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Structural Dynamics for EngineersSecond edition
Structural Dynamicsfor EngineersSecond edition
H.A. Buchholdt and S.E. Moossavi Nejad
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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.
Structural Dynamics for Engineers, 2nd edition
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
Causes and effects of structural vibration
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
Structural Dynamics for Engineers, 2nd edition
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
Causes and effects of structural vibration
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
Structural Dynamics for Engineers, 2nd edition
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
Causes and effects of structural vibration
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.
Structural Dynamics for Engineers, 2nd edition
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
Causes and effects of structural vibration
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
Structural Dynamics for Engineers, 2nd edition
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
Causes and effects of structural vibration
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.
Structural Dynamics for Engineers, 2nd edition
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.
Causes and effects of structural vibration
13
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.015
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
Structural Dynamics for Engineers, 2nd edition
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
Equivalent one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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
Equivalent one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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
Equivalent one degree-of-freedom systems
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
The weight of the equivalent lumped mass is therefore given by
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
The equivalent elastic spring stiffness is given by
KE ¼ðL0EI ’00 xð Þ� �2
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
Structural Dynamics for Engineers, 2nd edition
22
The equivalent geometrical spring stiffness is given by
KG ¼ðL0T ’0 xð Þ� �2
dx ¼ðL0T 3=2L3� �2
2Lx� x2� �2
dx ð2:52Þ
Hence
KG ¼ 6T=5L ð2:53Þ
The critical value for the axial force occurs when
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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
Equivalent one degree-of-freedom systems
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
The weight of the equivalent lumped mass is therefore given by
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
The equivalent elastic spring stiffness is given by
KE ¼ðL0EI ’00 xð Þ� �2
dx ¼ðL0EI 16=5L4� �2
12x2 � 12Lx� �2
dx ð2:65Þ
and hence
KE ¼ 6144EI=125L3: ð2:66Þ
The equivalent geometrical spring stiffness is given by
KG ¼ðL0T ’0 xð Þ� �2
dx ¼ðL0T 16=5L4� �2
4x3 � 6Lx2 þ L3� �2
dx ð2:67Þ
and hence
KG ¼ 4353T=875L: ð2:68Þ
Structural Dynamics for Engineers, 2nd edition
24
The critical value for the axial force occurs when
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�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKEg
W
� s¼ 1
2�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi6144EIg=L3
3968wL=7875
� sð2:71Þ
or
f ¼ 1:571919
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
q: ð2:72Þ
When the correct mode shape is used, the expression for the natural frequency is
1:5707963
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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
Equivalent one degree-of-freedom systems
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
The weight of the equivalent lumped mass is therefore given by
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
The equivalent elastic spring stiffness is given by
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
The equivalent geometrical spring stiffness is given by
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
The critical value for the axial force occurs when
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.
Structural Dynamics for Engineers, 2nd edition
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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
Equivalent one degree-of-freedom systems
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
The equivalent elastic spring stiffness is given by
KE ¼ðL0EI ’00 xð Þ� �2
dx ¼ðL0EI 32=L4� �2
6x2 � 6Lxþ L2� �2
dx ð2:96Þ
and hence
KG ¼ 1024EI=5L3: ð2:97Þ
The equivalent geometrical spring stiffness is given by
KG ¼ðL0T ’0 xð Þ� �2
dx ¼ðL0T 32=L4� �2
2x3 � 3Lx2 þ L2x� �2
dx ð2:98Þ
and hence
KG ¼ 512T=105L: ð2:99Þ
The critical value for the axial force occurs when
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�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKEg
W
� s¼ 1
2�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1024EIg=5L3
128wL=315
� sð2:102Þ
or
f ¼ 3:5730196ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
q: ð2:103Þ
Structural Dynamics for Engineers, 2nd edition
28
Using the correct mode shape,
f ¼ 3:5608213
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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
Equivalent one degree-of-freedom systems
29
The weight of the equivalent lumped mass is therefore given by
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
The equivalent elastic spring stiffness is given by
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
The equivalent geometrical spring stiffness is given by
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
The critical value for the axial force occurs when
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�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKEg
W
� s¼ 1
2�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi192EIg=L2
13wL=35
� sð2:118Þ
or
f ¼ 3:6185376ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
q: ð2:119Þ
The error resulting from this form of mode shape is therefore 1.62%.
Structural Dynamics for Engineers, 2nd edition
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
Equivalent one degree-of-freedom systems
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
The weight of the equivalent lumped mass is therefore given by
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
KE ¼ðL0EI ’00 xð Þ� �2
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
KG ¼ðL0T ’0 xð Þ� �2
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�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKEg
W
� s¼ 1
2�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi106:53028EIg=L3
0:4462246wL
� sð2:134Þ
or
f ¼ 2:4591211
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
q: ð2:135Þ
Structural Dynamics for Engineers, 2nd edition
32
Use of the correct mode shape yields
f ¼ 2:4509861ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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
The weight of the equivalent lumped mass is therefore given by
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
The equivalent elastic spring stiffness is given by
KE ¼ðL0EI ’00 xð Þ� �2
dx ¼ðL0EI 24=L4� �
�3Lx3 þ 4x3� �2
dx ð2:142Þ
and hence
KE ¼ 576EI=5L3: ð2:143Þ
The equivalent geometrical spring stiffness is given by
KG ¼ðL0T ’0 xð Þ� �2
dx ¼ðL0T 4=L4� �2
L3 � 9Lx2 þ 8x3� �2
dx ð2:144Þ
and hence
KG ¼ 192T=35L: ð2:145Þ
Equivalent one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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
The equivalent geometrical spring stiffness is given by
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
Equivalent one degree-of-freedom systems
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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
Structural Dynamics for Engineers, 2nd edition
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
The equivalent elastic spring stiffness is given by
KE ¼ðL0EI ’00 xð Þ� �2
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
KG ¼ðL0T ’0 xð Þ� �2
dx ¼ðL0T 6=L3� �2
Lx� x2� �2
dx ð2:173Þ
or
KG ¼ 6T=5L: ð2:174Þ
The critical value for the axial force occurs when
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
Equivalent one degree-of-freedom systems
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
The weight of the equivalent lumped mass is therefore given by
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
KE ¼ðL0EI ’00 xð Þ� �2
dx ¼ðL0EI 3=2L3� �2 �2xð Þ2 dx ð2:182Þ
or
KE ¼ 3EI=L3 ð2:183Þ
and the equivalent geometrical stiffness by
KG ¼ðL0T ’0 xð Þ� �2
dx ¼ðL0T 3=2L3� �2
L2 � x2� �2
dx ð2:184Þ
or
KG ¼ 6T=5L: ð2:185Þ
The critical value for the axial force occurs when
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
Structural Dynamics for Engineers, 2nd edition
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
Equivalent one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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.
Equivalent one degree-of-freedom systems
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
q:
The correct value is
f2 ¼ 6:4378174
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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:
Structural Dynamics for Engineers, 2nd edition
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
q:
The correct value is
f ¼ 0:613798
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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:
Equivalent one degree-of-freedom systems
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
q:
Structural Dynamics for Engineers, 2nd edition
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
Equivalent one degree-of-freedom systems
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wL4ð Þ
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:
Structural Dynamics for Engineers, 2nd edition
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:
Equivalent one degree-of-freedom systems
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:
The correct value is
f ¼ 0:0143625ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wð Þ
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
Structural Dynamics for Engineers, 2nd edition
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
f ¼ 0:0137832ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg=wð Þ
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
Equivalent one degree-of-freedom systems
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:
Structural Dynamics for Engineers, 2nd edition
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
Equivalent one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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 �
Equivalent one degree-of-freedom systems
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�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi411:30368
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.
Structural Dynamics for Engineers, 2nd edition
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:
Equivalent one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4766:938
28:149792
� s¼ 2:0711 Hz:
Percentage error ¼ 2:711� 1:9490
1:9490¼ þ6:265%
Equivalent one degree-of-freedom systems
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.
Structural Dynamics for Engineers, 2nd edition
58
REFERENCES
Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.
FURTHER READING
Buchholdt HA (1985) Introduction to Cable Roof Structures. Cambridge University Press,
Cambridge.
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
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
Equivalent one degree-of-freedom systems
59
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.061
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
∆
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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
Free vibration of one degree-of-freedom systems
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q
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
Structural Dynamics for Engineers, 2nd edition
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.
Free vibration of one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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%:
Free vibration of one degree-of-freedom systems
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:
Structural Dynamics for Engineers, 2nd edition
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�.
Free vibration of one degree-of-freedom systems
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:
Structural Dynamics for Engineers, 2nd edition
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:
Free vibration of one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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:
Free vibration of one degree-of-freedom systems
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.
Structural Dynamics for Engineers, 2nd edition
76
FURTHER READING
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
Craig RR Jr (1981) Structural Dynamics. Wiley, Chichester.
Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.
Irvine HM (1986) Structural Dynamics for the Practising Engineer. Allen & Unwin, London.
Paz M (1980) Structural Dynamics. Van Nostrand Reinhold, New York.
Stroud KA (1970) Engineering Mathematics. Macmillan, London.
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.
Free vibration of one degree-of-freedom systems
77
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.079
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
From Newton’s law of motion,
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q
: 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
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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
Forced harmonic vibration of one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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
Forced harmonic vibration of one degree-of-freedom systems
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:
Structural Dynamics for Engineers, 2nd edition
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�:
Forced harmonic vibration of one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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
Forced harmonic vibration of one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½1þ ð2�rÞ2�
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 2�rð Þ2� �q
2�¼
62:993982ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 2� 0:1� 1:0ð Þ2� �q
2� 0:1
¼ 321:20754 N:
Forced harmonic vibration of one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
92
where
F0 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½Cxg0!gÞ2 þ ðKxg0Þ2�
q¼ Kxg0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½1þ ð2�rÞ2�
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 2�rð Þ2� �q
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 2�rð Þ2� �q
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
Forced harmonic vibration of one degree-of-freedom systems
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.
Structural Dynamics for Engineers, 2nd edition
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
Forced harmonic vibration of one degree-of-freedom systems
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
ffiffiffiffiffiffiffiffiffiffiffiffiK
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:
Structural Dynamics for Engineers, 2nd edition
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
Forced harmonic vibration of one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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
Forced harmonic vibration of one degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
100
FURTHER READING
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
Craig Jr RR (1981) Structural Dynamics. Wiley, Chichester.
Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.
Irvine HM (1986) Structural Dynamics for the Practising Engineer. Allen & Unwin, London.
Paz M (1980) Structural Dynamics. Van Nostrand Reinhold, New York.
Stroud KA (1970) Engineering Mathematics. Macmillan, London.
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.
Forced harmonic vibration of one degree-of-freedom systems
101
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.103
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)
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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
Evaluation of equivalent viscous damping coefficients by harmonic excitation
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�
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� �2� �q
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
Structural Dynamics for Engineers, 2nd edition
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:
Evaluation of equivalent viscous damping coefficients by harmonic excitation
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
Structural Dynamics for Engineers, 2nd edition
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
Evaluation of equivalent viscous damping coefficients by harmonic excitation
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
Structural Dynamics for Engineers, 2nd edition
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
Evaluation of equivalent viscous damping coefficients by harmonic excitation
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.
Structural Dynamics for Engineers, 2nd edition
114
FURTHER READING
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.
Stroud KA (1970) Engineering Mathematics. Macmillan, London.
Evaluation of equivalent viscous damping coefficients by harmonic excitation
115
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.117
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)
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
120
xþ�x ¼ 16�€xx�t3 þ 1
2€xx�t2 þ _xx�tþ x ð6:10Þ
from which we obtain
�€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).
Response of linear and non-linear 1-DOF systems to random loading
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
Structural Dynamics for Engineers, 2nd edition
122
The acceleration at time (tþ�t) is therefore given by
€xx tþ�tð Þ ¼ 6
�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
_xx tþ�tð Þ ¼ 3
�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
Response of linear and non-linear 1-DOF systems to random loading
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
from which we obtain
�€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
Structural Dynamics for Engineers, 2nd edition
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
_xx tþ�tð Þ ¼ 2
�t�x� _xx ð6:42Þ
€xx tþ�tð Þ ¼ 4
�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
Response of linear and non-linear 1-DOF systems to random loading
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
Structural Dynamics for Engineers, 2nd edition
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
Response of linear and non-linear 1-DOF systems to random loading
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
Structural Dynamics for Engineers, 2nd edition
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
Response of linear and non-linear 1-DOF systems to random loading
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:
Structural Dynamics for Engineers, 2nd edition
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
Response of linear and non-linear 1-DOF systems to random loading
131
x tþ�tð Þ ¼ xþ�x
_xx tþ�tð Þ ¼ 2
�t�x� _xx ¼ 2
0:01�x� _xx ¼ 200�x� _xx
€xx tþ�tð Þ ¼ 4
�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
Structural Dynamics for Engineers, 2nd edition
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
Response of linear and non-linear 1-DOF systems to random loading
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.
Structural Dynamics for Engineers, 2nd edition
134
REFERENCES
Coates RC, Coutie MG and Kong FK (1972) Structural Analysis. Nelson, London.
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
FURTHER READING
Buchholdt HA (1985) Introduction to Cable Roof Structures. Cambridge University Press,
Cambridge.
Wood WL (1990) Practical Time-stepping Schemes. Oxford Applied Mathematics, Oxford.
Response of linear and non-linear 1-DOF systems to random loading
135
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.137
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
Structural Dynamics for Engineers, 2nd edition
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
ffiffiffiffiffiffiffiffiffiffiffiffiK
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)
Structural Dynamics for Engineers, 2nd edition
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:
Free vibration of multi-degree-of-freedom systems
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.
Structural Dynamics for Engineers, 2nd edition
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�
Free vibration of multi-degree-of-freedom systems
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg
w
� �s
!22 ¼ 41:1150� 10�3 EIg
w!2 ¼ 0:2027683
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg
w
� �s
f2 ¼ 0:0322715
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg
w
� �s
!23 ¼ 78:8793� 10�3 EIg
w!3 ¼ 0:2808547
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg
w
� �s
f3 ¼ 0:0446994
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg
w
� �s:
Structural Dynamics for Engineers, 2nd edition
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.
Free vibration of multi-degree-of-freedom systems
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.
Structural Dynamics for Engineers, 2nd edition
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
Free vibration of multi-degree-of-freedom systems
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:
Structural Dynamics for Engineers, 2nd edition
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
5th iteration: �3X1 ¼K
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
6th iteration: �3X1 ¼K
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
Free vibration of multi-degree-of-freedom systems
149
7th iteration: �3X1 ¼K
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
8th iteration: �3X1 ¼K
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
9th iteration: �3X1 ¼K
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
10th iteration: �3X1 ¼K
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
11th iteration: �3X1 ¼K
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
12th iteration: �3X1 ¼K
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
13th iteration: �3X1 ¼K
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
14th iteration: �3X1 ¼K
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
15th iteration: �3X1 ¼K
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:
Structural Dynamics for Engineers, 2nd edition
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg
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.
Free vibration of multi-degree-of-freedom systems
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.
Structural Dynamics for Engineers, 2nd edition
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEIg
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.
Free vibration of multi-degree-of-freedom systems
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
" #
Structural Dynamics for Engineers, 2nd edition
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
Free vibration of multi-degree-of-freedom systems
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
� �:
Structural Dynamics for Engineers, 2nd edition
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
Free vibration of multi-degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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=
ffiffiffiffiffiffiffi~MM2
q
Z3 ¼ X3=
ffiffiffiffiffiffiffi~MM3
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:
Free vibration of multi-degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
160
REFERENCES
Buchholdt HA (1988) Introduction to Cable Roof Structures, 2nd edn. Thomas Telford,
London.
FURTHER READING
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
Coates RC, Coutie MG and Kong FK (1972) Structural Analysis. Nelson, London.
Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.
Kreider DL, Kuller RG, Ostberg DR and Perkins FW (1966) An Introduction to Linear
Analysis. Addison-Wesley, London.
Paz M (1980) Structural Dynamics. Van Nostrand Reinhold, New York.
Stroud KA (1970) Engineering Mathematics. Macmillan, 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.
Free vibration of multi-degree-of-freedom systems
161
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.163
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
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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
Forced harmonic vibration of multi-degree-of-freedom systems
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;
Structural Dynamics for Engineers, 2nd edition
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
1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r21Þ
2 þ 4�2r21� �q
P0
K
1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið1� r21Þ
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
Forced harmonic vibration of multi-degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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.
Forced harmonic vibration of multi-degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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.
Forced harmonic vibration of multi-degree-of-freedom systems
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
Structural Dynamics for Engineers, 2nd edition
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.
Coates RC, Coutie MG and Kong FK (1972) Structural Analysis. Nelson, London.
Craig RR Jr (1981) Structural Dynamics. Wiley, Chichester.
Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.
Paz M (1980) Structural Dynamics. Van Nostrand Reinhold, New York.
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.
Forced harmonic vibration of multi-degree-of-freedom systems
175
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.177
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.
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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.
Damping matrices for multi-degree-of-freedom systems
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.
Structural Dynamics for Engineers, 2nd edition
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
Damping matrices for multi-degree-of-freedom systems
183
FURTHER READING
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
Craig RR Jr (1981) Structural Dynamics. Wiley, Chichester.
Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.
Paz M (1980) Structural Dynamics. Van Nostrand Reinhold, New York.
�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.
Structural Dynamics for Engineers, 2nd edition
184
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.185
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.
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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.
The nature and statistical properties of wind
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 �
where, from Equation 10.9,
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%
Structural Dynamics for Engineers, 2nd edition
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.
The nature and statistical properties of wind
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
Structural Dynamics for Engineers, 2nd edition
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
The nature and statistical properties of wind
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
Su nð Þ ¼ 4u2� f2
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.
Structural Dynamics for Engineers, 2nd edition
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
The nature and statistical properties of wind
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
Structural Dynamics for Engineers, 2nd edition
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.
The nature and statistical properties of wind
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
Structural Dynamics for Engineers, 2nd edition
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.
The nature and statistical properties of wind
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)
Structural Dynamics for Engineers, 2nd edition
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.
The nature and statistical properties of wind
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.
Structural Dynamics for Engineers, 2nd edition
202
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.203
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
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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.
Dynamic response to turbulent wind: frequency-domain analysis
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�
ffiffiffiffiffiffiffiffiffiffiffiffiK
M
� �s
� ¼ �st þ �a ¼ �st þ�CdAU
2!nM¼ �st þ
Fd
2�nUM
Su nð Þ ¼ 4u2� f2
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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.
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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
where, from Equation 11.37,
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
Structural Dynamics for Engineers, 2nd edition
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
From Equation 11.20,
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
From Equation 10.40,
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
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC2
xðxj � xkÞ2 þ C2yðyj � ykÞ2 þ C2
z ðzj � zkÞ2� �q
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
Dynamic response to turbulent wind: frequency-domain analysis
215
Sfknð Þ ¼ 4
F2 zkð ÞU2 zkð Þ
Ak nð ÞSuknð Þ ð11:61bÞ
’jk nð Þ ¼2n
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC2
xðxj � xkÞ2 þ C2yðyj � ykÞ2 þ C2
z ðzj � zkÞ2� �q
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
F2 zkð ÞU2 zkð Þ
Ak !ið ÞSuk!ið Þ ð11:68bÞ
’jk !ið Þ ¼!i
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiC2
xðxj � xkÞ2 þ C2yðyj � ykÞ2 þ C2
z ðzj � zkÞ2� �q
� 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
Structural Dynamics for Engineers, 2nd edition
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
nð Þ � Sf2nð Þ
� �q� e��12 nð Þ ð11:75Þ
and hence
Sfq1nð Þ ¼ Z2
11Sf1nð Þ þ 2Z11Z21
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1
nð Þ � Sf2nð Þ
� �q� e�’12 nð Þ þ Z2
21Sf2nð Þ: ð11:76aÞ
Similarly, the second modal force spectrum is given by
Sfq2nð Þ ¼ Z2
12Sf1nð Þ þ 2Z12Z22
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1
nð Þ � Sf2nð Þ
� �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
31Sf3nð Þ þ 2Z11Z21
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1
nð Þ � Sf2nð Þ
� �q� e�’12 nð Þ
þ 2Z11Z32
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1
nð Þ � Sf3nð Þ
� �q� e�’13 nð Þ þ 2Z21Z31
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf2
nð Þ � Sf3nð Þ
� �q� e�’23 nð Þ ð11:77aÞ
Sfq2nð Þ ¼ Z2
12Sf1nð Þ þ Z2
22Sf2nð Þ þ Z2
32Sf3nð Þ þ 2Z12Z22
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1
nð Þ � Sf2nð Þ
� �q� e�’12 nð Þ
þ 2Z12Z32
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1
nð Þ � Sf3nð Þ
� �q� e�’13 nð Þ þ 2Z22Z32
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf2
nð Þ � Sf3nð Þ
� �q� e�’23 nð Þ ð11:77bÞ
Sfq3nð Þ ¼ Z2
13Sf1nð Þ þ Z2
23Sf2nð Þ þ Z2
33Sf3nð Þ þ 2Z13Z23
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1
nð Þ � Sf2nð Þ
� �q� e�’12 nð Þ
þ 2Z13Z32
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf1
nð Þ � Sf3nð Þ
� �q� e�’13 nð Þ þ 2Z23Z33
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSf2
nð Þ � Sf3nð Þ
� �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).
Dynamic response to turbulent wind: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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:
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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:
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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:
Structural Dynamics for Engineers, 2nd edition
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:
Dynamic response to turbulent wind: frequency-domain analysis
225
where, from Equation 11.42,
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
� �2Aj !ið ÞSuj
!ið Þ ð11:92Þ
�2qi¼ 1
!3i
1
2�i
XNj¼ 1
2Fdj
Uj
� �2Aj !ið ÞSuj
!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:
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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.
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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,
Dynamic response to turbulent wind: frequency-domain analysis
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
From Equation 11.104,
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:
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to turbulent wind: frequency-domain analysis
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.
From Equation 11.113,
�2x ¼ 1
M
� �2
� �D4Lc
8�2S2
!2� !n
8�� SCL
!nð Þ
where, from Equation 2.30,
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:
Structural Dynamics for Engineers, 2nd edition
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:
Dynamic response to turbulent wind: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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 �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
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
Dynamic response to turbulent wind: frequency-domain analysis
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
xmax ¼ ��x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2:4� 3600ð Þ½ �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
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
xmax ¼ ��x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 2:4� 3600ð Þ½ �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 ln 2:4� 3600ð Þ½ �p
( )� 4:6443744� 10�3
¼ 0:0204 m:
Structural Dynamics for Engineers, 2nd edition
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
_xxi tþ�tð Þ ¼ 2
�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
Dynamic response to turbulent wind: frequency-domain analysis
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.
Structural Dynamics for Engineers, 2nd edition
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.
Harris CM (1988) Shock Vibration, 3rd edn. McGraw-Hill, London.
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
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
Dynamic response to turbulent wind: frequency-domain analysis
243
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.245
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)
Structural Dynamics for Engineers, 2nd edition
246
Figure 12.2 Seismic waves
P wave
S wave
Love wave
Rayleigh wave
Compressions
Dilatations
Double amplitudeWavelength
Undisturbed medium
The nature and properties of earthquakes
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).
Structural Dynamics for Engineers, 2nd edition
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
The nature and properties of earthquakes
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
Structural Dynamics for Engineers, 2nd edition
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
The nature and properties of earthquakes
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
Structural Dynamics for Engineers, 2nd edition
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
The nature and properties of earthquakes
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.
Key DE (1988) Earthquake Design Practice for Buildings. Thomas Telford, 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.
Structural Dynamics for Engineers, 2nd edition
254
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.255
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
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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:
Dynamic response to earthquakes: frequency-domain analysis
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.
From Newton’s law of motion,
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to earthquakes: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to earthquakes: frequency-domain analysis
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.
Structural Dynamics for Engineers, 2nd edition
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
_xxi tþ�tð Þ ¼ 2
�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
Dynamic response to earthquakes: frequency-domain analysis
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
S€xxg !ð Þ ¼S0 1þ 2�gr
� �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
Structural Dynamics for Engineers, 2nd edition
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
fgðtÞ ¼ fg sin !tð Þ ð13:38Þ
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
Dynamic response to earthquakes: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to earthquakes: frequency-domain analysis
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
€qqi þ 2�i!i _qqi þ !2i qi ¼ Z
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
Structural Dynamics for Engineers, 2nd edition
270
where
�! ¼ 12 �i!i
Mi !ð Þ ¼ 14 �
2i :
We therefore have
qi ¼ �i�qið13:54Þ
where, from Equation 10.48,
�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:
Dynamic response to earthquakes: frequency-domain analysis
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
where, from Equation 13.29,
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
2:0� 2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4� 0:62ð Þ
p ¼ 2783:3290m2=s3:
For !2¼ 42.390 rad/s,
S0 ¼0:141� 0:6� 280:942
2:0� 2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4� 0:62ð Þ
p ¼ 342:5934m2=s3:
For !3¼ 56.718 rad/s,
S0 ¼0:141� 0:6� 92:692
2:0� 2�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4� 0:62ð Þ
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:
Structural Dynamics for Engineers, 2nd edition
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
q2 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 42:390� 10=2�ð Þ½ �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 ln 42:390� 10=2�ð Þ½ �p
( )� 0:0887446 ¼ 0:2752m
q3 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 56:718� 10=2�ð Þ½ �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
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.
Dynamic response to earthquakes: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to earthquakes: frequency-domain analysis
275
then
fgðtÞ ¼ fg sin !tð Þ ð13:66Þ
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
From Equation 13.43,
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
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to earthquakes: frequency-domain analysis
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 �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
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 �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
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.
Structural Dynamics for Engineers, 2nd edition
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
€qqi þ 2�i!i _qqi þ !2i qi ¼ Z
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
Dynamic response to earthquakes: frequency-domain analysis
279
and hence
qi ¼ �i�qið13:85Þ
where, from Equation 10.48,
�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:
Structural Dynamics for Engineers, 2nd edition
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
5002ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 0:62ð Þ
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
5002ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 0:62ð Þ
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:
Dynamic response to earthquakes: frequency-domain analysis
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
q2 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 12:466� 10=2�ð Þ½ �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 ln 12:466� 10=2�ð Þ½ �p
( )� 0:39402 ¼ 1:05618m
q3 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 ln 18:025� 10=2�ð Þ½ �
pþ 0:577ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
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:
Structural Dynamics for Engineers, 2nd edition
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
Dynamic response to earthquakes: frequency-domain analysis
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,
S€xxg !ð Þ ¼S0 1þ 2�gr
� �2h i
1� r2ð Þ2 þ 2�rð Þ2
where
S0 ¼0:141�g€xx
2g;max
!g
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ 4�2g� �q
Structural Dynamics for Engineers, 2nd edition
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:
Dynamic response to earthquakes: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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:
Dynamic response to earthquakes: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
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.
Dynamic response to earthquakes: frequency-domain analysis
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
Structural Dynamics for Engineers, 2nd edition
290
REFERENCES
Blum JA, Newmark NM and Corning LH (1961) Design of Multistorey Reinforced Building for
Earthquake Motions. Portland Cement Association, Chicago.
Clough RW and Penzien J (1975) Dynamics of Structures. McGraw-Hill, London.
Coates RC, Coutie MG and Kong FK (1972) Structural Analysis. Nelson, London.
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.
Key DE (1988) Earthquake Design Practice for Buildings. Thomas Telford, London.
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.
Craig RR Jr (1981) Structural Dynamics. Wiley, Chichester.
Eiby GA (1980) Earthquakes. Heinemann, London.
Lomnitz F and Rosenbleuth E (1976) Seismic Risk and Engineering Decisions. Elsevier,
Amsterdam.
Paz M (1980) Structural Dynamics. Van Nostrand Reinhold, New York.
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.
Dynamic response to earthquakes: frequency-domain analysis
291
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.293
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
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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
Generation of wind and earthquake histories
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
Structural Dynamics for Engineers, 2nd edition
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
Generation of wind and earthquake histories
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
Structural Dynamics for Engineers, 2nd edition
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
Generation of wind and earthquake histories
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
Structural Dynamics for Engineers, 2nd edition
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
Generation of wind and earthquake histories
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
Structural Dynamics for Engineers, 2nd edition
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
Time lag τ: s, log scale
Frequency range
16
14
12
10
8
6
4
2
0
Generation of wind and earthquake histories
305
Box GEP and Jenkins CM (1977) Time Series Analysis: Forecasting and Control. Holden Day,
San Francisco.
Buchholdt HA (1985) Introduction to Cable Roof Structures. Cambridge University Press,
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.
Structural Dynamics for Engineers, 2nd edition
306
Structural Dynamics for Engineers, 2nd edition
ISBN: 978-0-7277-4176-9
ICE Publishing: All rights reserved
http://dx.doi.org/10.1680/sde.41769.293
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
Structural Dynamics for Engineers, 2nd edition
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
Structural Dynamics for Engineers, 2nd edition
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
Generation of wind and earthquake histories
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
Structural Dynamics for Engineers, 2nd edition
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
Generation of wind and earthquake histories
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
Structural Dynamics for Engineers, 2nd edition
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
Generation of wind and earthquake histories
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
Structural Dynamics for Engineers, 2nd edition
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
Generation of wind and earthquake histories
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
Structural Dynamics for Engineers, 2nd edition
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
Time lag τ: s, log scale
Frequency range
16
14
12
10
8
6
4
2
0
Generation of wind and earthquake histories
305
Box GEP and Jenkins CM (1977) Time Series Analysis: Forecasting and Control. Holden Day,
San Francisco.
Buchholdt HA (1985) Introduction to Cable Roof Structures. Cambridge University Press,
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.
Structural Dynamics for Engineers, 2nd edition
306