is expected

to exhibit non-linear behaviour. Such an analysis may be undertaken using one of the forward

step-by-step integration processes presented above by neglecting energy losses at impact and

assuming that the kinetic energy after impact is equal to the initial potential energy of the falling

load. This may be written as

mgH ¼ 12 M þmð Þ _xx20 ð6:58Þ

and yields

_xx0 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2mgH

M þm

� �s: ð6:59Þ

In this case, �P is zero and Equation 6.37 reduces to

K þ 2

�tC þ 4

�t2M

� ��x ¼ 2C _xxþM

4

�t_xxþ 2€xx

� �ð6:60Þ

where the initial value for the velocity _xx is given by Equation 6.59 and the initial value for €xx is

taken as zero.

Example 6.1

A portal frame is subjected to a horizontal impulse force P(t) at beam level. The specifications

for the portal frame and dynamic load are shown in Figures 6.7(a) and 6.7(b), respectively.

Use the linear acceleration method to predict the response of the frame.

The change of displacement during each time step is given by

�x ¼ K�1d �Pd

where the expressions for Kd and�Pd are given by Equations 6.15 and 6.16, respectively. The

periodic time of vibration of the frame is

T ¼ 2�

ffiffiffiffiffiffiffiffiffiffiffiffiM

K

� �s¼ 2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi40� 1000

1600� 1000

� �s¼ 0:9934588 s


127

Figure 6.7 Portal frame (a) subjected to shock load P(t) and (b) at beam level

P(t)

(a)

40 × 9.81 kN

c = 80 kN s/m

P(t)

: kN

k = 800kN/m

k = 800kN/m

(b)

Time: s

5

0.1 0.2 0.3 0.4 0.5 0.6

20

45 45

30

and hence a time step of �t¼ 0.1 s should be sufficient to describe the motion of response.

With reference to the diagram of the force P(t) in Figure 6.7, a time step of 0.1 s may be

slightly too large to describe the load history sufficiently accurately; for the purpose of this

example it is assumed that the chosen time step is small enough. By Equations 6.15, 6.16,

6.18 and 6.19, the following functions are obtained for Kd, �P, _xx and €xx:

Kd ¼ 1600þ 3

0:1� 80þ 6

0:12� 40 ¼ 28 000 kN=m

�Pd ¼ �Pþ 3� 80þ 6

0:1� 40

� �_xxþ 0:1

2� 80þ 3� 40

� �€xx ¼ �Pþ 2640 _xxþ 124€xx kNð Þ

_xx ¼ 3

0:1�x� 2 _xx� 0:1

2€xx ¼ 30�x� 2 _xx� 0:05€xx

€xx ¼ 6

0:12�x� 6

0:1_xx� 2€xx ¼ 600�x� 60 _xx� 2€xx:

Table 6.1 Response calculations using linear acceleration method to predict the response of portal

frame shown in Figure 6.7

t: Kd: �P: �Pd: �x: x: _xx: €xx:

s kN/m kN kN m m m/s m/s2

0.0 28000.0 0.0 0.00000 0.000000 0.0000000 0.000000 0.00000

0.1 28000.0 5.0 5.00000 0.000179 0.0001786 0.005357 0.10714

0.2 28000.0 15.0 42.42784 0.001515 0.0016939 0.029388 0.37346

0.3 28000.0 25.0 148.89388 0.005318 0.0070115 0.082080 0.68038

0.4 28000.0 0.0 301.05866 0.010752 0.0177636 0.124384 0.16569

0.5 28000.0 �15.0 333.91788 0.011926 0.0296900 0.100718 �0.63900

0.6 28000.0 �30.0 156.65791 0.005695 0.0352850 0.001637 �1.40809

0.7 28000.0 0.0 178.92592 �0.005390 0.0298950 �0.118030 �0.91970

0.8 28000.0 0.0 �425.63507 �0.015201 0.0146940 �0.174000 0.19900

0.9 28000.0 0.0 �484.11919 �0.017290 �0.0025960 �0.160720 0.46533

1.0 28000.0 0.0 �366.59374 �0.013093 �0.0156190 �0.094610 0.85682


128

The sequence of the iterative process is given in Table 6.1 and the time histories for the

displacement, velocity and acceleration of response are shown in Figure 6.8. As a rule of

thumb, it may be assumed that the maximum response of 1-DOF systems with zero damping

occurs when the ratio of the time of the impulse to the periodic time is 4 0.8 as indicated in

Figure 6.8 Displacement, velocity and acceleration histories for portal frame in Example 6.1

Time: s

0.2 0.4 0.6 0.8 1.0

0.2 0.4 0.6 1.0

0.2 0.4 0.8 1.0

40

30

20

10

0

–10

x: m

100

50

–50

–100

x: m

/s

1000

500

–500

–1000

x: m

/s2


129

Figure 6.10, where the dynamic magnification factor is plotted against the impulse length

ratio �/T. In this case, the ratio is 0.6/0.9934588¼ 0.604 which is less than 0.8, but the

damping is 22.36% of critical. The damping is therefore very high, and it is assumed that

the maximum response occurs within 1.0 s in order to keep the number of iterations to a

minimum.

From Table 6.1, it can be seen that the maximum displacement is 35.285mm. If the portal

frame had been subjected to a maximum static force of 45.0 kN, the displacement would

have been 45.0/16.0¼ 28.125mm. The dynamic magnification is therefore 25.46%.

The histories plotted in Figure 6.8 show, as expected, that the maximum displacement and

acceleration occur when the velocity is zero and the maximum velocity occurs when the

displacement and acceleration are zero.

Example 6.2

A weight of 1.0 kN is dropped from a height of 1.0 m onto the centre of a simply supported

beam having a span of 10.0 m. The beam supports a distributed load of 3.0 kN/m, which

includes a self-weight. The EI value for the beam is 28 000 kNm2. Neglecting the loss of

energy at impact and losses due to structural damping, calculate the initial velocity and

maximum displacement of the beam and the corresponding dynamic magnification factor.

Use the constant acceleration method to calculate the maximum central displacement, if

the structural damping of the beam is 2.0% of critical. Use a time step equal to approximately

1/20th of the natural period of the beam.

Treating the beam and 1.0 kN falling load as a mass–spring system, the equivalent lumped

weight is

We ¼ Pþ 17

35wL ¼ 1:0þ 17

35� 3:0� 10 ¼ 15:571429 kN:

The equivalent stiffness is given by

Ke ¼6144EI

125L3¼ 6144� 28 000:0

125� 103¼ 1376:256 kN=m

and the natural frequency of the beam plus load is therefore

f ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKeg

We

� �s¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1376:256� 1000� 9:81

15:571429� 1000

� �s¼ 4:6864071 Hz;

hence the time step

�t ¼ T

20¼ 1

20f¼ 1

20� 4:6864071¼ 0:0106691 s; say �t ¼ 0:01 s:


130

The damping coefficient is given by

C ¼ �Cc ¼ 2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKeMeð Þ

p¼ 2� 0:2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1376:256� 1 571 429

9:81

� �s

¼ 1:8695597 kN s=m

and the initial velocity of the beam after impact is given by

_xx0 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2mgH

M þm

� �s¼

ffiffiffiffiffiffiffiffiffiffiffiffiffi2PgH

We

s¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2� 1:0� 9:81� 1:0

15:571429

� �s

¼ 1:1224972 m=s:

The maximum displacement of the beam is found by equating the maximum potential energy

to the maximum strain energy, which yields

xmax ¼ xst 1þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 2H=xstð Þ

ph i

where

xst ¼ P=Ke ¼ 1:0=1376:256 ¼ 0:726609� 10�3 m

xmax ¼ 0:726609� 10�3 1þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 2� 1:0=0:726609� 10�3ð Þ

q� �¼ 38:85463� 10�3 m

K þ 2

�tC þ 4

�t2M

� ��x ¼ 2C _xxþM

4

�t_xxþ 2€xx

� �

or

Kd�x ¼ �Pd

where

K þ 2

�tC þ 4

�t2M

� �¼ 1376:256þ 2

0:011:8695597þ 4

0:01215:571429

9:81

� �

¼ 65 242:233 kN=m

2C _xxþM4

�t_xxþ 2€xx

� �¼ 2� 1:8695597 _xxþ 15:571429

9:81

4

0:01_xxþ 2€xx

� �

¼ 638:65977 _xxþ 3:1746033€xx:

We therefore have

�P0 ¼ 638:65977� 1:1224972þ 3:1746033� 0:0 ¼ 716:8938 kN


131

x tþ�tð Þ ¼ xþ�x


�t�x� _xx ¼ 2

0:01�x� _xx ¼ 200�x� _xx


�t2�x� 4

�t_xx� €xx ¼ 4

0:012�x� 4

0:01_xx� €xx

¼ 40 000�x� 400 _xx� €xx:

The response of the beam is calculated in time steps of 1/100ths of a second for 7 steps, where

the first trough is encountered. Table 6.2 shows the numerical values of response and

Figure 6.10 shows the deflection for the centre of the beam.

Figure 6.9 Variation of the central deflection of the beam in Example 6.2 caused by a falling load

with time

Time: s

Def

lect

ion:

m

0.0 0.01 0.02 0.03 0.04 0.05 0.06

0.00

0.01

0.02

0.03

0.04

Table 6.2 Response calculations using the constant acceleration method to predict the response of

the simply supported beam in Example 6.2 to the impact caused by a falling load

t: Kd: �x: x: _xx: €xx: �Pd:

s kN/m m m m/s m/s2 kN

0.00 65242.233 0.00000000 0.0000000 1.1224972 0.00000 716.89380

0.01 65242.233 0.01098810 0.0109881 1.0751228 �9.47488 656.55869

0.02 65242.233 0.01006340 0.0210515 0.9375572 �18.03824 541.51581

0.03 65242.233 0.00830008 0.0293515 0.7224588 �24.98144 382.09921

0.04 65242.233 0.00585662 0.0352081 0.4488652 �29.74896 192.23100

0.05 65242.233 0.00294641 0.0381545 0.1404168 �31.94072 �11.72055

0.06 65242.233 �0.00017965 0.0379748 �0.1763460 �31.41184 �212.34139


132

6.6. Response to impulse loadingTime-domain methods may also be used to study how the response of 1-DOF systems vary with

the duration of different forms of impulse, such as those shown in Figure 6.10 where the ratio of

the dynamic to the static response (the dynamic magnification factor) is plotted against the ratio

of the duration � of the impulse to the natural period T of the oscillator.

6.7. Incremental equations of motion for multi-DOF systemsThe method for predicting the response of linear and non-linear 1-DOF systems for random

loading may be extended to multi-DOF systems by writing the incremental equations of

motion in matrix form. Equation 6.37, in which the acceleration is assumed to remain constant

during the time step �t, therefore may be written as

Kþ 2

�tCþ 4

�t2M

� ��x ¼ �P þ 2C _xxþM

4

�t_xxþ 2€xx

� �ð6:61Þ

where K, C andM are the stiffness, damping and mass matrices for a multi-DOF structure,�x is

the incremental displacement vector, x, _xx and €xx are the displacement, velocity and acceleration

vectors at time t and �P is the incremental load vector.

The ith elements in vectors x, _xx and €xx at time tþ�t are given by Equations 6.41–6.43, and we

therefore have

xi tþ�tð Þ ¼ xi þ�xi ð6:62Þ

_xxi tþ�tð Þ ¼ 2

�t�xi � _xxi ð6:63Þ

Figure 6.10 Dynamic magnification factor versus impulse length ratio � /T for rectangular,

triangular and half-sinusoidal impulses of � s duration (after Clough and Penzien, 1975)

Impulse length ratio τ/T

Dyn

amic

mag

nific

atio

n fa

ctor

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

2.4

2.0

1.6

1.2

0.8

0.4

0.0


133

€xxi tþ�tð Þ 4

�t2�xi �

4

�t_xxi � €xxi: ð6:64Þ

From Equation 6.61, it follows that the dynamic stiffness matrix and incremental dynamic load

vector are

Kd ¼ Kþ 2

�tCþ 4

�t2M

� �ð6:65Þ

�Pd ¼ �P þ 2C _xxþM4

�t_xxþ 2€xx

� �ð6:66Þ

and hence

�x ¼ K�1d �P: ð6:67Þ

Similarly, the equations of motion for structures subjected to turbulent wind V(t) may be written

Kþ 2

�tCþ 4

�t2Mþ 4

�tFd V � _xxð Þ

� ��x ¼ 2Fd V � _xxð Þ �V þ 2 _xxð Þ þ 2C _xxþM

4

�t_xxþ 2€xx

� �ð6:68Þ

and the equations of motion for structures subjected to ground acceleration €xxg tð Þ:

Kþ 2

�tCþ 4

�t2M

� ��x ¼ M �€xxg þ

4

�t_xxþ 2€xx

� �þ 2C _xx: ð6:69Þ

Equations 6.61, 6.68 and 6.69 require the assembly of stiffness, mass and structural damping

matrices. The construction of stiffness matrices and mass matrices is considered in Chapters 7

and 8 and in most modern books on structural analysis such as Coates et al. (1972). The construc-

tion of damping matrices is dealt with in Chapter 9, which shows how damping matrices can be

constructed from modal damping ratios whose values need to be obtained from codes of practice

or published papers or by dynamic testing.

Problem 6.1

Use the constant acceleration method to predict the first 1 s response of the portal frame in

Example 6.1. Plot the time histories of displacement, velocity and acceleration response and

compare them to those shown in Figure 6.8.

Problem 6.2

A steel ball of mass 0.76 kg is dropped in turn from heights of 100 mm and 200 mm onto the

centre of a pre-stressed concrete plate. The equivalent mass of the plate is 230 kg and the

equivalent stiffness 3.77 kN/mm. The damping is 1.6% of critical. Use the constant accelera-

tion method to calculate the maximum response for each drop. The size of a suitable time step

may be assumed to be approximately 1/20th of the period of the plate. Compare the values

obtained with those obtained by equating the initial potential energy of the ball to the strain

energy stored in the plate.


134

REFERENCES

Coates RC, Coutie MG and Kong FK (1972) Structural Analysis. Nelson, London.


FURTHER READING


Cambridge.

Wood WL (1990) Practical Time-stepping Schemes. Oxford Applied Mathematics, Oxford.


135


ISBN: 978-0-7277-4176-9



Chapter 7

Free vibration of multi-degree-of-freedomsystems

7.1. IntroductionIn Chapter 1, it is mentioned that structures generally have an infinite number of degrees of

freedom which are usually approximated to N-DOF systems by replacing the distributed mass

of structures with an equivalent system of lumped masses and assuming the elastic members to

be weightless. A general preliminary introduction to the free and forced vibration of multi-

DOF systems entails an excessive amount of algebra and requires the use of computers to

study their behaviour if there are more than three degrees of freedom.

In order to avoid these difficulties, the dynamic analysis of large systems is introduced by studying

structures with only two and three degrees of freedom. This is quite feasible since the procedure

for determining free vibration, as well as response to dynamic loads, is exactly the same as for

structures with N degrees of freedom. At this stage it is merely noted that an N-DOF structure

has N eigenvalues and eigenvectors associated with the system of equations that defines its

motion, and that the square roots of the eigenvalues are equal to the natural angular frequencies

of the structure. The eigenvectors corresponding to the natural frequencies represent the natural

modes or mode shapes in which the structure can vibrate. The determination of eigenvalues of

eigenvectors is of fundamental importance to the frequency-domain method of analysis, in

which the distribution of energy of random forces such as wind, waves and earthquakes are

given as functions of their frequency content in terms of power spectra. Structural damping is

usually not included when formulating the eigenvalue problem, as it increases the numerical

effort considerably and only has a second-order effect on the calculated frequencies.

7.2. Eigenvalues and eigenvectorsThe mathematical concept arises from the solution of a set of N homogeneous equations

where two N�N matrices A and B are related by a set of vectors V and scalars � such that the

relationship

AX � �BX ¼ 0 ð7:1Þ

or

A� �Bð ÞX ¼ 0 ð7:2Þ

is valid for non-zero values ofX. For a set of homogeneous equations represented by Equation 7.1

or 7.2 to have a non-trivial solution, the determinant of the matrix (A� �B) must be zero, i.e.

A� �Bj j ¼ 0: ð7:3Þ

137

The matrix (A� �B) is called the characteristic matrix of the system, and its determinant is called

the characteristic function while |A� �B| is the characteristic equation. For a structure ofNDOF,

the characteristic equation is a polynomial of degree N in �. The equation therefore has N roots

(�1, �2, �3, . . . , �N) which are real if the system matrix

S ¼ B�1A ð7:4Þ

is symmetric. The roots of the characteristic equation are called the characteristic or latent roots

or eigenvalues of the matrix S. In structural engineering, the eigenvalues are associated with more

than 1-DOF. For associated matrices of order greater than 3, the numerical work involved in

solving the eigenvalue problem is too great and too time-consuming to be carried out manually;

computers are needed for its solution. A number of methods can be used for the manual

calculations of small problems; there are also approximate methods for the calculation of the

first few eigenvalues of larger problems. There are three basic approaches for solving the

eigenvalue problem

g direct solution of the characteristic polynomialg iterative optimisation of eigenvectorsg transformation of the system matrix.

The first two methods can be used relatively easily to determine the eigenvalues and eigenvectors

for structures of up to 3 DOF, while the third approach requires the use of computers. In the

following, the first two methods are applied to the solution of 2- and 3-DOFmass–spring systems.

7.3. Determination of free normal mode vibration by solution of thecharacteristic equation

Consider the 2-DOF mass–spring system shown in Figure 7.1 which, for example, could be

considered as the mass–spring model of a column with the mass lumped together at two points

along its length. From the free-body diagram, the equations of motion for the two masses are

M1€xx1 ¼ �K1x1 � Kcx1 þ Kcx2 þ P1 tð Þ ð7:5aÞ

M2€xx2 ¼ þKcx1 � Kcx2 � K2x2 þ P2 tð Þ: ð7:5bÞ

Equations 7.5a and 7.5b may be written in matrix form as

M1 0

0 M2

� �€xx1

€xx2

� �þ

K1 þ Kcð Þ �Kc

�Kc K2 þ Kcð Þ

� �x1

x2

� �¼

P1 tð ÞP2 tð Þ

� �: ð7:6Þ

Figure 7.1 2-DOF mass–spring system

P1(t) P2(t)

M1 M2

x1 x2

K1 Kc K2


138

In order to determine the natural frequencies and mode shapes of vibration, set P1(t)¼P2(t)¼ 0

which yields

M1 0

0 M2

� �€xx1

€xx2

� �þ

K1 þ KcÞð �Kc

�Kc K2 þ KcÞð

� �x1

x2

� �¼

0

0

� �: ð7:7Þ

It is assumed that the motion of each mass in free vibration is simple harmonic, then

x1 ¼ X1 sin !tð Þ ð7:8aÞ

x2 ¼ X2 sin !tð Þ ð7:8bÞ

€xx1 ¼ �X1!2 sin !tð Þ ð7:9aÞ

€xx2 ¼ �X2!2 sin !tð Þ: ð7:9bÞ

Substitution of the expressions for x and €xx into Equation 7.7 yields

K1 þ KcÞð �Kc

�Kc K2 þ Kcð Þ

� �X1

X2

� �� !2 M1 0

0 M2

� �X1

X2

� �¼

0

0

� �ð7:10Þ

or

K1 þ Kc � !2M1

� ��Kc

�Kc K2 þ Kc � !2M2

� �" #

X1

X2

� �¼

0

0

� �: ð7:11Þ

Equation 7.11 is satisfied only if the determinant

K1 þ Kc � !2M1

� ��Kc

�Kc K2 þ Kc � !2M2

� ��

�� ¼ 0: ð7:12Þ

Expansion of the above determinant yields the characteristic equation

!4 � K1 þ Kcð Þ=M1 þ K2 þ Kcð Þ=M2½ �!2 þ K1K2 þ K1 þ K2ð ÞKc½ �=M1M2 ¼ 0 ð7:13Þ

from which the two angular frequencies !1 and !2 can be determined. Substitution in turn of the

calculated values for !1 and !2 into Equation 7.10 yields

X1

X2

� �

1

¼ Kc

K1 þ Kc � !21M1

¼ K2 þ Kc � !21M2

Kc

ð7:14aÞ

X1

X2

� �

2

¼ Kc

K1 þ Kc � !22M1

¼ K2 þ Kc � !22M2

Kc

: ð7:14bÞ

With no forces applied to the system the amplitudes of vibration will have no absolute values, and

only the amplitude ratios that determine the mode shapes can be determined. The first and second

mode shapes for the system shown in Figure 7.1 are therefore determined by first calculating the

ratios (X1/X2)1 and (X2/X2)2 corresponding to !1 and !2, respectively, from Equations 7.14a and

7.14b, and then assigning a value of, say, 1 to either X1 or X2.

Free vibration of multi-degree-of-freedom systems

139

Example 7.1

Determine the natural frequencies and corresponding mode shapes of vibration for the

2-DOF mass–spring system shown in Figure 7.1 if K1¼KcþK2¼K and M1¼M2¼M.

Substitution for M1, M2, K1, Kc and K2 into Equation 7.12 yields

!4 � 4K

M!2 þ 3K2

M2¼ 0

and hence

!2 ¼ K

M2� 1ð Þ:

We therefore have

!1 ¼


M

� �s

!2 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi3K

M

� �s:

Substitution of the expressions for !1 and !2 into Equations 7.14a and 7.14b yields the

amplitude ratios

X1

X2

� �

1

¼ 2K �M K=Mð ÞK

¼ 1

X1

X2

� �

2

¼ 2K �M 3K=Mð ÞK

¼ �1:

Figure 7.2 (a) First and (b) second modes of vibration of the mass–spring system in Example 7.1

ω = √(k/m)

(a)

ω = √(3k/m)

(b)


140

7.4. Solution of cubic characteristic equations by the Newtonapproximation method

In the Newton approximation method, successive estimates of � are achieved through the iterative

procedure

�iþ 1 ¼ �1 �� if g�0 �if g ð7:15Þ

where �{�} is the characteristic polynomial

� �f g ¼ �3 þ a�2 þ b�þ c ¼ 0 ð7:16Þ

The above values for the amplitude ratios imply that in the first mode the two masses move

in the same direction as if connected by a solid rod; in the second mode they move in the

opposite direction such that the midpoint of the central spring is at rest at all times, as

shown in Figures 7.2(a) and 7.2(b), respectively.

Example 7.2

Write down the equations for free vibration of the structure in Example 2.5 (Figure 2.14) and

hence establish the characteristic equation for the structure.

The equations of motion are given by

3M 0 0

0 2M 0

0 0 M

264

375

€xx1

€xx2

€xx3

264

375þ

7K �3K 0

�3K 5K �2K

0 �2K 2K

264

375

x1

x2

x3

264

375 ¼

0

0

0

264

375:

If SHM is assumed, the corresponding eigenvalue equation is

7K �3K 0

�3K 5K �2K

0 �2K 2K

264

375

X1

X2

X3

264

375� �

3M 0 0

0 2M 0

0 0 M

264

375

X1

X2

X3

264

375 ¼

0

0

0

264

375:

The above eigenvalue equation will have a non-trivial solution only if

7K � 3M�ð Þ �3K 0

�3K 5K � 2M�ð Þ �2

0 �2K 2K �M�ð Þ

��

��¼ 0:

Evaluation of this determinant leads to the characteristic equation

� �ð Þ ¼ 6�3 � 41��2 þ 72�2�� 24�3 ¼ 0

where

� ¼ K=M:


141

where

�iþ 1 ¼ �i ��3i þ a�2

i þ b�i þ c

3�2i þ 2a�i þ b

: ð7:17Þ

When calculating the first eigenvalue let the initial value of � be �i¼ 1¼ 0. When calculating the

third eigenvalue, set the initial value of � equal to the trace of M�1K. The second eigenvalue

can then be calculated by applying Theorem 7.1 in Section 7.6.1 below.

7.5. Solution of cubic characteristic equations by the direct methodFor the general characteristic equation given by Equation 7.16, let

Q ¼ a2 � 3b

9

R ¼ 2a3 � 9abþ 27c

54: ð7:18Þ

If Q3 – R2< 0 the characteristic equation has only one root, but if Q3 – R25 0 the equation has

three real roots which can be found by first calculating

� ¼ cos�1 R=

ffiffiffiffiffiffiQ3

q� �: ð7:19Þ

The roots of the cubic equation are then found in terms of � and are

�1 ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiQð Þ cos �

3� a

3

rð7:20aÞ

�2 ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiQð Þ cos �þ 2�

3� a

3

rð7:20bÞ

�3 ¼ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiQð Þ cos

�þ 4�

3� a

3

r: ð7:20cÞ

7.6. Two eigenvalue and eigenvector theoremsThe following two theorems frommatrix theory are useful for checking calculated eigenvalues and

eigenvectors and, in the case of 2- and 3-DOF systems, for reducing the amount of calculations.

7.6.1 Theorem 7.1The sum of the elements along the leading diagonal of the system matrix S¼M�1K, referred to as

the trace of S, is equal to the sum of its eigenvalues.

From this theorem it follows that

XNi¼ 1

Sii ¼XNi¼ 1

�i ¼XNi¼ 1

!2i : ð7:21Þ

7.6.2 Theorem 7.2If Xi and Xj, i 6¼ j, are two of the eigenvectors of the eigenvalue equation KX� �MX¼ 0, then

XiTMXj¼ 0. If M�1

K is symmetric, then Xi�1Xj¼ 0.


142

In the case of structures with 3 DOF, the second angular frequency and mode shape vector may be

found by applying the two theorems from the matrix algebra stated above.

Example 7.3

Use the Newton approximation method to determine the eigenvalues � for the character-

istic equation established in Example 7.2. Hence determine also the natural frequencies

and mode shape vectors for the three-storey shear structure in Example 2.5 (Figure 2.14)

in terms of the flexural rigidity EI of the columns and the weight w per metre span of the

floors.

The Newton approximation formula for the characteristic equation developed in Example

7.2 is

�iþ 1 ¼ �i �� ð Þ�0 �ð Þ ¼ �i �

6�3i � 41��2

i þ 72�2�i � 24�3

18�2i � 82��i þ 72�2

which will always converge towards the nearest root. In order to determine the first

eigenvalue, it is therefore convenient to assume that �i¼ 1¼ 0.0�. To four decimal places,

this yields

�i¼ 2 ¼ 0:0000�þ 0:3333� ¼ 0:3333�

�i¼ 3 ¼ 0:333�þ 0:0929� ¼ 0:4262�

�i¼ 4 ¼ 0:4262�þ 0:0074� ¼ 0:4336�

�i¼ 5 ¼ 0:4336�þ 0:0000� ¼ 0:4336�

and hence

�1 ¼ !21 ¼ 0:4336K=M:

In order to determine the highest eigenvalue, bearing in mind that the Newton method

converges towards the nearest root, the initial value for � is assumed to be equal to the

trace of the system matrix. We therefore have

�i¼ 1 ¼7

3�þ 5

2�þ 2� ¼ 41

6� ¼ 6:3333�

which yields

�i¼ 2 ¼ 6:8333�� 1:3289� ¼ 5:5044�

�i¼ 3 ¼ 5:5044�� 0:7875� ¼ 4:7169�

�i¼ 4 ¼ 4:7169�� 0:3860� ¼ 4:309�

�i¼ 5 ¼ 4:3309�� 0:1138� ¼ 4:2171�

�t¼ 6 ¼ 4:2171�� 0:0101� ¼ 4:2070�


143

�i¼ 7 ¼ 4:070�� 0:0001� ¼ 4:2069�

�i¼ 8 ¼ 4:2069�� 0:0000� ¼ 4:2069�

and hence

�3 ¼ !23 ¼ 4:2069K=M:

The second eigenvalue can now easily be determined by applying Theorem 7.1, which states

that the sum of the eigenvalues is equal to the trace of the system matrix:

�1 þ �2 þ �3 ¼41

6

K

M:

Substitution of the values for �1 and �3 into the above equation yields

�2 ¼ !22 ¼ 2:1928K=M:

The eigenvectors can now be determined by assuming that, say, X1 = 1, and substituting the

different values for � one at a time into the eigenvalue equation, yielding

X1 ¼ f 1:0000 1:8997 2:4256 g

X2 ¼ f 1:0000 0:405 �1:4578 g

X3 ¼ f 1:0000 �1:8736 1:6979 g:

Finally, substitution of the values for K andM used in Example 2.4 gives the following values

for the eigenvalues, natural angular frequencies and frequencies

!21 ¼ 8:1300� 10�3 EIg

w!1 ¼ 0:0901665

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg

w

� �s

f1 ¼ 0:0143504


w

� �s

!22 ¼ 41:1150� 10�3 EIg

w!2 ¼ 0:2027683


w

� �s

f2 ¼ 0:0322715


w

� �s

!23 ¼ 78:8793� 10�3 EIg

w!3 ¼ 0:2808547


w

� �s

f3 ¼ 0:0446994


w

� �s:


144

Example 7.4

Determine the roots of the characteristic equation developed in Example 7.2 by the direct

method.

The characteristic equation may be written as

�3 � 41

6��2 þ 12�2�� 4�3 ¼ 0

and hence

Q ¼ a2 � 3b

9¼ �41=6ð Þ2�3� 12

9�2 ¼ 1:1882716�2

R ¼ 2a3 � 9abþ 27c

54¼ 2� ð�41=6Þ3 � 9� ð�41=6Þ � 12þ 27� ð�4Þ

54�3

¼ �0:1510631�3

Q3 � R2 ¼ 1:18827163 � �0:1510631ð Þ2

�6 ¼ 1:700647�6 > 0:

The characteristic equation therefore has three real roots:

� ¼ cos�1 R=

ffiffiffiffiffiffiQ3

q� �¼ cos�1 �0:1510631=

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:18827163

p=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:18827163

p� �

¼ 96:697255

�1 ¼ �2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiQð Þ cos �

3

� �� a

3

s

¼ �2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:1882716ð Þ cos 96:697255

3

� �s��41=6

3

( )� ¼ 0:4336011�

�2 ¼ �2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiQð Þ cos �þ 2�

3

� �� a

3

s

¼ �2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:1882716ð Þ cos 96:697255þ 360

3

� �s��41=6

3

( )�

¼ 4:2068786�

�3 ¼ �2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiQð Þ cos �þ 4�

3� a

3

� �s

¼ �2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:882716ð Þ cos 96:697255þ 7620

3

� �s��41=6

3

( )�

¼ 2:1928537�:

The direct method therefore yields the same values for the eigenvalues, to four decimal places,

as the Newton approximation method. The former method does not necessarily calculate the

roots in ascending order however, as is apparent from the above results.


145

7.7. Iterative optimisation of eigenvectorsEquation 7.10 may be written in general matrix notation as

KX � !2MX ¼ 0 ð7:22Þ

where K is the stiffness matrix and M is the mass matrix for the structure. When the mass of a

structure is lumped together at nodes, which is usually the case in manual calculations, the

mass matrix is diagonal.

In the iterative method, the eigenvalues !2 and eigenvectors X are determined by optimising an

assumed mode shape vector through an iterative procedure on either

!2X ¼ M�1KX ð7:23Þ

or

X=!2 ¼ K�1MX : ð7:24Þ

Iterations on Equation 7.23 will cause the assumed eigenvector to converge towards the mode

corresponding to the highest eigenvector and hence the highest frequency; iterations on Equation

7.24 will cause the assumed vector to converge towards the eigenvector corresponding to the

lowest frequency. Equation 7.23 involves the inversion of the mass matrix M which, when the

matrix is diagonal, is achieved by simply inverting each of the elements on the leading diagonal.

The calculation of the lowest eigenvalue using Equation 7.24 requires the inversion of the stiffness

matrix K. Because the stiffness matrix is banded, the inversion process takes more time than the

inversion of the mass matrix. The inversion of the stiffness matrix can, however, be avoided by

calculating the lowest eigenvalue and eigenvector as follows. Let

BX i ¼ �I�M�1K

�X i ð7:25Þ

where � is a constant larger than the highest eigenvalue, I is a unit matrix and B is a square matrix

of the same order as M and K. From Equation 7.23 it follows that

M�1KX i ¼ !2

i IX i: ð7:26Þ

Substitution of the expression for M�1KXi given in Equation 7.26 into Equation 7.25 yields

BX i ¼ �� !2i

�IX i ¼ �� !2

i

�X i: ð7:27Þ

Assuming an initial vector Xi, iterations on Equation 7.27 will yield the highest value of [�� !i2]

and hence the lowest possible value for !i2, and therefore

!2i ¼ !2

1

X i ¼ X1:ð7:28Þ

Iteration algorithms based on Equations 7.23 and 7.27 will yield the highest and lowest natural

frequency and corresponding mode shapes for any structure. In the following, the iterative

method for determining the natural frequencies and mode shapes is demonstrated by solving

first a 2- and then a 3-DOF system.


146

Example 7.5

Use two iterative optimisation procedures to determine the highest and lowest frequencies

and mode shapes for the mass–spring system specified in Example 7.1.

The eigenvalue equation for the mass–spring system is given by

2K �K

�K 2K

� �X1

X2

� �� !2 M 0

0 M

� �X1

X2

� �¼

0

0

� �

and hence

K ¼ K2 �1

�1 2

� �

M ¼ M1 0

0 1

� �:

The equation that will yield the highest natural angular frequency is therefore

!22X2 ¼

K

M

2 �1

�1 2

� �X12

X22

� �:

Assume the vector for starting the iterative process to be

X2 ¼X12

X22

� �¼

1:0

1:0

� �;

the iterative process then proceeds as follows.

1st iteration: !22

X12

X22

� �¼ K

M

2 �1

�1 2

� �1:000

0:000

� �¼ 2:00� K

M

1:000

�0:500

� �

2nd iteration: !22

X12

X22

� �¼ K

M

2 �1

�1 2

� �1:000

�0:500

� �¼ 2:500� K

M

1:000

�0:800

� �

3rd iteration: !22

X12

X22

� �¼ K

M

2 �1

�1 2

� �1:000

0:00

� �¼ 2:800� K

M

1:000

�0:929

� �

4th iteration: !22

X12

X22

� �¼ K

M

2 �1

�1 2

� �1:000

0:929

� �¼ 2:927� K

M

1:000

�0:976

� �

As the iterative process proceeds, the values for the product !22X2 will converge to

!22

X12

X22

� �¼ 3:0� K

M

1:0

�1:0

� �

and thus

!2 ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi3K

M

� �s

X2 ¼ 1:0 �1:0g:f


147

The first eigenvalue is now found by applying Theorem 7.1, which yields

!21 þ !2

2 ¼K

M2þ 2ð Þ:

Substitution of the expression for !2 into the above equation is determined by applying

Theorem 7.2. If we set X11¼ 1.0, then

1:0 X21½ �1:0

�1:0

� �¼ 0

which yields

X21 ¼ 1:0

and hence

X i ¼ 1:0 1:0f g:

Example 7.6

Use the iterative optimisation method to determine the first natural frequency and mode

shape vector for the structure given in Example 2.5 (Figure 2.14).

Let the shear stiffness 12EI/L3 of each column be K and the mass per span of each floor beM.

With this notation, the matrix formulation of the equation of motion is

M

3 0 0

0 2 0

0 0 1

264

375

€xx1

€xx2

€xx3

264

375þ K

7 �3 0

�3 5 �2

0 �2 2

264

375

x1

x2

x3

264

375 ¼

0

0

0

264

375:

Assuming SHM, substitution for x and €xx yields

K

7 �3 0

�3 5 �2

0 �2 2

264

375

X1

X2

X3

264

375� !2M

3 0 0

0 2 0

0 0 1

264

375

X1

X2

X3

264

375 ¼

0

0

0

264

375:

To determine the lowest eigenvalue, we iterate on Equation 7.27 which requires that the

matrix

B ¼ �I�M�1K

�

be established. This in turn requires that we first establish the system matrix M�1K and then

choose a value for �, i.e.

M�1K ¼ K

6M

14 �6 0

�9 15 �6

0 �12 12

264

375:


148

The value of � must be greater than the highest eigenvalue. Theorem 7.1 states that the trace

of the system matrix is equal to the sum of the eigenvalues. A value for � equal to the trace is

therefore satisfactory, i.e.

� ¼ K

6K14þ 15þ 12ð Þ ¼ 41K

6K

and hence

B ¼ �I�M�1K

�¼ K

6M

27 6 0

9 26 6

0 12 29

264

375:

It remains to assume a suitable initial vector X1; a simple choice would be

X1 ¼ 1:0 0:0 0:0 g:f

Alternatively, we can choose the vector used in Example 2.5 which assumes that the mode

shape is similar to the deflected form caused by a horizontal force applied at each level,

and proportional to the weight of the corresponding floor. If the latter is assumed, we have

X1 ¼ 1:0 1:67 2:0 gf

and the iterative procedure is as follows.

1st iteration: �3X1 ¼K

6M

27 6 0

9 26 6

0 12 29

264

375

1:000

1:670

2:000

264

375 ¼ 6:170� K

M

1:000

1:40

2:104

264

375

2nd iteration: �3X1 ¼K

6M

27 6 0

9 26 6

0 12 29

264

375

1:000

1:40

2:104

264

375 ¼ 6:240� K

M

1:000

1:86

2:187

264

375

3rd iteration: �3X1 ¼K

6M

27 6 0

9 26 6

0 12 29

264

375

1:000

1:786

2:87

264

375 ¼ 6:286� K

M

1:000

1:818

2:250

264

375

4th iteration: �3X1 ¼K

6M

27 6 0

9 26 6

0 12 29

264

375

1:000

1:818

2:0

264

375 ¼ 6:318� K

M

1:000

1:840

2:97

264

375


6M

27 6 0

9 26 6

0 12 29

264

375

1:000

1:840

2:297

264

375 ¼ 6:340� K

M

1:000

1:57

2:22

264

375


6M

27 6 0

9 26 6

0 12 29

264

375

1:000

1:857

2:322

264

375 ¼ 6:357� K

M

1:00

1:867

2:350

264

375


149


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:67

2:50

24

35 ¼ 6:367� K

M

1:000

1:875

2:370

24

35


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:875

2:370

24

35 ¼ 6:375� K

M

1:000

1:882

2:385

24

35


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:882

2:385

24

35 ¼ 6:382� K

M

1:000

1:887

2:396

24

35


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:887

2:396

24

35 ¼ 6:387� K

M

1:000

1:890

2:404

24

35


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:890

2:404

24

35 ¼ 6:390� K

M

1:000

1:893

2:410

24

35


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:893

2:410

24

35 ¼ 6:393� K

M

1:000

1:895

2:414

24

35


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:895

2:414

24

35 ¼ 6:395� K

M

1:000

1:896

2:417

24

35


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:896

2:417

24

35 ¼ 6:396� K

M

1:000

1:897

2:419

24

35


6M

27 6 0

9 26 6

0 12 29

24

35

1:000

1:897

2:419

24

35 ¼ 6:397� K

M

1:000

1:898

2:421

24

35:

From Equation 7.27,

�3 ¼ �� !21

and hence

!21 ¼

K 6:8333333� 6:397ð ÞM

¼ 0:4363333� K

M

where

K ¼ 12EI

4:03

M ¼ 10w

g:


150

7.8. The Rayleigh quotientThe eigenvalue equation for a general N-DOF system is given by Equation 7.22. Pre-multiplica-

tion of each term by XT yields

XTKX � !2XT

MX ¼ 0 ð7:29Þ

and hence

!2 ¼ XTKX

XTMX: ð7:30Þ

The expression for the square of the natural frequency given by Equation 7.30 is referred to as

the Rayleigh quotient. It has the property that, even for approximately correct values of the

eigenvectors or mode shape vectors, the values for the frequencies are reasonably correct as

demonstrated by Examples 2.1, 2.2, 2.5 and 2.6 (where the quotient is used without explicitly

stating so). That this is the case can be seen simply by pre-multiplying each term in Equation

7.29 by 1/2, which yields

XTKX ¼ !2XT

MX ð7:31Þ

which states that the maximum strain energy is equal to the maximum kinetic energy.

7.9. Condensation of the stiffness matrix in lumped mass analysisWhen the mass of a structure is assumed to be concentrated at the nodes, it is usual to consider

only the inertia due to translational movements and to neglect that is due to rotation. This assumes

that the lumped masses are concentrated as point masses with radii of gyration equal to zero. In

the case of flexible structures where the joints rotate, the elements on the leading diagonal of the

Substitution of the expressions for K and M into the expression for !21 yields

f1 ¼ 0:0143956


w

� �s:

This implies that the frequency, whose value is given in Example 2.4, has converged after 15

iterations to within 0.23% of the correct value, with a corresponding mode shape of

X1 ¼ 1:000 1:898 2:421 g:f

The highest natural frequency can be determined by iterations on Equation 7.23 and the

second frequency by applying Theorem 7.1. Alternatively, all the eigenvalues may be

determined by setting up and solving the characteristic equation, which in this case is a

cubic equation in �. This can be solved either graphically or by application of the theory

for solving cubic equations. Having determined one eigenvalue, the characteristic polynomial

can alternatively be reduced by factorisation. In this case, the resulting quadratic character-

istic equation can be solved by using the standard formula for determining the roots of such

equations.


151

mass matrix corresponding to the rotational degrees of freedom will therefore be zero. In such

cases, the mass matrix cannot be inverted. The elements related to rotation therefore need to

be eliminated by condensing the stiffness matrix. Condensation of the stiffness matrix may also

be desirable to reduce the overall DOF of structures with a very large number of DOF in order

to reduce the numerical problem. Assume that the degrees of freedom to be reduced or condensed

are the first � unknown rotations, and carry out a Gauss–Jordan elimination of these coordinates.

After this elimination process, the stiffness equation may be arranged in partitioned form as

follows

I �~TT

0 �KK

" #�

x

� �¼

0

P

� �ð7:32Þ

where � is the displacement vector corresponding to the � DOF to be reduced and x is the vector

corresponding to the remaining x independent DOF. It should be noted that in Equation 7.32 it is

assumed that at the dependent degrees of freedom �, the external forces are zero. Equation 7.32 is

equivalent to the following two relationships

� ¼ ~TTx ð7:33Þ

~KKx ¼ P: ð7:34Þ

Equation 7.33, which expresses the relationship between the displacement vectors x and �, may

also be written as

�

x

� �¼

~TT

I

" #x½ �: ð7:35Þ

In Equation 7.34, which shows the relationship between the displacement vector x and the

force vector P, ~KK is the reduced stiffness matrix. ~KK may also be expressed by the following trans-

formation of the system matrix

~KK ¼ TTKT ð7:36Þ

where

T ¼~TT

I

" #: ð7:37Þ

Similarly, the mass and the damping matrix (the latter is introduced in Chapter 8) may be reduced

by the transformations

~MM ¼ TTMT ð7:38Þ

~CC ¼ TTCT ð7:39Þ

where the transformation matrix T is given by Equation 7.37.


152

Example 7.7

Reduce the DOF of the three-storey shear structure in Example 2.5 (Figure 2.14) to a 1-DOF

system by eliminating the translational displacements at the first- and second-floor levels.

Hence calculate the natural frequency and compare the value obtained with those obtained

previously.

A Gauss–Jordan elimination of the elements in the stiffness matrix corresponding to the

displacements at the first- and second-floor levels results in the transformation

K ¼ K

7 �3 0

�3 5 �2

0 �2 2

264

375! K

1 0 � 3

13

0 1 � 7

13

0 012

13

266666664

377777775

and hence

~KK ¼ 12

13K :

The corresponding reduced mass matrix is found through the transformation

~MM ¼ TTMT ¼ M

3

13

7

131

� � 3 0 0

0 2 0

0 0 1

264

375

3

13

7

13

1

2666664

3777775¼ 294

169M

and hence

!2 ¼ 3� 294� K

13� 169�M¼ 0:4081153� K

M

where, as before,

K ¼ 12EI

L3

M ¼ 10w

g:

Substitution of the expressions for K and M into the expression for !2 yields

f ¼ 0:0139223


w

� �s

which represents an error of 3.96% as compared to the 4.28% error resulting from the much

simpler method of reduction used in Chapter 2. The above reduction can be checked by

substitution of the values for the matrices T and K into Equation 7.36 and carrying out

the implied matrix multiplications.


153

7.10. Consistent mass matricesThe modelling of structural mass by lumped mass matrices usually leads to satisfactorily accurate

values for the frequencies, and has the advantage of reducing the amount of computer storage and

calculations involved in solving the eigenvalue problem. In the case of buildings, the total mass will

vary with the usage and it is usually difficult to estimate the mass and mass distribution accurately.

This further justifies the lumpedmass approach. From a computational point of view, however, it is

probably equally convenient to set up a mass matrix that takes account of the distribution of the

mass in individual members by using consistent element mass matrices. For plane frames, it can

be shown that the relationship between the inertia force vector, mass matrix and acceleration

vector for an uniform element of length L and mass m per unit length is given by

Ix1

Iy1

I�1

Ix2

Iy2

I�2

2666666664

3777777775¼ mL

420

140 0 0 70 0 0

0 156 22L 0 54 �13L

0 22L 4L2 0 13L �3L2

70 0 0 140 0 0

0 54 13L 0 156 �22L

0 �13L �3L2 0 �22L 4L2

2666666664

3777777775

€xx

€yy€��

€xx

€yy€��

2666666664

3777777775; ð7:40Þ

from which it can be seen that the consistent mass matrix has the same banded form as the

stiffness matrix for the member. Similar 12� 12 mass matrices can also be set up for 3D

structures. Consistent mass matrices for both 2D and 3D structures can be reduced through the

transformation given by Equation 7.38 by, for example, eliminating the rotational coordinates.

Example 7.8

Construct the stiffness and mass matrices for the stepped antenna mast shown in Figure 7.3,

assuming that the flexural rigidity and mass of the lower half of the mast are 2EI and 2 m per

unit length, and the flexural rigidity and mass of the top half of the mast are EI andm per unit

length. Ignoring axial stiffness, condense the matrices to include horizontal translations only.

The general stiffness matrix for a plane frame member, ignoring the axial stiffness, is given by

K ¼ EI

12=L3 6=L2 �12=L3 6=L2

6=L2 4=L �6=L2 2=L

�12=L3 �6=L2 12=L3 �6=L2

6=L2 2=L �6=L2 4=L

26664

37775:

The general mass matrix for a plane frame member, ignoring the axial inertia forces, is

M ¼ mL

420

156 22L 54 �13L

22L 4L2 13L �3L2

54 13L 156 �22L

�13L �3L2 �22L 4L2

26664

37775:

The stiffness matrix for the mast is now constructed as indicated by

K ¼ K1ð Þ22 þ K

2ð Þ11 K

2ð Þ12

K2ð Þ21 K

2ð Þ22

" #


154

Figure 7.3 Stepped antenna mast

EIL

L2EI

and hence

Kx ¼ EI

36=L3 �6=L2 �12=L3 6=L2

�6=L2 12=L �6=L2 2=L

�12=L3 �6=L2 12=L3 �6=L2

6=L2 2=L �6=L2 4=L

266664

377775

x1

�1

x2

�2

266664

377775:

Rearrangement of K in order to reduce it by eliminating �1 and �2 yields

Kx ¼ EI

12=L 2=L �6=L2 �6=L2

2=L 4=L 6=L2 �6=L2

�6=L2 6=L2 36=L3 �12=L3

�6=L2 �6=L2 �12=L3 12=L3

266664

377775

�1

�2

x1

x2

266664

377775:

Finally, reduction of the matrix by the Gauss–Jordan elimination process leads to

Kx ¼ EI

1 0 �9=11L �3=11L

0 1 21=11L �15=11L

0 0 216=11L3 �60=11L3

0 0 �60=11L3 54=11L3

26664

37775

�1

�2

x1

x2

26664

37775


155

7.11. Orthogonality and normalisation of eigenvectorsBefore proceeding to determine the dynamic response of multi-DOF structures, it is necessary

to consider what are known as the orthogonality properties of the eigenvectors. Let �i and �jbe two of the eigenvalues corresponding to the eigenvectors or mode shape vectors Xi and Xj,

where i 6¼ j, of a multi-DOF system represented by the eigenvalue equation

KX � �MX ¼ 0: ð7:41Þ

We therefore have

KX i � �iMX i ¼ 0 ð7:42Þ

KX j � �jMX j ¼ 0 ð7:43Þ

and hence the condensed stiffness matrix ~KK and associated transformation matrix T are given

by

~KK ¼ 6EI

11L3

36 �10

�10 9

� �

T ¼

9=11L 3=11L

�21=11L 15=11L

1 0

0 1

26664

37775:

The mass matrix for the mast is assembled in exactly the same way as the stiffness matrix;

hence

M€xx ¼ mL

420

468 �22L 54 �13L

�22L 12L2 13L �3L2

54 13L 156 �22L

�13L �3L2 �22L 4L2

26664

37775

€xx1€��1

€xx2€��2

26664

37775:

Transformation of the above mass matrix to conform with the Gauss–Jordan elimination of

�1 and �2 yields

M€xx ¼ mL

420

12L2 �3L2 �22L 13L

�3L2 4L2 �13L �22L

�22L �13L 468 54

13L �22L 54 156

26664

37775

�1

�2

x1

x2

26664

37775:

The condensed mass matrix is now found through the transformation given by Equation 7.38

as

~MM ¼ TTMT

~MM ¼ mL

50 820

62 148 8880

8880 13 212

� �:


156

and transposition of each term in Equation 7.42 yields

XTi K

T � �iXTi M

T ¼ 0: ð7:44Þ

If both K and M are symmetric matrices, then

XTi K� �iX

Ti M ¼ 0: ð7:45Þ

Post-multiplication of each term in Equation 7.45 by Xj yields

XTi KX j � �iX

Ti MX j ¼ 0: ð7:46Þ

Pre-multiplication of each term in Equation 7.43 by XTi yields

XTi KX j � �jX

Ti MX j ¼ 0: ð7:47Þ

Finally, subtraction of Equation 7.47 from Equation 7.46 yields

�j � �i

� �XT

i MX j ¼ 0 ð7:48Þ

and since �j 6¼ �i, it follows that

XTi MX j ¼ 0: ð7:49Þ

If the zero value for XiTMXj is substituted into either Equation 7.46 or Equation 7.47, it also

follows that

XTi KX j ¼ 0: ð7:50Þ

The relationships given by Equations 7.49 and 7.50 still apply if the eigenvectors are normalised.

Let

XTi MX i ¼ ~MMi ð7:51aÞ

XTj MX j ¼ ~MMj ð7:51bÞ

and hence

Z i ¼ X i=

ffiffiffiffiffiffi~MMi

qð7:52aÞ

Z j ¼ X j=ffiffiffiffiffiffiffi~MMj

qð7:52bÞ

where Zi and Zj are the normalised eigenvectors of vectors Xi and Xj with respect to M. Hence

ZTi MZj ¼ XT

i MX j=ffiffiffiffiffiffiMi

p ffiffiffiffiffiffiffiMj

p¼ 0 ð7:53Þ

ZTi MZi ¼ XT

i MX i= ~MMi ¼ 1 ð7:54aÞ

ZTj MZj ¼ XT

j MX j= ~MMj ¼ 1: ð7:54bÞ


157

Pre-multiplication of Equation 7.42 by XTi and Equation 7.43 by XT

j and then substitution ofp(Mi)Zi for Xi and

p(Mj)Zj for Xj in the resulting equations yields

ZTi KZ i � �iZ

Ti MZ i ¼ 0 ð7:55aÞ

ZTj KZ j � �jZ

Tj MZ j ¼ 0: ð7:55bÞ

From Equation 7.53, it follows that

ZTj MZ j ¼ 0 ð7:56Þ

and since

ZTj MZ j ¼ ZT

j MZ j ¼ 1

it follows that

ZTj KZ i ¼ �i ¼ !2

i ð7:57aÞ

ZTj KZ j ¼ �j ¼ !2

j : ð7:57bÞ

The matrix Z in which the columns are the normalised eigenvectors

Z1;Z2; . . . ;Z i; . . . ;Z j ; . . . ;ZN

is referred to as the modal or mode shape matrix of the dynamic matrix M�1K. From Equations

7.55a, 7.55b, 7.57a and 7.57b it follows that

ZTMZ ¼ I ð7:58aÞ

ZTKZ ¼ � ð7:58bÞ

where I is the identity or unit matrix and � is the diagonal matrix

� ¼ diag �1; �2; . . . ; �i; . . . ; �j ; . . . ; �N

: ð7:59Þ

Example 7.9

Normalise the eigenvectors calculated in Example 7.3 with respect to the mass matrix, and

write down the normalised mode shape matrix. The weight of the floors is 20.0 kN/m.

~MM1 ¼ XT1MX1 ¼ 1:0000 1:8997 2:4256½ �

3M 0 0

0 2M 0

0 0 M

264

375

1:0000

1:8997

2:4256

264

375 ¼ 16:101256M

~MM2 ¼ XT2MX2 ¼ 1:0000 0:1405 �1:4578½ �

3M 0 0

0 2M 0

0 0 M

264

375

1:0000

0:1405

�1:4578

264

375 ¼ 5:1646613M

~MM3 ¼ XT3MX3 ¼ 1:0000 �1:8736 1:6979½ �

3M 0 0

0 2M 0

0 0 M

264

375

1:0000

�1:8736

1:6979

264

375 ¼ 12:902600M


158

7.12. Structural instabilityIn the case of structures, frequency analysis is mostly used to predict their response to various

forms of dynamic loading, but it can also be used to study their stability.

In Chapter 2, we introduced the concept of an equivalent structural spring stiffness and showed

that the critical value for the axial force of a column occurs when the equivalent spring stiffness

and hence the frequency was zero, and that this would happen when the sum of the elastic stiffness

and geometric stiffness was zero (Equation 2.38). This concept may be extended to investigate the

stability of multi-DOF structures by calculating their natural frequencies for increasing loading

by updating the geometry matrix for each increment of loading and extrapolating the calculated

lowest frequency to zero.

Frequency analysis is also a useful tool for checking the stability of structures, especially the rota-

tional stiffness of designs such as domes, circular cable beam roofs and guyed masts. Another

application is the investigation of stability of structures in which members need to be removed

and replaced for refurbishment, in which case their stiffness and hence frequencies need to be

calculated without the member or members removed for replacement. A very rotational mode

having a very low frequency could lead to rotational collapse (and has in fact done so).

7.12.1 Stiffness instabilityConsidering the two parts of the stiffness matrix KE and KG with respect to instability, we can

recognise two types of instability: geometrical and numerical. The first type, geometrical

instability, arises from large deflections normally associated with flexible structures such as

cable-stayed bridges and guyed masts. Updating the geometrical stiffness matrix KG in small

steps of loading should overcome this instability.

The numerical instability arises in multi-DOF systems when the elements of the stiffness matrix K

and/or mass matrix M are too unevenly distributed. In certain situations, when KE<KG then

M�1K< 0 and this produces negative eigenvalues. This is an indication of structural instability

of the numerical model, in particular when the mass matrix has zero values in its diagonal

elements. In such cases, a process of elimination of that degree of freedom in the system should

solve the problem. Another source of numerical instability is the time-stepping process for non-

linear time-domain analyses of structures. This has been discussed in detail by Buchholdt (1988).

Z1 ¼ X1=

ffiffiffiffiffiffiffi~MM1

q

Z2 ¼ X2=


q

Z3 ¼ X3=


q

and hence

Z ¼1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

264

375� 10�3:


159

7.12.2 Determination of elastic stability by eigenvalue analysisIn Chapter 2, it is shown how different types of linear beam elements can be represented as

single-DOF systems. Furthermore, it is shown that the total spring stiffness is the sum of the

elastic stiffness and a geometry stiffness, and that the critical load-causing instability can be

determined by setting K¼KEþKG to zero.

Since !2¼K/M where M is the equivalent lumped mass, it follows that !2¼ 0 when K¼ 0. The

square of the natural angular frequency !2 may be considered as the eigenvalue for the

single-DOF system.

Similarly for a multi-DOF system, a structure will be unstable if one or more of the eigenvalues are

zero. Theoretically, some matrices may yield negative eigenvalues; this is highly unlikely in the

case of a structure however, as it would indicate that some part of the structure has zero stiffness,

for example, be a mechanism. It is more likely that the lowest eigenvalue associated with a torsion

mode will be near to zero.

If that is the case, it may be advisable to undertake one or more eigenvalue calculations with

increased loading and updated stiffness matrix, and determine the critical load by extrapolation.

Such an analysis may also be worth considering when refurbishing structures in which elements

need to be replaced or removed altogether.

Problem 7.1

Figure 7.4 Three-storey shear structure

M

M

M

K/2

K/2

K/2

K/2

K/2

K/2

x3

x2

x1

Formulate the equations of motion for the free vibration of the three-storey shear structure

shown in Figure 7.4. Assume the mass of each floor to be M, the shear stiffness of each


160

REFERENCES

Buchholdt HA (1988) Introduction to Cable Roof Structures, 2nd edn. Thomas Telford,

London.

FURTHER READING




Kreider DL, Kuller RG, Ostberg DR and Perkins FW (1966) An Introduction to Linear

Analysis. Addison-Wesley, London.



column to be K/2 and the damping to be negligible. The weight of the columns may be

ignored. Establish the characteristic equation for the building. Solve the equation by plotting

the values of the characteristic polynomial versus increasing values of !2. Hence determine

the mode shapes of vibration by substitution of the obtained values for !2 into the equations

of motion. Normalise the mode shape vectors and provide the resulting mode shape matrix.

Problem 7.2

Use Newton’s approximation method to solve the characteristic polynomial established for

the structure in Problem 7.1.

Problem 7.3

Use iterative procedures to determine the first and third eigenvalues for the structure shown

in Figure 7.4. Hence determine the second eigenvalue and the natural frequencies of the

building. Finally, establish the eigenvectors and check the results by applying the orthogon-

ality properties of eigenvectors.

Problem 7.4

Determine the EI values for the two sections of the antenna-mast in Example 7.8 if each

section is 10.0 m long, the mast supports a disc of mass 500 kg at the top and the first natural

frequency has to be equal to or greater than 4.0 Hz. Assume the mass of the upper half of the

mast to be 1600 kg/m and that of the lower half to be 3200 kg/m. Determine also the first and

second mode shapes.


161


ISBN: 978-0-7277-4176-9



Chapter 8

Forced harmonic vibration ofmulti-degree-of-freedom systems

8.1. IntroductionWhen excited by random forces such as wind, waves and earthquakes, structures will respond in a

number of different modes although most civil engineering structures respond mainly in the first

mode. Particularly in the case of line-like structures such as towers and chimneys, responses in

higher modes will contribute to the maximum stresses and strain set up in the structure. It is there-

fore necessary to take account of the contribution from these modes.

In the case of linear structures, this can be done by calculating the response in the individual

modes and then applying the principle of superimposition. In order to present the method of

approach it is, as in the case of free vibration, only necessary to consider 2- and 3-DOF systems,

as the principles applied for the solution of the equations of motion for these systems are equally

applicable to structures with more degrees of freedom. This chapter not only presents methods for

solving the equations of motion, but also methods for constructing suitable damping matrices

that, as far as possible, will correctly model the damping in the different structural modes.

Before taking damping into account, it is convenient to first consider the problem of solving an

undamped 2-DOF system subjected to harmonic excitation.

8.2. Forced vibration of undamped 2-DOF systemsConsider the 2-DOF system shown in Figure 8.1, where the two masses are acted upon by the two

pulsating forces P1 sin(!1t) and P2 sin(!2t) as shown.

The equation of motion for this system is given by

M 0

0 M

� �€xx1

€xx2

� �þ

2K �K

�K 2K

� �x1

x2

� �¼

P1 sin !1tð ÞP2 sin !2tð Þ

� �ð8:1Þ

or

M€xxþ Kx ¼ P tð Þ: ð8:2Þ

Inspection of Equation 8.1 does not immediately indicate a straightforward method of solution.

In the following, it is therefore demonstrated how the use of the eigenvectors of the equations of

motion for free vibrations can be used to reduce the 2-DOF system shown in Figure 8.1 to

two equivalent 1-DOF systems by decoupling the equations of motion. In Example 7.1, it is

shown that the frequencies and mode shapes for the mass–spring system are !1¼p(K/M)

and !2¼p(3K/M) and X1¼ {1, 1} and X2¼ {1, �1}. The mode-shape matrix X is therefore

163

given by

X ¼1 1

1 �1

� �:

Let

x ¼ Xq;

then

€xx� X€qq: ð8:3Þ

Substitution of these expressions for €xx and x into Equation 8.2 yields

MX€qqþ KXq ¼ P tð Þ: ð8:4Þ

Pre-multiplication of each term in Equation 8.4 by XT yields

XTMX€qqþ X

TKXq ¼ X

TP tð Þ ð8:5Þ

and hence

1 1

1 �1

� �M 0

0 M

� �1 1

1 �1

� �€qq1

€qq2

� �þ

1 1

1 �1

� �2K �K

�K 2K

� �1 1

1 �1

� �q1

q2

� �¼

1 1

1 �1

� �P1 tð ÞP2 tð Þ

� �:

ð8:6Þ

The matrix multiplications yield

2M 0

0 2M

� �€qq1

€qq2

� �þ

2K 0

0 6K

� �q1

q2

� �¼

P1 tð Þ þ P2 tð ÞP1 tð Þ � P2 tð Þ

� �: ð8:7Þ

Since the stiffness matrix has been diagonalised as a result of this operation,

2M€qq1 þ 2Kq1 ¼ P1 tð Þ þ P2 tð Þ ð8:8aÞ

2M€qq2 þ 6Kq2 ¼ P1 tð Þ � P2 tð Þ: ð8:8bÞ

Figure 8.1 2-DOF lumped mass–spring system acted on by harmonic forces

P1 sin(ω1t) P2 sin(ω2t)

M M

x1 x2

K K K


164

If

P1 sin !1tð Þ ¼ P2 sin !2tð Þ ¼ P sin !tð Þ

then

2M€qq1 þ 2K€qq1 ¼ 2P sin !tð Þ ð8:9aÞ

2M€qq2 þ 6Kq2 ¼ 0: ð8:9bÞ

The equations of motion for the 2-DOF mass–spring system have therefore been transformed to

two decoupled equations, each having the same form as the equation of motion for a 1-DOF

system. It should be noted that the natural frequencies of the two equivalent 1-DOF systems

represented by Equations 8.8a and 8.8b or Equations 8.9a and 8.9b arep(K/M) and

p(3K/M)

respectively, and therefore are the same as the first and second natural frequencies of the original

2-DOF system.

From Equations 4.15 and 4.12, the response of an undamped 1-DOF system to harmonic

excitation (since �¼ 0) is

x ¼ P0

K

1

1� r2sin !tð Þ: ð8:10Þ

The solutions to Equations 8.9a and 8.9b are therefore

q1 ¼2P

K �M!2sin !tð Þ

q2 ¼ 0:

ð8:11Þ

Substitution of the expressions for q1 and q2 into Equation 8.3 yields

X ¼1 1

1 �1

� � 2P

K �M!2sin !tð Þ

0

264

375 ¼

2P

K �M!2sin !tð Þ

2P

K �M!2sin !tð Þ

2664

3775: ð8:12Þ

It should be noted that the decoupling of the equations of motion in the way shown is achieved as

a result of the orthogonality properties of the eigenvectors presented in Chapter 7. When the

eigenvectors are not normalised, this yields

XTKX ¼ ~XX ð8:13Þ

XTMX ¼ ~MM ð8:14Þ

where

~KK ¼ diag K11;K22; . . . ;Kii; . . . ;KNNf g ð8:15Þ

~MM ¼ diag M11;M22; . . . ;Mii; . . . ;MNNf g: ð8:16Þ

Forced harmonic vibration of multi-degree-of-freedom systems

165

The elements Kii andMii are referred to as the modal stiffness and modal mass in the ith mode. It

should also be noted that when the eigenvectors are normalised, Kii¼!i2 and Mii¼ 1.

An examination of Equation 8.3, which may be written as

X ¼X11 X12

X21 X22

� �q1

q2

� �¼

X11q1 þ X12q2

X21q1 þ X22q2

� �; ð8:17Þ

reveals that when the scalars q1 and q2 are multiplied by the first and second eigenvector, they yield

the contribution by each mode to the total response.

8.3. Forced vibration of damped 2-DOF systemsConsider the 2-DOF system shown in Figure 8.2, where the damping mechanism is represented by

two systems of equivalent viscous dampers. The first set (C1) represents the damping caused by

friction at the supports and any other forms of external damping forces, such as aerodynamic

and hydrodynamic forces. The second set (C2) represents the internal damping in the springs.

In a real structure, this would mainly be due to hysteresis losses and friction forces in member

joints as well as in the cladding.

The matrix formulation of the equations of motion for the system shown in Figure 8.2 is

M 0

0 M

� �€xx1

€xx2

� �þ

C1 þ 2C2ð Þ �C2

�C2 C1 þ 2C2ð Þ

� �_xx1

_xx2

� �þ

2K �K

�K 2K

� �x1

x2

� �¼

P1 tð ÞP2 tð Þ

� �ð8:18Þ

or

M€xxþ C _xxþ Kx ¼ P tð Þ: ð8:19Þ

As before, let

x ¼ Xq

And hence

_xx ¼ X _qq

€xx ¼ X€qq:ð8:20Þ

Figure 8.2 2-DOF damped lumped mass–spring systems acted on by harmonic forces

P1(t) = P1 sin(ω1t) P2(t) = P2 sin(ω2t)

M M

x1

C1 C1

C2 C2C2

x2

K K K


166

We therefore have

MX€qqþ CX _qqþ KXq ¼ P tð Þ: ð8:21Þ

Finally, pre-multiplication of each term in Equation 8.21 by XT yields

XTMX€qqþ X

TCX _qqþ X

TKXq ¼ X

TP tð Þ: ð8:22Þ

From Examples 7.1 and 7.5,

X ¼1 1

1 �1

� �:

Substitution of this matrix for X into Equation 8.22 and the implied matrix multiplications yield

2M 0

0 2M

� �q1

q2

� �þ

2 C1 þ C2ð Þ 0

0 2 C1 þ 3C2ð Þ

� �q1

q2

� �þ

2K 0

0 6K

� �q1

q2

� �¼

P1 tð Þ þ P2 tð ÞP1 tð Þ � P2 tð Þ

� �

ð8:23Þ

which may alternatively be written as

2M€qq1 þ 2 C1 þ C2ð Þ _qq1 þ 2Kq1 ¼ P1 tð Þ þ P2 tð Þ ð8:24aÞ

2M€qq2 þ 2 C1 þ 3C2ð Þ _qq2 þ 6Kq2 ¼ P1 tð Þ � P2 tð Þ: ð8:24bÞ

The equations of motion have (as in the case of the equations for the undamped system shown in

Figure 8.1) therefore been decoupled, although damping has been included. Inspection of

Equation 8.18 reveals that one part of the damping matrix is proportional to the mass matrix

and the other part to the stiffness matrix. The damping matrix C may therefore be written as

C ¼C1 0

0 C1

� �þ

2C2 �C2

�C2 2C2

� �¼ �0

M 0

0 M

� �þ �1

2K �K

�K 2K

� �ð8:25Þ

where �0 and �1 are coefficients of proportionality. If the damping mechanism can be represented

by a system of equivalent viscous dampers as shown in Figure 8.2, it is possible to model the

damping mechanism as a function of the mass and stiffness of the system. In such cases, the

damping may therefore be expressed as

C ¼ �0Mþ �1K ð8:26Þ

which is referred to as Rayleigh damping. Because the eigenvectors are orthogonal with respect to

both the mass matrix and the stiffness matrix, it follows that for this form of damping they are also

orthogonal with respect to the damping matrix.

The orthogonality property of the eigenvectors with respect to the damping matrix is also the

reason why the equations of motion for damped multi-DOF systems can be decoupled. Inspection

of Equations 8.24a and 8.24b shows that, as in the case of the undamped mass–spring system in

Figure 8.1, the natural angular frequencies arep(K/M) and

p(3K/M). This, together with the fact

that the values of q when multiplied by the eigenvalue matrix yield the contribution from each

mode to the total response, leads to the conclusion that the sum of the damping coefficients in


167

Equations 8.24a and 8.24b is equal to the damping coefficients in the first and second modes

respectively. It is therefore assumed (referring to Equation 3.22) that in Equations 8.24a and

8.24b:

2 C1 þ C2ð Þ ¼ 2�1!1 2Mð Þ ð8:27aÞ

2 C1 þ 3C2ð Þ ¼ 2�2!2 2Mð Þ: ð8:27bÞ

The elements in the damping matrix in Equation 8.19 can therefore be found by the matrix

multiplication

C ¼ X�T 2�!½ � ~MMX

�1 ð8:28Þ

where

~MM ¼ XTMX ð8:29Þ

2�!½ � ¼2�1!1 0

0 2�2!2

� �: ð8:30Þ

With the expressions for the damping coefficients given by Equations 8.27a and 8.27b, the

uncoupled Equations 8.24a and 8.24b may now be written as

M€qq1 þ 2�1!1M _qq1 þ Kq1 ¼ 12 P1 tð Þ þ P2 tð Þ½ � ð8:31aÞ

M€qq2 þ 2�2!2M _qq2 þ 3Kq2 ¼ 12 P1 tð Þ � P2 tð Þ½ �: ð8:31bÞ

If it is again assumed that

P1 sin !1tð Þ ¼ P2 sin !2tð Þ ¼ P0 sin !tð Þ

then

q1 ¼P0

K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½ð1� r21Þ

2 þ 4�2r21�q sinð!t� �1Þ ð8:32aÞ

q2 ¼ 0 ð8:32bÞ

where

r1 ¼!

!1

¼ !ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiK=Mð Þ

p

and, from Equation 4.12,

�1 ¼ tan�1 2�1r11� r21

:

The maximum response occurs when

sin !t� �ð Þ ¼ 1;


168

the maximum response vector is therefore given by

x1

x2

� �¼

1 1

1 �1

� � P0

K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r21Þ

2 þ 4�2r21� �q

0

264

375 ¼

P0

K


2 þ 4�2r21� �q

P0

K


2 þ 4�2r21� �q

2666664

3777775: ð8:33Þ

When using the method of mode superposition, it is generally more convenient to assume that

~CC ¼ XTCX ¼ 2�!½ � ~MM ð8:34Þ

where ~MM is given by Equation 8.29, rather than assuming Rayleigh damping which requires the

calculation of the coefficients �0 and �1. For the DOF system considered, we therefore have

~CC ¼ 2�1!1~MM1 0

0 2�2!2~MM2

" #ð8:35Þ

where

~MM1 ¼ 1 1½ �M 0

0 M

� �1

1

� �¼ 2M

~MM2 ¼ 1 1½ �M 0

0 M

� �1

�1

� �¼ 2M:

The products XTi MXi ¼ ~MMi and X

Ti CXi ¼ ~CCi are referred to as the modal mass and modal

damping in the ith mode; similarly, the product XTi KXi ¼ ~KKi is referred to as the modal stiffness.

8.4. Forced vibration of multi-DOF systems with orthogonal dampingmatrices

In the previous section, it is shown that the equations of motion for a 2-DOF system can be

decoupled provided that the damping matrix can be diagonalised by pre-multiplication and

post-multiplication of the mode-shape matrix. In the following, the process of decoupling of

the equations of motion is extended to a general N-DOF system. From Section 7.11, it is

known that pre-multiplication and post-multiplication of both the mass matrix and the stiffness

matrix by the mode-shape matrix will lead to diagonal matrices. The equations of motion can

therefore always be decoupled for systems with more degrees of freedom if the eigenvectors are

also orthogonal with respect to the damping matrix. The general theory for decoupling and

hence calculation of the response of multi-DOF systems to harmonic excitation can therefore

be easily presented.

Let the equations of motion for a general linear multi-DOF system be

M€xxþ C _xxþ Kx ¼ P tð Þ: ð8:36Þ

The corresponding eigenvalue equation for determination of the natural frequencies and mode

shapes is

KX� !2MX ¼ 0 ð8:37Þ


169

which, for an N-DOF system, will yield the eigenvalues and eigenvectors

!2 ¼ b!21; !

22; . . . ; !

2Nc ð8:38aÞ

X ¼ X1;X2; . . . ;XN½ �: ð8:38bÞ

In order to decouple the equations of motion, it is assumed that the damping matrix can be

diagonalised and that

~CC ¼ XTCX ¼ 2�!½ � ~MM ð8:39Þ

where

2�!½ � ¼ diag 2�1!1; 2�2!2; . . . ; 2�N!Nf g:

Now let

x ¼ Xq

_xx ¼ X _qq

€xx ¼ X €qq:

Substitution of the above expressions for x, _xx and €xx into Equation 8.39 and post-multiplication of

each term by XT yields

XTMX €qqþ XT

CX _qqþ XTKXq ¼ XTP tð Þ: ð8:40Þ

From the orthogonality properties of eigenvectors, we have

XTi MX i ¼ ~MMi and XT

i MX j ¼ 0 when j 6¼ i

XTi KX i ¼ ~KKi and XT

i KX j ¼ 0 when j 6¼ i;

Equation 8.40 therefore reduces to

~MM€qqþ ~CC _qqþ ~KKq ¼ XTP tð Þ: ð8:41Þ

Since ~MM, ~CC and ~KK are diagonal matrices, Equation 8.41 may also be written as

~MM1€qq1 þ 2�1!1~MM1 _qq1 þ ~KK1q1 ¼ XT

1P tð Þ

~MM2€qq2 þ 2�2!2~MM2 _qq2 þ ~KK2q2 ¼ XT

2P tð Þ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

~MMN €qqN þ 2�N!N~MMN _qqN þ ~KKNqN ¼ XT

NP tð Þ: ð8:42Þ

The elements in the vector q may now be determined by solving the N equivalent 1-DOF systems

given by Equation 8.42, in which each equation represents a mass–spring system that will vibrate

with the frequency and damping of the corresponding structural mode. Finally, the structural

response vector for theN-DOF system is found by pre-multiplying the vector q by the mode-shape


170

matrix X; we therefore have

x ¼ Xq: ð8:43Þ

The use of a normalised eigenvector to decouple the equations of motion will lead to a simplifica-

tion of Equations 8.41 and 8.42. Let

x ¼ Zq

_xx ¼ Z _qq

€xx ¼ Z€qq:

Substitution of the above expressions for x, _xx and €xx into Equation 8.36 and pre-multiplication of

each term by ZT yields

ZTMZ€qqþ Z

TCZ _qqþ Z

TKZq ¼ Z

TP tð Þ: ð8:44Þ

From the properties of orthogonal normalised eigenvectors presented in Chapter 7, and with the

assumptions made above with respect to the damping matrix,

ZTi MZi ¼ 1 and Z

Ti MZj ¼ 0 when j 6¼ i

ZTi KZi ¼ !2

i and ZTi KZj ¼ 0 when j 6¼ i

ZTi CZi ¼ 2�i!i and Z

Ti CZj ¼ 0 when j 6¼ i:

Equation 8.44 can therefore be written as

I€qqþ 2�! _qqþ !2q ¼ ZTP tð Þ ð8:45Þ

which represents, like Equation 8.41, a system ofN independent equations which can be written as

€qq1 þ 2�1!1 _qq1 þ !21q1 ¼ Z

T1P tð Þ

€qq2 þ 2�2!2 _qq2 þ !22q2 ¼ Z

T2P tð Þ

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

€qqN þ 2�N!N _qqN þ !2NqN ¼ Z

TNP tð Þ:

ð8:46Þ

The use of normalised eigenvectors therefore results in decoupled equations of motion in which

the mass is unity and the stiffness is equal to the eigenvalues. Finally, having determined the

elements in q by solving the equations in Equation 8.46, the total response of the system is

found from the transformation

x ¼ Zq: ð8:47Þ

When analysing a structure, the choice between Equations 8.41 and 8.45 is a matter of preference

as they will lead to the same results.

The numerical effort becomes considerable, even for the most trivial of problems, if each of the

elements in the forcing vector P(t) consists of a sum of harmonic functions such as

Pi tð Þ ¼XMi¼ 1

ai sin !i tð Þ; ð8:48Þ

calculations will normally require the use of a computer.


171

Example 8.1

Let the three-storey shear structure in Example 2.5 (Figure 2.14) be vibrated by a shaker

positioned on the top floor. Calculate the response if the vibrator exerts a force

P(t)¼ 0.6 sin(!1t) kN, where !1 is the first natural frequency of the structure. The weight

of each floor is 20.0 kN/m and the flexural rigidity EI of each column is 89 100.0 kNm2.

The distance between the columns is 10.0 m and the height of the columns is 4.0 m. The

damping in each mode is assumed to be 2.0% of critical.

The natural frequencies and mode shapes for the structure have been calculated in Example

7.3, and the normalised mode-shape matrix in Example 7.9. Hence

!21

!22

!23

264

375 ¼ 89100:0� 9:81� 10�3

20:0

8:1300

41:1150

78:8793

264

375 ¼

355:31

1796:87

3447:31

264

375 rad2=s2

!1

!2

!3

264

375 ¼

18:850

42:390

58:718

264

375 rad=s

Z ¼1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

264

375� 10�3

P1 tð ÞP2 tð ÞP3 tð Þ

264

375 ¼

0

0

600 sin 18:85tð Þ

264

375:

The decoupled equations of motion can therefore be written as

€qq1 þ 2� 0:02� 18:858 _qq1 þ 355:31q2 ¼ 4:2336� 10�3 � 600 sin 18:85tð Þ

€qq2 þ 2� 0:02� 42:390 _qq2 þ 1796:87q2 ¼ �4:4926� 10�3 � 600 sin 18:85tð Þ

€qq3 þ 2� 0:02� 58:718 _qq3 þ 3447:31q3 ¼ 3:3105� 10�3 � 600 sin 18:85tð Þ:

The general solution to these equations is

q ¼ Z3i � 600

!2i

MFi sin 18:85t� �ið Þ

where

MF1 ¼1=2�1 ¼ 25:0

MF2 ¼1=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r22Þ

2 þ 4�22r22

� �q¼ 1:2462

MF3 ¼1=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r23Þ

2 þ 4�23r23

� �q¼ 1:1148


172

8.5. Tuned mass dampersThe development of high tensile steel has made it possible to build increasingly taller buildings and

longer span bridges, both of which will respond to the buffeting of wind and earthquake tremors

without necessarily causing structural damage. In the case of tall buildings, the sway acceleration

may be so great as to make the upper part of a building too uncomfortable for use. In such cases,

the resulting vibration needs to be significantly reduced. This is done by installing tuned mass

dampers (TMDs) which may be in the form of large suspended blocks of concrete or metal,

large tanks in which water is sloshing through tuned by baffles, baffle-tuned liquid column

dampers (LCDs) or suspended circular tanks in which the water level can be adjusted

(TLCDs). All of these are placed in the part of a structure where the motion is greatest and

needs to be reduced.

TMDs are used extensively to reduce the vibration of electrical transmission cables, bridges and

tall buildings, and reduce the magnitude of the pulsating forces at the supports of car motors and

other rotating machinery.

�1 ¼ �=2 ¼ 1:57080 rad

�2 ¼ tan�1 2�2r21� r22

� �¼ 0:02217 rad

�3 ¼ tan�1 2�3r31� r23

� �¼ 0:01432 rad:

Since the contributions to the response from the second and third mode are obviously very

small, the maximum response occurs when sin (18.85t� �/2)� 1, i.e. when t¼ 0.1667 s.

Hence

q1 ¼4:2336� 10�3 � 600

355:31� 25:0� 1:0 ¼ 178:7284� 10�3 m

q2 ¼�4:4926� 10�3 � 600

1796:87� 1:55295 sin 18:85� 0:1667� 0:02217ð Þ ¼ �0:0500� 10�3 m

q3 ¼3:3105� 10�3 � 600

3447:31� 1:24274 sin 18:85� 0:1667� 0:01432ð Þ ¼ 0:0098� 10�3 m:

The maximum response is therefore

x1

x2

x3

264

375 ¼

1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

264

375

178:7284

�0:0500

0:0098

264

375� 10�6 ¼

0:3118

0:5926

0:7569

264

375� 10�3 m:

From the above calculations, it can be seen that the contributions from the response in the

second and third modes are negligible. This is to be expected, as the exciting frequency is

equal to the first mode frequency. To eliminate response completely in the two higher

modes would require a synchronised shaker system with a vibrator on each floor. Such

systems are available, but tend to be expensive.


173

The TMD is a device consisting of a mass, springs and dampers fitted at or near the point of

maximum response amplitude. The size of the mass itself can be very large and weigh many

hundred tons. It can be supported on rollers or as a pendulum and connected to the structure

by springs and hydraulic dampers. Large masses need large dampers to control the motion and

dissipate more energy than smaller masses. Masses vary in size from 2% to 10% of the lumped

mass of the structure. An average mass size is of the order 5%. Figure 8.3 depicts the general

arrangements of a variety of TMDs.

At the time of writing, the largest mass damper is deemed to be the one in Taipei 101, Taiwan, the

world’s tallest building. The building is 509.2 m tall and situated 201 m from a major fault. The

mass of the damper, which weighs 730 tons in the form of a globe, is supported in a sling

formed by eight steel cables. The resulting pendulum is damped by eight viscous dampers and

is limited to a movement of 1.52 m in any direction by a bumper ring.

Tuned mass dampers are most efficient when the mass moves in the opposite direction to that of

the structure. When that is the case, the springs and viscous dampers will oppose the motion at the

same time as the latter will dissipate some of the energy as heat. Thus, if the top of the structure in

Figure 8.3 sways to the left, the mass MD will swing to the right hence compressing the spring to

Figure 8.3 Equivalent mass-spring systems for reducing the amplitudes of vibration due to sinusoidal

excitation: (a) floor-supported mass damper; (b) pendulum mass damper; and (c) equivalent system

(b)(a)

(c)

KsKD

Cs

CD

Ms MD

Xs XD


174

the right and tensioning the spring to the left. Both springs will therefore oppose the movement of

the structure simultaneously, and the dampers will oppose the motion and dissipate some of the

sway energy. When the mass of the TMDmoves in phase with the structure, only the dampers will

act to reduce the amplitude of vibration. The equivalent mass–spring systems for reducing vertical

and horizontal amplitudes caused by sinusoidal forces are depicted in Figures 8.3a, b and c.

FURTHER READING

Clough RW (1975) Dynamics of Structures. McGraw-Hill, London.





Problem 8.1

The three-storey shear structure in Figure 7.4 has a first natural frequency of 2.0 Hz. The

mass of each floor is 4000 kg. Calculate the response of the structure if it is vibrated by a

harmonic force

P tð Þ ¼ 1:0 sin !3 tð ÞkN

at the second-floor level, where !3 is the third natural angular frequency. Assume the

damping in each mode to be 1.0% of critical.

Problem 8.2

Calculate the response of the stepped antenna mast in Example 7.8 (Figure 7.3) if it is excited

by a harmonic force of 100 sin(1.1�!1t) N at the top of the mast, where !1 is the first natural

frequency of the mast. Assume the damping in the first and second modes to be 1.5% and

1.0% of critical, respectively.


175


ISBN: 978-0-7277-4176-9



Chapter 9

Damping matrices formulti-degree-of-freedom systems

9.1. IntroductionThe parameters for dynamic analyses of free and forced vibration of multi-DOF systems are

discussed in Chapters 7 and 8. These include the stiffness and mass matrices and establishment

of eigenvalues and eigenvectors for such systems. In order to complete the required matrices

for the solution of Equation 8.36, the equation of motion, it is necessary to set up the damping

matrix C. This chapter provides the steps for the setting-up of damping matrices and their

evaluation.

9.2. Incremental equations of motion for multi-DOF systemsThe general incremental equations of motion for predicting the response of linear and non-linear

multi-DOF systems to load histories, assuming constant acceleration during a time step �t, are

given by Equation 6.61 as

Kþ 2

�tCþ 4

�t2M

� ��x ¼ �Pþ 2C _xxþM

4

�t_xxþ 2€xx

� �ð9:1Þ

where K, C andM are the stiffness, damping and mass matrices for a multi-DOF structure,�x is

the incremental displacement vector, x, _xx and €xx are the displacement, velocity and acceleration

vectors at time t and �P is the incremental load vector.

In the mode superposition method presented in Chapter 8, the variables are separated by replace-

ment of the displacement, velocity and acceleration vectors x, _xx and €xx in the equation of motion

with a new set of generalised vectors q, _qq and €qq, where

x ¼ Zq

_xx ¼ Z _qq

€xx ¼ Z€qq

ð9:2Þ

and then post-multiplication of each term in the resulting equation by ZT, the transpose of the

normalised mode-shape matrix Z. This operation yielded N independent equations representing

N 1-DOF systems, each with its own modal frequency and damping ratio. There was therefore

no need to assemble a damping matrix with the same dimensions as the stiffness and mass matrices.

A similar transformation of Equation 9.1 yields

!2 þ 2

�t2�!þ 4

�t2

� ��q ¼ Z

T�P þ 4�! _qqþ 4

�t_qqþ 2€qq

� �: ð9:3Þ

177

For linear structures, it is therefore possible to limit the forward integration or step-by-step

method to include only the response in significant modes. In the case of non-linear structures

this form of transformation is not permissible since, for such structures, the natural frequencies

and mode shapes vary with the amplitude of response. For non-linear structures, it is therefore

necessary to assemble not only the stiffness and mass matrices, but also the structural damping

matrices. This chapter presents two methods for modelling the structural damping in matrix

form in terms of modal damping ratios, natural frequencies, stiffness and damping matrices.

Theoretically, such damping matrices ought to be updated at the end of each time step as the

stiffness, frequencies and damping ratios are functions of the amplitude of response. In practice,

however, this is usually not necessary because the damping ratios used will in most cases only be

approximate values taken from codes of practice or the literature. However, before methods of

modelling structural damping by matrices are studied, an outline is given of how damping

ratios in higher modes are obtained.

9.3. Measurement and evaluation of damping in higher modesDamping in the first mode of multi-DOF systems can generally be evaluated as for 1-DOF

systems: from decay functions, frequency sweeps or by steady-state vibration at resonance. The

use of only one vibrator will in most cases cause structures to vibrate with a mode shape that

closely resembles the true shape, and will therefore lead to reasonable values for the first damping

ratio. In higher modes, there are however difficulties. Firstly, it is usually impossible to obtain

decay functions for higher modes as most structures will, when excitation of a higher mode is

stopped, revert to vibration in the first mode. Measurements of damping either by steady-state

vibration at resonance or by frequency sweeps is also generally unsatisfactory, because the use

of one vibrator will generally not be sufficient to cause a structure to vibrate in a pure mode.

This is particularly noticeable when attempting to excite a structure in an anti-symmetric mode

such as the second, fourth and sixth modes of a simply supported beam. This can be easily

seen or demonstrated by measuring the phase angle of response at different points on a structure.

If the phase angle is 908 at the point of excitation it will usually be different at points away from

the vibrator, with the difference increasing with increasing distance from the point of excitation.

To obtain reasonably accurate values for damping in higher modes it is therefore necessary to use

more than one vibrator, whose force and frequency must be adjusted so that the structure at all

points vibrates with phase angles of 908. To achieve this, the vibrators must be controlled by a

computer. As for 1-DOF systems, the damping can be measured by plotting the exciting force

versus the amplitude of response for one cycle for each vibrator (as described in Chapter 5)

and then summing the work done by each vibrator. The expression for the damping ratio for a

structure vibrated by N vibrators is given by

� ¼ 1

2� ~MM!2n

XNi¼ 1

Ani

x2n0ið9:4Þ

� ¼ 1

2� ~KK

XNi¼ 1

Ani

x2n0ið9:5Þ

where ~MM and ~KK are the modal mass and modal stiffness, respectively, and Ani is the area

encompassed by the force–displacement curve for vibrator n at the ith mode. Multi-point

shaker systems are expensive, and are all mainly used by research institutions and industrial

companies specialising in dynamic testing.


178

In general, values for damping ratios are obtained from codes of practice or the literature. The

former usually only give values for damping ratios to be used in the dominant mode, with no

guidance on values to be used in higher modes. It is therefore not uncommon to use the same

damping ratio for all modes.

9.4. Damping matricesIn Chapter 8, it is shown that the dynamic response of linear multi-DOF structures can be

determined by decoupling the equations of motion and summing the responses in each mode.

It is therefore only necessary to assign values to the damping ratios for the modes contributing

to the total response, without having to set up a damping matrix. The implied orthogonality of

the mode shapes with respect to the damping matrix enables realistic numerical modelling of

structural damping, provided it can be assumed that the damping does not couple the modes.

This assumption is usually correct providing that aerodynamic and hydrodynamic damping

(when significant) are modelled separately.

In general, it is only necessary to model the structural damping by a damping matrix when under-

taking the form of dynamic response analysis indicated by Equation 9.1, in which case damping

due to external forces such as those caused by air and water can be taken into account separately.

When this is the case, the construction of orthogonal damping matrices is a convenient method

of modelling the structural damping. In the case of non-linear structures, the principle of

orthogonality no longer applies as the mode shapes as well as the frequencies are functions of

the amplitude of response. For weakly non-linear structures, the non-linearity is not significant

as assumed damping ratios will at best only be approximately correct. A considerable amount

of experimental evidence indicates that modal damping ratios vary with the amplitude of

response.

9.5. Modelling of structural damping by orthogonal dampingmatrices

9.5.1 First methodIn Chapter 8, it is shown that the equations of motion for a multi-DOF structure can be

written as

M€xxþ C _xxþ Kx ¼ P tð Þ ð9:6Þ

and the equations can be uncoupled providing the damping matrix has the same orthogonal

properties with respect to the mode-shape vectors as the mass and stiffness matrices. When this

is the case,

ZTi CZi ¼ 2�i!i ð9:7Þ

ZTCZ ¼ 2�!½ � ð9:8Þ

and therefore

C ¼ Z�T 2�!½ �Z�1: ð9:9Þ

The inversion of the mode-shape matrix Z can be avoided. From Equation 7.57,

ZTMZ ¼ I ð9:10Þ

Damping matrices for multi-degree-of-freedom systems

179

and hence

Z�T ¼ MZ ð9:11Þ

Z�1 ¼ Z

TM: ð9:12Þ

Substitution of the expressions for Z�T and Z�1 into Equation 9.9 yields the following expression

for the damping matrix:

C ¼ MZ 2�!½ �ZTM: ð9:13Þ

9.5.2 Second methodAnother method of constructing a damping matrix with orthogonal properties from modal

damping ratios is to assume that the damping is a function of both the mass and the stiffness,

and to make use of the general orthogonal relationship

ZTi M M

�1K

� �qZj ¼ 0 ð9:14Þ

which is satisfied when i 6¼ j and q¼ . . . , �2, �1, 0, 1, 2, . . . .

When q¼ 0 and q¼ 1, Equation 9.12 yields the previously obtained orthogonality conditions of

the mode-shape vectors with respect to the mass matrix and the stiffness matrix. Inspection of

Equation 9.14 indicates that it is possible to formulate an orthogonal damping matrix of the form

C ¼XN�1

q¼ 0

�qM M�1K

� �q ð9:15Þ

which will contain the correct damping in N modes providing the corresponding values for � can

be determined. An expression for calculating values of � can be developed by first pre-multiplying

and post-multiplying both sides of Equation 9.15 by ZiT and Zi, respectively. This yields the i

mode contribution to the total damping as

Ci ¼ 2�i!i ¼XN � 1

q¼ 0

ZTi �qM M

�1K

� �qZi: ð9:16Þ

In order to simplify the right-hand side of Equation 9.16, the two sides of the frequency equation

for the ith mode

KZi ¼ !2i MZi ð9:17Þ

are multiplied by �q, transposed, post-multiplied by (M�1K)qZi and then written in reverse order.

This yields the relationship

!2i Z

Ti �qM M

�1K

� �qZi ¼ Z

Ti �qK M

�1K

� �qZi: ð9:18Þ

Substitution of !i2M for K in the right-hand side of Equation 9.18 gives

ZTi �qM M

�1K

� �qZi ¼ �q!

2qi ð9:19Þ


180

which finally, on substitution of the left-hand side of Equation 9.19 into Equation 9.16, yields

2�i!i ¼XN�1

q¼ 0

�q!2qi ð9:20Þ

from which as many values of � can be found as there are known damping ratios. Given the

damping ratios for, say, the first four modes of an N-DOF structure, only four values for �

(namely �0, �1, �2 and �3) can therefore be determined by using Equation 9.20. The modal

damping in the ith mode, given four values for �, is given by the polynomial

2�i!i ¼ �0 þ �1!2i þ �2!

4i þ �3!

6i ð9:21Þ

where the four values for � may be calculated from the matrix equation

2�1!1

2�2!2

2�3!3

2�4!4

266664

377775

¼

1 !21 !4

1 !61

1 !22 !4

2 !62

1 !23 !4

3 !63

1 !24 !4

4 !64

266664

377775:

�0

�1

�2

�3

266664

377775: ð9:22Þ

Assigning values to lower mode damping ratios only may result in values for damping ratios in

higher modes that are very different from the real values. This is not important, however,

providing the damping ratios for the modes in which a structure mainly responds are correct.

In practice, it is often assumed that only one or two values for � are different from zero. When

this is the case, Equation 9.15 is reduced to one of the following:

C ¼ �0M ð9:23aÞ

C ¼ �1K ð9:23bÞ

C ¼ �0Mþ �1K: ð9:23cÞ

This may result in adequate modelling of the damping for a large number of civil engineering

structures that vibrate only in a few of the lower modes, but will not suffice for structures such

as guyed masts, cable-stayed bridges and cable and membrane roofs that respond significantly

in a large number of modes. The expression for damping given by Equation 9.23 is referred to

as Rayleigh damping (mentioned in Chapter 8). When this form of damping is assumed, the

damping in any mode can be calculated from

2�i!i ¼ �0 þ �1!2i : ð9:24Þ

Rayleigh damping is a convenient form for modelling the damping of weakly non-linear

structures, as it leads to damping matrices with the same banding as the stiffness matrix.

From Equation 9.23a it can be seen that when the damping is assumed to be proportional to the

mass only (i.e. when q¼ 0), the damping ratios decrease with increasing mode frequencies; when

the damping is assumed proportional to the stiffness only (i.e. when q¼ 1), then damping ratios

increase with increasing mode frequencies. Equation 9.24 indicates that Equation 9.23c will model

the structural damping mechanism correctly if values of 2�!, when plotted against !2, yield a

straight line.


181

Example 9.1

Construct the damping matrix for the three-storey shear structure shown in Example 2.5

(Figure 2.14) by using Equation 9.13, if the weight of the floors is 20.0 kN/m and the damping

in each mode is assumed to be 1.0% of critical. The natural angular frequencies and the

normalised mode-shape matrix for the structure are given in Example 8.1 as

!1

!2

!3

264

375 ¼

18:850

42:390

58:718

24

35rad=s Z ¼

1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

24

35� 10�3:

The mass matrix for the structure is

M ¼ 20 387:36

3 0 0

0 2 0

0 0 1

24

35kg

2�!½ � ¼2� 0:01� 18:850 0 0

0 2� 0:01� 42:390 0

0 0 2� 0:01� 58:714

264

375

¼0:37700 0 0

0 0:84780 0

0 0 1:17428

264

375

and hence

C ¼ 415:64444�3 0 0

0 2 0

0 0 1

264

375

1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

264

375

�0:37700 0 0

0 0:84780 0

0 0 1:17428

264

375

1:7454 3:3157 4:2336

3:0818 0:4330 �4:4926

1:9498 �3:6531 3:3105

264

375

3 0 0

0 2 0

0 0 1

264

375Ns=m:

We therefore have

C ¼51:1165 �12:5967 �1:7114

�12:5967 33:2093 �8:7771

�1:7114 �8:7771 15:2700

24

35� 103 N s=m:

Note that the matrix is not only full, but also symmetric.

Example 9.2

Use Equations 9.15 and 9.20 to construct the damping matrix for the shear structure in

Example 2.5 (Figure 2.14). The EI value for the columns and the weight of the floor are as

for the previous examples: 89 100.00 kNm2 and 20.0 kN/m, respectively. The natural angular

frequencies for the structure are given in Examples 8.1 and 9.1.


182

From Equations 9.15 and 9.20,

C ¼XN�1

q¼ 0

�qM M�1K

� �q

2�i!i ¼XN�1

q¼ 0

�q!2qi

M ¼ 20 387:36

3 0 0

0 2 0

0 0 1

24

35kg

K ¼ 16 706:25

7 �3 0

�3 5 �2

0 �2 2

24

35kN=m

! ¼18:50

42:390

58:718

24

35rad=s:

Given the damping ratios in three modes, Equation 9.20 may be written in matrix form as

2�1!1

2�2!2

2�3!3

264

375 ¼

1 !21 !4

1

1 !22 !4

2

1 !23 !4

3

264

375

�0

�1

�2

264

375

�0

�1

�2

264

375 ¼

1 18:8502 18:8504

1 42:3902 42:3904

1 58:7184 58:7184

264

375�1

0:37700

0:84780

1:07436

264

375 ¼

0:2218644

4:58361� 10�4

�6:12287� 18�8

264

375:

For three values of �, Equation 9.15 may be written as

C ¼ �0Mþ �1Kþ �2M�1K

2

�0M ¼ 0:2218644� 20 387:36

3 0 0

0 2 0

0 0 1

264

375 ¼

13 569:688 0 0

0 9406:459 0

0 0 4523:229

264

375

�1K ¼ 4:58361� 10�4 � 16 706:25

7 �3 0

�3 5 �2

0 �2 2

2664

3775

¼

53 602:454 �22 972:480 0

�22 972:480 38 287:467 �15 314:87

0 �15 314:987 15 314:987

2664

3775

�2M�1K

2 ¼ �6:12287� 10�8 � 16 706:252

20 387:36�

1=3 0 0

0 1=2 0

0 0 1

264

375

7 �3 0

�3 5 �2

0 �2 2

264

3752


183

FURTHER READING





�2M�1K

2 ¼�16 205:362 10 058:501 �1676:417

15 087:571 �15 528:959 5867:459

�5029:250 11 734:917 �6705:667

264

375:

Substitution of the matrices for �2M, �1K and �2M�1K2 into the expression for C and their

addition yields

C ¼50 966:780 �12 913:979 �1676:417

�7884:909 32 164:968 �9447:528

�5029:250 �3580:070 13 132:549

264

375Ns=m:

Inspection of the above damping matrix reveals that it is not symmetric. The reason for this is

that the term M�1K is not symmetric unless all the elements in M are equal.

Problem 9.1

For the structure in Examples 9.1 and 9.2, assume the damping in the first two modes to be

1.0% of critical. Construct the damping matrix assuming Rayleigh damping and hence

calculate the implied damping ratio in the third mode.

Problem 9.2

Use the damping matrix calculated in Problem 9.1 to set up the dynamic matrix for the

structure in Examples 9.1 and 9.2. Assume the time step to be approximately equal to 1/

10th of the period of the highest frequency.


184


ISBN: 978-0-7277-4176-9



Chapter 10

The nature and statistical properties ofwind

10.1. IntroductionWind is unsteady and exhibits random fluctuations in both time and space domains. Because wind

can be considered to possess stationary characteristics, it is possible to describe its functions in

statistical terms. Advances in computational techniques have made it possible to carry out

statistical analysis of wind records and to determine their statistical characteristics such as

those described by the variance of fluctuations, auto-correlation and spectral density functions

(the latter also commonly referred to as power spectra). Further advances in computational

techniques have made it possible to generate wind histories and wind fields with the same

statistical characteristics as real wind.

For linear structures, reasonable estimates of the response to wind can be made through a

stochastic approach in which the statistical characteristics of the response are determined in

terms of the statistical properties of wind. This form of analysis is carried out in the frequency

domain, and is the method most used by practising engineers.

However, for non-linear structures suchasmembrane, cable and cable-stayed structureswhose struc-

tural characteristics vary with the amplitude of response and hence with time, reliable estimates of

response to wind can only be made using a determined approach in which the structural properties

are updated at the end of each time step. In deterministic analysis, single wind histories and wind

fields simulating real time are generated from spectral density functions for fluctuating wind

speeds. Basically, there are two distinctmethods for generatingwindhistories: first, the superposition

of harmonic waves and second, a method based on filtering sequences of white noise.

10.2. The nature of windWind is a phenomenon caused by the movement of air particles in the Earth’s atmosphere. The

movement of air in the atmospheric boundary layer, which extends to about 1 km above

the Earth’s surface, is referred to as surface wind. The wind derives its energy primarily from

the sun. Solar radiation accompanied by radiation away from Earth produces temperature

differences and consequently pressure gradients that cause acceleration of the air. Away from

the ground, the pressure system is relatively stationary because the pressure gradients are balanced

by the centripetal and Coriolis accelerations. The centripetal acceleration is due to the curvature

of the isobars and Coriolis acceleration is due to the Earth’s rotation. This balance of forces

results in a steady-state condition that causes the air to flow in a direction parallel to the isobars.

Near the ground, the balance of the pressure system is disturbed by drag forces caused by the

Earth’s surface roughness. Ground surface roughness, whether occurring naturally (e.g.

mountains, hills and forests) or as man-made obstructions (such as buildings, bridges and

dams), causes so much mechanical stirring of the air movement that

185

g the wind speed near the surface is retardedg the wind direction changes and is no longer parallel to the isobarsg the flow conditions become unsteady and the wind exhibits instantaneous random

variations in magnitude and direction.

The rougher the surface, the more prominent these effects are. The effects decrease with increasing

height above the ground. The height at which the effects have virtually vanished is referred to as

the gradient height, and ranges from 300 to 600 m depending on the degree of surface roughness.

Examination of wind records shows that the velocity of wind fluctuates and that the fluctuations

vary both with the wind speed and with the roughness of the ground. It has therefore been found

convenient to express the wind velocity as the sum of the mean velocity U(z, x) in the along-wind

direction at height z and the fluctuating time-dependent velocity components u(z, x, t), u(z, y, t)

and u(z, z, t), where x represents the along-wind, y the horizontal across-wind and z the vertical

across-wind directions at height Z. We therefore have

U z; x; tð ÞU z; y; tð ÞU z; z; tð Þ

264

375 ¼

U z; xð Þ0

0

264

375þ

u z; x; tð Þu z; y; tð Þu z; z; tð Þ

264

375 ð10:1Þ

or

U z; tð Þ ¼ U zð Þ þ u z; tð Þ: ð10:2Þ

In cases where the horizontal and vertical across-wind fluctuations are of secondary importance,

the instantaneous wind velocity can be treated as a scalar quantity. In this case, the instantaneous

velocity at height z is given by (omitting the direction indicator x)

U z; tð Þ ¼ U zð Þ þ u z; tð Þ: ð10:3Þ

Research has revealed that the long-term statistical properties of wind are general and indepen-

dent of type of terrain, wind strength and site location. This significant conclusion emerged

from power spectral analysis of wind recorded over several years and at different locations.

The resulting spectrum, in which the square of the amplitudes of each frequency was plotted

against the frequency, provides a measure of the distribution of the energy of the random fluctua-

tions of the wind velocity in the frequency domain. A typical spectrum, whose full line is known as

the van der Hoven power spectrum, is shown in Figure 10.1.

An examination of Figure 10.1 yields the following information.

g The energy is distributed in two main humps separated by a gap (the so-called spectral

gap), which exists for periods between 10 min and 2 h. This implies that the fluctuations in

the mean velocity of wind can be measured by calculating the mean velocities of wind

speed signals recorded over periods ranging from only 10 min to 2 h. In this way,

fluctuations due to the high-frequency components are eliminated so that only those due to

the long-term fluctuations can be observed. Thus, as mentioned above, the wind velocity

can be divided into two parts: an average steady-state velocity that varies with the

long-term fluctuations due to macrometeorological causes and a fluctuating velocity with

high-frequency components due to turbulence.


186

g The first peak is linked to the annual variation. The second peak is linked to the 4 day

period. This is the time of passage of a complete macrometeorological system, i.e. the

duration of an average storm. The third peak is due to day and night thermal fluctuations.

The fourth peak, which is in the micrometeorological range, is centred around a frequency

of nearly one cycle/min and is caused by ground roughness.

As a result of the properties of wind outlined above, the response calculations of structures can be

divided into two parts: (a) the calculation of the quasi-static response caused by the mean velocity

component of wind and due to the very low-frequency fluctuations in the macrometeorological

system; and (b) the calculation of the response due to the high-frequency components, which

are the source of dynamic excitation.

10.3. Mean wind speed and variation of mean velocity with heightIt has been established that recording periods between 10 min and 2 h provide reasonably stable

values for the mean component of the wind speed. A period of 1 h lies nearly in the middle of this

range and is the recording period adopted in the UK where meteorological stations in different

parts of the country record and summarise the maximum daily wind speeds. The hourly wind

speeds are recorded at a height of 10m, but wind speed increases with increasing altitude above

the ground until it reaches the velocity Vg at the gradient height. Several laws have been used to

describe the way in which the mean velocity varies with height. Today, the most generally adopted

law is the logarithmic law, which gives the mean speed U(z) at height z above the ground as

U zð Þ ¼ 2:5 u� ln z=z0ð Þ ð10:4Þ

which can be written

u� ¼U 10ð Þ

2:5 ln 10=z0ð Þ ð10:5Þ

or

u� ¼ U 10ð Þffiffiffik

pð10:6Þ

Figure 10.1 Spectrum of longitudinal wind fluctuations: the full line of the spectrum is after van der

Hoven (1957)

Cycles/h

10–4 10–2 10–1 0.5 1 2 5 10 50

Ener

gy s

pect

rum

Macrometeorological range Micrometeorological range

nlns

The nature and statistical properties of wind

187

where u� is the shear velocity or friction velocity, z0 is the roughness length (for values see

Table 10.1), k is the surface drag coefficient (for values see Table 10.1) and U(10) is the reference

mean velocity 10 m above ground level.

If the surface drag coefficient k is known, then the corresponding value for z0 can be found by

using Equation 10.4:

z0 ¼ z exp �U zð Þ=2:5u�½ �: ð10:7Þ

The logarithmic law is applicable to heights in excess of 10 m. Below this height, the velocity is

assumed to be constant and equal to U(10). In some of the more recent codes, the logarithmic

law has been modified and the mean velocity at height z is given by

UðzÞ ¼ 2:5u�½lnðz=z0Þ þ 5:75ðz=HÞ � 1:87ðz=HÞ2�1:33ðz=HÞ3 þ 0:25ðz=HÞ4� ð10:8Þ

where H, the gradient height, can be determined from

H ¼ u�=2�! sin� ð10:9Þ

and ! is the angular rotation of the Earth (7.2722� 10�5 rad/s), � is the local angle of latitude and

� is a constant (equal to 6).

For the lowest 200 m of the atmosphere, the contributions from the square, cubic and fourth-

order terms can be neglected; Equation 10.8 therefore reduces to

U zð Þ ¼ 2:5u� ln z=z0ð Þ þ 5:75 z=Hð Þ½ �: ð10:10Þ

When the expression for H given by Equation 10.9 is substituted into Equation 10.10 and

z¼ 10 m, the following relationship between u� and z0 is obtained when using the above values

for ! and �:

u� ¼U 10ð Þ � 0:1254454 sin’

2:5 ln 10=z0ð Þ : ð10:11Þ

Since �1.0< sin �< 1.0, Equation 10.11 can for most applications be simplified to

u� ¼U 10ð Þ

2:5 ln 10=z0ð Þ : ð10:12Þ

Table 10.1 Roughness lengths and surface drag coefficients for various types of terrain

Type of terrain z0: m k� 103

Sand 0.0001–0.001 1.2–1.9

Sea surface 0.005 0.7–2.6

Low grass 0.01–0.04 3.4–5.2

High grass 0.04–0.10 5.2–7.6

Pine forest 0.90–1.00 28.0–30.0

Suburban areas 0.20–0.40 10.5–15.4

Centres of cities 0.35–0.45 14.2–16.6

Centres of large cities 0.60–0.80 20.2–25.1


188

Example 10.1

A guyed mast is instrumented with anemometers at 10 m and 100 m above the ground. From

the analysis of the records of one set of hourly readings, it was found that for wind from the

northeast the mean velocity at 10 m was 19.6 m/s, while the corresponding mean velocity at

100m was 33.6 m/s. Assuming the variation of the mean wind speed with height as expressed

by Equation 10.4 to be correct, calculate the roughness length z0 and surface drag coefficient

k for the site for the given wind direction. Calculate also the gradient height if the latitude of

the site is 538.

From Equation 10.4,

U 10ð Þ ¼ 19:60 ¼ 2:5u� ln 10=z0ð Þ

U 100ð Þ ¼ 33:64 ¼ 2:5u� ln 100=z0ð Þ

from which

ln 100=z0ð Þ ¼ 1:7155 ln 10=z0ð Þ

and hence

100=z0ð Þ ¼ 10=z0ð Þ1:7155:

We therefore have

z1:71550 ¼ 0:5193976

z0 ¼ 0:400 m:

The surface drag coefficient is determined using Equation 10.6. This requires that the shear

velocity u� be determined first by substitution of the values for U(10) and z0 into Equation

10.5, yielding

u� ¼19:61

2:5 ln 10=0:4ð Þ ¼ 2:437m=s:

The value of the surface drag coefficient is therefore

k ¼ u2�=U2 10ð Þ ¼ 2:4372=19:612 ¼ 0:0154

which agrees with the value corresponding to z0¼ 0.4 m given in Table 10.1.

The gradient height for the site is found by using Equation 10.5; we therefore have

H ¼ 2:437=12� 7:2722� 10�5 sin 57� ¼ 3329:79 m:

This calculated value is considerably greater than the gradient height of 900m assumed in the

wind map issued by the Meteorological Office for the UK.


189

Example 10.2

For a site at longitude 578 the surface drag coefficient k¼ 0.01. The estimated maximumwind

speed occurring during a 50 year period at a height of 10 m is 25.0 m/s. Determine and

compare the corresponding mean wind profiles obtained from the ground and up to a

height of 100m, using Equations 10.4 and 10.10.

From Equation 10.6,

u� ¼ 25:0ffiffiffiffiffiffiffiffiffi0:01

p¼ 2:50 m=s:

The corresponding value for z0 is found by substituting the values U(10)¼ 25.0 m/s, z¼ 10

and u�¼ 2.5 m/s into Equation 10.7, yielding

z0 ¼ 10:0� exp �25:0=2:5� 2:50ð Þ ¼ 0:183 m:

From Equation 10.4,

U zð Þ ¼ 2:5u� ln z=z0ð Þ

and hence

U zð Þ ¼ 2:5� 2:50 ln z=0:183ð Þ:

From Equation 10.10,

U zð Þ ¼ 2:5u� ln z=z0ð Þ þ 5:75 z=Hð Þ½ �


H ¼ u�=2�! sin�

and hence

H ¼ 2:50=2� 6� 7:2722� 10�5 sin 57� ¼ 3415:872 m

and

U zð Þ ¼ 2:5� 2:50 ln z=0:183ð Þ þ 5:75 z=3415:872ð Þ½ �:

Table 10.2 Example 10.2 data

U(z) Equation 10.4 Equation 10.10 Difference

U(20) 29.34m/s 29.55 m/s 0.716%

U(40) 33.67m/s 34.09 m/s 1.247%

U(60) 36.21m/s 36.84 m/s 1.740%

U(80) 38.00m/s 38.84 m/s 2.211%

U(100) 39.40m/s 40.45 m/s 2.665%

U(150) 41.93m/s 43.51 m/s 3.768%

U(200) 43.73m/s 45.83 m/s 4.802%

U(250) 45.12m/s 47.75 m/s 5.829%


190

10.4. Statistical properties of the fluctuating velocity componentof wind

In the description of the nature of wind above, it is explained that the velocity of wind could be

considered to consist of a constant or mean wind speed component and a fluctuating velocity

component due to the turbulence or gusting caused by the ground roughness. Recordings of

wind have shown that the velocity of wind can be considered as a stationary random process,

therefore

1

T

ðT0U tð Þdt ¼ U ð10:13Þ

1

T

ðT0u tð Þdt ¼ 0: ð10:14Þ

Because of this, the characteristics of the fluctuating component of wind can be quantified by

statistical functions. The most important of these for the dynamic analyst are

g the variance �2 and the standard deviation �g the auto-covariance function Cu(�) for the fluctuating velocity component u(t)g the spectral density function or power spectrum Su(n)g the cross-covariance function Cuv(�) of the fluctuating velocity components u(t) and v(t)g the cross-spectral density function or cross-power spectrum Suv(n)g the coherence function cohuv(n)g the probability density function p(u) and peak factor � for u(t)g the cumulative distribution function P(U) of U(t)

where n is the frequency of a constituent harmonic wind component as opposed to f which (in this

book) is used to denote a structural mode-shape frequency. The definitions and mathematical

formulations of the above functions are given in the following sections.

10.4.1 Variance and standard deviationThe variance of the fluctuating or gust velocity component is defined as

�2 uð Þ ¼ 1

T

ðT0u tð ÞTu tð Þdt ¼ �2 uxð Þ þ �2 uy

� �þ �2 uzð Þ ð10:15Þ

where

u tð Þ ¼ux tð Þuy tð Þuz tð Þ

264

375: ð10:16Þ

Substitution in turn of the values 20 m, 40 m, 60 m, 80 m, 100 m, 150 m, 200 m and 250 m for

z into the above two expressions for U(z) yields the values given in Table 10.2. The use of

Equation 10.10 becomes more significant in the design of very tall structures such as

towers and guyed masts.


191

The variances along the x axis, y axis and z axis are therefore equal to the mean square value of the

fluctuations in these directions. From recorded data, it has been observed that the greatest part of

the variance is associated with the fluctuations of the velocity in the direction of the mean flow. If

the direction along the flow parallel to the ground is the x direction, the direction perpendicular to

the flow and parallel to the ground is the y direction and the direction perpendicular to the flow is

the z direction, then it can be stated that

�2 uxð Þ � 10�2 uy� �

�2 uy� �

> �2 uzð Þ:ð10:17Þ

In general, it is therefore assumed that

�2u � �2 uxð Þ ¼ 1

T

ðT0u2x tð Þdt: ð10:18Þ

The variance �2(u) is obviously a function of the ground roughness and may be expressed in terms

of the shear velocity u� as

�2u ¼ �u2�: ð10:19Þ

Previously, it was generally assumed that �u was independent of height and that (for engineering

purposes) the constant � � 6.0 when the averaging time was 1 h. The reader should however be

aware that, particularly over rough ground, values as low as �� 4.0 have been reported in the

literature. Nowadays, it is generally accepted that the variance varies with height and not only

with ground roughness and mean wind speed. An expression that takes this dependence on

height into account is

�u zð Þ ¼ 2:63u�� 0:538þ z=z0ð Þ�16 ð10:20Þ

where �¼ 1� z/H and the gradient height H is given by Equation 10.9.

The standard deviation at �(z) at height z provides a measure of the dispersion of the wind speed

around its mean valueU(z) and is used as a measure of the turbulence intensity I(z), which is given

by

Iu zð Þ ¼ �u zð Þ=Uu zð Þ: ð10:21Þ

10.4.2 Auto-correlation and auto-covariance functionsTwo other important statistical concepts are as follows: the so-called auto-correlation function

R(�), where

RU �ð ÞT!1¼ 1

T

ð1�1

U þ u tð Þ½ � U þ u tþ �ð Þ½ �dt ð10:22Þ

and the auto-covariance function Cu(�), where

Cu �ð ÞT!1¼ 1

T

ð1�1

u tð Þ u tþ �ð Þ½ � dt: ð10:23Þ


192

The function Cu(�) provides a measure of the interdependence of the fluctuating velocity

component u of the wind at times t and tþ � . From Equation 10.18, it follows that when � ¼ 0

Cu �ð Þ ¼ Cu 0ð Þ ¼ �2 uð Þ: ð10:24Þ

Because wind histories are considered as stationary random processes with statistical properties

independent of time, it follows that RU(�)¼RU(��) and Cu(�)¼Cu(0).

It has also been found convenient to define an auto-covariance coefficient which is defined as the

ratio of C(�) to C(0). The expression for the auto-covariance coefficient is given by

cu �ð Þ ¼ Cu �ð Þ=Cu 0ð Þ ¼ Cu �ð Þ=�2u ð10:25Þ

and when � ¼ 0, cu(�)¼ 1.0. In the limit when � !1, cu(�)! 0. The auto-covariance coefficient

can therefore be regarded as a measure of the extent to which the fluctuation of the wind at time t

is a function of the fluctuation at time tþ � . If the value of cu(�) is small then the two quantities are

almost independent, while if cu(�)¼ 1.0 they are completely dependent on each other. For wind

the auto-covariance coefficient decreases with increasing values of � as shown in Figure 10.2,

where cu(�) is plotted against the time lag � for a recorded along-wind and across-wind trajectory.

10.4.3 Spectral density functions of longitudinal velocity fluctuationsSpectral density functions, also referred to as power spectra, are important functions that define

the random nature of wind. A spectral density function is denoted by Su(n), where the variable n is

the frequency of the sinusoidal velocity components of the fluctuating part of the wind velocity.

Spectral density functions give a measure of the energy distribution of the harmonic velocity

components, and form the basis for dynamic response analysis of linear structures in the

frequency domain. They can be expressed as Fourier transforms of the auto-covariance

function Cu(�), that is

Su nð Þ ¼ 4

ð10Cu �ð Þ cos 2� n�ð Þ d� ð10:26Þ

Cu �ð Þ ¼ð10Su nð Þ cos 2� n�ð Þdn: ð10:27Þ

Figure 10.2 Auto-correlation functions for (a) along-wind and (b) across-wind components of recorded

wind

τ: s

(b)

τ: s

(a)

10 20 30 10 20 30

1.0 1.0

r: τ


193

When the time lag � ¼ 0, Cu(�)¼Cu(0)¼ �2u. Equation 10.27 therefore yields

ð10Su nð Þ dn ¼ �2

u: ð10:28Þ

Davenport suggested the following formulation for the spectral density function

Su nð Þ ¼ 4u2� f2

n 1þ f 2ð Þ4=3ð10:29Þ

where

f ¼ 1200n

U 10ð Þ :

Harris modified the formulation by Davenport and suggested the formulation


n 2þ f 2ð Þ5=6ð10:30Þ

where

f ¼ 1800n

U 10ð Þ :

Both the above expressions for the spectral density functions depend only on the mean wind

speed U(10) and the ground roughness z0 and are independent of the height z; this is contrary

to experimental evidence. The use of the constant length scales L¼ 1200m and L¼ 1800 m was

therefore doubted. As a result, Deaves and Harris introduced a length scale that varied with

height and developed the following expression for the spectral density function

Su nð Þ ¼ 0:115�2u zð ÞTu zð Þ

0:0141þ n2T2u zð Þ½ �5=6

ð10:31Þ

where the time scale Tu(z) is determined by integration of the auto-covariance coefficient cu(z). We

therefore have

Tu zð Þ ¼ð10cu z; �ð Þd� ð10:32Þ

and �2u(z) can be calculated from Equation 10.20. The time scale Tu(z) is related to the length scale

Lu(z) through

Lu zð Þ ¼ Tu zð Þ U zð Þ; ð10:33Þ

the dependence of the length scale on height is therefore implied in the expression for the spectral

density function given by Equation 10.31. The evaluation of Tu(z) by integration of cu(z) is not a

practical proposition for design purposes, and Lawson (1980) gives a method for calculating Lu(z)

from which Tu(z) can be calculated using Equation 10.33. The method is lengthy and considered

to be outwith the scope of this book.


194

A more convenient formulation of a spectral density function that varies with height is that

suggested by Kaimal, defined

Su z; nð Þ ¼ 200u2� f z; nð Þn 1þ 50f z; nð Þ½ � 5=3

ð10:34Þ

where

f z; nð Þ ¼ zn

U zð Þ :

In the higher frequency range in which structures are likely to respond, this function is a close

approximation of spectra of recorded wind histories. However, it is not as accurate in the

lower frequency range. Another spectral density function that also varies with height is based

on the current ESDU (European Statistical Data Unit) model, which is given by

Su z; nð Þ ¼U 10ð Þ4�4 1þ Stop�

� �2:662500z

1=250 n1=3U zð Þ2

ð10:35Þ

where Stop is a topographic factor and � is the hill slope. Values for the spectral density functions

given by Equations 10.29, 10.30, 10.34 and 10.35 are compared in Table 10.3 for a mean velocity

U(10)¼ 25.0 m/s and a roughness length z0¼ 0.3 m.

As can be seen from Table 10.3, the values of Su(z, n) obtained using Equations 10.34 and 10.35

decrease with increasing height. It can be observed that, for the lower frequencies, the spectrum

based on the ESDUmodel yields much lower values for the power spectral density function in the

lower frequency range than the other three spectra.

In Figure 10.3, Equations 10.29, 10.30 and 10.34 are plotted in non-dimensional form for

turbulent wind with U(10)¼ 30 m/s and z0¼ 0.08 m.

10.4.4 Cross-correlation and cross-covariance functionsThe cross-correlation and cross-covariance of two continuous records [Uþu(t)]j and [Vþv(t)]k,

recorded at two different stations j and k in space, are measures of the degree to which the two

Table 10.3 Variation in spectral density function values for U(10)¼ 25.0 m/s, z0¼ 0.3 m and Stop�¼ 0

Equation Height: m 0.1 Hz 0.5 Hz 1.0 Hz 2.0 Hz 3.0 Hz 4.0 Hz

10.29 108.0275 7.8016 2.4616 0.7757 0.3947 0.2244

10.30 84.5362 5.9598 1.8790 0.5920 0.3012 0.1865

10.34 100 54.1373 4.1139 1.3136 0.4166 0.2124 0.1317

10.34 200 38.9267 2.8276 0.8975 0.2838 0.1446 0.0896

10.32 300 31.6070 2.2567 0.7147 0.2257 0.1149 0.0712

10.35 100 1.5322 0.5499 0.3464 0.2182 0.1663 0.1375

10.35 200 1.2833 0.4389 0.2765 0.1742 0.1329 0.1097

10.35 300 1.1371 0.3889 0.2450 0.1543 0.1178 0.0972


195

records are correlated in the amplitude domain. The cross-correlation function is given by

RjkUV �ð ÞT!1¼ 1

T

ð1�1

U þ u tð Þ½ �j V þ v tþ �ð Þ½ �k dt ð10:36Þ

and the cross-covariance function by

Cjkuv �ð Þt!1¼ 1

T

ð1�1

uj tð Þvk tþ �ð Þdt: ð10:37Þ

When � ¼ 0,

Cjkuv �ð Þ ¼ Cjk

uv 0ð Þ ¼ �2 uj ; vk� �

¼ �2uv ð10:38Þ

where �2uv is the cross-variance.

10.4.5 Cross-spectral density and coherence functions for longitudinal velocityfluctuations

Having defined the cross-covariance function between the fluctuating velocity components of

wind at stations j and k at zero time lag, it can be shown that

Cjkuv 0ð Þ ¼

ð10Scrujvk

nð Þdn ¼ð10SCujvk nð Þdnþ

ð10SQujvk

nð Þdn ð10:39Þ

where ¼p�1 and Scr

ujvk nð Þ is the cross-spectral density function. The latter is a measure of the

degree to which two histories u(t) and v(t), recorded at stations j and k respectively, are correlated

Figure 10.3 Comparison of spectral density functions given by Equations 10.29, 10.30 and 10.34 for

U(10)¼ 30m/s and z0¼ 0.08m� Sn(z, u)

Kaimal, z = 10 m Kaimal, z = 500 m

Kaimal, z = 300 m

Kaimal, z = 100 m

HarrisDavenport

Frequency n: Hz

10.00

1.00

0.10

0.0110–3 10–2 10–1 1 10


196

in the frequency domain. The terms SCujvk

nð Þ and SQujvk

nð Þ are known as the co-spectrum and quad-

rature spectrum, respectively. In wind engineering, the quadrature spectrum is usually assumed to

be negligible compared to the co-spectrum. Equation 10.39 may therefore be reduced to

Cjkuv 0ð Þ ¼

ð10Scrujvk

nð Þdn ¼ð10SCujvk

nð Þdn: ð10:40Þ

On the basis of wind tunnel measurements, it has been suggested that it is reasonable to assume in

engineering calculations that

Scrujvk

nð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSuj

nð Þ � Svknð Þ

h ie��

rð10:41Þ

where e��, known as the narrow-band cross-correlation, is the square root of the coherence

function e�2�¼ coh2ujk(n), and

� ¼2n

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC2

xðxj � xkÞ2 þ C2yðyj � ykÞ2 þ C2

z ðzj � zkÞ2� �q

UðzjÞ þUðzkÞð10:42Þ

where the exponential decay coefficients Cz¼ 10 and Cy¼ 16. Full-scale measurements indicate

that Cy and Cz decrease with increasing height however, and increase with increasing wind

speed and increasing ground roughness. Different wind codes may therefore recommend other

values for Cy and Cz than those given above. In Chapter 11, it is shown that the response of

multi-DOF systems is a function of both spectral and cross-spectral density functions. It is there-

fore of interest to see how the value of e�� varies with the distance between two points. In general,

it can be observed that the value of e�� decreases with (a) increasing distance between two points,

(b) increasing frequencies and (c) decreasing wind speeds.

Substitution of different values for n, ( yj� yk), (zj� zk), U(zj) and U(zk) into Equation 10.42

indicates that, for values of U(10)� 25.0 m/s and frequencies greater than approximately

1.5 Hz, the correlation between two histories is negligible when the distance between two stations

is greater than, say, 5.0 m. For many civil engineering structures, this therefore seems to imply that

the effect of cross-correlation can frequently be ignored when undertaking dynamic analysis in the

frequency domain.

Generation of cross-correlated wind histories is a time-consuming task and there is little evidence

that it is worth the effort. The past research evidence shows that it will not make much of a

difference to the overall analysis results.

Example 10.3

A 45m tall mast, whose first natural frequency is 1.0Hz and second natural frequency is 2.0 Hz,

is subjected to amean wind speed of 25m/s 10m above ground level. Calculate the values of the

spectral density function corresponding to the two first natural frequencies of the mast at points

P1¼ 25.0m, P2¼ 35.0m, P3¼ 40.0 m and P4¼ 45.0m along the length of the mast. Hence

calculate the values of the coherence and the cross-spectral density functions for points P4and P3, points P4 and P2 and points P4 and P1. Assume the ground roughness length to be

0.3m and use Equation 10.34 when calculating the spectral density values.


197

Before calculating the spectral density, coherence and cross-spectral density functions, we

must first calculate the shear velocity u� and the velocities U(25), U(35), U(40) and U(45).

From Equation 10.5,

u� ¼U 10ð Þ

2 ln 10=z0ð Þ ¼25:0

2 ln 10=0:3ð Þ ¼ 2:8518 m=s

and hence

U 25ð Þ ¼ 2:5� 2:8518� ln 25=0:3ð Þ ¼ 31:327 m=s

U 35ð Þ ¼ 2:5� 2:518� ln 35=0:3ð Þ ¼ 33:9316 m=s

U 40ð Þ ¼ 2:5� 2:518� ln 40=0:3ð Þ ¼ 34:8836 m=s

U 45ð Þ ¼ 2:5� 2:8518� ln 45=0:3ð Þ ¼ 35:7233 m=s:

The spectral density function given by Equation 10.34 is

Su z; nð Þ ¼ 200u2� f z; nð Þn 1þ 50f z; nð Þ½ � 5=3

where

f z; nð Þ ¼ zn

U zð Þ

and hence

f 25:0; 1:0ð Þ ¼ 25:0� 1 � 031:5327

¼ 0:79283

Su 25:0; 1:0ð Þ ¼ 200� 2:5182 � 0:9283

1:0 1þ 50� 0:79283½ �5=3¼ 2:6843 m2=s

f 35:0; 1:0ð Þ ¼ 35:0� 1:0

33:9316¼ 1:03149

Su 35:0; 1:0ð Þ ¼ 200� 2:85182 � 1:03149

1:0 1þ 50� 1 � 03149½ �5=3¼ 2:2739 m2=s

f 40:0; 1:0ð Þ ¼ 50:00� 1:0

34:8836¼ 1:14667

Su 40:0; 1:0ð Þ ¼ 200� 2:85182 � 1:14667

1:0 1þ 50� 1:14667½ �5=3¼ 2:257 m2=s

f 45:0; 1:0ð Þ ¼ 45:0� 1:0

35:7233¼ 1:25968

Su 45:0; 1:0ð Þ ¼ 200� 2:85182 � 1:25968

1:0 1þ 50� 1:25968½ �5=3¼ 2:0017 m2=s

The cross-spectral density function and the square root of the coherence function e�� are

given by Equations 10.41 and 10.42. Because the points P1, P2, P3 and P4 lie on a vertical


198

line, the expression for e�� reduces to

e�’ ¼ exp�2nCz zj � zk

� �

U Zj

� ��U Zkð Þ

" #:

If it is assumed that Cz¼ 10, then

cohuv 45:0; 40:0; 1:ð Þ ¼ exp�2� 1:0� 10 45:0þ 40:0ð Þ

35:233þ 34:8836ð Þ

� �¼ 0:24261

cohuv 45:0; 35:0; 1:0ð Þ ¼ exp�2� 1:0� 10 45:0� 35:0ð Þ

35:7233þ 33:9316ð Þ

� �¼ 0:05663

cohuv 45:0; 25:0; 1:0ð Þ ¼ exp�2� 1:0� 10 45:0� 25:0ð Þ

35:7233þ 31:327ð Þ

� �¼ 0:0261:

The values for the different cross-spectral density functions at n¼ 1.0 Hz can now be calcu-

lated from Equation 10.42, which for two points along a vertical line may be written as

SCuv z1; z2; nð Þ ¼ e�� Su z1; nð ÞSv z2; nð Þ½ �1=2

and therefore

SCuv 45:0; 40:0; 1:0ð Þ ¼ 0:4261� 2:017� 2:1257½ �1=2¼ 0:5005 m2=s

SCuv 45:0; 40:0; 2:0ð Þ ¼ 0:038771 m2=s

SCuv 45:0; 25:0; 1:0ð Þ ¼ 0:00261� 2:0017� 2:6843½ �1=2¼ 0:0061 m2=s:

The above calculations repeated for n¼ 2.0 Hz yield

Su 25:0; 2:0ð Þ ¼ 0:8631 m2=s

Su 35:0; 2:0ð Þ ¼ 0:7277 m2=s

Su 40:0; 2:0ð Þ ¼ 0:6792 m2=s

Su 45:0; 2:0ð Þ ¼ 0:6388 m2=s

cohuv 45:0; 40:0; 2:0ð Þ ¼ 0:058861

cohuv 45:0; 35:0; 2:0ð Þ ¼ 0:003206

cohuv 45:0; 25:0; 2:0ð Þ ¼ 0:000007

SCuv 45:0; 40:0; 2:0ð Þ ¼ 0:038771 m2=s

SCuv 45:0; 35:0; 2:0ð Þ ¼ 0:002185 m2=s

SCuv 45:0; 25:0; 2:0ð Þ ¼ 0:000005 m2s:

The above calculations reveal that the values of the cross-spectral density functions, as well as

the ratios of the same functions to the spectral density functions, decrease with increasing

distance between two histories and with increasing frequency.


199

10.5. Probability density function and peak factor for fluctuatingcomponent of wind

Let the range of the amplitudes of the fluctuating velocity component of wind u(t) associated with

a given record be divided into equal intervals �u(t) and let the amplitude of u(t) lie within the

interval ui(t) to ui þ 1(t) a total of ni times. A graph in which the numbers of ni are plotted against

the interval ui(t) and ui þ 1(t) as shown in Figure 10.4(a) is called a histogram. If ni is divided by the

total number of readings n and the interval �u(t) is made so small that it may be written as du(t),

the histogram becomes a smooth curve as shown in Figure 10.4. The curve is referred to as the

probability density function or probability distribution function and is denoted p(u). Because

of its derivation, it follows that

ð1�1

p uð Þdu ¼ 1 ð10:43Þ

ð1�1

u tð Þp uð Þdu ¼ �2u: ð10:44Þ

In wind engineering, the fluctuating component of wind is considered as a normally distributed

stationary random signal with zero mean and standard deviation �u. The probability density func-

tion can therefore be assumed to be Gaussian, in which case it can be shown that

p zð Þ ¼ 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2�ð Þexp �z2=2½ �

p ð10:45Þ

where

z ¼ u tð Þ=�2u: ð10:46Þ

The magnitude of the amplitude of the maximum fluctuation that may occur within a given time

interval T of such a process is expressed as

u tð Þmax¼ ��u ð10:47Þ

where

� ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln Tð Þ

pþ 0:577

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln Tð Þ

pð10:48Þ

Figure 10.4 (a) Histogram, (b) probability density function and (c) cumulative distribution functions

(a)u(t)

ni P(u) Q(U0) P(U0)

u(t) U(t)(b) (c)


200

¼

ð10n2Su nð Þ dn

ð10Su nð Þdn

2664

3775

1=2

: ð10:49Þ

For weakly damped structures, v may be assumed to be equal to f¼ 2�!n. When this is the

case

u tð Þmax¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2�!nTð Þ½ �

pþ 0:577

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2�!nTð Þ½ ��u

p: ð10:50Þ

10.6. Cumulative distribution functionIn many codes, a design value for wind is defined as a value with a stated probability of being

exceeded. A useful tool for this purpose is the cumulative distribution function, which is usually

denoted P(U0). If p(U) is the probability distribution function for the total wind velocity U(T),

then

P U0ð Þ ¼ Prob U < U0½ � ¼ðU0

�1p Uð ÞdU: ð10:51Þ

Alternatively, a design value may be defined as a value that has a stated probability of not

being exceeded. If this is denoted Q(U0), the cumulative distribution function in this case is

given by

Q U0ð Þ ¼ Prob U > U0½ � ¼ð�1

U0

p Uð Þ dU: ð10:52Þ

P(U0) therefore yields the probability that the wind speed U(t) is less than U0, and Q(U0) the

probability that U(t) is greater than U0. Diagrams of both types of cumulative distribution

function are shown in Figure 10.4(c).

10.7. Pressure coefficientsThe fluctuating pressure caused by wind is given by

p tð Þ ¼ 12 �Cp U tð Þ � _xx½ �2 ð10:53Þ

where � is the density of air, Cp is the pressure coefficient for a given point and _xx is the velocity of

the structure at the same point and in the direction of the wind. The integration of the pressure p(t)

over the surface of a structure or structural element will yield the resultant force exerted by the

wind.

The force components parallel and perpendicular to the along-wind direction are given by

Fd ¼ 12 �CdAd U tð Þ � _xx½ �2 ð10:54aÞ

F1 ¼ 12 �C1A1 U tð Þ � _xx½ �2 ð10:54bÞ

respectively, where Cd is the drag coefficient, C1 is the lift coefficient, Ad is the area projected onto

a plane perpendicular to the direction of the wind and A1 is the area projected onto a plane in the

along-wind direction by a unit length of structure or structural element.


201

FURTHER READING

Davenport AG (1961) The spectrum of horizontal gustiness near the ground in high winds.

Quarterly Journal of the Royal Meteorological Society 87: 194.

Deaves DM and Harris IR (1978) A Mathematical Model of the Structure of Strong Winds.

Report No. 76, United Kingdom Construction Industry Research and Information

association.

Harris CM (1988) Shock vibration, 3rd edn. McGraw-Hill, London.

Kaimal JC, Wyngaard JC, Izumi Y and Cote OR (1972) Spectral characteristics of surface-

layer turbulence. Quarterly Journal of the Royal Meteriological Society 98: 563–589.

Lawson TV (1980) Wind Effects on Buildings, vols 1 and 2. Applied Science, London.

Simue E and Scanlan RH (1978) Wind Effects on Structures. Wiley, Chichester.

van der Hoven I (1957) Power Spectrum of Horizontal Wind Speed in the Frequency Range from

0.0007 to 900 Cycles Per Hour. US Weather Bureau.

Problem 10.1

Given that the wind speed 10m above the ground is 25.0 m/s and the surface drag coefficient

k¼ 28.0� 10�3, calculate the shear velocity u� and the roughness length z0.

Problem 10.2

A 100m tall transmission tower is situated in an area with pine forest for which the roughness

length may be assumed to be 1.0 m. The design wind speed 10m above the ground is 28 m/s.

Calculate the corresponding shear velocity and the wind velocities at the heights 50 m, 90 m

and 100 m.

Problem 10.3

Let the first natural frequency of the tower in Problem 10.2 be 1.0 Hz. Calculate the values of

the Davenport and Harris power spectra for this frequency and for the design wind speed

given in Problem 10.2. Compare the values obtained with those calculated for heights of

90 m and 100m using Kaimal’s spectrum.

Problem 10.4

Use the data obtained in Problems 10.2 and 10.3 to calculate the values of the square root

of the coherence function and of the cross-spectral density function at the first natural

frequency of the tower for heights 90 m and 100m. Assume the value of the exponential

decay coefficient Cz in Equation 10.32 to be 8.


202


ISBN: 978-0-7277-4176-9



Chapter 11

Dynamic response to turbulent wind:frequency-domain analysis

11.1. IntroductionWind acting on structures induces stresses and deflections. If the deformations caused alter the

boundary conditions of the incident wind to such an extent that they alter the flow pattern,

and this gives rise to succeeding deflections of an oscillating nature, a phenomenon referred

to as aeroelastic instability is said to occur. In this chapter, a variety of aeroelastic instabilities

due to wind are given with their use in dynamic response analysis of aeroelastically stable

structures.

11.2. Aeroelasticity and dynamic responseAll aeroelastic instabilities result from aerodynamic forces that are influenced by the motion of the

structure. The main types of aeroelastic instability are cross-wind galloping, torsional divergence

and flutter. Buffeting, which is defined as the random loading of structures due to velocity

fluctuations in the oncoming wind, may be aeroelastically stable or unstable.

Cross-galloping is mainly associated with slender sections having special cross-sections such as

rectangular or D-sections, or the effective sections of greased or ice-coated cables. Such structures

can exhibit amplitudes of vibration many times their cross-sectional dimensions, and at

frequencies much less than those of vortex shedding from the same sections.

Torsional divergence is divergence in which structures subjected to the lift, drag and torsional

forces due to wind will tend to twist, effectively increasing the angle of attack. As the wind velocity

increases, the structure will twist further until it may be twisted to destruction. The wind speed at

which structural collapse occurs is referred to as the critical divergence velocity of wind. In most

cases of interest to the structural engineer, the critical divergence velocities are extremely high and

much greater than the wind velocities considered in design.

The term ‘flutter’ covers a whole class of aeroelastic oscillations such as classical flutter, single-

DOF flutter and panel flutter. Classical flutter implies an aeroelastic phenomenon in which

wind causes and couples together oscillations of a structure in one vertical and one rotational

DOF. Single-DOF flutter is associated with bluff, which in un-streamlined bodies cause flow

separation. Notable examples are the decks of cable-suspended span bridges, which can exhibit

single-degree torsional instability. Panel flutter is sustained vibration of panels caused by the

passing of wind; the most prominent has been caused by the high-speed passage of air in

supersonic flows. In civil engineering, panel flutter has mainly been associated with membrane

structures and pre-stressed cable net roofs with insufficient local or global anticlastic

curvature.

203

Galloping, flutter and also vortex-induced vibration (when the motion of the structure controls

the vortex shedding) are referred to as self-excited vibration. If the flow of air results in an initial

disturbance, the oscillations will either diverge or decay according to whether the energy of

motion extracted from the flow is greater or less than the energy dissipated by the level of

structural damping. In the case of flutter, wind speeds that cause neither decaying nor diverging

oscillations are referred to as critical flutter velocities.

Buffeting, as mentioned above, is defined as the random loading due to the velocity fluctuations in

turbulent wind. Aeroelastic instability in the case of buffeting is mainly associated with non-linear

structures such as slender towers, the decks of cable-suspended span bridges and insufficiently

tensioned cables in cable beam and cable net roof structures.

A great deal of research has been undertaken in order to develop and improve methods for

predicting the response for the different types of aeroelastic instability. Many of the problems

are only partially understood, and the solution of a particular problem usually requires the use

of wind tunnels in order to generate numerical models with properties similar to the prototype.

Aeroelastic unstable problems are inherently non-linear and, although important, they are

outwith the scope of this book. For further information, the interested reader is referred to

books on wind engineering such as Lawson (1990) and Simue and Scalan (1978).

A unified viewpoint is that in general the solution of aeroelastic instability problems can be

tackled only by forward integration in time which enables wind speeds, structural deformations

and structural stiffness, aerodynamic coefficients and aerodynamic damping to be updated at

the end of successive time increments. In most textbooks, the theoretical solution of aeroelastic

instability problems is confined to 2D problems.

11.3. Dynamic response analysis of aeroelastically stable structuresThe dynamic response of aeroelastically stable structures to wind may be predicted by either

a time-domain or a frequency-domain approach. The former requires the generation of

spatially correlated wind histories. The latter is based on the use of spectral density or power

spectra for wind. Of the two, the frequency domain is more generally used although time-

domain methods are more powerful. The reason why the latter approach is not currently used

much is that although efficient methods are available and can provide correlated wind histories,

the time and cost of the analysis does not justify the differences in the results.

11.4. Frequency-domain analysis of 1-DOF systemsThe aim of this approach, originally proposed by Davenport (1961), is to predict the statistical

properties of the structural response starting from the knowledge of the statistical properties of

the forces due to wind. Assuming that the fluctuating nature of the wind velocity is stationary,

forces due to wind are fully defined by their mean values, their probability distributions and

their spectrum of fluctuations. The method is applicable only to structures whose response can

be assumed to be linear. When it is applied to non-linear structures it is assumed that the dynamic

response is small compared to the static response; the non-linearities are only taken into account

when calculating the latter. The total response is calculated by superimposing the dynamic

response on the static response.

The frequency-domain method is based on the following hypotheses


204

g the dynamic response of the structure is linearg the mean aerodynamic force due to turbulent wind is the same as that in a steady flow with

the same mean velocityg the relationships between the velocity and force fluctuations are linearg the probability distributions of the wind speed fluctuations are Gaussian.

The second hypothesis implies that the effect of the acceleration of the wind is negligible. If

required, this effect can be accounted for by an additional pressure term �Cm(A/B) du(t)/dt

where Cm is an additional mass coefficient, A is a reference area and B is a reference dimension.

The existence of this term follows from consideration of the dynamic equilibrium condition of the

wind. It represents the force that the wind flowing around a building exerts on the structure as a

consequence of the change in wind velocity. The third hypothesis requires that the velocity

fluctuations u should be negligible compared to the mean velocity U.

The prediction of statistical response requires knowledge of the mean response, the response

spectrum and the probability distribution of the response. The mean response is determined by

considering the load due to the mean wind speed U as a static load, while the response due to

the fluctuating component u(t) of wind is determined by first calculating the variance of response.

The reason for this is that the relationships between velocity, force and displacement fluctuations

are assumed to be linear and the distribution of the velocity fluctuations is assumed to be

Gaussian. The distribution of the amplitudes of the fluctuating wind force must also be Gaussian,

as must the distribution of the amplitudes of the fluctuating component of the response. From

Equation 10.28, the variance of the fluctuating component of wind is given by

�2u ¼

ð10Su nð Þdn: ð11:1Þ

Similarly, the variances of a drag force fd(t) and response x(t) are found from integration of the

force and response spectra, respectively. We therefore have

�2f ¼

ð10Sf nð Þdn ð11:2Þ

�2x ¼

ð10Sx nð Þdn: ð11:3Þ

11.5. Relationships between response, drag force and velocity spectrafor 1-DOF systems

The fluctuating along-wind drag force acting on the area A of a 1-DOF system vibrating with a

velocity _xx tð Þ is given by

fd tð Þ ¼ 12 �CdA U tð Þ � _xx tð Þ½ � 2 � 1

2 �CdAU2 ð11:4Þ

or

fd tð Þ ¼ 12 �CdA U2 þ u2 tð Þ þ _xx2 tð Þ þ 2Uu tð Þ � 2U _xx tð Þ � 2u tð Þ _xx tð Þ �U2

� �: ð11:5Þ

When it can be assumed that u(t) and _xx tð Þ are small compared to U, the terms u2(t), _xx2 tð Þ and2u tð Þ _xx tð Þ are neglected and the expression for fd(t) is written as

fd tð Þ ¼ 12 �CdA 2Uu tð Þ � 2U _xx tð Þ½ �: ð11:6Þ

Dynamic response to turbulent wind: frequency-domain analysis

205

The equation of motion for a 1-DOF system subjected to a fluctuating drag force may be written

as

M€xxþ 2�s!nM _xxþ Kx ¼ 12 �Cd 2Uu tð Þ � 2U _xx tð Þ½ �: ð11:7Þ

Since _xx tð Þ ¼ _xx, the terms in Equation 11.7 may be rearranged as

M€xxþ 2�s!nM _xxþ �CdAUð Þ _xxþ Kx ¼ 12 �CdA 2Uu tð Þ½ � ð11:8Þ

or

M€xxþ 2!nM �s þ �að Þ _xxþ Kx ¼ 12 �CdA 2Uu tð Þ½ � ð11:9Þ

where �a is the equivalent viscous aerodynamic damping ratio, which for light flexible structures

can contribute considerably to the total damping, and is defined

�a ¼�CdAU

2!nM: ð11:10Þ

Inspection of Equation 11.9 shows that the resulting dynamic force acting on the structure, when

the term �CdAU _xx tð Þ is considered as part of the total damping mechanism, is

fd tð Þ ¼ �CdAUu tð Þ ð11:11Þ

or

fd tð Þ ¼ 2Fd

Uu tð Þ ð11:12Þ

where

Fd ¼ 12 �CdAU

2: ð11:13Þ

In order to obtain a relationship between the spectrum of the fluctuating component of the drag

force and the spectrum of the fluctuating velocity component, the frequency spans of the fluctu-

ating wind and force components are divided into unit frequency intervals with each interval

centred at the frequency n. If only one frequency interval is considered, then

u tð Þ ¼ u sin 2�ntð Þ ð11:14Þ

fd tð Þ ¼ fd sin 2�ntð Þ ð11:15Þ

since fd varies linearly with u(t). Substitution of the expressions for u(t) and fd into Equation 11.12

yields

fd ¼ 2Fd u=Uð Þ: ð11:16Þ

The relationship between the amplitudes of force and velocity is therefore

fdFd

¼ 2u

Uð11:17Þ


206

or

f 2dF2d

¼ 4u2

U2: ð11:18Þ

As the coordinates of spectral density functions are proportional to the square of the amplitudes

and inversely proportional to the frequency of each of the constituent harmonics, it follows that

Sfdnð Þ

F2d

¼ 4Su nð ÞU2

ð11:19Þ

which may be written in non-dimensional form by multiplying each term by the frequency n.

The effects of the spatial variation in the wind velocity and the frequency dependence of the drag

coefficient, both of which are important for structures with large surfaces, may be taken into

account by introducing aerodynamic admittance function A(n). Equation 11.19 therefore may

be rewritten as

nSfdnð Þ

F2d

¼ 4A nð Þ nSu nð ÞU2

: ð11:20Þ

The literature provides little information on the proper values to be used for the aerodynamic

admittance function, and it appears that more research is required in this field. Experimental

values proposed by Davenport (1961) and Vickery (1965) are given in Figure 11.1.

Having developed an expression for the load spectrum in terms of the velocity spectrum, it remains

to express the response spectrum in terms of the load spectrum. From the theory of forced

vibrations of damped linear 1-DOF systems (see Equation 4.15), the response x(t) to a force

fd tð Þ ¼ fd sin 2�ntð Þ ð11:21Þ

Figure 11.1 Variation of the aerodynamic admittance factor A(n) with the reduced frequency nB/U(10):

the value B is a structural reference dimension, n corresponds to a structural mode frequency and U(10)

is the reference wind velocity

0.01 0.1 1.0 10.0

Davenport (1961)

Vickery(1965)

Reduced frequency nB/U(10)

A(n

)

2.0

1.0

0.5

0.1


207

is

x tð Þ ¼ fdK

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ ð2�rÞ2� �q sin 2�nt� �ð Þ ð11:22Þ

or

x tð Þ ¼ fdK

MF nð Þ sin 2� nt� �ð Þ: ð11:23Þ

The maximum value of x(t), which occurs when sin(2�nt)¼ 1, is therefore

x ¼ fdK

MF nð Þ ð11:24Þ

where x and fd are the amplitudes of the harmonic response and force components associated with

the unit frequency interval centred at the frequency n. SinceK¼Fd/xs, Equation 11.24 may also be

written in the form

x

xs¼ fd

Fd

MF nð Þ: ð11:25Þ

Squaring each term in Equation 11.25 yields

x2

x2s¼ f 2d

F2d

MF2 nð Þ ð11:26Þ

where MF2(n)¼M(n) is referred to as the mechanical admittance factor. Since the coordinates of

power spectra are proportional to the square of the amplitudes of the constituent harmonics, it

follows that

Sx nð Þx2s

¼Sfd

nð ÞF2d

M nð Þ ð11:27Þ

or, if each term in Equation 11.27 is multiplied by n, in non-dimensional form we have

nSx nð Þx2s

¼ 4M nð ÞA nð ÞSu nð ÞU2

: ð11:28Þ

The variance of the fluctuating component of the response is now determined by integration of

both sides of Equation 11.29 with respect to n:

�2x ¼

ð10Sx nð Þdn ¼ 4

x2sU2

ð10M nð ÞA nð ÞSu nð Þdn: ð11:29Þ

For weakly damped structures the expression for �2x can be approximated to

�2x ¼

ð10Sx nð Þdn � 4

x2sU2

M nð ÞA nð ÞSu nð Þ�n ð11:30Þ


208

where

�n ¼ ��n

M nð Þ ¼ 1=4�2 ð11:31Þ

in which case

�2x ¼ x2s

U2

�n

�A nð ÞSu nð Þ: ð11:32Þ

The maximum probable displacement is therefore given by

xmax ¼ ��x ð11:33Þ

where � is a peak factor for weakly damped structures (see Equation 10.50).

Example 11.1

Amotorway sign of dimensions shown in Figure 11.2 may be assumed to vibrate as a 1-DOF

system in the along-road direction. The supporting structure is designed as a portal frame

with a horizontal beam, which can be considered to be rigid. The EI value for each

column is 228 799.08 kNm2 and the equivalent lumped mass, 9.0 m above the ground, is

5.3 t. At the point where the sign is positioned the motorway runs through woodland, so

the roughness length z0 may be taken as 0.9 m. If the design wind speed U(10)¼ 30.0 m/s,

determine (i) the maximum dynamic and hence maximum total response; and (ii) the

maximum shear force and bending moment occurring at the foot of each column. Use the

power spectrum proposed by Davenport (1961) (Equation 10.26) and the curve for the

aerodynamic admittance factor proposed by Vickery (1965) (Figure 11.1) when calculating

the variance of response. The drag coefficient for the 20.0� 2.0 m motorway sign is

Cd¼ 2.03. The specific density of air is 1.226 kg/m3.

Figure 11.2 Motorway sign

To the West To the NorthM4

20 m

10 m

8 m

2 m

The prediction of the along-wind dynamic response to wind tends to be lengthy. At this stage,

therefore, the reader may find it helpful to have a listing of the expressions and equations

needed, as they form the framework for the required calculations.


209

The total response x is given by

x ¼ xs þ xd ¼ Fd=K þ ��x

where

Fd ¼ 12 �CdAU

2

K ¼ 2 3EI=L3� �

¼ 6EI=L3

�2x ¼ x2s

U2

�n

�A nð ÞSu nð Þ

n ¼ 1

2�


M

� �s

� ¼ �st þ �a ¼ �st þ�CdAU

2!nM¼ �st þ

Fd

2�nUM


n 1þ f 2ð Þ4=3

u� ¼U 10ð Þ

2:5 ln 10=z0ð Þ

f ¼ 1200n=U 10ð Þ

� ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln nTð Þ½ �

pþ 0:577=

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln nTð Þ½ �

p

where T¼ 3600 s, if the value of U(10) is based on records of 1 h duration.

Calculation of Fd, K and xs

Fd ¼ 12 � 1:226� 2:03� 40:� 30:02 ¼ 44 798:4 N

K ¼ 6� 228 799:08=9:03 ¼ 1883:12 kN=m

xs ¼ 44 789:04=1883:12� 1000 ¼ 0:238 m

Calculation of the along-wind natural frequency fn

fn ¼ n ¼ 1

2�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1883:12� 1000

5300:0

� �s¼ 3:0 Hz

Determination of the aerodynamic ratio �a and total damping ratio �

�a ¼44 798:04

2�� 3:0� 30:0� 5300:0¼ 0:0149

� ¼ 0:01þ 0:0149 ¼ 0:0249


210

It is worth noting that in the above example the air contributes significantly to the

total damping and that the dynamic response is 1.23 times the response due to the mean wind,

although the natural frequency of the structure lies within the part of the frequency spectrum

of the wind where the energy of the wind fluctuations is considerably reduced (see Figure 10.3).

The level of energy may be more fully appreciated by the following example, in which the

eccentricity and eccentric mass of a variable speed motor, which will produce the same

maximum amplitude of vibration as that caused by the wind, are calculated.

Calculation of the shear velocity u�

u� ¼30:0

2 ln 10=0:9ð Þ ¼ 4:984 m=s

Determination of the value of Davenport’s spectrum at n¼ fn

f ¼ 1200n=U 10ð Þ ¼ 1200� 3:0=30:0 ¼ 120:0

Su nð Þ ¼ 4� 4:9842 � 120:02

3:0 1þ 120:02ð Þ4=3¼ 1:3612 m2=s

Calculation of the aerodynamic admittance factor A(n)

reduced frequency nB=U 10ð Þ ¼ 3:0� 20:0=30:0 ¼ 2:0 Hz

Hence, from Figure 11.1,

A uð Þ ¼ A 3:0ð Þ ¼ 0:139

Evaluation of the variance �2x and standard variation of response �x

�2x ¼ 0:02382

30:02� �� 3:0

0:0259� 0:139� 1:3612 ¼ 4:3333� 10�5 m2

�x ¼ 0:006583 m

Determination of the peak factor �

� ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 3:0� 3600ð Þ½ �

pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln 3:0� 3600ð Þ½ �p ¼ 4:441

Calculation of the maximum response x, the maximum bending moment Mmax and the maximum

shear force SFmax in each column

x ¼ xs þ ��x ¼ 0:0238þ 4:441� 0:006583 ¼ 0:0530351 m

SFmax ¼ 12Kx ¼ 1

2 � 1883:12� 0:0530351 ¼ 49:936 kN

Mmax ¼ 12 Kxð ÞH ¼ 1

2 � 1883:12� 0:0530351� 9:0 ¼ 449:22 kN m


211

11.6. Extension of the frequency-domain method to multi-DOFsystems

The response of multi-DOF systems can now be calculated in a similar manner to that of 1-DOF

systems by first decoupling the equations of motion (see Chapter 8) and then considering each

modal equation as the equation of motion of a single-DOF system.

The equations of motion for a multi-DOF system subjected to the drag forces caused by the

fluctuating component wind can be written in matrix form as

M€xx ¼ C _xxþ Kx ¼ f d tð Þ: ð11:34Þ

In order to decouple the equations of motion, let

x ¼ Zq ð11:35Þ

where Z is the normalised mode-shape matrix and q is the principal coordinate vector of the

system. Substitution of the expression for x into Equation 11.34 and post-multiplication of

each term in the same equation by ZT yields the following system of decoupled equations

Example 11.2

Calculate the product value of the eccentric mass times the eccentricity of a variable speed

vibrator that will vibrate the motorway sign in Example 11.1 at resonance with the same

maximum amplitude as that caused by the wind, and hence calculate the value of the eccentric

mass at an eccentricity of 0.25 m.

The maximum dynamic response of a 1-DOF system to harmonic excitation caused by an

eccentric mass vibrator is

xmax ¼me!2

K

1

2�:

Neglecting the aerodynamic damping, which is a function of the mean wind velocity

me ¼ 2�Kxmax

!2n

¼ 2� 0:01� 1883:12� 1000� 4:441� 0:006583ð Þ2�� 3:0ð Þ2

¼ 3:099 kg m:

Assuming an eccentricity of 0.25 m, the size of the eccentric mass would be

m ¼ 3:099=0:25 ¼ 12:96 kg:

A very large vibrator would therefore be needed to produce the same maximum amplitudes of

vibration as those caused by the wind.


212

which govern the response

€qq1 þ 2�1!1 _qq1 þ !21q1 ¼ fq1 tð Þ

€qq2 þ 2�2!2 _qq2 þ !22q2 ¼ fq2 tð Þ

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

€qqi þ 2�i!i _qqi þ !2i qi ¼ fqi tð Þ

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

€qqN þ 2�N!N _qqN þ !2NqN ¼ fqN tð Þ

ð11:36Þ

where

fqi tð Þ ¼ ZTi f d tð Þ: ð11:37Þ

The relationship between the global and principal coordinates is given by Equation 11.35; we

therefore have

xj ¼XNi¼ 1

Zjiqi: ð11:38Þ

If the terms in Equation 11.38 are squared and the cross-coupling terms between the modes are

neglected,

x2j ¼XNi¼ 1

Z2jiq

2i : ð11:39Þ

The spectrum Sxj (n) can therefore be computed as the superposition of the spectra Sqi (n) as:

Sxjnð Þ ¼

XNi¼ 1

Z2jiSqi

nð Þ: ð11:40Þ

The spectrum associated with each principal coordinate qi is dependent on the spectrum Sfqinð Þ of

the corresponding force component fqi (t) in the modal force vector. For a 1-DOF system, the

response spectrum is given in terms of the force spectrum by Equation 11.27, which may be

rewritten as

Sx nð Þ ¼ 1

!2M nð ÞSf nð Þ: ð11:41Þ

Similarly for the ith principal coordinate qi,

Sqinð Þ ¼ 1

!2ið Þ2

Mi nð ÞSfqinð Þ: ð11:42Þ


213

Having obtained an expression for the spectrum of the generalised coordinate qi, it remains to

determine an expression for the spectra for the modal force component fqi. The integral of the

spectrum of the ith component fqi (t) in the modal force vector is

ð10Sfqi

nð Þdn ¼ �2fdi

¼ 1

T

ðT0fqi tð Þ fqi tð Þdt ð11:43Þ


fqi tð Þ ¼XNj¼ 1

Zji fdj tð Þ: ð11:44Þ

Substitution of this expression for fqi (t) into Equation 11.43 yields

ð10Sfqi

nð Þdn ¼ 1

T

ðT0

XNj¼ 1

Zji fdj tð Þ �XN

k¼ 1

Zki fdk tð Þdt: ð11:45Þ

Since in this equation only the global forces fj (t) and fk(t) vary with time, it may be written as

ð10Sfqi

nð Þdn ¼XNj¼ 1

XN

k¼ 1

ZjiZki �1

T0

ðT0fdj tð Þ � fdk tð Þdt

¼XNi¼ 1

XN

k¼ 1

ZjiZkiRikfd

0ð Þ

ð11:46Þ

whereRjkfd

0ð Þ is the cross-covariance between the fluctuating global loads at stations j and k at zero

time lag. Lawson (1990) shows that

Rjkfd

0ð Þ ¼ð10Scr

fdnð Þ dn ¼

ð10S c

fdnð Þdnþ �

ð10Sqfd

nð Þdn ð11:47Þ

where �¼p

�1 and Scrfjk

nð Þ is the cross-spectral density function or cross-power spectrum, S cfjk

nð Þis the co-spectrum and S

qfjk

nð Þ is the quadrature spectrum for the wind forces at stations j and k.

In wind engineering, the quadrature spectrum of the load is generally assumed to be negligible

compared to the co-spectrum and hence

Scrfjk

nð Þdn ¼ S cfjk

nð Þ: ð11:48Þ

Equation 11.46 can therefore be written as

ð10Sfjk

nð Þdn ¼XNj¼ 1

XN

k¼ 1

ZjiZki

ð10S c

fdnð Þ dn: ð11:49Þ

Differentiation of Equation 11.49 with respect to n yields the spectrum of the modal force fqi, and

we therefore have

Sfqinð Þ ¼

XNj¼ 1

XN

k¼ 1

ZjiZkiScfd

nð Þ: ð11:50Þ


214

Wind-tunnel testing and full-scale measurements indicate that for civil engineering purposes it is

sufficient to use the following formulation for the co-spectrum:

S cfjk

nð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSfj

nð Þ � Sfknð Þ

h ir� coh fd nð Þ: ð11:51Þ


Sfjnð Þ ¼ 4

F2 zj� �

U2 zj� �A nð ÞSuj

nð Þ ð11:52aÞ

Sfknð Þ ¼ 4

F2 zkð ÞU2 zkð Þ

A nð ÞSuknð Þ: ð11:52bÞ


coh fjk nð Þ ¼ e�’jk nð Þ ð11:53Þ

where �jk(n) (which is a function of the frequency n, the station coordinates xj, yj, zj and xk, yk, zk,

the mean wind speeds at stations j and k and the coefficients Cx,Cy and Cz) is defined by Equation

10.42 as

’jk nð Þ ¼2n




U zj� �

þU zKð Þ: ð11:54Þ

11.7. Summary of expressions used in the frequency-domain methodfor multi-DOF systems

The sequence of equations required to predict the response of a structure to the buffeting of turbu-

lent wind with a natural angular frequency vector !¼ [!1, !2, . . . , !i, . . . , !N]T and a normalised

mode-shape matrix Z¼ [Z1, Z2, . . . ,Zi, . . . ,ZN]T, is listed below.

x ¼ Zq ð11:55Þ

qi ¼ �qi�qi ð11:56Þ

�2qi¼ð10Sqi

nð Þ dn ð11:57Þ

Sqinð Þ ¼ 1

!2ið Þ2

Mi nð ÞSfqinð Þ ð11:58Þ

Sfqinð Þ ¼

XNj¼ 1

XN

k¼ 1

ZjiZkiScfjk

nð Þ ð11:59Þ

S cfjk

nð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSfj

nð Þ � Sfknð Þ

h ir� e��jk nð Þ ð11:60Þ

Sfjnð Þ ¼ 4

F2 zj� �

U2 zj� �Aj nð ÞSuj

nð Þ ð11:61aÞ


215

Sfknð Þ ¼ 4


Ak nð ÞSuknð Þ ð11:61bÞ

’jk nð Þ ¼2n




U zj� �

þU zkð Þ: ð11:62Þ

For structures that are weakly damped, as is normally the case, it is usually sufficient to assume

that

�2qi¼ð10Sqi

nð Þdn ¼ Sqinið Þ�n ¼ Sqi

!ið Þ�! ð11:63Þ

where

�! ¼ 12 �i!i; ð11:64Þ

in which case, Equations 11.58–11.62 may be written

Sqi !ið Þ 1

!2ið Þ2

Mi !ið ÞSfqi !ið Þ ð11:65Þ

Sfqi !ið Þ ¼XNj¼ 1

XN

k¼ 1

ZjiZkiScfjk

!ið Þ ð11:66Þ

S cfjk

!ið Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSfj

!ið Þ � Sfk!ið Þ

h ire��jk !ið Þ ð11:67Þ

Sfj!ið Þ ¼ 4

F2 zj� �

U2 zj� ��j !ið ÞSuj

!ið Þ ð11:68aÞ

Sfk!ið Þ ¼ 4


Ak !ið ÞSuk!ið Þ ð11:68bÞ

’jk !ið Þ ¼!i




� U zj� �

þU zkð Þ� � : ð11:69Þ

11.8. Modal force spectra for 2-DOF systemsFrom Equation 11.59, the expression for the modal force spectrum in the ith mode is given by

Sfqi nð Þ ¼XNj¼ 1

XN

k¼ 1

ZjiZkiScfjk

nð Þ: ð11:70Þ

The first modal force spectrum for a 2-DOF system having the mode-shape matrix

Z ¼Z11 Z12

Z21 Z22

" #ð11:71Þ


216

is therefore

S fq1 nð Þ ¼X2j¼ 1

X2

k¼ 1

Zj1Zk1Scfjk

nð Þ ð11:72Þ

or

S fq1 nð Þ ¼ Z211S

cf11

þ Z11Z21Scf12

nð Þ þ Z21Z11Scf21

nð Þ þ Z221S

cf22

nð Þ: ð11:73Þ

From Equation 11.60, it follows that

S cf11

nð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1 nð Þ � Sf1 nð Þ� �q

� e�0 ¼ Sf1 nð Þ ð11:74aÞ

S cf22

nð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf2

nð Þ � Sf2nð Þ

� �q� e�0 ¼ Sf2

nð Þ ð11:74bÞ

S cf12

nð Þ ¼ S cf21

nð Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1


� �q� e��12 nð Þ ð11:75Þ

and hence

Sfq1nð Þ ¼ Z2

11Sf1nð Þ þ 2Z11Z21

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1


� �q� e�’12 nð Þ þ Z2

21Sf2nð Þ: ð11:76aÞ

Similarly, the second modal force spectrum is given by

Sfq2nð Þ ¼ Z2




� �q� e�’12 nð Þ þ Z2

22Sf2nð Þ: ð11:76bÞ

11.9. Modal force spectra for 3-DOF systemsThe expressions for the first, second and third modal force spectra for a 3-DOF system can be

developed similarly:

Sfq1nð Þ ¼ Z2

11Sf1nð Þ þ Z2

21Sf2nð Þ þ Z2




� �q� e�’12 nð Þ

þ 2Z11Z32



� �q� e�’13 nð Þ þ 2Z21Z31



� �q� e�’23 nð Þ ð11:77aÞ

Sfq2nð Þ ¼ Z2

12Sf1nð Þ þ Z2

22Sf2nð Þ þ Z2




� �q� e�’12 nð Þ

þ 2Z12Z32



� �q� e�’13 nð Þ þ 2Z22Z32



� �q� e�’23 nð Þ ð11:77bÞ

Sfq3nð Þ ¼ Z2

13Sf1nð Þ þ Z2

23Sf2nð Þ þ Z2




� �q� e�’12 nð Þ

þ 2Z13Z32



� �q� e�’13 nð Þ þ 2Z23Z33



� �q� e�’23 nð Þ: ð11:77cÞ

The expressions for the force spectra for a multi-DOF system may therefore be quite lengthy.

Fortunately, in the case of practical engineering problems, most of the cross-spectral density

terms are negligible and may be ignored because of the distance between the load stations (see

Example 10.3).


217

11.10. Aerodynamic damping of multi-DOF systemsEquation 11.10 gives an expression for the aerodynamic damping of 1-DOF systems and in

Example 11.1 it is shown that the level of aerodynamic damping compared to that of structural

damping can be considerable. This is also the case with some multi-DOF structures such as

guyed masts, where the damping caused by the relative velocity of the structure to that of the

air flow is of greater importance than the structural velocity. It was shown how to construct

damping matrices that permit the equations of motion to be decoupled in Chapter 9; the difficulty

in including the damping due to air is that the aerodynamic damping terms couple the equations

of motion. This leads, as will be shown, to an iterative solution method of the modal equations,

unless unjustifiable assumptions are made.

The matrix equation of motion for a multi-DOF structure subjected to turbulent wind is given by

M€xxþ C _xxþ Kx ¼ 12�CdA Uþ u tð Þ � _xx tð Þ½ �2�U

2�

: ð11:78Þ

Let

Fd ¼ 12�CdAU

2; ð11:79Þ

substitution of Equation 11.79 into Equation 11.78, neglecting the terms with u2(t), _xx2 tð Þ and

u(t)x(t), yields

M€xxþ C _xxþ 2Fd=Uð Þ _xxþ Kx ¼ 2Fd=Uð Þu tð Þ ð11:80Þ

since _xx tð Þ ¼ _xx. Post-multiplication of each term in Equation 11.80 by ZT, where Z is the normal-

ised mode-shape matrix, and substitution of the following expressions for x, _xx and €xx

x ¼ Zq

_xx ¼ Z _qq

€xx ¼ Z€qq

into the resulting matrix equation yields

€qq1 þ 2�s1!1 _qq1 þ ZT1 2Fd=Uð ÞZ _qqþ !2

1q1 ¼XNi¼ 1

Zi1 2Fdi=Uið Þ ui tð Þ

€qq2 þ 2�s2!2 _qq2 þ ZT2 2Fd=Uð ÞZ _qqþ !2

2q2 ¼XNi¼ 1

Zi2 2Fdi=Uið Þ ui tð Þ

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

€qqn þ 2�sn!n _qqn þ ZTn 2Fd=U½ �Z _qqþ !2

nqn ¼XNi¼ 1

Zin 2Fdi=Uið Þ ui tð Þ:

ð11:81Þ

In the rth modal equation, the aerodynamic damping term is therefore given by

ZTr 2Fd=Uð ÞZ _qq ¼

XNi¼ 1

XNj¼ 1

Zjr 2Fdj=Uj

� �Zji _qqi ð11:82Þ


218

or

ZTr 2Fd=Uð ÞZ _qq ¼ �1r _qq1 þ �2r _qq2 þ . . .þ �rr _qqr þ . . .þ �nr _qqn ¼

XNi¼ 1

�irqi ð11:83Þ

where

�ir ¼XNj¼ 1

Zjr 2Fdj=Uj

� �Zji: ð11:84Þ

Equation 11.83 may also be written as

ZTr 2Fd=Uð ÞZ _qq ¼ �1r

_qq1_qqrþ �2r

_qq2_qqrþ . . .þ �rr þ . . .þ �nr

_qqn_qqr

� �_qqr ð11:85Þ

which shows that damping due to air couples the modal equations. These can therefore only be

solved by making certain assumptions. If the motion in each mode is assumed to be simple

harmonic or sinusoidal, then any of the terms in Equation 11.85, say term i, may be written as

�ir

_qqi_qqr¼ �ir

qi!i cos !it� �ið Þqr!r cos !rt� �rð Þ ð11:86Þ

where � is a random phase angle. Since the value of cos(!t� �) may vary between �1 and þ1, it

follows that the value of the ratio _qqi= _qqr may vary between �1 and þ1. To decouple the modal

equations it is therefore necessary to assume that the average values of the terms �ir _qqi= _qqrð Þ arezero. As the fluctuations in wind velocities are random the wind velocity itself is assumed to be

stationary; since the variance of response is only calculated in the frequency domain in the first

instant, this assumption does not seem to be unreasonable.

The modal Equation 11.80 therefore may be written as

€qq1 þ 2 �s1 þ �a1ð Þ!1 _qq1 þ !21q1 ¼

XNi¼ 1

Zi1 2Fdi=Uið Þui tð Þ

€qq2 þ 2 �s2 þ �a2ð Þ!2 _qq2 þ !22q2 ¼

XNi¼ 1

Zi2 2Fdi=Uið Þui tð Þ

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

€qqn þ 2 �sn þ �anð Þ!n _qqn þ !2nqn ¼

XNi¼ 1

Zin 2Fdi=Uið Þui tð Þ

ð11:87Þ

where the simplified modal aerodynamic damping ratio in the rth mode is given by either

�ar ¼�rr

2!r

¼ 1

2!r

ZTr 2F=Uð ÞZr ð11:88aÞ

or

�ar ¼�rr

2!r

¼ 1

2!r

XNj¼ 1

Zjr 2Fdj=Uj

� �Zjr: ð11:88bÞ


219

Example 11.2

The mast in Figure 11.3 supports two discs, one at 10 m and one at 20 m above the ground.

The diameter of each disc is 4.0 m and the drag coefficient Cd¼ 2.0. The mast is situated in an

area where the roughness length is assumed to be 1.0 m. Calculate the lateral response of each

disc when the mast is subjected to a mean wind of 30 m/s at a height of 10 m above ground

level. Assume the exponential decay coefficient for the wind speed and ground roughness to

be Cz¼ 8. Use the logarithmic law (Equation 10.11) to calculate the mean wind profile, and

Kaimal’s power spectrum in order to take account of the variation of the spectral density

function with height. The condensed stiffness matrix ~KK, the normalised mode-shape matrix

Z and the angular frequency vector ! for the tower are given below.

Figure 11.3 Tower supporting two discs

10 m

x2

x1

10 m

Assume the damping in the first and second modes to be 1.0% of critical and the aerodynamic

admittance factor to be 0.5 in the first mode and 0.25 in the second. The wind load on the

mast itself may be neglected.

~KK ¼ 765:79891�36 �10

�10 4

" #kN=m

Z ¼3:443 6:521

10:109 �10:753

" #� 10�3

! ¼25:133

119:098

" #rad=s:


220

Determination of the shear velocity u�

From Equation 10.5,

u� ¼30:0

2:0 ln 10:0=1:0ð Þ ¼ 5:212 m=s:

The mean wind velocity at a height of 20 m above ground level is found using Equation 10.4;

we therefore have

U 20ð Þ ¼ 2:5� 5:212 ln 20:0=1:0ð Þ ¼ 39:034 m=s:

The force vector due to the mean wind velocity is therefore

Fd1

Fd2

" #¼

12 � 1:226� 2:0� �� 2:02 � 30:0002

12 � 1:226� 2:0� �� 2:02 � 39:0342

" #� 10�3 ¼

13:866

23:474

�kN:

The inverse of the condensed stiffness matrix ~KK is

K�1 ¼

0:118711 0:296778

0:296778 1:068400

�� 10�3 m=kN

and hence the displacements due to the mean wind velocity are given by

xs1

xs2

" #¼

0:118711 0:296778

0:296778 1:068400

�13:866

23:474

�� 10�3 ¼

8:613

29:195

�� 10�3 m:

The decoupled equations of motion for the mast are given by

€qq1 þ 2�s1!1 _qq1 þ �11q1 þ !21q1 ¼ Z11 2Fd1=U1ð Þ u1 tð Þ þ Z21 2Fd2=U2ð Þ u2 tð Þ

€qq2 þ 2�s2!2 _qq2 þ �22q2 þ !22q2 ¼ Z12 2Fd1=U1ð Þ u1 tð Þ þ Z22 2Fd2=U2ð Þ u2 tð Þ

where, from Equation 11.88b,

�rr ¼XNj¼ 1

Zjr 2Fdj=Uj

� �Zjr

and hence

�11 ¼ Z11 2Fd1=U1ð ÞZ11 þ Z21 2Fd2=U2ð ÞZ21

�22 ¼ Z12 2Fd1=U1ð ÞZ12 þ Z22 2Fd2=U2ð ÞZ22

�11 ¼ 3:443 2� 13 866:0

30:000

� �3:443þ 10:109 2� 23 474:0

39:034

� �10:109

�� 10�6 ¼ 0:1339

�22 ¼ 6:521 2� 13 866:0

30:000

� �6:521þ 10:753

2� 23 474:0

39:034

� �10:753

�� 10�6 ¼ 0:1784


221

The damping force in the first mode is therefore

2�1!1 _qq1 þ �11 _qq1 ¼ 2 �1 þ�11

2!1

� �!1 _qq1 ¼ 2 0:01þ 0:1339

2� 25:133

� �!1 _qq1

and hence

�1 ¼ �s1 þ �a1 ¼ 0:01226:

The damping force in the second mode is

2�2!2 _qq2 þ �22 _qq2 ¼ 2 �2 þ�22

2!2

� �!2 _qq2 ¼ 2 0:01þ 0:1784

2� 119:098

� �!2 _qq2

and hence

�2 ¼ �s2 þ �a2 ¼ 0:01075:

The calculation of the principal coordinates q1 and q2 first requires the calculation of

the values of the power spectrum for wind velocities at heights 10 m and 20m for

!1¼ 25.133 rad/s and !2¼ 119.098 rad/s. The spectrum proposed by Kaimal given by

Equation 10.34 is

Su z; nð Þ ¼ 200u2� f

n 1þ 50fð Þ5=3

where

f ¼ zn

U zð Þ :

For H¼ 10m and !1¼ 25.133 rad/s,

f ¼ 10� 25:133=2�� 30:0 ¼ 1:3333471

Su 10; 25:133ð Þ ¼ 200� 5:2122 � 1:3333471

4:000 1þ 50� 1:3333471ð Þ5=3¼ 1:6117216 m2=s:

For H¼ 10m and !2¼ 119.098 rad/s,

f ¼ 10� 119:098=2�� 30:0 ¼ 6:3183451

Su 10; 119:098ð Þ ¼ 200� 5:2122 � 6:3183451

18:955 1þ 50� 6:3183451ð Þ5=3¼ 0:1229349 m2=s:

For H¼ 20m and !1¼ 25.133 rad/s,

f ¼ 20� 25:133=2�� 39:034 ¼ 2:0495164

Su 20; 25:133ð Þ ¼ 200� 5:2122 � 2:0495164

4:000 1þ 50� 2:0495164ð Þ5=3¼ 1:2205777 m2=s:


222

For H¼ 20m and !2¼ 119.098 rad/s,

f ¼ 20� 119:098=2�� 39:034 ¼ 9:7120641

Su 20; 119:098ð Þ ¼ 200� 5:2122 � 9:7120641

18:955 1þ 50� 9:7120641ð Þ5=3¼ 0:0924701 m2=s:

Having calculated the values of the velocity spectrum, the next step is to calculate the values

of the force spectra at stations H¼ 10m and H¼ 20m at frequencies !1¼ 25.133 rad/s and

!2¼ 119.098 rad/s. The expression for the force spectrum given by Equations 11.65 and

11.66, omitting the subscript q, is

Sfd¼ z; nð Þ ¼ 4

F2 zð ÞU2 zð Þ

A nð ÞSu z; nð Þ:

At H¼ 10m and !1¼ 25.133 rad/s,

Sf110; 25:133ð Þ ¼ 4� 13:8662 � 106

30:0002� 0:50� 1:6117216 ¼ 0:6886204� 106 N s:

At H¼ 10m and !2¼ 119.098 rad/s,

Sf110; 119:098ð Þ ¼ 4� 13:8662 � 106

30:0002� 0:25� 0:1229349 ¼ 0:0262624� 106 N s:

At H¼ 20m and !1¼ 25.133 rad/s,

Sf220; 25:133ð Þ ¼ 4� 23:4742 � 106

39:0342� 0:50� 1:2205777 ¼ 0:8828430� 106 N s:

At H¼ 20m and !2¼ 119.098 rad/s,

Sf220; 119:098ð Þ ¼ 4� 23:4742 � 106

39:0342� 0:25� 0:0924701 ¼ 0:0334419� 106 N s:

The square root of the coherence function is given by Equation 11.69. For the wind forces at

H¼ 10 m and H¼ 20 m, the function �(z1, z2, n) reduces to

’ z1; z2; nð Þ ¼ !iCz z2 � z1ð Þ� U z2ð Þ þU z1ð Þ½ � :

For !1¼ 25.133 rad/s,

’ 10; 20; 25:133ð Þ ¼ 25:133� 8� 20� 10ð Þ� 39:034þ 30:000½ � ¼ 9:2708896

e�’ 10;20;25:133ð Þ ¼ e�9:2708896 ¼ 9:41247� 10�5


223

and for !2¼ 119.098 rad/s,

’ 10; 20; 119:098ð Þ ¼ 119:098� 8� 20� 10ð Þ� 39:034þ 30:000½ � ¼ 43:932058

e�’ 10;20;119:098ð Þ ¼ e�43:932058 ¼ 8:32817� 10�20

We are now in a position to calculate the force spectra in the principal modes at frequencies

!1¼ 25.133 rad/s and !2¼ 63.369 rad/s.

For a 2-DOF system, the expression for the force spectrum in the first mode is given by

Equation 11.76a as

Sfqin1ð Þ ¼ Z2

11Sf1n1ð Þ þ 2Z11Z21

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1

n1ð Þ � Sf2n1ð Þ

� �q� e�’12 n1ð Þ þ Z2

21Sf2n1ð Þ

Sfq1ð25:133Þ ¼nð4:5152 � 0:6886204Þ

þh2� 4:515� 8:873

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið0:6886204� 0:8828430

p� 9:41247� 10�5

i

þ ð8:8732 � 0:8828430o� 10�3 � 10�3 � 106

¼ 83:549905 N s:

Similarly, the expression for the force spectrum in the second mode is given by Equation

11.76b as

Sfq2n2ð Þ ¼ Z2

12Sf1n2ð Þ þ 2Z12Z22

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1

n2ð Þ � Sf2n2ð Þ

� �q� e�’12 n2ð Þ þ Z2

22Sf2n2ð Þ

Sfq2ð119:098Þ ¼nð5:8382 � 0:0262624

�h2� 5:838� 12:453

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið0:0262624� 0:0334419Þ

p� 8:32817� 10�20

i

þ ð12:4532 � 0:0334419Þo� 10�3 � 10�3 � 106

¼ 6:081158 N s:

The expression for the response spectra for the principal coordinates q1 and q2 is given by

Equation 11.77:

Sqi!ið Þ ¼ 1

!2ið Þ2

Mi !ið ÞSfqi!ið Þ

Sq1 !1ð Þ ¼ 1

!41

M1 !1ð ÞSfq1!1ð Þ ¼ 1

25:1334� 1

0:012262� 83:549905 ¼ 348:27936� 10�3 m2 s

Sq2 !2ð Þ ¼ 1

!42

M2 !2ð ÞSfq2!2ð Þ ¼ 1

119:0984� 1

0:010752� 6:081158 ¼ 6:538700� 10�5 m2 s:


224

11.11. Simplified wind response analysis of linear multi-DOF structuresin the frequency domain

In Examples 11.2 and 10.3, it can be seen that if two stations are as much as 10 m apart, the values

of the cross-spectral density function for the wind forces at the two stations are negligible

compared to the direct spectral density functions. When the structures are heavy (as in the case

of the stepped mast), the aerodynamic damping is small compared to the structural damping.

It can also be noticed that in higher modes the reduced frequencies result in aerodynamic

admittance factors which, together with the fact that the energy of the wind at higher frequencies

is very much reduced, cause structures to respond mainly in the first mode. A simplified and less

time-consuming explorative response analysis can therefore be undertaken.

From Equation 11.31, for weakly damped structures

�2qi¼ð10Sqi

!ð Þ d! � 12 �i!iSqi

!ð Þ ð11:89Þ

For lightly damped structures, the variance of q1 is given by

�2qi ¼ð10Sqi

!ið Þd! ¼ Sqi!ið Þ�!

where

�! ¼ 12 �!i

and hence

�2q1 ¼ 1

2 � 348:27936� 10�3 � 0:01226� 25:133 ¼ 53:657761� 10�3 m2

�2q2 ¼ 1

2 � 6:53870� 10�5 � 0:01075� 119:098 ¼ 0:041858� 10�3 m2

�q1 ¼ 0:2316414

�q2 ¼ 0:0064697 m

and hence

q1 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi21n 4:0000� 3600ð Þ½ �

pnþ 0:577=

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi21n 4:0000� 3600ð Þ½ �

p o� 0:2316414 ¼ 1:0442 m

q2 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi21n 18:9550� 3600ð½ �

pnþ 0:577

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi21n 18:9550� 3600ð Þ½ �

p o� 0:0064697 ¼ 0:0313 m

xd1

xd2

" #¼

4:515 5:838

8:873 �12:543

�1:0442

0:0313

�� 10�3 ¼

4:897

8:873

�� 10�3 m:

Hence, the total displacements at H¼ 10m and H¼ 20m are

x1

x2

" #¼

xd1

xd2

" #þ

xd1

xd2

" #¼

8:613

29:915

�� 10�3 þ

4:897

8:873

�� 10�3 ¼

13:10

38:068

�� 10�3 m:


225


Sqi !ið Þ ¼ 1

~KK2i

M !ið ÞSfqi !ið Þ ð11:90Þ

and ~KKi ¼ !2i if the mode-shape vectors are normalised.

In cases where the load stations are so far apart that the cross-spectral density functions for the

wind forces can be ignored, Equation 11.50 may be written

Sfqi!ið Þ ¼

XNj¼ 1

Z2ji

2Fdj

Uj

� �2Aj !ið ÞSuj

!ið Þ ð11:91Þ

and hence

Sqi!ið Þ ¼ 1

!2ið Þ2

M !ið ÞXNj¼ 1

Z2ji

2Fdj

Uj


!ið Þ ð11:92Þ

�2qi¼ 1

!3i

1

2�i

XNj¼ 1

2Fdj

Uj


!ið Þ: ð11:93Þ

Example 11.3

Let the three-storey shear structure shown in Figure 11.4 represent a condensed numerical

model of a 30 m tall tower block, with width equal to depth equal to 10 m, with the floors

in the building lumped together in the numerical model as three floors 10 m apart. The

mass of each equivalent floor is 120 000 kg and the corresponding total shear stiffness of

the columns between each floor is 12.0� 106 h/m. Calculate the response to turbulent wind

having a mean velocity of 25 m/s at a height of 10 m above the ground, if the surface drag

coefficient for the area is 0.015. Assume the structural damping in the first, second and

third modes to be 1.5%, 1.0% and 1.0% of critical, respectively. The drag coefficient at

all levels of the building may be taken as Cd¼ 1.3. The density of air is 1.226 kg/m3.

Aerodynamic damping and the cross-correlation of wind may be ignored. Use the power

spectral density function proposed by Kaimal to take account of the variation of the

power spectrum of wind with height. The natural angular frequencies and normalised

mode-shape matrix for the model structures are

! ¼4:439

12:466

18:025

264

375 rad=s

!2 ¼19:70

155:40

324:90

264

375 rad2=s2

Z ¼0:947 2:128 1:703

1:706 0:950 �2:128

2:128 �1:703 0:953

264

375� 10�3:


226

Figure 11.4 Three-storey shear structure

10 m

10 m

10 m

10 m

Determination of mean wind speeds 20 m and 30 m above the ground

U zð Þ ¼ 2:5u� ln z=z0ð Þ

where

u� ¼ffiffiffiffiffiffiffikð Þ

pU 10ð Þ

and hence

u� ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:015ð Þ

p� 25:0 ¼ 3:062 m=s

z0 ¼ 10� exp �25:0=2:5� 3:062ð Þ ¼ 0:382 m:

We therefore have

U 10ð Þ ¼ U1 ¼ 25:000 m=s

U 20ð Þ ¼ U2 ¼ 2:5� 3:062 ln 20:0=0:382ð Þ ¼ 30:299 m=s

U 30ð Þ ¼ U3 ¼ 2:5� 3:062 ln 30:0=0:382ð Þ ¼ 33:403 m=s

F 10ð Þ ¼ F1 ¼ 12 � 1:226� 1:3� 10:0� 10:0� 25:0002 ¼ 49 806:250 N

F 20ð Þ ¼ F2 ¼ 12 � 1:226� 1:3� 10:0� 10:0� 30:2992 ¼ 73 157:763 N

F 30ð Þ ¼ F3 ¼ 12 � 1:226� 1:3� 10:0� 5:0� 33:4032 ¼ 44 457:474 N


227

Table 11.1 Data for Example 11.3

!i: rad/s H¼ 10m H¼ 20m H¼ 30m

!1¼ 4.439 8.10425 6.06436 5.02992

!2¼ 12.446 1.55880 1.13451 0.93013

!3¼ 18.025 0.85386 0.61838 0.50595

and

F1=U1ð Þ2¼ 49 806:250=25:000ð Þ2¼ 3:81128� 106 N2 s2=m2

F2=U2ð Þ2¼ 73 157:763=30:299ð Þ2¼ 5:82994� 106 N2 s2=m2

F3=U3ð Þ2¼ 44 457:474=33:403ð Þ2¼ 1:77141� 106 N2 s2=m2:

The Kaimal spectrum values in m2/s for angular frequencies !i at heights H(z) are given in

Table 11.1.

Determination of the reduced frequencies ~uuI and aerodynamic admittance factors A(!i)

corresponding to the natural angular frequencies !1, !2 and !3 (from the solid-line graph

in Figure 11.1) yields

~nn1 ¼4:439� 10:0

2�� 25:0¼ 0:2826 Hz

~nn2 ¼12:446� 10:0

2�� 25:0¼ 0:7923 Hz

~nn3 ¼18:025� 10:0

2�� 25:0¼ 1:1475 Hz

and hence

A !1ð Þ ¼ 0:6732

A !2ð Þ ¼ 0:3398

A !3ð Þ ¼ 0:2371:

The expression for the variance �2qi, which neglects the cross-spectral density function, is

given by Equation 11.91 and implies the transposition of Z and the evaluation of Z2ji,

which yield

~ZZT ¼Z2

11 Z221 Z2

31

Z212 Z2

22 Z232

Z213 Z2

23 Z233

264

375 ¼

0:8968 2:9104 4:5284

4:5284 0:9025 2:9002

2:9002 4:5284 0:9082

264

375� 10�6:

Equation 11.91 may be written in matrix form. For the structure concerned, the aerodynamic

admittance factors A(!i) are constants. The variances �2q2, �

2q2 and �2

q3 therefore may be


228

calculated as

�2q1 ¼

1

4:4393� 0:6732

2� 0:0150:8968 2:9104 4:5284� �

�

3:81128 0 0

0 5:82994 0

0 0 1:77141

2664

3775

8:10425

6:06436

5:02992

2664

3775

¼ 43:8555 m2

�2q2 ¼

1

12:4463� 0:3398

2� 0:014:5284 0:9025 2:9002� �

�

3:81128 0 0

0 5:82994 0

0 0 1:77141

2664

3775

1:55880

1:13451

0:93013

2664

3775

¼ 0:3318 m2

�2q3 ¼

1

18:0253� 0:2371

2� 0:012:9002 4:5284 0:9082� �

�

3:81128 0 0

0 5:82994 0

0 0 1:77141

2664

3775

0:85386

0:61838

0:50595

2664

3775

¼ 0:0538 m2:

Determination of the generalised coordinates qi¼�i�i

q1 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi21n 4:439� 3600ð Þ

2�

rþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

21n 4:439� 3600ð Þ=2�p �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi43:8555

p" #

¼ 27:190 m

q2 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi21n 12:446� 3600ð Þ

2�

rþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

21n 12:446� 3600ð Þ=2�p �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:3318

p" #

¼ 2:505 m

q3 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 18:025� 3600ð Þ

2�

rþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln 18:025� 3600ð Þ=2�p �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:0538

p" #

¼ 1:028 m:

Determination of maximum total displacements

x ¼ K�1Fþ Z��

Given the shear stiffness of the columns between floors, the stiffness matrix for the structure is

K ¼ 12� 1062 �1 0

�1 2 �1

0 �1 1

264

375N=m

and hence

K�1 ¼ 10�6

12�

1 1 1

1 2 2

1 2 3

264

375m=N


229

11.12. Concluding remarks on the frequency-domain methodThe frequency-domain method is convenient for predicting the dynamic response of structures. It

is limited to the analysis of linear structures, although in practice it is also applied to some non-

linear structures by taking only the non-linear response due to the mean wind speed component

into account. When the frequency-domain method is applied to determine the dynamic response

of non-linear structures such as cable roofs and guyed masts, whose stiffness and frequencies are

functions of the degree of deformation, the natural frequencies should be determined for the

deformed state due to the mean wind component and not for the case when there is no load on

the structure.

Apart from the assumptions with respect to the statistical characteristics of wind, the main

assumption made in order to make the method possible is that the amplitudes of the fluctuating

x1

x2

x3

264

375 ¼ 10�6

12�

1 1 1

1 2 2

1 2 3

264

375

49 806:250

73 157:763

44 457:474

264

375þ

0:947 2:128 1:703

1:706 0:950 �2:128

2:128 �1:703 0:953

264

375

27:190

2:505

1:028

264

375� 10�3:

We therefore have

x1

x2

x3

264

375 ¼

0:0140

0:0238

0:0275

264

375þ

0:0328

0:0466

0:0546

264

375 ¼

0:0468

0:0704

0:0821

264

375m:

Note that although all the three natural frequencies of the structure, i.e. f1¼ 0.7065 Hz,

f2¼ 1.9808 Hz and f3¼ 2.8688 Hz, lie within the part of the frequency spectrum in which

the wind is considered to have a considerable amount of energy, the structure responds

mainly in the first mode. It is therefore of interest to see to which extent the calculated

displacements alter if it is assumed that the structure responds only in the first mode. This

can easily be done by writing q2¼ q3¼ 0 in the transformation x¼Z, i.e.

x1

x2

x3

264

375 ¼ 10�6

12�

1 1 1

1 2 2

1 2 3

264

375

49 806:250

73 157:763

44 457:474

264

375þ

0:947 2:128 1:703

1:706 0:950 �2:128

2:128 �1:703 0:953

264

375

27:190

0

0

264

375� 10�3

and hence

x1

x2

x3

264

375 ¼

0:0140

0:0238

0:0275

264

375þ

0:0257

0:0464

0:0578

264

375 ¼

0:0397

0:0702

0:0854

264

375m:

As can be seen, the differences are marginal (except in the case of the displacements x1) and no

greater than those that can be caused by uncertainties in the assumed values of damping ratios

and the degree of accuracy of the spectral density function used. In many cases, especially for

buildings, it may therefore be sufficient (at least initially) to calculate the response in the first

mode only in order to see if a further, more rigorous, investigation is required.


230

component of the wind are sufficiently small compared to the mean wind speed; the terms

0:5�CdAu2 tð Þ, 0:5�Cd _xx

2 tð Þ and �Cdu tð Þ _xx tð Þ in Equation 11.5 can therefore be ignored. This

assumption is generally justified, but it may not be true for sites in mountainous areas where

fluctuations of the same order of magnitude as the mean wind speed have been observed.

Finally, inspection of Figure 11.3 and Table 10.3 highlight that the degree of accuracy to which

the dynamic response can be predicted by this method will vary with the type of spectral density

function used. For important structures, it may be advisable to construct spectral density

functions from recordings at the site concerned.

11.13. Vortex shedding of bluff bodiesSo far only the along-wind response caused by the natural turbulence in the flow approaching

the structure has been considered, and not the different types of response due to the change of

flow caused by the structure itself. Of these, the most important mechanism for wind-induced

oscillations is the formation of vortices in the wake flow behind certain types of structure such

as chimneys, towers, electrical transmission lines and suspended pipelines. Many failures due to

vortex shedding have been reported.

When bluff bodies are exposed to wind, vortices are shed from their sides creating a pattern in

their wake often referred to as the Karman vortex trail shown in Figure 11.5. The frequency of

the shedding depends on the shape of the body, the velocity of the flow and, to a lesser extent,

the surface roughness and the turbulence of the flow. The dominant frequency of vortex shedding

is given by

nv ¼SU

Dð11:94Þ

where S is a non-dimensional constant referred to as the Strouhal number, U is the mean wind

velocity and D is the width of the bluff body. The manner in which vortices are formed is a

function of the Reynolds number, which is given by

Re ¼ UL

ð11:95Þ

where U is the mean velocity of the flow, L is a representative dimension of the structural element

(which in the case of members with circular cross-sections is equal to the diameter D) and v is the

kinematic viscosity, which for air is equal to 1.5� 105 m2/s at 208C.

Figure 11.5 Regular periodic vortex shedding for flow past circular cylinder

30 < Re < 5000Karman vortex trail


231

The type of vortex shedding that is most important to civil engineers is when the shedding occurs

regularly and alternates from side to side. For bodies with rectangular or square cross-sections,

the Strouhal number is nearly independent of the Reynolds number. For a body with a circular

cross-section, the Strouhal number varies with the rate of flow and hence with the Reynolds

number. Three major regions are characterised by the Reynolds number: the subcritical region

for Re4 3� 105; the supercritical region for 3� 1054Re4 3� 106; and the transcritical

region for Re5 3� 106. Approximate values for the Strouhal number for circular and square

sections are given in Table 11.2.

Vortex shedding will give rise to lift or across-wind forces which, as a first approximation per unit

length, may be written as

PL tð Þ ¼ 12 �DU2CL tð Þ ð11:96Þ

where CL is a lift coefficient that fluctuates in a harmonic or random manner and depends on the

Reynolds number, the atmospheric turbulence and the surface roughness of the building. If the

vortex shedding frequency nv coincides with the natural frequency of a structure, such as a

chimney, quite large across-wind amplitudes of vibration will result unless sufficient damping is

present. Values for lift coefficients and Strouhal numbers for different types of sections are

given by ESDU (1978) and Simue & Scalan (1978).

If the vortex shedding is harmonic, Equation 11.96 may be written as

PL tð Þ ¼ P0 sin !vtð Þ ¼ 12 �DU2CL sin 2� nvð Þ: ð11:97Þ

From Equation 2.8, the equivalent modal mass of a prismatic member is given by

M ¼ðL0m xð Þ ’ xð Þ½ �2 dx ð11:98Þ

and from Equation 2.26, assuming a constant wind profile, the equivalent modal force due to the

fluctuating lift force given by Equation 11.97 is

P tð Þ ¼ PL sin 2� nvð Þ ¼ 12�DU2 sin 2�nvð Þ

ðL0CL xð Þ ’ xð Þ½ � dx: ð11:99Þ

Table 11.2 Data for prediction of vortex-induced oscillations in turbulent flow (data taken from

Davenport, 1961)

Cross-section Strouhal

number S

RMS lift

coefficient �L

Bandwidth

B

Correlation length

(diameters) L

Circular: region

Sub-critical 0.2 0.5 0.1 2.5

Super-critical Not marked 0.14 Not marked 1.0

Trans-critical 0.25 0.25 0.3 1.5

Square:

Wind normal to face 0.11 0.6 0.2 3.0


232

Since in Equation 4.14 xst¼P0K¼P0/M!2, the maximum response of a 1-DOF system subjected

to harmonic excitation may be written as:

xmax ¼PL

M!2� 1

2�; ð11:100Þ

it follows that, when the vortex shedding occurs with the same frequency as the natural frequency

of the structure,

xmax ¼12 �DU2

ðL0CL ’ xð Þ½ � dx

!2

ðL0m xð Þ ’ xð Þ½ �2 dx

� 1

2�ð11:101Þ

which can be simplified if it is assumed that the mass per unit length is constant (m(x)¼m) and

that the loss of span-wise correlation of the lift forces can be taken into account by assuming that

the lift coefficient CL(x) is proportional to the mode shape, i.e.

CL xð Þ ¼ CL ’ xð Þ½ �: ð11:102Þ

Substitution of the above expressions for m(x) and CL(x) into Equation 11.101 yields

xmax ¼12 �DU2CL

2�!2ð11:103Þ

and from Equation 11.94 we have that

!2s ¼

4�2S2U2

D2: ð11:104Þ

Substitution of the expression for !2 into Equation 11.103, remembering that !v¼! at resonance,

yields

xmax ¼�D3CL

16�2S2m�ð11:105Þ

for the maximum response. For the first mode of a cantilever structure, xmax occurs at the tip. In

higher modes, this amplitude occurs where the resonance takes place. For circular cylinders, a

design value for CL is aboutp(2�L); the maximum value for cylinders is 0.4. Approximate

values for �L are given in Table 11.2 and Equation 11.105 may be used as a first estimate of

likely response, yielding an upper bound solution.

Example 11.4

A 20m high industrial cable-stayed steel chimney has an external diameter of 1.0 m and a

natural frequency of 2.4 Hz. The mass is 150 kg/m. The Strouhal number for the circular

section of the chimney is S¼ 0.2 and the root mean square value of the lift coefficient

�1¼ 0.14. Calculate the wind velocity that will cause vortex shedding with a frequency

equal to the natural frequency of the chimney, the corresponding Reynolds number and,


233

Even when the vortex shedding appears to be regular, the lift force and hence CL(t) are random

rather than harmonic. From Equation 11.96, it follows that the spectral density function for the

lift force per unit length can be expressed as

SPL!ð Þ ¼ 1

2 �DU2Þ2 � SCL!ð Þ

�ð11:106Þ

where

ð10SCL

!ð Þd! ¼ 1

T

ðT0CL tð Þ � CL tð Þdt ¼ �2

L ð11:107Þ

is the variance of the lift coefficient CL(t) and SCL!ð Þ is the spectral density function of CL(t).

The spectral density function for the response of a 1-DOF system, assuming a correlation length

DLC, is therefore given by

Sx !ð Þ ¼ 1

K

� �2 1

2�D2LcU

2

� �2

�MF2 nð Þ � SCL!ð Þ: ð11:108Þ


U2 ¼ !2sD

2

4�2S2ð11:109Þ

and hence

Sx !ð Þ ¼ 1

K

� �2 !2v�D

4Lc

8�2S2

!2

�MF2 !ð Þ � SCL!ð Þ d! ð11:110Þ

finally, the maximum first mode amplitude of response at the tip. The specific density of air

�¼ 1.226 kg/m3 and the kinematic viscosity for air ¼ 1.5� 10�5 m2/s.

The velocity at which the frequency of vortex shedding is equal to the natural frequency of the

chimney is

U ¼ nvD

S¼ 2:4� 1:0

0:2¼ 12:0 m=s:

The Reynolds number for a flow of 12.0 m/s is

Re ¼ UD

¼ 12:0� 1:0

1:5� 10�5¼ 8:0� 105;

the Reynolds number is therefore just at the lower end of the supercritical range.

The maximum amplitude of response at the tip of the chimney is given by

xmax ¼�D3CL

16�2S2m�¼ 1:226� 1:03 �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ð Þ � 0:14

p

16� �2 � 0:22 � 150� 0:01¼ 0:0256 m:


234

which implies

�2x ¼

ð10Sx !ð Þ d! ¼ 1

K

� �2

� !2v�D

4Lc

8�2S2

!2

�MF2 !ð Þ �ð10SCL

!ð Þd!: ð11:111Þ

For weakly damped structures, Equation 11.111 may be written as

�2x ¼ Sx !ð Þ�! ¼ 1

K

� �2

� !2v�D

4Lc

8�2S2

!2

�MF2 !ð Þ � SCL!ð Þ�! ð11:112Þ

where

K ¼ !nM

MF !ð Þ ¼ 1=2�

�! ¼ 12 �!

!v � !n

�2x ¼ Sx !nð Þ�! ¼ 1

M

� �2

� �D4Lc

8�2S2

!2

� !n

8�

� � SCL

!nð Þ: ð11:113Þ

Approximate values for the correlation length L in diameters are given in Table 11.2. The

correlation length decreases with increasing turbulence intensity, increases with the ratio 2H/D

(where H is the height of the structure) and increases with the amplitude of the motion.

In the subcritical and transcritical range, the energy of the lift force acting on circular cylinders

is distributed closely on either side of the dominant shedding frequency and can be represented

by a Gaussian type distribution curve. Harris (1988) and Lawson (1990) give the spectral density

function for the lift coefficient for this type of distribution as

SCLnð Þ ¼ �2

L

nsBffiffiffiffiffiffiffiffiffiffiffi4�3ð Þ

p exp � 1� n=nvB

� �2" #

: ð11:114Þ

In the supercritical range, the spectral density function is broad and is given by Harris (1988)

as

SCLnð Þ ¼ 4:8�2

L

1þ 682:2 nD=Uð Þ2

1þ 227:4 nD=Uð Þ2� �2 �

D

Uð11:115Þ

or

SCLnð Þ ¼ 4:8�2

L

1þ 682:2 Sn=nvð Þ2

1þ 227:4 Sn=nvð Þ2� �2 �

S

nv: ð11:116Þ


235

Example 11.5

First use Equation 11.114 and then Equation 11.116 to calculate the maximum transverse tip

displacement of the 20 m high industrial steel chimney in Example 11.4 which has an external

diameterD¼ 1.0 m, a natural frequency nn¼ 2.4 Hz and a massm¼ 150 kg/m. The structural

damping �¼ 0.01. The Strouhal number for the circular section of the chimney S¼ 0.2, the

root mean square value of the lift coefficient �1¼ 0.14, the bandwidth B¼ 0.1, the correlation

length L¼ 2.5D and the specific density of air �¼ 1.226 kg/m3. Use the same values for S and

L when using Equations 11.114 and 11.116.


�2x ¼ 1

M

� �2

� �D4Lc

8�2S2

!2� !n

8�� SCL

!nð Þ


M ¼ 728=2835ð ÞmL ¼ 728=2835ð Þ � 150� 20 ¼ 770:37037 kg

and hence

�2x ¼ 1

770:37

� 2

� 1:226� 1:04 � 2:5

8�2 � 0:22

( )2

� 2�� 2:4

8� 0:01

� � SCL

!nð Þ

¼ 3:08235� 10�4 � SCL!nð Þ:

The expression for the spectral density function for lift coefficients given by Equation 11.114

yields

SCL!ð Þ ¼ �2

L

nsBffiffiffi�

p exp � 1� n=nsB

� �2" #

¼ 0:142

2� 0:1ffiffiffi�

p exp � 1� 2:4=2:4

0:1

� �2( )

¼ 0:0460754 m2 s=m

and hence

�2x ¼ 3:08235� 10�4 � 0:0460754 ¼ 14:20207� 10�6 m2

�x ¼ 3:76856� 10�3 m:

The maximum amplitude of lateral vibration due to vortex shedding, from Equation 11.114,

is therefore

xmax ¼ ��x

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2:4� 3600ð Þ½ �

pþ 0:577=

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2:4� 3600ð Þ½ �

pn o� 3:76856� 10�3

¼ 0:0166 m:


236

Comparison of the displacements calculated in Examples 11.4 and 11.5 indicates that the response

to vortex shedding in the supercritical region from Equation 11.116 is much less than in the

subcritical region by Equation 11.114, and both equations lead to a smaller displacement than

Equation 11.105.

11.14. The phenomenon of lock-inThe wind speed can be expressed in terms of a non-dimensional reduced velocity Ur as

U ¼ UrnnD ð11:117Þ

where Nn is the natural frequency of the structure and D is the width of the structure. Combina-

tion of Equations 11.117 and 11.94 by elimination of U yields

nv ¼ Snnð ÞUr: ð11:118Þ

As both S and nn are constants, it follows that the shedding frequency varies linearly with the

reduced velocity. Wind tunnel tests of flexible structural models have however shown that in a

region on either side of the reduced velocity, where this velocity is approximately equal to the

inverse of the Strouhal number (i.e. where Ur¼ 1/S), the shedding frequency remains constant

and is equal to the natural frequency of the structure. This phenomenon is referred to as a

lock-in, because the shedding frequency is locked into the natural frequency of the structure. In

steady flow the frequency of the structural vibration tends to be constant during a lock-in, with

the greatest amplitude occurring when nv¼ nn.

For a circular structure with Strouhal number s¼ 0.2, the extent of lock-in can be seen in

Figure 11.6 where the ratio nv/nn is plotted against the reduced velocity Ur.

The expression for the spectral density function for lift coefficients given by Equation 11.116

yields

SCLnð Þ ¼ 4:8� 0:142

1þ 682:2 0:2� 2:4=2:4ð Þ2

1þ 227:4 0:2� 2:4=2:4ð Þ2� �2 �

0:2

2:4

¼ 2:1758� 10�3 m2s=m

and hence

�2x ¼ 3:8235� 10�4 � 2:1758� 10�3 ¼ 6:70657 m2

�x ¼ 0:818936� 10�3 m:

The maximum amplitude of lateral vibration due to vortex shedding given by Equation

11.116 is

xmax ¼ ��x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2:4� 3600ð Þ½ �

pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln 2:4� 3600ð Þ½ �p

( )� 0:818936� 10�3

¼ 0:0036 m:


237

The lock-in phenomenon can also be observed in the behaviour of real structures. In turbulent

wind, however, the lock-in condition occurs only if the amplitudes of vibration are in excess of

approximately 2% of the width of the building. When this is the case, the motion will have

large amplitude and a regular frequency at a reduced mean wind velocity equal to 1/S. When

the amplitudes of across-wind oscillations are smaller, the magnitude of the amplitude varies

spasmodically with a lock-in occurring from time to time. The response of horizontal structures

such as the spans of pipeline bridges tends to correlate the vortex shedding along the span, and will

therefore cause the transverse amplitude displacement to be increased further. If either Equation

11.114 or Equation 11.116 yields a displacement greater than 2.0% of the diameter D, the

calculations need to be repeated with a larger correlation length (ESDU, 1978; Harris, 1988).

Example 11.6

Calculate the lengths of the 20 m high chimney in Examples 11.3 and 11.4 which are likely to

shed vortices if the lock-in of frequency shedding is assumed to last for wind velocities equal

to �20% of the wind velocity which first causes across-wind vibration at the tip of the

chimney. Assume roughness lengths of 0.2 m, 0.45 m and 0.9 m. Finally, assuming the

effective correlation length DL to be 1/3rd of the above lengths, calculate the maximum

tip displacements for each case using Equation 11.113 and 11.115. The specific density of

air �¼ 1.226 kg/m3.

The reduced velocity at which vortex shedding will occur is:

ur ¼ 1=S ¼ 1=0:2 ¼ 5:0:

From Equation 11.95, the corresponding wind velocity is

U ¼ UrnsD ¼ 5:0� 2:4� 1:0 ¼ 12:0 m=s

Figure 11.6 Variation of frequency ratio nvns with reduced wind velocity Ur, showing lock-in

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0

Reduced velocity Ur

n v/n

s

1.5

1.0

0.5


238

and a lock-in will therefore occur for wind velocities between

12:20� 0:8 ¼ 9:8 m=s and 12:0� 1:2 ¼ 14:4 m=s:

The maximum length of chimney shedding vortices will be found by assuming the maximum

shedding velocity at the top of the chimney to be 14.4 m/s and the minimum velocity further

down to be 9.6 m/s. In order to determine this distance it is first necessary to calculate the

shear velocity corresponding to the mean shedding velocity of 12.0 m/s.

When z0¼ 0.2 m,

u� ¼12:0

2 ln 20:0=0:2ð Þ ¼ 1:0423068 m=s:

The corresponding height at which the velocity is 9.6 m/s is

z ¼ z0 eU zð Þ=2:5u� ¼ 0:2 e9:6=2:5�1:0423068 ¼ 7:962 m:

Assuming the velocity fluctuations to be�20% of the initiating shedding velocity, the lock-in

lengths are therefore:

when z0 ¼ 0:20 mð Þ LL ¼ 20:0� 7 :962 ¼ 12:038 m

when z0 ¼ 0:45 mð Þ LL ¼ 20:0� 9:364 ¼ 10:636 m

when z0 ¼ 0:90 mð Þ LL ¼ 20:0� 9:364 ¼ 9:243 m:

The corresponding assumed correlation lengths are

when z0 ¼ 0:20 mð Þ LC ¼ 12:38=3� 1:0 ¼ 4:090 m

when z0 ¼ 0:45 mð Þ LC ¼ 10:636=3� 1:0 ¼ 3:545 m

when z0 ¼ 0:90 mð Þ LC ¼ 9:243=3� 1:0 ¼ 3:081 m :

When z0¼ 0.20 m,

�2x ¼ 8:2498974� 10�4 � 0:0460754 ¼ 38:01173� 10�6 m2

�x ¼ 6:16536� 10�3 m

xmax ¼ ��x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2:4� 3600ð Þ½ �


2 ln 2:4� 3600ð Þ½ �p

( )� 6:1653600� 10�3

¼ 0:0271 m:

When z0¼ 0.45 m,

�2x ¼ 6:1977551� 10�4 � 0:0460754 ¼ 28:556443� 10�6 m2


239

A comparison of the above displacements with that of 0.0256 m calculated in Example 11.5 using

Equation 11.105, which yields an upper bound solution, indicates that the assumed correlation is

not unreasonable.

11.15. Random excitation of tapered cylinders by vorticesTapered cylinders such as stacks also vibrate due to vortex shedding. However, less is known

about the mechanism of excitation. Experience seems to indicate that the lift forces are narrow

band random with a rather small correlation length, with the dominant frequency given by

Equation 11.94. As the diameter varies, local resonance between ns and the natural frequency

of the tapered cylinder takes place at different heights. As the wind speed increases, the resonance

first appears at the tip and then shifts downwards. The critical wind speed for each height occurs

when ns is equal to nn. An approximate method for calculating the mean standard deviation of

tapered cylinders is given by Harris (1988).

11.16. Suppression of vortex-induced vibrationVortex shedding can be prevented by

g destroying the span-wise correlation of the vorticesg bleeding air into the near-wake regiong preventing the interaction of the two shear layers.

One method used to prevent vortex shedding of chimneys where the level of structural damping is

insufficient is in the fitting of ‘stakes’. The most efficient stake known for destroying the span-wise

correlation of vortices is a three-start helix that makes one revolution in five diameters length of

chimney and extends over the top third of the height. The disadvantage of stakes is that they

increase the drag force.

Another method is the fitting of perforated shrouds. These prevent vortex shedding by bleeding

air into the near-wake region. Shrouds tend to be heavier than stakes, but increase the drag

flow less.

�x ¼ 5:3438229� 10�3 m



2 ln 2:4� 3600ð Þ½ �p

( )� 5:3438229� 10�3

¼ 0:0235 m:

When z0¼ 0.90 m,

�2x ¼ 4:6815033� 10�4 � 0:0460754 ¼ 21:570214� 10�6 m2

�x ¼ 4:6443744� 10�3 m



2 ln 2:4� 3600ð Þ½ �p

( )� 4:6443744� 10�3

¼ 0:0204 m:


240

A third method is the use of splitter plates. These are not generally a practical proposition because

they need to be aligned in the direction of the wind. As they need to extend four diameters

downwind, they tend to be heavy. However, they have the advantage that they do not increase

the drag forces to the same extent as stakes and perforated shrouds.

11.17. Dynamic response to the buffeting of wind usingtime-integration methods

If spatially correlated wind histories can be generated (e.g. by the method presented in Chapter

14), then the response of structures can be determined through step-by-step integration in the

time domain. In Chapter 6, three such integrated methods based on the Newmark -and

Wilson �-equations are presented. Experience indicates that schemes employing the Newmark

[¼ 1/4]-equations, i.e. assuming the accelerations to remain constant during the time step �t,

are the most efficient. For 1-DOF systems, the response to wind can be calculated using Equation

6.51. The response of multi-DOF systems can be calculated using Equation 6.68, given as

Kþ 2

�tCþ 4

�t2Mþ 4

�tFd V� _xxð Þ

�¼ 2Fd V� _xxð Þ �Vþ 2 _xxð Þ þ 2C _xxþM

4

�t_xxþ 2€xx

� �ð11:119Þ

where K, C and M are the stiffness, damping and mass matrices for a structure, �x is the change

in displacement vector x during a time step �t, _xx is a velocity vector and €xx is a acceleration

vector. Using the Newmark [¼ 1/4]-equations, from Equations 6.41–6.43 the ith elements in

the displacement, velocity and acceleration vectors at time (tþ�t) are

xi tþ�tð Þ ¼ xi tð Þ þ�xi ð11:120aÞ


�t�xi � _xxi tð Þ ð11:120bÞ

€xxi tþ�tð Þ ¼ 4

�t2�xi �

4

�t_xxi tð Þ � €xxi tð Þ: ð11:120cÞ

The size of the time step�t is important, as over-large as well as over-small time steps will lead to

inaccuracies in the calculated response. In the case of both wind and earthquakes, most of the

energy is contained within the part of the frequency spectrum that lies between 0 and 10Hz.

The period of the smallest frequency component that needs to be considered is therefore usually

approximately 0.1 s. Experience has shown that frequency components of that order of magnitude

can be sufficiently accurately modelled with time steps �t¼ 0.1/10¼ 0.01 s.

The forward integration process should be continued until the variance of response is constant.

Experience indicates that this will occur after approximately 120 s of real time. The maximum

response is found by multiplying the standard deviation of response by the peak factor �.

Problem 11.1

The tapering lattice tower shown in Figure 11.7 supports a circular disc 40 m above the

ground. The values of the lateral stiffness mass and damping coefficient of the equivalent

mass–spring system of the tower are 323.723 kN/m, 7200 kg and 1030.44 N s/m, respectively.

The disc weighs 1.0 t and has a diameter of 3.0 m and a drag coefficient of 1.3. Determine

the maximum displacement of the tower when the mean wind speed 10 m above the


241

ground, averaged from 10 min recording, is 40 m/s. Assume the surface drag coefficient for

the site to be equal to 0.006 and that the fluctuating component of the wind can be

represented by each of the Davenport, Harris and Kaimal power spectrums.

Figure 11.7 Lattice tower supporting 3.0m diameter disc

40 m

Problem 11.2

Use the Davenport spectrum to calculate the response of the structure in Example 11.3.

Include the effect of aerodynamic damping and comment on its effect on the calculated

dynamic response.

Problem 11.3

A cable-supported pipeline bridge has a span of 20 m. The mass and stiffness of an equivalent

mass–spring system are 4000 kg and 39 478.418 N/m, respectively. Make a preliminary esti-

mate of the maximum across-wind response by assuming the mode shape of vibration to

be similar to the deflected form of a built-in beam supporting a uniformly distributed load,

and by further assuming that the response to random alternate vortex shedding will give

rise to correlated vortex shedding along the span. The value of the Strouhal number for a

circular section is S¼ 0.2, the lift coefficient C1¼ 0.3 and the specific density of air

�¼ 1.226 kg/m3. Assume the first mode damping ratio to be 1% critical.


242

REFERENCES

Davenport AG (1961) The application of statistical concepts to the wind loading of structures.

Proceedings of Institution of Civil Engineers, 19 Aug, 449–472.

ESDU (1978) Across-wind response due to vortex shedding isolated cylindrical structures in wind

and gas flows. ESDU Data Item 75011, Oct. 1978.


Lawson TW (1990) Wind Effects on Buildings, vols 1 and 2. Applied Science, Barking.

Simue E and Scalan RH (1978) Wind Effects on Structures. Wiley, Chichester.

Vickery BJ (1965) Model for atmospheric turbulence for studies of wind on buildings.

Proceedings of 2nd Australasian Conference on Hydraulics and Fluid Mechanics, Auckland

University of Auckland.

FURTHER READING



243


ISBN: 978-0-7277-4176-9



Chapter 12

The nature and properties of earthquakes

12.1. IntroductionEarthquakes are normally experienced as a series of cyclic movements of the Earth’s surface and

are the result of the fracturing or faulting of the Earth’s crust. The source of the vibratory energy is

the release of accumulated strain energy resulting from sudden shear failures, which involve the

slipping of the boundaries of large rock masses tens or even hundreds of kilometres beneath

the Earth’s surface. On a global scale, these large rock masses are continental in size and comprise

the so-called tectonic plates into which the Earth’s crust is divided. The failure of the crust gives

rise to the propagation of two types of waves through the Earth: pressure or primary waves and

shear or secondary waves, referred to as P and S waves. The P waves travel faster than the S waves,

so that the waves arrive in alphabetical order. If the velocities of the two types of waves are

known, the distance from a focal point of observation can be calculated. Once P and S waves

reach the surface, a surface wave is generated. Figure 12.1 shows the principal geometrical

terms used to describe earthquakes and the travel paths of P and S waves.

12.2. Types and propagation of seismic wavesOnly the pressure and shear waves are propagated within the Earth’s body. The P waves, as

mentioned above, are the fastest of the two: their motion is the same as a sound wave that spreads

out and alternatively compresses and dilates the rock. Like sound waves, P waves can travel

through solid rock and water. The S waves, which travel more slowly than the P waves, shear

the rock sideways in a direction perpendicular to the direction of travel, and cannot propagate

through water. Surface waves, as their name implies, travel only on the surface of the Earth.

Seismic surface waves are divided into two types referred to as the Love wave and the Rayleigh

wave. The motion of Love waves is essentially the same as that of S waves with no vertical com-

ponents. They move from side to side on the Earth’s surface in a direction normal to the direction

of propagation. The Love waves are like rolling ocean waves, in which the disturbed material

moves both vertically and horizontally in a vertical plane in the along-direction of the quake.

The surface waves travel more slowly than the P and S body waves, and the Love waves generally

travel faster than the Rayleigh waves. The different forms of seismic waves described above are

depicted in Figure 12.2. When P and S waves are reflected or refracted at the interfaces between

rock types, some of the wave energy can be converted to waves of the other types. On land and in

strong earthquakes, after the first few shakes, two kinds of ground motion are usually felt

simultaneously.

12.3. Propagation velocity of seismic wavesThe wave velocity within an elastic homogeneous isotropic solid can be defined by two constants �

and �, where � is the modulus of incompressibility or bulk modulus and � is the modulus of

rigidity.

245

For granite,

� � 2� 1010 N=m2

� ¼ 1:6� 1010 N=m2:

For water,

� � 0:2� 1010 N=m2

� ¼ 0:

Figure 12.1 The principal terms used to describe earthquakes: (a) geometry and (b) transmission

Fault

Focal distance

Foca

l dep

th

Site EpicentreEpicentral distance

(a)

(b)Source

(magnitude)

Surface wave

Focus (source)

P and S waves P and S waves

Rock

Alluvium

Site(intensity)


246

Figure 12.2 Seismic waves

P wave

S wave

Love wave

Rayleigh wave

Compressions

Dilatations

Double amplitudeWavelength

Undisturbed medium


247

Within the body of an elastic solid with density �, the velocity of pressure and shear waves is given

by the following expressions. For P waves,

velocity � ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikþ 4

3�

� ��

� �s:

For granite, �¼ 5.5 km/s and for water, �¼ 1.5 km/s. For S waves,

velocity � ¼ffiffiffiffiffiffiffiffiffiffiffiffi�=�ð Þ

p:

For granite, �¼ 3.0 km/s and for water, �¼ 0.0 km/s.

The velocities of the Love and Rayleigh waves propagated along the surface of an elastic solid

body are given as follows. For Rayleigh waves,

velocity cr < 0:92 �

where � is the S wave velocity of the rock. For Love waves (layered solid),

velocity �1 < cL < �2

where �1 and �2 are S wave velocities of the surface and deeper layers, respectively.

12.4. Recording of earthquakesGround motion is measured by seismometers that can detect very small vibrations but go off

the scale in strong motion. Strong motion seismometers are usually set to operate only when

triggered by accelerations above a certain level. The results from ordinary seismometers are

used primarily in the study of earthquake mechanisms, while those from strong motion

seismometers are of importance in establishing design criteria and, when mounted on buildings,

the behaviour of structures during earthquakes. The basic design concepts of seismometers are

given in Chapter 12.

12.5. Magnitude and intensity of earthquakesAn earthquake disturbance is measured at its source by magnitude on the Richter scale, ranging

from 0 to 8.9 (which is the largest measured to date). The calculation of magnitude is based on

seismometer measurements and is a measure of the strain released at source. The Richter scale

is logarithmic, implying that a magnitude 5 event may be a minor one while a magnitude 6.5

event may be a major event with a release of energy at source 31.6 times that of an earthquake

of magnitude 5. The determination of magnitude is shown in Figure 12.3. To engineers, the

most important observation is that earthquakes of magnitude less than 5 are not likely to cause

any structural damage.

The effect of an earthquake diminishes with distance, so that the effect at a particular location

is not defined by the magnitude. This is measured in terms of intensity, commonly on the

modified Mercalli scale, although there are a number of other scales. The Mercalli scale is not

precise, being based on subjective factors such as the type of building damage and whether the

shock is felt by people in cars. The scale grades events from 1 (not felt) to 12 (damage nearly

total).


248

12.6. Influence of magnitude and surface geology on characteristicsof earthquakes

An earthquake may have a duration of up to a minute or so, with the interaction of various types

of waves depicted in Figure 12.3 together with the effect of refraction at discontinuities producing

extremely complex wave forms. Seismologists and engineers have developed formulae relating all

the principal parameters of earthquake transmission such as duration, dominant period and

attenuation. Earthquake waves, however, are affected by both soil conditions and topography;

practising engineers should bear in mind that real-life results show a considerable amount of

scatter on each side of these median rules. An extensive treatment of seismic risk is given by

Figure 12.3 Procedure for determination of magnitude on the Richter scale from seismometer

recording: first, the distance to the focus is measured using the time interval between the S and P waves

(S� P¼ 24 s); the height of the maximum wave is then measured on the seismogram (23mm); a

straight edge is placed between the appropriate points on the distance and amplitude scales to measure

magnitude ML¼ 5.0

50

40

30

20

4

2

1086

100

50

20

10

5

2

1

0.5

0.2

0.1

S – P = 24 s

Amplitude:mmMagnitude

Distance:km

S – P:s

Amplitude = 23 mm

1020

30

500

400

300

200

100

60

40

20

5

0

6

5

4

3

2

1

0

P S


249

Lomnitz and Rosenbleuth (1976). However, the following non-quantitative rules are worth

remembering

g the predominant period increases with increasing magnitude, distance and depth of

alluvium (Figures 12.4 and 12.5)g the peak acceleration increases with increasing magnitude and soil stiffness and decreases

with increasing distance (Figures 12.6 and 12.7)g the duration increases with increasing magnitude.

The dominant frequency of the ground varies between both different sites and regions and is a

function of the magnitude of the earthquake, the distance from the causative fault as shown in

Figure 12.4 and the depth of alluvium as shown in Figure 12.5. Soft surface material will

behave similarly to jelly on a shaking table and can demonstrate considerable amplification of

the base rock motion.

An amplification factor of 20, as well as considerable modification of the predominant period, has

been predicted for the San Francisco Bay mud. The dominant frequencies in California range

from 3.3 to 4.0 Hz. As in the Caribbean, the dominant frequencies are lower and range from

2.5 to 2.8 Hz. Earthquakes with much lower frequencies have however been recorded. The

dominant frequency during the San Salvador earthquake in 1986 was 1.48 Hz, and dominant

frequency of the Mexico City earthquake in 1985 was as low as 0.41 Hz. This wide variation in

Figure 12.4 Predominant period–distance relationship for maximum acceleration in rock (Seed, 1968)

0 80 160 240 320

0 50 100 150 200

7.5

6.5

5.5

6.0

7.0

M = 8

km

miles

Distance from causative fault

Pred

omin

ant

perio

d: s

1.2

0.8

0.4

0


250

Figure 12.5 Relationship between the natural period of the soil and alluvium depth (Seed and Idriss,

1970)

0 50 100 150 200 250 300

Fund

amen

tal p

erio

d of

dep

osit:

s

Depth of soil: m

2.0

1.5

1.0

0.5

0

Figure 12.6 Acceleration–magnitude–distance relationship (Seed et al., 1976)

0.50

0.40

0.30

0.20

0.10

00 40 80 120 160

0 25 50 75 100

7.5

6.5

7

M = 8

km

Max

imum

acc

eler

atio

n: g

miles

Distance from causative fault


251

the dominant frequency of the ground should be borne in mind by designers and writers of codes

of practice who, for economical reasons, attempt to simplify design procedures.

Although very weak soils can produce substantial amplification of the base rock vibration for

earthquakes of low intensity, in major shaking the effect is limited by shear failures in the soil.

This produces an effective cut-off point in the transmission of large shocks. Weak soils have a

bad reputation in earthquakes but this is due to consolidation, liquefaction and other effects

producing large displacements. The estimation of the effect of site geology on ground motion is

complex and the literature is extensive.

In the same way that a building may be regarded as a dynamic system shaken at its base, the

surface alluvium (extending from a few metres to hundreds of metres in depth) may be considered

as a dynamic system shaken by the motion of the underlying rock. This argument can obviously

be extended to a combination of two dynamic systems. This is desirable in the case of most

structures and necessary in the case of large rigid structures such as nuclear reactors. In the

case of medium relatively flexible buildings, however, the dynamics of the soil and the building

are usually considered separately. In the case of piled foundations, special considerations are

necessary as the piles modify the surface response.

Although soils can suffer damage such as consolidation, liquefaction, landslides, avalanches and

shear failures by earthquake motion, it is assumed in this book that the structures are sited on

ground that will substantially retain its integrity during an earthquake.

12.7. Representation of ground motionAt any point on the Earth’s surface, earthquake motion comprises three translational compo-

nents: two rocking components and one horizontal torsional component. Earthquakes are

Figure 12.7 Effect of local soil conditions on peak acceleration (Seed et al., 1976): the relationships

shown are based on a ground acceleration of 0.3 g and are extrapolated from a database

0.6

0.5

0.4

0.3

0.2

0.1

00 0.1 0.2 0.3 0.4 0.5 0.6

Max

imum

acc

eler

atio

n: g

Maximum acceleration in rock: g

Rock

Stiff soil conditions

Deep cohesionless soils

Soft to medium stiff clayand sand


252

commonly classified by their intensity and peak acceleration, although these are only an

approximate measure of their capacity for causing damage. Other important factors are the

frequency content, duration, peak velocity and peak displacement. Of these, the frequency content

relative to the natural frequencies of the structures is generally the most significant. Earthquakes

with the main energy concentration in frequency bands corresponding to dominant structural

frequencies can cause more damage than earthquakes with greater peak accelerations but with

energy concentrated in different frequency bands from the structural frequency bands.

Information about ground motion can be presented in the time domain in the form of accelera-

tion, velocity and displacement histories (Figure 12.8) and in the frequency domain in the form of

response or power spectra. Response spectra are commonly used in design and form a convenient

method of establishing suitable specifications for linear structures. Their values at any given

frequency represent the peak response of a single-DOF oscillator to a specific earthquake

record. In order to predict the response of non-linear structures, time histories are needed.

Strong motion histories, if not available, can be constructed from spectral density functions or

Figure 12.8 Strong motion earthquake records (from Earthquake Engineering Research Laboratory,

1980)

0 5 10 15 20 25 30 35 40Time: s

40

20

0

–20

–40

40

20

0

–20

–40

40

20

0

–20

–40

Acc

eler

atio

n: g

/100

Velo

city

: cm

/sD

ispl

acem

ent:

cm


253

auto-covariance functions for ground accelerations and require information on the variation of

acceleration with time. Observation of earthquake histograms and use of their statistical

properties as data for generation of earthquake histories can be beneficial. Many histograms of

recorded earthquakes are available in digital forms for this purpose.

Methods for generating earthquake histories and families of correlated earthquakes with similar

properties are presented in Chapter 14, together with methods for generating spatially correlated

wind histories.

REFERENCES

Earthquake Engineering Research Laboratory (1980) Earthquake strong motion records. EERL,

Pasadena, Report No. 80–01, 1980.

Lomnitz C and Rosenbleuth E (1976) Seismic Risk and Engineering Decisions. Elsevier,

Amsterdam.

Seed HB (1968) Characteristics of rock motion during earthquakes. University of California at

Berkeley, Earthquake Engineering Research Center, Report EERC 63–5.

Seed HB and Idriss IM (1970) Solid moduli and damping factors for dynamic response analysis.

University of California at Berkeley, Earthquake Engineering Research Center, Report

EERC 70–10.

Seed HB, Murarka R, Lysmer J and Idriss IM (1976) Relationships of maximum acceleration,

maximum velocity, distance from source, and local site conditions for moderately strong

earthquakes. Bulletin of Seismological Society of America 66(4), 1323–1342.

FURTHER READING

Bolt BA (1978) Earthquakes: A Primer. W. H. Freeman, San Francisco.

Eiby GA (1980) Earthquakes. Heinemann, London.


Seed HB and Idriss I (1968) Seismic response of horizontal layers. Journal of the Soil

Mechanics and Foundations Division 94(SM4), 1003–1031.

Seed HB and Idriss IM (1982) Ground motion and ground liquefaction during earthquakes.

Earthquake Engineering Research Institute, Berkeley.


254


ISBN: 978-0-7277-4176-9



Chapter 13

Dynamic response to earthquakes:frequency-domain analysis

13.1. IntroductionThe most common form of data bank used in the design of structures to resist earthquakes is

response spectra. As mentioned in Chapter 12, a response spectrum is a curve that shows how

the maximum response, velocity or acceleration of oscillators with the same damping ratio but

with different natural frequencies respond to a specified earthquake. Another approach which

is gradually gaining ground is the use of power spectral density functions for the ground accelera-

tion caused by earthquakes. Their construction and use are similar to those of wind engineering.

Both the above methods are of interest to the practising engineer because they are eminently very

useful, as demonstrated below, in connection with the mode superposition method introduced in

Chapter 8.

There are also time-domain methods, which are generally only used in the case of non-linear

structures. One such method is described in Chapter 6 and is briefly, for convenience and

completeness, repeated in this chapter. The problem with time-domain methods is that it is not

sufficient to calculate the response to only one earthquake, as no two earthquakes at the same

site are likely to be identical. It is therefore necessary to generate a family of earthquakes with

properties appropriate for a given area. How many such earthquakes need to be included in

any design calculation is determined by experience and from recommendations in design codes.

13.2. Construction of response spectraThe linear acceleration method, Wilson �-method and the Newton �-method given in Chapter 6,

as well as Duhamel’s integral (Clough and Penzien, 1975; Key, 1988), may be used to calculate the

maximum displacement, velocity and acceleration of a single oscillator for a given earthquake

record such as that shown in Figures 1.3 or 12.8.

From Equation 4.52, the equation of motion for a 1-DOF system (relative to the support) when

subjected to a ground acceleration €xxgðtÞ can be written as

M€xxþ C _xxþ Kx ¼ M€xxgðtÞ ð13:1Þ

or

€xxþ 2�!n _xxþ !2nx ¼ €xxgðtÞ: ð13:2Þ

By calculating the maximum response of oscillators with different frequencies but with the same

damping, it is therefore possible to construct a response spectrum in the frequency domain for

oscillators with the same damping ratio. By repeating this process for oscillators with different

255

damping ratios, it is possible to construct a number of response spectra for the same record. An

example of a response spectrum for the record is shown in Figure 13.1.

13.3. Tripartite response spectraConsider the single-DOF mass–spring system shown in Figure 13.2 when subjected to a support

displacement yg and a corresponding support velocity _yyg.

Figure 13.1 Displacement response spectra for elastic 1-DOF oscillator subjected to the ground motion

of the 1940 El Centro earthquake (from Blum et al., 1961, reproduced with permission of the Portland

Cement Association)

0.5 0.1 0.2 0.5 1.0 2.0 5.0 10.0 20.0 50.0Natural frequency: Hz

Max

imum

dis

plac

emen

t (r

elat

ive

to b

ase:

inch

es)

50.0

20.0

10.0

5.0

2.0

1.0

0.5

0.2

0.1

0.05

0.02

Figure 13.2 Single-DOF oscillator subjected to support motion

k

c

m (y – yg)

( y – yg)

yg

my


256

The equation of motion for this form of excitation is

M€yyþ C _yy� _yyg� �

þ K y� yg� �

¼ 0: ð13:3Þ

If the relative displacement and relative velocity are denoted u and _uu respectively, then Equation

13.3 may be written as

M€yyþ C _uuþ ku ¼ 0 ð13:4Þ

and, if the damping is neglected,

M€yyþ Ku ¼ 0 ð13:5Þ

or

€yyþ !2nu ¼ 0: ð13:6Þ

From Equation 13.6, the absolute acceleration is proportional to the relative displacement. The

maximum absolute acceleration €yymax is therefore proportional to the maximum relative

displacement umax, i.e.

€yymax ¼ !2numax: ð13:7Þ

If damping is taken into account, and it is assumed that the relative velocity _uu ¼ 0 when the

relative displacement is a maximum and equal to umax, Equation 13.7 is again obtained. This

expression for the maximum acceleration is, purely by coincidence, the same as for simple

harmonic motion (SHM). The fictitious velocity associated with an apparent SHM is referred

to as a pseudo-velocity. The maximum value of the pseudo-velocity is _uumax. Thus

_uumax ¼ !numax ¼ 2�fnumax ð13:8Þ_uumax ¼ €yymax=!n ¼ €yymax=2�fn: ð13:9Þ

Taking the logarithm of both sides of Equations 13.8 and 13.9 yields

log _uumax ¼ log fn þ log 2�umaxð Þ ð13:10Þlog _uumax ¼ �log fn þ log €yymax=2�ð Þ: ð13:11Þ

For a constant value of umax, Equation 13.10 is a straight-line plot of log _uumax against log fn with

a slope of 458; for a constant value of €yymax, Equation 13.11 represents a straight-line plot of

log _uumax against log fn with a slope of 1358. It is therefore possible to plot the maximum spectral

response umax, spectral acceleration €yymax and spectral pseudo-velocity _uumax on the same graph, as

shown in Figure 13.3. The graph shows the maximum predicted responses to the El Centro

earthquake of oscillators with four levels of damping and with increasing natural frequencies.

The spectra shown in Figure 13.3 are raw, and it is usual to smooth them for design purposes

as it is highly unlikely that the duration, peak acceleration, frequency content and energy

distribution of future earthquakes in the same area will be the same as those of previously

recorded earthquakes.

In design, it is usual to employ consolidated response spectra normalised to a peak acceleration of

1.0g with corresponding maximum values for ground displacement and velocity (Harris, 1988).

Dynamic response to earthquakes: frequency-domain analysis

257

One such set of spectra is shown in Figure 13.4, where the maximum ground displacement is 36

inches and the maximum pseudo-ground velocity is 48 inches/s. These values are consistent with a

motion that is more intense than those generally considered in earthquake engineering. They are,

however, of proportional magnitudes deemed satisfactory for the design of most linear elastic

structures.

Figure 13.4 shows an additional six curves to the curve assumed for the ground motion. These are

the corresponding response curves for single oscillators with damping ratios ranging from 0% to

10% of critical. For assumed peak ground accelerations different from 1.0g, the values obtained

from the graph need to be scaled linearly.

13.4. Use of response spectraValues for displacements, velocities and accelerations are obtained from Figure 13.4 by taking

the antilogarithm of the ratio of the coordinates of a variable (measured in millimetres,

centimetres or inches) and the appropriate scaling factor. The coordinate magnitude for each

variable is measured with the value 1.0 as origin. The position on the graph of a given frequency

is found by taking the logarithm of the frequency and multiplying it by the scaling factor for

frequencies.

The scaling factors for the variables in Figure 13.4 (with the coordinates in centimetres) are

determined as follows:

Figure 13.3 Response spectra for 1-DOF oscillators for the 1940 El Centro earthquake (from Blum et al.,

1961, reproduced with permission of the Portland Cement Association)

Max

imum

grou

nd ac

celer

ation

0.05 0.1 0.5 1.0 5.0 10.0

Period: s

S x: c

m/s

250.0

100.0

10.0

50.0

5.0

25.0

2.5

Maximum ground velocity

0.75

0.1 g

0.01

g

1 g

5 g

10g

20g

10

100Sx

S x

0.250.125

0.50

ξ = 0.2

ξ = 0.02

ξ = 0.0

ξ = 0.1

Maximum ground

displacement


258

frequencies: log 100� log 1¼ 2¼ 6.20 cm, scaling factor Sf¼ 3.15

displacements: log 100� log 1¼ 2¼ 4.40 cm, scaling factor Sd¼ 2.20

velocities: log 100� log 1¼ 2¼ 6.10 cm, scaling factor Sv¼ 3.05

accelerations: log 100� log 1¼ 2¼ 4.50 cm, scaling factor Sv¼ 2.25.

Figure 13.4 Basic design spectra normalised to 1.0g (based on Figures 5 and 11, Newmark and Hall,

1982)

Spectra for damping0%0.5%1%2%5%

10%

0.1 0.2 0.5 1 2 5 10 20 50 100

Ground motion maxima

Displacement

Accele

ratio

n: g

0.02

0.05

0.1

200

100

20

10

0.5

0.2

2

1

5

0.005

0.0050.01

0.1

0.2

0.5

10050

20

10

52

1

0.02

Velo

city

: inc

hes/

s

Frequency: Hz

500

200

100

50

20

10

5

2

1

Example 13.1

The top of a tall building, which has a first natural frequency of 1.0 Hz and a first modal

damping ratio of 1.0% of critical, is modelled as a mass–spring oscillator. Use the

appropriate response spectrum in Figure 13.4 to predict the maximum lateral displacement,

pseudo-velocity and acceleration of the roof of the structure that will be caused by an

earthquake having an assumed peak acceleration of 0.3g.

xmax ¼ 0:3� 103:05=2:20 ¼ 7:30 inches

_xxmax ¼ 0:3� 106:73=3:05 ¼ 48:27 inches=s

€xxmax ¼ 0:3� 100:86=2:25 ¼ 0:723g:


259

13.5. Response of multi-DOF systems to earthquakesLet a two-storey shear building subjected to a ground motion xg(t)¼xg be represented by the

mass–spring system shown in Figure 13.5, where the displacements y1 and y2 of the two masses

m1 and m2 are relative to a fixed point.


M1€yy1 þ C1 _yy1 � _xxg� �

þ C2 _yy1 � _yy2ð Þ þ K1 y1 � xg� �

þ K2 y1 � y2ð Þ ¼ 0 ð13:12aÞ

M2€yy2 � C2 _yy1 � _yy2ð Þ � K2 y1 � y2ð Þ ¼ 0: ð13:12bÞ

Now let

x1 ¼ y1 � xg

_xx1 ¼ _yy1 � _xxg

€xx1 ¼ €yy1 � �€xxgðtÞ

x2 ¼ y2 � xg

_xx2 ¼ _yy2 � _xxg

€xx2 ¼ €yy2 � �€xxgðtÞ

where €xxgðtÞ is the acceleration history of an earthquake normalised to a peak acceleration of

1.0g and � is a constant that defines the magnitude of the peak acceleration of the real quake.

Substitution for y, _yy and €yy into Equations 13.12a and 13.12b yields

M1€xx1 þ C1 þ C2ð Þ _xx1 � C2 _xx2 þ K1 þ K2ð Þx1 � K2x2 ¼ M1�€xxgðtÞ ð13:13aÞ

M2€xx2 � C2 _xx1 þ C2 _xx2 � K2x1 þ K2x2 ¼ M2�€xxgðtÞ: ð13:13bÞ

Equations 13.13a and 13.13b may be written in matrix form as

M1 0

0 M2

" #€xx1

€xx2

" #þ

C1 þ C2 �C2

�C2 C2

" #_xx1

_xx2

" #þ

K1 þ K2 �K2

�K2 K2

" #x1

x2

" #

¼M1 0

0 M2

" #�€xxgðtÞ

�€xxgðtÞ

" #ð13:14Þ

Figure 13.5 Mass–spring model of two-storey shear structure

c1

k1

c2

k2

m1 m2

y1

xgy2


260

or as

M€xxþ C _xxþ Kx ¼ M�€xxgðtÞ: ð13:15Þ

Equation 13.15 is obviously also the general form for the equation of motion for any linear

multi-DOF structure subjected to a support motion xg(t)¼xg with acceleration �€xxgðtÞ.

In order to solve the system of equations given by Equation 13.15 and to predict the response of an

N-DOF system to a given support motion let (as in Chapters 8 and 11)

x ¼ Zq

_xx ¼ Z _qq

€xx ¼ Z€qq

where

Z ¼ Z1;Z2; . . . ;Zi; . . . ;ZN½ �

is the normalised mode-shape matrix associated with Equation 13.15. Substitution of the above

expressions for x, _xx and €xx into Equation 13.15 and post-multiplication of each term in the

equation by ZT yields

ZTMZ€qqþ Z

TCZ _qqþ Z

TKZq ¼ Z

TM�€xxgðtÞ: ð13:16Þ

From the orthogonality properties of normalised eigenvectors considered in Chapter 7,

ZTMZ ¼ I

ZTCZ ¼ 2�!

ZTKZ ¼ !2:

Substitution of these expressions for the matrix products into Equation 13.16 will uncouple the

equations of motion and yield

€qqþ 2�! _qqþ !2q ¼ ZTM�€xxgðtÞ ð13:17Þ

where 2�! and !2 are diagonal matrices. Equation 13.17 may also be written as

€qq1 þ 2�1!1 _qq1 þ !21q1 ¼ Z

T1M�€xxgðtÞ

€qq2 þ 2�2!2 _qq2 þ !22q2 ¼ Z

T2M�€xxgðtÞ

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

€qqi þ 2�i!i _qqi þ !2i qi ¼ Z

Ti M�€xxgðtÞ

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

€qqN þ 2�N!N _qqN þ !2NqN ¼ Z

TNM�€xxgðtÞ

ð13:18Þ


261

where the product ZTi M is referred to as the ith participation vector. Since the mass in each of the

above equivalent 1-DOF systems is unity, it follows that the equivalent ground acceleration in the

generalised coordinate system €qqgi ðtÞ is equal to the ground acceleration vector�€xxg post-multiplied

by the participation vector ZTi M. The maximum value of €qqgi ðtÞ occurs when €xxgðtÞ is equal to 1.0g,

and we therefore have

€qqgi ðtÞ ¼ ZTi M�€xxgðtÞ ð13:19Þ

€qqgi ;max ¼ ZTi M�g: ð13:20Þ

13.6. Deterministic response analysis using response spectraThe ith generalised modal equation (Equation 13.18) can be considered as the equation of

motion of a 1-DOF oscillator with unit mass subjected to a maximum ground acceleration

€qqgi ;max ¼ ZTi M�g. The maximum response of this system can therefore be found by use of a

response spectrum based on a damping ratio with value �i from which the response ~qqi;max

corresponding to the frequency !i can be found. As the spectra in Figure 13.4 are normalised

to a peak acceleration of 1.0g, it follows that

qi;max ¼ ZTi M�~qqi;max: ð13:21Þ

We therefore have

qmax ¼ ZTi M�~qqmax ð13:22Þ

and hence

xmax ¼ Zqmax ¼ ZZTM�~qqmax: ð13:23Þ

Equation 13.23 assumes that the maximum responses in each of the modes will occur simulta-

neously and relative to each other as in the mode-shape matrix. As this is highly unlikely, the

above expression for the maximum response vector is modified for design purposes. Each element

in the response vector xmax is recalculated as the square root of the sum of the squares of the

contribution from each mode:

xr ¼ Zr1q1ð Þ2þ Zr2q2ð Þ2þ . . .þ ZrNqNð Þ2� �1=2 ð13:24Þ

or

xr ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiXNi¼ 1

Zriqið Þ2vuut : ð13:25Þ

Example 13.2

Use the response spectra in Figure 13.3 to calculate the maximum and maximum modified

displacements and accelerations of the floors in the three-storey shear structure shown in

Figure 2.14 if the building is subjected to an earthquake with a peak acceleration equal to

0.25g. Assume the damping in the first mode to be 2.0% of critical and that in the second

and third mode to be 1.0% of critical. The stiffness matrix, mass matrix, natural frequencies


262

and normalised mode-shape matrix for the structure are as follows

K ¼114 596:3 �49 112:7 0

�49 112:7 81 854:5 �32 741:8

0 �32 741:8 32 741:8

264

375 kN=m

M ¼61:16208 0 0

0 40:77472 0

0 0 20:38736

264

375� 103 kg

! ¼18:659513

41:961902

58:121498

264

375 rad=s

!2 ¼348:1774

1760:8012

3378:1085

264

375 rad2=s2

f ¼2:9697537

6:6784441

9:6869163

264

375Hz

Z ¼1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

264

375� 10�3:

From Equation 13.20, the peak acceleration vector in the generalised coordinate system is

given by

ZTM�g ¼

1:7454 3:3157 4:2336

3:0818 0:4330 �4:4926

1:9498 �3:6531 3:3105

264

375

61:16208 0 0

0 40:77472 0

0 0 20:38736

264

375

0:25g

0:25g

0:25g

264

375

and hence

ZTM�g ¼

106:7522 135:1967 86:3119

188:4893 17:6554 �91:5923

119:2538 �148:9541 67:4924

264

375

0:25g

0:25g

0:25g

264

375 ¼

82:0652g

28:6381g

9:4480g

264

375:

The decoupled equations of motion or generalised modal equations at the time when the peak

acceleration occurs may therefore be written as

€qq1 þ 2� 0:02� 18:659513 _qq1 þ 348:1774q1 ¼ 82:0652g

€qq2 þ 2� 0:01� 41:961902 _qq2 þ 1760:8012q2 ¼ 26:6381g

€qq3 þ 2� 0:01� 58:121498 _qq3 þ 3378:1085q3 ¼ 9:4480g:

From the response spectra in Figure 13.4 for oscillators with 1.0% and 2.0%, remembering

that 1.0 inches¼ 0.0254 m, the following values are calculated for the generalised


263

displacement coordinates:

q1;max ¼ 82:0652� 101:80=2:75 � 0:0254 ¼ 9:4089m

q2;max ¼ 28:6381� 100:09=2:75 � 0:0254 ¼ 0:7843m

q3;max ¼ 9:4480� 10�1:15=2:75 � 0:0254 ¼ 0:0916m

and therefore

x1

x2

x3

264

375 ¼

1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

264

375

9:4089

0:7843

0:0916

264

375� 10�3 ¼

0:0190

0:0312

0:0366

264

375m:

The corresponding modified displacement vector obtained applying Equation 13.25, used in

structural design, is

x1

x2

x3

264

375 ¼

0:0166

0:0312

0:0400

264

375m:

Comparison of the elements in the two displacement vectors reveals that the relative

displacement between the ground and first floor is greatest in the first vector, while the relative

displacements between the first and second floor and the second and third floor are greater in

the second vector.

The maximum acceleration of each floor is found by using the response spectra in Figure 13.4

once more. It should be noted that the acceleration is given in terms of acceleration due to

gravity g, and not in inches/s2. We therefore have

€qq1;max ¼ 82:0652g� 101:70=2:75 ¼ 340:675g

€qq2;max ¼ 28:6381g� 101:80=2:75 ¼ 129:268g

€qq3;max ¼ 9:4480g� 101:45=2:75 ¼ 31:814g:

This yields the following acceleration vector for the floors:

€xx1

€xx2

€xx3

264

375 ¼

1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

264

375

340:675

129:268

31:814

264

375� 9:81� 10�3 ¼

10:350

10:490

9:485

264

375m=s2:

The maximummodified acceleration vector, obtained by taking the square root of the sum of

the squares of the contribution from each mode, yields

€xx1

€xx2

€xx3

264

375 ¼

7:048

11:153

15:288

264

375m=s2:

In the case of accelerations, the modified solution therefore leads to a much lower

acceleration at the first floor level and a much greater acceleration at the top level.


264

13.7. Dynamic response to earthquakes using time-domainintegration methods

If earthquake histories are available or can be generated, the maximum response of a structure

can be determined through step-by-step integration in the time domain. In Chapter 6, three

such integration methods based on the Newmark �- and Wilson �-equations are presented. Of

these, experience indicates that schemes employing the Newmark [�¼ 1/4]-equations are the

most efficient. For 1-DOF systems, the response to earthquakes can be calculated using Equation

6.54. The response of multi-DOF systems can be calculated using Equation 6.69, given as

Kþ 2

�tCþ 4

�t2M

� ��x ¼ 2C _xxþM

�€xxg þ 4

�t_xxþ 2€xx

� �ð13:26Þ

whereK,C andM are the stiffness, damping and mass matrices for a structure,�x is the change in

the displacement vector x during a time step �t, _xx is a velocity vector and €xx is an acceleration

vector. By use of the Newmark [�¼ 1/4]-equations, from Equations 6.41–6.43 the ith elements

in the displacement, velocity and acceleration vectors at time (tþ�t) are

xi tþ�tð Þ ¼ xiðtÞ þ�xi ð13:27aÞ


�t�xi � _xxiðtÞ ð13:27bÞ

€xxi tþ�tð Þ ¼ 4

�t2�xi �

4

�t_xxiðtÞ � €xxiðtÞ: ð13:27cÞ

The size of the time step�t is important, as over-large as well as over-small time steps will lead to

inaccuracies in the calculated response. In the case of both wind and earthquakes, most of the

energy is contained within the part of the frequency spectrum that lies between 0 and 10Hz.

The period of the smallest frequency component that needs to be considered is therefore

approximately 0.1 s. Experience has shown that frequency components of this order of magnitude

can be sufficiently modelled with time steps �t¼ 0.1/10¼ 0.01 s.

As response spectra resulting from both recorded and generated earthquakes tend to be spiky, it is

usually recommended to carry out a time-domain analysis using different earthquakes normalised

to the same peak acceleration to ensure that the combined spectra approximate a consolidated

spectrum. This may need more computational effort than frequency domain method. However,

reasonable results can be obtained by generating only a suitable strong-motion history, calcu-

lating the variance of response to this history and then multiplying the resulting standard

deviation of response with a suitable peak factor.

13.8. Power spectral density functions for earthquakesThe mean amplitude, variance and frequency content of earthquakes vary with time; earthquakes

are therefore not stationary processes. If divided into sufficiently small segments, the process

within each segment may be considered to be approximately stationary. Each segmental process

may be modelled mathematically by the summation of harmonic components. The acceleration of

the ground motion may therefore be expressed as

€xxgðtÞ ¼XNi¼ 1

€xxi cos !itþ �ið Þ ð13:28Þ


265

where the values for €xxi and !i are found by Fourier analysis of real records and �i is a phase angle

that varies randomly between 0 and 2�.

Power spectral density functions or power spectra for the strong-motion part of earthquakes are

constructed by plotting values of €xx2i =!i against !i or values of €xx2i =ni against ni, where ni¼!i/2�.

Such spectra, however, tend to be spiky and require adjustments if needed for design purposes.

Power spectra used in design are, like wind spectra, averaged over a number of normalised earth-

quakes and smoothed. Kanai (1957) and Tajimi (1960) proposed the following formulation for

smoothed power spectra that are functions of the expected peak acceleration as well as the

damping and natural frequency of the ground:

S€xxg !ð Þ ¼S0 1þ 2�gr

� �2h i

ð1� r2Þ2 þ ð2�grÞ2ð13:29Þ

where

S0 þ0:141�g€xx

2g;max

!g

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4�2g� �q ð13:30Þ

and €xxg;max is the peak ground acceleration, !g is the natural angular frequency of ground, r¼!i/!g

and �g is the damping ratio for ground. For firm ground, Kanai (1957) suggested the values

!g¼ 12.7 rad/s and �g¼ 0.6.

For low frequencies, i.e. when !! 0, Equation 13.29 will lead to unbounded values for ground

velocity and ground displacements. Clough and Penzien (1975) therefore suggested the following

modification of the spectral density function


� �2h i

ð1� r2Þ2 þ ð2�grÞ2� r41

ð1� r21Þ2 þ ð2�r1Þ2

ð13:31Þ

where r1¼!/!1 and the frequency parameter !1 and damping parameter �1 are selected to give the

spectral density function the desired characteristic. Suggested values for !1 and �1 are given by

Key (1988) and Lin et al. (1989).

13.9. Frequency-domain analysis of single-DOF systems using powerspectra for translational motion

In Chapter 11, it is shown how frequency-domain analysis can be used to predict the variance of

dynamic response due to the buffeting of wind. In what follows, it will be shown how the same

approach can be extended to calculate the variance of the dynamic response to the strong-

motion part of an earthquake.

To obtain a relationship between the spectrum of the fluctuating force acting at a point on a

structure due to the acceleration of the ground and the spectrum of the ground acceleration, let

the frequency spans of both force and support motion be divided into unit frequency intervals

with each interval centred at the frequency !. From Equation 4.53, the force acting on a mass

M due to support acceleration €xxgðtÞ ¼ xg!2 sin !tð Þ is

fgðtÞ ¼ M€xxgðtÞ: ð13:32Þ


266

If

€xxgðtÞ ¼ €xxg sin !tð Þ ð13:33Þ

then

fgðtÞ ¼ fg sin !tð Þ ð13:34Þ

since it is assumed that fg(t) varies linearly with €xxgðtÞ. Substitution of the expressions for €xxgðtÞand fg(t) into Equation 13.32 yield

fg ¼ M€xxg ð13:35Þ

and hence

f 2g ¼ M2€xx2g: ð13:36Þ

As the coordinates of power spectra are proportional to the square of the amplitudes of the

constituent harmonics and inversely proportional to their frequencies, it follows that

Sfg!ð Þ ¼ M2S€xxg !ð Þ: ð13:37Þ

Having developed an expression for the force spectrum in terms of the ground acceleration

spectrum, it remains to express the response spectrum in terms of the force spectrum. From the

theory of forced vibrations of damped linear 1-DOF systems in Chapter 4 (Equation 4.15), the

response to a force


is

xðtÞ ¼fg

K

1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r2Þ2 þ ð2�rÞ2� �q sin !t� �ð Þ ð13:39Þ

or

xðtÞ ¼fg

KMF !ð Þ sin !t� �ð Þ: ð13:40Þ

The maximum value of x(t), which occurs when sin(!t� �)¼ 1, is therefore

x ¼fg

KMF !ð Þ ð13:41Þ

and squaring of each term in Equation 13.41 yields

x2 ¼f 2g

K2M !ð Þ ð13:42Þ

whereM(!)�MF2(!) is the mechanical admittance factor (Chapter 11). Because the coordinates

of the power spectrum are proportional to the square of the amplitudes of the constituent


267

harmonics, it follows that

Sx !ð Þ ¼ 1

K2M !ð ÞSfg

!ð Þ: ð13:43Þ

Finally, substitution of the expression for Sfg(!) given by Equation 13.37 into Equation 13.43 yields

Sx !ð Þ ¼ M2

K2!ð ÞS€xxg !ð Þ ð13:44Þ

and hence

�2x ¼

ð10Sx !ð Þd! ¼ M2

K2

ð10M !ð ÞS€xxg !ð Þ d!: ð13:45Þ

For weakly damped structures, and since !2n ¼K/M, the expression for �2

x can be approximated to

�2x ¼

ð10Sx !ð Þd! � 1

!4n

M !ð ÞS€xxg !ð Þ�! ð13:46Þ

where

�! ¼ 12 �!n

M !ð Þ ¼ 14 �

2:

Example 13.3

A tall building with a fundamental frequency of 1.0 Hz and a damping ratio of 1.0% of

critical is submitted to an earthquake with a peak acceleration of 0.3g. Use a probabilistic

method and Kanai’s power spectrum (1957) to determine the mean standard deviation or

root mean square response of the top of the building. Assume the dominant frequency of

the ground to be 2.0 Hz and a ground damping ratio �g¼ 0.6. Finally, assuming the duration

of the strong-motion part of the earthquake to be 10 s, calculate the maximum response.

Because the structural damping is only 1.0% of critical, the expression for the variance of

response given by Equation 13.46 may be used and written as

�2x ¼ 1

8!3n

� 1

�S€xxg !ð Þ

where

�g ¼ 0:6

!g ¼ 2:0� 2� rad=s

r ¼ 1:=2:0 ¼ 0:5

S0 ¼0:141� 0:6� 0:32 � 9:812

2:0� 2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4� 0:62ð Þ

p ¼ 0:0373289m2=s3

S€xxg !nð Þ ¼0:0373289� 1þ 2� 0:6� 0:5ð Þ2

� �

ð1� 0:52Þ2 þ ð2� 0:6� 0:5Þ2¼ 0:0550323m2=s3


268

13.10. Influence of the dominant frequency of the ground on themagnitude of structural response

In Figure 13.6, the Kanai power spectrum is used to show how the root mean square response

value or mean standard deviation of response of four different 1-DOF structures – with damping

and hence

�2x ¼ 1

8� 1:0� 2�ð Þ3� 1

0:010:0550323 ¼ 2:77324� 10�3 m2

�x ¼ 0:0527m:

The maximum probable response is obtained by multiplication of the mean standard devia-

tion by a peak factor which, since the structure is weakly damped, is given by

� ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln !nT=2�ð Þ½ �

pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln !nT=2�ð Þ½ �p

where T is the assumed duration for the strong-motion part of the earthquake. The maximum

response is therefore

xmax ¼ ��x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 1:0� 10ð Þ½ �

pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln 1:0� 10ð Þ½ �p

( )� 0:0527 ¼ 0:127m:

Figure 13.6 Root mean square response of 1-DOF structures sited on grounds with varying dominant

frequencies and assumed damping equal to 60% of critical to earthquakes with peak acceleration

€xxg ¼ 0:3g

fn: Hz ξ: %

0.50 1.0

0.75 1.01.00 1.02.00 1.0

0 1.0 2.0 3.0 4.0Dominant ground frequency: Hz

Root

mea

n sq

uare

res

pons

e: m

0.3

0.2

0.1

0.0


269

equal to 1.0% of critical but different natural frequencies – varies if the structures are sited on

grounds with the same damping but with increasing dominant frequencies and shaken by

earthquakes with peak accelerations equal to 0.3g. As can be seen from the graphs, the responses

tend to increase as the dominant frequency decreases and are greatest when the frequency of the

structure and that of the dominant frequency of the ground coincide.

The level of ground damping will vary with the type of alluvium, and it can be shown that the

response of a structure will increase with decreasing values of �g although the dominant frequency

of the ground and the peak acceleration of the earthquakes remain the same. A �g value equal to

0.3 will therefore result in root mean square responses of more than three times those shown in

Figure 13.6.

13.11. Extension of the frequency-domain method for translationalmotion to multi-DOF structures

It has previously been shown that the equations of motion for multi-DOF structures subjected to

ground acceleration can be written in matrix notation as

M€xxþ C _xxþ KX ¼ M�€xxgðtÞ ð13:47Þ

and that the decoupled equations of motion, obtained through the transformation x¼Zq, can be

written in matrix notation as

€qqþ 2�! _qqþ !2q ¼ ZTM�€xxgðtÞ: ð13:48Þ

The ith generalised modal equation is, as in Equation 13.18, given by


Ti M�€xxgðtÞ ð13:49Þ

where

ZTi M�€xxgðtÞ ¼ Z1iM1�þ Z2iM2�þ . . .þ ZNiMN�½ �€xxgðtÞ: ð13:50Þ

The spectral density function for qi is developed in exactly the same way as the spectral density

function for the response of a 1-DOF system, and is given by

Sqi!ð Þ ¼

ZTi M�

� �2!4i

Mi !ð ÞS€xxg !ð Þ ð13:51Þ

where S€xxg !ð Þ is the power spectral density function for an earthquake with a peak acceleration of

1.0g. Hence the variance of qi is given by

�2qi¼ð10Sqi

!ð Þd! ¼Z

Ti M�

� �2!4i

ð10Mi !ð ÞS€xxg !ð Þd! ð13:52Þ

and, for weakly damped structures, by

�2qi¼ð10Sqi

!ð Þd! ¼Z

Ti M�

� �2!4i

Mi !ð ÞS€zzg !ð Þ�! ð13:53Þ


270

where

�! ¼ 12 �i!i

Mi !ð Þ ¼ 14 �

2i :

We therefore have

qi ¼ �i�qið13:54Þ


�i ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln !iT=2�ð Þ½ �

pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln !iT=2�ð Þ½ �p ð13:55Þ

and finally

x ¼ Zq: ð13:56Þ

Example 13.4

Let the structure shown in Figure 2.14 be situated in an area where the dominant ground

frequency is 2.0 Hz and the ground damping is assumed to be 60% of critical. Calculate

the structural response to an earthquake, the strong-motion part of which can be represented

by Kanai’s spectrum, if the peak acceleration is 0.25g. Assume the duration of the strong

motion to be 10 s. The damping in the first mode is 2.0% and in the second and third

modes is 1.0% of critical. The mass matrix, angular frequencies and the normalised mode-

shape matrix for the structure are as follows:

M ¼600 0 0

0 400 0

0 0 200

264

375� 103

gkg

! ¼18:50

42:390

56:518

264

375 rad=s

Z ¼1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

264

375� 10�3:

From Equation 13.50, the generalised peak mode acceleration is given by

€qqg;max ¼ 0:5ZTM�g

and hence

€qqg;max ¼1:7454 3:3157 4:2336

3:0818 0:4330 �4:4926

1:9498 �3:6531 3:3105

264

375

600 0 0

0 400 0

0 0 200

264

375

0:25

0:25

0:25

264

375 ¼

805:06

280:94

92:69

264

375m=s2:


271

For weakly damped structures, the variance of the generalised mode response q is given by

Equation 13.53:

�2q ¼

1

8!3� 1

�S€qqg;max

!ð Þ


S€qqg !ið Þ ¼S0 1þ 2�gri

� �2h i

1� r2ið Þ2 þ 2�gri� �2

and from Equation 13.30

S0 ¼0:141�g€qq

2g;max

!g

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4�2g� �q :

For !1¼ 18.850 rad/s,

S0 ¼0:141� 0:6� 805:062


p ¼ 2783:3290m2=s3:

For !2¼ 42.390 rad/s,

S0 ¼0:141� 0:6� 280:942


p ¼ 342:5934m2=s3:

For !3¼ 56.718 rad/s,

S0 ¼0:141� 0:6� 92:692


p ¼ 37:0281m2=s3:

In the expression for S€qqg !ið Þ, the values for ri are

r1 ¼ 18:850=2:0� 2� ¼ 1:50004

r2 ¼ 42:390=2:0� 2� ¼ 3:37329

r3 ¼ 56:718=2:0� 2� ¼ 4:52348

and hence

S€qqg 18:850ð Þ ¼2783:3290 1þ 2� 0:6� 1:50004ð Þ2

� �

1� 1:500042ð Þ2 þ 2� 0:6� 1:50004ð Þ2¼ 2457:1857m2=s3

S€qqg 42:390ð Þ ¼342:5934 1þ 2� 0:6� 3:37329ð Þ2

� �

1� 3:373292ð Þ2 þ 2� 0:6� 3:37329ð Þ2¼ 47:9915m2=s3

S€qqg 56:718ð Þ ¼37:0281 1þ 2� 0:6� 4:52348ð Þ2

� �

1� 4:523482ð Þ2 þ 2� 0:6� 4:52348ð Þ2¼ 2:7633m2=s3:


272

The variances and mean standard deviations of response in the generalised coordinate system

are therefore:

�2q1 ¼1

8� 1

18:8503� 1

0:02� 2457:1857 ¼ 2:2928938m2

�q1 ¼ 1:5142304m

�2q2 ¼1

8� 1

42:3903� 1

0:01� 47:9915 ¼ 0:0078756m2

�q2 ¼ 0:0887446m

�2q3 ¼1

8� 1

56:7183� 1

0:01� 2:7633 ¼ 0:0001893m2

�q3 ¼ 0:0137586m:

We therefore have

q1 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 18:850� 10=2�ð Þ½ �

pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln 18:850� 10=2�ð Þ½ �p

( )� 1:5142304 ¼ 4:2843m



2 ln 42:390� 10=2�ð Þ½ �p

( )� 0:0887446 ¼ 0:2752m



2 ln 56:718� 10=2�ð Þ½ �p

( )� 0:0137586 ¼ 0:0439m

x1x2x3

264

375 ¼

1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

24

35

4:2843

0:2752

0:0439

24

35� 10�3 ¼

0:0084

0:0142

0:170

24

35m:

A comparison to the displacements calculated in Example 13.2 reveals that the use of Kanai’s

power spectrum leads to a much smaller response than for the Newmark response spectra.

The main reason for this is that the variances of the underlying power spectrum values for the

Newmark response spectra have been found to be much greater than those of the Kanai spectra.

Themodified displacement vector used in structural design, and calculated by taking the root of

the sum of the squares of the contribution from each mode as expressed by Equation 13.25, is

x1x2x3

264

375 ¼

0:0075

0:0142

0:0182

24

35m:

Finally, it is of interest to calculate the response on the assumption that the structure

responds mainly in the first mode. This yields

x1x2x3

264

375 ¼

1:7454 3:0818 1:9498

3:3157 0:4330 �3:6531

4:2336 �4:4926 3:3105

24

35

4:2843

0

0

24

35� 10�3 ¼

0:0075

0:0142

0:0181

24

35m:

The first mode response vector is therefore almost identical to the modified vector calculated

above.


273

13.12. Response of 1-DOF structures to rocking motionSo far, only the response of structures to the translational motion of earthquakes has been

considered. However, earthquakes also contain rocking components about the two horizontal

axes and one torsional component about the vertical axis, of which the former are caused by

the shear waves and Rayleigh waves depicted in Figure 12.2. Modern codes require that the

effect of these components be taken into account. Research in the USA has produced spectra

for the rocking motion of earthquakes which makes it possible to take this component into

account when generating earthquake loading and performing dynamic analysis of structures.

Consider the column shown in Figure 13.7, in which the rotational moment of inertia of the

equivalent lumped mass at the top is assumed to be zero.

Let the base of the column be subjected to a rocking motion

�gðtÞ ¼ �g sin !tð Þ: ð13:57Þ

Figure 13.7 Column considered as a 1-DOF system subjected to rocking excitation

θg

y

Hθg x

M

H


274

The translational equation of motion for the lumped mass is

M€yyþ C _yy� _xxgðtÞ� �

þ K y� xgðtÞ� �

¼ 0 ð13:58Þ

where

€yy ¼ €xxþ €xxgðtÞ

_yy ¼ _xxþ _xxgðtÞ

y ¼ xþ xgðtÞ

and hence

M€xx ¼ C _xxþ Kx ¼ �M€xxgðtÞ: ð13:59Þ

From Equation 13.57 it follows that

€xxgðtÞ ¼ H €��gðtÞ ¼ �H�g!2 sin !tð Þ ð13:60Þ

and finally that

M þ C _xxþ Kx ¼ MH�g!2 sin !tð Þ: ð13:61Þ

The response at the top relative to the position of the rotated but undeformed column, resulting

from the sinusoidal rocking motion �g(t)¼ �g sin(!t), is

x ¼MH�g!

2

K� 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1� r2ð Þ2 þ 2�rð Þ2h ir sin !t� �ð Þ ð13:62Þ

xmax ¼MH�g!

2

K� 1

2�: ð13:63Þ

13.13. Frequency-domain analysis of single-DOF systems using powerspectra for rocking motion

It has previously been shown how frequency-domain analysis can be used to predict the variance

of response due to the strong-motion translational components of earthquakes. In the following,

the same approach is extended to include the calculation of the variance of response due to

rocking components.

To obtain the relationship between the spectrum of the fluctuating force acting at a point on a

structure due to the angular acceleration of the ground and the spectrum of the angular

ground acceleration, let the frequency spans of both force and rocking motion be divided into

unit frequency intervals with each interval centred at the angular frequency !. From Equations

13.59 and 13.60, the force acting on a mass M due to support acceleration €��g!2 is

fgðtÞ ¼ MH €��gðtÞ: ð13:64Þ

If

€��gðtÞ ¼ €��g sin !tð Þ ð13:65Þ


275

then


since it is assumed that fg(t) varies linearly with €��gðtÞ. Substitution of the expressions for €��gðtÞ andfg(t) into Equation 13.64 yields

fg ¼ MH €�� ð13:67Þ

and hence

f 2g ¼ M2H2 €��2g: ð13:68Þ

As the coordinates of power spectra are proportional to the square of the amplitudes of the

harmonic components and inversely proportional to their frequencies, it follows that

Sfg!ð Þ ¼ M2H2S€��g

!ð Þ: ð13:69Þ


Sx !ð Þ ¼ 1

K2M !ð ÞSfg

!ð Þ: ð13:70Þ

Substitution of the expression for Sfg(!) given by Equation 13.69 into Equation 13.70 yields

Sx !ð Þ ¼ M2H2

K2M !ð ÞS€��g

!ð Þ ð13:71Þ

and hence

�2x ¼

ð10Sx !ð Þd! ¼ M2H2

K2

ð10M !ð ÞS€��g

!ð Þ d!: ð13:72Þ

For weakly damped structures, and since !2n ¼K/M, the expression for �2x can be approximated

to

�2x ¼

ð10Sx !ð Þd! � H2

!4n

M !ð ÞS€��g!ð Þ�! ð13:73Þ

where

�! ¼ 12 �!n

M !ð Þ ¼ 14 �

2:

13.14. Assumed power spectral density function for rocking motionused in examples

In order to present an example to illustrate the above theory, it is necessary to construct a function

that will yield reasonable response values. In Eurocode 8, part 3, response spectra for rocking

acceleration about the x, y and z axes have been proposed in terms of the lateral response


276

acceleration spectra. We therefore have

€��x !ð Þ ¼ 0:85!€xx !ð Þ=� ð13:74aÞ

€��y !ð Þ ¼ 0:85!€yy !ð Þ=� ð13:74bÞ

€��z !ð Þ ¼ 1:00!€zz !ð Þ=� ð13:74cÞ

where � is the shear wave velocity in m/s and ! is the frequency under consideration. By definition,

the coordinates of a spectral density function are equal to the square of the amplitudes of the

constituent frequency components divided by the frequencies of the same component. The

value of the spectral density function for rocking acceleration about the y axis at an angular

frequency !i is therefore

S€��gy!ið Þ ¼

€��2gy !ið Þ!i

¼0:852!2

i €xx2g !ið Þ

!i�2

¼ 0:852!2i

�2S€xxg !ið Þ: ð13:75Þ

Finally, substitution of the expression for the Kanai power spectrum given by Equations 13.30

and 13.31 into Equation 13.75 yields

S€��gy!ið Þ ¼

0:1019r2i !g�g€xx2p

�2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 þ 4�2g� �q �

ð1þ 2�gri�2

1� r2ið Þ2 þ 2�gri� �2 : ð13:76Þ

Example 13.5

Let the 40.0m tall tapering lattice tower shown in Figure 11.7 be subjected to an earthquake

with a peak acceleration of 0.3g. Calculate the acceleration and response at the top of the

tower due to the translation and rocking of the ground if the natural frequency of the mast

is 2.0 Hz, the equivalent mass of a mass–spring system depicting the movement of the disc

is 8200 kg, the structural damping is 2.0% of critical, the dominant frequency of the ground

is 2.0 Hz and the ground damping is 60.0% of critical. Assume the duration of the strong

motion of the earthquake to be 10 s and the velocity of the shear waves to be 500m/s.

For weakly damped structures, from Equation 13.73 the variance of response to rocking

motion is given by

�2x ¼ð10Sx !ð Þ d! � H2

!4n

M !ð ÞS€��gy!ð Þ�!

where

�! ¼ 12 �!n

M !ð Þ ¼ 14 �

2

and hence

�2x ¼ 40:02

8� 2�� 2:0ð Þ3� 1

0:02� S€��gy

!ð Þ ¼ 5:0393023S€��gy!ð Þ


277

where, from Equation 13.76 and since r¼ 2.0/2.0¼ 1.0,

S€��gy!ið Þ ¼ 0:1019� 1:02 � 2�� 2:0ð Þ � 0:6� 0:3� 9:81ð Þ2

500:02ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4� 0:62ð Þ

p � 1þ 2� 0:6� 1:0ð Þ2

1� 1:02ð Þ2 þ 2� 0:6� 1:0ð Þ2

S€��gy!ið Þ ¼ 28:7408� 10�6 m2=s3:

We therefore have

�2x ¼ 5:0393023� 28:87408� 10�6 ¼ 145:50522� 10�6 m2

�x ¼ 12:062554� 10�3 m

and hence

xmax; rocking ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2:0� 10ð Þ½ �


2 ln 2:0� 10ð Þ½ �p

( )� 12:062554� 10�3 ¼ 0:0324m:

The corresponding variance and hence displacement due to translational motion of the

ground is found by applying Equation 13.46 which, on substitution of the expressions for

M(!) and �!, is

�2x ¼

ð10Sx !ð Þd! � 1

8!3n

� 1

�� Sxg

!ð Þ

and hence

�2x ¼ 1

8� 2�� 2:0ð Þ3� 1

0:02� S€xxg !ð Þ ¼ 3:14956� 10�3S€xxg !ð Þ

where, from Equations 13.29 and 13.30,

S€xxg !ð Þ ¼ 0:141� 0:6� 0:3� 9:81ð Þ2

2�� 2:0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4� 0:2ð Þ

p � 1þ 2� 0:6� 1:0ð Þ2

1� 1:02ð Þ2 þ 2� 0:6� 1:0ð Þ2¼ 0:0632518m2=s3:

We therefore have

�2x ¼ 3:14956� 10�3 � 0:0632518 ¼ 1:99215� 10�4 m2

�x ¼ 0:0141143m

and hence

xmax; translation ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2:0� 10ð Þ½ �


2 ln 2:0� 10ð Þ½ �p

( )� 0:0141143 ¼ 0:0379m:

With the assumed spectral density function for the rocking motion and the assumed velocity

of the shear waves, the rocking contributes approximately 46% to the lateral motion at the

top of the tower.


278

13.15. Extension of the frequency-domain method for rocking motionto multi-DOF structures

It has previously been shown that the equations of motion for multi-DOF structures subjected to

translational ground acceleration can be written in matrix notation as

M€xxþ C _xxþ KX ¼ M�€xxgðtÞ: ð13:77Þ

With reference to Equations 13.59 and 13.60, it is clear that the equations of motion for structures

subjected to rocking as well as translational support motion can be established simply by addition

of the force vectorMH€��gðtÞ to the right-hand side of Equation 13.77, where €��gðtÞ is the history ofthe angular acceleration of the rocking motion corresponding to a translational motion with a

peak acceleration of �gm/s2 and � is a factor that defines the magnitude of the peak translational

acceleration. We therefore have

M€xxþ C _xxþ KX ¼ M�€xxgðtÞ þMH _��gðtÞ ð13:78Þ

where H€��gðtÞ is an acceleration vector in which the element Hi is the height of mass Mi above

the ground. Diagonalisation of the above equations, achieved through the transformation

x¼Zq and pre-multiplication of each term by ZT, yields

€qqþ 2�! _qqþ !2q ¼ ZTM�€xxgðtÞ þ Z

TMH€��gðtÞ: ð13:79Þ

Considering the rocking motion only, the ith generalised modal equation is given by


TMH€��gðtÞ ð13:80Þ

where

ZTi MH€��gðtÞ ¼ Z1iM1H1 þ Z2iM2H2 þ . . .þ ZNiMNHN½ �€��gðtÞ: ð13:81Þ

The spectral density function for the generalised coordinate qi is now found by following exactly

the same procedure as used for the 1-DOF system, and yields

Sqi!ð Þ ¼

ZTi MH

� �2!4i

Mi !ð ÞS€��g!ð Þ ð13:82Þ

where S€��g!ð Þ is the spectral density function for a rocking motion with a peak angular acceleration

of 1.0 rad/s. The variance of qi is therefore given by

�2qi¼ð10Sqi

!ð Þ d! ¼Z

Ti MH

� �2!4i

ð10Mi !ð ÞS€��g

!ð Þd!: ð13:83Þ

For weakly damped structures,

�2qi¼ð10Sqi

!ð Þ d! ¼Z

Ti MH

� �2!4i

Mi !ð ÞS€��g!ð Þ�! ð13:84Þ

where

�! ¼ 12 �i!i

Mi !ð Þ ¼ 14 �

2i


279

and hence

qi ¼ �i�qið13:85Þ


�i ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln !iT=2�ð Þ½ �

pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln !iT=2�ð Þ½ �p ð13:86Þ

and finally

x ¼ Zq: ð13:87Þ

Example 13.6

Let the shear structure shown in Figure 11.4 be subjected to an earthquake with a peak

acceleration of 0.3g. The dominant frequency, damping ratio of the ground, duration of

the strong motion and velocity of the shear waves may be taken as 2.0 Hz, 0.6, 10 s and

500 m/s, respectively. The mass of each equivalent floor is 120 000 kg. The modal damping

ratios in the first, second and third modes are 0.03, 0.02 and 0.01, respectively. The natural

frequencies and normalised mode-shape matrix for the structure are given below. Assume

that the foundation supporting the structure behaves as a rigid plate, and calculate the

response due to rocking motion.

We have

! ¼4:439

12:446

18:025

264

375 rads

!2 ¼19:70

155:40

324:90

264

375 rad2=s2

Z ¼0:947 2:128 1:703

1:706 0:950 �2:128

2:128 �1:703 0:953

264

375� 10�3:

From Equation 13.79, the decoupled equations for rocking motion are given by the matrix

equation

€qqþ 2�! _qqþ !2q ¼ ZTMH€��gðtÞ

where, from Equation 13.84, the variance of qi for weakly damped structures is given by

�2qi¼ð10Sqi

!ð Þd! ¼Z

Ti NH

� �28!3

i

� 1

�S€��g

!ð Þ:


280

The three values for ZTi MH are determined through evaluation of the matrix product:

ZTMH ¼

0:947 1:706 2:128

2:128 0:950 �1:703

1:703 �2:128 0:953

264

375

120:0 0 0

0 120:0 0

0 0 120:0

264

375

10:0

20:0

30:0

264

375 ¼

12 891:60

�1297:20

650:76

264

375 kgm:

The assumed expression for S€��g!ð Þ is given by Equation 13.76. Hence, when

r1 ¼ !1=!g ¼ 4:439=2�� 2:0 ¼ 0:3532;

we have

S€��gy!1ð Þ ¼ 0:1019� 0:35322 � 2�� 2:0ð Þ � 0:6� 0:3� 9:81ð Þ2

5002ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 0:62ð Þ

p

� 1þ 2� 0:6� 0:3532ð Þ2

1� 0:35322ð Þ2 þ 2� 0:6� 0:3532ð Þ2

¼ 2:12579� 10�6 � 1:2473694 ¼ 2:6516454� 10�6 rad2=s;

when

r2 ¼ !2=!g ¼ 12:4662�� 2:0 ¼ 0:9920;

we have

S _��gy!2ð Þ ¼ 0:1019� 0:99202 � 2�� 2:0ð Þ � 0:6� 0:3� 9:81ð Þ2


p

� 1þ 2� 0:6� 0:9920ð Þ2

1� 0:99202ð Þ2 þ 2� 0:6� 0:9920ð Þ2

¼ 16:768813� 10�6 � 1:7073847 ¼ 28:597277� 10�6 rad2=s

and when

r3 ¼ !3=!g ¼ 18:025=2�� 2:0 ¼ 1:4344

we have

S€��gy!3ð Þ ¼ 0:1019� 1:43442 � 2�� 2:0ð Þ � 0:6� 0:3� 9:81ð Þ2


p

� 1þ 2� 0:6� 1:4344ð Þ2

1� 1:43442ð Þ2 þ 2� 0:6� 1:4344ð Þ2

¼ 35:060614� 10�6 � 0:9710095 ¼ 34:044192� 10�6 rad2=s:


281

13.16. Torsional response to seismic motionTorsional response of buildings to ground motion is due to the nature of the motion itself, lack of

symmetry in the structure and/or lack of symmetry in the distribution of the total mass about the

shear centre of the building. In order to determine the contribution of the torsional vibration to

the lateral, vertical and rocking responses, it is necessary to analyse the structure as a 3D structure.

This requires the assembly of 3D stiffness and mass matrices. The construction of the former is

given by Coates et al. (1972) and that of the latter by Clough and Penzien (1975).

The total response may then be calculated either in the frequency domain by first calculating the

natural frequencies and modes of vibration and then applying the method of mode superposition,

using either translational and rotational response spectra or power spectra, or in the time domain

using either real or generated earthquake histories. For non-symmetric structures with large

overhangs, such as cable-stayed cantilever roofs (see Figure 1.1) and cantilevered cranes, the

Substitution of the given values for !i and �i and the calculated values for ZiTMHW and

S€��gy!ið Þ into Equation 13.84 yield

�2q1 ¼

12 891:602

8� 4:4293� 1

0:03� 2:6516454� 10�6 ¼ 21:13493m2

�2q2¼ 1297:202

8� 12:4663� 1

0:02� 28:597277� 10�6 ¼ 0:15525m2

�2q3 ¼

650:762

8� 18:0253� 1

0:01� 34:044192� 10�6 ¼ 0:03077m2;

and hence

�q1¼ 4:59727m

�q2¼ 0:39402m

�q3¼ 0:17541m:

We therefore have

q1 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 4:429� 10=2�ð Þ½ �

pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2 ln 4:429� 10=2�ð Þ½ �p

( )� 4:59727 ¼ 10:42780m



2 ln 12:466� 10=2�ð Þ½ �p

( )� 0:39402 ¼ 1:05618m



2 ln 18:025� 10=2�ð Þ½ �p

( )� 0:17541 ¼ 0:49354m

x1

x2

x3

264

375 ¼

0:947 2:128 1:703

1:706 0:950 �2:128

2:128 �1:703 0:953

264

375

10:42780

1:05618

0:49354

264

375� 10�3 ¼

0:013

0:018

0:021

264

375m:


282

torsional response modes could be the most significant. Even for symmetric structures, some

codes require that the effect of possible vibration in torsional modes be taken into account by

assuming the position of the centre of gravity of the structural mass to be eccentric to that of

the structure’s shear centre. Most cases, except the very simplest ones, will require the use of a

computer. One such problem is considered in Example 13.7 as an introduction to torsional

vibration, where the translational and rotational response of a platform considered as a shear

structure is calculated using the Kanai power spectrum (Equations 13.30 and 13.31).

The expression for the torsional response spectrum of a 1-DOF spectrum due to a torsional

moment

TðtÞ ¼ PðtÞ � e ¼ M€xxgðtÞ � e ð13:88Þ

where M is the equivalent mass of the structure, €xxgðtÞ is the ground acceleration at time t and e is

the eccentricity of the mass relative to the shear centre of the structure, can be developed in exactly

the same manner as for the translational response, and can be shown to be

S� !ð Þ ¼ M2e2

K2t

M !ð ÞS€xxg !ð Þ: ð13:89Þ

The variance of torsional response is therefore

�2� ¼

ð10S� !ð Þ d! ¼ M2e2

K2t

ð10M !ð ÞS€xxg !ð Þ d!: ð13:90Þ

For weakly damped structures, and since !2n ¼Kt/Ip¼Kt/Mk2 where k is the radius of gyration,

the expression for ��2 can be approximated to

�2 ¼ð10S� !ð Þ d! � 1

!4n

� e2

k4M !ð ÞS€xxg !ð Þ�! ð13:91Þ

where

�! ¼ 12 �!n

M !ð Þ ¼ 14 �

2:

For weakly damped structures, the expression for the variance of torsional response is therefore

�2� ¼

ð10S� !ð Þ d! � 1

8� 1

!3n

� e2

K4� 1

�S€xxg !ð Þ: ð13:92Þ

Example 13.7

Calculate the translational and rotational response of the platform structure shown in

Figure 4.15 when subjected to an earthquake with a peak acceleration of 0.25g. Assume

the dominant frequency of the ground to be 0.565 Hz and the ground damping ratio to be

60.0% of critical. The equivalent mass of the structure at platform level is 4.722� 106 kg.

The eccentricity of the centre of gravity of the mass relative to the shear centre of the structure

measured perpendicular to the direction of the quake is assumed to be 1.0m. The polar


283

moment of inertia of the mass is 1361.2421� 106 kg m2, the translational stiffness is

135.748� 103 kN/m and the rotational stiffness is 45 475.523� 103 kN/rad. The translational

and rotational frequencies are 0.8533Hz and 0.9199, Hz respectively. The damping in

both the translational and the rotational mode may be assumed to be 2.0% of critical and

the duration of the strong motion 10 s.

From Equation 13.46, the variance of the translational response of a weakly damped 1-DOF

system can be written as a function of an earthquake acceleration spectrum:

�2x ¼ 1

8� 1

!3n

� 1

�S€xxg !ð Þ

where

!n ¼ 2�� 0:8533 ¼ 5:361442 rad=s

and hence

�2x ¼ 1

8� 1

5:3614423� 1

0:02S€xxg !ð Þ ¼ 0:0405541S€xxg !ð Þ:

Similarly, from Equation 13.91 the variance of the rotational response of weakly damped

1-DOF systems can be written as

�2� ¼

1

8� 1

!3n

� e2

k4� 1

�S€xxg !ð Þ

where

!n ¼ 2�� 0:9199 ¼ 5:7799022 rad=s

k ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 361:2421� 106=4:722� 106ð Þ

q¼ 16:97871m

and hence

�2� ¼

1

8� 1

5:77990223� 1:02

16:978714� 1

0:02S€xxg !ð Þ ¼ 0:389493� 10�6S€xxg !ð Þ:

From Equations 13.29 and 13.30,


� �2h i

1� r2ð Þ2 þ 2�rð Þ2

where

S0 ¼0:141�g€xx

2g;max

!g

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4�2g� �q


284

and therefore

S0 ¼0:141� 0:6� 0:252 � 9:812

2�� 0:920ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4� 0:62ð Þ

p ¼ 0:0563541m2 s�4 rad�1:

When fn¼ 0.8533 Hz,

r ¼ 0:8533=0:920 ¼ 0:9275

S€xxg !ð Þ ¼0:0563541 1þ 2� 0:6� 0:9275ð Þ2

� �

1� 0:92752ð Þ2 þ 2� 0:6� 0:9275ð Þ2¼ 0:1002655m2 s�4 rad�1

and hence

�2x ¼ 0:0405541� 0:1002655 ¼ 4:06617� 10�3

�x ¼ 0:0637665m:

The corresponding peak factor is

� ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 0:8533� 10ð Þ½ �

pþ 0:577

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 0:8533� 10ð Þ½ �

p¼ 2:3493672

and the maximum translational amplitude is therefore

xmax ¼ ��x ¼ 2:3493672� 0:0637665 ¼ 0:1498m

When fn¼ 0.9199 Hz,

r ¼ 0:9199=0:920 ¼ 0:9998913

S€xxg !ð Þ ¼0:0563541 1þ 2� 0:6� 0:9998913ð Þ2

� �

1� 0:99989132ð Þ2 þ 2� 0:6� 0:9998913ð Þ2¼ 0:0954973m2 s�4 rad�1

and hence

�2� ¼ 0:389493� 10�6 � 0:0954973 ¼ 0:0371955� 10�6

�� ¼ 0:1928614� 10�3 rad:

The corresponding peak factor is

� ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 0:9199� 10ð Þ½ �

pþ 0:577

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 0:9199� 10ð Þ½ �

p¼ 2:380589

and the maximum angular rotational amplitude is therefore

�max ¼ �� ¼ 2:380589� 0:1928614� 10�3 ¼ 0:4591327� 10�3 rad:

The corresponding movements at the corners of the platform are therefore

xt ¼ 0:4591327� 10�3 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi202 þ 202ð Þ

q¼ 0:0123m:


285

13.17. Reduction of dynamic responseThe two most common techniques used for reducing the vibration caused by earthquakes are

isolation and energy absorption or damping. A third method involves active control in which

feedback from sensors recording the vibration of the structure is utilised to control the behaviour

of the structure. As an example, tuned mass dampers are used to reduce the amplitude of vibration

in buildings as given in Chapter 8.

Isolation involves the installation of springs or base isolators at the base of a structure, mainly in

order to limit the amount of horizontal ground acceleration transmitted to the building. Isolators

are therefore essentially soft springs, usually constructed from the lamination of steel and rubber.

Their effectiveness depends on correct anticipation of the frequency contents of future earth-

quakes. In practice, they are often designed to reduce the dominant frequency of a structure to

0.5 Hz or less.

Elastic absorption involves linking dampers between points with relative displacements, either

within the structure or between the structure and the ground. Such dampers may be fabricated

in the form of hydraulic dashpots but are more generally designed to behave elastically up to a

given maximum permitted relative displacement, above which they will yield. For further

information about isolation and energy absorption, the reader is referred to Harris (1988), Key

(1988) and Warburton (1992).

As a first attempt to establish the spring stiffness of isolators, a building may be reduced to a

1-DOF structure constructed on a base which again is supported on isolators. Such a system,

containing an internal energy absorber, is shown in Figure 13.8.

When subjected to a ground acceleration €xxgðtÞ, the equation of motion for a 1-DOF shear

structure with the base plate supported on isolators (Figure 13.8) is

M€xxþ C _xxþ Kx ¼ M€xxgðtÞ ð13:93Þ

Figure 13.8 1-DOF structure with isolators and energy absorber

M2

M1

C2

K2/2

K1/2 K1/2C1/2 C1/2

K2/2


286

or

M1

0 M2

" #€xx1

€xx2

" #þ

C1 þ C2ð Þ �C2

�C2 C2

" #_xx1

_xx2

" #þ

K1 þ K2ð �K2

�K2 K2

" #x1

x2

" #

¼M1 0

0 M2

" #€xxgðtÞ€xxgðtÞ

" #: ð13:94Þ

If the effect of damping is neglected, the corresponding eigenvalue equation is

KX� !2MX ¼ 0 ð13:95Þ

which is satisfied when

K1 þ K2 � !2M1

� ��K2

�K2 K2 � !2M2

� ��

��¼ 0: ð13:96Þ

Evaluation of the determinant and solving the resulting characteristic equation with respect to K1

yields the following expression for the shear stiffness of the isolators in terms of the required

frequency:

K1 ¼!2K2 M1 þM2ð Þ � !4M1M2

K2i!2M2

: ð13:97Þ

Example 13.8

The first natural frequency of the shear structure in Example 2.5 (Figure 2.14) obtained from

an eigenvalue analysis is 0.7419Hz. The shear stiffness of the columns at ground level is

25.0� 106 kN/m. Reduce the structure to a 1-DOF system and calculate the shear stiffness

of isolators required to reduce the first natural frequency to 0.5 Hz. Assume the mass of

the base slab supporting the columns at ground level to be 1700.0� 103 kg.

The equivalent mass of the 1-DOF system is

M2 ¼K2

!21

¼ 2500:0� 106

2�� 0:7419ð Þ2¼ 115051:77� 103 kg

and from Equation 2.92 the required combined shear stiffness of the isolators is

K1 ¼

2�� 0:5ð Þ2�2500:0� 106 1500:0þ 115 051:77ð Þ � 103� �

� 2�� 0:5ð Þ4�1500:0� 115 051:77� 106� �

2500:0� 106 � 2�� 0:5ð Þ2�115 051:77� 103� �

¼ 2095:28� 106 N=m:


287

13.18. Soil–structure interactionWhen constructing numerical dynamic models of structures, it is necessary to consider the

flexibility of the soil and also to what extent the weight of the structure is likely to reduce the

dominant frequency of the soil above the bedrock. The weight of most structures is fortunately

very small compared to the amount of soil, and will not significantly alter the dynamic character-

istics of the latter. Such changes usually only need to be considered in the case of exceptionally

heavy rigid structures such as nuclear containment buildings, when the weight of the structure

can affect the surface ground motion below and adjacent to the foundations. This may be

better appreciated by considering the lumped mass model of soil shown in Figure 13.9. When

modelling such structures, it is necessary to incorporate the supporting soil down to the rock

base. This type of analysis is sophisticated and specialised, and outwith the scope of this book.

For detailed work on the subject the reader is referred to Wolf (1985).

As stated above, the weight of most buildings will not alter the characteristics of the supporting

ground. The flexibility of the soil will however tend to reduce the overall stiffness of the structure

and thus reduce its frequencies and modify its modal response, as well as generating additional

damping through energy dissipation. At resonance the surrounding layers of certain types of

soil (such as wet clays) will also tend to vibrate in phase with the structure in the same

manner as water and air, and therefore add to the amount of the vibrating mass. For the

purpose of analysis the stiffness and damping properties of soil can be modelled as springs and

Figure 13.9 Lumped mass model of soil

Soil surface

Rock face


288

equivalent viscous dampers, as indicated in Figure 13.10. The modelling of such springs and

dampers is considered to be outwith the scope of this book; interested readers should refer to

Key (1988).

Figure 13.10 Numerical modelling of the stiffness and damping of soil by equivalent elastic springs and

viscous dampers

Problem 13.1

A pre-stressed concrete bridge which can be considered as a simply supported beam spans

40 m. The flexural rigidity and the mass of the bridge are 8.15887� 1010 N m2 and 35 t/m,

respectively. The structural damping is 2.0% of critical. Use the response spectra shown in

Figure 13.4 to calculate the maximum first mode response due to an earthquake with a

peak acceleration of 0.3g.

Problem 13.2

The bridge in Problem 13.1 is sited at a point where the depth of soil is approximately 21 m.

For this depth the dominant frequency of the ground is estimated to be approximately 2.0 Hz.

As the ground is firm, the ground damping is assumed to be 60% of critical. Use the Kanai

power spectrum to estimate the maximum first mode response due to an earthquake with a

peak acceleration of 0.3g.


289

Problem 13.3

The two-storey shear structure shown in Figure 13.11 is to be erected in a seismic zone on a

site where the dominant frequency is unknown and therefore conservatively assumed to be

equal to the first natural frequency of the building. Use the response spectra shown in

Figure 13.4 first and then the Kanai power spectrum to calculate the maximum response

of the structure to an earthquake with a peak acceleration of 0.35g. Assume the damping

in the first and second modes to be 1.5% and 1.0% of critical, respectively, the damping in

the ground to be 60% of critical and the duration of the strong-motion part of the quake

to be 10 s. The mass matrix, natural angular frequencies and normalised mode-shape

matrix for the structure are as follows:

M ¼6:0 0

0 6:0

� �� 104 kg

! ¼9:4248

24:6743

� �rad=s

Z ¼0:2146 0:4024

0:3473 �0:0687

� �� 10�2:

Figure 13.11 Two-storey shear structure

5 m

5 m

6 m6 m6 m

Problem 13.4

Calculate the response of the stepped tower in Problem 7.4 to the lateral and rocking motions

of an earthquake with a peak acceleration of 0.35g, if the duration of the strong-motion part

of the quake is 10 s. Assume the dominant frequency of the ground to be 2.0 Hz, the damping

in the ground to be 60% of critical and the shear velocity of the ground to be 500 m/s. The

structural damping may be taken as 3.0% of critical in the first mode and as 2.0% in the

second mode. The stiffness matrix, mass matrix, natural frequencies and normalised damping


290

REFERENCES

Blum JA, Newmark NM and Corning LH (1961) Design of Multistorey Reinforced Building for

Earthquake Motions. Portland Cement Association, Chicago.



Harris CM (1988) Shock Vibration Handbook, 3rd edn. McGraw-Hill, London.

Kanai K (1957) Semi-empirical formula for the seismic characteristics of the ground. University

of Tokyo Bulletin Earthquake Research Institute 35: 309–325.


Lin BC, Tadjbasksh IG, Papageorgiu AA and Ahmadi G (1989) Response of base-isolated

buildings to random excitation described by Clough–Penzien spectral model. Earthquake

Engineering and Structural Dynamics 18: 49–62.

Newmark NM and Hall WJ (1982) Earthquake Spectra and Design. Earthquake Engineering

Research Institute, Berkeley.

Tajimi H (1960) A statistical method of determining the maximum response of building

structures during an earthquake. Proceedings of 2nd International Conference on Earthquake

Engineering, Tokyo and Kyoto, vol. II: 781–798.

Warburton GB (1992) Reduction of Vibrations. Wiley, London.

Wolf JH (1985) Dynamic Soil–structure Interaction. Prentice-Hall, Englewood Cliffs.

FURTHER READING

Bolt BA (1978) Earthquakes: A Primer. WH Freeman, San Francisco.


Eiby GA (1980) Earthquakes. Heinemann, London.

Lomnitz F and Rosenbleuth E (1976) Seismic Risk and Engineering Decisions. Elsevier,

Amsterdam.


Seed HB and Idriss IM (1982) Ground motion and ground liquefaction during earthquakes.

Earthquake Engineering Research Institute, Berkeley.

matrix for the tower are given below:

~KK ¼27 568:761 �7657:989

�7657:989 3063:186

� �kN=m

~MM ¼20 066:47 2795:75

2795:75 4659:62

� �kg

! ¼25:133

119:098

� �rad=s

Z ¼3:443 6:521

10:109 �10:753

� �� 10�3

The given stiffness and mass matrices have been obtained by eliminating the rotational

degrees of freedom at stations 10 and 20m above the ground.


291


ISBN: 978-0-7277-4176-9



Chapter 14

Generation of wind and earthquakehistories

14.1. IntroductionA time-domain method was presented in Chapter 6 for predicting the linear and non-linear

response of 1-DOF systems to wind and earthquakes and to multi-DOF systems in general.

The equations developed are based on the incremental equation of motion, and arise from various

assumptions with respect to the change in acceleration during a time step �t. Other time-domain

methods, which are particularly suitable for highly non-linear structures such as guyed masts,

cable and membrane roofs, are those in which

g equilibrium of the dynamic forces at the end of each time step is sought by minimisation of

the gradient vector of the total potential dynamic energy by use of the Newton-Raphson or

conjugate gradient method, andg where increased convergence and stability are achieved through scaling and the calculation

of a step length in the descent direction to a point where the energy is a minimum

(Buchholdt, 1985; Buchholdt et al., 1986).

The prediction of response using any of the above methods requires the ability to generate

earthquake histories and single and spatially correlated wind histories. The problem with using

recorded earthquake histories is that no two earthquakes are the same. For the purpose of

design, it is therefore necessary to calculate the response to a family of simulated earthquakes

compatible with a given site. Because wind histories can be considered as stationary stochastic

processes, they are simpler to generate the earthquake histories; methods for simulating wind

histories are therefore presented first.

14.2. Generation of single wind histories by a Fourier seriesShinozuka and Jan (1952) have shown that it is possible to express the fluctuating velocity

component u(t) of wind at any time t as

uðtÞ ¼ffiffiffiffiffiffiffið2Þ

p Xni¼ 1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisu nið Þ ��nð Þ½ � cos 2�ni þ �ið Þ

pð14:1Þ

where Su(n) is the value of the power spectral density function for the fluctuating component of

wind at the frequency n, �n¼ ni þ 1� ni and �i is the phase angle with a uniform probability

distribution function that varies randomly between 0 and 2�.

The frequency band in Equation 14.1, which has been divided into N parts, must contain all the

significant natural frequencies of the structure. For non-linear structures the frequency step �n

293

needs to be small, as the natural frequencies of such structures vary with the amplitude of

response.

14.3. Generation of wind histories by the autoregressive methodAnother method for generating single wind histories that yields variances of response similar

to those of real wind is the autoregressive (AR) method. It is computationally more efficient

than the Fourier series (FS) method given by Equation 14.1, and can also be used to generate

earthquake histories. The AR method filters white noise and transforms it into a signal with a

specified variance and autocovariance function.

Mathematically, the method for transforming white noise may be expressed as

uðtÞ ¼ �ðBÞ � aðtÞ ð14:2Þ

where u(t) is the stochastic process to be generated, a(t) is the input white noise with zero mean and

variance �2a and �(B) is a transfer function or filter. The white noise a(t) may also be expressed as

aðtÞ ¼ �Nu �NðtÞ ð14:2Þ

where N(t) are random shocks with zero mean and unit variance. Substitution of this expression

for a(t) into Equation 14.2 yields

uðtÞ ¼ �ðBÞ � �Nu �NðtÞ: ð14:3Þ

The white noise process a(t) is transformed into the process u(t) by the filter or transfer function

�(B). One type of filter that has proved to be very suitable for modelling wind and earthquakes is

the so-called autoregressive filter, which regressively weights and sums previous values.

In an autoregressively simulated process of order p, the instantaneous values of u(t) are expressed

as a finite linear aggregate of the previous values of u(t) plus a random impulse with zero mean and

variance �2Nu. The expression for u(t) may therefore be written as

uðtÞ ¼Xp

s¼ 1

�su t� s�tð Þ þ �NuNðtÞ ð14:4Þ

where � is an autoregressive parameter, N(t) is a random impulse with zero mean and unit

variance and

�2Nu ¼

1

T

ðT0

�NuNðtÞ � uðtÞdt: ð14:5Þ

Alternatively, Equation 14.4 may be written as

uðtÞ ¼Xp

s¼ 1

�sBsuðtÞ þ �NuNðtÞ ð14:6Þ

where Bs is a backshift operator which is defined

BsuðtÞ ¼ u tþ s�tð Þ: ð14:7Þ


294

Solving Equation 14.6 with respect to u(t) yields

uðtÞ ¼ 1

1�Xp

s¼ 1

’sBs

� �NuNðtÞ: ð14:8Þ

Comparing Equation 14.8 to Equation 14.3 yields the following expression for an autoregressive

filter of order p

�ðBÞ ¼ 1

Xp

s¼ 1

’sBs

: ð14:9Þ

In order to obtain expressions for determining the values for the parameters � and the variance

�2Nu, both sides of Equation 14.4 are multiplied by u(t� kt) where k¼ 1, 2, . . . , p. Integration

and averaging over time T yields

1

T

ðT0uðtÞ � u t� ktð Þ dt ¼

Xp

s¼ 1

1

T

ðT0’su t� s�tð Þ u t� k�tð Þdtþ 1

T

ðT0�NuNðtÞ u t� k�tð Þ dt: ð14:10Þ

When k> 0, Equation 14.10 yields

Cu f�tð Þ ¼ Cu �k�tð Þ ¼Xp

s¼ 1

’sCu k� sð Þ�t½ � k ¼ 1; 2; . . . ; p ð14:11Þ

because of the symmetry of the autocovariance function and the randomness of the process N(t).

When k¼ 0, Equation 14.10 yields

�2u ¼

Xp

s¼ 1

’sCu s�tð Þ þ �2Nu: ð14:12Þ

Division of all the elements in Equations 14.11 and 14.12 by �2u yields

cu k�tð Þ ¼Xp

s¼ 1

�scu k� sð Þ�t½ � k ¼ 1; 2; . . . p ð14:13Þ

�2Nu ¼ �2u 1�

Xp

s¼ 1

’scu s�tð Þ" #

ð14:14Þ

where cu(k�t) is the autocovariance coefficient at time lag � ¼ k�t corresponding to the power

spectral density functions Su(n) and cu[(k� s)�t]¼ 1.0 when k¼ s.

Given an expression for the power spectrum Su(n), the values of the autocovariance coefficient cuare determined by dividing both sides of Equation 10.27 by �2

n and completing the integration:

cu k�tð Þ ¼ 1

�2u

ð10

SuðnÞ cos 2� nk�tð Þ dn: ð14:15Þ

Having determined the values of the autocovariance function using Equation 14.11, the

autoregressive parameters � can be determined by use of Equation 14.13 and the variances of

the impulses �2Nu from Equation 14.14.

Generation of wind and earthquake histories

295

Unlike the Fourier series model the autoregressive model is not unconditionally stationary, and

tends to become non-stationary when a short time step is chosen. In this case, the right-hand

side of Equation 14.11 may become negative. Another problem is concerned with the number

of parameters � to be used in the generation of the autoregressive model. Both problems have

been dealt with by Box and Jenkins (1977) in terms of the so-called partial autocorrelation

function which, for a process suitable for simulation by an autoregressive method of order q, is

nearly zero when p> q.

Figure 14.1 shows the variation of the partial autocorrelation function with the order of the model

and with the size of the time step. The curves are based on the Kaimal spectrum (Equation 10.34),

with V(10)¼ 30m/s and z0¼ 0.1 m. They indicate that, for the data used, a suitable number of �

Figure 14.1 Variation of the partial autocorrelation model with number of parameters � and size of

time step

5 2

2 4 6 8 10 12 14

1 1

1

0.5

0.2

0.1

0.05

Part

ial a

utoc

orre

latio

n fu

nctio

n (P

AC

F)

0.8

1.0

0.8

0.6

0.4

0.2

–0.2

–0.4

–0.6

∆t

∆t = 0.05 s


296

parameters is 3–5 and that the size of the time step should not be less than 0.1 s. If p¼ 3 and

�t¼ 0.1 s, from Equations 14.13 and 14.14 we have

’1

’2

’3

264

375 ¼

cu 0:0ð Þ cu 0:1ð Þ cu 0:2ð Þcu 0:1ð Þ cu 0:0ð Þ cu 0:2ð Þcu 0:2ð Þ cu 0:1ð Þ cu 0:0ð Þ

264

375�1 cu 0:1ð Þ

cu 0:2ð Þcu 0:3ð Þ

264

375 ð14:16Þ

�2Nu ¼ �2u 1� ’1cu 0:1ð Þ � ’2cu 0:2ð Þ � ’3cu 0:3ð Þ½ �: ð14:17Þ

It should be noted that a time step of�t¼ 0.1 s is of the order of ten times the size of the time step

used in the forward integration method when the acceleration is assumed to remain constant

during the time step. It is therefore necessary to interpolate to obtain the wind velocities required

at any time t during the dynamic analysis.

As the process u(t) is generated with � parameters that are functions of the autocovariance

coefficients, the simulated histories need to be multiplied by the ratio �u/�gu where �u is the

standard deviation of the required history and �gu is the standard deviation of the generated

process.

Of the two methods for generating wind histories, the FS method is more expensive in terms of

computer time. This is shown in Figure 14.2, where the time taken to generate wind histories

by the two methods is compared. An example of an FS model and AR models generated with

the same power spectrum and autocovariance functions as those of a recorded history is given

by Buchholdt et al. (1986).

14.4. Generation of spatially correlated wind historiesIn Chapter 10, the correlation between the fluctuating wind velocities at two points in space is

expressed in terms of the cross-covariance function (Equation 10.37) and the cross-spectrum

Figure 14.2 Comparison of computer time required to generate single FS and AR wind models: for the

latter the number of frequency steps N is used only to compute the autocovariance coefficients cu(k�t)

of Equation 14.15

0 400 800 1200 1600 2000Number of frequency steps N

Com

pute

r tim

e: s

AR(6)

AR(12)

60

40

20

0

FS


297

(Equation 10.41), the latter being expressed as a function of the square root of the product of the

power spectra of the individual histories and the coherence function (Equation 10.42). In Chapter

11, it is shown how power spectra and cross-spectra are used to establish model force spectra for

calculating the variance of response of multi-DOF structures in the frequency domain (Equations

11.52–11.59). The interdependence of the velocity fluctuations in space must also be included

when generating spatial wind fields. This can be achieved in different ways. Spinelli devised a

method with correlation at time lag � ¼ 0 (Buchholdt, 1985; Iannuzzi, 1987; Iannuzzi and Spinelli,

1989) and Iwatani (1982) devised a method with correlation at � 5 0. A third method is now

presented based on an eigenvalue analysis of the cross-covariance matrix Cuv(0) at � ¼ 0.

In Chapter 10, it is shown that the wind velocity vector at any time tmay be considered to consist

of a steady-state componentU(z), whose element can be determined using either Equation 10.4 or

Equation 10.10 and a fluctuating component u(z, t). We therefore have

U z; tð Þ ¼ UðzÞ þ u z; tð Þ: ð14:18Þ

Further, the vector u(z, t) may be expressed as D� v(z, t) where D is a correlation matrix whose

elements are evaluated from the cross-covariance of the elements in u(z, t) at � ¼ 0 and v(z, t) is a

fluctuating velocity vector in which the elemental time histories are uncorrelated and can be

modelled as either FS or AR series. Equation 14.11 may therefore be written as

U z; tð Þ ¼ UðzÞ þD� v z; tð Þ: ð14:19Þ

The elements in matrix D are determined as follows. From Equations 10.37 and 10.38, it is shown

that the cross-covariance matrix at time lag � ¼ 0 is

Cu z;tð Þ;u z;tð Þð0Þ � Cuð0Þ ¼1

T

ðT0u z; tð Þ � u z; tð ÞT dt ¼ �ij

� �ð14:20Þ

where [�ij] is a square symmetrical matrix in which the elements on the leading diagonal �ij¼�i2

are the variances and the off-diagonal elements are the cross-variances of the elemental processes in

u(z, t). The elements on the leading diagonal may therefore be calculated by use of Equation 10.28,

and the off-diagonal elements by use of Equation 10.40. To proceed, the eigenvalues and the

normalised eigenvectors ofCu(0) must be determined. Let the corresponding eigenvalue equation be

Cuð0ÞX ¼ �IX ð14:21Þ

where �¼ diag.{�1, �2, . . . , �n} is the eigenvalue matrix andX¼ [X1,X2, . . . ,Xn] is the eigenvector

matrix. In order to normalise an eigenvector Xi, let

XTi IXi ¼ L2

i : ð14:22Þ

The normalised eigenvector Zi is now found by dividing each element in Xi by Li, and therefore

Zi ¼ Xi=Li ð14:23Þ

ZTi IZi ¼ 1: ð14:24Þ

Writing the eigenvalue equation (Equation 14.21) in terms of the eigenvalue �i and the normalised

eigenvector Zi and post-multiplication of each term by ZiT yields

ZTi Cuð0ÞZi ¼ �iZ

Ti IZi ¼ �i ð14:25Þ


298

and hence

ZTCuð0ÞZ ¼ � ð14:26Þ

Cuð0Þ ¼ Z�T�Z�1: ð14:27Þ

Since Cu(0) is a symmetric positive definite matrix, Z1¼ZT and ZT¼Z and hence

Cuð0Þ ¼ Z�ZT: ð14:28Þ

If the uncorrelated wind histories in v(t) in Equation 14.19 are generated with power spectra

having p7 variances �1, �2, . . . , �n, then

Cv z;yð Þv z;tð Þð0Þ ¼ Cvð0Þ ¼1

T

ðT0v z; tð Þ � v z; tð Þ dt ¼ �: ð14:29Þ

From Equations 14.19 and 14.20, it follows that

1

T

ðT0u z; tð Þ � u z; tð ÞT dt ¼ 1

T

ðT0Dv z; tð Þ � Dv z; tð Þ½ �T dt

� �ð14:30Þ

or

1

T

ðT0u z; tð Þ � u z; tð ÞT dt ¼ D

1

T

ðT0v z; tð Þ � v z; tð ÞT dt

� �D

T ð14:31Þ

and hence

Cuð0Þ ¼ D�DT: ð14:32Þ

Comparison of Equations 14.28 and 14.32 reveals that D¼Z, from which it follows that

U z; tð Þ ¼ UðzÞ þ Z� v z; tð Þ ð14:33Þ

where the histories in v(t) are generated with different sets of random numbers and with variances

�1, �2, . . . , �n, and where �1 to �n are the eigenvalues of the cross-correlation matrix Cu(0)¼ [�ij],

whose elements are defined by Equation 14.20.

Examples of the use of spatially correlated wind histories to determine the dynamic response of

guyed masts are given by Buchholdt et al. (1986), Iannuzzi (1987) and Ashmawy (1991).

14.5. Generation of earthquake historiesIf the response to seismic excitation can be considered to be linear, then the analysis can be under-

taken in the frequency domain and the input excitation for the site under consideration can be

prescribed in the form of response or power spectra as shown in Chapter 13. If, on the other

hand, the structure is likely to behave non-linearly, the analysis should be carried out in the

time domain and the input prescribed in the form of earthquake accelerograms.

Figure 14.3 shows the accelerograms and Figure 14.4 the power spectral density functions for

three different earthquakes. The former are of relatively short duration and amplitude, and


299

Figure 14.3 Accelerograms of main horizontal earthquake components: (a) north–south component

of the San Salvador earthquake, 10 November 1986, duration 9.38 s, peak acceleration 0.69g;

(b) east–west component of the Friuli 1 earthquake, Italy, 6 May 1976, duration 41.5 s, peak

acceleration 0.16g; (c) Love Wave component of the Imperial Valley earthquake, USA, 15 May 1979,

duration 42.1 s, peak acceleration 0.81g

0.00

0.00 3.04 4.06 5.07 6.09 7.10 8.12 9.13 10.14

9.388.44

19.01 22.81 26.61 30.41 34.21 38.020.00

(a)

(b)

0.81

0.67

0.52

0.38

0.24

–0.10

–0.04

–0.18

–0.32

–0.46

–0.60

1.20

0.93

0.65

0.37

0.09

–0.19

–0.46

–0.74

–1.02

–1.30

–1.56

0.54

0.41

0.29

0.17

0.04

–0.08

–0.20

–0.33

–0.45

–0.57

–0.69

(c)

Time: s

Acc

eler

atio

n in

g m

/s2


300

Figure 14.4 Power spectral density functions of accelerograms of earthquake components in

Figure 14.3

0.05 1.04 2.04 3.04 4.03 5.03 6.02 7.02 8.01 9.01 10.00(a)

Pow

er s

pect

rum

× g

2

(b)

(c)

Frequency: Hz

0.11

0.10

0.09

0.08

0.07

0.06

0.05

0.03

0.02

0.01

0.00

0.00 0.60 1.20 1.80 2.40 3.00 3.60 4.20 4.80 5.40 6.00

0.24

0.21

0.19

0.17

0.14

0.12

0.10

0.07

0.05

0.03

0.00

0.03 0.62 1.22 1.82 2.41 3.01 3.61 4.21 4.80 5.40 6.00

0.05

0.05

0.04

0.04

0.03

0.03

0.02

0.02

0.01

0.01

0.00


301

hence the variance varies with time. The latter shows the distribution of the square of the ampli-

tudes of the acceleration histories of the frequency components in the frequency domain and also

(quite clearly) the values of the dominant frequencies of the ground.

A further spectrum analysis of adjacent time regions of each record would also reveal that the

frequency/amplitude contents change during the passages of earthquakes. The reason for this is

that there is a time difference between the arrivals of the P and S waves, and that the ground

tends to filter out some of the higher-frequency components. In order to take the non-stationarity

of earthquake histories into account, the duration of the underlying stochastic process needs to be

divided into separate contiguous time regions, each having a unique time-variable frequency/

amplitude content whose amplitudes can be varied by using a deterministic time envelope or

shaping function �(t).

An acceleration history for the ith time region, with zero mean and variance �2€xx, may be generated

from Equation 14.4. We therefore have

€xxiðtÞ ¼Xp

s¼ 1

�is€xxi t� s�tð Þ þ �N €xxiNðtÞ ð14:34Þ

or

€xxiðtÞ ¼Xp

s¼ 1

�isBis€xxiðtÞ þ �N €xxiNðtÞ ð14:35Þ

where Bis is the backshift operator for the ith time region, defined

Bis€xxiðtÞ ¼ €xxi tþ s�tð Þ: ð14:36Þ

Solving Equation 14.36 with respect to €xxiðtÞ yields

€xxiðtÞ ¼1

1�Pp

s¼ 1 �isBis

� �N€xxiNðtÞ ð14:37Þ

where the parameters �is and the variance �2N€xxi are determined in the same manner as for wind.

Generation of time histories for each time region with different autocovariance functions but

with the same variance �2€xx leads to

c€xx k�tð Þ ¼Xp

s¼ 1

’isci€xx k� sð Þ�t½ � k ¼ 1; 2; . . . ; p ð14:38Þ

�2N €xxi ¼ �2

€xx 1�Xp

s¼ 1

’isxi€xx s�tð Þ" #

ð14:39Þ

where Equation 14.39 can be written in matrix form (see Equation 14.16). The shape function for

the ith region �i(t) can be expressed as

�ðtÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�2i€xxðtÞ=�2€xx

� �qð14:40Þ


302

where �2xiis the variance for the ith time region; hence

€xxiðtÞ ¼ �iðtÞ � €xxtðtÞ: ð14:41Þ

Ashmawy (1991) investigated the above method and found that, when generating earthquakes, it

is better to use 10 � parameters rather than 3–5 (as indicated by Figure 14.1) in the case of wind,

together with a time step of 0.02 s. Using the autocovariance functions of a number of different

earthquakes, he showed that it is possible to generate families of earthquake histories with

power spectra similar to those of the parent spectra.

In the generation and use of single earthquakes and families of earthquakes based on recorded

earthquake histories, we are of the opinion that it is better to generate such histories using

constructed continuous spectra with a given peak acceleration containing the ground frequency

and the main structural frequencies. Although useful when analysing non-linear structures, this

is not the case in the analysis of linear structures.

14.6. Cross-correlation of earthquake historiesThe motion of earthquakes is usually recorded in the form of accelerograms along three mutually

perpendicular axes (two horizontal axes and one vertical axis). In order to generate families of

earthquakes with the same statistical properties as the parent earthquake, it is therefore necessary

to generate three accelerograms not only with similar power spectra, but also with similar cross-

covariance. Because the motion is non-stationary, each of the three recorded accelerograms and

the corresponding underlying stochastic processes needs to be divided into the same number of

separate contiguous time regions with the latter being correlated region by region; this may be

achieved as follows. Let the cross-covariance matrix for the rth time region of a recorded

quake at time lag � ¼ 0 be

C€xx;rð0Þ ¼ �i;j

� �r¼ Zr�rZ

Tr i ¼ 1; 2; 3 ð14:42Þ

where �r¼diag.[�1, �2, �3]r is the eigenvalue matrix and Zr¼ [Z1, Z2, Z3]r is the normalised

eigenvector matrix of C€xx;rð0Þ. If the uncorrelated acceleration histories are denoted by the

vector €xxrðtÞ and the correlated acceleration histories by €��rðtÞ, then it follows from Equation

14.42 that the correlated acceleration during the rth time region is given by

€��rðtÞ ¼ Zr€xxrðtÞ: ð14:43Þ

14.7. Design earthquakesThe number of actual strong earthquake records available is limited; even if they were available, it

is unlikely that they would form a basis for believing that future earthquakes occurring at the

same site would be similar to those previously recorded. There is therefore a need for a method

that enables the simulation of realistic earthquakes with different but defined statistical character-

istics. Ashmawy (1991) found that: (a) it is possible to generate realistic time histories by assuming

rectilinear autocovariance functions as shown in Figure 14.5 and (b) the magnitudes of the

dominant frequencies of the simulated quakes varied with the slope of the assumed auto-

covariance function. He therefore found that when the value of the time lag � increased in

steps from 0.2 to 5 s, the dominant frequency decreased from 3.6 to 0.48 Hz; the frequency spectra

of the simulated histories for � ¼ 0.2 s ranged from 3.6 to 15.5 Hz and for � ¼ 5 s from 0.05 to

0.38 Hz. Ashmawy’s results are summarised in Figures 14.6 and 14.7. Design earthquakes


303

Figure 14.5 Linearised autocovariance function for the simulation of design earthquakes

Time lag τ: s

c(0) = 1

c(k∆t)

Figure 14.6 Relationship between total time lag � of the autocovariance function and the maximum

dominant frequencies of the resulting design earthquakes (from Ashmawy, 1991)

0.01 0.1 1.0 10.0

Max

imum

dom

inan

t fr

eque

ncy:

Hz

Time lag τ: s, log scale

4

3

2

1

0


304

simulated with the same slope of the assumed autocovariance function but with different series of

random numbers will have different power spectra.

The curves in Figure 14.7 should therefore by taken as indicating trends and not exact

relationships. The smooth curves show median values derived from scattered points on a

graph, with the degree of scatter being a function of the underlying series of random numbers

used in generating wind histories.

It is therefore advisable to calculate the response to a family of design earthquakes as opposed to a

single earthquake. Ashmawy studied the validity of design earthquakes (simulated as described

above) by comparing the calculated responses of a 238.6 m tall guyed mast to recorded

earthquakes and to design earthquakes with the same peak acceleration and similar power

spectra, and found that the two responses were very similar.

The worst design scenario for sites where the dominant ground frequency and the cross-

correlation of the three acceleration components are unknown is when, given an assumed peak

acceleration, the dominant frequency of the simulated histories coincides with the first natural

frequency of the structure and the cross-correlations of all three ground acceleration components

are unity. The latter will be the case when these components are generated with the same set of

random numbers.

REFERENCES

Ashmawy MA (1991) Nonlinear dynamic analysis of guyed masts for wind and earthquake

loading. PhD thesis, Polytechnic of Central London.

Figure 14.7 Relationship between the total time lag � of the autocovariance function and the

frequency range of the resulting design earthquakes (from Ashmawy, 1991)

0.01 0.1 1.0 10.0

Freq

uenc

y: H

z


Frequency range

16

14

12

10

8

6

4

2

0


305

Box GEP and Jenkins CM (1977) Time Series Analysis: Forecasting and Control. Holden Day,

San Francisco.


Cambridge.

Buchholdt HA, Moossevinejad S and Iannuzzi A (1986) Non-linear dynamic analysis of guyed

masts subjected to wind and guy ruptures. Proceedings of Institution of Civil Engineers,

Part 2, September, 353–359.

Iannuzzi A (1987) Response of guyed masts to simulated wind. PhD thesis, Polytechnic of

Central London.

Iannuzzi A and Spinelli P (1989) Response of a guyed mast to real and simulated wind. IASS

Bulletin, No. 99, pp 38–45.

Iwatani Y (1982) Simulation of multidimensional wind fluctuations having any arbitrary power

spectra and cross spectra. Journal of Wind Engineering 11: 5–18.

Shinozuka M and Jan CM (1952) Digital simulation of random processes and its applications.

Journal of Aeronautical Science 19(12): 793–800.


306


ISBN: 978-0-7277-4176-9



Chapter 14

Generation of wind and earthquakehistories

14.1. IntroductionA time-domain method was presented in Chapter 6 for predicting the linear and non-linear

response of 1-DOF systems to wind and earthquakes and to multi-DOF systems in general.

The equations developed are based on the incremental equation of motion, and arise from various

assumptions with respect to the change in acceleration during a time step �t. Other time-domain

methods, which are particularly suitable for highly non-linear structures such as guyed masts,

cable and membrane roofs, are those in which

g equilibrium of the dynamic forces at the end of each time step is sought by minimisation of

the gradient vector of the total potential dynamic energy by use of the Newton-Raphson or

conjugate gradient method, andg where increased convergence and stability are achieved through scaling and the calculation

of a step length in the descent direction to a point where the energy is a minimum

(Buchholdt, 1985; Buchholdt et al., 1986).

The prediction of response using any of the above methods requires the ability to generate

earthquake histories and single and spatially correlated wind histories. The problem with using

recorded earthquake histories is that no two earthquakes are the same. For the purpose of

design, it is therefore necessary to calculate the response to a family of simulated earthquakes

compatible with a given site. Because wind histories can be considered as stationary stochastic

processes, they are simpler to generate the earthquake histories; methods for simulating wind

histories are therefore presented first.

14.2. Generation of single wind histories by a Fourier seriesShinozuka and Jan (1952) have shown that it is possible to express the fluctuating velocity

component u(t) of wind at any time t as

uðtÞ ¼ffiffiffiffiffiffiffið2Þ

p Xni¼ 1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisu nið Þ ��nð Þ½ � cos 2�ni þ �ið Þ

pð14:1Þ

where Su(n) is the value of the power spectral density function for the fluctuating component of

wind at the frequency n, �n¼ ni þ 1� ni and �i is the phase angle with a uniform probability

distribution function that varies randomly between 0 and 2�.

The frequency band in Equation 14.1, which has been divided into N parts, must contain all the

significant natural frequencies of the structure. For non-linear structures the frequency step �n

293

needs to be small, as the natural frequencies of such structures vary with the amplitude of

response.

14.3. Generation of wind histories by the autoregressive methodAnother method for generating single wind histories that yields variances of response similar

to those of real wind is the autoregressive (AR) method. It is computationally more efficient

than the Fourier series (FS) method given by Equation 14.1, and can also be used to generate

earthquake histories. The AR method filters white noise and transforms it into a signal with a

specified variance and autocovariance function.

Mathematically, the method for transforming white noise may be expressed as

uðtÞ ¼ �ðBÞ � aðtÞ ð14:2Þ

where u(t) is the stochastic process to be generated, a(t) is the input white noise with zero mean and

variance �2a and �(B) is a transfer function or filter. The white noise a(t) may also be expressed as

aðtÞ ¼ �Nu �NðtÞ ð14:2Þ

where N(t) are random shocks with zero mean and unit variance. Substitution of this expression

for a(t) into Equation 14.2 yields

uðtÞ ¼ �ðBÞ � �Nu �NðtÞ: ð14:3Þ

The white noise process a(t) is transformed into the process u(t) by the filter or transfer function

�(B). One type of filter that has proved to be very suitable for modelling wind and earthquakes is

the so-called autoregressive filter, which regressively weights and sums previous values.

In an autoregressively simulated process of order p, the instantaneous values of u(t) are expressed

as a finite linear aggregate of the previous values of u(t) plus a random impulse with zero mean and

variance �2Nu. The expression for u(t) may therefore be written as

uðtÞ ¼Xp

s¼ 1

�su t� s�tð Þ þ �NuNðtÞ ð14:4Þ

where � is an autoregressive parameter, N(t) is a random impulse with zero mean and unit

variance and

�2Nu ¼

1

T

ðT0

�NuNðtÞ � uðtÞdt: ð14:5Þ

Alternatively, Equation 14.4 may be written as

uðtÞ ¼Xp

s¼ 1

�sBsuðtÞ þ �NuNðtÞ ð14:6Þ

where Bs is a backshift operator which is defined

BsuðtÞ ¼ u tþ s�tð Þ: ð14:7Þ


294

Solving Equation 14.6 with respect to u(t) yields

uðtÞ ¼ 1

1�Xp

s¼ 1

’sBs

� �NuNðtÞ: ð14:8Þ

Comparing Equation 14.8 to Equation 14.3 yields the following expression for an autoregressive

filter of order p

�ðBÞ ¼ 1

Xp

s¼ 1

’sBs

: ð14:9Þ

In order to obtain expressions for determining the values for the parameters � and the variance

�2Nu, both sides of Equation 14.4 are multiplied by u(t� kt) where k¼ 1, 2, . . . , p. Integration

and averaging over time T yields

1

T

ðT0uðtÞ � u t� ktð Þ dt ¼

Xp

s¼ 1

1

T

ðT0’su t� s�tð Þ u t� k�tð Þdtþ 1

T

ðT0�NuNðtÞ u t� k�tð Þ dt: ð14:10Þ

When k> 0, Equation 14.10 yields

Cu f�tð Þ ¼ Cu �k�tð Þ ¼Xp

s¼ 1

’sCu k� sð Þ�t½ � k ¼ 1; 2; . . . ; p ð14:11Þ

because of the symmetry of the autocovariance function and the randomness of the process N(t).

When k¼ 0, Equation 14.10 yields

�2u ¼

Xp

s¼ 1

’sCu s�tð Þ þ �2Nu: ð14:12Þ

Division of all the elements in Equations 14.11 and 14.12 by �2u yields

cu k�tð Þ ¼Xp

s¼ 1

�scu k� sð Þ�t½ � k ¼ 1; 2; . . . p ð14:13Þ

�2Nu ¼ �2u 1�

Xp

s¼ 1

’scu s�tð Þ" #

ð14:14Þ

where cu(k�t) is the autocovariance coefficient at time lag � ¼ k�t corresponding to the power

spectral density functions Su(n) and cu[(k� s)�t]¼ 1.0 when k¼ s.

Given an expression for the power spectrum Su(n), the values of the autocovariance coefficient cuare determined by dividing both sides of Equation 10.27 by �2

n and completing the integration:

cu k�tð Þ ¼ 1

�2u

ð10

SuðnÞ cos 2� nk�tð Þ dn: ð14:15Þ

Having determined the values of the autocovariance function using Equation 14.11, the

autoregressive parameters � can be determined by use of Equation 14.13 and the variances of

the impulses �2Nu from Equation 14.14.


295

Unlike the Fourier series model the autoregressive model is not unconditionally stationary, and

tends to become non-stationary when a short time step is chosen. In this case, the right-hand

side of Equation 14.11 may become negative. Another problem is concerned with the number

of parameters � to be used in the generation of the autoregressive model. Both problems have

been dealt with by Box and Jenkins (1977) in terms of the so-called partial autocorrelation

function which, for a process suitable for simulation by an autoregressive method of order q, is

nearly zero when p> q.

Figure 14.1 shows the variation of the partial autocorrelation function with the order of the model

and with the size of the time step. The curves are based on the Kaimal spectrum (Equation 10.34),

with V(10)¼ 30m/s and z0¼ 0.1 m. They indicate that, for the data used, a suitable number of �

Figure 14.1 Variation of the partial autocorrelation model with number of parameters � and size of

time step

5 2

2 4 6 8 10 12 14

1 1

1

0.5

0.2

0.1

0.05

Part

ial a

utoc

orre

latio

n fu

nctio

n (P

AC

F)

0.8

1.0

0.8

0.6

0.4

0.2

–0.2

–0.4

–0.6

∆t

∆t = 0.05 s


296

parameters is 3–5 and that the size of the time step should not be less than 0.1 s. If p¼ 3 and

�t¼ 0.1 s, from Equations 14.13 and 14.14 we have

’1

’2

’3

264

375 ¼

cu 0:0ð Þ cu 0:1ð Þ cu 0:2ð Þcu 0:1ð Þ cu 0:0ð Þ cu 0:2ð Þcu 0:2ð Þ cu 0:1ð Þ cu 0:0ð Þ

264

375�1 cu 0:1ð Þ

cu 0:2ð Þcu 0:3ð Þ

264

375 ð14:16Þ

�2Nu ¼ �2u 1� ’1cu 0:1ð Þ � ’2cu 0:2ð Þ � ’3cu 0:3ð Þ½ �: ð14:17Þ

It should be noted that a time step of�t¼ 0.1 s is of the order of ten times the size of the time step

used in the forward integration method when the acceleration is assumed to remain constant

during the time step. It is therefore necessary to interpolate to obtain the wind velocities required

at any time t during the dynamic analysis.

As the process u(t) is generated with � parameters that are functions of the autocovariance

coefficients, the simulated histories need to be multiplied by the ratio �u/�gu where �u is the

standard deviation of the required history and �gu is the standard deviation of the generated

process.

Of the two methods for generating wind histories, the FS method is more expensive in terms of

computer time. This is shown in Figure 14.2, where the time taken to generate wind histories

by the two methods is compared. An example of an FS model and AR models generated with

the same power spectrum and autocovariance functions as those of a recorded history is given

by Buchholdt et al. (1986).

14.4. Generation of spatially correlated wind historiesIn Chapter 10, the correlation between the fluctuating wind velocities at two points in space is

expressed in terms of the cross-covariance function (Equation 10.37) and the cross-spectrum

Figure 14.2 Comparison of computer time required to generate single FS and AR wind models: for the

latter the number of frequency steps N is used only to compute the autocovariance coefficients cu(k�t)

of Equation 14.15

0 400 800 1200 1600 2000Number of frequency steps N

Com

pute

r tim

e: s

AR(6)

AR(12)

60

40

20

0

FS


297

(Equation 10.41), the latter being expressed as a function of the square root of the product of the

power spectra of the individual histories and the coherence function (Equation 10.42). In Chapter

11, it is shown how power spectra and cross-spectra are used to establish model force spectra for

calculating the variance of response of multi-DOF structures in the frequency domain (Equations

11.52–11.59). The interdependence of the velocity fluctuations in space must also be included

when generating spatial wind fields. This can be achieved in different ways. Spinelli devised a

method with correlation at time lag � ¼ 0 (Buchholdt, 1985; Iannuzzi, 1987; Iannuzzi and Spinelli,

1989) and Iwatani (1982) devised a method with correlation at � 5 0. A third method is now

presented based on an eigenvalue analysis of the cross-covariance matrix Cuv(0) at � ¼ 0.

In Chapter 10, it is shown that the wind velocity vector at any time tmay be considered to consist

of a steady-state componentU(z), whose element can be determined using either Equation 10.4 or

Equation 10.10 and a fluctuating component u(z, t). We therefore have

U z; tð Þ ¼ UðzÞ þ u z; tð Þ: ð14:18Þ

Further, the vector u(z, t) may be expressed as D� v(z, t) where D is a correlation matrix whose

elements are evaluated from the cross-covariance of the elements in u(z, t) at � ¼ 0 and v(z, t) is a

fluctuating velocity vector in which the elemental time histories are uncorrelated and can be

modelled as either FS or AR series. Equation 14.11 may therefore be written as

U z; tð Þ ¼ UðzÞ þD� v z; tð Þ: ð14:19Þ

The elements in matrix D are determined as follows. From Equations 10.37 and 10.38, it is shown

that the cross-covariance matrix at time lag � ¼ 0 is

Cu z;tð Þ;u z;tð Þð0Þ � Cuð0Þ ¼1

T

ðT0u z; tð Þ � u z; tð ÞT dt ¼ �ij

� �ð14:20Þ

where [�ij] is a square symmetrical matrix in which the elements on the leading diagonal �ij¼�i2

are the variances and the off-diagonal elements are the cross-variances of the elemental processes in

u(z, t). The elements on the leading diagonal may therefore be calculated by use of Equation 10.28,

and the off-diagonal elements by use of Equation 10.40. To proceed, the eigenvalues and the

normalised eigenvectors ofCu(0) must be determined. Let the corresponding eigenvalue equation be

Cuð0ÞX ¼ �IX ð14:21Þ

where �¼ diag.{�1, �2, . . . , �n} is the eigenvalue matrix andX¼ [X1,X2, . . . ,Xn] is the eigenvector

matrix. In order to normalise an eigenvector Xi, let

XTi IXi ¼ L2

i : ð14:22Þ

The normalised eigenvector Zi is now found by dividing each element in Xi by Li, and therefore

Zi ¼ Xi=Li ð14:23Þ

ZTi IZi ¼ 1: ð14:24Þ

Writing the eigenvalue equation (Equation 14.21) in terms of the eigenvalue �i and the normalised

eigenvector Zi and post-multiplication of each term by ZiT yields

ZTi Cuð0ÞZi ¼ �iZ

Ti IZi ¼ �i ð14:25Þ


298

and hence

ZTCuð0ÞZ ¼ � ð14:26Þ

Cuð0Þ ¼ Z�T�Z�1: ð14:27Þ

Since Cu(0) is a symmetric positive definite matrix, Z1¼ZT and ZT¼Z and hence

Cuð0Þ ¼ Z�ZT: ð14:28Þ

If the uncorrelated wind histories in v(t) in Equation 14.19 are generated with power spectra

having p7 variances �1, �2, . . . , �n, then

Cv z;yð Þv z;tð Þð0Þ ¼ Cvð0Þ ¼1

T

ðT0v z; tð Þ � v z; tð Þ dt ¼ �: ð14:29Þ

From Equations 14.19 and 14.20, it follows that

1

T

ðT0u z; tð Þ � u z; tð ÞT dt ¼ 1

T

ðT0Dv z; tð Þ � Dv z; tð Þ½ �T dt

� �ð14:30Þ

or

1

T

ðT0u z; tð Þ � u z; tð ÞT dt ¼ D

1

T

ðT0v z; tð Þ � v z; tð ÞT dt

� �D

T ð14:31Þ

and hence

Cuð0Þ ¼ D�DT: ð14:32Þ

Comparison of Equations 14.28 and 14.32 reveals that D¼Z, from which it follows that

U z; tð Þ ¼ UðzÞ þ Z� v z; tð Þ ð14:33Þ

where the histories in v(t) are generated with different sets of random numbers and with variances

�1, �2, . . . , �n, and where �1 to �n are the eigenvalues of the cross-correlation matrix Cu(0)¼ [�ij],

whose elements are defined by Equation 14.20.

Examples of the use of spatially correlated wind histories to determine the dynamic response of

guyed masts are given by Buchholdt et al. (1986), Iannuzzi (1987) and Ashmawy (1991).

14.5. Generation of earthquake historiesIf the response to seismic excitation can be considered to be linear, then the analysis can be under-

taken in the frequency domain and the input excitation for the site under consideration can be

prescribed in the form of response or power spectra as shown in Chapter 13. If, on the other

hand, the structure is likely to behave non-linearly, the analysis should be carried out in the

time domain and the input prescribed in the form of earthquake accelerograms.

Figure 14.3 shows the accelerograms and Figure 14.4 the power spectral density functions for

three different earthquakes. The former are of relatively short duration and amplitude, and


299

Figure 14.3 Accelerograms of main horizontal earthquake components: (a) north–south component

of the San Salvador earthquake, 10 November 1986, duration 9.38 s, peak acceleration 0.69g;

(b) east–west component of the Friuli 1 earthquake, Italy, 6 May 1976, duration 41.5 s, peak

acceleration 0.16g; (c) Love Wave component of the Imperial Valley earthquake, USA, 15 May 1979,

duration 42.1 s, peak acceleration 0.81g

0.00

0.00 3.04 4.06 5.07 6.09 7.10 8.12 9.13 10.14

9.388.44

19.01 22.81 26.61 30.41 34.21 38.020.00

(a)

(b)

0.81

0.67

0.52

0.38

0.24

–0.10

–0.04

–0.18

–0.32

–0.46

–0.60

1.20

0.93

0.65

0.37

0.09

–0.19

–0.46

–0.74

–1.02

–1.30

–1.56

0.54

0.41

0.29

0.17

0.04

–0.08

–0.20

–0.33

–0.45

–0.57

–0.69

(c)

Time: s

Acc

eler

atio

n in

g m

/s2


300

Figure 14.4 Power spectral density functions of accelerograms of earthquake components in

Figure 14.3

0.05 1.04 2.04 3.04 4.03 5.03 6.02 7.02 8.01 9.01 10.00(a)

Pow

er s

pect

rum

× g

2

(b)

(c)

Frequency: Hz

0.11

0.10

0.09

0.08

0.07

0.06

0.05

0.03

0.02

0.01

0.00

0.00 0.60 1.20 1.80 2.40 3.00 3.60 4.20 4.80 5.40 6.00

0.24

0.21

0.19

0.17

0.14

0.12

0.10

0.07

0.05

0.03

0.00

0.03 0.62 1.22 1.82 2.41 3.01 3.61 4.21 4.80 5.40 6.00

0.05

0.05

0.04

0.04

0.03

0.03

0.02

0.02

0.01

0.01

0.00


301

hence the variance varies with time. The latter shows the distribution of the square of the ampli-

tudes of the acceleration histories of the frequency components in the frequency domain and also

(quite clearly) the values of the dominant frequencies of the ground.

A further spectrum analysis of adjacent time regions of each record would also reveal that the

frequency/amplitude contents change during the passages of earthquakes. The reason for this is

that there is a time difference between the arrivals of the P and S waves, and that the ground

tends to filter out some of the higher-frequency components. In order to take the non-stationarity

of earthquake histories into account, the duration of the underlying stochastic process needs to be

divided into separate contiguous time regions, each having a unique time-variable frequency/

amplitude content whose amplitudes can be varied by using a deterministic time envelope or

shaping function �(t).

An acceleration history for the ith time region, with zero mean and variance �2€xx, may be generated

from Equation 14.4. We therefore have

€xxiðtÞ ¼Xp

s¼ 1

�is€xxi t� s�tð Þ þ �N €xxiNðtÞ ð14:34Þ

or

€xxiðtÞ ¼Xp

s¼ 1

�isBis€xxiðtÞ þ �N €xxiNðtÞ ð14:35Þ

where Bis is the backshift operator for the ith time region, defined

Bis€xxiðtÞ ¼ €xxi tþ s�tð Þ: ð14:36Þ

Solving Equation 14.36 with respect to €xxiðtÞ yields

€xxiðtÞ ¼1

1�Pp

s¼ 1 �isBis

� �N€xxiNðtÞ ð14:37Þ

where the parameters �is and the variance �2N€xxi are determined in the same manner as for wind.

Generation of time histories for each time region with different autocovariance functions but

with the same variance �2€xx leads to

c€xx k�tð Þ ¼Xp

s¼ 1

’isci€xx k� sð Þ�t½ � k ¼ 1; 2; . . . ; p ð14:38Þ

�2N €xxi ¼ �2

€xx 1�Xp

s¼ 1

’isxi€xx s�tð Þ" #

ð14:39Þ

where Equation 14.39 can be written in matrix form (see Equation 14.16). The shape function for

the ith region �i(t) can be expressed as

�ðtÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�2i€xxðtÞ=�2€xx

� �qð14:40Þ


302

where �2xiis the variance for the ith time region; hence

€xxiðtÞ ¼ �iðtÞ � €xxtðtÞ: ð14:41Þ

Ashmawy (1991) investigated the above method and found that, when generating earthquakes, it

is better to use 10 � parameters rather than 3–5 (as indicated by Figure 14.1) in the case of wind,

together with a time step of 0.02 s. Using the autocovariance functions of a number of different

earthquakes, he showed that it is possible to generate families of earthquake histories with

power spectra similar to those of the parent spectra.

In the generation and use of single earthquakes and families of earthquakes based on recorded

earthquake histories, we are of the opinion that it is better to generate such histories using

constructed continuous spectra with a given peak acceleration containing the ground frequency

and the main structural frequencies. Although useful when analysing non-linear structures, this

is not the case in the analysis of linear structures.

14.6. Cross-correlation of earthquake historiesThe motion of earthquakes is usually recorded in the form of accelerograms along three mutually

perpendicular axes (two horizontal axes and one vertical axis). In order to generate families of

earthquakes with the same statistical properties as the parent earthquake, it is therefore necessary

to generate three accelerograms not only with similar power spectra, but also with similar cross-

covariance. Because the motion is non-stationary, each of the three recorded accelerograms and

the corresponding underlying stochastic processes needs to be divided into the same number of

separate contiguous time regions with the latter being correlated region by region; this may be

achieved as follows. Let the cross-covariance matrix for the rth time region of a recorded

quake at time lag � ¼ 0 be

C€xx;rð0Þ ¼ �i;j

� �r¼ Zr�rZ

Tr i ¼ 1; 2; 3 ð14:42Þ

where �r¼diag.[�1, �2, �3]r is the eigenvalue matrix and Zr¼ [Z1, Z2, Z3]r is the normalised

eigenvector matrix of C€xx;rð0Þ. If the uncorrelated acceleration histories are denoted by the

vector €xxrðtÞ and the correlated acceleration histories by €��rðtÞ, then it follows from Equation

14.42 that the correlated acceleration during the rth time region is given by

€��rðtÞ ¼ Zr€xxrðtÞ: ð14:43Þ

14.7. Design earthquakesThe number of actual strong earthquake records available is limited; even if they were available, it

is unlikely that they would form a basis for believing that future earthquakes occurring at the

same site would be similar to those previously recorded. There is therefore a need for a method

that enables the simulation of realistic earthquakes with different but defined statistical character-

istics. Ashmawy (1991) found that: (a) it is possible to generate realistic time histories by assuming

rectilinear autocovariance functions as shown in Figure 14.5 and (b) the magnitudes of the

dominant frequencies of the simulated quakes varied with the slope of the assumed auto-

covariance function. He therefore found that when the value of the time lag � increased in

steps from 0.2 to 5 s, the dominant frequency decreased from 3.6 to 0.48 Hz; the frequency spectra

of the simulated histories for � ¼ 0.2 s ranged from 3.6 to 15.5 Hz and for � ¼ 5 s from 0.05 to

0.38 Hz. Ashmawy’s results are summarised in Figures 14.6 and 14.7. Design earthquakes


303

Figure 14.5 Linearised autocovariance function for the simulation of design earthquakes

Time lag τ: s

c(0) = 1

c(k∆t)

Figure 14.6 Relationship between total time lag � of the autocovariance function and the maximum

dominant frequencies of the resulting design earthquakes (from Ashmawy, 1991)

0.01 0.1 1.0 10.0

Max

imum

dom

inan

t fr

eque

ncy:

Hz


4

3

2

1

0


304

simulated with the same slope of the assumed autocovariance function but with different series of

random numbers will have different power spectra.

The curves in Figure 14.7 should therefore by taken as indicating trends and not exact

relationships. The smooth curves show median values derived from scattered points on a

graph, with the degree of scatter being a function of the underlying series of random numbers

used in generating wind histories.

It is therefore advisable to calculate the response to a family of design earthquakes as opposed to a

single earthquake. Ashmawy studied the validity of design earthquakes (simulated as described

above) by comparing the calculated responses of a 238.6 m tall guyed mast to recorded

earthquakes and to design earthquakes with the same peak acceleration and similar power

spectra, and found that the two responses were very similar.

The worst design scenario for sites where the dominant ground frequency and the cross-

correlation of the three acceleration components are unknown is when, given an assumed peak

acceleration, the dominant frequency of the simulated histories coincides with the first natural

frequency of the structure and the cross-correlations of all three ground acceleration components

are unity. The latter will be the case when these components are generated with the same set of

random numbers.

REFERENCES

Ashmawy MA (1991) Nonlinear dynamic analysis of guyed masts for wind and earthquake

loading. PhD thesis, Polytechnic of Central London.

Figure 14.7 Relationship between the total time lag � of the autocovariance function and the

frequency range of the resulting design earthquakes (from Ashmawy, 1991)

0.01 0.1 1.0 10.0

Freq

uenc

y: H

z


Frequency range

16

14

12

10

8

6

4

2

0


305

Box GEP and Jenkins CM (1977) Time Series Analysis: Forecasting and Control. Holden Day,

San Francisco.


Cambridge.

Buchholdt HA, Moossevinejad S and Iannuzzi A (1986) Non-linear dynamic analysis of guyed

masts subjected to wind and guy ruptures. Proceedings of Institution of Civil Engineers,

Part 2, September, 353–359.

Iannuzzi A (1987) Response of guyed masts to simulated wind. PhD thesis, Polytechnic of

Central London.

Iannuzzi A and Spinelli P (1989) Response of a guyed mast to real and simulated wind. IASS

Bulletin, No. 99, pp 38–45.

Iwatani Y (1982) Simulation of multidimensional wind fluctuations having any arbitrary power

spectra and cross spectra. Journal of Wind Engineering 11: 5–18.

Shinozuka M and Jan CM (1952) Digital simulation of random processes and its applications.

Journal of Aeronautical Science 19(12): 793–800.


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Documents

Structural Dynamics for Engineers, 2e