Dynamic Analysis of Structures
Dynamic Analysis of
Structures
John T. Katsikadelis
Professor Emeritus of Structural Analysis
School of Civil Engineering
National Technical University of Athens
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Dedication
To my wife Efi
for her loving patience and support,
to my children Stefan and Christina,
and to
my granddaughter Katharina
for patiently enduring and sharing the years of
preparation of the book with me.
This work is also dedicated to my numerous students
whose attendance of the Structural Dynamics
course over the past 40 years encouraged an
international publication of the book.
Preface
The statement of the laws of motion by Newton 334 years ago (1686)a was a
milestone in the evolution of mechanics and modern engineering. The relation
between force (cause) and motion (effect) was quantified as a relation between
the linear momentum of the body and the force exerted on it. Thus, this relation
from a subject of philosophy up to that time turned out to be a valuable tool of
science for the study of the natural world. The subsequent developments in the
sciences were rapid. Astronomy, mathematics, mechanics of fluid and deformable bodies, and in general, mechanics of continuous media reached their peaks
in the centuries that followed, with immense applications to all engineering disciplines. Nevertheless, the laws of motion, which were stated as an axiom
(Axiomata sive Leges Motus) by Newton because, apparently, he could not justify their derivation, was a consequence of the discoveries of great scientists
who preceded him such as Galileo,b Kepler,c Hook, etc.
The implementation of the laws of motion leads to mathematical models
described by differential equations, ordinary or partial, whose solution effort
has given a great impetus to the development of mathematics. Unfortunately,
analytical solutions are limited to simple problems such as vibrations of discrete
systems with a few degrees of freedom; linear vibrations of beams, membranes,
plates, and shells with simple geometry; and simple support conditions made
from materials, mostly with a linear behavior. These solutions, while useful
for extracting qualitative conclusions about the dynamic response of structures,
are not capable of solving realistic problems in engineering, where the geometry
and loads are complicated while the response is generally nonlinear. Although it
has been shown that Newton’s law of motion is of an integer-order derivative,d
in recent years, the fractional derivatives have been proven more suitable for
modeling the actual structures. However, the use of fractional calculus has
not been employed in mathematical physics for three centuries because the
a. I. Newton, Philosophiae Naturalis Principia Mathematica, Royal Society Press, London, 1686.
b. J.T. Katsikadelis, Derivation of Newton’s law of motion using Galileo’s experimental data, Acta
Mech. 226 (2015) 3195–3204, doi:10.1007/s00707-015-1354.
c. J.T. Katsikadelis, Derivation of Newton’s law of motion from Kepler’s laws of planetary motion,
Arch. Appl. Mech. 88 (2018) (2017) 27–38, doi:10.1007/s00419-017-1245-x.
d. J.T. Katsikadelis, Is Newton’s law of motion really of integer differential form? Arch. Appl.
Mech. 89 (2019) 639–647, doi:10.1007/s00419-018-1486-3.
xv
xvi Preface
fractional derivative, contrary to the integer-order derivative, has no direct
physical or geometrical meaning. Thus, it was not adopted by the famous scientists of earlier centuries who developed modern science and engineering, as
they could not visualize the significance of the fractional derivative. Thus, we
dare say that in a certain sense, Newton’s Principia has delayed the development
of mechanics and engineering.
Structural dynamics became a tool for the dynamic analysis of structures
only with the advent of computers in the middle of the last century. Nowadays,
the cheap, fast, and large-capacity computers as well as the development of efficient numerical methods to solve the equations of motion allow the safe study of
complicated structures under dynamic loads. Aircrafts, large bridges, ships,
high-rise buildings, dams, fast trains, etc., are now being safely analyzed and
designed. The finite element method (FEM) permits the dynamic analysis of
continuous systems by modeling them with equivalent discrete systems. The
boundary element method (BEM) analyzes with great accuracy the dynamic
response of systems described by ordinary or partial differential equations.
Besides, the coupling of the two methods paves the way to solving very difficult
dynamic problems such as the interaction of the structure with soil or fluids as
well as the large deformation analysis of systems with nonlinear constitutive
equations. Recently, meshless methods, a numerical tool beyond FEM, have
been able to solve efficiently the equations describing the dynamic response
of continuous systems.
The dynamic analysis of structures today is mainly done by using professional computer codes such as NASTRAN, SAP2000, ETABS, ADINA, etc.,
which have relatively unlimited capabilities. While these programs are
undoubtedly a useful and indispensable tool in the hands of the engineer, they
do not allow for an understanding of the dynamic behavior of the structures or
the limits of the validity of the results. This can be achieved only with a deep
theoretical background, which is constantly expanding with the development of
the computational methods in mechanics. The engineer without theoretical
knowledge is unable to formulate the structural model. The modeling of structures cannot be done by the computer but only by the engineer, who is the only
one responsible for performing proper dynamic analysis. The belief that theoretical knowledge is not necessary today because there are ready-to-use computer codes is, at the very least, dangerous. The engineer should never
accept the results of a computer program unless he/she can check their validity
and accuracy. Apparently, this is possible only if the engineer has a deep theoretical knowledge of the dynamics of structures. We emphasize that theoretical
knowledge today is more necessary than ever. Areas that a few decades ago
were covered by applied mathematics and theoretical or applied mechanics
have become the subject of the modern engineering praxis.
Dynamics of Structures for engineers, particularly for civil engineers, was
included as a regular course in the curricula of many universities long ago,
Preface xvii
particularly after the advent of computers. Therefore, many books have appeared
on this subject. However, these books provide a means only to understand the
response of simple and mostly unrealistic structures when subjected to dynamic
loads, especially to ground motion. The principles of dynamics are illustrated by
applying them to very simple models, which cannot describe actual structures
and therefore cannot be employed for dynamic analysis and design. We should
have in mind that the dynamic analysis of actual structures requires their modeling, the formulation of the governing equations of motion, their solution under
any dynamic load, and the physical interpretation of the results.
In the last 30 years, almost all seismic codes have encountered earthquake
ground motion as an effective dynamic load. The advent of computers in the
early 1960s encouraged engineers to develop methods of dynamic analysis
of structures, modeled first by the FEM and later by other advanced numerical
methods. Today, these methods constitute a powerful tool for dynamic engineering analysis. Thanks to the availability of cheap computer power, every
engineer can use them. The essential ingredients of a book on Dynamics of
Structures for Civil Engineers should be:
(a) The basic concepts and principles of structural dynamics as they are
applied to particles as well as rigid and deformable bodies, enabling the
student or the engineer to formulate the equations of motion of any structure, no matter how complex, once the dynamic model has been adopted.
(b) Realistic modeling of actual structures under dynamic loads.
(c) Analysis of the dynamic response of the structure represented by its
model under any specified load. The analysis should include single- and
multiple-degree-of-freedom systems for linear and nonlinear response
under any dynamic excitation.
(d) Approximation of real structures using computational methods such as the
FEM, which replaces the actual structure (distributed parameter system)
with an approximate discrete system for which analysis methods can be
applied.
(e) Effective present-day numerical methods for dynamic analysis, including the
numerical solution of eigenvalue problems and the direct solution of the equations of motion of large systems, namely, systems with a large number of
degrees of freedom such as those resulting from the employed discretization.
Students attending a course on Dynamics of Structures should be exposed at
least to the above subjects. However, not all of them can be found in a single
book. Therefore, people interested in structural dynamics should refer to more
than one book in order to retrieve the required knowledge. Apparently, these
books cannot be used as integrated textbooks in the sense described above.
The student should be acquainted with different symbols and approaches, which
complicate the acquisition of knowledge, an approach that is, at least, educationally inappropriate.
xviii Preface
Knowledge of the dynamics of structures is particularly necessary for engineers, who are studying the response of their structures subjected to seismic
ground motion. Modern earthquake codes, which as mentioned treat the ground
motion as a dynamic load, require a deep theoretical understanding of the
dynamics of structures. The aim of this book is to give the student as well as
the professional engineer the minimum knowledge necessary to understand
the dynamic response of the structures. The author has taught structural dynamics, both as an undergraduate and a graduate course, at the School of Civil Engineering of the National Technical University of Athens (NTUA) continuously
since 1981, when it was first introduced as a formal course by the late Prof. A.E.
Armenákas. The author’s teaching experience together with his long research in
this area have contributed to the presentation of material in a simple and comprehensible way. The book contains numerous illustrative examples that help to
understand the theory. Particular emphasis is given to numerical methods,
which are presented methodically and given in the form of pseudo-codes so that
interested readers can write their own computer codes in the language of their
choice. The list of the programs in FORTRAN and MATLAB are given in electronic form on this book’s companion website. Nevertheless, the numerical
results in the Examples have been obtained using the MATLAB programs
because it is easy for the students to master this language. The programs can
be used to solve many problems of engineering praxis.
The book is divided into two parts.
The first part deals with the single-degree-of-freedom (SDOF) systems
and contains nine chapters. In particular, Chapter 1 presents the general
concepts and principles of dynamics as they apply to structural dynamics.
Chapter 2 deals with the free vibrations of the SDOF systems while
Chapter 3 deals with their forced vibrations. Particular emphasis is given
to the study of the resonance phenomenon. Chapter 4 presents the most efficient numerical methods for solving the equation of motion. Chapter 5 deals
with the nonlinear response of the SDOF systems. Chapter 6 presents the
response of structures due to ground motion. Chapter 7 deals with the damping of structures while Chapter 8 approximates the continuous systems by
generalized SDOF systems and studies the dynamic response of beams treated as continuous systems. Finally, Chapter 9 describes the analysis of SDOF
systems in the frequency domain.
The second part deals with the multi-degree-of-freedom (MDOF) systems
together with the well-known computational methods for their analysis. It contains seven chapters. In particular, Chapter 10 describes the various methods of
modeling of the MDOF systems, which are classified into categories that facilitate the selection of the appropriate method for formulating the equations of
motion. Chapter 11 presents in detail the FEM for the dynamic analysis of skeletal structures (trusses, frames, grids) as an extension of the static matrix structural analysis to dynamics. The reduction of degrees of freedom due to the
omission of axial deformations in specific structural members as well as the
dynamic response of flexible structures containing rigid bodies are also
Preface
xix
discussed. The derivation of the equivalent element nodal quantities, that is,
mass and stiffness matrices and forces, are derived by using the Lagrange equations instead of the principle of virtual works. Although the principle of virtual works offers a handy tool for the derivation of these quantities, the use of
the Lagrange equations was preferred here. The reason is that the Lagrange
equations not only offer a straightforward method for the derivation of the
equivalent nodal quantities for all types of elements, especially in complex
systems with a nonlinear response, but they also allow understanding of their
physical significance. Chapter 12 studies the free vibrations of MDOF systems without and with damping. The linear eigenvalue problem is presented
from the mathematical point of view, aiming at drawing useful conclusions
about the eigenfrequencies and the mode shapes of the physical systems.
Chapter 13 presents the numerical methods for the computation of the eigenfrequencies and mode shapes, especially for systems with a large number of
degrees of freedom such as those derived from the application of finite elements. Chapter 14 studies the forced vibrations of MDOF systems. A large
part of this chapter is devoted to the mode superposition method. It also discusses the use of Ritz vectors to reduce the degrees of freedom. Particular
emphasis is given to the response spectrum method. The response of linear
systems when they are subjected to a synchronous and an asynchronous
motion of the supports are also discussed. This chapter ends with the presentation of the numerical methods, giving the respective pseudocodes for the
time integration of linear and nonlinear systems of equations of motion.
Chapter 15 discusses the approximation of multistory buildings by skeletal
structures and presents methods of formulating their equation of motion.
Finally, Chapter 16 discusses the response of seismically isolated buildings.
This chapter is introductory to the subject and aims primarily at understanding the impact of base isolation on structures.
The book is supplemented by an appendix. Therein, the basic theory of rigid
body dynamics is presented for large and small displacements and the relevant
equations are derived, which are employed in the development of the material of
the book.
In closing, the author wishes to express his sincere thanks to his former student and coworker Dr. A. J. Yiotis for carefully reading the manuscript as well
as for his suggestions, constructive recommendations and his overall contribution to minimizing the oversights in the text. Warm thanks also belong to Dr.
Nikos G. Babouskos, also a former student and coworker of the author, not only
for his careful reading of the manuscript and his apposite suggestions for
improvement of the book but also for his assistance in checking the computer
programs and in producing the numerical results of examples therein. Finally,
thanks belong to Dr. G. Dasios, professor of mathematics at the University of
Patras as well as to his former students, Dr. G. Tsiatas, associate professor of
mathematics at the University of Patras, and Dr. P. Tsopelas, associate professor
of mechanics at NTUA, for reading certain sections of the book and making
constructive suggestions.
xx Preface
It is a pleasure to make grateful acknowledgment of many helpful suggestions contributed by my students who have attended the course over the past
39 years.
J.T. Katsikadelis
Athens
April 2020
Chapter 1
General concepts and principles
of structural dynamics
Chapter outline
1.1
1.2
1.3
1.4
Introduction
Types of dynamic loads
Dynamic degrees of freedom
Dynamic model and
formulation of the equation
of motion of SDOF systems
1.5 Derivation of the equations
of motion using d’Alembert’s
principle
1.6 Principle of virtual
displacements
1.1
3
6
9
11
15
38
1.7 Hamilton’s principle
1.8 Lagrange’s equations
1.8.1 Derivation of
Lagrange’s equations
1.8.2 Lagrange multipliers
1.8.3 Small displacements
1.8.4 Raleigh’s dissipation
function
1.9 Influence of the gravity loads
1.10 Problems
References and further reading
42
54
54
64
68
72
73
74
82
Introduction
Apart from static loads, engineering structures may be subjected to dynamic
loads, that is, loads whose magnitude as well as direction of action and/or position vary with time. The analysis of stresses and deflections developed in a
given structure undergoing dynamic loads is the fundamental objective of the
dynamic analysis of structures. Between static and dynamic analysis of structures, there exist two substantial differences:
(a) In static analysis, the loads are assumed time-invariant, and the resulting
response is unique, at least in linear theory. On the other hand, in dynamic
analysis the loads are time-varying and the deformations and stresses
depend on time, that is, at each instant the response of the structure is
different.
(b) In dynamics analysis, the material points of the structure change position
with the time, hence they have velocity and acceleration. Inasmuch as
the structure has a mass, inertial forces are produced due to the accelerations
of the material points. These inertial forces constitute an additional loading
that cannot be ignored. To make it tangible, we consider the cantilever beam
of Fig. 1.1.1a. The beam has a mass per unit length m and a flexural rigidity
EI , both assumed constant along the length, and it is subjected to the timeDynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00001-7
© 2020 Elsevier Inc. All rights reserved.
3
4 PART
I Single-degree-of-freedom systems
varying position-dependent distributed loading pðx, t Þ (Fig. 1.1.1a). We can
derive the equation of motion by considering the dynamic equilibrium of a
beam element of length dx (Fig. 1.1.1b). In static consideration, the element
would be in equilibrium under the action of the external load pðx, t Þdx and
the internal forces, that is, the shear force Q, the bending moment M at the left
cross-section of the element, and the shear Q + dQ and moment M + dM at
its right cross-section.
(a)
(b)
(c)
FIG. 1.1.1 Vibrating cantilever beam.
General concepts and principles of structural dynamics Chapter
1
5
The transverse deflection is a function not only of the spatial variable x but
also of time t, namely it is u ¼ u ðx, t Þ. As the element has a mass mdx, an inertial
€ arises, which, according to d’Alembert’s principle (see Section 1.5)
force m udx
opposes the motion, that is, if the positive transverse displacement u ðx, t Þ in the
beam is directed downward, the inertial force is directed upward (see Fig. 1.1.1b
and c). Similarly, due to angular acceleration ∂u€ðx, t Þ=∂x of the cross-section, an
inertial moment is also developed, which we may neglect [1]. Thus, referring to
Fig. 1.1.1b, we obtain the equation of dynamic equilibrium of the beam element
in the y direction as
€ ¼0
Q + Q + dQ + pðx, t Þdx m udx
or
∂Q
∂2 u
¼ pðx, t Þ + m 2
∂x
∂t
(1.1.1)
From the Euler-Bernoulli beam theory we have
Q ¼ EI
∂3 u
∂x 3
(1.1.2)
Substituting Eq. (1.1.2) into Eq. (1.1.1) yields
EI
∂4 u
∂2 u
+ m 2 ¼ pðx, t Þ
4
∂x
∂t
(1.1.3)
Eq. (1.1.3) is known as the equation of the dynamic equilibrium or the equation of motion of the vibrating beam. It is apparent that if we omit the inertial
term m∂2 u=∂t 2 in Eq. (1.1.3), we obtain the equation of the deflection of the
beam under static loading, that is,
EI
d4u
¼ pðx Þ
dx 4
(1.1.4)
Fig. 1.1.1c shows the beam subjected to the inertial forces. These forces
resist the accelerations and they need to be accounted for in the solution. This
is the most important characteristic of the dynamic problem. Obviously, the
magnitude of the inertial forces depends on the magnitude of the acceleration.
When the produced accelerations are very small, as in the case of slow motion,
the inertial forces are very small too, and they can be neglected. In this case,
the time appears in the equation as a parameter and the response at any instant
6 PART
I Single-degree-of-freedom systems
can be obtained by the static structural analysis, even though the load and the
response are time-varying. This response is pseudodynamic and is referred to
as quasistatic. The inertial forces appear in the equation of motion of the structure with the second derivatives of the displacements with respect to time.
Therefore, the equations that must be solved in dynamic analysis in order
to establish the deformations and stresses in the structure are differential
equations, contrary to static analysis where the governing equations are algebraic. For this reason, the solution procedure in dynamic analysis is essentially
different from that used in static analysis.
1.2 Types of dynamic loads
As already mentioned, dynamic loads are time-varying. Such loads are those
due to an unbalanced rotating machinery, the motion of vehicles on structures,
blast loads, wind loads, etc. The motion of the support of a structure, even in the
absence of external dynamic loads, produces a dynamic response, too. This is
the case of seismic ground motion (see Chapter 6).
FIG. 1.2.1 Periodic load.
Dynamic loads can be classified into two great groups that characterize the
approach of evaluating the structural response: The deterministic dynamic loads
and the nondeterministic or random dynamic loads. In the first group are the
dynamic loads whose time variation is fully determined, regardless of the
complexity of their mathematical presentation. They are also referred to as prescribed dynamic loads. They can be represented by an analytic or a generalized
function (Dirac or Heaviside) as well as numerically by a set of their values at
discrete time instances. The second group includes the loads, whose time variation is not completely known but it can be defined in a statistical sense. In this
book, the dynamic response of structures only under deterministic loads is
studied.
General concepts and principles of structural dynamics Chapter
5
H = me
2
1
7
sin t
H(t)
2.5
0
–2.5
T
T
–5
0
2
4
T
t
6
8
10
FIG. 1.2.2 Harmonic loading due to an unbalanced rotating mass.
From the analytical point of view, it is convenient to divide the deterministic
loads into two basic categories, periodic and nonperiodic loads. Periodic loads
are those whose time variation profile repeats continually at regular time intervals T . Mathematically, they can be represented by a periodic function
pðt Þ ¼ pðt + nT Þ
(1.2.1)
where n is a natural number. The time interval T is called the period of the load.
A general type of periodic load is shown in Fig. 1.2.1, which also identifies the
period of the load. A usual type of periodic load is the harmonic load caused by
an unbalanced rotating machine (Fig. 1.2.2); H ¼ mew2 cos wt is the horizontal
component of the centrifugal force. Loads that do not show any periodicity are
called nonperiodic loads. They may be of long duration, such as those resulting
from an earthquake. Nonperiodic loads of short duration are called impulsive
loads. A windblast striking a building and the pressure of a bomb explosion
on a structure are typical impulsive loads (Fig. 1.2.3). The earthquake produces
a special type of load, which is due to the excitation of the support of the
8 PART
I Single-degree-of-freedom systems
structure and, as we will see later in this book (Chapter 6), it can be reduced to an
effective dynamic load if the accelerogram of the ground motion is known
(Fig. 1.2.4).
FIG. 1.2.3 Nonperiodic load due to explosion.
400
0
–400
0
10
20
30
FIG. 1.2.4 Effective dynamic load pðt Þ ¼ m u€g ðt Þ due to seismic ground motion.
If we examine static loading closer, we will see that even what we call static
loads are actually dynamic in nature. They are applied starting from a zero value
until the final prescribed value is reached within a time span. That is, they are
time-varying, thus dynamic. However, the duration of the application of the
static load is longer than the period of vibration of the structure. This produces
negligible accelerations and consequently the response under a “static load”
General concepts and principles of structural dynamics Chapter
1
9
could be considered as a special case of the dynamic problem with negligible
accelerations, that is, quasistatic.
1.3
Dynamic degrees of freedom
The displacement method is the most suitable method for the dynamic analysis
of structures. In this method, the unknowns are the displacements. For structures
with distributed mass, the displacements are functions of the spatial coordinates
and of the time as well. Their dynamic response is described by partial differential equations of the hyperbolic type, which must be solved in order to determine
these displacements. The solutions of such equations belong to the most difficult
problems of mathematics. The available solutions refer to simple structures, for
example, beams with a constant cross-section, which are not adequate to study
the dynamic response of engineering structures. Thanks to the development of
modern computational methods such as FEM, BEM and Meshless Methods, the
actual structure is approximated by discrete models in which the mass is localized at a finite number of points (nodal points). These models are adequate to
represent the effects of all significant inertial forces of a structure. At each
instant, the deformed configuration of the structure is determined from the displacementsa of the nodal points, which are functions only of time. The response
of the discretized structure is governed by ordinary differential equations, which
are easy to solve analytically or at least numerically. The number of the independent nodal displacements required to determine the deformed shape of the moving structure is called the number of degrees of freedom. It is apparent that
continuous systems have an infinite number of dynamic degrees of freedom.
Structures with one degree of freedom are called single-degree-of-freedom
(SDOF) systems. Accordingly, we have two-degree-of-freedom (2 DOF) systems, three-degree-of-freedom (3 DOF) systems, and generally multi-degreeof-freedom (MDOF) systems. Fig. 1.3.1 shows SDOF systems. Fig. 1.3.1a represents the idealization of a silo. It consists of two massless columns and a square
rigid plate of mass m. With the assumption that the axial deformation of the columns is negligible, the horizontal displacement u ðt Þ is adequate to completely
determine the motion of the system. Hence, the system has one degree of freedom. Likewise, under the same assumptions for the columns, the motion of
the water tower of Fig. 1.3.1b can be determined from the angle fðt Þ.
Fig. 1.3.1c represents the typical model of a SDOF. Fig. 1.3.2a represents the
model of the two-story shear frame. To determine its motion, it is necessary to
establish the two independent horizontal displacements u1 ðt Þ and u2 ðt Þ.
Fig. 1.3.2b represents a cantilever column with a mass atop. This may be
considered as the idealization of a water tower. The whole mass is lumped at
the top while the column is massless. During the motion, the mass undergoes
horizontal displacement u ðt Þ and rotation fðt Þ. These two geometrical quantities are independent. Hence, the system has two degrees of freedom and thus
a. The term displacent denotes both translation and rotation.
10 PART
I Single-degree-of-freedom systems
two differential equations of motion are necessary to determine these displacements. If, however, the mass m of the system is assumed to be concentrated at a
point, its rotational inertia I is equal to zero. Hence, the inertial moment I f€ðt Þ is
zero, too, and one of the equations of motion becomes algebraic. This permits
the elimination of the rotational displacement, leading to only one equation of
motion for u ðt Þ. Consequently, the system has only one dynamic degree of freedom, even though it has two static degrees of freedom. Apparently, the number
of static degrees of freedom is not necessarily equal to the number of dynamic
degrees of freedom. Concluding, we can say that in an MDOF system, the number of dynamic degrees of freedom is equal to the number of independent differential equations of motion that must be formulated to establish the dynamic
response of the system.
Massless
columns
Massless
columns
(a)
(b)
(c)
FIG. 1.3.1 Systems with one degree of freedom (SDOF).
Rigid
Rig
(a)
FIG. 1.3.2 Systems with two degrees of freedom (2 DOF).
(b)
General concepts and principles of structural dynamics Chapter
1
11
FIG. 1.3.3 Multi-degree-of-freedom (MDOF) system.
FIG. 1.3.4 System with infinite degrees of freedom. Continuous system.
The lumped mass idealization provides a simple means of reducing the number of degrees of freedom. Fig. 1.3.3 represents the discrete model of a cantilever column, whose mass has been localized at three points. Neglecting the
axial deformation of the column and considering plane motion, the system
has six degrees of freedom, the three transnational ui ðt Þ and the three rotational,
fi ðt Þ. If the masses are fully concentrated so that their rotational inertia can be
ignored, the inertial moments Ii f€i are zero and the number of dynamic degrees
of freedom reduces to three. Obviously, the number of degrees of freedom
increases with the number of nodal points, where the mass of the structure is
lumped. As the number of points becomes infinitely large, the discretized structure approaches the continuous system (Fig. 1.3.4).
1.4 Dynamic model and formulation of the equation
of motion of SDOF systems
The modeling of the real structure plays a fundamental role in the dynamic analysis of structures. It is the most difficult task in dynamic analysis because in this
stage of analysis, the experience and theoretical background of the engineer
intervene critically in approximating the structural response.
12 PART
I Single-degree-of-freedom systems
Spring
Damper
Frictionless rollers
FIG. 1.4.1 Model of a SDOF system.
Fig. 1.4.1 shows a simple dynamic model of a SDOF system. It consists of a
rigid body of mass m constrained to move along the x axis in the plane of the
paper, a weightless spring connecting the mass to the firm support, and a damper.
Center of mass
FIG. 1.4.2 Forces applied to the free body.
The forces applied to the body at time t are shown in the free body diagram
of Fig. 1.4.2. These are
(a)
(b)
(c)
(d)
The external load pðt Þ
The elastic force fS
The damping force fD
The inertial force fI .
The spring force fS depends on the displacement u ðt Þ and it is generally
expressed by a nonlinear function, fS ¼ fS ðu Þ. For linear response of the structure, the force fS is proportional to the displacement and is given by
fS ¼ ku
(1.4.1)
where k is the constant that represents the spring stiffness coefficient, that is, the
force required to change the length of the spring by a unit. The force fS represents the elastic force of the structure that resists the motion and tends to bring
the body to its initial undeformed position.
The damping force fD also resists the motion. It represents the energy loss
due to internal or external dissipative forces. Damping forces are complex in
nature. Their exact expression in terms of the parameters of motion and of
the geometrical and material properties of the structure is complicated and difficult to determine. The simplest form of damping is linear viscous damping.
This produces damping forces, which are the easiest to handle mathematically
General concepts and principles of structural dynamics Chapter
1
13
and provide analytical results for the response of a system close to the experimental ones. The linear viscous damping mechanism is indicated by a dashpot,
as shown in Fig. 1.4.1. In viscous damping, the resisting force is proportional to
the velocity
fD ¼ cu_
(1.4.2)
where c is a constant that can be established experimentally. Inasmuch as the
work done by this force is converted to heat, the damping force is a nonconservative force. It is the force that makes the amplitude of a vibrating
structure decay.
The inertial force fI depends on the mass m of the body and its acceleration
€ It also resists the motion. It is given by Newton’s second law of motionb
u.
fI ¼ m u€
(1.4.3)
A simple example of a structure that can be modeled as SDOF is the
one-story, one-bay frame of Fig. 1.4.3. It consists of two identical weightless
columns fixed on the ground and having height h, cross-sectional moment of
inertia Ι, and modulus of elasticity E. The cross-sectional moment of inertia
of the horizontal beam is assumed infinitely large. This means that the beam
behaves like a rigid body of mass m and hence the cross sections of the columns
at the roof level cannot rotate when the frame deforms. The frame is subjected to
an external horizontal force pðt Þ, as shown in Fig 1.4.3a, which forces the frame
to move. Neglecting the axial deformation of the beam and columns, an allowable assumption for frames, the only possible movement is the displacement
u ðt Þ at the roof level. The rotation of the beam as a rigid body is excluded
because this would cause a change in the length of columns.
(a)
(b)
(c)
FIG. 1.4.3 Two-column shear frame.
b. Actually, this form of Newton’s law of motion is attributed to L. Euler, who defined it independently as a mechanical principle [2, 3]. This law was recently derived from Kepler’s laws of planetary motion [4].
14 PART
I Single-degree-of-freedom systems
Referring to Fig. 1.4.3b, we see that the elastic forces are the shear forces Q
at the top cross-sections of the columns. These forces are given by the known
relation of statics
Q¼
12EI
u ðt Þ
h3
(1.4.4)
The quantity 12EI =h 3 represents the translational stiffness of the column.
This is the force required to produce a unit relative displacement between
the end cross-sections of the column. These shear forces tend to restore the
frame to the undeformed position. Therefore, they play the role of the spring
in the SDOF model with a stiffness coefficient
k ¼2
12EI
h3
(1.4.5)
_
The inertial force is given by fI ¼ m u€ while the damping force by fD ¼ cu.
Another convenient model to represent the single-story frame is shown in
Fig. 1.4.3c. It consists of a mass m placed at the top of a column with translational stiffness equal to the sum of the translational stiffness coefficients of the
columns of the frame. During the motion, the top cross-sections of columns
undergo only the translational displacement u ðt Þ. Models of this type are also
suitable to idealize multistory shear frames (see Fig. 1.4.4), in which the masses
are placed at the floor levels and the girders are assumed rigid.
FIG. 1.4.4 Four-story shear frame and its model without damping.
(a)
(b)
FIG. 1.4.5 Two-story, two-bay shear frame and its model without damping.
General concepts and principles of structural dynamics Chapter
1
15
Fig. 1.4.5a shows another two-story shear frame. The columns are assumed
weightless. Fig. 1.4.5b shows its dynamic model. The column 1-2 is represented
by a spring of stiffness k ¼ 12EI =h 3 . The stiffness coefficients k1 and k2 include
only the stiffness of the columns with heights h1 and h2 , respectively.
Given the dynamic model of the structure, the equation of motion of the system is formulated. For the SDOF system, the equation of motion can be formulated using Newton’s second law of motion as it is applied for the motion of a
particle
m u€ ¼ F
(1.4.6)
F ¼ pðt Þ fS fD
(1.4.7)
where
is the resultant of the external forces. Using Eqs. (1.4.1), (1.4.2), (1.4.7),
Eq. (1.4.6) is written
m u€ + cu_ + ku ¼ pðt Þ
(1.4.8)
Eq. (1.4.8) is the equation of motion of the SDOF system. The equation of
motion represents the dynamic equilibrium of the system. It is an ordinary differential equation of the second order with respect to the unknown variable u ðt Þ.
The solution of this equation yields the displacement as a function of time. For
MDOF systems, the number of equations of motion that must be formulated is
equal to the number of dynamic degrees of freedom. The use of Newton’s law of
motion is not always well suited to formulate the equations, especially for
MDOF systems or complex SDOF systems. It requires advanced knowledge
of the dynamics of the rigid and deformable body as well as mastering various
special methods. Generally, the equations of motion can be formulated using:
(a)
(b)
(c)
(d)
d’Alembert’s principle or method of equilibrium of forces.
Principle of virtual work.
Hamilton’s principle.
Lagrange’s equations.
These methods will be presented in the following and will be demonstrated by
appropriate examples. The acquaintance with the application of these methods
constitutes a fundamental presupposition for the analysis of the dynamic
response of structures.
1.5 Derivation of the equations of motion using
d’Alembert’s principle
Actually, d’Alembert’s principle is a different interpretation of Newton’s
second law of motion. Suppose that we write it in the form
F m€
u¼0
(1.5.1)
16 PART
I Single-degree-of-freedom systems
where F is the resultant of all external forces acting on the particle of mass m and
€ is its acceleration with respect to an inertial frame of reference.c If we consider
u
that the term m€
u is another force, known as inertial force, then Eq. (1.5.1) states
that the vector sum of all forces, external and inertial, is zero during the motion.
But this is the necessary and sufficient condition for the static equilibrium of the
particle. Thus, in a sense, the dynamic problem is reduced to a problem of statics
according to the following statement, known as d’Alembert’s principle.
The laws of static equilibrium can be applied also to a dynamic system with
respect to an inertial frame of reference if the inertial forces are considered as
applied forces on the system together with the actual external forces.
The motion of a rigid body of mass m with respect to an inertial frame of
reference X, Y ,Z is decomposed into a translational motion of its center of
mass, where the whole mass is considered to be concentrated, and a rotational
motion about it (Fig. 1.5.1).
FIG. 1.5.1 Rigid body moving with respect to the inertial. frame X,Y , Z .
If R ¼ X ðt Þi + Y ðt Þj + Z ðt Þk is the position vector of a particle A of
the body with respect to the inertial system of axes XYZ and
r ¼ x ðt Þe1 + y ðt Þe2 + z ðt Þe3 the position of the same point with respect to
the nonrotating system of axes xyz through the center of mass C (see
Fig. 1.5.1), then the equations of motion of the body can be written as
€c
F ¼ mR
(1.5.2a)
_c
Mc ¼ H
(1.5.2b)
where F ¼ Fx i + Fy j + Fz k is the resultant of the external forces,
€ c ¼ X€ ci + Y€ c j + Z€c k is the acceleration of the center of mass,
R
Mc ¼ Mx e1 + My e2 + Mz e3 is the resultant moment of the external forces with
c. In classical dynamics, an inertial frame of reference is a frame of reference in which a body with
zero force acting upon it is not accelerating; that is, the body is at rest or it is moving at a constant
velocity in a straight line [5].
General concepts and principles of structural dynamics Chapter
1
17
_ c is the rate of change of the angular momenrespect to the center to mass, and H
tum Hc of the body with respect to the same point given as
ZZZ
_ c¼
H
r r€rdV
(1.5.3)
V
in which r ¼ rðx, y, z Þ is the mass density of the body.
Eq. (1.5.2a) is the equation of the translational motion while Eq. (1.5.2b) is
the equation of the rotational motion.
For a plane body moving in its plane, Eqs. (1.5.2a), (1.5.2b) become (see
Appendix)
Fx ¼ m X€ c
(1.5.4a)
Fy ¼ m Y€ c
(1.5.4b)
Mc ¼ Ic w_
(1.5.4c)
where w is the angular velocity of the rotational motion about the center of mass
and Ic the polar moment of inertia of the body about the same point.
Path of P
FIG. 1.5.2 Plane body moving in the XY plane. The system of xy axes moves with P without
rotating.
It is often convenient to study the motion with reference to an arbitrary point
P, which is not the center of mass of the body (see Fig. 1.5.2). Special attention
should be paid in this case because Eqs. (1.5.2a), (1.5.2b) take the form
€ p ¼ m€rc
F mR
(1.5.5a)
€p¼H
_p
Mp rc m R
(1.5.5b)
where Rp is the position vector of point P moving with the body and rc the posi_ p are the moment of the
tion vector of the center of mass with respect to P. Mp , H
external forces and the rate of change of the angular momentum with respect to
P, respectively.
18 PART
I Single-degree-of-freedom systems
When small displacements are considered, as in the theory of linear vibrations, Eqs. (1.5.5a), (1.5.5b) become (see Appendix).
(1.5.6a)
Fx ¼ m X€ p yc w_
(1.5.6b)
Fy ¼ m Y€ p + xc w_
Mp ¼ Ip w_ + m xc Y€ p yc X€ p
(1.5.6c)
The kinetic energy of a plane body moving in its plane is given
(a) with respect to the center of mass
1
1 2
2
T ¼ m X_ c + Y_ c + Ic w2
2
2
(1.5.7)
(b) with respect to an arbitrary point P of the body (K€onig’s theorem)
1
1 2
2
T ¼ m X_ p + Y_ p + Ip w2 + m xc Y_ p yc X_ p w
2
2
(1.5.8)
We shall write now Eqs. (1.5.4a)–(1.5.4c) in terms of the displacement
vector. Apparently, the displacement vector from the beginning of the motion
is defined as
u ¼ Rðt Þ Rð0Þ ¼ u ðt Þi + v ðt Þj
(1.5.9)
u ¼ X ðt Þ X ð0Þ, v ¼ Y ðt Þ Y ð0Þ
(1.5.10)
where
€ Y€ ¼ v€. Moreover, if fðt Þ represents the change of the rotaHence, X€ ¼ u,
_ w_ ¼ f,
€ Eqs. (1.5.4a)–(1.5.4c) are
tion in the same time interval and set w ¼ f,
written in terms of displacements as
Fx ¼ m u€c
(1.5.11a)
Fy ¼ m v€c
(1.5.11b)
Mc ¼ Ic f€
(1.5.11c)
€c
Fc ¼ m c U
(1.5.12)
or in matrix form
where
8 9
8 9
2
3
m 0 0
< uc =
< Fx =
Fc ¼ Fx , Uc ¼ vc , mc ¼ 4 0 m 0 5
: ;
: ;
Mc
f
0 0 Ic
(1.5.13)
are the force vector, the displacement vector, and the mass matrix of the body,
respectively.
General concepts and principles of structural dynamics Chapter
Similarly, Eqs. (1.5.6a)–(1.5.6c) are written
Fx ¼ m u€p yc f€
Fy ¼ m v€p + xc f€
Mp ¼ Ip f€ + m xc v€p yc u€p
1
19
(1.5.14a)
(1.5.14b)
(1.5.14c)
or in matrix form
€p
Fp ¼ m p U
where
2
8
9
>
< Fx >
=
Fp ¼ F x
>
>
:
;
Mp
8 9
>
=
< up >
Up ¼ vp
>
;
: >
f
3
m
0
my c
mp ¼ 4 0
m
mx c 5
my c mx c Ic
(1.5.15)
(1.5.16a)
(1.5.16b)
(1.5.16c)
Note that the mass matrix is not diagonal when the point of reference is not
the center of mass.
Finally, Eqs. (1.5.7), (1.5.8) are written as
1
1 2
T ¼ m u_ 2c + v_ 2c + Ic f_
2
2
(1.5.17)
1_T
_c
¼ U
mc U
2 c
1
1 2
T ¼ m u_ 2p + v_ 2p + Ip f_ + m xc u_ p yc v_ p f_
2
2
(1.5.18)
1_T
_p
mp U
¼ U
2 p
The set of equations with reference to point P can also be derived from the
set of equations with reference to point C by transforming the displacements
and the forces from point C to P (see Section 10.7).
Example 1.5.1 Equation of motion of an elastically supported body
Consider the rigid plate of constant thickness and total mass m shown in
Fig. E1.1a. The plate is hinged at O and elastically supported at A. Formulate
20 PART
I Single-degree-of-freedom systems
the equation of motion of the system for small amplitude motion using the
method of equilibrium of forces.
Solution
The only possible motion of the plate is the rotation in its plane about the point
O. Hence, the system has one degree of freedom. The motion can be described
either by the rotation fðt Þ about O or by the translational displacement of
a point, for example, the displacement u ðt Þ of point B, which is related
to fðt Þ as
u ðt Þ ¼ a tan fðt Þ afðt Þ, u ðt Þ ¼ BB 0 cos f BB 0
(1)
because we have assumed small displacements. Moreover, AA0 ¼ BB 0 =2 ¼ u=2
The applied forces are shown in the free body diagram in Fig. E1.1b. These
are:
The weight of the body:
W ¼ mg
(2a)
2
fS ¼ k ðAA0 Þ ¼ ku
3
(2b)
The spring force:
The inertial force at the center of mass along x
fIx ¼ m
d2
1 b
ðCC 0 Þx ¼ m u€
2 a
dt 2
(2c)
The inertial force at the center of mass along y
fIy ¼ m
d2
1
ðCC 0 Þy ¼ m u€
2
2
dt
(2d)
The inertial moment about the center of mass
u€
MIc ¼ IC f€ ¼ IC
a
(2e)
pðt Þ
(2f)
The external force
The quantities ðCC 0 Þx and ðCC 0 Þy are the horizontal and the vertical displacements of the center of mass C due to rotation, respectively. They are
obtained from Fig. E1.1b as
General concepts and principles of structural dynamics Chapter
1
21
(a)
(b)
(c)
FIG. E1.1 Rigid plate in Example 1.5.1.
1b
u
2a
1
ðCC 0 Þy ¼ ðOC Þf cos b ¼ u
2
ðCC 0 Þx ¼ ðOC Þf sin b ¼
(3a)
(3b)
The equation of motion results from the dynamic equilibrium of moments
with respect to point O. Thus, we obtain
a
2a
b
a
W fS fIx fIy MIc + pðt Þa ¼ 0
2
3
2
2
(4)
22 PART
I Single-degree-of-freedom systems
which by virtue of Eq. (2) becomes
"
#
a 2 b2
m
+
+ IC
2
2
a2
4
W
+ pðt Þ
u€ + ku ¼
9
2
Using Steiner’s formula, we have
"
#
a 2 b2
a 2 + b2
IO ¼ m
+
+ IC ¼ m
3
2
2
(5)
(6)
Hence, Eq. (5) becomes
m
a 2 + b2
4
W
+ pðt Þ
u€ + ku ¼
2
3a
9
2
(7)
Eq. (7) can be also obtained if we consider the motion with reference to point
O and employ Eq. (1.5.14c) for
2a
a
u€
+ W + pðt Þa, MIo ¼ IO f€
u€O ¼ v€O ¼ 0, f€ ¼ IO , MO ¼ fS
3
2
a
The weight W can be eliminated from Eq. (7), if the total displacement u ðt Þ
is expressed as the sum of the static displacement ust caused by the weight plus
the additional dynamic displacement uðt Þ, as shown in Fig. E1.1c, that is,
u ðt Þ ¼ ust + uðt Þ
(8)
The static equilibrium of moments with respect to point O, when the plate is
loaded only by the weight yields (see Fig. E1.1c)
4
a
kau st ¼ W
9
2
(9)
Noting that u€st ¼ 0 because ust is a constant, and using Eqs. (8), (9), Eq. (7)
becomes
+ k ∗ u ¼ p ∗ ðt Þ
m ∗ u€
(10)
where
m∗ ¼ m
a 2 + b2
4
, k ∗ ¼ k, p ∗ ðt Þ ¼ pðt Þ
2
3a
9
Eq. (10) has the form of Eq. (1.4.8) and represents the equation of motion of
the system. The quantities m ∗ , k ∗ , having dimensions of mass and translational
stiffness, respectively, are referred to as the generalized mass and the generalized stiffness of the SDOF system.
If the rotation fðt Þ, measured from the position of static equilibrium, is
taken as the parameter of motion in place of uðt Þ, the equation of motion results
Thus, we have
from Eq. (10) using the relation u ¼ fa.
4
IO f€ + ka2 f ¼ apðt Þ
9
(11)
General concepts and principles of structural dynamics Chapter
1
23
Example 1.5.2 Equation of motion of a frame with a rigid column
Formulate the equation of motion of the plane frame shown in Fig. E1.2a for
small amplitude motion. The mass of the horizontal beam CD is negligibly small
while the column of height L and nonnegligible width h ¼ L=4 is assumed rigid
The elastic stiffness of the ground is simulated by the
with total mass m ¼ mL.
spring CR while its damping by the two dashpots with damping parameters c.
(a)
(b)
FIG. E1.2 Frame with a rigid column in Example 1.5.2.
Solution
The only possible motion of the system is the rotation fðt Þ of the column as a
rigid body about the hinged support at point A of its base. Because the rotation is
small, we have:
sin f f, cos f 1, f2 0
Hence
u ¼ L sin f ¼ Lf,
h
h
d ¼ sin f f
2
2
24 PART
I Single-degree-of-freedom systems
h
h
ð1 2Þ ¼ sin f + L cos f f + L,
2
2
h
h
ð3 4Þ ¼ L sin f + cos f Lf + ,
2
2
h
h
ð5 6Þ ¼ cos f 2
2
The forces applied on the column are shown in Fig. E1.2b. These are:
The elastic moment at the corner C
6EI
4EI
d + 1:5L
f
MS ¼ ð1:5L
Þ2
The elastic moment due to the rotational spring MR ¼ CR f ¼ EI
L f
The moment of inertia of the mass m
2
€
MIA ¼ IA f€ ¼ mL
3 f
The elastic shear force at the beam end C
12EI
6EI
QS ¼ ð1:5L
d + ð1:5L
f
Þ3
Þ2
The damping forces
fD ¼ c h2 f_
The external load
pðt Þ
The equilibrium of moments with respect to point A yields
MIA + MS + MR + QS ð3 4Þ + 2fD ð5 6Þ pðt Þ ð1 2Þ ¼ 0
which after substituting their exressions becomes
mL2 € cL2 _
79EI pL
28EI 2
f+
f+
f+
f ¼ Lpðt Þ
3
32
18L
8
9L
Further, linearizing (f2 0) gives
mL2 € cL2 _
79EI pL
f+
f+
f ¼ Lpðt Þ
3
32
18L
8
(1)
(2)
If the displacement u ¼ Lf at the level of the beam is taken as the parameter
of the motion, the equation of motion becomes
m
cL
79EI p
u€ + u_ +
u ¼ pðt Þ
(3)
3
32
18L3 8L
Example 1.5.3 Equation of motion of a system of rigid bodies
The rigid body assemblage shown in Fig. E1.3a consists of the rigid bar AF of
total mass m hinged at A, and the rigid square plane body supported rigidly at F.
The dynamic excitation of the bar is due to the uniformly distributed transverse
load pðt Þ. The motion is constrained by a spring at B and the damper at G. Formulate the equation of motion of the system for small amplitude displacements
using the method of equilibrium of forces. The mass per unit length of the bar is
¼ m=3L and the surface mass density of the body is g ¼ 2m=L2
m
General concepts and principles of structural dynamics Chapter
1
25
(a)
(b)
FIG. E1.3 System with two rigid bodies in Example 1.5.3.
Solution
As the bar AF is rigid, the only possible motion is its rotation about A. Hence,
the system has a SDOF. Its motion can be described either by the angle of rotation fðt Þ about the hinge at A or by the transverse displacement of any point
along the axis of the bar. We choose the upward displacement u ðt Þ at point
B as the parameter of the motion. For small amplitude motion, the forces acting
on the system are shown in Fig. E1.3b. These are:
The elastic force fS at B: As it opposes the motion, it is directed downward
and is expressed as
fS ¼ ku
(1)
The damping force fD at G: It is directed also downward and is expressed as
fD ¼ c
d
d
ðGG 0 Þ ¼ c ð1:625u Þ ¼ 1:625cu_
dt
dt
(2)
The inertial force fIK and the inertial moment MIK at the center of mass K of
are
the bar due the distributed mass m
Þ
fIK ¼ ðm3L
d2
ðKK 0 Þ ¼ 0:75m u€
dt 2
MIK ¼ IK f€
(3)
26 PART
I Single-degree-of-freedom systems
or taking into account that
ð3LÞ3 m ð3LÞ3
¼
¼ 0:75mL2
12
3L 12
u
u
u€
f ¼ ¼ 0:5 , f€ ¼ 0:5
2L
L
L
IK ¼ m
we obtain
MIK ¼ 0:375mLu€
(4)
The inertial force fIG and the inertial moment MIG at the center of mass G of
the rigid body due to the mass gL2 =2:
L d2
2m L2 d 2
ðGG 0 Þ ¼ 2
ð1:625u Þ ¼ 1:625m u€
(5)
fIG ¼ gL
2
2 dt
L 2 dt 2
"
#
LðL=2Þ3 ðL=2ÞL3
u€
G
€
+
¼ 0:052mLu€
(6)
M I ¼ IG f ¼ g
0:5
12
12
L
The external load is 2L
pðt Þ.
The equilibrium of the moments about A yields the equation of motion of the
system. Thus, we have
fS ð2LÞ fD ð3:25LÞ fIK ð1:5LÞ fIG ð3:25LÞ
MIK MIG + pðt Þ ð2LÞ L ¼ 0
(7)
or inserting Eqs. (1)–(6) into Eq. (7) we obtain
m ∗ u€ + c ∗ u_ + k ∗ u ¼ p ∗ ðt Þ
(8)
m ∗ ¼ 6:833m, c ∗ ¼ 5:281c, k ∗ ¼ 2k, p ∗ ðt Þ ¼ 2L
pðt Þ
(9)
where
The quantities defined by Eq. (9) are referred to as the generalized mass, the
generalized damping, the generalized stiffness, and the generalized load,
respectively.
Once the dynamic displacement u ðt Þ is established from the solution of
Eq. (8), the vertical reaction RA can be evaluated from the dynamic equilibrium
of forces in the direction of the y axis. This yields
RA + pðt Þ2L fIK fIG fS fD ¼ 0
or using Eqs. (1)–(3), (5) we obtain
RA ¼ ku + 1:625cu_ + 2:375m u€ 2L
pðt Þ
General concepts and principles of structural dynamics Chapter
1
27
Example 1.5.4 Equation of motion of a single-story shear building
Formulate the equation of motion of the single-story building shown in Fig. E1.4a.
The damping is neglected. The columns are fixed on the ground, are inextensible,
and their mass is assumed to be lumped at their ends. Moreover, the roof plate
is assumed rigid. The material of the structure is reinforced concrete having specific weight g ¼ 24kN=m3 and modulus of elasticity E ¼ 2:1 107 kN=m2 . The
total load of the plate (dead and live) is 20kN=m2 . The force pðt Þ acts in the direction of the x axis and is given by pðt Þ ¼ 20sin 13t. The acceleration of gravity is
g ¼ 9:81m=s2 and the dimensions of the rectangular cross-sections of columns are
k1 : 30 30cm2 and k2 : 30 20cm2 .
(b)
(a)
FIG. E1.4 Single-story shear building in Example 1.5.4.
Solution
Taking into account that the structure is symmetric with respect to the x axis,
the columns are inextensible, and the load pðt Þ acts on the axis of symmetry, the
only possible motion of the plate is the horizontal displacement u ðt Þ in the
direction of the x axis. The SDOF model of the structure is shown in Fig. E1.4b.
The total mass of the system is due to the load of the plate and to half the
weight of the columns
m¼
5 10 20 + ð4 0:3 0:3 + 2 0:3 0:2Þ 2 24
¼ 104:285
9:81
The stiffness of the system is equal to the sum of the translational stiffness
coefficients of all columns, which are given as
ki ¼
12EI i
hi3
where Ii is the moment of inertia of the cross-section of the i column with
respect to the y axis through its mass center and hi its height. Thus, we have
28 PART
I Single-degree-of-freedom systems
Columns 30 30:
k3030 ¼
12 2:1 107 43
0:304
12 ¼ 2657:8kN=m
Columns 30 20:
k3020 ¼
12 2:1 107 43
0:303 0:20
12
¼ 1771:9kN=m
Therefore the stiffness of the system is
k ¼ 4 2657:8 + 2 1771:9 ¼ 14175:0kN=m
The equation of motion results from the equilibrium of the forces shown in
Fig. E1.4b. This yields
fI fS + pðt Þ ¼ 0
or
m u€ + ku ¼ pðt Þ
Substituting the numerical values for m, k and the expression for pðt Þ, the
above equation of motion becomes
5:21u€ + 708:75u ¼ sin 13t
Example 1.5.5 Equation of motion of a two-story shear frame
Formulate the equations of motion of the two-story shear frame shown in
Fig. E1.5a using the method of equilibrium of forces. The damping is ignored.
Solution
The system has two degrees of freedom because the girders are rigid and the
axial deformation of columns is ignored. The model of the structure is shown
in Fig. E1.5b. The masses are lumped at the story levels. The motion of the
system can be fully determined from the horizontal displacements u1 ðt Þ and
u2 ðt Þ of the masses m1 and m2 , respectively.
Rigid
Rigid
(a)
(b)
FIG. E1.5 Two-story shear frame in Example 1.5.5.
(c)
General concepts and principles of structural dynamics Chapter
1
29
The equations of motion result from the dynamic equilibrium of forces
applied to the masses m1 and m2 . These forces are shown in the free body diagrams of Fig. E1.5c. Thus, we obtain
m1 u€1 + k1 ðu1 u2 Þ ¼ p1 ðt Þ
(1)
m2 u€2 k1 ðu1 u2 Þ + k2 u2 ¼ p2 ðt Þ
(2)
Eqs. (1), (2) are written in matrix form as
m€
u + ku ¼ pðt Þ
(3)
where
u¼
u1
m1 0
k1 k1
, m¼
, k¼
, pðt Þ ¼
u2
k1
k1 + k2
0 m2
p1 ðt Þ
p2 ðt Þ
Example 1.5.6 Equation of motion of a general single-story shear building
The rigid horizontal plate is supported by K columns as shown in Fig. E1.6. The
columns are fixed on the ground as well as on the plate. Their principal axes have
arbitrary directions in the xy plane. Formulate the equation of motion of the
ðt Þ through the point A.
plate when the plate is loaded by the horizontal load P
Solution
We choose O xy as the system of reference of the motion, whose origin coincides with point O at the beginning of motion. Let xi , yi represent the coordinates of the center of mass of the cross-section of i column and fi the angle
between its principal x axis and the x axis. The axes xy will be referred to
as the global axes of the system while the axes xy as the local axes of the
column.
Inasmuch as the axial deformation of columns is ignored, the only possible
motion of the plate is inside its plane, which can be determined by the two
translational displacements of a point and the rotation of the plate. We study
the motion of the plate with reference to point O and let U , V represent its
translational components with respect to the global axes xy, which are
the rotation of the plate. As a consequence
assumed fixed in the plane, and W
of this motion, the cross-section of the i column at the level of the plate
undergoes the displacements u i , v i , wi , with respect to its base. These displacements generate elastic forces X i , Y i , M i , which act on the plate. Thus,
we define the following vectors and matrices that will be used in the subsequent analysis.
30 PART
I Single-degree-of-freedom systems
FIG. E1.6 General single-story shear building in Example 1.5.6.
(a) In global axes:
8 9
< U =
¼ V displacements of point O of the plate
U
:;
W
8 9
< ui =
i ¼ vi displacements of i column
D
: i;
w
8 i9
< X =
i ¼ Y i elastic forces of i column
F
S
: i;
M
(b) In local axes
8 9
< ui =
i
D ¼ v i displacements of i column
: i;
w
8 9
< Xi =
FiS ¼ Y i elastic forces of i column
: i;
M
General concepts and principles of structural dynamics Chapter
1
31
The transformation matrix for the vector quantities related to i column from the
global axes to the local axes is given as
2
3
cos fi sin fi 0
Ri ¼ 4 sin fi cos fi 0 5
0
0
1
Hence, the vectors are transformed from one system of axes to the other as
i
D i ¼ Ri D
i ¼ Ri T D i
D
i
FiS ¼ Ri F
S
i ¼ Ri T F i
F
S
S
where
(1a)
(1b)
(2a)
(2b)
3
cos fi sin fi 0
R
¼ 4 sin fi
cos fi 0 5
0
0
1
i 1 i T
i
¼ R
because Ri is orthonormal.
is the transpose of R . Note that R
i
i
i
The elastic forces X , Y , M are related to the displacements u i , v i , wi by
i T
2
12EI y i
u
h3
12EI x i
v
Yi ¼
h3
Xi ¼
Mi ¼
GI t i
w
h
(3a)
(3b)
(3c)
where Ix ,Iy are the principal moments of inertia of the column cross-section and
It is the torsional constant, E and G are the material constants, and h is the
height of the column.
Setting
12EI y
12EI x
GI t
i
i
, k22
¼
, k33
¼
h3
h3
h
Eqs. (3a)–(3c) can be written in matrix form as
8 9 2 i
38 9
k11 0 0 < u i =
< Xi =
i
0 5 vi
Y i ¼ 4 0 k22
: i;
: i;
i
0 0 k33
M
w
i
¼
k11
(4)
or
Fi ¼ k i D i
(5)
32 PART
I Single-degree-of-freedom systems
The matrix
2
i
0
k11
6
ki ¼ 4 0
0
3
0
7
i
k22
0 5
i
0 k22
is the stiffness matrix of i column.
Eq. (5) is transformed in global axes using Eqs. (2b), (1a). Thus, we have
i ¼ Ri T F i
F
S
S
T
¼ Ri k i D i
T
i
¼ Ri k i R i D
or
i D
i ¼ k
i
F
(6)
i ¼ Ri T k i R i
k
(7)
where
is the stiffness matrix of the column in global axes, which becomes after performing the matrix multiplications
2 i
3
i
k11 k12 0
6
7
i ¼ 6 ki ki 0 7
(8)
k
4 21 22
5
i
0 0 k
33
where
i
i
i
k11 ¼ k11
cos 2 fi + k22
sin 2 fi
9
>
>
>
>
>
>
=
i
i
i
k22 ¼ k11
sin 2 fi + k22
cos 2 fi
i
i
i
i
>
k12 ¼ k21 ¼ k11
k22
sin fi cos fi >
>
>
>
>
;
i
i
k 33 ¼ k33
(9)
Inasmuch as the plate is rigid, the displacements ui , vi , wi of the i column
of point O. The geometrical reladepend on the plate displacements U , V , W
tions result from the following consideration.
General concepts and principles of structural dynamics Chapter
1
33
FIG. E1.7 Displacements of point i due to rotation of plate.
The point i of the plate undergoes translational displacements due to
(a) The translational displacements of point O
i
u t ¼ U
i
v ¼ V
t
(b) The rotation of the plate about O. Referring to Fig. E1.7 and observing that
¼ wi , we obtain
cos ai ¼ xi =ri , sin ai ¼ yi =ri , W
i
sin ai ¼ yi W
u r ¼ ri W
i
cosai ¼ xi W
v r ¼ ri W
Thus, we have
ui ¼ ui t + ui r ¼ U yi W
vi ¼ vi t + vi r ¼ V + xi W
(10b)
wi ¼ W
(10c)
The previous equations are written in matrix form as
8 9 2
38 9
1 0 y i < U =
< ui =
vi ¼ 4 0 1 xi 5 V
: i;
:;
0 0 1
w
W
or setting
2
3
1 0 yi
ei ¼ 4 0 1 xi 5
0 0 1
(10a)
(11)
34 PART
I Single-degree-of-freedom systems
we can write Eq. (11) as
i ¼ ei U
D
(12)
The matrix e defined by Eq. (12) is referred to as the translation matrix or
transformation matrix of the i column.
The equations of motion result from Eqs. (1.5.14a)–(1.5.14c) if they are
The external force
employed for point O, with P O, up ¼ U , vp ¼ V , f ¼ W.
ðt Þ ¼ P x ðt Þ, P y ðt Þ T . Thus,
F is equal to the sum of all elastic forces FiS plus P
we have
i
P x ðt Þ K
X
i
€
yc W
X ¼ m U€
(13a)
i
€
+ xc W
Y ¼ m V€
(13b)
i¼1
P y ðt Þ K
X
i¼1
ðt Þ M
K X
i
i
€
i ¼ m xc V€ yc U€ + Io W
xi Y yi X + M
(13c)
i¼1
ðt Þ with respect
ðt Þ ¼ xA P y ðt Þ yA P x ðt Þ is the moment of the force P
where M
to O, m is the mass of the plate, and Io its moment of inertia with respect to O.
Eqs. (13a)–(13c) are written in matrix form as
8
9
32
2
38 i 9 2
32
38
€
>
1 0 0
1 0 0 >
1 0 yc >
m 0 0
< P x ðt Þ >
= X
< X >
=
< U
K
6
7
6
7
6
7
6
7
i
¼ 4 0 1 0 54 0 m 0 54 0 1 xc 5 V€
P y ðt Þ 4 0 1 0 5 Y
>
>
>
>
: ; i¼1
: i>
;
: €
y c xc 1
y i xi 1
M ðt Þ
0 0 Ic
0 0 1
M
W
9
>
=
>
;
or
ðt Þ P
K X
T i
¼ ðec ÞT mðec ÞU
€
ei F
(14)
i¼1
where
9
8
>
=
< P x ðt Þ >
ðt Þ ¼ P y ðt Þ
P
>
;
: >
M ðt Þ
2
3
m 0 0
m ¼ 40 m 0 5
0 0 Ic
(15a)
(15b)
Finally, using Eqs. (6), (12), we obtain the equation of motion
U
€ + K
U
¼P
ðt Þ
M
(16)
General concepts and principles of structural dynamics Chapter
where
3
m
0
m yc
¼ ðec ÞT mec ¼ 4 0
m
m xc 5
M
m yc m xc
Io
1
35
2
¼
K
K X
T
T
ei
Ri ki Ri ei
(17a)
(17b)
i¼1
are the mass and stiffness matrices of the structure, respectively.
The equation of motion (16) can be transformed with reference to the center
of mass by working as follows.
Using Eq. (12), we relate the displacements of the center of mass C to the
displacements of O. Hence, we have
c ¼ ec U
U
(18)
¼ ðec Þ1 U
c
U
(19)
which can be inverted to give
We can readily show that
2
3
1 0 yc
ðec Þ1 ¼ 4 0 1 xc 5
0 0 1
(20)
Substituting Eq. (19) into Eq. (16) and premultiplying it byðec ÞT , we
obtaind
cU
ðt Þc
c ¼P
€ + K
mU
(21)
c ¼ ðec ÞT K
ðec Þ1
K
(22a)
ðt Þ
c ðt Þ ¼ ðec ÞT P
P
(22b)
c
where
Eq. (22a) represents the transformed stiffness matrix of the structure from
point O to the center of mass C .
The stiffness matrix with respect to an arbitrary point O is, in general, a full
3 3 matrix, namely
3
2
k11 k12 k13
7
¼6
(23)
K
4 k21 k22 k23 5
k31 k32 k33
T
d. The notation eT ¼ ðe1 Þ is employed.
36 PART
I Single-degree-of-freedom systems
The off-diagonal terms cause coupling between the elastic force in one
direction and the displacement in another direction. For example, the element
k12 represents the force acting in the direction of the x axis when the plate
undergoes a unit displacement in the direction of the y axis. Similarly, the element k31 represents the moment acting on the plate about the z axis, if the plate
undergoes a unit displacement, U ¼ 1, in the direction of the x axis. The elastic
center or center of resistance of the plate is defined as the point of the plate
where an applied force in any direction does not produce rotation. This implies
the vanishing of the elements k13 and k23 (hence also their symmetric k31 and
k32 ) in the stiffness matrix (23). This point can be established as follows.
is transformed from point O to the sought elastic cenThe stiffness matrix K
ter E ðxE , yE Þ according to Eq. (22a), if ec is replaced by eE . Namely
2
32
3
32
1
0 0
1 0 yE
k11 k12 k13
T
1
eE
E ¼ eE
¼ 40
1 0 54 k21 k22 k23 54 0 1 xE 5
K
K
0 0 1
yE xE 1
k31 k32 k33
or after performing the matrix multiplications
2 E E E3
k11 k12 k13
6
7
6 E E E7
E
6
K ¼ 6 k21 k22 k23 7
7
4
5
E E E
k 31 k 32 k 33
2
k12
k11
6
6
¼ 6 k21
k22
4
k11 yE k21 xE + k31 k12 yE k22 xE + k32
k11 yE k12 xE + k13
3
7
7
k21 yE k22 xE + k23 7
5
2
2
k 13 yE + k 23 xE + k 33
E
E
The vanishing of the elements k13 and k23 yields
k11 yE k12 xE + k13 ¼ 0
k21 yE k22 xE + k23 ¼ 0
from which we establish the coordinates of E
k11
k
xE ¼ 21
k 11
k13
k23
k12
k12
k
yE ¼ 22
k 11
k13
k23
k12
(24a)
k21 k22
k21 k22
(24b)
General concepts and principles of structural dynamics Chapter
1
37
Thus, the stiffness matrix with respect to the elastic center takes the form
2 E E
32
3
k11 k12 0
k11 k12 0
6
7
7
E ¼ 6 kE kE 0 76
K
5
4 21 22
54 k 21 k 22 0
2
2
E
0 0 k 13 yE + k 23 xE + k 33
0 0 k
33
The coupling is now limited between the translational displacements and
the corresponding elastic forces. They can also be decoupled if the matrix is
y by rotation
transformed into a new system of axes x 0 Ey 0 , resulting from xE
through an angle q and demanding the off-diagonal elements to vanish. The
i ¼ K
E and
stiffness matrix in the new system is obtained using Eq. (7) for k
2
3
cos q sin q 0
R ¼ 4 sin q cos q 0 5
0
0
1
Thus, we have
ER
^ E ¼ RT K
K
(25)
or after performing the matrix multiplications
3
k11 k22
7
6 k cos 2 q + k sin 2 q k sin 2q
0
sin 2q + k 12 cos2q
22
12
7
6 11
2
7
6
E
^
7
K ¼ 6 k 11 k 22
2
2
7
6
sin 2q + k12 cos 2q
k11 sin q + k22 cos q + k12 sin 2q 0
5
4
2
0
0
k13 y2 + k23 x 2 + k33
E
E
2
The vanishing of the off-diagonal elements yields
2k12
tan 2q ¼ k 22 k11
(26)
The axes defined by angle q are referred to as the principal directions of stiffness of the structure. The stiffness matrix becomes now diagonal
2
3
k^11 0 0
7
^E ¼6
K
4 0 k^22 0 5
0 0 k^33
where
k^11 ¼ k11 cos 2 q + k22 sin 2 q k12 sin 2q
(27a)
k^22 ¼ k11 sin 2 q + k22 cos 2 q + k12 cos 2q
(27b)
k^33 ¼ k13 y2E + k23 x2E + k33
(27c)
38 PART
I Single-degree-of-freedom systems
The previous analysis allows us to draw the following conclusions:
(a) In static analysis, the concepts of the elastic center and the principal directions permit the uncoupling of the three equations of static equilibrium and
give a better insight into the deformation of the structure. These concepts,
however, do not have any advantage in the dynamic analysis because, in
general, the center of mass does not coincide with the elastic center and
therefore the equations of motion remain coupled through the second derivatives of the displacements. Apparently, this fact complicates the study of
the dynamic analysis.
(b) The equations
of motion
can be decoupled with respect to the physical dis only if the center of mass coincides with the elastic
placements U , V , W
center. This uncoupling should not be confused with that achieved via
modal coordinates (see Chapter 12).
1.6 Principle of virtual displacements
D’Alembert’s principle allows the application of the principle of virtual displacements to formulate the equations of motion of structural systems, especially for complex assemblages comprising a number of interconnected
particles or rigid bodies of finite size. The principle of virtual displacements
or virtual works can be expressed as follows:
The necessary and sufficient condition for the dynamic equilibrium of a
system is the vanishing of the total work done by the set of all externally applied
forces (actual and inertial) when the system is subjected to a virtual displacement, that is, a displacement pattern compatible with the geometrical constraints of the system.
Thus, the equations of motion of the dynamic system can be derived by first
identifying all forces (imposed external forces, elastic forces, damping forces,
and inertial forces) acting on the masses. Then by introducing a virtual displacement corresponding to each degree of freedom, the equations of motion are
obtained by setting the virtual work produced by all forces equal to zero. A
major advantage of this approach is that the contribution of the work done
by the reactions of nonyielding bilateral supports as well as the internal interaction forces on the separated masses do not appear explicitly in the equations.
Moreover, the quantities we have to deal with are scalar and not vectorial, thus
they can be algebraically manipulated. An important provision for the application of the principle of virtual displacements is that the masses of the system are
subjected to small displacements. This implies that the geometry of the structure
remains essentially unchanged after the action of the displacements.
Example 1.6.1 Equation of motion of a complex SDOF system
Formulate the equation of motion of the system shown in Fig. E1.8a by using the
principle of virtual displacements for small amplitude motion. It is assumed that
the cable can undertake compression.
General concepts and principles of structural dynamics Chapter
1
39
(a)
(b)
FIG. E1.8 System in Example 1.6.1.
Solution
Because the cable is inextensible, the displaced configuration of the system can
be specified either by the angle of rotation of one of the bars or by the transverse
displacement of a point on it. Thus, the system has only one degree of freedom.
If the upward transverse displacement u ðt Þ of point C is taken as the basic
parameter of the motion, then all other displacements can be expressed in terms
of it. Fig. E1.8b shows the deformed system with all forces applied to it.
The elastic forces fS1 and fS2 are due to the deformation of the springs k1 and
k2 . They are directed downward as they oppose the motion. The force fD is
due to the viscous damping mechanism and is directed upward as it also
opposes the motion. The inertia moments MIA , MIO , and MIE are due to the rotation of the masses about A, O , and E, respectively. All forces are expressed in
terms of the single displacement u ðt Þ
fS1 ¼ k1 ðBB 0 Þ ¼ ku=2, fS2 ¼ k2 ðCC 0 Þ ¼ 2ku
fD ¼ c
d
ðDD 0 Þ ¼ cu_
dt
ð2LÞ3 u€
m
2 u€
MIA ¼ IA f€1 ¼
¼ 1:333mL
3 2L
MIE ¼ IE f€2 ¼
ð1:5LÞ3 u€
m
2 u€
¼ 0:750mL
3
1:5L
MIO ¼ IO f€3 ¼ mL
ð0:8LÞ2 u€
2 u€
¼ 0:200mL
8 0:4L
40 PART
I Single-degree-of-freedom systems
If point C is given a virtual displacement du, the forces ride the following
displacements
d ðCC 0 Þ ¼ du, dðBB 0 Þ ¼ du=2, dðDD0 Þ ¼ du
df1 ¼ du=2L, df2 ¼ du=1:5L, df3 ¼ du=0:4L
du ðx Þ ¼ xdf1 ¼ xdu=2L
The work done by the forces acting on the system due to the virtual displacement should be set equal to zero, that is,
fS1 d ðBB 0 Þ fS2 dðCC 0 Þ fD dðDD 0 Þ MIA df1
Z L
MIE df2 MIO df3 +
pðt Þdu ðx Þdx ¼ 0
(1)
0
Using the expressions for the forces and the displacements in terms of the
basic displacement derived previously, Eq. (1) yields
2 u=2L
€
2 u=1:5L
€
0:25ku 2ku cu_ 1:333mL
0:750mL
€
2 u=0:4L
+ pðt ÞL=4du ¼ 0
0:200mL
or, inasmuch as du 6¼ 0, the expression within the square brackets should vanish.
This yields the equation of motion
(2)
m ∗ v€ + c ∗ v_ + k ∗ v ¼ p ∗ ðt Þ
where
c ∗ ¼ c, k ∗ ¼ 2:25k, p ∗ ðt Þ ¼ 0:25
pðt ÞL
m ∗ ¼ 1:667mL,
Example 1.6.2 Equation of motion of a rigid body assemblage
Formulate the equations of motion of the rigid body assemblage shown inFig. E1.9a by using the principle of virtual displacements on the basis of small
amplitude motion.
Solution
Due to the spring k1 , the rigid bars can rotate independently from each other
about their hinged supports at A and F. Hence, the system has two degrees
of freedom. Its motion can be specified by the transverse downward displacements u1 ðt Þ and u2 ðt Þ of points C and E, respectively. The forces applied to the
displaced system are shown in Fig. E1.9b. They are
The elastic force fS1 ¼ k1 ðCC 0 Þ ¼ k ðu2 u1 Þ
The elastic force fS2 ¼ k2 ðDD 0 Þ ¼ 4ku 2
The damping force fD ¼ c dtd ðBB 0 Þ ¼ c u_21
u€1
2
€1
The inertial moment MIA ¼ IA f€1 ¼ IA 2a
¼ 4ma
3 u
u€2
8ma
2
F
€
The inertial moment M ¼ IF f2 ¼ IF ¼
u€2
I
a
3
The system is given a virtual displacement pattern du1 and du2 corresponding
to the two degrees of freedom. The forces ride the following displacements
d ðBB 0 Þ ¼
du1
du1
, dðCC 0 Þ ¼ du1 , df1 ¼
2
2a
General concepts and principles of structural dynamics Chapter
1
41
(a)
(b)
FIG. E1.9 System in Example 1.6.2.
d ðDD 0 Þ ¼ 2du2 , d ðEE 0 Þ ¼ du2 , df2 ¼
du2
a
According to the principle of virtual displacements, the work done by the
applied forces must be equal to zero, that is,
MIA df1 fD dðBB 0 Þ + fS1 dðCC 0 Þ fS1 dðEE 0 Þ fS2 dðDD 0 Þ
MIF df2 + pðt Þd ðCC 0 Þ ¼ 0
(1)
Introducing the expressions of the forces and virtual displacements into
terms of the basic displacements in Eq. (1) yields
2ma
u_ 1
u€1 c + k ðu2 u1 Þ + pðt Þ du1
4
3
(2)
8ma
+ u€2 k ðu1 u2 Þ 8ku 2 du2 ¼ 0
3
Inasmuch as the quantities du1 and du2 are arbitrary, Eq. (2) is valid
only if
2ma
u_ 1
u€1 + c k ðu2 u1 Þ pðt Þ ¼ 0
4
3
(3a)
8ma
u€2 + k ð9u2 u1 Þ ¼ 0
3
(3b)
42 PART
I Single-degree-of-freedom systems
Eqs. (3a), (3b) are the equations of motion of the system. In matrix form they
are written as
3
2
2 c 3( ) "
#( ) (
)
2ma
0
u_ 1
k k
u1
pðt Þ
0
7 u€1
6 3
44 5
+
¼
(4)
4
5 u€2 +
8ma
u2
u_ 2
k
9k
0
0
0 0
3
1.7 Hamilton’s principle
The development of dynamics and generally of mechanics has been accomplished through two different approaches. The first is based on Newton’s laws
of motion. These laws deal with the motion of a body under the action of forces
acting on it. The involved quantities are (i) the imposed forces, which may be
externally applied forces, forces of interaction between the masses, and reactions
of constraints and (ii) the momentum or the quantity of motion as defined by
Newton. Because both quantities are vector quantities, this approach of mechanics is called vectorial mechanics. The analysis of complicated systems by direct
application of Newton’s laws of motion becomes increasingly difficult. The principal reason is that the equations are vectorial in nature and the forces and accelerations are often difficult to determine. Moreover, the reactions of the constraints
and the interaction forces between bodies must be explicitly accounted in the
equations of motion and have to be evaluated even when there is no interest to
evaluate them. In addition, each problem seems to require its own particular insights and there are no general procedures for obtaining the equations of motion.
The second approach is based mainly on the work of Lagrange and Hamilton
and is called analytical mechanics. In this approach, the involved quantities are
scalar functions, and therefore the fundamental equations, in contrast to vectorial
mechanics, do not depend on the choice of the coordinates. Also, it is not necessary to include explicitly the forces of the constraints and the interaction forces. It
will be shown that this approach circumvents to some extent the difficulties found
in the direct application of Newton’s law of motion to complicated systems. Furthermore, the equations of motion are formulated in a standard convenient form.
Analytical dynamics is based on Hamilton’s principle and Lagrange’s equations.
Hamilton’s principle is presented in this section. Lagrange’s equations are presented in the next section resulting directly from Hamilton’s principle.
One of the most important principles of dynamics is Hamilton’s principle,
named after the famous Irish mathematician and physicist Sir William Rowan
Hamilton (1805–65). Inertial and elastic forces are not explicitly involved in
this principle; instead, variations of the kinetic and potential energy are utilized.
This formulation has the advantage of dealing only with purely scalar quantities. In the procedure of virtual displacements, even though the works themselves are scalar quantities, vector quantities, displacements, and forces are
utilized to represent them. Hamilton’s principle is presented here for discrete
parameter systems.
General concepts and principles of structural dynamics Chapter
1
43
FIG. 1.7.1 Particle moving in space.
Consider a particle of mass m moving in space under the action of a force
Fðt Þ as shown in Fig. 1.7.1. If r ¼ rðt Þ ¼ x ðt Þi + y ðt Þj + z ðt Þk represents the
position vector of the particle at time t, then according to Newton’s second
law of motion, the Newtonian path of the particle is governed by the differential
equation
m
d 2r
F¼0
dt 2
(1.7.1)
We confine our attention to an interval of time during which the particle
moves from point 1 at t ¼ t1 to point 2 at t ¼ t2 . We consider now a varied path,
specified by rðt Þ + drðt Þ, adjacent to the actual one. We will refer to the quantity drðt Þ ¼ dx ðt Þi + dy ðt Þj + dz ðt Þk as the variation of r. The only restriction is
that the two paths coincide at time t ¼ t1 and t ¼ t2 . This implies that the variation dr ¼ drðt Þ vanishes at these instants, that is,
drðt1 Þ ¼ drðt2 Þ ¼ 0
(1.7.2)
The first step to derive Hamilton’s principle is to take the inner product of
the left side of Eq. (1.7.1) with the vector dr and to integrate from time t1 to time
t2 . This gives
Z
t2 m
t1
d2r
dr
F
dr
dt ¼ 0
dt 2
(1.7.3)
Integrating by parts the first term in the above integral and knowing that the
operator d acts like the differential operator [6], we obtain
Z
t2
t1
d 2r
dr
m 2 drdt ¼ m dr
dt
dt
t2
t1
Z
t1
t2
dr
dr
m d
dt
dt
dt
44 PART
I Single-degree-of-freedom systems
The term outside the integral is equal to zero because of Eq. (1.7.2). Moreover, we can write the integrand as
" #
dr
dr
1 dr dr
1 dr 2 1
dr 2
¼ d m
¼m d
¼m d
¼ dT
m d
dt
dt
2 dt dt
2 dt
2
dt
where
2
1
dr
¼ x_ ðt Þ2 + y_ ðt Þ2 + z_ ðt Þ2
T¼ m
2
dt
(1.7.4)
is the kinetic energy of the particle. Hence, the integral (1.7.3) takes the form
Z t2
ðdT + F drÞdt ¼ 0
(1.7.5)
t1
The variation dr is a virtual displacement that leads from the actual path to
the varied one. Hence the term F dr in Eq. (1.7.5) is the virtual work done
by the force Fðt Þ. Eq. (1.7.5) is a statement of Hamilton’s principle as it is
applied to a particle. This equation can be transformed into a more convenient
form if the force Fðt Þ is separated in its conservative and nonconservative
components, that is
Fðt Þ ¼ Fc ðt Þ + Fnc ðt Þ
(1.7.6)
A potential function A ¼ Aðx, y, z, t Þ exists from which the conservative
force Fc ðt Þ is derived as its minus gradient
∂A
∂A
∂A
i+
j+
k
(1.7.7)
Fc ¼ ∂x
∂y
∂z
Hence
Fc dr ¼ ∂A
∂A
∂A
dx +
dy +
dz
∂x
∂y
∂z
or
Fc dr ¼ dA
Hence, Hamilton’s principle, Eq. (1.7.5), can be written as
Z t2
Z t2
d ðT AÞdt +
dWnc dt ¼ 0
t1
t1
where
dWnc ¼ Fnc dr
represents the virtual work of the nonconservative force.
(1.7.8)
(1.7.9)
General concepts and principles of structural dynamics Chapter
1
45
In the absence of nonconservative forces, Fnc ¼ 0, Eq. (1.7.9) becomes
Z t2
d ðT AÞdt ¼ 0
(1.7.10)
t1
The scalar quantity
L¼T A
(1.7.11)
is termed the Lagrangian or the kinetic potential. We should emphasize that
Hamilton’s principle depends upon the energies of the system and is invariant
under the coordinate transformation.
Eq. (1.7.10) states that of all possible paths of motion of the particle during
an interval of time from t1 to t2 , the actual path is that for which the integral
Z t2
Ldt ¼ 0
(1.7.12)
t1
has a stationary value. In fact, it can be shown that this value is the minimum
value of the integral.
The derivation of Hamilton’s principle for a particle can be extended to
MDOF systems as well as to continuous systems. The potential energy usually
arises from the gravity field. However, it may also arise from other sources such
as electrical and magnetic fields. The strain energy U ðt Þ should be included as
an additional potential energy. Thus, we can write
Z
t2
Z
d ðU T + AÞdt t1
t2
dWnc dt ¼ 0
(1.7.13)
t1
Hamilton’s principle is rather utilized to derive the equations of motion of
continuous systems. The equations of motion of discrete parameter systems can
result directly from Lagrange’s equations.
Example 1.7.1 Equation of motion of the SDOF system
Formulate the equation of motion of the SDOF system shown in Fig. 1.4.1 using
Hamilton’s principle.
Solution
The potential energy is due to the strain energy stored in the spring during deformation. It is expressed in terms of the spring stiffness coefficient k and the displacement u as
1
U ¼ ku 2
2
(1)
The kinetic energy is due to the motion of the mass m and is given as
1
T ¼ m u_ 2
2
(2)
46 PART
I Single-degree-of-freedom systems
_ as a dissipative force, is nonconservative. The
The damping force fD ¼ cu,
virtual work of this force is
D
_
¼ fD du ¼ cudu
dWnc
(3)
The negative sign results from the fact that fD is opposite to the virtual displacement du.
The external force is also treated as nonconservative and it does the
virtual work
p
¼ pðt Þdu
dWnc
(4)
Because no conservative external forces act on the system, it is A ¼ 0.
The variations dU and dT are obtained from Eqs. (1), (2)
_ u_
dU ¼ kudu, dT ¼ m ud
(5)
Introducing Eqs. (3)–(5) into Hamilton’s principle, Eq. (1.7.13), yields
Z t2
Z t2
_ u_ Þdt _ + pðt Þdu dt ¼ 0
ðkudu m ud
½cudu
(6)
t1
t1
The next step is to remove the variation d u_ of the velocity u_ from Eq. (6).
This is achieved using integration by parts as follows:
Z t2
Z t2
du
_ udt
_ ¼
_
m ud
m ud
dt
dt
t1
t1
Z t2
d
(7)
m u_ ðdu Þdt
¼
dt
t1
Z t2
€
_ tt21 m ududt
¼ ½m udu
t1
According to Hamilton’s principle it holds
du ðt1 Þ ¼ du ðt2 Þ ¼ 0
Thus, the term outside the integral vanishes and Eq. (7) becomes
Z t2
Z t2
_ udt
_ ¼
€
m ud
m ududt
t1
(8)
t1
Eq. (6) by virtue of Eq. (8) is written as
Z t2
½m u€ + cu_ + ku pðt Þdudt ¼ 0
(9)
t1
In order that the integral in Eq. (9) is equal to zero for any time interval
½t1 , t2 , its integrand should vanish, that is,
½m u€ + cu_ + ku pðt Þdu ¼ 0
General concepts and principles of structural dynamics Chapter
1
47
Moreover, because du is arbitrary, it must be
m u€ + cu_ + ku pðt Þ ¼ 0
or
m u€ + cu_ + ku ¼ pðt Þ
(10)
which is the equation of motion.
Example 1.7.2 Equation of motion of a two-story shear frame
Formulate the equations of motion of the frame in Example 1.5.5 using Hamilton’s principle.
Solution
Referring to Fig. E1.5b and c in Example 1.5.5, we have
1
1
U ¼ k1 ðu1 u2 Þ2 + k2 u22
2
2
1
1
T ¼ m1 u_ 21 + m2 u_ 22
2
2
Their variations are
dU ¼ k1 ðu1 u2 Þðdu1 du2 Þ + k2 u2 du2
¼ k1 ðu1 u2 Þdu1 k1 ðu1 u2 Þdu2 + k2 u2 du2
dT ¼ m1 u_ 1 d u_ 1 + m2 u_ 2 d u_ 2
(1)
(2)
Integrating by parts the variation dT in the interval ½t1 , t2 yields
Z t2
Z t2
dTdt ¼
ðm1 u_ 1 d u_ 1 + m2 u_ 2 d u_ 2 Þdt
t1
t1
Z
¼ ½m1 u_ 1 du1 + m2 u_ 2 du2 tt21 t2
ðm1 u€1 du1 + m2 u€2 du2 Þdt
t1
and taking into account that du1 ðt1 Þ ¼ du1 ðt2 Þ ¼ du2 ðt1 Þ ¼ du2 ðt2 Þ ¼ 0, we
obtain
Z t2
Z t2
dTdt ¼ ðm1 u€1 du1 + m2 u€2 du2 Þdt
(3)
t1
t1
Moreover, it is
p
dWnc
¼ p1 ðt Þdu1 + p2 ðt Þdu2 and A ¼ 0
(4)
Introducing Eqs. (1), (3), (4) into Hamilton’s principle, Eq. (1.7.13), we
obtain Z t2
½k1 ðu1 u2 Þdu1 k1 ðu1 u2 Þdu2 + k2 u2 du2 + m1 u€1 du1
t1
+ m2 u€2 du2 p1 ðt Þdu1 p2 ðt Þdu2 dt ¼ 0
48 PART
I Single-degree-of-freedom systems
or
Z
t2
f½m1 u€1 + k1 ðu1 u2 Þ p1 ðt Þdu1 + ½m2 u€2 k1 u1 + ðk1 + k2 Þu2
t1
p2 ðt Þdu2 gdt ¼ 0
(5)
Because Eq. (5) is valid for any interval ½t1 , t2 , its integrand must be equal
to zero, that is,
½m1 u€1 + k1 ðu1 u2 Þ p1 ðt Þdu1 + ½m2 u€2 k1 u1 + ðk1 + k2 Þu2 p2 ðt Þdu2 ¼ 0
(6)
Inasmuch as the quantities du1 and du2 are arbitrary, Eq. (6) is valid only if
the quantities in the square brackets are equal to zero, that is,
m1 u€1 + k1 ðu1 u2 Þ p1 ðt Þ ¼ 0
(7a)
m2 u€2 k1 u1 + ðk1 + k2 Þu2 p2 ðt Þ ¼ 0
(7b)
which give the equations of motion
m1 u€1 + k1 u1 k1 u2 ¼ p1 ðt Þ
(8a)
m2 u€2 k1 u1 + ðk1 + k2 Þu2 ¼ p2 ðt Þ
(8b)
Example 1.7.3 Equation of motion of a complex MDOF system
The system shown in Fig. E1.9 consists of the three rigid bars AB,BC , CD connected by hinges at points B and C , and it is supported by a roller at point D and
a hinge at point A. The relative rotations of the bars at the hinges B and C are
restrained by moment-resisting rotational springs with stiffness coefficients
k3 ¼ k4 ¼ 4kL2 and by the rotational dashpots with damping coefficients
c3 ¼ c4 ¼ 2cL2 . In the transverse direction, the motion is restrained by the
two springs at points E and Q with stiffness coefficients k1 ¼ k, k2 ¼ 2k, and
the two dashpots at points F and G with damping coefficients c1 ¼ c and
c2 ¼ 3c. A constant axial force P is applied at point D. The system is set in
motion by the transverse load pðx, t Þ ¼ ðpx=LÞf ðt Þ, linearly distributed along
while the
the bar CD. The mass per unit length of the bars AB and CD is m
bar BC is massless and supports the rigid body S at H having surface mass
density g ¼ m=L.
Assuming small amplitude displacements, formulate the
equations of motion of the system using Hamilton’s principle.
FIG. E1.10 System in Example 1.7.3.
General concepts and principles of structural dynamics Chapter
1
49
Solution
Inasmuch as the bars are assumed rigid, this system has only two degrees of
freedom. The displaced configuration of the system can be determined from
the two transverse displacements u1 ðt Þ and u2 ðt Þ of points B and C . Referring
to Fig. E1.11, we have
FIG. E1.11 Deformed configuration of the system.
9
f1 ¼ u1 =4L
=
f2 ¼ ðu2 u1 Þ=3L
;
f3 ¼ u2 =3L
(1)
The displacements of points of application of the forces and the changes of
angles are expressed in terms of the basic quantities u1 and u2 as
9
EE 0 ¼ u1 =4, FF 0 ¼ u1 =2, GG 0 ¼ 3u1 =4 >
>
=
HH 0 ¼ u1 + ðu2 u1 Þ=3, QQ 0 ¼ u2 =2
(2)
DfB ¼ f2 f1 ¼ ð4u2 7u1 Þ=12L
>
>
;
DfC ¼ f3 + f2 ¼ ð2u2 u1 Þ=3L
The potential energy U due to the deformation of the springs is
1
1
1
1
2
2
U ¼ k1 ðEE 0 Þ + k2 ðQQ 0 Þ + k3 ðDfB Þ2 + k4 ðDfC Þ2
2
2
2
2
which by virtue of Eqs. (2) becomes
1 2 1 2
1
2
ku 1 + ku 2 + k ð4u2 7u1 Þ2 + k ð2u2 u1 Þ2
32
4
72
9
¼ 0:934ku 21 + 1:361ku 22 1:667u1 u2
U¼
Its variation is
dU ¼ k ð1:868u1 1:667u2 Þdu1 + k ð1:667u1 + 2:722u2 Þdu2
(3)
The kinetic energy consists of the kinetic energies T1 and T2 of the bars ΑΒ
and CD, and of the kinetic energy T3 of the rigid body S. Thus, we have
1
1
1
d
2
2
ðHH 0 Þ
T ¼ IA f_ 1 + ID f_ 3 + m
2
2
2 dt
2
1
2
+ IH f_ 2
2
(4)
50 PART
I Single-degree-of-freedom systems
where
m ¼ mL,
IA ¼ m
ð4LÞ3
ð3LÞ3
L3
, ID ¼ m
, IH ¼ m
3
3
6
(5)
Introducing Eqs. (1), (2), (5) into Eq. (4) yields
2
1
1
1
T ¼ m u_ 21 + m u_ 22 + m ðu_ 2 + 2u_ 1 Þ2 +
m ðu_ 2 u_ 1 Þ2
3
2
18
108
¼ 0:898m u_ 21 + 0:565m u_ 22 + 0:204u_ 1 u_ 2
and its variation
dT ¼ m ð1:796u_ 1 + 0:204u_ 2 Þd u_ 1 + m ð0:204u_ 1 + 1:130u_ 2 Þd u_ 2
Rt
Using integration by parts in the integral t12 dTdt and taking into
account that
d
du1 ðt1 Þ ¼ du1 ðt2 Þ ¼ du2 ðt1 Þ ¼ du2 ðt2 Þ ¼ 0 and d u_ ¼ d du
dt ¼ dt ðdu Þ
we obtain
Z t2
Z
dTdt ¼ t1
t2
½m ð1:796u€1 + 0:204u€2 Þdu1 + m ð0:204u€1 + 1:130u€2 Þdu2 dt
t1
(6)
The nonconservative forces include the loading pðx, t Þ and the damping
forces. Their virtual work is expressed in terms of the basic quantities as follows:
Z 3L
x
p
¼
pðx, t Þ 1 du2 dx
dWnc
3L
0
(7)
Z 3L
x
x
pLf ðt Þdu2
¼
p f ðt Þ 1 du2 dx ¼ 1:5
L
3L
0
D
dWnc
¼ c1
d
d
d
ðFF 0 Þd ðFF 0 Þ c2 ðGG 0 ÞdðGG 0 Þ c3 ðDfB ÞdðDfB Þ
dt
dt
dt
(8)
d
c4 ðDfC Þd ðDfC Þ
dt
Using Eq. (2) and taking into account that c1 ¼ c, c2 ¼ 3c, c3 ¼ c4 ¼ 2cL2 ,
we can write
D
dWnc
¼ cð2:840u_ 1 0:833u_ 2 Þdu1 + cð0:833u_ 1 1:111u_ 2 Þdu2
Hence, we have
dWnc ¼ cð2:395u_ 1 0:833u_ 2 Þdu1 + cð0:833u_ 1 1:111u_ 2 Þdu2 + 1:5
pLf ðt Þdu2
(9)
General concepts and principles of structural dynamics Chapter
51
1
Finally, the potential A of the external conservative forces is due to the constant axial force P . Hence it is
A ¼ P ðDD 0 Þ ¼ Pe
and
dA ¼ Pde
(10)
The variation de is evaluated as follows.
Referring to Fig. E1.11, we have
e ¼ ðAD Þ ðAD0 Þ ¼ 10L 4L cos f1 3Lcos f2 3L cos f3
Therefore
de ¼ Lð4sin f1 df1 + 3 sin f2 df2 + 3sin f3 df3 Þ
¼ Lð4f1 df1 + 3f2 df2 + 3f3 df3 Þ
which is introduced into Eq. (10) to yield
7P
P
P
2P
u1 + u2 du1 +
u1 u2 du2
dA ¼ 12L
3L
3L
3L
(11)
(12)
Introducing the expressions for dU , dT , dWnc , and dA into Hamilton’s principle, Eq. (1.7.13), we obtain the following equations of motion
7P
u1
1:796m u€1 + 0:204m u€2 + 2:395cu_ 1 0:833cu_ 2 + 1:868k 12L
P
u2 ¼ 0
+ 1:667k +
3L
P
u1
0:204m u€1 + 1:130m u€2 0:833cu_ 1 + 1:111cu_ 2 + 1:667k +
3L
2P
+ 2:722k pLf ðt Þ
u2 ¼ 1:5
3L
or in the matrix form
"
#( )
"
#( )
2:395 0:833
u_ 1
1:796 0:204
u€1
+c
m
0:833 1:111
u€2
u_ 2
0:204 1:130
"
#( ) (
) (13)
1:868 0:583l 1:667 + 0:333l
u1
0
+k
¼
1:667 + 0:333l 2:722 0:667l
u2
1:5
pLf ðt Þ
where l ¼ P=kL.
The elastic forces of the system are
fS1 ¼ k ð1:868 0:583lÞu1 + k ð1:667 + 0:333lÞu2
fS2 ¼ k ð1:667 + 0:333lÞu1 + k ð2:722 0:667lÞu2
52 PART
I Single-degree-of-freedom systems
They may become zero if the system of equations
1:868 0:583l 1:667 + 0:333l u1
k
1:667 + 0:333l 2:722 0:667l
u2
¼
has a nontrivial solution. This occurs if
+1:868 0:583l 1:667 + 0:333l 1:667 + 0:333l +2:722 0:667l ¼ 0
0
0
(14)
(15)
Expanding the determinant yields
0:27797l2 1:7227l + 2:3058 ¼ 0
from which we obtain
l1 ¼ 1:9555 l2 ¼ 4:2419
The obtained values of l specify two critical values, Pcr1 ¼ 1:9555kL and
Pcr2 ¼ 4:2419kL, of the compressive axial force for which the structure exhibits
no resistance to deformation, that is, it has no stiffness and the structure buckles.
Therefore, these critical loads are the buckling loads of the structure (first and
second). The condition for buckling is the vanishing of the determinant of the
stiffness matrix, Eq. (15). The resulting equation is called the buckling equation.
It is apparent that if the axial force is tensile, the determinant cannot vanish for
real values of the parameter l. Concluding, we can state that tensile axial forces
increase the stiffness of the structure while compressive axial forces reduce it
and may lead to buckling.
Example 1.7.4 Equation of motion of the elastic cantilever beam
Derive the equation of motion of the cantilever beam shown in Fig. E1.12.
Solution
The mass of the beam element is equal to mdx. Its kinetic energy is
mdx ½∂u ðx, t Þ=∂t 2 =2, which is integrated along the beam length to yield the
kinetic energy of the beam, namely
FIG. E1.12 Cantilever in Example 1.7.4.
1
T¼
2
Z
L
0
∂u ðx, t Þ 2
m
dx
∂t
(1)
General concepts and principles of structural dynamics Chapter
1
53
The strain energy of the beam is obtained by integrating the strain energy
density over its volume V , namely
Z
1
sx ex dV
(2)
U¼
2 V
From the beam theory we have
sx ¼
M ðx Þ
sx
∂2 u ðx, t Þ
y, ex ¼ , M ðx Þ ¼ EI
E
I
∂x 2
Substituting the previous equations into Eq. (2) and integrating over the
cross-section of the beam yield
2
2
Z
1 L
∂ u ðx, t Þ
U¼
EI
dx
(3)
2 0
∂x 2
For the simplicity of the expressions, the differentiation with respect to time
t will be designated by an over-dot while that with respect to the spatial coordinated x by a prime. Moreover, the arguments will be dropped for the same
reason. Hence, expressions (1) and (3) can be rewritten as
Z
1 L
m u_ 2 dx
(4)
T¼
2 0
Z
1 L
2
U¼
EI ðu 00 Þ dx
(5)
2 0
Their variations are
Z
dT ¼
L
_ udx
_
m ud
(6)
EI u 00 du 00 dx
(7)
0
Z
L
dU ¼
0
Integrating twice by parts the integral representing dU yields
Z L
L
L
dU ¼
EI u 0000 dudx ½EI u 000 du 0 + ½EI u 00 du 0 0
(8)
0
The boundary conditions of the beam are
At x ¼ 0 u ¼ u 0 ¼ 0, hence du ¼ du 0 ¼ 0
At x ¼ L M ¼ EI u 00 ¼ 0 Q ¼ EI u 000 ¼ 0
Therefore, the quantities outside the integral vanish and Eq. (8) becomes
Z L
EI u 0000 dudx
(9)
dU ¼
0
54 PART
I Single-degree-of-freedom systems
Because no conservative loads act on the system, it is A ¼ 0. Moreover, the
virtual work of the external load is
Z L
p
pðx, t Þdudx
(10)
dWnc ¼
0
Introducing Eqs. (5), (9), (10) into Hamilton’s principle, Eq. (1.7.13), we
obtain
Z t2 Z L
Z L
Z L
_ udx
_ EI u 0000 dudx m ud
pðx, t Þdudx dt ¼ 0
(11)
t1
0
0
0
Interchanging the integration in the second term and performing integration
by parts with respect to time, yield
Z t2 Z L
Z L
Z t2
_ udx
_
€
m ud
dt ¼
½m ududt
+ ½mudu tt21 dx
0
0
t1
t1
(12)
Z t Z L
2
¼
t1
€
½m ududx
dt
0
On the basis of Eq. (12), Eq. (11) becomes
Z t2 Z L
½EI u 0000 + m u€ pðx, t Þdudx dt ¼ 0
(13)
0
t1
Because Eq. (13) is valid for any interval ½t1 , t2 , the integrand must vanish,
namely
Z L
½EI u 0000 + m u€ pðx, t Þdudx ¼ 0
(14)
0
Moreover, because du is arbitrary, Eq. (14) is valid only if
EI u 0000 + m u€ pðx, t Þ ¼ 0
(15)
which yields the equation of motion of the cantilever
EI u 0000 + m u€ ¼ pðx, t Þ
(16)
Apparently, Eq. 16 is identical to that obtained in Section 1.1.
1.8 Lagrange’s equations
1.8.1 Derivation of Lagrange’s equations
In a system with N degrees of freedom, the displaced configuration can be determined from a set of coordinates, which take a certain value at each instant. The
system of coordinates for the analysis of a given mechanical system is not necessarily unique. Many coordinate systems are possible. Furthermore, the number
of coordinates may vary, but it cannot be less than N . Anyhow, if the number of
coordinates is greater than N , then additional equations, referred to as equations
General concepts and principles of structural dynamics Chapter
1
55
of constraint, must relate the coordinates so that the number of coordinates is
equal to the number of degrees of freedom plus the independent equations of
constraint. The requirement that the equations of motion hold together with
the equations of constraint complicates the solution. For this reason, we seek,
if possible, to choose N independent coordinates, which can specify the configuration of the system. For example, we consider the simple pendulum shown in
Fig. 1.8.1a. The rod is rigid and weightless. Its length is L and it can rotate freely
about the hinge at O, such that the motion is confined in a single vertical plane.
The position of the mass can be specified by the angle q between the vertical
axis y and the rod. Hence, the system has a SDOF. However, the displaced
configuration can also be determined by the coordinates ðx, y Þ, which represent
the position of the mass m within the xy plane. These coordinates, however,
are not independent because they must satisfy the constraint equation
x 2 + y 2 ¼ L2
(a)
(b)
FIG. 1.8.1 Simple (a) and double (b) pendulum.
Similarly, the configuration of the double pendulum of Fig. 1.8.1b can be
specified by the two angles q1 and q2 . Hence the system has two degrees of freedom. On the other hand, the position of the masses m1 , m2 can be determined by
the coordinates ðx1 , y1 Þ and ðx2 , y2 Þ, which, however, are not independent
because they must satisfy the following two constraint equations
x12 + y12 ¼ L21
ðx2 x1 Þ2 + ðy2 y1 Þ2 ¼ L22
The quantities q in the simple pendulum or q1 , q2 in the double pendulum,
which would determine the configuration of the system, could be considered as
coordinates in a more general sense. Any set of quantities that serves to specify
the configuration of the system is referred to as generalized coordinates. The
geometrical significance of the generalized quantities is not always cognizable.
For systems in motion, the generalized coordinates vary with time and are treated as algebraic variables. The process of obtaining one set of generalized coordinates from another is known as a coordinate transformation.
56 PART
I Single-degree-of-freedom systems
We consider now a transformation from a set of N generalized coordinates
q1 ðt Þ, q2 ðt Þ,…, qN ðt Þ to a set K of ordinary (for example, Cartesian) coordinates
x1 ,x2 ,…, xK ðK N Þ. The transformation equations are of the form
x1 ¼ x1 ðq1 , q2 , …, qN Þ
x2 ¼ x2 ðq1 , q2 , …, qN Þ
⋯ ⋯
xK ¼ xK ðq1 , q2 , …, qN Þ
(1.8.1)
For example, the transformation equations of the generalized coordinates
q1 , q2 to the ordinary coordinates x1 , x1 , y2 , y2 of the double pendulum are
x1 ¼ L1 sin q1
y1 ¼ L1 cos q1
x2 ¼ L1 sin q1 + L2 sin q2
y2 ¼ L1 cos q1 L2 cos q2
The kinetic energy of a system with K degrees of freedom may also depend
on the generalized coordinates q1 , q2 , …, qN beside the generalized velocities
q_ 1 , q_ 2 ,…, q_ N , that is,
T ¼ T ðq1 , q2 , …, qN , q_ 1 , q_ 2 , …, q_ N Þ
(1.8.2)
In conservative systems, the potential energy A depends only on the position, namely, it is
A ¼ Aðq1 , q2 , …, qN Þ
(1.8.3)
The work done by the forces derivable from the potential energy A, when the
generalized coordinates qi are given a virtual displacement dqi , is expressed as
dA ¼ Q1 dq1 + Q2 dq2 + ⋯ + QN dqN
where
Q1 ¼ ∂A
∂A
∂A
¼ , Q2 ¼ ¼ , …, QN ¼ ∂q1
∂q2
∂qN
(1.8.4)
The quantity Qi dqi represents the work done through the displacement dqi .
Inasmuch as the quantity Qi may or may not represent a force, it is referred to as
generalized force. Hence, if qi represents a translational displacement then Qi is
a force, whereas if qi represents a rotation then Qi is a moment. In some problems, the quantities dqi may represent surfaces, volumes, etc. Therefore, the
nature of the corresponding Qi is defined so that the quantity Qi dqi has the physical dimension of work.
Lagrange’s equations may be derived by direct application of Hamilton’s
principle. Thus, when the applied forces are conservative, we write
Z t2
ðdT dAÞdt ¼ 0
(1.8.5)
t1
General concepts and principles of structural dynamics Chapter
1
57
The variations associated with the kinetic energy and the potential energy
defined by Eqs. (1.8.2), (1.8.3), respectively, are of the form
dT ¼
∂T
∂T
∂T
∂T
dq1 + ⋯ +
dqN +
d q_ + ⋯ +
d q_
∂q1
∂qN
∂q_ 1 1
∂q_ N N
dA ¼
∂A
∂A
dq1 + ⋯ +
dqN
∂q1
∂qN
Substituting these expressions into Eq. (1.8.5), integrating by parts the terms
including d q_ i and taking into account dq1 ¼ dq2 ¼ ⋯ ¼ dqN ¼ 0 at instants t1
and t2 , we obtain
Z t2 ∂T d ∂T
∂A
∂T d ∂T
∂A
dq1 + ⋯ +
dqN dt ¼ 0
∂q1 dt ∂q_ 1
∂q1
∂qN dt ∂q_ N
∂qN
t1
Because the time interval ½t1 , t2 as well as the virtual displacements dqi are
arbitrary, this previous equation results in the following equations
d ∂T
∂T ∂A
+
¼ 0 ði ¼ 1, 2, …, N Þ
(1.8.6)
dt ∂q_ i
∂qi ∂qi
which, using Eq. (1.8.4), become
d ∂T
∂T
¼ Qi ði ¼ 1, 2, …, N Þ
dt ∂q_ i
∂qi
(1.8.7)
Eq. (1.8.6) or (1.8.7) are the Lagrange equations of motion.
When nonconservative forces act on the system in addition to the conservative forces, we can include them in Lagrange’s equations, if the work done by
the nonconservative forces riding the virtual displacements is expressed in
terms of the generalized forces, that is,
dWnc ¼ Q1 dq1 + Q2 dq2 + ⋯ + QN dqN
(1.8.8)
Introducing Eq. (1.8.8) into Hamilton’s principle, Eq. (1.7.9), the Lagrange
equations (1.8.6) become
d ∂T
∂T ∂A
+
¼ Qi ði ¼ 1, 2, …, N Þ
(1.8.9)
dt ∂q_ i
∂qi ∂qi
The elastic force components, which are derivable from a potential U (strain
energy), can be also involved in Eq. (1.8.9). Noting that
U ¼ U ðq1 , q2 , …, qN Þ
(1.8.10)
the associated variation is
dU ¼
∂U
∂U
dq1 + ⋯ +
dqN
∂q1
∂qN
Therefore, the components ∂U =∂qi express generalized elastic forces and
Lagrange’s equations become
58 PART
I Single-degree-of-freedom systems
d ∂T
∂T ∂V
+
¼ Qi ði ¼ 1, 2, …, N Þ
dt ∂q_ i
∂qi ∂qi
(1.8.11)
V ¼U +A
(1.8.12)
where
is the total potential energy of the system.
The N generalized forces Qi can be evaluated from the set of K actual forces
Fk associated with the set of Cartesian coordinates. For this purpose, we consider the work done by the set of forces Qi when the coordinates qi are given an
increment, that is, a virtual displacement dqi
dW ¼
N
X
Qi dqi
(1.8.13)
i¼1
If dxk represent the ensuing virtual displacements of the coordinates xk then
the set of forces Fk do the work
dW ¼
K
X
Fk dxk
(1.8.14)
k¼1
From physical consideration, the work done by the two sets of forces is the
same. The only difference is that they are expressed in two different coordinate
systems. Therefore, we can write
N
X
i¼1
Qi dqi ¼
K
X
Fk dxk
(1.8.15)
k¼1
or in matrix form
QT dq ¼ FT dx
(1.8.16)
where
Q ¼ fQ1 Q2 ⋯ QN gT , dq ¼ f dq1 dq2 ⋯ dqN gT
(1.8.17a)
F ¼ f F1 F2 ⋯ FK gT , dx ¼ f dx1 dx2 ⋯ dxK gT
(1.8.17b)
The relation between dxk and dqi results from the transformation equations
(1.8.1) by considering the variation of dxk . Thus, we have
dxk ¼
∂xk
∂xk
dq1 + ⋯ +
dqN
∂q1
∂qN
(1.8.18)
or in matrix form
dx ¼ Jdq
(1.8.19)
General concepts and principles of structural dynamics Chapter
1
59
where J is the Jacobian matrix of the transformation (1.8.1), that is,
2 ∂x ∂x
∂x1 3
1
1
⋯
6 ∂q1 ∂q2
∂qN 7
6
7
6 ∂x ∂x
∂x2 7
2
6 2
7
⋯
6
7
(1.8.20)
J ¼ 6 ∂q1 ∂q2
∂qN 7
6
7
6 ⋮
7
⋮
⋱
⋮
6
7
4 ∂xK ∂xK
∂xK 5
⋯
∂q1 ∂q2
∂qN
Substituting Eq. (1.8.19) into (1.8.16) yields
QT dq ¼ FT Jdq
(1.8.21)
QT ¼ F T J
(1.8.22)
Q ¼ JT F
(1.8.23)
from which we obtain
or
Apparently, Eq. (1.8.23) represents the sought relation between Q and F.
Example 1.8.1 Equation of motion of the double pendulum
Formulate the equations of motion of the double pendulum shown in
Fig. 1.8.1b.
Solution
Because the bars are inextensional, the displaced configuration of the moving
system can be specified by the generalized coordinates q1 and q2 . Referring to
Fig. 1.8.1b, the Cartesian coordinates of the masses m1 and m2 are expressed in
terms of q1 and q2 by the geometrical relations
x1 ¼ L1 sin q1
y1 ¼ L1 cos q1
x2 ¼ L1 sin q1 + L2 sin q2
y2 ¼ L1 cos q1 L2 cos q2
(1)
The kinetic energy of the system is
1 1 T ¼ m1 x_ 21 + y_ 21 + m2 x_ 22 + y_ 22
2
2
which by virtue of Eqs. (1) becomes
1
1
2
2
T ¼ ðm1 + m2 ÞL21 q_ 1 + m2 L1 L2 q_ 1 q_ 2 cos ðq1 q2 Þ + m2 L22 q_ 2
2
2
The potential energy is
A ¼ m1 gy 1 + m2 gy 2
(2)
60 PART
I Single-degree-of-freedom systems
or using Eqs. (1)
A ¼ ðm1 + m2 ÞgL1 cos q1 m2 gL2 cos q2
(3)
Differentiating Eqs. (2), (3) yields
∂T
¼ ðm1 + m2 ÞL21 q_ 1 + m2 L1 L2 q_ 2 cos ðq1 q2 Þ
∂q_ 1
d ∂T
¼ ðm1 + m2 ÞL21 q€1 + m2 L1 L2 q€2 cos ðq1 q2 Þ
dt ∂q_ 1
m2 L1 L2 q_ 2 sin ðq1 q2 Þ q_ 1 q_ 2
∂T
¼ m2 L1 L2 q_ 1 q_ 2 sin ðq1 q2 Þ
∂q1
∂A
¼ ðm1 + m2 ÞgL1 sin q1
∂q1
∂T
¼ m2 L1 L2 q_ 1 cos ðq1 q2 Þ + m2 L22 q_ 2
∂q_ 2
d ∂T
¼ m2 L1 L2 q€1 cos ðq1 q2 Þ m2 L1 L2 q_ 1 sin ðq1 q2 Þ q_ 1 q_ 2 + m2 L22 q€2
_
dt ∂q 2
∂T
¼ m2 L1 L2 q_ 1 q_ 2 sin ðq1 q2 Þ
∂q2
∂A
¼ m2 gL2 sin q2
∂q2
Applying Eq. (1.8.6) for i ¼ 1, 2 and q1 ¼ q1 q2 ¼ q2 , we obtain the equations of motion of the double pendulum
2
ðm1 + m2 ÞL1 q€1 + m2 L2 q€2 cos a + q_ 2 sin a + ðm1 + m2 Þg sin q1 ¼ 0 (4a)
2
L1 q€1 cos a + L2 q€2 L1 q_ 1 sin a + g sin q2 ¼ 0
(4b)
where
a ¼ q1 q2
Example 1.8.2 Equation of motion of the “soft” pendulum
Formulate the equations of motion of the simple pendulum shown in Fig. E1.13,
taking into account the axial deformation of the rod (soft pendulum). The undeformed length of the rod is L, its cross-sectional area A, and the modulus of
elasticity of the material E.
General concepts and principles of structural dynamics Chapter
1
61
FIG. E1.13 “Soft” pendulum in Example 1.8.2.
Solution
Because the rod is no more inextensional, the system has two degrees of freedom.
Its displaced configuration can be specified either by the orthogonal coordinates x
and y of the mass or by the angle of the q and the axial deformation of the rod.
The kinetic energy of the system is
1 T ¼ m x_ 2 + y_ 2
2
(1)
The potential energy of the external force (gravitational force) is
A ¼ mgy
(2)
and the potential of the elastic force
1
U ¼ ke2
(3)
2
where k ¼ EA=L is the axial stiffness of the rod and e its elongation. The latter
is expressed in terms of x and y as
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
e ¼ x 2 + y2 L
(4)
Introducing Eq. (4) in the expression for the axial stiffness, Eq. (3), yields
2
ffi
1 EA pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
U¼
(5)
x 2 + y2 L
2 L
Differentiating the energies, we obtain
!
d ∂T
∂A
∂U EA
L
¼ m x€,
¼ 0,
¼
1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x
dt ∂x_
∂x
∂x
L
x 2 + y2
!
d ∂T
∂A
∂U EA
L
¼ m y€,
¼ mg,
¼
1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y
dt ∂y_
∂y
∂y
L
x 2 + y2
(6)
(7)
Introducing Eqs. (6), (7) into Lagrange’s equations (1.8.11), we obtain the
equations of motion of the soft pendulum
!
EA
L
1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x ¼ 0
(8a)
m x€ +
L
x 2 + y2
62 PART
I Single-degree-of-freedom systems
!
EA
L
m x€ +
1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y ¼ mg
L
x 2 + y2
(8b)
Example 1.8.3 Equation of motion of a general one-story shear building
Formulate the equations of motion of the one-story building in Example 1.5.6
using the method of Lagrange’s equations.
Solution
The system has three degrees of freedom. We choose the displacements U , V of
about the z axis as the genpoint O in the plane of the plate and its rotation W
eralized coordinates of the system, namely
q1 ¼ U , q2 ¼ V , q3 ¼ W
Because the point O is not the center of mass of the plate, the kinetic energy
is given by the expression (1.5.18) (K€
onig’s theorem)
1
1 _ 2 m y U_ W
_
_ + m xc V W
T ¼ m U_ 2 + V 2 + Io W
c
2
2
(1)
The potential energy U consists of the strain energy of all columns. For the i
column it is
i 2
i 2 i
1 h i i 2
i
i
u + k22
v + k33
w
Ui ¼ k11
2
or using matrix notation
8 9
i
k 0 0 < u i =
1 i i i 11 i
1 T
(2)
Ui ¼ u v w 0 k22 0 v i ¼ Di ki Di
;
:
2
2
0 0 k i wi
33
Taking into account that (see Eqs. 1a, 12 of Example 1.5.6)
i T
T e i T Ri T
¼ Ri ei U,
¼U
D i ¼ Ri D
D
Eq. (2) is written as
1 T i kU
Ui ¼ U
2
where
i ¼ ei T Ri T ki Ri ei
k
Thus, we have
U¼
K
X
1
Ui ¼ UT KU
2
i¼1
1
2
2
+ 2k23 V W
2 + 2k12 U V + 2k13 U W
¼ k11 U + k22 V + k33 W
2
(3)
General concepts and principles of structural dynamics Chapter
1
63
ij are the elements of the matrix
where k
¼
K
K
X
i ¼
k
i¼1
K X
T
T
ei
Ri k i Ri e i
(4)
i¼1
Because there are no external conservative forces, it is
A¼0
(5)
The generalized forces result from Eq. (1.8.20). In this case it is
F ¼ P x ðt Þ P y ðt Þ 0
T
The transformation relations between the displacements u, v, w of a point
y of the plate and the displacements of point O are
x,
v ¼ V + xW,
w¼W
u ¼ U yW,
which yields the Jacobian matrix, Eq. (1.8.20)
2 ∂u ∂u ∂u 3
6 ∂U
6
6 ∂v
J¼6
6 ∂U
6
4
∂w
∂U
7 2
3
∂V ∂W
7
1 0 0
∂v ∂v 7
7 ¼ 4 0 1 05
7
∂V ∂W
7
y x 1
5
∂w ∂w
∂V ∂W
Thus, for point A we have
Q ¼ JT F ¼ P x ðt Þ P y ðt Þ y A P x ðt Þ + xA P y ðt Þ
Differentiation of Eq. (1) yields
d ∂T
€
m yc W
¼ m U€
dt ∂U_
∂T
¼0
∂U
!
d ∂T
€
+ mx c W
¼ m V€
dt ∂ V
∂T
¼0
∂V
d ∂T
€ m y U€
+ m xc V€
¼ Io W
c
dt ∂W
_
∂T
¼0
∂W
Moreover, Eq. (3) yields
T
(6)
(7a)
(7b)
(7c)
(7d)
(7e)
(7f)
64 PART
I Single-degree-of-freedom systems
∂U
¼ k11 U + k12 V + k13 W
∂U
(8a)
∂U
¼ k21 U + k22 V + k23 W
∂V
∂U
¼ k 31 U + k 32 V + k 33 W
∂W
(8b)
(8c)
Introducing Eqs. (6)–(8) into the Lagrange
equations (1.8.11) with N ¼ 3,
¼ 0, yields the equation of motion
and taking into account that A U , V , W
of the structure
U
€ + K
U
¼P
ðt Þ
M
(9)
where
2
m
6
¼6 0
M
4
0
m
my c mx c
m yc
3
7
mx c 7
5,
Io
2
k 11
6
¼ 6 k21
K
4
k31
3
k12 k13
7
k22 k23 7
5,
k32 k33
9
8 P ðt Þ
>
>
=
< x
P ðt Þ ¼
P y ðt Þ
>
>
;
:
y A P x ðt Þ + xA P y ðt Þ
As was anticipated, Eq. (9) is identical to Eq. (16) of Example 1.5.6.
1.8.2 Lagrange multipliers
Lagrange’s equations result as a direct application of Hamilton’s principle provided that the energies (kinetic and potential) as well as the virtual work of the
nonconservative forces can be expressed in terms of the generalized coordinates
and velocities, as is indicated in Eqs. (1.8.2), (1.8.3), (1.8.8), (1.8.10).
Lagrange’s equations apply to linear as well as to nonlinear systems.
In certain cases, it is impossible or it is not convenient to choose N independent coordinates. Then, we choose K > N coordinates and we introduce
n ¼ K N constraint equations, which in general have the form
g1 ðq1 , q2 , …, qK Þ ¼ 0
g2 ðq1 , q2 , …, qK Þ ¼ 0
…
…
gn ðq1 , q2 , …, qK Þ ¼ 0
We distinguish two approaches to derive the equations of motion:
(1.8.24)
General concepts and principles of structural dynamics Chapter
1
65
(a) Eq. (1.8.24) can be solved in terms of n ¼ K N coordinates. Then the
redundant coordinates can be eliminated from Eqs. (1.8.2), (1.8.3),
(1.8.8), (1.8.10), and the equations of motion are formulated using
Lagrange’s equations (1.8.11). If the constraint equations are linear, the
technique presented in Section 1.8.1 can also be employed to reduce the
number of equations to N .
(b) Eq. (1.8.24) cannot be solved in terms of n ¼ K N coordinates. In this
case, the equations of motion can be derived by using the method of
Lagrange multipliers.
The second approach preserves the symmetry of the problem because there are
no preferred coordinates while others are eliminated. Though the method of
Lagrange multipliers deals with more coordinates than the degrees of freedom
of the system, quite often this procedure results in simpler equations.
To illustrate the method of Lagrange multipliers, we consider the variations
of the constraint functions given by Eq. (1.8.24)
dg1
dg1
dq1 + ⋯ +
dqK ¼ 0
dq1
dqk
dg2
dg2
dq1 + ⋯ +
dqK ¼ 0
dg2 ¼
dq1
dqk
…
…
dgn
dgn
dgn ¼
dq1 + ⋯ +
dqK ¼ 0
dq1
dqk
dg1 ¼
(1.8.25)
which we write as
K
X
aji dqi ¼ 0 ðj ¼ 1, 2, …, n Þ
(1.8.26)
dgj
dqi
(1.8.27)
i¼1
where
aji ¼
If we assume that the constraints are frictionless, then no work is done by the
constraint forces Ri when they ride any virtual displacement dqi , that is,
K
X
Ri dqi ¼ 0
(1.8.28)
i¼1
Now we multiply Eq. (1.8.26) by a factor known as the Lagrange multiplier.
Thus, we obtain
K
X
aji dqi ¼ 0 ðj ¼ 1, 2, …, n Þ
(1.8.29)
lj
i¼1
66 PART
I Single-degree-of-freedom systems
where we note that a separate equation is written for each of the constraints.
Next, we subtract the sum of equations of the form (1.8.29) from
Eq. (1.8.28) and, interchanging the order of summation, we obtain
!
K
n
X
X
Ri lj aji dqi ¼ 0
(1.8.30)
i¼1
j¼1
from which, noting that dqi are arbitrary, we conclude that
n
n
X
X
dgj
Ri ¼
lj aji ¼
lj
dqi
j¼1
j¼1
(1.8.31)
The constraint forces Ri constitute additional generalized forces, which
must be included in Lagrange’s equations. Thus Eq. (1.8.11) become
n
X
d ∂T
∂T ∂V
dgj
+
¼ Qi +
lj
ði ¼ 1, 2, …, K Þ
(1.8.32)
dt ∂q_ i
∂qi ∂qi
dqi
j¼1
What we have accomplished by this procedure is the inclusion of the constraint reactions in the equations of motion as additional generalized forces.
Therefore, the number of unknowns becomes K + n, namely the K generalized
coordinates Qi ðt Þ and the n functions lj ðt Þ. The available equations are also
K + n, that is Eq. (1.8.24) plus Eq. (1.8.32).
Eq. (1.8.32) can be derived from Hamilton’s principle if the potential energy
of the external forces is replaced by
n
X
¼A
lj gj
(1.8.33)
A
j¼1
is referred to as the reduced potential energy. For a more
The function A
advanced formulation, including nonholonomic constraints and a dynamic
treatment of the Lagrange multipliers, see Ref. [7].
Example 1.8.4 Equation of motion of the simple pendulum using Lagrange
multipliers
Formulate the equations of motion of the simple pendulum shown in Fig. E1.14
in terms of the Cartesian coordinates x ðt Þ, y ðt Þ, assuming that the rod is weightless and rigid.
FIG. E1.14 Pendulum in Example 1.8.4.
General concepts and principles of structural dynamics Chapter
1
67
Solution
The kinetic and the potential energies of the system are
1 T ¼ m x_ 2 + y_ 2
2
A ¼ mgy
U ¼0
Because the rod is rigid, the coordinates must satisfy the constraint equation
g1 ¼ x 2 + y 2 L2 ¼ 0
Differentiating the quantities T and A we obtain
d ∂T
∂T
∂A
∂g1
¼ 2x, Q1 ¼ px
¼ 0,
¼ m x€,
¼ 0,
∂x
dt ∂x_
∂qi
∂x
d ∂T
∂T
∂A
∂g1
¼ 2y, Q2 ¼ py
¼ m y€,
¼ 0,
¼ mg,
∂y
dt ∂y_
∂y
∂y
Applying Eq. (1.8.32) for q1 ¼ x and q2 ¼ y we obtain the equations of
motion
m x€ ¼ px + 2xl
(1a)
m y€ + mg ¼ py + 2yl
(1b)
x 2 + y2 l 2 ¼ 0
(1c)
Eqs. (1) must be solved for the three unknowns x, y, and l. It should be
noted that two of these equations are differential and one algebraic and therefore
special care is required for their solution. A convenient method is to differentiate the constraint equation twice with respect to time and then to solve the
€ y€ and the parameter
resulting linear system of equations for the accelerations x,
l. For the problem at hand, we obtain
x x€ + y y€ ¼ 2T
m
Eqs. (1a), (1b), (2) are combined and written in matrix form
9
2
38 9 8
m 0 2x < x€ = < px
=
4 0 m 2y 5 y€ ¼
py mg
;
: ; :
2T =m
x y
0
l
(2)
(3)
which are solved to yield
px
py mg
L2 x€ + x x_ 2 + y_ 2 ¼ y 2 xy
m
m
2
px
2
2
2 py mg
L y€ + y x_ + y_ ¼ xy + x
m
m
(4a)
(4b)
68 PART
I Single-degree-of-freedom systems
xpx y py mg
T
L l¼
2
2
2
(4c)
Eqs. (4a), (4b) are solved using techniques for nonlinear differential equations. Analytical solutions are in general out of the question. However, a numerical solution is always feasible using the methods presented in Chapter 5. Once
the coordinates x ðt Þ, y ðt Þ and the Lagrange multiplier l have been established,
they are utilized in Eq. (1.8.31) to evaluate the constraint forces, which are the
components of the axial force of the rod. Thus, we have
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
∂g
∂g
(5)
Rx ¼ l ¼ 2lx, Rx ¼ l ¼ 2ly, S ¼ R2x + R2y ¼ 2lL
∂x
∂y
1.8.3 Small displacements
So far, we have not imposed any restrictions on the magnitude of the displacements in our discussion on Lagrange’s equations. Therefore, the derived equations of motion hold equally for small and large displacements and they are in
general nonlinear differential equations. For small displacements, however,
about the position of static equilibrium, the equations of motions are highly simplified as they become linear. This is very important in structural dynamics,
where we usually deal with small displacements and the resulting linear differential equations can be readily solved and predict the response of the structure.
1.8.3.1 Potential energy and stiffness matrix
We shall consider a system of particles that is in static equilibrium under the
action of a set of conservative forces. If its configuration is specified by ordinary
coordinates x1 ,x2 , …, x3 N , then the condition for static equilibrium is that the
virtual work done by the applied forces Fi is zero, that is,
dW ¼
3N
X
Fi dxi ¼ 0
(1.8.34)
i¼1
for all virtual displacements dxi consistent with the constraints, which are
assumed workless and bilateral.
Inasmuch as the forces are conservative, they are derivable from a potential
function V ¼ V ðx1 , x2 , …, x3N Þ, V ¼ U + A, according to the relation
Fi ¼ ∂V
∂xi
(1.8.35)
Using Eq. (1.8.35), Eq. (1.8.34) is written as
dW ¼ 3N
X
∂V
i¼1
∂xi
dxi ¼ 0
(1.8.36)
General concepts and principles of structural dynamics Chapter
1
69
If the number of degrees of freedom is n < 3N , dx’s are not independent. It
is possible to find n independent generalized coordinates by considering transformation equations, that is,
x1 ¼ x1 ðq1 , q2 , …, qn Þ
x2 ¼ x2 ðq1 , q2 , …, qn Þ
…
…
x3N ¼ x3N ðq1 , q2 , …, qn Þ
(1.8.37)
Then we have
dxi ¼
n
X
∂xi
j¼1
∂qj
dqj
(1.8.38)
Substituting the previous expression for dxi into Eq. (1.8.36), we obtain
dW ¼ 3N X
n
X
∂V ∂xi
i¼1 j¼1
∂xi ∂qj
dqj ¼ 0
(1.8.39)
Noting that
3N
∂V X
∂V ∂xi
¼
∂qj
∂xi ∂qj
i¼1
(1.8.40)
and interchanging the order of summation, Eq. (1.8.39) becomes
dW ¼ n
X
∂V
j¼1
∂qj
dqj ¼ 0
(1.8.41)
Because dqj are assumed to be independent, the virtual work is zero only if
the coefficients of dqj are zero at the equilibrium condition, that is, if
∂V
¼ 0, j ¼ 1, 2, …, n
(1.8.42)
∂qj 0
The subscript zero denotes that the derivatives refer to the equilibrium
position.
Let us expand now the potential energy function V ðq1 , q2 , …, qn Þ in a Taylor series about the position of equilibrium
n n X
n 2
X
∂V
1X
∂ V
dqi +
dqi dqj + ⋯
(1.8.43)
V ¼ V0 +
∂qi 0
2 i¼1 j¼1 ∂qi ∂qj 0
i¼1
We can arbitrarily set the potential energy at the reference position equal to
zero, that is,
(1.8.44)
V0 ¼ 0
70 PART
I Single-degree-of-freedom systems
If we now assume that the displacements about the equilibrium position are
small, we can neglect terms of order higher than the second in Eq. (1.8.43).
Thus, using Eqs. (1.8.41), (1.8.44) the expression for the potential energy is simplified as
n X
n 2
1X
∂ V
qi qj
(1.8.45)
V¼
2 i¼1 j¼1 ∂qi ∂qj 0
or setting
kij ¼ kji ¼
∂2 V
∂qi ∂qj
(1.8.46)
0
we can write Eq. (1.8.45) in the form
V¼
n X
n
1X
kij qi qj
2 i¼1 j¼1
(1.8.47)
The quantities kij defined by Eq. (1.8.46) are the stiffness coefficients of the
system. Thus we see that the potential energy is expressed by a homogeneous
quadratic function of the generalized coordinates qi if small motions about the
position of equilibrium are examined.
Eq. (1.8.47) is written in matrix form
1
V ¼ qT kq
2
where
8 9
2
k11 k12
q1 >
>
>
>
< =
6 k21 k22
q2
, k ¼6
q¼
4 ⋮ ⋮
⋮>
>
>
>
: ;
qn
kn1 kn2
(1.8.48)
3
⋯ k1n
⋯ k2n 7
7
⋱ ⋮ 5
⋯ knn
(1.8.49)
The matrix k is called the stiffness matrix of the system.
The expression for the potential energy given in Eq. (1.8.47) is an example
of a quadratic form. For a system whose reference equilibrium configuration is
stable, the potential energy V is positive for all possible values of qi , except
q1 ¼ q2 ¼ … ¼ qn ¼ 0. In this case, the function V is referred to as positive definite. This condition, however, puts restrictions on the allowable values of kij . It
is clear that all diagonal elements must be positive. The necessary and sufficient
condition that V be positive definite is that
2
3
k11 k12 ⋯ k1n
k k 6 k21 k22 ⋯ k2n 7
7
k11 > 0, 11 12 > 0, …, 6
(1.8.50)
4 ⋮ ⋮ ⋱ ⋮ 5>0
k21 k22
kn1 kn2 ⋯ knn
General concepts and principles of structural dynamics Chapter
1
71
1.8.3.2 Kinetic energy and mass matrix
The kinetic energy is
2
3N
1X
∂xk
T¼
mk
∂t
2 k¼1
(1.8.51)
Differentiating Eq. (1.8.37) with respect to time yields
n
∂xk X
∂xk
¼
q_
∂t
∂qj j
j¼1
(1.8.52)
Consequently, it is about the position of the static equilibrium
∂xk
∂t
2
n X
n X
∂xk
∂xk
¼
q_ i q_ j
∂q
i 0 ∂qj 0
0
i¼1 j¼1
(1.8.53)
Introducing this expression into Eq. (1.8.51) we can write the kinetic energy
in the form
T¼
n X
n
1X
mij q_ i q_ j
2 i¼1 j¼1
(1.8.54)
where it was set
mij ¼ mji ¼
3N
X
k¼1
mk
∂xk
∂qi
∂xk
0 ∂qj 0
(1.8.55)
The quantities mij defined by Eq. (1.8.55) are the inertia coefficients of the
system.
Eq. (1.8.54) is written in matrix form
1
T ¼ q_ T mq_
2
(1.8.56)
where
8 9
2
m11 m12
q_ 1 >
>
>
>
< =
6 m21 m22
q_ 2
, m ¼6
q_ ¼
4 ⋮
⋮>
⋮
>
>
>
: ;
q_ n
mn1 mn2
3
⋯ m1n
⋯ m2n 7
7
⋱ ⋮ 5
⋯ mnn
(1.8.57)
The matrix m is called the mass matrix of the system. The kinetic energy is a
positive definite quadratic function because it is the sum of positive quantities,
that is, the kinetic energies of the masses of the individual particles.
72 PART
I Single-degree-of-freedom systems
The equations of motion are obtained by applying Eq. (1.8.11). Differentiating the kinetic energy, Eq. (1.8.54), and the potential energy, Eq. (1.8.47),
yields
∂T
¼0
∂qi
X
n
d ∂T
¼
mij q€j
dt ∂q_ i
j¼1
n
∂V X
¼
kij qj
∂qi j¼1
Substituting into Lagrange’s equations, we obtain the following equations of
motion
n
X
j¼1
mij q€j +
n
X
kij qj ¼ Qi ði ¼ 1, 2, …, n Þ
(1.8.58)
j¼1
or in matrix form
m€
q + kq ¼ pðtÞ
(1.8.59)
where p(t)¼Q
The matrices m and k are symmetric. It is an advantage of the Lagrange formulation of the equations of motion that it preserves the symmetry of the coefficient matrices for those cases where T and V are represented by quadratic
functions of the velocities and displacements, respectively.
1.8.4 Raleigh’s dissipation function
The dissipative forces arising in a mechanical system are nonconservative
forces. Therefore, they are not derivable from a potential function. They are
involved in the Lagrange equations with their virtual work. When the dissipative forces are due to such sources as air resistance or internal friction, they are
usually assumed to depend linearly on the velocities along with the physical
coordinates and opposed to the motion, that is
fDj ¼ n
X
cij q_ j
(1.8.60)
i¼1
where cij ¼ cji are the damping coefficients of the linear viscous damping.
Apparently, we can construct a quadratic function
R¼
n X
n
1X
cij q_ i q_ j
2 i¼1 j¼1
(1.8.61)
General concepts and principles of structural dynamics Chapter
which yields
fDj ¼ n
X
∂R
¼
cij q_ j
∂q_ j
i¼1
1
73
(1.8.62)
If these forces are introduced as generalized forces in Lagrange’s equations
(1.8.11) and shifted to the left side, we obtain
d ∂T
∂T ∂V
∂R
+
+
¼ Qi ði ¼ 1, 2, …, n Þ
(1.8.63)
dt ∂q_ i
∂qi ∂qi ∂q_ i
The function R defined by Eq. (1.8.61) is known as Raleigh’s dissipation
function.
The equations of motion, Eq. (1.8.58), become now
n
X
mij q€j +
j¼1
n
X
j¼1
cij q_ j +
n
X
kij qj ¼ pi ði ¼ 1, 2, …, n Þ
(1.8.64)
j¼1
where pi denotes the nonconservative external forces.
Eq. (1.8.64) is written in matrix form
m€
q + cq_ + kq ¼ p
(1.8.65)
The matrix c is called the damping matrix of the system.
1.9
Influence of the gravity loads
We consider the system of Fig. 1.9.1a, which can move in the vertical direction.
Apparently, the weight of the body must be added to the external forces. If the
vertical displacement from the undeformed position is designated by u ¼ u ðt Þ,
the equation of motion will read
m u€ + cu_ + ku ¼ pðt Þ + W
(a)
FIG. 1.9.1 Influence of the gravity load.
(b)
(1.9.1)
74 PART
I Single-degree-of-freedom systems
The elongation ust of the spring under its own weight will be
ust ¼ W =k ¼ constant
(1.9.2)
u ¼ ust + uðt Þ
(1.9.3)
Further we set
where uðt Þ represents the vertical displacement measured from the position of
the static equilibrium.
Differentiating Eq. (1.9.3) yields
_ u€ ¼ u€
u_ ¼ u,
(1.9.4)
Using Eqs. (1.9.3), (1.9.4), the equation of motion (1.9.1) becomes
+ cu_ + ku st + k u ¼ pðt Þ + W
m u€
or using Eq. (1.9.2) we obtain
+ cu_ + k u ¼ pðt Þ
m u€
(1.9.5)
The conclusion drawn from Eq. (1.9.5) states that, in the study of the
dynamic response of a system undergoing small displacements, the loads due
to gravity can be neglected. Of course, the total displacements will result as
the sum of the static plus dynamic displacements. That is, the superposition
principle is valid.
1.10 Problems
Problem P1.1 The plane square rigid body B of side length L and surface
mass density g is supported by two identical inclined columns having
cross-sectional moment of inertia I , modulus of elasticity E, and negligible
mass. Derive the equation of motion neglecting the axial deformation of
the columns (Fig. P1.1).
FIG. P1.1 Structure in problem P1.1
General concepts and principles of structural dynamics Chapter
1
75
Problem P1.2 Consider the structure of Fig. P1.2a. The square plate of constant
thickness h ¼ a=10 and mass density g is supported at its center by a flexible
column having a circular cross-section with diameter d ¼ a=10, height a, and
material constants E, n. The plate is loaded by a force P acting in the plane
of the plate at point (Aða=8, a=6Þ and in the direction ∡x, P ¼ b ¼ 30° as
shown in Fig. P1.2b. Derive the equations of motion of the plate when the mass
of the column is neglected.
(a)
(b)
FIG. P1.2 Structure in problem P1.2
FIG. P1.3 Structural model in problem P1.3
Problem P1.3 The semicircular rigid plate of constant thickness and total mass
m is supported as shown in Fig. P1.3. Taking into account that the support at
point O is a hinge, formulate the equation of motion of the plate using (i)
the method of equilibrium of forces, (ii) the principle of virtual displacements,
and (iii) the method of the Lagrange equations.
Problem P.1.4 Consider the system shown in Fig. P1.4. The bars AD and EG
are rigid with masses m and m=3 , respectively. The mass at end D is concentrated. The elastic supports at points at B, E , and D are simulated by springs
with a stiffness k while the end G is supported by a viscous damper with a
damping coefficient c. The rod CE is weightless and rigid. Derive the equation
of motion using the principle of virtual displacements.
76 PART
I Single-degree-of-freedom systems
FIG. P1.4 System in problem P1.4
Problem P1.5 Consider the system shown in Fig. P1.5. The mass m is supported at the top of the flexible and massless column 2 3, which is supported
The support 1 is
on the ground by means of the rigid body 1 2 of mass 2a m.
elastically restrained by the rotational spring CR . Formulate the equation of
¼ m=a.
motion of the structure using CR ¼ EI =2a, m
FIG. P1.5 Structure in problem P1.5
Problem P1.6 Consider the two-story frame of Fig. P1.6. The columns 1 2,
10 20 , and the beam 3 30 are rigid while the columns 2 3, 20 30 , and the
beam 2 20 are massless and flexible with cross-sectional moment of inertia
I and modulus of elasticity E. The supports at 1 and 10 are elastically restrained
by rotational springs with a stiffness CR . Formulate the equation of motion of
¼ m=a.
the structure taking CR ¼ EI =2a and m
FIG. P1.6 Two-story frame in problem P1.6
General concepts and principles of structural dynamics Chapter
1
77
Problem P1.7 The system of Fig. P1.7 consists of the beam AB and the rigid
body S interconnected at B. The beam AB has a negligible mass, modulus of
elasticity E , and cross-sectional moment of inertia I . The beam is fixed at A
while the rigid body is elastically restrained at C by a rotational spring with
a stiffness CR ¼ EI =10L. The total mass m is uniformly distributed. The system
is loaded by the concentrated moment M ðt Þ at point B. Derive the equation of
motion of the system using Lagrange’s equations.
FIG. P1.7 System in problem P1.7
10
10
FIG. P1.8 Frame in problem P1.8
Problem P1.8 The frame of Fig. P1.8 consists of the rigid beam BD of total
mass m and the two massless and flexible columns AB and CD with a
cross-sectional moment of inertia I and modules of elasticity E. The two massless cables FB and GD have cross-sectional area A and cannot undertake compressive force. Derive the equation of motion of the structure taking
¼ m=5a.
I =A ¼ a2 =25 and m
Problem P1.9 Consider the two-story frame of Fig. P1.9. The columns of the
frame are rigid and have a surface mass density g. Their elastic support on the
ground is simulated by the rotational springs with a stiffness CR ¼ EI =10a.
The horizontal beams are massless and flexible with a cross-sectional moment
of inertia I and modulus of elasticity E. Derive the equation of motion when the
structure is subjected to the horizontal loads pðt Þ at the beam levels.
78 PART
I Single-degree-of-freedom systems
FIG. P1.9 Two-story frame in problem P1.9
Problem P1.10 The hinge O of the soft pendulum of Fig. P1.10 is elastically
restrained by the rotational spring with a stiffness CR ¼ EAL=10. The length of
the rod is L, its cross-sectional area A, and the modulus of elasticity E. Formulate the equation of motion of the pendulum.
FIG. P1.10 Soft pendulum in problem P1.10
Problem P1.11 The rigid bar AB of circular cross-section and mass density
¼ m=a is hinged at point A (Fig. P1.11). The cables DB,FB have crossm
sectional area A and modulus of elasticity E. They are assumed massless
and are prestressed so that they can undertake compressive forces. Formulate
=
FIG. P1.11 Structure in problem P1.11
General concepts and principles of structural dynamics Chapter
1
79
the equation of motion of the structure taking into account that the load P is
removed suddenly at instant t ¼ 0. Evaluate the minimum prestressing force
of the cables DB, FB so that they can undertake compressive loads.
Problem P1.12 Consider the structure of Fig. P1.12. The column AC has a
¼ m=a; it is supported by
circular cross-section and a mass per unit length m
a spherical hinge on the ground and is kept in place by three elastic cables of
cross-sectional area A and modulus of elasticity E. The cables are assumed massless and are prestressed so that they can undertake compressive force. Derive the
equation of motion of the structure when it is loaded by the horizontal force P ðt Þ
acting at the top of the column in the direction ∡x,P ¼ b (Fig. P1.12b).
(a)
(b)
FIG. P1.12 Structure in problem P1.12
Problem P1.13 The silo of Fig. P1.13 is supported on its fundament by four
identical columns of a square cross-section. The silo is full of material of density
g. The ground yields elastically with a subgrade constant Ks . The silo and the
fundament are rigid. Derive the equation of motion of the structure when it is
loaded by the horizontal force P ðt Þ acting at the top of the silo in the direction
(a)
(b)
FIG. P1.13 Silo on elastic subgrade. (a) Vertical section. (b) Plan form.
80 PART
I Single-degree-of-freedom systems
∡x,P ¼ b (Fig. P1.13b) using the following data: Side of the columns a=4;
thickness of the bottom and walls of the silo a=8; density of the material of
the silo 1:5g; and soil constant Ks ¼ EI =1500a 3 .
Problem P1.14 Consider the one-story building of Fig. P1.14. The rigid plate is
an equilateral triangular with a side a and it is supported by three columns of
height a, rectangular cross-section a=10 a=20, and modulus of elasticity E.
The columns are fixed at both ends. Derive the equation of motion of the plate
when a horizontal force P ðt Þ acts at point Að0, a=5Þ in the direction ∡x, P ¼ b.
The dead weight of the plate is included in p (kN=m2 ).
Rigid plate
FIG. P1.14 One-story building in problem P1.14
Problem P1.15 The two one-story buildings of Fig. P1.15 are connected with a
beam as shown in the figure. All columns have a square cross-section with a
moment of inertia Ic ¼ 2I . The connecting beam has a square cross-section with
moment of inertia Ib ¼ I . The structure is loaded by the horizontal force F(t) at the
level of the plates as shown in Fig. P1.15b. Formulate the equations of motion
using Lagrange’s equations. Assume: Torsion constant It ¼ 2:25d 4 =16, d ¼side
length of the square cross-section of the beam.
Rigid plate
(a)
beam
(b)
FIG. P1.15 Structure in problem P1.15. (a) vertical section, (b) plan form.
Problem P1.16 The system of Fig. P1.16 consists of the block of mass m1 ,
which can slide without friction on the inclined surface, and the pendulum of
General concepts and principles of structural dynamics Chapter
1
81
length L and mass m2 , which is pivoted at the center of mass of the block. The
rod of the pendulum has a cross-sectional area A and modulus of elasticity E.
Assuming plane motion, derive the equation of motion of the system taking
EA=L ¼ 5k and m1 ¼ 5m2 .
.
FIG. P1.16 System in problem P1.16
Problem P1.17 Derive the equation of motion of the system shown in
Fig. P1.17. The mass of the case is m2 . Use m2 ¼ 5m1 , k2 ¼ 3k1 , c2 ¼ 1:5c1 .
FIG. P1.17 System in problem P1.17
Problem P1.18 Consider the crane of Fig. P1.18. The horizontal beam is
assumed rigid. The column is flexible with a cross-sectional moment of inertia
I and the cable axially deformable with cross-sectional area A. The mass of the
cable and column is negligible. Derive the equation of motion of the system
when it is loaded by the horizontal force pðt Þ in the plane of the structure using
I =A ¼ a 2 =100 and a common modulus of elasticity E.
82 PART
I Single-degree-of-freedom systems
FIG. P1.18 Crane in Problem P1.18.
References and further reading
[1] S. Timoshenko, D.H. Young, W. Weaver Jr., Vibration Problems in Engineering, fifth ed., John
Wiley, New York, 1990.
[2] R.L. Coelho, On the deduction of Newton’s second law, Acta Mech. 229 (5) (2018) 2287–2290.
[3] Euler, L. (1912). Decouverte d’un Nouveau Principe de Mecanique, Memoires de l’academie
des sciences de Berlin 6, 185–217, in: Opera Omnia, Series II, vol. 5, 81–108, Leipzig.
[4] J.T. Katsikadelis, Derivation of Newton’s Law of Motion from Kepler’s Laws of Planetary
Motion, Arch. Appl. Mech. 88 (2018) 27–38, https://doi.org/10.1007/s00419-017-1245-x.
[5] D.T. Greenwood, Principles of Dynamics, Prentice-Hall, Englewood Cliffs, NJ, 1965.
[6] J.T. Katsikadelis, The Boundary Element Method for Engineers and Scientists, Academic Press,
Elsevier, Oxford, UK, 2016.
[7] S. Natsiavas, E. Paraskevopoulos, A set of ordinary differential equations of motion for constrained mechanical systems, Nonlinear Dyn. 79 (2015) 1911–1938.
[8] E.N. Strømmen, Structural Dynamics, Springer Series in Solid and Structural Mechanics,
Springer, New York, 2014.
Chapter 2
Single-degree-of-freedom
systems: Free vibrations
Chapter outline
2.1 Introduction
2.2 Free undamped vibrations
2.3 Free damped vibrations
2.3.1 Critically damped system
2.3.2 Underdamped system
2.1
83
83
91
91
92
2.3.3 Overdamped system
2.4 Conservation of energy in an
undamped system
2.5 Problems
References and further reading
96
97
99
103
Introduction
In this chapter, the free vibrations of a single-degree-of-freedom system
(SDOF) are studied, that is, its response when it is not subjected to any external
force, pðt Þ ¼ 0, but it is excited by an initial displacement and/or initial velocity.
The dynamic model of the system is shown in Fig. 1.4.1 and the equation of
motion (1.4.8) takes the form
m u€ + cu_ + ku ¼ 0
(2.1.1)
Eq. (2.1.1) is an ordinary linear homogeneous differential equation of the
second order with constant coefficients and its solution can be obtained using
known mathematical methods. Inasmuch as we are interested in the physical
response of the system described by this equation, it is advisable to analyze
the free vibration response in two stages, first for c ¼ 0 and then c 6¼ 0. In the
first case, we speak of free undamped vibrations while in the second case we
speak of free damped vibrations. Illustrative examples analyzing the free vibrations of SDOF systems are presented. The pertinent bibliography with recommended references for further study is also included.
2.2
Free undamped vibrations
Although systems without damping do not exist in the real world, the undamped
response is studied because it provides an insight into the free vibration response
of damped systems. In the absence of damping, c ¼ 0, Eq. (2.1.1) becomes
m u€ + ku ¼ 0
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00002-9
© 2020 Elsevier Inc. All rights reserved.
(2.2.1)
83
84 PART
I Single-degree-of-freedom systems
We look for a solution in the form [1,2]
u ¼ elt
(2.2.2)
where l is an arbitrary constant to be determined. Substitution of Eq. (2.2.2)
into Eq. (2.2.1) gives
2
ml + k elt ¼ 0
(2.2.3)
Because for t 0 it is elt 6¼ 0, Eq. (2.2.3) holds only if
ml2 + k ¼ 0
(2.2.4)
Eq. (2.2.4) is the characteristic equation of the differential equation (2.2.1).
Its solutions are
pffiffiffiffiffiffiffi
l1 ¼ iw, l2 ¼ iw, i ¼ 1
(2.2.5)
where
rffiffiffiffiffi
k
w¼
> 0:
m
(2.2.6)
The general solution of Eq. (2.2.1) is
u ðt Þ ¼ A0 eiwt + B 0 eiwt
(2.2.7)
where A0 and B 0 are arbitrary constants. Using Euler’s formula
eiwt ¼ cos wt i sin wt
(2.2.8)
u ðt Þ ¼ ðA0 + B 0 Þcos wt + i ðA0 B 0 Þsin wt
(2.2.9)
Eq. (2.2.7) is written as
or introducing the new arbitrary constants A ¼ A0 + B 0 and B ¼ i ðA0 B 0 Þ
we can write
u ðt Þ ¼ A cos wt + B sin wt
(2.2.10)
The velocity is obtained by the differentiation of Eq. (2.2.10)
u_ ðt Þ ¼ wA sin wt + wB cos wt
(2.2.11)
The arbitrary constants A and B can be determined if the displacement u ðt Þ
and the velocity u_ ðt Þ of the moving system are known at a certain instant t0 .
Usually, it is taken t0 ¼ 0 and the quantities u ð0Þ ¼ u0 and u_ ð0Þ ¼ u_ 0 are
referred to as the initial conditions of the motion.
Single-degree-of-freedom systems: Free vibrations Chapter
2
85
Eqs. (2.2.10), (2.2.11) for t ¼ 0 give
A ¼ u0
(2.2.12a)
u_ 0
w
(2.2.12b)
B¼
and Eq. (2.2.10) becomes
u ðt Þ ¼ u0 cos wt +
u_ 0
sin wt
w
(2.2.13)
Obviously, it is u ðt Þ ¼ 0, when u0 ¼ u_ 0 ¼ 0. Hence, the system is set
to motion only if it is given an initial displacement and/or an initial
velocity.
We set
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ffi
u_ 0
r ¼ ðu 0 Þ2 +
(2.2.14)
w
and we write Eq. (2.2.13) in the form
u0
u_ 0
u ðt Þ ¼ r
cos wt +
sin wt
r
rw
(2.2.15)
Inasmuch as it is
u_ 0
1
rw
u0
1,
r
2 2
u0
u_ 0
and
+
¼1
r
rw
(2.2.16)
we can set
u0
¼ cos q
r
u_ 0
¼ sin q
rw
and
(2.2.17)
and Eq. (2.2.15) becomes
u ðt Þ ¼ rcos ðwt qÞ
(2.2.18)
where
q ¼ tan 1
u_ 0
wu0
(2.2.19)
Eq. (2.2.18) states that the motion of the system is a harmonic vibration
with amplitude ju ðt Þjmax ¼ r, angular velocity w, and phase angle q. The
86 PART
I Single-degree-of-freedom systems
geometrical meaning of Eq. (2.2.18) is shown in Fig. 2.2.1. The displacement
u ðt Þ of the system can be visualized as the projection OA0 of the radius r ¼ OA
on the horizontal diameter,
pffiffiffiffiffiffiffiffiffi when it rotates counterclockwise with a constant
angular velocity w ¼ k=m starting off at an angle q equal to the phase angle.
The quantity w is referred to as the natural frequency of the system. Its physical dimension is s1 and it is measured in radians per second. The natural
frequency is also called eigenfrequency as it results as an eigenvalue of the
eigenvalue problem of linear algebra (see Chapter 12).
FIG. 2.2.1 Geometrical meaning of the free undamped vibration.
The time required for the undamped system to complete one cycle of free
vibration is referred to as the natural period of vibration of the system, which
is denoted by T and measured in seconds. It is related to the natural frequency of
vibration through
T¼
2p
w
(2.2.20)
f¼
1
T
(2.2.21)
The inverse of the period
expresses the number of cycles that the system performs in 1 s. This is referred
to as the natural cyclic frequency. The unit of f is the hertz (Hz) (cycles per
second, cps) and it is related to w through
f¼
w
2p
(2.2.22)
The displacement versus time for a system with w ¼ 8s1 , u0 ¼ 0:05m, and
u_ 0 ¼ 1m=s is shown in Fig. 2.2.2.
Single-degree-of-freedom systems: Free vibrations Chapter
87
2
0.2
T = 2p /w
(du/dt)0
0.15
0.1
0.05
u0
0
–0.05
-
–0.1
–0.15
–0.2
T = 2p /w
0
0.5
1
1.5
2
FIG. 2.2.2 Response of an undamped SDOF system.
Example 2.2.1 Free undamped vibrations of an one-story shear building
Consider the one-story shear building shown in Fig. E2.1. The columns are
assumed massless, inextensible, and fixed on the base. Moreover, the slab is
assumed uniform and rigid. The material constants are E ¼ 2.1 107 kN/m2
and n ¼ 0:2. The total load of the plate (dead plus live) is 70kN=m2 . The dimensions of the column cross-sections are k1 30 30cm2 and k2 30 20cm2 . The
acceleration of gravity is g ¼ 9:81m=s2 . Determine:
(i) The natural frequencies of the building.
(ii) The motion of the slab as well as of the top cross-section of the columns
ki , if the horizontal force P ¼ 1:0 103 kN at point Að2:0, 1:0Þ of
the slab acting in the direction b ¼ ∡x,P ¼ p=6 is removed suddenly
at t ¼ 0.
(a)
(b)
FIG. E2.1 One-story shear building in Example 2.2.1. (a) Vertical section. (b) Plan form.
88 PART
I Single-degree-of-freedom systems
Solution
The motion of the slab is described by the two displacements U ,V of its
center O along the x,y axes, respectively, and its rotation W about O. The
stiffness and mass matrices of the structure can be established using
Eqs. (17a), (17b) for the single-story building in Example 1.5.6. However,
taking into account that the structure is symmetric with respect to both axes,
the three components of the motion are uncoupled and an ad hoc solution can
be readily obtained.
(i) The stiffness of the columns in the x and y directions are:
kx1 ¼ ky1 ¼
12EI 1y
¼ 1360:8kN=m
h3
(1a)
kx2 ¼
12EI 2y
¼ 907:2kN=m
h3
(1b)
ky2 ¼
12EI 2x
¼ 403:2kN=m
h3
(1c)
Hence the respective stiffnesses of the structure are
Kx ¼ 4kx1 + 2kx2 ¼ 7257:6kN=m
(2a)
Ky ¼ 4ky1 + 2ky2 ¼ 6249:6kN=m
(2b)
The torsional stiffness KW is equal to the moment produced by the elastic
forces of the columns for unit rotation of the slab. Referring to Fig. E2.2,
we have
FIG. E2.2 Displacements of point i due to rotation of slab.
u i ¼ ri W sin ai ¼ yi W
(3a)
v i ¼ ri W cos ai ¼ xi W
(3b)
Single-degree-of-freedom systems: Free vibrations Chapter
2
89
Hence, the elastic forces at the top cross-section are
fx i ¼ kxi yi W
(4a)
fy i ¼ kyi xi W
(4b)
Besides these forces, a torsional moment is produced due to the rotation
of the top cross-section of the column
M i ¼ kWi W,
kW i ¼
GI it
,
h
G¼
E
2ð 1 + n Þ
(5)
The moment of the elastic forces of column i with respect O is
M i ¼ yi2 kxi + xi2 kyi + kWi W
(6)
and its torsional stiffness results for W ¼ 1
kWi ¼ yi2 kxi + xi2 kyi + kWi
(7)
Taking into account that [3]
It1 0:141 0:34 ¼ 1:142 103 m4
It2 0:196 0:3 0:23 ¼ 4:7040 104 m4
kW1 ¼
GI 1t
GI 2
¼ 1998:7kNm, kW 2 ¼ t ¼ 823:2kNm
h
h
(8)
x1 ¼ x6 ¼ 4:85m, x2 ¼ x5 ¼ 0, x3 ¼ x4 ¼ 4:85m
y1 ¼ y2 ¼ y3 ¼ 2:85m, y4 ¼ y5 ¼ y6 ¼ 2:85m
we obtain the torsional stiffness of the structure
KW ¼
6
X
yi2 kxi + xi2 kyi + kWi ¼ 1:9663 105 kNm
(9)
i¼1
The mass of the slab and its moment of inertia with respect to O are
m ¼ 10 5 70=9:81 ¼ 356:78kNm1 s2
Io ¼
10 53 + 103 5
70=9:81 ¼ 3716:44kNs2
12
Therefore, the three natural frequencies w1 ,w2 ,w3 are
(10)
(11)
90 PART
I Single-degree-of-freedom systems
9
rffiffiffiffiffiffi
Kx
1 >
>
w1 ¼
¼ 4:510s >
>
>
m
>
>
>
rffiffiffiffiffiffi
=
Ky
1
w2 ¼
¼ 4:185s >
m
>
>
>
rffiffiffiffiffiffiffi
>
>
KW
1 >
;
w3 ¼
¼ 7:274s >
I0
(12)
(ii) The components of the force P are
Px ¼ P cos b ¼ 866:02kN Py ¼ P sin b ¼ 500:0kN
(13)
and its moment with respect to O
MO ¼ yA Px + xA Py ¼ 1866:02kNm
which produces the static displacements
(14)
9
>
=
U0 ¼ Px =Kx ¼ 0:1193m
V0 ¼ Py =Ky ¼ 0:0800m
W0 ¼ MO =KW ¼ 0:949 10
2
rad
(15)
>
;
The slab performs free vibrations with initial conditions U ð0Þ ¼ U0 ,
U_ ð0Þ ¼ 0, V ð0Þ ¼ V0 , V ð0Þ ¼ 0, Wð0Þ ¼ W0 , W_ ð0Þ ¼ 0. Hence, the displacements of the center of the slab are
U ðt Þ ¼ U0 cos w1 t,
V ðt Þ ¼ V0 cos w2 t,
Wðt Þ ¼ W0 cos w2 t
(16)
while the displacements of the i column
ui ðt Þ ¼ U ðt Þ yi Wðt Þ,
vi ðt Þ ¼ V ðt Þ + xi Wðt Þ,
wi ðt Þ ¼ Wðt Þ
(17)
Fig. E2.3 presents the displacements of the column at the upper right
corner (x1 ¼ 4:85m, y1 ¼ 2:85m)
0.2
u1(t)
0.15
v1(t)
w1(t)
0.1
0.05
0
–0.05
–0.1
–0.15
–0.2
0
1
2
3
4
5
t
FIG. E2.3 Displacements u ðt Þ,v ðt Þ and rotation wðt Þ of the top cross-section of column at
x1 ¼ 4:85m, y1 ¼ 2:85m in Example 2.2.1.
Single-degree-of-freedom systems: Free vibrations Chapter
2.3
2
91
Free damped vibrations
In this case, it is c 6¼ 0 and the equation of motion (2.1.1) becomes
m u€ + cu_ + ku ¼ 0
(2.3.1)
We look again for a solution of the differential equation (2.3.1) in the form
of Eq. (2.2.2). This produces the following characteristic equation
ml2 + cl + k ¼ 0
or
l2 +
c
l + w2 ¼ 0
m
The two roots of Eq. (2.3.2) are evaluated from the expression
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c
c 2
l1,2 ¼ w2
2m
2m
(2.3.2)
(2.3.3)
The type of the root depends on the discriminant of the quadratic polynomial, Eq. (2.3.2)
D¼
c
2m
2
w2
(2.3.4)
Eq. (2.3.2) has for:
D > 0 two discrete real roots
D ¼ 0 one double real root
D < 0 two complex conjugate roots
The type of root of Eq. (2.3.2) determines the form of the solution of the differential equation (2.3.1), and consequently the physical response of the system.
Therefore, we distinguish three types of motion with damping, according to
whether the discriminant is positive, negative, or zero. It is expedient to discuss
first the case when the discriminant vanishes, which expresses the condition for
critical damping.
2.3.1
Critically damped system
The value of the damping coefficient c for which the discriminant vanishes is
called critical damping and it is denoted by ccr . The system under critical damping is called a critically damped system. This designation is justified because, as
we shall see below, this value represents the transition threshold from oscillatory to nonoscillatory motion and vice versa. Eq. (2.3.4) for D ¼ 0 determines
this value as
c ¼ ccr ¼ 2mw
(2.3.5)
For this value of c, Eq. (2.3.3) gives the double root
l ¼ w
(2.3.6)
92 PART
I Single-degree-of-freedom systems
and the general solution of Eq. (2.3.1) is
u ðt Þ ¼ ðA + Bt Þewt
(2.3.7)
The arbitrary constants are evaluated from the initial conditions u ð0Þ ¼ u0
and u_ ð0Þ ¼ u_ 0 . Thus, we obtain
A ¼ u0 , B ¼ wu0 + u_ 0
(2.3.8)
u ðt Þ ¼ ½u0 + ðwu0 + u_ 0 Þt ewt
(2.3.9)
and Eq. (2.3.7) becomes
Fig. 2.3.1 shows the plot of the displacement given by Eq. (2.3.9) if
u0 ¼ 0:05m, u_ 0 ¼ 0:2ms1 , w ¼ 8s1 . We see that the motion of the critically
damped system is nonoscillatory, but the displacement vanishes exponentially,
that is, the system returns to static equilibrium in infinite time.
0.07
0.06
(du/dt)0
0.05
u(t)
0.04
0.03
u0
0.02
0.01
0
0
0.2
0.4
0.6
0.8
1
t
FIG. 2.3.1 Response of a system with critical damping.
2.3.2 Underdamped system
This case is the most interesting because in all structural systems and in the
majority of the mechanical systems, the damping is much less than the critical.
To facilitate the study of the underdamped systems, we introduce the damping
ratio x, which is defined as
c
c
<1
(2.3.10)
x¼
¼
ccr 2mw
Single-degree-of-freedom systems: Free vibrations Chapter
Using Eq. (2.3.10), we write Eq. (2.3.3) as
qffiffiffiffiffiffiffiffiffiffiffiffi
l1,2 ¼ xw w x 2 1
2
93
(2.3.11)
or
l1,2 ¼ xw iwD
where
wD ¼ w
qffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
(2.3.12)
(2.3.13)
Further, the equation of motion (2.3.1) can be written as
u€ + 2xwu_ + w2 u ¼ 0
(2.3.14)
u ðt Þ ¼ A0 eðxw + iwD Þt + B 0 eðxwiwD Þt
¼ exwt ðA0 eiwD t + B 0 eiwD t Þ
(2.3.15)
whose general solution is
or using Eq. (2.2.8), we obtain
u ðt Þ ¼ exwt ðA cos wD t + B sin wD t Þ
(2.3.16)
The arbitrary constants A,B are evaluated from the initial conditions
u ð0Þ ¼ u0 and u_ ð0Þ ¼ u_ 0 . These yield
A ¼ u0 , B ¼
u_ 0 + u0 xw
wD
and Eq. (2.3.16) is written as
u_ 0 + u0 xw
u ðt Þ ¼ exwt u0 cos wD t +
sin wD t
wD
(2.3.17)
(2.3.18)
We observe that in the absence of damping, that is, x ¼ 0, it is wD ¼ w and
Eq. (2.3.18), as anticipated, becomes identical to Eq. (2.2.13). Eq. (2.3.18) can
also be written in the form
u ðt Þ ¼ rexwt cos ðwD t qÞ
(2.3.19)
where the amplitude r and the phase angle q are evaluated from the relations
s
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
u_ 0 + u0 xw 2
2
(2.3.20a)
+ ðu0 Þ
r¼
wD
1 u_ 0 + u0 xw
q ¼ tan
(2.3.20b)
wD u0
Fig. 2.3.2 shows the plot of the displacement given by Eq. (2.3.19) with
u0 ¼ 1m, u_ 0 ¼ 1ms1 , w ¼ 6s1 , x ¼ 0:07. Moreover, Fig. 2.3.3 presents the
94 PART
I Single-degree-of-freedom systems
dynamic response of a SDOF system for various values of the damping ratio and
u0 ¼ 1, u_ 0 ¼ 7ms1 , w ¼ 6s1 . From the plot of Eq. (2.3.19), we conclude that
the motion of an underdamped system is harmonic vibration with frequency
wD and period T ¼ 2p=wD , whose amplitude, however, decays exponentially
with time and is bounded by the envelops rexwt . Hence, wD is called the
damped frequency and the respective period the damped period of the underdamped system.
2
T = 2p /wD
1.5
T
T
1
u(t)
0.5
re−xw t
u0
0
–0.5
−re-xw t
–1
–1.5
–2
0
1
2
3
4
5
t
FIG. 2.3.2 Response of an underdamped SDOF system.
2
x=0
x=0.03
x=0.06
x=0.1
1.5
1
u(t)
0.5
0
–0.5
–1
–1.5
–2
0
1
2
3
4
5
t
FIG. 2.3.3 Response of an underdamped SDOF system for various values of the damping ratio.
Single-degree-of-freedom systems: Free vibrations Chapter
2
95
If we denote the displacement at time t by u ðt Þ, then the displacement at
time t + nT , that is, after n complete cycles, will be
u ðt + nT Þ ¼ rexwðt + nT Þ cos ½wD ðt + nT Þ q
(2.3.21)
Because
T¼
2p
wD
(2.3.22)
we have
u ðt + nT Þ ¼ re
2p
xw t + n w
D
¼ rexwt e
cos ½ðwD t + 2npÞ q
p2npx
ffiffiffiffiffiffiffiffi
1x2 cos ðw
D t qÞ
(2.3.23)
2npx
pffiffiffiffiffiffiffi2ffi
¼ u ðt Þe 1x
from which we obtain
u ðt Þ
2npx
¼ pffiffiffiffiffiffiffiffiffiffiffiffi
d ¼ ‘n
u ðt + nT Þ
1 x2
(2.3.24)
The quantity d defined by Eq. (2.3.24) is called the logarithmic decrement
and can be employed to determine the damping ratio x experimentally, when the
displacements u ðt Þ and u ðt + nT Þ between n consecutive cycles are measured.
Example 2.3.1 Free damped vibrations of a silo
Fig. E2.4 presents the idealization of a silo. It consists of two massless and inextensional columns of cross-sectional area a a and a square rigid plate of mass
m. At time t ¼ 0, the silo is displaced horizontally by a force P. Then the force is
suddenly removed and the system starts to vibrate. For P ¼ 200 kN, h ¼ 5m,
a ¼ 0:3m, E ¼ 2.1 107 kN/m2, m ¼ 100kN=ms2 determine:
(i) The damping ratio x of the system, if the horizontal displacement is reduced
to u1 ¼ 0:1u0 after n ¼ 5 oscillations.
(ii) Determine the displacement at time t1 ¼ 2.
FIG. E2.4 Model of the silo in Example 2.3.1.
96 PART
I Single-degree-of-freedom systems
Solution
(i) The only possible motion of the structure is its horizontal displacement
u ðt Þ. Hence, the system has a single degree of freedom. The stiffness k
of the structure is due to the relative displacement of the column ends,
which yield the stiffness
k ¼2
12EI
12 2:1 107 0:34 =12
¼
2
¼ 2721:6kN=m
h3
53
(1)
The force P produces an initial displacement
u0 ¼ P=k ¼ 200=2721:6 ¼ 0:0735m
(2)
Eq. (2.3.24) for n ¼ 5, u1 ¼ 0:1u0 gives x ¼ 0:0731.
pffiffiffiffiffiffiffiffiffi
(ii) The natural frequency is w ¼ k=m ¼ 5:2169s1 . The motion of the
silo is given by Eq. (2.3.18) with u0 ¼ 0:0735m, u_ 0 ¼ 0,
pffiffiffiffiffiffiffiffiffiffiffiffi
wD ¼ w 1 x2 ¼ 5:2029
u ðt Þ ¼ exp e0:3813t ½0:07349 cos 5:203t + 0:005386 sin 5:203t
(3)
which for t ¼ t1 ¼ 2 gives u ðt1 Þ ¼ 0:0211m.
2.3.3 Overdamped system
When the damping is greater than the critical, that is, c > ccr , the system is said
to be overdamped. Hence
c
c
>1
(2.3.25)
x¼
¼
ccr 2mw
then Eq. (2.3.3) gives two distinct real roots
qffiffiffiffiffiffiffiffiffiffiffiffi
l1, 2 ¼ xw W, W ¼ w x2 1 x > 1
(2.3.26)
and Eq. (2.3.1) has the general solution
u ðt Þ ¼ exwt A0 eWt + B 0 eWt
(2.3.27)
Taking into account that
cosh Wt ¼
eWt + eWt
,
2
sinh Wt ¼
eWt eWt
2
(2.3.28)
Eq. (2.3.27) can be written in the alternative form
u ðt Þ ¼ exwt ðA cosh Wt + B sinh Wt Þ
(2.3.29)
Single-degree-of-freedom systems: Free vibrations Chapter
2
97
The arbitrary constants A,B are evaluated from the initial conditions
u ð0Þ ¼ u0 and u_ ð0Þ ¼ u_ 0 . Then, Eq. (2.3.29) becomes
u_ 0 + u0 xw
sinh Wt
(2.3.30)
u ðt Þ ¼ exwt u0 cosh Wt +
W
Eq. (2.3.30) has been plotted in Fig. 2.3.4 with u0 ¼ 1m, u_ 0 ¼ 10ms1 , and
w ¼ 6s1 . It becomes evident that the motion of the overdamped system is
nonoscillatory.
1.5
x =1
x =1.5
x =2
x =2.5
u(t)
1
0.5
0
0
0.5
1
t
1.5
2
FIG. 2.3.4 Response of an overdamped SDOF system for various values of the damping ratio.
2.4
Conservation of energy in an undamped system
The free undamped vibrations of the SDOF system are governed by Eq. (2.2.1),
namely
m u€ + ku ¼ 0
(2.4.1)
Multiplication of the previous equation by the velocity u_ yields
m u€u_ + ku u_ ¼ 0
(2.4.2)
1 2 1 2
md u_ + kd u ¼ 0
2
2
(2.4.3)
which may be written as
98 PART
I Single-degree-of-freedom systems
and after integration over the interval ½0, t gives
1
1
m u_ 2 + ku 2 ¼ E
2
2
(2.4.4)
where E denotes the constant
1
1
E ¼ m ½u_ 0 2 + k ½u0
2
2
2
(2.4.5)
The first term on the left side of Eq. (2.4.4) represents the kinetic energy
T of the system while the second term represents the potential energy U , which
in this case is the elastic energy of the spring. Thus, we may write
T +U ¼E
(2.4.6)
and make the following statement:
In a system that performs free undamped vibrations, the total energy, kinetic
plus potential, remains constant during the whole duration of the motion and is
equal to the sum of the kinetic and potential energy given to the system at the start
of the motion.
Eq. (2.4.4) for u_ ¼ 0 gives the maximum value of the potential energy
1
Umax ¼ ku 2max ¼ E
2
hence
rffiffiffiffiffiffi
2E
umax ¼
k
(2.4.7)
(2.4.8)
Similarly, the maximum value of the kinetic energy is obtained from the
same equation for u ¼ 0
1
Tmax ¼ m u_ 2max ¼ E
2
hence
rffiffiffiffiffiffi
2E
u_ max ¼
m
(2.4.9)
(2.4.10)
Eqs. (2.4.7), (2.4.9) imply
Umax ¼ Tmax
(2.4.11)
1 2
1
ku
¼ m u_ 2max ¼ E
2 max 2
(2.4.12)
or
Single-degree-of-freedom systems: Free vibrations Chapter
2
99
Using Eq. (2.2.18) the potential energy is written
1
1
U ¼ ku 2 ¼ kr2 cos 2 ðwt qÞ
2
2
(2.4.13)
which takes its maximum value when cos ðwt qÞ ¼ 1. Thus, we have
1
Umax ¼ kr2
2
(2.4.14)
Similarly, the kinetic energy is written
1
1
T ¼ m u_ 2 ¼ mw2 r2 sin 2 ðwt qÞ
2
2
(2.4.15)
which takes its maximum value when sin ðwt qÞ ¼ 1. Thus, we have
1
Tmax ¼ mw2 r2
2
(2.4.16)
Substituting Eqs. (2.4.14), (2.4.16) into Eq. (2.4.11) yields
1
1
mw2 r2 ¼ kr2
2
2
(2.4.17)
which provides the following relation for computation of the natural frequency
of the system
rffiffiffiffiffi
k
w¼
(2.4.18)
m
The method we just described for the evaluation of the natural frequency of a
system based on the conservation of energy is known as the Rayleigh method.
Actually, Eq. (2.4.18) is identical to Eq. (2.2.6). Obviously, the present method
does not seem to provide any advantage over the direct method. This is generally true. The Rayleigh method is particularly useful for the approximate determination of the natural frequency of continuous systems, for which an analytical
solution is either very difficult or impractical to obtain (see Chapters 8 and 12).
Nevertheless, modern numerical methods have significantly limited the use of
the Rayleigh method. Of course, this method is very useful in cases where we
want to check the correctness of numerical approaches in problems for which
analytical or other numerical solutions are not available for comparison.
2.5
Problems
Problem P2.1 The structure of Fig. P2.1 consists of two identical rigid bars BA
and BC , both having line density m (mass/length). The support A is a hinge
while the support C is a simple support. The bracing rod DF has a
100 PART
I Single-degree-of-freedom systems
cross-sectional area A, a modulus of elasticity E, and a negligible mass. A vertical force P applied at point B is suddenly removed at t ¼ 0. Determine the
motion of the structure taking the horizontal displacement u ðt Þ of C as the
parameter of motion. Data: m ¼ 5kg=m, a ¼ 3m, E ¼ 2.1 108 kN/m2,
A ¼ 12cm2 .
FIG. P2.1 Structure in Problem 2.1.
Problem P2.2 An equipment of weight 15kN is supported on the horizontal
base by three identical systems, each consisting of a spring and a damper placed
at the three vertices of an equilateral triangle. The projection of the weight center of the equipment coincides with that of the triangle. The weight of the equipment produces a static deflection ust ¼ 2:5cm. The system performs free
vibrations. The dampers are regulated so that the amplitude of vibration reduces
to 1=10 of the initial deflection after five complete cycles. Determine the
damping coefficient c and compare the frequencies w and wD .
Problem P2.3 The wooden body of mass m1 is constrained by the spring k and
the damper c as shown in Fig. P2.3. A projectile of mass m2 ¼ 0:2m1 is fired
into the body and becomes implanted in it. If the speed of the projectile is v,
determine
(i) the maximum displacement if x ¼ 0.
(ii) the displacement u ðt Þ of the body if x ¼ 0:1.
FIG. P2.3 SDOF system in Problem P2.3.
Problem P2.4 A SDOF system of mass m and stiffness k performs free vibrations. At the end of four complete cycles, the displacement is u ð0Þ=3. If the mass
Single-degree-of-freedom systems: Free vibrations Chapter
2
101
of the system is increased by 50%, determine the amplitude of the vibrations
after four complete cycles.
Problem P2.5 The packing of a sensitive instrument is modeled by the system
of Fig. P2.5. During transportation, the box of mass m2 ¼ 10m1 falls vertically
from a height h. Assuming that the box does not bounce after the collision with
the ground, determine the motion of the instrument. What is its maximum
acceleration?
FIG. P2.5 Structural system in Problem P2.5.
Problem P2.6 The horizontal force P applied to the structure of Fig. P2.6 is
suddenly removed at t ¼ 0. Determine the motion of the structure when (i)
the cables are free of any pretension and (ii) have been prestressed to withstand
compression, and compute the minimum required pretension forces. In both
cases, the cables are assumed massless. Data: a ¼ 2m, P ¼ 100kN,
m ¼ 100 kNm1 s2 , E ¼ 2.1 107 kN/m2, I ¼ 880cm4 , A ¼ 5cm2
FIG. P2.6 Structure in Problem P2.6.
Problem P2.7 In the structure of Fig. P2.7, the rigid rod AB of circular crosssection and line mass density m ¼ m=a is supported on the ground by a spherical hinge and held in vertical position by three identical elastic massless cables
of cross-sectional area A and modulus of elasticity E. The cables are prestressed
102 PART
I Single-degree-of-freedom systems
so that they can withstand compression. Their anchor points D,G,F form an
equilateral triangle. Determine the motion of the structure if the horizontal
force P applied at the top of the rod in the direction of the y axis is suddenly
removed at t ¼ 0. Evaluate the minimum values of the prestress forces, which
ensure the capability of the cables to withstand compression. Data: a ¼ 2m,
m ¼ 100 kN m1s2, A ¼ 5cm2 , E ¼ 2.1 108 kN/m2, P ¼ 100kN.
(a)
(b)
FIG. P2.7 Structure in Problem P2.7. (A) Vertical view. (B) Plan form.
Problem P2.8 In the pendulum of Fig. P2.8, the rigid rod suspending the concentrated mass m has a line mass density m ¼ 2m=L. The hinge O is elastically
restrained by the rotational spring with stiffness CR ¼ kL2 =2. The rod is
displaced by an angle q0 from the vertical position and then is left to move.
Considering small displacements, derive the equation of motion and compute
its period.
FIG. P2.8 Pendulum in Problem P2.8.
Problem P2.9 The rigid silo of Fig. P2.9 is supported on its fundament by four
identical columns of square cross-sections with a side length a=8. The silo is full
of material with density g. The bottom and the walls of the silo have a thickness
a=8 and material density 1:5g. Compute the frequencies and the periods of the
structure.
Single-degree-of-freedom systems: Free vibrations Chapter
(a)
2
103
(b)
FIG. P2.9 Silo on four columns. (a) Vertical section. (b) Plan form.
References and further reading
[1] F.B. Hildebrand, Advanced Calculus for Applications, Prentice-Hall, Inc, Englewood Cliffs,
NJ, 1962.
[2] E. Kreyszig, Advanced Engineering Mathematics, fourth ed., John Wiley & Sons, New York,
1979.
[3] S. Timoshenko, J.N. Goodier, Theory of Elasticity, McGraw-Hill, New York, 1951.
[4] S. Graham Kelly, Mechanical Vibrations, Schaum’s Outline Series, McGraw-Hill, New York,
1996.
[5] W.T. Thomson, Theory of Vibration with Applications, fifth ed., Prentice Hall, Upper Saddle
River, NJ, 1998.
[6] S.S. Rao, Mechanical Vibrations, fifth ed., Prentice Hall, Upper Saddle River, NJ, 2011.
Chapter 3
Single-degree-of-freedom
systems: Forced vibrations
Chapter outline
3.1 Introduction
105
3.2 Response to harmonic loading 106
3.2.1 Response of undamped
systems to harmonic
loading
106
3.2.2 Response of damped
systems to harmonic
loading
110
3.3 Response to arbitrary dynamic
loading—Duhamel’s integral 113
3.3.1 Undamped vibrations
113
3.3.2 Damped vibrations
116
3.4 Analytical evaluation of the
Duhamel integral-applications 117
3.4.1 Response to step
function load
117
3.4.2 Response to ramp
function load
120
3.4.3 Response to step function
load with finite
rise time
121
3.5 Response to impulsive loads
125
3.1
3.5.1 Rectangular pulse load
3.5.2 Triangular pulse load
3.5.3 Asymmetrical triangular
pulse load
3.5.4 Response to piecewise
linear loading
3.6 Response to a periodic loading
3.6.1 Periodic loads
3.6.2 Fourier series
3.6.3 Response of the SDOF
system to periodic
excitation
3.7 Response to unit impulse
3.7.1 The delta function or
Dirac’s delta function
3.7.2 Response to unit impulse
3.7.3 Response to arbitrary
loading
3.7.4 The reciprocal theorem
in dynamics
3.8 Problems
References and further reading
126
128
131
135
137
137
138
143
146
146
148
151
151
152
157
Introduction
In this chapter, the forced vibrations of the SDOF system are studied. The
dynamic model of the system is shown in Fig. 1.4.1 and the motion of the system
is governed by Eq. (1.4.8), namely
m u€ + cu_ + ku ¼ pðt Þ
(3.1.1)
where pðt Þ represents an arbitrary function of time. First, the response under a
harmonic load is examined. This type of loading is particularly important in the
dynamic analysis of structures because it allows understanding the major
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00003-0
© 2020 Elsevier Inc. All rights reserved.
105
106 PART I Single-degree-of-freedom systems
differences between the static and dynamic response and identifying phenomena such as resonance that are not conceived by the static consideration. Moreover, the harmonic load analysis allows studying the response of SDOF systems
under a general periodic load using the Fourier series representation of a periodic load. Then, the response of a SDOF system under an arbitrary load is studied using Duhamel’s integral. Finally, the response to a unit load is discussed by
exploiting the properties of the Dirac delta function. The chapter ends by presenting the dynamic reciprocal theorem. Throughout the chapter, illustrative
examples analyzing the forced vibrations of SDOF systems are presented.
The pertinent bibliography with recommended references for further study is
also included. The chapter is enriched with problems to be solved aiming at
better understanding the theoretical issues.
3.2 Response to harmonic loading
In this section, we study the motion of SDOF systems subjected to harmonic
excitations whose magnitudes may be represented by a sine or cosine function
of time, that is,
pðt Þ ¼ p0 sin wt
(3.2.1a)
pðt Þ ¼ p0 cos wt
(3.2.1b)
or
where p0 is a constant representing the amplitude of the force and w is the frequency known as the excitation frequency. As mentioned in Section 3.1, the study
of the response of a SDOF system under harmonic loads is particularly important
in the dynamic analysis of structures because it allows understanding the major
differences between the static and dynamic response and identifying phenomena
such as resonance that are not realized by the static consideration. In fact, any load
that is periodic can be treated by resolving it into its harmonic components using
the Fourier series representation of a periodic function. The response of the system is obtained as the superposition of individual responses to the harmonic
components of the external excitation. We defer this discussion to Section 3.6.
3.2.1 Response of undamped systems to harmonic loading
We study first the response of the system to harmonic loading in absence of
damping. In this case, the equation of motion (3.1.1) for the sine load becomes
m u€ + ku ¼ p0 sin wt
(3.2.2)
Eq. (3.2.2) is an ordinary linear nonhomogeneous differential equation of the
second order with constant coefficients and its solution can be obtained using any
established mathematical method [1, 2]. Here, we look for a solution in the form
u ¼ uh + up
(3.2.3)
Single-degree-of-freedom systems: Forced vibrations Chapter
3
107
where uh is the solution of the homogeneous equation and up a particular solution of the nonhomogeneous equation, that is any function that satisfies
Eq. (3.2.2). The solution of the homogeneous equation was obtained in
Section 2.2, where the free undamped vibrations were discussed. It is given
by Eq. (2.2.10), namely
pffiffiffiffiffiffiffiffiffi
(3.2.4)
uh ¼ A cos wt + B sin wt, w ¼ k=m
There are general methods for obtaining a particular solution [1, 2]. A simple
method that can be applied in this case is the method of undetermined coefficients. According to this method, we look for a solution up in the form
up ¼ C sin wt
(3.2.5)
where C is a constant to be determined. Substituting Eq. (3.2.5) into Eq. (3.2.2)
yields
m w2 + k C p0 ¼ 0
from which we obtain
C¼
p0 1
,
k 1 b2
(3.2.6a)
w
w
(3.2.6b)
b¼
and the particular solution (3.2.5) becomes
up ¼
p0 1
sin wt
k 1 b2
(3.2.7)
Hence the general solution of Eq. (3.2.2) reads
u ðt Þ ¼ Acos wt + B sin wt +
p0 1
sin wt
k 1 b2
(3.2.8)
The arbitrary constants A and B are evaluated from the initial conditions.
We examine the case u ð0Þ ¼ u_ ð0Þ ¼ 0. The expression for the velocity is
obtained by differentiation of Eq. (3.2.8) with respect to time t
u_ ðt Þ ¼ Aw sin wt + Bw cos wt +
p0 w
cos wt
k 1 b2
(3.2.9)
For the considered initial conditions, Eqs. (3.2.8), (3.2.9) give
A ¼ 0 and B ¼ p0 b
k 1 b2
and Eq. (3.2.8) becomes
u ðt Þ ¼
p0 1
b sin wt Þ
ð sin wt
k 1 b2
(3.2.10)
108 PART I Single-degree-of-freedom systems
Obviously
ust ¼ p0 =k
(3.2.11)
denotes the static displacement that would be produced by a load p0 , equal to
the amplitude of the harmonic excitation, if it were to be applied statically.
The time-dependent quantity
Rðt Þ ¼
u ðt Þ
1
b sin wt Þ
¼
ð sin wt
ust 1 b 2
(3.2.12)
is called the response ratio. It is dimensionless and expresses the number that
must multiply the static displacement at time t to obtain the respective dynamic
displacement. The response ratio provides a measure of the influence of the
dynamic loading.
The extreme value
D ¼ max jRðt Þj
(3.2.13)
is referred to as the dynamic magnification factor (DMF). It is a very useful
quantity in dynamic analysis because, if it has been established for a given
loading, the extreme state of deformation and stress can be obtained by static
analysis.
Eq. (3.2.12) for b ¼ 1 takes the indeterminate form
Rðt Þ ¼
0
0
whose limit can be determined using the L’H^
opital rule. Thus, on the basis of
Eq. (3.2.6b) we obtain
b sin wt
sin wt
b!1
1 b2
‘im Rðt Þ ¼ ‘im
b!1
sin bwt b sin wt
b!1
1 b2
¼ ‘im
wt cos bwt sin wt
¼ ‘im
b!1
2b
¼
(3.2.14)
wt cos wt sin wt
2
From the latter relation, it is concluded that when b tends to 1, that is, when
the excitation frequency w of the harmonic force approaches the natural frequency of the system, the dynamic displacement grows indefinitely with time,
although the amplitude of the harmonic loading is finite. This phenomenon is
called resonance. The growth of the amplitude of the displacement with time
due to resonance is shown in Fig. 3.2.1. The response is periodic with a period
Single-degree-of-freedom systems: Forced vibrations Chapter
3
109
2p=w. A measure of the growth rate can be obtained by taking the difference
of the amplitudes of two consecutive peaks.
The time peak occurs when
20
w t/2
15
2p /w
2p /w
10
R(t)
5
0
–5
p
p
–10
–15
–20
–w t/2
0
2
1
3
4
5
t
FIG. 3.2.1 Response ratio of an undamped system at resonance, D ¼ max jRðt Þj ! 1
when t ! 1 (w ¼ 7, b ¼ 1).
dRðt Þ
w2 sin wt
¼
¼0
dt
2
(3.2.15)
or
t¼
np
n ¼ 1, 2, …
w
Hence the difference between consecutive peaks is
np p
np
+
R
R
¼ p cos np
w
w
w
¼ p
(3.2.16)
(3.2.17)
the
When the excitation force is of the cosine type, pðt Þ ¼ p0 cos wt,
employed procedure yields the particular solution
up ¼
p0 1
cos wt
k 1 b2
(3.2.18)
and the general solution for zero initial conditions, u ð0Þ ¼ u_ ð0Þ ¼ 0, is
obtained as
u ðt Þ ¼
p0 1
b cos wt Þ
ð cos wt
k 1 b2
(3.2.19)
110 PART I Single-degree-of-freedom systems
3.2.2 Response of damped systems to harmonic loading
In this case, the equation of motion (3.1.1) for the sine load becomes
m u€ + cu_ + ku ¼ p0 sin wt
(3.2.20)
The general solution of Eq. (3.2.20) can be sought again in the form
u ¼ uh + up
(3.2.21)
where uh is the homogeneous solution and up a particular solution of the
nonhomogeneous equation. We will limit our discussion to underdamped systems. Thus, the homogeneous solution is given by Eq. (2.3.16), namely
uh ¼ exwt ðA cos wD t + B sin wD t Þ
(3.2.22)
which was derived in Section 2.3.2. The particular solution is established using
the method of undetermined coefficients. According to this method, up is sought
in the form [1, 2]
+ C2 cos wt
up ¼ C1 sin wt
(3.2.23)
which is introduced into Eq. (3.2.20) to yield
C1 ¼
p0
1 b2
p0
2xb
, C2 ¼ k 1 b 2 2 + ð2xbÞ2
k 1 b2 2 + ð2xb Þ2
(3.2.24)
Hence, the general solution of Eq. (3.2.20) becomes
u ðt Þ ¼ exwt ðA sin wD t + B cos wD t Þ
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
Transient response
+
p0
1
2xb cos wt
(3.2.25a)
1 b2 sin wt
2
k 1 b2 + ð2xb Þ2
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
Steady-state response
the previous procedure yields
If the excitation force is p0 cos wt,
u ðt Þ ¼ exwt ðA sin wD t + B cos wD t Þ
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
Transient response
+
p0
1
+ 1 b2 cos wt
(3.2.25b)
1 b2 2xb sin wt
2
k 1 b2 + ð2xbÞ2
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
Steady-state response
It is clear that the first term in Eq. (3.2.25a) decays rapidly with time because
of the exponential term exwt , so its contribution becomes negligible after a
short time, which, of course, depends on the damping ratio. Thus, the response
of the system is governed by the second term. For this reason, we say that the
Single-degree-of-freedom systems: Forced vibrations Chapter
3
111
first term expresses the transient response while the second term expresses the
steady-state response of the system. This is shown in Fig. 3.2.2.
FIG. 3.2.2 Response of an underdamped system to harmonic loading ðw ¼ 5, w ¼ 6, p0 =k ¼ 1,
x ¼ 0:1, u0 ¼ 1, u_ 0 ¼ 10Þ.
Referring to Eq. (3.2.25a) we see that the steady-state response can be written in the form
qÞ
u ðt Þ ¼ rsin ðwt
(3.2.26)
where
r¼
i1=2
2
p0 h
2xb
1 b 2 + ð2xbÞ2
, q ¼ tan 1
k
1 b2
(3.2.27)
Therefore, the steady-state response of the underdamped system subjected
to a harmonic loading is an undamped free vibration.
The DMF is
h
i1=2
2
r
¼ 1 b2 + ð2xbÞ2
(3.2.28)
D ¼ max jRðt Þj ¼
p0 =k
We observe that D ¼ D ðbÞ. Consequently the maximum value of D is
obtained when
dD
¼0
db
This condition gives
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b ¼ 1 2x 2
(3.2.29)
(3.2.30)
112 PART I Single-degree-of-freedom systems
6
x=0
5
x=0.1
D (b)
4
3
2
x=0.2
x=0.3
ξ=0.7
x=0.5
1
ξ=1.0
0
0
0.5
1
1.5
2
Frequency ratio, b
2.5
3
FIG. 3.2.3 Variation of the amplitude of the response ratio Dðb, x Þ.
and
Dmax ¼
1
pffiffiffiffiffiffiffiffiffiffiffiffi
2x 1 x2
(3.2.31)
Apparently, the maximum value of the dynamic factor does not occur for
b ¼ 1. Nevertheless, for a small value of x it is b 1, for example, for
x ¼ 0:05 Eq. (3.2.30) gives b ¼ 0:9975. Fig. 3.2.3 shows the variation of D versus the frequency ratio b for different values of the damping ratio x. Note that
if x ¼ 0, Dmax becomes infinite.
6
4
1/2x
R(t)
2
0
–2
–1/2x
–4
–6
0
5
t
10
15
FIG. 3.2.4 Response ratio of an underdamped system at resonance (w ¼ 5, x ¼ 0:1, b ¼ 1,
D ¼ max jRðt Þj ! 1=2x when t ! 1).
Single-degree-of-freedom systems: Forced vibrations Chapter
3
113
Let us study now the response of the system when b ¼ 1, which is the
conventional value for resonance. For this value of b it is w ¼ w and
Eq. (3.2.25a) becomes
u ðt Þ ¼ exwt ðAsin wD t + B cos wD t Þ p0 cos wt
k 2x
(3.2.32)
If we assume zero initial conditions, u ð0Þ ¼ u_ ð0Þ ¼ 0, Eq. (3.2.32) gives
"
!
#
1 xwt
x
pffiffiffiffiffiffiffiffiffiffiffiffi sin wD t + cos wD t cos wt
e
Rðt Þ ¼
2x
1 x2
(3.2.33)
The time variation of Rðt Þ is shown in Fig. 3.2.4.
We observe that the DMF D ¼ max jRðt Þj, for x 6¼ 0 tends asymptotically
to the value 1=2x. Namely, the presence of damping prevents the occurence
of infinite displacements when x 6¼ 0. The conclusion drawn from this analysis
is that damping limits considerably the consequences of resonance. However,
the resonance phenomenon should not be ignored in structural design because
it produces displacements much larger than the static ones.
3.3 Response to arbitrary dynamic loading—Duhamel’s
integral
3.3.1
Undamped vibrations
In this section, we study the response of undamped SDOF systems when the
external loading pðt Þ is neither harmonic nor periodic, but an arbitrary function
of time. In this case, the equation of motion is
m u€ + ku ¼ pðt Þ
(3.3.1)
which we write in the form
u€ + w2 u ¼
pffiffiffiffiffiffiffiffiffi
1
pðt Þ, w ¼ k=m
m
(3.3.2)
There are several methods to obtain the solution of Eq. (3.3.2). A convenient, straightforward, and rather simple method for solving ordinary differential equations with constant coefficients is the Laplace transform
method [1, 3]. This method is based on the Laplace transform, which for a function u ðt Þ, t 0 is commonly denoted by L and defined as
Z 1
u ðt Þest dt
(3.3.3)
L½u ðt Þ ¼ U ðs Þ ¼
0
where s is the variable of the transform.
114 PART I Single-degree-of-freedom systems
The inverse Laplace transform is defined as the function u ðt Þ, which results
as the solution of the integral equation (3.3.3), when the function U ðs Þ is given.
It is denoted by
u ðt Þ ¼ L1 ½u ðt Þ
(3.3.4)
and it is given by the Fourier-Mellin integral.
A method to evaluate the Laplace transform of a simple function as well as
its inverse is to use the Tables of Laplace transform [4]. On the other hand,
ready-to-use computer applications such as Wolfram ALFA, etc., are also available. These applications can be employed to obtain the Laplace transform and
its inverse of complicated functions.
Using integration by parts, we can derive the Laplace transform of the
derivatives of a function u ðt Þ. Thus, we have
L½u_ ðt Þ ¼ sU u ð0Þ
(3.3.5a)
L½u€ðt Þ ¼ s2 U ½su ð0Þ + u_ ð0Þ
(3.3.5b)
Application of the Laplace transform to both sides of Eq. (3.3.2) yields
1
L u€ + w2 u ¼ L½pðt Þ
m
or
L½u€ + w2 L½u ¼
1
L½pðt Þ
m
(3.3.6)
which by virtue of Eq. (3.3.5b) becomes
s2 U ðs Þ ½su ð0Þ + u_ ð0Þ + w2 U ðs Þ ¼
1
P ðs Þ
m
(3.3.7)
where P ðs Þ is the Laplace transform of pðt Þ.
Eq. (3.3.7) is an algebraic equation with respect to U ðs Þ giving
U ðs Þ ¼
s2
s
1
1
1
u ð 0Þ + 2
P ðs Þ
u_ ð0Þ +
2
2
2
+w
s +w
m s + w2
(3.3.8)
The function u ðt Þ is obtained by taking the inverse Laplace transform of
Eq. (3.3.8), namely
u ðt Þ ¼ L1 U ðs Þ
Single-degree-of-freedom systems: Forced vibrations Chapter
3
115
or
u ðt Þ ¼ u ð0ÞL1
s2
s
s
1
s
+ u_ ð0ÞL1 2
+ L1 2
P ðs Þ
2
2
+w
s +w
m
s + w2
(3.3.9)
From the table of the Laplace transforms we obtain
L1
L1
s2
s
¼ cos wt
+ w2
(3.3.10a)
s2
1
sin wt
¼
2
+w
w
(3.3.10b)
Now we focus our attention on the last term on the right side of Eq. (3.3.9).
Its inverse Laplace transform can be obtained using the convolution theorem.
The convolution of two functions f ðt Þ and g ðt Þ denoted by f ðt Þ∗ g ðt Þ or
ðf ∗ g Þt is defined as
Z t
f ðt τÞg ðτÞdτ
f ðt Þ∗ g ðt Þ ¼
0
(3.3.11)
Z t
g ðt τÞf ðτÞdτ
¼
0
The Convolution Theorem
Let f ðt Þ and g ðt Þ be functions with Laplace transforms F ðs Þ and G ðs Þ, respectively, that is, L½f ðt Þ ¼ F ðs Þ and L½g ðt Þ ¼ G ðs Þ, then
L½ðf ∗ g Þðt Þ ¼ F ðsÞG ðsÞ
or equivalently
Z
1
t
L ½F ðs ÞG ðs Þ ¼ ðf ∗g Þðt Þ ¼
f ðt τÞg ðτÞdτ
(3.3.12a)
(3.3.12b)
0
Eq. (3.3.12b) can be employed to obtain the inverse Laplace of the last term
on the right side of Eq. (3.3.9). Thus, for F ðs Þ ¼ 1=ðs2 + w2 Þ and G ðs Þ ¼ P ðs Þ,
we obtain
Z t
1
1
P
ð
s
Þ
¼
pðτÞ sin ½wðt τÞdτ
(3.3.13)
L1 2
s + w2
mw 0
Hence, substituting Eqs. (3.3.10a), (3.3.10b), (3.3.13) into Eq. (3.3.9) gives
the solution of Eq. (3.3.1)
u ðt Þ ¼
u_ ð0Þ
sin wt + u ð0Þcos wt
w Z
t
1
pðτÞsin ½wðt τÞdτ
+
mw 0
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
Duhamel’s integral
(3.3.14)
116 PART I Single-degree-of-freedom systems
It obvious that the first two terms in Eq. (3.3.14) express the contribution of
the initial conditions to the motion of the system. The third term, which
expresses the contribution of the external loading, is known as the Duhamel
integral for the undamped system.
3.3.2 Damped vibrations
When damping is taken into account, the motion is governed by Eq. (3.1.1),
namely
m u€ + cu_ + ku ¼ pðt Þ
(3.3.15)
which for an underdamped system is written as
u€ + 2xwu_ + w2 u ¼
1
pðt Þ
m
(3.3.16)
For an arbitrary loading function pðt Þ, the solution of Eq. (3.3.16) is
obtained using the Laplace transform method following the procedure presented in the previous section. Thus taking the Laplace transform of
Eq. (3.3.16), we obtain
L½u€ + 2xwL½u_ + w2 L½u ¼
1
L½pðt Þ
m
(3.3.17)
Using Eqs. (3.3.5a), (3.3.5b) and solving for U ðs Þ, we obtain
U ðs Þ ¼
s
1
u ð 0Þ + 2
½2xwu ð0Þ + u_ ð0Þ
2
+ 2xws + w
s + 2xws + w2
1
P ðs Þ
+
m s 2 + 2xws + w2
s2
(3.3.18)
The evaluation of the inverse Laplace transform using the table of the
Laplace transforms requires the factorization of the denominator polynomial
f ðs Þ ¼ s 2 + 2xws + w2
(3.3.19)
l1 ¼ xw + iwD , l2 ¼ xw iwD
(3.3.20)
s 2 + 2xws + iw2 ¼ ðs l1 Þðs l2 Þ
(3.3.21)
Its roots are
Hence
From the table of the Laplace transforms we obtain
L1
1
1 l1 t
¼
e e l2 t
l1 l2
ðs l1 Þðs l2 Þ
1
¼ exwt sin wD t
wD
(3.3.22a)
Single-degree-of-freedom systems: Forced vibrations Chapter
L1
L1
s
1 l1 t
¼
l1 e l2 el2 t
ðs l 1 Þð s l 2 Þ
l1 l2
P ðs Þ
1
¼
ðs l 1 Þ ð s l 2 Þ
wD
¼e
Z t
xwt
3
117
(3.3.22b)
cos wD t
pðτÞexwðtτÞ sin wD ðt τÞdτ
(3.3.22c)
0
The last expression was obtained using the convolution theorem.
By virtue of Eqs. (3.3.22a)–(3.3.22c), Eq. (3.3.18) gives
u ðt Þ ¼ L1 ½U ðsÞ
¼
u_ ð0Þ + u ð0Þxw
sin wD t + u ð0Þ cos wD t exwt
wD
Z t
1
+
pðτÞexwðtτÞ sin ½wD ðt τÞdτ
mwD 0
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
(3.3.23)
Duhamel’s integral
It is obvious that the first term in Eq. (3.3.23) expresses the contribution of
the initial conditions to the motion of the system. The last term represents the
Duhamel integral for the underdamped motion. It is obvious that Eq. (3.3.23)
for x ¼ 0 yields Eq. (3.3.14) as anticipated. The Duhamel integral can also be
derived using the method described in Section 3.7.3.
3.4
Analytical evaluation of the Duhamel integral-applications
The analytical evaluation of the Duhamel integral presents considerable difficulties, especially when damping is taken into account. In practical terms, the
analytical integration is recommended when the excitation function is simple.
In recent years, however, the development of symbolic languages, such as
MATHEMATICA and MAPLE, has boosted interest in the use of analytical
integration. This paragraph presents some applications of the Duhamel integral
that help in understanding the response of an SDOF system under certain characteristic forms of dynamic loading. The integration is performed analytically.
3.4.1
Response to step function load
A step load is a nonperiodic load applied suddenly at t ¼ t0 and remains constant
during the whole duration of the motion. Mathematically, this load can be represented using the Heaviside step function
pðt Þ ¼ p 0 H ðt Þ
where
H ðt t0 Þ ¼
0, t < t0
1, t > t0
(3.4.1)
(3.4.2)
118 PART I Single-degree-of-freedom systems
is the Heaviside step function and p0 the magnitude of the load. Fig. 3.4.1 shows
the step function load applied at t ¼ 0.
FIG. 3.4.1 The step function load.
For u ð0Þ ¼ u_ ð0Þ ¼ 0 the displacement is obtained from the Duhamel integral
in Eq. (3.3.14)
Z t
Z t
1
p0
p0 sin ½wðt τÞdτ ¼ sin ½wðt τÞd ½wðt τÞ
u ðt Þ ¼
mw2 0
mw 0
p0
¼
½ cos wðt τÞt0
mw2
or
u ðt Þ ¼
p0
ð1 cos wt Þ
k
(3.4.3)
Taking into account that ust ¼ p0 =k represents the static displacement, the
response ratio is
R ðt Þ ¼
u ðt Þ
¼ ð1 cos wt Þ
ust
(3.4.4)
and the DMF
D ¼ max jRðt Þj ¼ 2
(3.4.5)
Eq. (3.4.5) shows that the suddenly applied load produces a maximum displacement that is twice as large than the displacement that the load p0 would
produce if it were applied statically (slowly). This is an elementary but important result that illustrates the difference between static and dynamic loading of a
structure.
When the damping is taken into account, the displacement for zero initial
conditions, u ð0Þ ¼ u_ ð0Þ ¼ 0, is obtained from Eq. (3.3.23)
Z t
p0
exwðtτÞ sin ½wD ðt τÞdτ
(3.4.6)
u ðt Þ ¼
mwD 0
The evaluation of the Duhamel integral is more complicated. Nevertheless,
using MAPLE we obtain
!
"
#
p0
x
1 cos wD t + pffiffiffiffiffiffiffiffiffiffiffiffi sin wD t exwt
u ðt Þ ¼
(3.4.7)
k
1 x2
Single-degree-of-freedom systems: Forced vibrations Chapter
119
3
2.5
x=0
2
x=0.05
x=0.15
R(t)
1.5
1
0.5
0.
–0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
t/TD (TD=2p /w D)
FIG. 3.4.2 Response ratio of a system under step load.
We observe that Eq. (3.4.7) for x ¼ 0 becomes identical to Eq. (3.4.3).
The response ratio results from Eq. (3.4.7)
"
Rðt Þ ¼ 1 !
#
x
xwt
cos wD t + pffiffiffiffiffiffiffiffiffiffiffiffi sin wD t e
1 x2
(3.4.8)
Fig. 3.4.2 shows the plot of Eq. (3.4.8) for different values of the
damping ratio.
The extreme values of Rðt Þ occur when
"
#
2
dR
xwt ðwx Þ
¼e
+ wD sin wD t ¼ 0
wD
dt
(3.4.9)
which gives
tn ¼
np
wD
n ¼ 0,1, 2, …
(3.4.10)
Substituting tn into Eq. (3.4.8) yields
Rðtn Þ ¼ 1 ð1Þn enx=
pffiffiffiffiffiffiffi2ffi
1x
(3.4.11)
120 PART I Single-degree-of-freedom systems
The maximum value results for n ¼ 1. Hence
D ¼ 1 + ex=
pffiffiffiffiffiffiffi2ffi
1x
(3.4.12)
that is, the maximum displacement occurs at the first peak, where t1 ¼ p=wD .
This is shown in Fig. 3.4.2. Eq. (3.4.12) for x ¼ 0 gives D ¼ 2 as anticipated.
3.4.2 Response to ramp function load
The ramp function load is a load that increases linearly with time. Mathematically, it is defined by
pðt Þ ¼ lt
t0
(3.4.13)
where l is a nonzero constant denoting the rate of increase of the load
(Fig. 3.4.3).
FIG. 3.4.3 Ramp function load; l ¼ p0 =t1 .
For zero initial conditions, u0 ¼ u_ 0 ¼ 0, the response of an undamped SDOF
system to a ramp function load is obtained by substituting the load pðt Þ ¼ p0 t=t1
into the Duhamel integral in Eq. (3.3.14)
p0 =t1
u ðt Þ ¼
mw
Z
t
τ sin ½wðt τÞdτ
(3.4.14)
0
which after integration gives
u ðt Þ ¼
p0 t sin wt
k t1
wt1
(3.4.15)
Fig. 3.4.4 shows the response ratio Rðt Þ ¼ u ðt Þ=ust , ust ¼ p0 =k, of the
undamped system to a ramp function load. We see that it oscillates about the
line p0 t=t1 .
Single-degree-of-freedom systems: Forced vibrations Chapter
121
3
30
25
R(t)
20
15
10
5
0
0
0.5
1
1.5
t/T
2
2.5
3
FIG. 3.4.4 Response ratio of a system under ramp function load (t1 ¼ T =10).
For damped systems, the analytical evaluation is rather complicated. However, it can be evaluated using a symbolic language.
3.4.3
Response to step function load with finite rise time. Static load
The step function load with finite rise time is a constant load that, however, is
not applied suddenly, but rises linearly up to a value p0 within a time t1 , the rise
time, and thereafter remains constant, as shown in Fig. 3.4.5. Mathematically it
is defined by
(p
0
t t t1
(3.4.16)
pðt Þ ¼ t1
p0 t > t1
FIG. 3.4.5 Step function load with finite rise time.
For zero initial conditions, u0 ¼ u_ 0 ¼ 0, the response of the system is
obtained by studying the motion in two phases:
122 PART I Single-degree-of-freedom systems
Phase Ι: Forced vibration under a ramp function load.
This phase starts at t ¼ 0 and ends at t ¼ t1 . The response of the undamped system is given by Eq. (3.4.15)
u I ðt Þ ¼
p0 t sin wt
, 0 < t t1
k t1
wt1
(3.4.17)
Phase ΙI: Forced vibration under a step function load.
This phase starts at t ¼ t1 with initial conditions uII ð0Þ ¼ uI ðt1 Þ, u_ II ð0Þ ¼ u_ I ðt1 Þ
and loading pðt Þ ¼ p0 . The solution is given by Eq. (3.3.14). The value of the
Duhamel integral is given by Eq. (3.4.3). Thus, we have
uII ðt Þ ¼
p0
u_ I ðt1 Þ
sin wt+ uI ðt1 Þcos wt+ ð1 cos wtÞ
k
w
(3.4.18)
where t¼ t t1 > 0.
Note that for t1 ¼ nT , n ¼ 1, 2, …, Eq. (3.4.17) gives
uI ðt1 Þ ¼
p0
, u_ I ðt1 Þ ¼ 0
k
(3.4.19)
and substituting into Eq. (3.3.18) gives
p0
k
(3.4.20)
RII ðt Þ ¼ 1
(3.4.21)
uII ðt Þ ¼
or
which means that the motion in the constant load phase is not oscillatory but the
displacement is constant and equal to the static displacement.
Fig. 3.4.6 presents the response ratio Rðt Þ for different values of t1 .
We observe that for smaller values of t1 =T , the response is similar to that
of a step function load while for larger values, the response is similar to that of
a static load. Therefore, the loads in real structures should not be applied
suddenly but slowly rising, that is, in time much larger than the natural
period of the structure to avoid dynamic magnification effects. Due to this
property, the step function load with finite rise time is also called static
loading.
Single-degree-of-freedom systems: Forced vibrations Chapter
3
123
FIG. 3.4.6 Response ratio of a system under a step function load with finite rise time.
Example 3.4.1 Blast load on one-story building
A blast-induced pressure wave strikes the one-story building in Example 2.2.1
in the x direction. The time variation of the blast pressure is represented by
pðt Þ ¼ p0 ð1 t=t0 Þet=t0 (Fig. E3.1). Determine the stress resultants max Qx ,
max Mx of the columns. The peak positive pressure is p0 ¼ 12kN=m2 and
the duration of the positive phase t0 ¼ 1. The system is at rest at t ¼ 0.
FIG. E3.1 Blast-induced pressure in Example 3.4.1.
Solution
The mass of the structure, its stiffness in the x direction, and the corresponding
natural frequency were evaluated in Example 2.2.1
m ¼ 356:78kN m1 s2 , k ¼ Kx ¼ 7257:6kN=m, w ¼ w1 ¼ 4:51
124 PART I Single-degree-of-freedom systems
If A represents the area of the surface struck by the blast-induced pressure,
the peak force applied at the floor level of the building is
1
1
P ¼ A p0 ¼ 5 5 12 ¼ 150kN=m2
2
2
(1)
ust ¼ P=k ¼ 150=7257:6 ¼ 0:02067m
(2)
Hence
Because the system is at rest at t ¼ 0, the response is given by Eq. (3.3.14)
with u0 ¼ u_ 0 ¼ 0
Z t
P
ð1 τÞexp ðτÞsin ½wðt τÞdτ
(3)
u ðt Þ ¼
mw 0
The analytical evaluation of the Duhamel gives
(
P
2w3
w2 ð w2 1 Þ
u ðt Þ ¼
sin
wt
coswt
k ðw2 + 1Þ2
ð w2 + 1Þ 2
w2 ð w2 1 Þ
w2
+
t exp ðt Þ
exp
ð
t
Þ
ð w 2 + 1Þ
ð w 2 + 1Þ 2
)
(4)
1.5
D = 1.265
1
R(t)
0.5
0
tmax = 0.563
–0.5
–1
–1.5
0
0.5
1
1.5
2
2.5
t
3
3.5
4
4.5
5
FIG. E3.2 Response ratio in Example 3.4.1.
Fig. E3.2 shows the plot of the response ratio Rðt Þ ¼ u ðt Þ=ðP=k Þ. The DMF
was found at D ¼ max jRðt Þj ¼ 1:1265. It was determined as the maximum
value of the array used to plot Rðt Þ. Thus, we have
umax ¼ Du st ¼ 1:1265 0:02067 ¼ 0:02328m
(5)
Single-degree-of-freedom systems: Forced vibrations Chapter
3
125
The extreme values of the shear forces and bending moments are:
Columns 30 30
max Qx ¼ kx umax ¼ 1360:8 0:02328 ¼ 31:68kN
h
max Mx ¼ max Qx ¼ 31:68 5=2 ¼ 79:20kNm
2
Columns 30 20
max Qx ¼ kx umax ¼ 907:2 0:02328 ¼ 21:12kN
h
max Mx ¼ max Qx ¼ 21:12 5=2 ¼ 52:80kNm
2
3.5
Response to impulsive loads
p(t)
In this section, a special group of dynamic loads with similar characteristics will
be considered: the impulsive loads or shock loads. The main characteristics of
such loads are their short duration and their high intensity (large magnitude).
Their duration is of the order of the natural period of the structure. Impulsive
loads are of great importance in the design of certain structural systems, for example, buildings subject to aboveground wind blasts or explosions. Because of the
short duration, the maximum response is reached before damping starts to absorb
much energy. As a result, the maximum response is not significantly affected by
the presence of damping. Therefore, we can ignore damping in the study of the
SDOF systems under impulsive loads in the analysis that follows. The time function representing the impulsive load is arbitrary, e.g., Fig. 3.5.1. Nevertheless,
important conclusions can be drawn by considering impulsive loads with a regular shape. The response of a system to an impulsive load is studied in two phases.
The first phase is a forced vibration under the impulsive load with zero initial
conditions while the second phase is a free vibration with initial conditions being
the displacement and velocity that the system attains at the end of the first phase.
t
FIG. 3.5.1 Time variation of an impulsive load.
126 PART I Single-degree-of-freedom systems
The response of the system to an impulsive load can be analyzed using the
methods presented in previous sections for the solution of the differential equation of motion under an arbitrary loading, that is, either by solving directly the
differential equation or by evaluating Duhamel’s integral. Another method to
obtain the response is to express the pulse as the superposition of two or more
simpler pulses for which the response solution is available or simple to determine (Fig. 3.5.2). Nevertheless, the analytical methods, especially for arbitrary
impulse loads, have lost their importance because of the development of efficient numerical methods. For this reason, only two simple impulsive loads
are considered, the rectangular pulse load and the triangular pulse load.
p(t )
p0
O
p(t )
p 0H (t )
if t
t1
p(t )
0
if t
t1
t
t1
p(t )
p0
O
t1
p(t )
p0 sin t
if t
t1
p(t )
0
if t
t1
t
FIG. 3.5.2 Impulsive loads represented by simple functions.
3.5.1 Rectangular pulse load
The rectangular pulse load is shown in Fig. 3.5.3. Mathematically, this load is
defined as
t t1
p0
(3.5.1)
pðt Þ ¼
0
t > t1
FIG. 3.5.3 Rectangular pulse load.
Single-degree-of-freedom systems: Forced vibrations Chapter
3
127
Phase Ι: Forced vibration
The response of the system is a forced vibration due to the suddenly applied load
p0 at t ¼ 0 with zero initial conditions, u0 ¼ u_ 0 ¼ 0. The displacement is given
by Eq. (3.4.3), namely
p0
uI ðt Þ ¼ ð1 cos wt Þ, 0 < t t1
(3.5.2)
k
Phase ΙΙ: Free vibration
The displacement is given by Eq. (2.2.13), namely
uII ðt Þ ¼
u_ I ðt1 Þ
sin wt+ uI ðt1 Þcos wt,
w
t¼ t t1 0
(3.5.3)
where
p0
ð1 cos wt1 Þ
k
p0 w
u_ I ðt1 Þ ¼
sin wt1
k
uI ðt1 Þ ¼
(3.5.4a)
(3.5.4b)
We now evaluate the DMF D. First, we assume that the maximum displacement occurs in Phase I. Eq. (3.5.2) has a maximum when
p0 w
sin wt ¼ 0 or wt ¼ np n ¼ 1, 2, …
u_ I ðt Þ ¼
k
from which we obtain
np n 2p nT
¼
¼
(3.5.5)
w 2w
2
Therefore, the displacement can reach its maximum umax within Phase I
if t1 T =2. In this case, it is D ¼ 2.
We assume now that t1 < T =2, then the maximum displacement umax
occurs in Phase II. Hence, it will be
t¼
FIG. 3.5.4 Response spectrum for rectangular pulse load.
128 PART I Single-degree-of-freedom systems
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u_ I ðt1 Þ 2
umax ¼
+ ½ u I ðt1 Þ 2
w
(3.5.6)
Using Eqs. (3.5.4a), (3.5.4b) the previous relation becomes
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p0
2p
2p
2p
1 2cos
t1 + cos 2
t1 + sin 2
t1
umax ¼
k
T
T
T
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p0
2p
2 1 cos
¼
t1
k
T
or
umax ¼
2p0
t1
sin p
k
T
(3.5.7)
and the DMF D will be given by
D¼
umax
t1
¼ 2sin p
,
p0 =k
T
t1 T
2
(3.5.8)
Eq. (3.5.8) shows that the maximum response depends only on the ratio
t1 =T . The plot of the function D ¼ D ðt1 =T Þ shown in Fig. 3.5.4 is referred
to as the displacement response spectrum or simply the response spectrum of
the impulsive load (see also Chapter 6). It is evident that the response spectrum
of a pulse load serves to determine the maximum response of the system under
this load without solving the differential equation of motion for the particular
pulse load.
3.5.2 Triangular pulse load
The next impulsive load is the decreasing triangular pulse load shown in
Fig. 3.5.5a.
Mathematically, this load is defined as
8
t
<
t t1
p0 1 (3.5.9)
pðt Þ ¼
t
1
:
0
t > t1
(a)
(b)
FIG. 3.5.5 Decomposition of a triangular pulse load.
(c)
Single-degree-of-freedom systems: Forced vibrations Chapter
3
129
The response of the system is studied again in two phases:
Phase Ι: Forced vibration
Eq. (3.5.9) suggests that the pulse load can be considered as the superposition of
a rectangular pulse and a ramp function pulse (Fig. 3.5.5b and c) and use the
solutions we already found for the respective load functions, that is,
Eqs. (3.4.3), (3.4.15).
Thus, we obtain
uI ðt Þ ¼
p0
sin wt t
1 cos wt +
,
k
t1 w
t1
0 t t1
(3.5.10)
t¼ t t1 0
(3.5.11)
Phase ΙΙ: Free vibration
The displacement is given by Eq. (2.2.13)
uII ðt Þ ¼
u_ I ðt1 Þ
sin wt+ uI ðt1 Þcos wt,
w
where
u I ðt1 Þ ¼
p0 sin wt1
coswt1
k
t1 w
(3.5.12a)
u_ I ðt1 Þ ¼
p0 w cos wt1
1
+ sin wt1 wt1
k
wt1
(3.5.12b)
We now evaluate the DMF D.
The maximum displacement in Phase I is obtained when
u_ I ðt Þ ¼ w sin wt +
cos wt 1
¼0
t1
t1
(3.5.13)
which may be written as
2p
t1
t
t
sin 2p
+ cos 2p
1¼0
T
T
T
(3.5.14)
When t1 =T is specified, Eq. (3.5.14) is a nonlinear algebraic equation for the
ratio t=T . Solving this equation and introducing the obtained solution in
Eq. (3.5.10) yield the respective maximum displacement max uI . The curve
RI in Fig. 3.5.6 represents the function max uI =ust versus t1 =T .
130 PART I Single-degree-of-freedom systems
In Phase II, the maximum response is obtained from the expression
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u_ I ðt1 Þ 2
max uII ¼
+ ½uI ðt1 Þ2
(3.5.15)
w
where uI ðt1 Þ and u_ I ðt1 Þ are given by Eqs. (3.5.12a), (3.5.12b), which yield
when expressed in terms of t1 =T
2
uI ðt1 Þ ¼
p0 6
6
k4
sin 2p
2p
t1
T
t1
T
2
u_ I ðt1 Þ ¼
p0 w 6
6
k 4
cos 2p
t1
2p
T
t1
T
3
cos 2p
t1 7
7
T 5
(3.5.16a)
3
+ sin 2p
t1
1 7
7
t1 5
T
2p
T
(3.5.16b)
The curve RII in Fig. 3.5.6 represents the function max uII =ust versus t1 =T .
From this figure, we conclude that for t1 =T 0:4 the maximum response D of
the system to the triangular pulse load occurs in Phase II while for t1 =T > 0:4 it
occurs in Phase I.
R = maxu /ust
D(t1/T)
D(0.4)=1.0513
R = maxu /ust
t1/T
FIG. 3.5.6 DMF for the triangular pulse load.
Single-degree-of-freedom systems: Forced vibrations Chapter
3.5.3
3
131
Asymmetrical triangular pulse load
The next impulsive load is the asymmetrical triangular pulse load shown in
Fig. 3.5.7. The response can be determined in three phases using the solutions
obtained in Sections 3.4.2 and 3.5.2.
Phase Ι: Forced vibration
uI ðt Þ ¼
p0
sin wt
t
,
k t1
w
0 t t1
(3.5.17)
Phase II: Forced vibration
uII ðt Þ ¼
u_ I ðt1 Þ
sin wet + uI ðt1 Þ cos wet
w
p0
sin wet et
,
+
1 cos wet +
t2
k
t2 w
(3.5.18)
0 et ¼ t t1 < t2
Phase ΙΙΙ: Free vibration
uIII ðt Þ ¼
u_ II ðt1 + t2 Þ
sin wt+ uII ðt1 + t2 Þcos wt,
w
t¼ t ðt1 + t2 Þ 0
(3.5.19)
FIG. 3.5.7 Asymmetrical triangular pulse load.
132 PART I Single-degree-of-freedom systems
Table 3.5.1 summarizes the response ratios for various loadings.
TABLE 3.5.1 Response ratios Rðt Þ for various loading functions.
Loading pðt Þ
Response ratio Rðt Þ
Rðt Þ ¼ 1 cos wt
Rðt Þ ¼ 1 cos wt, t t1
Rðt Þ ¼ coswðt t1 Þ cos wt, t t1
Rðt Þ ¼
1
sin wt
t
, t t1
t1
w
Rðt Þ ¼ 1 +
1
½ sinwðt t1 Þ sin wt , t t1
wt1
Rðt Þ ¼ 1 cos wt +
Rðt Þ ¼
sin wt t
, t t1
wt1
t1
sinwt1 sinwðt t1 Þ
cos wt, t t1
wt1
1
sin wt
t1
t
, 0t 2
t1
w
2
1
t1
t1
sin wt ,
t t1
t1 t + 2sin w t Rðt Þ ¼
2
2
t1
w
Rðt Þ ¼
Rðt Þ ¼
2
t1
sinwt , t t1
sinwðt t1 Þ + 2sinw t 2
wt1
Rðt Þ ¼
b sinwt
sin wt
1 b2
Example 3.5.1 One-story building under an impulsive moment
The rigid slab of the one-story building of Fig. E3.3a,b is loaded by the impulsive moment M ðt Þ, whose time variation is shown in Fig. E3.3c. The columns
are assumed massless and fixed on the ground. The pulse data are:
M0 ¼ 300kNm, M1 ¼ 500kNm, t1 ¼ T , where T is the period of the structure.
The other data are the same as in Example 2.2.1.
Single-degree-of-freedom systems: Forced vibrations Chapter
3
133
y
k1
k2
M (t )
O
k1
(a)
(b)
k2
10 m
k1
5m
x
k1
(c)
FIG. E3.3 One-story building under impulsive load.
Determine the stress resultants max Qx , max Qy , max Mx , max My , and
max Mf of the corner columns using time history analysis.
Solution
Because the structure is symmetric with respect to both axes x and y, the only
possible motion of the slab due to the moment M ðt Þ is the rotation fðt Þ about its
center O. Hence the equation of motion of the slab is
IO f€ðt Þ + Kf fðt Þ ¼ M ðt Þ
(1)
where IO is the moment of inertia of the mass of the plate with respect to O and
Kf the torsional stiffness of the structure. These quantities have been computed
in Example 2.2.1. Thus we have
IO ¼ 3:7164 103 kNm s2
(2)
Kf ¼ 1:9663 105 kNm
(3)
rffiffiffiffiffiffi
Kf
¼ 7:2738
w¼
IO
(4)
2p
¼ 0:869s
w
(5)
t1 ¼ T ¼ 0:869s
(6)
T¼
134 PART I Single-degree-of-freedom systems
The motion is studied in two phases.
Phase Ι: 0 t t1 . Forced vibration
The solution is given by the Duhamel integral
Z t
1
fI ðt Þ ¼
M ðτÞsin wðt τÞdτ
IO w 0
(7)
Taking into account that
M ðt Þ ¼ M0 +
M1 M0
t
t1
(8)
we have
1
M0
fI ðt Þ ¼
IO w
Z
t
0
M1 M0
sin wðt τÞdτ +
t1
Z
t
τ sin wðt τÞdτ
(9)
0
Apparently, the response results as the superposition of a rectangular pulse
and a ramp function pulse. Hence, using Eqs. (3.4.2), (3.4.15), we obtain
fI ðt Þ ¼
1 M0
M1 M0
sin wt
ð1 cos wt Þ +
t
wt1
IO w w
w
(10)
M0
M1 M0
sin wt
ð1 cos wt Þ +
t
Kf
Kf t1
w
(11)
or
f I ðt Þ ¼
which for the adopted data becomes
fI ðt Þ ¼ 1:525 103 ð1 cos wt Þ + 1:177 103 t sin wt
w
(12)
Phase ΙI: t1 < t. Free vibration
The solution is
fII ðt Þ ¼
f_ I ðt1 Þ
sin wðt t1 Þ + fI ðt1 Þcos wðt t1 Þ
w
(13)
Eq. (12) for t ¼ t1 gives
fI ðt1 Þ ¼ 1:017 103 and f_ I ðt1 Þ ¼ 0
(14)
and Eq. (13) becomes
fII ¼ 1:017 103 cos wðt t1 Þ
(15)
Single-degree-of-freedom systems: Forced vibrations Chapter
135
3
The response of the structure is shown in Fig. E3.4.
–3
4 ×10
fmax = 3.594e-3
3
f(t)
2
1
0
tmax = 0.462
–1
–2
t1
0
0.5
1
1.5
2
2.5
3
3.5
4
t
FIG. E3.4 Response of the structure in Example 3.5.1.
The stress resultants of the columns are computed using Eqs. (4a), 4(b), (6)
of Example 2.2.1. Thus, for the upper right column we have
kx1 ¼ ky1 ¼ 1360:8kN=m, kf1 ¼ 1:9985 103 kNm
x1 ¼ 4:85m, y1 ¼ 2:35m
max Qx1 ¼ kx1 y1 fmax ¼ 11:49kN
(16a)
max Qy1 ¼ ky1 x1 fmax ¼ 23:72kN
(16b)
h
max Mx1 ¼ max Qx1 ¼ 28:73kNm
2
h
max My1 ¼ max Qy1 ¼ 59:30kNm
2
max Mf1 ¼ kf1 fmax ¼ 7:183kNm
3.5.4
(16c)
(16d)
(16e)
Response to piecewise linear loading
Certain types of loading can be represented by straight-line segments within
generally unequal time intervals, as shown in Fig. 3.5.8. The accelerogram, that
is, the employed recording of the acceleration of the ground motion during an
earthquake, is a representative example of such loading.
136 PART I Single-degree-of-freedom systems
FIG. 3.5.8 Piecewise linear loading.
The loading function in the i interval is given by the expression
pðt Þ ¼ ai + bi t,
ti1 t ti ,
i ¼ 1, 2, …, n
(3.5.20)
where
ai ¼ pi1 , bi ¼
pi pi1
, to ¼ 0
ti ti1
(3.5.21)
The response is obtained in n phases, where n is the number of intervals.
The solution in the interval ti1 t ti is given by Eq. (3.3.23). Hence
we have
ui ðtÞ ¼
u_ ðti1 Þ + u ðti1 Þxw
sin wD t+ u ðti1 Þcos wD t exwt
wD
Z t
1
+
ðai + bi τÞexwðt τÞ sin ½wD ðt τÞdτ
mwD 0
(3.5.22)
where t¼ t ti1 , 0 t ti ti1 .
The Duhamel integral can be evaluated as a sum of two integrals: one due to
the constant term ai of the loading and the other due to the linear term. The inteð1Þ
ð2Þ
grals are denoted by Di ðt Þ and Di ðt Þ, respectively.
The first integral is obtained from Eq. (3.4.7) for p0 ¼ ai
ai
ð1Þ
Di ðtÞ ¼
k
"
1
!
#
x
xwt
cos wD t + pffiffiffiffiffiffiffiffiffiffiffiffi sin wD t e
1 x2
(3.5.23a)
The analytical evaluation of the second integral is rather complicated.
Therefore, the recourse to a symbolic language is inevitable. Thus, using
MAPLE, we obtain
#
(" )
2x2 1
bi
2x
2x
ð2Þ sin wD t+ cos wD t exwt + t
(3.5.23b)
Di ð t Þ ¼
k
wD
w
w
Substituting Eqs. (3.5.23a), (3.5.23b) into Eq. (3.5.22) gives
ui ðtÞ ¼ Ai0 + Ai1 t+ exwt Ai2 cos wD t+ Ai3 sin wD t
(3.5.24)
Single-degree-of-freedom systems: Forced vibrations Chapter
3
137
in which
1
xbi
ai 2
w
k
bi
Ai1 ¼ , Ai2 ¼ u ðti1 Þ Ai0
k
1 i
u_ ðti1 Þ + xwAi2 Ai1
A3 ¼
wD
Ai0 ¼
(3.5.25)
Differentiating Eq. (3.5.24) with respect to time gives the velocity
u_ i ðt Þ ¼ Ai1 + exwt xwAi2 + wD Ai3 cos wD t xwAi3 + wD Ai2 sin wD t
(3.5.26)
The presented method is exact. However, numerical methods are more convenient to compute the response to a piecewise linear loading (see Chapter 4).
3.6
3.6.1
Response to a periodic loading
Periodic loads
A periodic load is one whose time variation profile repeats continually at regular intervals T . It can be represented by a periodic function
pðt Þ ¼ pðt nT Þ, n ¼ 0,1, 2, …
(3.6.1)
The smallest constant T that satisfies Eq. (3.6.1) is called the period of the
periodic function. Fig. 3.6.1 shows examples of periodic functions. Many loadings in nature are periodic or can be approximated by periodic loads.
(a)
(b)
(c)
FIG. 3.6.1 Examples of periodic loads.
138 PART I Single-degree-of-freedom systems
3.6.2 Fourier series
The periodic function can be represented by the Fourier series, that is, a
trigonometric series of the form [5]
pðt Þ ¼ a0 +
1
X
ðan cos nw0 t + bn sin nw0 t Þ
(3.6.2)
n¼1
where an ,bn are constant coefficients to be determined and w0 ¼ T =2p is the
fundamental frequency of the periodic function. The coefficients an , bn ,
n ¼ 0, 1, 2, … are known as the Fourier series coefficients. They can be determined using the orthogonality property of the sine and cosine functions
presented below.
In general, a set of functions F : ffn ðt Þg is called orthogonal in a interval
½t1 , t2 , if for any two functions fm , fn F, m, n N holds
Z t2
0 if m 6¼ n
(3.6.3)
fm ðt Þfn ðt Þdt ¼
cn if m ¼ n
t1
where cn is a constant.
The set F of functions fn is complete if no other function outside F exists
that satisfies the orthogonality condition (3.6.3).
It can be readily shown that the set
F ¼ f1, cos w0 t, cos 2w0 t, cos 3w0 t, …sin w0 t, sin 2w0 t, sin 3w0 t, …g (3.6.4)
is orthogonal in ½T =2, T =2 and complete. Indeed, have
Z T =2
1cos mw0 tdt ¼ 0
for each m
(3.6.5a)
T =2
Z
T =2
T =2
Z
T =2
T =2
Z
cos mw0 t cos nw0 tdt ¼
T =2
T =2
Z
1sin mw0 tdt ¼ 0
T =2
T =2
for each m
(3.6.5b)
0
if m 6¼ n
T=2 if m ¼ n
(3.6.5c)
cos mw0 t sin nw0 tdt ¼ 0 for each m, n
sin mw0 t sin nw0 tdt ¼
0
if m 6¼ n
T =2 if m ¼ n
(3.6.5d)
(3.6.5e)
The coefficients an are evaluated by multiplying Eq. (3.6.2) by cos mw0 t,
integrating over the interval ½T =2, T =2, and using the orthogonality relations
(3.6.5a)–(3.6.5e). Thus, we obtain
Single-degree-of-freedom systems: Forced vibrations Chapter
a0 ¼
1
T
2
an ¼
T
Z
T =2
T =2
Z
T =2
T =2
3
139
pðt Þdt
(3.6.6a)
pðt Þcos nw0 tdt
(3.6.6b)
Similarly, the coefficients bn are evaluated by multiplying Eq. (3.6.2) by
sin mw0 t, integrating over the interval ½T =2, T =2, and using the orthogonality relations (3.6.5a)–(3.6.5e). Thus, we obtain
bn ¼
2
T
Z
T =2
T =2
pðt Þ sin nw0 tdt
(3.6.6c)
The series (3.6.2) represents the function pðt Þ, that is, converges to pðt Þ for
n ! 1, provided that it satisfies the following conditions, known as Dirichlet
conditions:
(a) The function pðt Þ has a finite number of discontinuities in one period.
(b) The function pðt Þ has a finite number of maxima and minima in one period.
(c) The function pðt Þ is absolutely integrable over a period, that is,
Z T =2
(3.6.7)
jpðt Þjdt ¼ k < 1
T =2
We shall say that that the function pðt Þ is piecewise continuous in the finite
interval ½T =2, T =2, if it satisfies conditions (a) and (b). At the points of discontinuity, for example, point t1 in Fig. 3.6.1c, the Fourier series converges to
the mean value
1 p t1 + p t1+
2
(3.6.8)
where p t1 and p t1+ are the left and right limits of pðt Þ at t1 .
In practice, the periodic function pðt Þ is approximated by a finite number of
terms of the Fourier series, that is, by a finite Fourier series.
Let
k
X
ðan cos nw0 t + bn sin nw0 t Þ
(3.6.9)
Sk ð t Þ ¼ a 0 +
n¼1
be the sum of the first k + 1 terms of the Fourier series, which will represent the
function pðt Þ in the interval ½T=2, T =2. Then we will have
pðt Þ ¼ a0 +
k
X
n¼1
ðan cosnw0 t + bn sin nw0 t Þ + ek ðt Þ
(3.6.10)
140 PART I Single-degree-of-freedom systems
where
ek ðt Þ ¼ pðt Þ Sk ðt Þ
(3.6.11)
is the error between pðt Þ and its approximation. The mean square error Ek is
given by
Z
1 T =2
½ek ðt Þ2 dt
Ek ¼
T T =2
"
#2
(3.6.12)
Z
k
X
1 T =2
pðt Þ a0 ðan cos nw0 t + bn sin nw0 t Þ dt
¼
T T =2
n¼1
The error Ek is a function of a0 , an and bn . Therefore, it is minimized when
∂Ek
∂Ek
∂Ek
¼ 0,
¼ 0,
¼ 0, ðn ¼ 1, 2, …, k Þ
∂a0
∂an
∂bn
Differentiating Eq. (3.6.12) and interchanging differentiation with integration give
"
#
Z
k
X
∂Ek
2 T=2
¼
pðt Þ a0 ðan cos nw0 t + bn sin nw0 t Þ dt (3.6.13a)
∂a0
T T=2
n¼1
"
#
Z
k
X
∂Ek
2 T =2
¼
pðt Þ a0 ðan cos nw0 t + bn sin nw0 t Þ cos nw0 tdt
∂an
T T =2
n¼1
∂Ek
2
¼
∂bn
T
Z
T =2
T =2
"
pðt Þ a0 k
X
#
(3.6.13b)
ðan cos nw0 t + bn sin nw0 t Þ sin nw0 tdt
n¼1
(3.6.13c)
Using the orthogonality relations (3.6.5a)–(3.6.5e), the integrals (3.6.13a)–
(3.6.13c) become
Z
∂Ek
1 T =2
¼ a0 pðt Þdt ¼ 0
(3.6.14a)
∂a0
T T =2
Z
∂Ek
2 T =2
¼ an pðt Þcos nw0 tdt ¼ 0
(3.6.14b)
∂an
T T =2
Z
∂Ek
2 T =2
¼ bn pðt Þ sin nw0 tdt ¼ 0
(3.6.14c)
∂bn
T T =2
We observe that the values of the coefficients a0 , an , bn obtained by
Eqs. (3.6.14a)–(3.6.14c) are identical to those obtained by Eqs. (3.6.6a)–
(3.6.6c). Therefore, we may conclude that if a function is approximated by a
finite Fourier series, the mean square error is minimized.
Single-degree-of-freedom systems: Forced vibrations Chapter
3
141
Moreover, when a function is approximated by a finite Fourier series, the
error in the discontinuity region is considerable, even if the number of terms
of the series is very large. This is known as the Gibbs phenomenon. A technique
to overcome this problem is to adapt a suitable function at the region of discontinuity that restores continuity (Fig. 3.6.2a). Similarly, a concentrated load can
be replaced with a bell-shaped function (Fig. 3.6.2b). This, of course, can be
done if the physical problem under consideration is not altered by this substitution. For example, the functions
p t1 + p t1+
p t1 p t1+
pðt t 1 Þ
, jt t1 j e (3.6.15a)
sin
pðt Þ ¼
2e
2
2
pðt Þ ¼
P
pðt t1 Þ
1 + cos
2e
2e
(3.6.15b)
where e is a small number can play this role.
(a)
(b)
FIG. 3.6.2 Functions pðt Þ replacing pðt Þ in the interval ½t1 e, t1 + e.
Example 3.6.1 Fourier series expansion of a periodic function. Gibbs
phenomenon
Expand the periodic function of Fig. 3.6.1b in Fourier series if t1 ¼ T =2.
Solution
The function pðt Þ within a period is defined as
if
0 t < T =2
2p0 t=T
pðt Þ ¼
0
if
T=2 < t T
(1)
We readily prove that the Dirichlet conditions are satisfied. Hence, the function can be expanded in Fourier series. Eqs. (3.6.6a)–(3.6.6c) give
Z
1 T
1
pðt Þdt ¼ p0
(2a)
a0 ¼
T 0
4
Z
2 T
cos np 1
an ¼
pðt Þcos nw0 tdt ¼ p0
(2b)
T 0
n 2 p2
142 PART I Single-degree-of-freedom systems
2
bn ¼
T
Z
T
pðt Þ sin nw0 t ¼ p0
0
cos np
np
(2c)
Hence
pðt Þ ¼
N
N
p0 p0 X
cos np 1
p0 X
cos np
+ 2
sin nw0 t
cos
nw
t
+
0
4
p n¼1
p n¼1
n2
n
(3)
Fig. E3.5 shows the graphical representation of the finite Fourier series in
Eq. (3) for various values of N with p0 ¼ 1, T ¼ 1. We observe that the convergence is very slow at point t ¼ T =2 due to the Gibbs phenomenon.
N = 20
1.2
N = 100
1.2
1
0.8
0.8
0.6
0.6
p(t)
p(t)
1
0.4
0.4
0.2
0.2
0
0
–0.2
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
–0.2
1
N = 1000
0
0.1
0.2
0.3
0.8
0.8
0.6
0.6
0.6
0.7
0.8
0.9
1
0.7
0.8
0.9
1
p(t)
p(t)
1
0.5
t
N = 2000
1.2
1
0.4
0.4
0.4
0.2
0.2
0
0
–0.2
0
0.1
0.2
0.3
0.4
0.5
t
0.6
0.7
0.8
0.9
1
–0.2
0
0.1
0.2
0.3
0.4
0.5
t
0.6
FIG. E3.5 Gibbs’s phenomenon in Example 3.6.1.
Example 3.6.2 Fourier series expansion of a periodic load
Expand the periodic load of Fig. E3.6 in Fourier series if t1 ¼ T =3.
FIG. E3.6 Periodic load in Example 3.6.2.
Single-degree-of-freedom systems: Forced vibrations Chapter
3
143
Solution
The function pðt Þ within a period T ¼ 3t1 can be represented using the delta
function.a Thus we obtain
pðt Þ ¼ P ½d ðt T =3Þ d ðt 2T =3Þ
Eqs. (3.6.6a)–(3.6.6c) give
Z
1 T
a0 ¼
pðt Þdt ¼ 0
T 0
Z
2 T
2P
2np
4np
cos
cos
an ¼
pðt Þcos nw0 tdt ¼
T 0
T
3
3
Z
2 T
2P
2np
4np
bn ¼
pðt Þ sin nw0 t ¼
sin
sin
T 0
T
3
3
Inserting these values of the coefficients in Eq. (3.6.2) we obtain
(
1
2P X
2np
4np
cos
cos
cos nw0 t
pðt Þ ¼
T n¼1
3
3
)
2np
4np
+ sin
sin
sin nw0 t
3
3
3.6.3
(1)
(2a)
(2b)
(2c)
(3)
Response of the SDOF system to periodic excitation
As was shown in the previous section, any periodic function can be expressed as
a sum of harmonic functions. Theoretically, an infinite number of harmonic
terms are required to represent accurately the function as a Fourier series. However, the contribution of the terms decreases with increasing n. Consequently, a
few terms are adequate to approximate the loading with acceptable accuracy.
For periodic loading, each term of the series represents harmonic excitation.
Thus, the total response of an SDOF system to a periodic loading can be
obtained as the sum of the responses of the individual harmonic terms, provided
that the system is linear, which allows the application of the superposition principle. In subsequent analysis, we assume that the periodic excitation acts for a
sufficiently long time, by which the transient response due to the initial conditions has decayed. Thus, only the steady-state response will be considered.
The steady-state response to the constant load pðt Þ ¼ p0 can be obtained
from Eq. (3.2.25a) by setting p0 ¼ a0 , x ¼ 0, w ¼ 0 and neglecting the transient
response. This yields
ao
(3.6.16a)
u 0 ðt Þ ¼
k
a. The delta function d ðt t0 Þ and its properties are discussed in Section 3.7.1.
144 PART I Single-degree-of-freedom systems
For the sine term bn sin nw0 t the steady-state response results from
Eq. (3.2.25a) by setting p0 ¼ bn , w ¼ nw0 , and b ¼ b n ð¼ nw0 =wÞ
unsin ðt Þ ¼
bn
1
1 b2n sin nw0 t 2xbn cos nw0 t
2
2
2
k 1b
+ ð2xbn Þ
n
(3.6.16b)
while for the cosine term an cos nw0 t the steady-state response is obtained from
Eq. (3.2.25b) by setting p0 ¼ an , w ¼ nw0 , and b ¼ b n ð¼ nw0 =wÞ
uncos ðt Þ ¼
an
1
2xbn sin nw0 t + 1 b 2n cos nw0 t
k 1 b 2 2 + ð2xb Þ2
n
n
(3.6.16c)
The steady-state response of the damped system to periodic loading results
as the superposition of responses to individual terms of the Fourier series
u ðt Þ ¼ u0 ðt Þ +
n
X
unsin ðt Þ +
n¼1
n
X
uncos ðt Þ
(3.6.17)
n¼1
Inserting Eqs. (3.6.16a)–(3.6.16c) into Eq. (3.6.17) gives
u ðt Þ ¼
1
ao X
an
1
+
2xbn sin nw0 t + 1 b2n cos nw0 t
2
2
2
k
k 1b
+ ð2xb n Þ
n¼1
n
+
1
X
bn
n¼1
2 2
k 1b
n
1
+ ð2xb n Þ2
1 b2n sin nw0 t 2xbn cosnw0 t
(3.6.18)
or
u ðt Þ ¼
1
ao X
1
1
+
an ð2xbn Þ + bn 1 b2n sin nw0 t
2
k
k 1 b 2 + ð2xb Þ2
n¼1
n
n
2
+ an 1 b n bn ð2xbn Þ cos nw0 t
(3.6.19)
Theoretically, there is no transient response when x ¼ 0. However, for
small values of damping, which is the usual case in our structures, the
steady-state response can be obtained from Eqs. (3.6.16b), (3.6.16c) for
x 0. This yields
unsin ðt Þ bn
1
sin nw0 t
k 1 b2n
(3.6.20a)
uncos ðt Þ an
1
cos nw0 t
k 1 b 2n
(3.6.20b)
Single-degree-of-freedom systems: Forced vibrations Chapter
3
145
which are actually approximations of the particular solutions (3.2.7) and
(3.2.18) of the undamped motion due to harmonic loading.
On the basis of Eqs. (3.6.20a), (3.6.20b), (3.6.16a), the approximate steadystate response becomes
u ðt Þ 1
1
X
ao X
an 1
bn 1
+
cos nw0 t +
sin nw0 t
2
k
k
k 1 b 2n
1
b
n
n¼1
n¼1
(3.6.21)
The response is a periodic function with a period T ¼ 2p=w0 . The contribution of each harmonic term in the series (3.6.19) depends on: (i) the amplitudes
an and bn of the harmonic components of the load, and (ii) the value of b n . The
terms with b n near 1 have the greatest influence on the response. It should be
noted that in the case bn ¼ 1, x ¼ 0 is meaningless.
Example 3.6.3 Response of an SDOF system subjected to periodic loading
Determine the response of a SDOF system subjected to the periodic loading of
Fig. 3.6.1b. Adopt t1 ¼ T =2, x ¼ 0, T ¼ 1, b ¼ 1:25.
Solution
The function pðt Þ within a period is defined as
2p0 t if 0 < t < 1=2
pðt Þ ¼
0
if 1=2 < t 1
(1)
The Fourier series coefficients are obtained using Eqs. (3.6.6a)–(3.6.6c).
They have already been evaluated in Example 3.6.1. Then the response to
the periodic loading is obtained from Eq. (3.6.19). Fig. E3.7 shows the response
ratio Rðt Þ ¼ u ðt Þ=ðp0 =k Þ (a) for different numbers of series terms and (b) for
n ¼ 50 and various values of damping ratio x. We observe that a small number
of terms gives an acceptable accuracy of the response while the response for
small values of x is very close to that with x ¼ 0, which justifies the
expression (3.6.21).
(a)
(b)
FIG. E3.7 Response to periodic loading in Example 3.6.3.
146 PART I Single-degree-of-freedom systems
3.7 Response to unit impulse
3.7.1 The delta function or Dirac’s delta function
In problems of mechanics, we often come across concentrated loads, that is,
actions that are applied to a very small region, theoretically at a point in space
or an instant in time. In this section, we present a mathematical tool, the delta
function or Dirac’s delta function, that allows us to handle impulsive loads in a
simple way and understand better the dynamic response of SDOF systems to
such loads and in general to arbitrary dynamic loads.
Let P ¼ 1 be a unit force that acts on the SDOF system at time t ¼ 0. First,
we consider that the force is not imposed instantaneously but it is distributed
over a small time interval ½e, +e and let pðt Þ denote its distribution. This distribution, which most probably has the shape shown in Fig. 3.7.1, is not known.
Nevertheless, we know that it is sufficiently “concentrated” about t ¼ 0 and that
Z +1
pðt Þdt ¼ 1
(3.7.1)
1
FIG. 3.7.1 Distribution of a concentrated impulsive force over the interval ½e, +e.
namely, the total force is equal to unity. If we skip the problem of determining
analytically the function pðt Þ, we may assume a priori a prescribed shape for
this function, for example,
k=2, jt j < 1=k
pk ðt Þ ¼
(3.7.2a)
0,
jt j 1=k
or
pk ðt Þ ¼
where k is a positive number.
k
pð1 + k 2 t 2 Þ
(3.7.2b)
Single-degree-of-freedom systems: Forced vibrations Chapter
3
147
Fig. 3.7.2a and b show both functions pk defined in Eqs. (3.7.2a), (3.7.2b),
respectively. Moreover, they satisfy Eq. (3.7.1), which means that they are
equivalent to pðt Þ and can represent the actual force P.
(a)
(b)
FIG. 3.7.2 Functions pk sufficiently concentrated for large values of k.
The distribution of pðt Þ becomes more “concentrated” as the value of k in
Eqs. (3.7.2a), (3.7.2b) increases. The limiting case for k ! 1 yields a fictitious
distribution per unit time, which is denoted by d ðt Þ and defined by
d ðt Þ ¼ ‘im pk ðt Þ
k!1
(3.7.3)
The function d ðt Þ is known as the delta function or Dirac’s delta function.
When the concentrated force acts at instant t ¼ t1 , the delta function is denoted
by dðt t1 Þ. In mathematics, the delta function is treated in the theory of generalized functions [6, 7]. We give below the definition of the delta function as a
generalized function, and we mention some of its properties [8, 9].
For a point source applied at t ¼ 0, the one-dimensional delta function is
defined by the relation
Z +1
d ðt Þf ðt Þ dx ¼ f ð0Þ
(3.7.4)
1
or by the relation
Z
+1
1
dðt t1 Þf ðt Þ dt ¼ f ðt1 Þ
(3.7.5)
for a point source applied at t ¼ t1 . The function f ðt Þ is continuous in a finite
interval containing the source point t ¼ 0 or t ¼ t1 . The one-dimensional delta
function can also be described by the relations
0, t 6¼ 0
d ðt Þ ¼
(3.7.6a)
1, t ¼ 0
148 PART I Single-degree-of-freedom systems
and
Z
+1
1
Z
d ðt Þdt ¼
e
e
d ðt Þdt ¼ 1
(3.7.6b)
where e is a small positive number. According to this definition, the function
dðt Þ has zero value everywhere except at t ¼ 0, where it becomes infinite,
and satisfies Eq. (3.7.6b).
Eq. (3.7.4) may be obtained using the mean value theorem of integral calculus. Thus, referring to Fig. 3.7.2a and choosing e ¼ 1=k, we obtain
Z +e
Z 1
1
f ðt Þd ðt Þdt ¼ ‘im
f ðt Þpk ðt Þdt ¼ ‘im f ðt ∗ Þ
2e
e!0
e!0
2e
1
e
(3.7.7)
¼ ‘im ½f ðt ∗ Þ ¼ f ð0Þ
e!0
where t* is a value of t in the interval (e, þe)
Moreover, using integration by parts we can show
Z b
m
d m d ðt t1 Þ
m d f ðt1 Þ
f ðx Þ
dt
¼
ð
1
Þ
, a < t1 < b
dt m
dt m
a
(3.7.8)
3.7.2 Response to unit impulse
In dynamics, loads pðt Þ distributed over a very short interval ½e, +e are
referred to as impulsive loads (see Section 3.5). The integral of the distribution
of the impulsive load over the interval it acts is defined as the impulse of the load
and is denoted by I , namely
Z +e
pðt Þdt
(3.7.9)
I¼
e
If the impulsive load P is concentrated and acts at instant t ¼ t1 , it can be
denoted by
pðt Þ ¼ Pdðt t1 Þ
Obviously, Eq. (3.7.5) for P ¼ 1 gives
Z +e
Pdðt t1 Þdt ¼ 1
I¼
(3.7.10)
(3.7.11)
e
From Newton’s second law of motion, we have (see Appendix)
d ðm u_ Þ
¼ pðt Þ
dt
(3.7.12a)
or if the mass is constant
m
d u_
¼ pðt Þ
dt
(3.7.12b)
Single-degree-of-freedom systems: Forced vibrations Chapter
which after integration over the interval ½e, +e gives
Z +e
+e
m ½u_ e
¼
pðt Þdt
3
149
(3.7.13)
e
or for e ! 0
where
mDu_ ¼ 1
(3.7.14)
Du_ ¼ u_ t1+ u_ t1
(3.7.15)
Hence the impulsive load produces an abrupt change (discontinuity) of the
velocity at time
t1 . If the system is at rest before the action of the impulsive load,
then it is u_ t1 ¼ 0 and Eq. (3.7.14) gives
u_ ðt1 Þ ¼
1
m
(3.7.16)
where t1 designates t1+ . Therefore, Eq. (3.7.16) presents the initial velocity
given to the system by the impulsive load. However, the displacement remains
continuous, which means that
(3.7.17)
Du ¼ u t1+ u t1 ¼ 0
If the elastic and damping forces are taken into account, then Eq. (3.7.12b)
is written
m
d u_
du
¼ pðt Þ c ku
dt
dt
which after integration over the interval ½e, +e gives
Z +e
Z +e
+e
+e
pðt Þdt c½u e k
u ðt Þdt
m ½u_ e ¼
e
(3.7.18)
(3.7.19)
e
Applying the mean value theorem of integral calculus to the integral of elastic force, we write Eq. (3.7.19) as
Z +e
+e
+e
m ½u_ e
¼
pðt Þdt c½u e
ku ðt ∗ Þ2e, e < t ∗ < e
(3.7.20)
e
which by virtue of Eqs. (3.7.15), (3.7.17) becomes if e ! 0
mDu_ ¼ 1
(3.7.21)
This means that the elastic and damping forces do not influence the change
of the velocity
when
an impulsive load is applied.
If u t1 ¼ u_ t1 ¼ 0, then Eqs. (3.7.16), (3.7.17) imply that the unit
impulse produces free vibrations with initial conditions u ðt1 Þ ¼ 0,
u_ ðt1 Þ ¼ 1=m, where t1 designates t1+ .
h(t-t)
150 PART I Single-degree-of-freedom systems
t =1
t
FIG. 3.7.3 Response to unit impulse.
If τ denotes the instant t1 and h ðt τÞ
then we obtain from Eq. (2.3.18)
h ðt τ Þ ¼
u ðt Þ the produced displacement,
1 xwðtτÞ
e
sin ½wD ðt τÞ, t τ
mwD
(3.7.22a)
Obviously, for the undamped system, x ¼ 0, it is
h ðt τÞ ¼
1
sin ½wðt τÞ, t τ
mw
(3.7.22b)
Eqs. (3.7.22a), (3.7.22b) express the response of the SDOF system to a unit
impulse acting at time τ. Note that h ðt τÞ ¼ 0 if t < τ. Fig. 3.7.3 shows the
response to a unit impulse acting at t1 ¼ τ ¼ 1.
The derivation of the response function h ðt τÞ presented previously was
achieved by physical consideration. Mathematically, it expresses the solution
of the equation of motion under the external loading pðt Þ ¼ d ðt τÞ, that is,
u€ + 2xwu_ + w2 u ¼
1
d ðt τ Þ
m
(3.7.23)
The solution of the above equation can be obtained from Eq. (3.3.23) for
zero initial conditions and pðt Þ ¼ dðt t1 Þ. Thus, we have
1
u ðt Þ ¼
mwD
Z
0
t
dðτ t1 ÞexwðtτÞ sin ½wD ðt τÞdτ
(3.7.24)
Single-degree-of-freedom systems: Forced vibrations Chapter
3
151
which by virtue of Eq. (3.7.5) gives
u ðt Þ
h ð t t1 Þ ¼
1 xwðtt1 Þ
e
sin ½wD ðt t1 Þ, t t1
mwD
(3.7.25)
or by setting t1 ¼ τ, we recover Eq. (3.7.22a).
Eqs. (3.7.22a), (3.7.22b) represent the Green’s function of the equation of
motion of the damped and undamped SDOF systems, respectively.
3.7.3
Response to arbitrary loading
The arbitrary loading pðt Þ can be visualized as a sequence of pulse loads of
infinitesimal duration with magnitude P ¼ pðτÞdτ in the interval ½ 0, t . The
response of the system to this pulse load is
du ¼ ½pðτÞdτh ðt τÞ
(3.7.26)
Apparently, the response of the system at time t is the sum of all infinitesimal responses from t ¼ 0 τ ¼ t. This sum is expressed by the integral
Z t
Z t
du ¼
pðτÞh ðt τÞdτ
(3.7.27)
u ðt Þ ¼
0
0
The integral (3.7.27) is known as the convolution integral of the functions
pðt Þ and h ðt Þ, to which we referred as the Duhamel integral in Section 3.2.2.
3.7.4
The reciprocal theorem in dynamics
In statics, the reciprocity is expressed by Betti’s theorem, also known as the
Maxwell-Betti reciprocal work theorem, introduced by Enrico Betti in 1872.
This theorem is valid for systems with linear behavior and reads
If two sets of loads FI and FII act separately on a linearly elastic structure, the
work WI , II done by the first set of loads in acting through the displacements UII
produced by the second set of loads is equal to the work WII , I done by the second
set of loads in acting through the displacements UI produced by the first set of
loads, namely
WI , II ¼ WII , I
(3.7.28)
In dynamics of linear systems, the reciprocity is expressed by the dynamic
reciprocal theorem [10], known also as the dynamic Betti-Rayleigh theorem.
This theorem for SDOF systems reads
If two loadings pI ðt Þ and pII ðt Þ act separately on a linear dynamic system and
produce the responses uI ðt Þ and uII ðt Þ then the convolution CI , II of the loading
pI ðt Þ with the response uII ðt Þ is equal to the convolution CII , I of the loading pII ðt Þ
with the response uI ðt Þ
152 PART I Single-degree-of-freedom systems
CI , II ¼ CII , I
(3.7.29)
or using definition (3.3.11) for the convolution, we may write
pI ðt Þ∗ uII ðt Þ ¼ pII ðt Þ∗ uI ðt Þ
or in integral form
Z t
Z t
pI ðτÞuII ðt τÞdτ ¼
pII ðτÞuI ðt τÞdτ
0
(3.7.30a)
(3.7.30b)
0
The dynamic reciprocal theorem is readily proved by taking into account
that the convolution satisfies the following properties [8]:
ðiÞ f ðt Þ ∗g ðt Þ ¼ g ðt Þ ∗f ðt Þ
ðiiÞ f ðt Þ∗ ½g ðt Þ∗ q ðt Þ ¼ ½g ðt Þ ∗f ðt Þ∗ q ðt Þ
(3.7.31a)
(3.7.31b)
where f ðt Þ, g ðt Þ, and q ðt Þ are arbitrary functions.
If h ðt Þ represents the response to the unit impulse, Eqs. (3.7.22a), (3.7.22b),
we may write on the basis of Eq. (3.7.25)
uI ðt Þ ¼ pI ðt Þ ∗h ðt Þ
(3.7.32a)
uII ðt Þ ¼ pII ðt Þ∗ h ðt Þ
(3.7.32b)
We may further write
pI ðt Þ ∗uII ðt Þ ¼ pI ðt Þ∗ ½pII ðt Þ∗ h ðt Þ
¼ pI ðt Þ∗ ½h ðt Þ ∗pII ðt Þ
¼ ½pI ðt Þ ∗h ðt Þ∗ pII ðt Þ
¼ uI ðt Þ ∗pII ðt Þ
(3.7.33)
¼ pII ðt Þ∗ uI ðt Þ
which proves the dynamic reciprocal theorem.
The dynamic reciprocal theorem has many applications in mechanics, for
example, moving loads, the boundary integral equation method for dynamic
problems [10], etc.
3.8 Problems
Problem P3.1 A machine carrying a mass m0 is placed on the slab of the building of Fig. P3.1. The mass rotates eccentrically about the point ð2:5, 2:0Þ at a
distance s ¼ 1:0m with a frequency f ¼ 4Hz. The columns are massless, inextensible, and fixed on the base while the slab is assumed uniform and rigid.
Determine the time history of the shear forces Qx , Qy , the bending moments
Mx , My , and the torsion moment Mw at the top of the columns. The material
constants are E ¼ 1:2 107 kN=m2 , n ¼ 0:2. The total load of the plate (dead
plus live) is q ¼ 12kN=m2 ; m0 ¼ m=5, where m is the total mass of the slab.
The acceleration of gravity is g ¼ 9:81m=s2 .
Single-degree-of-freedom systems: Forced vibrations Chapter
3
153
FIG. P3.1 Building in problem P3.1
Problem P3.2 A SDOF system with parameters m, k, x is subjected to
Determine the expression of the steady-state
harmonic loading p ¼ p0 cos wt.
response and show that its amplitude is the same with that produced by the load
ing p ¼ p0 sin wt.
Problem P3.3 An undamped SDOF system is subjected to the loading
When w ¼ w1 , the system is set in resonance. Subsequently, a mass
p ¼ p0 sin wt.
Dm ¼ 0:20kN m1 s is added and the system is set in resonance, when
w ¼ 0:6w1 . Determine the mass m of the system.
Problem P3.4 The one-story building of Fig. P3.4a is supported by three columns placed at the vertices of an equilateral triangle. The slab is rigid and the
columns are fixed at both ends and have a rectangular cross-section, a modulus
of elasticity E, and negligible mass. The slab is subjected to the load
for 0 t t1 and pðt Þ ¼ 0 for t1 < t (Fig. P3.4b) acting in
pðt Þ ¼ p0 cos 3 wt
(a)
FIG. P3.4 One-story building in problem P3.4
(b)
154 PART I Single-degree-of-freedom systems
the x direction. Plot the response ratio Rðt Þ of the structure and the stress resultants at the base of the columns 1 and 3 using the following data: Height of columns a ¼ 4m, cross-section of columns a=10 a=20, side length of the
triangular slab a, p0 ¼ 10kN, E ¼ 2:1 107 kN=m2 , load of the plate, including
x ¼ 0:05, w ¼ 8s1 , acceleration of
the dead load, q ¼ 20kN=m2 , t1 ¼ p=2w,
2
gravity g ¼ 9:81m=s .
Problem P3.5 The structure of Fig. P3.5a consists of the rigid girder BC and
the two flexible columns AB and CD having a cross-sectional moment of inertia
I and a modulus of elasticity E. The cables FB and GC have cross-sectional area
A, cannot withstand compression, and are assumed massless. The structure is subjected to the impulsive loads pðt Þ shown in Fig. P3.5b and c. Study the response of
the structure and determine the maximum error when the impulsive loads are
substituted by equivalent concentrated forces. Plot the function D ðt1 =T Þ for the
two load cases. For which value of the ratio t1 =T is the maximum error less than
2%? Data: a ¼ 1:5m, I ¼ 33,740cm4 (IPE450), E ¼ 2:1 108 kN=m2 ,
¼ 1:0kNm1 s2 =m, p10 ¼ 10kN. The value p20 is deterA ¼ 3cm2 , t1 ¼ 0:1, m
mined so that both loads have the same impulse.
(a)
(b)
FIG. P3.5 Structure in problem P3.5
(c)
Single-degree-of-freedom systems: Forced vibrations Chapter
3
155
Problem P3.6 Consider the structure of Fig. P3.6. The rigid column AC of circu is supported by the three elastic cables
lar cross-section and mass per unit length m
of cross-sectional area A and modulus of elasticity E. The support on the ground is
a spherical hinge. The cables have been prestressed so that they can withstand compression. Three advertising panels are massless fixed at the top of the columns, as
shown in the figure. The structure is subjected to the wind blast load of Fig. P3.6c in
the y direction (see Fig. P3.6b). Determine the minimum prestress force of the
cables. The cables and the panels have negligible mass. Use the data a ¼ 5:0m,
¼ 0:5kNm1 s2 =m, E ¼ 2:1 108 kN=m2 , A ¼ 4cm2 ,
p0 ¼ 4kN=m2 , m
t1 ¼ 0:1s.
(a)
(b)
(c)
FIG. P3.6 Structure in problem P3.6
156 PART I Single-degree-of-freedom systems
Problem P3.7 The one-story building in Example 2.2.1 is subjected to a blast
pressure in the x direction. The time variation of the blast pressure is approximated as shown in Fig. P3.7. Determine the stress resultants max Qx and
max Mx of the columns. The peak positive pressure is p0 ¼ 10kN and the
peak negative pressure is p1 ¼ 0:2p0 ; t1 ¼ 0:2s and t2 ¼ 5t1 . The system is
at rest at t ¼ 0.
10
8
6
p
4
0
2
t1
0
t2
p1
–2
0
0.5
1
1.5
2
FIG. P3.7 Blast pressure in problem P3.7
Problem P3.8 Show that an impulsive load generated by the load pðt Þ
and acting over the interval ½t1 , t2 can be represented as pI ¼ ½H ðt t1 Þ
H ðt t2 Þpðt Þ, where H ðt ti Þ is the Heaviside step function. Write a
MATLAB program that constructs the impulsive load.
Problem P3.9 Determine the dynamic response of a SDOF system subjected to
w ¼ 1:1w,
x ¼ 0, and
the sine periodic loading of Fig. P3.9. Assume: T ¼ p=w,
x ¼ 0:1.
FIG. P3.9 Periodic load in problem P3.9
Single-degree-of-freedom systems: Forced vibrations Chapter
3
157
Problem P3.10 Determine the dynamic response of a SDOF system subjected
T ¼ p=w,
x ¼ 0,
to the periodic loading of Fig. P3.10. Assume w ¼ 1:1w,
and x ¼ 0:1.
FIG. P3.10 Periodic load in problem P3.10
References and further reading
[1] F.B. Hildebrand, Advanced Calculus for Applications, Prentice Hall, Englewood Cliffs, New
Jersey, 1962.
[2] E. Kreyszig, Advanced Engineering Mathematics, fourth ed., John Wiley & Sons, New York,
1979.
[3] R.V. Churchill, Operational Mathematics, second ed., MacGraw-Hill Book Company, Inc.,
New York, 1958.
[4] F. Oberhettinger, L. Badii, Tables of Laplace Transforms, Springer-Verlag, Berlin/Heidelberg/
New York, 1973.
[5] H.P. Hsu, Fourier Analysis, Simon and Schuster, New York, 1967.
[6] G.F. Roach, Green’s Functions, Van Nostrand Reinhold Company, London, 1970.
[7] Μ. Greenberg, Application of Green’s Functions in Science and Engineering, Prentice Hall,
Englewood Cliff, NJ, 1971.
[8] T. Myint-U, L. Debnath, Linear Partial Differential Equations for Scientists and Engineers,
fourth ed., Birkh€auser, Boston, 2007.
[9] J.T. Katsikadelis, The Boundary Element Method for Engineers and Scientists, Academic
Press, Elsevier, Oxford, UK, 2016.
[10] J. Dominguez, Boundary Elements in Dynamics, Computational Mechanics Publications,
Southampton, Boston, 1993.
Chapter 4
Numerical integration of the
equation of motion
Chapter outline
4.1 Introduction
4.2 The central difference method
4.3 The average acceleration
method
4.4 The analog equation method
4.5 Stability of the numerical
integration methods
4.5.1 Errors in the numerical
integration
4.5.2 Difference equations
4.5.3 Difference equations and
stability of the numerical
integration methods
4.1
159
160
164
169
174
174
175
4.5.4 Stability of the central
difference method
4.5.5 Stability of the average
acceleration method
4.5.6 Stability of the analog
equation method
4.6 Accuracy of the numerical
integration
4.7 Problems
References and further reading
180
182
185
185
188
189
178
Introduction
The previous analysis shows that an analytical solution of the equation of
motion for an single-degree-of-freedom (SDOF) system is possible only if
the external force is described by a simple function. If the excitation force varies
arbitrarily with time or is given by a set of its values, an analytical solution is out
of the question. However, such problems can be tackled numerically by time
step integration methods for differential equations. The literature about these
methods is vast. Extensive chapters and whole books cover this subject. They
present the mathematical development of these methods, their computer implementation, and their accuracy, convergence, and stability. Several computer
packages include ready-to-use subroutines for the solution of the differential
equation of motion.
Some of these methods have been specially developed for the study of the
dynamic response of systems. A survey of these methods is given in [1, 2]. The
central difference method (CDM), Houbolt’s method, Wilson’s q-Method, and
Newmark’s method are the most well known among them [3, 4]. Nevertheless,
with the increase of cheap computer power, some of them have lost their importance while others have taken dominating places in the computational arena.
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00004-2
© 2020 Elsevier Inc. All rights reserved.
159
160 PART I Single-degree-of-freedom systems
These methods determine the displacement u ðt Þ by numerical integration of
the equation of motion
m u€ + cu_ + ku ¼ pðt Þ
(4.1.1)
The displacement u ðt Þ is computed step by step. The numerical methods
start from the values u ð0Þ and u_ ð0Þ, which are known at t ¼ 0, and subsequently
marching in the integration time, these quantities are computed at instants
Dt, 2Dt, 3Dt,… where Dt denotes a small time interval. The development of
these methods is based on two assumptions. The first assumption is that the equation
of motion (4.1.1) is satisfied at discrete time instants Dt apart, contrary to the analytical methods in which the equation is satisfied at any instant t. This means that the
equilibrium of all forces, namely inertial, damping, elastic, and external, is satisfied
only at discrete instants within the time interval of the solution. The second assumption is that the parameters of the solution, namely the displacement, velocity, and
acceleration, vary within the time interval Dt according to a prescribed law. The
accuracy, stability, and efficiency of the solution method depend on the numerical
integration scheme as well as on the magnitude of Dt. In the following sections, only
some of these methods, which are usually employed for the study of the dynamic
response of systems, will be briefly discussed. We confine our presentation to the
central difference method (CDM), Newmark’s average acceleration method
(AAM) [5], and the analog equation method (AEM), a new efficient method based
on the principle of the analog equation [6].
The presentation that follows is intended to discuss only the basic concepts
underlying these methods and provide a few computational algorithms together
with the computer programs based on them. While this material would be sufficient for many practical problems and research applications, the reader should
recognize that a wealth of knowledge exists on the subject.
4.2 The central difference method
In this method, the values of u ðt + Dt Þ and u ðt Dt Þ of the function u ðt Þ at
instants t + Dt and t Dt (Fig. 4.2.1) are approximated by their Taylor series
1
1
u ðt + Dt Þ ¼ u ðt Þ + Dt u_ ðt Þ + Dt 2 u€ðt Þ + Dt 3 u
___ðt Þ + ⋯
2
6
1
1
___ðt Þ + ⋯
u ðt Dt Þ ¼ u ðt Þ Dt u_ ðt Þ + Dt 2 u€ðt Þ Dt 3 u
2
6
(4.2.1)
(4.2.2)
Subtracting Eq. (4.2.2) from Eq. (4.2.1) yields
2
___ðt Þ + ⋯
u ðt + Dt Þ u ðt Dt Þ ¼ 2Dt u_ ðt Þ + Dt 3 u
6
(4.2.3)
For small values of Dt, the terms of order higher than two can be neglected
and Eq. (4.2.3) yields the following relation for the approximation of the first
derivative at time t
u_ ðt Þ u ðt + Dt Þ u ðt Dt Þ
2Dt
(4.2.4)
Numerical integration of the equation of motion Chapter
4
161
FIG. 4.2.1 Discretization of the interval ½0, T into N equal intervals h ¼ T =N .
Further, adding Eqs. (4.2.1), (4.2.2) and neglecting the terms of order higher
than three, we obtain the following expression to approximate the second derivative of u ðt Þ at time t
u€ðt Þ u ðt + Dt Þ 2u ðt Þ + u ðt Dt Þ
Dt 2
(4.2.5)
Substitution of the derivatives u_ ðt Þ and u€ðt Þ into Eq. (4.1.1) with their
approximations (4.2.4) and (4.2.5) gives
m
u ðt + Dt Þ 2u ðt Þ + u ðt Dt Þ
u ðt + Dt Þ u ðt Dt Þ
+ ku ðt Þ ¼ pðt Þ
+c
2
Dt
2Dt
which is solved for u ðt + Dt Þ to yield
m
m
c 2m
c +
u
ð
t
+
Dt
Þ
¼
p
ð
t
Þ
k
u ðt Dt Þ
u
ð
t
Þ
Dt 2 2Dt
Dt 2
Dt 2 2Dt
(4.2.6)
The previous equation may be written as
K^ u ðt + Dt Þ ¼ P^
(4.2.7)
where
m
c
K^ ¼ 2 +
Dt
2Dt
2m
m
c P^ ¼ pðt Þ k 2 u ðt Þ u ðt Dt Þ
Dt
Dt 2 2Dt
(4.2.8)
(4.2.9)
The quantities K^ and P^ are referred to as the effective stiffness and the
effective load, respectively. Obviously, Eq. (4.2.7) allows the evaluation of
the displacement at instant t + Dt, if the displacements at the two preceding
instants t and t Dt are known. Because u ð0Þ is known from the initial
162 PART I Single-degree-of-freedom systems
conditions, the procedure starts at t ¼ Dt. Obviously, this requires the value of
u ðDt Þ, which is unknown in the first instance, but it can be determined from
Eq. (4.2.2) for t ¼ 0. Thus neglecting terms of order higher than two, we have
1
u ðDt Þ u ð0Þ Dt u_ ð0Þ + Dt 2 u€ð0Þ
(4.2.10)
2
In the above equation, the quantities u ð0Þ and u_ ð0Þ are known from the initial conditions while u€ð0Þ can be computed from the equation of motion,
Eq. (4.1.1), for t ¼ 0. Thus, we obtain
u€ð0Þ ¼ ½pð0Þ cu_ ð0Þ ku ð0Þ=m
(4.2.11)
The stability of the CDM requires that time step Dt is less than a certain
critical value, that is,
qffiffiffiffiffiffiffiffiffiffiffiffi
Dt Dtcr ¼ T 1 x2 =p
(4.2.12)
where T is the period of the system (see Eq. 4.5.30). Otherwise, the procedure
“blows up” with time and the solution makes no sense. This is discussed in
Section 4.5.
Because T is usually a small number, Dt should be small, which implies
that a large number of time steps are required to solve the equation of motion.
This has been a major drawback of the method, especially in older times when
the computer capabilities in terms of memory and speed were restricted. This
fact has led researchers to develop integration methods in which the size of
the time step is not restricted by a critical value. Table 4.2.1 presents the
TABLE 4.2.1 Central difference method (CDM).
A. Data
w (or k), m, x, u0 , u_ 0 , ttot , pðt Þ
B. Initial computations
B.1. k ¼ mw2 (or w ¼
pffiffiffiffiffiffiffiffiffi
k=m ), c ¼ 2mwx, T ¼ 2p=w, ust ¼ pmax =k
B.2. u€0 ¼ ðp0 cu_ 0 ku 0 Þ=m
pffiffiffiffiffiffiffiffiffiffiffiffi
B.3. Choose Dt < Dtcr ¼ T 1 x2 =p and compute
B.4. ao ¼ 1=Dt 2 , a1 ¼ 1=2Dt, a2 ¼ 2ao , a3 ¼ 1=a2
B.5. uDt ¼ u0 Dt u_ 0 + a3 u€0 , K^ ¼ ao m + a1 c
C. For each time step compute:
C.1. P^t ¼ pt ðk a2 m Þut ðao m a1 cÞutDt
C.2. ut + Dt ¼ P^t =K^ , Rt + Dt ¼ ut + Dt =ust
C.3. u_ t ¼ a1 ðut + Dt utDt Þ, u€ðt Þ ¼ ao ðut + Dt 2ut + utDt Þ
C.4. Increase time t ¼ t + D and check: If t ttot end. Else set utDt ¼ ut , ut ¼ ut + Dt
and go to C.1
Numerical integration of the equation of motion Chapter
4
163
algorithm for the numerical implementation of the CDM in pseudocode-type
notation so that the reader can write a computer code in the language of his/
her preference.
Adhering to the steps of Table 4.2.1, a computer program called centr_
diff_lin.m has been written in MATLAB for the numerical integration of the
equation of motion using the CDM. The program is available on this book’s
companion website. It computes the displacement u ðt Þ, the velocity u_ ðt Þ, the
acceleration u€ðt Þ, and the response ratio Rðt Þ ¼ u ðt Þ=ðpmax =k Þ and gives their
graphical representation. Moreover, it computes the dynamic magnification
factor D ¼ max jRðt Þj and the time tmax it occurs. The user of the program is
responsible for providing the function representing the excitation force.
Example 4.2.1 Response of a SDOF using the central difference method (CDM)
Using the CDM, determine the response of a SDOF system with
m ¼ 100kN m1 s2 , k ¼ 2500kN=m, x ¼ 0:05, and u ð0Þ ¼ u_ ð0Þ ¼ 0, and subjected to the load pðt Þ ¼ po exp ð1 0:5t Þ, po ¼ 100kN.
Solution
The solution is obtained using the program centr_diff_lin.m with Dt ¼ 0:01.
Fig. E4.1 gives the response of the SDOF system. Moreover, Fig. E4.2 shows
Displacement
0.2
Velocity
0.5
u(t)
u,t(t)
0.1
0
0
–0.1
0
5
10
–0.5
0
t
1
R(t)
u,tt (t)
Response ratio
2
2
0
0
–2
–4
10
t
Acceleration
4
5
0
5
10
–1
0
t
FIG. E4.1 Response of the SDOF system in Example 4.2.1.
5
t
10
164 PART I Single-degree-of-freedom systems
the displacement u ðt Þ as compared with the exact one together with the error
u ðt Þ uex ðt Þ. The exact solution was obtained by analytical evaluation of
Duhamel’s integral giving
u ðt Þ ¼
4po
ð1 2xwÞ
10:5t
1xwt
e
+
cos
w
t
+
sin
w
t
D
D e
2wD
m ð1 4xw + 4w2 Þ
(1)
0.2
1
u(t) computed
u(t) exact
0.15
x 10 -4
u-uex
0.8
0.6
0.4
0.2
0.05
–0.2
u(t)
0.1
0
–0.4
0
–0.6
–0.8
-0.05
0
2
4
6
8
10
–1
0
2
4
6
8
10
t
FIG. E4.2 Computed solution and error in Example 4.2.1.
4.3 The average acceleration method
In 1959, N. M. Newmark developed a family of time step methods for the
numerical integration of the equation of motion [5] based on the approximation of the acceleration in each time step. In the following, we describe
only the well-known AAM (Average Acceleration Method), which is also
known as Newmark’s method with b ¼ 1=4 or Newmark’s trapezoidal rule
method. The AAM is the most widely employed numerical method in structural dynamics as it is simple to implement numerically as well as being
unconditionally stable and accurate (see Section 4.5.5). It can also be
employed to solve nonlinear equations of short duration motion [6, 7].
In this method, the acceleration in the interval t to t + Dt is assumed constant and equal to its mean value (Fig. 4.3.1)
1
u€ðt + τÞ ¼ ½u€ðt Þ + u€ðt + Dt Þ
2
0 τ Dt
Integrating with respect to τ yields
τ
u_ ðt + τÞ ¼ ½u€ðt Þ + u€ðt + Dt Þ + C1
2
(4.3.1)
Numerical integration of the equation of motion Chapter
4
165
FIG. 4.3.1 Variation of the acceleration, velocity, and displacement in the average acceleration
method.
where C1 is an arbitrary constant. For τ ¼ 0 we obtain C1 ¼ u_ ðt Þ, hence
τ
u_ ðt + τÞ ¼ u_ ðt Þ + ½u€ðt Þ + u€ðt + Dt Þ
2
0 τ Dt
(4.3.2)
Integrating once more with respect to τ yields
u ðt + τÞ ¼ τu_ ðt Þ +
τ2
½u€ðt Þ + u€ðt + Dt Þ + C2
4
which for τ ¼ 0 yields C2 ¼ u ðt Þ. Thus we have
u ðt + τÞ ¼ u ðt Þ + τ u_ ðt Þ +
τ2
½u€ðt Þ + u€ðt + Dt Þ 0 τ Dt
4
(4.3.3)
Eqs. (4.3.2), (4.3.3) for τ ¼ Dt give
u_ ðt + Dt Þ ¼ u_ ðt Þ +
Dt
½u€ðt Þ + u€ðt + Dt Þ
2
u ðt + Dt Þ ¼ u ðt Þ + Dt u_ ðt Þ +
Dt 2
½u€ðt Þ + u€ðt + Dt Þ
4
(4.3.4)
(4.3.5)
166 PART I Single-degree-of-freedom systems
Setting
Du ¼ u ðt + Dt Þ u ðt Þ
(4.3.6a)
Du_ ¼ u_ ðt + Dt Þ u_ ðt Þ
(4.3.6b)
Du€ ¼ u€ðt + Dt Þ u€ðt Þ
(4.3.6c)
Eqs. (4.3.4), (4.3.5) are written as
Du_ ¼
Dt
½2u€ðt Þ + Du€
2
Du ¼ Dt u_ ðt Þ +
Dt 2
½2u€ðt Þ + Du€
4
(4.3.7)
(4.3.8)
€ we obtain
Solving Eq. (4.3.8) for Du,
Du€ ¼
4
½Du Dt u_ ðt Þ 2u€ðt Þ
Dt 2
(4.3.9)
Moreover, substituting Eq. (4.3.9) into Eq. (4.3.7) yields
Du_ ¼
2
Du 2u_ ðt Þ
Dt
(4.3.10)
We shall now express Du in terms of u_ ðt Þ and u€ðt Þ. For this purpose, we
apply Eq. (4.1.1) at time t + Dt and t. This yields
m u€ðt + Dt Þ + cu_ ðt + Dt Þ + ku ðt + Dt Þ ¼ pðt + Dt Þ
m u€ðt Þ + cu_ ðt Þ + ku ðt Þ ¼ pðt Þ
which after subtracting give
mDu€ + cDu_ + kDu ¼ Dp
(4.3.11)
where
Dp ¼ pðt + Dt Þ pðt Þ
Substituting Eqs. (4.3.9), (4.3.10) into Eq. (4.3.11) yields
K^ Du ¼ P^
(4.3.12)
where
2c
4m
+
K^ ¼ k +
Dt Dt 2
4m
^
+ 2c u_ ðt Þ + 2m u€ðt Þ
P ¼ Dp +
Dt
(4.3.13)
(4.3.14)
Numerical integration of the equation of motion Chapter
4
167
The quantities K^ and P^ are referred to as effective stiffness and effective
load, respectively.
Eq. (4.3.12) allows the computation of Du when u_ ðt Þ and u€ðt Þ are known at
time t. Then Du€ðt Þ and Du_ ðt Þ can be computed using Eqs. (4.3.9), (4.3.10).
Subsequently, the values of u ðt + Dt Þ, u_ ðt + Dt Þ, u€ðt + Dt Þ result from the
expressions
u ðt + Dt Þ ¼ u ðt Þ + Du
(4.3.15a)
u_ ðt + Dt Þ ¼ u_ ðt Þ + Du_
(4.3.15b)
u€ðt + Dt Þ ¼ u€ðt Þ + Du€
(4.3.15c)
It may be more convenient to compute the acceleration u€ðt Þ using the
equation of motion (4.1.1) than Eq. (4.3.15c), that is,
u€ðt + Dt Þ ¼ ½pðt + Dt Þ cu_ ðt + Dt Þ ku ðt + Dt Þ=m
(4.3.16)
Table 4.3.1 presents the algorithm for the numerical implementation of the
AAM in pseudocode-type notation so that the reader can write a computer code
in the language of his/her preference.
TABLE 4.3.1 Average acceleration method (AAM).
A. Data
w (or k), m, x, u0 , u_ 0 , ttot , pðt Þ
B. Initial computations
B.1. k ¼ mw2 (or w ¼
pffiffiffiffiffiffiffiffiffi
k=m ), c ¼ 2mwx, ust ¼ pmax =k
B.2. u€0 ¼ ðp0 cu_ 0 ku 0 Þ=m
B.3. Choose Dt (usually Dt ¼ 0:1T =p) and compute
B.4. K^ ¼ k + 2c=Dt + 4m=Dt 2
C. For each time step compute
C.1. Dp ¼ pðt + Dt Þ pðt Þ,
P^ ¼ Dp + ð4m=Dt + 2cÞu_ ðt Þ + 2m u€ðt Þ
^ K^
C.2. Du ¼ P=
Du_ ¼ 2Du=Dt 2u_ ðt Þ
Du€ ¼ 4½Du Dt u_ ðt Þ=Dt 2 2u€ðt Þ
C.3. u ðt + Dt Þ ¼ u ðt Þ + Du
u_ ðt + Dt Þ ¼ u_ ðt Þ + Du_
u€ðt + Dt Þ ¼ u€ðt Þ + Du€
C.4. Increase time t ¼ t + Dt and check: If t ttot end. Else set u_ ðt Þ ¼ u_ ðt + Dt Þ,
u€ðt Þ ¼ u€ðt + Dt Þ and go to C.1
168 PART I Single-degree-of-freedom systems
As will be shown, contrary to the CDM, the stability of the AAM does not
demand any restriction on the size of the time step Dt. The time step, however,
is influenced by the accuracy of the method and its capability to describe an
oscillatory motion. Therefore, it must be small enough. The selection of Dt
equal to 1/10 of the period of the system or of the period of the excitation force
produces accurate results.
Adhering to the steps of Table 4.3.1, a computer program called av_acc_lin.m
has been written in MATLAB for the numerical integration of the equation of
motion. The program is available on this book’s companion website. It computes
the displacement u ðt Þ, the velocity u_ ðt Þ, the acceleration u€ðt Þ, and the response
ratio Rðt Þ ¼ u ðt Þ=ðpmax =k Þ and gives their graphical representation. Moreover, it
computes the dynamic magnification factor D ¼ max jRðt Þj and the time tmax it
occurs. The user of the program is responsible for providing the function of the
excitation force.
Example 4.3.1 Response of an SDOF using the average acceleration
method (AAM)
Determine the response the SDOF system in Example 4.2.1 using the AAM.
Solution
The solution was evaluated using the program av_acc_lin.m with Dt ¼ 0:01.
Fig. E4.3 gives the graphical representation of the displacement together with
the error u ðt Þ uex ðt Þ. The computed error by the AAM is almost double the
error of the CDM. Moreover, Fig. E4.4 shows the response of the system under
the harmonic load p ¼ 2 sin 5t. Obviously, this excitation produces resonance
(w ¼ w ¼ 5).
0.2
2
u(t) computed
u(t) exact
x 10−4
1.5
u-uex
0.15
1
0.5
u(t)
0.1
0
0.05
–0.5
–1
0
–1.5
−0.05
0
2
4
6
8
10
–2
0
t
FIG. E4.3 Computed solution and error in Example 4.3.1.
2
4
6
t
8
10
Numerical integration of the equation of motion Chapter
Displacement
0.01
u,t(t)
u(t)
0.02
0
–0.005
–0.01
0
–0.02
0
5
10
15
–0.04
5
10
15
Response ratio
10
5
R(t)
0.1
u,tt (t)
0
t
t
Acceleration
0.2
0
–0.1
–0.2
169
Velocity
0.04
0.005
4
0
–5
0
5
10
–10
15
0
5
10
15
t
t
FIG. E4.4 Response of the system in Example 4.3.1 under harmonic load.
4.4
The analog equation method
The AEM (Analog Equation Method) is a general method for solving differential equations, ordinary or partial, linear or nonlinear, of the elliptic, parabolic,
or hyperbolic type. The AEM is based on the principle of the analog equation,
according to which a differential equation can be replaced by another equation,
the so-called analog equation, provided that the substitute equation retains the
principal mathematical symbol of the differential equation, that is, the highest
order derivative of the original differential equation. Thus, a differential equation whose solution cannot be obtained can be reduced to an equation with a
known solution. The AEM, which was first presented in 1994 [8], has been
employed as a computational method for the solution of numerous difficult
problems in engineering science and mathematical physics [9–11]. Here, the
AEM is presented for the solution of the equation of motion in structural dynamics as developed in [6]. It is self-starting, unconditionally stable, accurate, and
conserves energy. It performs well when large deformations and long-time
durations are considered and it can be used as a practical method for the integration of the equations of motion in cases where widely used time integration
procedures, for example, Newmark’s AAM, become unstable [6, 7].
We consider the IVP (initial value problem) for the SDOF system
m u€ + cu_ + ku ¼ pðt Þ
t ½0, ttot , ttot > 0
(4.4.1)
170 PART I Single-degree-of-freedom systems
u ð0Þ ¼ u0 ,
u_ ð0Þ ¼ u_ 0
(4.4.2)
Let u ¼ u ðt Þ be the sought solution. Then, if the operator d =dt is applied to
it, we have
2
2
u€ ¼ q ðt Þ
(4.4.3)
where q ðt Þ is a fictitious source, unknown in the first instance. Eq. (4.4.3) is the
analog equation of Eq. (4.4.1). It indicates that the solution of Eq. (4.4.1) can
be obtained by solving Eq. (4.4.3) with the initial conditions (4.4.2), if the q ðt Þ
is first established. This is achieved as follows.
Taking the Laplace transform of Eq. (4.4.3) we obtain
1
1
1
U ðs Þ ¼ u ð0Þ + 2 u_ ð0Þ + 2 Q ðs Þ
s
s
s
(4.4.4)
where U ðs Þ,Q ðs Þ are the Laplace transforms of u ðt Þ,q ðt Þ, respectively. The
inverse Laplace transform of Eq. (4.4.4) gives the solution in integral from
Z t
u ðt Þ ¼ u ð0Þ + u_ ð0Þt +
q ðτÞðt τÞdτ
(4.4.5)
0
Thus, the IVP of Eqs. (4.4.1), (4.4.2) is transformed into the equivalent
Volterra integral equation for q ðt Þ.
Eq. (4.4.5) is solved numerically within a time interval ½0, T . The interval
½0, T is divided into N equal intervals, Dt ¼ h, h ¼ T =N , in which q ðt Þ is
assumed to vary according to a certain law, for example, constant, linear,
etc. In this analysis, q ðt Þ is assumed to be constant and equal to the mean value
in the interval h (Fig. 4.4.1). That is
qrm ¼
qr1 + qr
2
(4.4.6)
FIG. 4.4.1 Discretization of the interval ½0, T into N equal intervals h ¼ T=N .
Hence, Eq. (4.4.5) at instant t ¼ nh can be written as
un ¼ u0 + nh u_ 0
" Z
Z
h
m
m
+ q1
ðnh τÞdτ + q2
0
h
2h
Z
ðnh τÞdτ + ⋯ + qnm
nh
ðn1Þh
#
ðnh τÞdτ
(4.4.7)
Numerical integration of the equation of motion Chapter
4
171
which after evaluation of the integrals yields
un ¼ u0 + nh u_ 0 + c1
n 1
X
½2ðn r Þ + 1qrm + c1 qnm
r¼1
n 1
X
¼ un1 + h u_ 0 + 2c1
qrm + c1 qnm
(4.4.8)
r¼1
where
c1 ¼
h2
2
(4.4.9)
The velocity is obtained by direct differentiation of Eq. (4.4.5) using
Leibnitz’ rule for integrals [12]. Thus, we have
Z t
u_ ðt Þ ¼ u_ ð0Þ +
q ðτÞdτ
(4.4.10)
0
Using the same discretization for the interval ½0, T to approximate the integral in Eq. (4.4.10), we have
u_ n ¼ u_ 0 + c2
n1
X
qrm + c2 qnm
r¼1
(4.4.11)
¼ u_ n1 + c2 qnm
where
Solving Eq. (4.4.11) for
n1
P
r¼1
c2 ¼ h
qrm
(4.4.12)
and substituting into Eq. (4.4.8) gives
un ¼ un1 + h u_ n c1 qnm
By virtue of Eq. (4.4.6), Eqs. (4.4.13), (4.4.11) are written as
c1
c1
qn h u_ n + un ¼ qn1 + un1
2
2
c2
c2
qn + u_ n ¼ u_ n1 + qn1
2
2
(4.4.13)
(4.4.14)
(4.4.15)
Moreover, Eq. (4.4.1) at time t ¼ nt is written as
mq n + cu_ n + ku n ¼ pn
Eqs. (4.4.14), (4.4.15), (4.4.16) can
2
2
3
0
m
c k 8 9
q
6 1
7< n = 6 1
c h 1 7 u_
6 c
6
¼6 2 1
6 2 1
7
n
4
5: ; 4
1
1
un
c2 1 0
c2
2
2
be combined as
3
0 0 8
9 8 9
7< qn1 = < 1 =
0 1 7 u_
+ 0 p
7
n1
; : ; n
5:
0
u
n1
1 0
(4.4.16)
(4.4.17)
172 PART I Single-degree-of-freedom systems
Because m 6¼ 0, the coefficient matrix in Eq. (4.4.17) is not singular for sufficient small h and the system can be solved successively for n ¼ 1, 2, … to yield
the solution un and the derivatives u_ n , u€n ¼ qn at instant t ¼ nh T . For n ¼ 1,
the value q0 appears in the right side of Eq. (4.4.17). This quantity can be readily
obtained from Eq. (4.4.1) for t ¼ 0. Thus, we have
q0 ¼ ðp0 cu_ 0 ku 0 Þ=m
(4.4.18)
Eq. (4.4.17) can be also written as
Un ¼ AUn1 + bpn ,
n ¼ 1, 2, …,N
(4.4.19)
in which
8 9
< qn =
Un ¼ u_ n
: ;
un
2
31 2
3
0
0 0
m
c k
6 1
7 6 1
7
6
6 c1 0 1 7
c1 h 1 7
6
7
6
7
A¼6 2
7 6 2
7
4 1
5 4 1
5
c2 1 0
c2 1 0
2
2
31
2
m
c k
8 9
7 <1=
6 1
7
6
c
h
1
1
7
b¼6
7 :0;
6 2
5
4 1
0
c2 1 0
2
(4.4.20a)
(4.4.20b)
(4.4.20c)
The recurrence formula (4.4.19) can be employed to construct the solution
algorithm. However, the solution procedure can be further simplified. Thus,
applying Eq. (4.4.19) for n ¼ 1, 2, … we have
U1 ¼ AUo + bp1
U2 ¼ AU1 + bp2
¼ AðAUo + bp1 Þ + bp2
¼ A2 Uo + Abp1 + bp2
⋯ ¼ ⋯⋯⋯⋯⋯⋯⋯⋯⋯
(4.4.21)
Un ¼ An Uo + An1 p1 + An2 p2 + ⋯A0 pn b
Obviously, the last of Eq. (4.4.21) gives the solution vector Un at instant
tn ¼ nh using only the known vector U0 at t ¼ 0. The matrix A and the vector
b are computed only once.
Table 4.4.1 presents the algorithm for the numerical implementation of
AEM in pseudocode-type notation so that the reader can write a computer code
in the language of his/her preference.
Numerical integration of the equation of motion Chapter
4
173
Adhering to the steps of Table 4.4.1, a computer program called aem_lin.m
has been written in MATLAB for the numerical integration of the equation of
motion using the AEM. The program is available on this book’s companion
website. It computes the displacement u ðt Þ, the velocity u_ ðt Þ, the acceleration
u€ðt Þ, and the response ratio Rðt Þ ¼ u ðt Þ=ðpmax =k Þ and gives their graphical representation. Moreover, it computes the dynamic magnification factor
D ¼ max jRðt Þj and the time tmax it occurs. The user is responsible for providing the function of excitation force.
TABLE 4.4.1 The analog equation method (AEM).
A. Data
w (or k), m, x, u0 , u_ 0 , ttot , pðt Þ, ttot
B. Initial computations
Choose: h ð¼ Dt Þ and compute ntot ,
Compute: k ¼ mw2 , c ¼ 2mwx
q0 ¼ ðp0 cu_ 0 ku 0 Þ=m, c1 ¼ h 2 =2, c2 ¼ h
Formulate: U0 :¼ fq0 u_ 0 u0 gT
3
2
31 2
2
31
0
0 0
m
c k
m
c k
8 9
7
6 c1
7 6 1
6 c1
7 <1=
6 c1 0 1 7
6
6
h 1 7
h 1 7
6
7
6
7
6
7
Compute: A ¼ 6 2
7, b ¼ 6 2
7 6 2
7 :0;
5
4 c2
5 4
4 c2
5
0
1
1 0
1 0
c2 1 0
2
2
2
C. Compute solution
for n :¼ 1 to ntot
Un ¼ AUn1 + bpn
end
Example 4.4.1 Response of an SDOF system using the analog equation method
Determine the response the SDOF system in Example 4.2.1 using the AEM.
Solution
The solution is obtained using the program aem_lin.m with Dt ¼ 0:01.
Fig. E4.5 gives the graphical representation of the displacement together with
the error u ðt Þ uex ðt Þ. Moreover, Fig. E4.6 shows the response of the system
under the static load p ¼ t, if 0 t 1 and p ¼ 1, if 1 < t.
174 PART I Single-degree-of-freedom systems
0.2
2
u(t) computed
u(t) exact
x 10−4
u-uex
1.5
0.15
1
0.5
u(t)
0.1
0
0.05
–0.5
–1
0
–1.5
–0.05
0
2
4
6
8
10
–2
t
0
2
4
6
8
10
FIG. E4.5 Computed solution and error in Example 4.4.1.
× 10
6
-4
Displacement
10
Velocity
du(t)
u(t)
0
2
0
5
–5
0
10
-3
2
× 10
Acceleration
1.5
10
1
R(t)
0
5
t
Response ratio
t
ddu(t)
-4
5
4
0
× 10
–2
–4
0
0.5
5
10
0
0
t
5
10
t
FIG. E4.6 Response of the system in Example 4.4.1 under static load.
4.5 Stability of the numerical integration methods
4.5.1 Errors in the numerical integration
The numerical integration of the differential equations, hence of the equation of
motion, introduces errors, which influence not only the accuracy of the obtained
solution but also its capability to produce a reliable solution, that is, a solution
that within certain accuracy represents the actual solution. Factors that may
Numerical integration of the equation of motion Chapter
4
175
contribute to errors in the results obtained from well-defined loadings are of
three types [13]:
(i) Round-off errors: They result from calculations being done using numbers
expressed by too few digits.
(ii) Truncation errors: They are introduced in representing u ðt + Dt Þ or
u_ ðt + Dt Þ by a finite number of terms in the Taylor series expansion.
(iii) Instability errors: They are introduced by amplification of the errors from
one step to the subsequent one during the computations. The stability of
any method is improved by reducing the length of the time step.
Round-off errors are random in nature and therefore must be treated by statistical methods. In computer calculations, it is often possible to reduce the roundoff errors by using a higher precision in the computations. Truncation errors are
accumulated locally at each step. These errors provide a useful criterion for
measuring the accuracy of the various methods of numerical integration, provided that the employed numerical scheme is stable.
To study the stability of the numerical method, it is important to know the
influence of the error introduced at one step on the computations at the next step.
If the error tends to increase, then the solution soon becomes unbounded and
ceases to make sense. In this case, we say that the method is unstable. The study
of the stability of numerical integration methods of the differential equation of
motion and general of IVPs of differential equations is facilitated by the use of
difference equations, elements of which are discussed below.
4.5.2
Difference equations
If tn represents the time nh ðh ¼ Dt Þ, n ¼ 0, 1, 2… and un the displacement at
the same time, then we may write Eq. (4.2.6) as
un + 1 ¼ 2a1 un a2 un1 + gn
(4.5.1)
where
a1 ¼
2 h 2 w2
,
2ð1 + hwx Þ
a2 ¼
1 hwx
,
1 + hwx
gn ¼
pn h 2
m ð1 + hwx Þ
Eq. (4.5.1) is a recursive relation, which permits the computation of un + 1
from its previous two values. Applying this relation successively for
n ¼ 0, 1, … we have
9
For n ¼ 0 u1 ¼ 2a1 uo a2 u1 + g0
>
>
>
>
For n ¼ 1 u2 ¼ 2a1 u1 a2 u0 + g1
=
(4.5.2)
For n ¼ 2 u3 ¼ 2a1 u2 a2 u1 + g2
>
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯ >
>
>
;
For n ¼ n un + 1 ¼ 2a1 un a2 un1 + gn
176 PART I Single-degree-of-freedom systems
We observe that un + 1 can be computed if the two initial values u1 and uo
are known. Eq. (4.5.1) may also be written as
un + 1 2a1 un + a2 un1 ¼ gn
(4.5.3)
Eq. (4.5.3) is a difference equation of the second order [14] whose solution
can yield un + 1 using the recursive procedure (4.5.2). Because Eq. (4.5.3) is
linear with respect to un1 , un , and un + 1 , it is called a linear difference equation. If gn ¼ 0 the difference equation is called homogeneous while if gn 6¼ 0 it is
called nonhomogeneous. In general, an equation of the form
un + k + a1 un + k1 + a2 un + k2 + … + ak un ¼ gn
(4.5.4)
where a1 ,a2 , …,ak are constants is a linear difference equation of order k
with constant coefficients. The solution of the difference equation is a
sequence of uk values for which it is true. The order of the difference equation
is the difference between the largest and the smallest value of the argument k
appearing in it.
First, we examine the homogeneous linear difference equation ðgn ¼ 0Þ. The
solution of this equation is sought in the form un ¼ rn , which is inserted in
Eq. (4.5.4) to give
rn + k + a1 rn + k1 + a2 rn + k2 + … + ak rn ¼ 0
or dividing by rn we obtain the characteristic equation of the difference
equation (4.5.4)
pðrÞ ¼ rk + a1 rk1 + a2 rk2 + … + ak ¼ 0
(4.5.5)
which is polynomial of order k.
If we assume that all roots r1 ,r2 ,…,rk of pðrÞ are distinct, then
rn1 ,rn2 ,…, rnk are solutions of Eq. (4.5.4). Moreover, because of the linearity
of the equation
un ¼ c1 rn1 + c2 rn2 + … + ck rnk
(4.5.6)
is also a solution for all n for which the difference equation is defined with
c1 ,c2 ,…, ck being arbitrary constants. Eq. (4.5.6) is the general solution of
the difference equation (4.5.4). The arbitrary constants c1 ,c2 ,…, ck are determined from the k initial conditions.
If two roots of the polynomial (4.5.5) are complex conjugate, say r1 ¼
a + ib and r2 ¼ a ib, then we can write them in exponential form
r1 ¼ reiq ,
where
r¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 + b 2 ,
r2 ¼ reiq
(4.5.7)
q ¼ tan 1 ðb=aÞ
(4.5.8)
Numerical integration of the equation of motion Chapter
4
177
Using Euler’s identity (2.2.8), we obtain
rn1 ¼ r n einq ¼ r n ð cos nq + i sin nqÞ:
rn2 ¼ r n einq ¼ r n ð cos nq i sin nqÞ
thus we can write
c1 rn1 + c2 rn2 ¼ r n ½ðc1 + c2 Þcos nq + iðc1 c2 Þsin nq
¼ r n c10 cos nq + c20 sin nq
(4.5.9)
where c10 ¼ c1 + c2 and c20 ¼ iðc1 c2 Þ. We conclude from Eq. (4.5.9) that the
contribution of the complex roots to the solution is oscillatory.
Finally, we examine the case where the polynomial (4.5.5) has a double root,
say r1 . Then, the second solution will be nrn1 . This is verified by direct substitution of un ¼ nrn1 into Eq. (4.5.4). Thus we obtain
ðn + k Þrn1 + k + a1 ðn + k 1Þrn1 + k1 + … + ak1 ðn + 1Þrn1 + 1 + ak nrn1
¼ nrn1 rk1 + a1 rk1
+ … + ak1 r1 + ak
1
+ rn1 + 1 krk1
+ a1 ðk 1Þrk2 + … + ak1
1
which yields
nrn1 pðr1 Þ + rn1 + 1 p0 ðr1 Þ ¼ 0
The latter relation results because not only pðr1 Þ ¼ 0 but also p0 ðr1 Þ ¼ 0.
The solution of the nonhomogeneous equation, gn 6¼ 0, is obtained as the
sum of the homogeneous solution uno and a particular solution of the nonhomogeneous equation unp , namely
un ¼ uno + unp
(4.5.10)
In the special case where gn ¼ g ¼ constant, the particular solution is readily obtained. Indeed, introducing un ¼ G ¼ constant in Eq. (4.5.4) yields
ð1 + a1 + a2 + … + ak ÞG ¼ g
from which we obtain
G¼
g
1 + a1 + a2 + … + ak
(4.5.11)
provided that 1 + a1 + a2 + … + ak does not vanish.
Example 4.5.1 Solution of a second-order difference equation
Find the solution of the IVP
un + 2 2un + 1 + 2un ¼ 0
(1)
u0 ¼ 0, u1 ¼ 1
(2)
Solution
The characteristic equation of (1) is
r2 2r + 2 ¼ 0
(3)
178 PART I Single-degree-of-freedom systems
Its roots are r1 ¼ 1 + i, r2 ¼ 1 i and Eq. (4.5.8) give
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi
r ¼ 12 + 12 ¼ 2, q ¼ tan 1 ð1Þ ¼ p=4
Hence, Eq. (4.5.9) gives
np
np
+ c2 sin
un ¼ 2n=2 c1 cos
4
4
(4)
Applying the initial conditions gives c1 ¼ 0, c2 ¼ 1 and the solution becomes
np
(5)
un ¼ 2n=2 sin
4
Example 4.5.2 Solution of a third-order difference equation
Find the solution of the IVP
un + 3 2un + 2 un + 1 + 2un ¼ 0
(1)
u0 ¼ 0, u1 ¼ 1, u2 ¼ 1
(2)
Solution
The difference equation is of the third order. Its characteristic equation is
r3 2r2 r + 2 ¼ 0
(3)
Its roots are þ1, 1, 2. Hence the general solution of (1) is
un ¼ c1 ð1Þn + c2 ð1Þn + c3 ð2Þn
¼ c1 + ð1Þn c2 + 2n c3
(4)
Applying the initial conditions gives the following system for the arbitrary
constants c1 ,c2 , c3
c1 + c2 + c3 ¼ 0
c1 c2 + 2c3 ¼ 1
c1 + c2 + 4c3 ¼ 1
(5)
which is solved to give c1 ¼ 0, c2 ¼ 1=3, c3 ¼ 1=3. Therefore, we obtain the
closed form solution for the difference equation
1
2n
un ¼ ð1Þn +
3
3
(6)
4.5.3 Difference equations and stability of the numerical
integration methods
Along with the widespread use of computers for solving differential equations,
it was observed that some known integration schemes lead to errors in the
solution that are much greater than expected due to discretization. Moreover,
these errors for a certain value of the variable t increased, although the time step
Numerical integration of the equation of motion Chapter
4
179
Dt ¼ h was reduced. To understand this behavior, we examine the solution of
the differential equation
u_ ¼ 2u + 1, u ð0Þ ¼ 1
(4.5.12)
Using the central difference to approximate the derivative, we obtain the
difference equation
un + 1 + 4hu n un1 ¼ 2h
(4.5.13)
where h is the constant time step of the integration.
The solution of the previous difference equation is
un ¼ c1 rn1 + c2 rn2 +
1
2
(4.5.14)
where r1 and r2 are the roots of the characteristic equation
r2 + 4hr 1 ¼ 0
These roots are
r1, 2 ¼ 2h pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 + 4h 2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Expanding 1 + 4h 2 in the Taylor series and keeping only the linear terms,
the roots are expressed as
r1 ¼ 1 2h
(4.5.15a)
r2 ¼ ð1 + 2h Þ
(4.5.15b)
and Eq. (4.5.14) is written
un ¼ c1 ð1 2h Þn + c2 ð1Þn ð1 + 2h Þn +
1
2
(4.5.16)
It is known from differential calculus that
‘im ð1 + eÞ1=e ¼ e
e!0
Using this relation and taking into account that n ¼ tn =h, we obtain for a
given tn
‘im ð1 + 2h Þn ¼ ‘im ð1 + 2h Þð1=2hÞð2tn Þ ¼ e2tn
h!0
h!0
In the same way, we obtain
‘im ð1 2h Þn ¼ e2tn
h!0
Consequently, for h ! 0 the solution (4.5.16) becomes
1
2tn
+
+ c2 ð1Þn e2tn
u n ¼ c1 e
2
(4.5.17)
180 PART I Single-degree-of-freedom systems
On the other hand, the exact solution of Eq. (4.5.12) is
un ¼ c1 e2tn +
1
2
(4.5.18)
Evidently, the first term in Eq. (4.5.17) is the exact solution. The second
term is spurious (extraneous) and results from the fact that the first-order differential equation is substituted by a second-order difference equation. The
application of the initial conditions would give c2 ¼ 0 if the computations were
exact. In practice, however, errors are introduced, which are mainly due to the
rounding of numbers or the inaccuracy of the starting value. Therefore, the constant c2 is not exactly zero and consequently, a small error is introduced in each
integration step. This is magnified because it is multiplied by the factor
ð1Þn e2tn , which increases exponentially. Because the first term of the solution
(4.5.17) diminishes exponentially, the introduced error due to the spurious solution dominates the exact solution and leads to a totally wrong result. We
describe a method as unstable if the error increases exponentially with tn .
For the first order differential equations, the one-step integration methods do
not exhibit instability for small values of h. The multistep methods, however,
which lead to difference equations of order greater than one, introduce spurious
solutions and they may be unstable either for all values of h or for a certain
region of the values of h. In order to decide whether a multistep method is stable, we work as follows.
If the multistep method leads to a difference equation of order k, we find the
roots of the characteristic equation. If ri (i ¼ 1,2, …,kÞ are these roots, the general solution will be
un ¼ c1 rn1 + c2 rn2 + … + ck rnk
One of the solutions, say rn1 , will tend to the actual solution of the differential
equation. The remaining roots are spurious. We will say that a multistep method
is strongly stable if for h ! 0 the spurious roots satisfy the condition
jri j < 1, i ¼ 2, 3, …, k
(4.5.19)
Because we do not know which is the actual solution, the above condition
should apply to all roots ri . Apparently, this condition ensures that the error
diminishes as n increases. On the contrary, the error increases exponentially
if jri j > 1.
4.5.4 Stability of the central difference method
As was shown in Section 4.5.2, the CDM leads to the difference equation
(4.5.3), namely
un + 1 2a1 un + a2 un1 ¼ gn
(4.5.20)
Numerical integration of the equation of motion Chapter
4
181
where
a1 ¼
2 h 2 w2
1 hwx
pn h 2
, gn ¼
, a2 ¼
1 + hwx
2ð1 + hwx Þ
m ð1 + hwx Þ
(4.5.21)
The characteristic equation of Eq. (4.5.20) is
r2 2a1 r + a2 ¼ 0
whose roots are
r1, 2 ¼ a1 (4.5.22)
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a21 a2
The type of root depends on the sign of the discriminant D ¼ a21 a2 , which
may be written as
pffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffi
s2 s 2 1 x2 s + 2 1 x2
, s ¼ hw
D ðs Þ ¼
4ð1 + sxÞ2
Because
pffiffiffiffiffiffiffiffiffiffiffiffis > 0, the sign of the discriminant depends only on the factor
s 2 1 x2 . Hence, we distinguish the following two cases
pffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffi
2 1 x2 T
2
ðiÞ Dðs Þ > 0 s > 2 1 x
¼
1 x2
or h >
(4.5.23)
p
w
pffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffi
2 1 x2 T
2
¼
1 x 2 (4.5.24)
ðiiÞ Dðs Þ 0 s 2 1 x
or h p
w
In case (i), the characteristic equation has two real roots, r1 , r1 , The stability
of the solution requires that jr1 j < 1 and jr2 j < 1. But it can be shown that
jr2 j > 1, hence jr1 j > 1. Consequently, the solution is unstable in this case.
In case (ii), the characteristic equation has two complex conjugate roots
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r1, 2 ¼ a1 i a2 a21
(4.5.25)
This gives
pffiffiffiffiffi
j r1 j ¼ j r2 j ¼ a2 ¼
sffiffiffiffiffiffiffiffiffiffiffiffiffi
1 sx
1
1 + sx
(4.5.26)
Consequently, the solution is stable in case (ii).
The stability criterion (4.5.26) is also explained by writing the complex roots
in exponential form
r1 ¼ reiqt ,
where
pffiffiffiffiffi
r ¼ a2 ¼
r2 ¼ reiqt
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffiffiffi
1 hwx
a2
1 and q ¼ tan 1
1
1 + hwx
a21
(4.5.27)
(4.5.28)
182 PART I Single-degree-of-freedom systems
Then solution (4.5.20) is written as
pffiffiffiffiffi n
un ¼ ð a2 Þ ðc1 cos nq + c2 sin nqÞ + unp
(4.5.29)
pffiffiffiffiffi
which is bounded because a2 1.
The previous investigation leads to the conclusion that the CDM is stable if
the time step satisfies the condition.
qffiffiffiffiffiffiffiffiffiffiffiffi
T
T
(4.5.30)
Dt <
1 x2 p
p
This condition imposes a constraint on the size of the time step. In such a
case, we say that the numerical integration method is conditionally stable. It
should be noted that this condition is never a restriction for SDOF systems
because the accuracy of the results requires choosing a much smaller time step,
usually Dt ¼ 0:1T =p is adequate.
4.5.5 Stability of the average acceleration method
We consider the equation of motion (4.4.1) at time t ¼ nt
u€n + 2xwu_ n + w2 un ¼ pn ,
pn ¼ pn =m
(4.5.31)
and the approximate expressions (4.3.5) and (4.3.4) for the displacement and
velocity at the same instant, namely
un ¼ un1 + h u_ n +
h2
½u€n1 + u€n 4
h
u_ n ¼ u_ n1 + ½u€n1 + u€n 2
The previous three equations can be combined as
2
3
2
3
1 2xw w2 8 9
0 0 0 8
9 8 9
7< u€n1 = < 1 =
6 h2
7< u€n = 6 h 2
6
7
6
7
h
1
0
1 7 u_
6
7 u_ n1 + 0 pn
6
7:
6 4
7: n ; ¼ 6 4
4h
5 un1 ; : 0 ;
4 h
5 un
1 0
1
0
2
2
(4.5.32)
(4.5.33)
(4.5.34)
Solving for the vector Un ¼ f u€n u_ n u_ n gT , we obtain
Un ¼ AUn1 + b
pn ,
n ¼ 1, 2, …,N
(4.5.35)
3
31 2
1 2xw w2
0 0 0
7
6 h2
7 6 h2
7
6
7 6
h
1
0
1
6
7
6
7
A¼6 4
7
7 64
5
4 h
5 4h
1 0
1
0
2
2
(4.5.36a)
where
2
Numerical integration of the equation of motion Chapter
2
1
2xw w2
6
6 h2
6
b ¼ 6 4 0
6
4 h
1
2
31
7
7
1 7
7
7
5
0
8 9
1>
>
>
=
< >
0
>
>
>
;
: >
0
4
183
(4.5.36b)
Applying Eq. (4.5.35) for n ¼ 1, 2, … we have
U1 ¼ AUo + b
p1
U2 ¼ AU1 + b
p2
¼ AðAUo + b
p1 Þ + b
p2
(4.5.37)
¼ A2 Uo + Ab
p1 + b
p2
⋯ ¼ ⋯⋯⋯⋯⋯⋯⋯⋯⋯
Un ¼ An Uo + An1 p1 + An2 p2 + …A0 pn b
The matrix A is known as the amplification matrix. The stability of the
method requires that An is bounded. This is true if the spectral radius of A satisfies the condition [15]
rðAÞ ¼ max fjr1 j, jr2 j, jr3 jg 1
(4.5.38)
where ri (i ¼ 1, 2, 3) are the eigenvalues of the matrix A.
Using a symbolic language (here MATLAB) we find
pffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffi
4 s2 + 4s x2 1
4 s 2 4s x2 1
r1 ¼
, r2 ¼
, r3 ¼ 0
4 + 4sx + s 2
4 + 4sx + s 2
(4.5.39)
where s ¼ xw.
The type of the roots r1 , r2 depends on the sign of the discriminant
DðxÞ ¼ x2 1. Hence, we distinguish the following two cases
(i) If D ðx Þ > 0, both eigenvalues are real. It can be shown that jr1 j < 1, hence
jr2 j < 1. Therefore, the method is stable
(ii) If DðxÞ 0, the eigenvalues are complex conjugate
pffiffiffiffiffiffiffiffiffiffiffiffi
4 s 2 i4s 1 x2
r1,2 ¼
4 + 4sx + s 2
and
j r1 j ¼ j r2 j ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð4 s 2 Þ2 + 16s2 1 x 2
ð4 + 4sx + s 2 Þ
The equality is valid for x ¼ 0.
¼
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðs 2 + 4Þ2 16s 2 x2
½ðs 2 + 4Þ + 4sx 2
1 (4.5.40)
184 PART I Single-degree-of-freedom systems
The conclusion is that the AAM is stable without imposing any constraint on
the size of the time step. We say in this case that the integration method is
unconditionally stable.
Essentially, the procedure based on the condition (4.5.38) to prove the stability of the AAM is not different from that presented in Section 4.5.3, where the
stability results from the response of the difference equation. This is shown in
what it follows.
We write Eq. (4.5.35) for tn ,tn1 , tn2
Un AUn1 ¼ b
pn
Un1 AUn2 ¼ b
pn1
Un2 AUn3 ¼ b
pn2
(4.5.41)
or in matrix form
9
8
8
9
3> Un >
>
>
I A 0
0 <
=
< pn =
U
n1
4 0 I A 0 5
¼ b pn1
> Un2 >
:
;
>
0 0
I A >
pn2
;
:
Un3
2
(4.5.42)
Eq. (4.5.42), beside the displacements un ,un1 ,un2 , un3 , contains the
velocities u_ n , u_ n1 , u_ n2 , u_ n3 and the accelerations u€n , u€n1 , u€n2 , u€n3 . Reordering these equations and eliminating the velocities and accelerations yield the
equation
un 2a1 un1 + a2 un2 + a3 un3 ¼ c1 pn + c2 pn1 + c3 pn2
(4.5.43)
where
a1 ¼
4 w2 h 2
4 4sx + w2 h 2
, a2 ¼
,
4a
4a
1 + xwh
1 xwh
, c2 ¼
,
c1 ¼
a
a
a3 ¼ 0
(4.5.44)
c3 ¼ 0
(4.5.45)
where
a ¼ 1 + xwh +
w2 h 2
4
Eq. (4.5.43) is a difference equation whose characteristic equation is
r3 2a1 r2 + a2 r ¼ 0
The roots of Eq. (4.5.46) are
pffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffi
4 s 2 + 4s x 2 1
4 s 2 4s x2 1
r1 ¼
, r2 ¼
, r3 ¼ 0
4 + 4sx + s 2
4 + 4sx + s 2
that is, they are identical to the eigenvalues of the matrix A.
(4.5.46)
(4.5.47)
Numerical integration of the equation of motion Chapter
4.5.6
4
185
Stability of the analog equation method
The stability of the AEM is studied by considering the eigenvalues of the amplification matrix A given by Eq. (4.4.20b). Hence, the stability requires the validity of Eq. (4.5.38). Using a symbolic language (here MATLAB) we obtain the
three eigenvalues
pffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffi
4 s2 + 4s x2 1
4 s 2 4s x2 1
r1 ¼
, r2 ¼
, r3 ¼ 0
(4.5.48)
4 + 4sx + s 2
4 + 4sx + s 2
where s ¼ xw. Obviously, they are identical to those given by Eq. (4.5.39).
Consequently, we can state that the AEM is unconditionally stable.
4.6
Accuracy of the numerical integration
The methods for numerical integration of the equation of motion found in the
literature are numerous. They are either unconditionally stable or conditionally
stable. In the latter case, wDt must be less than a certain value, which is greater
than
one, for example, in the method of central differences this value is
pffiffiffiffiffiffiffiffiffiffiffiffi
2 1 x2 . However, the accuracy of the results requires values of wDt much
smaller than one. Therefore, the limitation of ensuring the stability of the solution is not decisive for the choice of the integration step, at least for the SDOF
systems. Hence, the choice of method should be based on the relative accuracy
of the numerical results. Actually, the accuracy of the numerical method
depends on the truncation error in the Taylor series representing the derivatives
(for example, CDM) or the approximation of the second derivative within the
time step (for example, AAM, AEM).
For free vibrations, as a measure of the relative accuracy, we define the
quantities
T T
T
(4.6.1)
rn rn + 1
rn
(4.6.2)
PE ¼
and
AD ¼
where T and T are the exact and the approximate periods, respectively, and
rn ,rn + 1 are the amplitudes at the consecutive times tn ,tn + 1 .
The first relation expresses the period elongation (PE) while the second the
amplitude decay (AD) over time.
The numerical solution obtained by the AAM or the AEM can be written in
terms of the eigenvalues
un ¼ c10 rn1 + c20 rn2
ð r 3 ¼ 0Þ
(4.6.3)
186 PART I Single-degree-of-freedom systems
or
un ¼ r n ðc1 sin nq + c2 cos nqÞ
n + c2 cos wt
nÞ
¼ r n ðc1 sin wt
(4.6.4)
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
where r ¼ a 2 + b2 , q ¼ tan 1 ðb=a Þ, a ¼ Reðr1 Þ, b ¼ Imðr1 Þ, w ¼ q=h,
tn ¼ nh.
The corresponding exact solution is
qffiffiffiffiffiffiffiffiffiffiffiffi
(4.6.5)
un ¼ exwtn ðc1 sin wD tn + c2 cos wD tn Þ, wD ¼ 1 x2 , tn ¼ nh
Comparison of Eqs. (4.6.4), (4.6.5) could show the accuracy of the numerical scheme. Thus, the period elongation is
pffiffiffiffiffiffiffiffiffiffiffiffi
T T h 1 x2
¼
PE ¼
1
(4.6.6)
T
q
For the amplitude decay, we can define an equivalent damping ratio x from
the relation
n
¼ exqn
r n ¼ exwt
(4.6.7)
x ¼ ln r=q
(4.6.8)
which gives
The difference Dx ¼ x x can be employed as a measure for the amplitude
decay. The dependence of the period elongation and amplitude decay on h=T is
shown in Figs. 4.6.1 and 4.6.2, respectively. Obviously, for small values of h=T
the scheme is accurate. Note that for x ¼ 0 it is jr2 j ¼ jr3 j ¼ r ¼ 1 and Eq. (4.6.8)
yields x ¼ 0. That is, there is no amplitude decay.
Period elongation %
0.25
x=0
x = 0.1
x = 0.2
0.2
0.15
0.1
0.05
0
0
0.05
0.1
0.15
0.2
h/T
FIG. 4.6.1 Period elongation versus h=T for different values x.
0.25
0.3
Numerical integration of the equation of motion Chapter
4
187
−3
2
x 10
x=0
x = 0.1
x = 0.2
0
Δx
−2
−4
−6
−8
0
0.05
0.1
0.15
h/T
0.2
0.25
0.3
FIG. 4.6.2 Amplitude decay Dx ¼ x x versus h=T for different values of x.
Example 4.6.1 Stability of Houbolt’s method
Houbolt’s method for the numerical integration of the equation of motion in the
absence of damping requires the solution of the difference equation [4]
2 + s 2 un + 1 5un + 4un1 un2 ¼ 0, s ¼ wh
(1)
Investigate the stability of the method.
Solution
The characteristic equation of Eq. (1) is
2 + s 2 r3 5r2 + 4r 1 ¼ 0
(2)
Its roots are
1 b
12a 25
2
+5
3a 2
b
pffiffiffi 3 b
1
b 12a 25
12a 25
+
+5 i
+2
r2,3 ¼
3a
4
b
b
6a 2
r1 ¼
(2)
(3)
where
a ¼ 2 + s2
(4)
pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=3
b ¼ 720a + 108a 2 + 1000 + 12 3a 104a + 100 + 27a 2
(5)
The stability condition requires that jr1 j < 1 and jr2 j ¼ jr3 j < 1. This
condition is satisfied, as is shown in Fig. E4.7.
188 PART I Single-degree-of-freedom systems
1
|r1|
0.9
|r2|
0.8
|r3|
0.7
0.6
0.5
|r2|=|r3|
0.4
0.3
0.2
0.1
0
| r1| (spurious)
1
2
3
4
5
6
7
8
9
10
h/t
FIG. E4.7 Houbolt’s method. Roots of the characteristic equation.
4.7 Problems
Problem P4.1 Find the solution of the IVPs for the difference equations
un + 1 3un ¼ 5, u0 ¼ 1
un + 2 4un + 1 + 3un ¼ 2n , u0 ¼ 0, u1 ¼ 1
Hint: The particular solution of (ii) will be sought in the form gn ¼ 2n c,
where c is a constant.
Problem P4.2 A SDOF system is subjected to pulse the load
t pðt Þ ¼ p0 1 1 + et=t1 , 0 t t1
t1
pðt Þ ¼ 0, t > t1
Study the response of the system using
(i) The central difference method.
(ii) The average acceleration method.
(iii) The analog equation method.
Compare the results with the exact solution.
Data: u ð0Þ ¼ u_ ð0Þ ¼ 0, m ¼ 50kNm1 s2 , x ¼ 0:05, w ¼ 5s1 , p0 ¼ 100kN,
t1 ¼ 0:5s and ttot ¼ 10s.
Problem P4.3 A SDOF system is subjected to the piecewise linear load pðt Þ
shown in Fig. P4.3. Use the three discussed numerical methods to establish
its response.
Numerical integration of the equation of motion Chapter
4
189
Compare the results with the exact solution. Data: u ð0Þ ¼ u_ ð0Þ ¼ 0,
m ¼ 50kN m1 s2 , x ¼ 0:05, w ¼ 5s1 , ti ¼ 0:01i, pi ¼ 20 1 + ð1Þi ði + 5Þ=
ði + 1Þ , i ¼ 0, 2, …,100.
FIG. P4.3 Piecewise linear load in Problem P4.3.
References and further reading
[1] M.A. Dokainish, K. Subbaraj, A survey of direct time-integration method in computational
structural dynamics. I. Explicit methods, Comput. Struct. 32 (1989) 1371–1386.
[2] K. Subbaraj, M.A. Dokainish, A survey of direct time-integration methods in computational
structural dynamics. II. Implicit methods, Comput. Struct. 32 (1989) 1387–1401.
[3] C.H. Norris, Structural Design for Dynamic Loads, McGraw-Hill, New York, 1959.
[4] K.J. Bathe, E.L. Wilson, Numerical Methods in Finite Elements, Prentice-Hall, Englewood
Cliffs, NJ, 1976.
[5] N.M. Newmark, A method of computation of structural dynamics, J. Eng. Mech. 85 (1959)
67–94.
[6] J.T. Katsikadelis, A new direct time integration method for the equations of motion in structural dynamics, ZAMM Z. Angew. Math. Mech. 94 (9) (2014) 757–774, https://doi.org/
10.1002/zamm.20120024.
[7] K.J. Bathe, Conserving energy and momentum in nonlinear dynamics: a simple implicit time
integration scheme, Comput. Struct. 85 (2007) 437–445.
[8] J.T. Katsikadelis, The analog equation method–a powerful BEM-based solution technique
for solving linear and nonlinear engineering problems, in: Transactions on Modelling and
Simulation, vol. 7, WIT Press, pp. 167–182, www.witpress.com, 1743-355X.
[9] J.T. Katsikadelis, The analog boundary integral equation method for nonlinear static and
dynamic problems in continuum mechanics, J. Theor. Appl. Mech. 40 (4) (2002).
[10] J.T. Katsikadelis, The Boundary Element Method for Plate Analysis, Academic Press,
Elsevier, Oxford, UK, 2014.
[11] J.T. Katsikadelis, The Boundary Element for Engineers and Scientists, Academic Press,
Elsevier, Oxford, UK, 2016.
[12] F.B. Hildebrand, Advanced Calculus for Applications, Prentice-Hall, Inc., Englewood Cliffs,
NJ, 1962.
[13] J.L. Humar, Dynamics of Structures, second ed., A.A. Balkema Publishers, Lisse, NL, 2002.
[14] F. Scheid, Numerical Analysis, second ed., Schaum’s Outline Series, McGraw-Hill, New
York, 1998.
[15] U.N. Faddeeva, Computational Methods in Linear Algebra, Dover Publications, New York,
1959.
Chapter 5
Nonlinear response:
Single-degree-of-freedom
systems
Chapter outline
5.1 Introduction
191
5.2 The central difference method 195
5.3 The average acceleration
method
197
5.1
5.4 The analog equation
method
5.5 Problems
References and further reading
203
211
215
Introduction
The equation of motion of a vibrating system expresses the equilibrium condition of all forces applied to the system, namely the external excitation force, the
inertial force, the damping force, and the elastic force. The equilibrium condition reads
fI ðt Þ + fD ðt Þ + fS ðt Þ ¼ pðt Þ
(5.1.1)
The forces fI , fD , and fS depend on the physical properties of the system.
In the systems we analyzed, the physical properties are not time-dependent
and the dependence of these forces on the cause that produces them is linear,
that is,
fI ¼ m u€
(5.1.2a)
fD ¼ cu_
(5.1.2b)
fS ¼ ku
(5.1.2c)
where m, c,k are constant quantities. Systems with such a physical response are
referred to as linear systems.
In general, however, Eq. (5.1.2a)-(5.1.2c) may be of the form
fI ¼ m ðt Þu€
(5.1.3a)
_ tÞ
fD ¼ fD ðu, u,
(5.1.3b)
_ tÞ
fS ¼ fS ðu, u,
(5.1.3c)
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00005-4
© 2020 Elsevier Inc. All rights reserved.
191
192 PART
I Single-degree-of-freedom systems
That is, the mass may vary with time and the damping and elastic forces may
_ t Þ, fS ðu, u,
_ t Þ. In this case,
_ and t, that is, fD ðu, u,
be nonlinear functions of u, u,
Εq. (5.1.1) takes the form
_ t Þ + fS ðu, u,
_ t Þ ¼ pðt Þ
m ðt Þu€ + fD ðu, u,
(5.1.4)
which is a nonlinear differential equation of the second order. A dynamic system, whose response is described by Eq. (5.1.4), is referred to as nonlinear.
Although systems with variable mass are not unusual [1], our discussion will
be limited to systems with constant mass. Besides, the forces fD and fS will
be considered of the form fD ðu_ Þ and fD ðu Þ. Thus, Eq. (5.1.4) becomes
m u€ + fD ðu_ Þ + fS ðu Þ ¼ pðt Þ
(5.1.5)
We distinguish two types of nonlinearity: the geometric nonlinearity, which
is due to large displacements implying large deformations of the structure, and
the material nonlinearity, which is due to nonlinear constitutive equations (e.g.,
hyperelastic or elastoplastic materials). Of course, both types of nonlinearity
can simultaneously characterize the response of a system.
The analytical solution of the nonlinear equations of motion is a difficult and
complicated mathematical problem. Exact solutions are available only for a few
cases and for differential equations of a specific form [2, 3]. The existing solutions aim rather at a qualitative study of the response of the system described by
a nonlinear equation than at offering a computational means for practical analyses. The knowledge of the nonlinear response of the single-degree-of-freedom
(SDOF) systems comes from approximate methods, and mainly from numerical
methods. Therefore, the recourse to numerical methods to solve the nonlinear
equations of motion is inevitable. The step-by-step methods play a dominant
role. The Runge-Kutta methods, usually employed for the solution of nonlinear
equations, belong to these methods [4].
The dynamic response of nonlinear systems can be studied effectively by
demanding the fulfillment of equation motion (5.1.5) at discrete time instants
Dt apart by the use of the step-by-step integration methods we discussed in
Chapter 4. These methods as developed for nonlinear equations of motion
are presented directly below while for the analytical methods, the reader is
advised to look in the vast related literature [2, 3, 5].
Example 5.1.1 Systems with a geometrical nonlinearity
Derive the equation of motion of the system shown in Fig. E5.1. The supports at
A,B and the interconnection at C are hinges. The mass m at C is concentrated.
The system is set to motion by the initial conditions u0 , u_ 0 and/or the vertical
external force pðt Þ. The bars are assumed massless.
Nonlinear response: Single-degree-of-freedom systems Chapter
l
A
5
193
l
C
φ
u(t )
E, A
B
E, A
m
p(t )
fI
fS
S
S
p(t )
FIG. E5.1 SDOF system with nonlinear response.
Solution
During the motion, the force pðt Þ, the inertial force fI , and the elastic forces of
the bars are in equilibrium
The inertial force is given by the relation
fI ¼ m u€
(1)
The total elastic force is caused by the elongation of the bars and is given by
fS ¼ 2S sin f
(2)
where
S¼
EA
d
l
(3)
E is the modulus of elasticity of the material of the bar, A its cross-sectional area,
and d the elongation of the bars.
The elongation of the bars at time t is
d¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
l 2 + u2 l
(4)
and
u
sin f ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
l 2 + u2
(5)
194 PART
I Single-degree-of-freedom systems
Hence, the elastic force in Eq. (2) by virtue of Eqs. (3)–(5) is expressed as
2
3
u6
1
7
fS ðu Þ ¼ 2EA 41 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5
l
2
1 + ðu=l Þ
(6)
and the equation of motion (5.1.1) in the absence of damping becomes
2
3
u6
1
7
m u€ + 2EA 41 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 ¼ pðt Þ
l
1 + ðu=l Þ2
(7)
Obviously, the equation of motion is nonlinear. Fig. E5.2 shows the graph
of fS ðu Þ. Because d 2 fS =du 2 > 0 the curve is concave upward. In this system,
the slope kT increases continually, which implies that the elastic force
increases with increasing u. In this case, we say that the system exhibits hardening, in contrast to other systems that exhibit softening. In the latter systems,
the curve fS ðu Þ is concave downward (d 2 fS =du 2 < 0), that is, the slope
kT decreases continually, which implies that elastic force decreases with
increasing u, for example, a system with stiffness fS ¼ 40u u 3 exhibits softening; see Fig. E5.3.
FIG. E5.2 System with hardening.
Nonlinear response: Single-degree-of-freedom systems Chapter
5
195
FIG. E5.3 System with softening.
5.2
The central difference method
We consider the case where the damping force depends linearly on the velocity
(viscous damping) and the elastic force depends nonlinearly on the displacement, then Eq. (5.1.5) becomes
m u€ + cu_ + fS ðu Þ ¼ pðt Þ
(5.2.1)
The central difference method (CDM) developed for linear equations can be
readily adjusted to solve this equation explicitly. Thus, using Eqs. (4.2.4),
(4.2.5) to replace the derivatives in Eq. (5.2.1), we obtain
m
un + 1 2un + un1
un + 1 un1
+ fS ðun Þ ¼ pn
+c
2
h
2h
or solving for un + 1 gives
K^ un + 1 ¼ P^
(5.2.2)
(5.2.3)
where
m
c
K^ ¼ 2 +
h
2h
m c 2m
P^ ¼ pn fS ðun Þ + 2 un 2 un1
h
h 2h
(5.2.4)
(5.2.5)
Eq. (5.2.3) is used to compute un + 1 from the displacements at the two preceding
instants. The value of u1 is computed from Eq. (4.2.10), namely
1
u ðDt Þ u ð0Þ Dt u_ ð0Þ + Dt 2 u€ð0Þ
2
(5.2.6)
196 PART
I Single-degree-of-freedom systems
in which u€0 results now from Eq. (5.2.1) for t ¼ 0
u€0 ¼ ½p0 cu_ 0 fS ðu0 Þ=m
(5.2.7)
Obviously, the numerical algorithm presented in Table 4.2.1 can be employed
to write a computer code for the nonlinear response of the SDOF systems after
modifying appropriately to include the nonlinear restoring force fS ðu Þ. On the
basis of this algorithm, a computer program called centr_diff_nlin.m has been
written in MATLAB for the numerical integration of the nonlinear equation of
motion using the CDM. The program is available on this book’s companion
website.
Example 5.2.1 Simple pendulum. CDM solution
Solve the initial value problem (IVP) describing the motion of the simple pendulum (Fig. E5.4) under its own weight, namely
g
(1)
q€ + sin q ¼ 0
l
qð0Þ ¼ q0
(2a)
q_ ð0Þ ¼ q_ 0
(2b)
where qðt Þ represents the angle of the pendulum from the vertical position, l is
its length, and g the acceleration of gravity. The equation of motion (1) can be
readily derived using the method of Lagrange’s equations (see Section 1.8).
Eq. (1) admits an exact solution [6]
npffiffiffi hpffiffiffiffiffiffiffi
io
k sn
g=l ðt + T0 Þ; k
(3)
qðt Þ ¼ 2 sin1
where k ¼ sin 2 ðq0 =2Þ and T0 is the quarter of the period; sn represents the
sn-Jacobean elliptic function [7].
Solution
The response of the pendulum for l ¼ g, q0 ¼ 0:40p, and q_ 0 ¼ 0 is obtained using
the program centr_diff_nlin.m with Dt ¼ 0:01. It is shown in Fig. E5.5 as compared with the exact one.
FIG. E5.4 Simple pendulum.
Nonlinear response: Single-degree-of-freedom systems Chapter
t
5
197
t
FIG. E5.5 Response of the simple pendulum in Example 5.2.1.
5.3
The average acceleration method
The average acceleration method (AAM) presented in Section 4.3 can be modified to solve the nonlinear equation of motion, Eq. (5.2.1). Although this
method becomes unstable for long-duration motions [9], it is presented here
because it is still widely used for the solution of the nonlinear equation of
motion.
At time t + Dt this equation is written
m u€ðt + Dt Þ + fD ðt + Dt Þ + fS ðt + Dt Þ ¼ pðt + Dt Þ
(5.3.1)
Subtracting Eq. (5.2.1) from the foregoing equation yields
mDu€ + DfD + DfS ¼ Dp
(a)
(5.3.2)
(b)
FIG. 5.3.1 Tangent and secant of (a) damping and (b) stiffness curve.
Expanding fD ðt + Dt Þ and fS ðt + Dt Þ in the Taylor series gives
DfD ¼ fD ðt + Dt Þ fD ðt Þ
¼ fD ðu_ + Du_ Þ fD ðu_ Þ
df
1 d 2 fD
ðDu_ Þ2 + ⋯
¼ D Du_ +
d u_
2 d u_ 2
(5.3.3)
198 PART
I Single-degree-of-freedom systems
DfS ¼ fS ðt + Dt Þ fS ðt Þ
¼ fS ðu + Du Þ fS ðu Þ
df
1 d 2 fS
¼ S Du +
ðDu Þ2 + ⋯
du
2 du 2
(5.3.4)
For small values of Dt, the quantities Du, Du_ are also small. Thus, neglecting the nonlinear terms in Eqs. (5.3.3), (5.3.4), we obtain
DfD cT Du_
(5.3.5)
DfS kT Du
(5.3.6)
Obviously, cT and kT express the slope of the tangent to the curves fD ðu_ Þ
and fS ðu Þ, respectively, at time t.
Referring to Fig. 5.3.1a, we have
fD ðt + Dt Þ ¼ fD ðt Þ + DfD
However, the exact value of DfD is
DfD ¼ Du_ tan fc
where fc is the angle of the secant. Hence, approximating DfD by Eq. (5.3.5)
introduces the error (see Fig. 5.3.1a)
eD ¼ ðcT tan fc ÞDu_
(5.3.7)
because cT is the slope of the tangent.
Similarly, the use of kT to approximate DfS introduces the error (see
Fig. 5.3.1b)
eS ¼ ðkT tanfk ÞDu
(5.3.8)
The errors eD and eS cannot be avoided because u_ ðt + Dt Þ and u ðt + Dt Þ
are not known at instant t + Dt. However, as we will show, they can be kept
under a given bound, which specifies the accuracy of the solution procedure.
By virtue of Eqs. (5.3.5), (5.3.6), Eq. (5.3.2) is written in incremental form
mDu€ + cT Du_ + kT Du ¼ Dp
(5.3.9)
The previous equation is of the form (4.3.11). Hence, the AAM is suitable to
_ we obtain
solve it. Thus, using Eqs. (4.3.9), (4.3.10) to express Du€ and Du,
2
Du 2u_ ðt Þ
Dt
(5.3.10)
4
½Du Dt u_ ðt Þ 2u€ðt Þ
Dt 2
(5.3.11)
Du_ ¼
Du€ ¼
Then Eq. (5.3.9) becomes
2cT
4m
4m
+ 2 Du ¼ Dp +
+ 2cT u_ ðt Þ + 2m u€ðt Þ
kT +
Dt
Dt
Dt
(5.3.12)
Nonlinear response: Single-degree-of-freedom systems Chapter
5
199
or
k ∗ Du ¼ Dp∗
(5.3.13)
where
k ∗ ¼ kT +
and
Dp∗ ¼ Dp +
2cT
4m
+ 2
Dt
Dt
4m
+ 2cT u_ n + 2m u€n
Dt
(5.3.14)
(5.3.15)
The value of Du obtained from Eq. (5.3.13) is used in Eq. (5.3.10) to eval_ Then we obtain
uate Du.
un + 1 ¼ un + Du
(5.3.16a)
u_ n + 1 ¼ u_ n + Du_
(5.3.16b)
The acceleration u€n + 1 is evaluated directly from Eq. (5.3.1)
u€n + 1 ¼
1
½pn + 1 fD ðu_ n + 1 Þ fS ðun + 1 Þ
m
(5.3.16c)
Apparently, the value u€n + 1 obtained from the above equation reduces the
error introduced by cT and kT . The study of the stability of the time step integration schemes for nonlinear differential equation of motion is an issue beyond
the scope of this book.
The described method gives good results if the slope of the tangent is close
to that of the secant. For this reason, it is recommended to check the following
errors at the end of each step
kT ksec cT csec a
a, ec ¼ (5.3.17)
ek ¼ kT cT where
ksec ¼
DfS
DfD
, csec ¼
Du
Du_
(5.3.18)
and a is a small specified number defining the upper bound of the error, for
example, a ¼ 0:01. If it is
min fek , ec g > a
(5.3.19)
then we must reduce the time step Dt and repeat the computations. Table 5.3.1
presents the algorithm for the numerical solution of the nonlinear equation of
motion (5.1.5) using the AAM.
200 PART
I Single-degree-of-freedom systems
TABLE 5.3.1 Average acceleration method. Nonlinear equation of motion.
A. Data
m, fD ðu_ Þ, fS ðu Þ, pðt Þ, u ð0Þ, u_ ð0Þ, ttot , a
B. Initial computations
u€ð0Þ ¼ fpð0Þ fD ½u_ ð0Þ fS ½u ð0Þg=m
Select: Dt and set t ¼ 0 and
C. In each step compute
1. kT ¼ ðdf S =du Þt cT ¼ ðdf D =d u_ Þt
2. k ∗ ¼ kT + 2cT =Dt + 4m=Dt 2 , Dp ¼ pðt + Dt Þ pðt Þ
3. p∗ ¼ Dp + ð4m=Dt + 2cT Þu_ ðt Þ + 2m u€ðt Þ
4. Du ¼ p∗ =k ∗ , Du_ ¼ 2Du=Dt 2u_ ðt Þ
5. u_ ðt + Dt Þ ¼ u_ ðt Þ + Du_
u€ðt + Dt Þ ¼ fpðt + Dt Þ fD ½u_ ðt + Dt Þ fS ½u ðt + Dt Þg=m
c Df =Du_ S =Du ek ¼ kT Df
, ec ¼ T cTD kT
6. Check: if max fec , ek g < a go to the next time step. Else set t ¼ t + Dt and check
if t > ttot end. Else set u ðt Þ ¼ u ðt + Dt Þ, u_ ðt Þ ¼ u_ ðt + Dt Þ, u€ðt Þ ¼ u€ðt + Dt Þ, and
go to C.1
When the damping force depends linearly on the velocity, then the response
is governed by Eq. (5.2.1). In this case, the previously presented incremental
method can be improved by employing an iterative procedure within each step,
which minimizes the error introduced by the tangent stiffness kT .
The starting point is Eq. (5.3.13), which we write as
k ∗ du ð1Þ ¼ Dp∗
(5.3.20)
The quantity du ð1Þ is the first approximation to Dun within the time step
from tn to tn + 1 . That is
Dunð1Þ ¼ du ð1Þ
(5.3.21)
FIG. 5.3.2 Graph of the function p*(u).
The index n denotes the number of the step that brings us from the displacement un to un + 1 . Fig. 5.3.2 presents the graph of the function p∗ ðu Þ. Apparently,
Nonlinear response: Single-degree-of-freedom systems Chapter
5
201
the displacement du ð1Þ resulting from Eq. (5.3.20) decreases Dp∗ by dpð1Þ .
Hence, there is a remaining force
DF ð2Þ ¼ Dp∗ dpð1Þ
(5.3.22)
which must be equilibrated. The change dpð1Þ is computed using Eq. (5.3.20), if
the tangential slope kT is replaced with the slope of the secant
" #
fS un + du ð1Þ fS ðun Þ 2c
4m
ð1Þ
+
dp ¼
+
du ð1Þ
Dt Dt 2
du ð1Þ
2c
4m
¼ fS un + du ð1Þ fS ðun Þ +
+ 2 du ð1Þ
Dt Dt
which by virtue of Eq. (5.3.14) becomes
dpð1Þ ¼ fS un + du ð1Þ fS ðun Þ + ðk ∗ kT Þdu ð1Þ
(5.3.23)
The remaining force DF ð2Þ produces an additional displacement du ð2Þ ,
which is computed from the relation
k ∗ du ð2Þ ¼ DF ð2Þ
(5.3.24)
Hence, the new approximation to Dun is
Dunð2Þ ¼ du ð1Þ + du ð2Þ
¼ Dunð1Þ + du ð2Þ
(5.3.25)
Then, it is used to compute the new remaining force
DF ð3Þ ¼ DF ð2Þ dpð2Þ
(5.3.26)
where
dpð2Þ ¼ fS un + Dunð2Þ fS un + Dunð1Þ + ðk ∗ kT Þdu ð2Þ
(5.3.27)
The force DF ð3Þ produces the additional displacement du ð3Þ , which is computed from the relation
k ∗ du ð3Þ ¼ DF ð3Þ
(5.3.28)
Thus, the new approximation of Dun is
Dunð3Þ ¼ Dunð2Þ + du ð3Þ
(5.3.29)
Consequently, for the i + 1 approximation it is
DF ði + 1Þ ¼ DF ðiÞ dpðiÞ
dpðiÞ ¼ fS un + DunðiÞ fS un + Dunði1Þ + ðk ∗ kT Þdu ðiÞ
(5.3.30)
(5.3.31)
202 PART
I Single-degree-of-freedom systems
k ∗ du ði + 1Þ ¼ DF ði + 1Þ
(5.3.32)
Dunði + 1Þ ¼ DunðiÞ + du ði + 1Þ
(5.3.33)
Note that for i ¼ 1 it must be set Dunð0Þ ¼ 0 and DF ð1Þ ¼ Dp∗ .
The iteration procedure is terminated after I iterations, if
du ði + 1Þ e
DunðI Þ (5.3.34)
where e is a specified small number. Then we assume that the convergence has
been achieved. The value Dun ¼ DunðI Þ is considered exact and it is used to
compute un + 1 and u_ n + 1 , u€n + 1 . Subsequently, the procedure continues to
the next step. The iterative procedure within the time step from tn to tn + 1 is
summarized in Table 5.3.2. This procedure is known as the modified
Newton-Raphson method.
TABLE 5.3.2 Modified Newton-Raphson method for the minimization
of the error in the average acceleration method.
A. Initial values
ð0Þ
ð0Þ
un + 1 ¼ un , fS ¼ fS ðun Þ, DRð1Þ ¼ Dp∗n
B. In each iteration i ¼ 1,2,3,… compute:
1. du ðiÞ ¼ DRðiÞ =k ∗
ði Þ
ði1Þ
2. un + 1 ¼ un + 1 + du ðiÞ
ði Þ
ði1Þ
DpðiÞ ¼ fS fS
ði + 1Þ
ði Þ
3.
+ ðk ∗ kT Þdu ðiÞ
4. DR
¼ DR DpðiÞ
5. du ði + 1Þ ¼ DRði + 1Þ =k ∗
6. If du ði + 1Þ =DunðI Þ > e set i ¼ i + 1 and go to B.2
Adhering to the steps in Table 5.3.1, a computer program called av_acc_
nlin.m has been written in MATLAB for the solution of the nonlinear equation
of motion. The electronic version of the program is available on this book’s
companion website.
Example 5.3.1 Simple pendulum. AAM solution
Solve the IVP describing the motion of the simple pendulum (Fig. E5.4) under
its own weight. The motion is described by Eqs. (1), (2a), (2b) of Example 5.2.1.
Solution
The solution is obtained using the program av_acc_nlin.m with Dt ¼ 0:01. The
response is shown in Fig. E5.5 as compared with the exact one.
Nonlinear response: Single-degree-of-freedom systems Chapter
5.4
5
203
The analog equation method
The solution procedure developed in Section 4.4 for the linear equation of
motion can be straightforwardly extended to the nonlinear equation. The nonlinear IVP treated here is described by the equations
m u€ + fD ðu_ Þ + fS ðu Þ ¼ pðt Þ
u ð0Þ ¼ u0 ,
u_ ð0Þ ¼ u_ 0
(5.4.1)
(5.4.2)
where fD ðu_ Þ and fS ðu Þ are nonlinear functions of their arguments.
Applying Eq. (5.4.1) for t ¼ nDt, we obtain
mq n + fD ðu_ n Þ + fS ðun Þ ¼ pn ,
qn ¼ u€n
Moreover, Eqs. (4.4.14), (4.4.15) hold also, that is,
2 c1 3
2 c1 3
"
#( ) "
#(
)
+
h 1
u_ n
0 1
u_ n1
6 27
6 2 7
qn + 4
q
¼
+4
5
c2
c2 5 n1
1 0
un
un1
1 0
+
2
2
(5.4.3)
(5.4.4)
TABLE 5.4.1 The analog equation method. Nonlinear equation of motion.
A. Data
Read: m, fD ðu_ Þ, fS ðu Þ, u0 , u_ 0 , pðt Þ, ttot
B. Initial computations
1. Select: h ¼ Dt and compute ntot
2. Compute: c1 ¼ h 2 =2, c2 ¼ h, q0 ¼ ½pðt Þ fD ðu_ 0 Þ + fS ðu0 Þ=m
C. Compute solution
for n ¼ 1 to ntot solve for fqn u_ n un gT the system of the nonlinear algebraic equations:
mq n + fD ðu_ n Þ + fS ðun Þ ¼ pn
2 c 3
2 c 3
1
1
+
h 1 u_ n
0 1 u_ n1
6 27
6 2 7
¼
+ 4 c 5qn + 4 c 5qn1
2
2
1 0 un
1 0 un1
+
2
2
Eqs. (5.4.3), (5.4.4) constitute a system of three algebraic equations, one
nonlinear and the other two linear. They can be solved successively for
n ¼ 1, 2, … to yield the solution un and the derivatives u_ n , u€n ¼ qn at instant
t ¼ nh T . For n ¼ 1, the value q0 appears in the right side of Eq. (5.4.4). This
quantity is readily obtained from Eq. (5.4.1) for t ¼ 0. This yields
q0 ¼ ½pð0Þ fD ðu_ 0 Þ + fS ðu0 Þ=m
(5.4.5)
Several procedures can be applied to solve this system of algebraic equations. For example, a simple procedure is to substitute un and u_ n from
Eq. (5.4.4) into Eq. (5.4.3) and solve the resulting nonlinear equation for qn .
This can be achieved by employing any ready-to-use subroutine for nonlinear
algebraic equations, for example, MATLAB function fsolve. The steps of the
204 PART
I Single-degree-of-freedom systems
solution procedure are presented in Table 5.4.1. Adhering to the steps of this
table, a computer program called aem_nlin.m has been written in MATLAB
for the numerical integration of the nonlinear equation of motion using the
AEM. The program is available on this book’s companion website. It computes
the displacement u ðt Þ, the velocity u_ ðt Þ, and the acceleration u€ðt Þ.
Example 5.4.1 The Duffing equation. AEM solution
Use the AEM to solve the IVP for the Duffing equation
u€ + 0:2u_ + u + u 3 ¼ pðt Þ
(1)
u ð 0Þ ¼ 0
(2a)
u_ ð0Þ ¼ 1
(2b)
For
pðt Þ ¼ e0:1t ½ð0:01sin t 0:2cost sin t Þ 0:2ð0:1 sin t cos t Þ
+ sin t + e0:2t ð sin t Þ3 Eq. (1) admits an exact solution uexact ðt Þ ¼ e0:1t sin t.
Solution
The solution is obtained using the program aem_nlin.m with Dt ¼ 0:01. The
graph of the solution is shown in Fig. E5.6 as compared with the exact one.
1
Computed
Exact
0.8
error X103
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
0
5
10
15
t
FIG. E5.6 Solution u and error u uexact in Example 5.4.1.
20
25
Nonlinear response: Single-degree-of-freedom systems Chapter
5
205
Example 5.4.2 Response of a nonlinear system with hardening
Use the AEM to study the dynamic response of the SDOF system shown in
Fig. E5.1, when it is subjected to:
(i) An initial displacement u0 ¼ 0:05m from the position of static equilibrium.
(ii) External load p0 ¼ mg suddenly applied at time t ¼ 0.
(iii) External load pðt Þ ¼ 10 sin Wt.
Data: m ¼ 3:8722kN m1 s2 , E ¼ 2:1 108 kN=m2 , A ¼ 3:142 104 m2 ,
g ¼ 9:81m=s2 , l ¼ 3:00m.
Solution
(i) The static displacement ust produced by the weight of the body is calculated
from Eq. (7) of Example 5.1.1 for u€ ¼ 0 and pðt Þ ¼ mg, namely
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(1)
2EAðust =l Þ 1 1= 1 + ðust =l Þ2 ¼ mg
which is solved for the data of the problem to yield ust ¼ 0:25m.
The external force is the constant weight of the body, that is, pðt Þ ¼ mg,
while the elastic force
(2)
fS ðust + u Þ
where u ¼ u ðt Þ denotes the additional displacement due to the dynamic
response. Thus, the equation of motion becomes
m u€ + fS ðust + u Þ ¼ mg
(3)
u_ 0 ¼ 0
(4)
with initial conditions
u0 ¼ 0:05m,
It should be noted that due to the nonlinearity of the elastic force, the superposition of the displacements does not apply. The computed response of the system with Dt ¼ 0:01 is shown in Fig. E5.7.
(ii) In this case the equation of motion reads
m u€ + fS ðu Þ ¼ mg
(5)
u0 ¼ 0, u_ 0 ¼ 0
(6)
with initial conditions
where u ¼ u ðt Þ denotes the total dynamic displacement from the undeformed
position. The computed response of the system with Dt ¼ 0:01 is shown in
Fig. E5.8.
(iii) In this case, the IVP becomes
m u€ + fS ðu Þ ¼ mg + 10sin Wt
(7)
u0 ¼ 0, u_ 0 ¼ 0
(8)
The computed response
with Dt ¼ 0:01 is shown in Fig. E5.9 for the ratio
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
W=w ¼ 0:1, where w ¼ EA=lm . In all cases, the results are compared with
those obtained by the CDM
206 PART
u(t)
I Single-degree-of-freedom systems
t
u(t)
FIG. E5.7 Response in Example 5.4.2 (i).
t
FIG. E5.8 Response in Example 5.4.2 (ii).
FIG. E5.9 Response in Example 5.4.2 (iii).
Nonlinear response: Single-degree-of-freedom systems Chapter
5
207
Example 5.4.3 Response of a nonlinear system with softening
Study the dynamic response of a SDOF system with the following data:
m ¼ 1kN m1 s2 , u ð0Þ ¼ 1m, u_ ð0Þ ¼ 1m=s, fS ¼ 2u 3 , and ttot ¼ 1s. Because
d 2 fS=du 2 ¼ 12u < 0 the system exhibits softening. The integration of the
equation of motion will be performed using all three methods, that is, CDM,
AAM, and AEM. The problem admits an exact solution u ðt Þ ¼ 1=ð1 + t Þ.
Solution
The IVP is
u€ 2u 3 ¼ 0
(1)
u0 ¼ 1, u_ 0 ¼ 1
(2)
The solution in the interval 0 t 1 for Dt ¼ 0:001 is shown in Fig. E5.10.
FIG. E5.10 Response of the SDOF system in Example 5.4.2.
208 PART
I Single-degree-of-freedom systems
Example 5.4.4 Response of an elastoplastic system
In this example, the dynamic response of a system with material nonlinearity is
studied. The one-story building (Fig. E5.11a) is modeled as a SDOF system
with mass m. The structure is subjected to the horizontal rectangular pulse load
pðt Þ of Fig. E5.11b. The material is elastoplastic in nature. This yields the forcedisplacement relation shown in Fig. E5.12 due to the elastoplastic hinges produced by the horizontal load. Neglecting damping, study the response of the
system using both the AAM and AEM. Data: t1 ¼ 0:25s, fSy ¼ 69:68kN,
k ¼ 5360kN=m, m ¼ 36kN m1 s2 , and p0 ¼ 45kN.
(b)
(a)
FIG. E5.11 System with a material nonlinearity in Example 5.4.4.
FIG. E5.12 Restoring force in Example 5.4.4.
Solution
It is
fSy
69:68
¼ 0:013m ¼ yield displacement
uy ¼ ¼
k 5360:00
um ¼ maximum displacement where the velocity changes sign
uR ¼ um uy ¼ remaining plastic deformation
The equation of motion is
m u€ + fS ðu Þ ¼ pðt Þ
Nonlinear response: Single-degree-of-freedom systems Chapter
5
209
where fS is the stiffness of the structure. It is
fS ¼ ku
if u < uy ¼ 0:013m
if u
fS ¼ fSy
y u um
fS ¼ fSy k ðum u Þ if um 2uy < u < um , u_ < 0
Moreover, we have
rffiffiffiffiffi
k
w¼
¼ 12:20s1 ,
m
T¼
2p
¼ 0:515s
w
(i) Solution by the AAM.
We employ the AAM following the displacement step by step.
At instant t we have
pðt Þ fS
¼ 0:0278½pðt Þ fS m
4m
4 36
k ∗ ¼ kT + 2 ¼ k +
¼ kT + 1, 440, 000, kT ¼ 0, u > uy
Dt
0:012
The numerical results obtained by the AAM with Dt ¼ 0:01 < T =10 are
shown in Table E5.1 for two different time steps.
u€ ¼
(ii) Solution by AEM.
A computer program has been written in MATLAB for the evaluation of the
elastoplastic response of the SDOF system. The obtained numerical results
are shown in Table E5.1 for two different time steps.
Finally, the time history of response of the system as well as of the restoring
force are shown in Fig. E5.13. and Fig. E5.14.
TABLE E5.1 Numerical solution of elastoplastic system in Example 5.4.4.
u ðt Þ (cm)
Dt ¼ 0:01
t (s)
AEM
AAM
Dt ¼ 0:001
AEM
AAM
0.00
0.000
0.000
0.000
0.000
0.05
0.124
0.151
0.149
0.151
0.10
0.503
0.550
0.546
0.551
0.15
1.003
1.053
1.050
1.055
0.20
1.446
1.483
1.480
1.484
0.25
1.721
1.745
1.743
1.745
0.30
1.643
1.709
1.677
1.677
Continued
TABLE E5.1 Numerical solution of elastoplastic system in
Example 5.4.4.—cont’d
u ðt Þ (cm)
Dt ¼ 0:01
t (s)
AEM
AAM
Dt ¼ 0:001
AEM
AAM
0.35
1.134
1.234
1.177
1.175
0.40
0.378
0.487
0.423
0.421
0.45
0.353
0.262
0.312
0.313
0.50
0.796
0.744
0.763
0.763
0.55
0.791
0.786
0.768
0.766
0.60
0.341
0.374
0.324
0.321
0.65
0.393
0.343
0.408
0.412
0.70
1.147
1.107
1.164
1.167
0.75
1.649
1.642
1.671
1.673
0.80
1.718
1.757
1.747
1.747
0.85
1.330
1.410
1.363
1.362
0.90
0.624
0.726
0.659
0.658
0.95
0.145
0.049
0.111
0.113
1.00
0.701
0.637
0.663
0.663
FIG. E5.13 Displacement in Example 5.4.4.
Nonlinear response: Single-degree-of-freedom systems Chapter
5
211
FIG. E5.14 Restoring force in Example 5.4.4.
5.5
Problems
Problem P5.1 Show that the motion of the simple pendulum is governed by
the IVP
g
(1)
q€ + sin q ¼ 0, qð0Þ ¼ q0 , q_ ð0Þ ¼ q_ 0
l
Solve the equation of motion numerically when g=l ¼ 1, q0 ¼ 0:1p, q_ 0 ¼ 0
and compute the period T of the pendulum. Give the graphical representation
of the function T ¼ T ðq0 Þ, 0:1 q0 1 for the two time steps Dt ¼ 0:1 and
Dt ¼ 0:0001. Compare with the exact expression
sffiffiffi Z
l p=2
df
sin ðq=2Þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , k ¼ sin ðq0 =2Þ, sin f ¼
(2)
T ðq 0 Þ ¼ 4
g 0
sin ðq0 =2Þ
1 k 2 sin 2 f
Use numerical integration to evaluate the Legendre elliptic integral of first
kind in Eq. (2).
Problem P5.2 The support O of the simple pendulum in Fig. P5.2 is subjected
The rotation about the
to the horizontal harmonic motion u ðt Þ ¼ u0 sin wt.
support is elastically restrained by the spring CR . Study the motion of the
system if g=l ¼ 1, u0 w2 =l ¼ 5, CR ¼ kl 2 =2, q0 ¼ p=6, q_ 0 ¼ 0, u0 ¼ 0:1m,
w ¼ 5s1 .
212 PART
I Single-degree-of-freedom systems
FIG. P5.2 Pendulum in problem P5.2.
Problem P5.3 Study the response of the system shown in Fig. P5.3, when
(i) u ð0Þ ¼ 0:05m, u_ ð0Þ ¼ 0, pðt Þ ¼ 0. The initial displacement will be taken
from the position of the static equilibrium.
(ii) u ð0Þ ¼ u_ ð0Þ ¼ 0, pðt Þ ¼ mg ð1 t=t1 ÞH ðt1 t Þ:pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
where w ¼ kmin =m .
(iii) u ð0Þ ¼ u_ ð0Þ ¼ 0, P ðt Þ ¼ 15sin wt,
(iv) Plot of the dependence of the period T as a function of u0 , T ¼ T ðu0 Þ, if
u_ ð0Þ ¼ 0 and pðt Þ ¼ 0. The initial displacement will be taken from the
position of the static equilibrium.
Data: t1 ¼ 1s, ttot ¼ 5s, m ¼ 10kN m1 s2 , E ¼ 2:1 108 kN=m2 , A ¼ 1cm2 ,
L ¼ 4:0m, and g ¼ 9:81 ms2 . The cables are assumed massless to avoid the
sag due to self-weight [10–12].
FIG. P5.3 System in problem P5.3.
Problem P5.4 Study the response of the system shown in Fig. P5.3 if the elastic
force is approximated by the first three terms of its Tailor series. Compare the
results with those in Problem P5.3 when pðt Þ ¼ mg ð1 t=t1 ÞH ðt1 t Þ and
u ð0Þ ¼ u_ ð0Þ ¼ 0.
Problem P5.5 The response of a nonlinear system is governed by the following
equation, known as the van der Pol equation
u€ m 1 u 2 u_ + u ¼ 0
Plot the solution u ðt Þ, if u ð0Þ ¼ 0, u_ ð0Þ ¼ 0:1 and for (i) m ¼ 0:2 (ii) m ¼ 1:2.
Problem P5.6 The horizontal beams of the frame in Fig. P5.6 have negligible
mass and are flexible while the shear walls having a uniform mass density with
Nonlinear response: Single-degree-of-freedom systems Chapter
5
213
specific weight g are assumed rigid. Their support on the ground is elastic and it
is expressed by the relation
1
(a)
MR ¼ CR f f2
4
The system is set in motion by the initial conditions u0 ¼ 5cm, u_ 0 ¼ 0. Plot
the dependence of the ratio T=T0 (T0 is the period of the structure resulting
when the nonlinear term in Eq. a is ignored) on the initial displacement and discuss the influence of the nonlinearity of the elastic supports on the response of
the structure. Data: a ¼ 5m, g ¼ 24kN=m3 , E¼2.1107kN/m2, CR ¼ EI =5a,
and cross-sectional dimensions of the beam 0:20 0:40m2 .
FIG. P5.6 Frame in problem P5.6.
Problem P5.7 The buoy of Fig. P5.7 consists of two massless cones with a base
dimeter 2R and a height h. A concentrated mass m attached at the bottom of the
body keeps the buoy floating at the position shown in the figure. Study the
FIG. P5.7 Floating buoy in problem P5.7.
214 PART
I Single-degree-of-freedom systems
vertical motion of the buoy if it is displaced vertically downward from the equilibrium position by u0 . Data: m ¼ 10kNm1 s2 h ¼ 5m, R ¼ 2m,
u ð0Þ ¼ 0:30m, u_ ð0Þ ¼ 0, and specific weight of the liquid g ¼ 2kN=m3 .
Problem P5.8 The water tower of Fig. P5.8a is subjected to the blast load of
Fig. P5.8b. The response of the structuren is elastoplastic.
The
h
iorestoring force in
the elastic branch is given by fS ¼ ku 1 + 1= 1 + ð10u Þ6
=2 (Fig. P5.8c).
The structure is modeled by a SDOF system. Study the response of the structure
in the interval of t 0 ¼ 1s. Data: m ¼ 50kNm1 s2 , k ¼ 2000kN=m, x ¼ 0:07,
p0 ¼ 25kN, t1 ¼ 0:1s, and yield displacement uy ¼ 0:1m.
(b)
(a)
{
(c)
FIG. P5.8 Water tower in problem P5.8.
}
Nonlinear response: Single-degree-of-freedom systems Chapter
5
215
References and further reading
[1] J.T. Katsikadelis, Derivation of Newton’s law of motion using Galileo’s experimental data,
Acta Mech. 226 (9) (2015) 3195–3204, https://doi.org/10.1007/s00707-015-1354-y.
[2] H. Nayfeh, D.T. Mook, Nonlinear Oscillations, John Wiley & Sons, Inc, New York, 1995.
[3] F. Verhulst, Nonlinear Differential Equations and Dynamical Systems, Springer, SpringerVerlag, Berlin Heidelberg, 1996.
[4] J.C. Butcher, Numerical Methods for Ordinary Differential Equations, second ed., John Wiley
& Sons Ltd, 2008.
[5] H.T. Davis, Introduction to Nonlinear Differential and Integral Equations, Dover Publications
Inc, New York, 2010.
[6] A. Belendez, C. Pascual, D.I. Mendez, T. Belendez, C. Neipp, Exact solution for the nonlinear
pendulum, Rev. Bras. Ensimo Fis. 29 (4) (2007) 645–648.
[7] M. Abramowitz, I.A. Stegun (Eds.), Handbook of Mathematical Functions, Dover Publications, New York, 1970.
[8] K.J. Bathe, Conserving energy and momentum in nonlinear dynamics: a simple implicit time
integration scheme, Comput. Struct. 85 (2007) 437–445.
[9] J.T. Katsikadelis, A new direct time integration method for the equations of motion in structural dynamics, Angew. Math. Mech. 94 (9) (2014) 757–774, https://doi.org/10.1002/
zamm.20120024.
[10] J.W. Leonard, Tension Structures, McGraw-Hill, New York, 1988.
[11] J.T. Katsikadelis, Finite deformation of cables under 3-D loading: an analytic solution, in:
D.E. Beskos, D.L. Karabalis, A.N. Kounadis (Eds.), Proc. of the 4th National Congress on
Steel Structures, Patras, May 24–25, vol. II, 2002, pp. 526–534.
[12] C.G. Tsiatas, J.T. Katsikadelis, Nonlinear analysis of elastic cable-supported membranes, Eng.
Anal. Bound. Elem. 35 (2011) 1149–1158.
Chapter 6
Response to ground motion
and vibration isolation
Chapter outline
6.1 Introduction
6.2 Equation of motion: Relative
displacement
6.2.1 Response spectra
6.3 Equation of motion in terms
of the total displacement
6.1
217
217
220
6.4 Vibration isolation
6.4.1 Transmission of force
6.4.2 Transmission of motion
6.5 Problems
References and further reading
235
235
237
240
243
230
Introduction
Structural systems are often excited by the motion of their support. The
response of a structure to support excitation is dynamic even though no external
dynamic loads act on it. The seismic motion of the ground represents a typical
example of support excitation of structures. The study of the response of structures to earthquake-induced motion is a specific but very important subject of
structural dynamics. It is discussed in depth in books on earthquake engineering
as well as in books on structural dynamics, preparing engineers to design structures for earthquake-induced motion [1,2]. This book treats the dynamic
response of structures when the excitation force is known. Therefore, the discussion in this chapter is limited only to the study of the dynamic response
of the SDOF system due to support excitation. Besides, some basic concepts such
as the response spectrum concept, which facilitates the dynamic analysis of structures due to ground motion, are presented. The general problem of the support
excitation of structures will be examined later when the MDOF (multi-degreeof-freedom) systems are studied. The transmission of vibrations from the
structure to the fundament and vice versa are also discussed. Illustrative examples
analyzing the response of SDOF systems due to ground motion are presented. The
pertinent bibliography with recommended references for further reading is also
included. The chapter is enriched with problems to be solved.
6.2
Equation of motion: Relative displacement
Fig. 6.2.1 shows a simplified dynamic model of a one-story shear building
whose support moves horizontaly according to a known law described by the
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00006-6
© 2020 Elsevier Inc. All rights reserved.
217
218 PART
I Single-degree-of-freedom systems
function ug ðt Þ. The usual method to analyze the motion of the system is to
decompose the total displacement, namely the displacement with respect to a
fixed reference frame, utot , in two components (Fig. 6.2.1)
utot ðt Þ ¼ ug ðt Þ + u ðt Þ
(6.2.1)
where u ðt Þ is the relative displacement of the top of the column with respect to
its base.
The deformation of the structure is caused by the relative displacement u ðt Þ.
Therefore, the elastic force fS ðt Þ and the damping force fD ðt Þ depend only on
u ðt Þ and u_ ðt Þ, respectively. However, the inertial force fI ðt Þ depends on the
total acceleration u€tot ðt Þ. Thus, we will have
fI ðt Þ ¼ m u€tot ðt Þ,
fS ðt Þ ¼ ku ðt Þ, fD ðt Þ ¼ cu_ ðt Þ
(6.2.2)
utot
ug
Fixed axis of reference
m
m
c
(a)
fI
u
k
k
ug (t )
(b)
m
mutot
fS
fD
c
cu
ku
(c)
FIG. 6.2.1 SDOF system subjected to ground motion (a), Dynamic model (b), Forces on the
free body (c).
The dynamic equilibrium of the system (see Fig. 6.2.1c) requires
fI ðt Þ + fD ðt Þ + fS ðt Þ ¼ 0
which by virtue of Eq. (6.2.2) becomes
m u€tot + cu_ + ku ¼ 0
(6.2.3)
m u€ + cu_ + ku ¼ m u€g
(6.2.4)
or using Eq. (6.2.1) we have
Response to ground motion and vibration isolation Chapter
6
219
Eq. (6.2.4) is an equation of forced motion, where the excitation function
is m u€g . In this case, the excitation function is called effective load and is
denoted by
pðt Þ ¼ m u€g ðt Þ
(6.2.5)
We observe that the influence of the ground motion on the structure does not
depend directly on the ground displacement ug ðt Þ, but on its acceleration u€g ðt Þ.
Therefore, the dynamic response of the structure due to ground motion can be
studied if the accelerogram of the seismic motion is known, namely if there is a
record of the function u€g ðt Þ during the earthquake (see Fig. 6.2.2). It must be
emphasized that the initial conditions u ð0Þ and u_ ð0Þ for the relative displacement do not vanish if ug ð0Þ 6¼ 0 or u_ g ð0Þ 6¼ 0, even though the system is at rest
at time t ¼ 0. Indeed, if
utot ð0Þ ¼ 0,
u_ tot ð0Þ ¼ 0
ug,tt (cm/s2)
max ug,tt=258.59
–
–
–
t
FIG. 6.2.2 Accelerogram from Athens earthquake, Sept. 7, 1999 (Recorded by ITSAK, Α399-1.
V2, longitudinal component, max u€g ¼ 258:59cm=s2 ).
then Eq. (6.2.1) gives
u ð0Þ ¼ ug ð0Þ, u_ ð0Þ ¼ u_ g ð0Þ
(6.2.6)
Eq. (6.2.4) can also be written as
m u€ + 2mwx u_ + ku ¼ m u€g ðt Þ
(6.2.7a)
u€ + 2xwu_ + w2 u ¼ u€g ðt Þ
(6.2.7b)
or
220 PART
I Single-degree-of-freedom systems
Apparently, Eq. (6.2.7b) states that the deformation of the system due to
given ground acceleration u€g ðt Þ depends only on the natural frequency w, hence
on the natural period T ¼ 2p=w, and on the damping ratio x, that is,
u ¼ u ðt, T , x Þ. Consequently, two systems with the same natural period T
and the same damping ratio x will undergo the same displacement u ðt Þ under
the same ground motion, in spite of the fact that the two systems may have different masses or different stiffnesses.
The negative sign in the effective load pðt Þ ¼ m u€g ðt Þ affects only the
direction of the displacement and not its magnitude. In practice, this has little
significance inasmuch as the engineer is usually interested in the maximum
absolute value of u ðt Þ. Therefore, the sign can be omitted in this case. This
assumption allows us to write the Duhamel integral in the forma
1
u ðt Þ ¼ U ðt Þ
(6.2.8)
wD
where
Z t
(6.2.9)
U ðt Þ ¼
u€g ðτÞexwðtτÞ sin wD ðt τÞdτ
0
For a given ground motion and a fixed damping ratio, we can evaluate the
largest absolute value of the function U ðt Þ, hence of u ðt Þ, for an interval of
values of the natural period T of the damped SDOF system and plot the curves
u ðT , xÞ ¼ max t ju ðt, T , xÞj for discrete values of the damping ratio x. Fig. 6.2.3
shows the curves u ðT , xÞ ¼ U ðT , x Þ=wD , 0 < T 2 for different values of x,
when the ground motion is induced by the Athens earthquake, Sept. 7, 1999.
The respective accelerogram is shown in Fig. 6.2.2. The curves u ðT , x Þ were
obtained by direct solution of Eqs. (6.2.7a), (6.2.7b). The solution can be
obtained using either the analytic solution presented in Section 3.5.4 or numerically using any of the methods presented in Chapter 4.
6.2.1 Response spectra
The curve u ðT , x Þ ¼ max t ju ðt, T , x Þj is called the response spectrum of the
relative displacement or the deformation response spectrum. The response
spectrum, introduced by M. A. Biot in 1932 [3], can be used as a practical means
to study the effect of the ground motion on structures. It is clear that the deformation response spectrum permits the direct evaluation of the absolute maximum value of the relative displacement of the SDOF system for a specified
excitation without solving the equation of motion, if its natural period T and
damping ratio x are given.
pffiffiffiffiffiffiffiffiffiffiffiffi
a. In Eqs. (6.2.8) and (6.2.9), we can set wD ¼ w 1 x 2 w, because in real structures the value of
the damping ratio is small (x ¼ 3% 15%, hence x2 ≪1) and the error due to this approximation is
much smaller than that due to the uncertainty of the determination of u€g ðt Þ.
Response to ground motion and vibration isolation Chapter
6
221
u T
T (s)
FIG. 6.2.3 Deformation response spectrum u ðT , x Þ for Athens earthquake, Sept. 7, 1999.
Similarly, we define the relative velocity spectrum and the acceleration
response spectrum as the graphs of the curves u_ ðT , xÞ ¼ max t ju_ ðt, T , x Þj
and u€tot ðT , xÞ ¼ max t ju€tot ðt, T, x Þj, respectively.
The expression of the relative velocity results by direct differentiation of
Eq. (6.2.9) with respect to time t. Because t appears as a parameter in the limits
of the integral, the derivative can be obtained using Leibnitz’s rule [4].
Thus, we obtain
Z t
(6.2.10)
u€g ðτÞexwðtτÞ cos wD ðt τÞdτ
u_ ðt Þ ¼ wxu ðt Þ +
0
which by virtue of Eq. (6.2.8) is written
x
u_ ðt Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi U ðt Þ + U ðt Þ
1 x2
(6.2.11)
where it was set
U ðt Þ ¼
Z
t
u€g ðτÞexwðtτÞ cos wD ðt τÞdτ
(6.2.12)
0
Differentiation of the expression (6.2.10) gives the relative acceleration
and the total acceleration is obtained from Eq. (6.2.1), which gives
u€tot ðt Þ ¼ u€g ðt Þ + u€ðt Þ. However, the total acceleration can be obtained directly
from Eq. (6.2.7b), that is
u€tot ðt Þ ¼ 2xwu_ ðt Þ w2 u ðt Þ
222 PART
I Single-degree-of-freedom systems
which by virtue of Eqs. (6.2.8), (6.2.11) becomes
w u€tot ðt Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi 2x2 1 U ðt Þ 2xwU ðt Þ
1 x2
(6.2.13)
The graphs of the absolute maximum of the functions (6.2.11) and (6.2.13),
that is, u_ ðT , xÞ ¼ max t ju_ ðt, T , x Þj and u€tot ðT , xÞ ¼ max t ju€tot ðt, T , x Þj, in an
interval of the undamped natural period T for a fixed value of the damping ratio
x of the SDOF systems give the response spectra of the relative velocity and the
(total) acceleration. Figs. 6.2.4 and 6.2.5 show these response spectra for the
accelerogram of Fig. 6.2.2. The curves were obtained by direct solution of
Eq. (6.2.7b). It should be noted that the quantities u ðT , xÞ, u_ ðT , xÞ, and
u€ðT , xÞ do not occur at the same time instant.
The shear force at the base of the SDOF system of Fig. 6.2.1 is equal to the
elastic force, that is,
Qo ¼ fS ¼ ku
or taking into account that k ¼ mw2 and using Eq. (6.2.8) we obtain
w
Qo ¼ m pffiffiffiffiffiffiffiffiffiffiffiffi U ðt Þ
(6.2.14)
1 x2
Obviously, the maximum value of the base shear force is
w
max Qo ¼ m pffiffiffiffiffiffiffiffiffiffiffiffi U ðT , xÞ
1 x2
(6.2.15)
T (s)
FIG. 6.2.4 Response spectrum of the relative velocity u_ ðT , xÞ for the Athens Earthquake,
Sept. 7, 1999.
Response to ground motion and vibration isolation Chapter
6
223
T (s)
FIG. 6.2.5 Response spectrum of the total acceleration u€tot ðT , xÞ for the Athens earthquake,
Sept. 7, 1999.
For small values of the damping ratio (say, 0 x 0:15), we may set x 2 0.
Thus, the last two equations become
Qo ¼ mwU ðt Þ
(6.2.16)
max Qo ¼ mwU ðT , x Þ
(6.2.17)
We observe that the quantity wU ðT , x Þ in Eq. (6.2.17) has dimensions of
acceleration. In earthquake engineering, this quantity is designated by
Spa ðT , x Þ and it is known as spectral pseudoacceleration. It is an important
quantity because it allows direct evaluation of the maximum elastic force (base
shear force) in the SDOF system from the graph of Spa ðT , xÞ.
The quantity Spv ðT , x Þ ¼ U ðT , x Þ has dimensions of velocity and it is
known as spectral pseudovelocity. Actually, the quantities Spv ðT , x Þ,
Spa ðT , x Þ are different from u_ ðT , xÞ and u€tot ðT , x Þ. Therefore, they should
not be confused. Nevertheless, it is Spa ðT , xÞ ¼ u€tot ðT , xÞ, if x ¼ 0. Indeed,
Eq. (6.2.3) for x ¼ 0 becomes
u€tot ¼ w2 u ¼ wU ðt Þ
(6.2.18)
u€tot ðT , x Þ ¼ Spa ðT , x Þ
(6.2.19)
from which we obtain
The deviation of the pseudoacceleration from the extreme value of the total
acceleration is small for small values of x (say 0 x 0:1). Thus we may set
Spa ðT , x Þ u€tot ðT , x Þ. This is shown in Fig. 6.2.6.
224 PART
I Single-degree-of-freedom systems
(utot),tt(T,0)
Spa(T,0)
T (s)
FIG. 6.2.6 Response spectra Spa ðT , x Þ and u€tot ðT , x Þ, ðx ¼ 0Þ.
For the sake of uniform notation, we set Sd ðT , x Þ ¼ u ðT , xÞ. It is obvious
that if one of the quantities Sd ðT , x Þ, Spv ðT , xÞ, Spa ðT , xÞ is known, the remaining quantities result immediately. Thus, if Sd ðT , xÞ is known then we obtain
Spv ðT , xÞ ¼ wSd ðT , x Þ
(6.2.20)
Spa ðT , x Þ ¼ wSpv ðT , xÞ
(6.2.21)
The above relations are valid because the resulting values refer to the same
time and even at the instant where the maximum displacement Sd ðT , xÞ occurs.
Eq. (6.2.17) can be written as
max Qo ¼ mS pa ðT , x Þ
¼W
Spa ðT , xÞ
g
(6.2.22)
where W is the weight of the SDOF system and g the acceleration of gravity.
The ratio
e¼
Spa ðT , x Þ
g
(6.2.23)
referred to as the elastic seismic coefficient is a quantity specified in the earthquake codes and it is used to evaluate the maximum equivalent shear force at the
base of a SDOF shear building.
Response to ground motion and vibration isolation Chapter
6
225
The response spectra of the displacement, pseudovelocity, and pseudoacceleration contain the same information for the dynamic response of the system
because each of them results from another by multiplying or dividing it by a
constant number. The reason for using the three spectra is that each of them
refers to a different natural quantity. We saw that Sd ðT , x Þ expresses the
extreme value of the relative displacement while Spa ðT , x Þ is employed to
evaluate the shear force at the base of the one-story shear building. Finally,
Spv ðT , x Þ is used to calculate the extreme value of the energy of the
system during the earthquake. Indeed, the maximum value of the energy of
deformation is
1
E ðT , xÞ ¼ kSd ðT , x Þ2
2
1 Spv ðT , xÞ2
¼ k
w2
2
1
¼ mS pv ðT , xÞ2
2
(6.2.24)
Fig. 6.2.7 presents the graphs of Sd ðT , x Þ, Spv ðT , x Þ, and Spa ðT , xÞ for the
Athens earthquake, Sept. 7, 1999. In earthquake engineering, the three plots are
usually presented in a single graphical representation using a logarithmic scale
for the four axes [2].
All previous results have been obtained by direct solution of the equation of
motion. For this reason, a computer program has been written in MATLAB and
given the name response_spectrum_aem.m. The program evaluates the spectra of the deformation, relative velocity, and acceleration and makes their
graphs. The program uses the numerical method presented in Section 4.4.
The user should provide the one-dimensional array containing the values of
the accelerogram at the respective times. The electronic version of the program
is given on this book’s companion website.
Example 6.2.1 Fig. E6.1a shows a cement silo of square plan form supported on
four identical columns. The silo is made of reinforced concrete. When the silo is
empty, a horizontal load P in the x direction is applied, which causes a horizontal displacement of the silo equal to 5cm. Then, the force is suddenly removed
and the system starts to vibrate. After time to the silo has performed six complete vibrations and the amplitude was reduced to 1cm. Determine:
(i) The damping ratio x.
(ii) The damping coefficient c, the damped natural frequency wD , and the
time to
(iii) The displacement after 40 and 60s.
(iv) The maximum shear force max Q and the maximum bending moment
max M of the columns, when the silo is full and is subjected to the support
of duration t1 ¼ 3s in the absence of damping.
excitation ug ¼ uo sin wt
Determine also the dynamic magnification factor D.
226 PART
I Single-degree-of-freedom systems
Sd
(a)
(b)
Spa
(c)
FIG. 6.2.7 Response spectra for the Athens earthquake, Sept. 7, 1999 (x ¼ 0:1).
Response to ground motion and vibration isolation Chapter
6
227
5m
2.5 m
z
x
1m
utot
P
ug
u
m
c
10 m
k
x
(a)
(b)
FIG. E6.1 Silo in Example 6.2.1
Data: Specific weight of cement g c ¼ 17kN=m3 , specific weight of reinforced
concrete g b ¼ 24kN=m3 , uo ¼ 1:7mm, w ¼ 2:5s1 , modulus of elasticity of
the reinforced concrete E ¼ 2:1 107 kN=m2 , cross-sectional area of columns
0:35 0:35m2 , and thickness of the silo walls and bottom 0:20m. The axial deformation of the columns is ignored while their mass is assumed lumped at their ends.
Solution
The walls of the silo do not deform inside their plane. Therefore, the silo
behaves in actual fact as a rigid body. Because the axial deformation of the columns is neglected, the only possible motion of the structure in the xz plane is the
horizontal one along the x axis. This is shown in the dynamic model of the system in Fig. E6.1b.
(i) Τhe damping ratio x is computed from the relation (2.3.24) for n ¼ 6, t ¼ 0,
u ð0Þ ¼ 5cm, u ðnT Þ ¼ 1cm. This yields
x
1
5
pffiffiffiffiffiffiffiffiffiffiffiffi ¼
‘n
2
2
6p
1
1x
which gives x ¼ 4:265%.
(ii) The computation of c and wD requires the determination of the mass m and
the stiffness k of the structure
2
24
5:0 4:62 2:5 + 4:62 0:2=0:928 + 4 0:352 10=2 9:81
¼ 40:6kNm1 s2
m¼
12EI
12 2:1 107 0:354 =12
¼
4
¼ 1260:5kN=m
h3
103
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
w ¼ 1260:5=40:6 ¼ 5:57s1
k ¼4
228 PART
I Single-degree-of-freedom systems
Hence
c ¼ 2mwx ¼ 2 40:6 5:57 0:04265 ¼ 19:3kNm1 s
qffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
wD ¼ w 1 x 2 ¼ 5:57 1 0:042652 ¼ 5:56s1
to ¼ 6T ¼ 6
2p
¼ 6:78s
wD
(iii) The system performs free damped vibrations. The displacement u ðt Þ is
computed from Eq. (2.3.18) for u ð0Þ ¼ 0:05m, u_ ð0Þ ¼ 0, x ¼ 0:04265,
w ¼ 5:57s1 , wD ¼ 5:56s1 . Thus we have
u ðt Þ ¼ e0:238t ð5cos 5:56t + 0:214 sin 5:56t Þ 102
(1)
which yields
u ð40Þ ¼ 0:282 103 cm,
u ð60Þ ¼ 0:268 105 cm
(iv) When the silo is filled with cement, we have
17
¼ 142:1kNm1 s2
m ¼ 40:6 + 4:62 2:5 + 4:62 0:8=3 9:81
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
w ¼ 1260:5=142:1 ¼ 2:97834s1
The motion is examined in two phases.
Phase Ι (0 t 3). The motion is forced vibration. The equation of motion
is given by Eq. (6.2.4) for c ¼ 0, that is
m u€ + ku ¼ pðt Þ
(2)
The effective force is
¼ 142:1 1:7 103 2:52 sin wt
¼ 1:5098 sin wt
pðt Þ ¼ mu o w2 sin wt
The displacement is given by Eq. (3.2.8). Because the system is at rest at
t ¼ 0, it is utot ð0Þ ¼ u_ tot ð0Þ ¼ 0. Hence
u ð0Þ ¼ utot ð0Þ ug ð0Þ ¼ 0
and
o ¼ 2:5 1:7 103 ¼ 4:25 103 m=s
u_ ð0Þ ¼ u_ tot ð0Þ u_ g ð0Þ ¼ wu
These initial conditions yield
A¼
u_ ð0Þ Po b
and B ¼ 0
k 1 b2
w
Hence
u I ðt Þ ¼
po 1
u_ ð0Þ
b sin wt Þ
ð sin wt
sin wt +
k 1 b2
w
Response to ground motion and vibration isolation Chapter
6
229
¼ 2:5=2:98 ¼ 0:8389, po ¼ 1:5098kN,
The above relation for b ¼ w=w
k ¼ 1260:5kN=m, and u_ ð0Þ ¼ 4:25 103 m=s yields
Þ
uI ðt Þ ¼ 103 ð4:8303 sin wt 4:0545sin wt
Þ
u_ Ι ðt Þ ¼ 102 ð1:4386 cos wt 1:0136 cos wt
Phase II (t 3). The motion is free vibration and the displacement is given
by Eq. (2.2.13). Thus we have
uII ðt Þ ¼
u_ I ð3Þ
sin wet + uI ð3Þ cos wet , et ¼ t 3 0
w
(3)
The initial conditions for this phase are
uI ð3Þ ¼ 1:5309 103 m, u_ I ð3Þ ¼ 1:620884630 102 ms1
which give
ffi
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u_ ð3Þ 2
max uII ¼ r ¼
+ u_ ð3Þ2 ¼ 5:65341 103 m
w
(4)
R(t)
Fig. E6.2 shows the time history of the response ratio Rðt Þ ¼ u ðt Þ=ðpo =k Þ.
We observe that the max u takes place in phase II. Hence the dynamic load factor is
max uII
¼ 4:7199
(5)
D¼
p0 =k
t (s)
FIG. E6.2 Time history of the response ratio Rðt Þ in Example 6.2.1.
230 PART
I Single-degree-of-freedom systems
and
12EI
k
max u ¼ max uII ¼ 1:781kN
3
h
4
h
max M ¼ max Q ¼ 8:907kNm
2
max Q ¼
(6a)
(6b)
6.3 Equation of motion in terms of the total displacement
In certain cases, it is convenient to formulate the equation of motion in terms of
the total displacement utot ðt Þ. Then, the elastic force and damping force will be
expressed as
(6.3.1)
fS ¼ k utot ug , fD ¼ c u_ tot u_ g
The dynamic equilibrium yields the following equation of motion
(6.3.2)
m u€tot + c u_ tot u_ g + k utot ug ¼ 0
or
m u€tot + cu_ tot + ku tot ¼ pðt Þ
(6.3.3)
pðt Þ ¼ cu_ g + ku g
(6.3.4)
where now
Eq. (6.3.3) is suitable to study the dynamic response when the u€g does not
exist in the classical sense or the supports of the structure undergo different
excitations (see Example 6.3.2).
Example 6.3.1 The support of a SDOF system, whose dynamic model is shown
in Fig. E6.3a, undergoes a sudden constant displacement ug ¼ u0 . Determine the
dynamic response of the system when utot ð0Þ ¼ u_ tot ð0Þ ¼ 0.
(a)
FIG. E6.3 SDOF system in Example 6.3.1
(b)
Response to ground motion and vibration isolation Chapter
6
231
Solution
The function representing the ground motion is shown in Fig. E6.3b. The problem can be solved by the following two ways:
(a) Formulating the equation of motion in terms of the total displacement, that
is, Eq. (6.3.3). This yields
m u€tot + ku tot ¼ ku 0
(1)
This implies that an effective constant excitation force pðt Þ ¼ ku 0 is
suddenly applied. The solution of Eq. (1) is obtained from Eq. (3.4.3)
for p0 ¼ ku 0 . Hence
pffiffiffiffiffiffiffiffiffi
(2)
utot ðt Þ ¼ u0 ð1 cos wt Þ, w ¼ k=m
We observe that
ðutot Þmax ¼ 2u0
The relative displacement is
u ðt Þ ¼ utot ug ¼ u0 cos wt
(3)
which yields an elastic force
fS ¼ ku ¼ ku 0 cos wt
(4)
max jfS j ¼ ku 0
(5)
and
(b) Formulating the equation of motion in terms of the relative displacement
u ðt Þ. In this case, we have
ug ð0Þ ¼ u0 , u_ g ð0Þ ¼ 0, u€g ð0Þ ¼ 0
Consequently
pðt Þ ¼ 0, u ð0Þ ¼ utot ð0Þ ug ð0Þ ¼ u0 , u_ ð0Þ ¼ u_ tot ð0Þ u_ g ð0Þ ¼ 0
and the equation of motion becomes
m u€ + ku ¼ 0
(4)
with initial conditions u ð0Þ ¼ u0 , u_ ð0Þ ¼ 0.
The solution of Eq. (4) is obtained from Eq. (2.2.13) as
u ðt Þ ¼ u0 cos wt
(5)
which is identical with that given by Eq. (3).
Example 6.3.2 The supports of the columns of the one-story frame of Fig. E6.4
qÞ and ug2 ¼ uo sin wt.
are subjected to the displacements ug1 ¼ uo sin ðwt
Determine the equation of motion of the structure and give the expressions
of the relative displacement u ðt Þ and the stress resultants Q ðt Þ, M ðt Þ at the
top cross-sections of the columns. The dead load of the rigid beam is included
232 PART
I Single-degree-of-freedom systems
utot (t )
p
utot (t )
EI
h
40 × 80
35 × 70
ug 1
L
ug 2
FIG. E6.4 One-story frame in Example 6.3.2
in the load p. The material of the columns is reinforced concrete. Data: specific
weight of concrete g ¼ 24kN=m3 , u0 ¼ 0:02m, x ¼ 0:05, w ¼ 2:5rad=s,
u ð0Þ ¼ 0, u_ ð0Þ ¼ 0, L ¼ 15m, h ¼ 7m, E ¼ 2:1 107 kN=m2 , and
p ¼ 200kN=m. The mass of the columns is assumed lumped at their ends.
Solution
The system has one degree of freedom. The equation of motion with respect to
the total displacement utot ðt Þ is
m u€tot + c1 u_ tot u_ g1 + c2 u_ tot u_ g2 + k1 utot ug1 + k2 utot ug2 ¼ 0 (1)
or
m u€tot + cu_ tot + ku tot ¼ pðt Þ
(2)
c ¼ c1 + c2 , k ¼ k1 + k2
(3)
p ¼ c1 u_ g1 + c2 u_ g2 + k1 ug1 + k2 ug2
(4)
where
The mass of the system is
m¼
200 15 + ð0:35 0:70 + 0:40 0:80Þ ð7=2Þ 24
¼ 310:6kNm1 s2
9:81
The stiffness of the columns
k1 ¼
12EI 1 12 2:1 107 0:70 0:353
¼
¼ 1837:5kN=m
h3
12 73
k2 ¼
3EI 2 3 2:1 107 0:80 0:403
¼
¼ 783:7kN=m
h3
12 73
Hence the stiffness of the system is
k ¼ k1 + k2 ¼ 2621:2kN=m
Response to ground motion and vibration isolation Chapter
6
233
The natural frequency is
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
w ¼ 2621:2=310:6 ¼ 2:905s1
and the natural period
T ¼ 2p=w ¼ 2:163s
The damping coefficient c is
c ¼ 2mwx ¼ 2 310:6 2:905 0:05 ¼ 90:23kNm1 s
and the damped frequency
wD ¼ w
qffiffiffiffiffiffiffiffiffiffiffiffi
1 x2 ¼ 2:901s1
The damping coefficients c1 and c2 that contribute to the effective force cannot be directly determined. However, they can be estimated by setting
c1 ¼ ac and c2 ¼ ð1 a Þc, 0 a 1
Apparently, the cases a ¼ 0 and a ¼ 1 are not realistic.
For the data of the problem, Eq. (2) becomes
310:6u€tot + 90:23u_ tot + 2621:2utot ¼ pðt Þ
(5)
qÞ + 15:67sin wt
pðt Þ ¼ 36:75sin ðwt
qÞ + ð1 a Þ cos wt
+ 1:80w½a cos ðwt
(6)
where
utot(t) (m)
Eq. (5) can be solved using the analytic method presented in Chapter 3.
Figs. E6.5 and E6.6 show the graphs of utot ðt Þ for various values of the phase
angle q and the allocation coefficient a. Fig. E6.7 shows the influence of a on
the maximum displacement max |utot ðt Þ|.
t
FIG. E6.5 Total displacement for different values of the phase angle q in Example 6.3.2 (a ¼ 0:2).
I Single-degree-of-freedom systems
utot(t) (m)
234 PART
t
max|utot(t)|
FIG. E6.6 Total displacement for different values of the allocation coefficient a in Example 6.3.2
(q ¼ 3p=4).
a
FIG. E6.7 Extreme value max|utot ðt Þ| versus the allocation coefficient a in Example 6.3.2
(x ¼ 0:05).
From the study of the numerical results, we may draw the following
conclusion:
The percentage allocation of the damping to the two columns does not affect
significantly the dynamic response of the structure. Consequently, an arbitrary
but reasonable allocation, for example, 0:4 < a < 0:6, allows treating practical
cases of asynchronous support excitations.
Response to ground motion and vibration isolation Chapter
6
235
The stress resultants Q ðt Þ and M ðt Þ at the top cross-sections of the columns
are evaluated from the relations.
12EI 1 utot ug1 ¼ 1837:5 utot ug1
3
h
3EI 2 Q2 ðt Þ ¼ 3 utot ug2 ¼ 783:7 utot ug2
h
6EI 1 M1 ðt Þ ¼ 2 utot ug1 ¼ 6431:2 utot ug1
h
3EI 2 M2 ðt Þ ¼ 2 utot ug2 ¼ 5485:9 utot ug2
h
Q1 ðt Þ ¼
6.4
Vibration isolation
Very often, machines that generate vibrations are mounted on our structures.
These vibrations, which are transferred to the supporting structure (soil, foundation), are annoying or even harmful and need to be minimized. This is
achieved by inserting spring and damping devices between the machine and
its foundation. An inverse problem arises when vibrations from the environment
(ground support) are transferred to the structure. This problem occurs when sensitive instruments are placed on top or inside moving structures (earthquake
motion of a building, motion of a car on a rough road, spacecrafts, airplanes,
ships, etc.) or special structures near vibration generation sources (monuments).
In the first situation, we talk about the transmission of force while the second is
about the transmission of motion.
6.4.1
Transmission of force
Let us consider first the force isolation problem. Fig. 6.4.1 presents a SDOF
system consisting of a mass m mounted on the fundament by means of a system
of springs and dampers having total stiffness k and total damping c. A vertical
is applied to the mass. The transfer of force is
harmonic force p ¼ p0 sin wt
FIG. 6.4.1 Force transmitted to the foundation.
236 PART
I Single-degree-of-freedom systems
examined when the motion is in the steady state phase. Hence, the vertical displacement is given by Eq. (3.2.26), that is,
qÞ
u ðt Þ ¼ rsin ðwt
(6.4.1)
where
i 1
2
p0 h
2
1 b2 + ð2xb Þ2
k
2xb
q ¼ tan 1
, b ¼ w=w
1 b2
r¼
(6.4.2a)
(6.4.2b)
The total force transmitted to the foundation is
f ¼ fS + fD
¼ ku + cu_
qÞ + crw cos ðwt
qÞ
¼ krsin ðwt
(6.4.3)
q fÞ
¼ fT sin ðwt
where
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
k 2 + ðcwÞ2
1 cw
f ¼ tan
k
fT ¼ r
(6.4.4)
(6.4.5)
Using Eqs. (6.4.2a), (6.4.2b) for the expression of r and c=k ¼ 2x=w, the
previous relations are written as
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u
1 + ð2xbÞ2
(6.4.6)
fT ¼ p0 t
2
1 b2 + ð2xb Þ2
f ¼ tan 1 ð2xb Þ
(6.4.7)
The ratio TR ¼ fT =p0 is called the transmission ratio or transmissibility. It
provides a measure of the force transmitted from the vibrating machine to the
fundament. Fig. 6.4.2 shows the graph of the function TRðb, x Þ for various
values of
pffiffix.ffi We observe that the reduction of the transmitted force is achieved
for b > 2. It is also apparent that the reduction of the ratio TR increases with
decreasing x. Theoretically, the value x ¼ 0 gives the minimum value of the
transmission ratio. However, this is not realistic for two reasons: first, because
the actual isolation systems have damping and second, some damping is desirable to avoid adverse results at the startup of the machine as it passes through the
resonance frequency.
6
237
TR=fT /p0
Response to ground motion and vibration isolation Chapter
–
–
FIG. 6.4.2 Transmissibility for harmonic excitation.
6.4.2
Transmission of motion
In this case, the problem is to determine the motion that is transmitted from the
ground to the machine and in general to the structure supported on the ground.
Fig. 6.4.3 shows a mass supported on the ground by means of a system of
springs and dampers with total stiffness k and damping c. The vertical ground
motion is harmonic
ug ¼ u0 sin wt
(6.4.8)
FIG. 6.4.3 Ground motion transmitted to the structure.
The equation of motion with respect to the relative displacement is
m u€ + cu_ + ku ¼ m w2 u0 sin wt
(6.4.9)
The relative displacement in the steady state is given by Eq. (3.2.25) for
p0 ¼ m w2 u0 . That is
238 PART
I Single-degree-of-freedom systems
m w2 u0
1
2xb cos wt
1 b2 sin wt
2
2
2
k
1 b + ð2xbÞ
u0
2xb3 cos wt
¼
1 b2 b 2 sin wt
2
2 2
1 b + ð2xbÞ
u ðt Þ ¼
(6.4.10)
and the total displacement
utot ¼ ug + u
u0
+
2xb3 cos wt
¼ u0 sin wt
1 b2 b 2 sin wt
2
1 b2 + ð2xbÞ2
nh
o
i
u0
2xb 3 cos wt
¼
1 b2 + ð2xb Þ2 sin wt
2
1 b2 + ð2xbÞ2
Þ
¼ uT sin ðwt
(6.4.11)
where
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u
1 + ð2xbÞ2
uT ¼ u0 t
2
1 b2 + ð2xb Þ2
!
2xb3
1
¼ tan
1 b 2 + ð2xb Þ2
(6.4.12)
(6.4.13)
From Eqs. (6.4.8), (6.4.12), we obtain
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
ðu€tot Þmax ðutot Þmax uT u
1 + ð2xbÞ2
TR ¼
¼
¼ ¼ t
2
u0
ug max
u€g max
1 b2 + ð2xb Þ2
(6.4.14)
We observe that the transmissibility of the amplitude of the ground harmonic motion is the same as that of the force transmissibility. The same is valid
for the extreme values of the accelerations.
Example 6.4.1 A sensitive instrument of weight W ¼ 30kN is installed at a
place where the environment exerts vertical harmonic motion with a cyclic frequency f ¼ 15Hz and maximum amplitude u0 ¼ 0:02m. For the insulation of
the motion, the instrument is placed on an elastic layer with stiffness
k ¼ 28kN=m and damping x ¼ 0:1. Determine (i) the acceleration transmitted
to the instrument and (ii) If the smooth functioning of the instrument requires
that the acceleration applied to it is less than 0:1g, indicate how to ensure that by
using the same elastic layer.
Solution
(i) Computation of the transmission ratio TR
The mass and the natural frequency of the instrument are
Response to ground motion and vibration isolation Chapter
6
239
W
30kN
¼
¼ 3:058kNm1 s2
g 9:81m=s2
rffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi
k
28
w¼
¼
¼ 3:026s1
m
3:058
m¼
The circular frequency of the harmonic motion and the maximum
acceleration are
w ¼ 2pf ¼ 94:248s1
u€g max ¼ w2 u0 ¼ 177:653 ¼ 18:109g g ¼ 9:81ms2
hence
w
b ¼ ¼ 31:147
w
and
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u
1 + ð2xbÞ2
¼ 0:0065
TR ¼ t
2
1 b 2 + ð2xbÞ2
Consequently, the maximum acceleration transmitted to the
instrument is
ðu€tot Þmax ¼ TR u€g max ¼ 0:118g
(ii) Modification of the system so that ðu€tot Þmax < 0:1g
It must be
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u
ðu€tot Þ
1 + ð2xbÞ2
0:1g
¼ max <
¼ 0:0055
TR ¼ t
2
2 2
u€g max 18:109g
1 b + ð2xbÞ
For the same elastic layer, the stiffness k and the damping coefficient c
are unaltered. Consequently, the transmission ratio TR can be modified if
the ratio b is changed. But because w is prescribed, this ratio can be changed, only if the natural frequency w is changed. This is possible if the mass
of the system (instrument) is changed by an increment Dm.
Let m 0 ¼ m + Dm be the new mass and x 0 , w0 the new damping ratio and
the frequency of the system (instrument), respectively. Then we will have
rffiffiffiffiffiffi
w0
m
(1)
¼
w
m0
c ¼ 2mwx ¼ 2m 0 w0 x0
240 PART
I Single-degree-of-freedom systems
which yield
mw
x ¼x 0 0 ¼x
mw
0
Moreover, we have
w
b ¼ 0¼b
w
0
rffiffiffiffiffiffi
m
m0
rffiffiffiffiffiffi
m0
m
(2)
(3)
From Eqs. (2), (3) we obtain
x0 b0 ¼ xb ¼ 3:115
Consequently, it must be
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
u
2
u
u
1 + ð2x 0 b0 Þ
1 + ð2 3:115Þ2
u
TR ¼ u
< 0:0055
¼
t
t
2 2
2 2
2
1 b0
+ ð2x0 b 0 Þ
1 b0
+ ð2 3:115Þ2
which holds if
33:885 < b0 ¼ b
pffiffiffiffiffiffiffiffiffiffiffiffi
m 0 =m or m 0 > 3:619kNm1 s2
Hence Dm 0:561kNm1 s2 .
6.5 Problems
Problem P6.1 The supports 1 and 2 of the structure shown in Fig. P6.1 are sub and ug2 ðt Þ ¼ u0 sin ðwt
p=3Þ, respecjected to the motions ug1 ðt Þ ¼ u0 sin wt
tively. A plane square rigid body of side a and density g is attached to
node 3. The mass of the column and the beam is neglected. Determine the
response of the structure and the reactions at the supports as well as the forces
that produce the support excitations. Data: L ¼ 8m, a ¼ 2m, cross section of the
FIG. P6.1 Structure in problem P6.1
Response to ground motion and vibration isolation Chapter
6
241
beam and the column IPE360, w ¼ 10s1 , u0 ¼ 3cm, g ¼ 10kN=m2 , and
E ¼ 2:1 108 kN=m2 .
ug,tt
Problem P6.2 A SDOF system with parameters m, x,k is subjected
to ground motion whose accelerogram is given by u€gi ¼ 100ð1Þi
½ði + 5Þ=ði + 1Þ ði + 1Þ=ði + 5Þ with duration t1 ¼ 2:5s (Fig. P6.2). Compute
and plot the response spectra of the displacement, velocity, and acceleration
for the interval of the natural period 0 < T 2 and x ¼ 0:05, 0:1.
Data: u ð0Þ ¼ u_ ð0Þ ¼ 0, u€g ð0Þ ¼ 0, m ¼ 4, and k ¼ 100.
(ug)i
(ug)i
t
FIG. P6.2 Accelerogram in problem P6.2
Problem P6.3 The rigid vertical column AC (Fig. P6.3) of a circular cross is supported on the ground by means of a spherical hinge
section, line density m
at A and three elastic cables of cross-sectional area A and modulus of elasticity E.
(a)
FIG. P6.3 Structure in problem P6.3
(b)
242 PART
I Single-degree-of-freedom systems
The cables have been prestressed so that they can undertake compressive forces.
which are
The column carries three advertising panels of total mass 10ma,
arranged as in Fig. P6.3b. Their support on the column extends to a length
1:25a. The structure is subjected to ground motion in the y direction, whose accelerogram u€g ðt Þ is given in Problem P6.2. Determine the minimum prestressing
force of the cable GB using the results of Problem P6.2. The cables are assumed
¼ 0:5kNm1 s2 =m, E ¼ 2:1 108 kN=m2 , and
massless. Data: a ¼ 5m, m
2
A ¼ 4cm .
Problem P6.4 The one-story building of Fig. P6.4 is subjected to ground
motion whose accelerogram u€g ðt Þ is given in Problem P6.2. The motion takes
place in the direction of angle b with respect to the x axis. Determine the
dynamic response of the structure and plot the relative displacement u ðb Þ
of the top cross-section of columns 1 and 2 as a function of the angle b,
0 b 2p. Moreover, compute the extreme normal stresses of the same
cross-sections due to bending. Data: x ¼ 0:07 and E ¼ 2:1 107 kN=m2 . The
mass of the columns is neglected. The load q includes also the dead weight
of the slab.
FIG. P6.4 One-story building in problem P6.4
Problem P6.5 The vertical columns of the frame in Fig. P6.5 have specific
weight g b and are assumed rigid. The elastic support on the ground is simulated
by the rotational springs CR . The horizontal beams are flexible with crosssectional moment of inertia I and modulus of elasticity E while their mass
and axial deformation are assumed negligible. The structure is subjected to
of total duration ttot ¼ 3s. Give
the horizontal ground motion ug ðt Þ ¼ u0 sin wt
the graph of the response spectrum of the rotation of the structure and compute
the extreme values of the shear force and the bending moment of the beams.
Data: Cross-sectional area of the beams A ¼ a=10 a=10, CR ¼ EI =a,
E ¼ 2:1 107 kN=m, g b ¼ 24kN=m3 , u0 ¼ 0:03m, w ¼ 3, 5 and 7s1 .
Response to ground motion and vibration isolation Chapter
6
243
=∞
FIG. P6.5 Frame in problem P6.5
Problem P6.6 A vehicle traveling with a velocity v¼80km/h on a multispan
bridge (Fig. P6.6) is idealized by the SDOF system of Fig. P6.6a. The length
of each span is L ¼ 20m. The deck of the bridge has been permanently
deformed due to creep so that each span can be simulated by a sinusoidal curve
with amplitude h ¼ 5cm. The stiffness of the system is k ¼ 10kN=m and its
damping ratio x ¼ 0:15. The tires of the wheels are assumed undeformable.
The total weight of the vehicle is W ¼ 20kN. (i) Determine the vertical motion
of the vehicle, (ii) Compute the maximum acceleration applied to the passenger
of the vehicle, and (iii) Compute the velocity that causes resonance to the vehicle and gives its extreme amplitude.
(a)
h
h
(b)
L
FIG. P6.6 Vehicle traveling on a bridge in problem P6.6
References and further reading
[1] R.W. Clough, J. Penzien, Dynamic of Structures, second ed., McGraw-Hill, Inc., New York, 1993.
[2] A.K. Chopra, Dynamics of Structures: Theory and Applications to Earthquake Engineering,
Prentice Hall, Englewood Cliffs, NJ, 1995.
[3] M.A. Biot, Vibrations of buildings during earthquake, in: Transient Oscillations in Elastic
System, Aeronautics Department, Calif. Inst. of Tech., Pasadena, CA, 1932. Chapter II in
Ph.D. Thesis No. 259.
244 PART
I Single-degree-of-freedom systems
[4] F.B. Hildebrand, Advanced Calculus for Applications, Prentice Hall, Englewood Cliffs, NJ,
1962.
[5] R.R. Craig Jr., J. Andrew, A.J. Kurdila, Fundamentals of Structural Dynamics, second ed., John
Wiley, New Jersey, 2006.
[6] J.W. Leonard, Tension Structures, McGraw-Hill, New York, 1988.
[7] J.T. Katsikadelis, Finite deformation of cables under 3-D loading: an analytic solution, in:
D.E. Beskos, D.L. Karabalis, A.N. Kounadis (Eds.), Proc. of the 4th National Congress on Steel
Structures, Patras, May 24–25, vol. II, 2002, pp. 526–534.
Chapter 7
Damping in structures
Chapter outline
7.1
7.2
7.3
7.4
7.5
Introduction
Loss of energy due to damping
Equivalent viscous damping
Hysteretic damping
Coulomb damping
7.5.1 Free vibrations with
Coulomb damping
7.5.2 Forced vibrations with
Coulomb damping
7.6 Damping modeling via
fractional derivatives
7.1
245
246
249
250
252
252
255
257
7.6.1 Introduction
7.6.2 The fractional derivative
7.7 Measurement of damping
7.7.1 Free vibration decay
method
7.7.2 Resonance amplitude
method
7.7.3 Width of response curve
method
7.8 Problems
References and further reading
257
258
260
261
262
263
265
267
Introduction
In this chapter, the damping of structures is discussed. Damping appears in all
mechanical systems that perform vibrations. It is the dissipation of energy in a
vibrating structure. The type of energy into which the mechanical energy is
transformed depends on the system and the physical mechanism that causes
the dissipation. The energy is lost either in the form of heat or is radiated into
the environment. For example, the loss of energy in the form of heat is perceived
when an iron rod is subjected to alternating bending. The sound produced by a
body that is hit represents the loss of energy dissipated into the environment. In
the study of vibrations, we are interested in the damping related to the response
of the structure. Damping is due to different energy dissipation mechanisms acting simultaneously. In spite of the age-long detailed studies on the damping of
structures, the understanding of damping mechanisms is quite primitive. A
well-known method to get rid of this problem is to use so-called viscous damping. This approach was first introduced by Rayleigh [1] via his famous dissipation function (see Section 1.8.4).
The loss of energy of a vibrating system reduces the amplitude of the free
vibration. When a system undergoing forced vibrations reaches the phase of
the steady-state response, the loss of energy is balanced by the energy input into
the system by the excitation force.
In vibrating systems, we distinguish different types of damping forces,
which may be due to the internal molecular friction, the sliding friction, or the
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00007-8
© 2020 Elsevier Inc. All rights reserved.
245
246 PART
I Single-degree-of-freedom systems
resistance of the fluid in which the system moves. Sophisticated mathematical
models of damping forces are complex and are not convenient to study the vibrations. For this reason, simplified damping models have been developed that
allow an adequate study of the dynamic response of structures. We have already
used the model of viscous damping, which leads to equations that can be solved
analytically. The presented examples facilitate the comprehension of all concepts. The pertinent bibliography with recommended references for further
study is also included. The chapter is enriched with problems to solved.
7.2 Loss of energy due to damping
The energy loss is determined in the phase of the steady-state response. The graphical representation of the relationship fD ðu Þ between the damping force and the
displacement varies greatly for different types of damping. In all cases, however,
the curve is closed in a complete oscillation and includes a region called the hysteresis loop, whose area is equal to the energy lost per cycle (see also Refs. [2–5]).
In general, the loss of energy is expressed by the integral
Z T
WD ¼
fD ðu Þdu
(7.2.1)
0
where T ¼ 2p=
w the period of the vibration in the steady-state response. The
quantity WD depends on several factors such as temperature, frequency, or
amplitude of the vibration.
In this section, we consider the simplest form of energy loss, that is, the loss
due to viscous damping. The damping force, in this case, is given by the relation
_ Moreover, the displacement in the steady-state phase, due to the
fD ¼ cu.
t, is given by Eq. (3.2.26), namely
harmonic force pðt Þ ¼ p0 sin w
t qÞ
u ðt Þ ¼ rsin ðw
(7.2.2)
p0
1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
k 2 2
1 b + ð2xbÞ2
(7.2.3)
where
r¼
tan q ¼
2xb
1 b2
(7.2.4)
Consequently
t qÞ
u_ ðt Þ ¼ r
w cos ðw
The integral (7.2.1) yields
Z
WD ¼
T
0
(7.2.5)
cu_ 2 dt
¼ c
w2 r 2
Z
¼ pc
wr2
2p=
w
0
t qÞdt
cos 2 ðw
(7.2.6)
Damping in structures Chapter
7
247
Of particular
loss at resonance. Then it is
pffiffiffiffiffiffiffiffiffi interest is the
pffiffiffiffiffiffienergy
ffi
¼ w ¼ k=m , c ¼ 2mxw ¼ 2x km and Eq. (7.2.6) becomes
w
WD ¼ 2xpkr2
(7.2.7)
Moreover, the energy input into the system is due to the work that the force
t produces in a complete oscillation. Namely
pðt Þ ¼ p0 sin w
Z T
pðt Þdu
Wp ¼
0
Z
¼
2p=
w
0
t qÞ sin w
tdt
p0 r
w cos ðw
(7.2.8)
¼ p0 rp sin q
Taking into account that (see Eq. 3.2.27)
2xb
sin q ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 2xbrk=p0
2
1 b2 + ð2xbÞ2
(7.2.9)
we can write Eq. (7.2.8) as
wr2 ¼ WD
Wp ¼ pc
(7.2.10)
Indeed, during the steady-state response, the input into the system by the
excitation force is equal to the dissipated energy due to damping. We come
to the same conclusion if we examine the work produced by the elastic and
inertial forces.
The work produced by the elastic force in a cycle is
Z T
fS ðu Þdu
WS ¼
0
Z
¼
2p=
w
0
r
¼ kw
2
_
ku udt
Z
2p=
w
0
(7.2.11)
t qÞ cos ðw
t qÞdt ¼ 0
sin ðw
Similarly, the work produced by the inertial force is
Z T
fI ðu Þdu
WI ¼
0
Z
¼
0
2p=
w
_
m u€udt
¼ m
w3 r2
Z
2p=
w
0
(7.2.12)
t qÞ cos ðw
t qÞdt ¼ 0
sin ðw
248 PART
I Single-degree-of-freedom systems
Thus, it must be
WI + WD + W S ¼ Wp
(7.2.13)
which by virtue of Eqs. (7.2.11), (7.2.12) yields Eq. (7.2.10).
Eq. (7.2.5) yields
2 r2 cos 2 ðw
t qÞ
u_ ðt Þ2 ¼ w
2 2
t qÞ
r 1 sin 2 ðw
¼w
2 r2 u 2
¼w
(7.2.14)
fD2 ¼ c2 u_ ðt Þ2
2 r2 u 2
¼ c2 w
(7.2.15)
hence
which is readily transformed into
2
fD 2
u
+
¼1
r
c
wr
(7.2.16)
FIG. 7.2.1 Hysteresis loop in a system with viscous damping.
The above relation represents an ellipse with semiaxes r and c
wr in the
plane u, fD , Fig. 7.2.1. The area of the ellipse is pc
wr2 , that is, it is equal to
the energy lost in a cycle. Therefore, the hysteresis loop of a system with viscous
damping is an ellipse.
The properties of damping in the material systems are expressed in
different ways depending on the scientific area where they are applied. Here
we mention two units for measuring the relative dissipated energy that are most
commonly encountered: the specific damping capacity and the specific damping factor.
The specific damping capacity is defined as
wD ¼
WD
U
where U ¼ kr2 =2 is the elastic energy of the system.
(7.2.17)
Damping in structures Chapter
7
249
The specific damping factor is defined as
¼
WD
2pU
(7.2.18)
For viscous damping the above quantities result as
wD ¼
pc
wr2 2pc
w 2pð2mwx Þ
w
¼
¼
¼ 4pxb
mw2
kr2 =2 mw2
¼ 2xb
7.3
(7.2.19)
(7.2.20)
Equivalent viscous damping
When the damping of the structure is not inherently viscous, the curve fD ðu Þ is
not generally an ellipse. It is, however, possible to determine an equivalent viscous damping coefficient ceq by equating the area WD of the experimentally
obtained damping loop in a cycle of harmonic vibration (Fig. 7.3.1) with the
theoretical value of the area of the hysteresis loop representing dissipation of
energy due to viscous damping.
FIG. 7.3.1 Energy dissipated in a cycle of harmonic vibration determined experimentally.
On the basis of Eq. (7.2.6) we write
WD ¼ ceq p
wr2
(7.3.1)
from which we define the equivalent damping coefficient
ceq ¼
WD
p
wr2
(7.3.2)
and the respective damping ratio
ceq
xeq ¼ pffiffiffiffiffiffiffi
2 km
(7.3.3)
250 PART
I Single-degree-of-freedom systems
7.4 Hysteretic damping
When the structure is subjected to cyclic loading, the dissipation of energy
occurs in the interior of the material of the structure. Experiments performed
by many researchers have shown that in most structural materials, such as steel
and aluminum, the energy loss per cycle does not depend on the frequency of the
external excitation force, at least for a wide range of frequencies, but it is proportional to the square of the amplitude of the oscillation. This damping is called
structural damping or hysteretic damping.
The equation of motion for forced vibrations with hysteretic damping is of
the form
m u€ + f ðu Þ ¼ pðt Þ
(7.4.1)
where f ðu Þ is in general a nonlinear function of the displacement. The solution
of Eq. (7.4.1) can be obtained numerically using any of the methods presented in
Chapter 5.
The force f ðu Þ can be expressed as the sum of two forces, an elastic force
with a mean value fS ¼ ku and a damping force
fD ¼
k
u_
w
(7.4.2)
where is the constant of damping. This is the simplest model that can be used
to represent the hysteretic damping.
The loss of energy in a cycle due to hysteretic damping results from
Eq. (7.2.6), if the coefficient c is replaced with k=
w. This yields
WD ¼ kpr2h
(7.4.3)
where rh is the amplitude of the vibration with hysteretic damping. We observe
that the dissipation of energy per cycle does not depend on the excitation fre , but on the square of the amplitude of the vibration.
quency w
Apparently, the function f ðu Þ representing the resisting force can be written
f¼
k
u_ + ku
w
(7.4.4)
and the equation of motion (7.4.1) for harmonic excitation takes the form
m u€ +
k
t
u_ + ku ¼ p0 sin w
w
(7.4.5)
w, which yields
We define the damping coefficient ch ¼ k=
x h ¼ ch =2mw ¼ =2b
(7.4.6)
that is, the damping ratio changes with b.
The solution of Eq. (7.4.5) for the steady-state response is obtained from
Eq. (7.2.2), if x is replaced with xh . Thus
Damping in structures Chapter
t qh Þ
u ðt Þ ¼ rh sin ðw
7
251
(7.4.7)
where
rh ¼
p0
1
ffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
k 1 b2 + 2
tan qh ¼
1 b2
(7.4.8)
(7.4.9)
Working as in Section 7.2, we can determine the geometry of the hysteresis
loop for this type of damping. Thus, we obtain from Eq. (7.4.7)
2 cos 2 ðw
t qh Þ
u_ ðt Þ2 ¼ r2h w
2 2
1 sin 2 ðw
t qh Þ
¼ rh w
2 r2h u 2
¼w
(7.4.10)
Moreover, it is
fD2 ¼ ðf ku Þ2 ¼
k
w
2
u_ 2
(7.4.11)
ðf ku Þ2 ¼ 2 k 2 r2h u 2
(7.4.12)
2
f ku 2
u
+
¼1
krh
rh
(7.4.13)
which by virtue of (7.4.10) is written as
or
Therefore, in the plane u, f , Eq. (7.4.13) represents a rotated ellipse
(Fig. 7.4.1). The area of the ellipse is pkr2h and expresses the loss of energy
in a cycle. We observe that Eq. (7.4.13) does not involve the excitation frequency. This implies that the hysteresis loop can be determined experimentally
using a low excitation frequency, that is, quasistatic, by plotting the loaddisplacement curve.
For f ¼ 0, we obtain the abscissa of the ellipse on the u axis
d ¼ rh pffiffiffiffiffiffiffiffiffiffiffiffiffi
(7.4.14)
1 + 2
which can be used to evaluate the damping coefficient . Hence
d
¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r2h d 2
(7.4.15)
252 PART
I Single-degree-of-freedom systems
FIG. 7.4.1 Hysteresis loop in a system with hysteretic damping.
7.5 Coulomb damping
As mentioned in Section 7.1, damping is due to various energy dissipation
mechanisms acting simultaneously. Although the model of viscous damping
or hysteretic damping in most cases can adequately approximate the behavior
of the structure, it is not suitable to describe the response of the structure when
forces due to Coulomb friction are present. These forces arise on the dry contact
interface of two bodies with a relative sliding. The friction force is expressed by
the relation F ¼ mN , where N is the force normal to the contact surface and m
the friction coefficient, which is determined experimentally and is greater
before the sliding. The direction of the frictional force is opposite to that of
motion, that is, its sign is opposite to the sign of the velocity. Therefore, we
can set
F ¼ signðu_ ÞmN
(7.5.1)
_ ju_ j
signðu_ Þ ¼ u=
(7.5.2)
where
7.5.1 Free vibrations with Coulomb damping
We consider the SDOF system of Fig. 7.5.1a, which is sliding on a dry surface.
When the body is moving to the right (Fig. 7.5.1b) the equilibrium of forces
yields the equation of motion
m u€ + ku ¼ F
(7.5.3)
while when the body is moving to the left (Fig. 7.5.1c) the equilibrium of forces
yields the equation of motion
m u€ + ku ¼ F
(7.5.4)
Damping in structures Chapter
7
253
(a)
(b)
(c)
FIG. 7.5.1 Motion with Coulomb friction.
On the basis of Eq. (7.5.1), Eqs. (7.5.3), (7.5.4) can be combined as
m u€ + ku ¼ F signðu_ Þ
(7.5.5)
m u€ + F signðu_ Þ + ku ¼ 0
(7.5.6)
or
Eq. (7.5.6) is nonlinear because of the term of damping and can be solved by
any of the methods developed in Chapter 5. In the following, an analytical solution of Eq. (7.5.6) is presented with zero initial conditions, u ð0Þ ¼ u0 , u_ ð0Þ ¼ 0.
The body is displaced from the equilibrium position to the right by u0 and we
study the subsequent motion. The body will move to the left according to
Eq. (7.5.4). A particular solution to this equation is
up ðt Þ ¼
F
k
(7.5.7)
Hence the general solution of Eq. (7.5.4) is
u ðt Þ ¼ A cos wt + B sin wt +
F
k
(7.5.8)
Applying the initial conditions
u ð0Þ ¼ u0 , u_ ð0Þ ¼ 0
(7.5.9)
F
A ¼ u0 , B ¼ 0
k
(7.5.10)
gives
and the displacement (7.5.8) becomes
F
F
u ðt Þ ¼ u 0 cos wt +
k
k
(7.5.11)
254 PART
I Single-degree-of-freedom systems
Eq. (7.5.11) holds until the velocity vanishes, namely t ¼ p=w. At that
instant, the body is at the extreme left position, where the displacement is
u ðp=wÞ ¼ ðu0 2F=k Þ. The body will now start moving to the right with initial conditions
p
F
p
u
¼ u0 2
, u_
¼0
(7.5.12)
w
k
w
The motion is now described by Eq. (7.5.3), which has a particular solution
up ðt Þ ¼ F
k
(7.5.13)
and general solution
u ðt Þ ¼ A cos wt + B sin wt F
k
(7.5.14)
The initial conditions (7.5.12) give
A ¼ u0 3F
, B¼0
k
(7.5.15)
and Eq. (7.5.14) becomes
F
F
u ðt Þ ¼ u 0 3
cos wt k
k
FIG. 7.5.2 Free vibrations with Coulomb damping.
(7.5.16)
Damping in structures Chapter
7
255
Eq. (7.5.16) holds until the body reaches the extreme right position, namely
until the instant t ¼ 2p=w. At that time, the body has completed a full oscillation
and the displacement is u ð2p=wÞ ¼ ðu0 4F=k Þ. This solution procedure continues to obtain the response of the next oscillations. The graphical representation of the displacement versus time is shown in Fig. 7.5.2. The curve was
obtained by numerical integration of the equation of motion (7.5.6) with
m ¼ 10kNm1 s2 , T ¼ 0:5s, F ¼ 23:685kN, u0 ¼ 0:55m, u_ 0 ¼ using the program aem_nlin.m developed in Chapter 5. The numerical results coincide with
those obtained by the above-presented analytical solution. The motion is a
vibration with a period T ¼ 2p=w, which means that the Coulomb friction does
not affect the frequency or the period of vibration. The amplitude of vibration is
reduced in each cycle by 4F=k. A consequence of this is that the envelopes of
the curve are straight lines, unlike in the cases of viscous or hysteretic damping
where the envelopes are exponential functions. The motion of the system continues until the elastic force ku becomes smaller than the force F of the friction.
Until now, no difference was made between static friction Fs ¼ ms N and
dynamic friction Fd ¼ md N . The first occurs when the body is stationary and
the second when the body moves. Generally, it is md < ms , hence the dynamic
friction coefficient md will be used in the equation of motion while the static
friction coefficient ms is used for the control of the motion. In viscous or hysteretic damping, theoretically, the body does not stop moving because the
amplitude of the vibration reduces exponentially. Nevertheless, real structural
systems stop after a finite time. This is due to the fact that the Coulomb friction
coexists with other forms of damping and forces the moving systems to stop.
7.5.2
Forced vibrations with Coulomb damping
In this case, the equation of motion becomes
m u€ + F signðu_ Þ + ku ¼ pðt Þ
(7.5.17)
An analytical solution can be achieved by splitting the above equation into
two equations. Thus the equation describing the motion to the right is
m u€ + ku ¼ F + pðt Þ
whose general solution is
u ðt Þ ¼ Acos wt + B sin wt F
1
+
k mw
Z
t
0
pðτÞ sin wðt τÞdτ
(7.5.18)
(7.5.19)
On the other hand, the equation describing the motion to the left is
m u€ + ku ¼ F + pðt Þ
whose general solution is
F
1
u ðt Þ ¼ A cos wt + B sin wt + +
k mw
Z
0
t
pðτÞsin wðt τÞdτ
(7.5.20)
(7.5.21)
256 PART
I Single-degree-of-freedom systems
t the general solution is (see Eq. 3.2.8)
For a harmonic load pðt Þ ¼ p0 sin w
u ðt Þ ¼ A cos wt + B sin wt F p0 1
w
t, b ¼
+
sin w
k
w
k 1 b2
(7.5.22)
Apparently, the derivation of the solution for forced vibrations becomes
quite complicated if we follow the procedure applied for the free oscillations,
that is, by splitting the equation of motion into two equations. Nevertheless, for
small damping, we can approximate the solution in the phase of the steady-state
response. Thus, the solution is given by Eq. (7.2.2).
The graph of the friction force versus displacement in one cycle takes the
form of the rectangle of Fig. 7.5.3. Hence, the work produced by the friction
force in a complete oscillation is
WD ¼ 4Fr
(7.5.23)
We can determine an equivalent coefficient of viscous damping by equating
the loss of energy (Eq. 7.5.23) with that of the viscous damping given by
Eq. (7.3.1). Namely,
FIG. 7.5.3 Hysteresis loop in a system with Coulomb damping.
ceq p
wr2 ¼ 4Fr
(7.5.24)
which gives
4F
p
wr
(7.5.25)
ceq
2F
¼
2mw pkrb
(7.5.26)
ceq ¼
and an equivalent damping ratio
xeq ¼
Damping in structures Chapter
7
257
Substituting this value of the damping ratio in Eq. (7.2.3) yields
r¼
p0
1
ffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
k 2
2
1 b + ð4F=pkrÞ
(7.5.27)
which is solved for r to give
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u
p0 u1 ð4F=pp0 Þ2
r¼ t 2
k
1 b2
(7.5.28)
Eq. (7.5.28) holds if 1 ð4F=pp0 Þ2 > 0, that is, when F=p0 < p=4. Obviously, for F=p0 > p=4 r becomes imaginary and this method for determining
an equivalent damping coefficient does not apply.
The phase angle results from Eq. (7.2.4) by setting x ¼ xeq and taking the
value of r from Eq. (7.5.28). Thus, we have
4F=pp0
tan q ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 ð4F=pp0 Þ2
(7.5.29)
The plus sign is valid when b < 1 while the minus sign is valid when b > 1.
7.6
7.6.1
Damping modeling via fractional derivatives
Introduction
In the modeling of the linear elastic behavior of structures, damping has been
classically modeled as being linearly dependent on velocity. This damping
model is adequate for the dynamic analysis of lightly damped structures. However, this model, while mathematically straightforward, fails to describe the
damping response of heavily damped structures. This drawback is more pronounced in materials used in devices aimed at increasing the damping of structures, such as seismic isolators.
Historically, the need for more refined models to control damping has
pushed researchers to use viscoelastic models in the dynamic analysis of structures. Applicable viscoelastic models are expressed as a series of ordinary time
derivatives relating time-dependent stress to strain fields. These models yield
acceptable plots of material properties, but they have drawbacks. Normally,
these models contain many terms, making them mathematically cumbersome
and increasing the order of the differential equations describing the response
of the system. An alternative representation of these models is realized via
fractional derivatives, that is, derivatives of noninteger order [6]. The real
strength of this approach is that the fractional derivative models describe
258 PART
I Single-degree-of-freedom systems
damping effects using fewer material parameters than the integer order differential models but with equal precision. This approach involves fractional differential equations, which needs acquaintance with fractional calculus.
7.6.2 The fractional derivative
The fractional (noninteger order) derivatives are as old as calculus. The theory
of derivatives of noninteger order goes back to G. W. Leibnitz. After Leibnitz
defined the derivative of integer order, d n y=dx n with n being an integer,
L’H^
opital asked: “What if n is a fraction, say n ¼ 1=2?” Leibnitz gave an
answer to this question (Sept. 30, 1695) concluding “It will be a paradox”
and added prophetically “From this apparent paradox, one day useful consequences will be drawn” [7]. For three centuries, the fractional derivative
inspired pure theoretical mathematical developments useful only for mathematicians. The integer order derivative allows giving geometrical interpretations to
the proposed physical models resulting from Newton’s law of motion. Apparently, this made the then-revolutionary concepts accessible to contemporary
scientists, who were well experienced in geometry. In a sense, the long delay
to apply fractional calculus may be attributed to this fact. We recall that fractional derivatives lack the straightforward geometrical interpretation of their
integer counterparts. However, research carried out in recent years has pointed
out that fractional order derivatives provide an effective tool to reliably model
many complex physical and engineering systems. In recent years, this fact gave
a great boost to the study of fractional differential equations. In this context,
many books on fractional calculus as well as numerous publications on the
study of physical and engineering systems via fractional derivatives have been
written; see for example [8] and the references therein.
There are several definitions of the fractional derivative [8]. The RiemannLiouville and Caputo fractional derivatives are among the most widely used.
Nevertheless, as the scope of this section is not to give a detailed account of
this subject, we will restrict our discussion to the Caputo derivative, defined as
Z t
1
u ðmÞ ðτÞ
DCa u ðt Þ ¼
dτ, m 1 < a < m
(7.6.1)
Gðm aÞ 0 ðt τÞa + 1m
where Gðz Þ is the gamma function with argument z, a is the order of the fractional derivative, and m an integer. It can be shown that
lim DCa u ðt Þ ¼ u ðmÞ ðt Þ
a!m
lim DCa u ðt Þ ¼ u ðm1Þ ðt Þ u ðm1Þ ð0Þ
a!m1
(7.6.2)
Obviously, for m ¼ 1, the fractional derivative interpolates the first-order
derivative [9]. Fig. 7.6.1 shows the fractional derivative of the function
u ¼ t t 3 =6 + t 5 =120 for various orders of the fractional derivative.
Damping in structures Chapter
7
259
1.2
D 1u
1
c
D 0.99999u
0.8
c
D 0.9u
0.6
c
D 0.5u
c
0.4
D 0.1u
c
0.2
D 0.00001u
0
u
c
−0.2
−0.4
−0.6
0
0.5
1
1.5
2
2.5
3
FIG. 7.6.1 The fractional derivative of the function u ¼ t t 3 =6 + t 5 =120 for various orders.
If damping is modeled via the fractional derivative, the equation of motion
of a SDOF system is written as
m u€ + cD aC u + ku ¼ pðt Þ
(7.6.3)
Obviously, for a ! 1, we obtain the equation of motion with viscous
damping
m u€ + cu_ + ku ¼ pðt Þ
(7.6.4)
while for a ! 0 the equation of motion becomes
m u€ + ðc + k Þu ¼ pðt Þ + cu 0
(7.6.5)
that is, it yields the elastic response with stiffness k ∗ ¼ c + k and excitation
force p∗ ¼ pðt Þ + cu 0 . The Caputo derivative is employed because, contrary
to other types of fractional derivatives, it allows the application of initial conditions having a direct physical significance.
Eq. (7.6.3) represents a fractional differential equation. Analytical solutions
of such equations are difficult or impossible to obtain. This reason has recently
boosted the development of efficient numerical methods for solving fractional
differential equations [10, 11].
Fig. 7.6.2 shows the free vibration response of an DOF system for various
values of the order a
260 PART
I Single-degree-of-freedom systems
FIG. 7.6.2 Free vibration response of a SDOF system for various values of the order a.
The solution of Eq. (7.6.3) has been obtained using the method presented
in [11]. The developed MATLAB program has been given the name
three_term_FD.m and is included on this book’s companion website. Note
that for a 1 and a 0 the solutions are identical to the corresponding
analytical ones.
The fractional calculus has allowed the definition of any order of fractional
derivative, real or imaginary. This fact enables us to consider the fractional
derivative to be a function of time (explicit variable-order fractional derivative)
or of some other time-dependent variable (implicit variable-order fractional
derivative). Thus, the variable-order Caputo derivative for m ¼ 1 reads
Z t
1
u_ ðτÞ
aðt Þ
DC u ð t Þ ¼
dτ ðExplicitÞ
(7.6.6a)
Gð1 aðt ÞÞ 0 ðt τÞaðt Þ
Z t
1
u_ ðτÞ
aðu Þ
dτ ðImplicitÞ
(7.6.6b)
DC u ð t Þ ¼
Gð1 aðu ÞÞ 0 ðt τÞaðu Þ
Z t
1
u_ ðτÞ
aðu_ Þ
DC u ð t Þ ¼
dτ ðImplicitÞ
(7.6.6c)
Gð1 aðu_ ÞÞ 0 ðt τÞaðu_ Þ
The concept of a variable-order fractional derivative exhibits notable advantages over the constant order derivative and it has been recently used to model
the dynamic response of actual structures [12, 13].
7.7 Measurement of damping
As stated in Chapter 1, the mass and stiffness of a dynamic system can be determined from its physical characteristics. The mass can be determined from the
geometry and the mass density of the structural elements. The stiffness is also
Damping in structures Chapter
7
261
determined if the geometry and material properties of the structural elements
are known. However, it is difficult or at least impractical to relate damping
to known or measurable physical characteristics of the system. Therefore, the
damping of a given structure cannot be measured precisely during the design
phase, but only experimentally after its construction. There are several experimental techniques to measure the damping of a structure. Most of them are
based on the assumption that the damping is viscous. When damping is not viscous in nature, an equivalent viscous damping is usually determined.
7.7.1
Free vibration decay method
This is the simplest and most frequently used method of determining the viscous
damping ratio x through experimental measurements. This method was already
presented in Chapter 2 when we discussed the free damped vibrations. The system is excited by means of an appropriate experimental instrumentation and is
left to perform free oscillations. Subsequently, the peak amplitudes ui and ui + n
over n consecutive cycles are measured. As shown in Section 2.3.2, the damping ratio can be calculated using Eq. (2.3.24), namely
2npx
ui
pffiffiffiffiffiffiffiffiffiffiffiffi ¼ ln
(7.7.1)
ui + n
1 x2
which yields
dn
x ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
4p n 2 + d n 2
where
dn ¼ ln
ui
ui + n
(7.7.2)
(7.7.3)
This method was illustrated in Example E6.2.1.
When the damping is hysteretic, the damping force is given by Eq. (7.4.2). In
w, provided that the damping is small. Hence,
free vibrations, it can be set as w
the equivalent damping coefficient ch and the damping ratio are obtained from
the relations
k
w
xh ¼
2
ch ¼
(7.7.4a)
(7.7.4b)
The hysteretic damping coefficient is obtained by using Eq. (7.7.2), if it is
set x ¼ xh . This yields
262 PART
I Single-degree-of-freedom systems
2dn
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
4p n 2 + dn 2
(7.7.5)
Example 7.7.1 Evaluation of damping in a SDOF system
In a test, the silo in Example 2.3.1 is subjected to harmonic excitation at a frequency 15s1 and the force-displacement relation is obtained at the steady state.
The amplitude of the displacement is measured as 10 cm, and the energy loss per
cycle 50kNm. If damping is considered to be viscous, determine c and x.
Solution
The mass of the system is m ¼ 100kNm1 s2 , the stiffness k ¼ 2721:6kN=m,
and frequency w ¼ 5:2169s1 , taken from Example 2.3.1.
(i) For viscous damping, the energy loss per cycle is given by Eq. (7.2.6),
namely
WD ¼ pc
wr2
(1)
1
¼ 15s , r ¼ 0:1m, WD ¼ 50kNm gives
which for w
WD
50
¼
¼ 106:10
p
wr2 p 15 0:12
(2)
c
106:10
¼
¼ 0:102
2mw 2 100 5:2169s1
(3)
c¼
x¼
7.7.2 Resonance amplitude method
t. In the phase of
The system is subjected to harmonic excitation pðt Þ ¼ p0 sin w
the steady-state response, the amplitude and the phase angle are given by
Eq. (3.2.27), namely
i
2
p0 h
2
1 b2 + ð2xb Þ2
k
2xb
q ¼ tan 1
1 b2
1
r¼
(7.7.6a)
(7.7.6b)
For b ¼ 1 Eq. (7.7.6b) yields q ¼ 90° regardless of the value of x. If the
employed experimental instrumentation allows the measurement of the phase
angle q, then we adjust the excitation frequency so that q ¼ 90° and we measure
the amplitude of vibration r. Besides, setting b ¼ 1 in Eq. (3.2.28) we obtain
D ¼ max jRðt Þj ¼
which yields
r
1
¼
p0 =k 2x
(7.7.7)
Damping in structures Chapter
x¼
p0 =k
2r
7
263
(7.7.8)
This previous method requires knowledge of the stiffness k of the structure,
which is determined either from the physical characteristics of the structure or
experimentally, for example, by imposing a load and measuring the resulting
displacement.
If the measurement of the phase difference is not easy, then we measure
experimentally the amplitude of the vibration in the range of resonance. Subsequently, we plot the curve Dðb Þ ¼ rðb Þ=ðp0 =k Þ (see Fig. 7.7.1) and determine
its maximum. Then Eq. (3.2.31), namely
Dmax ¼
1
pffiffiffiffiffiffiffiffiffiffiffiffi
2x 1 x 2
(7.7.9)
FIG. 7.7.1 Graphical representation of the curve D ðbÞ in the range of resonance.
allows the evaluation of x. For small values of x, it is x 2 0, hence
x 1=2Dmax . This method also requires knowledge of the stiffness k of the
structure.
7.7.3
Width of response curve method
The width of the response curve DðbÞ in the range of resonance can be used to
determine the damping of the structure. In this method, the frequencies corresponding to the phase angles 45° are measured. One of these frequencies is
264 PART
I Single-degree-of-freedom systems
below the resonance frequency while the other is above it (see Fig. 7.7.2). The
respective values of b are obtained from Eq. (7.7.6b). Thus, we have
2xb1
¼1
1 b21
(7.7.10a)
2xb2
¼ 1
1 b22
(7.7.10b)
1 b21 2xb1 ¼ 0
(7.7.11a)
1 b22 2xb2
(7.7.11b)
which can be written as
¼0
D
FIG. 7.7.2 Response curve DðbÞ in the neighborhood of resonance to determine bI and b II .
Subtracting the previous equations gives
1
2 w
1
1w
x ¼ ðb 2 b 1 Þ ¼
2
2 w
(7.7.12)
1 and w
2 have
Eq. (7.7.12) can be used to calculate x when the frequencies w
been measured and the natural frequency w is known or can be determined.
The previously described method is based on the capability of measuring the
phase angle, which may not be simple because it requires complex instrumentation. However, another property of the response curve may be used to determine the damping ratio. This method is not based on the measurement of the
phase angle, but on experimental measurements. The response curve D ðb Þ is
plotted in the range of resonance and the values b corresponding to
Damping in structures Chapter
7
265
pffiffiffi
1= 2Dmax are determined. It is apparent from Fig. 7.7.2 that these values are
two, which are denoted by bI and bII . For small values of the damping ratio, it is
Dmax 1=2x. Besides, using Eq. (3.2.28) we can write
1 1
1
pffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2x
2
1 b2 + ð2xb Þ2
The solution of the above equation gives two values of b
qffiffiffiffiffiffiffiffiffiffiffiffi
2
2
b I ¼ 1 2x 2x 1 + x2
(7.7.13)
(7.7.14a)
qffiffiffiffiffiffiffiffiffiffiffiffi
b 2II ¼ 1 2x2 + 2x 1 + x2
(7.7.14b)
pffiffiffiffiffiffiffiffiffiffiffiffi
Using the binomial theorem to expand 1 + x 2 and neglecting terms of
order higher than the second, the above relations reduce to
b2I 1 2x 2x 2
(7.7.15a)
b2II 1 + 2x 2x 2
(7.7.15b)
or after expanding in Taylor series
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3
bI 1 2x 2x2 1 x x 2
2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3
b II 1 + 2x 2x2 1 + x x 2
2
(7.7.16a)
(7.7.16b)
which give
1
x ¼ ðbII b I Þ
2
(7.7.17)
Eq. (7.7.17) is similar to Eq. (7.7.12), but avoids the measurement of the
phase angles.
7.8
Problems
Problem 7.1 The damping force applied to a system moving in a fluid is given
by the relation fD ¼ lu_ a , where l and a are real constants. Give:
(i) The graphical representation of the displacement u ðt Þ when the system
performs free vibrations with initial conditions u ð0Þ ¼ 0:02m, u_ ð0Þ ¼ 0.
(ii) The graphical representation of the displacement u ðt Þ when the system is
t with zero initial conditions.
subjected to the harmonic load pðt Þ ¼ p0 sin w
Data: m ¼ 10kNm1 s2 , k ¼ 1500kN=m, l ¼ 100, a ¼ 3, p0 ¼ 300kN, and
¼ 2s1 .
w
266 PART
I Single-degree-of-freedom systems
Problem 7.2 The system of Fig. P7.2 is subjected to Coulomb damping. Give:
(i) The graphical representation of the displacement u ðt Þ when the system
performs free vibrations with initial conditions (a) u ð0Þ ¼ 0:40m,
u_ ð0Þ ¼ 0 and (b) u ð0Þ ¼ 0, u_ ð0Þ ¼ 2:8ms1 .
(ii) The graphical representation of the displacement u ðt Þ when the system is
t with zero initial conditions.
subjected to the harmonic load pðt Þ ¼ p0 sin w
(iii) Calculate the equivalent damping ratio xeq of the equivalent viscous
damping.
=w ¼ 0:4, N ¼ 70kN,
Data: m ¼ 10:132kNm1 s2 , k ¼ 1600kN=m, b ¼ w
m ¼ 0:32, and p0 ¼ 1:57F.
FIG. P7.2 System subjected to Coulomb damping.
Problem P7.3 Two bodies B1 and B2 with masses m1 and m2 , respectively, are
placed on two inclined planes whose angles are f1 and f2 , as shown in
Fig. P7.3. The bodies are connected by a massless cable of length L and axial
stiffness k ¼ EA=L. The friction coefficient between the bodies and the inclined
planes is m while between the cable and the pulley it is zero. Determine the
equation of motion of the system.
FIG. P7.3 System in problem P7.3.
Problem P7.4 The damping of a SDOF system is expressed by the Caputo fractional derivative of order a ¼ 0:5. Compare the response of the system with that
of viscous damping. Data: m ¼ 10kNm1 s2 , x ¼ 0:1, k ¼ 500kN=m,
p ¼ p0 sin 5t, and p0 ¼ 2kN. Hint: Use the program three_term_FD.m available on this book’s companion website.
Damping in structures Chapter
7
267
References and further reading
[1] L. Rayleigh, Theory of Sound, second ed., Dover Publications, New York, 1877; reissued
1945.
[2] R.W. Clough, J. Penzien, Dynamics of Structures, second ed., McGraw-Hill, New York, 1993.
[3] A.K. Chopra, Dynamics of Structures: Theory and Applications to Earthquake Engineering,
Prentice Hall, Englewood Cliffs, NJ, 1995.
[4] J.L. Humar, Dynamics of Structures, second ed., A.A. Balkema Publishers, Lisse, NL, 2002.
[5] S. Adhikari, Damping Models for Structural Vibration, Dissertation submitted to the University of Cambridge for the Degree of Doctor of Philosophy, Trinity College, Cambridge, 2000.
[6] R.L. Bagley, P.J. Torvik, Fractional calculus—a different approach to the analysis of viscoelastically damped structures, AIAA J. 27 (1998) 1412–1417.
[7] G.W. Leibniz, S€amtliche Schriften und Briefe, 3te Reihe, Mathematischer, Naturwissenschaftlicher und Technischer Briefwechsel, Band VI, Letter 163, 2004, p. 510. BerlinBrandenburgischen Akademie der Wissenschaften und Akademie der Wissenschaften in
G€
ottingen.
[8] I. Podlubny, Fractional Differential Equations, Academic Press, New York, 1999.
[9] J.T. Katsikadelis, Generalized fractional derivatives and their applications to mechanical systems. Arch. Appl. Mech. 85 (2015) 1307–1320, https://doi.org/10.1007/s00419-014-0969-0.
[10] K. Diethelm, N.J. Ford, A.D. Freed, Y. Luchko, Algorithms for fractional calculus: a selection
for numerical Methods, Comput. Methods Appl. Mech. Eng. 194 (2005) 743–773.
[11] T. Katsikadelis, Numerical Solution of multi-term fractional differential equations, ZAMM,
Z. Angew. Math. Mech. 89 (7) (2009) 593–608.
[12] C.F.M. Coimbra, Mechanics with variable-order differential operators, Ann. Phys. (Leipzig)
12 (11–12) (2003) 692–703, https://doi.org/10.1002/andp.200310032.
[13] J.T. Katsikadelis, Numerical solution of variable order fractional differential equations,
arXiv:1802.00519 [math.NA] (2018).
Chapter 8
Generalized single-degree-offreedom systems—Continuous
systems
Chapter outline
8.1 Introduction
8.2 Generalized single-degree-offreedom systems
8.3 Continuous systems
8.3.1 Introduction
8.3.2 Solution of the beam
equation of motion
8.1
269
275
284
284
285
8.3.3 Free vibrations of beams
8.3.4 Orthogonality of the
free-vibration modes
8.3.5 Forced vibrations of
beams
8.4 Problems
References and further reading
286
291
293
295
297
Introduction
In this chapter, the method of global shape functions is employed to approximate the response of continuous systems by SDOF systems, which we call
generalized SDOF systems. The example that follows helps in understanding
the basic ingredients of the method as well as the error introduced by the
lumped mass assumption.
In order to study the dynamic response of the frame shown in Fig. 8.1.1a, we
approximate it by the model of Fig. 8.1.1b. As mentioned, in formulating this
model, it was assumed that the mass of the columns is concentrated at their ends,
that is, half at the top and half at the foot of the column. The consequence of this
assumption is that the elastic curve of the columns has the form of an unloaded
beam fixed at both ends, whose end cross-sections undergo a relative displacement u ðt Þ. This assumption is, however, not entirely correct because the mass of
the columns is actually distributed along their length. This fact, apparently, pro u€
ðx, t Þ, where uðx, t Þ is the actual elastic curve
duces inertial forces fI ðx, t Þ ¼ m
of the columns (see Fig. 8.1.1c). Thus, the problem would be correctly
addressed if the columns were treated as systems of infinite degrees of freedom,
that is, continuous systems whose top cross-sections are connected by the rigid
beam of mass m. Such an approach, however, would be quite difficult because
the analysis leads to partial differential equations (see Section 1.1). Nevertheless, it is possible to approximate the system by another model, which is closer
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00008-X
© 2020 Elsevier Inc. All rights reserved.
269
270 PART
I Single-degree-of-freedom systems
to the actual system than that of Fig. 8.1.1b. In this model, we consider that the
mass is distributed along the length of columns, but their deformed shape is chosen so that it is close to the actual one. We observe that the end cross-sections of
the columns in Fig. 8.1.1a do not rotate during the motion. Hence the shape function ðx Þ ( see Eq. (8.1.1)) representing the shape of the elastic curve should be
chosen so that it satisfies the geometrical boundary conditions at the ends of the
columns, which demand ð0Þ ¼ 0, 0 ð0Þ ¼ 0, ðh Þ 6¼ 0, and 0 ðh Þ ¼ 0. The functions ðx Þ are not unique. They constitute an infinite set of functions called
geometrically admissible functions.
(a)
(b)
(c)
FIG. 8.1.1 Model of a two-column frame with rigid beam.
After this consideration, we may set
uðx, t Þ ¼ ðx Þu ðt Þ
(8.1.1)
where u ðt Þ is a time-dependent function denoting the displacement of a
certain cross-section of the column, say at x ¼ a, 0 < a h. This yields
uða, t Þ ¼ ða Þu ðt Þ ¼ u ðt Þ, which implies that ða Þ ¼ 1. If it is taken a ¼ h, then
u ðt Þ ¼ uðh, t Þ represents the displacement at the top of the column.
Eq. (8.1.1) states that the shape of the elastic curve remains the same
during motion while its values are multiplied by the common factor u ðt Þ.
The function ðx Þ, which approximates the elastic curve, is referred to as shape
function.
In the example of Fig. 8.1.1, a shape function for the columns is the elastic
curve of a beam with constant stiffness EI , whose ends are subjected to a relative displacement equal to one. Therefore, the shape function is obtained from
the solution of the differential equation of the elastic curve of the beam in the
absence of loading (Fig. 8.1.2a)
EI
d 4 ðx Þ
¼0
dx 4
(8.1.2)
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 271
(a)
(b)
FIG. 8.1.2 Shape function (a) and and deflection curve (b) in Example 8.1.1
with boundary conditions
ð0Þ ¼ 0,
0
ð0Þ ¼ 0,
ðh Þ ¼ 1,
0
ðh Þ ¼ 0
(8.1.3)
Integrating Eq. (8.1.2) yields
ðx Þ ¼ c 1
x3
x2
+ c2 + c3 x + c4
6
2
(8.1.4)
where ci (i ¼ 1, 2, 3, 4) are arbitrary constants.
Applying now the boundary conditions (8.1.3), we obtain
c1 ¼ 12
,
h3
c2 ¼
6
,
h2
c3 ¼ 0, c4 ¼ 0
(8.1.5)
Then substituting these values for the arbitrary constants into Eq. (8.1.4)
yields the shape function
x 2
x 3
ðx Þ ¼ 3
2
¼ 3x2 2x3 , x ¼ x=h
(8.1.6)
h
h
After the selection of the shape function, the motion of the system is determined by the time-dependent function u ðt Þ. The equation of motion can be
derived using different methods such as Hamilton’s principle, Lagrange equations, or the principle of virtual work. To implement these methods, it is necessary to determine the elastic energy U and the kinetic energy T of the system
as well as the virtual work of the external force pðt Þ.
(a) Evaluation of the elastic energy U . The elastic energy is due to the strain
energy of the two columns because of bending. The strain energy per unit
volume is
1
U0 ¼ s x e x
2
(8.1.7)
272 PART
I Single-degree-of-freedom systems
where sx is the normal stress on the cross-section of the column and ex the corresponding strain. For a linearly elastic material with modulus of elasticity E,
it is ex ¼ sx =E. Taking into account that the bending stress in a beam is
sx ¼ Mx y=I ðx Þ, we obtain
1 Mx y 2
U0 ¼
(8.1.8)
2E I ðx Þ
where I ðx Þ is the moment of inertia of the, in general, variable cross-section
and Mx the bending moment. Hence, the total elastic energy of the one
column is
Z
U
¼ U0 dV
2
V
(8.1.9)
Z
1 Mx y 2
¼
dV
V 2E I ðx Þ
where V is the volume of the column. It is known from the beam theory
that
Mx ¼ EI ðx Þu00 ðx, t Þ
which is introduced into Eq. (8.1.9) to yield
Z
U 1 h
2
¼
EI ðx Þ½u00 ðx, t Þ dx
2 2 0
(8.1.10)
(8.1.11)
For I ðx Þ ¼ I ¼ constant and uðx, t Þ ¼ ðx Þu ðt Þ, the previous equation
becomes
Z h
U EI
2
2
¼ ½u ðt Þ
½ 00 ðx Þ dx
(8.1.12)
2
2
0
Differentiating Eq. (8.1.6) twice with respect to x gives
00
ðx Þ ¼
6
ð1 2xÞ
h2
(8.1.13)
Substituting Eq. (8.1.13) into Eq. (8.1.12) and integrating give
U 1 12EI 2
¼
u
2 2 h3
(8.1.14)
Therefore, the elastic energy of the two columns is
U¼
12EI 2
u
h3
(8.1.15)
12EI
udu
h3
(8.1.16)
and its variation
dU ¼ 2
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 273
(b) Evaluation of the kinetic energy T . The kinetic energy is due to the velocity
u_ ðt Þ of the mass m of the horizontal beam as well as to the velocity u_ ðx, t Þ
at points x of the column axis. Thus, we have
of the mass m
Z h
1
1
2
2
½u_ ðx, t Þ dx
m
(8.1.17)
T ¼ m u_ + 2
2
2
0
By virtue of Eq. (8.1.1), the above relation is written as
Z h
1
u_ 2
½ ðx Þ2 dx
T ¼ m u_ 2 + m
2
0
(8.1.18)
Then using Eq. (8.1.6) and integrating give
1
u_ 2
T ¼ m u_ 2 + 0:371mh
2
(8.1.19)
Þud
_ u_
dT ¼ ðm + 0:742h m
(8.1.20)
and
(c) Evaluation of the virtual work dWnc . This is due to the nonconservative
external force pðt Þ. This is
dWnc ¼ pðt Þdu
(8.1.21)
Now we proceed to the derivation of the equation of motion using:
1. Hamilton’s principle. Substituting Eqs. (8.1.16), (8.1.20), (8.1.21) into
Eq. (1.7.13) and taking into account that A ¼ 0, we obtain
Z t2 EI
Þud
_ u_ pðt Þdu dt ¼ 0
24 3 udu ðm + 0:742h m
(8.1.22)
h
t1
which after elimination of du_ using integration by parts yields the equation of motion
m ∗ u€ + k ∗ u ¼ p∗ ðt Þ
(8.1.23)
where
m ∗ ¼ m + 0:741mh,
k∗ ¼
24EI
, p∗ ðt Þ ¼ pðt Þ
h3
(8.1.24)
2. Lagrange equations. The equation of motion will result from
Eq. (1.8.11) by setting qi ¼ u, A ¼ 0 and Qi ¼ pðt Þ. This gives
d ∂T
∂T ∂U
+
¼ pðt Þ
(8.1.25)
dt ∂u_
∂u
∂u
which by virtue of Eqs. (8.1.15), (8.1.19) yields Eq. (8.1.23).
274 PART
I Single-degree-of-freedom systems
3. The principle of virtual work. The principle of virtual work stated in
Section 1.6 for rigid bodies can also be extended to deformable bodies,
but now we must take account of the work done by the internal forces in
riding through the virtual deformation. Thus, for a deformable body, the
principle of virtual work becomes
dWex ¼ dWin
(8.1.26)
where dWex and dWin denote the virtual work of the external and internal forces, respectively.
For the system of Fig. 8.1.1, the external forces are the excitation
force pðt Þ and the inertial forces. Hence
Z
h
€ 2
dWex ¼ m udu
u€
ðx, t Þd uðx, t Þdx + pðt Þdu
m
0
which by virtue of Eqs. (8.1.1), (8.1.6) becomes
€ 0:742mhdu
dWex ¼ m udu
+ pðt Þdu
(8.1.27)
The virtual work of the internal forces for the two columns is
obtained from the relationa
Z
dWin ¼ 2 sx dex dV
(8.1.28)
V
where V is the volume of the column.
Taking into account that sx ¼ Mx y=I , sx ¼ Eex , Mx ¼ EI u00 ðx, t Þ,
we obtain from Eq. (8.1.28)
dWin ¼ 2
12EI
udu
h3
(8.1.29)
Substituting Eqs. (8.1.27), (8.1.29) into Eq. (8.1.26) yields the equation
of motion (8.1.23).
a
The expression for the strain energy of a beam due to bending is derived from the strain energy
of the beam by taking its variation ([1], Chap. 1):
Z
Z
1
1
Win ¼
sx ex dV ¼
Es2 dV
(a)
2 V
2 V x
which by taking the variation gives
Z
dWin ¼
Z
Eex dex dV ¼
V
sx dex dV
V
(b)
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 275
FIG. 8.1.3 Dependence of T*/T on the ratio of the column mass over total beam mass in the frame
of Fig. 8.1.1
If the mass of the column is assumed concentrated at the ends of the columns, the coefficients of the corresponding equation of motion result as
m ∗ ¼ m + mh,
k∗ ¼
24EI
, p∗ ðt Þ ¼ pðt Þ
h3
(8.1.30)
Fig. 8.1.3
We observe that the generalized mass is less by 0:259mh.
where T ∗
shows the variation of the ratio T ∗ =T versus the ratio mh=m,
is the natural period of the generalized single-degree-of-freedom system
and T the the period of the model in Fig. 8.1.1b. We observe that the lumped
mass assumption has a small influence on the natural period when the mass
of the columns with respect to the mass of the horizontal beam is small.
Illustrative examples facilitating the comprehension of all concepts are presented and the pertinent bibliography with recommended references for further study is also included. The chapter is enriched with problems to be
solved.
8.2
Generalized single-degree-of-freedom systems
After the introductory example in the previous section, we can extend the discussed method to approximate the response of a more complex system with a generalized SDOF system. When the members of the system are undeformed, the
substitute SDOF system represents the actual response of the system. Such examples were discussed in Chapter 1. Example 1.7.3 is a representative case. The
shape function of that example is shown in Fig. E1.11. However, when the members connecting the lumped masses are deformable, then the generalized SDOF
system simulating the actual one can be used to approximate its response. The
accuracy of the approximation depends on the selection of the shape function.
276 PART
I Single-degree-of-freedom systems
Nevertheless, the method allows us to approximate easily the dynamic characteristics of a complex system, thus circumventing exact solution methods.
To illustrate the application of the method to complex systems, we consider
the beam of Fig. 8.2.1a, which is fixed at point 1 and simply supported at point 2.
Moreover, the beam rests on the nonhomogeneous Winkler’s type elastic foundation with variable reaction modulus k ðx Þ. The foundation reacts also with distributed linear damping having coefficient cðx Þ. The cross-section of the beam
is variable, hence I ¼ I ðx Þ and m ¼ m ðx Þ. The beam caries the lumped masses
mi with rotational inertia Ii at points x ¼ xi , i ¼ 1, 2, …,N . The beam is loaded
by the distributed load pðx, t Þ and the axial load P at the end cross-section 2.
The beam may also be subjected to concentrated loads and/or concentrated and
distributed moments. However, for the convenience of the illustration of the
method, these loads are not considered.
To treat the structure as a generalized SDOF, we will seek the transverse
deflection in the form of Eq. (8.1.1), that is, uðx, t Þ. The shape function ðx Þ
will be selected so that
ð0Þ ¼ 0,
0
ð0Þ ¼ 0,
ðLÞ ¼ 0,
0
ðLÞ 6¼ 0
(8.2.1)
The deflection curve of the beam fixed at one end and simply supported at
the other is a geometrically admissible function and thus can be used as a shape
function. Taking ðL=2Þ ¼ 1, we may write
uðx, t Þ ¼ ðx Þu ðt Þ
(8.2.2)
Obviously, it is uðL=2, t Þ ¼ u ðt Þ, hence u ðt Þ represents the displacement at
the middle of the beam.
We will use Hamilton’s principle to derive the equation of motion of
the generalized SDOF system. For this purpose, we evaluate the quantities
U , K , dWnc , and A.
(a)
(b)
(c)
(d)
FIG. 8.2.1 Beam resting on foundation with nonhomogeneous elastic and damping reaction.
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 277
(a) Evaluation of the elastic energy U . This energy consists of the strain energy
of the beam due to bending and the elastic energy of the springs of Winkler’s
model. That is,
Z
Z
1 L
1 L
2
00
EI ðx Þ½u ðx, t Þ dx +
k ðx Þ½uðx, t Þ2 dx
(8.2.3)
U¼
2 0
2 0
which yields
Z L
Z
dU ¼
EI ðx Þu00 ðx, t Þd u00 ðx, t Þdx +
0
L
k ðx Þuðx, t Þd uðx, t Þdx
(8.2.4)
0
Eq. (8.1.1) gives duðx, t Þ ¼ ðx Þdu, du00 ðx, t Þ ¼
Eq. (8.2.4) becomes
00
ðx Þdu. Thus,
dU ¼ k ∗ udu
(8.2.5)
where
Z
k∗
L
¼
00
EI ðx Þ½
Z
2
L
ðx Þ dx +
0
k ðx Þ½ ðx Þ2 dx
(8.2.6)
0
(b) Evaluation of the kinetic energy T . This energy consists of the kinetic
energy of the distributed mass m ðx Þ and the kinetic energy of the lumped
masses mi .That is
Z
N
N
h
i2
1 L
1X
1X
2
2
0
T¼
m ðx Þ½u_ ðx, t Þ dx +
mi ½u_ ðxi , t Þ +
Ii u_ ðxi , t Þ
2 0
2 i¼1
2 i¼1
(8.2.7)
0
where u_ ðx, t Þ and u_ ðx, t Þ are the transverse velocity and the angular velocity, respectively, at the cross-section x. Eq. (8.2.7) gives
Z L
N
X
dT ¼
m ðx Þu_ ðx, t Þdu_ ðx, t Þdx +
mi u_ ðxi , t Þdu_ ðxi , t Þ
0
i¼1
(8.2.8)
N
X
0
0
+
Ii u_ ðxi , t Þdu_ ðxi , t Þ
i¼1
which by virtue of Eq. (8.1.1) results in
_ u_
dT ¼ m ∗ ud
(8.2.9)
where
Z
m∗
L
¼
0
m ðx Þ½ ðx Þ2 dx +
N
X
i¼1
mi ½ ðxi Þ2 +
N
X
i¼1
Ii ½ 0 ðxi Þ
2
(8.2.10)
278 PART
I Single-degree-of-freedom systems
(c) Evaluation of the virtual work dWnc of the nonconservative forces. This is
due to the damping force fD ðx, t Þ ¼ cðx Þu_ ðx, t Þ and the external force
pðx, t Þ. Hence we have
Z L
Z L
cðx Þu_ ðx, t Þd uðx, t Þdx +
pðx, t Þd uðx, t Þdx
(8.2.11)
dWnc ¼ 0
0
which by virtue of Eq. (8.1.1) gives
dWnc ¼ ½c∗ u_ + p∗ ðt Þdu
(8.2.12)
where
Z
L
c∗ ¼
cðx Þ ðx Þdx
(8.2.13a)
pðx, t Þ ðx Þdx
(8.2.13b)
0
Z
L
p∗ ðt Þ ¼
0
(d) The potential A of the conservative forces. This is due to the work of the
axial force P. Obviously, if the axial deformation is neglected, it is
A ¼ 0. But if we consider shortening of the beam due to bending, that is,
if we adopt large displacements, then the shortening is expressed by the
nonlinear term of the strain-displacement relation. Thus, according to the
nonlinear theory of elasticity, we have [1, 2]
1 ∂u 2
ex ¼
(8.2.14)
2 ∂x
and the total shortening is
DL ¼
1
2
Z
L
½u0 ðx, t Þ dx
2
(8.2.15)
0
Consequently
1
A ¼ PDL ¼ P
2
Z
L
½u0 ðx, t Þ dx
2
(8.2.16)
0
and
Z
dA ¼ P
L
u0 ðx, t Þd u0 ðx, t Þdx
(8.2.17)
0
which by virtue of Eq. (8.1.1) gives
dA ¼ k∗ udu
(8.2.18)
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 279
where
k∗ ¼ P
Z
L
½ 0 ðx Þ dx
2
(8.2.19)
0
Substituting Eqs. (8.2.5), (8.2.9), (8.2.12), (8.2.18) into Hamilton’s principle, Eq. (1.7.13), results in
Z t2
Z t2
_ u_ k∗ udu dt k ∗ udu m ∗ ud
½c∗ u_ + p∗ ðt Þdudt ¼ 0 (8.2.20)
t1
t1
or after integration by part to eliminate the derivative from du_
Z
t2
m ∗ u€ + c∗ u_ + k ∗ k∗ u p∗ ðt Þ dudt ¼ 0
(8.2.21)
t1
from which we obtain the equation of motion of the generalized SDOF
system
m ∗ u€ + c∗ u_ + k ∗ k∗ u ¼ p∗ ðt Þ
(8.2.22)
Looking at the equation of motion (8.2.22), we draw a useful conclusion
regarding the stability of the structure. It is evident that the stiffness of the
system vanishes when the axial load P takes the critical value
Z
Pcr ¼ k ∗ =
h
½ 0 ðx Þ dx
2
(8.2.23)
0
This value of the axial force is the buckling load of the structure.
A consequence of this is the vanishing of the natural frequency,
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
w¼
k ∗ k∗ =m ∗ ¼ 0. Therefore, a method of determining the buckling
load is to find the value of the compressive axial load, which annuls the natural frequency. This method is known as the dynamic criterion for buckling.
The method of approximating the continuous systems by a generalized
SDOF system can be successful with regard to the displacement. However,
we should be careful when we need to determine the stress resultants from
the obtained deflection curve using the known relations
M ðx, t Þ ¼ EI u00 ðx, t Þ
000
Q ðx, t Þ ¼ EI u ðx, t Þ
(8.2.24)
(8.2.25)
The stress resultants resulting from the above relations may deviate considerably from the actual ones. This is illustrated by the following example.
We consider the cantilever beam of Fig. 8.2.2. The function
px
(8.2.26)
ðx Þ ¼ 1 cos
2L
280 PART
I Single-degree-of-freedom systems
is a geometrically admissible function, that is,
ð0Þ ¼ 0,
0
ð0Þ ¼ 0,
ðLÞ ¼ 1,
0
ðLÞ 6¼ 0
(8.2.27)
FIG. 8.2.2 Cantilever beam as generalized SDOF system
Therefore, it can be used as a shape function for the cantilever beam and
the resulting displacement is
px uðx, t Þ ¼ 1 cos
u ðt Þ
(8.2.28)
2L
Eq. (8.2.25) gives
Q ðx, t Þ ¼ EI
p 2
2L
sin
px
u ðt Þ
2L
(8.2.29)
which results in
Q ð0, t Þ ¼ 0
(8.2.30)
This result is absurd. Nevertheless, this problem can be circumvented
if the stress resultant is evaluated using the procedure described in
Example 8.2.1.
Example 8.2.1 The industrial chimney of length L ¼ 75m shown in Fig. E8.1
consists of the outer reinforced concrete shell, which supports the linings. The
thickness of the thermal insulation layer is ti ¼ 0:10m and that of the refractory
layer tr ¼ 0:10m. The chimney is fixed on the ground.
1. Determine the natural frequency of the structure using two different shape
functions: (i) the elastic curve of a cantilever with constant cross-section
under a uniformly distributed load, and (ii) the first vibration mode of a
cantilever with constant cross-section.
2. Study the dynamic response of the chimney subjected to the impulsive wind
pressure shown in Fig. E8.1b. The analysis will be done using the shape
function that produces more accurate results.
3. Compute the bending moment and shear force at the base of the chimney and
give their expressions as a function of time.
4. Compute the dynamic magnification factor D ¼ max jRðt Þj for the
displacement.
Data:
Specific weight of reinforced concrete: g b ¼ 24kN=m3
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 281
Specific weight of thermal insulation: g i ¼ 0:7kN=m3
Specific weight of refractory bricks: g r ¼ 20kN=m3
Peak value of wind pressure: pw ¼ 1kN=m2
Time duration of wind pressure: t1 ¼ 2s
(a)
(b)
FIG. E8.1 Industrial chimney and loading.
Modulus of elasticity of reinforced concrete: E ¼ 2:1 107 kN=m2
Solution
Computation of m ∗ ,k ∗ ,p∗ ðt Þ.
Mean radius of reinforced concrete shell: r ðx Þ ¼ 3:05 0:0243x
Mean radius of thermal insulation: ri ðx Þ ¼ 2:85 0:0233x
Mean radius of the refractory lining: rr ðx Þ ¼ 2:75 0:0233x
Thickness of reinforced concrete: t ðx Þ ¼ 0:30 0:002x
The mass and the moment of inertia
m ðx Þ ¼ ½2pr ðx Þt ðx Þg b + 2pri ðx Þti g i + 2prr ðx Þtr g r =g
(1)
I ðx Þ pr ðx Þ3 t ðx Þ
(2)
1. The natural frequency of the chimney
(i) The shape function is the elastic curve of a cantilever with constant
cross-section under a uniformly distributed load p ¼ 8EI =L4 , that is,
282 PART
I Single-degree-of-freedom systems
ðx Þ ¼
1 2
6x 4x3 + x4 , x ¼ x=L
3
Computation of the integrals using MATLAB gives
Z L
2
∗
EI ðx Þ½ 00 ðx Þ dx ¼ 2977:6218
k ¼
(3)
(4)
0
Z
m∗
L
¼
m ðx Þ½ ðx Þ2 dx ¼ 120:9955
(5)
0
Hence
w¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi
k ∗ =m ∗ ¼ 4:9608
(6)
(ii) The shape function is the first vibration mode of a cantilever with a
uniform cross-section (see Section 8.3.3.2)
1
ðx Þ ¼ ½ cosh lx cos lx 0:7341ð sinh lx sinh lx Þ, l ¼ 1:8751=L (7)
3
Z L
2
k∗ ¼
EI ðx Þ½ 00 ðx Þ dx ¼ 2715:2000
(8)
0
Z
L
m∗ ¼
m ðx Þ½ ðx Þ2 dx ¼ 116:8332
0
w¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi
k ∗ =m ∗ ¼ 4:8208
(9)
(10
Obviously, case (ii) is more accurate because it yields a smaller
natural frequency (see Chapter 12).
2. The dynamic response of the chimney
The outer diameter of the chimney is
d ðx Þ ¼ 6:40 0:0507x
(11)
hence the wind load per unit length is given by
where
pðx, t Þ ¼ pw d ðx Þf ðt Þ
(12)
8
t
>
>
2
if 0 t t1 =2
>
>
< t1 t
f ðt Þ ¼
2 1
if t1 =2 t t1
>
>
>
t
1
>
:
0
if
t1 t
(13)
hence the peak load of the generalized SDOF system is
Z L
p∗w ¼
pw d ðx Þ ðx Þdx ¼ 106:81
0
(14)
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 283
The dynamic response will be obtained from the solution of the equation
of motion
m ∗ u€ + k ∗ u ¼ p∗w f ðt Þ
(15)
The solution of Eq. (15) is obtained analytically in three phases.
Phase Ι. 0 t 1. The solution is given by Eq. (3.4.15)
2p∗w
sin wt
t
¼ 0:039337t 0:008160 sin 4:8208t
(16a)
uI ¼
w
k ∗ t1
Phase ΙΙ. 1 t 2, et ¼ t 1. The solution is given by Eqs. (3.3.14),
(3.5.10)
p∗w
sin wet e
u_ I ð1Þ
sin wet + uI ð1Þcos wet +
t
(16b)
1 cos wet +
uII ðt Þ ¼
w
w
k∗
where
uI ð1Þ ¼ 0:04745m, u_ I ð1Þ ¼ 0:03508m=s
Phase IΙΙ. 2 t, ^t ¼ t 2. The solution is given by Eq. (2.2.13)
uIII ðt Þ ¼
u_ II ð2Þ
sin w^t + uII ð2Þcos w^t
w
(16c)
where
uII ð2Þ ¼ 0:01447m, u_ II ð2Þ ¼ 0:00759m=s
Fig. E8.2 shows the graphical representation of the response ratio Rðt Þ,
from which we conclude that max jRðt Þj occurs in phase ΙI. The numerical
solution gives D ¼ 1:2673 occurring at t ¼ 1:12s.
FIG. E8.2 Graphical representation of the response ratio in Example 8.2.1
284 PART
I Single-degree-of-freedom systems
The stress resultants at the base of the chimney are obtained by considering
the equilibrium of all external forces, that is, Q ð0, t Þ, M ð0, t Þ, the wind pressure pðx, t Þ, and the inertia force fI ðx, t Þ. Thus, referring to Fig. E8.3, we have
FIG. E8.3 External forces in Example 8.2.1
Z
L
Q ð0, t Þ +
Z
0
Z
L
M ð0, t Þ L
fI ðx, t Þdx pðx, t Þdx ¼ 0
0
Z
L
xf I ðx, t Þdx +
0
xpðx, t Þdx ¼ 0
0
By setting in the previous equilibrium equations
ðx, t Þ ¼ m ðx Þ ðx Þu€ðt Þ and pðx, t Þ ¼ pw d ðx Þf ðt Þ
fI ðx, t Þ ¼ m ðx Þu€
we obtain after the evaluation of the integrals
Z
Z L
m ðx Þ ðx Þdx + pw f ðt Þ
Q ð0, t Þ ¼ u€ðt Þ
0
Z
L
M ð0, t Þ ¼ u€ðt Þ
Z
L
L
m ðx Þ ðx Þxdx pw f ðt Þ
0
d ðx Þdx
0
d ðx Þxdx
0
which yield
Q ð0, t Þ ¼ 211:6454u€ðt Þ + 337:4063f ðt Þ
(17)
M ð0, t Þ ¼ 10421:7751u€ðt Þ 10870:3125f ðt Þ
(18)
8.3 Continuous systems
8.3.1 Introduction
The modeling of a structure as a continuous system can accurately express its
real response, provided that the employed constitutive relations represent
the actual physical law that relates the stresses and strains. As we saw in
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 285
Section 1.1, this modeling leads to partial differential equations of the
hyperbolic type, namely, for the vibrations of the beam [3]
EI
∂4 u
∂2 u
+
m
¼ pðx, t Þ
∂x 4
∂t 2
(8.3.1)
where u ¼ u ðx, t Þ represents the transverse displacement, EI the bending stiff the mass per unit length, and pðx, t Þ the transverse load
ness of the beam, m
density.
Similarly, for the vibrations of the thin plate [1]
∂4 u
∂4 u
∂4 u r ∂2 u pðx, y, t Þ
+
2
+
+
¼
∂x 4
∂x 2 ∂y 2 ∂y 4 D ∂t 2
D
(8.3.2)
where u ¼ u ðx, y, t Þ represents the transverse displacement, D the bending
stiffness of the plate, r the surface mass density, and pðx, y, t Þ the transverse
load density. The above equations are more complicated for the shell and
become even more complicated if the thickness of the structural element is variable or the response is nonlinear [1, 4]. The effort to obtain analytical solutions
of the equations governing the response of continuous systems has inspired
mathematicians and applied physicists to develop the theory of partial differential equations [5, 6]. This, however, did not satisfy people who wanted to use the
derived solutions in engineering praxis because these solutions treat a very limited number of problems. The development of approximate and especially of
modern numerical solutions (e.g., FEM, BEM, MM) has provided us with
efficient tools to solve problems of mathematical physics and engineering
described by complicated ordinary and partial differential equations. The question that arises nowadays and demands investigation is how reliably these equations describe the actual response of physical systems. The system identification
based on experimental data can answer this question. Though the available professional computer codes efficiently analyze engineering problems, they cannot
give adequate insight into the dynamic response of structures. Therefore, the use
of analytical methods is necessary to understand the response of structures and
to develop simple and desired-accuracy solutions, which are very useful to
check new numerical methods. Because the scope of this book is the analysis
of structures consisting of straight-line structural elements, the discussion will
be limited to the solution of the equation of the vibrating beam. Results of this
analysis are used in SDOF generalized systems.
8.3.2
Solution of the beam equation of motion
The transverse flexural vibrations of the beam with constant stiffness are governed by Eq. (8.3.1) subjected to the specified boundary and initial conditions.
The solution u ¼ u ðx, t Þ can be obtained in the form
u ¼ uh + up
(8.3.3)
286 PART
I Single-degree-of-freedom systems
where uh is the solution of the homogeneous equation
∂4 uh
∂2 uh
2 ¼0
+m
4
∂x
∂t
and up a particular solution of the nonhomogeneous equation
EI
EI
(8.3.4)
∂4 up
∂2 up
+
m
¼ pðx, t Þ
∂x 4
∂t 2
(8.3.5)
8.3.3 Free vibrations of beams
For pðx, t Þ ¼ 0, Eq. (8.3.1) yields the homogeneous equation, which represents
the equation of free vibrations of the beam. Assuming constant mass,
¼ constant, and dropping the subscript for convenience in Eq. (8.3.4),
m
we write this equation as
∂4 u
∂2 u
+
m
¼0
(8.3.6)
∂x 4
∂t 2
Eq. (8.3.6) is solved using the method of separation of variables for partial
differential equations [6], that is, the solution is sought in the form
EI
u ðx, t Þ ¼ fðx ÞY ðt Þ
(8.3.7)
This equation indicates that the free vibration motion is the product of a
specific deflection shape fðx Þ and an amplitude Y ðt Þ depending on time.
Introducing the above expression for the deflection into Eq. (8.3.6) gives
ðx ÞY€ ðt Þ ¼ 0
EI fðivÞ ðx ÞY ðt Þ + mf
(8.3.8)
EiI fðivÞ ðx Þ
Y€ ðt Þ
¼
fðx Þ
m
Y ðt Þ
(8.3.9)
which is written as
For the obvious problem associated with the possibility of the vanishing of
the product fðx ÞY ðt Þ, we refer to [7, 8]. Here, we just state that the roots of the
denominator in Eq. (8.3.9) coincide with those of the nominator, an assumption
that permits this division.
Because the left side of Eq. (8.3.9) is independent of t and the right side
independent of x, this equation is valid only if both sides are equal to a
constant l. Hence
EI fðivÞ ðx Þ
Y€ ðt Þ
¼
¼l
fðx Þ
m
Y ðt Þ
or
Y€ ðt Þ + lY ðt Þ ¼ 0
(8.3.10a)
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 287
m
f ðx Þ ¼ 0
(8.3.10b)
EI
If l 0, the solution of Eq. (8.3.10a) does not represent an oscillatory
motion. Therefore, l must be a positive constant, l ¼ w2 . Thus, the solution
of Eq. (8.3.10a) is (see Section 2.2)
fðivÞ ðx Þ l
Y ðt Þ ¼ A cos wt + B sin wt
(8.3.11)
where A,B are arbitrary constants depending on the initial conditions Y ð0Þ and
Y_ ð0Þ. Thus we have
Y ðt Þ ¼ Y ð0Þcos wt +
Y_ ð0Þ
sin wt
w
(8.3.12)
Apparently, w is the natural frequency of the vibration, unknown in the first
instance.
Similarly, Eq. (8.3.10b) is written as
fðivÞ ðx Þ b 4 fðx Þ ¼ 0
(8.3.13)
where
b 4 ¼ w2
m
EI
(8.3.14)
The solution of Eq. (8.3.13) is sought in the form
fðx Þ ¼ C 0 ekx
(8.3.15)
which is introduced into Eq. (8.3.13) to give the characteristic equation
k 4 b4 ¼ 0
(8.3.16)
k1, 2 ¼ ib, k3, 4 ¼ b
(8.3.17)
whose roots are
Using each of these roots in Eq. (8.3.15) yields four terms, which are added
to give the general solution
fðx Þ ¼ C10 eibx + C20 eibx + C30 ebx + C40 ebx
(8.3.18)
where C 0 i ði ¼ 1, 2, 3, 4Þ are arbitrary complex constants.
Using Euler’s formula (2.2.8) and the expressions of the hyperbolic sine and
cosine, Eq. (8.3.18) becomes
fðx Þ ¼ C1 cos bx + C2 sin bx + C3 cosh bx + C4 sinh bx
(8.3.19)
in which Ci ði ¼ 1, 2, 3, 4Þ are new arbitrary constants related to
C 0 i ði ¼ 1, 2, 3, 4Þ, and can be determined from the boundary (support) conditions of the one-span beam.
288 PART
I Single-degree-of-freedom systems
In the following, the above-described procedure for free-vibration analysis
is illustrated by analyzing the simply supported and the cantilever beams.
For further reading, the reader is advised to look in Refs. [9–13].
8.3.3.1 The simply supported beam
For a simply supported beam of length L (Fig. 8.3.1) the support conditions are
realized as:
u ð0, t Þ ¼ 0, M ð0, t Þ ¼ EI u 00 ð0, t Þ ¼ 0
(8.3.20a)
u ðL, t Þ ¼ 0, M ðL, t Þ ¼ EI u 00 ðL, t Þ ¼ 0
(8.3.20b)
which by virtue of Eq. (8.3.7) become
fð0ÞY ðt Þ ¼ 0, f00 ð0ÞY ðt Þ ¼ 0
(8.3.21a)
fðLÞY ðt Þ ¼ 0, f00 ðLÞY ðt Þ ¼ 0
(8.3.21b)
FIG. 8.3.1 Vibration modes and natural frequencies of a uniform simply supported beam.
Inasmuch as Eqs. (8.3.21a), (8.3.21b) are valid for all values of t, they are
satisfied only if
fð0Þ ¼ 0, f00 ð0Þ ¼ 0
(8.3.22a)
fðLÞ ¼ 0, f00 ðLÞ ¼ 0
(8.3.22b)
Introducing Eq. (8.3.19) into Eq. (8.3.22a) yields
C1 + C3 ¼ 0
(8.3.23a)
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 289
C1 + C3 ¼ 0
(8.3.23b)
C1 ¼ C3 ¼ 0
(8.3.24)
from which we obtain
Further, introducing Eq. (8.3.19) into Eq. (8.3.22b), we obtain
C2 sin bL + C4 sinh bL ¼ 0
(8.3.25a)
b2 ½C2 sin bL + C4 sinh bL ¼ 0
(8.3.25b)
Eqs. (8.3.25a), (8.3.25b) provide the system of two homogeneous algebraic
equations for the evaluation of C2 ,C4 , that is,
sin bL sinh bL C2
0
¼
(8.3.26)
sin bL sinh bL C4
0
The system has a nontrivial solution if its determinant vanishes, that is,
sin bL sinh bL ¼ 0
(8.3.27)
Nevertheless, because sinh bL 6¼ 0, it must be
sin bL ¼ 0
(8.3.28)
np
, n ¼ 1, 2, …
L
(8.3.29)
which is satisfied if
bn ¼
Introducing this value of b n into Eq. (8.3.14) gives the corresponding natural
frequencies
rffiffiffiffiffiffiffiffiffiffi
EI
2 2
wn ¼ n p
(8.3.30)
4
mL
Substituting Eq. (8.3.28) into Eq. (8.3.26), we obtain
"
#( ) ( )
0
0 sinh bL
C2
¼
0
0 sinh bL
C4
(8.3.31)
which yields C4 ¼ 0 and C2 ¼ arbitrary, therefore
fðx Þ ¼ C2 sin bx
On the basis of Eq. (8.3.29), we obtain the mode shapes
np x , n ¼ 1, 2, …
fn ðx Þ ¼ C2 sin
L
(8.3.32)
(8.3.33)
The first three of these mode shapes are shown in Fig. 8.3.1 along with their
natural frequencies
290 PART
I Single-degree-of-freedom systems
8.3.3.2 The cantilever beam
For a cantilever beam of length L the support conditions are realized as:
u ð0, t Þ ¼ 0, u 0 ð0, t Þ ¼ 0
00
(8.3.34a)
000
M ðL, t Þ ¼ EI u ðL, t Þ ¼ 0, Q ðL, t Þ ¼ EI u ðL, t Þ ¼ 0
(8.3.34b)
which by virtue of Eq. (8.3.7) become
fð0ÞY ðt Þ ¼ 0, f0 ð0ÞY ðt Þ ¼ 0
(8.3.35a)
f00 ðLÞY ðt Þ ¼ 0, f000 ðLÞY ðt Þ ¼ 0
(8.3.35b)
Because Eqs. (8.3.35a), (8.3.35b) are valid for all values of t, they are
satisfied only if
fð0Þ ¼ 0, f0 ð0Þ ¼ 0
(8.3.36a)
f00 ðLÞ ¼ 0, f000 ðLÞ ¼ 0
(8.3.36b)
Introducing Eq. (8.3.19) into Eqs. (8.3.36a), (8.3.36b) yields the system of
equations for the evaluation of the coefficients Ci
2
38 9 8 9
1
0
1
0
>
>
>0>
> >
> C1 >
6 0
7 < C2 = < 0 =
1
0
1
6
7
(8.3.37)
¼
4 cos bL sin bL coshbL sinh bL 5> C3 > > 0 >
>
;
: >
; >
: >
0
sin bL cos bL sinh bL cosh bL
C4
Eq. (8.3.37) has a nontrivial solution if the
matrix vanishes, that is,
1
0
1
0
1
0
cos bL sin bL cosh bL
sin bL cos bL sinh bL
determinant of the coefficient
0
1
¼0
sinh bL cosh bL (8.3.38)
or after evaluation of the determinant
cos bL cosh bL + 1 ¼ 0
(8.3.39)
The roots bn , n ¼ 1, 2, … of Eq. (8.3.39) are used in Eq. (8.3.14) to obtain
the natural frequencies of the vibrating cantilever beam.
Substituting Eq. (8.3.39) into Eq. (8.3.37) and solving the resulting
homogeneous system of equations, we obtain
C3 ¼ C1 , C4 ¼ C2 , C2 ¼ cos b n L + cosh bn L
C1
sin b n L + sinh bn L
(8.3.40)
which are introduced into Eq. (8.3.19) to give the mode shapes
cos b n L + cosh bn L
ð sin bn x sinh bn x Þ
fn ðx Þ ¼ C1 cos bn x cosh b n x sin b n L + sinh bn L
(8.3.41)
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 291
Table 8.3.1 gives the first five roots of bn L of Eq. (8.3.39). They have been
obtained numerically using the function fsolve of MATLAB. Note that for n > 3
they can be obtained from the relation
p
(8.3.42)
bn L ð2n 1Þ
2
The first three of these mode shapes are shown in Fig. 8.3.2 along with their
natural frequencies
TABLE 8.3.1 First five roots of Eq. (8.3.39).
n
bn L
1
1.8751040688
2
4.6940911329
3
7.8547574382
4
10.995540734
5
14.137168391
FIG. 8.3.2 Vibration modes and natural frequencies of a uniform cantilever beam.
8.3.4
Orthogonality of the free-vibration modes
The infinite set of free-vibration mode shapes F : ffn ðx Þg, n ¼ 1, 2, … has a
nice property. They are orthogonal in the interval ½0, L, that is, they satisfy
the orthogonality condition (see Section 3.6.2)
292 PART
I Single-degree-of-freedom systems
Z
L
0 if r 6¼ n
cn if r ¼ n
fr ðt Þfn ðt Þdt ¼
0
(8.3.43)
for any two functions fn ,fr F, m, nN . In the language of partial differential
equations, the free-vibration modes are called the eigenfunctions of the eigenvalue
problem described by the differential equation (8.3.13) and its boundary conditions. The orthogonality condition is readily proved by proceeding as follows.
The mode shapes fn ,fr satisfy Eq. (8.3.13), that is,
fðnivÞ ðx Þ b4n fn ðx Þ ¼ 0
(8.3.44a)
fðrivÞ ðx Þ b 4r fr ðx Þ ¼ 0
(8.3.44b)
Multiplication of Eq. (8.3.44a) by fr ðx Þ and integrating over the interval
½0, L gives
Z L
Z L
fn ðx Þfr ðx Þdx ¼
fðnivÞ ðx Þfr ðx Þdx
(8.3.45)
b4n
0
Z
0
0
Further, integrating the right side of the above equation twice by parts gives
Z L
L
L
L
4
00
00
00
0
000
fn ðx Þfr ðx Þdx fn ðx Þfr ðx Þ 0 + fn ðx Þfr ðx Þ 0 ¼ bn
fn ðx Þfr ðx Þdx
0
(8.3.46)
Obviously, the terms in square brackets in the above equation vanish if
either end of the beam is simply supported, fixed, or free. Thus, we have
Z L
Z L
4
bn
fn ðx Þfr ðx Þdx ¼
f00n ðx Þf00r ðx Þdx
(8.3.47a)
0
0
Similarly, multiplying Eq. (8.3.44b) by fn ðx Þ, integrating over the interval
½0, L, and performing the integrations by parts yields the symmetric relation
Z L
Z L
fn ðx Þfr ðx Þdx ¼
f00n ðx Þf00r ðx Þdx
(8.3.47b)
b4r
0
0
Subtracting Eq. (8.3.47b) from Eq. (8.3.47a) gives
Z L
0 ¼ b 4n b4r
fn ðx Þfr ðx Þdx
(8.3.48)
0
which for b 4n 6¼ b4r results in the orthogonality condition for the mode shapes
Z L
fn ðx Þfr ðx Þdx ¼ 0
(8.3.49)
0
It can also be shown that the set F : ffn ðx Þg is complete, that is there is no
other function outside the set F, which satisfies the condition (8.3.49).
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 293
8.3.5
Forced vibrations of beams
The method of separation of variables can be employed to study the forced
vibrations of one-span beams by solving Eq. (8.3.1). The beam may be
subjected to initial conditions
u ðx, 0Þ ¼ f ðx Þ, u_ ðx, 0Þ ¼ g ðx Þ
The total flexural displacement of the beam is obtained as the superposition
of all modal contributions, namely
u ðx, t Þ ¼
∞
X
fn ðx ÞYn ðt Þ
(8.3.50)
n¼1
Substituting the previous equation into Eq. (8.3.1), multiplying by fr ðx Þ,
and integrating over the interval ½0, L, we obtain
X
Z L
Z L
∞ ∞ X
ðivÞ
€
n ðx Þfr ðx Þdx
Yn ð t Þ
EI fn ðx Þfr ðx Þdx +
Y n ðt Þ
mf
n¼1
0
Z
L
¼
0
n¼1
fr ðx Þpðx, t Þ
(8.3.51)
0
Using now the orthogonality condition (8.3.49) and taking into account
Eqs. (8.3.45), (8.3.14), we obtain
Mn Y€ n ðt Þ + Kn Yn ðt Þ ¼ Pn ðt Þ
(8.3.52)
where
Z
L
Mn ¼ m
0
Z
f2n ðx Þdx, Kn ¼ w2n Mn , Pn ¼
L
fn ðx Þpðx, t Þdx
(8.3.53)
0
denote the modal mass, the modal stiffness, and the modal force. These quantities are also referred to as the generalized mass, the generalized stiffness, and
the generalized force, respectively.
The solution of Eq. (8.3.52) is given by Eq. (3.3.14), that is,
Z t
1
Y_ n ð0Þ
Y n ðt Þ ¼
sin wn t + Yn ð0Þcos wn t +
Pn ðτÞsin ½wn ðt τÞdτ
wn
M n wn 0
(8.3.54)
The initial conditions Yn ð0Þ, Y_ n ð0Þ for the time function result from
Eq. (8.3.50). This yields
u ðx, 0Þ ¼
∞
X
fn ðx ÞYn ð0Þ ¼ f ðx Þ
(8.3.55)
n¼1
Multiplying the previous equations by fn ðx Þ, integrating over the interval
½0, L, and using the orthogonality condition (8.3.49), we obtain
294 PART
I Single-degree-of-freedom systems
Z
Yn ð 0Þ ¼
L
f ðx Þfn ðx Þdx
0
Z
L
0
(8.3.56a)
f2n ðx Þdx
Similarly, we obtain
Z
Y_ n ð0Þ ¼
0
L
g ðx Þfn ðx Þdx
Z
L
0
(8.3.56b)
f2n ðx Þdx
Example 8.3.1 A simply supported beam of length L is subjected to a suddenly
applied uniform p0 under zero initial conditions, f ðx Þ ¼ g ðx Þ ¼ 0. Determine the
expressions of the displacement, bending moment, and shear force.
Solution
1. Determine the natural frequencies and mode shapes of the vibration.
They are obtained from Eqs. (8.3.30), (8.3.33):
rffiffiffiffiffiffiffiffiffiffi
np EI
2 2
x , n ¼ 1, 2, …
(1)
wn ¼ n p
ð
x
Þ
¼
sin
,
f
n
4
mL
L
2. Determine the modal mass, modal force, and initial conditions. They are
obtained from Eqs. (8.3.53), (8.3.56a), (8.3.56b):
Z L
np mL
Mn ¼ m
x dx ¼
(2)
sin 2
L
2
0
Z L
np L
x dx ¼ p0 ½1 ð1Þn Pn ¼ p 0
sin
(3)
L
np
0
Yn ð0Þ ¼ 0, Y_ n ð0Þ ¼ 0
(4)
3. Determine the time-varying amplitudes Yn ðt Þ. Substituting Eqs. (2)–(4)
into Eq. (8.3.54) and using Eq. (3.4.3) give
Yn ð t Þ ¼
4p0 L4
ð1 cos wn t Þ, n ¼ 1, 3, 5, …
n 4 p4 EI
(5)
Hence
u ðx, t Þ ¼
∞
∞
np X
4p0 L4 X
1
x ð1 cos wn t Þ,
fn ðx ÞYn ðt Þ ¼ 4
sin
4
L
p EI n¼1 n
n¼1
n ¼ 1, 3, 5, …
(6)
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 295
The bending moment and the shear force are evaluated from the
expressions
M ðx, t Þ ¼ EI u 00 ðx, t Þ ¼ EI f00 ðx ÞY ðt Þ
000
000
Q ðx, t Þ ¼ EI u ðx, t Þ ¼ EI f ðx ÞY ðt Þ
8.4
(7)
(8)
Problems
Problem 8.1 The television tower of Fig. P8.1 is subjected to seismic ground
motion ug ðt Þ. Derive the equation of motion if the structure is approximated by
a SDOF system. The reaction moment
of the elastic ground is represented by the
nonlinear expression MR ¼ CR f + 14 f2 , where CR ¼ KI f ; If is the moment of
inertia of the planform of the fundament and K ¼ E=10h the foundation modulus with E being the modulus of elasticity of the material of the structure. The
cross-section of the flexible column, the planform of the fundament, and the
body B are circular with diameters D, Df ¼ 8D, and DB ¼ 5D, respectively.
The density of the material is r. The fundament and the body B are
assumed rigid.
FIG. P8.1 Television tower in problem P8.1
Problem P8.2 The continuous beam of Fig. P8.2 rests on Winkler’s elastic
foundation with variable modulus k ðx Þ. The beam is axially subjected to the
load P. Determine the value of the load that produces buckling.
FIG. P8.2 Continous beam on nonhomogeneous Winkler’s elastic foundation in problem P8.2
296 PART
I Single-degree-of-freedom systems
Problem P8.3 A vehicle of weight W ¼ 30kN is traveling on a simply supported bridge with velocity v ¼ 80km=h as shown in Fig. P8.3a. The vehicle
is simulated by a single wheel of negligible mass. Determine the dynamic magnification factor for the bridge when the beam is approximated by a generalized
SDOF system. The material of the bridge is prestressed concrete with modulus
of elasticity E ¼ 2:1 107 kN=m2 and specific weight g ¼ 25kN=m3 . The
cross-section of the bridge is shown in Fig. P8.3b.
(a)
(b)
FIG. P8.3 Simple supported bridge in Problem P8.3
Problem P8.4 Use the method of separation of variables to solve the equation
of free flexural vibrations of the:
(a) Fixed-fixed beam.
(b) Fixed-simply supported beam
In both cases, determine the natural mode shapes and the frequency equation.
Compute the first three natural frequencies.
Problem P8.5 Analyze the free flexural vibrations of the two-span continuous
beam of Fig. P8.5 by solving the equation of motion of the beam. Hint: Consider
the continuity condition at support 2, that is, u 0 I ðL, t Þ ¼ u 0 II ð0, t Þ.
FIG. P8.5 Two-span continuous beam in problem P8.5
Generalized single-degree-of-freedom systems—Continuous systems Chapter 8 297
References and further reading
[1] J.T. Katsikadelis, The Boundary Element Method for Plate Analysis, Academic Press,
Elsevier, Oxford, UK, 2014.
[2] D.O. Brush, B.O. Almroth, Buckling of Bars, Plates and Shells, McGraw-Hill, New York,
1975.
[3] S. Timoshenko, D.E. Young, W. Weaver Jr., Vibration Problems in Engineering, fourth ed.,
John Wiley & Sons, New York, 1974.
[4] J.T. Katsikadelis, G.C. Tsiatas, Nonlinear dynamic analysis of beams with variable stiffness,
J. Sound Vib. 270 (2004) 847–863.
[5] E. Zauderer, Partial Differential Equations of Applied Mathematics, second ed., John Wiley &
Sons, Singapore, 1989.
[6] T. Myint-U, L. Debnath, Linear Partial Differential Equations for Scientists and Engineers,
fourth ed., Birkh€auser, Boston, 2007.
[7] E.C. Titchmarsh, Eigenfunction Expansions, Part I (1958) & Part II (1962), Oxford University
Press, Oxford, UK, 1958.
[8] G.L. Cain, G.H. Meyer, Separation of Variables for Partial Differential Equations: An Eigenfunction Approach, in: Studies in Advanced Mathematics, CRC Press, London, 2006.
[9] R.W. Clough, J. Penzien, Dynamics of Structures, McGraw-Hill, New York, 1993.
[10] J.L. Humar, Dynamics of Structures, second ed., A.A. Balkema Publishers, Lisse, NL, 2002.
[11] A.K. Chopra, Dynamics of Structures: Theory and Applications to Earthquake Engineering,
Prentice Hall, Englewood Cliffs, NJ, 1995.
[12] S.S. Rao, Vibration of Continuous Systems, John Wiley & Sons, New Jersey, 2007.
[13] L. Meirovitch, Analytical Methods in Vibration, Macmillan, London, 1967.
Chapter 9
Analysis in the frequency
domain
Chapter outline
9.1 Introduction
9.2 Complex form of the Fourier
series
9.3 Complex dynamic response
to periodic load
9.4 Fourier integral
representation of a
nonperiodic load
9.5 Response to a nonperiodic
load
9.1
299
301
303
304
9.6 Discrete Fourier transform
9.7 Application of the discrete
Fourier transform
to dynamic analysis
9.8 Fast Fourier transform
9.8.1 The Sande-Tukey
algorithm
9.9 Problems
References and further reading
310
313
314
315
320
322
307
Introduction
The methods we discussed in the previous chapters for solving the equation of
motion of a SDOF system were accomplished using time as an independent variable or, said differently, the solution was obtained in the time domain. These
methods are either analytical, which in the general case lead to the evaluation of
the Duhamel integral, or numerical, implemented by the step-by-step integration methods. Occasionally, simpler or more convenient analytical or numerical
solutions may be possible for certain types of dynamics problems, using integral
transforms such as the Laplace transform or the Fourier transform. The integral
converts the linear differential equation into a linear algebraic equation, from
which the integral transform of the unknown function is obtained. Then the
inverse transform results in the solution in the time domain.
We have already discussed the Laplace transform in Section 3.3, where it
was employed to solve the equation of motion of a SDOF system under an arbitrary external excitation. The Laplace transform uses a parameter that does not
have a direct physical meaning. Instead, the parameter in the Fourier transform
has the physical meaning of frequency [1–3]. The method of analyzing dynamical systems using the Fourier transform is known as the analysis in the frequency domain. It plays an important role in studying the dynamic response
of linear systems, that is, systems described by linear differential equations.
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00009-1
© 2020 Elsevier Inc. All rights reserved.
299
300 PART
I Single-degree-of-freedom systems
The analysis in the frequency domain is employed in many fields of engineering
science and physics. Therefore, the available related literature is extensive and
reference to it is made with regard to the specific scientific discipline under
treatment.
As we have seen in Section 3.4, the drawback of time-domain analysis is the
evaluation of the Duhamel integral, which can be achieved analytically only for
very simple loading functions. Instead, the Fourier transform converts the convolution integral, such as the integral of Duhamel, into the product of the transforms of the functions involved in the convolution. Then the inverse transform
of the product gives the response of the system in the time domain.
The efficiency of the method depends on the possibility of finding the Fourier transform of the functions of the convolution as well as the inverse of their
product. There are tables [4] that give the Fourier transform and its inverse for a
variety of functions. Unfortunately, these functions are rather simple while the
use of tables requires tedious analytical work with the risk of a possible error.
Certainly, the symbolic languages such as MAPLE, MATHEMATICA, and
WOLFRAM ALPHA offer new possibilities for the application of the method.
Nevertheless, the Fourier transform method cannot always meet the practical needs, especially when the load function is specified by a set of values, as for
example the accelerogram of an earthquake. For this reason, the Fourier transform for a long time has not been equally applied to the dynamic analysis of
structures as it has to other areas of engineering science. The use of numerical
methods of computing the Fourier transform has brought to light this method in
its discrete form, referred to as the discrete Fourier transform. Of course, the
discrete Fourier transform is not simpler or more functioning than the numerical
methods of evaluating the Duhamel integral. Therefore, its use would have been
limited if a specific algorithm for the numerical computation of the Fourier
transform had not been developed, the so-called Fast Fourier transform, which
limits the numerical computations by several orders and makes dynamic analysis in the frequency domain quite effective.
Besides its effectiveness, the analysis in the frequency domain has other
important advantages. It shows clearly the frequencies of the loading, providing
thus the possibility of detecting those frequencies that unfavorably excite the
structure. It is also suitable for the study of systems involving large or infinite
regions, such as the interaction of dams in large reservoirs with waves, soil
structure interaction in earthquake engineering, and, in general, problems where
the physical characteristics of the system such as stiffness or damping depend
on the frequency of the oscillation. For these reasons, it was considered appropriate to include a brief discussion of the analysis in the frequency domain in
this book. Apparently, this would enable the engineers involved in earthquake
analysis to understand the capabilities of the method.
The method could be presented starting from the mathematical development
of the Fourier transform [4], but in order to understand its physical significance,
the complex Fourier series and the response of the SDOF system to periodic
Analysis in the frequency domain Chapter
9
301
loading are preceded. Then, the method extends to nonperiodic loads by introducing the Fourier integral and the Fourier transform. The chapter closes with
the presentation of the discrete Fourier transform and the fast Fourier transform.
9.2
Complex form of the Fourier series
In Section 3.6, the dynamic response of a structure under a general periodic load
was obtained in the time domain by expanding the periodic load in Fourier
series, that is, by representing it through an infinite sum of harmonic loads. This
procedure was discussed in detail there. However, the study of the response of a
system in the frequency domain is facilitated by representing the Fourier series
(3.6.2) in exponential form. We recall that
pðt Þ ¼ a0 +
1
X
t + bn sin n w
t Þ
ðan cos n w
(9.2.1)
n¼1
a0 ¼
an ¼
2
T
bn ¼
2
T
1
T
Z
Z
T =2
T =2
T =2
Z
T =2
T =2
T =2
pðt Þdt ¼
1
T
Z
T
0
pðt Þdt
(9.2.2a)
tdt, n ¼ 1, 2, 3…
pðt Þcos n w
(9.2.2b)
tdt,
pðt Þ sin n w
(9.2.2c)
n ¼ 1, 2, 3…
From Euler’s formula (2.2.8), we have
t ¼
cos n w
einwt + einwt
2
(9.2.3a)
t ¼
sin n w
einwt einwt
2i
(9.2.3b)
Substituting Eq. (9.2.3a) into Eq. (9.2.2b) gives
Z
1 T =2
an ¼
pðt Þ einwt + einwt dt
T T =2
Replacing n with n in the previous equation gives
Z
1 T =2
an ¼
pðt Þ einwt + einwt dt ¼ an
T T =2
Similarly, substituting Eq. (9.2.3b) into Eq. (9.2.2c) yields
Z
1 T =T
ein wt einwt
dt:
bn ¼
pðt Þ
T T =2
i
(9.2.4a)
(9.2.4b)
(9.2.5a)
302 PART
I Single-degree-of-freedom systems
Hence
1
bn ¼
T
Z
T =2
T =2
pðt Þ
einwt einwt
dt ¼ bn
i
(9.2.5b)
The Fourier series (9.2.1) by virtue of Eqs. (9.2.3a), (9.2.3b) becomes
pðt Þ ¼ a0 +
1
1
X
einwt + einwt X
ein wt einwt
+
an
bn
2
2i
n¼1
n¼1
1
1
1X
1X
¼ ao +
ðan ibn Þeinwt +
ðan + ibn Þeinwt
2 n¼1
2 n¼1
(9.2.6)
Further, using Eqs. (9.2.4a), (9.2.5a), the previous equation becomes
pðt Þ ¼ ao +
¼
1
1
1X
1 X
ðan ibn Þeinwt +
ðan ibn Þein wt
2 n¼1
2 n¼1
1
X
cn e
(9.2.7)
t
in w
n¼1
in which the series coefficients are given as
1
cn ¼ ðan ibn Þ
2
Z T =2
1
¼
pðt Þ einwt + einwt ein wt einwt dt
2T T =2
Z
1 T =2
pðt Þeinwt dt
¼
T T =2
(9.2.8a)
and
1
¼ ðan + ibn Þ
Z2
1 T =2
¼
pðt Þeinwt dt
T T =2
Z
1 T =2
c0 ¼
pðt Þdt ¼ a0
T T =2
cn
(9.2.8b)
(9.2.8c)
Eq. (9.2.7) represents the complex form or exponential form of the Fourier
series.
From Eq. (9.2.7), the reasonable question arises as to how it is possible for a
real function to be expressed as a sum of complex terms. It is easy, however, to
prove that the right side of this equation is a real function as long as we think that
to each term cn einwt the term cn ein wt corresponds, whose sum yields a real
function.
Analysis in the frequency domain Chapter
9.3
9
303
Complex dynamic response to periodic load
In Section 3.6.3, the method of determining the steady-state response of a SDOF
system subjected to a periodic load was presented. In this method, the periodic
load was analyzed into harmonic terms, both sine and cosine, by expanding it in
a Fourier series. After establishing the steady-state response for each term, the
total response was obtained as the sum of all responses. Following a similar procedure, it is easy to determine the steady-state response to a periodic load when
it is has been expanded in a complex Fourier series. To this end, we must first
determine the steady-state response of the system when subjected to the load
pðt Þ ¼ p0 eiwt . In this case, the equation of the SDOF system is written
m u€ + cu_ + ku ¼ p0 eiwt
(9.3.1)
The solution will be obtained as a sum of the homogeneous solution uh and a
particular solution up of the nonhomogeneous equation. The homogeneous
solution is given by Eq. (3.2.22), namely
uh ¼ exwt ðA cos wD t + B sin wD t Þ
(9.3.2)
The particular solution is sought in the form
up ¼ Ceiwt
(9.3.3)
Substituting the previous expression into Eq. (9.3.1) gives
p0
C¼
2
+ icw
+ kÞ
ðm w
which is inserted into Eq. (9.3.3) to yield
p0
eiwt
up ¼
2 + icw
+k
m w
(9.3.4)
(9.3.5)
Hence, the general solution of Eq. (9.3.1) is
u ðt Þ ¼ exwt ðA cos wD t + B sin wD t Þ +
p0
eiwt
2
+k
m w + icw
(9.3.6)
The first term in Eq. (9.3.6) becomes negligible with increasing time and it
represents the transient response of the system. The second term expresses the
steady-state response and can be written as
Þp0 eiwt
u ðt Þ ¼ H ðw
(9.3.7)
where it was set
1
2 + icw
+k
m w
1
, b ¼ w
=w
¼ k 1 b 2 + 2ixb
Þ ¼
H ðw
(9.3.8)
304 PART
I Single-degree-of-freedom systems
Þ is called the complex frequency response function, also
The function H ðw
known as the transfer function. In summary, setting p0 ¼ cn we may write the
steady-state response to a periodic load as
u ðt Þ ¼
1
X
Þeinwt
c n H ðw
(9.3.9)
n¼1
where
cn ¼
1
T
Z
T =2
T =2
pðt Þeinwt dt ¼
1
T
Z
0
T
pðt Þein wt dt
(9.3.10)
9.4 Fourier integral representation of a nonperiodic load
When the load is nonperiodic, it cannot be represented as a Fourier series. But it
is possible to express the nonperiodic function in the form of an integral over the
interval ð1, 1Þ, which one might regard as a Fourier series whose period is
infinitely large. In Section 3.6.2, we saw that a function periodic in the interval
½0, 1Þ or ð1, 1Þ is expanded in a Fourier series, Eq. (9.2.1). The coefficients
in this equation are written as
Z
1 T =2
a0 ¼
pðτÞdτ
(9.4.1a)
T T =2
Z
2 T =2
τdτ, n ¼ 1, 2, 3…
an ¼
pðτÞcos n w
(9.4.1b)
T T =2
Z
2 T =2
τdτ, n ¼ 1, 2, 3…
bn ¼
pðτÞ sin n w
(9.4.2)
T T =2
The previous equations resulted from Eq. (9.2.2a), Eq. (9.2.2b), Eq. (9.2.2c)
by substituting t with τ in the integrand. Note that it is allowed because τ is a
dummy variable.
Substituting Eqs. (9.4.1a), (9.4.1b), (9.4.2) in Eq. (9.2.1) yields
"
#
Z
1 Z T =2
1 T =2
2X
τ cosn w
tdτ
pðτÞdτ +
pðτÞ cos n w
pðt Þ ¼
T T =2
T n¼1 T =2
"
#
(9.4.3)
1 Z T =2
2X
τ sin n w
tdτ
+
pðτÞ sin n w
T n¼1 T =2
or
1
pðt Þ ¼
T
Z
1
2X
pðτÞdτ +
T n¼1
T =2
T =2
"Z
T=2
T =2
#
ðτ t Þdτ
pðτÞcos n w
(9.4.4)
Analysis in the frequency domain Chapter
9
305
If the function pðt Þ is not periodic, we may set
( Z
"
#)
1 Z T =2
1 T=2
2X
ðτ t Þdτ
pðτÞdτ +
pðτÞcos n w
pðt Þ ¼ lim
T !1 T T=2
T n¼1 T =2
(9.4.5)
in which t does not change in passing to the limit. It is assumed that the function
pðt Þ satisfies the Dirichlet conditions (see Section 3.6.2). Therefore, it is absolutely integrable over the interval ½T =2, T =2, that is,
Z
1 Z T =2
1 T =2
k
(9.4.6)
j a0 j ¼ pðτÞdτ jpðτÞjdτ <
T T =2
T T =2
T
which vanishes for T ! 1. Hence
(
"
#)
1 Z T =2
2X
ðτ t Þdτ
pðt Þ ¼ lim
pðτÞ cos n w
T !1 T
T =2
n¼1
(9.4.7)
If we set now
n + 1 w
n ¼
¼w
n , D
w¼w
nw
2p
T
(9.4.8)
we may write Eq. (9.4.7) as
pðt Þ ¼ lim
T !1
1
X
n ÞD
P ðw
w
(9.4.9)
n¼1
n Þ is the value of the function
where P ðw
Z
1 T =2
ðτ t Þdτ
Þ ¼
pðτÞcos w
P ðw
p T =2
(9.4.10)
is understood as a continuous variable (Fig. 9.4.1)
¼w
n , where w
for w
FIG. 9.4.1 Fourier transform of p(t).
306 PART
I Single-degree-of-freedom systems
Þ, we readily deduce that the sum
From the graphical representation of P ðw
1
X
n ÞD
P ðw
w
(9.4.11)
n¼1
Þ. When T ! 1, then D
approaches the area under the curve y ¼ P ðw
w ! 0 and
the sum (9.4.11) becomes a definite integral. Consequently, we may write
Eq. (9.4.9) as
Z 1
Þd w
pðt Þ ¼
P ðw
(9.4.12)
0
or by virtue of Eq. (9.4.10)
Z 1 Z 1
1
ðτ t Þdτ d w
pðt Þ ¼
pðτÞcos w
p 1
0
(9.4.13)
The integral (9.4.13) is known as the Fourier integral. The Fourier integral converges to the function if it is piecewise continuous in every finite interval and
absolutely integrable over ð1, 1Þ. At points of discontinuity, the Fourier
integral converges to the mean value.
The Fourier integral can be expressed in a complex form using Euler’s formula, Thus, writing
i
1h
ðτ t Þ ¼ eiwðτt Þ + eiwðτt Þ
cos w
(9.4.14)
2
and inserting into Eq. (9.4.13) splits it into two integrals, that is,
Z Z 1
Z Z 1
1 1
1 1
ðτt Þ
ðτt Þ
iw
i w
+
pðτÞe
dτ d w
pðτÞe
dτ d w
pðt Þ ¼
2p 0
2p 0
1
1
to Changing the integration variable from w
w in the second integral, we
obtain
Z Z 1
1 1
pðτÞeiwτ dτ eiwt d w
(9.4.15)
pðt Þ ¼
2p 1 1
The forgoing relation allows writing
Z 1
Þ ¼
P ðw
pðτÞeiwτ dτ
1
or returning to the variable t from τ
Z
Þ ¼
P ðw
1
1
pðt Þeiwt dt
and inserting it into Eq. (9.4.15), gives
Z
1 1
Þeiwt d w
pðt Þ ¼
P ðw
2p 1
(9.4.16)
(9.4.17)
Analysis in the frequency domain Chapter
9
307
Þ defined by Eq. (9.4.16) is called the (direct) Fourier
The function P ðw
transform of pðt Þ while the function pðt Þ resulting from Eq. (9.4.17) is called
Þ.
the inverse Fourier transform of P ðw
In the time domain, a function will be denoted by a small letter while its
Fourier transform is by the same capital letter. The relationship between them
will be symbolized by
Þ
pðt Þ , P ðw
(9.4.18)
Usually, we denote the Fourier transform of a function pðt Þ by F ½pðt Þ while
its inverse is by F 1 ½pðt Þ, namely
Þ ¼ F ½pðt Þ
P ðw
(9.4.19)
Þ
pðt Þ ¼ F 1 ½P ðw
(9.4.20)
The Fourier transform of the derivative of a function pðt Þ is readily established by applying integration by parts to Eq. (9.4.16). Generally, for a function
of which the ðn 1Þ order derivatives are continuous and the nth order derivative is piecewise continuous, it can be shown that
Z 1
Þn P ðw
Þ
pðnÞ ðt Þeiwt dt ¼ ðiw
(9.4.21)
1
Example 9.4.1 The Fourier transform of a function
Find the Fourier transform of the function pðt Þ ¼ ejt j .
Z
Solution
Þ ¼
P ðw
¼
1
1
Z 0
ejt j eiwt dt
eð1iwÞt dt +
1
1
1
+
1 + iw
1 iw
2
¼
2
1+w
Z
1
eð1 + iwÞt dt
0
¼
9.5
Response to a nonperiodic load
In Section 3.7.3, we have seen that the response of a SDOF system to an arbitrary load is given by the convolution integral
Z t
u ðt Þ ¼ pðt Þ∗ h ðt Þ ¼
pðτÞh ðt τÞdτ
(9.5.1)
0
where h ðt τÞ is the response to the unit impulse given by
h ðt τ Þ ¼
exwðtτÞ
sin wD ðt τÞ, t > τ
mwD
(9.5.2)
308 PART
I Single-degree-of-freedom systems
for x 6¼ 0 and
h ðt τ Þ ¼
1
sin wðt τÞ, t > τ
mw
for x ¼ 0.
The integral (9.5.1) can be also written as
Z 1
u ðt Þ ¼ pðt Þ∗ h ðt Þ ¼
pðτÞh ðt τÞdτ
(9.5.3)
(9.5.4)
1
because pðτÞ ¼ 0, when τ < 0 and h ðt τÞ ¼ 0, when τ > t.
The establishment of the response to an arbitrary load in the frequency
domain is achieved by taking the Fourier transform of the convolution
(9.5.4). Thus, we have
Z 1 Z 1
Þ ¼
U ðw
pðt Þh ðt τÞdτ eiwt dt
(9.5.5)
1
1
If we set t τ ¼ s, then t ¼ τ + s and Eq. (9.5.5) gives
Z 1 Z 1
Þ ¼
h ðs Þeiws ds pðτÞeiwτ dτ
U ðw
1
Z1
1
ÞpðτÞeiwτ dτ
H ðw
¼
1
Z 1
Þ
¼ H ðw
pðτÞeiwτ dτ
(9.5.6)
1
ÞH ðw
Þ
¼ P ðw
From Eq. (9.5.6) we deduce that the Fourier transform of the response to an
arbitrary load, namely of the convolution integral, is equal to the product of the
Fourier transforms of the functions in the convolution. Hence, we may write
symbolically
Þ
pðt Þ , P ðw
(9.5.7)
Þ
h ðt Þ , H ðw
(9.5.8)
Þ
u ðt Þ ¼ pðt Þ ∗h ðt Þ , U ðw
(9.5.9)
The Fourier transform of the function u ðt t0 Þ is obtained as
Z 1
F½uðt t0 ¼
u ðt t0 Þeiwt dt
1
Z 1
¼
u ðxÞeiwðx + t0 Þ dt
1
Z 1
¼ eiwt0
u ðx Þeiwx dt
1
¼ eiwt0 F ½u ðt Þ
(9.5.10)
Analysis in the frequency domain Chapter
9
309
The forgoing equation represents the shifting property of the Fourier
transform.
Þ of the response h ðt τÞ to the unit impulse can
The Fourier transform H ðw
be obtained as follows.
The equation of motion of the response h ðt τÞ to the unit impulse results by
setting pðt Þ ¼ d ðt τÞ in the equation of the SDOF system. That is,
m h€ + ch_ + kh ¼ dðt τÞ
By virtue of Eqs. (9.5.10), (9.4.21), (3.7.6b), Eq. (9.5.11) gives
2 + ci w
+ k H ðw
Þ ¼ 1
m w
(9.5.11)
(9.5.12)
from which we obtain
Þ ¼
H ðw
2
ðm w
1
+ kÞ
+ ci w
(9.5.13)
or
1
Þ ¼ 2
H ðw
k b 1 + 2ixb
(9.5.14)
Eq. (9.5.14) is identical to Eq. (9.3.8), that is, the complex response function
is the Fourier transform of the response of the SDOF system to the unit impulse.
Example 9.5.1 The Fourier transform method for the dynamic response of a
SDOF system
Determine the response of a SDOF (k,w,x) system subjected to the load
pðt Þ ¼ p0 , 0 < t using the frequency domain analysis.
Solution
The Fourier transform of the load is
Z
Z
1 1
1 1
p0
t
i w
Þ ¼
P ðw
p0 e
dt ¼
p0 eiwt dt ¼
2p 1
2p 0
2pi w
(1)
The Fourier transform of the response to the unit impulse is
1
, b ¼ w
Þ ¼ 2
=w
H ðw
k b 1 + 2ixb
Consequently, by virtue of Eq. (9.5.9), we obtain
p0
2
Þ ¼ P ðw
ÞH ðw
Þ ¼
U ðw
k b 1 + 2ixb
2pi w
(2)
(3)
The response in the time domain results as the inverse Fourier transform of
Þ, that is,
U ðw
Z 1
p0
eiwt
2
dw
(4)
u ðt Þ ¼
2pik 1 w
b + 2ixb 1
310 PART
I Single-degree-of-freedom systems
which is further written as
p0
u ðt Þ ¼
2pikw
Z
1
eiwbt
db
1 b ðb b 1 Þðb b 2 Þ
where b 1 , b2 are the roots of the polynomial b2 + 2ixb 1, namely
qffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffi
b 1 ¼ ix + 1 x2 , b 2 ¼ ix 1 x 2
(5)
(6)
The integral (5) is evaluated using the method of closed line integrals in the
complex domain of b. This method yields
u ðt Þ ¼ 0, t 0
"
!#
p0
x
xwt
1e
cos wD t + pffiffiffiffiffiffiffiffiffiffiffiffi sin wD t , 0 < t
u ðt Þ ¼
k
1 x2
From the last example, we observe that the dynamic analysis in the frequency domain requires the evaluation of complicated integrals, even for the
simplest load cases. This problem is circumvented by applying numerical
methods for the evaluation of the Fourier transform, such as the discrete Fourier
transform (DFT), and the fast Fourier transform (FFT). These methods are discussed in the next sections.
9.6 Discrete Fourier transform
In the previous paragraph, we developed the method of dynamic analysis in the
frequency domain. The steps we followed are summarized as follows:
Þ ¼ F ½pðt Þ and
1. We take the Fourier transforms of the loading function, P ðw
Þ ¼ F ½h ðt Þ.
the response function to the unit pulse H ðw
Þ ¼ P ðw
ÞH ðw
Þ obtained
2. We evaluate the product of the two transforms U ðw
in the previous step.
3. We evaluate the inverse Fourier transform, which gives the response of the
Þ in the time domain.
system u ðt Þ ¼ F 1 ½U ðw
The success of the method is based on the capability of finding the Fourier transform of a given function and its inverse. However, this is not always easy. With
the exception of simple functions whose Fourier transforms are obtained from
tables [4], their establishment requires the computation of complicated integrals, a task that is difficult, tedious, or even impossible.
Very often, the loading function is specified by a set of values at distinct time
instants, for example, the accelerogram of an earthquake. As we mentioned, the
previous difficulties can be overcome by developing methods for the numerical
computation of the Fourier transform (direct and inverse). The discrete
Fourier transform (DFT) is among them. Below, we present the numerical
Analysis in the frequency domain Chapter
9
311
implementation of the DFT and its application for dynamic analysis in the frequency domain.
We consider the function pðt Þ of Fig. 9.6.1, which is defined in the interval
0 t ttot .
tot
FIG. 9.6.1 Nonperiodic function expanded perodically.
Then we assume that the function is periodically extended from 1 to +1
with a period T0 ttot to include a number of zero values of pðt Þ in T0 .
If this function is expanded in Fourier series, we obtain by virtue of
Eqs. (9.2.7), (9.2.8a), (9.2.8b), (9.2.8c)
1
X
pðt Þ ¼
cn ein wt
(9.6.1)
,
(9.6.2)
n¼1
1
cn ¼
T0
1
¼
T0
Z
T0 =2
T0 =2
Z T0
0
pðt Þeinwt dt
pðt Þe
t
in w
n ¼ 0, 1, 2, …
dt
The integral (9.6.2) can be evaluated numerically. For this purpose, we
divide the interval T0 into N equal subintervals of length Dt ¼ T0 =N and
approximate the integral with a sum of N rectangles.
If we set
s ðt Þ ¼ pðt Þeinwt
(9.6.3)
then we have
cn ¼
1
1 NX
s ðtk ÞDt
T0 k¼0
(9.6.4)
where s ðtk Þ is the value of the integrand at instants tk ¼ kDt. That is,
s ðtk Þ ¼ pðtk ÞeinwkDt
(9.6.5)
312 PART
I Single-degree-of-freedom systems
or taking into account that
¼ 2p=T0 ¼ 2p=N Dt
w
T0 ¼ N Dt,
(9.6.6)
we obtain
s ðtk Þ ¼ pðtk Þe2pikn=N
(9.6.7)
and Eq. (9.6.4) becomes
cn ¼
1
1 NX
pðtk Þe2pikn=N
N k¼0
(9.6.8)
As was shown in Section 3.6, the Fourier series can be approximated by a
sum of finite terms, Eq. (3.6.9), with very good accuracy even in cases of a discontinuous function, as demonstrated by Example 3.6.2. Therefore, on the basis
of Eq. (9.6.6) we can write Eq. (9.6.1) as
M
X
pðtk Þ ’
cn e2pikn=N
(9.6.9)
n¼M
In the foregoing equation, we split the sum into two sums
1
X
pðtk Þ ¼
cn e2pikn=N +
M
X
cn e2pikn=N
(9.6.10)
n¼0
n¼M
then taking into account that the function pðt Þ is periodic with a period
T0 ¼ N Dt, we can write the first sum as
1
X
cn e2pikn=N ¼
n¼M
1
X
cn + N e2pik ðn + N Þ=N
(9.6.11)
n¼M
Subsequently, setting M ¼ ðN 1Þ=2, n~ ¼ n + N and taking into account
the periodicity of pðt Þ, we can write the right side of the previous equation as
2M
X
cn~ e2pik n~ =N ¼
n~ ¼M + 1
M
X
cn e2pikn=N
(9.6.12)
n¼1
Substituting Eq. (9.6.11) into Eq. (9.6.10), and taking into account
Eq. (9.6.12), we obtain
pðtk Þ ¼
2M
X
cn e2pikn=N
(9.6.13)
cn e2pikn=N
(9.6.14)
n¼0
or because 2M ¼ N 1, it follows:
pðtk Þ ¼
N
1
X
n¼0
Analysis in the frequency domain Chapter
9
313
In order to match the expressions between the continuous Fourier transform
and the discrete Fourier transform, we set cn ¼ Pn and Eqs. (9.6.8), (9.6.14) are
written
Pn ¼
1
1 NX
pk e2pikn=N ,
N k¼0
pk ¼
N
1
X
Pn e2pikn=N ,
k ¼ 0, 1, 2, …,N 1
k ¼ 0, 1, 2, …,N 1
(9.6.15)
(9.6.16)
n¼0
The foregoing relations express the discrete Fourier transform (DFT), direct
and inverse, respectively.
The DFT approximates numerically the continuous Fourier transform,
defined by Eqs. (9.4.16), (9.4.17). The accuracy of the DFT is very good if
Dt is selected small. However, there is a fundamental difference between the
continuous Fourier transform and the discrete Fourier transform. The first provides the exact transform of the actual function while the second assumes a periodic extension of the function. This means that the discrete transform is
applicable when the interval T0 is finite. It holds only within the period. Outside
it, the two transforms are completely different unless the function happens to be
periodic.
9.7 Application of the discrete Fourier transform
to dynamic analysis
As we mentioned in Section 9.5, the response in the time domain of a SDOF
system subjected to an arbitrary load is given by the convolution integral of
the load function pðt Þ and the response function to the unit impulse h ðt Þ.
Namely,
u ðt Þ ¼ pðt Þ ∗h ðt Þ
(9.7.1)
Eq. (9.7.1) can be used as the basis to obtain the response in the frequency
domain by the DFT adhering to the following steps:
1. We compute the DFT of the function pðt Þ, 0 t ttot , which is assumed
extended periodically with a period T0 ttot so that pðtN Þ ¼ 0.
Pn ¼
1
1 NX
pðtk Þe2pikn=N
N k¼0
(9.7.2)
2. We compute the DFT of the response function h ðt Þ. This requires the confinement of h ðt Þ in an interval equal or smaller than T0 .
Hn ¼
1
1 NX
h ðtk Þe2pikn=N
N k¼0
(9.7.3)
314 PART
I Single-degree-of-freedom systems
3. We compute the DFT of the product
U n ¼ Pn H n
(9.7.4)
4. We compute the inverse DFT of the product
u ðtk Þ ¼
N
1
X
Un e2pikn=N ,
k ¼ 0, 1, 2, …,N 1
(9.7.5)
n¼0
which yields the response in the time domain.
Details about the application of the DFT method to dynamic analysis can be
found in the relevant literature, for example, Refs. [3, 5].
9.8 Fast Fourier transform
The discrete convolution of two functions pðt Þ, h ðt Þ is defined as
u ðtk Þ ¼
N
1
X
pðtm Þh ðtkm ÞDt, tk ¼ kDt, tkm ¼ ðk m ÞDt
(9.8.1)
m¼0
where both functions pðt Þ and h ðt Þ are periodic, that is,
pðtm + rN Þ ¼ pðtm Þ
(9.8.2a)
h ðtm + rN Þ ¼ h ðtm Þ, r ¼ 0, 1, 2, …
(9.8.2b)
Eq. (9.8.1) gives the dynamic response u ðtk Þ directly in the time domain at
time tk if pðt Þ represents the load and h ðt Þ the response to the unit
impulsive load.
We observe that the time-domain analysis based on Eq. (9.8.1) requires N 2
multiplications between real numbers. Instead, for the dynamic analysis in the
frequency domain, the required multiplications are N 2 between real and complex numbers as dictated by each of Eqs. (9.7.2), (9.7.3), N of complex numbers
as dictated by Eq. (9.7.4), and N 2 multiplications of complex numbers as dictated by Eq. (9.7.5).
It is clear that the dynamic analysis in the frequency domain using the DFT
requires significantly more computations, a fact that does not encourage its use.
However, it is possible to reduce the number of computations drastically by taking advantage of the harmonic properties of the involved functions. The algorithm that reduces the calculations is referred to as the fast Fourier transform
(FFT). The appearance of the FFT gave an impetus to dynamic analysis in the
frequency domain. The number of operations (multiplications) required by the
FFT decreases from N 2 to N log 2 N . Fig. 9.8.1 illustrates the advantage of FFT
Analysis in the frequency domain Chapter
9
315
over DFT. The first FFT algorithm was developed by Gauss in the early 19th
century [6]. Also, the contributions of Runge, Danielson, Lanczos, and others
in the early 20th century were significant. However, its use did not attract the
interest of many researchers because the calculations had to be performed by
hand. It was only with the advent of computers that the FFT came to the foreground. In 1965, J. W. Cooley and J. W. Tukey published an algorithm for calculating the FFT [7, 8]. This algorithm is similar to that of Gauss and others and
is named after them as the Cooley-Tukey algorithm. Today, there are several
algorithms for FFT based on this algorithm. Below we present the Sande-Tukey
algorithm that is a variation of the Cooley-Tukey.
FIG. 9.8.1 Number of operations in DFT and FFT.
9.8.1
The Sande-Tukey algorithm
In this algorithm, we assume that N is a power of 2, that is,
N ¼ 2M
(9.8.3)
where M is an integer. This constraint is introduced to simplify the algorithm.
In general, the DFT can be represented as
Pn ¼
N
1
X
k¼0
~k ¼ pk =N , n ¼ 0, 1, 2, …,N 1
~k e2pikn=N , p
p
(9.8.4)
316 PART
I Single-degree-of-freedom systems
Eq. (9.8.4) can also be written in the form
Pn ¼
N
1
X
~k W nk
p
(9.8.5)
k¼0
where W is the complex weight function defined as
W ¼ e2pi=N
(9.8.6)
We divide now the interval into two subintervals, and we express
Eq. (9.8.4) as
Pn ¼
ðNX
=2Þ1
N
1
X
~
pk e2pikn=N +
~k e2pikn=N , n ¼ 0, 1, 2, …,N 1 (9.8.7)
p
k¼N =2
k¼0
Then we introduce a new variable m ¼ k N =2, so that the total number
of the indices is the same in both sums. Thus, we may write the foregoing
equation as
Pn ¼
ðNX
=2Þ1
~k e2pikn=N +
p
ðNX
=2Þ1
~m + N =2 e2pinðm + N =2Þ=N
p
(9.8.8)
m¼0
k¼0
or
Pn ¼
ðNX
=2Þ1
~k + epin p
~k + N =2 e2pikn=N
p
(9.8.9)
k¼0
n
We observe that eipn ¼ ðeip Þ ¼ ð1Þn . Consequently, for points with
even n this factor is equal to one while with odd n it is equal to 1. The next
step is to separate the terms of Eq. (9.8.9) into two sums corresponding to the
even and odd values of n. Hence, for even values, we have
P2n ¼
ðNX
=2Þ1
~k + N =2 e2pik ð2nÞ=N
~k + p
p
k¼0
¼
ðNX
=2Þ1
k¼0
(9.8.10)
~k + N =2 e2pikn=ðN =2Þ
~k + p
p
Analysis in the frequency domain Chapter
9
317
while for odd values
P2n + 1 ¼
ðNX
=2Þ1
~k + N =2 e2pik ð2n + 1Þ=N
~k p
p
k¼0
¼
ðNX
=2Þ1
(9.8.11)
~k + N =2 e2pik=N e2pikn=ðN =2Þ
~k p
p
k¼0
where n ¼ 0, 1, 2, …, ðN =2Þ 1.
By virtue of Eq. (9.8.6), Eqs. (9.8.10), (9.8.11) may be written
P2n ¼
NX
=21
~k + N =2 W 2kn
~k + p
p
(9.8.12)
~k ~
pk + N =2 W k W 2kn
p
(9.8.13)
k¼0
P2n + 1 ¼
NX
=21
k¼0
We can now make an important observation, which is the key to the method.
The even and the odd expressions can be considered as two DFTs of N =2 points
each. We further set
~k + p
~k + N =2
gk ¼ p
~k p
~k + N =2 W k , k ¼ 0, 1, 2, …, ðN =2Þ 1
hk ¼ p
(9.8.14)
(9.8.15)
Hence
P2n ¼ Gn
P2n + 1 ¼ Hn
)
, n ¼ 0, 1, 2, …, ðN =2Þ 1
(9.8.16)
In other words, a computation at N points has been replaced by two computations at N =2 points each. Because each of the latter computations requires
ðN =2Þ2 complex calculations against N 2 required by the initial computation,
the number of multiplications is reduced to 2ðN =2Þ2 ¼ N 2 =2. It is obvious,
therefore, that the process of bisecting the transforms can be repeated in a second stage. We can thus compute 4 DFT with N =4 points each using the first and
last point of each sequence of N =4 points. This technique continues and ends in
318 PART
I Single-degree-of-freedom systems
DFT with two points (see Fig. 9.8.2). The total number of complex multiplication for a given DFT reduces to N log 2 N . The importance of FFT over DFT is
demonstrated in Fig. 9.8.1.
FIG. 9.8.2 Flow chart of the first substitution stage of the DFF with N points by two DFF of N =2
points each when N ¼ 8.
On the basis of the previous analysis, a computer program has been
written in MATLAB that evaluates the dynamic response of a SDOF. The program uses the MATLAB functions fft.m and ifft.m. This program, which is
given the name FFT_dymamic.m, is available on this book’s companion
website.
Example 9.8.1 The Fourier transform for the dynamic response of a SDOF
system
Determine the response of SDOF system subjected to the Athens earthquake
in 1999 using the FFT method. Data: m ¼ 1, x ¼ 0:1, k ¼ 25, and pðt Þ ¼ u€g ðt Þ
where u€g ðt Þ is the accelerogram of the earthquake.
Solution
The solution is obtained using the program FFT_dymamic.m. The response
of the system is shown in Fig. E9.1 as compared with the numerical solution
in the time domain using the program aem.lin.m given in Section 4.4. It
is obvious that the computed responses by both methods are graphically
identical.
u(t) (m)
Analysis in the frequency domain Chapter
u,t(t) (m/s)
t
u,tt(t) (m/s2)
t
t
FIG. E9.1 Dynamic response of the SDOF system using FFT in Example 9.8.1.
9
319
320 PART
I Single-degree-of-freedom systems
Example 9.8.2 Amplitude spectrum of an accelerogram
Compute and plot the amplitude spectrum of the 1999 Athens earthquake
using the FFT method.
Solution
The amplitude spectrum is established by computing the magnitude of the
values of the Fourier transform of the accelerogram and it is plotted versus
the frequency (Hz) in the interval of the duration of the excitation. Fig. E9.2
shows the amplitude spectrum of the Athens 1999 earthquake as computed
using the program FFT_Ampl_Spectrum.m, which utilizes the function fft.m
of MATLAB. The program FFT_Ampl_Spectrum.m is available on this
book’s companion website.
FIG. E9.2 Amplitude spectrum of the 1999 Athens earthquake in Example 9.8.2.
9.9 Problems
Problem P9.1 Write a computer program for the evaluation of DFT and compute the DFT of the function shown in Fig. P9.1. Then compute the inverse DFT
and compare the results with the exact function.
FIG. P9.1 Function p(t) in problem P9.1
Analysis in the frequency domain Chapter
9
321
Problem P9.2 The SDOF system m, x,k is subjected to a ground motion ug ðt Þ.
Study its dynamic response using:
(a) The exact analytic method in the time domain.
(b) The numerical computation of the convolution integral in the time domain.
(c) The DFT in the frequency domain.
h
i
Data: x ¼ 0:1, u€gi ¼ 20 1 + ð1Þi ði + 5Þ=ði + 1Þ , ti ¼ 0:05i, i ¼ 1, 2, …,20,
u ð0Þ ¼ u_ ð0Þ ¼ 0, and w ¼ 4ps1 .
Þ versus the freProblem P9.3 Give the graph of the Fourier transform Aðw
quency (Hz) of the accelerogram of the Mexico City earthquake using the
FFT. The file Mexico_Earthquake.txt including the values of the accelerogram
is available on this book’s companion website.
Problem P9.4 The water tower of Fig. P9.4a is subjected to the load
pðt Þ ¼ 150sin pt=t1 , t1 ¼ 0:5 (Fig. P9.4b), which acts at the center of mass of
the tank. Approximating the structure by a generalized SDOF system, study
the dynamic response in the frequency domain for ttot ¼ 1s. The material of
the structure is reinforced concrete with a specific weight g ¼ 25kN=m3 . The column and the tank have a circular cross-section. The interior of the tank is divided
into small compartments to eliminate the liquid-structure interaction phenomena.
Use the first mode shape of the cantilever with a constant cross-section as a shape
function, that is, ψðx Þ ¼ ½ coshlx cos lx 0:7341ð sinh lx sinhlx Þ=3,
l ¼ 1:8751=L, where L is the length of the cantilever.
(b)
(a)
FIG. P9.4 Water tower (a) and load (b) in problem P9.4
322 PART
I Single-degree-of-freedom systems
References and further reading
[1] R. Bracewell, The Fourier Transform and Its Applications, McGraw-Hill, New York, 1978.
[2] T. Myint-U, L. Debnath, Linear Partial Differential Equations for Scientists and Engineers,
fourth ed., Birkh€auser, Boston, 2007.
[3] J.L. Humar, Dynamics of Structures, second ed., A.A. Balkema Publishers, Lisse, NL, 2002.
[4] F. Oberhettinger, Tabellen zur Fourier Transformation, 1957 Sringer-Verlag, Berlin, 1957.
[5] R.R. Craig Jr., A.J. Kurdila, Fundamentals of Structural Dynamics, second ed., John Wiley,
New Jersey, 2006.
[6] M.T. Heideman, D.H. Johnson, C.S. Burrus, Gauss and the history of the fast Fourier transform,
IEEE ASSP Mag. 1 (4) (1984) 14–21.
[7] J.W. Cooley, J.W. Tukey, An algorithm for the machine calculation of complex Fourier series,
Math. Comput. 19 (1965) 297–301.
[8] J.F. Hall, A FFT algorithm for structural dynamics, Earthq. Eng. Struct. Dyn. 10 (1982)
797–811.
[9] W.H. Press, B.P. Flannery, S.A. Teukolsky, W.T. Vetterlin, Numerical Recipes in FORTRAN,
second ed., Cambridge University Press, New York, 1992.
Chapter 10
Multi-degree-of-freedom
systems: Models and equations
of motion
Chapter outline
10.1 Introduction
10.2 Systems with localized mass
and localized stiffness
10.3 Systems with distributed
mass and localized stiffness
10.4 Systems with localized mass
and distributed stiffness
10.4.1 The method of
influence coefficients
325
327
328
330
334
10.5 Systems with distributed
mass and distributed stiffness
10.5.2 The method of global
shape functions
10.6 Mixed systems
10.7 Transformations of the
equations of motion
10.8 Problems
References and further reading
341
342
347
351
354
358
10.1 Introduction
So far, we have studied the dynamic response of SDOF systems. We have also
shown how a system with infinite degrees of freedom can be approximated by
a SDOF system. The trustworthiness of this approximation depends on various
issues. If the actual distribution of the physical properties of the structure, that
is, mass and stiffness, and that of the external force produce deformation during
the motion similar to the assumed, then the approximation with a SDOF system
gives acceptable results. A key shortcoming of this approximation is the difficulty in determining the degree of reliability of the obtained results. In general,
however, the study of the dynamic response of structures requires their modeling with MDOF systems, especially when the deformation shapes are complicated. In engineering structures, the mass, though distributed to all its members,
is usually lumped at certain points or regions. For example, in buildings the
mass is lumped at the levels of the stories or in a water tower at the top of
the column that supports the tank. This fact allows describing the motion of
a structure with that of a MDOF system with deformation parameters the displacements of the points where the dynamic characteristics (mass and moment
of inertia) are concentrated.
Fig. 10.1.1a shows a three-story frame whose horizontal beams are virtually
rigid. In this structure, the mass of the columns is negligible compared to that of
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00010-8
© 2020 Elsevier Inc. All rights reserved.
325
326 PART
II Multi-degree-of-freedom systems
the beams. Hence, the mass is lumped at the level of the beams and the structure
can be approximated by the model of Fig. 10.1.1b. The deformation shape during motion is shown in Fig. 10.1.1c. Obviously, its motion can be determined by
establishing the displacements u1 ðt Þ, u2 ðt Þ, and u3 ðt Þ, that is, the system has
three degrees of motion.
(a)
(b)
(c)
FIG. 10.1.1 Three-story frame (a), dynamic model (b) and deformation shape (c).
(a)
(b)
FIG. 10.1.2 Water tower (a) and its deformed dynamic model (b).
Fig. 10.1.2a shows a water tower. With the assumption that the mass of the
column is negligible compared with that of the tank, we can model the water
tower with the system of Fig. 10.1.2b. That is, the water tower is simulated
by a flexible column, which is fixed at the ground and has a mass m at its
top with a moment of inertia Io . The mass m can move horizontally and rotate
within the plane. The determination of the motion requires the establishment of
the displacement u ðt Þ and the rotation fðt Þ of the top cross-section of the column, that is, the system has two degrees of freedom.
The MDOF systems can be categorized as follows:
1. Systems with localized mass and localized stiffness.
2. Systems with distributed mass and localized stiffness.
MDOF systems: Models and equations of motion Chapter
10
327
3. Systems with localized mass and distributed stiffness.
4. Systems with distributed mass and distributed stiffness.
5. Mixed systems resulting from a combination of the systems of previous
categories.
In what it follows, the term localized mass denotes a lumped mass that may have
a moment of inertia while the term concentrated mass denotes a mass that is
concentrated at a point and has a zero moment of inertia. The above classification is useful as it dictates a convenient method to formulate the equations of
motion of the MDOF system. Its effectiveness is explained by illustrative examples. The pertinent bibliography with recommended references for further study
is also included. The chapter is enriched with problems to be solved.
10.2 Systems with localized mass and localized stiffness
The system of Fig. 10.2.1a consists of two rigid bodies with masses m1 and m2 ,
whose horizontal motion is elastically constrained by two massless springs k1 and
k2 . The system has localized masses and stiffnesses. The degrees of freedom are
two, the displacements u1 ðt Þ and u2 ðt Þ. The equations of motion can be derived
using the method of the equilibrium of the forces (D’Alembert’s principle) presented in Section 1.5. The forces applied to the masses are shown in Fig. 10.2.1b.
The equilibrium of mass m1 gives
fI 1 + fD1 + fS1 ¼ p1
(10.2.1)
m1 u€ 1 + c1 ðu_ 1 u_ 2 Þ + k1 ðu1 u2 Þ ¼ p1 ðt Þ
(10.2.2)
or
Similarly, the equilibrium of mass m2 gives
fI 2 + fD2 fD1 + fS2 fS1 ¼ p2
(10.2.3)
or
m2 u€2 c1 u_ 1 + ðc2 + c1 Þu_ 2 k1 u1 + ðk2 + k1 Þu2 ¼ p2 ðt Þ
or
(10.2.4)
Eqs. (10.2.2), (10.2.4) are written in matrix form
m1 0
u€1
c1 c1
u_ 1
k1 k1
u1
+
+
u€2
u_ 2
c1
c2 + c1
k1
k2 + k1
u2
0 m2
p1
(10.2.5)
¼
p2
M€
u + Cu_ + Ku ¼pðt Þ
where
c1 c1
k1 k1
m1 0
, C¼
, K¼
M¼
0 m2
c1
c1 + c2
k1
k1 + k2
(10.2.6)
(10.2.7)
328 PART
II Multi-degree-of-freedom systems
are the mass, the damping, and stiffness matrices of the system, respectively, and
p1 ð t Þ
u1
(10.2.8)
, pðt Þ ¼
u¼
u2
p2 ðt Þ
the displacement and load vectors.
(a)
(b)
FIG. 10.2.1 System with localized mass and localized stiffness (a). Forces applied to the masses
m1 and m2 (b).
10.3 Systems with distributed mass and localized stiffness
The system of Fig. 10.3.1a consists of the rigid bar AC , whose mass is uni and the circular
formly distributed along its length with line mass density m,
also uniformly distributed. The motion of
rigid body of total mass m ¼ mL,
the system is elastically constrained by the two massless springs, k1 ¼ k and
k2 ¼ 2k. A damper with damping coefficient c acts at point C . The system is
loaded by the moment M ðt Þ and the force pðt Þ. This structure, which represents a system with distributed mass and localized stiffness, has two degrees
of freedom. We choose the vertical displacement u ðt Þ of the end C of the bar
and the rotation fðt Þ of the circular body to determine the motion of the system. Referring to Fig. 10.3.1b, the equilibrium of the rigid bar with respect to
point A gives:
MDOF systems: Models and equations of motion Chapter
10
329
(a)
(b)
FIG. 10.3.1 System with distributed mass and localized stiffness (a). Forces acting on the two
bodies (b).
MIA + 2Lf D + Lf S1 + 2Lf S2 ¼ pðt ÞL
(10.3.1)
Obviously, we have
MIA ¼ IA
ð2LÞ3 u€ 4
u€
_ fS1 ¼ 0:5ku
2 u€ , fD ¼ cu,
¼m
¼ mL
3 2L 3
2L
fS2 ¼ 2k ðu 0:5LfÞ
which are substituted into Eq. (10.3.1) to yield
2
9
1
u€ + cu_ + ku kLf ¼ pðt Þ
mL
3
4
2
(10.3.2)
Similarly, the equilibrium of the circular body with respect to point O gives
MIO 0:5Lf S2 ¼ M ðt Þ
(10.3.3)
Taking into account that
MIO ¼ IO f€ ¼
mL
3€
1
mL
4€
f, fS2 ¼ 2k ðu 0:5LfÞ
f
¼
p
ð
0:5L
Þ
8
pð0:5LÞ2 2
and substituting into Eq. (10.3.3), we obtain
3€
mL
f kLu + 0:5L2 kf ¼ M ðt Þ
8
(10.3.4)
330 PART
II Multi-degree-of-freedom systems
Eqs. (10.3.2), (10.3.4) are the equations of motion, which in matrix form
become
M€
u + Cu_ + Ku ¼pðt Þ
(10.3.5)
where
3
2
"
#
9
0
mL
c 0
7
63
k kL
, C¼
, K¼
M¼4
4
35
mL
0 0
kL 0:5L2 k
0
8
2
(10.3.6)
represent the mass, damping, and stiffness matrices of the system and
9
8
< 1 pðt Þ =
u
(10.3.7)
u¼
, pðt Þ ¼ 2
f
;
:
M ðt Þ
the displacement and load vectors.
10.4 Systems with localized mass and distributed stiffness
The system of Fig. 10.4.1 a represents an example of this case. It consists of the
flexible column AO, whose mass is assumed negligible, and the plane square
body of side length a and mass m uniformly distributed.
(a)
(b)
FIG. 10.4.1 System with localized mass and distributed stiffness (a). Deformed dynamic model (b).
The equations of motion will results from the motion of the plane rigid body in
its plane. We examine the motion with respect to the point O, which does not coincide with the center of mass of the body. In general, the system has three degrees of
freedom, namely the horizontal displacement, the vertical displacement, and the
rotation about O. Because the column is flexible, the horizontal displacement and
the rotation are due to the bending deformation. The vertical displacement is
caused by (i) the axial deformation of the column, which is very small and thus
MDOF systems: Models and equations of motion Chapter
10
331
neglected, and (ii) the shortening of the chord of the deflection curve, which is
also neglected in the linear theory. Therefore, the parameters of the motion are
the horizontal displacement u ðt Þ and the rotation fðt Þ, Fig. 10.4.1b. In the following, the equations of motion are derived using two different methods.
1. The method of the Lagrange equations
(i) Elastic energy: This is due to the bending deformation of the column. The
deflection curve can be set in the form
uðx, t Þ ¼
1 ðx Þu ðt Þ +
2 ðx Þfðt Þ
(10.4.1)
where 1 ðx Þ and 2 ðx Þ are the elastic curves of the column for u ðt Þ ¼ 1,
fðt Þ ¼ 0 and fðt Þ ¼ 1, u ðt Þ ¼ 0, respectively. They can be obtained from
the solution of the following two boundary value problems
d4 1
¼ 0,
dx 4
1 ð0Þ ¼ 0,
1
0
ð0Þ ¼ 0,
1 ðh Þ ¼ 1,
1
0
ðh Þ ¼ 0
(10.4.2a)
and
d4 2
¼ 0,
dx 4
2 ð0Þ ¼ 0,
2
0
ð0Þ ¼ 0,
2 ðh Þ ¼ 0,
2
0
ðh Þ ¼ 1
(10.4.2b)
Integrating of the differential equation (10.4.2a) gives
1 ðx Þ ¼
1 3 1 2
c1 x + c2 x + c3 x + c4
6
2
(10.4.3)
After evaluation of the arbitrary constants by applying the boundary
conditions, we obtain
x 2
x 3
+2
¼ 3x2 + 2x3 , x ¼ x=h
(10.4.4)
1 ðx Þ ¼ 3
h
h
Similarly, we obtain
x 2 x
1
¼ hx 2 ðx 1Þ, x ¼ x=h
ð
x
Þ
¼
h
2
h
h
(10.4.5)
The elastic energy is given (see Eq. 8.1.11)
Z
1 h
2
EI ½u00 ðx, t Þ dx
U¼
2 0
which by virtue of Eq. (10.4.1) becomes
Z
1 h 00
2
EI 1 ðx Þu ðt Þ + 002 ðx Þfðt Þ dx
U¼
2 0
(10.4.6)
Differentiating Eq. (10.4.6) with respect to u ðt Þ and fðt Þ gives
332 PART
II Multi-degree-of-freedom systems
∂U
¼ EI
∂u
Z
h
0
00
1 ðx Þu ðt Þ +
Z
¼ EIu ðt Þ
0
h
2
00
1 ðx Þ dx
00
2 ðx Þfðt Þ
00
1 ðx Þdx
Z
h
+ EI fðt Þ
00
00
2 ðx Þ 1 ðx Þdx
0
12EI
6EI
u+ 2 f
h3
h
Z h
00
∂U
00
00
¼ EI
1 ðx Þu ðt Þ + 2 ðx Þfðt Þ
2 ðx Þdx
∂f
0
Z h
Z h
00
00
¼ EIu ðt Þ
1 ðx Þ 2 ðx Þdx + EI fðt Þ
(10.4.7a)
¼
0
¼
0
2
00
2 ðx Þ dx
(10.4.7b)
6EI
4EI
f
u+
h2
h
(ii) Kinetic energy: The kinetic energy with respect to point O is evaluated
from Eq. (1.5.8). Taking the origin of the coordinates at point O we have:
_ xc ¼ 0, yc ¼ a=2, IP ¼ IO and Eq. (1.5.8)
XP ¼ u, YP ¼ 0, w ¼ f,
becomes
1
1
a
T ¼ m u_ 2 + IO f_ 2 m u_ f_
2
2
2
(10.4.8)
Differentiating Eq. (10.4.8) with respect to u_ and f_ gives
∂T
a
¼ m u_ m f_
∂u_
2
∂T
a
¼ IO f_ m u_
2
∂f_
(10.4.9a)
(10.4.9b)
(iii) Generalized forces: They result from Eq. (1.8.23) by setting q1 ¼ u,
q2 ¼ f. Thus, we obtain
Q1 ¼ pðt Þ, Q2 ¼ pðt Þa=2
(10.4.10a,b)
(iv) The potential of the external conservative forces: Because there are no
conservative forces, it is A ¼ 0.
Substituting Eqs. (10.4.7a), (10.4.7b), (10.4.9a) (10.4.9b), (10.4.10a,b)
into the Lagrange equation (1.8.11) gives
a
12EI
6EI
m u€ m f€ + 3 u + 2 f ¼ pðt Þ
2
h
h
a
6EI
4EI
a
f ¼ pðt Þ
IO f€ m u€ + 2 u +
2
h
h
2
(10.4.11a)
(10.4.11b)
MDOF systems: Models and equations of motion Chapter
10
333
or in matrix form
M€
u + Ku ¼pðt Þ
where
2
u
u¼
,
f
2
3
12EI 6EI
6 h3
h2 7
7
K¼6
4 6EI 4EI 5
h2
h
(
)
pðt Þ
a
pðt Þ ¼
pðt Þ
2
ma 3
m 2 5,
M ¼ 4 ma
IO
2
(10.4.12)
(10.4.13a,b)
(10.4.13c,d)
2. The method of equilibrium of forces
The forces acting on the square rigid body during the motion are shown in
Fig. 10.4.2. The equilibrium of forces may be considered either with respect
to point O or to the center of mass C .
FIG. 10.4.2 Free body diagram.
a. Equilibrium with respect to point O.
Because point O does not coincide with the mass center, the equations
of motion will be obtained from Eqs. (1.5.14a), (1.5.14b), (1.5.14c) by
taking the origin of axes at the point O and setting: P O, uP ¼ u,
vP ¼ 0, xc ¼ 0, yc ¼ a=2, Fx ¼ pðt Þ fS , Fy ¼ 0, MO ¼ pðt Þa=2 MS .
Thus, we obtain
a
(10.4.14a)
m€
u m f€ + fS ¼ pðt Þ
2
a
a
(10.4.14b)
m u€ + IO f€ + MS ¼ pðt Þ
2
2
The elastic forces fS and MS are the shear force and the bending
moment of a beam fixed at end A and subjected to the displacement u
and rotation f at end O. Thus, we have
fS ¼
12EI
6EI
u+ 2 f
3
h
h
(10.4.15a)
334 PART
II Multi-degree-of-freedom systems
MS ¼
6EI
4EI
f
u+
2
h
h
(10.4.15b)
Obviously, they are identical to those given by Eqs. (10.4.7a),
(10.4.7b).
Substituting the previous expressions for fS and MS in
Eqs. (10.4.14a), (10.4.14b) yields
a
12EI
6EI
m€
u m f€ + 3 u + 2 f ¼ pðt Þ
2
h
h
a
6EI
4EI
a
f ¼ pðt Þ
IO f€ m u€ + 2 u +
2
h
h
2
(10.4.16a)
(10.4.16b)
which are identical to Eqs. (10.4.11a), (10.4.11b).
b. Equilibrium with respect to the mass center C .
The equations of motion will be obtained from Eqs. (1.5.11a),
(1.5.11b), (1.5.11c) by setting uc ¼ u af=2, Fx ¼ fS + pðt Þ,
MC ¼ MS fS a=2. Thus, we obtain
a
m€
u m f€ + fS ¼ pðt Þ
(10.4.17a)
2
a
Ic f€ + MS + fS ¼ 0
(10.4.17b)
2
The first of the above equations is identical to (10.4.14a). The second
equation, however, looks different from (10.4.14b). Nevertheless, multiplying Eq. (10.4.17a) by a=2 and adding it to (10.4.17b) gives
a 2 a
a
f€ + MS ¼ pðt Þ
u + IC + m
(10.4.18a)
m€
2
2
2
Because IC + m ða=2Þ2 ¼ IO (Steiner’s formula), the previous equation becomes
a
a
u + IO f€ + MS ¼ pðt Þ
(10.4.18b)
m€
2
2
which is identical to (10.4.14b).
The above transformation of the equations of motion from the center
of mass to point O is rather occasional. A formal method to transform the
equations of motion when we change the point of reference is presented in
Section 10.7.
10.4.1 The method of influence coefficients
The equations of motion of linear systems with localized masses and distributed
stiffnesses can be derived using the method of influence coefficients, which is
based on the superposition principle. A frame structure can be idealized as an
assemblage of beam elements interconnected at nodal points. We illustrate this
MDOF systems: Models and equations of motion Chapter
10
335
method with the plane frame of Fig. 10.4.3. The masses are localized at the
nodes, where the external loads are applied. In the general case in which the
axial deformation of the beams and columns is not neglected, each node i
has three degrees of freedom with respect to the global system of axes X Y ,
Y axes, respectively, and a rotatwo translations ui , vi in the directions of the X,
tion fi about the Z axes. Hence, the frame of Fig. 10.4.3 with n ¼ 6 free nodes
has in total N ¼ 3n ¼ 18 degrees of freedom.
FIG. 10.4.3 Frame with n ¼ 6 nodes.
Referring to Fig. 10.4.4, we represent the displacement vector by
T ¼ u1 v1 f1 u2 v2 f2 ⋯ ui vi fi ⋯ un vn fn
u
(10.4.19)
or if they are labeled from 1 to N ¼ 3n, we may write
uT ¼ f u1 u2 u3 u4 u5 u6 ⋯ u3i2 u3i1 u3i ⋯ uN g
(10.4.20)
where
u3i2 ¼ ui ,
u3i1 ¼ vi ,
u3i ¼ fi ,
i ¼ 1, 2, …,n
FIG. 10.4.4 Nodal displacements of the frame. N ¼ 3n ¼ 18.
(10.4.21)
336 PART
II Multi-degree-of-freedom systems
Similarly, we formulate the vector of the external nodal loads (Fig. 10.4.5)
FIG. 10.4.5 External loads acting on the nodes of the frame.
pðt ÞT ¼ f px1 py1 M1 px2 py2 M2 ⋯ pxi pyi Mi ⋯ MN g
¼ f p1 p2 p3 p4 p5 p6 ⋯ p3i2 p3i1 p3i ⋯ pN g
(10.4.22)
where
p3i2 ¼ pxi ,
p3i1 ¼ pyi ,
p3i ¼ Mi ,
i ¼ 1, 2, …,n
(10.4.23)
The number of equations will be equal to the number of degrees of freedom.
The equations of motion can be formulated using the method of equilibrium of
forces. If fIi , fDi , fSi represent the inertia, damping, and elastic forces, respectively, and pi ðt Þ the external load in the direction of the displacement ui , then
it should be (see Fig. 10.4.6)
fIi + fDi + fSi ¼ pi ðt Þ,
(a)
i ¼ 1, 2, …,N
(10.4.24)
(b)
FIG. 10.4.6 Equilibrium of forces (a) or moments (b) in the direction of the displacement ui .
Eq. (10.4.24) holds in the direction of all displacement components
u1 , u2 ,…,uN , N ¼ 3n. The expressions of the forces fIi , fDi , fSi can be determined using the following method, which is referred to as the method of influence coefficients.
MDOF systems: Models and equations of motion Chapter
10
337
10.4.1.1 Elastic forces
In a SDOF system, the elastic force depends only on one displacement. Apparently, in a MDOF system this force will depend on all the displacements. Thus,
for a linear system, the superposition principle allows writing
fSi ¼ ki1 u1 + ki2 u2 + … + kij uj + … + kiN uN
(10.4.25)
FIG. 10.4.7 Stiffness influence coefficients ki4 .
where kij are constants expressing the stiffness influence coefficients or simply
the stiffness coefficients. They relate the elastic force fSi to the displacement uj .
The physical meaning of the coefficient kij results by setting uj ¼ 1 and uk6¼j ¼ 0
in Eq. (10.4.25), that is, by applying a unit displacement along the degree of
freedom j, holding all other displacements zero, as shown in Fig. 10.4.7. This
yields
fSi ¼ kij
(10.4.26)
Eq. (10.4.26) states that the stiffness coefficient kij expresses the elastic
force applied along the degree of freedom i for a unit displacement along the
degree of freedom j, that is, the displacement uj , while all other displacements
are zero.
Fig. 10.4.7 shows the deformation of the frame when the displacement u4 ¼ 1
is applied at node 4 while all other displacements are zero, that is, uj ¼ 0
( j ¼ 1, 2, …, N , j 6¼ 4). Obviously, the stiffness influence coefficients are equal
to the forces required to maintain the deformed shape of the frame when subjected to the displacement u4 ¼ 1.
For i ¼ 1, 2, …, N , Eq. (10.4.25) yields N equations, which we write in
matrix form
9 2
8
38 9
k k ⋯ k1N >
f >
>
>
>
> u1 >
= 6 11 12
< S1 >
7< u2 =
fS2
k
k
⋯
k
21
22
2N
7
¼6
(10.4.27)
4 ⋯ ⋯ ⋯ ⋯ 5> ⋮ >
⋮ >
>
>
>
>
;
;
:
: >
fSN
kn1 kn2 ⋯ kNN
uN
338 PART
II Multi-degree-of-freedom systems
or
f S ¼ ku
where
9
8
fS1 >
>
>
>
>
>
<
fS1 =
,
fS ¼
>
> ⋮ >
>
>
>
;
:
fSN
8 9
u1 >
>
>
>
> >
<
u2 =
u¼
,
>
> ⋮ >
>
>
>
: ;
uN
(10.4.28)
2
3
k11 k12 ⋯ k1N
6k k ⋯ k 7
2N 7
6 21 22
k¼6
7
4⋯ ⋯ ⋯ ⋯ 5
kn1 kn2 ⋯ kNN
(10.4.29)
The vector f S with dimensions N 1 represents the vector of the elastic
forces, the vector u with dimensions N 1 represents the vector of the displacements, and the matrix k with dimensions N N represents the stiffness matrix
of the structure. Obviously, this method, which explains the physical meaning
of the stiffness coefficients, is by no means suitable for their evaluation because
it requires N static analyses of the fixed structure successively for u1 ¼ 1,
u2 ¼ 1, …, uN ¼ 1. The stiffness matrix, however, can be established using
other methods of structural analysis, for example, the flexibility method or
the direct stiffness method. Anyhow, the establishment of the stiffness matrix
of a structure is a subject of the static structural analysis and it is evaluated using
the most suitable method for a particular structure.
10.4.1.2 Damping forces
The concept of the influence coefficients can be employed to express the damping forces fDi . Thus, assuming viscous damping we may write
fDi ¼ ci1 u_ 1 + ci2 u_ 2 + … + cij u_ j + … + ciN u_ N
(10.4.30)
Applying Eq. (10.4.30) for i ¼ 1, 2, …, N yields N equations, which we
write in matrix form
9 2
8
38 9
c11 c12 ⋯ c1N > u_ 1 >
fD1 >
>
>
>
> >
>
>
=
< f = 6 c c ⋯ c 7>
< u_ >
D2
2N 7
2
6 21 22
¼6
(10.4.31)
7
>
4 ⋯ ⋯ ⋯ ⋯ 5>
⋮ >
⋮ >
>
>
>
>
>
>
>
>
;
:
: ;
fDN
cn1 cn2 ⋯ cNN
u_ N
or
f D ¼ cu_
where
9
8
fD1 >
>
>
>
>
>
<
fD2 =
,
fD ¼
>
> ⋮ >
>
>
>
;
:
fDN
8 9
u_ 1 >
>
>
>
> >
<
u_ 2 =
u_ ¼
,
>
> ⋮ >
>
>
>
: ;
u_ N
(10.4.32)
2
c11 c12 ⋯ c1N
6c c ⋯ c
2N
6 21 22
c¼6
4⋯ ⋯ ⋯ ⋯
cn1 cn2 ⋯ cNN
3
7
7
7
5
(10.4.33)
MDOF systems: Models and equations of motion Chapter
10
339
FIG. 10.4.8 Damping influence coefficients ci4 .
The vector f D with dimensions N 1 represents the vector of the damping
forces, the vector u_ with dimensions N 1 represents the vector of the velocities, and the matrix c with dimensions N N represents the damping matrix of
the structure. The elements cij of the damping matrix are the damping influence
coefficients. Their physical meaning is analogous to that of the stiffness influence coefficients, namely cij expresses the damping force applied along the
degree of freedom i for unit velocity along the degree of freedom j, that is,
u_ j ¼ 1, while all other velocities are zero. Fig. 10.4.8 show the damping influence coefficients at the nodes of the frame, when u_ 4 ¼ 1, u_ j ¼ 0 (j ¼ 1, 2, …, N ,
j 6¼ 4). Note that the damping coefficients, contrary to the stiffness coefficients,
cannot be established from the geometrical data of the structure and the physical
properties of its elements. In Section 12.11, we will show how the damping
matrix of the structure can be established by assuming known damping ratios
for each mode shape estimated from experimental data obtained from similar
structures.
10.4.1.3 Inertial forces
Similarly, we can relate the inertial force fIi to the acceleration u€ j . Thus, we may
write
fIi ¼ mi1 u€ 1 + mi2 u€ 2 + … + mij u€ j + … + miN u€ N
(10.4.34)
For i ¼ 1, 2, …, N , Eq. (10.4.34) yields N equations, which we write in
matrix form
8 9 2
38 9
m11 m12 ⋯ m1N >
fI 1 >
u€ 1 >
>
>
>
>
=
< = 6
< >
7
€2
fI 2
m
m
⋯
m
u
21
22
2N 7
¼6
(10.4.35)
⋮ > 4 ⋯ ⋯ ⋯ ⋯ 5>
⋮ >
>
>
>
;
;
: >
: >
fIN
mn1 mn2 ⋯ mNN
u€ N
or
f I ¼ m€
u
(10.4.36)
340 PART
II Multi-degree-of-freedom systems
where
8 9
fI 1 >
>
>
=
< >
fI 2
,
fI ¼
⋮ >
>
>
;
: >
fIN
8 9
u€ 1 >
>
>
=
< >
u€ 2
€¼
u
,
>
>
> ⋮ >
;
:
u€ N
2
m11 m12
6 m21 m22
m¼6
4 ⋯ ⋯
mn1 mn2
3
⋯ m1N
⋯ m2N 7
7
⋯ ⋯ 5
⋯ mNN
(10.4.37)
The vector f I with dimensions N 1 represents the vector of the inertial
€ with dimensions N 1 represents the vector of the accelforces, the vector u
erations, and the matrix m with dimensions N N represents the mass matrix
or inertial matrix of the structure. The elements mij of the mass matrix are the
mass influence coefficients. Their physical meaning is analogous to that of the
stiffness and damping influence coefficients, namely mij expresses the inertial
force applied along the degree of freedom i for unit acceleration along the
degree of freedom j, that is, u€ j ¼ 1, while all other accelerations are zero.
Fig. 10.4.9 shows the mass influence coefficients at the nodes of the frame,
when u€ 4 ¼ 1, u€ j ¼ 0 (j ¼ 1, 2, …, N , j 6¼ 4).
FIG. 10.4.9 Mass influence coefficients mi4 .
In actual structures, the mass is distributed. The model that considers the
mass lumped at certain points of the structure, for example, at the nodes of a
frame, approximates adequately the dynamic response of the structure. When
it is assumed that the lumped mass has no geometrical dimensions, that is, it
is simulated by a material particle, then its rotational inertia is zero and the
respective influence coefficients vanish.
Writing now Eq. (10.4.24) for all directions i ¼ 1, 2, …, N , we obtain
fI 1 + fD1 + fS1 ¼ p1 ðt Þ
fI 2 + fD2 + fS2 ¼ p2 ðt Þ
… … … …
fIN + fDN + fSN ¼ pN ðt Þ
(10.4.38)
f I + f D + f S ¼ p ðt Þ
(10.4.39)
or
MDOF systems: Models and equations of motion Chapter
10
341
which by virtue of Eqs. (10.4.28), (10.4.32), (10.4.36) gives the equation of
motion of the structure
m€
u + cu_ + ku ¼ pðt Þ
(10.4.40)
10.5 Systems with distributed mass and distributed stiffness
Systems with distributed mass and distributed stiffness are also referred to as
continuous systems or distributed parameter systems. Theoretically, they have
infinite degrees of freedom and their motion is described by partial differential
equations. The systems of Fig. 10.5.1 belong to this category.
(a)
(b)
FIG. 10.5.1 Continuous systems: (a) Cantilever beam, (b) Chimney fixed on the ground.
Continuous systems can be approximated by MDOF systems, even if the
mass distribution does not necessarily have a particular concentration at some
points. This approximation can be realized using two methods. The first method
analyzes the structure by expressing its deformed shape as a superposition of a
series of global shape functions of the spatial coordinates, each multiplied by its
own generalized coordinate. This method is known as the method of global
shape functions. On the contrary, the second method treats the structure as a
discrete MDOF system in which the mass and stiffness are concentrated at certain points, but with interacting displacements. The discretization can be performed either by the flexibility method or the stiffness method. The latter is
realized by the finite element method (FEM) [1], in which the structure is
approximated by a set of discrete elements with known deformation and inertial
properties. Both methods, that is, the method of the global shape functions and
the FEM, convert the governing partial differential equation into a system of
ordinary differential equations of motion, which can be solved by wellestablished numerical methods. Consequently, the dynamic analysis of continuous systems through a direct analytical solution of partial differential
equations, a very difficult and in most cases insurmountable mathematical
342 PART
II Multi-degree-of-freedom systems
problem, is circumvented. Nevertheless, the research is ongoing and new efficient methods have been developed for solving time-dependent structural problems, for example, the boundary element method (BEM) [2] or the meshless
methods beyond the element methods [3]. The method of global shape functions
is presented below while the FEM is presented in this Chapter 11.
10.5.1 The method of global shape functions
In this method, it is assumed that the continuous function u ðx, t Þ, which represents the displacement of the points of the structure, can be approximated by the
finite superposition series
uðx, t Þ ¼
1 ðx Þu1 ðt Þ +
2 ðx Þu2 ðt Þ + … +
N ðx ÞuN ðt Þ ¼
N
X
i ðx Þui ðt Þ
i¼1
(10.5.1)
The functions i ðx Þ express shapes of deformation of the structure and are
referred to as global shape functions while ui ðt Þ are functions of time that
express the generalized coordinates, for example, displacements or rotations
of certain points of the structure. The minimum requirements for the
expression (10.5.1) to approximate the actual deformation u ðx, t Þ are: (i) the
shape functions must be at least geometrically admissible, that is, satisfy
the geometric (essential) boundary (support) conditions of the structure, and
(ii) must be linearly independent, that is, any of them cannot result as a linear
combination of two or more of the others.
The generalized coordinates are the unknown time functions, which must be
determined by the solution of equations of motion. The number of generalized
coordinates is equal to the degrees of freedom of the substitute MDOF system,
which approximates the actual continuous system. The equations of motion are
usually derived using the principle of virtual work, the Lagrange equations, or
the Hamilton principle. Here, the latter approach, as presented for the generalized SDOF system in Section 8.2, is employed to illustrate the method of global
shape functions.
FIG. 10.5.2 Cantilever beam.
MDOF systems: Models and equations of motion Chapter
10
343
Suppose we are going to apply the method of global shape functions to
approximate the dynamic response of the cantilever beam of Fig. 10.5.2. The
cantilever has variable stiffness I ðx Þ and mass m ðx Þ. It is loaded by the transverse load pðx, t Þ and the axial force P at the free end.
(i) Elastic energy: The elastic energy is equal to the strain energy of the cantilever and it is given (see Eq. 8.1.11)
Z
1 L
2
EI ðx Þ½u00 ðx, t Þ dx
(10.5.2)
U¼
2 0
Its variation is
Z
dU ¼
L
EI ðx Þu00 ðx, t Þd u00 ðx, t Þdx
(10.5.3)
0
or using Eq. (10.5.1)
Z L
N
X
dU ¼
EI ðx Þ
0
Z
¼
00
i ui
i¼1
L
N X
N
X
EI ðx Þ
0
¼
!
N
X
!
00
j duj
j¼1
00 00
i j ui duj
dx
!
dx
(10.5.4)
i¼1 j¼1
N X
N
X
kij ui duj
i¼1 j¼1
where
Z
L
kij ¼
0
EI ðx Þ
00 00
i j dx
(10.5.5)
(ii) Kinetic energy: The kinetic energy is due to the transverse displacements
and rotations of the mass elements. Thus, we have
Z
Z
h
i2
1 L
1 L
2
0
m ðx Þ½u_ ðx, t Þ dx +
I ðx Þ u_ ðx, t Þ dx
(10.5.6)
T¼
2 0
2 0
The second term in the above expression is due to the rotation of the
cross-sections and its contribution is small. In the following development,
without limiting the generality, we omit this term for the sake of simplicity.
The variation of the kinetic energy is
Z L
m ðx Þu_ ðx, t Þd u_ ðx, t Þdx
(10.5.7)
dT ¼
0
or using Eq. (10.5.1)
344 PART
II Multi-degree-of-freedom systems
Z
dT ¼
L
m ðx Þ
0
Z
¼
i u_ i
i¼1
L
m ðx Þ
!
N
X
j d u_ j
j¼1
N X
N
X
dx
!
j u_ i d u_ j
i
0
¼
!
N
X
dx
(10.5.8)
i¼1 j¼1
N X
N
X
mij u_ i d u_ j
i¼1 j¼1
where
Z
L
mij ¼
m ðx Þ
i
j dx
(10.5.9)
0
(iii) Virtual work of the nonconservative forces:
The nonconservative forces are the load pðx, t Þ and the damping force.
The damping may be external or internal. As in the case of generalized
SDOF systems (see Fig. 8.2.1), we assume that the external damping force
is viscous and is distributed along the length of the beam, that is,
fD ðx, t Þ ¼ cðx Þu_ ðx, t Þ, where cðx Þ is the damping coefficient. The internal
damping is due to the deformation of the element of the beam, resists it, and
depends on the velocity of the strain. Hence, if the produced stress is denoted
by sD , we may write
∂ex
(10.5.10)
sD ¼ cs
∂t
where cs is the coefficient of the internal damping and ex the stain. Then,
the virtual work of the internal nonconservative forces is
Z
in
(10.5.11)
dWnc ¼ sD dex dV
V
Taking into account that
sx
M
y, M ¼ EI ðx Þu00 ðx, t Þ
ex ¼ , sx ¼
E
I ðx Þ
Eq. (10.5.11) becomes
Z L
00
in
dWnc
¼
cs I ðx Þu_ ðx, t Þd u00 ðx, t Þdx
(10.5.12)
(10.5.13)
0
or using Eq. (10.5.1)
Z L
N
X
in
¼
cs I ðx Þ
dWnc
0
Z
¼
i¼1
L
cs I ðx Þ
0
¼
!
00
i u_ i
N X
N
X
i¼1 j¼1
N X
N
X
i¼1 j¼1
cijin u_ i duj
N
X
!
00
j duj
j¼1
00 00
i j u_ i duj
dx
!
dx
(10.5.14)
MDOF systems: Models and equations of motion Chapter
10
345
where
Z
cijin
L
¼
00 00
i j dx
c s I ðx Þ
0
(10.5.15)
The virtual work of the external damping force is
Z L
ex
dWnc
¼
cðx Þu_ ðx, t Þd uðx, t Þdx
(10.5.16)
0
or using Eq. (10.5.1)
!
!
Z L
N
N
X
X
ex
c ðx Þ
dWnc ¼ i u_ i
j duj dx
0
Z
i¼1
L
¼
c ðx Þ
0
¼
j¼1
N X
N
X
i
!
j u_ i duj
dx
(10.5.17)
i¼1 j¼1
N X
N
X
cijex u_ i duj
i¼1 j¼1
where
Z
cijex ¼
L
c ðx Þ
i
j dx
(10.5.18)
0
Finally, the virtual work due the external nonconservative load pðx, t Þ is
Z L
p
dWnc ¼
pðx, t Þd uðx, t Þdx
(10.5.19)
0
or using Eq. (10.5.1)
Z
p
dWnc
¼
L
pðx, t Þ
0
¼
N
X
j duj dx
j¼1
N
X
(10.5.20)
pj ðt Þduj
j¼1
where
Z
L
pj ðt Þ ¼
pðx, t Þ j dx
(10.5.21)
0
Consequently, the virtual work due to all nonconservative forces is
dWnc ¼ N X
N
X
i¼1 j¼1
cij u_ i duj +
N
X
j¼1
pj ðt Þduj
(10.5.22)
346 PART
II Multi-degree-of-freedom systems
where it was set
cij¼ cijin + cijex
(10.5.23)
(iv) The potential energy of the conservative forces: The potential energy of the
conservative forces is due to the constant axial force P. Thus, we have
A ¼ Pe
(10.5.24)
where e is the shorting of the elastic curve due to bending and is given as
Z
Z L
1 L 0
2
ex dx ¼
½u ðx, t Þ dx
(10.5.25)
e¼
2
0
0
The variation of the potential energy is
dA ¼ Pde
Z L
¼ P
u0 ðx, t Þd u0 ðx, t Þdx
(10.5.26)
0
or using Eq. (10.5.1)
Z
dA ¼ P
L
0
Z
¼ P
!
0
i ui
i¼1
L
0
¼ P
N
X
N
X
j¼1
N X
N
X
0
i
!
0
j duj
dx
!
0
j ui duj
dx
(10.5.27)
i¼1 j¼1
N X
N
X
kGij ui duj
i¼1 j¼1
where
Z
kGij ¼
0
L
0
i
0
j dx
(10.5.28)
Substituting the previous expressions for dU ,dK , dWnc , and dA in
Hamilton’s principle, Eq. (1.7.13), produces the following equations of
motion
N
X
j¼1
mij u€ j +
N
X
cij u_ j +
j¼1
N
X
kij Pk Gij uj ¼ pi , i ¼ 1, 2…, N
(10.5.29)
j¼1
or in matrix form
M€
u + Cu_ + ðK PKG Þu ¼pðt Þ
(10.5.30)
The matrices M, C, K represent the mass, damping, and stiffness matrices, respectively. Their elements are evaluated from Eqs. (10.5.9),
MDOF systems: Models and equations of motion Chapter
10
347
(10.5.23), (10.5.5). The matrix KG , whose elements are evaluated from
Eq. (10.5.28), is referred to as the geometric stiffness matrix. It expresses
here the influence of the axial force on the dynamic response of the structure. It is obvious that the compressive axial force reduces the stiffness of the
system while the tensile force increases it. The vanishing of the determinant
of the total stiffness matrix K PKG , namely
det ðK PKG Þ ¼ 0
(10.5.31)
yields the N values of the buckling load.
The method of global shape functions is also known as the Ritz method. The
shape functions are, as we have already mentioned, geometrically admissible
and linearly independent functions representing deformation patterns of the
entire structure. They are chosen arbitrarily but appropriately, based on the
experience of the user of the method. The success of the method depends on
the choice of shape functions, but this is still a difficult problem [4]. The use
of the Ritz method has been substantially reduced with the emergence of modern computational methods, such as the finite element method (FEM) [1], the
boundary element method (BEM) [2], and the meshless methods [3]. However,
it is useful either as the theoretical background for developing new computational methods or for testing them.
10.6 Mixed systems
Mixed systems are those resulting from the combination of the previously discussed categories. As an example of this case, we refer to the structure of
Fig. 10.6.1a. Indeed, this system has localized masses (m1 ,m2 ), distributed
masses (mass of bars), localized stiffness (stiffness kR of the rotational spring),
and distributed stiffnesses (stiffness of bars). We assume: m1 ¼ m2 ¼ m,
¼ m=a, kR ¼ EI =4a.
m
(a)
(b)
FIG. 10.6.1 Mixed system (a) and motion parameters (b).
348 PART
II Multi-degree-of-freedom systems
For the determination of motion, we take as degrees of freedom the displace
ments of the masses, namely the rotation f1 of the mass m1 , the displacement u,
and the rotation f2 of the mass m2 ; see Fig. 10.6.1b. The equations of motion
will be derived using the Lagrange equations.
(i) Elastic energy: If u1 , u2 , u3 , u4 denote the displacements and the rotations at
the ends of the beam element as in Fig. 10.6.2, its elastic curve can be set in
the form
uðx, t Þ ¼
1 ðx Þu1
+
2 ðx Þu2
+
3 ðx Þu3
+
4 ðx Þu4
(10.6.1)
The shape functions i express the elastic curves of the beam for ui ¼ 1,
when uj ¼ 0, j 6¼ i. Following the procedure presented in Section 10.4, we
obtain.
FIG. 10.6.2 Degrees of freedom of the beam element.
1 ðx Þ ¼ 1 3
x 2
x 3
¼ 1 3x 2 + 2x 3
L
2
x x
1 ¼ Lx ðx 1Þ2
2 ðx Þ ¼ L
L L
x 2
x 3
2
¼ 3x2 2x3
3 ðx Þ ¼ 3
L
L
x 2 x
1
¼ Lx2 ðx 1Þ
ð
x
Þ
¼
L
4
L
L
L
+2
(10.6.2a)
(10.6.2b)
(10.6.2c)
(10.6.2d)
where
x ¼ x=h
(10.6.3)
For the element 2–3 holds: u1 ¼ af1 , u2 ¼ f1 , u3 ¼ u a f2 =2, u4 ¼ f2 ,
L ¼ 3a.
For the element 4–5 holds: u1 ¼ u + a f2 =2, u2 ¼ f2 , u3 ¼ 0, u4 ¼ 0,
L ¼ 3a.
The elastic energy is given by the expression
Z
Z
1 3
1 5
1
2
2
2
U¼
EI ½u00 ðx, t Þ dx +
EI ½u00 ðx, t Þ dx + kR f1
(10.6.4)
2 2
2 4
2
or using Eq. (10.6.1), we obtain
MDOF systems: Models and equations of motion Chapter
1
f1 , f2 ¼
U u,
2
Z
+
3a
EI
0
1
2
Z
00
1 ðx Þa f1 +
3a
EI
0
00
1 ðx Þ
00
2 ðx Þf1 +
u + a f2 =2 +
00
3 ðx Þ
u a f2 =2 +
00
2
2 ðx Þf2 dx
10
349
00
2
4 ðx Þf2 dx
1
2
+ kR f1
2
(10.6.5)
Differentiating the above relation yields after evaluation of the integrals
∂U
¼ k11 u + k12 f1 + k13 f2
∂u
∂U
¼ k21 u + k22 f1 + k23 f2
∂f1
∂U
¼ k31 u + k32 f1 + k33 f2
∂f2
(10.6.6a)
(10.6.6b)
(10.6.6c)
where
8 EI
10 EI
, k12 ¼ k21 ¼ , k13 ¼ k31 ¼ 0
9 a3
9 a2
121 EI
17 EI
38 EI
, k23 ¼ k32 ¼
, k33 ¼
k22 ¼
36 a
9 a
9 a
Therefore the stiffness matrix of the system is
2
3
32 40a
0
EI 4
k¼
40a
121a2 17a 2 5
36a 3
0
17a 2 152a 2
k11 ¼
(ii) Kinetic energy: The kinetic energy results from the expression
Z
Z
1 3
1 5
2
2
½u_ ðx, t Þ dx +
½u_ ðx, t Þ dx
T¼
m
m
2 2
2 4
(10.6.7)
(10.6.8)
1
1
1
+ I1 f_ 12 + m u_ 2 + IC f_ 22
2
2
2
or using Eq. (10.6.1), we obtain
1 Z 3a h
_ f_ 1 , f_ 2 ¼
1 ðx Þa f_ 1 + 2 ðx Þf_ 1 + 3 ðx Þ u_ a f_ 2 =2 +
m
T u,
2 0
Z
i2
1 3a h
1 ðx Þ u_ + a f_ 2 =2 + 2 ðx Þf_ 2 dx
m
+
2 0
_
4 ð x Þf 2
i2
dx
2
2
1
1
1
2
+ I1 f_ + m u_ + IC f_
2
2
2
1
2
(10.6.9)
350 PART
II Multi-degree-of-freedom systems
Differentiating the above relation yields after evaluation of the integrals
d ∂T
+ m12 f€1 + m13 f€2
(10.6.10a)
¼ m11 u€
dt ∂u_
!
d ∂T
+ m22 f€1 + m23 f€2
(10.6.10b)
¼ m21 u€
dt ∂f_
1
!
d ∂T
+ m32 f€1 + m33 f€2
(10.6.10c)
¼ m31 u€
dt ∂f_
2
where
2712
558
m, m12 ¼ m21 ¼
ma, m13 ¼ m31 ¼ 0
840
840
2294 2
675 2
1832 2
m22 ¼
ma , m23 ¼ m32 ¼ ma , m33 ¼
ma
840
840
840
m11 ¼
Therefore, the mass matrix of the system is
2
3
2712
558a
0
m 4
m¼
558a
2294a 2 675a 2 5
840
0
675a 2
1832a 2
(10.6.11)
(iii) Generalized forces: The virtual work of the nonconservative forces is due
to the load pðt Þ. Thus, we have
Q1 d u + Q2 df1 + Q3 df2 ¼ pðt Þd u + a f2 =2
¼ pðt Þd u pðt Þadf2 =2
(10.6.12)
from which we obtain
Q1 ¼ pðt Þ
Q2 ¼ 0
Q3 ¼ pðt Þa=2
Hence the equation of motion of the mixed system is
38 9
€
2712 558a
0
>
=
< u >
m6
€
2
2 7 f
4 558a 2294a 675a 5
1
>
840
;
: € >
0
675a 2
1832a 2
f2
9 (10.6.13)
2
38 9 8
32 40a
0
pðt Þ
u >
>
>
>
<
=
<
=
EI 6
2
2 7
+
40a
121a
¼
0
17a
f
4
5
1
>
>
36a3
: >
; >
:
;
pðt Þa=2
f2
0
17a 2 152a 2
2
MDOF systems: Models and equations of motion Chapter
10
351
10.7 Transformations of the equations of motion
In the previous sections, we presented various methods for the formulation of
the equations of motions of a MDOF system with N degrees of freedom. In each
case, a set of independent displacements is selected, which, as stated in
Section 1.8, expresses the generalized coordinates or simply the coordinates,
as we will now be referring to, of the system and are denoted by the vector
of the displacements u in the equation of motions. It is often necessary or con
venient to express the equations in a system of also independent coordinates u
other than that originally selected. The transformation of the coordinates from u
to u is a linear relation of the form
u ¼R
u
(10.7.1)
where R is the N N square matrix referred to as the transformation matrix. If
M, C, and K are the mass, damping, and stiffness matrices, respectively, and
pðt Þ the vector of the external loads in the original system of coordinates u,
K
C,
and the vector p
ðt Þ in the transwe need to determine the matrices M,
. This can be achieved using the principle of virformed system of coordinates u
tual work as follows.
The elastic forces in the coordinates u are given by the relation
f S ¼Ku
(10.7.2)
by the relation
while in the coordinates u
f S ¼K
u
(10.7.3)
a virtual displacement d
We give u
u. Then the corresponding virtual displacement du results from the relation (10.7.1)
du ¼Rd
u
(10.7.4)
The virtual work done by the elastic forces f S due to the virtual displacement
du is
dWS ¼duT f S
¼ duT Ku
(10.7.5)
or using Eqs. (10.7.1), (10.7.4)
uT RT KR
u
dWS ¼d
(10.7.6)
Similarly, the virtual work of the elastic forces in the system of the coordi is
nates u
S ¼d
dW
uT f S
u
¼ d
uT K
(10.7.7)
352 PART
II Multi-degree-of-freedom systems
S ¼ dWS , which gives
Obviously, it must be d W
u
u ¼ d
uT K
d
uT RT KR
or
RT KR u
¼0
d
uT K
(10.7.8)
and d
Because the vectors u
u are nonzero, it must be
¼ RT KR
K
(10.7.9)
Similarly, by considering the virtual work of the inertial forces f I ¼ M€
u and
f I ¼ M
u
€ in the two systems of coordinates u and u
, we obtain
¼ RT MR
M
(10.7.10)
Further, by considering the virtual work of the damping forces f D ¼ Cu_ kai
f D ¼ C
u
_ in the two systems of coordinates, we prove that
¼ RT CR
C
(10.7.11)
Finally, by expressing the virtual work of the external forces in the two systems of coordinates, we have
dWp ¼ duT pðt Þ ¼ ðRd
uÞT pðt Þ
(10.7.12)
¼ d
uT RT pðt Þ
and
p ¼ d
ðt Þ
dW
uT p
(10.7.13)
p , we obtain
Because these two virtual works are equal, dWp ¼ d W
ðt Þ ¼ RT pðt Þ
p
(10.7.14)
Example 10.7.1 The equation of motion of the system shown in Fig. 10.4.1 with
respect to point O is (see Eqs. 10.4.11a, 10.4.11b):
2
ma 3 ( pðt Þ )
m u
6
2 7 u€ + k11 k12
a
¼
(1)
4 ma
5 €
pðt Þ
f
k21 k22
f
IO
2
2
where
k11 ¼
12EI
,
h3
k12 ¼ k21 ¼
6EI
,
h2
k22 ¼
4EI
h
Transform Eq. (1) with respect to the mass center C of the body.
(2)
MDOF systems: Models and equations of motion Chapter
10
353
Solution
If u and f denote the displacement of the mass center and its rotation about it,
then the transformation relations result by geometrical consideration as
a
(3)
u ¼ u + f
2
f ¼ f
(4)
or
Hence
" a # u
u
1
2
¼
f
f
0 1
"
a#
2 ,
R¼
0 1
1
"
1 0
RT ¼ a
1
2
(5)
#
(6)
Then Eq. (10.7.10) gives
2
a 32 a 3
#
m m
1 0
2 74 1 2 5
6
¼ a
M
4
5
a
1
0 1
IO
m
2
2
"
#
m 0
¼
0 IC
"
(7)
where
a2
6
is the moment of inertia with respect to the center of mass C .
Further, Eq. (10.7.9) gives
3
"
#"
#2
1 0 k11 k12 1 a
¼ a
4 25
K
1 k21 k22
0 1
2
"
#
k11 k12
¼
k21 k22
IC ¼ m
(8)
(9)
where
12EI
k11 ¼ k11 ¼ 3
h
a
6EI
k 12 ¼ k 21 ¼ k11 + k12 ¼ 3 ða + h Þ
2
h
a 2
12EI a 2 ah h 2
+
k11 + ak 12 + k22 ¼ 3
+
k 22 ¼
4
3
2
h
2
(10)
354 PART
II Multi-degree-of-freedom systems
Finally, Eq. (10.7.14) gives
9
38
1 0 < pðt Þ =
ðt Þ ¼ 4 a 5
p
a
1 : pðt Þ ;
2
2
(
)
pðt Þ
¼
0
2
Therefore, the transformed equation of motion is
2
3( ) (
"
#( )
)
2
a+h
m 0
u
pðt Þ
u€
6EI 4
+ 3
3a 2 + 6ah + 2h 2 5 ¼
h
0 IC
f
0
a+h
f€
6
(11)
(12)
Note that we obtain the same equations if the equations are derived directly
with respect to center of mass C .
10.8 Problems
Problem P10.1 Formulate the equations of motion of the system shown in
Fig. P10.1. Data: k1 ¼ 3k, k2 ¼ 2k, k3 ¼ k, c1 ¼ c3 ¼ c, c2 ¼ 2c, m1 ¼ m2 ¼ m,
and m3 ¼ 2m.
FIG. P10.1 System in problem P10.1.
Problem P10.2 A square plate with a side length a, constant thickness
d ¼ a=10, and mass density g is supported by three columns of height
h ¼ a=2 and square cross-section of a side length b ¼ a=20 (Fig. P10.2).
Neglecting the axial deformation and the mass of the columns, formulate the
equations of motion of the structure using the Lagrange equations. The module
of elasticity is E and the shear modulus G ¼ 0:4E.
MDOF systems: Models and equations of motion Chapter
10
355
FIG. P10.2 Structure in problem P10.2.
Problem P10.3 A rigid bar of total mass m is supported by the three springs
k1 , k2 , k3 and the damper c as shown in Fig. P10.3. The circular disc of mass
0:5m at the end D of the bar has a diameter 0:2L. Formulate the equation of
motion. Data: k2 ¼ 1:5k1 and k3 ¼ 2k1 .
FIG. P10.3 System in problem P10.3.
Problem P10.4 Formulate the equations of motion of the plane frame of
Fig. P10.4 when the supports 1 and 2 are subjected to the horizontal displacements ug1 ðt Þ and ug2 ðt Þ, respectively. Give the expressions for the evaluation of
the support reactions. The beams are assumed rigid (I ¼ 1) while the columns
have the same cross-section and modulus of elasticity E.
FIG. P10.4 Frame in problem P10.4.
356 PART
II Multi-degree-of-freedom systems
Problem P10.5 The beam of Fig. P10.5 consists of the flexible and massless
per unit length. The beam is clamped
part AB and the rigid part BC of mass m
at A while the hinged support at C is elastically restrained by the rotational
spring CR ¼ EI =10L. The beam is loaded by a concentrated moment M ðt Þ
applied at B. Formulate the equation of motion of the structure and give the
expressions for the evaluation of the support reactions.
FIG. P10.5 Beam in problem P10.5.
Problem P.10.6 The structure of Fig. P10.6 consists of the two bars AB and
BC . The bar AB is massless and flexible and is simply supported at A while
per unit length and is elastically restrained at
the bar BC is rigid with mass m
the hinged end C by the rotational spring CR ¼ EI =10L. The structure is
loaded by a concentrated moment M ðt Þ applied at B. Formulate the equation
of motion of the structure and give the expressions for the evaluation of the
support reactions.
FIG. P10.6 Structure in problem P10.6.
Problem P10.7 Formulate the equation of motion of the plane frame of
Fig. P10.7. The massless cables FB and GC have cross-sectional area A and
are prestressed so that they can undertake compressive forces. Data:
¼ m=a (kg=m), and CR ¼ EI =10L.
I =A ¼ a2 =25, m
MDOF systems: Models and equations of motion Chapter
10
357
FIG. P10.7 Plane frame in problem P10.7.
Problem P10.8 Formulate the equations of motion of the system shown in
per unit length while the cable
Fig. P10.8. The bar AB is rigid and has mass m
pffiffiffi
and the pulley are assumed massless. The total length of the cable is 1:5a 2.
The hinged support at A is elastically restrained by the rotational spring
is moving downward with
CR ¼ EAa=250. The hanging body of mass m ¼ ma
a velocity v ¼ v0 sin wt.
FIG. P10.8 System in problem P10.8.
Problem P10.9 The thin spherical tank of Fig. P10.9 has a diameter R ¼ a=3
and is full with a liquid of density g. It is supported on the ground through nine,
hinged at both ends, massless rods of cross-sectional area A and modulus of
elasticity E. The points 10 ,20 ,30 lie on a horizontal circular ring of the spherical
tank at a height a from the ground and at a distance R=3 from the center of the
sphere. The mass of the tank is neglected.
358 PART
II Multi-degree-of-freedom systems
Formulate the equation of motion when
1. The structure performs free vibrations.
2. The structure is subjected to ground motion ug ðt Þ in the direction b with
respect to the x axis.
FIG. P10.9 Spherical tank in problem P10.9.
Problem P10.10 A vehicle traveling along a bridge is idealized as shown in
Fig. P10.10. The unsprung mass mc is connected to the sprung mass through
the suspension system (k,c). The bridge has a span L and is simply supported.
Using the Ritz method, formulate the equation of motion of the system bridgevehicle when the vehicle is traveling with a constant velocity v. The bridge is
simulated by a two-degree-of-freedom system with global shape functions
n ¼ sin ðnpx=LÞ, n ¼ 1, 2. The mass mc maintains its contact to the bridge dur
ing the motion. Data: mc ¼ m0 =10, m0 ¼ mL=20,
and k ¼ EI =100L.
FIG. P10.10 Vehicle traveling on a bridge in problem P10.10.
References and further reading
[1] O. Zienkiewicz, R. Taylor, The finite element method, seventh ed., Butterworth-Heinemann,
Oxford, UK, 2013.
[2] J.T. Katsikadelis, The boundary element method for engineers and scientists, second ed.,
Academic press, Elsevier, Oxford, UK, 2016.
[3] G.R. Liu, Meshfree methods: moving beyond the finite element method, second ed., CRC Press,
Boca Raton, FL, 2010.
[4] J.T. Katsikadelis, A generalized ritz method for partial differential equations in domains of arbitrary geometry using global shape functions. Eng. Anal. Bound. Elem. 32 (5) (2008) 353–367,
https://doi.org/10.1016/j.enganabound.2007.001.
Chapter 11
The finite element method
Chapter outline
11.1 Introduction
359
11.2 The finite element method
for the plane truss
360
11.2.1 Properties of the plane
truss element
360
11.2.2 Transformation of the
nodal coordinates of
the truss element
374
11.2.3 Equation of motion of
the plane truss
377
11.2.4 Steps to formulate the
equations of motion
for a plane truss by the
finite element method 381
11.2.5 Modification of the
equations of motion
due to the supports
of the structure
382
11.3 The finite element method
for the plane frame
392
11.3.1 Properties of the
plane frame element 392
11.3.2 Transformation of the
nodal coordinates of the
plane frame element 410
11.4 Static condensation: Guyan’s
reduction
424
11.5 Flexural vibrations of a
plane frame
429
11.6 Reduction of the degrees
of freedom due to
constraints
439
11.7 Axial constraints in the
plane frame
442
11.8 The finite element method
for the plane grid
453
11.8.1 Properties of the
plane grid element
453
11.8.2 Transformation of the
nodal coordinates
of the plane grid
element
463
11.9 The finite element method
for the space frame
476
11.9.1 Properties of the
space frame element 476
11.9.2 Transformation of the
nodal coordinates of
the space frame
element
482
11.10 The finite element method
for the space truss
494
11.10.1 Properties of the
space truss element 494
11.10.2 Transformation of
the nodal coordinates
of the space truss
element
498
11.11 Rigid bodies within flexible
skeletal structures
503
11.11.1 Rigid bodies in
spaces frames
503
11.11.2 Rigid bodies in
spaces trusses, plane
grids, plane frames,
and plane trusses
510
11.12 Problems
517
References and further reading
521
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00011-X
© 2020 Elsevier Inc. All rights reserved.
359
360 PART
II Multi-degree-of-freedom systems
11.1 Introduction
The distributed mass is the usual case we encounter in actual structures, that is,
the actual structures should be modeled as continuous systems. This modeling
leads to partial differential equations with prescribed initial and boundary conditions. The solution of such equations is a difficult mathematical problem, even
for individual components of structures (e.g., beams, plates, walls, etc.). On the
other hand, the method of global shape functions, which can provide an acceptable solution, has limited capabilities due to the difficulty in choosing the global
shape functions [1]. In practice, this method cannot cope with conventional
structures such as a multistory frame or a grid. Besides, it cannot be programmed for automatic use on a computer. Therefore, neither method is
suitable for the dynamic analysis of realistic structures in engineering. At this
dead end, the finite element method (FEM) method gives a way out.
The FEM represents a major breakthrough in the field of computational
mechanics. Extensive literature on the general formulation of the FEM [2] as
well as on its application to dynamic problems is now available [3].
In the finite element approach, the structure being analyzed (e.g., frame,
wall, plate, shell, three-dimensional deformable body, or a combination of
them) is divided into a finite number of small subregions, elements, which
are interconnected at discrete points, the nodes, where the compatibility
condition for the displacements and the equilibrium of equivalent nodal forces
are ensured. After the discretization, we assume that the field function (the displacement in this case) within each element varies according to a known law,
which is expressed as a superposition of shape patterns, the shape functions,
corresponding to unit values of the element nodal displacements. After that,
the elastic and kinetic energy as well as the virtual work of the nonconservative
nodal forces in terms of the nodal displacements, viewed as generalized coordinates, are established. The equivalent nodal forces (elastic, inertial, damping,
and external forces) may result by the use of the Lagrange equations or the principle of virtual work. Finally, the compatibility of the nodal displacements and
the equilibrium of the nodal forces give the differential equations of motion of
the structure. The FEM is presented below only for skeletal structures, that is,
structures consisting of straight-line elements (trusses, frames, grids) because
surface structures (walls, plates, shells) as well as three-dimensional (3D)
deformable bodies do not fall within the scope of this book.
11.2 The finite element method for the plane truss
11.2.1 Properties of the plane truss element
We consider a plane truss element with a variable cross-section Aðx Þ, mass per
unit length m ðx Þ, and modulus of elasticity E, whose local axis in the undeformed state coincides with the x axis. The ends j, k of the element are called
nodes of the element, thus any quantity (force, displacement) associated with
them is referred to as nodal quantity (e.g., nodal force, nodal displacement).
The finite element method Chapter
11
361
Fig. 11.2.1 Degrees of freedom of a plane truss element.
In the deformed state at time t, the ends of the elements are displaced to points
j 0 , k 0 and the element occupies the position j 0 k 0 (Fig. 11.2.1).
The element has four degrees of freedom, the displacements u1 ðt Þ, u2 ðt Þ of
node j and the displacements u3 ðt Þ, u4 ðt Þ of node k. These quantities constitute
the nodal coordinates of the plane truss element. A typical point x of the element axis undergoes two displacements, the axial displacement u ðx, t Þ and the
transverse one v ðx, t Þ. We consider the element as a generalized system and
apply the Ritz method presented in Section 10.5.1. Thus, we set the axial
displacement in the form
u ðx, t Þ ¼ u1 ðt Þ
1 ðx Þ + u 3 ðt Þ 3 ðx Þ
(11.2.1)
where 1 ðx Þ and 3 ðx Þ are shape functions expressing the axial deformation for
u1 ¼ 1, u3 ¼ 0 and u1 ¼ 0, u3 ¼ 1, respectively. The functions 1 ðx Þ and 3 ðx Þ
can be established as follows.
The equilibrium of the element dx (see Fig. 11.2.2) yields
dN
¼0
dx
(11.2.2)
N ¼ Aðx Þsx
¼ EAðx Þex
d i
¼ EAðx Þ
dx
(11.2.3)
d
d i
¼0
EAðx Þ
dx
dx
(11.2.4)
or taking into account that
we write Eq. (11.2.2) as
and after integration
Z
i ¼ c1
Fig. 11.2.2 Beam element dx.
dx
+ c2
Aðx Þ
(11.2.5)
362 PART
II Multi-degree-of-freedom systems
The arbitrary constants c1 and c2 are evaluated from the boundary conditions. It is obvious that for a variable cross-section, the determination of the
shape functions requires the evaluation of the integral in Eq. (11.2.5). This
relation, although it gives shape functions expressing the exact static axial
deformation, is not suitable for the automation of the method because different
shape functions have to be determined for elements with a different law of
variation of the cross-section. This difficulty is surpassed if we accept the
same shape functions for all elements, regardless of the variation law of the
cross-section and, indeed, those resulting from Eq. (11.2.4) for a constant
cross-section. In that case, we obtain
i
which for
1 ð0Þ ¼ 1
and
¼ c1 x + c2
1 ð LÞ ¼ 0
gives
1 ðx Þ ¼ 1 x,
and for
3 ð 0Þ ¼ 0
and
x ¼ x=L
(11.2.6a)
3 ð LÞ ¼ 1
3 ðx Þ ¼ x,
x ¼ x=L
(11.2.6b)
The shape functions (11.2.6a), (11.26b) can also result if we arbitrarily
accept a linear law of variation of axial displacement within the element.
The assumption of linear variation of the displacement is at the expense of accuracy. However, the error is acceptable when the element is small and the crosssectional variation is not intense. This becomes clear from Example 11.2.1.
Taking into account that the element is a straight line in the deformed state,
we can write the transverse displacement asa
v ðx, t Þ ¼ u2 ð1 x Þ + u4 x
¼ u2 2 ðx Þ + u4 4 ðx Þ
(11.2.7)
a. Due to the axial deformartion we should write
v ðx, t Þ ¼ u2 ð1 x 0 Þ + u4 x 0 , x0 ¼ x 0 =L0
But it is
x0 ¼
x0
x + u ðx, t Þ
x=L + u ðx, t Þ=L
u ðx, t Þ
¼
x+
¼
L0 L + ðu3 u1 Þ 1 + ðu3 u1 Þ=L
L
(a)
(b)
because ðu3 u1 Þ=L≪1:
Eq. (a) by virtue of Eq. (b) becomes
u ðx, t Þ
u ðx, t Þ
+ u4 x +
v ðx, t Þ u2 1 x L
L
u4 u2
u ðx, t Þ
u2 ð1 x Þ + u4 x L
u2 ð1 x Þ + u4 x
2
because the nonlinear term u4 u
L u ðx, t Þ is much smaller compared to the linear ones and thus it can
be neglected.
The finite element method Chapter
11
363
Obviously, it is 2 ðx Þ ¼ 1 ðx Þ and 4 ðx Þ ¼ 3 ðx Þ.
The equivalent nodal forces in the directions of the degrees of freedom can
be established using the method of the Lagrange equations or the principle of
virtual work. Both methods are presented in the following.
11.2.1.1 The method of the Lagrange equations
We refer to Eq. (1.8.11), which we write for the truss element
d ∂T
∂T ∂U ∂A
+
+
¼ Qi ,
i ¼ 1, 2, 3, 4
dt ∂u_ i
∂ui ∂ui ∂ui
(11.2.8)
The forces ∂A=∂ui , derived from the potential A, are gravitational forces. As
shown in Section 1.9, they are ignored in the linear dynamic analysis but the
deformation produced by them should be superimposed to that resulting from
the dynamic analysis in order to obtain the total deformation of the structure.
Obviously, we deduce from Eq. (11.2.8) that the terms
d ∂T
∂T
(11.2.9)
fIi ¼
dt ∂u_ i
∂ui
fSi ¼
∂U
∂ui
(11.2.10)
express the nodal inertial force and the nodal elastic force, respectively, in the
direction of the generalized coordinate ui . Finally, the force Qi in Eq. (11.2.8)
includes the nonconservative generalized forces. In our case, these are the
damping forces fDi and the external forces pi . Regarding the sign convention
for the nodal forces, they are positive when their direction coincides with the
positive direction of the axes they refer.
(i) Nodal elastic forces and stiffness matrix of the truss element
Eq. (11.2.7) expresses the displacement of the element as a rigid body.
Therefore, the elastic energy is due only to the axial deformation of the element
given by
Z
1
sx ex dV
(11.2.11)
U¼
2 V
Taking into account that sx ¼ Eex , ex ¼ ∂u=∂x, the previous relation
becomes
Z Z 2
1 L
∂u
E
dxdydz
U¼
2 0 A
∂x
(11.2.12)
2
Z L
1
∂u
¼
EAðx Þ
dx
2 0
∂x
or using Eq. (11.2.1)
364 PART
II Multi-degree-of-freedom systems
U ðu1 , u3 Þ ¼
1
2
Z
L
EAðx Þ u1
0
2
0
0
1 ðx Þ + u3 3 ðx Þ dx
(11.2.13)
The nodal elastic forces result from Eq. (11.2.10) for i ¼ 1, 2, 3, 4. Thus,
we obtain
fS1 ¼
∂U
¼ k11 u1 + k13 u3
∂u1
(11.2.14a)
∂U
¼0
∂u2
(11.2.14b)
∂U
¼ k31 u1 + k33 u3
∂u3
(11.2.14c)
∂U
¼0
∂u4
(11.2.14d)
fS2 ¼
fS3 ¼
fS4 ¼
where
Z
L
kij ¼
0
EAðx Þ 0i ðx Þ 0j ðx Þdx,
Eqs. (11.2.14a)–(11.2.14d) are written in
8 9 2
k11 0 k13
fS1 >
>
>
> 6
> >
>
>
>
>
< fS2 = 6 0 0 0
6
¼6
>
>
6
>
> 4 k31 0 k33
> fS3 >
>
>
>
>
: ;
fS4
0 0 0
i, j ¼ 1, 3
matrix form as
38 9
0 > u1 >
>
>
> >
>
7>
>
=
< >
07
7 u2
7
> u3 >
07
>
> >
5>
>
>
>
;
: >
u4
0
(11.2.15)
(11.2.16)
or
f eS ¼ ke ue
(11.2.17)
u are the vectors of the nodal elastic forces and the nodal displacewhere
ments, respectively, and ke the stiffness matrix of the e truss element. Hence, the
element matrix ke for the plane truss element is defined as
2
3
k11 0 k13 0
6
7
60 0 0 07
6
7
e
k ¼6
(11.2.18)
7
6 k31 0 k33 0 7
4
5
f eS ,
e
0
0 0
0
Obviously, we deduce from Eq. (11.2.15) that kij ¼ kji . Hence, the stiffness
matrix is symmetric.
For an element with a constant cross-section Aðx Þ ¼ Ae and a length Le ,
Eq. (11.2.15) is integrated analytically and Eq. (11.2.18) yields
The finite element method Chapter
2
1
e6
EA
0
ke ¼ e 6
L 4 1
0
3
0 1 0
0 0 07
7
0 1 05
0 0 0
365
11
(11.2.19)
It should be noted that Eq. (11.2.10) expresses the Castigliano theorem.
Therefore, the previous method for establishing the stiffness matrix is identical
to the so-called energy method.
(ii) Nodal inertial forces and mass matrix of the truss element
In the FEM, the equivalent inertial forces are obtained by two different assumptions of the mass distribution on the element: the consistent mass assumption,
which assumes a continuous distribution of the mass on the element, and the
lumped mass assumption, which lumps the mass at its nodes. The inertial mass
matrices resulting from both assumptions are derived below.
(a) Consistent mass matrix
During the motion, the infinitesimal mass m ðx Þdx undergoes the two displacements u ðx, t Þ and v ðx, t Þ. Therefore, the kinetic energy of the truss element will
be given by the expression
Z
n
o
1 L
m ðx Þ ½u_ ðx, t Þ2 + ½v_ ðx, t Þ2 dx
(11.2.20)
T¼
2 0
or using Eqs. (11.2.1), (11.2.7)
Z
n
1 L
T ðu_ 1 , …, u_ 4 Þ ¼
m ðx Þ ½u_ 1
2 0
2
1 ðx Þ + u_ 3 3 ðx Þ
+ ½u_ 2
1 ðx Þ + u_ 4 3 ðx Þ
2
o
dx
(11.2.21)
The inertial forces result from Eq. (11.2.9) for i ¼ 1, 2, 3, 4.
performing the differentiations we obtain
d ∂T
∂T
¼ m11 u€1 + m13 u€3
fI 1 ¼
dt ∂u_ 1
∂u1
d ∂T
∂T
¼ m22 u€2 + m24 u€4
fI 2 ¼
dt ∂u_ 2
∂u2
d ∂T
∂T
¼ m31 u€1 + m33 u€3
fI 3 ¼
dt ∂u_ 3
∂u3
d ∂T
∂T
fI 4 ¼
¼ m42 u€2 + m44 u€4
dt ∂u_ 4
∂u4
Thus, after
(11.2.22a)
(11.2.22b)
(11.2.22c)
(11.2.22d)
where
Z
mij ¼ mi + 1, j + 1 ¼
0
L
m ðx Þ i ðx Þ j ðx Þdx, i, j ¼ 1, 3
(11.2.23)
366 PART
II Multi-degree-of-freedom systems
Eqs. (11.2.22a)–(11.2.22d) are written in matrix form
8 9 2
38 9
m11 0
u€
fI 1 >
m13 0
>
>
>
>
>
> 1>
>
>
> >
>
>
>
7>
< fI 2 = 6
<
m22 0
m24 7 u€2 =
60
7
6
¼6
>
u€3 >
fI 3 >
m33 0 7
>
>
>
>
5>
4 m31 0
>
>
>
>
>
> ;
> >
;
:
:
fI 4
m44
0
m42 0
u€4
(11.2.24)
or
€e
f eI ¼ me u
(11.2.25)
€e are the vectors of the nodal inertial forces and the nodal accelerwhere f eI , u
ations, respectively, and me the mass matrix of the e truss element. Therefore,
the element mass matrix me is defined as
2
3
m13 0
m11 0
60
m22 0
m24 7
7
me ¼ 6
(11.2.26)
4 m31 0
m33 0 5
m44
0
m42 0
Obviously, we deduce from Eq. (11.2.23) that mij ¼ mji . Hence, the mass
matrix is symmetric.
Eq. (11.2.23) is integrated
For an element with a constant mass, m ðx Þ ¼ m,
analytically and Eq. (11.2.26) gives
2
3
2 0 1 0
e6
m 60 2 0 17
7
me ¼
(11.2.27)
6 41 0 2 05
0 1 0 2
e Le is the total mass of the e element.
where m e ¼ m
(b) Lumped mass matrix
According to this assumption, the mass of the element is concentrated at its
nodes by static considerations, that is, they are obtained as the reactions of a
simply supported beam under the load m ðx Þ (see Fig. 11.2.3). Thus, we have
Z L
m ðx Þð1 x Þdx
(11.2.28a)
m1 ¼
0
Z
L
m2 ¼
m ðx Þxdx
(11.2.28b)
0
Therefore, the kinetic energy of the truss element is given by the expression
1 1 T ¼ m1 u_ 1 2 + u_ 2 2 + m2 u_ 3 2 + u_ 4 2
2
2
(11.2.29)
The finite element method Chapter
11
367
Fig. 11.2.3 Lumped mass assumption.
The inertial forces result from Eq. (11.2.9) for i ¼ 1,2, 3, 4. Hence, after
performing the differentiations, we obtain
d ∂T
∂T
¼ m11 u€1
(11.2.30a)
fI 1 ¼
dt ∂u_ 1
∂u1
d ∂T
∂T
¼ m22 u€2
(11.2.30b)
fI 2 ¼
dt ∂u_ 2
∂u2
d ∂T
∂T
¼ m33 u€3
(11.2.30c)
fI 3 ¼
dt ∂u_ 3
∂u3
d ∂T
∂T
fI 4 ¼
¼ m44 u€4
(11.2.30d)
dt ∂u_ 4
∂u4
where
m11 ¼ m22 ¼ m1 , m33 ¼ m44 ¼ m2
Eqs. (11.2.30a)–(11.2.3d) are written in matrix form as
8 9 2
38 9
m11 0
fI 1 >
0
0
u€
>
>
>
>
>
> 1>
>
>
> >
>
>
>
7>
< fI 2 = 6
<
m22 0
0 7 u€2 =
60
6
7
¼6
>
0
m33 0 7
fI 3 >
u€3 >
>
>
>
>
40
5>
>
>
>
>
>
> ;
>
;
:
: >
fI 4
0
0
0
m44
u€4
Therefore, the element mass matrix me is defined as
2
3
m11 0
0
0
60
m22 0
0 7
6
7
me ¼ 6
7
40
0
m33 0 5
0
0
0
(11.2.31)
(11.2.32)
(11.2.33)
m44
We observe that the lumped mass assumption results in a diagonal mass matrix.
Eqs. (11.2.28a),
For an element with constant mass, m ðx Þ ¼ m
(11.2.28b) give
e Le =2
m11 ¼ m22 ¼ m33 ¼ m44 ¼ m
(11.2.34)
368 PART
II Multi-degree-of-freedom systems
and the mass matrix, Eq. (11.2.33), becomes
2
1 0 0
1 e6
0 1 0
e
m ¼ m 6
2 40 0 1
0 0 0
3
0
07
7
05
1
(11.2.35)
e Le is the total mass of the truss element e.
where m e ¼ m
(iii) Nodal damping forces and damping matrix of the truss element
We consider only internal damping. It is therefore due to the deformation of the
truss element; it resists it and depends on the velocity of the strain. If we denote
with sD the axial stress due to damping, we can write
sD ¼ cs
∂ex
∂t
(11.2.36)
where cs is the coefficient of the internal damping and ex the axial strain.
The virtual work of the internal damping force in the truss element is
Z
in
(11.2.37)
dWnc ¼ sD dex dV
V
Taking into account that ex ¼ ∂u=∂x, the previous relation becomes
Z L
∂2 u ðx, t Þ ∂u ðx, t Þ
in
d
dx
(11.2.38)
¼
c s Að x Þ
dWnc
∂x∂t
∂x
0
or using Eq. (11.2.1)
Z L
in
dWnc
¼
cs Aðx Þ u_ 1
0
d u1
0
0
1 ðx Þ + u_ 3 3 ðx Þ
0
0
1 ðx Þ + u 3 3 ðx Þ
dx
¼ ðc11 u_ 1 + c13 u_ 3 Þdu1 ðc31 u_ 1 + c33 u_ 3 Þdu3
(11.2.39)
where
Z
L
cij ¼
0
cs Aðx Þ 0i ðx Þ 0j ðx Þdx, i, j ¼ 1, 3
(11.2.40)
The damping forces are obtained as generalized forces from Eqs. (11.2.8),
(1.8.8) for i ¼ 1, 2, 3, 4. Thus we have
fD1 ¼ Q1 ¼ c11 u_ 1 + c13 u_ 3
(11.2.41a)
fD2 ¼ Q2 ¼ 0
(11.2.41b)
fD3 ¼ Q3 ¼ c31 u_ 1 + c33 u_ 3
(11.2.41c)
fD4 ¼ Q4 ¼ 0
(11.2.41d)
The finite element method Chapter
11
369
The minus sign in the above relations is due to the fact that they are shifted
to the left side of Eq. (11.2.8) in order that their positive direction is the same
with that of fIi and fSi .
Eqs. (11.2.41a)–(11.2.41d) are written in matrix form as
8 9 2
38 9
c11 0 c13 0 > u_ 1 >
fD1 >
>
>
>
>
>
>
> >
> >
>
>
7>
< fD2 = 6
6 0 0 0 0 7< u_ 2 =
6
7
¼6
(11.2.42)
7 u_ >
>
fD3 >
>
>
>
4 c31 0 c33 0 5>
3>
>
>
>
>
>
>
> ;
> ;
:
:
0 0 0 0
fD4
u_ 4
or
f eD ¼ ce u_ e
(11.2.43)
where f eD , u_ e are the vectors of the nodal damping forces and nodal velocities,
respectively, and ce is the damping matrix of the e truss element. Therefore, the
element damping matrix ce is defined as
2
3
c11 0 c13 0
60 0 0 07
6
7
ce ¼ 6
(11.2.44)
7
4 c31 0 c33 0 5
0
0 0
0
Obviously, we deduce from Eq. (11.2.40) that cij ¼ cji . Hence, the damping
matrix is symmetric.
For an element with a constant cross-section, Aðx Þ ¼ Ae , Eq. (11.2.40) are
integrated analytically and Eq. (11.2.44) becomes
2
3
1 0 1 0
6
7
cs A e 6 0 0 0 0 7
7
ce ¼ e 6
(11.2.45)
7
L 6
4 1 0 1 0 5
0 0 0 0
The evaluation of the matrix ce requires knowledge of the coefficient cs , the
determination of which is practically not feasible. The damping matrix in engineering praxis, as we will see in Section 12.11.3, is constructed by assuming
known values of the damping ratio for each vibration mode, which are estimated
by experiments on similar structures. Nevertheless, we presented this approach
for the completeness of FEM, which is structured through the element approach.
(iv) Equivalent nodal loads of the truss element
The truss element can also be subjected to axial external loading, which may
consist of a distributed load pðx, t Þ and a finite number of concentrated loads
Pk ðt Þ, k ¼ 1, 2, …, K , acting at the points xk of the element, Fig. 11.2.4.
370 PART
II Multi-degree-of-freedom systems
Fig. 11.2.4 Loading and equivalent nodal loads of a plane truss element.
Obviously, equivalent nodal loads will act only in directions of u1 and u3 .
The loads pðx, t Þ and Pk ðt Þ are considered as nonconservative, thus the equivalent loads can be established as generalized forces in the directions of u1 and
u3 . The virtual work due to axial displacement is
Z L
K
X
p
dWnc ¼
pðx, t Þdu ðx, t Þdx +
Pk du ðxk , t Þ
0
Z
¼
0
k¼1
L
pðx, t Þ½du1
1 ðx Þ + du3 3 ðx Þdx
+
K
X
Pk ðt Þ½du1
1 ðxk Þ + du3 3 ðxk Þ
k¼1
¼ p1 ðt Þdu1 + p3 ðt Þdu3
(11.2.46)
where it was set
Z
pi ðt Þ ¼
0
L
pðx, t Þ i ðx Þdx +
K
X
Pk ðt Þ i ðxk Þ, i ¼ 1, 3
(11.2.47)
k¼1
The nodal load forces are obtained from Eq. (11.2.8) for i ¼ 1, 2, 3, 4, that is,
Q1 ¼ p1 , Q2 ¼ 0, Q3 ¼ p3 , Q4 ¼ 0. Hence, the vector of the equivalent nodal load
forces for the e element is
p e ð t Þ ¼ f p 1 ð t Þ 0 p 3 ð t Þ 0g T
(11.2.48)
Example 11.2.1 Compute the stiffness, mass, and damping matrices as well as
the vector of the nodal load forces of an element with variable cross-section and
compare them to those of the element with a constant cross-section equal to the
mean value of the variable
cross-section. Data: Aðx Þ ¼ A0 1 + x x2 ,
x ¼ x=L, pðx, t Þ ¼ p0 ðt Þ 1 + x x 2 . Modulus of elasticity E, material density
r, and damping coefficient cs .
Solution
The mean cross-sectional area is
Z L
Z L
7A0
¼1
Aðx Þdx ¼ A0
1 + x x 2 dx ¼
A
6
L 0
0
(1)
The finite element method Chapter
11
371
Moreover, from Eq. (11.2.6a), we obtain
1
0
1 ðx Þ ¼ ,
0
3 ðx Þ ¼
L
1
L
(2)
Stiffness matrix. The elements of the stiffness matrix are computed from
Eq. (11.2.15), which gives
Z L
EAðx Þ 0i ðx Þ 0j ðx Þdx
kij ¼
0
(3)
EA
¼
L
Obviously, the stiffness matrix of the element with a variable cross-section
is the same as that of an element with a constant cross-section equal to the mean
cross-section. Hence, we may write
2
3
1 0 1 0
6 0 0 0 07
EA
6
7
ke ¼
(4)
L 4 1 0 1 0 5
0 0 0 0
Mass matrix. The elements of the
from Eq. (11.2.23), which gives
2
23
me 6
e
60
m ¼
70 4 12
0
consistent mass matrix are computed
0
23
0
12
12
0
23
0
3
0
12 7
7
0 5
23
(5)
where m e ¼ 7rA0 L=6 is the total mass of the element. Comparison with the
mass matrix of the element of constant cross-section, Eq. (11.2.27), shows that
the deviation is negligible, ð23=70Þ=ð2=6Þ 0:99, ð12=70Þ=ð1=6Þ 1:03.
The elements of the mass matrix for lumped mass assumption are computed
from Eq. (11.2.28), which give
Z 1
7
rA0 L 1 + x x 2 xdx ¼ rLA0
(6)
m11 ¼ m22 ¼
12
0
Z 1
7
m33 ¼ m44 ¼
rA0 L 1 + x x 2 ð1 x Þdx ¼ rLA0
(7)
12
0
2
3
1 0 0 0
1 60 1 0 07
7
(8)
me ¼ m e 6
2 40 0 1 05
0 0 0 1
where m e ¼ 7rA0 L=6 is the total mass of the element.
372 PART
II Multi-degree-of-freedom systems
Damping matrix. The elements of the damping matrix are computed from
the Eq. (11.2.40), which gives
2
3
1 0 1 0
6 0 0 0 07
cs A
6
7
ce ¼
(9)
L 4 1 0 1 0 5
0 0 0 0
Nodal Load forces. They are computed from Eq. (11.2.47), which gives
pe ðt Þ ¼
pe ðt Þ
f 1 0 1 0 gT
2
(10)
where pe ðt Þ ¼ 7p0 ðt ÞL=6 is the total load of the element.
11.2.1.2 The method of virtual work
The method of virtual work is the usual method employed in FEM to derive the
pertinent nodal forces and matrices of the element. This principle states that if a
deformable system being in equilibrium under a set of forces is given virtual
displacements that are consistent with the constraints on the system, the virtual
work dWex done by the external (applied) forces riding the virtual displacements equals the virtual work done by the internal forces, dWin . We illustrate
this method by deriving the equivalent elastic nodal and inertial forces as well as
the stiffness and mass matrices of the truss element. For the other element
equivalent nodal forces matrices, we refer to the method of the Lagrange equations, where the virtual work of the damping and external forces was evaluated,
cases (iii) and (iv).
(i) Nodal elastic forces and stiffness matrix of the truss element
The virtual work due to the axial elastic deformation of the element when it is
given a virtual displacement du is equal to the virtual strain energy
Z
Z L
Aðx ÞEex dex dx
(11.2.49)
dWin ¼ sx dex dx ¼
0
V
where V is the volume of the element.
Taking into account Eq. (11.2.1), we obtain
ex ¼
dex ¼ d
∂u
¼ u1
∂x
∂u
¼ du1
∂x
+ u3
0
3
¼½
0
1
0
3
u1
u3
+ du3
0
3
¼½
0
1
0
3
du1
du3
0
1
0
1
¼ N0 u
¼ duT N0
(11.2.50)
T
(11.2.51)
where
N0 ¼ ½
0
1
0
3 ,
u¼
u1
, du ¼
u3
du1
du3
(11.2.52)
The finite element method Chapter
Substituting Eqs. (11.2.50), (11.2.51) into Eq. (11.2.49) yields
Z L
T
0T 0
EAðx ÞN N dx u
dWin ¼ du
11
373
(11.2.53)
0
Besides, the virtual work of the equivalent elastic nodal forces
f S ¼ ½ fS1 fS3 T is
dWex ¼ duT f
(11.2.54)
Inasmuch as dWin ¼ dWex we obtain
fS1
k k
u1
¼ 11 13
fS3
k31 k33
u3
(11.2.55)
where
Z
kij ¼
0
L
EAðx Þ 0i ðx Þ 0j ðx Þdx,
i, j ¼ 1, 3
(11.2.56)
Further, taking into account that fS2 ¼ fS4 ¼ 0, we can write Eq. (11.2.55)
8 9 2
38 9
k11 0 k13 0 > u1 >
fS1 >
>
>
>
> >
> 6
> =
=
<
< >
7>
fS2
6 0 0 0 0 7 u2
¼6
(11.2.57)
7
>
4 k31 0 k33 0 5>
fS3 >
u3 >
>
>
>
>
>
>
>
>
: ;
: ;
0 0 0 0
fS4
u4
which is identical to Eq. (11.2.16).
(ii) Nodal inertial forces and mass matrix of the truss element
The internal inertial force of mass element m ðx Þdx is
fI ðx, t Þ ¼ m ðx Þ½u€ðx, t Þ + v€ððx, t Þdx
(11.2.58)
and the virtual work produced when the end displacements are given a virtual
displacement ðdu1 , du2 , du3 , du4 Þ is
Z L
m ðx Þ½u€ðx, t Þdu + v€ðx, t Þdv dx
(11.2.59)
dWin ¼
0
which by virtue of Eqs. (11.2.1), (11.2.7) is written as
Z L
€
dWin ¼ duT
m ðx ÞNT Ndx u
(11.2.60)
0
where
N¼½
1
2
3
4¼
1
½ ð1 xÞ ð1 x Þ x x L
(11.2.61)
The integral in Eq. (11.2.60) gives the mass matrix of the truss element
374 PART
II Multi-degree-of-freedom systems
2
Z
me ¼
0
L
m11 0
m13 0
3
6
7
m24 7
m22 0
60
7
m ðx ÞNT Ndx ¼ 6
6m 0
m33 0 7
4 31
5
0
m42 0
(11.2.62)
m44
where
Z
mij ¼ mi + 1, j + 1 ¼
L
m ðx Þ i ðx Þ j ðx Þdx, i, j ¼ 1, 3
(11.2.63)
0
Consequently, Eq. (11.2.60) becomes
dWin ¼ duT m€
u
(11.2.64)
Besides, the virtual work of the equivalent nodal inertial forces is
dWex ¼ duT f I
Inasmuch as dWin ¼ dWex we obtain
8 9 2
m11 0
fI 1 >
m13
>
>
>
>
> 6
> >
=
<
m22 0
fI 2
60
¼6
6
>
fI 3 >
m33
>
>
> 4 m31 0
>
>
;
: >
fI 4
0
m42 0
(11.2.65)
38 9
u€
>
>
> 1>
>
> >
7>
<
m24 7 u€2 =
7
u€3 >
0 7
>
>
5>
>
>
>
;
: >
m44
u€4
0
(11.2.66)
which is identical to (11.2.32).
From the previous discussion, it becomes clear that the method of the
Lagrange equations is more convenient to derive the equivalent element
quantities (nodal forces and matrices), provided that the reader is familiar with
this method.
11.2.2 Transformation of the nodal coordinates of the truss element
In order to formulate the equilibrium equations at the nodes of the truss, all element nodal quantities, forces, and displacements should be referred to a single
system of axes. The key to achieving this is the transformation matrix of the
components of a vector with respect to a rotated system of axes. Thus, we consider two systems of axes in the plane, O
x y and Oxy. The second system results
by rotating the first through an angle f about the z axis (Fig. 11.2.5) in the positive sense, that is, counterclockwise. If ða1 , a2 Þ and ða1 , a2 Þ denote the components of a vector a with respect to these systems of axes, the following
transformation relation is valid
( ) "
#( )
cos f sin f
a1
a1
¼
(11.2.67)
sin f cos f
a2
a2
The finite element method Chapter
11
375
Fig. 11.2.5 Transformation of the coordinates of a vector.
The previous relation is readily obtained by projecting the broken line OAP
successively onto the axes Ox and Oy and taking into account that OA ¼ a1 ,
AP ¼ a2 , OC ¼ a1 , and CP ¼ a2 (see Fig. 11.2.5).
Eq. (11.2.67) is written as
a ¼L
a
where
a¼
a1
,
a2
a ¼
a1
,
a2
(11.2.68)
L¼
cos f sin f
sin f cos f
(11.2.69)
The matrix L is referred to as the transformation matrix of the components
of a plane vector.
It can be readily shown that
L1 ¼LT
(11.2.70)
that is, the matrix L is orthonormal. Thus Eq. (11.2.68) is readily inverted to
a ¼LT a
(11.2.71)
!
We now consider the element e, whose longitudinal axis j k defines the
right-handed system of axes jxy. Let O xy
be another arbitrarily chosen righthanded system of axes and let f be the angle between x and x axes, that is,
the angle that the axis x sweeps when it is rotated counterclockwise to coincide
with the x axis, Fig. 11.2.6. The axes x, y are called the global axes while the
axes x, y defined by the truss element are called the local axes. Any quantity,
force, or displacement referred to the global axes will be henceforth designated
by an overbar.
Fig. 11.2.6 Nodal displacements of the truss element in global and local axes.
376 PART
II Multi-degree-of-freedom systems
Empoying Eq. (11.2.67) for the nodal
u1
cos f
¼
u2
sin f
u3
cos f
¼
u4
sin f
displacements, we obtain
sin f
u1
(11.2.72a)
cos f u2
sin f
u3
(11.2.72b)
cos f u4
The previous two relations can be combined as
8 9 2
38 9
cos f sin f
0
0
u1 >
>
>
>
>
>
> u1 >
< = 6
7< u2 =
sin
f
cos
f
0
0
u2
6
7
¼
0
cos f sin f 5>
u > 4 0
u >
>
>
>
;
;
: 3>
: 3>
0
0
sin f cos f
u4
u4
(11.2.73)
or
ue ¼Re ue
(11.2.74a)
ue ¼ðRe ÞT ue
(11.2.74b)
and after inverting
where
2
cos fe
6
sin fe
Re ¼ 6
4 0
0
sin fe 0
cos f 0
0
cos fe
0
sin fe
3
0
7
0
7
e5
sin f
cos fe
(11.2.75)
represents the transformation matrix of the plane truss element e.
The vectors of the equivalent nodal forces are defined with respect to
the global axes as shown in Fig. 11.2.7. It is obvious that they also obey the
transformation law (11.2.74b). Thus, we have
f e ¼ðRe ÞT f e
S
S
(11.2.76)
f e ¼ðRe ÞT f e
I
I
(11.2.77)
f e ¼ðRe ÞT f e
D
D
(11.2.78)
pe ðt Þ ¼ðR Þ p ðt Þ
(11.2.79)
e T e
Fig. 11.2.7 Nodal forces of the truss element in global and local axes.
The finite element method Chapter
11
377
Eq. (11.2.76) is further written as
f e ¼ðRe ÞT f e
S
S
¼ ðRe ÞT ke ue
(11.2.80)
e T e
¼ ðR Þ k Re ue
or
f e ¼ke ue
S
(11.2.81)
e
k ¼ ðRe ÞT ke Re :
(11.2.82)
where
is the element stiffness matrix with respect to the global axes.
Similarly, we obtain
f e ¼m
e ue
I
(11.2.83)
f D ¼
ce ue
(11.2.84)
e ¼ ðRe ÞT me Re
m
(11.2.85)
ce ¼ ðRe ÞT ce Re
(11.2.86)
where
11.2.3 Equation of motion of the plane truss
The formulation of the equations of motion in FEM can be viewed as a process
of assembling of the element stiffness, mass, and damping matrices as well as of
the external nodal forces into the corresponding global matrices and global load
nodal vector of the structure. This process involves first the formulation of the
global displacement vector u. This vector contains all nodal displacements of
the truss referred to the global axes including those of the supports. If the truss
has n nodes, then the vector u will have N ¼ 2n elements. The displacements of
the i node will take the places 2i 1 and 2i in the global vector. Thus, if j, k
denote the numbers of origin and the end of the e element, the displacements
ue will take the places u2j1 , u2j , u2k1 , u2k in u. This can be realized computationally by means of the linear transformation
ue ¼ae u
(11.2.87)
where ae is an 4 N Boolean matrix, that is, a matrix consisting of zeros and ones.
ð11:2:88Þ
378 PART
II Multi-degree-of-freedom systems
The above transformation is inverted asb
u ¼ðae ÞT ue
(11.2.89)
^f e
S
If
denotes the enlarged vector with dimension N , then the assemblage
e
of the elements of f S can be realized according to Eq. (11.2.89)
^f e ¼ ðae ÞT f e
S
S
e
The next step is to write the relation f S
(11.2.90)
¼k u as a relation between ^f S and u.
Starting with Eq. (11.2.90) and using the transformations (11.2.87), we obtain
e e
^f e ¼ ðae ÞT f e
S
S
e T e e
¼ ða Þ k u
¼ ðae ÞT k ae u
^ e u
¼K
e
e
(11.2.91)
where
^ e ¼ ðae ÞT ke ae
K
(11.2.92)
denotes the enlarged stiffness matrix with dimensions N N . Obviously, the
e
^ e are:
places of the elements of k in K
k^2j1,2j1 ¼ k11 , k^2j1,2j ¼ k12 , k^2j1, 2k1 ¼ k13 , k^2j1, 2k ¼ k14
k^2j, 2j1 ¼ k21 , k^2j,2j ¼ k22 , k^2j,2k1 ¼ k23 , k^2j, 2k ¼ k24
k^2k1,2j1 ¼ k31 , k^2k1,2j ¼ k32 , k^2k1, 2k1 ¼ k33 , k^2k1, 2k ¼ k34
k^2k, 2j1 ¼ k41 , k^2k,2j ¼ k42 , k^2k,2k1 ¼ k43 , k^2k, 2k ¼ k44
Following the same process, we obtain the relations
^f e ¼M
^ e u€
I
(11.2.93)
^f e
D
(11.2.94)
^ e u_
¼C
b. The matrix a is not square, therefore it cannot be inverted. The inverse transformation (11.2.89)
can be proved either by inspection or by considering the equality of the work produced by the nodal
forces in the two systems of diplacements ue and u. Indeed, according to Eq. (11.2.87) we have
f e ¼ae^f e
where f and ^f e are element force vectors with a dimension 4 and N , respectively.
e
Equating the works gives
^f e
T
e
u ¼ f ue
¼ ae ^f e
¼ ^f e
which results in Eq. (11.2.89).
T
T
ue
ðae ÞT ue
The finite element method Chapter
^e ðt Þ ¼ ðae ÞT pe ðt Þ
p
^f e ,
I
11
379
(11.2.95)
^f e ,
D
^e ðt Þ denote the enlarged nodal inertial, damping matrices,
and p
where
and load vector, respectively, while
^ e ¼ ðae ÞT m
e ae
M
^e
e T e e
C ¼ ða Þ ca
(11.2.96)
(11.2.97)
ðt Þ with a dimenFinally, all nodal loads are assembled into a load vector P
sion N , corresponding to u. Hence, the loads applied to node i will take the
ðt Þ.
places P2i1 and P2i in P
The equations of motion can be derived using the method of the Lagrange
equations with generalized coordinates the components of the vector u. Thus,
we have:
Elastic energy. The elastic energy of the e element is
e
1
Ue ¼ uT ^f S
2
1 ^e
u
¼ uT K
2
(11.2.98)
and the total elastic energy is
U¼
Ne
1X
^ e u
uT K
2 e¼1
1 ¼ uT K
u
2
N X
N
1X
ij ui uj
K
¼
2 i¼1 j¼1
(11.2.99)
where
¼
K
Ne
X
^e
K
(11.2.100)
e¼1
represents the stiffness matrix of the structure with Ne denoting the number of
all elements.
Kinetic energy. Similarly, the kinetic energy of the structure results as the
sum of the kinetic energies of all elements
T¼
Ne
1X
T ^e
u_
u_ M
2 e¼1
1 T
¼ u_ M
u_
2
N X
N
1X
ij u_ i u_ j
¼
M
2 i¼1 j¼1
(11.2.101)
380 PART
II Multi-degree-of-freedom systems
where
¼
M
Ne
X
^e
M
(11.2.102)
e¼1
represents the mass matrix of the structure.
Virtual work of the damping forces. For the e element, we have
e
dWDe ¼ d
uT ^f D
^ e u_
¼ d
uT C
(11.2.103)
and the total virtual work of the damping forces is
dWD ¼ Ne
X
^ e u_
d
uT C
e¼1
u_
¼ d
uT C
¼
N X
N
X
ij u_ j d
ui
C
(11.2.104)
i¼1 j¼1
where
¼
C
Ne
X
^e
C
(11.2.105)
e¼1
represents the damping matrix of the structure.
Virtual work of the external loads. It results as the sum of all external loads,
ðt Þ applied
^e ðt Þ and the loads P
namely of the equivalent element nodal loads p
directly to the nodes of the truss. Therefore, we have
uT Pðt Þ +
dWp ¼ d
Ne
X
^ ðt Þe
d
uT p
e¼1
¼ d
u
T
Ne
X
ðt Þ +
^ e ðt Þ
p
P
!
(11.2.106)
e¼1
¼ d
u pðt Þ
T
where
ðt Þ +
pðt Þ ¼ P
Ne
X
^ e ðt Þ
p
(11.2.107)
e¼1
represents the vector of all external loads.
Applying Eq. (1.8.11), we obtain
∂T
¼0
∂
ui
(11.2.108a)
The finite element method Chapter
11
381
X
N
d ∂T
¼
Mij u€j
dt ∂u_ i
j¼1
(11.2.108b)
N
∂U X
¼
Kij uj
∂
u i j¼1
(11.2.108c)
ij u_ j + pi ðt Þ
Qi ¼ C
(11.2.108d)
and the equation of motion becomes
N
X
j¼1
ij u€j +
M
N
X
j¼1
ij u_ j +
C
N
X
ij uj ¼ pi ðt Þ
K
(11.2.109)
j¼1
or
u_ + K
u€ + C
u ¼
M
pðt Þ
(11.2.110)
Obviously, Eq. (11.2.110) has the same form as that obtained using the
methods discussed in Chapter 10.
11.2.4 Steps to formulate the equations of motion for a plane truss
by the finite element method
The formulation of the equations of motion for a structure (here the plane truss)
by the FEM based on the analytical procedure described in the previous section
can be summarized as a sequence of the following steps:
Fig. 11.2.8 Plane truss (n ¼ 5, N ¼ 2n ¼ 10, Ne ¼ 8).
(i) Idealization of the truss by a set of finite elements interconnected at its
nodes. This step includes (Fig. 11.2.8):
a. Selection of the system of global axes x y.
b. Numbering the nodes of the truss ð1, 2, …, n Þ and determining their
coordinates with respect to the global axes.
c. Determination of the vector of the N ¼ 2n degrees of freedom u.
ðt Þ of the external loads directly applied to
d. Formulation of the vector P
the nodes of the truss.
382 PART
II Multi-degree-of-freedom systems
e. Numbering of the elements ð1, 2, …, Ne Þ and determination of their
orientation by selecting the positive direction of the local x axis.
(ii) For each element ðe ¼ 1, 2, …, Ne Þ
a. Compute the matrices:
– ke using Eq. (11.2.18)
– me using Eq. (11.2.26) or (11.2.33) depending on the mass
assumption
– ce using Eq. (11.2.44), if the coefficient cs is given
– pe ðt Þ using Eq. (11.2.48)
– Re using Eq. (11.2.75)
– ae using Eq. (11.2.88)
b. Compute the matrices and load vector:
e
k ¼ ðRe ÞT ke Re
e ¼ ð Re Þ T m e Re
m
ce ¼ ðRe ÞT ce Re
pe ðt Þ ¼ ðRe ÞT pe ðt Þ
c. Using Eqs. (11.2.92), (11.2.96), (11.2.97), (11.2.95), formulate the
enlarged matrices and load vector
^ e , and p
^ e, C
^ e, M
^ e ðt Þ
– K
(iii) Compute the matrices and the load vector of the structure
¼
K
Ne
X
e¼1
¼
^ e, M
K
Ne
X
e¼1
¼
^ e, C
M
Ne
X
e¼1
^ e , pðt Þ ¼ P
ðt Þ +
C
Ne
X
^ e ðt Þ
p
e¼1
11.2.5 Modification of the equations of motion due to the supports
of the structure
Eq. (11.2.110) describes the motion of the free structure. However, the truss
may be supported at some of its nodes. Supporting the structure means constraining certain degrees of freedom or, in other words, specifying certain nodal
displacements. The minimum number of degrees of freedom that should be constrained is three in order to prevent the movement of the structure as a rigid
body. These degrees of freedom are selected so that an infinitesimal movement
is excluded.
Let s be the number of the specified degrees of freedom, hence s elements of
the vector u are known while f ¼ N s are unknown. We reorder the elements
of u to separate the known displacements from the unknown ones and we denote
by us the vector of the known (support) displacements and by uf the vector of
the unknown (free) displacements. The elements of us are equal to zero for
The finite element method Chapter
11
383
nonyielding supports. In general, the elements of us may depend on time,
us ¼ us ðt Þ, as in the case of support excitation. The nodal quantities ui and
pi ðt Þ are referred to as dual quantities. For a well-posed problem, the dual quantities cannot be specified simultaneously. Either ui is specified and pi ðt Þ is
unknown or pi ðt Þ is specified and ui is unknown.
Consequently, the vector ps ðt Þ, which corresponds to us , contains the
unknown nodal forces, namely the support reactions plus the element nodal
contributions, while the vector pf ðt Þ, which corresponds to uf , contains the
known nodal forces.
The N equations of motion (11.2.110) will be solved to determine the
unknown vectors uf and ps ðt Þ. The solution procedure requires first the rearrangement of the equations of motion, which is done as follows.
We rearrange the elements of u so that the known elements occupy the last
positions. This rearrangement can be done computationally by the transformation
u ¼Ve
u
(11.2.111)
where e
u is the rearranged vector and V is a square Boolean matrix. As an
example, we consider the truss of Fig. 11.2.8. The specified displacements
are u1 ¼ u2 ¼ u4 ¼ 0. Therefore the rearranged displacement vector is
e
e1 u
e2 u
e3 u
e4 u
e5 u
e6 u
e7 u
e8 u
e9 u
e10 g
uT ¼ f u
¼ f u3 u5 u6 u7 u8 u9 u10 u1 u2 u4 g
and the matrix V is defined from the
9 2
8
0 0 0 0 0 0 0 1
u1 >
>
>
>
>
>
>
>
6
>
>
u2 >
>
60 0 0 0 0 0 0 0
>
>
>
>
6
>
>
>
>
61 0 0 0 0 0 0 0
>
>
u
3 >
>
>
>
> 6
>
60 0 0 0 0 0 0 0
>
>
>
u4 >
>
6
>
>
>
>
>
< u = 6
60 1 0 0 0 0 0 0
5
¼6
60 0 1 0 0 0 0 0
>
>
u6 >
>
6
>
>
>
>
6
>
>
>
>
60 0 0 1 0 0 0 0
u
>
7 >
>
>
6
>
>
>
>
60 0 0 0 1 0 0 0
>
u8 >
>
>
6
>
>
>
>
6
>
>
>
>
u
>
>
9
> 40 0 0 0 0 1 0 0
>
>
>
;
:
0 0 0 0 0 0 1 0
u10
relation
9
38
u3 >
0 0 >
>
>
>
>
>
>
> u5 >
1 07
>
7>
>
>
>
>
7>
>
>
>
0 0 7>
u
6 >
>
>
7>
>
>
>
7
>
>
0 1 7>
u
7 >
>
>
>
>
>
7>
0 0 7< u8 =
7
or u ¼ Ve
u
0 07
u9 >
>
>
7>
>
>
>
7>
>
>
> u10 >
0 07
>
>
>
7>
>
>
>
>
7
>
0 0 7>
u
>
1 >
>
>
>
7>
>
>
>
>
5
0 0 >
u
>
2
>
>
>
>
;
:
0 0
u4
(11.2.112)
(11.2.113)
K
C,
and the vector pðt Þ
We can readily show that the matrices M,
are rearanged by virtue of matrix V as
e ¼ VT MV
M
(11.2.114a)
e ¼ VT CV
C
(11.2.114b)
e ¼VT KV
K
(11.2.114c)
e
pðtÞ ¼VT pðtÞ
(11.2.114d)
384 PART
II Multi-degree-of-freedom systems
Thus, the rearranged equations of motion become
ee
eu
e u ¼e
e€ + C
M
u_ + Ke
pðt Þ
which are partitioned after separating the known from the unknown
displacements
ð11:2:115Þ
or
"
e ff M
e fs
M
e
e
Msf Mss
#
e€f
u
e
u€s
"
e ff C
e fs
C
+
e
e
Csf Css
#
e_ f
u
e
u_ s
"
e ff K
e fs
K
+
e
e
Ksf Kss
#
e
uf
e
us
¼
e
p f ðt Þ
e
p s ðt Þ
(11.2.116)
Performing the matrix multiplications in the previous equation yields
e ff e
e ff e
e ff e
M
u€f + C
u_ f + K
uf ¼ e
pf∗ ðt Þ
e sf e
e ss e
e ss e
e sf e
e ss e
e sf e
u€f + M
u€s + C
u_ f + C
u_ s + K
uf + K
us ¼ e
p s ðt Þ
M
(11.2.117)
(11.2.118)
where it was set
e fs e
e fs e
e fs e
e
u€s C
u_ s K
us
pf ðt Þ M
pf∗ ðt Þ ¼ e
(11.2.119)
Eq. (11.2.117) is the equation of motion of the supported truss. Its solution
gives the vector of the unknown displacements e
uf , which is then inserted in
Eq. (11.2.118) to obtain the vector of nodal forces e
ps ðt Þ. Note that this vector contains the equivalent element nodal loads contributing to the supported node. Obviously, when the support displacements do not depend on time, Eq. (11.2.119)
becomes
e fs e
e
us
pf ðt Þ K
pf∗ ðt Þ ¼ e
(11.2.120)
The finite element method Chapter
11
385
and Eq. (11.2.118) is written
e sf e
e sf e
e sf e
e ss e
M
u€f + C
u_ f + K
uf + K
us ¼ e
ps ðt Þ
(11.2.121)
Finally, when the supports do not yield, the foregoing equations are further
simplified and become
e
p f ðt Þ
pf∗ ðt Þ ¼ e
(11.2.122)
e sf e
e sf e
e sf e
u€f + C
u_ f + K
uf ¼ e
p s ðt Þ
M
(11.2.123)
Example 11.2.2 Formulate the equations of motion of the truss in Fig. E11.1a
when (i) It is subjected to the external loading. (ii) The support 1 is
subjected to the vertical motion ug ðt Þ. Assumed data: Nodal coordinates:
1ð0, 0Þ, 2ð3, 3:5Þ, 3ð6, 0Þ; cross-sectional areas of the bars: A1 ¼ 1:5A,
A2 ¼ A3 ¼ A; distributed load on the element 1: pðx, t Þ ¼ P ðt Þ=L1 ; damping
coefficient: cs ¼ 0, modulus of elasticity E; material density: r, lumped mass
assumption.
(a)
Fig. E11.1a Plane truss in Example 11.2.2.
Solution
(i) Equation of motion under the external loading.
(b)
Fig. E11.1b Nodal displacements of the plane truss in Example 11.2.2.
386 PART
II Multi-degree-of-freedom systems
The truss has n ¼ 3 nodes, hence the free structure has N ¼ 2n ¼ 6 degrees of
freedom (see Fig. E11.2b). The numbering of the elements and the positive
direction of the local axes are shown in Fig. E11.1b.
TABLE E11.1 Geometrical data of the elements in Example 11.2.2.
e
xj
xk
D
x
yj
yk
1
0
6
6
0
0
2
0
3
3
0
3.5
3
3
6
3
3.5
0
D
y
Le
cos fe
sinfe
0
6
1
3.5
4.61
0.651
þ0.759
3.5
4.61
0.651
0.759
0
1. Computation of ke , me , ce , pe ðt Þ, Re , e ¼ 1, 2, 3
Matrices ke . The elements have constant cross-section. Thus Eq. (11.2.19) gives
2
3
2
3
0:25 0 0:25 0
0:217 0 0:217 0
6 0
6 0
0 0
07
0 0
07
3
7 2
6
7
k1 ¼ EA6
4 0:25 0 0:25 0 5, k ¼ k ¼ EA4 0:217 0 0:217 0 5
0
0 0
0
0
0 0
0
Matrices me . The elements have constant cross section. Thus, we obtain
from Eq. (11.2.35)
2
2
3
3
4:5 0 0 0
2:305 0
0
0
6 0 4:5 0 0 7
60
7
2:305 0
0
2
3
6
7
7
m1 ¼ rA6
4 0 0 4:5 0 5, m ¼ m ¼ rA4 0
5
0
2:305 0
0 0 0 4:5
0
0
0
2:305
Matrices ce . It is given cs ¼ 0. Thus, Eq. (11.2.45) gives ce ¼ 0.
Vectors pe ðt Þ. They are obtained from Eq. (11.2.47), which gives
9
8
8 9
0:5P ðt Þ >
0>
>
>
>
>
>
=
=
<
< >
0
0
, p2 ðt Þ ¼ p3 ðt Þ ¼
p1 ðt Þ ¼
0:5P ðt Þ >
0>
>
>
>
>
>
;
;
:
: >
0
0
Matrices Re . The direction cosines cos fe and sin fe of the element axes are
calculated from the global coordinates of the element nodes and are given in
Table E11.1. Thus, we have
2
3
2
3
1 0 0 0
0:651 0:759 0
0
60 1 0 07
6 0:759 0:651 0
7
0
2
6
7
7
R1 ¼ 6
4 0 0 1 0 5, R ¼ 4 0
0
0:651 0:759 5
0 0 0 1
0
0
0:759 0:651
The finite element method Chapter
11
387
2
3
0:651 0:759 0
0
6 0:759 0:651 0
7
0
7
R3 ¼ 6
40
0
0:651 0:759 5
0
0
0:759 0:651
e
e , pe ðt Þ, e ¼ 1, 2, 3
2. Computation of k , m
e
Matrices k . Eq. (11.2.82) gives
2
0:25 0 0:25
6
0
0 0
k1 ¼ EA6
4 0:25 0 0:25
0
0 0
2
0:092 0:107
6 0:107 0:125
2
6
k ¼ EA4
0:092 0:107
0:107 0:125
2
0:092 0:107
6
0:107
0:125
k3 ¼ EA6
4 0:092 0:107
0:107 0:125
3
0
07
7,
05
0
3
0:092 0:107
0:107 0:125 7
7
0:092 0:107 5
0:107 0:125
3
0:092 0:107
0:107 0:125 7
7
0:092 0:107 5
0:107 0:125
e . Eq. (11.2.85) gives
Matrices m
2
3
2
4:5 0 0 0
2:305
6 0 4:5 0 0 7
60
1
2
3
6
6
7
m ¼ rA4
, m ¼ m ¼ rA4
0 0 4:5 0 5
0
0 0 0 4:5
0
0
2:305
0
0
0
0
2:305
0
3
0
7
0
7
5
0
2:305
0
0
0
0
0
0
1
0
Vectors pe ðt Þ. Eq. (11.2.79) gives
9
8
8 9
0:5P ðt Þ >
0>
>
>
>
>
>
=
=
<
< >
0
0
1
2
3
, p ðt Þ ¼ p ðt Þ ¼
p ðt Þ ¼
0:5P ðt Þ >
0>
>
>
>
>
>
;
;
:
: >
0
0
^ e, M
^ e, p
^e ðt Þ, e ¼ 1, 2, 3
3. Computation K
The element
2
1 0
6
0 1
a1 ¼ 6
40 0
0 0
assembly matrices ae are
3
2
0 0 0 0
1 0
6
0 0 0 07
7, a2 ¼ 6 0 1
40 0
0 0 1 05
0 0 0 1
0 0
0
0
1
0
0
0
0
1
0
0
0
0
3
2
0
0
6
07
7, a3 ¼ 6 0
40
05
0
0
1
0
0
0
0
1
0
0
3
0
07
7
05
1
388 PART
II Multi-degree-of-freedom systems
They yield the enlarged element matrices:
^ e . Eq. (11.2.92) gives
Matrices K
2
3
0:25 0 0 0 0:25 0
6 0
0 0 0 0
07
6
7
6
0 0 0 0
07
^ 1 ¼ EA6 0
7,
K
6 0
0 0 0 0
07
6
7
4 0:25 0 0 0 0:25 0 5
0
0 0 0 0
0
2
3
0:092 0:107 0:092 0:107 0 0
6 0:107 0:125 0:107 0:125 0 0 7
6
7
6
7
^ 2 ¼ EA6 0:092 0:107 0:092 0:107 0 0 7
K
6 0:107 0:125 0:107 0:125 0 0 7
6
7
4 0
0
0
0
0 05
0
0
0
0
0 0
2
3
0 0 0
0
0
0
60 0 0
7
0
0
0
6
7
6
7
0
0
0:092
0:107
0:092
0:107
3
^
6
7
K ¼ EA6
7
0
0
0:107
0:125
0:107
0:125
6
7
4 0 0 0:092 0:107 0:092 0:107 5
0 0 0:107 0:125 0:107 0:125
^ e . Eq. (11.2.96) gives
Matrices M
2
3
4:5 0 0 0 0 0
6 0 4:5 0 0 0 0 7
6
7
6
7
^ 1 ¼ rA6 0 0 0 0 0 0 7
M
60 0 0 0 0 0 7
6
7
4 0 0 0 0 4:5 0 5
0 0 0 0 0 4:5
2
3
2:305 0
0
0
0 0
60
2:305 0
0
0 07
6
7
6
0
0
2:305
0
0
07
2
^
6
7
M ¼ rA6
0
0
2:305 0 0 7
60
7
40
0
0
0
0 05
0
0
0
0
0 0
2
3
0 0 0
0
0
0
60 0 0
7
0
0
0
6
7
6
7
0
0
^ 3 ¼ rA6 0 0 2:305 0
7
M
60 0 0
7
2:305 0
0
6
7
40 0 0
5
0
2:305 0
0 0 0
0
0
2:305
The finite element method Chapter
11
389
^e ðt Þ. Eq. (11.2.95) gives
Vectors p
9
8
8 9
0:5P ðt Þ >
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0>
>
>
>
>
>
>
>
=
=
<
< >
0
0
1
2
3
^ ðt Þ ¼ p
^ ðt Þ ¼
^ ðt Þ ¼
, p
p
0
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
0:5P ðt Þ >
0>
>
>
>
>
>
>
>
;
;
:
: >
0
0
M
and total load vector pðt Þ of
4. Computation of the total matrices K,
the truss
2
0:342
0:107 0:092 0:107 0:25
0
0
0
3
6
7
6 0:107 0:125 0:107 0:125 0
7
0
6
7
6
7
3
X e
0:092 0:107 7
6 0:092 0:107 0:184 0
¼
^ ¼ EA6
7
K
K
6 0:107 0:125 0
0:25
0:107 0:125 7
6
7
e¼1
6
7
6 0:25
0
0:092 0:107 0:342 0:107 7
4
5
2
6:805
6
60
6
3
60
X e
¼
^ ¼ rA6
M
M
6
60
e¼1
6
6
40
0:107 0:125 0:107
0
0
0
0
0
6:805 0
0
0
0
0
4:610 0
0
0
0
0
4:610 0
0
0
0
0
6:805 0
0:125
3
7
7
7
7
7
7,
7
7
7
5
0
0
0
0
0
6:805
9 8
9 8
9
8
P1 + 0:5P ðt Þ >
0:5P ðt Þ >
P1 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
P2 >
P2
>
>
>
>
>
>
>
>
>
>
>
>
=
3
<
=
=
<
<
X
0
P
ð
t
Þ
P
ð
t
Þ
ðt Þ +
^ e ðt Þ ¼
¼
+
p
pðt Þ ¼ P
0
>
>
>
> 3P ðt Þ >
> 3P ðt Þ
>
>
>
>
>
e¼1
> >
>
>
>
>
>
>
>
>
>
0:5P ðt Þ >
0:5P ðt Þ
>
>
> >
>
> 0
>
>
>
>
>
>
>
; :
; :
;
:
0
P6
P6
Note that P1 , P2 , P6 denote the unknown support reactions
e M
e and load vector e
5. Computation of the modified matrices K,
pðt Þ
Due to the supports, the displacement vector should be modified as
e
e2 u
e3 u
e4 u
e5 u
e6 g ¼ f u3 u4 u5 u1 u2 u6 g
e1 u
uT ¼ f u
390 PART
II Multi-degree-of-freedom systems
Therefore, the matrix V defined by Eq. (11.2.111) is
2
3
0 0 0 1 0 0
60 0 0 0 1 07
6
7
61 0 0 0 0 07
7
V¼6
60 1 0 0 0 07
6
7
40 0 1 0 0 05
0 0 0 0 0 1
Then, applying Eqs. (11.2.114a), (11.2.114c), (11.2.114d) gives
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
e ff , M
e fs , M
e sf , M
e ss , K
e ff , K
e fs , K
e sf , K
e ss and
6. Formulation of the matrices M
ps ðt Þ
vectors e
pf ðt Þ, e
e K,
e e
These matrices are obtained after partitioning M,
pðt Þ on the
previously indicated partition. Thus, we have
2
3
2
4:610 0
0
0 0
e fs ¼ M
e sf ¼ rA4 0 0
e ff ¼ rA4 0
5, M
4:610 0
M
0
0
6:805
0 0
2
3
6:805 0
0
e ss ¼ rA4 0
5
6:805 0
M
0
0
6:805
basis of the
3
0
05
0
The finite element method Chapter
2
0:184 0
e ff ¼ EA6
K
4 0
0:25
0:092 0:107
0:092
3
2
0:092 0:107
11
391
0:107
3
7 e
6
7
0:107 5, K
fs ¼ EA4 0:107 0:125 0:125 5
0:342
0:25
0
0:107
2
3
2
3
0:092 0:107 0:25
0:342 0:107 0
e ss ¼ EA4 0:107 0:125 0
e sf ¼ EA4 0:107 0:125 0
5, K
5
K
0:107 0:125 0:107
0
0
0:125
9
9
8
8
< P ðt Þ =
< P1 + 0:5P ðt Þ =
e
, e
ps ðt Þ ¼ P2
pf ðt Þ ¼ 3P ðt Þ
;
;
:
:
0:5P ðt Þ
P6
7. Equation of motion of the truss
Because e
us ¼ 0, it is also e
u€s ¼ 0 and Eq. (11.2.119) gives e
pf∗ ðt Þ ¼ e
pf ðt Þ.
Hence, Eq. (11.2.117) results in the following equation of motion
2
38 9
2
38 € 9
>
e
0:184 0
0:092 > u
4:610 0
0
e1 >
u
>
>
>
>
>
> 1>
=
<
6
7< =
6
7
6
7
7 u
€
e
+
EA
0
0:250
0:107
rA6
0
4:610
0
u
4
5> 2 >
4
5> e 2 >
>
>
>
>
;
>
: >
;
:
e3
0:092 0:107 0:342
0
0
6:805
u
e€3
u
9
8
P ðt Þ >
>
>
>
=
<
¼ 3P ðt Þ
>
>
>
>
;
:
0:5P ðt Þ
The solution of the equation of motion yields the vector e
uf and
Eq. (11.2.118) the nodal forces at the supports
8 9
2
38 9
e1 =
0:125 0
0:107 < u
ps1 =
<e
e
e2
0:092 0:092 5 u
ps2 ¼ EA4 0
: ;
: ;
e
e3
0:125 0:107 0
ps3
u
Hence, the reactions at the supports are
8 9 8 9 8
9
ps1 = < 0:5P ðt Þ =
< P1 = < e
P
¼ e
0
p
: 2 ; : s2 ; :
;
e
P6
ps3
0
(ii) Equation of motion due to the vertical motion ug ðt Þ of the support 1.
In this case it is
e
pf ðt Þ ¼ 0,
8 9
< P1 =
e
ps ðt Þ ¼ P2 ,
: ;
P6
8
9
<0
=
e
us ¼ ug ðt Þ
:
;
0
392 PART
II Multi-degree-of-freedom systems
and Eq. (11.2.119) gives
e fs e
e fs e
e fs e
e
u€s C
u_ s K
us
pf∗ ðt Þ ¼ M
or
9
8
< 0:107 =
e
pf∗ ðt Þ ¼ EA 0:125 ug ðt Þ
;
:
0
Hence, the equation of motion becomes
9
2
38 u
2
38 € 9
>
e1 >
0:184 0
0:092 >
4:610 0
0
<u
= E
< e1 >
=
40
5 u
e2
0:250 0:107 5 u
4:610 0
e€2 + 4 0
> >
: >
; r 0:092 0:107 0:342 >
;
0
0
6:805 : u
e3
u
e€3
9
8
0:107 =
E<
¼
0:125 ug ðt Þ
;
r:
0
and Eq. (11.2.123) gives the reactions
9
8
8 9
2
38 9
e1 =
0:092 0:107 0:250 < u
=
<0
< P1 =
5 u
e2 + rA 6:805 u€g
P2 ¼ EA4 0:107 0:125 0
;
:
: ;
: ;
e3
0:107 0:125 0:107
0
u
P6
9
8
< 0:107 =
+ EA 0:125 ug
;
:
0
ef ¼ f u
e1 u
e2 u
e3 gT can be converted to
Remark: The displacement vector u
the original vector of the free nodal displacements, that is, u ¼ f u3 u4 u5 gT
ef .
by inverting the transformations that have created u
11.3 The finite element method for the plane frame
11.3.1 Properties of the plane frame element
We consider a plane frame element with variable cross-section Aðx Þ, crosssectional moment of inertia I ðx Þ, mass per unit length m ðx Þ, and modulus of
elasticity E. The element local axis in the undeformed state coincides with the
x axis. The frame element is a beam, therefore it is also called a beam element.
The ends j,k are the nodes of the element, thus any quantity (force, displacement) associated with them is referred to as nodal quantity (i.e., nodal force,
nodal displacement). In the deformed state at time t, the ends of the element
are displaced to the points j 0 , k 0 and the element occupies the position j 0 k 0
(Fig. 11.3.1). The element, besides the axial deformation, undergoes a flexural
deformation due to the end transverse displacements and rotations. The element
has six degrees of freedom: the displacements u1 ðt Þ, u2 ðt Þ and the rotation
u3 ðt Þ ¼ q1 of node j as well as the displacements u3 ðt Þ, u4 ðt Þ and rotation
The finite element method Chapter
11
393
u6 ðt Þ ¼ q2 of node k. These six quantities represent the nodal coordinates of the
plane frame element. A typical point x of the element axis undergoes two displacements: the axial displacement u ðx, t Þ and the transverse displacement
v ðx, t Þ. The rotation of the cross-section v 0 ðx, t Þ has a small influence on the
dynamic properties of the beam so it is ignored in the presentation that follows.
We consider the element as a generalized system and apply the Ritz method
presented in Section 10.5.1. Thus, we set the axial displacement in the form
u ðx, t Þ ¼ u1
1 ðx Þ + u 4 4 ðx Þ
(11.3.1)
where 1 ðx Þ and 4 ðx Þ are shape functions expressing the axial deformation for
u1 ¼ 1, u4 ¼ 0 and u1 ¼ 0, u4 ¼ 1, respectively. These functions for an element
with constant cross-section were obtained in Section 11.2.1 as
1 ðx Þ ¼ 1 x
(11.3.2a)
4 ðx Þ ¼ x
(11.3.2b)
where x ¼ x=L.
Fig. 11.3.1 Degrees of freedom of a plane frame element.
The transverse displacement can be set in the form
v ðx, t Þ ¼ u2
2 ðx Þ + u3 3 ðx Þ + u5 5 ðx Þ + u6 6 ðx Þ
(11.3.3)
where i ðx Þ are shape functions expressing the elastic curve of the element for
ui ¼ 1, ði ¼ 2, 3, 5, 6Þ and uj ¼ 0, j 6¼ i.
These functions can be determined from the solution of the beam equation
under bending. Without restricting the generality, we assume that the Bernoulli
assumption is valid for the bending of the element. Therefore, the transverse
deformation results only from bending. In this case, the elastic curve is obtained
by successive integration of the equation
d2
d 2 ðx Þ
¼0
(11.3.4)
EI
ð
x
Þ
dx 2
dx 2
The solution of this equation, although it gives shape functions expressing
the exact static transverse deformation of the beam element, is not suitable for
the automation of the method because different shape functions have to be
determined for elements with a different law of variation of the moment of
394 PART
II Multi-degree-of-freedom systems
inertia. As we pointed out in Section 11.2.1, this difficulty is surpassed if we
accept the same shape functions for all elements, regardless of the variation
law of the cross-section. Apparently, the shape functions resulting from
Eq. (11.3.4) for a constant cross-section are suitable. In this case, the solution
of Eq. (11.3.4) is given by the cubic polynomial
ðx Þ ¼ c 1
x3
x2
+ c2 + c3 x + c4
6
2
(11.3.5)
x2
+ c2 x + c3
2
(11.3.6)
and its first derivative by
0
ðx Þ ¼ c1
The arbitrary constants are evaluated from the boundary conditions.
Thus, the shape function
2 ðx Þ
results
3 ðx Þ results
5 ðx Þ results
6 ðx Þ results
for
for
for
for
2 ð0Þ ¼ 1,
3 ð0Þ ¼ 0,
5 ð0Þ ¼ 0,
6 ð0Þ ¼ 0,
0
2 ð0Þ ¼ 0,
0
3 ð0Þ ¼ 1,
0
5 ð0Þ ¼ 0,
0
6 ð0Þ ¼ 0,
2 ðLÞ ¼ 0,
3 ðLÞ ¼ 0,
5 ðLÞ ¼ 1,
6 ðLÞ ¼ 0,
0
2 ðLÞ ¼ 0
0
3 ðLÞ ¼ 0
0
5 ðLÞ ¼ 0
0
6 ðLÞ ¼ 1
This procedure yields
2 ðx Þ ¼ 1 3x
2
+ 2x 3
2
3
3 ðx Þ ¼ L x 2x + x
5 ðx Þ ¼ 3x
6 ðx Þ ¼ L
(11.3.7a)
(11.3.7b)
2x3
(11.3.7c)
x + x
(11.3.7d)
2
2
3
where x ¼ x=L.
Fig. 11.3.2 Shape functions of the transverse deformation of the beam element.
The finite element method Chapter
11
395
The cubic polynomials given by Eqs. (11.3.7a)–(11.3.7d) are known as
Hermitian polynomials. Their graphic representation is shown in Fig. 11.3.2.
Obviously, they satisfy the differential equation of the elastic curve of a beam
with a constant cross-section. However, as mentioned, they can be used as shape
functions of elements with a variable cross-section, provided that the elements
are small. The accuracy of the solution is improved by subdividing the element
into smaller elements. This technique increases the number of elements.
The equivalent nodal forces of the element will be obtained using the
methods employed for the truss element. It should be recalled here that, unlike
the classic notation of statics, the nodal forces are positive when they act in
the direction of the axis they refer. This is shown in Fig. 11.3.3. The actions
in the directions of the coordinates u1 , u2 , u4 , and u5 are forces while in the
directions of u3 and u6 are moments. In the following, the nodal elastic forces
are established using both methods, the method of the Lagrange equations and
that of the virtual work.
Fig. 11.3.3 Positive direction of the nodal forces af the plane frame element.
11.3.1.1 The method of the Lagrange equations
(i) Nodal elastic forces and stiffness matrix of the beam element
The elastic energy is due to the axial deformation of the element given by the
Eq. (11.2.12), that is,
Z
1 L
2
EAðx Þ½u 0 ðx, t Þ dx
(11.3.8)
Ua ¼
2 0
as well as to the bending deformation of the element given by Eq. (10.5.2),
that is,
Z
1 L
2
Ub ¼
EI ðx Þ½v 00 ðx, t Þ dx
(11.3.9)
2 0
Hence, the total elastic energy is
U ¼ Ua + U b
Z
Z
1 L
1 L
2
2
¼
EAðx Þ½u 0 ðx, t Þ dx +
EI ðx Þ½v 00 ðx, t Þ dx
2 0
2 0
(11.3.10)
396 PART
II Multi-degree-of-freedom systems
which by virtue of Eqs. (11.3.1), (11.3.3) becomes
Z
2
1 L
EAðx Þ u1 01 ðx Þ + u4 04 ðx Þ dx
U ðu1 , u2 , …, u6 Þ ¼
2 0
Z
1 L
+
EI ðx Þ u2 002 ðx Þ + u3 003 ðx Þ + u5
2 0
00
5 ðx Þ + u6
2
00
6 ðx Þ dx
(11.3.11)
The nodal elastic forces result from Eq. (11.2.10) for i ¼ 1, 2, …,6.
Thus, after performing the differentiation, we obtain
fS1 ¼
∂U
¼ k11 u1 + k14 u4
∂u1
(11.3.12a)
fS2 ¼
∂U
¼ k22 u2 + k23 u3 + k25 u5 + k26 u6
∂u2
(11.3.12b)
fS3 ¼
∂U
¼ k32 u2 + k33 u3 + k35 u5 + k36 u6
∂u3
(11.3.12c)
fS4 ¼
∂U
¼ k41 u1 + k44 u4
∂u4
(11.3.12d)
fS5 ¼
∂U
¼ k52 u2 + k53 u3 + k55 u5 + k56 u6
∂u5
(11.3.12e)
fS6 ¼
∂U
¼ k62 u2 + k63 u3 + k65 u5 + k66 u6
∂u6
(11.3.12f)
where
Z
L
kij ¼
Z
kij ¼
0
0
L
EAðx Þ 0i ðx Þ 0j ðx Þdx, i, j ¼ 1, 4
EI ðx Þ
00
00
i ðx Þ j ðx Þdx,
i, j ¼ 2, 3, 5, 6
Eqs. (11.3.12a)–(11.3.12f) are written in matrix form as
38 9
9 2k
8
0 k14 0 0 > u1 >
fS1 >
11 0
>
> 6
>
>
>
>
> u2 >
> 6 0 k22 k23 0 k25 k26 7
>
>
f >
>
>
7>
= 6
=
< S2 >
< >
7
0
k
k
0
k
k
fS3
u
32
33
35
36
3
7
¼6
6
7 u4 >
>
> 6 k41 0 0 k44 0 0 7>
> fS4 >
>
>
>
>
>u >
>
>
>
4
0 k52 k53 0 k55 k56 5>
>
>
;
;
: fS5 >
: 5>
u6
fS6
0 k62 k63 0 k65 k66
(11.3.13a)
(11.3.13b)
(11.3.14)
or
f eS ¼ ke ue
f eS ,
(11.3.15)
where
ue are the vectors of the nodal elastic forces and the nodal displacements, respectively, and ke is the stiffness matrix of the e beam element.
Therefore, the matrix ke for the plane beam is defined as
The finite element method Chapter
2
k11
60
6
60
e
k ¼6
6 k41
6
40
0
0
k22
k32
0
k52
k62
0
k23
k33
0
k53
k63
k14
0
0
k44
0
0
0
k25
k35
0
k55
k65
3
0
k26 7
7
k36 7
7
0 7
7
k56 5
k66
11
397
(11.3.16)
Obviously, we deduce from Eqs. (11.3.13a), (11.3.13b) that kij ¼ kji . Hence,
the stiffness matrix is symmetric.
For an element with a constant cross-section Aðx Þ ¼ Ae , I ðx Þ ¼ I e and
length Le , Eqs. (11.3.13a), (11.3.13b) are integrated analytically and
Eq. (11.3.16) becomes
3
2
EA
EA
0
0
0
0 7
6 L
L
7
6
12EI
6EI
12EI
6EI 7
6
7
6 0
0
6
L3
L2
L3
L2 7
6
6EI
4EI
6EI
2EI 7
7
6 0
0
7
6
2
2
e
L
L
L
L
7
6
(11.3.17)
k ¼6
7
EA
EA
7
6
0
0
0
0
7
6 L
L
7
6
12EI
6EI
12EI
6EI 7
6
7
6 0
0
3 2
6
L
L
L3
L2 7
4
6EI
2EI
6EI
4EI 5
0
0
L2
L
L2
L
Note that the superscript e in Ae , I e , and Le has been dropped from the elements of the matrix in Eq. (11.3.17) for the sake of simplicity of the expressions.
(ii) Nodal inertial forces and mass matrix of the beam element
As in the case of the truss, the equivalent inertial nodal forces for the beam element are obtained by two different assumptions of the mass distribution on
the element: the consistent mass assumption, which assumes a continuous distribution of the mass on the element, and the lumped mass assumption, which
lumps the mass at its nodes. The inertial mass matrices resulting from both
assumptions are derived below.
(a) Consistent mass matrix
During the motion, the infinitesimal mass m ðx Þdx undergoes the two displacements u ðx, t Þ and v ðx, t Þ. Therefore, the kinetic energy of the beam element will
be given by the expression.c
c. Due to bending, the cross-sections of the beam rotate by an angle v 0 ðx, t Þ. Under this rotation the
mass element m ðx Þdx exhibits a kinetic energy rðx ÞI ðx Þv_ 0 ðx, t Þdx=2, where rðx Þ is the mass
density of the material. Therefore, Eq. (11.3.18) should also include the term
Z
Z
1 L
1 L
rðx ÞI ðx Þv_ 0 ðx, t Þdx ¼
rðx ÞI ðx Þ u_ 2 02 ðx Þ + u_ 3 03 ðx Þ + u_ 5 05 ðx Þ + u_ 6 06 ðx Þ dx
2 0
2 0
However, it can be shown that the contribution of this term is small and it can be neglected.
398 PART
II Multi-degree-of-freedom systems
T¼
1
2
Z
L
n
o
m ðx Þ ½u_ ðx, t Þ2 + ½v_ ðx, t Þ2 dx
(11.3.18)
0
or using Eqs. (11.3.1), (11.3.3), the previous equation becomes
Z
h
i
1 L
m ðx Þ ½u_ 1 1 ðx Þ + u_ 4 4 ðx Þ2 dx
T ðu_ 1 , u_ 2 , …, u_ 6 Þ ¼
2 0
Z
1 L
m ðx Þ½u_ 2 2 ðx Þ + u_ 3 3 ðx Þ + u_ 5 5 ðx Þ + u_ 6
+
2 0
6 ðx Þ
2
dx
(11.3.19)
The nodal inertial forces result from Eq. (11.2.9) for i ¼ 1, 2, …, 6. Thus,
after performing the differentiations, we obtain
d ∂T
∂T
fI 1 ¼
¼ m11 u€1 + m14 u€4
(11.3.20a)
dt ∂u_ 1
∂u1
d ∂T
∂T
¼ m22 u€2 + m23 u€3 + m25 u€5 + m26 u€6
(11.3.20b)
fI 2 ¼
dt ∂u_ 2
∂u2
d ∂T
∂T
fI 3 ¼
¼ m32 u€2 + m33 u€3 + m35 u€5 + m36 u€6
(11.3.20c)
dt ∂u_ 3
∂u3
d ∂T
∂T
¼ m41 u€1 + m44 u€4
(11.3.20d)
fI 4 ¼
dt ∂u_ 4
∂u4
d ∂T
∂T
¼ m52 u€2 + m53 u€3 + m55 u€5 + m56 u€6
(11.3.20e)
fI 5 ¼
dt ∂u_ 5
∂u5
d ∂T
∂T
fI 6 ¼
¼ m62 u€2 + m63 u€3 + m65 u€5 + m66 u€6
(11.3.20f)
dt ∂u_ 6
∂u6
where
Z
mij ¼
L
m ðx Þ i ðx Þ j ðx Þdx, i, j ¼ 1, 4 and i, j ¼ 2, 3, 5, 6
(11.3.21)
0
Eqs. (11.3.20a)–(11.3.20f)
8 9 2m
fI 1 >
11 0
>
>
>
>
>
60
m22
>
>
f
I
2
>
= 6
< >
60
m
fI 3
32
¼6
6 m41 0
f
>
>
I
4
> 6
> >
>
>
m52
> 40
> fI 5 >
;
:
fI 6
0
m62
are written in matrix form as
38 9
0
m14 0
0
u€1 >
>
>
>
>
m23 0
m25 m26 7
>
>
u€2 >
>
7>
=
<
m33 0
m35 m36 7
€
u
3
7
0
m44 0
0 7
>
> u€4 >
7>
>
>
>
5
m53 0
m55 m56 >
>
;
: u€5 >
€
u
m63 0
m65 m66
6
(11.3.22)
or
€e
f eI ¼ me u
(11.3.23)
The finite element method Chapter
11
399
€e are the vectors of the nodal inertial forces and the nodal accelerwhere f eI , u
ations, respectively, and me the mass matrix of the e beam element. Therefore,
the mass matrix me is defined as
2
3
m11 0
0
m14 0
0
60
m25 m26 7
m22 m23 0
6
7
60
m32 m33 0
m35 m36 7
6
7
e
(11.3.24)
m ¼6
7
6 m41 0
0
m44 0
0 7
6
7
40
m55 m56 5
m52 m53 0
0
m62 m63 0
m65 m66
Obviously, we deduce from Eq. (11.3.21) that mij ¼ mji . Hence, the mass
matrix is symmetric.
Eq. (11.3.21) are integrated
For an element with constant mass, m ðx Þ ¼ m,
analytically and Eq. (11.3.24) becomes
2
3
140 0
0
70
0
0
60
156
22L 0
54 13L 7
6
7
2
27
e 60
0
13L
3L
22L
4L
m
6
7
(11.3.25)
me ¼
6
7
420 6 70 0
0
140 0
0 7
6
7
40
54
13L 0
156 22L 5
2
22L 4L2
0 13L 3L 0
e Le is the total mass of the e element. Note that the superscript
where m e ¼ m
e
e in L has been dropped from the elements of the matrix in Eq. (11.3.25) for the
sake of simplicity of the expressions.
The shape functions employed to derive the mass matrix are the same as
those employed to derive the stiffness matrix. The mass matrix resulting in this
way is referred to as the consistent mass matrix.
(b) Lumped mass matrix
According to this assumption, the mass of the element is concentrated at its
nodes by static considerations, that is, they are obtained as the reactions of a
simply supported beam under the load m ðx Þ (see Fig. 11.2.3). Thus, we have
Z L
m ðx Þð1 x Þdx
(11.3.26a)
m1 ¼
0
Z
m2 ¼
L
m ðx Þxdx
(11.3.26b)
0
Therefore, the kinetic energy of the beam element will be given by the
expression
1 1 T ¼ m1 u_ 1 2 + u_ 2 2 + m2 u_ 4 2 + u_ 5 2
2
2
(11.3.27)
400 PART
II Multi-degree-of-freedom systems
The nodal forces result from Eq. (11.2.9) for i ¼ 1, 2, 3, 4. Thus, after
performing the differentiations, we obtain
d ∂T
∂T
¼ m11 u€1
(11.3.28a)
fI 1 ¼
dt ∂u_ 1
∂u1
d ∂T
∂T
fI 2 ¼
¼ m22 u€2
(11.3.28b)
dt ∂u_ 2
∂u2
d ∂T
∂T
¼0
(11.3.28c)
fI 3 ¼
dt ∂u_ 3
∂u3
d ∂T
∂T
¼ m44 u€4
(11.3.28d)
fI 4 ¼
dt ∂u_ 4
∂u4
d ∂T
∂T
fI 5 ¼
¼ m55 u€5
(11.3.28e)
dt ∂u_ 5
∂u5
d ∂T
∂T
¼0
(11.3.28f)
fI 6 ¼
dt ∂u_ 6
∂u6
where
m11 ¼ m22 ¼ m1 , m44 ¼ m55 ¼ m2
Eqs. (11.3.28a)–(11.3.28f) are written in matrix form as
8 9 2
38 9
fI 1 >
>
m11 0
0 0
0
0 >
>
>
>
>
> u€1 >
>
> 6
> fI 2 >
>
> u€2 >
>
7>
0
0
0
0
0
m
>
>
>
22
>
>
>
7>
=
= 6
<
<f >
6
7
€
0
0
0
0
0
0
u
I3
3
7
¼6
60
07
0
0 m44 0
>
fI 4 >
>
>
> u€4 >
>
6
7>
>
>
>
>
>
>
4
5>
0
0
0
0
0
m
u€5 >
>
>
>
>
55
>
>
>
fI 5 >
;
>
:
>
: ;
0
0
0 0
0
0
u€6
fI 6
(11.3.29)
(11.3.30)
Eqs. (11.3.26a), (11.3.26b)
For an element with constant mass, m ðx Þ ¼ m,
are integrated analytically to give
e Le =2
m11 ¼ m22 ¼ m33 ¼ m44 ¼ m
and the mass matrix (11.3.30) becomes
2
1 0
60 1
6
e 60 0
m
6
me ¼
6
2 60 0
6
40 0
0 0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
3
0
07
7
07
7
7
07
7
05
0
(11.3.31)
(11.3.32)
e Le is the total mass of the e element.
where m e ¼ m
We observe that the lumped mass assumption results in a diagonal mass
matrix.
The finite element method Chapter
11
401
(iii) Nodal damping forces and damping matrix of the beam element
As in the case of the truss element, we assume only internal damping. It is therefore due to the deformation of the beam element, it resists it, and depends on the
velocity of the strain. If we denote by sD the axial stress due to damping, we
may write
sD ¼ cs
∂ex
∂t
(11.3.33)
where cs is the coefficient of the internal damping and ex the axial strain.
The virtual work of the internal damping force in the beam element is
Z
in
¼ sD dex dV
(11.3.34)
dWnc
V
Taking into account that
ex ¼
sx
N ðx Þ M ðx Þ
, sx ¼
+
y, N ðx Þ ¼ EAðx Þu 0 , M ðx Þ ¼ EI ðx Þv 00 ðx, t Þ
E
Aðx Þ
I ðx Þ
the previous relation gives
dWnc ¼ Z
L
Z0
L
cs Aðx Þu_ 0 ðx, t Þdu 0 ðx, t Þdx
(11.3.35)
cs I ðx Þv_ 00 ðx, t Þdv 00 ðx, t Þdx
0
or using Eqs. (11.3.1), (11.3.3)
Z L
in
dWnc ¼ cs Aðx Þ u_ 1 01 ðx Þ + u_ 4
Z
0
0
4 ðx Þ
d u1
0
0
1 ðx Þ + u 4 4 ðx Þ
dx
cs I ðx Þ u_ 2 002 ðx Þ + u_ 3 003 ðx Þ + u_ 5 005 ðx Þ + u_ 6 006 ðx Þ
0
d u2 02 ðx Þ + u3 03 ðx Þ + u5 05 ðx Þ + u6 06 ðx Þ dx
¼ ðc11 u_ 1 + c14 u_ 4 Þdu1 ðc41 u_ 1 + c44 u_ 4 Þdu4
L
ðc22 u_ 2 + c23 u_ 3 + c25 u_ 5 + c26 u_ 6 Þdu2
ðc32 u_ 2 + c33 u_ 3 + c35 u_ 5 + c36 u_ 6 Þdu3
ðc52 u_ 2 + c53 u_ 3 + c55 u_ 5 + c56 u_ 6 Þdu5
ðc62 u_ 2 + c63 u_ 3 + c65 u_ 5 + c66 u_ 6 Þdu6
(11.3.36)
where
Z
cij ¼
0
L
cs Aðx Þ 0i ðx Þ 0j ðx Þdx, i, j ¼ 1, 4
(11.3.37a)
402 PART
II Multi-degree-of-freedom systems
Z
cij ¼
0
L
c s I ðx Þ
00
00
i ðx Þ j ðx Þdx,
i, j ¼ 2, 3,5, 6
(11.3.37b)
Writing Eq. (1.8.11) for the beam element, we obtain the nodal damping
forces as generalized forces for i ¼ 1, 2, …,6. Thus we have
fD1 ¼ Q1 ¼ c11 u_ 1 + c14 u_ 4
(11.3.38a)
fD2 ¼ Q2 ¼ c22 u_ 2 + c23 u_ 3 + c25 u_ 5 + c26 u_ 6
(11.3.38b)
fD3 ¼ Q3 ¼ c32 u_ 2 + c33 u_ 3 + c35 u_ 5 + c36 u_ 6
(11.3.38c)
fD4 ¼ Q4 ¼ c41 u_ 1 + c44 u_ 4
(11.3.38d)
fD5 ¼ Q5 ¼ c52 u_ 2 + c53 u_ 3 + c55 u_ 5 + c56 u_ 6
(11.3.38e)
fD6 ¼ Q6 ¼ c62 u_ 2 + c63 u_ 3 + c65 u_ 5 + c66 u_
(11.3.38f)
Eqs. (11.3.38a)–(11.3.38f) are written in matrix form as
8 9 2
38 9
c11 0 0 c14 0 0 >
fD1 >
>
>
>
>
> u_ 1 >
>
>
7>
> 6
>
> >
>
>
>
>
>
6
7
>
>
>
_
f
c
0
c
c
0
c
u
D2 >
22
23
25
26 > 2 >
>
>
> 6
>
> >
7>
>
>
>
>
>
7>
= 6
< fD3 >
<
6 0 c32 c33 0 c35 c36 7 u_ 3 =
6
7
¼6
7 u_ >
>
fD4 >
>
>
> 4>
6 c41 0 0 c44 0 0 7>
>
>
>
>
>
>
>
6
7>
>
>
>
>
>
>
>
>
6
7
>
>
>
>
_
0
c
f
c
0
c
c
u
>
>
>
>
D5
52
53
55
56
5
4
5
>
>
>
>
>
>
>
;
: ;
: >
fD5
0 c62 c63 0 c65 c66
u_ 6
(11.3.39)
or
f eD ¼ ce u_ e
(11.3.40)
where f eD , u_ e are the vectors of the nodal damping forces and nodal velocities,
respectively, and ce is the damping matrix of the beam element e. Hence, the
damping matrix ce is defined as
2
3
c11 0 0 c14 0 0
6
7
6 0 c22 c23 0 c25 c26 7
6
7
6
7
0
c
c
0
c
c
6
7
32
33
35
36
e
6
7
(11.3.41)
c ¼6
7
6 c41 0 0 c44 0 0 7
6
7
60 c c 0 c c 7
52
53
55
56 5
4
0
c62 c63 0
c65 c66
For an element with constant cross-section Aðx Þ ¼ Ae , I ðx Þ ¼ I e and length
L , Eqs. (11.3.37a), (11.3.37b) are integrated analytically and Eq. (11.3.41)
becomes.
e
The finite element method Chapter
2
cs A
cs A
0
0
6 L
L
6
6
12c
I
6c
I
s
s
6 0
0
6
L3
L2
6
6
6cs I
4cs I
6
0
6 0
2
6
L
L
ce ¼ 6
6 cs A
cs A
6
0
0
6 L
L
6
6
12c
I
6c
I
s
s
6 0
0
3 2
6
L
L
6
4
6cs I
2cs I
0
0
2
L
L
11
403
3
0
0
7
7
12cs I
6cs I 7
7
3
L
L2 7
7
6cs I
2cs I 7
7
2
7
L
L 7
7
7
0
0 7
7
7
12cs I
6cs I 7
2 7
L3
L 7
7
6cs I
4cs I 5
2
L
L
(11.3.42)
Note that, the superscript e in Ae , I e , and Le has been dropped from the
elements of the matrix in Eq. (11.3.42) for the sake of simplicity of the
expressions.
(iv) Equivalent nodal loads of the beam element
The equivalent nodal forces can be evaluated in two ways: (a) as generalized
forces in the direction of the nodal displacements ue and (b) as static equivalent
forces. In the first case, the shape functions utilized to derive elastic forces are
employed too. The so-resulting vector of the equivalent nodal forces is referred
to as the consistent load vector.
Fig. 11.3.4 Loading and consistent equivalent nodal loads of a plane beam element.
(a) Consistent nodal load vector
Referring to Fig. 11.3.4, the virtual work produced by the displacements of the
beam element dx is
Z L
Z L
p
px ðx, t Þdu ðx, t Þdx +
py ðx, t Þdv ðx, t Þdx
(11.3.43)
dWnc ¼
0
0
which by virtue of Eqs. (11.3.1), (11.3.3) becomes
p
dWnc
¼ p1 ðt Þdu1 + p2 ðt Þdu2 + p3 ðt Þdu3 + p4 ðt Þdu4 + p5 ðt Þdu5 + p6 ðt Þdu6
(11.3.44)
404 PART
II Multi-degree-of-freedom systems
where it was set
Z
L
pi ðt Þ ¼
Z
pi ðt Þ ¼
px ðx, t Þ i ðx Þdx, i ¼ 1, 4
(11.3.45a)
0
L
py ðx, t Þ i ðx Þdx, i ¼ 2, 3, 5, 6
(11.3.45b)
0
The nodal load vector results from Eq. (11.3.44)
pe ðt Þ ¼ f p1 ðt Þ p2 ðt Þ p3 ðt Þ p4 ðt Þ p5 ðt Þ p6 ðt Þ gT
(11.3.46)
(b) Statically equivalent nodal load vector
In this case, the bending deformation of the beam element is ignored and its
transverse deformation is assumed linear, hence it is given by the relation
v ðx, t Þ ¼ u2
in which
2 ðx Þ ¼
1 ðx Þ ¼ 1 x,
2 ðx Þ + u 5 5 ðx Þ
5 ðx Þ ¼
4 ðx Þ ¼ x
and Eq. (11.3.44) becomes
p
¼ p1 ðt Þdu1 + p2 ðt Þdu2 + p4 ðt Þdu4 + p5 ðt Þdu5
dWnc
(11.3.47)
where it was set
Z
L
pi ðt Þ ¼
px ðx, t Þ i ðx Þdx, i ¼ 1, 4
(11.3.48a)
py ðx, t Þ i ðx Þdx, i ¼ 2, 5
(11.3.48b)
0
Z
pi ðt Þ ¼
L
0
Consequently
pe ðt Þ ¼ f p1 ðt Þ p2 ðt Þ 0 p4 ðt Þ p5 ðt Þ 0 gT
(11.3.49)
Inserting the expressions of the linear shape function in Eqs. (11.3.48a),
(11.3.48b) yields
Z L
Z L
px ðx, t Þð1 xÞdx, p4 ðt Þ ¼
px ðx, t Þxdx
(11.3.50)
p1 ðt Þ ¼
0
Z
p 2 ðt Þ ¼
0
0
L
Z
py ðx, t Þð1 xÞdx, p5 ðt Þ ¼
L
py ðx, t Þxdx
(11.3.51)
0
namely, the equivalent nodal loads are obtained as the reactions of a simply supported beam under the loads px ðx, t Þ and py ðx, t Þ. Hence, their name statically
equivalent nodal loads.
The finite element method Chapter
11
405
The forces resulting in this way are less accurate because the transverse
displacement is approximated by a linear polynomial instead of a cubic one.
The linear shape functions are not geometrically admissible, which means that
the conditions of the Ritz method are violated. Nevertheless, the accuracy of the
results is acceptable, especially when the elements are small. The statically
equivalent nodal load vector is particularly useful when inertial forces due
to the rotations of the cross-sections are ignored, as in the lumped mass
assumption.
(v) Geometric stiffness matrix of the beam element
When the produced rotation of the cross-sections is significant, there may be
shortening of the length of the element even if the axial deformation is ignored
(∂u=∂x ¼ 0). This shortening is due to the nonlinear term of the straindisplacement relation [4,5], that is,
1 ∂v ðx, t Þ 2
(11.3.52)
ex ¼
2
∂x
In the presence of an axial force N ðx Þ, the elastic energy due to its
shortening is
Z
1 L
N ðx Þex dx
UG ¼
2 0
(11.3.53)
Z
1 L
∂v ðx, t Þ 2
N ðx Þ
dx
2 0
∂x
which by virtue of Eq. (11.3.3) becomes
Z
1 L
UG ¼
N ðx Þ u2 02 ðx Þ + u3 03 ðx Þ + u5
2 0
2
0
0
5 ðx Þ + u6 6 ðx Þ dx
Differentiation of the previous expression for UG with respect to ui ,
i ¼ 1, 2, …,6, results in the additional elastic nodal forces
fG1 ¼
∂UG
¼0
∂u1
(11.3.54a)
fG2 ¼
∂UG
¼ kG22 u2 + kG23 u3 + kG25 u5 + kG26 u6
∂u2
(11.3.54b)
fG3 ¼
∂UG
¼ kG32 u2 + kG33 u3 + kG35 u5 + kG36 u6
∂u3
(11.3.54c)
fG4 ¼
∂UG
¼0
∂u4
(11.3.54d)
406 PART
II Multi-degree-of-freedom systems
fG5 ¼
∂UG
¼ kG52 u2 + kG53 u3 + kG55 u5 + kG56 u6
∂u5
(11.3.54e)
fG6 ¼
∂UG
¼ kG62 u2 + kG63 u3 + kG65 u5 + kG66 u6
∂u6
(11.3.54f)
where
Z
kGij ¼
0
L
N ðx Þ 0i ðx Þ 0j ðx Þdx, i, j ¼ 2, 3, 5, 6
Eqs. (11.3.54a)–(11.3.54f)
8 9 2
0 0
fG1 >
>
>
>
> >
>
>
6 0 kG22
f
>
>
G2
>
= 6
< >
6 0 kG32
fG3
¼6
6
>
> 60 0
> fG4 >
>
>
>
>
4 0 kG52
>
>
> fG5 >
;
:
fG6
0 kG62
are written in matrix form as
38 9
0
0 0
0
u1 >
>
>
>
> >
>
>
kG23 0 kG25 kG26 7
u
>
2>
>
7>
=
<
7
kG33 0 kG35 kG36 7 u3
0
0 0
0 7
>
> u4 >
7>
>
>
>
kG53 0 kG55 kG56 5>
>
>
> u5 >
;
:
kG63 0 kG65 kG66
u6
(11.3.55)
(11.3.56)
or
f eG ¼ keG ue
(11.3.57)
ue are the vectors of the elastic nodal forces and the nodal displacewhere
ments, respectively, and keG is the geometric stiffness matrix of the e beam
element. Therefore, the matrix keG for the plane frame is defined as
2
3
0 0
0
0 0
0
6 0 kG22 kG23 0 kG25 kG26 7
6
7
6 0 kG32 kG33 0 kG35 kG36 7
7
(11.3.58)
keG ¼ 6
60 0
0
0 0
0 7
6
7
4 0 kG52 kG53 0 kG55 kG56 5
0 kG62 kG63 0 kG65 kG66
f eG ,
When the axial force is constant, N ðx Þ ¼ N , Eq. (11.3.55) are integrated
analytically and Eq. (11.3.58) becomes
3
2
0 0
0 0 0
0
7
6
12
12
60
1 0 1 7
7
6
L
L
7
6
7
6
4L
L
60 1
0 1
7
6
N
3
3 7
7
(11.3.59)
keG ¼ 6
6
10 6 0 0
0 0 0
0 7
7
7
6
6 0 12 1 0 12 1 7
7
6
L
L
7
6
5
4
L
4L
0 1
0 1
3
3
The axial force may be a priori known, as it happens when the structure is
under stress before the dynamic loads are applied. The total stiffness of the
The finite element method Chapter
11
407
element is the sum ke + keG . Thus, we conclude from Eq. (11.3.59) that the stiffness increases if the axial force is tensile while it decreases if it is compressive.
The vanishing of the stiffness leads to the buckling of the structure.
The axial force is given by the relationd
N ðx Þ ¼ EA
∂u
∂x
EA
ðu1 + u4 Þ
L
¼
(11.3.60)
namely, it depends on the displacements. Therefore, if the geometric stiffness is
taken into account in the analysis, the problem becomes nonlinear.
11.3.1.2 The method of virtual work
The application of this method is illustrated with the derivation of the stiffness
matrix of the beam element, under both axial and bending deformation.
The virtual work due to the axial deformation is written as
Z
Z L
AEeax deax dx
(11.3.61)
dWina ¼ sax deax dx ¼
0
V
Using Eq. (11.3.1) and taking into account that
8 9
u1 >
>
>
>
>
>
>
u2 >
>
>
>
>
=
<
∂u
u3
a
0
0
0
0
¼ N0a u
ex ¼ ¼ u 1 1 + u 4 4 ¼ ½ 1 0 0 4 0 0 u4 >
>
∂x
>
>
>
>
>
u >
>
>
>
;
: 5>
u6
deax ¼ N0a du
(11.3.62)
(11.3.63)
where
Na ¼ ½
1
0 0
4
0 0
u ¼ f u1 u2 u3 u4 u5 u6 gT
(11.3.64a)
(11.3.64b)
Substituting now Eqs. (11.3.62), (11.3.63) into Eq. (11.3.61) yields
Z L
T
dWina ¼ duT
EAðx ÞN0 a N0a dx u
(11.3.65)
0
The virtual work due to the bending deformation is
Z
b
dWin ¼ sbx debx dx
V
"
#
∂u 1 ∂v 2
[4].
d. The exact expression for the axial force is N ðx Þ ¼ EA
+
∂x 2 ∂x
(11.3.66)
408 PART
II Multi-degree-of-freedom systems
Using Eq. (11.3.3), we obtain
sbx ¼
M
y ¼ Eyv 00 ¼ Ey ½ 0
I
00
2
00
3
00
5
0
8 9
u1 >
>
>
>
>
>
>
u >
>
=
< 2>
u3
00
¼ EyN00b u
6 u
>
>
4
>
> >
>
>
>
>
;
: u5 >
u6
debx ¼ yN00b du
where
N00b ¼ ½ 0
00
2
00
3
00
5
0
00
6
(11.3.67)
Substituting the foregoing expressions for sbx and debx into Eq. (11.3.66)
and integrating over the cross-section, we obtain
Z L
T
EI ðx ÞN00 b N00b dx u
(11.3.68)
dWinb ¼ duT
0
The total virtual work of the internal forces is
Z L
Z
T
EAðx ÞN0 a N0a dx +
dWin ¼ dWina + dWinb ¼ duT
0
L
0
T
EI ðx ÞN00 b N00b dx u
(11.3.69)
which is set equal to the virtual work of the equivalent nodal forces, that is,
dWex ¼ duT f S
to yield
9 2k
8
fS1 >
11
>
> 6
>
>
> 60
> fS2 >
>
>
= 6
<
0
fS3
¼6
6 k41
fS4 >
>
>
>
6
>
>
> 40
>
>
;
: fS5 >
fS6
0
0
k22
k32
0
k52
k62
0
k23
k33
0
k53
k63
k14
0
0
k44
0
0
0
k25
k35
0
k55
k65
38 9
0 > u1 >
>
>
> u2 >
k26 7
>
>
7>
=
< >
7
k36 7 u3
u4 >
0 7
>
>
7>
>
>
>
>
5
k56 >
;
: u5 >
u
k66
6
(11.3.69)
(11.3.70)
where
Z
L
kij ¼
Z
kij ¼
0
0
L
EAðx Þ 0i ðx Þ 0j ðx Þdx, i, j ¼ 1, 4
EI ðx Þ
00
00
i ðx Þ j ðx Þdx,
i, j ¼ 2, 3, 5, 6
(11.3.71a)
(11.3.71b)
Obviously, Eq. (11.3.70) is identical to Eq. (11.3.14). A similar procedure
will give the remaining element matrices.
The finite element method Chapter
11
409
Example 11.3.1 Compute the stiffness and mass matrices and the nodal load
vector
of a beam element, which is subjected to the loads px ðx, t Þ ¼ pðt Þ
1 + x x 2 and py ðx, t Þ ¼ 3pðt Þð1 x Þ, x ¼ x=L. The height of the crosssection varies linearly, h ðx Þ ¼ h ð1 + 0:5x Þ, while its width b is constant.
The material density is r. Consider the lumped mass assumption.
Solution
i. Stiffness matrix: We set A ¼ bh, I ¼ bh 3 =12. Hence Aðx Þ=A ¼ ð1 + 0:5x Þ,
I ðx Þ=I ¼ ð1 + 0:5x Þ3 . Then we obtain from Eq. (11.3.13)
Z
L
k11 ¼
EAðx Þ
0
0
1 ðx Þ 1 ðx Þdx
c
¼ 1:25k11
EAðx Þ
0
0
1 ðx Þ 4 ðx Þdx
c
¼ 1:25k14
0
Z
L
k14 ¼
0
Z
L
k22 ¼
0
EI ðx Þ
Z
L
k23 ¼
EI ðx Þ
0
Z
L
k25 ¼
EI ðx Þ
0
Z
L
k26 ¼
Z
L
k33 ¼
00
00
3 ðx Þ 3 ðx Þdx
c
¼ 1:4875k33
L
EI ðx Þ
0
L
EI ðx Þ
0
L
k55 ¼
EI ðx Þ
0
Z
L
k56 ¼
EI ðx Þ
0
k66 ¼
0
L
c
¼ 2:09375k25
EI ðx Þ
k35 ¼
Z
00
00
2 ðx Þ 5 ðx Þdx
c
¼ 1:7k23
c
¼ 2:4875k26
Z
k36 ¼
00
00
2 ðx Þ 3 ðx Þdx
00
00
2 ðx Þ 6 ðx Þdx
0
Z
c
¼ 2:09375k22
EI ðx Þ
0
Z
00
00
2 ðx Þ 2 ðx Þdx
EI ðx Þ
00
00
3 ðx Þ 5 ðx Þdx
00
00
3 ðx Þ 6 ðx Þdx
00
00
5 ðx Þ 5 ðx Þdx
c
¼ 2:125k36
c
¼ 2:09375k55
00
00
5 ðx Þ 6 ðx Þdx
00
00
6 ðx Þ 6 ðx Þdx
c
¼ 1:7k35
c
¼ 2:4875k56
c
¼ 2:66875k66
The superscript c refers to the element of constant cross-section A ¼ bh.
410 PART
II Multi-degree-of-freedom systems
m ðx Þ ¼ rAðx Þ ¼ rAð1 + 0:5x Þ
ii. Mass
matrix:
Substituting
Eq. (11.3.26), we obtain
Z
L
m11 ¼ m22 ¼
m ðx Þð1 xÞdx ¼
0
Z
m44 ¼ m55 ¼
L
m ðx Þxdx ¼
0
into
7
rAL
12
8
rAL
12
iii. Nodal load vector: Using Eqs. (11.3.50), (11.3.51), we obtain
Z
L
p1 ðt Þ ¼
0
px ðx, t Þð1 x Þdx ¼
Z
L
p4 ðt Þ ¼
px ðx, t Þxdx ¼
0
Z
L
p2 ðt Þ ¼
7
pðt ÞL
12
7
pðt ÞL
12
py ðx, t Þð1 x Þdx ¼ pðt ÞL
0
Z
p5 ðt Þ ¼
0
L
1
py ðx, t Þxdx ¼ pðt ÞL
2
11.3.2 Transformation of the nodal coordinates
of the plane frame element
The translational components u1 and u2 are transformed from the local to the
global system of axes according to Eq. (11.2.67), while the component u3 ,
which expresses a rotation about the z axis, remains unaltered in the rotated system because the local axis z and the global axis z are identical, Fig. 11.3.5.
Hence, we can write
cos f sin f
u1
u1
¼
(11.3.72a)
u2
sin f cosf u2
u3 ¼ u3
which may be combined as
3( )
( ) 2
cos f sin f 0
u1
u1
u2 ¼ 4 sin f cos f 0 5 u2
u3
u3
0
0
1
(11.3.72b)
(11.3.73a)
The finite element method Chapter
11
411
Fig. 11.3.5 Nodal displacements of the beam element in global and local axes.
Similarly, we obtain
3( )
( ) 2
cosf sin f 0
u4
u4
u5 ¼ 4 sin f cos f 0 5 u5
u6
u6
0
0
1
(11.3.73b)
Eqs. (11.3.73a), (11.3.73b) are further combined as
8 9 2
38 9
cos f sin f 0
0
0
0 >
u1 >
>
>
>
>
> u1 >
> 6
> >
>
>
> u2 >
7>
sin
f
cos
f
0
0
0
0
u
>
>
>
2
>
>
>
7>
= 6
=
< >
<
6
7
0
0
1
0
0
0
u3
u
3
7
¼6
6
0
0
cos f sin f 0 7
>
> 6 0
>
> u4 >
> u4 >
7>
>
>
>
>
>
>
4
5>
0
0
0
sin
f
cos
f
0
u >
u
>
>
>
>
5
>
> ;
>
;
:
: 5>
0
0
0
0
0
1
u6
u6
(11.3.74)
or
ue ¼ Re ue
(11.3.75a)
ue ¼ ðRe ÞT ue
(11.3.75b)
hence
where
2
cos fe
6 sin fe
6
6 0
e
R ¼6
6 0
6
4 0
0
sin fe
cos fe
0
0
0
0
0
0
0
0
1
0
0
cos fe
0 sin fe
0
0
0
0
0
sin fe
cos fe
0
3
0
07
7
07
7
07
7
05
1
(11.3.76)
represents the transformation matrix of the e plane frame element.
Eq. (11.3.75b) holds because the matrix Re is orthonormal.
e
e
e
The global vectors of the nodal forces f S , f I , f D and pe ðt Þ are defined in
relation to ue . Their transformation obeys the same law, Eq. (11.3.75b). Thus,
we have
412 PART
II Multi-degree-of-freedom systems
f e ¼ðRe ÞT f e
S
S
(11.3.77)
f e ¼ðRe ÞT f e
I
I
(11.3.78)
f e
D
(11.3.79)
¼ðRe ÞT f eD
pe ðt Þ ¼ðR Þ p ðt Þ
e T e
(11.3.80)
Using Eqs. (11.3.15), (11.3.23), (11.3.40), (11.3.75a), the first three of the
foregoing equations are transformed in the global axes as
f e ¼ ke ue
S
(11.3.81)
f e
I
u
¼m
(11.3.82)
f D ¼ c u
(11.3.83)
e
k ¼ ðRe ÞT ke Re
(11.3.84)
e ¼ ð Re Þ T m e Re
m
(11.3.85)
ce ¼ ðRe ÞT ce Re
(11.3.86)
e €e
e _e
where
e
e , and ce representing the stiffness, mass, and damping matrices of the
with k , m
e beam element in global axes, respectively. Note that the lumped mass matrix
remains unaltered under this transformation.
Applying the procedure presented in Section 11.2.3 for the plane truss,
we obtain the equation of motion of the plane frame
u_ + K
u€ + C
u ¼
M
pð t Þ
(11.3.87)
which gives after applying the support conditions and partitioning
e ff e
e ff e
e ff e
u€f + C
u_ f + K
uf ¼ e
pf∗ ðt Þ
M
e sf e
e ss e
e sf e
e ss e
e sf e
e ss e
M
u€f + M
u€s + C
u_ f + C
u_ s + K
uf + K
us ¼ e
p s ðt Þ
(11.3.88)
(11.3.89)
where
e fs e
e fs e
e fs e
e
u€s C
u_ s K
us
pf ðt Þ M
pf∗ ðt Þ ¼ e
(11.3.90)
Apparently, Eqs. (11.3.88), (11.3.89), (11.3.90) are the counterpart of
Eqs. (11.2.117), (11.2.118), (11.2.119), which hold for the plane truss.
For the sake of convenience, we write Eq. (11.3.88) as
M€
u + C€
u + Ku ¼ pðt Þ
(11.3.91)
The finite element method Chapter
413
11
Example 11.3.2 Formulate the equation of motion of the frame shown in
Fig. E11.2.
Fig. E11.2 Frame in Example 11.3.2
Assumed data:
Coordinates of frame nodes: 1ð0, 0Þ, 2ð0, 5Þ, 3ð4, 8Þ, 4ð10, 0Þ.
Properties of the elements: A1 ¼ A, A2 ¼ A3 ¼ 2A, I1 ¼ I , I2 ¼ I3 ¼ 8I ,
I =A ¼ ðL1 =50Þ2 . Load pðx, t Þ ¼ P ðt Þ=L2 . Damping coefficient cs ¼ 0,
modulus of elasticity E, material density r. Lumped mass assumption.
Solution
The system has n ¼ 4 nodes, hence the free structure has N ¼ 3n ¼ 12 degrees
of freedom, (u3i2 , u3i1 , u3i , i ¼ 1, 2, 3, 4). The numbering of the elements and
the positive direction of the local axes are chosen as in Fig. E11.2.
TABLE E11.2 Geometrical data of the elements in Example 11.3.2.
e
xj
xk
D
x
yj
yk
1
0
0
0
0
5
2
0
4
4
5
3
4
10
6
8
D
y
Le
cos fe
sinfe
5
5
1
0
8
3
5
0.8
0.6
0
8
10
0.6
0.8
1. Computation of ke , me , ce , pe ðt Þ, Re for e ¼ 1, 2, 3
Matrices ke . The elements have constant cross-section. Therefore, Eq. (11.3.17)
is employed, which gives
414 PART
II Multi-degree-of-freedom systems
2
3
2500
0
0 2500
0
0
6
0 12
30
0 12
30 7
6
7
6
0 30
100
0 30
50 7
EI 6
7
1
k ¼
6
7
125 6 2500
0
0 2500
0
07
6
7
4
0 12 30
0 12 30 5
0 30
50
0 30
100
3
2
5000
0
0 5000
0
0
6
0
96 240
0 96 240 7
7
6
6
0
240 800
0 240 400 7
EI 6
7
2
k ¼
7
6
125 6 5000
0
0 5000
0
07
7
6
4
0
96 240
0
96 240 5
0
240 400
0 240 800
2
3
2500
0
0 2500
0
0
6
0 12
60
0 12
60 7
6
7
6
7
EI
0
60
400
0
60
200
3
6
7
k ¼
7
2500
0
0
2500
0
0
125 6
6
7
4
0 12
60
0 12
60 5
0 60
200
0 60
400
Matrices me . The elements have constant cross-section. Therefore,
Eq. (11.3.32) is employed for lumped mass assumption, which gives
2
3
2
3
2:5 0 0 0 0 0
5:0 0 0 0 0 0
6 0 2:5 0 0 0 0 7
6 0 5:0 0 0 0 0 7
6
7
6
7
6
7
60 0 0 0 0 07
0 0 0 0 0 07
2
6
7
m1 ¼ rA6
¼
rA
,
m
6 0 0 0 2:5 0 0 7
6 0 0 0 5:0 0 0 7
6
7
6
7
4 0 0 0 0 2:5 0 5
4 0 0 0 0 5:0 0 5
0 0 0 0 0 0
0 0 0 0 0 0
2
3
10:0 0 0 0
0 0
6 0 10:0 0 0
0 07
6
7
6
0
0 0 0
0 07
3
6
7
m ¼ rA6
0 0 10:0 0 0 7
6 0
7
4 0
0 0 0 10:0 0 5
0
0 0 0
0 0
Matrices ce . It is cs ¼ 0 and Eq. (11.3.42) gives ce ¼ 0.
Vectors pe ðt Þ. To be consistent to the lumped mass assumption, the statically
equivalent nodal load vector will be employed. Therefore, Eqs. (11.3.50), (11.3.51)
give for the load-free elements 1 and 3, p1 ðt Þ ¼ p2 ðt Þ ¼ p3 ðt Þ ¼ p4 ðt Þ ¼ 0 while
for element 2, we obtain
p1 ðx, t Þ ¼ PLð2t Þ cos fð2Þ sin fð2Þ ,
p2 ðx, t Þ ¼ PLð2t Þ cos 2 fð2Þ
The finite element method Chapter
11
415
where cos fð2Þ ¼ 0:8, sin fð2Þ ¼ 0:6. Thus, we have
9
8
8 9
0:048P ðt Þ >
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
0:064P
ð
t
Þ
0>
>
>
>
>
>
>
>
=
=
<
< >
0
0
2
1
3
, p ðt Þ ¼ p ðt Þ ¼
p ðt Þ ¼
0:048P ðt Þ >
0>
>
>
>
>
>
>
>
>
>
>
>
>
0:064P ðt Þ >
0>
>
>
>
>
>
>
>
;
;
:
: >
0
0
Matrices Re . The direction cosines cos fe and sin fe are computed from the
Cartesian coordinates of the element nodes and are given in Table E11.2. Thus
employing Eq. (11.3.76), we obtain
2
3
2
3
0 1 0 0 0 0
0:8 0:6 0 0 0 0
6 1 0 0 0 0 0 7
6 0:6 0:8 0 0 0 0 7
6
7
6
7
6
7
6 0 0 1 0 0 07
0 0 1 0 0 07
2
6
7
¼
R1 ¼ 6
,
R
6 0 0 0 0 1 07
6 0 0 0 0:8 0:6 0 7
6
7
6
7
4 0 0 0 1 0 0 5
4 0 0 0 0:6 0:8 0 5
0 0 0 0 0 1
0 0 0 0 0 1
2
3
0:6 0:8 0 0
0 0
6 0:8 0:6 0 0
0 07
6
7
6
0
0
1
0
0 07
3
6
7
R ¼6
0 0 0:6 0:8 0 7
60
7
40
0 0 0:8 0:6 0 5
0
0 0 0
0 1
e
e , pe ðt Þ for e ¼ 1, 2, 3
2. Computation of k , m
e
Matrices k . They are computed using Eq. (11.3.84)
2
3
12
0 30 12
0
30
6 0
2500
0
0 2500
07
6
7
6
EI
30
0 100
30
0
50 7
1
6
7
k ¼
0
30
12
0
30 7
125 6
6 12
7
4 0 2500
0
0
2500
05
30
0
50
30
0
100
2
3
3234:6 2353:9 144:0 3234:6 2353:9 144:0
6 2353:9 1861:4
192:0 2353:9 1861:4 192:0 7
6
7
6
EI
144:0
192:0
800:0
144:0 192:0 400:0 7
2
6
7
k ¼
144:00 3234:6 2353:9 144:0 7
125 6
6 3234:6 2353:9
7
4 2353:9 1861:4 192:0
2353:9 1861:4 192:0 5
144:0
192:0
400
144:0 192:0
800:0
416 PART
II Multi-degree-of-freedom systems
2
3
907:7 1194:2 48:0 907:6
1194:2 48:0
6 1194:2 1604:3 36:0
1194:2 1604:3
36:0 7
6
7
6
EI
48:0
36:0 400:0
48:0
36:0 200:0 7
6
7
k3 ¼
907:7 1194:2 48:0 7
125 6
6 907:6 1194:2 48:0
7
4 1194:2 1604:3 36:0 1194:2
1604:3 36:0 5
48:0
36:0 200:0
48:0
36:0 400:0
e . They are computed using Eq. (11.3.85)
Matrices m
2
3
2
2:5 0 0 0
0 0
5:0
60
7
60
2:5
0
0
0
0
6
7
6
60
60
0 0 0
0 07
2
7, m
6
1 ¼ rA6
¼
rA
m
60
7
60
0 0 2:5 0 0 7
6
6
40
5
40
0 0 0
2:5 0
0
0 0 0
0 0
0
2
10:0
6 0
6
6 0
3 ¼ rA6
m
6 0
6
4 0
0
0
10:0
0
0
0
0
0
0
0
0
0
0
0
0
0
10:0
0
0
0
0
0
0
10:0
0
0
5:0
0
0
0
0
0
0
0
0
0
0
0
0
0
5:0
0
0
0
0
0
0
5:0
0
3
0
07
7
07
7
07
7
05
0
3
0
07
7
07
7
07
7
05
0
Vectors pe ðt Þ. They are computed using Eq. (11.3.80).
9
8
8 9
0
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0:08P
ð
t
Þ
0>
>
>
>
>
>
>
>
=
=
<
< >
0
0
2
1
3
, p ðt Þ ¼ p ðt Þ ¼
p ðt Þ ¼
0
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0:08P
ð
t
Þ
0>
>
>
>
>
>
>
>
;
;
:
: >
0
0
^ e, M
^ e, p
^e ðt Þ for e ¼ 1, 2, 3
3. Computation of the enlarged matrices K
Assembly matrices ae . By inspection, we obtain
2
1 0 0 0 0 0 0 0 0 0 0 0
60
6
6
60
6
1
a ¼6
60
6
6
40
3
2
0 0 0 1 0 0 0 0 0 0 0 0
60
1 0 0 0 0 0 0 0 0 0 07
6
7
6
7
60
0 1 0 0 0 0 0 0 0 0 07
6
7
2
7, a ¼ 6
60
0 0 1 0 0 0 0 0 0 0 07
6
7
6
7
40
0 0 0 1 0 0 0 0 0 0 05
0 0 0 0 0 1 0 0 0 0 0 0
3
0 0 0 1 0 0 0 0 0 0 07
7
7
0 0 0 0 1 0 0 0 0 0 07
7
7
0 0 0 0 0 1 0 0 0 0 07
7
7
0 0 0 0 0 0 1 0 0 0 05
0 0 0 0 0 0 0 0 1 0 0 0
The finite element method Chapter
2
0 0 0 0 0 0 1 0 0 0 0 0
60
6
6
60
6
3
a ¼6
60
6
6
40
11
417
3
0 0 0 0 0 0 1 0 0 0 07
7
7
0 0 0 0 0 0 0 1 0 0 07
7
7
0 0 0 0 0 0 0 0 1 0 07
7
7
0 0 0 0 0 0 0 0 0 1 05
0 0 0 0 0 0 0 0 0 0 0 1
^ e . They are computed using the relation K
^ e ¼ ðae ÞT ke ae
Matrices K
2
12
6 0
6
6
6 30
6
6 12
6
6
6 0
6
6 30
EI
6
^1 ¼
K
6
125 6 0
6
6 0
6
6
6 0
6
6 0
6
6
4 0
0
0
6
60
6
6
60
6
6
60
6
6
60
6
6
60
^ 2 ¼ EI 6
K
6
125 6 0
6
6
60
6
6
60
6
6
60
6
6
60
4
0
30 0 0 0 0 0 0
0
0
2500
0
0
100
30
30
12
0
0
2500
0
0
50
0
30
2500
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0 0 0 0 0 07
7
7
50 0 0 0 0 0 0 7
7
30 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 07
7
100 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 07
7
0 0 0 0 0 0 07
7
7
0 0 0 0 0 0 07
7
0 0 0 0 0 0 07
7
7
0 0 0 0 0 0 05
0
0
0
0
0 0 0 0 0 0 0
0 0
0
0
0
0
0
0
0 0
0
0
0
0
0
0
0 0
0
0
0
0
0
0
3234:6
2353:9 144:0 3234:6 2353:9 144:0
0 0
2353:9
1861:4
0 0
144:0
192:0
800:0
0 0 3234:6 2353:9
0 0
192:0 2353:9 1861:4
192:0
144:0
192:0
400:0
144:0
3234:6
2353:9
144:0
0 0 2353:9 1861:4 192:0
2353:9
1861:4 192:0
0 0
3
2500
0
2
30 12
144:0
192:0
192:0
400:0
144:0
0 0
0
0
0
0
0
800:0
0
0 0
0
0
0
0
0
0
0 0 0
0
0
0
0
0
0
0 0 0
3
7
0 0 07
7
7
0 0 07
7
7
0 0 07
7
7
0 0 07
7
7
0 0 07
7
7
0 0 07
7
7
0 0 07
7
7
0 0 07
7
7
0 0 07
7
7
0 0 07
5
0 0 0
418 PART
2
II Multi-degree-of-freedom systems
0 0 0 0 0 0
0
0
0
0
0
0 0 0 0 0
0
0
0
0
0
0 0 0 0 0
0
0
0
0
0
0 0 0 0 0
0
0
0
0
0
0 0 0 0 0
0
0
0
0
0
0
0
6
60
6
60
6
6
60
6
6
60
6
6
60
EI
6
^3 ¼
K
125 6
60
6
60
6
6
60
6
6
60
6
6
40
0 0 0 0 0
0 0 0 0 0
907:6 1194:2
0 0 0 0 0 1194:2
0
48
1604:3
36
0 0 0 0 0
48
36
400
0 0 0 0 0
907:6
1194:2
48
0 0 0 0 0
1194:2 1604:3
0 0 0 0 0 0
48
36
0
907:6
200
1604:3
36
400
0
1194:2
36
907:6 1194:2
36 1194:2
48
3
7
07
7
07
7
7
07
7
7
07
7
7
07
7
7
48 7
7
36 7
7
7
200 7
7
7
48 7
7
7
36 5
1194:2 1604:3
48
0
^ e . They are computed using the relation M
^ e ¼ ðae ÞT m
e ae
Matrices M
3
2
2:5 0 0 0
0 0 0 0 0 0 0 0
60
2:5 0 0
0 0 0 0 0 0 0 07
7
6
60
0 0 0
0 0 0 0 0 0 0 07
7
6
7
6
0 0 2:5 0 0 0 0 0 0 0 0 7
60
7
6
60
0 0 0
2:5 0 0 0 0 0 0 0 7
7
6
60
0 0 0
0 0 0 0 0 0 0 07
7
6
1
^
M ¼ rA6
7,
60
0 0 0
0 0 0 0 0 0 0 07
7
6
60
0 0 0
0 0 0 0 0 0 0 07
7
6
60
0 0 0
0 0 0 0 0 0 0 07
7
6
7
6
60
0 0 0
0 0 0 0 0 0 0 07
7
6
40
0 0 0
0 0 0 0 0 0 0 05
0
0 0 0
0 0 0 0 0 0 0 0
2
3
0 0 0 0 0 0 0 0 0 0 0 0
60 0 0 0 0 0 0 0 0 0 0 07
6
7
60 0 0 0 0 0 0 0 0 0 0 07
6
7
6
7
6 0 0 0 5: 0 0 0 0 0 0 0 0 7
6
7
6 0 0 0 0 5: 0 0 0 0 0 0 0 7
6
7
60 0 0 0 0 0 0 0 0 0 0 07
6
7
2
^ ¼ rA6
M
7
6 0 0 0 0 0 0 5: 0 0 0 0 0 7
6
7
6 0 0 0 0 0 0 0 5: 0 0 0 0 7
6
7
60 0 0 0 0 0 0 0 0 0 0 07
6
7
6
7
60 0 0 0 0 0 0 0 0 0 0 07
6
7
40 0 0 0 0 0 0 0 0 0 0 05
0 0 0 0 0 0 0 0 0 0 0 0
The finite element method Chapter
2
0
60
6
60
6
60
6
60
6
6
60
3
^
M ¼ rA6
60
6
60
6
60
6
60
6
40
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
10:
0
0
0
0
0
0
0
0
0
0
0
0
10:
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
10:
0
0
0
0
0
0
0
0
0
0
0
0
10:
0
11
419
3
0
07
7
07
7
07
7
07
7
7
07
7
07
7
07
7
07
7
07
7
05
0
^e ðt Þ. They are computed using the relation p
^ e ðt Þ ¼ ðae ÞT pe ðt Þ
Vectors p
9
8
8 9
0
0>
>
>
>
>
>
>
>
>
>
>
>
>
> 0
>
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> 0
>
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0:08P
ð
t
Þ
0
>
>
>
>
>
>
>
=
=
<
< >
0
0
2
1
3
^ ðt Þ ¼ p
^ ðt Þ ¼
^ ðt Þ ¼
, p
p
0
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0:08P
ð
t
Þ
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0
>
>
>
>
>
>
>
>
>
>
>
>
>
> 0
>0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0
>
>
>
;
;
:
: >
0
0
K,
pðt Þ of the frame
4. Computation of the total matrices M,
2
2:5
60
6
60
6
6
60
6
60
6
3
X
6
¼
^ e ¼ rA6 0
M
M
60
6
e¼1
60
6
6
60
6
60
6
40
0
0
2:5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7:5
0
0
0
0
0
0
0
0
0
0
0
0
7:5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
15:
0
0
0
0
0
0
0
0
0
0
0
0
15:
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
10:
0
0
0
0
0
0
0
0
0
0
0
0
10:
0
3
0
07
7
07
7
7
07
7
07
7
07
7
07
7
07
7
7
07
7
07
7
05
0
¼
K
3
X
^e
K
e¼1
2
12
0
6 0 2500
6
6
6 30
0
6
6 12
0
6
6
6 0 2500
6
0
EI 6
6 30
¼
6
125 6 0
0
6
6 0
0
6
6
6 0
0
6
6 0
0
6
6
4 0
0
0
0
30
12
0
30
0
0
0
0
0
0
100
0
30
2500
0
0
50
0
0
0
0
0
0
0
0
0
0
30
3246:6
2353:9 114
3234:6
2353:9
144
0
0
0
50
2353:9
114
4361:4
192
2353:9
144
1861:4
192
192
400
0
0
0
0
1159:7
3465:7
192
156
192
900
0 3234:6 2353:9 144
0 2353:9 1861:4 192
0
144
192
400
0
0
0
0
0
0
0
0
0
0
0
0
4142:28
1159:7
192
156
907:68 1194:2
1194:2 1604:3
48
36
9
8
0
>
>
>
>
>
>
>
>
0
>
>
>
>
>
>
>
>
0
>
>
>
>
>
>
>
>
2P
ð
t
Þ
>
>
>
>
>
>
>
>
0:08P
ð
t
Þ
>
>
>
>
3
=
<
X
0
ðt Þ +
^ e ðt Þ ¼
pðt Þ ¼ P
p
0
>
>
>
>
e¼1
>
>
>
1:08P ðt Þ >
>
>
>
>
>
>
>
>
>
> 0
>
>
>
>
>
>
>
> 0
>
>
>
>
>
>
>
>
0
>
>
;
:
0
1200
48
36
200
907:68 1194:2
1194:2 1604:3
48
36
907:68 1194:2
1194:2
1604:3
48
36
0
3
07
7
7
07
7
07
7
7
07
7
07
7
7
48 7
7
36 7
7
7
200 7
7
48 7
7
7
36 5
400
The finite element method Chapter
11
421
K,
pðt Þ due to support conditions of
5. Modification of the matrices M,
the frame
Referring to Fig. E11.2, the displacement vector should be rearranged as
e
e1 u
e2 u
e3 u
e4 u
e5 u
e6 u
e7 u
e8 u
e9 u
e10 u
e11 u
e12 g
uT ¼ f u
¼ f u4 u5 u6 u7 u8 u9 u12 u1 u2 u3 u10 u11 g
Hence, the matrix V is
9 2
8
0
e1 >
u
>
>
>
>
>u
60
>
>
e
2
>
>
> 6
>
>
60
>
>
e3 >
u
>
>
6
>
>
>
>
>
>
e4 >
u
> 6
>
61
>
>
>
>
60
>
e
u5 >
>
>
>
= 6
<
60
e6
u
¼6
60
e
u
>
7 >
>
>
6
>
>
>
>u
60
e
>
>
8 >
>
6
>
>
>
>u
60
>
>
e
9 >
>
6
>
>
>
>
60
>
>
e
u
10 >
>
6
>
>
>
>
40
>
>
e
u
11
>
>
;
:
0
e12
u
defined from the relation
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
38
9
0 > u4 >
>
>
>
>
> u5 >
07
>
7>
>
>
>
>
>
>
07
u
>
>
6
7>
>
>
>
>
>
07
u
>
7
7>
>
>
>
>
>
07
>
> u8 >
>
7>
=
<
07
7 u9
07
> u12 >
>
7>
>
>
> u1 >
07
>
>
>
7>
>
>
>
>
>
07
u
2 >
>
7>
>
>
> u >
>
07
3 >
>
7>
>
>
>
>
>
1 5>
u
10
>
>
;
:
0
u11
Applying Eqs. (11.2.114a), (11.2.114c), (11.2.114d), we obtain
422 PART
II Multi-degree-of-freedom systems
e ff , M
e fs , M
e sf , M
e ss , K
e ff , K
e fs , K
e sf , K
e ss ,
6. Formulation of the matrices M
e
pf ðt Þ, e
p s ðt Þ
e M,
e e
These matrices result directly from K,
pðt Þ taking into account the partitioning dictated by the separation of the free displacements from the specified
(support) ones. Thus, we have
2
3
2
3
7:5 0 0 0
0 0 0
0 0 0 0 0
60
60 0 0 0 07
7:5 0 0
0 0 07
6
7
6
7
60
7
60 0 0 0 07
0 0 0
0 0 07
6
6
7
e ff ¼ rA6 0
e
6
7
0 0 15: 0 0 0 7
M
6
7, Mfs ¼ rA6 0 0 0 0 0 7
60
7
6
7
0 0 0
15: 0 0 7
6
60 0 0 0 07
40
5
4
0 0 0
0 0 0
0 0 0 0 05
0
0 0 0
0 0 0
0 0 0 0 0
3
3
2
2
0 0 0 0 0 0 0
2:5 0 0 0
0
60 0 0 0 0 0 07
60
2:5 0 0
0 7
7
7
6
6
e ss ¼ rA6 0
e sf ¼ rA6 0 0 0 0 0 0 0 7, M
0
0
0
0 7
M
7
7
6
6
40 0 0 0 0 0 05
40
0 0 10: 0 5
0 0 0 0 0 0 0
0
0 0 0
10:
2
3
3246:6 2353:9 114 3234:6 2353:9 144
0
6 2353:9 4361:4 192 2353:9 1861:4 192
07
6
7
6 114
7
192
900
144
192
400
0
6
7
EI
e
6
Kff ¼
3234:6 2353:9 144 4142:2 1159:7 192 48 7
6
7
125 6
7
6 2353:9 1861:4 192 1159:7 3465:7 156 36 7
4 144
192
400
192
156
1200 200 5
0
0
0
48
36
200 400
3
2
12
0 30
0
0
6 0 2500
0
0
0 7
7
6
6 30
0 50
0
0 7
7
6
EI
e fs ¼
6 0
0
0 907:6 1194:2 7
K
7
125 6
6 0
0
0 1194:2 1604:3 7
7
6
4 0
0
0 48
36 5
0
0
0 48
36
The finite element method Chapter
11
423
2
3
12
0 30
0
0
0
0
6 0 2500
0
0
0
0
07
6
7
e sf ¼ EI 6 30
K
0
50
0
0
0
07
6
7
125 4
0
0
0 907:6 1194:2 48 48 5
0
0
0 1194:2 1604:3 36 36
2
3
12
0 30
0
0
6 0 2500
0
0
0 7
7
EI 6
e
6
30
0
100
0
0 7
Kss ¼
6
7
125 4
0
0
0
907:6 1194:2 5
0
0
0 1194:2 1604:3
9
8
2:0P >
>
9
8
>
>
>
>
>
0:08P >
P1 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
=
=
< P2 >
< 0
e
, e
ps ðt Þ ¼ P3
0
pf ðt Þ ¼
>
>
>
>
>
>
>
>
> P10 >
> 1:08P >
>
>
>
>
;
>
:
>
>
>
>
>
P11
0
>
>
;
:
0
7. Equation of motion
In as much as e
us ¼ 0, it is also e
u€s ¼ 0 and Eq. (11.3.90) yields e
pf∗ ðt Þ ¼ e
pf ðt Þ.
The equation of motion results from Eq. (11.3.88). Thus we obtain
2
7:5
60
6
60
6
6
rA6 0
6
60
6
40
0
2
0
7:5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
15:0
0
0
0
0
0
0
0
15:0
0
0
0
0
0
0
0
0
0
38 9
0 >
e€1 >
u
>
>
>
>
>
>
>
07
u
7>
> e€2 >
>
>
>
>
7
0 7>
e€3 >
<u
=
7 €
07 u
e4
7> € >
>
>
0 7>
e5 >
>u
>
>
7>
>
>
€6 >
0 5>
>
>
e
u
>
:€ >
;
0
e7
u
38 9
e1 >
3246:6 2353:9 114:0 3234:6 2353:9 144:0
0
>
>u
>
>
6 2353:9 4361:4 192:0 2353:9 1861:4 192:0
7>
>
>
>
e2 >
u
0
6
7>
>
>
>
>
6 114:0
7
>
e3 >
192:0 900:0
144:0 192:0 400:0
0 7>
u
<
=
6
EI 6
7
e4
+
6 3234:6 2353:9 144:0 4142:2 1159:7 192:0 48:0 7 u
7> >
125 6
>u
>
6 2353:9 1861:4 192:0 1159:7 3465:7 156:0 36:0 7>
e5 >
>
>
>
6
7>
>
>
>
>
4 144:0
5
e
192:0 400:0
u
192:0 156:0 1200:0 200:0 >
6
> >
>
:
;
e7
u
0
0
0
48:0
36:0 200:0 400:0
8
9
2:0P >
>
>
>
>
> 0:08P >
>
>
>
>
>
>
>
>
>
>
>
0
<
=
¼
0
>
>
>
>
>
1:08P >
>
>
>
>
>
>
>
>
0
>
>
>
>
:
;
0
424 PART
II Multi-degree-of-freedom systems
The solution of the equation of motion gives the vector e
uf . Then Eq. (11.3.89)
yields the support reactions because there are no nodal element loads in the
direction of the supports, hence
e sf e
e
uf
p s ðt Þ ¼ K
or
8 9
e1 >
u
>
>
2
3>
8
9
>
>
>
e2 >
12
0
30
0
0
0
0
u
P
>
>
1 >
>
>
>
>
>
>
>
>
>
>
>
6
7
>
>
>
e
0
2500
0
0
0
0
0
u
P
< 2 = EI 6
=
7< 3 >
6
7
e4
50
0
0
0
0 7 u
P3 ¼
6 30 0
>
>
7> >
>
> 125 6
>
>
>
4 0
e5 >
0
0 907:68 1194:2 48 48 5>
u
> P10 >
>
>
>
:
;
>
>
>
>
>
e
0
0
0
1194:2 1604:3 36 36 >
u
P11
>
6>
>
: >
;
e7
u
11.4 Static condensation: Guyan’s reduction
Often the inertial forces in the directions of some coordinates are small and can
be omitted. This allows reducing significantly the number of equations of
motion to be solved. Such a case is common in the FEM, where the kinetic
energy of the rotational degrees of freedom is neglected because it is small
compared to that of the translational degrees.
If the elements of the displacement vector are reordered so that the vector ut
includes the translational degrees of freedom and the vector uq the rotational
ones, then for C ¼ 0 and M diagonal Eq. (11.3.91) is written as
€t
u
Ktt Ktq
ut
pt
Mtt 0
+
¼
(11.4.1)
€q
Kqt Kqq
uq
pq
0 0 u
where the vector pq ðt Þ includes the actions (moment) applied directly to
the nodes.
Obviously, Eq. (11.4.1) yields after performing the multiplications
€t + Ktt ut + Ktq uq ¼ pt ðt Þ
Mtt u
(11.4.2a)
Kqt ut + Kqq uq ¼ pq ðt Þ
(11.4.2b)
Eq. (11.4.2b) is quasistatic and can be used to eliminate uq from
Eq. (11.4.2a). Τhe matrix Kqq is square and can be inverted. Thus solving
Eq. (11.4.2b) for uq gives
uq ¼ K1
qq ðpq ðt Þ Kqt ut Þ
which is substituted into Eq. (11.4.2a) to yield
€t + Ktt∗ ut ¼ pt∗ ðt Þ
Mtt u
(11.4.3)
The finite element method Chapter
11
425
where
Ktt∗ ¼ Ktt Ktq K1
qq Kqt
(11.4.4a)
pt∗ ¼ pt Ktq K1
qq pq
(11.4.4b)
Ktt∗
The matrix
is called the statically condensed stiffness matrix while the
procedure producing it is referred to as the static condensation. Obviously,
when pq ¼ 0, pt∗ ¼ pt .
The reduction of the degrees of freedom can also be achieved when the
mass matrix is not diagonal or the inertial forces in the rotational degrees
of freedom are not negligible. In the employed technique, a subset of the coordinates is selected arbitrarily as the set of active (or “master”) coordinates, say
ut , and the remaining coordinates are dependent (or “slave”) coordinates, say
uq . In this case, the mass matrix of Eq. (11.4.1) is written in the partitioned
form
Mtt Mtq
(11.4.5)
M¼
Mqt Mqq
where now the matrices Mtq , Mqt , Mqq may not vanish.
The starting point is Eq. (11.4.2b), which for pq ðt Þ ¼ 0 becomes
uq ¼ K1
qq Kqt ut
(11.4.6)
In the foregoing equation ut represents the independent coordinates and
uq the dependent ones. Thus, we can write
I
ut
¼
ut
(11.4.7)
K1
uq
qq Kqt
or
u ¼ Tut
where
I
T¼
K1
qq Kqt
(11.4.8)
(11.4.9)
represents the transformation matrix of the independent coordinates ut to u.
By virtue of the transformation (11.4.9), the elastic energy is written
1
U ¼ uT Ku
2
1
¼ uTt TT KTut
2
1
¼ uTt K ut
2
(11.4.10)
426 PART
II Multi-degree-of-freedom systems
where
K ¼TT KT
¼Ktt Ktq K1
qq Kqt
(11.4.11)
Similarly, the kinetic energy is written
1
K ¼ u_ T Mu_
2
1
¼ u_ Tt TT MTu_ t
2
1
¼ u_ Tt M u_ t
2
(11.4.12)
where
M ¼TT MT
"
#"
#
I
Mtt Mtq
1
¼ I Kqt Kqq
Mqt Mqq K1
qq Kqt
(11.4.13)
1
1
1
¼Mtt Mtq K1
qq Kqt + Kqt Kqq Mqq Kqq Kqt Kqt Kqq Mqt
Obviously, if M is diagonal then Mtq ¼ Mqt ¼ 0 and the foregoing equation
becomes
1
M ¼ Mtt + Kqt K1
qq Mqq Kqq Kqt
(11.4.14)
Finally, the virtual work of the external nodal loads is written
dWp ¼duT pðt Þ
¼duTt TT pðt Þ
(11.4.15)
¼duTt p ∗ ðt Þ
where
p ∗ ðt Þ ¼TT pðt Þ
¼pt ðt Þ K1
qq Kqt pq ðt Þ
(11.4.16)
This reduction of the degrees of freedom is known as the Guyan reduction [6]. Obviously, in this reduction, the stiffness of the structure is not
affected, as all elements of the original stiffness matrix contribute to the reduced
stiffness matrix. On the contrary, the reduced mass matrix is affected as it
involves combinations of elements of the mass and stiffness matrix. This technique reduces the number of equations. Nevertheless, it introduces approximations, and the reliability of the results depends on the selection of the eliminated
degrees of freedom.
The finite element method Chapter
11
427
Example 11.4.1 Formulate the equation of motion of the frame in Example
11.3.2 after the static condensation of the rotational degrees of freedom.
Solution
e3 , u
e6 , u
e7
The rotational degrees of freedom are u6 , u9 , u12 , will take the places u
in the modified displacement vector e
uf . Therefore, the elements of the vector
e
uf must be reordered as follows:
38 9
8 9 2
e1 >
1 0 0 0 0 0 0 >
u
u1 >
>
>
>
>
> 6
>
> >
> >
7>
>
>
>
>
>
0
1
0
0
0
0
0
e
u
u
2>
2>
6
7>
>
>
>
>
>
>
>
>
>
>
6
7
> 6 0 0 0 0 1 0 0 7>
>
>
>
>
e4 >
u
= 6
=
< u3 >
<
7
7 u
0
0
1
0
0
0
0
e
u4 ¼ 6
5
6
7
>
>
> 6 0 0 0 1 0 0 0 7>
>
>u >
>u
>
>
>
e
>
>
>
>
6
7
5
3>
>
>
>
> 6
>
> >
> >
7>
>
>
>
>
>
4
5
e
0
0
0
0
0
1
0
u
u
6>
6>
>
>
>
>
>
;
;
: >
: >
e7
0 0 0 0 0 0 1
u
u7
which defines the transformation matrix
2
1 0 0
60 1 0
6
6
60 0 0
6
6
VC ¼ 6 0 0 1
6
60 0 0
6
6
40 0 0
0 0 0
3
0 0 0 0
0 0 0 07
7
7
0 1 0 07
7
7
0 0 0 07
7
1 0 0 07
7
7
0 0 1 05
0 0 0 1
e ff and the vector e
e ff , K
pf as
and modifies the matrices M
2
7:5 0
0
0
6 0
7:5
0
0
6
6
6 0
0
15:0
0
6
6
T e
M ¼ VC Mff VC ¼ 6 0
0
0
15:0
6
6 0
0
0
0
6
6
0
0
0
4 0
0
0
0
0
3
0 0 0
0 0 07
7
7
0 0 07
7
7
0 0 07
7
0 0 07
7
7
0 0 05
0 0 0
428 PART
II Multi-degree-of-freedom systems
The implied partitioning of the matrices M, K, and vector pðt Þ in the foregoing
equations yields
2
2
7:5
60
Mtt ¼ rA6
40
0
3
0
0 0
7:5 0 0 7
7
0
15 0 5
0
0 15
3
3246:6 2353:9 3234:6 2353:9
7
EI 6
6 2353:9 4361:4 2353:9 1861:4 7,
Ktt ¼
4
125 3234:6 2353:9 4142:2 1159:7 5
2353:9 1861:4 1159:7 3465:7
2
3
114 144 0
7
EI 6
6 192 192 0 7
Ktq ¼
4
5
144
192
48
125
192 156 36
2
3
2
3
114 192 144 192
900 400
0
EI 4
EI
4 400 1200 200 5
144 192 192 156 5, Kqq ¼
Kqt ¼
125
125
0
0 48
36
0 200 400
2
3
2 3
2P ðt Þ
0
6 0:08P ðt Þ 7
7, pq ðt Þ ¼ 4 0 5
pt ðt Þ ¼ 6
4 0
5
0
1:08P ðt Þ
Applying Eqs. (11.4.4a), (11.4.4b) gives (11.4.4a)
2
3
3222:71 2389:02 3205:83 2387:19
7
EI 6
6 2389:02 4308:10 2396:65 1810:18 7,
Ktt∗ ¼
4
5
3205:83
2396:65
4101:75
1196:10
125
2387:19 1810:18 1196:10 3412:98
9
8
2:0P ðt Þ >
>
>
>
>
>
<
0:08P ðt Þ =
∗
pt ðt Þ ¼
>
>
0
>
>
>
>
;
:
1:08P ðt Þ
The finite element method Chapter
11
429
Hence, the equation of motion after the static condensation is
2
7:5
6
60
rA6
60
4
0
2
38 9
> u€1 >
>
> >
7>
7:5 0 0 7< u€2 =
7
0
15 0 7
u€4 >
>
>
5>
>
;
: >
0
0 15
u€5
0
0
0
38 9 8
9
2389:02 3205:83 2387:19 > u1 > > 2:0 >
>
>
>
>
>
6
7> > >
2389:02 4308:10 2396:65 1810:18 7< u2 = < 0:08 =
EI 6
6
7
+
¼
7 u > > 0 >P ð t Þ
125 6
>
>
4 3205:83 2396:65 4101:75 1196:10 5>
4> >
>
>
;
:
; >
: >
1:08
2387:19 1810:18 1196:10 3412:98
u5
3222:71
11.5 Flexural vibrations of a plane frame
In a frame structure, the work due to the axial deformation of its elements is
small compared to that due to bending deformation. Thus, in the usual cases,
the axial deformation is omitted in order to reduce the number of equations
of motion to be solved, which is an important issue in the dynamic analysis.e
This omission is particularly convenient in small plane frames with elements
parallel to the global axes (e.g., rectangular frames), where the equations of
motion can be easily derived by hand without recourse to the FEM. This is illustrated with the examples below. However, the general problem of neglecting the
axial deformations in frames is discussed in Section 11.7.
In flexural vibrations, the displacement vector results by omitting the
degrees of freedom u1 and u4 in the direction of the element local x axis.
The remaining degrees of freedom are shown in Fig. 11.5.1 after the renumbering. Thus, the displacement vector is written now
u ¼ f u1 u2 u3 u4 gT
(11.5.1)
where u1 , u3 are the translational degrees of freedom and u2 , u4 the rotational
ones. Then the reduced stiffness, mass, and damping matrices and the vectors
of the nodal forces of the element will result from the corresponding original
ones by omitting the nodal forces in these directions. Thus, for an element
with a constant cross-section, the stiffness matrix results from Eq. (11.3.17)
by omitting the first and fourth line as well as the first and fourth column. This
yields
e. The omission of the axial deformation is not always permitted because it may lead to considerable
mistakes in certain structures such as high-rise buildings, whose overall deformation pattern resembles that of a cantilever when subjected to horizontal loads. Obviously, this deformation cannot be
realized when the axial deformation of the columns is neglected.
430 PART
II Multi-degree-of-freedom systems
Fig. 11.5.1 Degrees of freedom of a beam element with bending deformation.
2
12EI
6EI
3
6
L
L2
6
4EI
6 6EI
6
2
6
L
ke ¼ 6 L
6EI
6 12EI
6 3 2
6
L
L
4 6EI
2EI
L2
L
3
12EI
6EI
L3
L2 7
7
6EI
2EI 7
7
2
L
L 7
7
12EI
6EI 7
7
L3
L2 7
6EI
4EI 5
2
L
L
(11.5.2)
Similarly, the consistent mass matrix is obtained from Eq. (11.3.25) as
2
3
156
22L 54 13L
4L2 13L 3L2 7
me 6
6 22L
7
(11.5.3)
me ¼
6
7
420 4 54
13L 156 22L 5
13L 3L2 22L
4L2
and the lumped mass matrix from Eq. (11.3.32) as
2
3
1 0 0 0
7
me 6
60 0 0 07
me ¼
6
7
2 40 0 1 05
(11.5.4)
0 0 0 0
The damping matrix is obtained from Eq. (11.3.42) as
2
3
12cs I
6cs I
12cs I
6cs I
3
6
L3
L2
L
L2 7
6
7
6 6cs I
4c
I
6c
I
2c
s
s
sI 7
6
7
2
6 L2
7
L
L
L
e
6
7
c ¼6
12cs I
6cs I 7
6 12cs I 6cs I
2 7
6
L3
L2
L3
L 7
6
7
4 6cs I
2cs I
6cs I
4cs I 5
2
L2
L
L
L
(11.5.5)
The finite element method Chapter
while the geometric stiffness matrix for
Eq. (11.3.59)
2
12
1
6 L
6
4L
6 1
N6
e
3
6
kG ¼ 6 12
10 6 6 L 1
4
L
1
3
11
431
constant axial force results from
12
L
1
12
L
1
3
1
7
L 7
7
7
3 7
7
1 7
7
4L 5
3
(11.5.6)
Finally, the consistent nodal load vector is obtained from Eq. (11.3.46),
which for a constant load py ðx, t Þ ¼ pe becomes
pe ðt Þ ¼
p e Le
2
1
L
L
1 6
6
T
(11.5.7)
and the static equivalent nodal load vector is obtained from Eq. (11.3.49) as
p e Le
(11.5.8)
f 1 0 1 0 gT
2
In the following examples, the formulation of the equation of motion of
certain frames with elements parallel to the global axes is presented. The method
of influence coefficients is employed to establish the matrices of the structure.
pe ðt Þ ¼
Example 11.5.1 Formulate the equation of motion of the structure shown in
Fig. E11.3a. Consider the lumped mass assumption for the column. Examine
the case I0 ¼ 0. Assumed material density r ¼ m=10AL.
Solution
The structure may be viewed as a plane frame consisting of a single element, the
beam column. Obviously, the system has two free degrees of freedom in global
axes: the nodal displacement u1 and the rotation u2 of the top cross-section
(see Fig. E11.3b). Hence, the stiffness matrix of the structure is
(a)
(b)
(c)
(d)
(e)
Fig. E11.3 Structure in Example 11.5.1 (a), parameters of motion (b), deformation patterns (c), (d),
and beam element degrees of freedom (e)
¼ k 11
K
k 21
k12
k22
(1)
432 PART
II Multi-degree-of-freedom systems
are evaluated by considering the
The elements of the stiffness matrix K
deformation of the system in global axes. The matrix element kij expresses
the elastic force in the direction of ui , when uj ¼ 1 while the remaining displacements are zero (i, j, ¼ 1, 2). The deformation patterns are shown in Fig. E11.3c
and d. The nodal elastic forces can be evaluated using the stiffness matrix of the
beam element with a constant cross-section, Eq. (11.5.2), when matching
u1 ¼ u1 , u2 ¼ u2 , u3 ¼ u4 ¼ 0 (see Fig. E11.3e). Consequently, we obtain:
With reference to Fig. E11.3c
12EI
ð1Þ
k11 ¼ k11 ¼ 3 ,
L
6EI
ð1Þ
k21 ¼ k21 ¼ 2
L
With reference to Fig. E11.3d
6EI
ð1Þ
k12 ¼ k12 ¼ 2 ,
L
4EI
ð1Þ
k22 ¼ k22 ¼
L
Obviously, it is not necessary to evaluate the element k21 because the stiffness matrix is symmetric. Thus, we have
2 12EI 6EI 3
L3
L2 7
¼6
K
4
5
6EI 4EI
L2
L
The mass matrix is
m11 0
M¼
0
m22
(2)
(3)
where
ð1Þ
11 ¼ m11 + m ¼ rAL=2 + m ¼ 1:05m and m22 ¼ I0
m
Hence, the mass matrix of the structure is
¼ 1:05m 0
M
0
I0
(4)
The vector of the nodal loads is
pðt Þ ¼
Therefore, the equation
11 0
m
22
0
m
pðt Þ
M ðt Þ
of motion of the structure is
pðt Þ
u1
u€1
k11 k12
+ ¼
€
M ðt Þ
u2
u2
k 21 k 22
(5)
(6)
or after performing the operations
11 u€1 + k11 u1 + k12 u2 ¼ pðt Þ
m
(7a)
22 u€2 + k21 u1 + k22 u2 ¼ M ðt Þ
m
(7b)
The finite element method Chapter
11
433
22 ¼ I0 ¼ 0 static condensation is applied and Eq. (7b) becomes
When m
k21 u1 + k22 u2 ¼ M ðt Þ
(8)
from which we obtain
1
1
u2 ¼ k22 M ðt Þ k22 k21 u1
(9)
Then substituting into Eq. (7a) gives
∗
11 u€1 + k11
u1 ¼ p ∗ ðt Þ
m
(10)
3EI
1
∗
k11
¼ k11 k12 k22 k21 ¼ 3
L
(11)
where
1
p ∗ ðt Þ ¼ pðt Þ k12 k22 M ðt Þ ¼ pðt Þ 3
M ðt Þ
2L
(12)
∗
Inserting the values for k11
and p ∗ ðt Þ into Eq. (10) yields the condensed
equation of motion
1:2m u€1 +
3EI
M ðt Þ
u1 ¼ pðt Þ 1:5
3
L
L
(13)
We observe that the moment M ðt Þ in the direction of u2 adds a force in the
direction of u1 .
Example 11.5.2 Formulate the equation of motion of the plane frame shown
Fig. E11.4 by neglecting the axial deformation and the damping of the elements.
Column 1 is elastically supported on the ground, whose stiffness is simulated by
the rotational spring kR ¼ EI =2L. All elements have a constant cross-section.
Their properties are shown in the figure. The mass matrix will be obtained
using the lumped mass assumption. The rotational degrees of freedom will
be condensed. Assume M ðt Þ ¼ 2P ðt ÞL.
Fig. E11.4 Frame in Example 11.5.2.
434 PART
II Multi-degree-of-freedom systems
Solution
Inasmuch as the axial deformation of the elements is neglected, the degrees of
freedom of the structure are reduced to seven, two translational, u1 , u2 , and five
rotational, u3 , u4 , u5 , u6 , u7 . The degrees of freedom are shown in Fig. E11.5.
Fig. E11.5 Degrees of freedom of the frame in Example 11.5.2.
Stiffness matrix. The stiffness matrix will have the form
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Fig. E11.6 Deformation patterns of the frame in Example 11.5.2 due to unit nodal displacements.
The finite element method Chapter
2
k11
6
6 k 21
6
6 k31
6
¼6
K
6 k41
6
6 k51
6
6
4 k 61
k12
k22
k32
k13
k23
k33
k42 k43
k52 k53
k62 k63
k14
k24
k34
k15
k25
k35
k44 k45
k54 k55
k64 k65
k16
k26
k36
k46
k56
k66
k 71 k 72 k 73 k 74 k 75 k76
11
3
k17
7
k27 7
7
k37 7
7
7
k47 7
7
k57 7
7
7
k67 5
k77
435
(1)
Its elements will be computed as the elastic forces at the nodes by giving a
unit value to each displacement and keeping the other displacements equal to
zero, following the method of the influence coefficients discussed in
Section 10.4.1. For the economy of the computation, we will take advantage
of the symmetry of the matrix and restrict the computations to the elements
of the main diagonal and those below it.
are computed by referring to
The elements of the 1st column of K
Fig. E11.6a. The displacement u1 ¼ 1 “activates” (deforms) the elements
1 and 3. Their matching with the beam element is shown in Fig. E11.5, namely:
Element 1: u1 ¼ u1 , u2 ¼ 0, u3 ¼ 0, u4 ¼ 0
Element 3: u1 ¼ u1 , u2 ¼ 0, u3 ¼ 0, u4 ¼ 0
Then, with reference to Eq. (11.5.2), we obtain
12EI 12EI 27 EI 12EI
ð1Þ
ð3Þ
ð3Þ
k11 ¼ k11 + k11 ¼
+ 3 ¼
, k 21 ¼ k31 ¼ 3
L
2 L3
L
ð2LÞ3
6EI
3 EI 6EI
3 EI
ð1Þ
ð1Þ
k31 ¼ k41 ¼
¼
, k 41 ¼ k21 ¼
¼
ð2LÞ2 2 L2
ð2LÞ2 2 L2
6EI
6EI
ð3Þ
ð3Þ
k51 ¼ k21 ¼ 2 , k61 ¼ k41 ¼ 2 , k71 ¼ 0
L
L
The elements of the 2nd column of K are computed by referring to
Fig. E11.6b. The displacement u2 ¼ 1 activates (deforms) the elements 3, 4,
and 6. Their matching with the beam element is:
Element 3: u1 ¼ 0, u2 ¼ 0, u3 ¼ u2 , u4 ¼ 0
Element 4: u1 ¼ u2 , u2 ¼ 0, u3 ¼ 0, u4 ¼ 0
Element 6: u1 ¼ u2 , u2 ¼ 0, u3 ¼ 0, u4 ¼ 0
Then, with reference to Eq. (11.5.2), we obtain
12EI 12EI 12EI 36EI
ð3Þ
ð4Þ
ð6Þ
k22 ¼ k33 + k11 + k11 ¼ 3 + 3 + 3 ¼ 3
L
L
L
L
6EI
ð3Þ
k32 ¼ 0,
k42 ¼ 0,
k52 ¼ k23 ¼ 2
L
6EI
6EI
6EI
ð3Þ
ð4Þ
ð6Þ
k62 ¼ k43 + k21 ¼ 2 + 2 ¼ 0,
k72 ¼ k21 ¼ 2
L
L
L
436 PART
II Multi-degree-of-freedom systems
are computed by referring to
The elements of the 3rd column of K
Fig. E11.6c. The rotation u3 ¼ 1 activates (deforms) only element 1. Its matching with the beam element is:
Element 1: u1 ¼ 0, u2 ¼ 0, u3 ¼ u2 , u4 ¼ 0
Then, with reference to Eq. (11.5.2), we obtain
4EI EI 5 EI 2EI EI
ð1Þ
ð1Þ
k33 ¼ k44 + kR ¼
+
¼
, k 43 ¼ k24 ¼
¼
2L
2L 2 L
2L
L
k 63 ¼ 0,
k 73 ¼ 0
k 53 ¼ 0,
are computed by referring to
The elements of the 4th column of K
Fig. E11.6d. The rotation u4 ¼ 1 activates (deforms) the elements 1 and 2. Their
matching with the beam element is:
Element 1: u1 ¼ 0, u2 ¼ u4 , u3 ¼ 0, u4 ¼ 0
Element 2: u1 ¼ 0, u2 ¼ u4 , u3 ¼ 0, u4 ¼ 0
Then, with reference to Eq. (11.5.2), we obtain
4EI 4ð8EI Þ 34EI
ð1Þ
ð2Þ
k44 ¼ k22 + k22 ¼
+
¼
2L
L
L
2ð8EI Þ 16EI
ð2Þ
¼
,
k54 ¼ k42 ¼
L
L
k64 ¼ 0,
k74 ¼ 0
are computed by referring to
The elements of the 5th column of K
Fig. E11.6e. The rotation u5 ¼ 1 activates (deforms) the elements 2 and 3. Their
matching with the beam element is:
Element 2: u1 ¼ 0, u2 ¼ 0, u3 ¼ 0, u4 ¼ u5
Element 3: u1 ¼ 0, u2 ¼ u5 , u3 ¼ 0, u4 ¼ 0
Then, with reference to Eq. (11.5.2), we obtain
4ð8EI Þ 4EI 36EI
ð2Þ
ð3Þ
k55 ¼ k44 + k22 ¼
+
¼
L
L
L
2EI
ð3Þ
,
k75 ¼ 0
k65 ¼ k42 ¼
L
are computed by referring to
The elements of the 6th column of K
Fig. E11.6f. The rotation u5 ¼ 1 activates (deforms) the elements 3, 4, and 5.
Their matching with the beam element is:
Element 3: u1 ¼ 0, u2 ¼ 0, u3 ¼ 0, u4 ¼ u6
Element 4: u1 ¼ 0, u2 ¼ u6 , u3 ¼ 0, u4 ¼ 0
Element 5: u1 ¼ 0, u2 ¼ u6 , u3 ¼ 0, u4 ¼ 0
The finite element method Chapter
11
437
Then, with reference to Eq. (11.5.2), we obtain
4EI 4EI 4ð8EI Þ 88 EI
ð3Þ
ð4Þ
ð5Þ
k66 ¼ k44 + k22 + k22 ¼
+
+
¼
L
L
1:5L
3 L
2ð8EI Þ 32 EI
ð5Þ
k76 ¼ k42 ¼
¼
1:5L
3 L
are computed by referring to
Finally, the elements of the 7th column of K
Fig. E11.6g. The rotation u7 ¼ 1 activates (deforms) the elements 5 and 6. Their
matching with the beam element is:
Element 5: u1 ¼ 0, u2 ¼ 0, u3 ¼ 0, u4 ¼ u7
Element 6: u1 ¼ 0, u2 ¼ u6 , u3 ¼ 0, u4 ¼ 0
Then, with reference to Eq. (11.5.2), we obtain
4ð8EI Þ 4EI 76 EI
ð5Þ
ð6Þ
k77 ¼ k44 + k22 ¼
+
¼
1:5L
L
3 L
Therefore, the stiffness matrix is
ð2Þ
Mass matrix. The lumped mass assumption for the elements, Fig E11.7a,
concentrates the mass at the nodes of the frame as shown in Fig. E11.7b.
Obviously, inertial forces arise only in the directions of the translational degrees
of freedom u1 and u2 . Thus, we have
ð1Þ
ð2Þ
ð2Þ
ð3Þ
m11 ¼ m1 + m1 + m2 + m1 ¼ 11:5mL
ð3Þ
ð4Þ
ð5Þ
ð5Þ
ð6Þ
m22 ¼ m2 + m1 + m1 + m2 + m1 ¼ 16:5mL
Therefore the mass matrix of the structure reads
ð3Þ
438 PART
(a)
II Multi-degree-of-freedom systems
(b)
Fig. E11.7 Lumped masses of the elements (a) and nodal masses of the frame (b) in Example
11.5.2.
Vector of the external nodal loads. This includes the nodal loads that are
directly applied to the nodes of the frame as well as those resulting from
the contribution of the equivalent nodal loads due to the element loads. In
our problem, only element 6 is loaded. To be consistent with the mass
matrix, the equivalent nodal load will be obtained from Eq. (11.3.51).
9
8
Hence
2
>
>
>
=
< P ðt Þ >
3
p1 ðt Þ
ð6Þ
¼
p ðt Þ ¼
p3 ðt Þ
>
>
>
;
: 1 P ðt Þ >
3
Therefore, the vector of the external loads of the structure reads
Static condensation. This includes the elimination of the five rotations
M
, and pðt Þ is indicated in
u3 , u4 ,…, u7 . The required partitioning of K,
Eqs. (2), (3), (4). Thus, we have
2
3
2
3
27
3 3
tq ¼ EI 4 2 L 2 L 6L 6L 0 5
tt ¼ EI 4 2 12 5, K
K
L3
L3
12
36
0 0 6L 0 6L
3
2
2
3
5 2 2
3
0
0
0
7
62L L
L 0
7
6
62
7
7
6 2
6
7
34L2 16L2 0
0
7
6L
63
7
7
6
6 L 0 7
7
60
62
7
2
2
2
EI
EI
16L
36L
2L
0
7
6
6
7
, Kqq ¼ 3 6
Kqt ¼ 3 6
7
7
L 6 6L 6L 7
L 6
7
88
32
60
6
7
0
2L2
L2
L2 7
6
6 6L 0 7
3
3 7
7
6
4
5
4
32 2 76 2 5
L
L
0
0
0
0
6L
3
3
The finite element method Chapter
11
439
9
8
0
>
>
9
8
>
>
>
>
>
>
P ðt Þ =
0
=
<
<
mL
33
0
, pq ðt Þ ¼ 0
, pt ðt Þ ¼ 2
Mtt ¼
>
: P ðt Þ ;
>
2 0 23
>
>
0
>
>
>
>
3
;
:
2P ðt ÞL
Applying Eqs. (11.4.4a), (11.4.4b) give
"
#
10:209 10:344
EI
0:546
∗
tt ¼
K
, p ∗ t ðt Þ ¼ P ðt Þ
0:837
L3 10:344 33:131
Therefore the equation of motion reads
33 0
EI
mL
10:209 10:344
0:546
¼ P ðt Þ
+ 3
0:837
L 10:344 33:131
2 0 23
11.6 Reduction of the degrees of freedom due to constraints
The number of degrees of freedom for the dynamic analysis of plane and space
frames may be considerably reduced by introducing axial constraints as a
consequence of ignoring the axial deformation of the elements. Before studying
this particular problem, we will discuss the following problem.
We assume that the equations of motion have been derived with respect to
N generalized coordinates u, that is,
M€
u + Cu_ + Ku ¼pðt Þ
(11.6.1)
and that the coordinates u are not independent, but they are constrained
by K < N linear constraint equations
Du ¼0
(11.6.2)
where D is a K N rectangular matrix.
The problem is to determine L ¼ N K independent equations of motion
with respect to L independent generalized coordinates, which will be solved
to obtain the dynamic response of the system.
We cope with this problem by proceeding as follows.
We choose L independent displacements (coordinates) and we partition the
displacement vector as
u ¼
ua
ub
(11.6.3)
where ua are the independent and ub the dependent displacements. Then we
write Eq. (11.6.2) as
440 PART
II Multi-degree-of-freedom systems
½ Da Db ua
ub
¼0
(11.6.4)
or
Da ua + Db ub ¼0
(11.6.5)
a
ub ¼ D1
b Da u
(11.6.6)
which gives
From Eq. (11.6.6), we conclude that the selection of the independent
coordinates cannot be arbitrary, but they must be chosen so that the inversion
of the matrix Db is ensured. This is accomplished, as will be shown in
Section 11.7, by the Gauss-Jordan elimination method. Generally, after selecting the independent coordinates, it is necessary to rearrange the elements of
the vector u to comply with the partition suggested by Eq. (11.6.3). Then
Eq. (11.6.1) is written
u€ + K
u€ + C
u ¼
M
pðt Þ
(11.6.7)
K,
C,
u, and pðt Þ are the modified matrices and vectors after the
where M,
rearrangement.
Eq. (11.6.3) can be further written
u ¼
I
ua
D1
b Da
(11.6.8)
u ¼T
ua
(11.6.9)
or
where I is the unit matrix with dimensions L L and
T¼
I
D1
b Da
(11.6.10)
Eq. (11.6.9) expresses a linear transformation, which transforms the
independent coordinates ua into u. This transformation can be employed to
reduce the number of Eq. (11.6.1).
Indeed, the kinetic energy of the system is
1 T
u_
T ¼ u_ M
2
(11.6.11)
or using Eq. (11.6.9)
1 T
u_ a
T ¼ u_ a TT MT
2
1 T^
¼ u_ a M
u_ a
2
(11.6.12)
The finite element method Chapter
11
441
where
^ ¼ TT MT
M
(11.6.13)
is the L L reduced transformed mass matrix.
is partitioned as indicated by the vectors ua and ub , we have
If the matrix M
"
¼
M
ab
aa M
M
ba M
bb
M
#
(11.6.14)
and Eq. (11.6.13) gives
n
^ ¼ I D1 D
M
a
b
T
"
#(
aa M
ab
o M
I
)
ba M
bb
D1
M
b Da
ab D1 Da D1 Da T M
aa M
ba + D1 Da
¼M
b
b
b
(11.6.15)
T
bb D1 Da
M
b
Similarly, the elastic energy of the system is
1 u
U ¼ uT K
2
1
ua
¼ uTa TT KT
2
(11.6.16)
1 ^
¼ uTa K
ua
2
where
^ ¼ TT KT
K
(11.6.17)
is the L L transformed reduced stiffness matrix.
as M
in Eq. (11.6.14) we obtain from Eq. (11.6.17)
Partitioning K
^¼ I
K
D1
b Da
"
#(
aa K
ab
I
K
)
ba K
bb
D1
K
b Da
T
1
1
ba + D1 Da
aa K
ab D Da D Da K
¼K
b
b
b
(11.6.18)
T
bb D1 Da
K
b
442 PART
II Multi-degree-of-freedom systems
The transformed damping matrix is established by considering the virtual
work of the nodal damping forces
uT f D
dWD ¼ d
¼ d
uT Cu_
u_ a
¼ d
uTa TT CT
(11.6.19)
^ u_ a
¼ d
uTa C
Hence, we have
^ ¼ TT CT
C
(11.6.20)
Similarly, we establish the transformed vector of the external loads that is,
dWp ¼d
uT pðt Þ
¼d
uTa TT pðt Þ
(11.6.21)
^ ðt Þ
¼d
uTa p
Hence
^ðt Þ ¼TT pðt Þ
p
¼
pa ðt Þ D1
b Da
T
pb ðt Þ
(11.6.22)
11.7 Axial constraints in the plane frame
In Section 11.5, we examined the flexural vibrations of plane frames. We saw
that the mass, stiffness, and damping matrices of the elements have dimensions
4 4 if the axial deformation is ignored. Using these matrices in Example
11.5.2, we formulated the equations of motion of a plane frame with elements
parallel to the global axes. Obviously, the procedure we used is tedious even for
simple frames having elements parallel to the global axes because its implementation requires the inspection of the user, which, however, does not facilitate the
automation of the method. The difficulty becomes practically insurmountable
when the frame has elements inclined with respect to the global axes. Neglecting the axial deformation implies the equality of the displacements u1 and u4
in local axes (see Fig. 11.7.1). This annuls the nodal elastic and damping forces
in those directions, ðfS1 ¼ fS4 ¼ 0, fD1 ¼ fD4 ¼ 0Þ, but not the inertial forces
ðfI 1 , fI 4 6¼ 0Þ. This becomes evident by setting u1 ¼ u4 in Eqs. (11.3.14),
(11.3.22), (11.3.39), and taking into account Eqs. (11.3.13a), (11.3.21),
(11.3.37a), respectively. The inertial behavior of the element in the direction
of its local x axis can be simulated with that of a one-dimensional rigid body
within the frame. To tackle this problem, we must start by considering the frame
elements with the six degrees of freedom.
The finite element method Chapter
11
443
Fig. 11.7.1 Degrees of freedom of an element with flexural vibrations, u4 ¼ u1 .
We consider the beam element with its six degrees of freedom with respect
to global axes, as shown in Fig. 11.7.1. If we want to study only the flexural
vibrations of the frame, we must neglect the axial deformation of the elements.
This entails the imposition of constraints on the translational degrees of freedom. These constraints result from the fact that the projections of end displacements on the element axis must be equal [7,8], that is,
u1 cos f + u2 sin f ¼ u4 cos f + u5 sin f
(11.7.1)
The degrees of freedom of a frame with n nodes are N ¼ 3n, of which
Nt ¼ 2n are translational degrees of freedom. However, these degrees are not
independent because they are subject to constraints of the form (11.7.1). The
maximum number of constraints is at first sight Kmax ¼ Ne , where Ne is the
number of the elements of the frame. Therefore the independent translational
degrees of freedom are 2n Ne 0. Obviously, this is true if the layout of
the elements is such that each of them constrains only one degree of freedom.
Nevertheless, this is not always true as one can readily understand from the
frames of Fig. 11.7.2. More specifically, the frame of Fig. 11.7.2a has Nt ¼ 4
translational degrees of freedom. The three elements impose three axial constraints, that is, it is Nc ¼ Ne ¼ 3, where Nc denotes the number of constraints.
Thus the remaining translational degrees are: Nt Nc ¼ 4 3 ¼ 1. In the frame
of Fig. 11.7.2b, the four elements impose four axial constraints, that is
Nc ¼ Ne ¼ 4, hence Nt Nc ¼ 4 4 ¼ 0, which means that all translational
degrees of freedom are constrained. Further, in the frame of Fig. 11.7.2c, the
fifth element is redundant because the translational degrees of freedom are
again four, Nc ¼ 4, while it is Ne ¼ 5. In this case, we say that the frame is overconstrained. The frame of Fig. 11.7.2d has four elements, that is, Ne ¼ 4. However, the number of constraints is three, Nc ¼ 3, because, in this layout, the
elements 3 and 4 lie on a straight line and consequently cannot constrain the
normal displacement un . Finally, the number of constraints in the frame of
Fig. 11.7.2e is Nc ¼ 4, even it is Ne ¼ 5.
From the previous examples, it becomes apparent that the number of elements Ne cannot determine the number of active axial constraints. In simple
444 PART
II Multi-degree-of-freedom systems
frames, this can be done by inspection. However, a general procedure should be
developed that would allow determining automatically the number of
active constraints, rejecting the redundant ones and separating the independent
degrees of freedom from the dependent ones. This procedure is presented
right below.
The constraint equation (11.7.1) for the typical element e starting at the point
j and ending at the point k is written
u3j2 cos fe + u3j1 sin fe u3k2 cos fe u3k1 sin fe ¼ 0
(11.7.2)
which for e ¼ 1, 2, …, Ne yields
D
ut ¼ 0
(11.7.3)
where D is an Ne Nt matrix and ut the vector of the translational degree of
freedom. Matrix D can be formulated by establishing appropriate assembly
matrices.
As we will show below, the rank of the matrix D will determine the active
constraints.
Eq. (11.7.3) is written
½ Da Db ua
ub
¼0
(11.7.4)
or
Da ua + Db ub ¼0
(11.7.5)
where ua the independent and ub the dependent displacements.
Eq. (11.7.5) yields
a
ub ¼ D1
b Da u
(11.7.6)
The foregoing equation points out that the selection of ub cannot be arbitrary, but such to ensure invertibility of the matrix Db , that is, det ðDb Þ 6¼ 0.
Hence the dimensions of Db , which is a submatrix of D, cannot be greater
than the rank of D. The determination of the rank can be achieved using the
Gauss-Jordan elimination [9]. This method is a variation of the Gauss elimination. The main difference is that by the elimination procedure, zeros are generated both below and above each pivot element. Besides, all rows are normalized
via division by the pivoting elements. Therefore, the resulting matrix that
determines the rank of the constraint matrix is a unit matrix and not an upper
diagonal matrix. This matrix is also known as the reduced row echelon form.
There are ready-to-use computer functions that transform a matrix to this form,
for example, the function R ¼ rref (A) of MATLAB transforms the matrix A
into a reduced row echelon matrix R.
The finite element method Chapter
(a)
(b)
(c)
(d)
11
445
(e)
Fig. 11.7.2 Frames with axial constraints.
This method leads to an automatic selection of the independent node translations. The procedure is better understood by applying it to the frames of Fig. 11.7.2.
Frame of Fig. 11.7.2a
The Cartesian coordinates of the nodes are 1 ð0, 0Þ, 2 ð0, 3Þ, 3 ð4, 4Þ, 4 ð6, 0Þ.
The computed geometrical data of the elements are shown in Table 11.7.1.
446 PART
II Multi-degree-of-freedom systems
TABLE 11.7.1 Geometrical data of the elements of the frame in Fig. 11.7.2a.
Number of
element e
Node j
Node k
D
x
1
1
2
0
2
2
3
3
3
4
D
y
Le
cosfe
3
3
0
1
4
1
4.123
0.970
0.243
2
4
4.472
0.447
0.894
The constraint equations are three, namely
8 9
2
3> u1 > 8 9
>
>
0
1
0
0
>
= <0=
< >
6
7 u2
¼ 0
4 0:970 0:243 0:977 0:243 5
>
> : ;
> u4 >
0
0
0
0:447 0:894 >
;
: >
u5
sin fe
(11.7.7)
Hence
3
0
1
0
0
7
6
D ¼ 4 0:970 0:243 0:970 0:243 5
0
0
0:447 0:894
2
(11.7.8)
Gauss-Jordan elimination. Before starting the elimination, it is convenient
to interchange the first with second row. Thus, we have
2
3
0:970 0 0:970 0:243
40
5
1 0
0
0
0 0:447 0:894
Dividing the second row by 0:970 gives
2
3
1 0 1
0:250
40 1 0
5
0
0 0 0:447 0:894
Then, dividing the third row by 0:447 gives
2
3
1 0 1 0:250
40 1 0 0
5
0 0 1 2:000
The finite element method Chapter
Finally, adding the third row to the first row gives
2
3
1 0 0 2:225
40 1 0 0
5
0 0 1 2:000
447
11
(11.7.9)
Therefore, the rank of the matrix D is 3, hence Nt ¼ 4, Nc ¼ 3, N ¼ 4 3 ¼ 1.
We can take u1 as the independent. Thus, referring to Eq. (11.7.7) we have
2
3
1:000 0
0
Db ¼ 4 0:243 0:977 0:243 5
(11.7.10a)
0
0:447 0:894
2
3
0
Da ¼ 4 0:970 5
(11.7.10b)
0
Note that we can also use Eq. (11.7.9) instead of Eq. (11.7.7) to obtain
2
3
1 0 0
(11.7.11a)
Db ¼ 4 0 1 0 5
0 0 1
2
3
0
Da ¼ 4 0:856 5
(11.7.11b)
0:572
because Eq. (11.7.7) are homogeneous and the Gauss-Jordan elimination does
not change the order of the unknowns.
Apparently, performing Gauss-Jordan elimination by inspection is a tedious
task and it becomes too complicated for large matrices. However, computer
codes have been developed to apply this technique and ready-to-use programs
are available.
Frame of Fig. 11.7.2b
The Cartesian coordinates of the nodes are 1 ð0, 0Þ, 2 ð0, 3Þ, 3 ð4, 4Þ, 4 ð6, 0Þ,
5ð4, 6Þ. The computed geometrical data of the elements are shown in
Table 11.7.2.
TABLE 11.7.2 Geometrical data of the elements of the frame in Fig. 11.7.2b.
Number of
element e
Node j
Node k
D
x
1
1
2
0
2
2
3
3
3
4
3
D
y
Le
cosfe
sinfe
3
3
0
1
4
1
4.123
0.970
0.243
4
2
4
4.472
0.447
0.894
5
0
2
2
0
1
448 PART
II Multi-degree-of-freedom systems
The constraint equations are four, namely
2
38 9 8 9
0
1:000 0
0
>
>
> >
>0>
> u1 >
6 0:970 0:243 0:970 0:243 7< u2 = < 0 =
6
7
¼
40
0>
0
0:447 0:894 5>
> >
>
> u4 >
;
;
: >
:
0
u5
0
0
0
1
Hence
2
0
6 0:970
D¼6
40
0
(11.7.12)
3
1:000 0
0
0:243 0:970 0:243 7
7
0
0:447 0:894 5
0
0
1
The Gauss-Jordan elimination gives
2
1 0
60 1
D¼6
40 0
0 0
0
0
1
0
(11.7.13)
3
0
07
7
05
1
(11.7.14)
the rank of which is 4, thus Nt ¼ 4, Nc ¼ 4, N ¼ 4 4 ¼ 0. Therefore, there is no
translational independent variable, which means that the number of active constrains is equal to the number of degrees of freedom.
Frame of Fig. 11.7.2c
The Cartesian coordinates of the nodes are 1 ð0, 0Þ, 2 ð0, 3Þ, 3 ð4, 4Þ, 4 ð6, 0Þ,
5ð4, 6Þ, 6ð6, 4Þ. The computed geometrical data of the elements are shown in
Table 11.7.3.
TABLE 11.7.3 Geometrical data of the elements of the frame in Fig. 11.7.2c.
Number of
element e
Node j
Node k
D
x
1
1
2
0
2
2
3
3
3
4
5
D
y
Le
cos fe
3
3
0
1
4
1
4.123
0.970
0.243
4
2
4
4.472
0.447
0.894
3
5
0
2
2
0
1
3
6
2.5
0
2.5
1
0
The constraint equations are five, namely
8 9
2
3
8 9 >0>
0
1:000 0
0
> >
>
>0>
> >
> u1 >
6 0:970 0:243 0:970 0:243 7>
=
< >
= >
< u >
6
7>
2
6
7
¼ 0
0
0:447 0:894 7
60
6
7> u4 >
>
>0>
> >
>
40
0
0
1:000 5>
>
> >
; >
: >
>
;
: >
u5
0
0
0
1:000 0
sinfe
(11.7.15)
The finite element method Chapter
Hence
2
0
1:000
0
0
11
449
3
6
7
6 0:970 0:243 0:970 0:243 7
6
7
D¼6
0
0:447 0:894 7
60
7
6
7
0
0
1:000 5
40
0
0
1:000
0
(11.7.16)
The Gauss-Jordan elimination gives
2
1
60
6
D¼6
60
40
0
0
1
0
0
0
0
0
1
0
0
3
0
07
7
07
7
15
0
(11.7.17)
the rank of which is 4. Therefore, there is no translational independent variable.
Apparently, the fifth constraint is redundant.
Frame of Fig. 11.7.2d
The Cartesian coordinates of the nodes are 1 ð0, 0Þ, 2 ð0, 3Þ, 3 ð4, 4Þ, 4 ð6, 0Þ,
5ð3, 6Þ. The computed geometrical data of the elements are shown in
Table 11.7.4.
TABLE 11.7.4 Geometrical data of the elements of the frame in Fig. 11.7.2d.
Number of
element e
Node j
Node k
1
1
2
D
x
D
y
2
0
3
3
0
1
2
3
4
1
4.123
0.970
0.243
3
3
4
2
4
4.472
0.447
0.894
4
3
5
1
2
2.236
0.447
0.894
Le
The constraint equations are four, namely
2
38 9 8 9
0
u
0
1:000 0
0
>
>
> >
> >
> 1>
=
<0>
= >
< u >
6 0:970 0:243 0:970 0:243 7>
2
6
7
¼
6
7
40
u4 >
0:447 0:894 5>
0
>
>0>
> >
>
>
;
: >
; >
: >
0
u5
0
0
0:447 0:894
cosfe
sin fe
(11.7.18)
450 PART
II Multi-degree-of-freedom systems
Hence
2
3
0
1:000 0
0
6 0:970 0:243 0:970 0:243 7
6
7
D¼6
7
40
0
0:447 0:894 5
0
0
0:447 0:894
(11.7.19)
The Gauss-Jordan elimination gives
2
3
1 0 0 2:250
60 1 0 0
7
6
7
D¼6
7
4 0 0 1 2:000 5
0 0 0
(11.7.20)
0
the rank of which is 3. Therefore, there is one independent displacement, which
means that the number of active constraints is three.
Frame of Fig. 11.7.2e
The Cartesian coordinates of the nodes are 1 ð0, 0Þ, 2 ð0, 3Þ, 3 ð4, 4Þ, 4 ð6, 0Þ,
5ð3, 6Þ, 6(6,4). The computed geometrical data of the elements are shown
in Table 11.7.5.
TABLE 11.7.5 Geometrical data of the elements of the frame in Fig. 11.7.2e.
Number of
element e
Node j
Node k
1
1
2
D
x
D
y
2
0
3
3
0
1
2
3
4
1
4.123
0.970
0.243
3
3
4
2
4
4.472
0.447
0.894
4
3
5
1
2
2.236
0.447
0.894
5
3
6
2
0
2
Le
cos fe
1
sinfe
0
The constraint equations are five, namely
8 9
0>
38 9 >
>
>
>
>
u
>
>
>
>
1
>
>
>
>
>
>
>
>
>
>
>
7>
6
0
>
>
>
>
>
>
>
6 0:970 0:243 0:970 0:243 7>
>
>
>
=
7< u2 = < >
6
7
6
¼ 0
0:447 0:894 7
0
60
>
7>
6
>
> >
> >
>
>
>
60
>
>
> >
> u4 >
0
0:447 0:894 7
>
>
>
5>
4
>
>
>
0
>
>
>
>
>
>
: ; >
>
>
>
u5
0
0
1
0
;
: >
0
2
0
1:000
0
0
(11.7.21)
The finite element method Chapter
Hence
2
0
6 0:970
6
D¼6
60
40
0
3
1:000 0
0
0:243 0:970 0:243 7
7
0
0:447 0:894 7
7
0
0:447 0:894 5
0
1
0
The Gauss-Jordan elimination gives
2
1 0
60 1
6
D¼6
60 0
40 0
0 0
0
0
1
0
0
3
0
07
7
07
7
15
0
11
451
(11.7.22)
(11.7.23)
the rank of which is 4. Therefore, there is no translational independent variable.
Apparently, the fifth constraint is redundant.
Example 11.7.1 Formulate the equations of motion of the frame in Example
11.3.2, when the axial constraints of the elements are taken into account.
Fig. E11.8 The translational degrees of freedom.
Solution
We note that in this example, two elements of the frame are not parallel to the
global axes. Therefore, the procedure for flexural vibrations described in
Example 11.5.2 is not convenient to formulate the equation of motion of the
structure. Consequently, the general procedure for the FEM analysis of a plane
frame should be employed.
First, the mass and stiffness matrices and the vector of the nodal loads
with respect to the seven free degrees of freedom are computed. Then, the rotational degrees of freedom u3 , u6 , u7 are eliminated by static condensation. This
results in a system with four translational degrees of freedom (Fig. E11.8).
452 PART
II Multi-degree-of-freedom systems
Subsequently, the three axial constraints are imposed as described in
Section 11.7, which leads to the elimination of three translational displacements. Thus, the final equation will have only one degree of freedom. The relevant matrices have been computed in Example 11.4.1, namely
2
7:5
60
6
Mtt ¼ rA6
40
0
0
7:5
0
0
0
0
15
0
3
2
3
0
3222:71 2389:02 3205:83 2387:19
7
0 7
EI 6
7
6 2389:02 4308:10 2396:65 1810:18 7
7, Ktt∗ ¼
6
7
5
4
125
0
3205:83 2396:65 4101:75 1196:10 5
15
2387:19 1810:18 1196:10 3412:98
8
9
2:0P ðt Þ >
>
>
>
>
< 0:08P ðt Þ >
=
∗
pt ðt Þ ¼
>
>
0
>
>
>
>
:
;
1:08P ðt Þ
The translational displacements are shown in Fig. E11.8. Obviously, the
active constraints are three. Nevertheless, we apply the procedure described
in Section 1.7.
TABLE E11.3 Geometrical data of the elements.
Number of
Element e
Node j
Node k
D
x
1
1
2
2
2
3
3
D
y
Le
0
5
5
0
1
3
4
5
5
0.8
0.6
4
6
8
10
0.6
0.8
cos fe
sinfe
On the base of the geometrical data shown in Table E11.3, the constraint
equations are
8 9 8 9
3> u1 > > 0 >
2
>
> >
>
0 1:0 0
0
>
=
<0>
= >
< >
7 u2
6
¼
0:8
0:6
0:8
0:6
5
4
>
>
>0>
> >
> u4 >
0 0
0:6 0:8 >
;
: >
; >
: >
0
u5
hence
2
0
1:0
0
0
0
0
0:6 0:8
3
6
7
D ¼ 4 0:8 0:6 0:8 0:6 5
Using the Gauss-Jordan Elimination, we find out that the rank of the
matrix D is 3. Thus, the number of independent degrees of freedom is equal
The finite element method Chapter
11
453
to 4 3 ¼ 1. Taking u1 as the independent displacement, we can write the
constraint equations as
2
38 9 8 9
2 3
e2 = < 0 =
1:0 0
0
0
<u
4 0:8 5 u
e4 ¼ 0
e1 + 4 0:6 0:8 0:6 5 u
: ; : ;
e5
0
0:6 0:8
u
0
0
hence
2
3
2
3
0
1:0 0
0
Da ¼ 4 0:8 5, Db ¼ 4 0:6 0:8 0:6 5
0
0
0:6 0:8
Consequently, Eq. (11.6.10) gives
T¼
I
D1
b Da
9
8
1 >
>
>
>
>
=
<0 >
¼
>
0:64 >
>
>
>
>
;
:
0:48
The matrices Mtt , Ktt∗ , pt∗ ðt Þ are modified due to the axial constraints
by virtue of Eqs. (11.6.13), (11.6.17), (11.6.22). Thus, we have
M ¼ TT Mtt T ¼ 17:10rA
K ¼ TT Ktt∗ T ¼ 0:2306EI
pðt Þ ¼ TT pt∗ ðt Þ ¼ 1:4816P ðt Þ
Therefore the equation of motion of the frame is
17:10rAu€1 + 0:2306EI u1 ¼ 1:4816P ðt Þ
From the solution of the foregoing equation, we determine the displacement
e1 . Subsequently, the remaining translational displacements are computed from
u
the relation
e2 u
e4 u
e5 gT ¼ T
u1
ub ¼ f u
Then the rotations are computed from Eq. (11.4.2b)
uq ¼ K1
qq Kqt ut
11.8 The finite element method for the plane grid
11.8.1 Properties of the plane grid element
We consider the grid of Fig. 11.8.1, whose plane coincides with the global x y
plane. The axis of the e element defines its local x axis. The local y axis lies in
the x y plane while the local z axis coincides with the global axis z
454 PART
II Multi-degree-of-freedom systems
(Fig. 11.8.2). The element cross-sections are symmetric with respect to the xz
plane. The nodal loads consist of forces in the z direction and of moments
about x and y axes (Fig. 11.8.1). The elements may be subjected to loads
normal to the xy plane and moments about x and y axes. Moreover, the axial
deformation of the elements is ignored. Under these assumptions, the nodes of
the grid undergo rotations about the global axes x, y and a translational
displacement normal to the x y plane. These global displacements produce
the rotations u1 , u4 about the local x axis, the rotations u2 , u5 about the local
y axis, and the translational displacements u3 , u6 in the direction of the z axis
at the ends of the element (see Fig. 11.8.2).
Thus the element displacement vector is
u ¼ f u1 u2 u3 u4 u5 u6 g
(11.8.1)
The displacements u1 , u4 produce torsion of the element while the displacements u2 , u3 , u5 , u6 produce bending in x z plane. The bending deformation of the element is given by Eq. (11.3.3) after changing appropriately the
role of the shape functions i ðx Þ to correspond to the nodal displacements.
Thus, we have
v ðx, t Þ ¼ u2
2 ðx Þ + u3 3 ðx Þ + u5 5 ðx Þ + u6 6 ðx Þ
(11.8.2)
where now
2 ðx Þ ¼ L
x 2x 2 + x 3
3 ðx Þ ¼ 1 3x
5 ðx Þ ¼ L
+ 2x 3
(11.8.3b)
x 2 + x3
(11.8.3c)
6 ðx Þ ¼ 3x
Fig. 11.8.1 Plane grid.
2
2
(11.8.3a)
2x3
(11.8.3d)
The finite element method Chapter
11
455
Fig. 11.8.2 Nodal displacements of the grid element.
The sign of 2 ðx Þ and 5 ðx Þ has changed because the directions of the
corresponding displacements have changed with reference to Fig. 11.3.1.
In order to determine the torsional deformation of the element, we consider elements of constant cross-section. Regarding elements of variable cross-section,
we approximate them by a series of rigidly interconnected elements of constant
cross-section. This avoids the use of advanced methods employed for torsional
analysis to elements of variable cross-section.
The rotation qðx, t Þ about the x due to the rotations u1 and u4 is written in the
form
qðx, t Þ ¼ u1
1 ðx Þ + u 4 4 ðx Þ
(11.8.4)
The shape functions 1 ðx Þ and 4 ðx Þ can be obtained from the solution of
the equilibrium equation of beam with constant cross-section subjected to the
end rotations u1 , u4 . Thus, referring to Fig. 11.8.3, we have
Fig. 11.8.3
dM t
¼0
dx
(11.8.5)
where Mt is the twisting moment given by
Mt ¼ GI t
dq
dx
(11.8.6)
in which G ¼ E=2ð1 + n Þ is the shear modulus and It the torsional constant.
456 PART
II Multi-degree-of-freedom systems
Substituting Eq. (11.8.6) into Eq. (11.8.5) yields
d 2q
¼0
dx 2
(11.8.7)
q ¼ c1 x + c2
(11.8.8)
from which we obtain
The boundary conditions for Eq. (11.8.7) are qð0Þ ¼ u1 and qðLÞ ¼ u4 , which
are applied to Eq. (11.8.8) and give the arbitrary constants
u4 u1
c1 ¼
c2 ¼ u4 ,
L
Introducing these values in Eq. (11.8.8) yields
qðx, t Þ ¼ u1 ð1 xÞ + u4 x, x ¼ x=L
(11.8.9)
Therefore, the shape functions defined by Eq. (11.8.4) are
1 ðx Þ ¼ 1 x,
4 ðx Þ ¼ x
(11.8.10)
The value of It obtained on the base of the Saint-Venant theory depends on
the shape of the cross-section of the bar. For a circular cross-section, It is equal
to the polar moment of inertia with respect to its center. For a cross-section of
arbitrary shape, It is calculated from the expression [4]
Z
2
It ¼
y + z 2 zfy + yfz dydz
(11.8.11)
A
where f ¼ fðy, z Þ is the warping function of the cross-section with respect to
the twist center and is obtained from the solution of the following boundary
value problem
∂2 f ∂2 f
+
¼0
∂y 2 ∂z 2
y, zA
(11.8.12a)
∂f
¼ zn y yn z
∂n
y, zG
(11.8.12b)
in which A is the cross-section of the bar and G its boundary; ny , nz are the
components of the unit vector normal to the boundary of the cross-section.
For the solution of the torsion problem, the reader is advised to look in the
literature. A detailed study of the Saint-Venant torsion problem is presented
in [4]. The Saint-Venant theory under certain support conditions, which
approach the Saint-Venant assumptions, gives acceptable results. This, however, does not always hold in actual structures. It is well known that the response
of a beam under general twisting loading and general boundary conditions leads
to the problem of nonuniform torsion. This problem has been extensively examined in the literature, for example, [10, 11].
The finite element method Chapter
11
457
Fig. 11.8.4 Positive directions of the nodal forces of the grid element.
The equivalent nodal forces of the grid element will be obtained using the
Lagrange equations. Fig. 11.8.4 shows the positive directions of element nodal
forces. The actions in the directions of u1 , u2 , u4 , and u5 are moments while in
the directions of u3 and u6 are forces.
(i) Nodal elastic forces and stiffness matrix of the grid element
The elastic energy is due to the torsional deformation of the element given
by the expressionf
1
2
Ut ¼ GI t ðq0 Þ L
2
(11.8.13)
as well as to the bending deformation of the element given by Eq. (10.5.2),
that is,
f. The elastic energy due to torsion is given by
Ut ¼
Taking into account that
1
2
Z
V
G g 2yx + g 2zx dV , y,zA
g yx ¼ q0 fy z , g zx ¼ q0 ðfz + y Þ
(1)
(2)
and that q0 ¼ dq=dx is constant, Eq. (1) becomes
1
2
Ut ¼ Gq0 L
2
Z
Z
A
f2y + f2z zfy + yfz dA +
zfy + yfz + y 2 + z 2 dA
(3)
A
The second integral is equal to It while the first one is equal to zero. Indeed, the first integral is
transformed as
Z h
Z
i
f2y + f2z dA +
ðzfÞy + ðyfÞz dA
R¼
A
A
Z
Z
Z
(4)
f fyy + fzz dA + ffn dA f zn y yn z dA
¼
A
A
A
¼0
The foregoing result is obtained by applying the Gauss-Green theorem to the first term and the
Gauss divergence theorem to the second term and taking into account that fyy + fzz ¼ 0 and
fn ¼ zn y yn z . Hence, Eq. (3) becomes
1
1
2
Ut ¼ GI t ðq0 Þ L ¼ Mt q0 L
2
2
(5)
458 PART
II Multi-degree-of-freedom systems
Ub ¼
1
2
Z
L
EI ½v 00 ðx, t Þ dx
2
(11.8.14)
0
Thus, the total elastic energy is
U ¼Ut + Ub
1
1
2
¼ GI t ðq0 Þ L +
2
2
Z
L
EI ½v 00 ðx, t Þ dx
2
(11.8.15)
0
which by virtue of Eqs. (11.8.2), (11.8.4) becomes
1
U ðu1 , u2 , …, u6 Þ ¼ GI t u1 01 ðx Þ + u4 04 ðx Þ L
2
Z
1 L +
EI u2 002 ðx Þ + u3 003 ðx Þ + u5
2 0
2
00
00
5 ðx Þ + u6 6 ðx Þ dx
(11.8.16)
Differentiation of the foregoing equation with respect to ui , i ¼ 1, 2, …, 6
yields the nodal elastic forces. Thus, we obtain
9 2
8
38 9
k11 0 0 k14 0 0 > u1 >
fS1 >
>
>
>
> >
>
>
6 0 k k 0 k k 7>
>
>
>u >
>
>
fS2 >
22
23
25
26 7>
>
>
>
>
6
2>
>
>
>
>
>
>
>
7
6
=
< f = 6 0 k k 0 k k 7< u >
32
33
35
36
S3
3
7
¼6
(11.8.17)
7
6
>
u >
f > 6 k41 0 0 k44 0 0 7>
>
>
> 6
>
> S4 >
> 4>
>
>
>
>
7
>
>
>
>
>
4 0 k52 k53 0 k55 k56 5>
u5 >
fS5 >
>
>
>
>
>
>
>
;
;
:
: >
u6
fS6
0 k62 k63 0 k65 k66
in which
Z
L
kij ¼
Z
kij ¼
GI t
0
L
EI
0
0
0
i ðx Þ j ðx Þdx,
00
00
i ðx Þ j ðx Þdx,
i, j ¼ 1, 4
(11.8.18a)
i, j ¼ 2, 3, 5, 6
(11.8.18b)
or
f eS ¼ ke ue
(11.8.19)
u are the vectors of the nodal elastic forces and nodal displacements,
where
respectively, and ke the stiffness matrix of the e grid element. Further, substituting Eqs. (11.8.3a)–(11.8.3d) and (11.8.10) into Eqs. (11.8.18a), (11.8.18b) and
performing the integration gives
f eS ,
e
The finite element method Chapter
2
GI t
0
6 L
6
6
4EI
6 0
6
L
6
6
6EI
6 0
2
6
L
ke ¼ 6
6 GI t
6
0
6 L
6
6
2EI
6 0
6
L
6
4
6EI
0
L2
0
6EI
2
L
12EI
L3
0
6EI
L2
12EI
3
L
GI t
L
0
0
GI t
L
0
0
11
459
3
0
0
7
7
2EI
6EI 7
7
L
L2 7
7
6EI
12EI 7
2 3 7
L
L 7
7
7
7
0
0
7
7
4EI
6EI 7
7
7
L
L2 7
6EI
12EI 5
L2
L3
(11.8.20)
The superscript e has been dropped from Ite , I e , and Le for the sake of the
simplicity of the expressions.
(ii) Nodal inertial forces and mass matrix of the grid element
The equivalent inertial nodal forces result again on the base of two different
assumptions regarding the mass distribution along the length of the element,
namely the consistent mass assumption and the lumped mass assumption. The
inertial mass matrices resulting from both assumptions are derived right below.
(a) Consistent mass matrix
During the motion, the infinitesimal mass m ðx Þdx of the element undergoes the
two displacements, that is, the rotation qðx, t Þ and the transverse displacement
v ðx, t Þ. Therefore, the kinetic energy of the grid element will be given by the
expression
Z
o
2
1 Ln
I0 ðx Þ q_ ðx, t Þ + m ðx Þ½v_ ðx, t Þ2 dx
(11.8.21)
T¼
2 0
where I0 ðx Þ is the polar moment of inertia of the mass m ðx Þ ¼ rAðx Þ with
respect to the axis of the element, which is assumed to coincide with the
twist axis.
Using Eqs. (11.8.2), (11.8.4), Eq. (11.8.21) becomes
Z
h
i
1 L
I0 ðx Þ ½u_ 1 1 ðx Þ + u_ 4 4 ðx Þ2 dx
T ðu_ 1 , u_ 2 , …, u_ 6 Þ ¼
2 0
Z
1 L
m ðx Þ½u_ 2 2 ðx Þ + u_ 3 3 ðx Þ + u_ 5 5 ðx Þ + u_ 6 6 ðx Þ2 dx
+
2 0
(11.8.22)
460 PART
II Multi-degree-of-freedom systems
The nodal inertial forces result from Eq. (11.2.9) for i ¼ 1, 2, …, 6. Thus,
after performing the differentiations, we obtain
38 9
8 9 2m 0
0
m14 0
0
fI 1 >
11
>
> u€1 >
>
>
>
6
7>
>
>
>
>
>
>
>
>
6
7
>
>
>
>
0
m
m
0
m
m
f
€
>
>
>
u
22
23
25
26
I2 >
2>
>
>
>
6
7
>
>
>
> 6
>
> >
> >
7>
>
>
< fI 3 = 6 0
<
7
m32 m33 0
m35 m36 7 u€3 =
6
¼6
(11.8.23)
7
>
>
6 m41 0
u€4 >
fI 4 >
0
m44 0
0 7
>
>
>
>
>
>
>
>
6
7
>
>
>
>
>
>
>
7>
> 6
>
>
>
>
>
> 6
>
> fI 5 >
> u€5 >
0
m52 m53 0
m55 m56 7
>
>
>
>
4
5
>
>
>
;
: ;
: >
u€6
fI 6
0
m62 m63 0
m65 m66
where
Z
L
mij ¼
Z
mij ¼
I0 ðx Þ i ðx Þ j ðx Þdx
i, j ¼ 1, 4
(11.8.24a)
i, j ¼ 2, 3, 5, 6
(11.8.24b)
0
L
m ðx Þ i ðx Þ j ðx Þdx
0
Eq. (11.8.23) is written in matrix form
€e
f eI ¼ me u
(11.8.25)
€ are the vectors of the nodal inertial forces and the nodal acceleru
where
ations, respectively, and me the mass matrix of the e grid element.
For an element with constant mass, m ðx Þ ¼ rA, rI ðx Þ ¼ rI 0 , Eqs. (11.8.24a),
(11.8.24b) are integrated analytically and the mass matrix becomes
3
2
140rg2
0
0
70rg2
0
0
7
6
60
22L 0
3L2 13L 7
4L2
7
6
7
6
7
6
e 60
22L
156 0
13L 54 7
m
7
6
(11.8.26)
me ¼
7
420 6
6 70rg2
0
0
140rg2
0
0 7
7
6
7
6
2
2
60
4L 22L 7
3L 13L 0
5
4
f eI ,
e
0
13L
54
0
22L
156
p
ffiffiffiffiffiffiffiffiffiffi
where m e ¼ rALe is the mass of the element and rg ¼ I0 =A the radius of
gyration of the cross-section.
(b) Lumped mass matrix
According to this assumption, the mass of the element is concentrated at its
nodes by static considerations, that is, they are obtained as the reactions of a
simply supported beam under the load m ðx Þ (see Fig. 11.2.3). Thus, we have
The finite element method Chapter
Z
m1 ¼
0
L
m ðx Þð1 x Þdx
Z
m2 ¼
L
m ðx Þxdx
11
461
(11.8.27a)
(11.8.27b)
0
Therefore, the kinetic energy of the element is
1
1
T ¼ m1 u_ 3 2 + m2 u_ 6 2
2
2
(11.8.28)
which by virtue of Eq. (11.2.9) for i ¼ 1, 2, …,6 yields
8 9 2
0
fI 1 >
>
>
>
>
> 60
>f >
>
>
I2 >
>
> 6
>
>
= 6
<f >
60
I3
¼6
> 6
> fI 4 >
>
60
>
>
>
>
6
>
>
>
>
fI 5 > 4 0
>
>
;
: >
0
fI 6
0
0
0 0
0
0
0 0 0
m1 0 0
0
0
0 0
0
0
0
0
0 0
0 0
38 9
u€1 >
>
>
>
>
>
> u€ >
>
0 7
2>
>
7>
>
>
>
7>
<
0 7 u€3 =
7
0 7
>
> u€4 >
7>
>
>
>
7>
>
>
>
>
5
€
0 >
u
5
>
> >
;
:
m2
u€6
0
(11.8.29)
For an element of constant cross-section, m ðx Þ ¼ rA, the mass matrix
becomes
3
2
0 0 0 0 0 0
60 0 0 0 0 07
7
6
7
6
e 60 0 1 0 0 07
m 6
7
me ¼
(11.8.30)
7
2 6
60 0 0 0 0 07
7
6
40 0 0 0 0 05
0 0 0 0 0 1
where m e ¼ rAe Le is the mass of e the element.
(iii) Nodal damping forces and damping matrix of the grid element
Following a procedure analogous to that for the plane frame, we can construct a damping matrix for the grid element. However, as we will see in
Section 12.11, the practical significance of the damping matrices constructed
in this way is limited and therefore, for reasons of saving space, we avoid its
derivation here.
(iv) Equivalent nodal loads of the grid element
The equivalent nodal loads can be evaluated in two ways: (a) as generalized
forces in the direction of the nodal displacements ue and (b) as static equivalent
forces. In the first case, the shape functions of the stiffness matrix are used and
462 PART
II Multi-degree-of-freedom systems
the resulting vector of the equivalent nodal forces is referred to as a consistent
nodal load vector.
(a) Consistent nodal load vector
We assume that the element is loaded by the transverse load pz ðx, t Þ and the
twisting moment mx ðx, t Þ. Referring to Fig. 11.8.5, the virtual work produced
during the deformation of a beam element dx is
Fig. 11.8.5 Loading and equivalent nodal loads of a plane grid element.
Z
p
dWnc
¼
L
Z
L
mx ðx, t Þdqðx, t Þdx +
0
pz ðx, t Þdv ðx, t Þdx
(11.8.31)
0
which by virtue of Eqs. (11.8.2), (11.8.4) gives
p
¼ p1 ðt Þdu1 + p2 ðt Þdu2 + p3 ðt Þdu3 + p4 ðt Þdu4 + p5 ðt Þdu5 + p6 ðt Þdu6
dWnc
(11.8.32)
where
Z
L
pi ðt Þ ¼
mx ðx, t Þ i ðx Þdx,
i ¼ 1, 4
(11.8.33a)
i ¼ 2, 3, 5, 6
(11.8.33b)
0
Z
L
pi ðt Þ ¼
pz ðx, t Þ i ðx Þdx,
0
Consequently, the resulting vector of equivalent nodal loads of the e
element is
9
8
p1 ðt Þ >
>
>
>
>
>
>
>
>
p2 ðt Þ >
>
>
>
>
>
=
< p ðt Þ >
3
e
(11.8.34)
p ðt Þ ¼
>
>
p
ð
t
Þ
4
>
>
>
>
>
>
>
>
>
p5 ðt Þ >
>
>
>
>
;
:
p6 ðt Þ
The finite element method Chapter
11
463
(b) Statically equivalent nodal load vector
In this case, we assume that the grid element is a simply supported beam
under the load pz ðx, t Þ and the torsional moment mx ðx, t Þ. Actually, the components p2 ðt Þ and p5 ðt Þ are zero. The remaining components are obtained as the
reactions of the simply supported beam
Z L
mx ðx, t Þ i ðx Þdx, i ¼ 1, 4
(11.8.35a)
pi ðt Þ ¼
0
Z
p i ðt Þ ¼
L
pz ðx, t Þ i ðx Þdx,
i ¼ 3, 6
(11.8.35b)
0
where now
3 ðx Þ ¼ 1 x,
6 ðx Þ ¼ x.
Hence, we have
9
8
p1 ðt Þ >
>
>
>
>
>
>
>
0
>
>
>
>
=
<
p3 ðt Þ
e
p ðt Þ ¼
>
>
> p4 ðt Þ >
>
>
>
>
>
>
>
>0
;
:
p6 ðt Þ
(11.8.36)
The statically equivalent vector of the nodal loads is particularly useful
when the rotational inertial nodal forces of the element are ignored.
11.8.2 Transformation of the nodal coordinates of the plane
grid element
The displacements u1 , u2 are transformed into the global system of axes according to Eq. (11.2.67), while the component u3 , which expresses the displacement
in the direction of the z axis, remains unaltered in the rotated system because the
local z axis and the global z axis are identical. Hence, we have
u1
cos f sin f
u1
¼
(11.8.37a)
u2
sin f cos f u2
u3 ¼ u3
The previous relations can be written
8 9 2
cos f sin f
< u1 =
u2 ¼ 4 sin f cos f
: ;
0
0
u3
(11.8.37b)
as
38 9
0 < u1 =
0 5 u2
: ;
1
u3
Similarly, we obtain
8 9 2
38 9
cosf sin f 0 < u4 =
< u4 =
u
¼ 4 sin f cos f 0 5 u5
: 5;
: ;
u6
u6
0
0
1
(11.8.38a)
(11.8.38b)
464 PART
II Multi-degree-of-freedom systems
Eqs. (11.8.38a), (11.8.38b) are combined
8 9 2
cos f sin f 0
0
u1 >
>
>
>
>
>
>
>
6 sin f cos f 0
0
u
>
>
2
>
= 6
< >
6 0
0
1
0
u3
¼6
0
0
cos f
> 6
> u4 >
>
6 0
>
>
>
0
0 sin f
> 4 0
> u5 >
>
>
: ;
0
0
0
0
u6
as
0
0
0
sin f
cos f
0
38 9
0 >
u1 >
>
>
>
>
>
>
07
>
>
> u2 >
7<
=
7
0 7 u3
07
>
> u4 >
7>
> >
0 5>
u5 >
>
>
>
;
: >
1
u6
(11.8.39)
or
where
2
ue ¼ Re ue
(11.8.40a)
ue ¼ ðRe ÞT ue
(11.8.40b)
3
cos fe sin fe 0
0
cos fe 0
0
0
1
0
0
0
cos fe
0
0 sin fe
7
07
7
0
07
7
7
e
sin f 0 7
7
7
cos fe 0 5
0
0
0
6
e
6 sin f
6
6 0
6
Re ¼ 6
6 0
6
6
4 0
0
0
0
0
0
(11.8.41)
1
is the transformation matrix of the plane grid element e.
e e e
The global vectors of the nodal forces f S , f I , f D , and pe ðt Þ are defined in relae
tion to u . Their transformation obeys the same law, that is, Eq. (11.8.40b).
Thus, we obtain
f e ¼ðRe ÞT f e
S
S
(11.8.42)
f e ¼ðRe ÞT f e
I
I
(11.8.43)
f e ¼ðRe ÞT f e
D
D
(11.8.44)
pe ðt Þ ¼ðRe ÞT pe ðt Þ
(11.8.45)
Using Eqs. (11.8.19), (11.8.25), (11.3.41), (11.8.40a), the first three of the
foregoing equations are transformed into the global axes as
f e ¼ke ue
S
(11.8.46)
f e
I
e €e
(11.8.47)
e _e
f e ¼
D c u
(11.8.48)
u
¼m
e
e , and where k , m
ce represent the stiffness, mass, and damping matrices of the e
grid element in global axes, respectively, and are given by
e
k ¼ ðRe ÞT ke Re
(11.8.49)
The finite element method Chapter
11
465
e ¼ ð Re Þ T m e Re
m
(11.8.50)
ce ¼ ð Re Þ T ce Re
(11.8.51)
Note that the lumped mass matrix remains the same under this transformation.
Applying the procedure presented in Section 11.2.3 for the plane truss, we
obtain the equation of motion of the plane grid, which, if damping is taken into
account, is written as
ee
e u ¼e
ee
u_ + Ke
pð t Þ
u€ + C
M
(11.8.52)
or after applying the support conditions
e ff e
e ff e
e ff e
M
u€f + C
u_ f + K
uf ¼ e
pf∗ ðt Þ
e sf e
e ss e
e ss e
e sf e
e ss e
e sf e
u€f + M
u€s + C
u_ f + C
u_ s + K
uf + K
us ¼ e
ps ðt Þ
M
(11.8.53)
(11.8.54)
where
e fs e
e fs e
e fs e
e
u€s C
u_ s K
us
pf ðt Þ M
pf∗ ðt Þ ¼ e
(11.8.55)
For convenience we will write Eq. (11.8.53) Check this as
M€
u + C€
u + Ku ¼ pðt Þ
(11.8.56)
Example 11.8.1 Formulate the equation of motion of the plane grid of Fig. E11.9.
The grid is fixed at nodes 3 and 4. The element 1 is loaded by the transverse
t. A moment M ðt Þ ¼ 3p0 L2 sin w
t acts at the node 2
load pðx, t Þ ¼ p0 sin w
t. The cross-section
while the support 1 is subjected to the rotation qg ðt Þ ¼ q0 sin w
of all elements is rectangular with aspect ratio h=b ¼ 2. Consider lumped mass
assumption for the elements. Data: L ¼ 3:0m, h ¼ 0:50m, a ¼ p=4,
E ¼ 2:1 107 kN=m2 , n ¼ 0:2, q0 ¼ 0:5p0 , mass density r.
Solution
The system has n ¼ 4 nodes. Hence, the free structure has N ¼ 3n ¼ 12 degrees
of freedom. The numbering of the nodes and the positive direction of the
elements are shown in Fig. E11.9
Fig. E11.9 Plane grid in Example 11.8.1.
466 PART
II Multi-degree-of-freedom systems
1. Computation of ke , me , ce , pe ðt Þ, Re for e ¼ 1,2, 3.
Matrices ke . All elements have a rectangular cross-section with side
lengths h ¼ 0:50 m and b ¼ 0:25 m. Their cross-sectional moment of inertia
is I ¼ bh 3 =12. The torsional constant of the elements is given by the relation
It ¼ b3 h [12]. For h=b ¼ 2 it is ¼ 0:229. Hence, It ¼ 0:229hb3 ¼ 0:6866I .
Moreover, G ¼ E=2ð1 + n Þ ¼ 0:4166E. The stiffness matrices of the elements
are computed using Eq. (11.8.20)
2
3
2:572
0
0 2:572
0
0
6 0
36:0 18:0 0
18:0 18:0 7
6
7
6
EI
0
18:0 12:0 0
18:0 12:0 7
7
k1 ¼ k 2 ¼ k 3 ¼ 3 6
0
0
2:572
0
0 7
L 6
6 2:572
7
4 0
18:0 18:0 0
36:0 18:0 5
0
18:0 12:0 0
18:0 12:0
Matrices me . For lumped mass assumption, the mass matrices are computed
using Eq. (11.8.30)
2
3
0 0 0 0 0 0
60 0 0 0 0 0 7
6
7
6 0 0 1:5 0 0 0 7
7
m1 ¼ m2 ¼ m3 ¼ rA6
60 0 0 0 0 0 7
6
7
40 0 0 0 0 0 5
0 0 0 0 0 1:5
Vectors pe ðt Þ. To be consistent with the lumped mass assumption, the
vectors pe ðt Þ will be computed as the static equivalent nodal loads. Thus, using
Eq. (11.8.36), we have
9
8
0 >
>
>
>
> 0 >
>
>
>
>
>
=
< 1:5 >
1
t, p2 ðt Þ ¼ p3 ðt Þ ¼ 0
p0 sin w
p ðt Þ ¼
>
>
0
>
>
>
>
>
>
>
>
>
;
: 0 >
1:5
TABLE E11.4 Geometrical data of the elements.
Element
number e
xj
1
0
2
3
D
x
yj
yk
D
y
Le
0
0
0
3
3
3
0
1
0
2.12
2.12
3
5.12
2.12
3
0.707
0.707
0
2.12
2.12
3
5.12
2.12
3
0.707
0.707
xk
cosfe
sinfe
The finite element method Chapter
11
467
On the base of the geometrical data given in Table E.11.4, we obtain
2
0
6 1
6
6
6 0
R1 ¼ 6
6 0
6
6
4 0
3
2
0 0 0
0:707
7
6
0 0 07
6 0:707
7
6
7
6 0
0 1 0 0 0
7 , R2 ¼ 6
7
6 0
0 0 0 1 07
6
7
6
4 0
0 0 1 0 0 5
1 0
0 0
0 0 0
0 0 1
0
3
0
07
7
7
1 0
0
07
7
0 0:707 0:707 0 7
7
7
0 0:707 0:707 0 5
0
0
0:707 0
0:707 0
0
0
0
0
0
0
0
0
0
2
0:707 0:707 0 0
0
6 0:707 0:707 0 0
0
6
6
6 0
0
1 0
0
R3 ¼ 6
6 0
0
0 0:707 0:707
6
6
4 0
0
0 0:707 0:707
0
0
0
0
0
1
3
0
07
7
7
07
7
07
7
7
05
1
e
e , pe ðt Þ for e ¼ 1, 2, 3
2. Computation of k , m
e
Matrices k . They are computed using Eq. (11.8.49)
2
3
36:0 0
18:0 18:0 0
18:0
6 0
2:57
0
0 2:57
0 7
6
7
6
7
6
7
18:0
0
12:0
18:0
0
12:0
EI
1
6
7
k ¼ 36
L 6 18:0 0
18:0 36:0 0
18:0 7
7
6
7
4 0 2:57
0
0
2:57
0 5
18:0
0
12:0 18:0 0
12:0
3
2
19:29 16:71 12:73
7:71 10:29 12:73
6 16:71 19:29 12:73 10:29
7:71 12:73 7
7
6
7
6
7
6
12:73
12:73
12:0
12:73
12:73
12:0
EI
2
7
6
k ¼ 36
L 6 7:71 10:29 12:73 19:29 16:71 12:73 7
7
7
6
4 10:29
7:71 12:73 16:71 19:29 12:73 5
2
12:73
12:73 12:0
12:73
12:73
12:0
19:29
16:71 12:73
7:71
10:29
12:73
12:73
12:73 12:0
12:73
12:73
12:0
3
6 16:71 19:29 12:73 10:29
7:71 12:73 7
6
7
6
7
6
7
12:73
12:73
12:0
12:73
12:73
12:0
EI
7
k3 ¼ 3 6
6
L 6 7:71 10:29 12:73 19:29 16:71 12:73 7
7
6
7
4 10:29
7:71 12:73 16:71 19:29 12:73 5
468 PART
II Multi-degree-of-freedom systems
e . They are computed using
Matrices m
2
0 0
60 0
6
6
60 0
1
2
3
m ¼ m ¼ m ¼ rA6
60 0
6
6
40 0
0 0
Eq. (11.8.50)
0
0 0
0
3
0
0
0 0
0 0
7
7
7
7
7
0 7
7
7
0 5
0
0 0
1:5
0 0 0
1:5 0 0
0
0
Vectors pe ðt Þ. They are computed using Eq. (11.8.45)
9
8
0 >
>
>
>
>
>
>
0 >
>
>
>
>
=
<
1:5
1
t, p2 ðt Þ ¼ p3 ðt Þ ¼ 0
p0 sin w
p ðt Þ ¼
0 >
>
>
>
>
>
>
0 >
>
>
>
>
;
:
1:5
^ e, M
^ e, p
^e for e ¼ 1, 2, 3
3. Computation of the enlarged matrices K
Assembly matrices ae .
2
1 0 0 0 0 0 0 0 0 0 0 0
60
6
6
60
1
a ¼6
60
6
6
40
2
2
1 0 0 0 0 0 0 0 0 0 07
7
7
0 1 0 0 0 0 0 0 0 0 07
7,
0 0 1 0 0 0 0 0 0 0 07
7
7
0 0 0 1 0 0 0 0 0 0 05
0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 0 0
60
6
6
60
2
a ¼6
60
6
6
40
3
0 0 0 1 0 0 0 0 0 0 07
7
7
0 0 0 0 1 0 0 0 0 0 07
7
0 0 0 0 0 1 0 0 0 0 07
7
7
0 0 0 0 0 0 1 0 0 0 05
0 0 0 0 0 0 0 0 1 0 0 0
0
60
6
6
60
3
a ¼6
60
6
6
40
3
3
0 0 1 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 07
7
7
0 0 0 0 1 0 0 0 0 0 07
7
0 0 0 0 0 0 0 0 1 0 07
7
7
0 0 0 0 0 0 0 0 0 1 05
0 0 0 0 0 0 0 0 0 0 0 1
The finite element method Chapter
11
469
^ e . They are computed using the relation K
^ e ¼ ðae ÞT ke ae (see
Matrices K
Eq. 11.2.92)
2
3
36:0 0
18:0 18:0 0
18:0 0 0 0 0 0 0
6 0
2:572
0
0 2:572
0 0 0 0 0 0 07
6
7
6 18:0 0
12:0 18:0 0
12:0 0 0 0 0 0 0 7
6
7
6
7
18:0 36:0 0
18:0 0 0 0 0 0 0 7
6 18:0 0
6
7
6 0 2:572
0
0
2:572
0 0 0 0 0 0 07
6
7
6 18:0
7
0
12:0
18:0
0
12:0
0
0
0
0
0
0
EI
6
7
^1 ¼ 6
K
7
L3 6 0
0
0
0
0
0 0 0 0 0 0 07
6
7
6 0
0
0
0
0
0 0 0 0 0 0 07
6
7
6 0
0
0
0
0
0 0 0 0 0 0 07
6
7
6
7
6 0
0
0
0
0
0 0 0 0 0 0 07
6
7
4 0
0
0
0
0
0 0 0 0 0 0 05
0
0
0
0
0
0 0 0 0 0 0 0
2
0
60
6
6
60
6
60
6
6
60
6
EI 6
60
2
^
K ¼ 36
L 60
6
60
6
6
60
6
60
6
6
40
0
2
0
60
6
6
60
6
60
6
60
6
6
60
EI
3
^
K ¼ 36
L 6
60
6
60
6
60
6
60
6
6
40
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
19:29 16:71 12:73
7:71 10:29 12:73
16:71
19:29 12:73 10:29
7:71 12:73
12:73 12:73
12:0
12:73 12:73 12:0
7:714 10:29 12:73
19:29 16:71 12:73
10:29
7:71 12:73 16:71
19:29 12:73
12:73 12:0
12:73
12:73 12:0 12:73
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0
0
0
0
0
0 0
0
0
0
0
0 0
0 19:29
16:71 12:73 0 0
0 16:71
19:29 12:73 0 0
0 12:73 12:73 12:0 0 0
0
0
0
0
0 0
0
0
0
0
0 0
0
0
0
0
0 0
0
7:714 10:29 12:73 0 0
0 10:29
7:71 12:73 0 0
0 12:73
12:73 12:0 0 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
3
0
07
7
7
07
7
07
7
7
07
7
07
7
7
07
7
07
7
7
07
7
07
7
7
05
0
3
0
0
0
0
7
0
0
0
0
7
7
0
0
0
0
7
7
0
7:71 10:29 12:73 7
7
0 10:29
7:71 12:73 7
7
7
0 12:73 12:73 12:0 7
7
7
0
0
0
0
7
7
7
0
0
0
0
7
7
0
0
0
0
7
7
0 19:29 16:71 12:73 7
7
0 16:71 19:29 12:73 5
0 12:73 12:73 12:0
470 PART
II Multi-degree-of-freedom systems
^ e . They are computed
Matrices M
(see Eq. 11.2.96)
2
0 0 0 0 0
60 0 0 0 0
6
6 0 0 1:5 0 0
6
60 0 0 0 0
6
6
60 0 0 0 0
6
60 0 0 0 0
1
^
M ¼ rA6
60 0 0 0 0
6
60 0 0 0 0
6
6
60 0 0 0 0
6
60 0 0 0 0
6
40 0 0 0 0
0 0 0 0 0
2
0
60
6
60
6
60
6
60
6
6
^ 2 ¼ rA6 0
M
60
6
60
6
60
6
60
6
40
0
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
^ e ¼ ð ae Þ T m
e ae
using the relation M
0
0
0
0
0
1:5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1:5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1:5
0
0
0
0 0 0 0 0 0
60
6
6
60
6
60
6
6
60
6
60
^ 3 ¼ rA6
M
6
60
6
60
6
6
60
6
60
6
6
40
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
3
0
07
7
07
7
07
7
7
07
7
07
7,
07
7
07
7
7
07
7
07
7
05
0
3
0
07
7
07
7
07
7
07
7
07
7
07
7
07
7
07
7
07
7
05
0
0 0 0 0 0 0
3
0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
0 0 0 0 1:5 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 5
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 1:5
The finite element method Chapter
11
471
^e ðt Þ. They are computed using the relation p
^e ðt Þ ¼ ðae ÞT pe ðt Þ
Vectors p
(see Eq. 11.2.95)
9
8
8 9
0 >
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
1:5
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
=
< 1:5 =
<0>
p0 sin w
^ 2 ðt Þ ¼ p
^ 3 ðt Þ ¼
^ 1 ðt Þ ¼
t, p
p
>
>
0 >
0>
>
>
>
>
>
>
>
>
>
>
> >
>
>
>
>
>
>
>
>
>
>
>
>
> 0 >
>0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> 0 >
>0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
0
>
>
>
>
>
>
>
;
;
:
: >
0
0
3
P
¼
4. Computation of M
¼
^ e, K
M
e¼1
2
3
P
^ e , and p ¼ Pðt Þ +
K
e¼1
0 0 0
0 0 0
0 0
0 0 0
0 0
0 0 0
6
60
6
6
60
6
60
6
6
60
6
6
60
¼ rA ¼ 6
M
6
60
6
6
60
6
6
60
6
6
60
6
60
4
3
P
^ e ðt Þ
p
e¼1
0 0 0 0 0 0
3
7
0 0 0 0 0 0 7
7
7
0 1:5 0 0 0 0 0 0 0 0 0 7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
7
0 0 0 0 4:5 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
7
0 0 0 0 0 0 0 0 0 0 0 7
7
0 0 0 0 0 0 0 0 0 0 0 7
5
0 0 0 0 0 1:5
3
36:0 0
18:0 18:0
0
18:0
0
0
0
0
0
0
6 0
2:572
0
0
2:57
0
0
0
0
0
0
0 7
7
6
6 18:0 0
12:0
18:0
0
12:0
0
0
0
0
0
0 7
7
6
7
6
6 18:0 0
18:0 74:57 0
18:0
7:71 10:29 12:73
7:71 10:29 12:73 7
7
6
6 0 2:572
7:71 12:73 10:29
7:71 12:73 7
0
0
41:14 25:46 10:29
7
6
7
6 18:0 0
25:46
36:0
12:73
12:73
12:0
12:73
12:73
12:0
12:0
18:0
EI
7
6
¼ 6
K
7
L3 6 0
0
0
0 7
0
0
7:71 10:29 12:73 19:29 16:71 12:73
7
6
6 0
0
0
0 7
0
0 10:29 7:714 12:73 16:71 19:29 12:73
7
6
6 0
0
0
0 7
0
0 12:73 12:73 12:0 12:73 12:73 12:0
7
6
7
6
6 0
0
0
0
19:29 16:71 12:73 7
0
0
7:71 10:29 12:73
7
6
4 0
0
0
0
16:71 19:29 12:73 5
0
0
10:29 7:714 12:73
0
0
0
12:73 12:73 12:0
0
0
0
12:73 12:73 12:0
2
ðt Þ +
pðt Þ ¼ P
3
X
i¼1
^ i ðt Þ ¼
p
9
8
Mg
>
>
>
>
>
>
>
>
>
>
P
2
>
>
>
>
>
>
>
>
>
>
P3
>
>
>
>
>
>
2
>
>
3p
L
sin
w
t
>
>
0
>
>
>
>
>
>
>
>
0
>
>
>
>
>
>
=
<0
>
P7
>
>
>
>
P
>
8
>
>
>
>
P
>
9
>
>
>
>
P
>
10
>
>
>
>
P
11
>
>
:
P12
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
;
+
9
8
0 >
>
>
>
>
>
>
>
>
>
0
>
>
>
>
>
>
>
>
>
1:5 >
>
>
>
>
>
>
> 0 >
>
>
>
>
>
>
>
>
>
>
0
>
>
>
>
>
=
< 1:5 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
0
0
0
0
0
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
;
p0 sin w
t ¼
9
8
Mg
>
>
>
>
>
>
>
>
>
>
P
2
>
>
>
>
>
>
>
>
>
t >
P3 1:5p0 sin w
>
>
>
>
>
>
>
>
sin
w
t
27p
>
>
0
>
>
>
>
>
>
>
>
0
>
>
>
>
>
>
=
< 1:5p sin w
t
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
0
P7
P8
P9
P10
P11
P12
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
;
The finite element method Chapter
11
473
Note that Mg denotes the moment that produces the rotation qg ðt Þ. It is
unknown and will be established as a reaction force.
M,
pðt Þ due to support conditions of the
6. Modification of the matrices K,
grid.
Referring to Fig. E11.9, the displacement vector should be modified as
e
e1 u
e2 u
e3 u
e4 u
e5 u
e6 u
e7 u
e8 u
e9 u
e10 u
e11 u
e12 g
uT ¼ f u
¼ f u4 u5 u6 u1 u2 u3 u7 u8 u9 u10 u11 u12 g
Hence, the matrix V that rearranges the
2
0 0 0 1 0 0 0
6
60 0 0 0 1 0 0
6
6
60 0 0 0 0 1 0
6
61 0 0 0 0 0 0
6
6
60 1 0 0 0 0 0
6
6
60 0 1 0 0 0 0
6
V¼6
60 0 0 0 0 0 1
6
6
60 0 0 0 0 0 0
6
6
60 0 0 0 0 0 0
6
6
60 0 0 0 0 0 0
6
60 0 0 0 0 0 0
4
matrices of the structure is
3
0 0 0 0 0
7
0 0 0 0 07
7
7
0 0 0 0 07
7
0 0 0 0 07
7
7
0 0 0 0 07
7
7
0 0 0 0 07
7
7
0 0 0 0 07
7
7
1 0 0 0 07
7
7
0 1 0 0 07
7
7
0 0 1 0 07
7
0 0 0 1 07
5
0 0 0 0 0 0 0 0 0 0 0 1
e ¼ VT KV,
e ¼ VT MV,
e
Thus, using the equations K
M
pðt Þ ¼ VT pðt Þ,
we obtain
474 PART
II Multi-degree-of-freedom systems
Following the indicated partitioning, we obtain
2
18:0
6 0
6
6
6 18:0
6
6
2
3
6 7:71
74:57
0
18:0
6
EI
EI
6
7
e ff ¼ 4 0
e sf ¼ 6
K
41:14 25:46 5, K
6 10:29
3
3
L
L 6
6 12:73
18:0 25:46 36:0
6
6
6 7:71
6
6
4 10:29
2:57
6
e fs ¼ EI 6
K
3
L 4
18:0
0
18:0
0
2:57
0
18:0
0
12:0
2
36:0
6 0
6
6
6 18:0
6
6
6 0
6
EI
e
0
Kss ¼ 3 6
L 6
6
0
6
6
6 0
6
6
4 0
0
0
18:0
2:57 0
0
0
0
0
0
0
0
7:71
10:29
10:29
12:73
7:714
12:73
12:73 12:73
0
0
0
0
12:0
0
0
0
12:0
0
10:29
12:73
7:71
12:73
12:73
12:0
10:29
12:73
12:73
7:714
12:73
2
18:0
0
12:0
12:73
7:71
18:0
10:29
0
7
10:29 7
5
12:73 12:0
0
0
18:0
0
3
7:71
12:73
0
0
3
7
7
7
7
12:0
0
0
0
0
12:0
0
7
7
0
19:29 16:71 12:73 0
0
19:29 7
7
0 16:71 19:29 12:73 0
0 16:71 7
7
7
0 12:73 12:73 12:0
0
0 12:73 7
7
7
0
0
0
0
19:29 0
0
7
7
0
0
0
0
16:71 0
0
5
0
0
0
0
12:73 0
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
The finite element method Chapter
2
0
60
6
60
6
2
3
60
0 0 0
6
e
e
4
5
, Mss ¼ rA6
Mff ¼ rA 0 0 0
60
60
0 0 4:5
6
60
6
40
0
0
0
0
0
0
0
0
0
0
0
0
1:5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1:5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
11
475
3
0
0 7
7
0 7
7
0 7
7
0 7
7
0 7
7
0 7
7
0 5
1:5
e sf ¼ 0
e fs ¼ M
M
9
9
8
8
Mg
q g ðt Þ >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
P2
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
P
1:5p
sin
w
t
0
>
>
> 3
>
0
9
8
>
>
>
>
>
>
>
>
>
>
>
>
27
P
0
=
=
< 7
=
<
<
es ð t Þ ¼ 0
e
, u
t, e
p0 sin w
pf ðt Þ ¼
0
p s ð t Þ ¼ P8
>
>
>
;
>
:
>
>
>
>
P9
1:5
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
P
0
10
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
P
0
11
>
>
>
>
;
;
:
:
P12
0
e ff ¼ 0
The equation of motion results from Eq. (11.8.53) with C
e ff e
e ff e
u€f + K
M
uf ¼ e
pf∗ ðt Þ
Taking into account that q0 ¼ 0:5p0 , Eq. (11.8.55) gives
8 9
< 18 =
e fs u
e sf us ¼ 0
€s K
e
t
p sin w
pf ðt Þ M
p ∗ f ðt Þ ¼ e
: ; 0
7:5
Hence, the equation of motion is
5. Static condensation
The rotational degrees of freedom u4 , u5 are statically condensed. Obviously,
the displacements should be rearranged as u6 , u4 , u5 . This defines the
modification matrix
2
3
0 1 0
VC ¼ 4 0 0 1 5
1 0 0
476 PART
II Multi-degree-of-freedom systems
which yields
Ktt ¼ 36
EI
EI
EI 18:0
,
K
¼
½
18:0
25:45
,
K
¼
tq
qt
L3
L3
L3 25:45
EI 74:57
0
Kqq ¼ 3
0 41:14
L
Mtt ¼ 4:5rA, Mtq ¼ Mqt ¼ Mqq ¼ 0
t, pq ¼
pt ¼ 1:5p0 sin w
27
t
p0 sin w
0
Hence
Ktt∗ ¼ Ktt Ktq K1
qq Kqt ¼ 15:91
EI
L3
t
pt∗ ¼ pt Ktq K1
qq pq ¼ 5:017p0 sin w
which are inserted into Eq. (11.4.3) to give the final form of the equation of
motion
4:5rA u€6 + 15:91
EI
t
u6 ¼ 5:017p0 sin w
L3
11.9 The finite element method for the space frame
11.9.1 Properties of the space frame element
We consider the space frame of Fig. 11.9.1, whose nodes are determined by
their Cartesian coordinates with respect to the global system of axes x y z.
The loads applied to the nodes may consist of forces or moments while the
elements may be loaded by distributed and/or concentrated loads (forces or
moments). The axis of the e element determines its local x axis while the y
and z axes lie on the principal planes of bending (see Fig. 11.9.2). The nodes
of the frame, hence the ends of the elements, undergo six displacements (three
translations and three rotations). Therefore, the nodal displacements of the
element are 12, six for each end, and are numbered as shown in Fig. 11.9.2.
They are represented by the vector
u ¼ f u1 u2 u3 u4 u5 u6 u7 u8 u9 u10 u11 u12 g
(11.9.1)
The finite element method Chapter
11
477
Fig. 11.9.1 Space frame.
Fig. 11.9.2 Nodal displacements of a space frame element.
The displacements u1 , u7 produce axial deformation of the element, the
displacements u4 , u10 produce torsion, and the displacements u2 , u6 , u8 , u12
and u3 , u5 u9 , u11 produce bending in the planes x y and x z, respectively.
The elastic energy of the element is
U ¼ Ua + Ut + Uby + Ubz
(11.9.2)
where Ua ,Ut ,Uby , Ubz represent the elastic energies due to axial and torsional
deformation as well as to bending deformation in the planes xy and xz, respectively. Using appropriately the shape functions developed for the plane frame
and grid elements, we can express the displacements of a cross-section x of the
space frame element as
u ðx, t Þ ¼ u1
qðx, t Þ ¼ u4
1 ðx Þ + u 7 7 ðx Þ
(11.9.3a)
4 ðx Þ + u10 10 ðx Þ
(11.9.3b)
v ðx, t Þ ¼ u2
2 ðx Þ + u6 6 ðx Þ + u8 8 ðx Þ + u12 12 ðx Þ
(11.9.3c)
w ðx, t Þ ¼ u3
3 ðx Þ + u5 5 ðx Þ + u9
(11.9.3d)
9 ðx Þ + u11 11 ðx Þ
478 PART
II Multi-degree-of-freedom systems
where u, v,w are the translations in the directions of the x, y, z axes, respectively, and q the rotation about the x while i ðx Þ (i ¼ 1,2, …,12) are given by
1 ðx Þ ¼
4 ðx Þ ¼ ð1 x Þ
7 ðx Þ ¼
2 ðx Þ ¼
(11.9.4a)
10 ðx Þ ¼ x
3 ðx Þ ¼ 1 3x
6 ðx Þ ¼ 5 ðx Þ ¼ L
2
(11.9.4b)
+ 2x3
x 2x 2 + x3
2
3
12 ðx Þ ¼ 11 ðx Þ ¼ L x + x
9 ðx Þ ¼
8 ðx Þ ¼ 3x
2
2x3
(11.9.4c)
(11.9.4d)
(11.9.4e)
(11.9.4f)
On the base of Eqs. (11.3.8), (11.3.9), (11.8.13), Eq. (11.9.2) yields
Z
Z
1 L
1 L
2
2
0
EA½u ðx, t Þ dx +
GI t ½q0 ðx Þ dx
U ðu1 , …, u12 Þ ¼
2 0
2 0
(11.9.5)
Z
Z
1 L
1 L
2
2
00
00
+
EI z ½v ðx, t Þ dx +
EI y ½w ðx, t Þ dx
2 0
2 0
Further, for the consistent mass assumption, the kinetic energy is given by
Z
n
o
1 L
T ðu_ 1 , …, u_ 12 Þ ¼
m ðx Þ ½u_ ðx, t Þ2 + ½v_ ðx, t Þ2 + ½w_ ðx, t Þ2 dx
2 0
(11.9.6)
Z
2
1 L
_
I0 ðx Þ q ðx, t Þ dx
+
2 0
while for the lumped mass assumption by
1 1 T ðu_ 1 , …, u_ 12 Þ ¼ m1 u_ 21 + u_ 22 + u_ 23 + m2 u_ 27 + u_ 28 + u_ 29
2
2
(11.9.7)
where
Z
L
m1 ¼
0
Z
L
m ðx Þð1 xÞdx, m2 ¼
m ðx Þxdx
(11.9.8)
0
The element mass and stiffness matrices as well as the equivalent nodal
forces are obtained using the method of the Lagrange equations as described
for the plane frame and the grid element. In the following, we consider an element with constant cross-section.
(i) Nodal elastic forces and stiffness matrix of the space frame element
For an element with a constant cross-section, the stiffness matrix having dimensions 12 12 results as
e e k k
e
(11.9.9)
k ¼ jje jk
kkj kekk
The finite element method Chapter
11
479
where
2
L2 A 0
6
60
6
6
60
6
E
kejj ¼ 3 6
L 6
60
6
6
6
40
2
0
0
0
12Iz
0
0
0
0
12Iy 0
0
0
0
6LI z
0
0
0
0
0
0
12Iz 0
0
0
L2 A
6
60
6
6
60
6
E
e
kkk ¼ 3 6
L 6
60
6
6
60
4
2
6
6
6
6
6
6
E
kekj ¼ 3 6
L 6
6
6
6
6
4
0
12Iy 0
0
0
0
6LI y 0
0
4L2 Iz
0
6LI y
4L2 Iy
0
L2 A
0
0
0
0
0
12Iz
0
0
0
0
0
12Iy
0
6LI y
0
0
0
0
0
0
6LI z
6LI y
0
G 2
L It 0
E
0
2L2 Iy
0
0
kejk ¼ kekj
3
4L2 Iz
0
(11.9.10a)
7
6LI z 7
7
7
7
0
7
7
7
7
0
7
7
7
0
5
G 2
L It 0
E
6LI z 0
0
4L2 Iy
0
6LI y 0
0
7
6LI z 7
7
7
7
0
7
7
7
7
0
7
7
7
0
5
6LI y
G 2
L It
E
3
0
0
(11.9.10b)
3
7
6LI z 7
7
7
7
0
7
7
7
7
0
7
7
7
0
5
(11.9.10c)
2L2 Iz
T
(11.9.10d)
In the previous matrices, A is the area of the cross-section of the element e,
L its length, Iy , Iz are the cross-sectional moments of inertia with respect to the
y and z axes, respectively, and It the torsional constant of the cross-section.
Note that the superscript e has been omitted from Ite , Iye , Ize , and Le for the
simplicity of the expressions.
(ii) Nodal inertial forces and mass matrix of the space frame element
(a) Consistent mass matrix
The consistent matrix results as
"
m ¼
e
mejj mejk
mekj mekk
#
(11.9.11)
480 PART
where
II Multi-degree-of-freedom systems
2
3
140 0
0
0
0
0
60
156
0
0
0
22L 7
6
7
e 6
m 60
0
156 0
22L 0 7
7
mejj ¼
0 7
0
0
140rg2 0
420 6
60
7
40
0
22L 0
4L2 0 5
0
22L 0
0
0
4L2
2
3
140 0
0
0
0
0
60
156 0
0
0
22L 7
6
7
e 6
m
0
0
156
0
22L
0 7
e
6
7
mkk ¼
0 7
0
0
140rg2 0
420 6
60
7
40
0 5
0
22L 0
4L2
0
22L 0
0
0
4L2
2
3
70 0
0
0
0
0
60
54 0
0
0
13L 7
6
7
e 6
m
0
0
54
0
13L
0 7
e
6
7
mkj ¼
0 7
0
0
70rg2 0
420 6
60
7
40
0 5
0
13L 0
3L2
0 13L 0
0
0
3L2
mejk ¼ mekj
(11.9.12a)
(11.9.12b)
(11.9.12c)
T
m e ¼ rAL is the total mass of the e element and rg ¼
gyration of the cross-section.
(11.9.12d)
pffiffiffiffiffiffiffiffiffiffi
I0 =A is the radius of
(b) Lumped mass matrix
According to this assumption, the mass of the element is concentrated at its
nodes, that is, they are obtained as the reactions of a simply supported beam
under the load m ðx Þ. Thus, for an element with constant mass, the mass lumped
assumption yields
2
3
m1 0 0 0 0 0
6 0 m1 0 0 0 0 7
6
7
6 0 0 m1 0 0 0 7
e
6
7
mjj ¼ 6
(11.9.13a)
7
60 0 0 0 0 07
40 0 0 0 0 05
0 0 0 0 0 0
2
3
m2 0 0 0 0 0
6 0 m2 0 0 0 0 7
6
7
6 0 0 m2 0 0 0 7
e
6
7
mkk ¼ 6
(11.9.13b)
7
60 0 0 0 0 07
40 0 0 0 0 05
0 0 0 0 0 0
mejk ¼ mekj ¼ 0
(11.9.13c)
The finite element method Chapter
11
481
(iii) Equivalent nodal loads of the space frame element
(a) Consistent nodal load vector
The element is subjected to the axial load px ðx, t Þ, the transverse loads
py ðx, t Þ, pz ðx, t Þ, and the torsional moment mx ðx, t Þ. The resulting equivalent
load vector is
pe ð t Þ ¼
pej
pek
(11.9.14)
where
pej ¼ f p1 p2 p3 p4 p5 p6 gT
(11.9.15a)
pek ¼ f p7 p8 p9 p10 p11 p12 gT
(11.9.15b)
The components pi (i ¼ 1, 2, …, 12) are given by
Z L
px ðx, t Þ i ðx Þdx, i ¼ 1, 7
pi ðt Þ ¼
(11.9.16a)
0
Z
L
pi ðt Þ ¼
py ðx, t Þ i ðx Þdx, i ¼ 2, 6, 8, 12
(11.9.16b)
pz ðx, t Þ i ðx Þdx, i ¼ 3, 5, 9, 11
(11.9.16c)
0
Z
L
pi ðt Þ ¼
0
Z
L
pi ðt Þ ¼
mx ðx, t Þ i ðx Þdx, i ¼ 4, 10
(11.9.16d)
0
(b) Statically equivalent nodal load vector
In this case, we assume that the element is a simply supported beam under
the loads px ðx, t Þ, py ðx, t Þ, pz ðx, t Þ and the torsional moment mx ðx, t Þ.
Apparently, the components p5 ðt Þ, p6 ðt Þ, p11 ðt Þ, and p12 ðt Þ are zero. The
remaining components result as
Z L
px ðx, t Þ i ðx Þdx, i ¼ 1, 7
(11.9.17a)
p i ðt Þ ¼
0
Z
L
pi ðt Þ ¼
py ðx, t Þ i ðx Þdx, i ¼ 2, 8
(11.9.17b)
pz ðx, t Þ i ðx Þdx, i ¼ 3, 9
(11.9.17c)
0
Z
pi ðt Þ ¼
Z
pi ðt Þ ¼
0
L
0
L
mx ðx, t Þ i ðx Þdx, i ¼ 4, 10
(11.9.17d)
482 PART
where
II Multi-degree-of-freedom systems
i ðx Þ ¼ 1 x,
i + 6 ðx Þ ¼ x,
i ¼ 1, 2, 3, 4. Therefore
pej ðt Þ ¼ f p1 p2 p3 p4 0 0 gT
(11.9.18a)
pek ¼ f p7 p8 p9 p10 0 0 gT
(11.9.18b)
11.9.2 Transformation of the nodal coordinates of the space
frame element
We consider the two systems of axes: the global system x y z and the local system
xyz. The latter is defined by the axis x of the element and the principal axes y, z of
its cross-section, Fig. 11.9.3. The base unit vectors of the two systems are denoted
by ei and ei i ¼ 1, 2, 3, respectively. The projection of ei on the axes x, y z gives
e1 ¼ l11 e1 + l12 e2 + l13 e3
(11.9.19a)
e2 ¼ l21 e1 + l22 e2 + l23 e3
(11.9.19b)
e3 ¼ l31 e1 + l32 e2 + l33 e3
(11.9.19c)
The quantities lij are the direction cosines of the xyz axes with respect
to x y z. Because ei are unit vectors and normal to each other, we have
6 j.
jei j2 ¼ l2i1 + l2i2 + l2i3 ¼ 1 and ei ej ¼ li1 lj1 + li2 lj2 + li3 lj3 ¼ 0, i ¼
For the matrix of the direction cosines
2
3
l11 l12 l13
(11.9.20)
L ¼ 4 l21 l22 l23 5
l31 l32 l33
we can readily show that det ðLÞ ¼ 1. Hence, the matrix L is orthonormal and
consequently
L1 ¼ LT
Fig. 11.9.3 Global and local systems of axes of the space frame element.
(11.9.21)
The finite element method Chapter
11
483
A vector a is written with respect to the two systems of axes
a ¼ a1 e1 + a2 e2 + a3 e3
(11.9.22)
a ¼ a1 e1 + a2 e2 + a3 e3
(11.9.23)
where a1 , a2 ,a3 and a1 , a2 , a3 denote the components of vector a with respect to
the local and global axes, respectively.
Substituting Eqs. (11.9.19a)–(11.9.19c) into Eq. (11.9.23) gives
e1 + ðl12 a1 + l22 a2 + l32 a3 Þ
e2
a ¼ðl11 a1 + l21 a2 + l31 a3 Þ
+ ðl13 a1 + l23 a2 + l33 a3 Þ
e3
(11.9.24)
which, when compared with Eq. (11.9.22), yield
a1 ¼ l11 a1 + l21 a2 + l31 a3
a2 ¼ l12 a1 + l22 a2 + l32 a3
a3 ¼ l13 a1 + l23 a2 + l33 a3
(11.9.25)
a ¼ LT a
(11.9.26)
a ¼ L
a
(11.9.27)
or
and by virtue of Eq. (11.9.21)
The matrix L is the transformation matrix of the components of a vector
from the axes x y z to xyz. In two dimensions, the matrix L is determined from
the Cartesian coordinates of the end points of the element. These are, however,
not adequate in three dimensions because an infinite number of systems may
have the common x axis. In order to define the system of local axes, we must
also know one of the principal axes of the cross-section. The simplest
way to
accomplish it is to specify the global coordinates of a point P xp , yp , zp on
one of the principal planes, say of xy.
The direction cosines of the x axis are computed from the relations
l11 ¼
yk yj
xk xj
zk zj
, l12 ¼
, l13 ¼
L
L
L
(11.9.28)
where xj , yj , zj , ðxk , yk , zk Þ are the global coordinates of the end points of the
h
i
2
2
2 1=2
. The points
e element with length L ¼ xk xj + yk yj + zk zj
j and P define the vector
r ¼ xp xj e1 + yp yj e2 + zp zj e3
(11.9.29)
The vector e3 is obtained from the vector product
e3 ¼
e1 r
je1 rj
(11.9.30)
484 PART
II Multi-degree-of-freedom systems
This is a unit vector and as a vector product, it is normal to the plane defined
by the vectors e1 and r, that is, it is in the direction of the local z axis. Its components give the direction cosines l31 , l32 , l33 .
Finally, the direction cosines l21 , l22 , l23 will result from the components
of the vector
e2 ¼ e3 e1
(11.9.31)
Apparently, the end displacements (translations and rotations) of the
element are transformed according to Eq. (11.9.27). Thus, we have
8 9 2
38 9
l11 l12 l13 < u1 =
< u1 =
u
¼ 4 l21 l22 l23 5 u2
(11.9.32a)
: 2;
: ;
u3
l31 l32 l33
u3
8 9 2
38 9
l11 l12 l13 < u4 =
< u4 =
(11.9.32b)
u5 ¼ 4 l21 l22 l23 5 u5
: ;
: ;
u6
l31 l32 l33
u6
8 9 2
38 9
l11 l12 l13 < u7 =
< u7 =
¼ 4 l21 l22 l23 5 u8
(11.9.32c)
u
: 8;
: ;
u9
l31 l32 l33
u9
9
8 9 2
38
l11 l12 l13 < u10 =
< u10 =
¼ 4 l21 l22 l23 5 u11
(11.9.32d)
u
;
: 11 ;
:
u12
l31 l32 l33
u12
which are combined to yield
ue ¼ Re ue
where
2
L
60
e
R ¼6
40
0
0
L
0
0
0
0
L
0
(11.9.33)
3
0
07
7
05
L
(11.9.34)
Obviously, the matrix Re with dimensions 12 12 represents the transformation matrix of the e space frame element. This vector is orthonormal, hence
Eq. (11.9.33) is inverted as
ue ¼ ðRe ÞT ue
(11.9.35)
The vectors of the equivalent nodal forces obey the same transformation
law. Namely
f e ¼ðRe ÞT f e
S
S
(11.9.36)
f e
I
(11.9.37)
¼ðRe ÞT f eI
The finite element method Chapter
f e ¼ðRe ÞT f e
D
D
11
485
(11.9.38)
pe ðt Þ ¼ðRe ÞT pe ðt Þ
(11.9.39)
e
It can be readily shown that the stiffness matrix k , damping matrix ce , and
e of the element e with respect to the global axes are computed
mass matrix m
from the relations
e
k ¼ ðRe ÞT ke Re
(11.9.40)
e T e
(11.9.41)
e T
(11.9.42)
ce ¼ ð R Þ c Re
e ¼ ð R Þ m e Re
m
Applying the procedure presented in Section 11.2.3 for the plane truss,
we obtain the equation of motion of the space frame, which, if damping is taken
into account, is written as
u_ + K
u€ + C
u ¼
M
pðt Þ
(11.9.43)
or after applying the support conditions and partitioning
e ff e
e ff e
e ff e
u€f + C
u_ f + K
uf ¼ e
pf∗ ðt Þ
M
e sf e
e ss e
e ss e
e sf e
e ss e
e sf e
u€f + M
u€s + C
u_ f + C
u_ s + K
uf + K
us ¼ e
ps ðt Þ
M
(11.9.44)
(11.9.45)
where
e fs e
e fs e
e fs e
e
u€s C
u_ s K
us
pf ðt Þ M
pf∗ ðt Þ ¼ e
(11.9.46)
For convenience, we will write Eq. (11.9.44) as
M€
u + Cu_ + Ku ¼ pðt Þ
(11.9.47)
Example 11.9.1 Compute the transformation matrix L of the space frame
element, which is determined from the end coordinates j ð2, 1, 1Þ, k ð5, 2, 2Þ
and the point P ð7, 0, 0Þ of the principal xy plane.
Solution
First, we compute
Dx ¼ xk xj ¼ 3, Dy ¼ yk yj ¼ 1, Dz ¼ zk zj ¼ 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
L ¼ 32 + 11 + 12 ¼ 3:317
and using Eq. (11.9.28), we obtain
l11 ¼
Dx
Dy
Dz
¼ 0:904, l12 ¼
¼ 0:301, l13 ¼
¼ 0:301
L
L
L
(1)
486 PART
II Multi-degree-of-freedom systems
Then we compute
rp ¼ xp xj e1 + yp yj e2 + zp zj e3
2
¼5
e1 e2 e3
e1
6
e1 rp ¼ 6
4 0:904
e2
e3
0:301
3
7
0:301 7
e1 + 2:412
e2 2:412
e3
5 ¼ 0
1
1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
je1 rj ¼ 02 + 2:4122 + ð2:412Þ2 ¼ 3:141
5
(2)
and Eq. (11.9.30) gives
e3 ¼
e1 r
¼ 0
e1 + 0:707
e2 0:707
e3
je1 rj
(3)
Finally, using Eq. (11.9.31), we obtain
2
3
e2
e3
e1
6
7
e2 ¼ e3 e1 ¼ 6
0:707 0:707 7
e1 + 0:639
e2 0:639
e3 (4)
40
5 ¼ 0:426
0:904 0:301 0:301
Hence
2
0:904 0:301
0:301
3
6
7
7
L¼6
4 0:426 0:639 0:639 5
0
0:707 0:707
(5)
Example 11.9.2 Formulate the equation of motion of the space frame
of Fig. E11.10a. The nodal coordinates are 1 ð0:886L, 0, 0:5LÞ,
2 ð0, 0:5L, 0:5LÞ, 3 ð0, 0:5L, 0:5LÞ, 4 ð0:886L, 0, 0:5LÞ. The frame is
fixed at nodes 1 and 4. The local z axis of all elements is parallel to the
zx plane. The elements have a rectangular cross-section A with dimensions
h and b in the directions of y and z axes, respectively. The axial deformation
of the elements is ignored and the lumped mass assumption is adopted. The
element 2 is loaded by the distributed load py ðx, t Þ ¼ 5f ðt Þ while a
moment Mðt Þ ¼ f 0 0 200Lf ðt Þ gT is applied at node 3. Assume:
h=b ¼ 2, A ¼ 300Iy , elastic constants E, n ¼ 0:2, and mass density r.
The finite element method Chapter
487
11
(a)
Fig. E11.10a Space frame in Example 11.9.2.
(b)
Fig. E11.10b Translational and rotational degrees of freedom in Example 11.9.2.
Solution
The system has n ¼ 4 nodes. Thus, the free structure has N ¼ 6n ¼ 24
degrees of freedom. We set Iy ¼ I , hence A ¼ 300I , Iz ¼ 4I , and
It ¼ 0:229hb3 ¼ 2:748I . Moreover, it is G ¼ 0:4166E.
TABLE E11.5 Geometrical data of the elements.
e
xj
xk
D
x
yj
yk
1
0.886L
0
0.886L
0
0.5L
0.5L
0.5L
0.5L
0
0
0.5L
0.5L
L
0.5L
0.5L
0
2
0
0
0
0.5L
0.5L
3
0
0.886L
0.886L
0.5L
0
D
y
0.5L
zj
zk
D
z
488 PART
II Multi-degree-of-freedom systems
Computation of ke , me , pe , Re for e ¼ 1, 2, 3. Applying Eqs. (11.9.10a)–
(11.9.10d) we obtain
2
7500
60
6
EI 6 0
k1jj ¼ 3 6
L 6
60
40
0
0
0
0
0
48
0
0
0
0
12 0
30
0
0 28:62 0
0
30 0
100
120 0
0
0
2
7500 0
60
48
6
EI 6
0
0
1
kkk ¼ 3 6
0
0
L 6
6
40
0
0
120
3
0
120 7
7
0 7
7,
0 7
7
0 5
400
3
0
0
0
0
0
0
0
120 7
7
12 0
30
0 7
7
0 28:62 0
0 7
7
30 0
100 0 5
0
0
0
400
2
3
7500 0
0
0
0 0
6 0
48
0
0
0 120 7
6
7
6
EI
0
0
12
0
30
0 7
1
6
7, k 1 ¼ k1
kjk ¼ 3 6
kj
jk
0
0 28:62 0 0 7
L 6 0
7
4 0
0
30 0
50 0 5
0
120 0
0
0 200
T
k2 ¼ k3 ¼ k1
The lumped mass assumption results in the matrices me
2
0:5
60
6
60
1
mjj ¼ rAL6
60
6
40
0
0
0:5
0
0
0
0
0
0
0:5
0
0
0
0
0
0
0
0
0
m1jk ¼ m1kj
0
0
0
0
0
0
T
3
2
0
0:5
60
07
7
6
6
07
7, m1 ¼ rAL6 0
kk
60
07
7
6
40
05
0
0
0
0:5
0
0
0
0
¼ 0 and m2 ¼ m3 ¼ m1
0
0
0:5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
3
0
07
7
07
7
07
7
05
0
The finite element method Chapter
11
489
Because the loading is referred to the global axes, we obtain directly the
equivalent nodal loads pe ðt Þ
9
9
8
8
8 9
0 >
0 >
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
2:5
2:5
0>
>
>
>
>
>
>
>
>
>
>
>
=
=
=
<
<
< >
0
0
0
2
2
1
1
3
3
Lf ðt Þ, pk ¼
Lf ðt Þ, pj ¼ pk ¼ pj ¼ pk ¼
pj ¼
0 >
0>
>
> 0 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0 >
0 >
0>
>
>
>
>
>
>
>
>
>
>
>
;
;
;
:
:
: >
0
0
0
Taking the positive direction of the local x to coincide with increasing node
number and the points P1 20 , P2 O, P3 30 to define the principal planes xy
of the elements 1, 2, 3, respectively, and Table E11.5, we obtain:
Element 1:
e1 ¼ ½ 0:866 0:500 0 , rP1 ¼ f 4:33 0 0 gT ,
e3 ¼
e1 rP1
¼ ½ 0 0 1 , e2 ¼ e3 e1 ¼ ½ 0:5 0:866 0 je1 rP1 j
2 1
3
L
0:866 0:500 0
60
1
1
6
4
5
L ¼ 0:5
0:866 0 , R ¼ 4
0
0
0
1
0
2
0
L1
0
0
0
0
L1
0
3
0
0 7
7
0 5
L1
Element 2:
e1 ¼ ½ 0 0 1 , rP2 ¼ ½ 0 2:5 2:5 , e3 ¼
e1 rP2
¼ ½ 1 0 0 je1 rP2 j
e2 ¼ ½ 0 1 0 2
3
2
L2
0 0 1
60
L2 ¼ 4 0 1 0 5, R2 ¼ 6
40
1 0 0
0
0
L2
0
0
0
0
L2
0
3
0
0 7
7
0 5
L2
490 PART
II Multi-degree-of-freedom systems
Element 3:
e1 ¼ ½ 0:866 0:500 0 , rP3 ¼ ½ 0 2:5 0 , e3 ¼
e1 rP3
¼ ½ 0 0 1 je1 rP3 j
e2 ¼ ½0:50 0:866 0 2 3
3
L
0:866 0:500 0
60
3
3
L ¼ 4 0:500 0:866 0 5, R ¼ 6
40
0
0
1
0
2
0
L3
0
0
0
0
L3
0
3
0
0 7
7
0 5
L3
K
and total load vector p of the strucComputation of total matrices M,
ture. First, we transform the element matrices and load vectors from the
local to global axes using Eqs. (11.9.39), (11.9.40), (11.9.42). Then we
determine the element assembly matrices ae ðe ¼ 1, 2, 3Þ, which are used
to evaluate the enlarged element matrices and load vectors. Finally, we evaluate the total matrices and the total load vector of the free structure from the
relations
¼
K
3
X
e¼1
e
¼
ðae ÞT k ae , M
3
X
e ae , pðt Þ ¼ Pext +
ða e ÞT m
e¼1
3
X
ðae ÞT pe
e¼1
where Pext ¼ f0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 200Lf ðt Þ 0 0 0 0 0 0gT is the vector
of the loads directly applied to the nodes. Subsequently, the transformation
matrix V is constructed by considering the support conditions. After
e ff and
e ff , K
partitioning the transformed matrices, we compute the matrices M
∗
e
the load vector pf using Eqs. (11.9.44) and (11.9.45). This procedure yields
the equation of motion in terms of the 12 free degrees of freedom
ef ¼ f u7 u8 u9 u10 u11 u12 u13 u14 u15 u16 u17 u18 gT
u
e ff e
e ff e
u€f + K
uf ¼ pt∗
M
Then, reordering the displacements to separate the translations from the
rotations and eliminating the six rotational degrees of freedom yields the following equation with respect to the six translational degrees of freedom
f u7 u8 u9 u13 u14 u15 gT
2
38 € 9
u7 >
>
>
>
>
>
>
>
7>
>
>
€
7
>
>
8 >
u
1:0 0 0 0 0 7>
>
>
>
>
>
7>
>
>
>
<
7
0 1:0 0 0 0 7 u€9 =
7
> u€13 >
>
0 0 1:0 0 0 7
>
>
7>
>
>
>
7>
>
>
>
>
7
€
u
0 0 0 1:0 0 5>
> 14 >
>
>
>
>
>
>
:€ >
;
u15
0 0 0 0 1:0
1:0 0
6
60
6
6
60
6
rAL6
60
6
6
60
4
0
0
0
0
0
9
38
u7 >
225L2 + 7:68 130L2 6:17
3:17
4:68
0:976
3:17
>
>
>
>
>
7>
6 130L2 6:17 75L2 + 13:6
>
u8 >
1:68
0:976
1:26
1:68
>
>
7>
6
>
>
<
=
7
6
2
2
EI 6
6:17
1:68
300L + 7:66
3:17
1:68
300L 0:88 7 u9
+ 36
7
7>
u13 >
4:68
0:976
3:17
225L2 + 7:68 130L2 + 6:17
3:17
L 6
>
>
>
7>
6
>
>
2
>
5
4 0:976
u14 >
1:68
1:26
1:68
1:68
75L + 13:6
>
>
>
>
:
;
u15
3:17
1:68
300L2 0:88
3:17
1:68
300L2
2
8
9
9:39
>
>
>
>
>
>
>
>
>
16:26 2:5L >
>
>
>
>
<
=
0
¼
f ðt Þ
>
>
140:61
>
>
>
>
>
>
>
> 243:54 2:5L >
>
>
>
:
;
0
492 PART
II Multi-degree-of-freedom systems
To save space, the intermediate computed matrices are not given here
because they are very large. The reader can find them on this book’s companion
website.
Reduction of the degrees of freedom due to axial constraints. The equations
of the constraints are
l111 u7 + l112 u8 + l113 u9 ¼ 0
l211 u7 + l212 u8 + l213 u9 ¼ l211 u13 + l212 u14 + l213 u15
l311 u13 + l312 u14 + l313 u15 ¼ 0
Using the computed values for lij the constraint matrix is
2
3
0:866 0:5 0 0
0 0
6
7
D ¼ 40
0 1 0
0 15
0
0
0 0:866 0:5 0
The Gauss-Jordan elimination yields
2
3
1 0:577 0 0 0
0
6
7
1 0 0
1 5
4 0 0
0 0
0 1 0:577 0
We readily conclude that the rank of the constraint matrix is 3. Therefore,
the independent displacements are 3. In the following, we take u8 u9 u14 as
independent displacements. This requires the rearrangement of the equations
using the transformation matrix
2
3
0 0 0 1 0 0
61 0 0 0 0 07
6
7
6
7
60 1 0 0 0 07
^
6
7
V¼6
7
60 0 0 0 1 07
6
7
40 0 1 0 0 05
0 0 0 0 0 1
which transforms the constraint equations to
The finite element method Chapter
11
Thus the constraint matrix becomes
Hence
2
3
2
3
0:5 0 0
0:866 0
0
0
1 5,
Da ¼ 4 0 1 0 5, Db ¼ 4 0
0
0 0:5
0
0:866 0
2
3
1
0 0
6 0
7
1 0
6
7
6
7
I
0
0 1
6
7
¼
T¼
1
6
7
Db Da
6 0:577 0 0
7
4 0
0 0:577 5
0
1 0
2
^ ¼ TT
M
3
1:33 0 0
^TM
e tt V
^ T ¼ rAL4 0
V
2 0 5
0
0 1:33
2
^ ¼ TT
K
23:3
^TK
e ∗V
^ T ¼ EI 4 3:35
V
tt
L3
1:43
3:35
13:5
3:35
3
1:43
3:35 5
23:3
9
8
21:68
2:5L
=
<
T ∗
^ p ðt Þ ¼
^ðt Þ ¼ TT V
f ðt Þ
0
p
t
;
:
324:73 2:5L
and the sought equation of motion reads
9
9
38
2
38
1:33 0 0
< u€8 = EI 23:3 3:35 1:43 < u8 =
4 3:35 13:5 3:35 5 u9
rAL4 0
2 0 5 u€9
+
;
: € ; L3
:
0
1:33
u14
1:43 3:35 23:3
u14
2
9
8
< 21:68 2:5L =
f ðt Þ
¼
0
;
:
324:73 2:5L
493
494 PART
II Multi-degree-of-freedom systems
11.10 The finite element method for the space truss
11.10.1 Properties of the space truss element
In a space truss (Fig. 11.10.1), the rods are interconnected at the nodes, which
are spherical hinges and as such allow freely the relative rotation of the end
cross-sections. In the deformed state at time t, the ends of the element j, k
are displaced to points j 0 , k 0 and the element occupies the position j 0 k 0 .
Therefore, the vector of nodal coordinates is (see Fig. 11.10.2)
u ¼ f u1 u2 u3 u4 u5 u6 gT
(11.10.1)
According to what we mentioned for the plane truss, a point on the x
local axis of the element will undergo displacements along the x, y and z axes,
that is,
u ðx, t Þ ¼ u1
1 ðx Þ + u4 4 ðx Þ
(11.10.2a)
v ðx, t Þ ¼ u2
2 ðx Þ + u 5 5 ðx Þ
(11.10.2b)
w ðx, t Þ ¼ u3
3 ðx Þ + u6 6 ðx Þ
(11.10.2c)
Fig. 11.10.1 Space truss.
Fig. 11.10.2 Nodal displacements of a space truss element.
The finite element method Chapter
11
495
where
1 ðx Þ ¼
2 ðx Þ ¼
3 ðx Þ ¼ 1 x,
4 ðx Þ ¼
5 ðx Þ ¼
x ¼ x=L
6 ðx Þ ¼ x
(11.10.3a)
(11.10.3b)
Because the element does not undergo bending, the axes on the cross-section
of the rod may have any orientation. Nevertheless, it is recommended to take
them in the directions of the cross-sectional principal axes.
The elastic energy of the element is due only to the axial deformation.
Therefore, for an element with variable cross-section Aðx Þ and mass m ðx Þ, it is
Z
1 L
2
EAðx Þ½u 0 ðx, t Þ dx
(11.10.4)
U ðu1 , u2 , …, u6 Þ ¼
2 0
The kinetic energy for consistent mass assumption is
Z
n
o
1 L
T ðu_ 1 , u_ 2 , …, u_ 6 Þ ¼
m ðx Þ ½u_ ðx, t Þ2 + ½v_ ðx, t Þ2 + ½w_ ðx, t Þ2 dx
2 0
(11.10.5)
while for lumped mass assumption is
1 1 T ðu_ 1 , u_ 2 , …, u_ 6 Þ ¼ m1 u_ 21 + u_ 22 + u_ 23 + m2 u_ 24 + u_ 25 + u_ 26
2
2
(11.10.6)
where
Z
m1 ¼
L
Z
m ðx Þð1 x Þdx and m2 ¼
0
L
m ðx Þxdx
(11.10.7)
0
Applying the procedure developed for the plane truss, we obtain the
stiffness matrix, the mass matrix, and the vector of equivalent nodal forces
of the space truss.
(i) Nodal elastic forces and stiffness matrix of the space truss element
The relation between the elastic nodal forces and nodal displacements is
in which
f eS ¼ ke ue
(11.10.8)
8 9
fS1 >
>
>
>
>
>
>f >
>
S2 >
>
>
>
>
>
=
<f >
S3
e
fS ¼
>
> fS4 >
>
>
>
>
>
>
>
>
>
f
>
>
S5
>
;
: >
fS6
(11.10.9a)
496 PART
II Multi-degree-of-freedom systems
2
8 9
u1 >
>
>
>
> >
>
>
>
u2 >
>
>
>
>
>
=
<u >
3
e
u ¼
>
> u4 >
>
>
>
>
>
>
>
>
>
u
>
>
5
>
;
: >
u6
(11.10.9b)
k11 0 0 k14 0 0
60
6
6
60
ke ¼ 6
6k
6 41
6
40
0
3
0 0 k44
0 0 0
0 07
7
7
0 07
7
0 07
7
7
0 05
0 0 0
0 0
0 0 0
0 0 0
(11.10.9c)
where
Z
kij ¼
L
EAðx Þ
0
0
i
0
j dx,
i, j ¼ 1, 4
(11.10.10)
For Aðx Þ ¼ A ¼ constant it is
2
3
1 0 0 1 0 0
6 0 0 0 0 0 07
7
6
7
e6
6
0 0 0 0 0 07
EA
7
ke ¼ e 6
7
L 6
6 1 0 0 1 0 0 7
7
6
4 0 0 0 0 0 05
0 0 0
(11.10.11)
0 0 0
(ii) Nodal inertial forces and mass matrix of the truss element
The relation between the inertial nodal forces and nodal displacements is
€e
f eI ¼ me u
(11.10.12)
8 9
fI 1 >
>
>
>
>
>
>f >
>
I2 >
>
>
>
>
>
=
<f >
I3
e
fI ¼
>
> fI 4 >
>
>
>
>
>
>
>
>
>
f
>
>
I
5
>
;
: >
fI 6
(11.10.13a)
in which
The finite element method Chapter
8 9
u€1 >
>
>
>
> >
>
>
>
€2 >
u
>
>
>
>
>
=
< u€ >
3
e
€ ¼
u
>
> u€4 >
>
>
>
>
>
>
>
>
>
€
u
>
>
5
>
;
: >
u€6
and me is the mass matrix.
For the consistent mass assumption, we obtain
2
3
m11 0
0
m14 0
0
60
0
m25 0 7
m22 0
6
7
6
0
0
m36 7
0
0
m
33
e
6
7
m ¼6
0
m44 0
0 7
6 m41 0
7
40
0
m54 0 5
m52 0
0
m66
0
0
m63 0
11
497
(11.10.13b)
(11.10.14)
where
Z
mij ¼ mi + 1, j + 1 ¼ mi + 2, j + 2 ¼
L
m ðx Þ i ðx Þ j ðx Þdx, i, j ¼ 1, 4
0
(11.10.15)
¼ constant the matrix
For m ðx Þ ¼ m
2
2 0
60 2
6
me 6
e
60 0
m ¼
6 6
61 0
40 1
0 0
in Eq. (11.10.14) becomes
3
0 1 0 0
0 0 1 07
7
2 0 0 17
7
(11.10.16)
0 2 0 07
7
0 0 2 05
1 0 0 2
¼ rAL is the total mass of the element.
where m e ¼ mL
For lumped mass assumption, the mass matrix results as
2
3
m11 0
0
0
0
0
60
0
0
0 7
m22 0
6
7
6
0
0 7
0
0
m33 0
7
me ¼ 6
60
0 7
0
0
m44 0
6
7
40
0
0
0
m55 0 5
0
0
0
0
0
m66
where m11 ¼ m22 ¼ m33 ¼ m1 , m44 ¼ m55 ¼ m66 ¼ m2 and
Z L
Z L
m ðx Þð1 xÞdx, m2 ¼
m ðx Þxdx
m1 ¼
0
0
(11.10.17)
(11.10.18)
498 PART
II Multi-degree-of-freedom systems
¼ constant, Eq. (11.10.17) becomes
When m ðx Þ ¼ m
2
3
1 0 0 0 0 0
60 1 0 0 0 07
6
7
60 0 1 0 0 07
1
7
me ¼ m e 6
7
2 6
60 0 0 1 0 07
40 0 0 0 1 05
0 0 0 0 0 1
(11.10.19)
(iii) Equivalent nodal loads of the space frame element
If the element is subjected to the axial load pðx, t Þ, the equivalent nodal load
vector is
9
8
p1 ðt Þ >
>
>
>
>
>
>
>
0
>
>
>
>
=
<
0
e
(11.10.20)
p ðt Þ ¼
p4 ðt Þ >
>
>
>
>
>
>
>
0
>
>
>
>
;
:
0
where
Z
L
pi ðt Þ ¼
px ðx, t Þ i ðx Þdx, i ¼ 1, 4
(11.10.21)
0
11.10.2 Transformation of the nodal coordinates of the space
truss element
By selecting the local axes y, z to coincide with the principal axes of the crosssection, we can compute the matrix of the direction cosines of the element
exactly as for the space frame element. Then, the nodal coordinates at the
end points are transformed according to Eqs. (11.9.32a), (11.9.32b). Hence,
we will have
8 9 2
38 9
l11 l12 l13 < u1 =
< u1 =
¼ 4 l21 l22 l23 5 u2
(11.10.22a)
u
: 2;
: ;
u3
l31 l32 l33
u3
8 9 2
38 9
l11 l12 l13 < u4 =
< u4 =
¼ 4 l21 l22 l23 5 u5
(11.10.22b)
u
: 5;
: ;
u6
l31 l32 l33
u6
which are combined to
ue ¼ ðRe ÞT ue
(11.10.23)
The finite element method Chapter
where
Re ¼
L 0
0 L
11
499
(11.10.24)
is the transformation matrix of the e element of the space truss with dimensions
6 6. The remaining procedure for formulating the equation of motion is the
same as for the plane truss.
Example 11.10.1 Formulate the equation of motion of the space truss of
Fig. E11.11. All bars have the same cross-sectional area A. The distributed load
on element 1 is pðx, t Þ ¼ ð1 x=LÞP ðt Þ=L. Assume: modulus of elasticity E,
material density r, lumped mass assumption.
Solution
Fig. E11.11 Space truss in Example 11.10.1.
The free structure has n ¼ 4 nodes and N ¼ 3n ¼ 12 degrees of freedom. We
assume that the principal planes xy of all elements are normal to the plane
x y. Thus, we take the projection of the node 4 on the plane x y as point P, that
is, P ð1:5, 2, 0Þ, which is common for all elements. Then, taking into account the
geometrical data of Table E11.6, we obtain
TABLE E11.6 Geometrical data of the elements.
e
xj
xk
D
x
yj
yk
D
y
zj
zk
D
z
L
1
3
1.5
1.5
0
2
2
0
4
4
4.717
2
0
1.5
1.5
0
2
2
0
4
4
4.717
3
3
1.5
1.5
4
2
2
0
4
4
4.717
500 PART
II Multi-degree-of-freedom systems
Computation of the transformation matrices Re, e¼1,2,3
Element 1
e1 ¼ f 0:318 0:424 0:848 gT ,
e3 ¼
rP1 ¼ f 1:5 2 0 gT
e1 rP1
T
T
¼ f 0:8 0:6 0 g , e2 ¼ e3 e1 ¼ f 0:508 0:678 0:530 g
je1 rP1 j
2
3
"
#
1
L
0
7
0:678 0:530 5, R1 ¼
0 L1
0:800 0:600
0
0:318
6
L1 ¼ 4 0:508
0:424
0:848
Element 2
e1 ¼ f 0:318 0:424 0:848 gT ,
e3 ¼
rP2 ¼ f 1:5 2 0 gT
e1 rP2
¼ f 0:8 0:6 0 gT , e2 ¼ e3 e1 ¼ f 0:508 0:678 0:530 gT
je1 rP2 j
2
3
2
0:318 0:424 0:848
L 0
L ¼ 4 0:508 0:678 0:530 5, R2 ¼
0 L2
0:800 0:600 0
2
Element 3
e1 ¼ f 0:318 0:424 0:848 gT , rP3 ¼ f 1:5 2 0 gT
e3 ¼
e1 rP3
¼ f 0:8 0:6
je1 rj
2
0:318
L3 ¼ 4 0:508
0:800
0 gT , e2 ¼ e3 e1 ¼ f 0:508 0:678 0:530 gT
3
3
0:424 0:848
L 0
3
5
0:678 0:530 , R ¼
0 L3
0:600 0
Computation of ke , me , pe ðt Þ, e ¼ 1, 2, 3
2
3
1 0 0 1 0 0
6 0 0:5 0
0 0 0 7
6
7
6
7
0
0
0:5
0
0
0
EA
6
7
k1 ¼ k2 ¼ k3 ¼
6
7
L 6 1 0 0
1 0 0 7
6
7
4 0 0 0
0 0:5 0 5
0 0 0
0 0 0:5
2
3
0:5 0 0 0 0 0
60
0:5 0 0 0 0 7
6
7
6
0
0 0:5 0 0 0 7
7
m1 ¼ m2 ¼ m3 ¼ rAL6
60
0 0 0:5 0 0 7
6
7
40
0 0 0 0:5 0 5
0
0 0 0 0 0:5
The finite element method Chapter
Z
p11 ðt Þ ¼
L
1 ðx Þdx
¼ 0:333P ðt Þ,
pðx, t Þ
4 ðx Þdx
¼ 0:167P ðt Þ
0
Z
p14 ðt Þ ¼
pðx, t Þ
11
L
0
9
8
0:333 >
>
>
>
>
>
>
>
0
>
>
>
>
=
<
0
1
P ðt Þ,
p ðt Þ ¼
0:167 >
>
>
>
>
>
>
>
>
>
>
>0
;
:
0
p2 ðt Þ ¼ p3 ðt Þ ¼ 0
e
e , pe ðt Þ, e ¼ 1, 2, 3
Computation of k , m
T
k1 ¼ R1 k1 R1
3
2
0:101 0:135 0:27 0:101 0:135 0:27
6 0:135 0:18
0:36
0:135 0:18 0:36 7
7
6
7
6
0:27
0:36
0:719 0:27 0:36 0:719 7
EA 6
7
6
¼
L 6
0:101 0:135 0:27 7
7
6 0:101 0:135 0:27
7
6
4 0:135 0:18 0:36 0:135 0:18
0:36 5
0:27
T
k2 ¼ R2 k2 R2
2
0:101
6 0:135
6
6
0:27
EA 6
6
¼
6
L 6 0:101
6
4 0:135
0:27
T
k3 ¼ R3 k3 R3
2
0:101
6 0:135
6
6
0:27
EA 6
6
¼
6
L 6 0:101
6
4 0:135
0:27
0:36
0:719 0:27
0:135 0:27
0:36
0:101 0:135
0:719
0:27
3
0:36 7
7
7
0:36
0:719 0:27
0:36 0:719 7
7
0:135 0:27
0:101 0:135 0:27 7
7
7
0:18
0:36
0:135 0:18 0:36 5
0:36 0:719 0:27 0:36
0:719
0:18
0:36
0:135 0:18
0:101 0:135 0:27
0:135
0:27
0:18
0:36
0:36 0:135 0:18
0:719 0:27 0:36
0:135 0:27
0:101
0:18
0:36
0:135
0:27
0:36
0:719
3
0:36 7
7
7
0:719 7
7
0:135 0:27 7
7
7
0:18
0:36 5
0:36
0:719
501
502 PART
II Multi-degree-of-freedom systems
2
0:5 0
0
0
0
3
0
6 0 0:5 0 0 0 0 7
7
6
7
6
6
0 0 0:5 0 0 0 7
7
1 ¼m
2 ¼m
3 ¼ rAL6
m
6 0 0 0 0:5 0 0 7,
7
6
7
6
4 0 0 0 0 0:5 0 5
0 0 0 0 0 0:5
9
8
0:106 >
>
>
>
>
>
>
>
>
0:141 >
>
>
>
>
>
=
< 0:283 >
P ðt Þ, p2 ðt Þ ¼ p3 ðt Þ ¼ 0
p1 ðt Þ ¼
>
>
0:053
>
>
>
>
>
>
>
>
>
0:070 >
>
>
>
>
;
:
0:141
Subsequently, the assembly matrices ae , ðe ¼ 1, 2, 3Þ are formulated
2
1
60
6
60
a1 ¼ 6
60
6
40
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
0
60
6
60
a3 ¼ 6
60
6
40
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
3
2
0
0
60
07
7
6
6
07
7, a2 ¼ 6 0
60
07
7
6
40
05
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0 0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
3
0
07
7
07
7
07
7
05
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
3
0
07
7
07
7
07
7
05
1
and the total matrices of the structure are computed, namely
¼
K
3
X
e¼1
e
¼
ðae ÞT k ae , M
3
X
e¼1
e ae , pðt Þ ¼ Pðt Þ +
aT m
3
X
ðae ÞT pðt Þe
e¼1
e M
e and load vector p
eðt Þ due to support conditions
Modified matrices K,
The equation of motion will be derived with respect to the three free degrees
of freedom of node 4. Due to the support conditions, the displacement vector
should be modified as
e
e1 u
e2 u
e3 u
e4 u
e5 u
e6 u
e7 u
e8 u
e9 u
e10 u
e11 u
e12 g
uT ¼ f u
¼ f u10 u11 u12 u1 u2 u3 u4 u5 u6 u7 u8 u9 g
The finite element method Chapter
Therefore, the matrix V
2
0
60
6
61
6
60
6
61
6
60
V¼6
60
6
60
6
60
6
60
6
40
0
11
503
defined by Eq. (11.2.111) is
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
1
0
1
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
3
0
07
7
07
7
07
7
07
7
07
7
07
7
07
7
07
7
07
7
05
0
After partitioning the modified matrices, the equation of motion results from
Eq. (10.2.117) which in terms of the original numeration reads
9
9
2
38
38
0:303 0:135 0:270 < u10 =
1:5 0 0
< u€10 = EA
4 0:135 0:539 0:359 5 u11
rAL4 0 1:5 0 5 u€11 +
;
:€ ;
:
L
0:270 0:359 2:157
u12
0 0 1:5
u12
9
8
< 0:053 =
¼ 0:929 P ðt Þ
;
:
1:859
2
11.11 Rigid bodies within flexible skeletal structures
Skeletal structures may often embrace bodies of finite dimensions whose deformation is very small compared to that of the elements of the structure. These
bodies may be considered as rigid bodies. Fig. 11.11.1 shows examples of such
structures. During motion, these bodies oblige the ends of the elements connected with them to ride the displacements of the rigid body, which are three
for the plane body and six for the three-dimensional body.
11.11.1 Rigid bodies in spaces frames
First, we will derive the equations of motion of a space frame embracing rigid
bodies. We consider an element e, whose ends j and k are connected to the bodies ðJ Þ and ðK Þ, respectively, Fig 11.11.2. Points J and K of the bodies, which
in general do not coincide with the center of mass of the bodies, are taken as
points of reference of their motion. The axes x y z are parallel to the global axes
of the structure. All points of reference of the motion are considered as virtual
(fictitious) nodes of the frame, the displacements of which will be the
504 PART
II Multi-degree-of-freedom systems
Fig. 11.11.1 Rigid bodies B in skeletal structures.
Fig. 11.11.2 Element e of a space frame with rigid bodies at its ends.
coordinates of the motion. This imposes the transfer of the element end
forces from its ends j and k to J and K , respectively. This is achieved by working as follows.
The positive direction of the element is from point j to point k. We denote by
y j ¼ yj yJ , D
z j ¼ zj zJ the coordinates of point j with
D
x j ¼ xj xJ , D
respect to the system J x y z (see Fig. 11.11.3). The position vector defined
by the relation
x j i + D
y j j + D
zj k
rj ¼ D
(11.11.1)
will be referred to as the translation (transfer) vector or the offset vector and is
directed from point J to point j.
(i) Transfer of the nodal forces from j to J
The finite element method Chapter
11
505
Fig. 11.11.3 Rigid body (J ).
The nodal element forces at the end j and those transferred to point J are
given by
ð11:11:2a; bÞ
in which f aj , f aJ denote forces while f bj , f bJ moments all referred to the global
system of axes. Nevertheless, the overbars are omitted here for the sake of simplifying the symbols. Therefore, the vector f j has dimensions 6 1 and may
e
e
represent any of the vectors f I , f S , pe ðt Þ at j. The damping is not considered
here but it will be introduced as proportional damping in the final equation
of motion (see Section 12.11).
Due to transfer, we have
(11.11.3a)
f aJ ¼ f aj
f bJ ¼ f bj + rj f aj
(11.11.3b)
After expanding the vector product using vector components, the foregoing
equations are written as
ð11:11:4Þ
or
f J ¼ Tj f j
(11.11.5)
506 PART
II Multi-degree-of-freedom systems
where
Tj ¼
I 0
ej I
(11.11.6)
represents the transformation matrix of the end forces from point j to J .
I represents 3 3 unit matrix and
2
3
0
D
zj
D
yj
zj
0
D
xj 5
(11.11.7)
ej ¼ 4 D
D
yj
D
xj 0
is the offset (or eccentricity) matrix, which is equal to zero in the absence of a
rigid body because rj ¼ 0. Obviously, the matrix ej is antisymmetric, namely
eTj ¼ ej
(11.11.8)
Eq. (11.11.5) can be readily inverted by taking point j as a reference point
and repeating the previous procedure. This yields
f j ¼ TTj f J
where
TTj ¼
I ej
I eTj
¼
¼
0 I
0 I
(11.11.9)
(11.11.10)
(ii) Transfer of the displacements from J to j
The nodal displacements at J and those transferred to point j are denoted in
correspondence to Eq. (11.11.2a) , Eq. (11.11.2b) by
ð11:11:11a; bÞ
in which uaj ,uaJ denote translations while ubj , ubJ rotations.
The displacements at j in terms of the displacements J are given by the
expressions
uaj ¼ uaJ
(11.11.12a)
ubj ¼ ubJ + uaJ rj
(11.11.12b)
The term uaJ rj in Eq. (11.11.12b) results from Eq. (Α.2.10) of Appendix Α
by setting w ¼uaJ and R ¼ rj .
After expanding the vector product using vector components, the foregoing
equations yield
The finite element method Chapter
uj ¼ TTj uJ
11
507
(11.11.13)
Similar relations are obtained for the nodal displacements and forces at end
k. Namely
uk ¼ TTk uK
(11.11.14)
f k ¼ TTk f K
(11.11.15)
where
TTk
2
I eTk
¼
0 I
(11.11.16)
3
0
D
zk
D
yk
zk
0
D
xk 5
ek ¼ 4 D
D
y k D
xk 0
(11.11.17)
y k , D
z k are the coordinates of the point k with respect
The quantities D
x k , D
to the system of axes K x y z (see Fig. 11.11.2). Thus, the offset vector is
x k i + D
y k j + D
zk k
rk ¼ D
(11.11.18)
Eqs. (11.11.5), (11.11.15) are combined as
f 0 e ¼ Te f e
where
f e ¼
fj
0e
, f ¼
fk
(11.11.19)
fJ
T 0
, Te ¼ j
fK
0 Tk
(11.11.20)
Similarly, Eqs. (11.11.13), (11.11.14) are combined as
u0 ¼ Te ue
e
(11.11.21)
where
ue ¼
uj
e
, u0 ¼
uk
uJ
uK
(11.11.22)
The quantities provided with a prime are referred to the virtual nodes
J and K .
(iii) Transformation of the stiffness matrix, mass matrix, and equivalent nodal
load vector due to the offset
Eq. (11.11.19) is written as
f 0 e ¼ Te f e
S
S
e
¼ Te k ue
e e
(11.11.23)
e T 0e
¼ T k ðT Þ u
hence the transformed stiffness matrix is
0e
e
k ¼ Te k ðTe ÞT
(11.11.24)
508 PART
II Multi-degree-of-freedom systems
Applying the same procedure yields the transformed mass matrix
0 ¼ Te m
e ðTe ÞT
m
e
(11.11.25)
The vector of the equivalent nodal forces is transformed according to
Eq. (11.11.19). Thus, we have
p0 ðt Þe ¼ Te pðt Þe
(11.11.26)
The remaining procedure to formulate
the
equation is known. That is,
e
e
e
^ 0 , and p
^0 , M
^0 ðt Þ using the assembly
we formulate the enlarged matrices K
matrices and we establish the equation
0 u0 ¼ p0 ðt Þ
0 u€0 + K
M
(11.11.27)
where
0¼
M
Ne
X
e¼1
0 ¼
^0 , K
M
e
Ne
X
^ 0 , p0 ðt Þ ¼ P0 ðt Þ +
K
e
e¼1
Ne
X
^ 0 ðt Þe
p
(11.11.28)
e¼1
The components of the vector P0 ðt Þ represent the loads that are applied
directly to the virtual nodes. But in real structures, the external loads are usually
applied to points not coinciding with the virtual nodes, therefore they must
previously be transferred to these points. The procedure is the same as that
employed for the element nodal forces. Thus, if P is the point of application
of the load with an offset vector.
x p i + D
y p j + D
zpk
rp ¼ D
(11.11.29)
its transfer from P to J is performed using the relation
PJ ðt Þ ¼ Tp Pp
(11.11.30)
Eq. (11.11.27) is not the equation of motion of the structure because so far
the inertial forces produced by the motion of the rigid bodies have not been
included in it. Apparently, these forces are the most important in the considered
type structures.
This is achieved if the motion of the rigid bodies is taken into account.
In Appendix, it is shown that the motion of a rigid body with respect to its center
of mass is described by Eqs. (A.2.2), (A.2.8), namely
€c
F ¼ mR
(11.10.31a)
_ c
Mc ¼ H
(11.10.31b)
where F is the resultant force, Mc the resultant moment with respect to the
center of mass, and Hc the angular momentum.
Introducing Eq. (A.2.14) in Eq. (11.11.31b), using (Α.2.11) and neglecting
the nonlinear terms (the products of rotations) because we examine small
displacements, we obtain the linearized Euler’s equations with respect to the
principal axes of the body
The finite element method Chapter
€c
F ¼ mR
11
509
(11.11.32a)
Mc ¼ Ix w_ x + Ixy w_ y + Ixz w_ z e1
+ Iyx w_ x + Iy w_ y + Iyz w_ z e2
+ Izx w_ x + Izy w_ y + Iz w_ z e3
(11.11.32b)
The quantities Ix , Ixy , …, Iz are the moments and the products of inertia,
which are given by Eq. (Α.2.15). Using the notation of the current section,
we write the forgoing equations
€c
f c ¼ Mc u
(11.11.33)
where
8 9
8 9
2
m
fc1 >
uc1 >
>
>
>
>
>
>
>
> >
>
>
>
>
>
60
f
u
>
>
>
>
c2
c2
>
>
6
=
=
< >
< >
60
f
uc3
, Mc ¼ 6
f c ¼ c3 , uc ¼
60
uc4 >
>
>
>
>
> fc4 >
>
6
>
>
>
>
>
>
>
>
40
u >
>
>
>
> fc5 >
>
;
;
:
: c5 >
0
fc6
uc6
0
m
0
0
0
0
0
0
m
0
0
0
0
0
0
Ix
Iyx
Izx
0
0
0
Ixy
Iy
Izy
3
0
0 7
7
0 7
7
Ixz 7
7
Iyz 5
Iz
(11.11.34)
f c , uc , and Mc the vector of the inertia forces applied at the center of mass C , the
displacement vector of the center of mass and the mass (inertia) matrix of
the body with respect to center of mass, respectively. If the point of reference
of the motion is not the center of mass, but an arbitrary point P, then Mc must
be transformed from C to P. The transformation of f c and uc is achieved by
Eqs. (11.11.5), (11.11.13). Thus, we have
where
f p ¼ Tc f c
(11.11.35)
uc ¼ TTc up
(11.11.36)
3
0
D
zc
D
yc
I 0
, ec ¼ 4 D
zc 0
D
xc 5
Tc ¼
ec I
D
y c D
xc 0
2
(11.11.37)
y c ¼ yc yp , D
z c ¼ zc zp are the components of the
and D
x c ¼ xc xp , D
offset vector rc from the center of mass to the reference point P.
After that, we may write
f p ¼ Tc f c
€c
¼ Tc M c u
€p
¼ Tc Mc TTc u
Hence
Mp ¼ Tc Mc TTc
(11.11.38)
0
of
Then the mass matrices of all rigid bodies are added to mass matrix M
Eq. (11.11.27) after they are enlarged by appropriate assembly matrices. Therefore, we obtain the equation of motion
510 PART
II Multi-degree-of-freedom systems
0 u0 ¼ p0 ðt Þ
∗ u€0 + K
M
(11.11.39)
where now
∗¼
M
Nb
X
^i +
M
p
i¼1
Ne
X
^0
M
e
(11.11.40)
e¼1
^ i are the enlarged mass matrices of the Nb rigid bodies. The procein which M
p
∗ is illustrated with Example 11.11.1.
dure of determining M
11.11.2 Rigid bodies in spaces trusses, plane grids, plane frames,
and plane trusses
Rigid bodies enclosed in the other types of skeletal structures can be studied
following the procedure developed for the space frame. They differ only in
the transformation matrix Tj and in the inertia matrix Mc . These matrices
for all types of structures are summarized in Table 11.11.1. The transformation
matrices can be readily found from that valid for the space frame (Type 5 in
Table 11.11.1) by deleting appropriate rows and columns. For example, point
TABLE 11.11.1 Transformation matrices Tj and rigid body mass-inertia
matrix Mc of skeletal structures.
Type of structure
1. Plane truss
(structure in
xy-plane)
2. Plane frame
(bending in
xy-plane)
3. Grid (structure
in xy-plane)
4. Space truss
5. Space frame
Tj
2
Mc
3
1
0
4 0
1 5
D
yj
D
xj
2
3
1
0
0
4 0
1
05
D
xj 1
D
yj
2
3
1 0 D
yj
4 0 1 D
xj 5
0 0 1
3
2
1
0
0
6 0
1
0 7
7
6
6 0
0
1 7
7
6
7
6 0
D
z
D
y
j
j7
6
4 D
zj
0
D
xj 5
D
yj
D
xj
0
2
1
0
0
0
6 0
1
0
0
6
6 0
0
1
0
6
6 0
D
zj
D
yj 1
6
4 D
zj
0
D
xj 0
D
yj
D
xj
0
0
2
0
0
0
0
1
0
3
0
07
7
07
7
07
7
05
1
m
40
0
2
m
40
0
2
Ix
4 Iyx
0
2
m
60
6
60
6
60
6
40
0
2
m
60
6
60
6
60
6
40
0
3
0 0
m 05
0 Iz
3
0 0
m 05
0 Iz
3
Ixy 0
Iy 0 5
0 m
0
m
0
0
0
0
0
0
m
0
0
0
0
0
0
Ix
Iyx
Izx
0
0
0
Ixy
Iy
Izy
0
m
0
0
0
0
0
0
m
0
0
0
0
0
0
Ix
Iyx
Izx
0
0
0
Ixy
Iy
Izy
3
0
0 7
7
0 7
7
Ixz 7
7
Iyz 5
Iz
3
0
0 7
7
0 7
7
Ixz 7
7
Iyz 5
Iz
The finite element method Chapter
11
511
J for a plane body in a plane truss has three displacements, two translations in
the x and y directions and a rotation about the z axis. However, a joint j of the
truss has only two displacements, the translations in the x and y directions.
Therefore, we keep only the first, second, and sixth rows and the first and second columns in the 6 6 matrix. Thus, a 3 2 transformation matrix results for
the plane truss. It should be noted that the 3 3 matrix pertaining to the grid
requires not only the deletion of rows and columns but also their rearrangement.
The inertia matrix Mc results from the inertia properties of the rigid body moving within the structure. It can also be found from that valid for the space frame
(Type 5 in Table) by deleting appropriate rows and columns.
Example 11.11.1 Formulate the equation of motion of the truss of Fig. E11.12.
Coordinates of nodes: 1ð0, 0Þ, 2ð4, 0Þ, 3ð0, 3Þ, 4ð4, 3Þ, 5ð8, 3Þ; cross-section of
bars: A1 ¼ A2 ¼ 1:5A, A3 ¼ A5 ¼ A, A4 ¼ A6 ¼ A7 ¼ 2:1A, A ¼ 27cm2 ; damping coefficient: cs ¼ 0; modulus of elasticity: E ¼ 2:1 108 kN=m2 ; acceleration of gravity g ¼ 9:81m=s2 material density r ¼ 7:55kNm1 s2 =m3 ;
consistent mass assumption. The plane rigid body B has dimensions 3:0 3:0 0:02 m3 and specific weight 24kN=m3 .
Fig. E11.12 Plane truss in Example 11.5.2.
Solution
The structure has n ¼ 5 nodes. We take node 1 as the point of reference of the
motion of the rigid body. The free structure has N ¼ 9 degrees of freedom, namely
the translational displacements u1 , u2 and the rotation u3 of the rigid body about
point 1 and the six displacements of nodes 2, 4, 5 (see Fig. E11.12). Thus, it is
u ¼ f u1 u2 u3 u4 u5 u6 u7 u8 u9 gT
Computation of ke , me , pe ðt Þ for e ¼ 1, …,7
2
0:375
6
0
k1 ¼ k2 ¼ EA6
4 0:375
0
3
2
0 0:375 0
0:250
6
0 0
07
7, k3 ¼ EA6 0
4 0:250
0 0:375 0 5
0 0
0
0
3
0 0:250 0
0 0
07
7
0 0:250 0 5
0 0
0
512 PART
II Multi-degree-of-freedom systems
2
0:420
6 0
4
7
6
k ¼ k ¼ EA4
0:420
0
3
2
0
0:20
6
07
7, k5 ¼ EA6 0
4 0:20
05
0
0
0 0:420
0 0
0 0:420
0 0
2
0:70
6
0
k6 ¼ EA6
4 0:70
0
2
2
6
0
m1 ¼ m2 ¼ rA6
41
0
0
2
0
1
2
3:50
6
0
m4 ¼ m7 ¼ rA6
4 1:75
0
0 0:70
0 0
0 0:70
0 0
3
2
0
1:333
6
17
7, m3 ¼ rA6 0
4 0:666
05
1
0
2
0
2
0
3:50
0
1:75
0
1:75
0
3:50
0
2
2:10
6
0
m6 ¼ rA6
4 1:05
0
0 0:20
0 0
0 0:20
0 0
3
0
07
7
05
0
0
1:333
0
0:666
3:50
0
1:05
0
2:10
0
3
0
0:666 7
7
5
0
0:666
0
1:333
0
3
2
0
1:667
6
1:75 7
7, m5 ¼ rA6 0
4 0:833
0 5
0
2:10
0
1:05
3
0
07
7
05
0
1:333
0
1:667
0
0:833
0:833
0
1:667
0
3
0
1:05 7
7
0 5
2:10
p1 ðt Þ ¼ p2 ðt Þ ¼ p3 ðt Þ ¼ p4 ðt Þ ¼ p5 ðt Þ ¼ p6 ðt Þ ¼ p7 ðt Þ ¼ 0
Computation of Re
2
1
60
1
2
3
R ¼R ¼R ¼6
40
0
0
1
0
0
0
0
1
0
3
2
0
0:8
6 0:6
07
4
7
7, R ¼ R ¼ 6
4 0
05
1
0
2
3
2
0:8 0:6 0
0
0
6 0:6 0:8 0
7
6 1
0
5
6
7, R ¼ 6
R ¼6
4 0
4 0
0 0:8 0:6 5
0
0 0:6 0:8
0
e
e , pe ðt Þ
Computation of k , m
2
0:375
6 0
1
2
k ¼ k ¼ EA6
4 0:375
0
0 0:375
0 0
0 0:375
0 0
0:6 0
0:8 0
0
0:8
0 0:6
1 0
0 0
0 0
0 1
3
2
0
0:250
6 0
07
3
7, k ¼ EA6
4 0:250
05
0
0
3
0
0 7
7
0:6 5
0:8
3
0
07
7
15
0
0 0:250
0 0
0 0:250
0 0
3
0
07
7
05
0
3
0
0:833 7
7
5
0
1:667
The finite element method Chapter
513
11
2
2
3
0:268 0:202 0:268 0:202
0 0
6 0:202 0:151 0:202 0:151 7
6 0 0:7
4
7
6
6
6
7
k ¼ k ¼ EA4
, k ¼ EA4
0:268 0:202
0:268 0:202 5
0 0
0:202 0:151
0:202 0:151
0 0:7
3
0 0
0 0:7 7
7
0 0 5
0 0:7
2
3
0:128 0:096 0:128
0:096
6 0:096
0:072 0:096 0:072 7
7
k5 ¼ EA6
4 0:128
0:096 0:128 0:096 5
0:096 0:072 0:096
0:072
2
2
3
2 0 1 0
1:333 0
0:666
60 2 0 17
60
1:333
0
1
2
3
7 ¼ rA6
¼ rA6
¼m
m
4 1 0 2 0 5, m
4 0:666 0
1:333
0 1 0 2
0
0:666 0
2
2
3
3:50 0
1:75 0
1:667 0
6
6
7
0
3:50
0
1:75
0
1:667
7, m
7 ¼ rA6
5 ¼ rA6
4 ¼ m
m
4 1:75 0
4 0:833 0
3:50 0 5
0
1:75 0
3:50
0
0:833
2
2:10
60
6
¼ rA6
m
4 1:05
0
0
2:10
0
1:05
1:05
0
2:10
0
3
0
0:666 7
7
5
0
1:333
0:833
0
1:667
0
3
0
0:833 7
7
5
0
1:667
3
0
1:05 7
7
0 5
2:10
Computation of Te
Because point 1 has been chosen as the point of reference, only the elements 2
and 5 are affected by the motion of the rigid body. Referring to Table 11.11.1
and Eqs. (11.11.24), (11.11.25), we obtain
2
3
1 0 0 0
2
3
6 0 1 0 07
1 0
6
7
7
y 3 ¼ 3, T2j ¼ 4 0 1 5 and T2 ¼ 6
Element 2. D
x 3 ¼ 0, D
6 3 0 0 0 7
4 0 0 1 05
3 0
0 0 0 1
2
3
0:375 0 1:125 0:375 0
6 0
0 0
0
07
6
7
02
1:125
0
3:375
1:125
07
k ¼ EA6
6
7
4 0:375 0 1:125 0:375 0 5
0
0 0
0
0
514 PART
II Multi-degree-of-freedom systems
2
1 0 0
60 1 0
1 0
6
Element 5. D
y 3 ¼ 3, D
x 3 ¼ 0, T5k ¼ 4 0 1 5 and T5 ¼ 6
60 0 1
40 0 0
3 0
0 0 3
2
3
0:128 0:096 0:128 0:096 0:384
6 0:096 0:072 0:096 0:072 0:288 7
6
7
05
7
k ¼ EA6
6 0:128 0:096 0:128 0:096 0:384 7
4 0:096 0:072 0:096 0:072 0:288 5
0:384 0:288 0:384 0:288 1:152
2
3
0e
3
0
07
7
07
7
15
0
0e ¼ m
e , p0 e ¼ pe .
For the remaining elements e ¼ 1, 3, 4, 6, 7 it is k ¼ k , m
e
Computation of the mass matrix M1
It is convenient to express the area AB and the density rB of the plane
body in terms of A, r. Thuswe obtain AB ¼ 3333:33A, rB ¼ 0:324r. This yields
m ¼ 21:6rA, Ic ¼ 21:6rA 32 + 32 =12 ¼ 32:4rA. Then from Table 11.11.1,
we have
2
3 2
3
2
3
1
0 0
21:6 0
0
1
0
0
1 05
1
05 ¼ 4 0
Mc ¼ rA4 0 21:6 0 5 Tc ¼ 4 0
xc 1
1:5 1:5 1
0
0 32:4
D
y c D
and by virtue of Eq. (11.11.38), we obtain
2
21:6
0
M1 ¼ rA4 0
21:6
32:4 32:4
3
32:4
32:4 5
128:9
Finally, the load P ðt Þ is transferred from point A to 1
8
9
2
3
1 0
< 0=
0
¼ 1 P ðt Þ
P1 ðt Þ ¼ 4 0 1 5
P ðt Þ
:
;
3 3
3
The finite element method Chapter
Assembly matrices ae ðe ¼ 1, 2, …, 7Þ. Referring to Fig.
2
2
3
1 0 0 0 0
1 0 0 0 0 0 0 0 0
60 1 0 0 0
60 1 0 0 0 0 0 0 07 2 6
6
7
a1 ¼ 6
4 0 0 0 1 0 0 0 0 0 5, a ¼ 6 0 0 1 0 0
40 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0
2
3
2
0 0 0 0 0 1 0 0 0
1 0 0 0 0
6
7
6
0
0
0
0
0
0
1
0
0
7 4 60 1 0 0 0
a3 ¼ 6
4 0 0 0 0 0 0 0 1 0 5, a ¼ 4 0 0 0 0 0
0 0 0 0 0 0 0 0 1
0 0 0 0 0
2
3
2
0 0 0 1 0 0 0 0 0
0 0 0 1 0
60 0 0 0 1 0 0 0 07
6
7 6 60 0 0 0 1
5
6
7
a ¼6
6 1 0 0 0 0 0 0 0 0 7, a ¼ 4 0 0 0 0 0
40 1 0 0 0 0 0 0 05
0 0 0 0 0
0 0 1 0 0 0 0 0 0
2
3
0 0 0 1 0 0 0 0 0
60 0 0 0 1 0 0 0 07
7
a7 ¼ 6
40 0 0 0 0 0 0 1 05
0 0 0 0 0 0 0 0 1
The assembly matrix for the
2
1
ap ¼ 4 0
0
11
515
E11.12, we have
3
0 0 0 0
0 0 0 07
7
0 0 0 07
7
1 0 0 05
0 1 0 0
3
0 0 0 0
0 0 0 07
7
1 0 0 05
0 1 0 0
0
0
1
0
0
0
0
1
0
0
0
0
3
0
07
7
05
0
plane rigid body is
3
0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 05
0 1 0 0 0 0 0 0
Note that the dimensions of a2 and a5 are 5 9 while the dimensions of ap
are 39.
The enlarged element matrices and load vectors are computed using
Eqs. (11.2.92), (11.2.95), (11.2.96), namely
e
e
e
^ 0 ¼ ðae ÞT k0 e ae , p
^ 1 ¼ ðap ÞT M1 ap
^ 0 ¼ ðae ÞT m
0 ae , K
^0 ðt Þ ¼ ðae ÞT p0 ðt Þ, M
M
e
e
and load vector pðt Þ are computed using
The total stiffness matrix K
Eqs. (11.2.100), (11.2.107), respectively, namely
¼
K
Ne
X
e¼1
^ 0 , pðt Þ ¼ P
ðt Þ +
K
e
Ne
X
^ 0 ðt Þ
p
e
e¼1
ðt Þ is the vector of the external loads, that is,
where P
ðt Þ ¼ P ðt Þf 0 1 3 0 0 0 1 0 5 gT
P
516 PART
II Multi-degree-of-freedom systems
Regarding the total mass matrix, it is computed using Eq. (11.11.40)
∗ ¼M
^1+
M
7
X
^0
M
e
e¼1
Taking into account the support conditions (
u 2 ¼ u4 ¼ u5 ¼ 0), the following
transformation matrix V is constructed
2
3
1 0 0 0 0 0 0 0 0
60 0 0 0 0 0 1 0 07
6
7
60 1 0 0 0 0 0 0 07
6
7
60 0 0 0 0 0 0 1 07
6
7
7
V¼6
60 0 0 0 0 0 0 0 17
60 0 1 0 0 0 0 0 07
6
7
60 0 0 1 0 0 0 0 07
6
7
40 0 0 0 1 0 0 0 05
0 0 0 0 0 1 0 0 0
and the load vector pðt Þ yielding the
∗, K
which modifies the matrices M
equation of motion
e ff e
e ff e
u€f + K
uf ¼ e
pf∗ ðt Þ
M
where
2
30:77
6 43:40
6
6
e ff ¼ rA6 2:75
M
6 0
6
4 0
0
43:40 2:75 0
162:60 3:00 0
3:00 8:93 0
0
0
8:93
0
0:67 0
0
0
0:67
0
0
0:67
0
4:83
0
3
0
0 7
7
0 7
7:
0:67 7
7
0 5
4:83
2
3
1:15 1:51 0:64 0:20 0
0
6 1:51 4:53 0:12 0
7
0
0
6
7
6 0:64 0:12 0:89 0:20 0:25 0
7
e ff ¼ EA6
7,
K
6 0:20 0
7
0:20 0:85 0
0
6
7
4 0
0
0:25 0
0:52 0:20 5
0
0
0
0
0:20 0:15
e
uf ðt Þ ¼ f u3 u4 u6 u7 u8 u9 gT
pf∗ ðt Þ ¼ P ðt Þf 0 3 0 1 0 5 gT , e
Remark. The solution of the equation of motion gives the components of the
displacement e
uf ðt Þ due to the dynamic loads. However, the weight Wc ¼ 3 3 0:02 24 ¼ 4:32kN of the rigid body produces a static deformation, which
should be added to the dynamic one in order to obtain the total deformation. This
is achieved as follows.
The finite element method Chapter
First, the weight Wc is transferred from point C to point 1
8
9
2
3
1
0
< 0 =
0
¼ 4:32
1 5
W1 ¼ 4 0
Wc
:
;
6:48
1:5 1:5
11
517
T
This yields the load vector Pst ¼ 0 4:32 6:48 0 0 0 0 0 0 , which
is
e st ¼ VPst ¼ 0 6:48 0 0 0 0 4:32 0 0 T .
modified due to the supports
to
P
e stf ¼ 0 6:48 0 0 0 0 T and finally
Then we obtain P
e ff
ust ¼ K
1
Pstf ¼
1
f 2:34 3:07 3:45 1:37 3:45 4:60 gT
EA
11.12 Problems
Problem
P11.1 A plane truss element of length L has variable cross-section
A ¼ A0 1 0:05x2 , x ¼ x=L. The element is loaded by the distributed axial load
t and the suddenly applied concentrated load P ¼ P0 at
pðx, t Þ ¼ p0 ð1 + xÞ sin w
the cross-section x ¼ 0:25L. Compute: (a) the matrices ke , me and the vector pe .
For the mass matrix consider (i) consistent mass assumption and (ii) lumped
mass assumption. Compare ke , me with those resulting when the exact shape
functions are employed. (b) Transform ke , me , pe to the
pffiffiffi system of axes,
global
if the coordinates of element end points j and k are 1 + 3 L=2, 3L=2 and
ðL=2, LÞ, respectively.
Problem P11.2 The height of the cross-section Aðx Þ ¼ bh ðx Þ of a plane frame
element of length L varies linearly so that h ðLÞ ¼ 0:5h ð0Þ. The element is
t, the distributed bendloaded by the distributed load
py ðx, t Þ ¼ p0 ð1 x Þsin w
t as well as by the concentrated
ing moment mz ðx, t Þ ¼ m0 1 x + x 2 cos w
load Py ðL=3Þ and the concentrated moment Mz ð3L=4Þ, which are applied suddenly. Compute the matrices ke , me and the vector pe ðt Þ and compare them with
those resulting when the exact shape functions are employed. Consider the
lumped mass assumption for the mass matrix.
Problem P11.3 Considering only flexural vibrations, formulate the equation of
motion of the frame of Fig. P11.3. Use lumped mass assumption for the
element mass.
Fig. P11.3 Frame in problem P11.3.
518 PART
II Multi-degree-of-freedom systems
Problem P11.4 Considering only flexural vibrations, formulate the equation of
motion of the frame of Fig. P11.4. Assume CT ¼ EI =L and lumped mass
assumption for the element mass.
Fig. P11.4 Frame in problem P11.4.
Problem P11.5 Find the number of the active axial constraints of the frame of
Fig. P11.5 and determine its degrees of freedom after their reduction due to
static condensation and axial constraints. The coordinates of the nodes are:
Að0, 0Þ, B ð7, 0Þ, 1ð0, 3Þ, 2ð0, 8Þ, 3ð4, 8Þ, 4ð6, 8Þ.
Fig. P11.5 Frame in problem P11.5
Problem P11.6 Neglecting the axial deformation, formulate the equation of
motion of the frames of Fig. P11.6.
The finite element method Chapter
11
519
Fig. P11.6 Frames in problem P11.6.
Problem P11.7 Neglecting the axial deformation of the beam elements, formulate the equation of motion of the frame of Fig. P11.7. The elements have a
length L ¼ 4a, cross-sectional area A, moment of inertia I , modulus of elasticity
The rigid bodies B1 and B2 have dimensions a a
E, and line density m.
and 1:5a 0:5a, respectively, and unit thickness. Their material density is r.
Consider lumped mass assumption for the elements.
Fig. P11.7 Frame in problem P11.7.
Problem P11.8 Formulate the equation of motion of the grid of Fig. P11.8.
Assume the data: coordinates of nodes 1ð0, 0Þ, 2ð2, 0Þ, 3ð3, 2Þ, 4ð3, 2Þ;
rectangular cross-section of all elements h b ¼ 0:50 0:30m2 ; material constants E ¼ 2:1 107 , G ¼ 0:4E, r ¼ 2:4kNm1 s2 /m3; loading pðx, t Þ ¼ p0 f ðt Þ,
M ðt Þ ¼ p0 L2 f ðt Þ. Consider lumped mass assumption for the elements.
520 PART
II Multi-degree-of-freedom systems
Fig. P11.8 Grid in problem P11.8.
Problem P11.9 Formulate the equation of motion of the space frame of
Fig. P11.9. The principal plane xy of the elements is vertical. Assume data:
t, G ¼ 0:4E, 1 ð2a, 0, 0Þ, 2 ð2a, 1:5a, a Þ, 3 ða, 1:5a, a Þ,
pðx, t Þ ¼ p0 sin w
4 ð0, 1:5a, a Þ, 5 ða, 5a, 0Þ. The elements have a square cross-section with side
length a=15.
Fig. P11.9 Space frame in problem P11.9.
Problem P11.10 A rigid circular plate of radius R and thickness 0:1a is supported
on the ground by truss bars as shown in Fig. P11.10. The plate is horizontal
and stands off at a distance h ¼ 0:5a from the ground. The support points on the
ground lie on the vertices of an equilateral triangle. A horizontal force
P ¼ 100kN produced by a rotating mass is applied at the center of the plate.
Formulate the equation of motion of the structure. Assume: a ¼ 20m, R ¼ 4m,
cross-sectional area of the bars A; density of the plate r; modulus of elasticity
E; load of the plate q ¼ 10kN=m2 ; lumped mass assumption for the elements.
Fig. P11.10 Structure in problem P11.10.
The finite element method Chapter
11
521
Problem P11.11 Formulate the equation of motion of the space frame of
Fig. P11.11. The axial deformation of the elements is neglected and the lumped
mass assumption is adopted. All elements have a rectangular cross-section with
dimensions h and b in the directions of the principal axes y and z, respectively.
The plane xy is vertical. The shape of the rigid body is a rectangular parallelepiped with dimensions a 0:5a 0:4a. The force P ðt Þ is applied at the middle
300 of the edge. The material constants are: modulus of elasticity E, Poisson’s
ratio n ¼ 0:2, and density r.
Fig. P11.11 Space frame in problem P11.11.
References and further reading
[1] J.T. Katsikadelis, A generalized Ritz method for partial differential equations in domains of
arbitrary geometry using global shape functions. Eng. Anal. Bound. Elem. 32 (5) (2008)
353–367, https://doi.org/10.1016/j.enganabound.2007.001.
[2] O. Zienkiewicz, R. Taylor, The Finite Element Method, seventh ed., Butterworth-Heinemann,
Oxford, UK, 2013.
[3] T.J.R. Hughes, The Finite Element Method: Linear Static and Dynamic Finite Element Analysis, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1987
[4] J.T. Katsikadelis, The Boundary Element Method for Engineers and Scientists, second ed.,
Academic Press, Elsevier, Oxford, UK, 2016.
[5] D.O. Brush, B.O. Almroth, Buckling of Bars, Plates and Shells, McGraw-Hill, New York,
1975.
[6] R.G. Guyan, Reduction of stiffness and mass matrices, AIAA J. 3 (1965) 380.
522 PART
II Multi-degree-of-freedom systems
[7] W. Weaver Jr., P.R. Johnston, Structural Dynamics by Finite Elements, Prentice Hall,
Englewood Cliffs, NJ, 1987.
[8] W. Weaver Jr., M. Eisenberger, Dynamics of frames with axial constraints, ASCE J. Struct.
Eng. 109 (3) (1983) 773–784.
[9] S. Lipschutz, M. Lipson, Linear Algebra (Fourth Edition), Schaum’s Outline Series, McGrawHill Companies, Inc., New York, 2009
[10] V.Z. Vlasov, Thin-walled Elastic Beams, second ed., Israel Program for Scientific Translations, Jerusalem, Israel, 1961.
[11] E.J. Sapountzakis, V.G. Mokos, Warping shear stresses in nonuniform torsion, Comput.
Methods Appl. Mech. Eng. 192 (2003) 4337–4353.
[12] S. Timoshenko, G.N. Goodier, Theory of Elasticity, second ed., McGraw-Hill, NY, 1951.
[13] R. Debasish, G.V. Rao, Elements of Structural Dynamics: A New Perspective (1st Edition),
John Wiley & Sons, 2012.
Chapter 12
Multi-degree-of-freedom
systems: Free vibrations
Chapter outline
12.1 Introduction
523
12.2 Free vibrations without
damping
524
12.3 Orthogonality of eigenmodes 532
12.4 Eigenmodes of systems with
multiple eigenfrequencies
534
12.5 The linear eigenvalue
problem
541
12.5.1 The standard
eigenvalue problem
of linear algebra
541
12.5.2 Properties of the
eigenvalues and
eigenvectors
544
12.5.3 The generalized
eigenvalue problem 552
12.6 The Rayleigh quotient
558
12.7 Properties of
eigenfrequencies and
modes of MDOF systems
without damping:
A summary
563
12.8 Solution of the vibration
problem without damping
563
12.9 The method of mode
superposition
565
12.10 Solution of the vibration
problem with damping
570
12.10.1 Direct solution of
the differential
equation
571
12.10.2 Linearization of
the quadratic
eigenvalue problem 575
12.10.3 The use of a
proportional viscous
damping matrix
578
12.11 Construction of a
proportional damping
matrix
584
12.11.1 Rayleigh damping 584
12.11.2 Additional
orthogonality
conditions: Caughey
damping matrix
587
12.11.3 Construction of
theo proportional
damping matrix
using the modal
matrix
590
12.12 Problems
597
References and further reading
600
12.1 Introduction
In this chapter, we study the dynamic response of multi-degree-of-freedom
(MDOF) systems in the absence of external forces. Like the SDOF systems,
the MDOF systems perform oscillations caused by initial displacements and/
or initial velocities. In accordance with the SDOF systems, we call these
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00012-1
© 2020 Elsevier Inc. All rights reserved.
523
524 PART
II Multi-degree-of-freedom systems
oscillations free vibrations of MDOF systems. In this case, it is pðt Þ ¼ 0 and the
equation of motion (see Eq. 10.2.6) becomes
M€
u + Cu_ + Ku ¼0
(12.1.1)
where M is the mass matrix, C the damping matrix, K the stiffness matrix, and
u ¼ uðt Þ the displacement vector. When damping is neglected, C ¼ 0, we talk
about free vibrations without damping or free undamped vibrations. Their study
plays an important role in the analysis of the MDOF systems because, in addition to facilitating the understanding of their dynamic response by introducing
the physical concepts of the eigenfrequencies, periods, and mode shapes, it also
serves to develop methods for solving free vibration problems with damping as
well as forced vibration problems.
As mentioned, the equation describing the free undamped vibrations results
from Eq. (12.1.1) by setting C ¼ 0. Thus, we have
M€
u + Ku ¼0
(12.1.2)
The solutions of Eqs. (12.1.1) and (12.1.2) under the specified initial conditions uð0Þ and u_ ð0Þ are the subject that will be discussed in this chapter.
All presented material is illustrated by appropriately chosen examples. The pertinent bibliography with recommended references for further study is also
included. The chapter is enriched with problems to be solved.
12.2 Free vibrations without damping
We look for a solution of Eq. (12.1.2) in the form
u1 ¼ b1 T ðt Þ, u2 ¼ b 2 T ðt Þ, …,uN ¼ bN T ðt Þ
(12.2.1)
where b1 , b2 ,…,bN are constants and T ðt Þ a function of time, which is common for all displacements. If we define the vector b ¼ fb 1 , b2 , …, b N gT , then
we can write Eq. (12.2.1) in the form
u ¼bT ðt Þ
(12.2.2)
Substituting Eq. (12.2.2) into Eq. (12.1.2) yields
MbT€ ðt Þ + KbT ðt Þ ¼0
(12.2.3)
Τ
Premultiplying the foregoing equation by b gives
bΤ MbT€ ðt Þ + bΤ KbT ðt Þ ¼0
(12.2.4)
Taking into account that both quantities bΤ Mb, bΤ Kb are constants and
excluding the value T ðt Þ ¼ 0 (see Section 8.3.3), we may write Eq. (12.2.4) as
T€ ðt Þ bΤ Kb
¼
T ðt Þ bΤ Mb
(12.2.5)
Because the right side of Eq. (12.2.5) is a constant, say l, it implies that also
the left side is a constant, namely
Multi-degree-of-freedom systems: Free vibrations Chapter
T€ ðt Þ
¼l
T ðt Þ
12
525
(12.2.6)
or
T€ ðt Þ + lT ðt Þ ¼ 0
(12.2.7)
The constant l is unknown in the first instance and will be determined
subsequently.
Eq. (12.2.7) allows us to determine the time-dependent function T ðt Þ. To
this end, we look for solutions representing oscillatory motion. We distinguish
three cases:
(i) l ¼ 0. In this case, Eq. (12.2.7) becomes
T€ ðt Þ ¼ 0
(12.2.8)
T ðt Þ ¼ C 1 t + C 2
(12.2.9)
which is integrated to yield
where C1 , C2 are arbitrary integration constants. Obviously, the function
T ðt Þ obtained from Eq. (12.2.9) implies that Eq. (12.2.2) does not represent an oscillatory motion. Therefore, the value l ¼ 0 must be excluded.
(ii) l ¼ w2 < 0. In this case, Eq. (12.2.7) becomes
T€ ðt Þ w2 T ðt Þ ¼ 0
(12.2.10)
whose solution is (see Chapter 2)
T ðt Þ ¼ C1 ewt + C2 ewt
(12.2.11)
From the foregoing expression of T ðt Þ, it follows that Eq. (12.2.2)
does not represent an oscillatory motion. Consequently, the negative
values of the constant l must also be excluded.
(iii) l ¼ w2 > 0. In this case, Eq. (12.2.7) becomes
T€ ðt Þ + w2 T ðt Þ ¼ 0
(12.2.12)
whose solution is (see Chapter 2)
T ðt Þ ¼ c cos wt + d sin wt
¼ a cos ðwt qÞ
(12.2.13)
where c,d or a,q are arbitrary integration constants.
From the foregoing expression of T ðt Þ, it follows that the displacements
obtained by Eq. (12.2.2) are bounded and the function T ðt Þ expresses a harmonic oscillation with the natural frequency w.
Eq. (12.2.12) gives T€ ðt Þ ¼ w2 T ðt Þ, which is substituted into Eq. (12.2.3)
to yield
K w2 M bT ðt Þ ¼ 0
(12.2.14)
526 PART
II Multi-degree-of-freedom systems
Because the foregoing equation must hold for all values of t > 0, it must be
K w2 M b ¼0
(12.2.15)
Eq. (12.2.15) represents a system of N linear algebraic equations, which can
be solved to determine the vector b ¼ fb1 , b 2 , …, bN gT , that is, the amplitude
of the displacements. The system of Eq. (12.2.15) is homogeneous; therefore it
has a nontrivial solution only if the determinant of the coefficient matrix vanishes, namely
det K w2 M ¼0
(12.2.16)
or
k11 w2 m11
k21 w2 m21
⋯
k N 1 w2 m N 1
k12 w2 m12
k22 w2 m22
⋯
k N 2 w2 m N 2
⋯
⋯
⋯
⋯
k1N w2 m1N k2N w2 m2N ¼0
⋯
kNN w2 mNN (12.2.17)
The expansion of the determinant in Eq. (12.2.17) produces a polynomial of
the N degree with respect to w2 . Therefore Eq. (12.2.16) may hold true for its N
values w2 . These values are the eigenfrequencies or natural frequencies of the
system. Eq. (12.2.16), from which the eigenfrequencies are determined, is
referred to as the frequency equation of the system. As we will show below,
the roots of Eq. (12.2.16) are real and positive for the problems of dynamics
we examine. The eigenfrequencies are arranged in the order of magnitude,
w1 < w2 … < wN , the smallest of which is called the fundamental eigenfrequency. The eigenfrequencies may be all distinct or some of them may be multiple. First, we assume that the system has discrete eigenfrequencies. For each
value wi , we obtain a system of linear algebraic equations of the form (12.2.15),
which allows the determination of the corresponding vector bðiÞ . Due to the fact
that the determinant is equal to zero, the number of independent equations is
ði Þ ði Þ
ði Þ
N 1, which implies that one of the components b 1 , b2 ,…,b N of the vector
ði Þ
ði Þ
b can be determined arbitrarily. In this respect, we may take b 1 ¼ 1, and
write the system of Eq. (12.2.15) as
ð12:2:18Þ
ði Þ
Partitioning the matrix A
and the vector b
Eq. (12.2.18) and defining the matrices
ði Þ
A11 ¼ k11 w2i m11 11
¼ K w2i M
ði Þ
as indicated in
(12.2.19a)
Multi-degree-of-freedom systems: Free vibrations Chapter
ði Þ
A12 ¼ k12 w2i m12 ⋯ k1N w2i m1N 1ðN 1Þ
2
3
k21 w2 m21
ði Þ
5
A21 ¼ 4
⋮
2
kn1 wi mN 1 ðN 1Þ1
2
3
k22 w2i m22 ⋯ k2N w2i m2N
ði Þ
5
A22 ¼ 4
⋮
⋮
⋮
2
2
kN 2 wi mN 2 ⋯ kNN wi mNN ðN 1ÞðN 1Þ
8 ðiÞ 9
< b2 =
ði Þ
b2 ¼
⋮
: ðiÞ ;
bN ðN 1Þ1
we can write Eq. (12.2.18) as
"
#
ði Þ
ði Þ 1
0
A11 A12
¼
ði Þ
ði Þ
ði Þ
b
0
A21 A22
2
12
527
(12.2.19b)
(12.2.19c)
(12.2.19d)
(12.2.19e)
(12.2.20)
which after performing the multiplication gives
A11 + A12 b2 ¼ 0
ði Þ
ði Þ ði Þ
(12.2.21a)
ði Þ
A21
ði Þ ði Þ
+ A22 b2
(12.2.21b)
¼0
The second of the foregoing equations, that is, Eq. (12.2.21b), can be used to
ði Þ
establish the vector b2 . Thus we have
h
i1
ði Þ
ði Þ
ði Þ
A21
(12.2.22)
b2 ¼ A22
ði Þ
Eq. (12.2.21a) must hold true if the value of b2 obtained from Eq.
(12.2.21b) is inserted in it. Therefore, this equation can be employed to verify
ði Þ
the computed value of b2 .
Eq. (12.2.18) states that for each value wi , we obtain a vector bðiÞ , which
henceforth will be denoted by bi for convenience. These vectors are called
eigenvectors and are displayed in a square matrix B of dimensions N N , each
column of which is an eigenvector
2
3
1
1
⋯ 1
6 b21 b 22 ⋯ b2N 7
7
(12.2.23)
B ¼ ½ b1 b2 ⋯ bN ¼ 6
4⋯ ⋯ ⋯ ⋯ 5
bN 1 b N 2 ⋯ bNN
For each value of wi , Eq. (12.2.2) gives a solution of the form
ui ¼ bi ðci cos wi t + d sin wi t Þ
¼ bi ai cos ðwt qi Þ
which is called an eigensolution.
(12.2.24)
528 PART
II Multi-degree-of-freedom systems
The superposition of the solutions given by Eq. (12.2.24) for i ¼ 1, 2…, N ,
gives the general solution for the free vibration problem of the MDOF system
with N degrees of freedom. Thus we have
u¼
N
X
bi ðci cos wi t + di sin wi t Þ
(12.2.25)
i¼1
or
u¼
N
X
bi ai cos ðwi t qi Þ, ai ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ci2 + di2 , qi ¼ tan 1 ðdi =ci Þ
(12.2.26)
i¼1
The 2N arbitrary constants ci ,di or ai ,qi are established from the initial conditions using the method we will describe after discussing the properties of the
eigenvectors.
From Eq. (12.2.24), we conclude that the deformation pattern of the system
due to an eigensolution is given by a certain shape, which is expressed by the
components of the eigenvector bi multiplied each instant by the value of the
corresponding common time function T ðt Þ ¼ ai cos ðwi t qi Þ. The deformation
shape defined by vector bi is called the i-eigenmode of the free vibration.
As previously mentioned, the components of the eigenvector are determined
relative to one of its components, which can be arbitrarily chosen. Therefore,
the absolute magnitude of the eigenvector is not determined, that is, the eigenvectors define only directions in the N -dimensional space while their measure
remains undetermined. In view of this fact, it is appropriate to adopt a standard
scale-setting procedure for measuring it. This process is called normalization.
A common method of normalization is the use of the nondimensional
vectors
2
3
2
3
b1i
f1i
6 f2i 7
7
1 6
7
6 b2i 7
fi ¼ 6
(12.2.27)
4 ⋯ 5 ¼ max b 4 ⋯ 5
ki
fNi
bNi
The vectors fi (i ¼ 1,2…,N ) are called the normal modes or normal
eigenmodes of the vibration of the system. The shape of deformation defined
by the eigenmode fi is also referred to as the i-mode shape or simply the i-mode
of the free vibration.
The N normal modes can then be displayed in a single square matrix, each
column of which is a normal mode, namely
2
3
f11 f12 ⋯ f1N
6 f21 f22 ⋯ f2N 7
7
F ¼ ½ f1 f2 ⋯ fN ¼ 6
(12.2.28)
4⋯ ⋯ ⋯ ⋯ 5
fN 1 fN 2 ⋯ fNN
The matrix F is called the modal matrix.
Multi-degree-of-freedom systems: Free vibrations Chapter
12
529
Another method of normalization of the eigenvectors, which is usual in theoretical discussions and computer programs, is to normalize the modes with
respect to the mass. In this method, the scale-setting for measuring the components of the eigenvector is chosen so that the following relation is satisfied
fTi Μfi ¼ 1
(12.2.29)
where fi represents the i normalized vector and Μ the mass matrix of the
system. This is achieved as follows:
Let mi be the number that must multiply the elements of bi in order to
normalize it. Thus, we will have
fi ¼ mi bi
Substituting it into Eq. (12.2.29) gives
1 ¼ fTi Μfi ¼ m2i bTi Μbi
from which we obtain
1
mi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
bTi Μbi
(12.2.30)
Hence
bi
ffi
(12.2.31)
fi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
bTi Μbi
qffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffi
If it is M ¼ I, then the scalar quantity bTi Ιbi ¼ bTi bi is the magnitude of
bi , that is, the normalized eigenvector fi is a unit vector. The described process
is also called orthonormalization with respect to mass.
The establishment of the natural frequencies wi and mode shapes fi can be
achieved when the flexibility matrix F ¼ K1 is known without inverting it to
obtain the stiffness matrix.
For this purpose, Eq. (12.2.15) is premultiplied by the matrix F to yield
FMw2 FK b ¼0
(12.2.32)
then taking into account that FK ¼ I, Eq. (12.2.32) becomes
FM 1
I b ¼0
w2
(12.2.33)
which allows the establishment of the inverse of the eigenfrequencies, 1=w2i ,
and the eigenvectors bi . It should be noted that the matrix FM is not in general
symmetric.
Example 12.2.1 Compute the eigenfrequencies and mode shapes of the twostory shear frame of Fig. E12.1a. Assume data: h1 ¼ 3:5m, h2 ¼ 4:0m,
q1 ¼ 40kN=m, q2 ¼ 50kN=m, and L ¼ 6:0m. The cross-sectional area of the
530 PART
II Multi-degree-of-freedom systems
columns of the second floor is 25 25cm2 while that of the first floor is
30 30cm2 . The material of the structure is reinforced concrete with a specific
weight g b ¼ 24kN=m3 and modulus of elasticity E ¼ 2:1 107 kN=m2 . The
dead weight of the beams is included in the loads q1 ,q2 . Consider lumped mass
assumption of the columns.
Solution
(a)
(b)
FIG. E12.1 Two-story shear frame.
The system has two degrees of freedom. Its dynamic model is shown in
Fig. E12.1b.
(i) Computation of the mass matrix M.
6:0 40 + 2 0:25 0:25 m11 ¼
3:5
24
2
9:81
¼ 25kNm1 s2
3:5
4:0
+ 0:30 0:30 24
2
2
9:81
6:0 50 + 2 0:25 0:25 m22 ¼
¼ 32 kNm1 s2
Hence
M¼
25 0
0 32
(ii) Computation of the stiffness matrix K.
Employing the method used in Example 11.5.2, we obtain
ð1Þ
ð2Þ
k11 ¼ k11 + k11 ¼ 2
¼2
12EI 1
h13
12 2:1 107 0:254
¼ 3826:5kN=m
3:53 12
(1)
Multi-degree-of-freedom systems: Free vibrations Chapter
ð1Þ
ð2Þ
k21 ¼ k31 + k31 ¼ 2
12
531
12EI 1
h13
12 2:1 107 0:254
¼ 3826:5kN=m
3:53 12
12EI 1
12EI 3
ð1Þ
ð2Þ
ð3Þ
ð4Þ
k22 ¼ k33 + k33 + k11 + k11 ¼ 2 3 + 2 3
h1
h2
¼ 2
¼2
12 2:1 107 0:254
12 2:1 107 0:304
+
2
¼ 9142:1kN=m
3:53 12
43 12
Hence
K¼
3826:5 3826:5
3826:5 9142:1
(2)
(iii) Computation of the eigenfrequencies.
The frequency equation results from Eq. (12.2.16) as
3826:5 25w2 3826:5
det K w2 M ¼ 2¼0
3826:5
9142:1 32w
(3)
Expansion of the determinant yields the polynomial
w4 438:7506w2 + 25425:1792 ¼ 0
(4)
whose roots are
w21 ¼ 68:7089, w22 ¼ 370:0416
and the eigenfrequencies
w1 ¼ 8:289, w2 ¼ 19:236
(5)
(iv) Computation of the eigenmodes.
For N ¼ 2, Eqs. (12.2.21) become
3826:5 25w2i 3826:5b2i ¼ 0
3826:5 + 9142:1 32w2i b2i ¼ 0
The first of the foregoing equation
gives:
For i ¼ 1, b21 ¼ 3826:50=9142:1 32 8:2892 ¼ 0:5511
For i ¼ 2, b22 ¼ 3826:50= 9142:1 32 19:2362 ¼ 1:41751
Therefore, the matrix of the eigenvectors is
B¼
1
1
0:5511 1:41751
(6)
532 PART
II Multi-degree-of-freedom systems
FIG. E12.2 Mode shapes of the shear frame in Example 12.2.1.
The normalization based on Eq. (12.2.27) gives the modal matrix
F¼
1
0:7054
0:5511 1
(7)
while the normalization based on Eq. (12.2.31) gives
b1
f1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
bT1 Μb1
1
0:1697
1
¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi 0:5511 ¼ 0:0935
25 0
1
f 1 0:5511 g
0 32 0:5511
b2
f2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffi
bT2 Μb
1
0:1061
1
ffi
¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼
1:41276
0:1499
25 0
1
f 1 1:41276 g
0 32 1:41276
Hence the modal matrix is
F¼
0:1697 0:1061
0:0935 0:1499
(8)
Fig. E12.2 presents the normal modes of the dynamic model of the frame
resulting from the first normalization, Eq. (7).
12.3 Orthogonality of eigenmodes
As we have already pointed out and will prove in Section 12.5, the eigenfrequencies w2i resulting from the frequency equation (12.2.16) are real and positive for the problems we encounter in the dynamics of structures. In the
Multi-degree-of-freedom systems: Free vibrations Chapter
12
533
following, we will derive the orthogonality conditions satisfied by the eigenmodes. For the simplicity of the proof, we will assume that the eigenfrequencies
are distinct.
We write Eq. (12.2.15) for the two eigenfrequencies wn and wi , where
wn 6¼ wi , respectively
Kbn ¼w2n Mbn
(12.3.1)
Kbi ¼w2i Mbi
(12.3.2)
Kfn ¼w2n Mfn
(12.3.3)
Kfi ¼w2i Mfi
(12.3.4)
or after normalization
Premultiplying now both sides of Eq. (12.3.3) by fTi gives
fTi Kfn ¼w2n fTi Mfn
(12.3.5)
Transposing both sides of the foregoing equation, we obtain
fTn KT fi ¼w2n fTn MT fi
(12.3.6)
Further, using the symmetry of the matrices K and M, that is, KT ¼ K and
M ¼ M, we obtain
T
fTn Kfi ¼w2n fTn Mfi
Next, premultiplying both sides of Eq. (12.3.4) by
(12.3.7)
fTn
gives
fTn Kfi ¼w2i fTn Mfi
(12.3.8)
Finally, by subtracting Eq. (12.3.8) from (12.3.6), we obtain
2
wn w2i fTn Mfi ¼ 0
(12.3.9)
fTn Mfi ¼ 0,
(12.3.10)
or
wi 6¼ wn
Eq. (12.3.10) states that the eigenmodes are orthogonal with respect to the
mass matrix. This relation expresses the first orthogonality condition of the
eigenmodes. It is evident from Eq. (12.3.8) that
fTn Kfi ¼ 0,
wi 6¼ wn
(12.3.11)
Eq. (12.3.11) states that the eigenmodes are also orthogonal with respect to
the stiffness matrix. This relation expresses the second orthogonality condition
of the eigenmodes.
534 PART
II Multi-degree-of-freedom systems
Example 12.3.1 Verify the orthogonality conditions for the eigenmodes in
Example 12.2.1:
0:1697
0:1061
, f2 ¼
f1 ¼
0:0935
0:1499
Solution
The computed mass and stiffness matrices are
M¼
25 0
3826:5 3826:5
, K¼
0 32
3826:5 9142:1
(a) The first orthogonality condition gives
fT1 Mf2
¼ f 0:1697 0:0935 g
T
25 0
0 32
0:1061
0:1499
(b) The second orthogonality condition gives
fT1 Kf2 ¼ f 0:1697 0:0935 gT
3826:5 3826:5
3826:5 9142:1
¼0
0:1061
0:1499
(1)
¼0
(2)
Because the eigenmodes have been orthonormalized with respect to the
mass, we may write
0:1697 0:1061
0:0935 0:1499
T
25 0
0 32
0:1697 0:1061
1 0
¼
0:0935 0:1499
0 1
or
FT MF ¼ I
(3)
12.4 Eigenmodes of systems with multiple eigenfrequencies
In complex systems with many degrees of freedom, it is possible that a number
of eigenfrequencies are equal or differ very little from each other. In these cases,
we say that the system has multiple eigenfrequencies. Systems including a rigid
body motion are treated as systems with multiple eigenfrequencies (see
Example 12.4.2).
Let wn be the eigenfrequency, which has a multiplicity p, namely
wn ¼ wn + 1 ¼ wn + 2 ¼ ⋯ ¼ wn + p1
(12.4.1)
As we will show in the next section, in a system with p equal eigenfrequencies, there are p linearly independent eigenmodes corresponding to the multiple
eigenfrequency, provided that its algebraic multiplicity is equal to its geometric
Multi-degree-of-freedom systems: Free vibrations Chapter
12
535
multiplicity, that is, if the eigenvalue problem has no degenerate eigenvalues.a
These eigenmodes can be determined as follows.
The rank of the matrix Aðwn Þ defined by Eq. (12.2.18) for w ¼ wn is N p.
We partition this matrix as
"
#(
) ðn Þ
ðn Þ
ðn Þ
0
b1
A11 A12
¼
(12.4.2)
ðn Þ
ðn Þ
ðn Þ
0
A21 A22
b2
where
ðn Þ
A11 : has dimensions : p p
ðn Þ
A12 :
ðn Þ
A21 :
ðn Þ
A22 :
ðn Þ
b1 :
ðn Þ
b2 :
(12.4.3a)
has dimensions : p ðN pÞ
(12.4.3b)
has dimensions : ðN pÞ p
(12.4.3c)
has dimensions : ðN pÞ ðN pÞ
(12.4.3d)
has dimensions : p 1
(12.4.3e)
has dimensions : ðN pÞ 1
(12.4.3f)
Eq. (12.4.2) after performing the multiplication gives
ðn Þ ðn Þ
ðn Þ ðn Þ
ðn Þ ðn Þ
A21 b1
ðn Þ ðn Þ
+ A22 b2
A11 b1 + A12 b2 ¼ 0
Because
ðn Þ
A22
(12.4.4a)
¼0
is not singular, Eq. (12.4.4b) is solved for
1
ðn Þ
ðn Þ
ðn Þ ðn Þ
b2 ¼ A22
A21 b1
(12.4.4b)
ðn Þ
b2
(12.4.5)
If we now choose arbitrarily p linearly independent vectors, each of them
with dimension p
ðn Þ
ðn + 1Þ
b1 ,b1
ðn + p1Þ
, ⋯,b1
then Eq. (12.4.5) can be used to determine the p vectors
ðn Þ
ðn + 1Þ
b2 ,b2
ðn + p1Þ
, ⋯,b2
For example, a set of p arbitrary linearly independent vectors is
8 9
8 9
8 9
1>
0>
0>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
1
=
< =
< =
<0>
ðn Þ
ðn + 1Þ
ðn + p1Þ
b1 ¼ 0
,b1
¼ 0
,⋯, b1
¼ 0
>
>
>
>
>
>
>
>
>
⋮>
⋮>
⋮>
>
>
>
>
>
>
>
>
>
;
;
;
: >
: >
: >
0 p1
0 p1
1 p1
(12.4.6)
The so-obtained arbitrary p vectors bðnÞ ,bðn + 1Þ , ⋯,bðn + p1Þ corresponding
to the multiple eigenfrequency wn are obviously linearly independent; however,
a. If the eigenvalue problem has degenerate
eigenvalues,
the eigenvectors
are established from the
p
solution of the eigenvalue problem K w2n M b ¼0, instead of K w2n M b ¼0 [1].
536 PART
II Multi-degree-of-freedom systems
they do not satisfy in general the orthogonality condition between each other.
As we will see in Section 12.8, The fulfillment of the orthogonality condition is
necessary for the solution of the free vibration problem of the MDOF systems.
This can be achieved using the Gram-Schmidt orthogonalization process, which
is described right below
FIG. 12.4.1 Geometrical interpretation of Gramm-Schmidt orthogonalization.
Let x1 ,x2 ,…, xN be a set of N linear independent but not orthogonal vectors.
These vectors constitute the basis of a space with N dimensions. We can readily
show that the vectors defined by the relations
e
x1 ¼ x1
e
x2 ¼ x2 e
x3 ¼ x3 e
xT1 x2
x1 j2
je
e
xT1 x3
e
x1
e
xN ¼ xN e
x1 e
xT2 x3
e
x2
x1 j
x2 j2
je
je
⋯ ¼ ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
2
e
xT1 xN
x1 j2
je
e
x1 e
xT2 xN
x2 j2
je
e
x2 ⋯ (12.4.7)
e
xTN 1 xN
xN 1 j2
je
e
xN 1
are orthogonal to each other. From the geometrical point of view, the vector e
xk
is equal to xk minus its projection on the space defined by the vectors
e
x1 ,e
x2 ,…,e
xk1 . For example, the vector e
x2 is equal to x2 minus its projection
on the vector e
x1 . This is shown in Fig. 12.4.1.
The orthogonality of the eigenvectors with respect to the mass, which is also
needed, is achieved using the following relations.
e
x1 ¼ x1
e
x2 ¼ x2 e
xT1 Mx2
e
x1
e
x1
xT1 Me
e
x3 ¼ x3 e
e
xT1 Mx3
xT2 Mx3
e
e
x
x2
1
e
e
x1
x2
xT1 Me
xT2 Me
⋯ ¼ ⋯⋯⋯⋯⋯⋯⋯⋯⋯
e
xN ¼ xN e
e
e
xT1 MxN
xT MxN
xT MxN
e
e
e
x1 2T
x2 ⋯ TN 1
xN 1
T
e
e
e
x1
x2
xN 1
x1 Me
x2 Me
xN 1 Me
(12.4.8)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
537
Example 12.4.1 Compute the eigenfrequencies and eigenmodes of a structure
whose stiffness and mass matrices are:
2
3
2
3
3826 1913 7652
25 0 25
5, M ¼ 4 0 8
K ¼ 4 1913
2285:5 6484
16 5
(1)
7652
6484 21307:08
25 16 67
Solution
(i) Computation of the eigenfrequencies.
The frequency equation is given by the determinant
3826 25w2 1913
7652 + 25w2
2
2
¼0
det ½AðwÞ ¼ 1913
2285:5 8w
6484 16w
2
7652 + 25w2 6484 16w2
21307:08 67w
(2)
which after expanding yields the characteristic polynomial
w6 507:4355w4 + 55567:85905w2 1746816:390 ¼ 0
(3)
from which we obtain the roots
w21 ¼ w22 ¼ 68:7088, w23 ¼ 370:0179
(4)
and the eigenfrequencies
w1 ¼ w2 ¼ 8:2891, w3 ¼ 19:2358
(5)
We observe that the structure has two equal eigenfrequencies, that is,
p ¼ 2.
(ii) Computation of the eigenmodes.
For w1 ¼ w2 ¼ 8:2891 we have
Because the rank of the matrix is N p ¼ 3 2 ¼ 1, the nonvanishing
determinants have dimensions 1 1 and Að1Þ is partitioned as shown above.
Thus partitioning Að1Þ as indicated, we have
ðn Þ
A11 ¼
ðn Þ
A12 ¼
2108:2705 1913:0000
1913:0000 1735:8266
(6)
5934:2705
5384:6531
(7)
ðn Þ
A21 ¼ ½ 5934:2705 5384:6531 ðn Þ
A22 ¼ ½16703:5650
(8)
(9)
538 PART
II Multi-degree-of-freedom systems
For n ¼
1,we obtain
1
ð1Þ
ð1Þ
b1 ¼
and Eq. (12.4.5) gives b2 ¼ f0:3553g. Hence
0
8
9
<1
=
b1 ¼ 0
:
;
0:3553
For n ¼
2,we obtain
0
ð2Þ
ð2Þ
and Eq. (12.4.5) gives b2 ¼ f0:322365g. Hence
b1 ¼
1
8
9
< 0
=
1
b2 ¼
:
;
0:322365
(10)
(11)
For n ¼ 3 the eigenfrequency is discrete, w3 ¼ 19:2358, the rank of the
matrix is 3 1 ¼ 2, and the matrix Að3Þ is partitioned as
ð12Þ
Thus we have
ð3Þ
A11 ¼ ½ 5424:4 ð3Þ
(13)
A12 ¼ ½ 1913 1598:4 (14)
1913
1598:4
(15)
ð3Þ
A21 ¼
ð3Þ
A22 ¼
674:6280
563:7440
563:7440 3483:9921
and Eq. (12.2.22) gives
h
i1
2:8357
ð3Þ
ð3Þ
ð3Þ
ð3Þ
A21 ¼
. Hence
b1 ¼ f1g, b2 ¼ A22
8
90
< 1
=
b3 ¼ 2:8357
:
;
0
(16)
(17)
(iii) Orthogonalization of the eigenmodes.
The computed eigenvectors b1 , b2 , and b3 are linearly independent.
This is readily verified by computing the determinant of the eigenvectors,
which apparently does not vanish
1
0
1
(18)
det ðBÞ ¼ 0
1
2:8356 ¼ 1:26936
0:3553 0:322365 0
Multi-degree-of-freedom systems: Free vibrations Chapter
12
539
However, they do not satisfy the orthogonality condition. They are made
orthogonal using the Gram-Schmidt process described previously. Thus
applying Eq. (12.4.8), we obtain
e1 ¼ b1 ¼ f 1 0 0:3553 gT
b
eT Mb2
b
e
e1 ¼ f 0:3868 1 0:4598 gT
b2 ¼ b2 1T
b
e1
e Mb
b
(19)
1
e3 ¼ b3 ¼ f 1 2:8357 0 gT
b
and after their orthonormalization with respect to the mass matrix, we obtain
2
3
0:2524 0:2551 0:1058
F ¼ 4 0:0000 0:6595 0:3000 5
(20)
0:0897 0:3032 0:0000
Example 12.4.2 Determine the eigenfrequencies and eigenmodes of the free
moving flexible bar of Fig. E12.3a by considering only flexural vibrations.
The bar is approximated by two equal elements. The element properties
r, A,E,I are assumed constant. Consider lumped mass assumption.
FIG. E12.3a Moving flexible bar in Example 12.4.2.
Solution
For lumped mass assumption, the dynamic model of the system is shown in
Fig. E12.3b. The system has six degrees of freedom, which are reduced to three,
u1 , u2 ,u3 , after the static condensation of the rotations.
FIG. E12.3b Dynamic model in Example 12.4.2.
Thus the mass matrix results as
2
3
1 0 0
rAL 4
M¼
0 2 05
4
0 0 1
(1)
540 PART
II Multi-degree-of-freedom systems
and the stiffness matrix as
2
3
1 2 1
12EI
K ¼ 3 4 2 4 2 5
L
1 2 1
The frequency equation is
1 l 2
1
¼ l 2 ð l 4Þ ¼ 0
det ½AðlÞ ¼ 2
4 2l 2
1
2
1l
(2)
(3)
where l ¼ w2 rAL4 =48EI .
The roots of Eq. 3 are
l1 ¼ l2 ¼ 0, l3 ¼ 4
(4)
Obviously, there is a double eigenfrequency equal to zero, which is due to
the rigid body motion of the bar. The eigenvectors corresponding to the double
eigenfrequency are established using the procedure presented in Section 12.4.
Thus, for l1 ¼ l2 ¼ 0 it is
ð5Þ
Because the rank of the matrix is 3 2 ¼ 1, the nonvanishing determinants
have dimensions 1 1.
Assuming the partitioning of Aðl1 Þ and Aðl1 Þ shown in Eq. 5, we have
ð1Þ
A11 ¼
1 2
1
ð1Þ
ð1Þ
ð1Þ
, A12 ¼
, A21 ¼ ½ 1 2 , A22 ¼ ½1
2 4
2
For n ¼ 1
1
ð1Þ
ð1Þ
and Eq. (12.4.5) gives b2 ¼ f1g. Hence
l1 ¼ 0, b1 ¼
0
8 9
< 1=
0
bð1Þ ¼
: ;
1
For n ¼ 2
1
ð2Þ
ð2Þ
and Eq. (12.4.5) gives b2 ¼ f1g. Hence
l2 ¼ 0, b1 ¼
1
8 9
<1=
bð2Þ ¼ 1
: ;
1
(6)
(7)
(8)
For n ¼ 3 it is l3 ¼ 4, the rank of the matrix is 3 1 ¼ 2, and the matrix
Aðl3 Þ is partitioned as
Multi-degree-of-freedom systems: Free vibrations Chapter
12
541
Therefore
ð3Þ
ð3Þ
ð3Þ
A11 ¼ ½3, A12 ¼ ½ 2 1 , A21 ¼
2
4 2
ð3Þ
, A22 ¼
1
2 3
and Eq. (12.2.22) give
h
i1
1
ð3Þ
ð3Þ
ð3Þ
b2 ¼ A22
A21 ¼
. Hence
1
8
9
< 1=
bð3Þ ¼ 1
:
;
1
(9)
The graphical representations of the eigenmodes are shown in Fig. E12.3c.
FIG. E12.3c Eigenmodes in Example 12.4.2.
We readily find out that the orthogonality condition with respect to the mass
matrix is satisfied.
12.5 The linear eigenvalue problem
12.5.1 The standard eigenvalue problem of linear algebra
As already mentioned, the problem of determining the eigenfrequencies and
eigenmodes is directly related to the linear eigenvalue problem of the linear
algebra [1, 2],
is stated as follows.
which
Let A ¼ aij be a square N N matrix and x an N 1 vector, whose components with respect to a certain base are x1 ,x2 ,…, xN . The relation
y ¼Ax
(12.5.1)
represents a linear transformation, which transforms the vector x to the vector y
with components
y1 ¼ a11 x1 + a12 x2 + ⋯ + a1N xN
y2 ¼ a21 x1 + a22 x2 + ⋯ + a2N xN
⋯ ¼ ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
yN ¼ aN 1 xN 1 + aN 2 x2 + ⋯ + aNN xN
(12.5.2)
In general, the matrix A is complex. However, in the problems we are studying, it is real. For example, we consider the matrix
A¼
2
1
5:5 3
(12.5.3)
542 PART
II Multi-degree-of-freedom systems
Eq. (12.5.1) transforms the vector x ¼ f 2 3 gT to
y1
2
1
2
1
¼
¼
y2
5:5 3 3
2
(12.5.4)
The vectors x and y are shown in Fig. 12.5.1.
FIG. 12.5.1 Transformation of a vector x to a vector y.
In general, the direction of y is different than that of x. The question now
arises whether for the given square matrix A a vector x exists, which when premultiplied by the matrix produces a vector that is proportional to the original
vector. The answer is affirmative. The constant of proportionality is called
an eigenvalue of the matrix. Mathematically, this is expressed as
Ax ¼ lx
(12.5.5)
ðA lIÞx ¼ 0
(12.5.6)
or
where I is the N N unit matrix and l is the corresponding eigenvalue.
Eq. (12.5.6) is written in expanded form as
2
38 9 8 9
0>
A11 l A12
⋯ A1N
x1 > >
>
>
=
< >
= >
< >
6 A21
7
0
A
l
⋯
A
x
22
2N
2
6
7
(12.5.7)
¼
4⋯
5> ⋯ > > ⋯ >
⋯
⋯ ⋯
>
;
: >
; >
: >
0
AN 1
xN
AN 2
⋯ ANN l
The foregoing relation represents a system of N linear algebraic equations,
which can be solved to determine the vector x. Because the system is homogeneous, a nontrivial solution can be obtained only if the determinant of coefficient matrix is equal to zero, that is,
A11 l A12
⋯ A1N
A21
A22 l ⋯ A2N
¼0
det ðA lIÞ ¼ (12.5.8)
⋯
⋯ ⋯
⋯
AN 1
AN 2
⋯ ANN l Multi-degree-of-freedom systems: Free vibrations Chapter
12
543
Expanding the determinant yields a polynomial of N degree with respect to
l, namely
PðlÞ ¼ a0 lN + a1 lN 1 + ⋯ + aN 1 l + aN
(12.5.9)
which is referred to as the characteristic polynomial of the matrix A.
The equation
det ðA lIÞ ¼ PðlÞ ¼ 0
(12.5.10)
is called the characteristic equation of A. Because it was assumed that A is a
real matrix, the coefficients of PðlÞ are also real. It is known from the fundamental theorem of algebra that PðlÞ has N roots l1 ,l2 ,…,lN , which may
be real or complex conjugate while some of them may be multiple. The roots
lk ðk ¼ 1, 2, …, N Þ are called the eigenvalues of A. Understandably, for each
lk , the system of Eq. (12.5.7) yields a vector xk , which is real or complex
depending on whether lk is real or complex. The vectors xk (k ¼ 1, 2, …, N )
are called the eigenvectors of A. Obviously, if xk is an eigenvector then rxk ,
with r being a scalar, is also an eigenvector. This means that only the line of
action of the eigenvector is determined while its magnitude is undetermined.
The pairs ðlk , xk Þ, k ¼ 1, 2, …,N , are referred to as eigenpairs.
Example 12.5.1 Determine the eigenvalues and the eigenvectors of the matrix
given by Eq. (12.5.3).
Solution
The characteristic equation is
2 l 1
2
5:5 3 l ¼ l + 5l + 0:5 ¼ 0
(1)
The degree of the characteristic equation is 2, thus two eigenvalues are
obtained
l1 ¼ 0:1021, l2 ¼ 4:8980
(2)
The corresponding equation will be obtained from Eq. (12.5.7) for N ¼ 2.
Thus for l1 ¼ 0:1021, we have
2 ð0:1021Þ
1
x1
0
¼
(3)
5:5
3 ð0:1021Þ x2
0
which gives
x1 ¼
1
1:8779
and for l2 ¼ 4:8980, we have
2 ð4:8980Þ
1
5:5
3 ð4:8980Þ
(4)
x1
x2
0
¼
0
(5)
544 PART
II Multi-degree-of-freedom systems
which gives
x2 ¼
1
2:8978
(6)
12.5.2 Properties of the eigenvalues and eigenvectors
In this section, we will present some properties of the eigenvalues and eigenvectors that are useful for the development of numerical methods for calculating
the eigenfrequencies and eigenmodes of the free vibrations of the MDOF systems, but also in general for the study and understanding of their dynamic
response.
Let x ¼ fx1 , x2 , …, xN gT and y ¼ fy1 , y2 , …, yN gT be two, in general, complex eigenvectors of the N dimensional space. Their inner or scalar product is
defined by the relation
N
X
xk yk ¼ xT y
(12.5.11)
k¼1
is the vector with components yk , the complex conjugate of yk .
where y
We readily verify that the inner product obeys the following rules:
(i)
(ii)
(iii)
(iv)
> 0 if x 6¼ 0 and xT x
¼ 0 if x ¼ 0
xT x
T
¼ yT x
x y
¼ a x T y
ðaxÞT y
Þ ¼ a xT y
where a is a complex scalar and a its conjugate.
xT ðay
¼ x, y
¼ y and therefore xT y ¼ yT x.
If x and y are real then x
1. The matrix A and its transpose AT have the same eigenvalues.
This follows directly from the fact that the determinant of the matrix
does not change if its lines are interchanged with its columns. Therefore,
the matrices A and AT have the same characteristic polynomial, that is,
(12.5.12)
det ðA lIÞ ¼ det AT lI ¼ PðlÞ ¼ 0
hence the same eigenvalues.
2. If lr and ls are two distinct eigenvalues, lr 6¼ ls , of a real matrix A and ls
the complex conjugate of ls , then the eigenvectors xr of A corresponding to
0s ¼ 0.
lr and x0s of AT corresponding to ls are orthogonal, xTr x
To prove this property, we formulate the inner product of the vectors
yr ¼ Axr and x0s in two ways:
Besides, Taking into account that Axr ¼ lr xr and property (iii) of the
inner product, we have
0s
0s ¼ðAxr ÞT x
yTr x
0s
¼ ðlr xr ÞT x
0s
¼ lr xTr x
(12.5.13a)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
545
T ¼ AT ,
Taking into account that the matrix A is real, hence A
0
¼ ls xs , we obtain on the basis of property (iv) of the inner product
AT x0s
0s ¼ ðAxr ÞT x
0s
yTr x
Tx
0s
¼ xTr A
0 0
s s
¼ xTr AT x
0
s x
s
¼ xTr l
(12.5.13b)
¼ ls xTr x
Subtracting Eq. (12.5.13b) from Eq. (12.5.13a) yields
0s ¼ 0
ðlr ls ÞxTr x
Apparently, because it was assumed that lr 6¼ ls , we obtain
0s ¼ 0
xTr x
(12.5.14)
When all eigenvalues are distinct, this property produces N 2 N
orthogonality relations between the eigenvectors of the matrices A and AT .
3. If the eigenvalues l1 ,l2 ,…,lN of the matrix A are distinct, then the eigenvectors x1 ,x2 , …,xN are linearly independent.
Let k < N eigenvectors x1 ,x2 , …,xk be linear independent. Then the
eigenvectors xk + 1 ,xk + 2 ,…,xN can be expressed as linear combinations
of the independent ones, namely
xj ¼
k
X
ci xi , j ¼ k + 1, k + 2, …,N
(12.5.15)
i¼1
which when multiplied by A, gives
Axj ¼
k
X
ci Axi
i¼1
k
X
¼
ci li xi
(12.5.16)
i¼1
On the other hand, we have
Axj ¼ lj xj
¼
k
X
lj ci xi
(12.5.17)
i¼1
Subtracting Eq. (12.5.17) from (12.5.16), we obtain by virtue of
Eq. (12.5.15)
k X
i¼1
lj li ci xi ¼ 0
(12.5.18)
546 PART
II Multi-degree-of-freedom systems
Because it was assumed that lj 6¼ li and xi are linearly independent, it
implies that all coefficients ci are zero, thus Eq. (12.5.15) gives xj ¼ 0. This,
however, is absurd because xj as an eigenvector cannot be equal to zero.
Therefore, x1 , x2 ,…,xN are linearly independent.
4. If the matrix A is symmetric, then its eigenvalues are real.
To prove this property, we formulate the inner product of the vectors
y ¼ Ax and x in two ways as previously.
First, we obtain on the basis of Eq. (12.5.13a)
¼lxT x
yT x
(12.5.19)
Then taking into account that the matrix A is real and symmetric, it holds
T ¼ AT ¼ A, AT x ¼ l
s x0 , Ax ¼ lx, and on the basis of property (iv) of
A
s
the inner product, we obtain
¼ðAxÞT x
yT x
¼ x T AT x
x
¼ xT A
T
¼ x Ax
¼ xT lx
Tx
¼ lx
(12.5.20)
Subtracting Eq. (12.5.20) from Eq. (12.5.19) yields
xT x
¼0
ll
(12.5.21)
> 0 and
and on the basis of property (i) it is xT x
l¼l
(12.5.22)
hence the eigenvalue l is real.
The orthogonality properties are simplified significantly for real and
symmetric matrices because, in addition to real eigenvalues, they have real
eigenvectors, too. Moreover, the matrices A and AT , in addition to having
the same eigenvalues, have the same eigenvectors, x0 ¼ x. Thus
Eq. (12.5.14) can be written as
xTr xs ¼ 0
(12.5.23)
5. If the real matrix A is symmetric and positive definite, then its eigenvalues
are positive.
Consider the matrix A and an arbitrary vector x ¼ fx1 , x2 , …, xN gT . The
homogeneous polynomial of the second degree defined by
Pðx1 , x2 , …, xN Þ ¼
N X
N
X
i¼1 k¼1
aik xi xk
(12.5.24)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
547
or
PðxÞ ¼ xT Ax
(12.5.25)
is referred to as the quadratic form of the matrix A. For a real and symmetric
matrix A, the coefficients of the polynomial PðxÞ are real and aik ¼ aki .
The quadratic form PðxÞ is called positive definite, if
T
x Ax > 0, if x 6¼ 0
PðxÞ ¼
(12.5.26)
xT Ax ¼ 0, if x ¼ 0
A matrix A is called positive definite if its quadratic form is positive.
A direct consequence of this is that the diagonal elements of A are positive.
Indeed, if we take xi ¼ 1 and xk ¼ 0, k 6¼ i, then Eq. (12.5.24) gives
Pðx1 , x2 , …, xN Þ ¼ aii > 0
(12.5.27)
Let x be an eigenvector of A. Because A is real and symmetric, the
eigenpair ðl, xÞ is real according to the previous property. To prove that
l > 0 we write Eq. (12.5.25) as
T
xT Ax ¼ AT x x
¼ ðAxÞT x
(12.5.28)
¼ lxT x
Because xT x > 0, hence xT Ax > 0, the foregoing relation gives
l¼
xT Ax
>0
xT x
(12.5.29)
6. To each eigenvalue of a symmetric matrix there correspond as many linearly independent eigenvectors as the multiplicity of the eigenvalue.
Let A be an N N symmetric matrix. We assume that we have computed the eigenvalue lk and the corresponding eigenvector xk . Note that
the eigenvalue problem has no degenerate eigenvalues because the matrix
is assumed symmetric. We construct now an N N orthonormal matrix Q,
whose first column is the vector xk , namely
^ , QT Q ¼ I
Q ¼ xk Q
(12.5.30)
^
Apparently, the dimensions of Q are N ðN 1Þ. The construction of Q
is always possible because the vectors in Q provide a basis of the N dimensional space defined by A. This is understood from Example 12.5.2, which
is presented below.
First we evaluate
"
#
xTk
T
^
Q AQ ¼
A xk Q
T
^
Q
"
#
(12.5.31)
^
xTk Axk xTk AQ
¼
^
^ T Axk Q
^ T AQ
Q
548 PART
II Multi-degree-of-freedom systems
^ T Axk ¼
Then taking into account that Axk ¼ lk xk , xTk xk ¼ 1, Q
T
T T
T ^
^
^
lk Q xk ¼ 0, xk AQ ¼ Q A xk ¼ 0 because it was assumed AT ¼ A,
we write the foregoing relation as
QT AQ ¼
lk 0
^
0 A
(12.5.32)
where
^ ¼Q
^
^ T AQ
A
(12.5.33)
^ ¼ c, that is, a conis a fully populated ðN 1Þ ðN 1Þ. If N ¼ 2, then A
T
stant, and the matrix Q AQ becomes diagonal
QT AQ ¼
lk 0
0 c
(12.5.34)
Premultiplying the previous equation by Q gives
lk 0
0 c
lk 0
^
¼ x Q
0 c
AQ ¼ Q
(12.5.35)
^ is another eigenvector of A and c is the other eigenvalue
Hence Q
regardless of whether lk is a multiple or a distinct eigenvalue. For arbitrary
N the proof is achieved by mathematical induction. Taking into account the
previous proof, the Gramm-Schmidt orthogonalization, and property (4), we
deduce that an N N symmetric matrix has N orthonormal eigenvectors.
Example 12.5.2 Construct the orthonormal matrix Q ¼ ½ x1 x2 x3 , from
xT1 ¼ f 1:5 3 2 g.
Solution
We normalize the vector x1 with respect to its magnitude jx1 j ¼ 3:9051. Thus
we obtain
xT1 ¼ f 0:3841 0:7682 0:5121 g
The vector
relation
xT2
(1)
¼ f x12 x22 x32 g can be computed from the orthogonality
xT1 x2 ¼ 0:3841x12 0:7682x22 + 0:5121x32 ¼ 0
(2)
Thus, by determining the two components arbitrarily, then the third component is computed from Eq. (2). In this respect, we take arbitrarily
x12 ¼ 5, x22 ¼ 1. Then Eq. (2) gives x32 ¼ 5:25 and
xT2 ¼ f 5 1 5:25 g
(3)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
549
which is normalized with respect to its magnitude jx2 j ¼ 7:3186 and becomes
xT2 ¼ f 0:6832 0:1366 0:7173 g
(4)
The vector xT3 ¼ f x13 x23 x33 g can be computed from the orthogonality
relations
xT1 x3 ¼ 0:3841x13 0:7682x23 + 0:5121x33 ¼ 0
(5)
xT2 x3 ¼ 0:6832x13 0:1366x23 0:7173x33 ¼ 0
(6)
Thus, by determining x13 arbitrarily, say x13 ¼ 1, the foregoing relations
yield the linear system of equations
0:7682x23 + 0:5121x33 ¼ 0:3841
(7)
0:1366x23 0:7173x33 ¼ 0:6832
(8)
from which we obtain x23 ¼ 1:0070, x33 ¼ 0:7606, hence
x3 ¼ f 1 1:0070 0:7606 gT
(9)
which is normalized with respect to its magnitude jx3 j ¼ 1:6102 and becomes
x3 ¼ f 0:6211 0:6254 0:4724 gT
Therefore the requested orthonormal matrix is
2
3
0:3841 0:6832 0:6211
Q ¼ 4 0:7682 0:1366 0:6254 5
0:5121 0:7173 0:4724
(10)
(11)
The vectors x1 ,x2 , x3 provide a new basis in the three-dimensional space.
7. Reduction of a square matrix to diagonal form
From the previous properties, it can be concluded that the real and symmetric matrix A with dimensions N N has always N eigenvectors, which
are linearly independent and orthogonal to each other. Therefore, the eigenvectors can provide a new basis in the N dimensional space. A vector x
represented with respect to the new basis will have components
x^1 , x^2 ,…, x^N . On this basis, we can write
x ¼ x^1 x1 + x^2 x2 + … + x^N xN
(12.5.36)
x ¼ X^
x
(12.5.37)
or
where
2
X ¼ ½ x1 x2
x11
6 x21
⋯ xN ¼ 6
4⋯
xN 1
x12
x22
⋯
xN 2
⋯
⋯
⋯
⋯
3
x1N
x2N 7
7
⋯ 5
xNN
(12.5.38)
550 PART
II Multi-degree-of-freedom systems
represents the transformation matrix. It should be noted that matrix X can
be inverted,because the vectors xi are linearly independent, that is,
det ðXÞ 6¼ 0.
Next, we transform the relation y ¼Ax with respect to the new base.
According to Eq. (12.5.37), the vector y is written
^ ¼ X1 y
y
¼ X1 Ax
¼ X1 AX^
x
^x
¼ A^
(12.5.39)
where
^ ¼ X1 AX
A
(12.5.40)
But it holds
AX ¼ A½ x1 x2 ⋯ xN ¼ ½ Ax1 Ax2 ⋯ AxN ¼ ½ l1 x1 l2 x2 ⋯ lN xN (12.5.41)
¼ XL
where
2
l1
60
L ¼6
4⋯
0
0
l2
⋯
⋯
⋯
⋯
⋯
0
3
0
0 7
7
⋯ 5
lN
(12.5.42)
is a diagonal matrix with elements the N eigenvalues of A.
Finally, Eq. (12.5.40) by virtue of Eq. (12.5.41) becomes
^ ¼ X1 XL ¼ L
A
(12.5.43)
L ¼ X1 AX
(12.5.44)
or using Eq. (12.5.40)
that is, the matrix A in the space defined by the eigenvectors takes a
diagonal form.
Further, solving Eq. (12.5.44) with respect to A gives
A ¼ XLX1
(12.5.45)
The forgoing representation of A is called the spectral decomposition of
the matrix.
Taking into account the orthogonality property of the eigenvectors,
we obtain
Multi-degree-of-freedom systems: Free vibrations Chapter
2
x1 T x1
6
0
XT X ¼ XXT ¼ 6
4⋯
0
0
x2 T x2
⋯
0
⋯
⋯
⋯
⋯
3
0
7
0
7
5
0
T
xN xN
12
551
(12.5.46)
which,
ffiffiffiffiffiffiffiffiffiffiffiif the eigenvectors are normalized with respect to their magnitude,
p
xi T xi ¼ 1, becomes
XT X ¼ XXT ¼ I
(12.5.47)
X1 ¼ XT
(12.5.48)
A ¼ XLXT
(12.5.49)
Hence
and Eq. (12.5.45) becomes
8. Similar matrices have the same eigenvalues.
We consider that the change of the base is performed by the
transformation
x ¼T^
x
(12.5.50)
where T is a nonsingular matrix.
The matrix A is transformed in the space defined by the new base
according to Eq. (12.5.40), when we set X ¼ T. Hence
^ ¼ T1 AT
A
(12.5.51)
The above transformation is referred to as a similarity transformation
^ are similar matrices.
while A and A
^ is
The eigenvalue problem for the matrix A
^ lI x
^¼0
A
(12.5.52)
and its characteristic polynomial
^ lI
^ ðlÞ ¼ det A
P
¼ det T1 ðA lIÞT
(12.5.53)
¼ det T1 det ðA lIÞ det ðTÞ
¼ det T1 det ðTÞPðlÞ
Because det T1 and det ðTÞ are nonzero constants, we obtain
^ ðlÞ ¼ PðlÞ ¼ 0. Hence the characteristic polynomials P
^ ðlÞ and PðlÞ
P
^
have the same roots, which implies that the matrices A and A have the same
eigenvalues. Note that the property of the product of determinants has been
552 PART
II Multi-degree-of-freedom systems
employed to reach the foregoing conclusion, that is, det ðABCÞ ¼
det ðAÞdet ðBÞ det ðCÞ. Regarding the eigenvectors, they are transformed
according to Eq. (12.5.50).
12.5.3 The generalized eigenvalue problem
The problem of determining the eigenvalues and eigenvectors of a matrix as
stated above represents the typical or standard eigenvalue problem. However,
the eigenvalue problem for determining the eigenfrequencies and mode shapes
has a more general form
ðA lBÞx ¼ 0
(12.5.54)
This problem is known as the generalized eigenvalue problem of linear algebra. In the literature, it is also referred to as the linearized eigenvalue problem.
The study of the properties of the eigenvalues and eigenvectors of the generalized eigenvalue problem, Eq. (12.5.54), is facilitated if it is transformed to the
standard eigenvalue problem, Eq. (12.5.6). Thus, the properties that apply to the
standard eigenvalue problem can be transferred to the generalized eigenvalue
problem. Without excluding the generality, the discussion will be restricted
to real, symmetric, and positive definite matrices A and B because in free vibrations, they represent the stiffness and mass matrices, that is, A ¼ K and B ¼ M.
Their positive definiteness results from the fact that the elastic energy U ðuÞ and
the kinetic energy T ðu_ Þ are expressed by positive definite quadratic forms,
that is,
8
1
>
< uT Ku > 0, if u 6¼ 0
U ðuÞ ¼ 2
(12.5.55)
>
: 1 uT Ku ¼ 0, if u ¼ 0
2
8
1
>
< u_ T Mu_ > 0, if u_ 6¼ 0
T ðu_ Þ ¼ 2
(12.5.56)
>
: 1 u_ T Mu_ ¼ 0, if u_ ¼ 0
2
Applying Eq. (12.5.49) to matrix B we have
eX
eL
eT
B¼X
(12.5.57)
e is the diagonal matrix of the eigenvalues of B and X
e the matrix
where L
of its eigenvectors. Obviously, when the matrix B is diagonal, as in the case
e ¼ I.
of concentrated masses, then X
Further, we can set
B ¼ SST
(12.5.58)
Multi-degree-of-freedom systems: Free vibrations Chapter
where
e
S¼X
pffiffiffiffi
e
L
12
553
(12.5.59)
e are positive because B
The matrix S is real on account that the elements of L
was assumed positive definite.
Substituting Eq. (12.5.59) into Eq. (12.5.54), yields
Ax ¼ lSST x
(12.5.60)
^ ¼ ST x
x
(12.5.61)
^
x ¼ ST x
(12.5.62)
We define now the vector
thenb
1
Premultiplying Eq. (12.5.60) by S and using Eq. (12.5.62) yield
^ lI x
^¼0
A
(12.5.63)
where
^ ¼ S1 AST
A
(12.5.64)
^ is real and symmetric. Therefore, according to
Obviously, the matrix A
^ are real. Moreover, taking into
property 3, its eigenvalues and eigenvectors x
^ satisfy the orthogonality condition, x
^Ti x
^j ¼ 0, i 6¼ j,
account that the vectors x
we obtain.
T
^Ti x
^j ¼ ST xi ST xj
x
¼xTi SST xj
¼xTi Bxj
,
i 6¼ j
(12.5.65)
¼0
which implies that the eigenvectors of the generalized eigenvalue problem are
orthogonal with respect to the matrix B.
The eigenvalue problems (12.5.54) and (12.5.63) have the same eigenvalues. Indeed, we can set I ¼ S1 SST ST and write Eq. (12.5.63) by virtue
of Eqs. (12.5.64) and (12.5.58) as
1
^¼0
S ðA lBÞST x
(12.5.66)
^ ðlÞ are the characteristic polynomials of the eigenvalue
If PðlÞ and P
problems (12.5.54) and (12.5.63), respectively, we obtain
1
b. The notation ST ¼ ST
is employed.
554 PART
II Multi-degree-of-freedom systems
^ lI
^ ðlÞ ¼ det A
P
¼ det S1 ðA lBÞST
(12.5.67)
¼ det S1 det ðA lBÞdet ST
¼ det S1 PðlÞ det ST
Because det S1 6¼ 0 and det ST 6¼ 0, it implies that both eigenvalue
problems have the same characteristic equation, hence the same
eigenvalues.
The spectral decomposition of matrix B requires the complete solution of
the eigenvalue problem. Therefore, the transformation of the generalized eigenvalue problem on the basis of Eq. (12.5.59) is not the most convenient one.
A usual method to determine the matrix S is the Cholesky decomposition
method, or the square root method, in which the matrix B is written in the form
of a product, that is,
B ¼ UT U
(12.5.68)
where U is an upper triangular matrix. Hence
S ¼ UT
(12.5.69)
9. If the matrices A and B real, symmetric, and positive definite, then the
generalized eigenvalue problem has positive eigenvalues.
This is readily proved by premultiplying Eq. (12.5.54) by xT . This
gives
xT Ax ¼ lxT Bx
(12.5.70)
Because A and B are positive definite, we obtain
l¼
xT Ax
>0
xT Bx
(12.5.71)
10. The eigenvectors xi of the generalized eigenvalue problem are orthogonal
with respect to the matrices A and B.
It was previously shown that
xTi Bxj ¼ 0,
i 6¼ j
(12.5.72)
Consequently, we obtain
xTi Axj ¼ lxTi Bxj
¼0
(12.5.73)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
555
Example 12.5.3 Transform the generalized eigenvalue problem into a standard
eigenvalue problem using the spectral decomposition method, when
A¼
50 40
2:5 1
, B¼
40 90
1
3:2
Solution
The characteristic equation of B is
2:5 l
e 1
¼l
e2 5:7l
e + 7:0 ¼ 0
1
e
3:2 l (1)
from which we obtain
e2 ¼ 3:909
e1 ¼ 1:790, l
l
(2)
Hence
e ¼ 1:790 0
L
0
3:909
(3)
The eigenvectors are computed from the solution of the homogeneous linear
system
e 1
x1
0
2:5 l
(4)
e x2 ¼ 0
1
3:2 l
e1 ¼ 1:790 and l
e2 ¼ 3:909.
for l
Thus, we obtain the matrix of the eigenvectors normalized with respect to
their magnitude
"
#
0:8157 0:5784
e¼
X
(5)
0:5785 0:8158
Using Eq. (12.5.59) we obtain
pffiffiffiffi
e ¼ 1:0913 1:1435 , S1 ¼ 0:6097 0:4323
e L
S¼X
0:2926 0:4126
0:7740 1:6129
(6)
and on the basis of Eq. (12.5.64)
^ ¼ S1 AST ¼
A
14:3213 2:1288
2:1288 29:2564
(7)
11. If the real and symmetric matrix A is singular, then the generalized eigenvalue problem has at least one zero eigenvalue and the corresponding
eigenvector is different from zero.
556 PART
II Multi-degree-of-freedom systems
First, we will show that this property holds for the standard eigenvalue
problem Ax ¼ lx. For this purpose, we write A in the form of its spectral
decomposition,
A ¼ XLXT
(12.5.74)
Because it was assumed that A is
it implies that det ðAÞ ¼ 0.
singular,
Moreover, Eq. (12.5.47) gives det XT det ðXÞ ¼ 1, hence
det ðAÞ ¼ det XT det ðLÞdet ðXÞ
¼ det ðLÞ
¼l1 l2 ⋯lN
(12.5.75)
¼0
from which we conclude that at least one of the li is zero and the eigenvalue problem for this value becomes
Axi ¼ 0
(12.5.76)
which yields xi 6¼ 0 because det ðAÞ ¼ 0.
The generalized eigenvalue problem Ax ¼ lBx is transformed to the
^ x ¼ l^
standard eigenvalue problem A^
x, hence it is
^ ¼ l1 l2 ⋯lN
det A
(12.5.77)
which by virtue of Eq. (12.5.64) gives
^ ¼ det S1 det ðAÞdet ST
det A
¼0
(12.5.78)
because it was assumed det ðAÞ ¼ 0. Eqs. (12.5.77) and (12.5.78) imply
^i 6¼ 0 and
that at least one of the eigenvalues li is zero. Moreover, it is x
^i 6¼ 0.
by virtue of Eq. (12.5.62), we obtain xi ¼ ST x
If the matrices A and B represent the stiffness and mass matrices of
the structure, that is, A ¼ K, B ¼ M, then the eigenvector xi corresponding to the zero eigenvalue represents rigid body motion. This is shown
right below.
If we set xi ¼ ui , then Kui represents the vector of the elastic force f Si
corresponding to the displacement ui , that is
f Si ¼ Kui
(12.5.79)
Kui ¼ li Mui
¼0
(12.5.80)
or because li ¼ 0
Hence f Si ¼ 0 while Eq. (12.5.76) yieldsui 6¼ 0, which is due to the
motion of the structure as a rigid body.
Multi-degree-of-freedom systems: Free vibrations Chapter
12
557
12. If the real and symmetric matrix B is singular, then the generalized eigenvalue problem has at least one infinite eigenvalue.
This is shown if the eigenvalue problem is written in the form
Bx ¼ mAx
(12.5.81)
where m ¼ 1=l.
13. Any vector u with a dimension N can be represented as the superposition
of the eigenvectors of the eigenvalue problem.
We showed that the set of the eigenvectors x1 ,x2 ,…, xN of an N N
symmetric matrix is linearly independent and can be employed as a base of
the N dimensional space to represent an arbitrary vector u in that space.
Thus we may set
u ¼ a1 x1 + a2 x2 + ⋯ + aN xN
(12.5.82)
u ¼ Xa
(12.5.83)
or
where X is the matrix of the eigenvectors and a ¼ fa1 , a2 , …, aN gT the
vector of the coefficients. The matrix X is not singular because the eigenvectors xi are linearly independent. Hence
a ¼ X1 u
(12.5.84)
a ¼ XT u
(12.5.85)
or using Eq. (12.5.48)
or
ai ¼ xi T u,
i ¼ 1, 2, …,N
(12.5.86)
Similarly, we can use the eigenvectors of the generalized eigenvalue
problem to represent the vector u. The establishment of the coefficients
ai in that case is established by premultiplying both sides of
Eq. (12.5.82) by xTi B and noting that xTi Bxj ¼ 0 for i 6¼ j. This yields
ai ¼
xTi Bu
xTi Bxi
(12.5.87)
If the eigenvectors xi are normalized with respect to B so that
xTi Bxi ¼ 1, then Eq. (12.5.87) gives
ai ¼ xTi Bu, i ¼ 1, 2, …,N
(12.5.88)
a ¼ XT Bu
(12.5.89)
or
The representation of a vector as a superposition of the eigenvectors as
in Eq. (12.5.82) is known as the expansion theorem. As we will see in
558 PART
II Multi-degree-of-freedom systems
Section 14.5, this theorem is a special case of a Ritz vector representation
when the eigenvectors are used as Ritz vectors.
14. If a real and symmetric matrix A is singular, then the quadratic form
PðuÞ ¼ uT Au is positive semidefinite, that is, PðuÞ ¼ uT Au 0 for u 6¼ 0.
We write the vector u in the form of Eq. (12.5.83) and the matrix A in
the form of Eq. (12.5.49). Thus we have
PðuÞ ¼uT Au
¼aT XT AXa
¼aT La
¼
(12.5.90)
N
X
li ai2
i¼1
Because A is singular, at least on of its eigenvalues is zero, say lk ¼ 0.
If we take u ¼ xk , it will be ak 6¼ 0, ai6¼k ¼ 0 and Eq. (12.5.90) becomes
PðuÞ ¼lk ak2
¼0
(12.5.91)
Therefore
PðuÞ ¼ uT Au 0 for u 6¼ 0
(12.5.92)
12.6 The Rayleigh quotient
In Section 12.5, we discussed the standard and the generalized eigenvalue problems and we studied the properties of the eigenvalues and eigenvectors. In this
section, we will supplement these properties with Rayleigh’s quotient, which
yields further important properties that are very useful for the numerical computation of eigenvalues and eigenvectors.
The Rayleigh quotient for an N N symmetric matrix A is the scalar
quantity defined by the relation
rðuÞ ¼
uT Au
uT u
(12.6.1)
where u is a vector with dimension N . We will show that the following
properties hold.
1. If l1 l2 l3 ⋯ lN are the eigenvalues of A, then
l1 rðuÞ lN
(12.6.2)
and it is rðuÞ > 0, if A is positive definite and rðuÞ 0 if A is positive
semidefinite. Moreover, it is rðuÞ ¼ l1 if u ¼ x1 and rðuÞ ¼ lN if u ¼ xN .
Multi-degree-of-freedom systems: Free vibrations Chapter
12
559
On the basis of the expansion theorem, Eq. (12.5.82), we may set
u¼
N
X
ak xk
(12.6.3)
k¼1
where the eigenvectors xk of A are assumed normalized with respect to their
magnitude, that is
xTi xj ¼ d ij
(12.6.4)
where dij is the Kronecker delta defined as d ij ¼ 1 if i ¼ j and dij ¼ 0 if i 6¼ j.
Substituting Eq. (12.6.3) into Eq. (12.6.1) and taking into account
Eq. (12.6.4) and Axk ¼ lk xk yield
rðuÞ ¼
l1 a12 + l2 a22 + ⋯ + lN aN2
a12 + a22 + ⋯ + aN2
(12.6.5)
which for l1 6¼ 0 becomes
rðuÞ ¼ l1
a12 + ðl2 =l1 Þa22 + ⋯ + ðlN =l1 ÞaN2
a12 + a22 + ⋯ + aN2
(12.6.6)
Because lk =l1 1 (k ¼ 2, 3, …, N ), it follows that the nominator of
the foregoing fraction is greater or equal to the denominator, hence
rðuÞ l1 . If u ¼ x1 we have a1 ¼ 1, a2 ¼ a3 ¼ ⋯ ¼ aN ¼ 0 and Eq.
(12.6.5) gives rðuÞ ¼ l1 .
Similarly, for lN 6¼ 0 Eq. (12.6.5) is written as
rðuÞ ¼ lN
a12 ðl1 =lN Þ + ðl2 =lN Þa22 + ⋯ + aN2
a12 + a22 + ⋯ + aN2
(12.6.7)
The nominator of the forgoing fraction is less or equal to the denominator
because lk =lN 1 (k ¼ 1, 2, …,N 1), hence rðuÞ lN . If u ¼ xN we
have a1 ¼ a2 ¼ ⋯ ¼ aN 1 ¼ 0, aN ¼ 1 and Eq. (12.6.7) gives rðuÞ ¼ lN .
2. If the vector u is chosen from a subset of vectors that is orthogonal to the
first k 1 eigenvectors of A, then
rðuÞ lk
(12.6.8)
Applying the expansion theorem, Eq. (12.5.82), we have
u ¼ a1 x1 + a2 x2 + ⋯ + ak1 xk1 + ak xk + ⋯ + aN xN
(12.6.9)
Due to the assumed orthogonality, it is uT xj ¼ 0, j ¼ 1, 2, …,k 1
and Eq. (12.5.86) gives aj ¼ 0, j ¼ 1, 2, …,k 1. Hence Eq. (12.6.9) is
written as
u¼
N
X
j¼k
aj xj
(12.6.10)
560 PART
II Multi-degree-of-freedom systems
Substituting the previous expression for u into Rayleigh’s quotient,
Eq. (12.6.1), yields
rðuÞ ¼
lk ak2 + lk + 1 ak2 + 1 + ⋯ + lN aN2
ak2 + ak2 + 1 + ⋯ + aN2
a 2 + ðlk + 1 =lk Þak2 + 1 + ⋯ + ðlN =lk ÞaN2
¼lk k
ak2 + ak2 + 1 + ⋯ + aN2
(12.6.11)
Inasmuch as lk + 1 =lk 1, ⋯,lN =lk 1, it follows that the nominator
of the foregoing fraction is greater or equal to the denominator, hence
rðuÞ lk . If u ¼ xk we have ak ¼ 1, ak + 1 ¼ ⋯ ¼ aN ¼ 0 and Eq.
(12.6.11) gives rðuÞ ¼ lk .
3. If the vector u deviates from the eigenvector xk by an error of order e, then
Rayleigh’s quotient approximates the eigenvalue lk with an error of order
e2 , namely
rðuÞ ¼ lk + O e2
(12.6.12)
To prove this statement we set
u
u ¼ xk + ee
(12.6.13)
where ee
u is the deviation of u from xk and expresses the contribution of the
remaining eigenvectors.
Substituting Eq. (12.6.13) into Eq. (12.6.1) gives
xTk + ee
uT Aðxk + ee
uÞ
rðuÞ ¼ uT ðxk + ee
uÞ
xTk + ee
(12.6.14)
T
xTk Axk + ee
u Axk + exTk Ae
u + e2 e
u
uT Ae
¼
e + e2 u
eT u
e
uT xk + exTk u
xTk xk + ee
Further, by virtue of the expansion theorem, Eq. (12.5.82), we can write
e
u¼
N
X
aj x j
(12.6.15)
j ¼1
j 6¼ k
Substituting the foregoing expression for e
u into Eq. (12.6.14) and taking
into account the orthogonality of the eigenvectors, we obtain
lk + e2
rðuÞ ¼
N
X
j¼1, j6¼k
1 + e2 S
lj aj2
(12.6.16)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
561
where it was set
S¼
N
X
j ¼1
j 6¼ k
aj2
(12.6.17)
Expanding the denominator in Eq. (12.6.16) in series using the
polynomial theorem for a negative exponent, we obtain
1 + e2 S
1
¼ 1 e2 S + e4 S 2 e6 S 3 + ⋯
(12.6.18)
which is introduced into Eq. (12.6.16) to give
0
1
B
C
N
X
B
C
2
2C
2
4 2
6 3
rðuÞ ¼B
l
+
e
a
l
j j C 1e S +e S e S +⋯
B k
@
A
j ¼1
j 6¼ k
0
1
(12.6.19)
B X
C
B N
C
2
¼lk + e B
lj aj lk S C
C + higher order terms
@j ¼1
A
2B
j 6¼ k
Obviously, Eq. (12.6.19) proves Eq. (12.6.12).
The Rayleigh’s quotient for the standard eigenvalue problem and the
properties resulting thereof can be extended to the generalized eigenvalue
problem as well. This becomes obvious if the generalized eigenvalue prob^ x ¼ l^
lem Ax ¼ lBx is transformed into the standard one A^
x. Thus writing
Rayleigh’s quotient in the form
^Þ ¼
rðu
^u
^T A^
u
T
^ u
^
u
(12.6.20)
and taking into account that (see Eqs. 12.5.58, 12.5.61, and 12.5.64)
^ ¼ S1 AST
^ ¼ ST u, A
ST S ¼ B, u
Eq. (12.6.20) becomes
rðuÞ ¼
uT Au
uT Bu
(12.6.21)
562 PART
II Multi-degree-of-freedom systems
Example 12.6.1 Determine the eigenfrequencies and eigenmodes of the twostory building in Example 12.2.1 using the properties of Rayleigh’s quotient.
Solution
The mass and stiffness matrices of the building were computed in Example
12.2.1, that is,
M¼
25 0
,
0 32
K¼
We take
u¼
3826:5 3826:5
3826:5 9142:1
1
x
(1)
(2)
where x is an unknown. We will determine x so that Rayleigh’s quotient takes
an extreme value. For this purpose, we formulate Rayleigh’s quotient,
Eq. (12.6.21), for A ¼ K and B ¼ M. Thus we obtain
rðx Þ ¼
uT Ku
uT Mu
½1 x 3826:5 3826:5
1
3826:5 9142:1 x
25 0
1
½1 x 0 32 x
9142:1x 2 7653:0x + 3826:5
¼
32x 2 + 25
¼
(3)
The value of x is obtained by establishing the extreme values of the function
rðx Þ. Thus we obtain
drðx Þ 244896x 2 + 212209x 191325
¼
dx
ð32x 2 + 25Þ2
(4)
¼0
Eq. (4) has two roots: x1 ¼ 0:5511 kai x2 ¼ 1:4176. For these values, the
second derivative
d 2 rðx Þ 107 ð1:567x 3 + 2:037x 2 3:673x 0:530Þ
¼
dx 2
ð32x 2 + 25Þ3
(5)
gives
d 2 rðx 1 Þ
d 2 rðx 2 Þ
¼
399:98
>
0,
¼ 60:4469 < 0
(6)
dx 2
dx 2
Therefore, x1 yields the minimum and x2 the maximum. On the base of
Eq. (3), we obtain
1
rðx1 Þ ¼ l1 ¼ w21 ¼ 68:7089, w1 ¼ 8:289, u1 ¼
(7)
0:5511
Multi-degree-of-freedom systems: Free vibrations Chapter
rðx2 Þ ¼ l2 ¼ w22 ¼ 370:0416,
w2 ¼ 19:236,
u2 ¼
12
1
1:4127
563
(8)
Normalizing the eigenmodes with respect to the mass, we obtain
F¼
0:1697 0:1061
0:0935 0:1499
(9)
Obviously, the obtained eigenfrequencies and mode shapes are identical to
those obtained by the method in Section 12.2.
12.7 Properties of eigenfrequencies and modes of MDOF
systems without damping: A summary
The properties of the generalized eigenvalue problem are applied to the problem
of the free vibrations of MDOF systems if the matrices A and B in Eq. (12.5.54)
represent the stiffness and mass matrices, respectively, that is, A ¼ K, B ¼ M.
Then the eigenvalues express the squares of the eigenfrequencies, li ¼ w2i , and
the eigenvectors the normalized eigenmodes of the vibrations. In Table 12.7.1,
these properties are summarized using the terminology employed in free vibration analysis.
TABLE 12.7.1 Properties of the eigenfrequencies and eigenmodes in free
vibrations of MDOF systems without damping.
1. When K and M are real, symmetric, and positive definite, then all eigenfrequencies are
positive.
2. When K is singular, det ðKÞ ¼ 0, at least one eigenfrequency is zero and the corresponding
eigenmodes express rigid body motion while the elastic energy is positive semidefinite,
U ðuÞ ¼ 12 uT Ku 0 for u 6¼ 0.
3. When M is singular, det ðMÞ ¼ 0, at least one eigenfrequency is infinite and the kinetic
energy is positive semidefinite, T ðu Þ ¼ 12 u_ T Mu_ 0 for u_ 6¼ 0.
4. To each eigenfrequency there correspond as many linearly independent eigenvectors
as the multiplicity of the eigenfrequency.
5. The eigenmodes, including those corresponding to a repeated eigenfrequency, are
linearly independent and orthogonal with respect to M and K, that is
fTm Mfn ¼ 0,
m 6¼ n
fTm Kfn
m 6¼ n
¼ 0,
6. Any arbitrary vector u with dimension N can be expressed as a superposition of the
N eigenmodes
u ¼ Fa, a ¼ FT Mu
12.8 Solution of the vibration problem without damping
In Section 12.2, it was shown that the solution of Eq. (12.1.2) is given by
Eq. (12.2.25) or Eq. (12.2.26), which we rewrite here for convenience
564 PART
II Multi-degree-of-freedom systems
uðt Þ ¼
N
X
fi ðci cos wi t + di sin wi t Þ
(12.8.1)
i¼1
or
uðt Þ ¼
N
X
fi ai cos ðwi t qi Þ
(12.8.2)
i¼1
in which wi and fi are the eigenfrequencies and eigenmodes, respectively,
obtained from the solution of the eigenvalue problem
Kf ¼ lMf
(12.8.3)
The use of the eigenmodes fi instead of the eigenvectors bi is permitted
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
because the normalization coefficient mi ¼ 1= fTi Mfi can be incorporated
in the arbitrary constants ci, di or ai ,qi . The solution of the vibration problem
is accomplished when the arbitrary constants are determined. This is achieved
by using the 2N initial conditions specified by the displacement vector uð0Þ and
velocity vector u_ ð0Þ, each of dimension N , at t ¼ 0.
Eq. (12.8.1) for t ¼ 0 gives
or
uð0Þ ¼ f1 c1 + f2 c2 + ⋯ + fN cN
(12.8.4)
uð0Þ ¼ Fc
(12.8.5)
T
where F is the modal matrix and c ¼ fc1 , c2 , ⋯, cN g the vector of the
unknown coefficients ci. Eq. (12.8.5) provides a system of N linear algebraic
equations with respect to N unknowns, whose solution results in the vector c,
namely
c ¼ F1 uð0Þ
(12.8.6)
The constants di are determine by differentiating Eq. (12.8.1) and using an
analogous procedure.
The foregoing solution requires the inversion of the matrix F, which is not a
simple task. This can be avoided if the orthogonality of the eigenmodes with
respect to the mass is employed. For this purpose both sides of Eq. (12.8.4)
are premultiplied by fTi M. Thus we obtain
fTi Muð0Þ ¼ fTi Mfi ci
because the terms
fTi Mfj
(12.8.7)
¼ 0, i 6¼ j, vanish. Therefore
fTi Muð0Þ
fTi Mfi
(12.8.8)
ci ¼ fTi Muð0Þ
(12.8.9)
ci ¼
which is further written as
Multi-degree-of-freedom systems: Free vibrations Chapter
12
565
if the eigenmodes have been orthonormalized with respect to the mass, that is,
fTi Mfi ¼ 1 (see Eq. 12.2.29).
The same procedure is used to evaluate the coefficients di . Differentiating
Eq. (12.8.1) with respect to time yields
u_ ðt Þ ¼
N
X
fi wi ðci sin wi t + di cos wi t Þ
(12.8.10)
i¼1
and for t ¼ 0
u_ ð0Þ ¼ f1 w1 d1 + f2 w2 d2 + ⋯ + fN wN dN
(12.8.11)
Premultiplying the foregoing equation by fTi M yields
di ¼
1 fTi Mu_ ð0Þ
wi fTi Mfi
(12.8.12)
di ¼
1 T
f Mu_ ð0Þ
wi i
(12.8.13)
and if fTi Mfi ¼ 1
The constants ai ,qi are evaluated from the relations
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ai ¼ ci2 + di2 , qi ¼ tan 1 ðdi =ci Þ
(12.8.14)
12.9 The method of mode superposition
The solution to the problem of free vibrations can be obtained in another way,
which is based on the expansion theorem. Thus according to Eq. (12.5.82) the
displacement vector can be represented as
uðt Þ ¼f1 Y1 ðt Þ + f2 Y2 ðt Þ + ⋯ + fN YN ðt Þ
¼FYðt Þ
(12.9.1)
The foregoing equation transforms the components of the vector
u ¼ f u1 u2 ⋯ uN gT referred to a system with base vectors e1 ð1, 0, …, 0Þ,
e2 ð0, 1, …, 0Þ, …,eN ð0, 0, …, 1Þ into the components Y ¼ f Y1 Y2 ⋯ YN gT
referred to the system with base vectors the eigenmodes f1 , f2 ,⋯,fN . The components Yn , n ¼ 1, 2, …,N are generalized coordinates and are referred to as modal or
normal coordinates.
Substituting Eq. (12.9.1) into Eq. (12.1.2) yields
M f1 Y€ 1 + f2 Y€ 2 + ⋯ + fN Y€ N + Kðf1 Y1 + f2 Y2 + ⋯ + fN YN Þ ¼ 0
(12.9.2)
fTn
and taking into account
Then, premultiplying the foregoing equation by
the orthogonality conditions fTn Mfj ¼ 0 and fTn Kfj ¼ 0 for n 6¼ j, we obtain
566 PART
II Multi-degree-of-freedom systems
fTn Mfn Y€ n + fTn Kfn Yn ¼0,
n ¼ 1, 2, …, N
(12.9.3)
Setting further
Mn ¼ fTn Mfn
(12.9.4)
Kn ¼ fTn Kfn
(12.9.5)
Eq. (12.9.3) is written
Mn Y€ n + Kn Yn ¼0,
n ¼ 1, 2, …,N
(12.9.6)
The quantities Mn and Kn express the generalized mass and the generalized
stiffness of the nth eigenmode, respectively, and are called the generalized or
modal mass and generalized or modal stiffness. Further, taking into account that
Kfn ¼ w2n Mfn (see Eq. 12.3.3), we obtain
Kn ¼ w2n Mn
(12.9.7)
and observing that Mn 6¼ 0, Eq. (12.9.6) becomes
Y€ n + w2n Yn ¼ 0,
n ¼ 1, 2, …,N
(12.9.8)
The transformation of the natural coordinates un to modal coordinates
reduces the system of the N coupled equations to a set N uncoupled equations
of a SDOF system. Hence, the solution of the nth equation is given by
Eq. (2.2.13), namely
Y_ n ð0Þ
sin wn t + Yn ð0Þcoswn t, n ¼ 1, 2, …,N
(12.9.9)
wn
The quantities Yn ð0Þ, Y_ n ð0Þ represent the initial conditions of the transformed equations and can be obtained from the specified initial conditions
for the physical displacements, that is, uð0Þ, u_ ð0Þ, as follows.
Premultiplied Eq. (12.9.1) by fTn M yields
Yn ¼
Y n ðt Þ ¼
fTn Muðt Þ
,
Mn
n ¼ 1, 2, …,N
(12.9.10)
and differentiating it with respect to t gives
f Mu_ ðt Þ
,
Y_ n ðt Þ ¼ n
Mn
T
n ¼ 1, 2, …,N
(12.9.11)
For t ¼ 0, Eqs. (12.9.10) and (12.9.11) give the sought initial conditions
fTn Muð0Þ
,
Mn
n ¼ 1, 2, …,N
(12.9.12)
f Mu_ ð0Þ
,
Y_ n ð0Þ ¼ n
Mn
n ¼ 1, 2, …,N
(12.9.13)
Y n ð 0Þ ¼
T
Multi-degree-of-freedom systems: Free vibrations Chapter
12
567
The foregoing relations are further simplified if the eigenmodes are normalized on the basis of Eq. (12.2.29). This gives
Mn ¼ fTn Μfn ¼ 1
(12.9.14)
Kn ¼ w2n
(12.9.15)
that is, the modal masses become equal to the unity and the modal stiffnesses
equal to the square of the corresponding eigenfrequency. Thus, we have
Yn ð0Þ ¼ fTn Muð0Þ,
n ¼ 1, 2, …,N
(12.9.16)
Y_ n ð0Þ ¼ fTn Mu_ ð0Þ,
n ¼ 1, 2, …,N
(12.9.17)
The vector un ¼ fn Yn , n ¼ 1, 2, …,N expresses the contribution of the nth
eigenmode to the displacement vector u and is referred to as the nth modal
component of the displacement vector. Hence, Eq. (12.9.1) is written as
u ¼ u 1 + u2 + ⋯ + u N
(12.9.18)
which implies that the displacement vector is the superposition of the modal
components or, in other words, the displacement vector is the superposition
of the eigenmodes multiplied by a time-dependent weight coefficient.
This method of solving the equation of motion for the free vibration problem
is known as the modal superposition method or the method of the superposition
of the eigenmodes. It is used not only for free undamped vibrations but also for
damped vibrations as well as for forced vibrations. The superposition is
illustrated in Fig. 12.9.1, where the mode shapes correspond to those of the
two-story shear frame in Example 12.2.1.
FIG. 12.9.1 Modal superposition.
Example 12.9.1 Determine the displacements of the frame in Example 12.2.1
using the modal superposition method to solve the equation of motion. Assumed
initial conditions uð0Þ ¼ f0:03 0:01gT , u_ ð0Þ ¼ 0.
Solution
In Example 12.2.1, it was found
M¼
25 0
3826:5 3826:5
, K¼
0 32
3826:5 9142:1
(1)
568 PART
II Multi-degree-of-freedom systems
w1 ¼ 8:289,
0:1697
f1 ¼
,
0:0935
w2 ¼ 19:236
0:1061
f2 ¼
0:1499
(2)
Because the eigenmodes are orthonormalized with respect to the mass, the
modal matrices resulting from Eq. (12.9.14) are
M1 ¼ fT1 Μf1 ¼ 1,
M2 ¼ fT2 Μf2 ¼ 1
Applying Eqs. (12.9.12) and (12.9.13) gives
25 0
0:03
Y1 ð0Þ ¼ ½ 0:1697 0:0935 ¼ 0:15720
0 32 0:01
25 0
0:03
¼ 0:0316
Y2 ð0Þ ¼ ½ 0:1061 0:1499 0 32 0:01
(3)
(4)
(5)
Y_ 1 ð0Þ ¼ Y_ 2 ð0Þ ¼ 0
(6)
Y1 ¼ 0:15720cos 8:289t
(7a)
Y2 ¼ 0:0316cos 19:236t
(7b)
Thus Eq. (12.9.9) gives
Then Eq. (12.9.1) gives the displacement vector
2:6677
0:3353
u 1 ðt Þ
cos 8:289t + 102 cos 19:236t
¼ 102 1:4698
0:4700
u 2 ðt Þ
(8)
Example 12.9.2 The loads P1 and P2 applied to the structure in Fig.E12.4a are
removed suddenly at t ¼ 0. Determine the ratio P1 =P2 so that the structure
vibrates in the form of the second eigenmode. The axial deformation of the col ¼ 1,
umn is neglected. Consider lumped mass assumption. Assume a ¼ 1, m
E ¼ 2:1 107 kN=m2 , and column cross-section 0:20 0:20 m2 .
(a)
(b)
FIG. E12.4 Structure in Example 12.9.2 (a); Parameters of motion (b).
Multi-degree-of-freedom systems: Free vibrations Chapter
12
569
Solution
The system has two degrees of freedom, the horizontal displacement u and the
rotation f at the top of the column shown in Fig. E12.4b. The equation of
motion can be obtained either by applying Eqs. (A.4.8) or by considering the
structure as a plane frame consisting of one element and including a rigid body
at its free end (see Section 11.11). Here, Eqs. (A.4.8) are employed. Thus,
€ Fx ¼ fS , Fy ¼ 0, MP ¼ MS , IP ¼ IO ,
setting yc ¼ 0, XP ¼ u, YP ¼ 0, w_ ¼ f,
we obtain
m u€ + fS ¼ 0
(1)
Fy ¼ mx c f€
(2)
IO f€ + MS ¼ 0
(3)
The elastic forces are applied at point O and are given by the relations
fS ¼ k11 u + k12 f
MS ¼ k21 u + k22 f
Eqs. (1) and (3) are combined in matrix form
M€
u + Ku ¼ 0
(4)
where
M¼
m 0
k k
, K ¼ 11 12
0 Io
k21 k22
Note that Eq. (2) will be used to evaluate the unknown force Fy once u,f are
established from the solution of Eq. (4).
Stiffness matrix: The stiffness matrix will result from Eq. (11.5.2) for
u3 ¼ u4 ¼ 0. Thus, we have
2
3
2 3 3a 3
12EI 6EI
6 ð4a Þ3 ð4a Þ2 7 EI
6
7
6 16 8 7
K¼6
(5)
7¼ 4
5
4 6EI 4EI 5 a 3 3a 2
a
8
ð4a Þ2 4a
Mass matrix: We have
1
+ ð3a Þ10m
¼ 32a m
m ¼ ð4a Þm
2
Z 2a
2 dx ¼ 30ma
3
Io ¼
10mx
a
hence
M¼
m 0
32 0
¼ ma
0 Io
0 30a 2
(6)
570 PART
II Multi-degree-of-freedom systems
Eigenvalues and eigenmodes: They are obtained from the eigenvalue
problem
4 =EI
ðK lMÞb ¼ 0, l ¼ w2 ma
(7)
l1 ¼ 3:606, l2 ¼ 106:133
(8)
which yields
F¼
0:1654 0:0625
0:0645 0:1708
(9)
Initial conditions: There are only initial displacements, which are equal to
the static displacements due to the loads P1 , P2 . They are evaluated from the
relation
P1
0:7619P1 + 0:5714P2
¼ 102
uð0Þ ¼ K1
(10)
2P2 a
0:2857P1 0:2857P2
In order that the structure vibrates in the form of the second eigenmode, it
must be Y1 ðt Þ 0, which holds if Y1 ð0Þ ¼ Y_ 1 ð0Þ ¼ 0. Inasmuch as u_ ð0Þ ¼ 0,
Eq. (12.9.17) gives Y_ 1 ð0Þ ¼ 0. If we take uð0Þ ¼ kf2 , with k being an arbitrary
constant, then Eq. (12.9.16) gives
Y1 ð0Þ ¼fT1 Muð0Þ
¼kfT1 Mf2
(11)
¼0
Therefore, it must be
0:7619P1 + 0:5714P2
0:0625
2
¼k
10
0:2857P1 0:2857P2
0:1708
(12)
from which we obtain
P1 ¼ 212:1k, P1 ¼ 271:9k
P1 =P2 ¼ 0:7801
(13)
12.10 Solution of the vibration problem with damping
In the case of free vibrations with damping, the damping matrix C does not
vanish. Thus the equation that must be solved is
M€
u + Cu_ + Ku ¼0
(12.10.1)
subject to the specified initial conditions uð0Þ, u_ ð0Þ.
In the following, three methods are presented for the solution of this
problem.
Multi-degree-of-freedom systems: Free vibrations Chapter
12
571
12.10.1 Direct solution of the differential equation
We look for a solution of the form
u ¼ belt
which is substituted into Eq. (12.10.1) to yield
2
l M + lC + K b ¼0
(12.10.2)
(12.10.3)
The foregoing equation represents a homogeneous system of N linear
algebraic equations, which can be solved to give a nontrivial vector b, provided
that the determinant of the coefficient matrix
is zero. Namely
SðlÞ ¼ l2 M + lC + K
(12.10.4)
det l2 M + lC + K ¼ 0
(12.10.5)
Expansion of the determinant yields a real polynomial of 2N degree in l,
whose vanishing gives the characteristic equation of the differential equation
of motion Eq. (12.10.1), namely
PðlÞ ¼ a0 l2N + a1 l2N 1 + ⋯ + a2N ¼ 0
(12.10.6)
According to the fundamental theorem of algebra, Eq. (12.10.6) has 2N
roots l1 ,l2 ,…,l2N . Because the coefficients of the polynomial are real, the
roots are either real or complex conjugate pairs. Eq. (12.10.3) represents an
eigenvalue problem referred to as a quadratic eigenvalue problem. For each
value of ln a vector bn is obtained, which is real or complex depending on
whether ln is real or complex. Assuming that the roots are discrete, the general
solution of Eq. (12.10.1) is given by the superposition
u¼
2N
X
an bn eln t
(12.10.7)
n¼1
where an are 2N arbitrary constants, which are determined from the initial
conditions.
The response of the system depends on the type of roots ln . We distinguish
the following cases:
(i) Real roots
Let ln be a real root, the corresponding solution (12.10.2) is
un ¼ bn eln t
(12.10.8)
Obviously, for ln > 0 the solution diverges exponentially while for
ln < 0 it converges exponentially. In both cases, the motion of the system
is not an oscillation. If the root has a multiplicity k, then the solution
(12.10.2) becomes [3]
572 PART
II Multi-degree-of-freedom systems
un ¼ bn a0 + a1 t + ⋯ + t k1 eln t
(12.10.9)
which again diverges or converges exponentially depending on whether
ln > 0 or ln < 0, respectively, and the motion of the system is not an
oscillation.
(ii) Complex roots.
m is also a root.
Let lm be complex, then its complex conjugate l
We set
m ¼ m iwm
lm ¼ mm + iwm and l
m
(12.10.10)
m are also complex conjugate and their
The respective vectors bm and b
contribution to the general solution is
m e lm t
um ¼am bm elm t + am∗ b
m eiwm t
(12.10.11)
¼emm t am bm eiwm t + am∗ b
m cos wm t + i am bm a ∗ b
¼emm t am bm + am∗ b
m m sin wm t
The constants am , am∗ are arbitrary constants, hence they can be chosen so that
they are complex conjugates, that is
am ¼ cm + id m and am∗ ¼ cm id m
(12.10.12)
where cm and dm are also arbitrary constants.
Further, we set
m ¼ pm iqm
bm ¼ pm + iqm and b
(12.10.13)
Substituting Eqs. (12.10.12) and (12.10.13) into Eq. (12.10.11) gives
um ¼ 2emm t ½ðcm pm dm qm Þ cos wm t ðcm qm + dm pm Þsin wm t (12.10.14)
The factor within the square bracket expresses a harmonic vibration. Its
amplitude diverges exponentially if mm > 0 and converges exponentially if
mm < 0. Finally, if mm ¼ 0, the roots are imaginary and the amplitude of vibration remains constant. The vibration with mm > 0 is known as flutter or negative
damping. The solution is obtained as a superposition of the terms given by Eq.
(12.10.14), namely
u¼
N
X
um
(12.10.15)
m¼1
Eq. (12.10.14) implies that the natural components of the displacements are
real, and as would be anticipated, the general solution is real.
Multi-degree-of-freedom systems: Free vibrations Chapter
12
573
Eq. (12.10.14) is further written
3
0 8
8
8
8
9
91
9
91
p1m >
q1m >
p1m >
>
>
>
>
> p1m >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
7
C
C
6B >
B
>
>
>
>
>
>
>
>
>
>
>
>
>
>
7
C
C
6B >
B >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
7
C
C
6B >
B
>
>
>
>
>
>
>
>
p
p
q
p
>
>
>
>
>
>
>
>
2m >
2m >C
2m >
2m >C
7
6B >
B >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
7
C
C
6B >
B >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
7
C
C
6B >
B
>
>
>
< ⋮ >
<
< ⋮ >
< ⋮ >
=
=C
=
=C
7
6B >
B >
7
C
C
6B
B
m
t
m
C cos wm t Bcm
C sin wm t 7
6Bcm
um ¼2e
dm
+ dm
>
>
>
>
>
>
7
C
C
6B >
B >
>
>
>
>
>
>
>
p
q
p
p
>
>
>
>
>
>
>
>
7
C
C
6B >
B
im
im
im
im
>
>
>
>
>
>
>
>
>
>
>
>
>
>
7
C
C
6B >
B >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
7
C
C
6B >
B >
>
>
>
>
>
>
> ⋮ >
>
>
>
>
>
7
C
C
6B >
B >
>
>
>
>
>
>
>
⋮
⋮
>
>
>
>
>
>
7
C
C
6B >
B >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
5
A
A
4@ >
@ >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
:
:
:
;
;
;
;
pNm
qNm
pNm
pNm
9
8
r1m cos ðwm t q1m Þ >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
r
cos
ð
w
t
q
Þ
>
>
m
2m
2m
>
>
>
>
>
>
>
>
>
>
>
>
=
<
⋮
¼2emm t
>
>
>
r cos ðwm t qim Þ >
>
>
>
>
>
> im
>
>
>
>
>
>
>
>
>
>
⋮
>
>
>
>
>
>
>
>
>
>
;
:
rNm cos ðwm t qNm Þ
20
(12.10.16)
where
rim ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðcm pim dm qim Þ2 + ðcm qim + dm pim Þ2
qin ¼ tan 1
ðcm qim + dm pim Þ
cm pim dm qim
(12.10.17a)
(12.10.17b)
It becomes obvious from Eq. (12.10.17b) that the displacement components
in a vibration mode have different phase angles, contrary to the undamped
vibrations where the phase angles are the same within a vibration mode (see
Eq. 12.8.2).
Example 12.10.1 Determine the displacements of the frame in Example 12.2.1
in the presence of damping by direct solution of the equation of motion.
Assume:
C¼
30 5
, uð0Þ ¼ f0:03 0:01gT , u_ ð0Þ ¼ 0
5
40
Solution
The matrices M kai K were computed in Example 12.2.1. Hence, we have
M¼
25 0
,
0 32
K¼
3826:5 3826:5
3826:5 9142:1
(1)
574 PART
II Multi-degree-of-freedom systems
The characteristic equation results from Eq. (12.10.5)
det l2
25 0
30 5
3826:5 3826:5
+l
+
0 32
5
40
3826:5 9142:1
¼0
(2)
which after expanding gives
l4 + 2:45l3 + 440:219375l2 + 486:3225l + 25425:17925 ¼ 0
The roots of the polynomial (3) are
9
8 9 8
0:52764 + 8:27232i >
l1 >
>
>
>
>
>
>
=
< = <
0:52764 8:27232i
l2
¼
0:69736 + 19:22372i >
l > >
>
>
>
;
:
; >
: 3>
0:69736 19:22372i
l4
and Eq. (12.10.3) gives
b1 ¼
b2 ¼
b3 ¼
b4 ¼
+0:87585 0:69134 103 i
+0:48259 + 0:12547 102 i
+0:87585 + 0:69134 103 i
+0:48259 0:12547 102 i
+0:57638 + 0:30308 102 i
0:81719 + 0:21377 102 i
+0:57638 0:30308 102 i
0:81719 0:21377 102 i
(3)
(4)
(5)
The general solution is given by Eq. (12.10.7) for N ¼ 2
u¼
4
X
an bn eln t
(6)
n¼1
The previous equation for t ¼ 0 gives
uð0Þ ¼
4
X
an bn
(7)
an ln bn
(8)
n¼1
u_ ð0Þ ¼
4
X
n¼1
which are further written in matrix form
B
uð0Þ
a¼
BL
u_ ð0Þ
(9)
where B is the matrix of the vectors bn with dimensions 2 4, L the diagonal
matrix of ln with dimension 4 4, and a the vector of the coefficients an .
Multi-degree-of-freedom systems: Free vibrations Chapter
12
Eq. (9) for the data of the problem gives
9
8 9 8
a1 > >
+0:15232 101 0:99900 103 i >
>
>
>
=
= >
<
< >
a2
+0:15232 101 + 0:99900 103 i
¼
a > >
>
+0:28787 102 0:93751 104 i >
>
>
;
; >
:
: 3>
a4
+0:28787 102 + 0:93751 104 i
575
(10)
Finally, applying Eq. (12.10.15) for N ¼ 1, 2, and taking into account
Eq. (12.10.14) give
u1
+0:26681 101
0:52764t
¼e
cos 8:27232t
u2
+0:14704 101
+0:17710 102
+
sin 8:27232t
+0:92599 103
+0:33190 102
+e0:69736t
cos 19:22372t
0:47044 102
+0:90624 104
+
sin 19:22372t
0:16553 103
12.10.2 Linearization of the quadratic eigenvalue problem
As mentioned in the previous section, Eq. (12.10.3) represents the second-order
eigenvalue problem, commonly known as the quadratic eigenvalue problem.
The 2N values ln , which make the determinant SðlÞ equal to zero, are its eigenvalues. Moreover, the 2N vectors bn obtained from Eq. (12.10.3) are its eigenvectors. In the previous section, we presented a direct method for determining
these quantities. Another method is to transform the quadratic eigenvalue problem to a standard eigenvalue problem. This method is of great interest because it
allows methods applied to undamped vibrations to be applied also to damped
vibrations. It is based on the transformation of the equation of motion to two
first-order equations. To this end, we set
u_ ¼ x1 and u ¼ x2
(12.10.18)
x_ 2 ¼ x1
(12.10.19)
Mx_ 2 Mx1 ¼ 0
(12.10.20)
hence
or premultiplying by M
Moreover, Eq. (12.10.1) is written as
Mx_ 1 + Cx_ 2 + Kx2 ¼0
(12.10.21)
576 PART
II Multi-degree-of-freedom systems
Eqs. (12.10.20) and (12.10.21) are combined to
0 M x_ 1
M 0
x1
0
+
¼
x_ 2
M C
0 K x2
0
(12.10.22)
or
^ x_ + Kx
^ ¼0
M
where
^¼
M
0 M ^
M 0
, K¼
, x¼
M C
0 K
(12.10.23)
x1
x2
¼
u_
u
(12.10.24)
Eq. (12.10.23) accepts a solution of the form
x ¼ belt
which is substituted into Eq. (12.10.23) to give
^ lM
^ b¼0
K
(12.10.25)
(12.10.26)
namely, the quadratic eigenvalue problem of order N is transformed into a gen^ and K
^ are real
eralized linear eigenvalue problem of order 2N . The matrices M
and symmetric. Thus, according to the first property in Table 12.7.1, the eigenvalue problem in Eq. (12.10.26) has real eigenvalues but not positive because
these matrices are not positive definite. The transformation to a standard eigenvalue problem using the method presented in Section 12.5.3 produces a matrix
^ in Eq. (12.5.64) that is not real. Therefore, if the matrix M
^ is not singular, it is
A
convenient to transform Eq. (12.10.26) as
^ lI b ¼ 0
A
(12.10.27)
where
^ ¼M
^
^ 1 K
A
(12.10.28)
Eq. (12.10.27) gives 2N eigenvalues and the respective eigenvectors bn that
^ and K
^ (see Section 12.5.2,
satisfy the orthogonality condition with respect to M
property 2). The eigenvalues of the problems we encounter are complex with a
negative real part and express convergent vibrations.
^ we will have
If the eigenvectors are orthonormalized with respect to M,
^ n ¼1
fTn Mf
(12.10.29)
^ n ¼ ln
fTn Kf
(12.10.30)
The general solution of Eq. (12.10.23) is obtained as a superposition of the
solutions of Eq. (12.10.25), namely
x¼
2N
X
n¼1
an fn eln t
(12.10.31)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
577
where an are arbitrary constants determined from the following initial
conditions
u_ ð0Þ
xð0Þ ¼
(12.10.32)
uð0Þ
Eq. (12.10.31) for t ¼ 0 gives
2N
X
xð0Þ ¼
an fn
(12.10.33)
n¼1
^ and by using Eq. (12.10.29) becomes
which if premultiplied by fTn M
^ ð 0Þ
an ¼ fTn Mx
(12.10.34)
The coefficients an , the eigenvalues ln , and the eigenvectors fn are
complex conjugate. Hence we may set
an ¼ cn + id n ,
an ¼ cn id n
(12.10.35)
ln ¼ mn + iwn ,
n ¼ m iwn
l
n
(12.10.36)
fn ¼ pn + iqn ,
n ¼ pn iqn
f
(12.10.37)
Apparently, for each pair of eigensolutions, we will have
n
xm ¼ xn + x
¼ 2emn t ½ðcn pn dn qn Þcos wn t ðcn qn + dn pn Þ sin wn t (12.10.38)
and the general solution is obtained as
x¼
N
X
xm
(12.10.39)
m¼1
The solution of Eq. (12.10.23) may result by considering the transformation
of xðt Þ in terms of the modal coordinates Yðt Þ
x ¼ FY
(12.10.40)
If L represents the matrix of the eigenvalues and F the matrix of the
^ we have
eigenvectors orthonormalized with respect to M,
^ ¼I
FT MF
(12.10.41)
^ ¼L
FT KF
(12.10.42)
Substituting Eq. (12.10.40) into Eq. (12.10.23) gives
^ Y_ + KFY
^
MF
¼0
(12.10.43)
which is further premultiplied by F and gives by virtue of Eqs. (12.10.41) and
(12.10.42)
T
Y_ LY ¼ 0
(12.10.44)
578 PART
II Multi-degree-of-freedom systems
or
Y_ n ln Yn ¼ 0, n ¼ 1, 2, …,2N
(12.10.45)
that is, the system of the 2N coupled equations is transformed to 2N
SDOF equations with respect to the modal coordinates. Integration of
Eq. (12.10.45) gives
Yn ¼ an elt , n ¼ 1, 2, …,2N
(12.10.46)
Thus the general solution is given by Eq. (12.10.40). The vector
a ¼ fa1 , a2 , …, a2N gT of the arbitrary constants is obtained from the relation
^ ð 0Þ
a ¼ FT Mx
(12.10.47)
In summary, the solution of the vibration problem with damping presented
in this section can be obtained by adhering to the following steps:
^ and K
^ defined by Eq. (12.10.24).
1. Formulation of the matrices M
2. Computation of the eigenvalues and eigenvectors of the eigenvalue
problem, Eq. (12.10.26).
^
3. Orthonormalization of the eigenvectors with respect to M.
4. Computation of the arbitrary constants using Eq. (12.10.34).
5. Computation of the solution using Eq. (12.10.31).
6. Separation of the vectors u and u_
u_ ¼
N
X
n¼1
an fðn1Þ eln t ,
u¼
N
X
an fðn2Þ eln t
(12.10.48)
n¼1
where fðn1Þ and fðn2Þ are the upper and lower half of fn , respectively.
12.10.3 The use of a proportional viscous damping matrix
It becomes obvious from the previous example that the solution to the problem
of free vibrations with viscous damping through directly solving the quadratic
eigenvalue problem exhibits significant computational difficulties. On the other
hand, transforming it to a linear eigenvalue problem increases the number of
computations in addition to the complexity of handling complex numbers.
For these reasons, the previously discussed methods are rather theoretical
and, except in special cases, they do not represent the common praxis in the
dynamic analysis of structures. In practice, the modal damping assumption is
adopted, according to which the damping affects separately each eigenmode
by a specified damping coefficient. This is, of course, true when the modes
of the free vibration without damping are orthogonal with respect to the
damping matrix, that is,
Cn if n ¼ m
T
fn Cfm ¼
(12.10.49)
0 if n 6¼ m
Multi-degree-of-freedom systems: Free vibrations Chapter
12
579
Using Eq. (12.9.1), which represents the transformation from the physical
coordinates to modal coordinates of the undamped vibrations, we may write
Eq. (12.10.1) as
€ + CFY_ + KFY ¼0
MFY
(12.10.50)
which is premultiplied by F to yield
T
€ + FT CFY_ + FT KFY ¼0
FT MFY
(12.10.51)
Taking into account the orthogonality of the eigenmodes with respect to M,
K, and assuming that Eq. (12.10.49) holds, we obtain
Mn Y€ n + Cn Y_ n + Kn Yn ¼ 0, n ¼ 1, 2, …,N
(12.10.52)
namely, the uncoupling of equations of motion is achieved when damping is
present.
Further, setting
Cn ¼ 2x n Mn wn
and taking into account that
Kn ¼ Mn w2n
(12.10.53)
Eq. (12.10.52) becomes
Y€ n + 2x n wn Y_ n + w2n Yn ¼ 0, n ¼ 1, 2, …, N
(12.10.54)
whose solution is given by Eq. (2.3.18), namely
Yn ðt Þ ¼ exn wn t
Y_ n ð0Þ + Yn ð0Þxn wn
sin wDn t + Yn ð0Þcos wDn t
wDn
(12.10.55)
The quantity xn defined by Eq. (12.10.53) expresses the damping ratio of the
n-eigenmode. The initial conditions Yn ð0Þ, Y_ n ð0Þ are evaluated from the vectors uð0Þ and u_ ð0Þ using Eqs. (12.9.12) and (12.9.13). After determining
Yn ðt Þ,n ¼ 1, 2, …,N , the solution of Eq. (12.10.1) is obtained from
Eq. (12.9.1), namely
uðt Þ ¼
N
X
f n Yn ð t Þ
(12.10.56)
n¼1
This solution is referred to as the mode superposition method with modal
damping.
Eq. (12.10.55) by virtue of Eq. (2.3.19) is written as
Yn ðt Þ ¼ rn exn wn t cos ðwDn t qn Þ
where
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
Y_ n ð0Þ + Yn ð0Þx n wn
rn ¼
+ ½Yn ð0Þ2
wDn
qn ¼ tan 1
Y_ n ð0Þ + Yn ð0Þxn wn
wD Y ð 0Þ
(12.10.57)
(12.10.58a)
(12.10.58b)
580 PART
II Multi-degree-of-freedom systems
From the last equation, one concludes that the displacements in each mode
are in phase.
The above solution resulted by assuming that the transformation u ¼ FY
diagonalizes also the damping matrix, that is,
^
FT CF ¼ C
(12.10.59)
where
2
2x1 w1
6⋯
6
6⋯
^ ¼6
C
6
6⋯
6
4⋯
0
0
2x2 w2
⋯
⋯
⋯
0
⋯
⋯
⋯
⋯
⋯
⋯
0
0
⋯
2x n wn
⋯
0
⋯
⋯
⋯
⋯
⋯
⋯
3
0
7
0
7
7
⋯
7
7
7
⋯
7
5
⋯
2xN wN
(12.10.60)
The damping matrix satisfying Eq. (12.10.49) is called proportional.
The damping that satisfies the orthogonality condition of the eigenmodes of
the undamped free vibrations is referred to in the literature as classical damping
versus nonclassical damping where this condition is not satisfied. Methods of
constructing proportional damping matrices will be presented in the next
section.
Example 12.10.2 Determine the displacements of the frame in Example 12.2.1
in the presence of damping by the method of linearization of the quadratic
eigenvalue problem. Assume:
C¼
30 5
, uð0Þ ¼ f0:03 0:01gT , u_ ð0Þ ¼ 0
5
40
Solution
The matrices M and K were computed in Example 12.2.1. Thus we have
M¼
25 0
,
0 32
Eq. (12.10.24) gives
K¼
3826:5 3826:5
3826:5 9142:1
3
0
0 25 0
6 0
0 M
0
0 32 7
^¼
7
M
¼6
4 25 0 30 5 5
M C
0 32 5 40
2
3
25 0
0
0
6
7
0
^ ¼ M 0 ¼ 6 0 32 0
7
K
4
0 K
0
0
3826:5 3826:5 5
0
0 3826:5 9142:1
(1)
2
(2)
(3)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
581
The characteristic equation is
31
2
3
0
0 25 0
25 0
0
0
C
6 0
7
B6
0
0 32 7
0
7C
6
7
6 0 32 0
^ lM
^ ¼B
det K
B6
7C ¼ 0
7 l6
@4 0
4 25 0 30 5 5A
0
3826:5 3826:5 5
0 32 5 40
0
0 3826:5 9142:1
02
(4)
which after expanding gives
l4 + 2:45l3 + 440:219375l2 + 486:3225l + 25425:17925 ¼ 0
(5)
We observe that the characteristic polynomial (5) is identical to that in
Example 12.10.1. Hence, it gives the same eigenvalues c
9
8 9 8
0:52764 + 8:27232i >
l1 >
>
>
>
>
>
>
>
=
< 0:52764 8:27232i >
= >
<l >
2
(6)
¼
>
>
0:69736 + 19:22372i >
l3 >
>
>
>
>
>
>
> :
> ;
;
:
0:69736 19:22372i
l4
The eigenvectors are computed from Eq. (12.10.26) and are then normalized
^ Thus
with respect to mass M.
For l1 ¼ 0:52764 + 8:27232i we obtain
9
8
+0:259428 0:228755i >
>
>
>
>
=
< +0:143371 0:125558i >
(7)
f1 ¼
>
0:029533 0:029477i >
>
>
>
>
;
:
0:016218 0:016297i
For l2 ¼ 0:52764 8:27232i we obtain
9
8
+0:259428 + 0:228755i >
>
>
>
>
=
< +0:143371 + 0:125558i >
f2 ¼
>
0:029533 + 0:029477i >
>
>
>
>
;
:
0:016218 + 0:016297i
For l3 ¼ 0:69736 + 19:22372i we obtain
9
8
+0:241638 0:222178i >
>
>
>
>
>
<
0:340096 + 0:317685i =
f3 ¼
>
0:011998 0:012135i >
>
>
>
>
;
:
+0:017145 + 0:017070
(8)
(9)
^ K
^ are not positive definite.
c. The matrices M, C,K are positive definite. However, the matrices M,
This justifies that the eigenvalue problem has complex eigenvalues and complex eigenvectors.
582 PART
II Multi-degree-of-freedom systems
For l4 ¼ 0:69736 19:22372i we obtain
9
8
+0:241638 + 0:222178i >
>
>
>
>
=
< 0:340096 0:317685i >
f4 ¼
>
0:011998 + 0:012135i >
>
>
>
>
;
:
+0:017145 0:017070
(10)
The vector xð0Þ representing the initial conditions is computed from
Eq. (12.10.32)
9
8
0 >
>
>
>
=
<
u_ ð0Þ
0
(11)
x ð 0Þ ¼
¼
uð0Þ
0:03 >
>
>
>
;
:
0:01
and Eq. (12.10.47) gives the coefficients an
9
8
0:211292 + 0:240875i >
>
>
>
>
=
< 0:211292 0:240875i >
T ^
a ¼ F Mxð0Þ ¼
>
0:066486 + 0:071021i >
>
>
>
>
;
:
0:066486 0:071021i
(12)
Finally, Eqs. (12.10.48) give
8 9
<u =
1
:u ;
¼
2N
X
ð2Þ
an fn eln t
n¼1
2
9
08
< +0:26681 101 =
¼e0:52764t @
8
9
< +0:17710 102 =
1
cos 8:27232t +
sin 8:27232t A
: +0:14704 101 ;
: +0:92599 103 ;
9
8
9
08
1
< +0:33190 102 =
< +0:90624 104 =
0:69736t
@
+e
cos 19:22372t +
sin 19:22372t A
: 0:47044 102 ;
: 0:16553 103 ;
8 9
< u_ =
1
: u_ ;
2
¼
2N
X
ð1Þ
an f n e ln t
n¼1
8
9
9
08
1
< 0:22165 =
< +0:57263 103 =
sin 8:27232t A
cos 8:27232t +
¼e0:52764t @
: 0:12213 ;
: 0:98528 104 ;
9
8
9
08
1
< 0:57241 103 =
< 0:63866 101 =
0:69736t
@
+e
cos 19:22372t +
sin 19:22372t A
: +0:98450 104 ;
: +0:90551 101 ;
Note that the obtained solution by this method is identical to that obtained in
Example 12.10.1.
Example 12.10.3 Determine the displacements of the frame in Example 12.2.1
in the presence of classical damping. Assume:
Multi-degree-of-freedom systems: Free vibrations Chapter
C¼
12
583
30 5
, uð0Þ ¼ f0:03 0:01gT , u_ ð0Þ ¼ 0
5
40
Solution
The eigenfrequencies and mode shapes were computed in Example 12.2.1,
namely
w1
8:289
0:1697 0:1061
¼
, F¼
(1)
w2
19:236
0:0935 0:1499
The transformation (12.10.49) gives
FT CF ¼
¼
0:1697
0:1061
T
0:0935 0:1499
30 5
5
40
0:1697
0:1061
0:0935 0:1499
1:05496 0:05712
0:05712 1:39556
(2)
The orthogonality assumption with respect to the damping matrix requires
that the off-diagonal terms are omitted. This is permitted here because these
terms are small compared to the diagonal ones. Thus, taking into account that
M1 ¼ M2 ¼ 1, we can set
^ 1 ¼ 2x w1 ¼ 1:05496
C
1
(3a)
^ 2 ¼ 2x w2 ¼ 1:39556
C
2
(3b)
which yield
x1 ¼ 0:0636,
x2 ¼ 0:0363,
qffiffiffiffiffiffiffiffiffiffiffiffi
wD1 ¼ w1 1 x21 ¼ 8:272
qffiffiffiffiffiffiffiffiffiffiffiffi
wD2 ¼ w2 1 x22 ¼ 19:223
(4a)
(4b)
Eqs. (12.9.16) and (12.9.17) give
Y1 ð0Þ ¼ fT1 Muð0Þ ¼ 0:157195
(5a)
Y2 ð0Þ ¼ fT2 Muð0Þ ¼ 0:031607
(5b)
Y_ 1 ð0Þ ¼ Y_ 2 ð0Þ ¼ 0
(6)
and Eq. (12.10.55) gives
Y1 ðt Þ ¼ e0:52748t ð0:010018 sin 8:272t + 0:157195 cos 8:272t Þ
(7a)
Y2 ðt Þ ¼ e0:69778t ð0:001148 sin 19:223t + 0:031607 cos 19:223t Þ
(7b)
584 PART
II Multi-degree-of-freedom systems
Finally, we obtain from Eq. (12.9.1)
(
u1
u2
)
¼
2
X
f n Yn
n¼1
¼e
0:52748t
(
0:02667
)
(
0:1697 102
)
!
sin 8:272t
cos 8:272t +
0:01469
0:9366 103
(
)
(
)
!
0:3353 102
1:218 104
0:69778t
+e
cos 19:223t +
sin 19:223t
0:4737 102
0:172 103
(8)
Comparing the above solution with that obtained in Example 12.10.1, we
conclude that omitting the diagonal terms in Eq. (2) negligibly influences the
solution.
12.11 Construction of a proportional damping matrix
12.11.1 Rayleigh damping
When the solution is obtained by the method of superposition of eigenmodes
and the modal damping ratios are known (see Eq. 12.10.54), it is obvious that
the knowledge of a proportional damping matrix is not required. However,
when other methods for solving the equation of motion are employed, for example, numerical integration methods, the knowledge of the damping matrix is
necessary. Thus, the problem of constructing the damping matrix arises when
all or some of the modal damping ratios are specified. This problem will be the
subject of this section.
It is obvious that the damping matrix satisfies Eq. (12.10.49) if it is proportional to the mass or stiffness of the structure.
For a mass-proportional damping matrix it is
C ¼ a0 M
(12.11.1)
where a0 is a constant.
Premultiplying the foregoing relation by fTn and postmultiplying it by fm
yield
a0 Mn if n ¼ m
(12.11.2)
fTn Cfm ¼ a0 fTn Mfm ¼
0
if n 6¼ m
Hence
Cn ¼ a0 Mn
(12.11.3)
a0 ¼ 2x n wn
(12.11.4)
or by virtue of Eq. (12.10.53)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
585
The foregoing relation allows the determination of a0 if xn is specified. After
that, the damping ratios xk of the remaining eigenmodes (k 6¼ n) can be established from the relation
a0
xk ¼
2wk , k 6¼ n
(12.11.5)
wn
¼ xn
wk
For a stiffness-proportional damping matrix, we may set
C ¼ a1 K
Following the previous procedure we obtain
a1 w2n Mn if n ¼ m
fTn Cfm ¼ a1 fTn Kfm ¼
if n 6¼ m
0
(12.11.6)
(12.11.7)
Hence
a1 w2n Mn ¼ Cn
¼ 2xn Mn wn
(12.11.8)
which yields
a1 ¼
2xn
wn
(12.11.9)
The foregoing relation allows the determination of a1 if xn is specified.
After that, the damping ratios x k of the remaining eigenmodes (k 6¼ n) can be
established from the relation
a 1 wk
xk ¼
2 , k 6¼ n
(12.11.10)
wk
¼ xn
wn
Obviously, the linear combination
C ¼ a0 M + a1 K
(12.11.11)
satisfies also the orthogonality condition (12.10.49). Inasmuch as this relation
involves two arbitrary constants, a0 and a1 , it is possible to establish matrix C
so that two of its eigenmodes have specified damping ratios. Indeed,
Eq. (12.11.11) gives
fTn Cfn ¼ 2x n Mn wn ¼ a0 Mn + a1 Mn w2n
(12.11.12a)
fTm Cfm ¼ 2x m Mm wm ¼ a0 Mm + a1 Mm w2m
(12.11.12b)
which provide the linear system of equations
586 PART
II Multi-degree-of-freedom systems
2 1
3
w
n
1 6 wn
xn
7 a0
¼
4 1
5
a1
xm
2
wm
wm
(12.11.13)
from which the coefficients a0 and a1 are obtained
a0 ¼
2wn wm ðwn xm wm xn Þ
w2n w2m
(12.11.14a)
2 ð wn x n wm x m Þ
w2n w2m
(12.11.14b)
a1 ¼
If xn ¼ xm ¼ x, which is a reasonable assumption, the foregoing relations
yield
a0 ¼
2xwn wm
wn + wm
(12.11.15a)
a1 ¼
2x
wn + wm
(12.11.15b)
Once the coefficients a0 and a1 are established, the damping ratios of the
remaining eigenmodes (k 6¼ m, n) can be evaluated from the relation
xk ¼
1
1
a0 + a1 wk , k 6¼ m,n
2
wk
xk = xk (wk )
xk = a0 /2wk
xk = (a0 /wk + a1wk)/2
xk = a1wk /2
wk
FIG. 12.11.1 Damping ratio versus frequency for Rayleigh damping.
(12.11.16)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
587
The proportional damping determined by Eq. (12.11.11) is known as Rayleigh damping. Fig. 12.11.1 shows the variation of xk ðwk Þ, as obtained from
Eqs. (12.11.5), (12.11.10) and (12.11.16).
Example 12.11.1 Construct a proportional damping matrix of the frame in
Example 12.2.1, so that x1 ¼ 0:08 and x2 ¼ 0:06.
Solution
The mass and stiffness matrices as well as the natural frequencies were
computed in Example 12.2.1, namely
25 0
3826:5 3826:5
w1
8:289
M¼
, K¼
,
¼
(1)
w2
0 32
3826:5 9142:1
19:236
Using these data, Eq. (12.12.13) becomes
2 1
3
8:289
0:08
1 6 8:289
7 a0
¼
4
5
1
2
0:06
a1
19:236
19:236
(2)
which yields a0 ¼ 1:1023 and a1 ¼ 0:003259. Then using Eq. (12.12.11) we
obtain
25 0
3826:5 3826:5
+ 0:003259 0 32
3826:5 9142:1
40:029 12:4717
¼
12:4717 65:0705
C ¼ 1:1023 (3)
12.11.2 Additional orthogonality conditions: Caughey
damping matrix
Rayleigh damping enables us to construct a proportional damping matrix by
specifying the damping ratio of only two eigenmodes. However, if we are interested in constructing a proportional damping matrix by specifying the damping
ratio of more than two eigenmodes, we must use matrices beyond the mass and
stiffness matrices that satisfy the orthogonality condition. This is achieved by
formulating orthogonality conditions more than those defined by Eqs. (12.3.10)
and (12.3.11). These conditions are called additional orthogonality conditions
and can be derived as follows [4–6].
We assume that the eigenmodes fn have been orthonormalized with respect
to the mass, hence Mn ¼ 1. Starting from the relation
Kfn ¼ ln Mfn
(12.11.17)
where ln ¼ w2n , and assuming that M is not singular, the multiplication of both
sides of the foregoing equation by fTm KM1 yields
588 PART
II Multi-degree-of-freedom systems
fTm KM1 Kfn ¼ ln fTm Kfn
0 n 6¼ m
¼
l2n n ¼ m
(12.11.18)
Further, multiplying Eq. (12.11.17) by fTm KM1 KM1 gives
fTm KM1 KM1 Kfn ¼ ln fTm KM1 Kfn
which by virtue of Eq. (12.11.18) yields
2
0 n 6¼ m
fTm KM1 Kfn ¼
l3n n ¼ m
(12.11.19)
Repeating the previous procedure results in the orthogonality relations
p
0
n 6¼ m p ¼ 1, 2, …1
fTm KM1 Kfn ¼
(12.11.20)
lpn + 1 n ¼ m p ¼ 1, 2, …1
Observing that MM1 ¼ I, we obtain
p
p
KM1 K ¼ MM1 KM1 K
¼ MM1 KM1 KM1 ⋯ KM1 K
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
p
1
¼ M M K M1 K ⋯ M1 K
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
1
¼M M K
and Eq. (12.11.20) becomes
fTm M
1
M K
p + 1
p + 1
fn ¼
p+1
(12.11.21)
0
n 6¼ m p ¼ 1, 2, …1
lpn + 1 n ¼ m p ¼ 1, 2, …1
(12.11.22)
Because the above relation is valid for p ¼ 1 we can write it as
1 q
0 n 6¼ m q ¼ 0, 1, 2, …1
T
fm M M K fn ¼
(12.11.23)
lqn n ¼ m q ¼ 0, 1, 2, …1
If matrix K is not singular, it holds ln 6¼ 0, n ¼ 1, 2, …,N and Eq.
(12.11.17) is written as
K1 Mfn ¼ l1
n fn
(12.11.24)
or
1 1
M K fn ¼ l1
n fn
Multiplication of both sides of the foregoing equation by
1
T
fTm M M1 K fn ¼ l1
n fm Mfn
0
n 6¼ m
¼
1
ln n ¼ m
(12.11.25)
fTm M
yields
(12.11.26)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
589
1
Further, multiplying Eq. (12.11.25) by fTm M M1 K
yields
2
1
T
1
fn
(12.11.27)
fTm M M1 K fn ¼ l1
n fm M M K
which by virtue of Eq. (12.11.26) gives
2
fTm M M1 K fn ¼
0
n 6¼ m
l2
n¼m
n
(12.11.28)
Repeating the previous procedure results in the orthogonality relations
1 q
0
n 6¼ m q ¼ 1, 2, …1
T
(12.11.29)
fm M M K fn ¼
lq
n ¼ m q ¼ 1, 2, …1
n
Eqs. (12.11.22) and (12.11.29) are combined to
q
0 n 6¼ m q ¼ 0, 1, 2, … 1
fTm M M1 K fn ¼
lqn n ¼ m q ¼ 0, 1, 2, … 1
Therefore, the matrices defined by the relation
q
Cq ¼ M M1 K , q ¼ 0, 1, 2, … 1
(12.11.30)
(12.11.31)
satisfy the orthogonality condition. Evidently, any linear combination of k N
terms
C¼
k
X
aq Cq ¼
X
q
q
aq M M1 K
(12.11.32)
q
can be employed as a proportional damping matrix with k specified modal
damping ratios. Eq. (12.11.32) is known as the Caughey series and the damping
matrix C obtained from this equation is known as the Caughey damping
matrix [7]. Obviously, for k ¼ 2 and q ¼ 0,1 the Rayleigh damping matrix is
obtained. The summation index k may be arbitrary. Let us assume that
q ¼ 0, 1, 2, …, k 1. Then
C¼
k1
X
q
aq M M1 K
(12.11.33)
q¼0
The coefficients aq are evaluated from the relation
fTn Cfn ¼
k 1
X
q
aq fTn M M1 K fn
(12.11.34)
q¼0
which by virtue of Eqs. (12.10.53) and (12.11.30) becomes
2xn wn ¼
k 1
X
q¼0
aq lqn
(12.11.35a)
590 PART
II Multi-degree-of-freedom systems
or because ln ¼ w2n
xn ¼
1
1
a0 + a1 wn + a2 w3n + ⋯ + ak1 w2k3
n
2
wn
Applying the
linear equations
2
1
6 w1
6
61
16
6
w
26
6 ⋯2
6
41
wk
(12.11.35b)
foregoing equation for n ¼ 1, 2,…,k provides the system of
3
w1 w31 ⋯ w2k3
1
9 8 9
78
7 > a0 > > x 1 >
> > >
7>
7 < a1 = < x 2 =
w2 w32 ⋯ w2k3
7
2
7> ⋮ > ¼ > ⋮ >
> > >
>
⋯ ⋯ ⋯ ⋯ 7
7: ak1 ; : xk ;
5
wk w3k ⋯ w2k3
k
(12.11.36)
which yields the coefficients ak when xk is specified.
12.11.3 Construction of the proportional damping matrix
using the modal matrix
An alternative way of constructing a proportional damping matrix follows
directly from the orthogonality relation (12.10.59) if it is premultiplied by
FT and postmultiplied by F1 . Thus, we obtain
^ 1
C ¼ FT CF
(12.11.37)
The foregoing formulation requires the inversion of the modal matrix. This
can be achieved using the relation
FT MF ¼I
(12.11.38)
FT ¼MF
(12.11.39a)
F1 ¼FT M
(12.11.39b)
which gives
and
The previous equations are substituted into Eq. (12.11.37) to yield
^ TM
C ¼ MFCF
We can further write
2
2x 1 w1 ⋯ 0
6
6⋯
6
6
^ T ¼ f1 ⋯ fn ⋯ fN 6 ⋯
FCF
6
6
6⋯
4
0
¼
N
X
n¼1
2x n wn fn fT
n
⋯ ⋯
(12.11.40)
⋯ 0
⋯ ⋯
⋯ 2x n wn ⋯ ⋯
⋯ ⋯
⋯ ⋯
⋯ 0
⋯ 2x N wN
3
7
7
7h
i
7 T
7 f ⋯ fT ⋯ fT
n
7 1
N
7
7
5
Multi-degree-of-freedom systems: Free vibrations Chapter
and Eq. (12.11.40) becomes
C¼M
N
X
12
591
!
2xn wn fn fTn
M
(12.11.41)
n¼1
Eq. (12.11.41) shows clearly the contribution of each eigenmode to the proportional damping matrix.
When using a proportional damping constructed by the previous methods,
we should take into account the following:
1. The damping matrix C obtained from Eq. (12.11.41) is in general complete.
The same holds for the damping matrix obtained using the additional
orthogonality conditions for k > 2. Apparently, the use of a complete damping matrix increases the computational cost.
2. The linear system of Eq. (12.11.36) exhibit ill conditioning because the
coefficients 1=wn ,wn ,w3n ,⋯,w2k3
may differ significantly. As a remedy,
n
it is recommended that the values of series exponents in Eq. (12.11.33)
are taken near zero. For example, it is advisable to take q ¼ 2, 1,
0, 1, 2, when k ¼ 5.
3. In Rayleigh or Caughey damping, it is possible to obtain negative values for
x n . These values should be excluded; otherwise, the response of the structure will be divergent.
4. It is inappropriate when the system consists of parts with significantly different damping, for example, in the case of soil-structure interaction where
soil damping is much greater than that of the structure. The same is true
when energy-absorbing dampers are incorporated into the structure or
base isolation systems are used. The resulting damping matrix does not
satisfy the orthogonality condition; therefore, the method presented in
Section 12.10.2 should be employed. Another method to cope with this
problem is to use the method of substructures with a different proportional
damping matrix for each of them.
5. The use of Rayleigh damping, which offers a reasonable solution to the
damping problem, yields increased values of the damping ratios for
higher-order eigenmodes.
Example 12.11.2 The computed mass and stiffness matrices of a single-story
building are
2
3
2
3
52 0
0
1472 407 5553
M ¼ 4 0 52
0 5, K ¼ 4 407 1001 4154 5
0 0 441
5553 4154 43516
Determine a proportional damping matrix for the structure using (i) additional orthogonality conditions, (ii) the alternative method based on the modal
matrix, and (iii) Rayleigh damping. Assume: x 1 ¼ 0:06, x2 ¼ 0:08, x 3 ¼ 0:10
592 PART
II Multi-degree-of-freedom systems
Solution
To obtain reliable results, the computations were performed using arithmetic
with 10 significant figures.
The solution of the eigenvalue problem ðK w2 MÞf ¼ 0 gives
8 9 8
9
2
3
0:07127 0:10731 0:05133
< w1 = < 2:9290 =
¼
3:9429 , F ¼ 4 0:10257 +0:08574 0:03682 5
(1)
w
: 2; :
;
11:0502
+0:02068 +0:00653 0:04239
w3
The eigenmodes are orthonormalized with respect to the mass.
(i) Additional orthogonality conditions
The damping matrix will be computed using Eq. (12.11.32) for
q ¼ 0, 1, 2
2
C ¼ a0 M + a1 K + a2 M M1 K
(2)
¼ a0 M + a1 K + a2 KM1 K
The coefficients a0 ,a1 ,a2
2
1=w1 w1
14
1=w2 w2
2
1=w3 w3
result from Eq. (12.11.36) for N ¼ 3
9
38 9 8
w31 < a0 = < 0:06 =
w32 5 a1 ¼ 0:08
: ; :
;
0:10
a2
w33
(3)
Substituting the values of the eigenfrequencies from Eq. (1) and solving
the system give
a0 ¼ 0:22227 101 , a1 ¼ 0:45470 101 , a2 ¼ 0:22666 103
and Eq. (2) gives
3
40:219 2:549 88:247
C ¼ 4 2:549 30:647 70:130 5
88:247 70:130 806:837
(4)
2
(ii) The alternative method based on the modal matrix
^ using Eq. (12.10.60) for
We compute the matrix C
2
3 2
2x1 w1 0
0
0:351 0
^ ¼ 40
5 ¼ 40
C
2x 2 w2 0
0:631
0
0
0
0
2x 3 w3
(5)
N ¼3
3
0
5
0
2:210
which is substituted into Eq. (12.11.40) to give
2
3
40:219 2:549 88:247
^ T M ¼ 4 2:549 30:647 70:130 5
C ¼ MFCF
88:247 70:130 806:837
(6)
(7)
Multi-degree-of-freedom systems: Free vibrations Chapter
12
593
(iii) Rayleigh damping
The damping matrix will be computed from Eq. (12.11.11) while the
coefficients a0 ,a1 are obtained from the solution of the system
2
3
1
w
n 7
a
xn
16
6 wn
7 0 ¼
(8)
5 a
24 1
x
1
m
wm
wm
and the third eigenfrequency from Eq. (12.11.16), that is,
xk ¼
1
1
a0 + a1 wk , k 6¼ m,n
2
wk
(9)
There are three possibilities for computing the damping matrix.
(a) n ¼ 1, m ¼ 2. The system (8) is written
a0
0:06
1 1=w1 w1
¼
2 1=w2 w2
0:08
a1
(10)
which gives
a0 ¼ 0:74714 102 , a1 ¼ 0:40098 101
(11)
Then substituting these values in Eq. (12.11.11) yields the damping
matrix
2
3
59:41 16:32 222:66
6
7
C ¼ a0 M + a1 K ¼ 4 16:32 40:52 166:56 5
222:66 166:56 1748:22
and Eq. (9) gives x3 ¼ 0:221.
(b) n ¼ 1, m ¼ 3. The system (8) is written
a0
0:06
1 1=w1 w1
¼
2 1=w3 w3
0:10
a1
(12)
which gives
a0 ¼ 0:21103, a1 ¼ 0:16371 101
Then Eq. (12.11.11) gives the damping matrix
2
3
35:07 6:66 90:90
6
7
C ¼ a0 M + a1 K ¼ 4 6:66 27:36 68:00 5
90:90 68:00 805:46
and Eq. (9) gives x 2 ¼ 0:059
(13)
(14)
594 PART
II Multi-degree-of-freedom systems
(c) n ¼ 2, m ¼ 3. The system (8) is written
1 1=w2 w2
a0
0:08
¼
a1
0:10
2 1=w3 w3
(15)
which gives
a0 ¼ 0:40047, a1 ¼ 0:14819 101
Then Eq. (12.11.11) gives the damping matrix
2
3
42:63 6:03 82:29
C ¼ a0 M + a1 K ¼ 4 6:03 35:65 61:56 5
82:29 61:56 821:49
(16)
(17)
and Eq. (9) gives x 1 ¼ 0:090.
We observe that the damping ratio computed from Eq. (9) deviates significantly from the corresponding predetermined. One technique to restrict the
deviation is by best fitting a straight line x ¼ a
w + b through the pairs ðwi , xi Þ,
i ¼ 1, 2, …, k using the least square method and then taking the values xi .
Example 12.11.3 Determine the eigenfrequencies and eigenmodes of the truss
in Fig. E12.5 and construct a proportional damping matrix based on the modal
matrix. The truss is loaded by the static (gravity) load W ¼ 400 kN. Assumed
data: Nodal coordinates: 1ð0, 0Þ, 2ð3, 3:5Þ, 3ð6, 0Þ; Cross-sectional areas of
the bars: A1 ¼ 1:5A, A2 ¼ A3 ¼ A, and A ¼ 27 cm2 ; modulus of elasticity
E ¼ 2:1 108 kN=m2 ; material density: r ¼ 7:55 kNm1 s2 =m3 ; lumped
mass assumption; and modal damping ratios x1 ¼ 0:1, x2 ¼ 0:08, x3 ¼ x4 ¼
… ¼ xN ¼ 0:06.
FIG. E12.5 Plane truss in Example 12.12.3.
Solution
1. Global mass and stiffness matrices
The geometrical data of the truss and the element properties were computed in Example 11.2.2 except for the mass of the gravity load W , which
constitutes an additional nodal mass in the directions 3, 4 of the global mass
matrix. Therefore, we have m33 ¼ m44 ¼ W =g ¼ 40:775. Hence
Multi-degree-of-freedom systems: Free vibrations Chapter
2
W
M
0 0 0
0
0
0
0
0 0 0
0
0 0
12
595
3
7
0 07
7
7
0 2000:0 0
0 07
7
7
0 0
2000:0 0 0 7
7
7
0 0
0
0 07
5
6
60
6
6
60
6
¼ rA6
60
6
6
60
4
0 0
and
2
¼ MW
M
6:805
60
6
6
3
X
60
^ e ¼ rA6
+
M
60
6
e¼1
6
40
0
0
0
6:805 0
0
0
0
0
0
0
0
0
2004:610 0
0
0
2004:610 0
0
0
0
0
0
6:805 0
0
0
0
0
3
7
7
7
7
7
7
7
7
5
6:805
3
0:342 0:107 0:092 0:107 0:25
0
6 0:107 0:125 0:107 0:125 0
7
0
6
7
6
7
3
X
6
7
0:092
0:107
0:184
0
0:092
0:107
e
^
6
7
K¼
K ¼ EA6
7
0:107
0:125
0
0:25
0:107
0:125
6
7
e¼1
6
7
4 0:25
0
0:092
0:107 0:342 0:107 5
2
0
0
0:107 0:125 0:107
0:125
M
due to support conditions
2. Modification of the matrices K,
The truss is supported as in Example 11.2.2. Thus the matrix V that
M
is the same. This yields
reorders the matrices K,
596 PART
II Multi-degree-of-freedom systems
3. Eigenfrequencies and eigenmodes.
The eigenfrequencies and eigenmodes result from the solution of the
eigenvalue problem
e ff lM
e ff f ¼0
K
(1)
where l ¼ w2 r=E. Taking into account the above-indicated partitioning
we have
2
3
10 4:610 0
0
7
e ff ¼ rA6
M
104:610 0
40
5,
0
0
6:805
2
0:184 0
e ff ¼ EA6
K
4 0
0:25
0:092 0:107
0:092
3
7
0:107 5
0:342
The solution of the eigenvalue problem (1) gives
9
8 9 8
2
3
0:0206 0:0086 0:0004
>
=
< 45:20 >
= >
< w1 >
6
7
56:30 , F ¼ 4 0:0086 0:0206 0:0004 5
w2 ¼
>
>
>
>
;
: ; :
0:0082 0:0041 0:3832
1182:66
w3
Fig. E12.6 shows the eigenmodes
FIG. E12.6 Eigenmodes of the truss in Example 12.12.3.
Multi-degree-of-freedom systems: Free vibrations Chapter
12
597
4. Construction of the proportional damping matrix.
^ is
The matrix C
2
3
0:1 45:20 0
0
^ ¼ 24 0
5
C
0:8 56:30 0
0
0
0:06 1182:66
2
3
9:0392 0
0
5
¼ 40
9:0084 0
0
0
141:9195
and Eq. (11.12.40) gives
3
181:77 0:98 2:43
^
C ¼ MFCF
M ¼ 10 4 0:98
81:56 2:83 5
2:43
2:83
9:65
T
2
2
Note that, as was anticipated,
2
9:0392
FT CFT ¼ 4 0
0
it is
3
0
0
^
5¼C
9:0084 0
0
141:9195
12.12 Problems
Problem P12.1 Compute the eigenfrequencies and eigenmodes of the truss
shown in Fig. P12.1. W1 ¼ 400 kN and W2 ¼ 250 kN are gravity dead loads.
Assumed data: Coordinates of the nodes:1ð0, 0Þ, 2ð4, 0Þ, 3ð0, 4Þ,4ð4, 4Þ,
5ð8, 4Þ; cross-sectional area of bars:A1 ¼ A2 ¼ A3 ¼ 1:5A, A4 ¼ A5 ¼ A7 ¼ A,
A6 ¼ 2A, and A ¼ 27 cm2 ; modulus of elasticity E ¼ 2:1 108 kN=m2 ; and
material density r ¼ 7:55 kNs2 =m4 . Consistent mass assumption. The rigid plane
body B has dimensions 4 4 m2 , thickness h ¼ 0:10 m, and specific weight
24 kN=m3 .
FIG. P12.1 Plane truss in problem P12.1.
598 PART
II Multi-degree-of-freedom systems
Problem P12.2 A single-story building whose plan form is an equilateral triangle of side a is supported by three columns of a rectangular cross as shown in
Fig. P12.2. The floor plate is assumed rigid and the columns fixed at both ends.
The mass of the columns is neglected. Construct a proportional damping matrix
using (i) Rayleigh damping, (ii) three additional orthogonality conditions, and
(iii) the modal matrix method. Assume: a ¼ 10 m, height of columns h ¼ 5 m;
cross-sectional area of columns 30 60 cm2 ; modulus of elasticity
E ¼ 2:1 107 kN=m2 and Poisson’s ratio n ¼ 0:2; and total load of the plate
(including dead weight) p ¼ 12 kN=m2 . Damping ratios: x1 ¼ 0:10,
x2 ¼ x3 ¼ 0:08.
FIG. P12.2 Plan form of the single-story building in problem P12.2.
Problem P12.3 Formulate the equation of motion of the two-story shear building of Fig P12.3. Then (i) compute the eigenfrequencies and eigenmodes, and
(ii) construct a proportional damping matrix of the form C ¼ a0 M + a1 K with
x1 ¼ 1:2, x2 ¼ 0:1 and formulate the equation of motion with damping. Assumed
data: E ¼ 2:1 107 kN=m2 and g ¼ 25 kN=m3 . All columns have a square
cross-section 0:30 0:30m2 . The load of the slabs also includes their dead load.
FIG. P12.3 Two-story building in problem P12.3.
Multi-degree-of-freedom systems: Free vibrations Chapter
12
599
Problem P12.4 The industrial chimney of length L ¼ 75 m shown in Fig. P12.4
consists of the outer reinforced concrete shell, which supports the linings.
The thickness of the thermal insulation layer is ti ¼ 0:10 m and that of the
refractory layer tr ¼ 0:10 m. The chimney is fixed on the ground. The structure
is modeled by three constant elements. Adopting lumped mass assumption:
(i) formulate the equation of motion and compute the eigenfrequencies and
eigenmodes, and (ii) construct a proportional damping if x 1 ¼ 0:15, x 2 ¼ 0:1
and xk ¼ 0:8 for k > 2 using all discussed methods. Assume the following data:
Specific weight of reinforced concrete g b ¼ 24 kN=m3
Specific weight of thermal insulation g i ¼ 0:7 kN=m3
Specific weight of refractory bricks g r ¼ 20 kN=m3
Modulus of elasticity of reinforced concrete E ¼ 2:1 107 kN=m2
FIG. P12.4 Industrial chimney in problem P12.4.
Problem P12.5 Formulate the Rayleigh proportional damping matrix C for the
three-degree-of-freedom system with modal damping ratios x1 ¼ 1:5, x2 ¼ 0:10,
and x3 ¼ 0:08. Consider all different cases and compare them with that obtained
using the modal matrix. Data
2
3
2
3
1000 0
5000
120000 40000 0
6
7
6
7
M ¼ 40
500 4000
5, K ¼ 4 40000 40000 0
5,
5000 4000 77833:33
8
9
< 1:120 =
w ¼ 11:553
:
;
13:541
0
0
100000
600 PART
II Multi-degree-of-freedom systems
References and further reading
[1] G. Strang, Linear Algebra and Its Applications, fourth ed., Cengage Learning, United States,
2005.
[2] S. Lipschutz, M.L. Lipson, Linear Algebra, fourth ed., Schaum’s Outline SeriesMcGraw-Hill
Companies, Inc., New York, 2009
[3] E.L. Ince, Ordinary Differential Equations, Dover Publications, New York, 1956.
[4] J.L. Humar, Dynamics of Structures, second ed., A.A. Balkema Publishers, Lisse, NL, 2002.
[5] R.W. Clough, J. Penzien, Dynamics of Structures, second ed., McGraw-Hill, New York, 1993.
[6] K.-J. Bathe, E.L. Wilson, Numerical Methods in Finite Element Analysis, Prentice-Hall Inc.,
Englewood Cliffs, NJ, 1976
[7] T.K. Caughey, Classical normal modes in damped linear dynamic systems, J. Appl. Mech.
27 (1960) 269–271. ASME.
Chapter 13
Numerical evaluation of the
eigenfrequencies and
eigenmodes
Chapter outline
13.1 Introduction
601
13.2 The vector iteration method 603
13.2.1 The inverse vector
iteration method
604
13.2.2 Convergence of the
inverse vector iteration
method
606
13.3 Computation of higher-order
eigenpairs
609
13.3.1 The vector purification
method
13.3.2 The inverse vector
iteration method with
shifts
13.4 Free or partially supported
structure
13.5 Problems
References and further reading
609
610
612
617
618
13.1 Introduction
From the study of the problem of free vibrations of MDOF systems presented in
Chapter 12, it becomes obvious that the computation of the eigenfrequencies
and eigenmodes, that is, the solution of the eigenvalue problem, plays a crucial
role in the dynamic analysis of structures. In the examples presented there, the
computation of the eigenvalues li was made by determining the roots of the
characteristic polynomial
PðlÞ ¼ a0 lN + a1 lN 1 + ⋯ + aN 1 l + aN
(13.1.1)
and subsequently the eigenvectors fi were obtained from the solution of the
linear algebraic system
ðK li MÞfi ¼ 0
(13.1.2)
This method requires the computation of the coefficients of PðlÞ. As soon as
this is achieved, its roots are computed by a numerical method because an analytical determination of the roots is possible only for polynomials up to fourth
degree (N ¼ 4). The determination of the polynomial coefficients requires a large
number of computations because the roots of PðlÞ are very sensitive to small
changes of the coefficients, leading to a considerable inaccuracy of the roots.
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00013-3
© 2020 Elsevier Inc. All rights reserved.
601
602 PART
II Multi-degree-of-freedom systems
Thus, this method is rather theoretical in nature and not suitable for solving the
eigenvalue problem, particularly for large degree-of-freedom systems.
The numerical solution of the eigenvalue problem is a difficult problem of
computational linear algebra. Great efforts have been made to develop reliable
methods for solving this problem during the last two centuries. The emergence
of computers in the 1950s gave a colossal impetus for research in this area.
Thus, since that time, many new methods have been developed for the solution
of the eigenvalue problem. Generally, these methods can be grouped into the
following three categories:
1. Transformation methods.
2. Iteration methods.
3. Determinant search method.
The transformation methods are applied when the size of the eigenvalue problem is comparatively small and the matrices are fully populated or banded with
large bandwidth. The underlying idea of these methods is the use of consecutive
transformations, which render the matrix A of the standard eigenvalue problem
ðA lIÞx ¼ 0 diagonal. These methods compute all the eigenvalues simultaneously while the eigenmodes are computed either by adhering to the inverse
transformations or by iteration procedures. The known transformation methods
are (i) Jacobi diagonalization, named after Carl Gustav Jacob Jacobi, who first
proposed the method in 1846, although it became widely used only in the 1950s
with the advent of computers, (ii) Givens triangularization, (iii) Householder
transformation, and (iv) QR and QL transformations.
The iteration methods are suitable for solving large eigenvalue problems
with banded or sparse matrices as those in the finite elements. Particularly, they
are applied in cases where it is necessary to compute only a few eigenvalues, the
smallest or the largest ones. The most known iteration methods that are effectively applied to the solution of the vibration problem in engineering are: (i)
vector iteration, (ii) vector iteration with shift, (iii) subspace iteration, and
(iv) the Lanczos method.
Finally, the determinant search method is based on the search for the roots of
the function det ðA lBÞ. It is employed when the size of the eigenvalue
problem is comparatively large while the matrices are banded with a small
bandwidth and the computation of only a few first eigenpairs is required.
The previous methods can be combined to give the best solution for a particular problem. It should be emphasized that all methods are by nature iterative
because the solution of the eigenvalue problem is equivalent to that of determining the roots of a polynomial, which, as already mentioned, are established only
by numerical iterative methods for orders of the characteristic polynomial N > 4.
Although an iterative process is necessary for the computation of the eigenpair ðli , fi Þ, it should be noted that as soon as one of its members has been
computed, the other member can be computed without further iterative process.
Suppose that li is computed by an iteration method, then fi can be computed by
solving the linear system (13.1.2). On the other hand, if fi has been computed
Numerical evaluation of the eigenfrequencies and eigenmodes Chapter
13
603
by an iteration method, then li can be computed from the Rayleigh quotient
using the relations
li ¼ fTi Kfi , fTi Mfi ¼ 1
(13.1.3)
Therefore, when selecting the appropriate method, the question arises as to
whether it is computationally less expensive to compute li first and then fi , or
vice versa, or both at the same time. The answer to this question depends on the
requirements of the particular problem and the properties of the matrices K and
M, for example, the number of the desired eigenpairs, the dimensions and the
density of K and M, etc.
From the above discussion, we realize that there is not always a single algorithm leading to an effective solution to the eigenvalue problem. Indeed, an
algorithm that can be effective for a particular problem may be totally inadequate for another one. The criteria for choosing the solution algorithm are
two: (a) the reliability of the method, and (b) its computational cost. Reliability
implies the computation of the solution with the desired precision without the
method collapsing. The computational cost is determined by the number of
numerical operations and the required computer memory.
All methods mentioned above are described in detail in many books of structural dynamics published earlier, for example, Refs. [1–4]. The reason is clear.
At that time, the ready-to-use computer codes and functions for the numerical solution of the eigenvalue problem were limited, if not unavailable, and the engineers
had to write their own computer codes to cope with the largely increasing needs for
determining the eigenfrequencies and mode shapes. The full and detailed development of all the above methods requires extensive work, the presentation of which
goes beyond the objectives of this book. Besides, this issue is the subject of many
books on numerical analysis to which the interested reader may refer [4]. On the
other hand, mathematical libraries, for example, IMSL, MAPLE, MATLAB,
Numerical Recipes, etc., offer many ready-to-use subroutines and functions for
the solution of the eigenvalue problem. In the following, for educational purposes,
we limit our presentation only to the vector iteration method, which can be readily
used to solve small-sized eigenvalue problems by hand or by writing simple
computer code. The chapter closes with the discussion of the response of the
free or partially supported structures. The examples facilitate the understanding
of the presented material. References and related bibliography for further study
are also included. The chapter is enriched with problems to be solve.
13.2 The vector iteration method
In the vector iteration method, the starting point is the equation
Kf ¼ lMf
(13.2.1)
aiming at approximating the vector f so that this equation is satisfied. Thus, we
assume arbitrarily a vector f, which we designate as x1 , and a value of l, say
l ¼ 1. Then we evaluate the right side of Eq. (13.2.1). This yields
R1 ¼ 1 Mx1
(13.2.2)
604 PART
II Multi-degree-of-freedom systems
Inasmuch as x1 is arbitrary, it will not, in general, satisfy Kx1 ¼ R1 .
Nevertheless, if this happens, then x1 is an eigenvector. Therefore, a vector
x2 6¼ x1 should be determined to satisfy the equation
Kx2 ¼ R1
(13.2.3)
Eq. (13.2.3) can be considered as the static equilibrium of a system subjected
to load R1 . From intuition, we understand that the vector x2 resulting from the
solution of Eq. (13.2.3) is closer to the eigenvector f than x1 . Repeating
the above procedure with an initial vector x2 gives a better approximation of
the eigenvector. Thus, we can develop a simple iterative procedure that yields
f, when it converges. The assumption of an arbitrary value for l does not affect
the method because, as is known, the eigenvectors are determined to an arbitrary
multiplier. The iteration method resulting in this way is known as the direct
vector iteration method. The procedure we have described is the underlying
idea of the inverse vector iteration method, in which Eq. (13.2.1) is employed
inversely. Apparently, in the direct vector iteration method, the vector
R1 ¼ Kx1 is first computed and then a better approximation x2 is obtained from
the solution of the linear system Mx2 ¼ R1 .
13.2.1 The inverse vector iteration method
This method, also known as the Stodola-Vianello method, is very efficient for
the computation of the eigenvectors and the respective eigenvalues. Stodola
used it for the solution of the vibration problem of a rotating shaft in 1904. Earlier in 1898, Vianello used it for determining the critical load in buckling for a
rotating shaft. It applies when K is not singular while M may be diagonal
including zeros elements, banded or fully populated. If K is singular, the inverse
vector iteration method with shifts is employed (Section 13.3.2). This happens
when the structure is free or partially supported (Section 13.4). In this method,
as mentioned, we start with an initial vector x1 and then adhere to the iteration
steps k ¼ 1, 2,… until convergence is achieved:
k + 1 by solving the linear system
1. We determine the vector x
K
xk + 1 ¼ Mxk
(13.2.4)
2. We compute the approximate value of the eigenvalue lðk + 1Þ corresponding
k + 1 using the Rayleigh quotient, Eq. (12.6.21),
to x
k + 1 Þ ¼
rðx
Tk+ 1 K
xk + 1 ðk + 1Þ
x
k + 1 Þ
, l
¼ rðx
T
k + 1 M
xk + 1
x
(13.2.5)
3. We check for convergence by comparing two consecutive values of l
ðk + 1Þ
l
lðk Þ <e
(13.2.6)
lðk + 1Þ
where e is a specified tolerance
Numerical evaluation of the eigenfrequencies and eigenmodes Chapter
13
605
k + 1 with respect
4. If the convergence criterion is not satisfied, we normalize x
to mass
xk + 1 ¼ k + 1
x
Tk+ 1 M
xk + 1
x
1=2
(13.2.7)
and go back to step 1.
5. If the criterion is satisfied, the procedure ends and we set
f1 xk + 1 , l1 lðk + 1Þ
(13.2.8)
The normalization does not affect the convergence. However, if the vector xk + 1
is not normalized, its elements increase in each step, a fact that can cause numerical problems. If the desired precision of l1 is 2d digits, then the tolerance should
be taken as e ¼ 102d . The resulting precision of the eigenvector will be d digits.
Computationally, it is more effective to formulate the previous algorithm as
follows.
We take an arbitrary vector x1 and compute the vector z1 ¼ Mx1 .
Subsequently, we compute for k ¼ 1, 2, … until convergence is achieved
k + 1 Þ¼
rðx
K
xk + 1 ¼ zk
(13.2.9)
zk + 1 ¼ M
xk + 1
(13.2.10)
Tk+ 1 zk
x
,
T
k + 1zk + 1
x
zk + 1 ¼ k + 1 Þ
lðk + 1Þ ¼ rðx
zk + 1
Tk+ 1zk + 1
x
1=2
(13.2.11)
(13.2.12)
If convergence is achieved after k iterations, we take
f1 ’ k + 1
x
1=2
Tk+ 1zk + 1
x
,
l1 lðk + 1Þ
(13.2.13)
The method works provided that zT1 f1 6¼ 0.
Example 13.2.1 Compute the eigenpair ðl1 , f1 Þ of the eigenvalue problem
Kf ¼ lMf using the inverse vector iteration method, when
2
3
2
3
32 24 18
0:5 0 0
K ¼ 4 68 71 38 5, M ¼ 4 0 0:5 0 5
36 27 20
0 0 1
Solution
A convenient trial vector to start the procedure is the vector x1 ¼ f1 1 1gT . The
computed eigenpair for different values of k is shown in Table E13.1.
606 PART
II Multi-degree-of-freedom systems
Apparently, after eight iterations, it coincides with that obtained by the
MATLAB function [V,D] ¼ eig(A,B).
TABLE E13.1 Computation of the first eigenpair using
the inverse vector iteration method.
k
1
l1
fT1
1
0.1522
0.6875
0.8672
2
13.2829
0.0327
0.6113
0.9015
3
6.4215
0.0460
0.6103
0.9015
4
6.6418
0.0438
0.6112
0.9013
5
6.6439
0.0442
0.6110
0.9013
6
6.6416
0.0441
0.6110
0.9013
7
6.6421
0.0442
0.6110
0.9013
8
6.6420
0.0442
0.6110
0.9013
MATLAB
6.6420
0.0442
0.6110
0.9013
13.2.2 Convergence of the inverse vector iteration method
In the preceding section, we presented the algorithm of inverse vector iteration
assuming that the method converges to the eigenvector corresponding to the
smallest eigenvalue. We will now present a formal proof of the convergence
because it allows us to understand the extension of the method in determining
the higher-order eigenpairs.
To prove the convergence, we consider Eq. (13.2.4), which we write as
Kxk + 1 ¼ Mxk
(13.2.14)
We assume that the eigenmodes have been computed and let
F ¼ ½f1 f2 ⋯fN be the modal matrix. Applying the expansion theorem
(12.5.82) for u ¼ xk ,xk + 1 , X ¼ F and a ¼ yk , yk + 1 yields
xk ¼ Fyk
(13.2.15a)
xk + 1 ¼ Fyk + 1
(13.2.15b)
Substituting Eqs. (13.2.15a), (13.2.15b) into Eq. (13.2.14) and premultiplying by FT gives
FT KFyk + 1 ¼ FT MFyk
(13.2.16)
Numerical evaluation of the eigenfrequencies and eigenmodes Chapter
13
607
Using the orthogonality relations FT MF ¼ I and FT KF ¼ L, the forgoing
equation becomes
Lyk + 1 ¼ yk
(13.2.17)
where L is the diagonal matrix including the eigenvalues.
Eqs. (13.2.14), (13.2.17) are equivalent, therefore the convergence properties of Eq. (13.2.14) are those of Eq. (13.2.17). The latter, however, simplifies
the proof of convergence because the elements of L are the eigenvalues li and
the eigenvectors are unit vectors ei , that is,
ð13:2:18Þ
In the inverse vector iteration method, the starting vector x1 should not be
M-orthogonal to f1 . Now y1 should not be M-orthogonal to e1 . Apparently, the
vector
y 1 ¼ f 1 1 ⋯ 1 gT
satisfies this requirement.
Applying Eq. (13.2.17) for k ¼ 1, 2, …l gives
1 1
1 T
⋯
yl + 1 ¼
ll1 ll2
llN
(13.2.19)
(13.2.20)
Because the eigenvector does not change if it is multiplied by a number,
we can write
(
¼
l + 1 ¼ ll1 yl + 1
y
l
l )T
l1
l1
1
⋯
l2
lN
(13.2.21)
Assuming that l1 < l2 < ⋯lN , we obtain
l + 1 ¼ e1
lim y
l!1
(13.2.22)
Therefore, for l ! 1, yl + 1 converges to a multiple of e1 .
Using the Euclidian norm, the convergence of a sequence of vectors
y1 , y2 ,…,yl to a vector e1 is defined by the relation
y
l + 1 e1 lim
¼r
(13.2.23)
l!1 jy
l e1 jp
where p is the number characterizing the rate of convergence, for example, for
p ¼ 1 the rate is linear, for p ¼ 2 quadratic and so forth. In the present case it is
608 PART
II Multi-degree-of-freedom systems
lim
y
l + 1 e1 l!1
l e1 j
jy
¼
l1
l2
(13.2.24)
Hence, convergence is linear and the rate of convergence is r ¼ l1 =l2 . The
smaller this ratio, the faster the rate of convergence.
In the preceding proof, we have accepted that the eigenvalues are distinct.
Let us now consider the case of multiple eigenvalues, for example,
l1 ¼ l2 ⋯ ¼ lm . Then Eq. (13.2.21) becomes
(
l )T
l1 l
l1
l + 1 ¼ 1 1 ⋯
y
(13.2.25)
⋯
lm + 1
lN
from which we conclude that the iteration procedure converges with rate of convergence l1 =lm + 1 . Hence, we can generally state that the rate of convergence
is given by the ratio of l1 to the nearest distinct eigenvalue.
The Rayleigh quotient for the vector yl + 1 is obtained by Eq. (12.6.1) for
A ¼ L and u ¼ yl + 1 . Thus we have
yT Ly
r yl + 1 ¼ l +T 1 l + 1
yl + 1 yl + 1
which by virtue of Eq. (13.2.17) becomes
yT y
r yl + 1 ¼ T l + 1 l
yl + 1 yl + 1
(13.2.26)
Using Eq. (13.2.21) to express yl and yl + 1 , and substituting into the
foregoing equation gives
r yl + 1 ¼
l1
N
X
ðl1 =li Þ2l1
i¼1
N
X
(13.2.27)
ðl1 =li Þ
2l
i¼1
which implies
lim r yl + 1 ¼ l1
l!1
(13.2.28)
The preceding proof was made by taking x1 ¼ f 1 1 ⋯ 1 gT as a starting
vector. However, we can show that the convergence properties apply for any
starting vector provided that it is not M-orthogonal to f1 . Nevertheless, in some
cases, the convergence may require a large number of iterations.
The convergence rate depends, as we have shown, on the ratio l1 =lm + 1 .
The convergence is very slow if l1 is almost equal to lm + 1 , but it can be
accelerated using the inverse vector iteration method with shifts presented in
Section 13.3.2.
Numerical evaluation of the eigenfrequencies and eigenmodes Chapter
13
609
13.3 Computation of higher-order eigenpairs
13.3.1 The vector purification method
When the starting vector x1 is selected to be M-orthogonal to f1 , that is
xT1 Mf1 ¼ 0, the inverse vector iteration method converges, at list theoretically,
to an eigenvector other than f1 , and indeed to that corresponding to the next
eigenvalue l2 . The selection of the starting vector may be again arbitrary
and become orthogonal to f1 by the Gram-Schmidt orthogonalization process
presented in Section 12.4. Thus, on the basis of the second of Eqs. (12.4.8),
we obtain
e
x1 ¼ x1 fT1 Mx1
f1
fT1 Mf1
(13.3.1)
which is orthogonal to f1 . Obviously, the contribution of the eigenmode f1
has been removed from the trial vector e
x1 . The vector that remains is said to
be purified. Now, Eq. (13.3.1) can be further written as
!
f1 fT1 M
e
x1 ¼ I T
x1
f1 Mf1
(13.3.2)
¼ S 1 x1
where
S1 ¼ I f1 fT1 M
fT1 Mf1
(13.3.3)
The matrix S1 is referred to as the sweeping matrix. It is used to purify the
vector x1 from f1 . Theoretically, the iterations start with the purified vector e
x1
and we expect convergence to the second eigenvector f2 . Unfortunately, during
the iterations, small arithmetic rounding errors are introduced, which result in
pollution of the vector x1 by f1 . This demands the purification procedure in
each iteration, which, in fact, increases the computational cost. For the computation of the nth order eigenpair ðln , fn Þ, the starting vector x1 must be free
from the contribution of the previous n 1 eigenmodes and M-orthogonal to
them. This is achieved by the Gram-Schmidt orthogonalization procedure.
Thus, by virtue of Eq. (12.4.8), we obtain
fT1 Mx1
fT2 Mx1
fTn Mx1
f
f
⋯
fn
1
2
fT1 Mf1
fT2 Mf2
fTn Mfn
!
f1 fT1 M f2 fT2 M
fn fTn M
¼ I T
⋯ T
x1
f1 Mf1 fT2 Mf2
fn Mfn
e
x1 ¼ x1 ¼ Sn x1
(13.3.4)
610 PART
II Multi-degree-of-freedom systems
where Sn represents the sweeping matrix
Sn ¼ I f1 fT1 M f2 fT2 M
fn fTn M
⋯
fT1 Mf1 fT2 Mf2
fTn Mfn
(13.3.5)
or
Sn ¼ Sn1 fn fTn M
,
fTn Mfn
n ¼ 1, 2, …, S0 ¼ I
(13.3.6)
Although Gram-Schmidt orthogonalization combined with the inverse vector iteration may yield the higher-order eigenpairs, it is not suitable for writing a
general computer program for the solution of the eigenvalue problem because
convergence becomes slow as the order of the eigenvector increases. Instead,
we prefer the vector iteration method with shifts, which is described in the next
section.
13.3.2 The inverse vector iteration method with shifts
The study of the convergence of the inverse iteration method presented in
Section 13.2.2 showed that for l1 < l2 , the iteration process converges to the
eigenvector f1 with a convergence rate l1 =l2 . Thus, the convergence is fast
when l1 is very small compared to l2 , for example, l1 =l2 ¼ 0:01, or extremely
slow when l1 is near l2 , for example, l1 =l2 ¼ 0:9999. Therefore, a problem
of improving or even controlling the rate of convergence arises. This is
accomplished by the vector iteration method with shifts, which can also be used
to determine any other eigenpair ðli , fi Þ. This method also addresses the case
where the matrix K is positive semidefinite, hence the solution of Eq. (13.2.4) is
not possible.
A shift m in K in the eigenvalue problem Kf ¼ lMf defines the matrix
^ ¼ K mM. The matrices K
^ and K have the same eigenvectors while the
K
^ and l satisfy the relation
eigenvalues l
^ ¼lm
l
(13.3.7)
^ is shifted by m with respect to the origin
That is, the eigenvalue l
(Fig. 13.3.1). Indeed
^ ¼ ðK mMÞf
Kf
¼ Kf mMf
¼ ðl mÞMf
^
¼ lMf
(13.3.8)
Numerical evaluation of the eigenfrequencies and eigenmodes Chapter
13
611
Π( ) = det (K − M)
|
|
1
m
1
− m|
2
− m|
|
3
− m|
2
3
4
FIG. 13.3.1 Graph of the characteristic polynomial and shifts
^ i of the eigenvalue problem (13.3.8) are
We assume that all eigenvalues l
distinct. In this case, Eq. (13.2.20) becomes
T
1
1
1
⋯
(13.3.9)
yl + 1 ¼
ðl1 mÞl ðl2 mÞl
ðlN mÞl
where it was assumed that li m is nonzero, that is, it is a positive or a negative
^ i ¼ li m has the smallest absolute value and we
number. We assume that l
formulate the product
ð13:3:10Þ
In the foregoing relation, it is jli mj=lp m < 1 for all p 6¼ i. Hence,
l + 1 ! ei as l ! 1 and the iteration vector converges to fi . Moreover, we
y
^ i + m. The rate of convergence is determined by the element
obtain li ¼ l
having the largest absolute value, that is,
li m r ¼ max (13.3.11)
p6¼i lp m
Because li is closer to m, the rate of convergence of the iteration vector to fi
will be
li m li m r ¼ max ,
(13.3.12)
li1 m li + 1 m
For the data shown in Fig 13.3.1, we have jl2 mj < jl1 mj < jl3 mj.
Therefore, the iterative procedure will converge to f2 with rate of convergence
r ¼ jl2 mj=jl1 mj.
Evidently, the selection of the shift very close to the eigenvalue we are
interested in permits, at least theoretically, achieving the desired rate of
612 PART
II Multi-degree-of-freedom systems
convergence. In practice, however, this is difficult because the eigenvalue is not
known in advance. Often, in certain problems, we need to compute a few of the
first eigenpairs. In such cases, the vector iteration method with shifts is combined with that without shifts. Namely, a few, say five, eigenpairs are computed
by applying directly the inverse vector iteration method in conjunction with the
Gram-Schmidt orthogonalization. The same procedure is employed for the
computation of the next eigenpair, that is, the sixth one. However, instead of
completing the iteration process until convergence is achieved, we stop it after
some iterations and obtain a first estimation of the sixth eigenvalue from the
Rayleigh quotient, Eq. (13.2.5). This value is used as a new shift and the next
five eigenpairs are computed using the inverse vector iteration method. The rate
of convergence can be improved if in each iteration, the shift m takes the value of
k Þ in the k + 1 iteration. Thus
the Rayleigh quotient, namely if we set mk ¼ rðx
the algorithm described by Eqs. (13.2.9)–(13.2.13) is modified as
k ÞM
½K rðx
x k + 1 ¼ zk
(13.3.13)
zk + 1 ¼ M
xk + 1
(13.3.14)
k + 1 Þ ¼ x
Tk+ 1 zk
rðx
T
k Þ
k + 1zk + 1 + rðx
x
(13.3.15)
zk + 1 ¼ zk + 1
Tk+ 1zk + 1
x
1=2
(13.3.16)
where now
k + 1 Þ ! li when k ! 1
zk + 1 ! Mfi and rðx
(13.3.17)
The inverse vector iteration method is very useful when the system is subjected to a harmonic excitation with a specified frequency and we want to see if
there are frequencies close to the excitation frequency. This allows us to prevent
resonance by changing the dynamic characteristics of the structure.
13.4 Free or partially supported structure
When the rigid body motion of the structure is not restrained, that is, the structure is free to move or the restrained degrees of freedom are less than those
required to prevent rigid body motion, then the stiffness matrix K contains zero
eigenvalues, hence it is singular and cannot be inverted. Therefore, the methods
of computation of the eigenpairs that involve the inversion of the stiffness
matrix, for example, the inverse vector iteration, do not apply. This troubleshooting can be overcome in the following two ways:
(a) Shifting of the eigenvalue
This method is mathematical in nature and it is accomplished by virtue
of Eq. (13.3.8)
^
^ ¼ lMf
Kf
(13.4.1)
Numerical evaluation of the eigenfrequencies and eigenmodes Chapter
13
613
where
^ ¼lm
^ ¼ K mM and l
K
(13.4.2)
^ is generally not singular and can be inverted. If the
Obviously, K
selected value of m is negative and the matrix M is diagonal, this shift is
equivalent to adding a positive number to the diagonal elements of K,
which from a physical point of view expresses restraining each degree of
freedom by a spring. The advantage of this method is that the
eigenmodes remain unchanged.
(b) Restraining the free degrees using springs
The simplest method to treat the unrestrained structure that is also computationally easy to implement is by restraining the degrees of freedom corresponding to rigid body motion by springs with a very small stiffness.
These fictitious springs are placed between the structure and the supports
and inhibit the rigid body motion. Computationally, these springs represent
additive terms to the diagonal elements of K corresponding to the unrestrained degrees of freedom. The addition of a small stiffness, while making the stiffness matrix nonsingular, modifies the structure. However, the
small stiffness of the springs negligibly affects the eigenmodes and the
eigenfrequencies while at the same time the eigenmodes corresponding
to the rigid body of motion are also computed. As is anticipated, the
computed eigenfrequencies corresponding to the rigid body motion are
very small (approximately zero). This is illustrated with the example that
follows.
Example 13.4.1 Compute the mode shapes and eigenfrequencies of the plane
structure in Fig. E13.1a. The density of the rigid bodies is r while the mass of
the column is negligible.
f
f
′
′
′
(a)
(b)
(c)
FIG. E13.1 Plane structure in Example 13.4.1 (a), degrees of freedom (b), and free body diagram (c).
614 PART
II Multi-degree-of-freedom systems
Solution
The system has three degrees of freedom, u1 ,f, u3 (Fig. E13.1b). The equation
of motion can be formulated by different methods. Here, the method of equilibrium of forces with respect to the center of mass C is employed. Referring
to Fig. E13.1c, we obtain
m1 u€1 + fS1 ¼ 0
(1a)
a
Ic f€ + fS2 + fS1 ¼ 0
(1b)
2
(1c)
m3 u€3 + fS3 ¼ 0
where
fS1 ¼ k11 ðu2 u3 Þ + k12 f
(2a)
fS2 ¼ k21 ðu2 u3 Þ + k22 f
(2b)
fS3 ¼ k31 ðu2 u3 Þ + k32 f
(2c)
The coefficients kij in the previous equations are computed using
Eq. (11.5.2). Moreover, we have
1
m1 ¼ ra 2 , m3 ¼ 3ra 2 , Ic ¼ ra 4
6
Substituting Eqs. (2a), (2b), (2c) into Eqs. (1a), (1b), (1c) gives
M€
u + Ku ¼ 0
(3)
where
8 9
3
2
3
1 0
0
1:5
2:25a
1:5
< u1 =
EI
M ¼ ra 2 4 0 a 2 =6 0 5, K ¼ 3 4 2:25a 3:875a2 2:25a 5, u ¼ f
: ;
a
0 0
3
1:5 2:25a
1:5
u3
2
(4a--c)
It is convenient to eliminate the length a and make the matrices dimensionless. To this end, the rotational coordinate f is replaced by the translational
coordinate u2 ¼ u1 + a2 f (see Fig. E13.1b). This is accomplished by the
transformation (see Section 10.7).
8 9 2
38 9
1
0
0 < u1 =
< u1 =
f ¼ 4 2=a 2=a 0 5 u2
: ;
: ;
0
0
1
u3
u3
which yields
2
1:667 0:667 0
3
7
^ ¼ RT MR ¼ ra 2 6
M
4 0:667 0:667 0 5,
0
0
3
2
3
8:0 11:0 3:0
7
^ ¼ RT KR ¼ EI 6
K
4 11:0 15:5 4:5 5
a3
3:0 4:5 1:5
2
3
1
0
0
where R ¼ 4 2=a 2=a 0 5
0
0
1
Numerical evaluation of the eigenfrequencies and eigenmodes Chapter
13
615
Therefore, the eigenvalues and eigenmodes will be obtained from the solution of the eigenvalue problem
^ lM
^ f ¼ 0, l ¼ w2 ra5 =EI
K
^ ¼ 0. Hence the solution is obtained either by the inverse vector
It is det K
iteration method with shift or by restraining u3 using a spring of small stiffness.
1. The inverse vector iteration with shift
a. Computation of the first eigenpair.
We employ the algorithm presented in Section 13.3.2. We take
arbitrarily x1 ¼ f 1 2 1 gT as the starting vector, which gives
m1 ¼ rðx1 Þ ¼ 3:321. After five iterations we obtain l1 ¼ 0:868e 16 0
and f1 ¼ f 1 1 1 gT . The obtained values of the eigenpair were anticipated
because the structure has a rigid body motion.
b. Computation of the second eigenpair
We apply the vector purification method described in Section 13.3.1.
We take arbitrarily x1 ¼ f 1 1 2 gT as the trial vector. This gives
2
3
0:75 0:472 1015 0:75
T
f f M 6
7
S1 ¼ I 1T 1 ¼ 4 0:25 1:000
0:75 5,
f1 Mf1
0:25 0:482 1015 0:25
9
8
>
=
< 0:749 >
e
x1 ¼ S1 x1 ¼ 0:749 , m1 ¼ rðx1 Þ ¼ 2:0:
>
>
;
:
0:249
After four iterations we obtain
l2 ¼ 0:239 and f2 ¼ f 0:834 0:508 0:278 gT .
c. Computation of the third eigenpair.
We take arbitrarily x1 ¼ f 2 2 1 gT as the trial vector. This gives
2
3
0:128 0:182 0:053
T
f f M 6
7
S2 ¼ S1 2T 2 ¼ 4 0:784 1:111 0:325 5,
f2 Mf2
0:043 0:061 0:018
9
8
>
=
< 0:054 >
e
0:325 , m1 ¼ rðe
x1 Þ ¼ 25:0096
x 1 ¼ S2 x 1 ¼
>
>
;
:
0:018
After five iterations we obtain
l3 ¼ 24:998 and f3 ¼ f 0:231 1:411 0:077 gT .
2. Restraining u3 using a spring of small stiffness
0
Adding the small stiffness k33
¼ 0:0001EI =a 3 to the element k33 yields
2
3
8:0 11:0 3:0
^ ¼ EI 4 11:0 15:5 4:5 5
K
a3
3:0 4:5
1:5001
616 PART
II Multi-degree-of-freedom systems
a. Computation of the first eigenpair.
It is now det ðKÞ ¼ 0:0003 6¼ 0, therefore we can directly apply the
inverse vector iteration method using the algorithm described in
Section 13.2.1. We take arbitrarily x1 ¼ f 1 2 1 gT as a trial vector.
After five iterations we obtain l1 ¼ 0:2499 104 0 and
f1 ¼ f 1 0:999 0:998 gT f 1 1 1 gT
b. Computation of the second eigenpair.
We take x1 ¼ f 1 1 2 gT as the trial vector, which gives
8
9
2
3
0:75 0:00 0:75
< 0:75 =
f1 fT1 M 4
¼ 0:25 1:00 0:75 5, e
x1 ¼ S1 x1 ¼ 0:75
S1 ¼ I T
:
;
f1 Mf1
0:25 0:00 0:25
0:25
After four iterations we obtain
l2 ¼ 0:2399 and f2 ¼ f 0:834 0:508 0:278gT .
c. Computation of the third eigenpair.
We take x1 ¼ f 2 2 1 gT as the trial vector, which gives
8
9
2
3
0:128 0:181 0:053
< 0:054 =
f2 fT2 M 4
¼ 0:785 1:111 0:323 5, e
x1 ¼ S 2 x1 ¼
0:325
S2 ¼ S1 T
:
;
f2 Mf2
0:043 0:061 0:018
0:018
After five iterations we obtain
l3 ¼ 25:009 and f3 ¼ f 0:231 1:411 0:077gT .
Table E13.2 summarizes the eigenpairs computed by the previous two methods
and the MATLAB function [V,D] ¼ eig(A,B). The obtained results illustrate
that the method of restraining the degrees corresponding to rigid body motion
by springs of small stiffness, besides being simple to implement, gives good
results.
TABLE E13.2 Eigenvalues and eigenmodes in Example 13.3.1.
Eigenpair
Iteration with shift
n
ln
Restraining with a spring
fn
16
1
0:868 10
2
0:239
0:834
0:508
0:278
3
24:998
0:231
1:411
0:077
0:999
0:999
1:000
ln
fn
4
0:249 10
MATLAB
ln
fn
1:000
0:999
0:998
0.0000
1:000
1:000
1:000
0:2399
0:834
0:508
0:278
0.2399
0:834
0:508
0:278
25:009
0:231
1:411
0:077
24.998
0:231
1:411
0:077
Numerical evaluation of the eigenfrequencies and eigenmodes Chapter
13
617
13.5 Problems
Problem P13.1 The one-story building of Fig. P13.1 is subjected to the harmonic
excitation p ¼ p0 sin20t. Examine whether the structure is at resonance risk.
Assume: E ¼ 2:1 107 kN=m2 , n ¼ 0:2: Columns 1, 3, 6, 8 0:35 0:35 m2 ,
Columns 2, 4, 5, 7 0:20 0:40 m2 . The dead weight of the plate is included in q.
FIG. P13.1 One-story building in problem P13.1
Problem P13.2 We consider the plane structure of Fig. P13.2. At time t the support A becomes a hinge. Compute the eigenfrequencies and eigenmodes of the
structure at t + .
FIG. P13.2 Plane structure in problem P13.2
Problem P13.3 We consider the plane structure of Fig. P13.3 whose base is isolated. Compute the eigenfrequencies (i) if the stiffness of the isolators is small
compared with that of the columns, that is, kb =k ¼ 0:1, (ii) if the isolators break
at time t, that is, kb ¼ 0. Assume; m1 ¼ m2 , mb ¼ 0:8m1 .
FIG. P13.3 Plane structure in problem P13.3
618 PART
II Multi-degree-of-freedom systems
References and further reading
[1] K.J. Bathe, E.L. Wilson, Numerical Methods in Finite Element Analysis, Prentice-Hall,
Englewood Cliffs, NJ, 1976.
[2] J.L. Humar, Dynamics of Structures, second ed., A.A. Balkema Publishers, Lisse, NL, 2002.
[3] J.T. Katsikadelis, Dynamic of Structures, Vol. II, Symmetria Publications, Athens, 2004 (in
Greek).
[4] W.H. Press, B.P. Flannery, S.A. Teukolsky, W.T. Vetterlin, Numerical Recipes in Fortran,
second ed., Cambridge University Press, New York, 1992.
[5] T.J.R. Hughes, The Finite Element Method: Linear Static and Dynamic Finite Element
Analysis, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1987.
Chapter 14
Multi-degree-of-freedom
systems: Forced vibrations
Chapter outline
14.1 Introduction
14.2 The mode superposition
method
14.3 Modal contribution in the
mode superposition method
14.3.1 Modal participation
14.3.2 Static correction
method
14.3.3 Error in mode
superposition method
due to truncation of
higher modes
14.4 Reduction of the dynamic
degrees of freedom
14.4.1 Static condensation
14.4.2 Kinematic constraints
14.5 Rayleigh-Ritz method
14.5.1 Ritz transformation
14.5.2 Approximation using
Ritz vectors
14.6 Selection of Ritz vectors
14.6.1 Method of natural
mode shapes
14.6.2 The method of
derived Ritz vectors
14.7 Support excitation
14.7.1 Multiple support
excitation
14.7.2 Uniform support
excitation
14.8 The response spectrum
method
620
620
629
629
641
647
649
649
650
651
651
654
655
656
659
663
663
665
14.9 Comparison of mode
superposition method
and Rayleigh-Ritz method
672
14.10 Numerical integration
of the equations of
motions—Linear MDOF
systems
675
14.10.1 The central
difference method
(CDM)—Linear
equations
675
14.10.2 The average
acceleration method
(AAM)—Linear
equations
677
14.10.3 The analog equation
method (AEM)—
Linear equations
679
14.11 Numerical integration of the
equations of motions—
Nonlinear MDOF systems
681
14.11.1 The average
acceleration method
(AAM)—Nonlinear
equations
681
14.11.2 The analog equation
method (AEM)—
Nonlinear equations 684
14.12 Problems
688
References and further reading
692
668
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00014-5
© 2020 Elsevier Inc. All rights reserved.
619
620 PART
II Multi-degree-of-freedom systems
14.1 Introduction
In this chapter, we study the dynamic response of MDOF systems under the
action of external forces. As was shown in Chapter 10, the equation of motion
of a linear system with viscous damping is
M€
u + Cu_ + Ku ¼pðt Þ
(14.1.1)
where M is the mass matrix, C the damping matrix, K the stiffness matrix, u the
displacement vector, and pðt Þ the vector of the external forces in the directions
of the displacements. Besides, the system is subjected to the initial conditions
uð0Þ ¼u0 , u_ ð0Þ ¼u_ 0
(14.1.2)
where u0 and u_ 0 are known vectors.
The time-dependent forces pðt Þ may be due to moving loads, machine
excitations, explosions, wind pressure, aerodynamic actions, ground motions,
etc. Their determination, which depends on the type of structure and its use, constitutes a special problem that does not concern the discussion that follows.
The methods that can be used to solve the initial value problem (14.1.1),
(14.1.2) are:
1.
2.
3.
4.
The mode superposition method.
The response spectrum method.
The direct time integration method of the equations of motion.
The analysis in the frequency domain.
None of the above methods is purely analytical because their implementation
includes numerical techniques, at least at a certain stage of their application.
Indeed, the mode superposition method and the response spectrum method
require the determination of the eigenfrequencies and eigenmodes, which is
accomplished, as we saw in Chapter 13, numerically. The analysis in the frequency domain requires the use of numerical methods for computing the Fourier transform and its inverse. Finally, the direct time integration method of the
equations of motion is predominantly a numerical method. Therefore, the
description of a method as analytical is only theoretical because the real problems of structural dynamics encountered by the engineer in practice require
numerical processes and their programming on a computer.
14.2 The mode superposition method
As was shown in Section 12.9, the displacement vector is written by virtue of the
expansion theorem as
uðt Þ ¼ f1 Y1 ðt Þ + f2 Y2 ðt Þ + ⋯ + fN YN ðt Þ
(14.2.1)
where f1 ,f2 ,…, fN are the eigenmodes of the free undamped vibrations.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
621
Substituting Eq. (14.2.1) into Eq. (14.1.1) gives
M f1 Y€ 1 + f2 Y€ 2 + ⋯ + fN Y€ N + C f1 Y 1 + f2 Y 2 + ⋯ + fN Y N
+ Kðf1 Y1 + f2 Y2 + ⋯ + fN YN Þ ¼ pðt Þ
(14.2.2)
fTn
Premultiplying the previous equation by
and assuming that the damping
matrix is proportional, we obtain by virtue of the orthogonality properties
Mn Y€ n + Cn Y n + Kn Yn ¼Pn ðt Þ, n ¼ 1, 2,…,N
(14.2.3)
where
Mn ¼ fTn Mfn
(14.2.4a)
Cn ¼ fTn Cfn
(14.2.4b)
Kn ¼ fTn Kfn
(14.2.4c)
Pn ¼ fTn pðt Þ
(14.2.4d)
The quantities defined previously are referred to as generalized or modal
mass, generalized or modal damping, generalized or modal stiffness, and
generalized or modal force of the nth mode shape, respectively. Taking into
account that
Kn ¼ w2n Mn
(14.2.5)
Cn ¼ 2x n Mn wn
(14.2.6)
Eq. (14.2.3) is written as
1
Pn ðt Þ,
Y€ n + 2xn wn Y n + w2n Yn ¼
Mn
n ¼ 1,2, …,N
whose solution is given by Eq. (3.3.23)
"
#
x n wn t Y n ð0Þ + Yn ð0Þx n wn
sin wDn t + Yn ð0Þcos wDn t
Y n ðt Þ ¼ e
wDn
Z t
1
+
Pn ðτÞexn wn ðtτÞ sin ½wDn ðt τÞdτ
Mn wDn 0
(14.2.7)
(14.2.8)
The initial conditions Yn ð0Þ, Y n ð0Þ are computed using Eqs. (12.9.12),
(12.9.13), namely
Y n ð 0Þ ¼
Y n ð 0Þ ¼
fTn Muð0Þ
,
Mn
n ¼ 1, 2, …, N
(14.2.9a)
fTn Mu_ ð0Þ
,
Mn
n ¼ 1, 2, …, N
(14.2.9b)
622 PART
II Multi-degree-of-freedom systems
Once the modal coordinates Yn ðt Þ have been computed, the solution of
Eq. (14.1.1) is obtained from the superposition (14.2.1). The Duhamel integral
in Eq. (14.2.8) can be evaluated analytically using a symbolic language. Nevertheless, this is not always possible or the obtained expressions are too complicated for numerical computation, therefore it is preferable to solve
Eq. (14.2.7) directly numerically using any of the numerical methods presented
in Chapter 4.
The study of the dynamic response of a MDOF system under an external
force pðt Þ via the mode superposition method is accomplished by adhering
to the following steps:
1. Determine the dynamic model of the structure, hence the degrees of freedom, and define the nodal displacement vector uðt Þ and the nodal load vectors pðt Þ.
2. Compute the mass matrix M and stiffness matrix K of the structure (see
Chapters 10 and 11).
3. Solve the eigenvalue problem ðK w2 MÞf ¼0 to compute the eigenfrequencies wn and mode shapes fn (Chapter 13).
4. Compute the modal masses Mn and modal loads Pn ðt Þ using Eqs. (14.2.4a),
(14.2.4d) and formulate Eq. (14.2.7) for the modal coordinate Yn ðt Þ.
5. Compute the initial conditions Yn ð0Þ, Y n ð0Þ from the natural initial conditions uð0Þ, u_ ð0Þ using Eqs. (14.2.9a), (14.2.9b).
6. Compute the modal coordinates Yn ðt Þ either using Eq. (14.2.8) with analytical evaluation of the Duhamel integral or by direct numerical solution of
Eq. (14.2.7).
7. Compute the displacement vector uðt Þ from the modal contributions
fn Yn ðt Þ using the superposition relation (14.2.1).
8. Compute the nodal elastic forces f S ¼ Ku.
Example 14.2.1 Determine the dynamic response of the two-story shear frame
in Example 12.2.1 for the following two load cases:
(a) The horizontal loads p1 ¼ 190 kN and p2 ¼ 300 kN are suddenly applied
at t ¼ 0 while the structure is at rest.
(b) The ground undergoes the displacement ug ðt Þ ¼ 0:02 sin 10t.
More specifically, in case (a), compute the extreme values of the displacements,
shear forces, and bending moments at the top of the columns while in case (b),
compute the maximum shear force at the base of the frame and the maximum
overturning moment. Consider proportional damping with modal damping
ratios x1 ¼ 0:06, x 2 ¼ 0:04.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
623
Solution
The structure properties, eigenfrequencies, and mode shapes are taken from
Example 12.2.1, namely
25 0
3826:5 3826:5
M¼
,
K¼
,
0 32
3826:5 9142:1
8:289
0:1697 0:1061
w¼
,
F¼
19:236
0:0935 0:1499
Case (a)
The modal matrices are computed using Eq. (14.2.4a)
25 0
0:1697
¼1
M1 ¼ ½0:1697 0:0935
0 32 0:0935
25 0
0:1061
M2 ¼ ½ 0:1061 0:1499 ¼1
0 32 0:1499
The load vector is
pðt Þ ¼
190
300
and Eq. (14.2.4d) gives the modal loads
190
¼ 4:1930
300
190
¼ 65:1290
P2 ¼ ½ 0:1061 0:1499 300
P1 ¼ ½ 0:1697 0:0935 The equations of motion for the modal coordinates Y1 ðt Þ, Y2 ðt Þ result from
Eq. (14.2.7), that is
Y€ 1 + 0:9947Y 1 + 68:7089Y1 ¼4:1930
Y€ 2 + 1:5389Y 2 + 370:0416Y2 ¼65:1290
Because the structure is at rest at t ¼ 0, we have uð0Þ ¼ 0, u_ ð0Þ ¼ 0 and
Eqs. (14.2.9a), (14.2.9b) give
Yn ð0Þ ¼
fTn Muð0Þ
¼0
Mn
Y n ð0Þ ¼
fTn Mu_ ð0Þ
¼0
Mn
n ¼ 1, 2
The damped eigenfrequencies are
qffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffi
wD1 ¼ w1 1 x21 ¼ 8:2741, wD2 ¼ w2 1 x22 ¼ 19:2206
624 PART
II Multi-degree-of-freedom systems
Thus, Eq. (14.2.8) gives
Z t
Pn
Yn ð t Þ ¼
exn wn ðtτÞ sin ½wDn ðt τÞdτ
Mn wDn 0
0
2
¼
1
3
Pn 6
xn
B
C
7
41 @ cos wDn t + qffiffiffiffiffiffiffiffiffiffiffiffi sin wDn t Aexn wn t 5
2
M n wn
2
1x
n
The Duhamel integral was evaluated using MAPLE (see Eq. 3.4.7). Hence,
using the data of the problem, we obtain
Y1 ðt Þ ¼ 0:06125 0:06125 cos 8:2741t + 0:36817 102 sin 8:2741t e0:4973t
Y2 ðt Þ ¼ 0:17629 0:17629cos 19:2206t + 0:70572 102 sin 19:2206t e0:7694t
The displacements result from Eq. (14.2.1) for N ¼ 2
(
)
(
) (
)
0:1061
0:1697
u 1 ðt Þ
Y2 ðt Þ
¼
Y1 ðt Þ +
0:1499
u 2 ðt Þ
0:0935
Fig. E14.1 shows the graphical representations of u1 ðt Þ and u2 ðt Þ in the
interval 0 t 5, from which we obtain
max u1 ¼ 0:0473 m, min u2 ¼ 0:04555 m
The shear forces at the top of columns 1 and 3 are
12EI 1
ðu1 u2 Þ,
h13
Q2 ¼
12EI 2
u2
h23
u2(t)
u1(t)
Q1 ¼
t
FIG. E14.1 Time history of u1(t) and u2(t) in Example 14.2.1.
t
14
625
Q2(t)
Q1(t)
Multi-degree-of-freedom systems: Forced vibrations Chapter
t
t
FIG. E14.2 Time history of Q1(t) and Q2(t) in Example 14.2.1
Fig. E14.2 shows the graphical representations of Q1 ðt Þ and Q2 ðt Þ in the
interval 0 t 5, from which we obtain
max |Q1 | ¼ 168:66 kN,
max |Q2 | ¼ 11:3082 kN
The respective bending moments are
max M1 ¼
h1
max |Q1 | ¼ 295:16 kN m
2
max M2 ¼
h2
max |Q2 | ¼ 22:61 kN m
2
Case (b)
The equation of motion in terms of the relative displacements u 1 ¼ u1 ug ,
u 2 ¼ u2 ug (see Chapter 6) is
€ + Cu
_ + K
Mu
u ¼
pð t Þ
where
ðt Þ ¼ M
p
1
u€ ¼
1 g
50
sin 10t
64
The modal loads are
50
sin 10t ¼ 14:4690 sin 10t
64
50
sin 10t ¼ 4:2886 sin 10t
P2 ¼ ½ 0:1061 0:1499 64
P1 ¼ ½ 0:1697 0:0935 626 PART
II Multi-degree-of-freedom systems
The initial conditions with respect to the relative displacements are
1
0
ð0Þ ¼ utot ð0Þ u
ug ð0Þ ¼
1
0
1
0:2
_ ð0Þ ¼ u_ tot ð0Þ u
u_ ð0Þ ¼
1 g
0:2
The equations of motion for the modal coordinates result from
Eq. (14.2.7) as
Y€ 1 + 0:9947Y 1 + 68:7089Y1 ¼14:4690 sin 10t
Y€ 2 + 1:5389Y 2 + 370:0416Y2 ¼ 4:2886 sin 10t
The solution of the foregoing equations is obtained from Eq. (14.2.8). The
τ
Duhamel integral is evaluated using MAPLE. Thus, setting Pn ðτÞ ¼ Pn sin w
we can obtain
Yn ð t Þ ¼
Pn
Mn wDn
Z
t
τ sin ½wDn ðt τÞdτ
exn wn ðtτÞ sin w
0
Pn 20x n wn cos 10t + w2Dn + x2n w2n 100 sin 10t
¼
Mn x2 w2 + ðw 10Þ2 x2 w2 + ðw + 10Þ2
n n
+
Pn
Mn
Dn
n n
Dn
x 2 w2 1000
20x n wn cos wDn t + 10wDn + 10 n n +
sin wDn t exn wn t
wDn
wDn
x 2n w2n + ðwDn 10Þ2 x 2n w2n + ðwDn + 10Þ2
The initial conditions for the modal coordinates are
fT2 Muð0Þ
¼0
M2
25 0
0:2
¼ 1:4469
Y 1 ð0Þ ¼ ½ 0:1697 0:0935 0 32 0:2
25 0
0:2
¼ 0:42886
Y 2 ð0Þ ¼ ½ 0:1061 0:1499 0 32 0:2
Y1 ð 0Þ ¼
fT1 Muð0Þ
¼ 0,
M1
Y 2 ð 0Þ ¼
Therefore, applying Eq. (14.2.8) for the data of the problem yields
Y1 ðt Þ ¼ 0:13348cos 10t 0:41996 sin 10t
+ ð0:1334 cos 8:2741t + 0:34063 sin 8:2741t Þe0:4973t
Y2 ðt Þ ¼ 0:9022 103 cos 10t 0:01583 sin 10t
+ 0:9022 103 cos 19:2206t + 0:03051 sin 19:2206t e0:7694t
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
627
The displacements result from Eq. (14.2.1) for N ¼ 2
0:1697
0:1061
u 1 ðt Þ
¼
Y 1 ðt Þ +
Y2 ð t Þ
0:0935
u 2 ðt Þ
01499
The elastic forces applied at the roof levels are computed from
u
f S ¼ K
¼ KFYðt Þ
3826:5 3826:5 0:1697 0:1061 Y1 ðt Þ
¼
3826:5 9142:1 0:0935 0:1499 Y2 ðt Þ
291:5793Y1 ðt Þ + 979:584Y2 ðt Þ
¼
205:4293Y1 ðt Þ 1776:39244Y2 ðt Þ
Hence
fS1 ¼ 291:5793Y1 ðt Þ + 979:584Y2 ðt Þ
fS2 ¼ 205:4293Y1 ðt Þ 1776:3924Y2 ðt Þ
A particularly useful response parameter in earthquake analysis of buildings
is the base shear. This is the total horizontal force Q0 ðt Þ that must be resisted at
the foundation, and is found by summing the inertia (D’Alambert) forces over
the structure. These forces are equal to the elastic forces if damping is neglected.
Thus, we have
Q0 ðt Þ ¼ fS1 + fS2 ¼ 497:0Y1 ðt Þ 796:8084Y2 ðt Þ
and the overturning moment
M0 ðt Þ ¼ fS1 ðh1 + h2 Þ + fS2 h2 ¼ 3008:5619Y1 ðt Þ + 241:3104Y2 ðt Þ
Q0(t)
M0(t)
The graphical representations of Q0 ðt Þ and M0 ðt Þ are shown in Fig. E14.3,
from which we obtain max jQ0 j ¼ 283:28 kN and max jM0 j ¼ 1783:83 kN m.
t
FIG. E14.3 Time history of Q0(t) and M0(t) in Example 14.2.1.
t
628 PART
II Multi-degree-of-freedom systems
Example 14.2.2 Determine the steady-state response of the two-story shear
t at
frame in Example 12.2.1, when it is subjected to the load p1 ðt Þ ¼ po sin w
¼ 10 s1 and (i)
the roof level of the second floor. Assume: po ¼ 5 kN, w
x1 ¼ 0:06, x2 ¼ 0:04, and (ii) x 1 ¼ x 2 ¼ 0.
Solution
The load vector is
t0gT
pðt Þ ¼ f p0 sin w
Eq. (14.2.4d) gives the modal loads
t
P1 ¼ fT1 pðt Þ ¼ f11 p0 sin w
t
P2 ¼ fT2 pðt Þ ¼ f12 p0 sin w
and Eq. (14.2.7) becomes
t,
Y€ n + 2xn wn Y n + w2n Yn ¼pn0 sin w
pn0 ¼
f1n p0
,
Mn
n ¼ 1, 2
(1)
The steady-state response is obtained from Eq. (3.2.26) by setting u ¼ Yn
t qn Þ
Yn ðt Þ ¼ rn sin ðw
where now
i
Pn0 h
2
2
2 2
1
b
+
ð
2x
b
Þ
,
rn ¼
n
n
n
Mn w2n
1
qn ¼ tan 1
(2)
!
2x n b n
,
1 b2n
bn ¼
w
wn
(3)
The displacements of the steady state response result from Eq. (14.2.1).
t q1 Þ + f2 r2 sin ðw
t q2 Þ
u ¼ f1 r1 sin ðw
(4)
The eigenfrequencies and mode shapes are taken from Example 12.2.1,
namely
f1 ¼ f 0:1697 0:0935 gT ,
w1 ¼ 8:289,
f2 ¼ f 0:1061 0:1499 gT
w2 ¼ 19:236,
M1 ¼ M2 ¼ 1
(i) x 1 ¼ 0:06, x2 ¼ 0:04. We have
b1 ¼
w
¼ 1:2064,
w1
b2 ¼
w
¼ 0:5198
w2
and Eqs. (3) give
r1 ¼ 0:02584,
r2 ¼ 0:001961,
q1 ¼ 0:3076,
q2 ¼ 0:0569
Multi-degree-of-freedom systems: Forced vibrations Chapter
Hence
u1 ðt Þ
u2 ðt Þ
14
629
0:438
¼ 102 sin ð10t + 0:3076Þ + 103
2:416
0:208
sin ð10t 0:0569Þ
0:294
(ii) x1 ¼ x2 ¼ 0. Eqs. (3) give
r1 ¼ 0:2561 102 ,
Hence
u 1 ðt Þ
u 2 ðt Þ
r2 ¼ 0:7636 103 ,
¼ 103 q1 ¼ q2 ¼ 0
0:5156
sin 10t
0:1250
14.3 Modal contribution in the mode superposition method
In the mode superposition method, the displacement vector uðt Þ results as a sum
of the individual modal displacements un , namely
uðt Þ ¼
N
X
un
(14.3.1a)
un ¼ fn Yn ðt Þ
(14.3.1b)
n¼1
The exact solution is obtained when the sum includes all the modal displacements. However, the economy of the analysis requires the use of a few modal
displacements corresponding to the lower eigenmodes, at least those whose
contribution is not negligible, while cutting off higher order modal displacements. Besides, the truncation is necessary for another important reason. The
modeling of the structure with finite elements reveals deviations between the
discretized and the actual structure, which become pronounced in higher-order
eigenmodes. Therefore, the additional error due to truncated higher eigenmodes
does not concern us very much. In order to estimate this error, we first need to
understand the contribution of each eigenmode to the overall response.
14.3.1 Modal participation
The forces produced by the modal component un ¼ fn Yn ðt Þ of the displacement u are:
un ¼ Mfn Y€ n
The inertial force: ðf I Þn ¼ M€
The damping force: ðf D Þn ¼ Cu_ n ¼ Cfn Y n
The elastic force: ðf S Þn ¼ Kun ¼ Kfn Yn
630 PART
II Multi-degree-of-freedom systems
Obviously, the resultant of the above forces equilibrates only a part pn ðt Þ of the
external force pðt Þ. Therefore, we have
pn ðt Þ ¼ ðf I Þn + ðf D Þn + ðf S Þn
¼ MY€ n + CY n + KYn fn
(14.3.2)
¼ Aðt Þfn
and consequently
pðt Þ ¼
N
X
pn ðt Þ
(14.3.3)
n¼1
If the damping matrix C is proportional, for example, C ¼ a0 M + a1 K, then
the matrix Aðt Þ satisfies the relation
Aðt Þfn ¼ MY€ n + CY n + KYn fn
h
i
(14.3.4)
¼ Y€ n + a0 + a1 w2n Y n + w2n Yn Mfn
¼ ðt ÞMfn
and Eq. (14.3.2) is written as
pn ðt Þ ¼ ðt ÞMfn
¼ ðt Þen
(14.3.5)
From the forgoing equation, it is evident that the vectors en ¼ Mfn
(n ¼ 1, 2, …, N ) do not change with time, therefore they provide a constant vector base to express the vector pðt Þ. Therefore, we can write
pðt Þ ¼
N
X
an ðt Þen
(14.3.6)
n¼1
The coefficients an ðt Þ are evaluated using the orthogonality conditions
(12.3.10)
0 i 6¼ n
(14.3.7)
fTi en ¼ fTi Mfn ¼
Mn i ¼ n
Premultiplying Eq. (14.3.6) by fTn gives
an ðt Þ ¼
fTn pðt Þ
Mn
(14.3.8)
hence
fTn pðt Þ
en
Mn
fT pðt Þ
¼ n
Mfn
Mn
pn ðt Þ ¼
(14.3.9)
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
631
The component pn ðt Þ of the load vector defined by the foregoing relation
produces the generalized forces according to Eq. (14.2.4d)
ðpn Þk ¼ fTk pn ðt Þ
¼
fTn pðt Þ T
fk Mfn ,
Mn
k ¼ 1, 2, …, N
(14.3.10)
which by virtue of the orthogonality condition yields
Pn ð t Þ k ¼ n
ðpn Þk ¼
0
k 6¼ n
That is, only the component Pn ðt Þ does not vanish. Hence, we deduce that
pn ðt Þ excites only the nth eigenmode.
We shall examine now the case where the external load is of the form
pðt Þ ¼ Rf ðt Þ
(14.3.11)
where R is a constant vector expressing the spatial distribution of the loading
and f ðt Þ a scalar function specifying the amplitude of the components pk ðt Þ
ðk ¼ 1, 2, 3, …, N Þ of pðt Þ with respect to the global system of coordinates
at time t. This type of loading is usual in structures, for example, in the case
of support excitation produced by seismic ground motion (see Section 14.7).
Substituting Eq. (14.3.11) into Eq. (14.3.3), gives
pðt Þ ¼ f ðt Þ
N
X
Rn
(14.3.12)
n¼1
and by virtue of Eq. (14.3.9)
fTn R
Mfn
Mn
¼ Gn Mfn
Rn ¼
(14.3.13)
The quantity
Gn ¼
fTn R
Mn
(14.3.14)
provides a measure for determining the degree to which the nth eigenmode participates in the dynamic response and it is called the modal participation factor
or simply the participation factor. However, this definition of Gn is not useful in
the study of the dynamic response of structures as it depends on how the modes
are normalized. Besides, it cannot be used as a measure of the modal contribution to a response quantity. Both drawbacks are overcome by introducing the
modal contribution described next [1].
When pðt Þ ¼ Rf ðt Þ, the equation of motion (14.2.7) for the modal coordinate Yn is written as
632 PART
II Multi-degree-of-freedom systems
Y€ n + 2xn wn Y n + w2n Yn ¼ Gn f ðt Þ
(14.3.15)
y€n + 2xn wn y_ n + w2n yn ¼ f ðt Þ
(14.3.16)
Yn ¼ G n y n
(14.3.17)
or
where
Eq. (14.3.16) is identical to Eq. (3.3.16) and expresses the motion of the
SDOF system when m ¼ 1, w ¼ wn , and pðt Þ ¼ f ðt Þ. The solution of
Eq. (14.3.16) gives yn ðt Þ and Eq. (14.3.1b) gives the modal displacement
un ¼ Gn fn yn ðt Þ
(14.3.18)
The corresponding elastic force is
ðf S Þn ¼ Kun
¼ KGn fn yn ðt Þ
which, taking into account that Kfn ¼ w2n Mfn , becomes
ðf S Þn ¼ Gn Mfn w2n yn ðt Þ
¼ Rn w2n yn ðt Þ
(14.3.19)
(14.3.20)
The dynamic response at time t is obtained as the static responsea when the
structure is loaded by the elastic forces. We denote by qn ðt Þ the contribution of
the nth mode to a certain quantity q ðt Þ (representing deformation or stress).
Hence qn ðt Þ will be obtained as the static response of the structure produced
by the load ðf S Þn . If qnst denotes the static response due to static load Rn , then
we may write
(14.3.21)
qn ðt Þ ¼ qnst w2n yn ðt Þ
The foregoing equation leads to the conclusion that the nth mode contribution qn ðt Þ is obtained as a result of two analyses: (1) the static analysis of the
structure subjected to the external load Rn , and (2) the dynamic analysis of the
nth mode SDOF system under the excitation force f ðt Þ. Therefore, the dynamic
analysis requires the static analysis of the structure by N loadings Rn ,
n ¼ 1, 2, …,N , and the dynamic analysis of N different SDOF systems under
the same excitation force f ðt Þ.
a. The term static response is used in the sense that time is involved as a parameter and no inertial
forces or damping forces are produced. This response is called also quasistatic.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
633
The superposition of all response contributions gives the total response
N
X
q ðt Þ ¼
qn ðt Þ
n¼1
N
X
¼
w2n yn ðt Þ
qnst
(14.3.22)
n¼1
The previous modal analysis procedure provides a means to understand the
contribution of modes to the dynamic response of a quantity and it can be measured by defining the quantity
g qn ¼
qnst
,
q st
q st ¼
N
X
qnst
(14.3.23)
n¼1
Note that q st results as the static analysis of the structure subjected to the
load R. Evidently, this definition of g qn does not require the computation of
all qnst when the response is approximated by the first few mode shapes.
The quantity g qn defined by Eq. (14.3.23) is called the nth modal contribution
factor of the quantity q. This modal contribution factor has three important
properties, which show its advantage over the modal participation factor Gn
defined by Eq. (14.3.14), namely
(1) It is by definition dimensionless.
(2) It does not depend on how the modes are normalized because qnst is due to
the load Rn , which does not depend on the mode normalization method,
and the modal properties do not enter into q st .
(3) The sum of all participation factors is equal to unity
N
X
g qn ¼ 1
(14.3.24)
n¼1
On the basis of Eqs. (14.3.21), (14.3.23), the extreme value of qn ðt Þ is
qn0 ¼ g qn q st w2n yn0
(14.3.25)
where yn0 ¼ max t jyn ðt Þj.
The static response ynst ðt Þ is obtained from Eq. (14.3.16), if the inertial and
damping forces are neglected, that is,
ynst ðt Þ ¼
f ðt Þ
w2n
(14.3.26)
f0
w2n
(14.3.27)
Hence its extreme value is
where f0 ¼ max t jf ðt Þj.
ynst
0
¼
634 PART
II Multi-degree-of-freedom systems
The quantity
Dn ¼ yn0
ynst 0
(14.3.28)
defines the dynamic magnification factor of the nth mode. Thus, Eq. (14.3.25)
by virtue of Eqs. (14.3.27), (14.3.28) becomes
qn0 ¼ g qn q st f0 Dn
(14.3.29)
Eq. (14.3.29) shows that the extreme value of the nth mode contribution to
the quantity q ðt Þ is the product of four quantities:
(1)
(2)
(3)
(4)
The dimensionless modal contribution factor g qn
the static value of q due to the loading R
the maximum value of the function f ðt Þ
the nth mode dynamic magnification factor Dn .
It is emphasized that the quantities q st and g n depend only on the spatial
distribution R of the external forces while Dn and f0 only on f ðt Þ. When the
excitation is due to seismic ground motion, the quantity w2n yn0 , according to
Eq. (6.2.21), represents the pseudoacceleration, Spa ðTn , xn Þ ¼ w2n yn0 , hence
its value can be taken from the respective response spectrum (see Fig. 6.2.6).
Example 14.3.1 The chimney of the variable cross-section shown in
Fig. E14.4a is modeled with six beam elements of constant cross-section.
Compute the modal contribution factor of (i) the base shear force Qb , (ii)
the overturning moment Mb , and (iii) the top displacement u6 when the excita t. Assume: m1 ¼ 6m, m2 ¼ 5m,
tion is due to ground motion ug ðt Þ ¼ uo sin w
m3 ¼ 4m, m4 ¼ 3m, m5 ¼ 2m, m6 ¼ m, I1 ¼ 16I , I2 ¼ 11I , I3 ¼ 7I , I4 ¼ 4I ,
I5 ¼ 2I , and I6 ¼ I :
q
q
q
q
q
q
(a)
(b)
(c)
FIG. E14.4 Chimney in Example 14.2.2 (a); FEM model (b); Degrees of freedom (c).
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
635
Solution
The equation of motion is (see Section 14.7).
M€
u + Cu_ + Ku ¼ M1u€g
T
(1)
t
where 1 ¼ f 1 1 1 1 1 1 g , u€g ðt Þ ¼ w u0 sin w
2
1. Formulation of the matrices M, K and vector R
The structure has 12 degrees of freedom, 6 translational, and 6 rotational
(Fig. E14.4c). After the static condensation of the rotational degrees of freedom,
we obtain:
2
3
6 0 0 0 0 0
60 5 0 0 0 07
6
7
60 0 4 0 0 07
7
M ¼ m6
(2)
60 0 0 3 0 07
6
7
40 0 0 0 2 05
0 0 0 0 0 1
2
3
232:80 121:78 37:89 7:11
1:21 0:15
6 121:78 122:97 69:99 21:01 3:57 0:44 7
6
7
6 37:89 69:99 72:03 39:04 10:73 1:34 7
6
7
(3)
K ¼ k6
21:01 39:04 38:38 19:04 3:88 7
6 7:11
7
4
1:21 3:57 10:73 19:04 16:00 5:00 5
0:15
0:44 1:34
3:88 5:00 2:12
where
k¼
Moreover, it is
EI
ðL=6Þ3
8 9
6>
>
>
>
> >
>
5>
>
>
>
=
< >
4
, f ðt Þ ¼ u€g ðt Þ
R ¼ M1 ¼ m
3>
>
>
>
>
>
>
>
>
>
>2>
;
:
1
2. Computation of eigenfrequencies and eigenmodes
Using the MATLAB function [V, D]¼eig (K,M), we obtain
pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
w1 ¼ 0:2030 k=m , w2 ¼ 0:7570 k=m , w3 ¼ 1:7928 k=m
pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
w4 ¼ 3:3067 k=m , w5 ¼ 5:1964 k=m , w6 ¼ 7:9089 k=m
(4)
636 PART
II Multi-degree-of-freedom systems
2
0:0161
6 0:0645
6
1 6
6 0:1469
F ¼ pffiffiffiffiffi 6
m 6 0:2640
6
4 0:4133
0:5835
0:0547
0:1747
0:2736
0:2310
0:0662
0:6010
3
0:1123 0:1665 0:2106 0:2806
0:2478 0:1858 0:0205 0:2626 7
7
0:1305 0:1863 0:2569 0:1893 7
7
7
0:2348 0:1831 0:3320 0:1067 7
7
0:2981 0:4140 0:2491 0:0498 5
0:4539 0:2807 0:1148 0:0178
(5)
3. Computation of the vectors Rn
The eigenmodes have been orthonormalized with respect to the mass. Hence,
Mn ¼ 1, n ¼ 1, 2, …, 6. Then Eq. (14.3.14) gives
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
G1 ¼ 3:2094 m , G2 ¼ 2:2555 m , G3 ¼ 1:5881 m
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
G4 ¼ 1:1813 m , G5 ¼ 0:9509 m , G6 ¼ 0:8893 m
The vectors Rn are computed from Eq. (14.3.13)
R1 =m R2 =m R3 =m R4 =m R5 =m R6=m
0:311 0:739 1:070 1:180 1:201 1:497
1:036 1:970 1:967 1:098 0:097 1:168
1:886 2:468 0:829 0:880 0:977 0:673
2:542 1:563 1:119 0:649 0:947 0:284
2:653 0:298 0:947 0:978 0:474 0:088
1:873 1:355 0:721 0:333 0:109 0:016
st
st
st
, Mbn
, and u6n
4. Computation of Qbn
The base shear is obtained as the sum of all the elastic forces while the overturning moment as the sum of the moments of all the elastic forces with respect
to the base of the chimney. Thus, we have
st
¼
Qbn
6
X
Rkn ¼ 1T Rn
k¼1
st
Mbn
¼
6
X
hk Rkn ¼ hT Rn ,
k¼1
L
hT ¼ f 1 2 3 4 5 6 g
6
The vector ust
n is computed by considering the static analysis of the structure
Gn
1
subjected to the external force Rn , that is, ust
n ¼ K Rn ¼ w2 fn . Hence
n
st
u6n
¼
Gn
f :
w2n 6n
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
637
Finally, the modal contribution factors are computed from Eq. (14.3.23):
b
gQ
n ¼
st
Qbn
,
6
X
st
Qbk
b
gM
n ¼
k¼1
st
Mbn
,
6
X
st
Mbk
g un6 ¼
st
u6n
6
X
st
u6k
k¼1
k¼1
Their values are given in Table E14.1. From this table, we deduce that each
quantity requires a different number of contributing modes to achieve a specified degree of approximation. For example, four modes for the base shear, two
modes for the overturning moment, and one mode for the top displacement
should contribute in order to achieve an approximation over 90%.
TABLE E14.1 Modal contribution factors in Example 14.3.1.
Base shear force
Number
of mode
shape n
b
gQ
n
n¼1
1
0.4905
2
K
P
b
gQ
n
Overturning moment
K
P
b
gM
n
Top displacement
K
P
g un6
g un6
b
gM
n
n¼1
0.4905
0.7627
0.7627
1.0500
1.0500
0.2423
0.7328
0.1556
0.9183
0.0549
0.9954
3
0.1201
0.8529
0.0466
0.9648
0.0052
1.0006
4
0.0664
0.9193
0.0186
0.9834
0.0007
0.9999
5
0.0430
0.9623
0.0096
0.9930
0.0000
1.0000
6
0.0377
1.0000
0.0070
1.0000
0.0000
1.0000
n¼1
Example 14.3.2 For the structure in Example 14.3.1, compute the extreme
¼ 1:5w1 , m ¼ 31:056 kN m1 s2 ,
values Qbn0 , Mbn0 , u6n0 when w
k ¼ 5486:208 kN=m, L ¼ 75 m, x1 ¼ 0:06, and x2 ¼ x3 ¼ ⋯ ¼ x6 ¼ 0:04,
u0 ¼ 1.
Solution
The extreme value qn0 of a quantity q is obtained from Eq. (14.3.29), namely
qn0 ¼ g qn q st f0 Dn
The modal contribution factors g qn (q Qb , Mb , u6 ) were computed in
t, hence
2 sin w
Example 14.3.1 and are given in Table E14.1. It is f ðt Þ ¼ u0 w
2
st
st
. The quantities Qb , Mb are obtained from the relations:
f0 ¼ max jf ðt Þj ¼ u0 w
Qbst ¼
6
X
k¼1
Rk ¼ 1T R ¼ 652:1760 kN
(1)
638 PART
II Multi-degree-of-freedom systems
Mbst ¼
6
X
hk Rk ¼ hT R ¼ 21739:2 kN m
(2)
k¼1
The vector ust is obtained from the relation
ust ¼ K1 R ¼ f 0:0088 0:0331 0:0711 0:1209 0:1802 0:2451 gT
Hence
u6st ¼ 0:2451
(3)
The dynamic magnification factors Dn are computed from Eq. (14.3.28).
This requires the solution of Eq. (14.3.16), which is given by Eq. (14.2.8).
The initial conditions for the relative displacements are:
wu0 1:
uð0Þ ¼ utot ð0Þ ug ð0Þ ¼ 0 u_ ð0Þ ¼ u_ tot ð0Þ u_ g ð0Þ ¼ Therefore, the initial conditions yn ð0Þ, y_ n ð0Þ result from Eqs. (14.2.9a),
(14.2.9b) with Mn ¼ 1, n ¼ 1, 2, …,6:
yn ð0Þ ¼
y_ n ð0Þ ¼ Y n ð0Þ ¼
Yn ð0Þ fTn Muð0Þ
¼
¼ 0,
Gn
Gn
T
_ g ð0Þ
f n M u
Gn
¼ wu0
n ¼ 1, 2, …, 6
fTn M1
,
Gn
n ¼ 1, 2, …,6
(4a)
(4b)
or taking into account that Gn ¼ fTn R ¼ fTn M1, we obtain
wu0 , n ¼ 1, 2, …,6
y_ n ð0Þ ¼ (5)
t
2 sin w
Then Eq. (14.2.8) gives for Mn ¼ 1, Pn ðt Þ ¼ u0 w
yn ðt Þ xn wn t
w
¼e
sin wDn t
u0
wDn
Z
2 t
w
τexn wn ðtτÞ sin ½wDn ðt τÞdτ
sin w
wDn 0
(6)
The eigenfrequencies wn are taken from Example 14.3.1, which for data of
this problem become
w1 ¼ 2:6977,
w2 ¼ 10:0608,
w3 ¼ 23:8283
w4 ¼ 43:9498,
w5 ¼ 69:0658,
w6 ¼ 105:1181
14
639
R2(t)
R1(t)
Multi-degree-of-freedom systems: Forced vibrations Chapter
t
R4(t)
R3(t)
t
t
R6(t)
R5(t)
t
t
t
FIG. E14.5 Time history of the response ratios and modal magnification factors in Example 14.4.2.
Eq. (6) demands the analytical evaluation of the Duhamel integral, which
leads to a complicated expression. To avoid this, the solution is obtained numerically using the program called aem_lin.m described in Section 4.4 while the
modal dynamic magnification factors are obtained from the relation
Dn ¼ max ðjRn jÞt , where Rn ðt Þ ¼ yn =ðyn Þst ðn ¼ 1, 2, …, 6Þ represents the
respective response ratio within the integration interval. Their graphical
640 PART
II Multi-degree-of-freedom systems
representation is shown in Fig. E14.5, in which the dynamic magnification factors are annotated. Table E14.2 shows the extreme values Qbn0 , Mbn0 , u6n0
TABLE E14.2 Extreme values of Qbn0 , Mbn0 , and u6n0 in Example 14.3.2.
Qbn0 ¼
st
b
gQ
n Qb f0 Dn
Mbn0 ¼
st
b
gM
n Mb f0 Dn
2.084
10916.50
565819.0
8.78
0.171
2.575
6663.12
142631.0
0.56
3
0.067
5.628
7218.45
93361.1
0.11
4
0.036
10.263
7277.62
67953.7
0.03
5
0.023
16.082
7385.09
54958.8
0.00
6
0.016
24.435
9837.86
60888.6
0.00
n
tn
1
1.819
2
Dn
u6n0 ¼
g un6 u6st f0 Dn
Example 14.3.3 The chimney of Example 14.3.1 is subjected to horizontal
loading pðt Þ ¼ Rf ðt Þ, where R ¼ lfk , that is, the spatial distribution is proportional to the kth eigenmode when l is a constant. Compute the modal contribution factors of the base shear when k ¼ 1, 2.
Solution
The vector Rn is obtained from Eq. (14.3.13), namely
Rn ¼ Gn Mfn
where
Gn ¼ fTn R ¼ lfTn fk
n ¼ 1, 2, …,6
In general it is
6¼ 0. Hence, Rn 6¼ 0 and the loading pðt Þ excites all the
mode shapes, in contrast to the free vibrations where only the corresponding
mode shape is excited when the initial conditions are proportional to a certain
of its mode shapes (see Example 12.9.2). The mass matrix, the stiffness matrix,
and the mode shapes are taken from Example 14.3.1.
fTn fk
(i) R ¼ lf1
Computation of Rn and g Q
n
The mode shapes have been orthonormalized with respect to the mass, hence
Mn ¼ 1, n ¼ 1, 2, …, 6. Eq. (14.3.14) gives:
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
G1 ¼ 0:6071l m , G2 ¼ 0:2647l m , G3 ¼ 0:1167l m
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
G4 ¼ 0:0537l m , G5 ¼ 0:0186l m , G6 ¼ 0:0025l m
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
641
The vectors Rn are obtained from Eq. (14.3.13):
lR1 =m lR1 =m lR1 =m lR1 =m lR1 =m lR1 =m
0:0588 0:0868 0:0786 0:0536 0:0235 0:0043
0:1959 0:2312 0:1445 0:0499 0:0019 0:0034
0:3568 0:2897 0:0609 0:0400 0:0191 0:0019
0:4808 0:1835 0:0822 0:0295 0:0185 0:0008
0:5018 0:0351 0:0695 0:0445 0:0093 0:0002
0:3542 0:1591 0:0529 0:0151 0:0021 0:0000
The contribution factors are obtained from Eq. (14.3.23)
gQ
n ¼ ½1:30894, 0:40112, 0:12445, 0:04263, 0:01188, 0:00153
(ii) R ¼ lf2
Computation of Rn and g Q
n
Eq. (14.3.14) gives:
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
G1 ¼ 0:2647l m , G2 ¼ 0:5273l m , G3 ¼ 0:22218l m
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
G4 ¼ 0:0895l m , G5 ¼ 0:0309l m , G6 ¼ 0:0040l m
Eq. (14.3.13) gives:
lR1 =m
0:0256
0:0854
0:1555
0:2096
0:2188
0:1544
lR2 =m lR2 =m lR2 =m lR2 =m lR2 =m
0:1729 0:1497 0:0894 0:0391 0:0067
0:4605 0:2752 0:0832 0:0031 0:0052
0:5770 0:1159 0:0667 0:0317 0:0030
0:3654 0:1565 0:0491 0:0308 0:0012
0:0698 0:1324 0:0741 0:0154 0:0003
0:3168 0:1008 0:0251 0:0035 0:0000
and using Eq. (14.3.23), we obtain
gQ
n ¼ ½12:7194, 17:8053, 5:2824, 1:5837, 0:4405, 0:0532
14.3.2 Static correction method
When the time function f ðt Þ in Eq. (14.3.11) is harmonic, that is,
t, the dynamic load magnification factor Dn approaches the unity
f ðt Þ ¼ f0 sin w
, where Tn ¼ 2p=wn is the nat =wn ¼ Tn =T
for small values of the ratio b n ¼ w
ural period of the nth eigenmode and T ¼ 2p=
w is the period of the loading. This
becomes evident from Fig. 3.2.3. The same applies when the ratio Tn =ttot is
very small or the ratio ttot =Tn is too large, where ttot is the duration of loading.
This is shown in Fig. 14.3.1, which represents the variation of the dynamic magnification factor Dn versus the ratio ttot =Tn when the loading is the acceleration
of the Athens earthquake, September 7, 1999, f ðt Þ ¼ u€g ðt Þ, of total duration
642 PART
II Multi-degree-of-freedom systems
ttot ¼ 39 s. We observe that for ttot =Tn > 1000, hence for periods Tn < 0:039 s
or eigenfrequencies wn > 160s1 , we may ignore the dynamic effect and
assume Dn 1 for eigenmodes of order n > Nd , where Nd is the number of
eigenmodes with Tn > 0:039 s and Dn noticeably greater than the one. This
observation is the quintessence of the static correction method, presented next.
Suppose that we use N eigenmodes to approximate the dynamic response of
a quantity q ðt Þ. We split the contribution of eigenmodes into two parts. The first
Nd eigenmodes with Dn greater than one and the remaining N Nd with Dn
close to one. Then we can write
q ðt Þ ¼
Nd
X
N
X
qnst w2n yn ðt Þ +
n¼1
qnst w2n ynst ðt Þ
(14.3.30)
n¼Nd + 1
where yn ðt Þ is taken from the solution of Eq. (14.3.16) and ynst ðt Þ from
Eq. (14.3.26). Substituting Eq. (14.3.26) into Eq. (14.3.30) and using
Eq. (14.3.23) we obtain
q ðt Þ ¼
Nd
N
X
X
qnst wn2 yn ðt Þ + f ðt Þ
qnst
n¼1
(
¼q
st
n¼Nd + 1
Nd
X
gqn
wn2 yn ðt Þ
+ f ðt Þ
n¼1
)
N
X
(14.3.31)
gqn
n¼Nd + 1
3.5
x =0.05
x =0.1
x =0.15
3
Dn
2.5
2
1.5
1
0.5
0
0
200
400
600
800 1000
t tot /Tn
1200
1400
1600
FIG. 14.3.1 Variation of the dynamic magnification factor Dn versus the ratio ttot/Tn
Eq. (14.3.24) gives
N
X
n¼Nd + 1
g qn ¼ 1 Nd
X
n¼1
g qn
(14.3.32)
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
643
which is substituted into Eq. (14.3.31) to give
(
q ðt Þ ¼ q st
Nd
X
g qn w2n yn ðt Þ + f ðt Þ 1 n¼1
Nd
X
!)
g qn
(14.3.33)
n¼1
or
(
q ðt Þ ¼ q st f ðt Þ +
Nd
X
g qn w2n yn ðt Þ f ðt Þ
)
(14.3.34)
n¼1
It becomes obvious from the foregoing equation that the computation of q ðt Þ
requires the determination of only Nd eigenfrequencies and eigenmodes.
In Eq. (14.3.33), the term
!
Nd
X
q
gn
(14.3.35)
f ðt Þ 1 n¼1
represents the static correction that should be added to the dynamic solution in
order to obtain the total dynamic response. This method is known as the static
correction method.
This static correction method is very useful when the contribution of
higher eigenmodes to a quantity q ðt Þ cannot be neglected in dynamic analysis,
but the corresponding dynamic magnification factors produced by the excitation force f ðt Þ are close to one. In such cases, the combination of the dynamic
response of a few lower eigenmodes together with the static correction will
give results, which adequately approximate the response obtained by the mode
superposition method. The advantage of the static correction method offers a
significant reduction in the computational task, especially when the equations
are solved numerically. The example that follows illustrates the merits of this
method.
Example 14.3.4 The chimney in Example 14.3.1 is subjected to the wind blast
load pðt Þ ¼ Rf ðt Þ R ¼ f 1:5 6:0 15:0 25:0 40:0 60:0 gT . The time function
f ðt Þ is given in Fig. E14.6 (p0 ¼ 1 t1 ¼ 0:1). Compute the response of the base
shear force, the overturning moment, and the top displacement using the exact
mode superposition method as well as the static correction method by including
the dynamic response of the eigenmodes with Dn > 1:15. Compare the extreme
values obtained from the two methods. The equation of motion will be solved
numerically to obtain yn ðt Þ. L ¼ 75 m, x1 ¼ 0:06, x2 ¼ x3 ¼ ⋯ ¼ x6 ¼ 0:04,
m ¼ 31:056 kN m1 s2 , and k ¼ 5486:208 kN=m.
II Multi-degree-of-freedom systems
f(t)
644 PART
t
FIG. E14.6 Time variation of the blast load in Example 14.3.4.
Solution
The mass matrix, stiffness matrix, eigenfrequencies, and mode shapes are taken
from Example 14.3.1 for the given values of k and m. The vectors Rn are
obtained from Eq. (14.3.13) with Mn ¼ 1. Thus, we obtain
R1
5:888
19:604
35:708
48:124
50:225
35:456
R2
9:085
24:191
30:311
19:197
3:6693
16:644
R3
R4
R5
R6
8:7984 6:284 2:279 0:096
16:170 5:843 0:184 0:075
6:8147 4:685 1:853 0:043
9:196
3:454 1:796 0:018
7:782 5:206 0:899 0:005
5:925
1:765 0:207 0:001
st
st
st
The quantities Qbn
, Mbn
, and u6n
are computed from the relations
st
¼
Qbn
6
X
Rkn ¼ 1T Rn
k¼1
st
Mbn
¼
6
X
hk Rkn ¼ hT Rn ,
k¼1
st
u6n
¼
L
hT ¼ f 1 2 3 4 5 6 g
6
Gn
f
w2n 6n
The modal contribution factors of Qb , Mb are obtained from Eq. (14.3.23). The
computed values are given in Table E14.3. The dynamic magnification factors
are computed from the relation Dn ¼ max jRðt Þj, where Rðt Þ ¼ yn ðt Þ= ynst 0 is
t
the response ratio of the nth mode and ynst 0 ¼ f0 =w2n . For the given loading it
is f0 ¼ 1. Hence
Multi-degree-of-freedom systems: Forced vibrations Chapter
st y1 0 ¼ 0:1374,
y2st
st y4 0 ¼ 0:5170 103 ,
0
¼ 0:9870 102 ,
y3st
st y5 0 ¼ 0:2096 103 ,
0
645
14
¼ 0:1761 102
y6st
0
¼ 0:9045 104
TABLE E14.3 Modal contribution factors in Example 14.3.4.
Base shear force Qb
Mode
shape n
K
P
b
gQ
n
n¼1
b
gQ
n
Overturning moment
Mb
K
P
b
gM
n
n¼1
b
gM
n
Top
displacement u6
K
P
g un6
n¼1
g un6
1
1.3221
1.3221
1.1254
1.1254
0.9652
0.9652
2
0.4235
0.8985
0.1489
0.9765
0.0325
0.9977
3
0.1405
1.0391
0.0298
1.0063
0.0021
0.9998
4
0.0503
0.9887
0.0077
0.9986
0.0002
0.9999
5
0.0116
1.0003
0.0014
1.00004
0.0000
0.9999
6
0.0003
1.0000
0.00004
1.0000
0.0000
1.000
Qbst
¼ 147:5 kN
Mbst
¼ 8981:25 kN m
u6st
¼ 0:156 m
The values max jyn ðt Þj are obtained from the solution of Eq. (14.3.16),
t
where f ðt Þ is obtained from Fig. E14.6. The initial conditions are
uð0Þ ¼ u_ ð0Þ ¼ 0, which yields yn ð0Þ ¼ y_ n ð0Þ ¼ 0. The solution of the equation
of motion is obtained numerically using the program aemlin.m developed in
Section 4.4. Thus, we obtain
D1 ¼ 0:2461,
D2 ¼ 0:8803,
D3 ¼ 1:4990
D4 ¼ 1:4787,
D5 ¼ 1:1482,
D6 ¼ 1:1169
We observe that the dynamic magnification factors of the last two modes
are close to one, so we can include the two higher modes in the static correction and obtain the total response from Eq. (14.3.34) with Nd ¼ 4. That is
(
)
4
X
st
q
2
g n wn yn ðt Þ f ðt Þ
q ðt Þ ¼ q f ðt Þ +
n¼1
646 PART
II Multi-degree-of-freedom systems
Qb (t)
Static correction
Exact
t
Mb (t)
Static correction
Exact
t
u6(t)
Static correction
Exact
t
FIG. E14.7 Time history of Qb , Mb , u6 .
where q Qb ,Mb , u6 . Fig. E14.7 gives the graphical representation of
Qb ,Mb ,u6 . They were obtained from the foregoing equation as well as from
the exact method, that is, by including all eigenmodes. The extreme values
max jQb j, max jMb j, and max ju6 j together with the instants tmax they occur
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
647
are given in Table E14.4. It becomes obvious that the static correction method
provides very good results.
TABLE E14.4 Extreme values max jQb j, max jMb j, max ju6 j in Example 14.3.4.
Base shear force
max jQb j
Extreme
value
tmax
Overturning moment
max jMb j
Top displacement
max ju6 j
Exact
Static
correction
Exact
Static
correction
Exact
Static
correction
92.4876
91.500
3341.2
3341.4
0.0384
0.0384
0.634
0.634
0.751
0.751
0.620
0.620
14.3.3 Error in mode superposition method due to truncation
of higher modes
From the results in Examples 14.3.1 and 14.3.3, we conclude that for a given
spatial distribution R of the external load, the contribution of eigenmodes is
not the same for all quantities q ðt Þ. For instance, in Example 14.3.1, the contribution of one eigenmode of the top displacement is sufficient to reach
100% of the total response while the contribution of five eigenmodes of the base
shear force and three eigenmodes of the overturning moment are required to
cover 95% of their total response. Similar findings result from Example
14.3.3, that is, more eigenmodes of the base shear force than of the top displacement or the overturning moment should contribute in order to obtain the same
approximation of the response. However, we cannot reach decisive conclusions
because the above findings apply to structures of a specific geometry and support conditions (here, cantilever). But even in such cases, as can be seen from
Eq. (14.3.22), the modal contribution depends also on the factor w2n yn ðt Þ, that is,
the pseudoacceleration, which is influenced by the dynamic characteristics of
the eigenmode and the time function f ðt Þ with an extreme value f0 Dn . The
results in Example 14.3.2 illustrate this influence. Therefore, the answer to
the question as to how many eigenmodes must be included in the dynamic analysis by the mode superposition method to obtain an accurate solution is: Include
all of them at the first instance.
However, the need to reduce the computational task requires the truncation
of eigenmodes. We can adequately approximate the value of a quantity q ðt Þ by
including modes with a large modal contribution factor g qn and a dynamic magnification factor Dn appreciably greater than one. For the eigenmodes with large
values of g qn but with Dn 1, the method of static correction limits the dynamic
analysis to a few lower eigenmodes. It should be noted that the extreme value
648 PART
II Multi-degree-of-freedom systems
max jq ðt Þj cannot be obtained as a superposition of max jqn ðt Þj ¼ q st g qn f0 Dn
t
t
because the values Dn do not occur at the same time. A measure to estimate
the truncation error produced by omitting higher-order eigenmodes may result
from the following consideration.
Let K N be the number of eigenmodes that we include in the superposition. This means that the displacement vector is approximated by the sum
u¼
K
X
un
n¼1
¼
K
X
(14.3.36)
fn Yn
n¼1
Substituting u into Eq. (14.1.1) gives
K X
MY€ n + CY n + KYn fn ¼ e
pðt Þ
(14.3.37)
n¼1
Obviously, it is e
pðt Þ 6¼ pðt Þ, when K < N . Taking into account Eq. (14.3.2),
Eq. (14.3.37) becomes
K
X
pn ðt Þ ¼ e
pðt Þ
(14.3.38)
n¼1
Namely, the K < N modal forces equilibrate a part of the external force
pðt Þ. This results in an error
e¼pe
p
(14.3.39)
Then the error norm eK for e can be defined as
eK ¼
pT e
pT p
(14.3.40)
This gives eK ¼ 0, if K ¼ N , and e ¼ 1, K ¼ 0.
Obviously, when the loading is given by Eq. (14.3.11), the error takes the form
eK ¼
RT e
RT R
(14.3.41)
where now
e¼R
K
X
Rn
n¼1
K
X
¼R
Gn Mfn
(14.3.42)
n¼1
Table E14.5 gives the mode truncation error in Examples 14.3.1 and 14.3.3.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
649
TABLE E14.5 Truncation error e ¼ RT e=RT R.
Example 14.3.3
Number
of modes K
Example 14.3.1
k ¼1
1
0.6770
3.275e5
2
0.3814
2.325e5
5.409e5
3
0.2160
2.000e5
8.891e5
4
0.1201
2.008e5
9.377e5
5
0.0565
2.057e5
9.505e5
6
3.59e11
2.048e5
9.485e5
k ¼2
1.000
14.4 Reduction of the dynamic degrees of freedom
The modeling of structures with finite elements leads to MDOF systems with a
large number of degrees of freedom. Although in the static analysis, a fine discretization is necessary to determine accurately the deformation and stress of
the structure, in dynamic analysis reliable results can be obtained without a fine
discretization. On the contrary, the large number of degrees of freedom makes
the dynamic analysis laborious and complicated because it requires the determination of a large number of eigenfrequencies and eigenmodes, whose contribution to the response may be negligible.
We can use the discretization employed for the static analysis also for the
dynamic analysis, but after reducing considerably the degrees of freedom.
The usual methods of achieving that are: (a) by the static condensation,
(b) by imposing kinematic constraints, and (c) by the Rayleigh-Ritz method.
14.4.1 Static condensation
In the directions of some displacements of a structure, the inertial forces are zero
or very small, and therefore they can be neglected. The equations of motion in
these directions degenerate into static equations and are used to eliminate the
corresponding displacements. This is common when simulating structures with
finite elements. Because of the lumped assumption, the rotational inertia of the
concentrated masses is zero, hence the corresponding inertial quantities are also
zero. Therefore, the rotational degrees of freedom, though being necessary to
approximate accurately the stiffness of the structure, have a negligible contribution to the dynamic response. Apparently, static condensation significantly
reduces the degrees of freedom. The static condensation and its implementation
procedure have already been described in detail in Section 11.4.
650 PART
II Multi-degree-of-freedom systems
14.4.2 Kinematic constraints
The geometry of a structure and the properties of parts or members of it allow
for the introduction of the kinematic constraints, which express the displacements of certain degrees of freedom in terms of a smaller set of them. We
encountered this problem in Section 11.7, where we discussed the axial constraints in frames, or in Section 11.11, where the general problem of rigid bodies
in flexible structures was studied. The importance of this reduction is illustrated
with the following two examples.
We consider the frame plane of Fig. 14.4.1, in which the infill wall B contributes to the stiffness of the frame. The frame without the wall has 9 3 ¼ 27
degrees of freedom. The response of the wall is that of a plane elastic body.
Because its deformation is very small compared to that of the frame, it can
be considered as a plane rigid body. Therefore, the nodes a,b,c, d are constrained by the rigid body B and the 4 3 degrees of freedom are reduced to
3, thus the remaining degrees of freedom are 18ð¼ 5 3 + 3Þ. The static condensation of the 5 unconstrained rotational degrees of freedom reduces the
degrees of freedom to 18 5 ¼ 13. Finally, the axial constraints of the 11 beam
elements reduce the degrees of freedom to 13 11 ¼ 2. Hence, the remaining
degrees of freedom are the two horizontal displacements u1 ð¼ u2 Þ and u3 .
FIG. 14.4.1 Plane frame with an infill wall.
As a second example, we consider the 3-story building of Fig. 14.4.2 consisting of 6 plane frames, 3 in the x-direction and 3 in the y-direction, and 3 horizontal slabs (plates). The 9 vertical columns together with the beams
supporting the plates are simulated with a space frame of total nodes n ¼ 27,
each having 6 degrees of freedom, 3 translational and 3 rotational. Thus the
number of degrees of freedom are N 0 ¼ 27 6 ¼ 162. The floor slabs, also
called diaphragms, although flexible in the vertical direction, are usually very
stiff in their own plane and can be assumed as rigid plane bodies. This
assumption restrains the displacements in the xy-plane and rotations about
the vertical axis at each floor level and reduces their number to 3, that is,
equal to the degrees of freedom of a plane rigid body moving in its plane
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
651
(see Example 1.5.6). Therefore, 3 9 ¼ 27 degrees of freedom within the xy
plane are reduced to 3 and the total degrees of freedom are reduced to
N 00 ¼ 162 3 3 9 + 3 3 ¼ 90. Further, neglecting the axial deformation
of the columns, an assumption valid for a building of small height compared
to the other two dimensions, introduces additional 3 9 ¼ 27 constraints, which
reduce the degrees of freedom to N 000 ¼ 90 3 9 ¼ 63. Finally, static condensation allows the elimination of the rotational degrees of freedom about x and y,
which amass 2 9 ¼ 18 degrees of freedom at each floor. Hence, the active
degrees of freedom are limited to N ¼ 63 3 2 9 ¼ 9, that is, equal to the
number of degrees of freedom of the three slabs. The procedures of reducing
the degrees of freedom were presented in Section 11.6.
FIG. 14.4.2 Three-story building with diaphragms.
14.5 Rayleigh-Ritz method
14.5.1 Ritz transformation
In Section 10.5.1, we described the Ritz method as the method of global shape
functions to approximate the dynamic response of continuous systems. In that
method, the function u ðx, t Þ representing the displacement of the points of the
structure was approximated by the superposition
u ðx, t Þ K
X
i ðx Þui ðt Þ
(14.5.1)
i¼1
where i ðx Þ are functions representing deformation patterns of the structure and
ui ðt Þ time functions that play the role of generalized coordinates and can represent displacements of certain points or rotations of cross-sections. In order
that function u ðx, t Þ approximates the actual deformation of the structure,
the functions i ðx Þ must be (i) geometrically admissible and (ii) linearly independent. Besides, they must be continuously differentiable up to the degree
required by the equation in which they are involved.
The Ritz method can also be applied to MDOF systems. Indeed, its application to these systems does not encounter the difficulties of continuous systems,
where the establishment of shape functions is a very difficult mathematical
652 PART
II Multi-degree-of-freedom systems
problem, especially for bodies of arbitrary shape [2]. In MDOF systems, the
shape functions are replaced by a set of discrete displacements, which define
admissible deformation patterns of the structure. Ritz applied this method in
1909 as an extension of Rayleigh’s method to establish the lower eigenfrequencies, therefore it is also known as the Rayleigh-Ritz method.
For that reason, in the Ritz method for the MDOF systems, the displacement
vector is expressed as the superposition of a number of linearly independent
vectors representing possible deformation patterns of the structure. These vectors are referred to as Ritz vectors. Thus, we can write
u ¼ y1 z1 ðt Þ + y2 z2 ðt Þ + ⋯ + yK zK ðt Þ
¼ xz
(14.5.2)
where z represent the vector of K generalized coordinates and C the matrix of
the Ritz vectors with dimensions N K ðK N Þ. The transformation defined
by Eq. (14.5.2) is known as the Ritz transformation. Apparently, the transformation u ¼ FY we met in Section 12.9 represents a special case of the Ritz
transformation when the vibration modes are used as the Ritz vectors.
Substituting the vector u from Eq. (14.5.2) into Eq. (14.1.1) gives
Mx€z + Cx z_ + Kxz ¼pðt Þ
(14.5.3)
which, when premultiplied by x , gives
T
e z + Kz
e z + C_
e ¼e
M€
pðt Þ
(14.5.4)
where
e ¼ x T Cx, K
e ¼ x T Kx, e
e ¼ x T Mx, C
pðt Þ ¼ x T pðt Þ
M
(14.5.5)
Eq. (14.5.4) represents a system of K differential equations for the generalized coordinates zk ðt Þ. The Ritz vectors generally do not satisfy the orthogonality conditions, so consequently the matrices defined by Eq. (14.5.5) are not
diagonal; hence Eq. (14.5.4) is coupled. Nevertheless, their solution can be
obtained using the method of superposition presented in Section 14.2. Thus,
e n and eigenmodes zn from
we have first to establish the eigenfrequencies w
the solution of the eigenvalue problem
e w
e z¼0
e2 M
K
(14.5.6)
Then the transformation
e ðt Þ
z ¼ ΖY
(14.5.7)
reduces Eq. (14.5.4) to K SDOF equations, whose solution gives the modal
e 2 ðt Þ ⋯ Y
e K ðt ÞgT .
e ð t Þ ¼ fY
e 1 ðt Þ Y
coordinates Y
The natural displacements result from Eq. (14.5.2), which by virtue of
Eq. (14.5.7) gives
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
653
e ðt Þ
u ¼ xΖY
eY
e ðt Þ
¼F
(14.5.8)
e ¼ xΖ
F
(14.5.9)
e n ¼ xz
f
n
(14.5.10)
where it was set
or
The latter equation is analogous to Eq. (14.5.2) and expresses the Ritz transformation of the mode shapes of the eigenvalue problem (14.5.6). We can reade n satisfy the orthogonality
ily show that the transformed eigenmodes f
conditions with respect to the original mass and stiffness matrices M and K,
namely
e i ¼ 0,
e T Mf
f
n
e T Kf
e i ¼ 0, n 6¼ i
f
n
(14.5.11)
Indeed, we have
e T Mf
e i ¼ xzn T M xzi
f
n
¼ zTn x T Mx zi
e i
¼ zTn Mz
¼0
(14.5.12)
because zn and zi are mode shapes of the eigenvalue problem (14.5.6). Similarly, we prove the second orthogonality condition. The orthogonality with
respect to the damping matrix holds only if it is a proportional damping matrix
as discussed in Section 12.11. If the eigenmodes are orthonormalized with
respect to mass, we have
e n ¼ 1,
zTn Mz
e n ¼w
e 2n
zTn Kz
(14.5.13)
e T Kf
en ¼ w
e 2n
f
n
(14.5.14)
A consequence of this is
e n ¼ 1,
e T Mf
f
n
It is reasonable to raise the question of why we use the Rayleigh-Ritz method
because it also demands the solution of an eigenvalue problem. The reason is
because the Rayleigh-Ritz method has important advantages over the mode
superposition method, the most noteworthy of which is the significant reduction
in the number of equations of motion if the number K of the Ritz vectors is
taken sufficiently smaller than N . The obtained solution by this method is
approximate in nature. Therefore, to be able to use the Rayleigh-Ritz method,
we have to answer first the following questions:
654 PART
II Multi-degree-of-freedom systems
1. How good is the dynamic solution approximated when it is obtained using
the Ritz method?
2. How can we derive Ritz vectors for a given structure?
The answers are given immediately below.
14.5.2 Approximation using Ritz vectors
We assume that the displacement vector u defined by Eq. (14.5.2) represents a
deformation configuration of the structure. The Rayleigh quotient corresponding to this vector is given by Eq. (12.6.21). Hence
uT Ku
uT Mu
zT x T Kxz
¼ T T
z x Mxz
e
zT Kz
¼
e
zT Mz
rðuÞ ¼
or
r ð zÞ ¼
e
zT Kz
T
e
z Mz
(14.5.15)
The value of the Rayleigh quotient is altered by changing the selected deformation pattern, which occurs when one or more of the components of the vector
e and M
e are positive definite, hence rðzÞ has finite
z change. The matrices K
values and according to Eq. (12.6.2), we have
0 < l1 rðuÞ lN < 1
(14.5.16)
We know that in the neighborhood of an eigenvalue, rðzÞ takes a stationary
value (see Section 12.6), which is a minimum near all eigenvalues except near
lN , where it is a maximum. The condition ensuring this is
∂r
¼ 0,
∂zk
or
k ¼ 1, 2, …, K
Te
e
∂ zT Mz
e
1 ∂ z Kz
zT Kz
¼0
2
e
∂zk
∂zk
zT Mz
e
zT Mz
or by virtue of Eq. (14.5.15)
e
∂ zT Kz
∂zk
r
e
∂ zT Mz
∂zk
¼0
(14.5.17)
(14.5.18)
(14.5.19)
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
655
e and M
e are symmetric and that
Taking into account that the matrices K
T
∂z=∂zk ¼ ∂z =∂zk ¼ I, we obtain
e
∂ zT Kz
∂zT e
e ∂z
¼
Kz + zT K
(14.5.20a)
∂zk
∂zk
∂zk
e
¼ 2Kz
and similarly
e
∂ zT Mz
∂zk
¼
∂zT e
e ∂z
Mz + zT M
∂zk
∂zk
(14.5.20b)
e
¼ 2Mz
Substituting Eqs. (14.5.20a), (14.5.20b) into Eq. (14.5.19) gives
e rM
e z¼0
K
(14.5.21)
which proves that the stationary values of r are the eigenvalues of the transformed eigenvalue problem (14.5.21). The eigenvalue problem is solved using
any of the methods presented in Chapter 13. The eigenvalues give the
eigenfrequencies
pffiffiffiffiffi
en ¼ rn , n ¼ 1, 2, …, K
(14.5.22)
w
and the eigenvector zn gives the natural mode
e n ¼ xz
f
n
(14.5.23)
e n are the best approximation to the corresponding
e n and f
The quantities w
actual ones. It is observed (see Example 14.6.1) that the accuracy of the approximation is better for the lower eigenfrequencies and mode shapes. According to
what we stated in Section 12.6, it holds
e1 ,
w1 w
e2 , …
w2 w
eK 1
wK 1 w
and
e K wK
w
(14.5.24)
14.6 Selection of Ritz vectors
The efficiency of the Rayleigh-Ritz method depends on the appropriate selection of the Ritz vectors so that they can approximate the natural vibration
modes. Various methods have been developed for their selection. Two of them
are presented below. The first method is based on the knowledge of the shapes
of natural modes while in the second method the Ritz vectors are derived by a
computational process.
656 PART
II Multi-degree-of-freedom systems
14.6.1 Method of natural mode shapes
In this method, we select the shapes of natural modes as Ritz vectors. Thus n is
selected so that it has the shape of fn . The eigenmode fn is not computed by
solving the eigenvalue problem, but its shape is known from the analysis of
similar structures. For example, we know that the first two natural modes of
a multistory shear frame have the shapes shown in Fig. 14.6.1. The components
i1 and
i2 are arbitrary but with the limitation that their graphs give the
shapes of 1 and 2 shown in Fig. 14.6.1. For example, we can take
T
T
1 ¼ f 0:25 0:5 0:75 1:00 g and 2 ¼ f 0:3 0:5 0:20 1:00 g .
FIG. 14.6.1 Shapes of natural modes of a multistory shear frame.
Apparently, the method of natural mode shapes requires the experience of its
user. It is clear that this method becomes difficult for complicated structures and
practically impossible for three-dimensional structures. The following example
illustrates the implementation of the method of natural mode shapes.
Example 14.6.1 Determine the eigenfrequencies and mode shapes of the chimney in Example 14.3.1 using the Rayleigh-Ritz with (i) two, (ii) three, and (iii)
four Ritz vectors. Use the mode shapes of the cantilever with a constant crosssection as Ritz vectors.
(a)
(b)
(c)
FIG. E14.8 Ritz vectors in Example 14.6.1.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
657
Solution
The mode shapes of the cantilever with a constant cross-section are obtained
from Eq. (8.3.41). Because C1 is arbitrary, we may write
n ðx Þ ¼
1
½ cosh an x cos an x Cn ð sinh an x sin an xÞ
2
where
an ¼ bn L,
x ¼ x=L,
Cn ¼
cosh an + cos an
sinh an + sin an
The values of an for n ¼ 1, 2, 3, 4 are taken from Table 8.3.1, That is
a1 ¼ 1:875,
a2 ¼ 4:694,
a3 ¼ 7:855,
a4 ¼ 10:995
TABLE E14.6 Ritz vectors in Example 14.6.1.
xi
1
2
3
4
0.167
0.05
0.23
0.49
0.70
0.333
0.16
0.59
0.72
0.22
0.500
0.34
0.71
0.02
0.72
0.667
0.55
0.42
0.64
0.17
0.833
0.77
0.22
0.22
0.55
1.000
1.00
1.00
1.00
1.00
The values of the Ritz vectors are given in Table E14.6 and their graphs in
Fig. E14.8.
1. Mass and stiffness matrices
They are obtained from Example 14.3.1.
3
2
6 0 0 0 0 0
60 5 0 0 0 07
7
6
60 0 4 0 0 07
7
M ¼ m6
6 0 0 0 3 0 0 7,
7
6
40 0 0 0 2 05
0 0 0 0 0 1
2
3
232:80 121:78 37:89 7:11
1:21 0:15
6 121:78 122:97 69:99 21:01 3:57 0:44 7
6
7
6 37:89 69:99 72:03 39:04 10:73 1:34 7
7
K ¼ k6
6 7:11
21:01 39:04 38:38 19:04 3:88 7
6
7
4
1:21
3:57 10:73 19:04 16:00 5:00 5
0:15
0:44 1:34
3:88 5:00 2:12
658 PART
II Multi-degree-of-freedom systems
e and K
e for two Ritz vectors
(a) Computation of the matrices M
In this case, it is
2
3
0:05 0:23
6 0:16 0:59 7
6
7
6 0:34 0:71 7
7
x¼6
6 0:55 0:42 7
6
7
4 0:77 0:22 5
1:00 1:00
Hence
e ¼ x T Mx ¼ m 3:6987 0:8608 ,
M
0:8608 5:7003
e ¼ x T Kx ¼ k 0:1911 0:3397
K
0:3397 4:5018
e rM
e z ¼ 0 becomes
Thus, the reduced eigenvalue problem K
m
0
0:1948 0:3305
3:6987 0:8608
z1
¼
, l¼r
l
z2
0
0:3305 4:4985
0:8608 5:7003
k
Using any of the methods presented in Chapter 12, we obtain l1 ¼ 0:0459,
l2 ¼ 0:7974. Hence
k
k
1:0000 0:1257
r1 ¼ 0:0459 , r2 ¼ 0:7974 , z ¼
0:0708 1:0000
m
m
(b) Computation of the approximate eigenfrequencies and mode shapes. They
are obtained from the relations
pffiffiffiffiffi
e n ¼ rn , fn ¼ xzn
w
The obtained results are given in Table E14.7 in juxtaposition to the exact ones.
Moreover, Tables E14.8 and E14.9 present the results obtained using three and
four Ritz vectors, respectively. We note that the use of four Ritz vectors produces very good results for the first two eigenfrequencies and mode shapes.
A practical rule is to use 2n Ritz vectors to achieve a good approximation of
n eigenfrequencies and mode shapes.
TABLE E14.7 Results obtained using two Ritz vectors in Example 14.6.1.
Rayleigh-Ritz
Exact
pffiffiffiffiffiffiffiffiffi
e1 ¼ 0:2143 k=m
w
pffiffiffiffiffiffiffiffiffi
w
e2 ¼ 0:8930 k=m
3
2
0:0178 0:0950
6 0:0623 0:2421 7
7
6
7
6
e ¼ p1ffiffiffi 6 0:1526 0:2834 7
F
7
m 6 0:2740
0:1490
7
6
4 0:4137 0:1346 5
0:5639 0:4782
pffiffiffiffiffiffiffiffiffi
w1 ¼ 0:2030 k=m
pffiffiffiffiffiffiffiffiffi
w2 ¼ 0:7570 k=m
2
0:0162
6 0:0645
6
6 0:1469
6
F ¼ p1ffiffiffi
m 6 0:2640
6
4 0:4133
0:5835
3
0:0547
0:1747 7
7
0:2736 7
7
0:2310 7
7
0:0662 5
0:6010
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
659
TABLE E14.8 Results obtained using three Ritz vectors in Example 14.6.1.
Rayleigh-Ritz
Exact
pffiffiffiffiffiffiffiffiffi
e 1 ¼ 0:2130 k=m
w
pffiffiffiffiffiffiffiffiffi
w
e 2 ¼ 0:7787 k=m
pffiffiffiffiffiffiffiffiffi
w
e 3 ¼ 2:1869 k=m
3
2
0:0186 0:0548 0:1951
6 0:0618 0:1859 0:2872 7
7
6
6 0:1478 0:2910 0:0060 7
1
e
7
6
F ¼ pffiffiffi
m 6 0:2680 0:2149 0:2665 7
7
6
4 0:4145 0:1058 0:1181 5
0:5765 0:5563 0:3437
pffiffiffiffiffiffiffiffiffi
w1 ¼ 0:2030 k=m
pffiffiffiffiffiffiffiffiffi
w2 ¼ 0:7570 k=m
pffiffiffiffiffiffiffiffiffi
w3 ¼ 1:7928 k=m
2
0:0161 0:0547
6 0:0645 0:1747
6
6 0:1469 0:2736
6
F ¼ p1ffiffiffi
m 6 0:2640 0:2310
6
4 0:4133 0:0662
0:5835 0:6010
3
0:1123
0:2478 7
7
0:1305 7
7
0:2348 7
7
0:2981 5
0:4539
TABLE E14.9 Results obtained using four Ritz vectors in Example 14.6.1.
Rayleigh-Ritz
Exact
pffiffiffiffiffiffiffiffiffi
w
e 1 ¼ 0:2091 k=m
pffiffiffiffiffiffiffiffiffi
w
e 2 ¼ 0:7648 k=m
pffiffiffiffiffiffiffiffiffi
e 3 ¼ 1:8840 k=m
w
pffiffiffiffiffiffiffiffiffi
e 4 ¼ 4:0680 k=m
w
pffiffiffiffiffiffiffiffiffi
w1 ¼ 0:2030 k=m
pffiffiffiffiffiffiffiffiffi
w2 ¼ 0:7570 k=m
pffiffiffiffiffiffiffiffiffi
w3 ¼ 1:7928 k=m
pffiffiffiffiffiffiffiffiffi
w4 ¼ 3:3067 k=m
2
0:0171
6 0:0623
6
6
6
e ¼ p1ffiffiffiffi 6 0:1482
F
m6
6 0:2616
6
4 0:4110
0:5898
0:0574
0:1759
0:2796
0:2340
0:0838
0:5784
3
0:1166 0:2800
0:2742 0:1070 7
7
7
0:1056
0:2745 7
7
0:2867 0:0412 7
7
7
0:2105 0:2419 5
0:4035
0:2215
2
0:0161
6 0:0645
6
6
6 0:1469
ffi6
F ¼ p1ffiffiffi
m6
6 0:2640
6
4 0:4133
0:5835
0:0547
0:1747
0:2736
0:2310
0:0662
0:6010
3
0:1123 0:1665
0:2478 0:1858 7
7
7
0:1305
0:1863 7
7
0:2348 0:1831 7
7
7
0:2981 0:4140 5
0:4539
0:2807
14.6.2 The method of derived Ritz vectors
We consider the case where the external load is of the form given by
Eq. (14.3.11), namely
pðt Þ ¼ Rf ðt Þ
(14.6.1)
The constant vector R may be used to derive a sequence of Ritz vectors,
which are orthonormal with respect to mass using the following procedure.
The first Ritz vector 1 is derived as the vector of the static displacements
produced by the load R, that is
y1 ¼ K1 R
which is normalized with respect to mass to give
(14.6.2)
660 PART
II Multi-degree-of-freedom systems
1
¼
y1
1=2
T
y1 My1
(14.6.3)
The second vector is derived from the vector y2 , which is obtained as the
static displacement vector under the load R ¼ M 1 , that is, the inertial forces
associated with the first Ritz vector. Thus, we have
y2 ¼ K1 M
(14.6.4)
1
The vector y2 , which is not orthogonal to 1 , is mass-orthogonalized to it
using the Gram-Schmidt method presented in Section 12.4. Thus, taking into
account that T1 M 1 ¼ 1, the second of Eq. (12.4.8) gives
e
(14.6.5)
y2 ¼ y2 yT2 M 1 1
which is normalized with respect to mass to give
2
Similarly, the vector
k
e
y2
¼
1=2
T
e
y2
y2 Me
(14.6.6)
is obtained as
yk ¼ K1 M
e
yk ¼ yk k 1
X
(14.6.7)
k1
ai
i
(14.6.8)
i¼1
where
ai ¼ yTk M
i
(14.6.9)
Hence
k
e
yk
¼
1=2
T
e
yk
yk Me
(14.6.10)
The procedure is repeated until we obtain the desired number of Ritz vectors.
The set of vectors 1 , 2 , …, k is orthogonal with respect to mass and linearly
independent, thus satisfying the requirement of the Ritz method. The Ritz
vectors derived by this procedure are also called load-dependent Ritz vectors.
Example 14.6.2 Determine the eigenfrequencies and mode shapes of the chimney in Example 14.3.1 using the Rayleigh-Ritz with (i) two, (ii) three, and (iii)
four derived Ritz vectors.
Solution
The mass and stiffness matrices were computed in Example 14.3.1. We adopt a
load vector proportional to the mass of the structure, that is,
R ¼ m f 6 5 4 3 2 1 gT
Multi-degree-of-freedom systems: Forced vibrations Chapter
a. Computation of the first Ritz vector
Taking m ¼ 31:056 kN m
14.3.2), Eq. (14.6.2) gives
1
14
661
1
s and k ¼ 5486:208 kN=m (see Example
2
y1 ¼ K1 R ¼ f 0:0088 0:0331 0:0711 0:1209 0:1802 0:2451 gT
and substituting into Eq. (14.6.3)
1
¼
y1
1=2
T
y1 My1
¼ f 0:0036 0:0135 0:0289 0:0491 0:0732 0:0996 gT
b. Computation of the second Ritz vector
Eq. (14.6.4) gives
y2 ¼ K1 M
1
¼ f 0:0402 0:1605 0:3643 0:6521 1:0171 1:4316 gT 102
which is orthogonalized with respect to
a1 ¼ yT2 M 1
1
¼ 0:1371
e
y2 ¼ y2 a1 1
¼ f 0:0885 0:2408 0:3182 0:2113 0:1351 0:6653 gT 103
which after normalization gives
2
e
y2
¼
1=2
T
e
y2
y2 Me
¼ f 0:0136 0:0371 0:0491 0:0326 0:0208 0:1026 gT
c. Computation of the third Ritz vector
Adhering to the previous steps we obtain
y3 ¼ K1 M
2
¼ f 0:0808 0:2363 0:3004 0:0638 0:6461 1:7681 g 103
a 1 ¼ yT
3M
T
1 ¼ 0:00649,
e
y3 ¼ y 3 a 1
a 2 ¼ yT
3M
2 ¼ 0:010059
1 a2 2
¼ f 0:3343 0:5016 0:05419 0:5528 0:3853 0:9141 g 104
T
e
y3
1=2
3¼ T
ey3 Mey3
¼ f 0:03003 0:04510 0:00493 0:04965 0:03476 0:08186 g
T
662 PART
II Multi-degree-of-freedom systems
d. Computation of the fourth Ritz vector
Similarly we obtain
y4 ¼ K1 M
3
¼ f 0:0255 0:0376 0:0298 0:1270 0:0612 0:2817 gT 103
a1 ¼ yT4 M
1
¼ 5:8861 1016 , a2 ¼ yT4 M
a3 ¼ yT4 M
3
¼ 0:0018245
e
y4 ¼ y4 a1
2
¼ 0:0011036
1 a2 2 a3 3
¼ f 0:14213 0:037082 0:15304 0:0050278 0:20841 0:19202 g 104
T
e
y4
1=2
T
e
y4 Me
y4
4¼
¼ f 0:04338 0:01131 0:04671 0:00153 0:06361 0:05861 g
Hence
2
6
6
1 6
x ¼ pffiffiffiffiffi 6
m6
6
4
0:0200
0:0751
0:1610
0:2737
0:4080
0:5550
T
3
0:0760 0:1674 0:2418
0:2069 0:2514 0:0631 7
7
0:2733 0:0275 0:2603 7
7
0:1816 0:2767 0:0086 7
7
0:1161 0:1937 0:3545 5
0:5715 0:4562 0:3266
Applying the Rayleigh-Ritz method as described in Example 14.6.1, we
obtain the approximate eigenfrequencies and mode shapes. The obtained results
are given in Tables E14.10–E14.12 in juxtaposition to the exact ones.
TABLE E14.10 Results obtained using two derived
Ritz vectors in Example 14.6.2.
Rayleigh-Ritz
Exact
pffiffiffiffiffiffiffiffiffi
e 1 ¼ 0:2030 k=m
w
pffiffiffiffiffiffiffiffiffi
e 2 ¼ 0:7946 k=m
w
3
2
0:0161 0:0769
6 0:0645 0:2104 7
7
6
7
6
e ¼ p1ffiffiffi 6 0:1470 0:2811 7
F
m 6 0:2642 0:1952 7
7
6
4 0:4134 0:0929 5
0:5832 0:5427
pffiffiffiffiffiffiffiffiffi
w1 ¼ 0:2030 k=m
pffiffiffiffiffiffiffiffiffi
w2 ¼ 0:7570 k=m
3
2
0:0162 0:0547
6 0:0645 0:1747 7
7
6
6 0:1469 0:2736 7
7
6
F ¼ p1ffiffiffi
m 6 0:2640 0:2310 7
7
6
4 0:4133 0:0662 5
0:5835 0:6010
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
663
TABLE E14.11 Results obtained using three derived Ritz vectors in
Example 14.6.2.
Rayleigh-Ritz
Exact
pffiffiffiffiffiffiffiffiffi
w
e 1 ¼ 0:2030 k=m
pffiffiffiffiffiffiffiffiffi
e 2 ¼ 0:7571 k=m
w
pffiffiffiffiffiffiffiffiffi
e 3 ¼ 1:9777 k=m
w
3
2
0:0162 0:0542 0:1761
6 0:0645 0:1754 0:2769 7
7
6
6 0:1469 0:2751 0:0643 7
1
e
7
6
F ¼ pffiffiffi
m 6 0:2640
0:2301 0:2486 7
7
6
4 0:4133 0:0687 0:2048 5
0:5835 0:5979 0:3804
pffiffiffiffiffiffiffiffiffi
w1 ¼ 0:2030 k=m
pffiffiffiffiffiffiffiffiffi
w2 ¼ 0:7570 k=m
pffiffiffiffiffiffiffiffiffi
w3 ¼ 1:7928 k=m
2
0:0161 0:0547
6 0:0645 0:1747
6
6 0:1469 0:2736
6
F ¼ p1ffiffiffi
m 6 0:2640 0:2310
6
4 0:4133 0:0662
0:5835 0:6010
3
0:1123
0:2478 7
7
0:1305 7
7
0:2348 7
7
0:2981 5
0:4539
TABLE E14.12 Results obtained using four derived Ritz vectors in
Example 14.6.2.
Rayleigh-Ritz
Exact
pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
e1 ¼ 0:2030 k=m , w
e 2 ¼ 0:7570 k=m
w
pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
e3 ¼ 1:7958 k=m , w
e 4 ¼ 3:8410 k=m
w
pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
w1 ¼ 0:2030 k=m , w2 ¼ 0:7570 k=m
pffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffi
w3 ¼ 1:7928 k=m , w4 ¼ 3:3067 k=m
2
3
0:2772
0:2540
0:1281 7
7
7
0:1272 0:2378 7
7
0:2311 0:2421 0:0532 7
7
7
0:0663 0:2854
0:2941 5
0:5835
0:6009
0:4457 0:2218
0:0161
6 0:0645
6
6
6 0:1469
e ¼ p1ffiffiffiffi 6
F
m6
6 0:2640
6
4 0:4133
0:0547
0:1120
0:1747
0:2736
2
0:0161
6 0:0645
6
6 0:1469
ffi6
F ¼ p1ffiffiffi
6
m 6 0:2640
6
4 0:4133
0:5835
0:0546
0:1747
0:2736
0:2310
0:0662
0:6010
3
0:1123 0:1665
0:2478 0:1858 7
7
0:1305 0:1863 7
7
7
0:2348 0:1831 7
7
0:2981 0:4140 5
0:4539 0:2807
14.7 Support excitation
14.7.1 Multiple support excitation
When the motion of an MDOF system is due only to support excitation, the
external forces applied to the free nodes are zero, that is, is pf ðt Þ ¼ 0. As
was shown in Section 11.2.5, the equations of motion of the supported
structure are
€f + Cff u_ f + Kff uf ¼ Mfs u
€s Cfs u_ s Kfs us
Mff u
(14.7.1)
€f + Mss u
€s + Csf u_ f + Css u_ s + Ksf uf + Kss us
ps ðt Þ ¼ Msf u
(14.7.2)
664 PART
II Multi-degree-of-freedom systems
where uf ¼ uf ðt Þ is the vector of the f unknown free displacements, us ¼ us ðt Þ
the vector of the s known support displacements, and ps ðt Þ the vector of the s
support reactions. We first examine the case where the components of us are s
time functions, different from each other. We refer to this type of support excitation as multiple support excitation to distinguish it from the special case where
us f ðt Þ, with e
us being a constant vector and f ðt Þ a
the vector us has the form us ¼ e
common time function, to which we refer as uniform support excitation.
Eqs. (14.7.1), (14.7.2) are transformed to another form, which is more practical in dynamic analysis. To this end we set
f
uf ¼ ust
f +u
(14.7.3)
The displacements ust
f are obtained as the static solution of Eq. (14.7.1), that
is, when the accelerations and the velocities are zero. Thus, setting
€s ¼ u_ f ¼ u_ s ¼ 0 and uf ¼ ust
€f ¼ u
u
f into Eqs. (14.7.1), (14.7.2), we obtain
Kff ust
f + Kfs us ¼ 0
(14.7.4)
st
Ksf ust
f + Kss us ¼ ps
(14.7.5)
Assuming
that the supports do not permit rigid body motion, it is
det Kff 6¼ 0 and Eq. (14.7.4), gives
1
ust
f ¼ Sus , S ¼ Kff Kfs
(14.7.6)
which is substituted into Eq. (14.7.5) to yield the support reaction due to the
static application of the displacements us
Ksf S + Kss us ¼ pst
(14.7.7)
s
Moreover, if the matrix C is proportional to K, we can readily show that it
also holds
_s ¼0
Cff u_ st
f + Cfs u
(14.7.8)
Indeed, if C ¼ cK, where c is a constant, Eq. (14.7.8) holds true due to
Eq. (14.7.4). If C is not proportional to K, the contribution of the damping force
to the excitation force does not vanish. However, it is small compared with the
inertial force and it can be neglected.
Substituting Eq. (14.7.3) into Eq. (14.7.1) gives
f ¼ Mff u
€st
€s Cff u_ st
_s
Mff u€f + Cff u_ f + Kff u
f Mfs u
f + Cfs u
Kff ust
+
K
u
(14.7.9)
fs s
f
which by virtue of Eqs. (14.7.4), (14.7.6), and (14.7.8) becomes
f ¼ Mff S + Mfs u
€s
Mff u€f + Cff u_ f + Kff u
(14.7.10)
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
665
f .
The foregoing equations permit the establishment of the displacements u
On the basis of Eq. (14.7.1), Eq. (14.7.3) and Eq. (14.7.4), we can write the
elastic forces at the free nodes as
f Sf ¼ Kff uf + Kfs us
f + Kff ust
¼ Kff u
f + Kfs us
(14.7.11)
f
¼ Kff u
which states that the elastic forces are produced only by the dynamic disf , as they are defined by Eq. (14.7.3) and computed from
placements u
Eq. (14.7.10). Further, using Eq. (14.7.2) and Eq. (14.7.7) we can write
the elastic forces at the supports
f Ss ¼ Ksf uf + Kss us
f + Ksf ust
¼ Ksf u
f + Kss us
f + Ksf S + Kss us
¼ Ksf u
(14.7.12)
f + pst
¼ Ksf u
s
To simplify the expressions, we use the notation Mff ¼ M, Cff ¼ C,
f ¼ u
and Eq. (14.7.10) is written as
Kff ¼ K, u
€s
Mu€ + Cu_ + K
u ¼ MS + Mfs u
(14.7.13)
If the mass matrix is diagonal, then Mfs ¼ 0 and Eq. (14.7.13) reduces to
u ¼ MS€
us
Mu€ + Cu_ + K
(14.7.14)
14.7.2 Uniform support excitation
We now examine the case where the displacement vector of the supports has
the form
us f ðt Þ
us ¼ e
(14.7.15)
es is a constant vector specifying the
where, as stated in the previous section, u
spatial distribution of the support displacements, and f ðt Þ a common time function specifying their amplitude at time t.
Substituting Eq. (14.7.15) into Eq. (14.7.14) gives
u ¼ Mrf€ðt Þ
Mu€ + Cu_ + K
(14.7.16)
r ¼ Se
us
(14.7.17)
where it was set
666 PART
II Multi-degree-of-freedom systems
That is, the vector r represents the static displacements of the free nodes
produced by the vector e
us .
When the supports are subjected to a uniform motion ug ðt Þ, as in the
seismic ground motion, the vector of the support displacements can be
written as
us ¼ 1ug ðt Þ
(14.7.18)
and inserting it into Eq. (14.7.17) gives
r ¼ S1
(14.7.19)
T
in which 1 ¼ f 1 1 1 ⋯ 1 g is the vector with dimension s. In this case, the
vector r represents rigid body motion, therefore it is
ust
f ¼ 1ug ðt Þ
(14.7.20)
Eqs. (14.7.4), (14.7.5) by virtue of Eqs. (14.7.18), (14.7.20) are written as
( )
0
^ g ðt Þ ¼
K1u
(14.7.21)
pst
s
where
"
^¼
K
Kff Kfs
Ksf Kss
#
(14.7.22)
^ is the stiffness matrix of the free structure, hence det K
^ ¼0
The matrix K
st
and ps ¼ 0, because Eq. (14.7.21) has a nontrivial solution. Hence, it is not
necessary to use Eq. (14.7.7) to compute pst
s .
Example 14.7.1 Formulate the equation of motion of the bridge in Fig. E14.9
when the supports 1 and 2 are subjected to the displacements ug1 ¼ ug ðt Þ and
ug2 ¼ 2ug ðt Þ. The cross-sectional area of the cable is A and its modulus of
elasticity Es while the modulus of elasticity of the remaining structure
is E ¼ 2:1 107 kN=m2 . Assume: I1 ¼ I2 ¼ I3 ¼ 2:055 m4 , I4 ¼ 52 m4 ,
¼ 11:2 kN=m, and L ¼ 50 m. Adopt the lumped
I5 ¼ 31 m4 , Es A ¼ 0:4EI 1 , m
mass assumption.
Solution
By neglecting the axial deformations of members undergoing bending,
the structure has 9 degrees of freedom including those corresponding to
ground motion. The degrees of freedom are shown in Fig. E14.9. Their numbering was based on the reasoning to avoid renumbering after static condensation. The lumped mass assumption is adopted. The cable is treated as a
truss member while its sag due to self-weight is neglected. Possible compression forces are encountered by appropriate prestress. The stiffness matrix
is formulated using the procedure described in Example 11.5.2. Thus,
we obtain
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
667
FIG. E14.9 Bridge with support excitations in Example 14.7.1
The static condensation (see Section 11.4) yields
2
3
2
3
105 0
0 0
201 134 226
24:7
60
6
560 0 0 7
104
132
160 7
7, Ktt∗ ¼ 103 6 134
7
Mtt ¼ 6
40
4 226
0
m3 0 5
132
308 82:3 5
0
0
0 m4
24:7 1:59 82:3 57:6
The displacements u3 , u4 are known. Thus, we can write
1
u1
u3
, us ¼
¼
u ðt Þ
uf ¼
u2
u4
2 g
201 134
105 0
3
3 226 24:7
, Kfs ¼ 10
, Kff ¼ 10
Mff ¼
134 104
132 160
0
560
668 PART
II Multi-degree-of-freedom systems
The displacements of free nodes are written in the form of Eq. (14.7.3), that
f , and Eq. (14.7.6) gives the static solution
is, uf ¼ ust
f +u
1
1
ust
u g ðt Þ
¼
K
K
fs
f
ff
2
2:01 0:961 1
¼
u g ðt Þ
1:25 1:250 2
0:088
¼
ug ðt Þ
1:250
f is obtained from Eq. (14.7.14), that is
while u
201 134 u 1
105 0
u€1
3
+
10
134 104 u 2
0
560 u€2
¼
10
u€ ðt Þ
674 g
14.8 The response spectrum method
The mode superposition method gives the dynamic response of a certain
quantity q ðt Þ as a superposition of its modal responses qn ðt Þ, namely
q ðt Þ ¼
N
X
qn ðt Þ
(14.8.1)
n¼1
where
qn ðt Þ ¼ qnst w2n yn ðt Þ
(14.8.2)
Eq. (14.8.1) expresses the time history of q ðt Þ. From Eq. (14.8.2), we deduce
that the peak value qn0 of the modal response qn ðt Þ occurs simultaneously with
the extreme value of the pseudoacceleration w2n yn ðt Þ, That is
qn0 ¼ qnst max w2n yn ðt Þ
(14.8.3)
t
or as stated in Section 6.2
qn0 ¼ qnst Spa ðxn , Tn Þ
(14.8.4)
It should be noted that the quantity Spa ðxn , Tn Þ is by definition positive
while the quantity qnst is algebraic. Because the value of the pseudoacceleration
depends on the period of the system (see Fig. 6.2.6), we can deduce from
Eq. (14.8.4) that the peak values of the modal quantities do not occur at the same
time, nor do they have the same phase. Therefore, the peak value of q ðt Þ cannot
be obtained by superposition. Nevertheless, because the quantities qn0 for a
given excitation are readily computed from the response spectrum without
dynamic analysis, researchers earlier sought methods of exploiting these quantities to compute the peak value of q ðt Þ.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
669
A first estimate is obtained from the relation
jq jmax N
X
jqn0 j
(14.8.5)
n¼1
which provides an upper bound to jq jmax . Eq. (14.8.5) expresses the absolute
sum (ABSSUM) modal combination rule. It assumes that all modal peak values
occur at the same instant. It gives a too conservative estimate of the peak value.
Therefore, it is not popular in structural design.
Another estimate is obtained from the relation
!1=2
N X
N
X
rnm qn0 qm0
(14.8.6)
jq jmax n¼1 m¼1
where rmn is the correlation coefficient of the peak modal values qn0 and qm0 . Its
value varies between 0 and 1, with rmn ¼ 1 if m ¼ n. The foregoing equation is
written in matrix form
qffiffiffiffiffiffiffiffiffiffiffiffiffi
(14.8.7)
jq jmax qT0 rq0
where q0 ¼ f q10 q20 ⋯ qN 0 gT and r the matrix of the correlation
coefficients.
Various mathematical expressions for computing the correlation coefficients
are reported in the literature when the excitation is due to seismic ground motion.
Here, we quote only Der Kiureghian’s relation [3], which is widely used today
pffiffiffiffiffiffiffiffiffiffi
8 xn xm ðxn + b nm xm Þb3=2
nm
(14.8.8)
rmn ¼ 2
2
2
1 b nm + 4xn x m b nm 1 + b nm + 4 x 2n + x 2m b 2nm
where b nm ¼ wm =wn , and rmn ¼ rnm . The above relationship results from a
stochastic process starting from the mean square response, which we do not present here because it goes beyond the scope of this book. The interested reader
should consult the related literature, for example, [4]. The method of computing
jq jmax using Eq. (14.8.7) is known as the complete quadratic combination
(CQC) rule. Eq. (14.8.7) is greatly simplified when the eigenfrequencies are
well separated, that is, there are no multiple eigenfrequencies nor are they close
to each other. In this case, the off-diagonal elements of the correlation matrix,
rmn ðm 6¼ n Þ, are negligible and we can set r I. Then Eq. (14.8.7) becomes
qffiffiffiffiffiffiffiffiffiffiffi
(14.8.9)
jq jmax qT0 q0
or
jq jmax N
X
m¼1
!1=2
2
qn0
(14.8.10)
670 PART
II Multi-degree-of-freedom systems
The method of computing jq jmax using Eq. (14.8.10) developed by Rosenblueth [4] is known as the square-root-of-sum-of-squares (SRSS) rule. It should
be emphasized that Eq. (14.8.10) can lead to unacceptable results when the
eigenfrequencies are not well separated.
Example 14.8.1 Using the response spectrum method, compute the peak values
of the base shear force Qb , the overturning moment Mb , and the top displacement u6 of the chimney in Example 14.3.1 when subjected to the Athens earthquake, Sept. 7, 1999. Assume: m ¼ 31:056 kN m1 s2 , k ¼ 5486:208 kN=m,
L ¼ 75 m, x 1 ¼ 0:06, x2 ¼ x3 ¼ ⋯ ¼ x6 ¼ 0:04.
Solution
The eigenfrequencies and mode shapes were computed in Example 14.3.1.
The peak modal values will be computed using Eq. (14.8.4).
st
st
st
, Mbn
, and u6n
Computation Qbn
From Example 14.3.1, we obtain for m ¼ 31:056 kN m1 s2
R1
R2
R3
R4
R5
R6
9:66 22:98 33:23 36:65 37:32 46:49
32:17 61:18 61:09 34:10
3:02 36:27
58:57 76:65 25:75 27:33 30:34 20:92
78:94 48:54 34:75 20:15 29:41 8:86
82:39 9:28 29:40 30:37 14:71
2:75
58:17 42:08 22:39 10:30
3:39 0:49
st
st
st
, Mbn
, u6n
are computed from the relations (see Example
The quantities Qbn
14.3.1)
st
st
st
¼ 1T Rn , Mbn
¼ hT Rn , u6n
¼
Qbn
Gn
f
w2n 6n
The computed values are given in Table E14.13.
st
st
st
TABLE E14.13 Modal quantities Qbn
, Mbn
, u6n
in
Example 14.8.1.
st
u6n
¼ Gwn2 f6n
Mode n
st
Qbn
¼ 1T Rn
st
Mbn
¼ hT Rn
1
319.89
16580.0
0.257
2
157.99
3381.8
0.013
3
78.32
1012.4
1.269e3
4
43.33
403.8
0.172e3
5
28.08
209.1
0.023e3
6
24.56
151.9
0.001e4
n
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
671
Computation of the modal pseudoacceleration Spa ðxn , Tn Þ
According to Eq. (14.8.4), we have
Spa ðxn , Tn Þ ¼ w2n max jyn ðt Þj
t
The peak value max t jyn ðt Þj is obtained from the solution of Eq. (14.3.16)
by setting f ðt Þ ¼ u€g ðt Þ, where u€g ðt Þ is given by the accelerogram of the specified earthquake. The computation of Spa ðx n , Tn Þ is performed numerically
using any of the methods described in Section 6.2. The computed values are
given in Table E14.14 together with the modal values of Qb , Mb , u6 .
Computation of the correlation matrix
Eq. (14.8.8) is used to compute the correlation coefficients. For the data of
the problem, it gives
3
r ¼ 102
3
1000:0
3:53
0:716
0:262
0:128
0:067
6
3:53 1000:0
7:34
1:91
0:85
0:423 7
6
7
6
7
0:716
7:34
1000:0
15:6
4:47
1:88
6
7
6
7
0:262
1:91
15:6
1000:0
29:1
7:15
6
7
4
0:128
0:85
4:47
29:1
1000:0
33:8 5
0:067
0:423
1:88
7:15
33:8
1000:0
Table E14.15 presents the peak values jQb jmax , jMb jmax , and ju6 jmax as computed using the three modal combination rules in juxtaposition with the exact
values obtained by the response history analysis (RHA) using the relation
X
6
st
2
qn wn yn ðt Þ jq jmax ¼ max t n¼1
with numerical integration of Eq. (14.3.16).
TABLE E14.14 Modal quantities Qbn0 , Mbn0 , and u6n0 in Example 14.8.1.
Spa ðx n , Tn Þ
Mode
n
Tn
xn
1
2.2861
0:06
31.9908
102.33
5304.0
2
0.6242
0:04
437.4272
691.10
14793.0
3
0.2634
0:04
412.2019
322.84
4173.0
0.523
4
0.1429
0:04
506.3626
219.42
2044.6
0.087
5
0.0909
0:04
540.8229
151.85
1131.0
0.012
6
0.0597
0:04
507.2341
124.58
ðcm=s2 Þ
Qbn0
ðkNÞ
Mbn0
ðkN mÞ
770.52
u6n0
ðcmÞ
8.2324
5.858
0.0007
672 PART
II Multi-degree-of-freedom systems
TABLE E14.15 Peak values in Example 14.8.1.
Method
jQb jmax
jMb jmax
ju6 jmax
ABSSUM
1612.12
28216.12
0.147
CQC
830.73
16511.35
0.1013
SRSS
824.04
16444.95
0.1012
RHA
(exact)
922.50
18545.48
0.1157
The results in Table E14.15 lead us to certain conclusions. First, the
ABSSUM rule gives extremely conservative estimations (þ75%, þ52%, and
þ27% larger than the respective quantities jQb jmax , jMb jmax , and ju6 jmax u6
obtained by the RHA). Second, the SRSS and CQC rules give essentially the
same estimations for the extreme values. This is due to the fact that the eigenfrequencies of the structure are well separated (w1 ¼ 2:69, w2 ¼ 10:06,
w3 ¼ 23:82, w4 ¼ 43:94, w5 ¼ 69:06, w6 ¼ 105:11), which produces negligible
off-diagonal terms of the correlation matrix. Third, the SRSS and CQC rules
give estimations of the extreme values smaller than the exact resulting from
RHA. Consequently, the estimation of the extreme values using the SRSS
and CQC is not in favor of safety for the analyzed structure. However, this is
not a general rule because larger values may result by SRSS and CQC for other
types of structures, for example, buildings. Nevertheless, the response spectrum
method is the most convenient method for dynamic analysis of structures under
seismic excitation and has been introduced in all earthquake codes. It should
not, however, be applied to any structure without a previous consideration
because, as we have shown, the estimated peak values may not ensure safety.
Because of the great interest in the response spectrum method, the search for
a rule combining extreme values to obtain a better approximation of the exact
values is still of interest for researchers. The reader may find improved rules in
the literature, mainly based on the theory of probability, which ensure a better
correlation of mode shapes. Nevertheless, increased and cheap computer power
facilitates the exact evaluation of the peak values, a fact that may reduce noticeably the importance of the combination rules.
14.9 Comparison of mode superposition method
and Rayleigh-Ritz method
The transformation of the equations of motion using K < N Ritz vectors
reduces the number of equations from N to K (Eq. 14.5.4), which, however,
e K
e C,
e (Eq. 14.5.5) are not in
are coupled because the resulting matrices M,
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
673
e ¼I
general diagonal, except when the Ritz vectors are load-dependent. Then M
because the Ritz vectors are orthonormalized with respect to mass.
According to Eqs. (14.3.12)–(14.3.14), we can write
e
pðt Þ ¼ f ðt Þ
K
X
en M
Rn , Rn ¼ G
n
en ¼
where G
T
nR
(14.9.1)
n¼1
The number K of the Ritz vectors that should be used in dynamic analysis
must yield a load e
pðt Þ sufficiently close to pðt Þ. Hence, according to
K
P
Rn from R is
Eq. (14.3.42), the deviation of
n¼1
e ¼R K
X
Rn
n¼1
¼R K
X
(14.9.2)
en M
G
n
n¼1
and the truncation error
eK ¼
RT e
RT R
(16.9.3)
Before stating some general conclusions about the preference between
the mode superposition method and the Rayleigh-Ritz method, it is advisable
to examine the truncation error eK as a function of the number of eigenmodes
fn and Ritz vectors n ðn ¼ 1, 2, …, K Þ, respectively, for the chimney in of
Example 14.3.1, when it is subjected to loads pi ðt Þ ¼ Ri f ðt Þ (i ¼ 1, 2, 3, 4) for different spatial distributions of R: f 1 1 1 1 1 1 gT , f 0 0 0 0 1:5 0:5 gT ,
f 0 0 1:5 1 1 0:5 gT , and f 0 0 0 0 0 1 gT . The results are presented
in Fig. 14.9.1. We observe that the truncation error is noticeably smaller when
the analysis is done using Ritz vectors than mode shapes in all four cases. This
is because the Ritz vectors result from the spatial distribution R of the external
load. Although the Rayleigh-Ritz method is preferable on the base of the truncation error, however, the use of eigenmodes still has advantages because it leads to
uncoupled equations of motion and allows the use of the response spectrum
method for determining the peak values. Another advantage of the Rayleigh-Ritz
method is that static correction is unnecessary because this correction is contained
in the first Ritz vector resulting from the static analysis under the load R. As a
conclusion to the preceding analysis, we can state the following in terms of advantages and disadvantages of the two approaches:
(a) Rayleigh-Ritz method
Advantages: It leads to a considerable reduction of the equations of
motion, thus it reduces the computational task, especially when the
674 PART
II Multi-degree-of-freedom systems
FIG. 14.9.1 Error eK as a function of the number K of mode shapes fn and Ritz vectors c n .
equations of motion are solved numerically. The truncation error is noticeably smaller. The static correction is unnecessary.
Disadvantages: The transformed equations are coupled. Thus, the solution of the eigenvalue problem is necessary to determine the eigenfrequencies and eigenmodes of the reduced equations. Only the lower
eigenfrequencies and mode shapes are accurately determined, so to obtain
K eigenfrequencies and eigenmodes with acceptable accuracy, we need
about a double number of Ritz vectors. It is not suitable for the response
spectrum method.
(b) The mode superposition method
Advantages: Leads to uncoupled equations of motion. Therefore, the
solution of a MDOF system with N degrees of freedom results from the
solution of N SDOF systems. It is suitable for the response spectrum
analysis.
Disadvantages: It requires the solution of large eigenvalue problems.
The truncation error is larger. Static correction is necessary. It is less suitable for numerical analysis.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
675
14.10 Numerical integration of the equations
of motions—Linear MDOF systems
The method of mode superposition presented in Section 14.2 requires a proportional damping matrix to uncouple the equations of motion. In this case, it is
possible to obtain an analytical solution when the load functions are simple.
However, when those functions are complicated or described by a set of discrete
values, such as for example by the seismic accelerogram, it is necessary to use
numerical methods (see Chapter 4) for the computation of the modal coordinates Yn ðt Þ. In the general case where the damping matrix is not proportional
or the equations are nonlinear, recourse to direct time integration numerical
methods is inevitable.
The nonlinear equation of motion of a SDOF system is given by Eq. (5.1.5).
Apparently, for MDOF systems, this equation takes the form
M€
u + f D ðu_ Þ + f S ðuÞ ¼ pðt Þ
(14.10.1)
For the linear response of the system, the foregoing equation becomes (see
Section 10.2)
M€
u + Cu_ + Ku ¼ pðt Þ
(14.10.2)
When the number of equations is large, the numerical solution is simplified
if the degrees of freedom are previously reduced. In linear systems, this is
accomplished by the Rayleigh-Ritz method as discussed in Section 14.5. This
reduction procedure applies also to nonlinear system [5].
As in SDOF systems, the direct numerical integration method yields the
solution uðt Þ at time instants Dt apart starting from uð0Þ. Next, for convenience, we denote the displacement vector at time tn ¼ nDt, n ¼ 0, 1, 2, … by
un . The literature on numerical integration of equations of motion is extensive.
Here, the discussion will be limited to the methods presented in Chapter 4,
which are modified to apply to MDOF systems.
14.10.1 The central difference method (CDM)—Linear equations
At time tn , Eq. (14.10.2) is written
M€
un + Cu_ n + Kun ¼ pn
(14.10.3)
Applying the procedure described in Section 4.2, the derivatives of the vector uðt Þ are approximated by the central differences
un + 1 un1
u_ n (14.10.4)
2Dt
un + 1 2un + un1
€n (14.10.5)
u
Dt 2
676 PART
II Multi-degree-of-freedom systems
which are substituted into Eq. (14.10.3) to yield
1
1
2
1
1
C un + 1 ¼ pn K 2 M un C un1
M+
M
Dt 2
2Dt
Dt
Dt 2
2Dt
(14.10.6)
or
^1 un k
^2 un1 , n ¼ 0, 1, 2, …
^ n + 1 ¼ pn k
Ku
(14.10.7)
where
^ ¼ 1 M+ 1 C
K
Dt 2
2Dt
^1 ¼ K 2 M
k
Dt 2
^2 ¼ 1 M 1 C
k
Dt 2
2Dt
(14.10.8a)
(14.10.8b)
(14.10.8c)
Eq. (14.10.7) provides the recursive formula to compute un + 1 in terms of
un and un1 . For n ¼ 0 the vector u1 appears in the right side of the formula,
which can be computed by the truncated Taylor series
1
€0
u1 u0 Dt u_ 0 + Dt 2 u
2
(14.10.9)
€0 is obtained from
in which u0 , u_ 0 are known from the initial conditions while u
the equation of motion for t ¼ 0
€0 ¼ M1 p0 Cu_ 0 Ku0
u
(14.10.10)
The CDM is conditionally stable. The stability condition requires
Dt < TN =p, where TN is the smallest natural period. Obviously, for systems
with a large number of degrees of freedom, TN is very small so that the stability
criterion of CDM demands a very small time step Dt, which renders the method
inappropriate. This disadvantage is circumvented if the use of the method is preceded by a significant reduction of the degrees of freedom as described in
Section 14.5. The steps of the numerical scheme are given in Table 14.10.1
in the form of a pseudocode so that the reader can write a computer code in
the language of his/her preference. Adhering to the steps of Table 14.10.1, a
computer program called centr_diff_lin_MDOF.m has been written in
MATLAB for the numerical integration of the equation of motion using the central difference method. The program is available on this book’s companion
website.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
677
TABLE 14.10.1 Central difference method. Linear equations.
A. Data
1. Read: M, C, K, u0 , u_ 0 , pðt Þ, ttot
B. Initial computation
1. Select: Dt ¼ 0:1Tmin , Tmin ¼ 2p=wN
€0 ¼ M1 ðp0 Cu_ 0 Ku0 Þ
2. Compute: u
€0
3. Compute: u1 ¼ u0 Dt u_ 0 + Dt2 u
^ ¼ 12 M + 1 C
4. Compute: K
2
Dt
2Dt
^2 ¼ 1 2 M 1 C
^1 ¼ K 2 2 M, k
5. Compute: k
Dt
Dt
2Dt
C. For each time step
^1 un k
^2 un1
^n ¼ pn k
1. Compute: p
^
^n
2. Solve: Kun + 1 ¼ p
1
€n ¼ Dt1 2 ðun + 1 2un + un1 Þ
ðun + 1 un1 Þ, u
3. Compute: u_ n ¼ 2Dt
4. Set n ¼ n + 1 and check:
If tn ttot end. Else set un1 ¼ un , un ¼ un + 1 and go to step C.1.
14.10.2 The average acceleration method (AAM)—Linear
equations
The basic concept of the method is to approximate the acceleration vector in the
interval Dt with its mean value, that is,
1
€ ðt + τ Þ ¼ ½ u
€ ðt Þ + u
€ðt + Dt Þ
u
2
0 τ Dt
(14.10.11)
Following the procedure in Section 4.3, we obtain identical relations
where, however, the displacement, velocity, and acceleration are replaced
with the respective vectors of the MDOF system. Thus, the vectors un + 1 ,
€n + 1 at instant tn + 1 ¼ ðn + 1ÞDt, n ¼ 0, 1, 2, … are given by the
u_ n + 1 , and u
relations
un + 1 ¼ Du + un
(14.10.12a)
u_ n + 1 ¼ Du_ + u_ n
(14.10.12b)
€n + 1 ¼ D€
€n
u
u+u
(14.10.12c)
678 PART
II Multi-degree-of-freedom systems
TABLE 14.10.2 Average acceleration method. Linear equations.
A. Data
1. Read: M, C, K, u0 , u_ 0 , pðt Þ, ttot
B. Initial computations
1. Select: Dt
€0 ¼ M1 ðp0 Cu_ 0 Ku0 Þ
2. Compute: u
^ ¼ K + 2 C + 42 M
3. Compute: K
Dt
Dt
4
^ ¼ 2M
M + 2C, m
4. Compute: ^c ¼ Dt
C. For each step
^ un
pn ¼ Dpn + ^cu_ n + m€
1. Compute: Dpn ¼ pn + 1 pn , D^
^
2. Solve: KDu
¼ D^
pn
2
4 _
u n 2€
3. Compute: Du_ ¼ Dt
Du 2u_ n , D€
u ¼ Dt4 2 Du Dt
un
€n + 1 ¼ D€
€n
4. Compute: un + 1 ¼ Du + un , u_ n + 1 ¼ Du_ + u_ n , u
u+u
5. Set n ¼ n + 1 and check:
€n ¼ u
€n + 1 and go to step C.1.
If tn ttot end. Otherwise set un ¼ un + 1 , u_ n ¼ u_ n + 1 , u
_ D€
The difference vectors Du,
u are computed from the relations
2
Du 2u_ n
Dt
4
4
un
D€
u ¼ 2 Du u_ n 2€
Dt
Dt
Du_ ¼
(14.10.13)
(14.10.14)
while Du from
^
KDu
¼ D^
pn
(14.10.15)
^ ¼K+ 2 C+ 4 M
K
Dt
Dt 2
(14.10.16)
^ un
D^
pn ¼ Dpn + ^cu_ n + m€
(14.10.17)
where
and
in which
^c ¼
4
^ ¼ 2M, Dpn ¼ pn + 1 pn
M + 2C, m
Dt
(14.10.18)
The AAM, contrary to CDM, is unconditionally stable. Therefore, the time
step Dt can be chosen arbitrarily. Its size, however, is influenced by the accuracy of the method and its capability to describe an oscillatory motion. Therefore, it must be small enough. The selection of Dt equal to 1/10 of the smallest
natural period of the system produces accurate results. Table 14.10.2 presents
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
679
the algorithm for the numerical implementation of the AAM in pseudocodetype notation so that the reader can write a computer code in the language of
his/her preference.
Following the steps in Table 14.10.2, a computer program called av_acc_lin_MDOF.m has been written in MATLAB for the numerical integration
of the equations of motion using the average acceleration method. The program
is available on this book’s companion website.
14.10.3 The analog equation method (AEM)—Linear equations
The underlying concepts of this method are presented in Section 4.4, where the
AEM was developed for a SDOF system. Apparently, the solution procedure
applies also to MDOF systems provided that the coefficients m, c, k in
Eq. (4.4.1) and the quantities u0 , u_ 0 , un , u_ n , qn , pn are replaced with the coefficient
matrices M,C,K and the vectors u0 , u_ 0 , un , u_ n , qn , pn , respectively, and the scalar
operations with matrix operations. Thus, Eqs. (4.4.17), (4.4.20a)–(4.4.20c) read
3
2
2
3
M
C K 8 9
0
0 0 8
9 8 9
7< qn = 6 c1
6 c1
7< qn1 = < I =
7
6
I 0 I7
6 2 I hI I 7 u_ n ¼ 6
7 u_ n1 + 0 pn
2
7: ; 6
6
; : ;
4
5:
5 un
4 c2
c2
0
un1
I I 0
I I 0
2
2
(14.10.19a)
q0 ¼ M1 p0 Cu_ 0 Ku0 , det ðMÞ 6¼ 0
(14.10.19b)
€n . Eq. (14.10.19a), when solved successively for n ¼ 1, 2, …, gives
where qn ¼ u
the solution of the equation of motion (14.10.2). The AEM is self-starting and,
as shown in Section 4.4, unconditionally stable and accurate while conserving
energy. Therefore, it can be used as a practical method for integration of the
equations of motion in cases where widely used time integration procedures
become unstable [6]. Table 14.10.3 presents the algorithm for the numerical
implementation of the AEM in pseudocode-type notation so that the reader
can write a computer code in the language of his/her preference. Following
the steps in that table, a computer program called aem_lin_MDOF.m has been
written in MATLAB for the numerical integration of the equations of motion
using the AEM. The program is available on this book’s companion website.
Example 14.10.1 Solve the discretized equations of motion of the chimney in
Example 14.3.4 using the three numerical methods, CDM, AAM, and AEM.
Give the graphical representation of the base shear force Qb ðt Þ, overturning
moment Mb ðt Þ, and top displacement u6 ðt Þ. Assume: k ¼ 5486:208 kN=m,
m ¼ 31:056 kN m1 s2 , x1 ¼ 0:06, x2 ¼ x3 ¼ ⋯ ¼ x6 ¼ 0:04, pðt Þ ¼ Rf ðt Þ.
The time function f ðt Þ is given in Fig. E14.6 (t1 ¼ 0:1, p0 ¼ 1),
R ¼ f 1:5 6:0 15:0 25:0 40:0 60:0 gT .
680 PART
II Multi-degree-of-freedom systems
TABLE 14.10.3 Analog equation method (AEM). Linear equations.
A. Data
1. Read: M, C, K, u0 , u_ 0 , pðt Þ, ttot
B. Initial computations
1. Select: h ¼ Dt and compute ntot ¼ bttot =h c
2. Compute: q0 ¼ M1 ðp0 Cu_ 0 Ku0 Þ, c1 ¼ h 2 =2, c2 ¼ h
3. Formulate: U0 ≔f q0 u_ 0 u0 gT
2
M
C
6 c1
I hI
6
4. Compute: A ¼ 6 2
4
c2
I I
2
2
M
C
6 c1
I hI
6
b¼6 2
4 c
2
I I
2
3
0 0 0
7 6 c1
7
I 7 6 I 0 I 7
7 6 2
7
5 4
5
c2
0
I I 0
2
3
K 1 8 9
I
< >
=
7 >
I 7
0
7
5 >
: >
;
0
0
K
31 2
C. Compute solution
for n≔1 to ntot
Un ¼ AUn1 + bpn
end
Solution
The mass and stiffness matrices, eigenfrequencies, and mode shapes are taken
from Example 14.3.1, which for the given values of k and m give.
2
3
186:336 0
0
0
0
0
60
7
155:280 0
0
0
0
6
7
60
7
0
124:224
0
0
0
6
7
M¼6
7
0
0
0
93:168
0
0
6
7
40
5
0
0
0
62:112 0
0
0
0
0
0
31:056
2
3
1277189:2 668110:4
207872:4 39006:9
6638:3
822:9
6 668110:4
674639:0 383979:7
115265:2 19585:7
2413:9 7
6
7
6 207872:4 383979:7
395171:5 214181:5
58867:0 7351:5 7
6
7
K¼6
7
6 39006:9
115265:2 214181:5
210560:6 104457:4
21286:4 7
6
7
4
6638:3 19585:7
58867:0 104457:4
87779:3 27431:0 5
822:9
2413:9
7351:5
21286:4 27431:0
11630:7
The obtained eigenfrequencies and eigenmodes are
w1 ¼ 2:697,
w2 ¼ 10:060,
w3 ¼ 23:828
w4 ¼ 43:949,
w5 ¼ 69:065,
w6 ¼ 105:118
Multi-degree-of-freedom systems: Forced vibrations Chapter
2
0:0028
6 0:0115
6
6 0:0263
F¼6
6 0:0473
6
4 0:0741
0:1047
0:0098
0:0313
0:0491
0:0414
0:0118
0:1078
14
681
3
0:0201 0:0298 0:0377 0:0503
0:0444 0:0333 0:0036 0:0471 7
7
0:0234 0:0334 0:0460 0:0339 7
7
0:0421 0:0328 0:0595 0:0191 7
7
0:0534 0:0742 0:0447 0:0089 5
0:0814 0:0503 0:0205 0:0031
The numerical methods require the damping matrix. This is constructed as a
proportional damping matrix from the eigenfrequencies, eigenmodes, and
modal damping ratios. Thus applying Eq. (12.11.40), we obtain
^ TM
C ¼ MFCF
2
3
1152:72 396:84
59:28
6:35
1:97
0:19
6 396:84 657:43 289:32
37:54
2:82
1:69 7
6
7
6
7
6
59:28 289:32 441:10 192:30
23:00
0:12 7
7
¼6
6
6:35
37:54 192:30 277:56 113:97 13:23 7
6
7
6
7
4
1:97
2:82
23:00 113:97 148:39 44:83 5
0:19
1:69
0:12
13:23
44:83
35:59
The graphical representation of Qb ðt Þ, Mb ðt Þ, and u6 ðt Þ, obtained using the
three numerical methods with Dt ¼ 0:01, are shown in Fig. E14.10. Apparently,
the obtained results are graphically identical.
14.11 Numerical integration of the equations
of motions—Nonlinear MDOF systems
The solution to Eq. (14.10.1) is obtained by the step-by-step integration technique as already developed for the SDOF systems in Chapter 5. This technique
demands the fulfillment of the equation of motion at discrete time instants
Dt apart. Numerous methods using this concept are available in the literature
[7,8]. Here, we will limit our discussion to the average acceleration method
(AAM) in conjunction with the modified Newton-Raphson method for the control of the error in the tangential stiffness, and the analog equation method
(AEM) [6]. The AAM is very popular in numerical dynamic analysis while
the AEM, besides its efficiency and accuracy, performs well where other
well-known methods fail.
14.11.1 The average acceleration method (AAM)—Nonlinear
equations
The average acceleration method (AAM) developed for linear systems in
Section 14.10.2 can also be applied to nonlinear systems. We limit our presentation of the AAM as applied to equation
M€
u + Cu_ + f S ðuÞ ¼ pðt Þ
(14.11.1)
682 PART
II Multi-degree-of-freedom systems
FIG. E14.10 Time histories of Qb ðt Þ, Mb ðt Þ, and u6 ðt Þ in Example 14.10.1.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
683
TABLE 14.11.1 Average acceleration method (AAM). Nonlinear equations.
A. Data
1. Read: M, C, f S ðuÞ, u0 , u_ 0 , ttot , a
B. Initial Computations
1. Select: Dt
€0 ¼ M1 ½p0 Cu_ 0 f S ðu0 Þ
2. Compute: u
4
^ ¼ 2M
^
M + 2C m
3. Compute: c ¼ Dt
C. In each step
1. Compute: ðkT Þn ¼ d ðfSi Þ=du j n ,
^n ¼ ðkT Þ + 2 C + 4 2 M,
2. Compute: k
n
Dt
Dt
Dp ¼ pn + 1 pn
^n ¼ Dp + ^cu_ n + m€
^ un
3. Compute: p
^n and p
^n using the modified Newton-Raphson method
4. Compute: Du from k
(see Table 14.11.2)
2
Du 2un
5. Compute: Du_ ¼ Dt
6. Compute: un + 1 ¼ un + Du, u_ n + 1 ¼ u_ n + Du_
€n + 1 ¼ M1 ½pn + 1 Cu_ n + 1 f S ðun + 1 Þ
u
7. Set n ¼ n + 1 and check:
If tn
ttot end. Else set un ¼ un + 1 , u_ n ¼ u_ n + 1 , u_ n ¼ u_ n + 1 and go to step C.1.
This equation represents a special case of Eq. (14.10.1), that is, only when the
stiffness vector is a nonlinear function of the displacements while the damping
vector is linear. The steps of the method are given in Table 14.11.1. Moreover,
Table 14.11.2 shows the steps for the modified Newton-Raphson method as
applied to MDOF systems. Following these steps, a computer program called
av_acc_nlin_MDOF.m has been written in MATLAB for the numerical integration of the nonlinear equations of motion using the average acceleration method.
The program list is available on this book’s companion website.
TABLE 14.11.2 Modified Newton-Raphson method.
A. Data
ð0Þ
ð0Þ
pn
1. Read: un + 1 ¼ un , f S ¼ f S ðun Þ, DRð1Þ ¼ D^
B. For each iteration Compute i ¼ 1,2,3…
^1 DRðiÞ
1. duðiÞ ¼ k
n
ði Þ
ði1Þ
2. un + 1 ¼ un + 1 + duðiÞ h
i
ði Þ
ði1Þ
ði Þ
^n ðkT Þ duðiÞ
3. Dp ¼ f S f S + k
n
4. DRði + 1Þ ¼ DRðiÞ DpðiÞ
^1 DRði + 1Þ
5. Compute
duði + 1Þ ¼ k
n
6. If duði + 1Þ =DuðnI Þ > a Set i ¼ i + 1 and go to step Β.2
684 PART
II Multi-degree-of-freedom systems
14.11.2 The analog equation method (AEM)—Nonlinear equations
The solution procedure developed in Section 5.3 for nonlinear SDOF systems
can be straightforwardly extended to MDOF systems. The nonlinear initial
value problem for MDOF systems reads
_ uÞ ¼ pðt Þ
M€
u + Fðu,
(14.11.2a)
uð0Þ ¼ u0 , u_ ð0Þ ¼ u_ 0
(14.11.2b)
_ uÞ is an
where M is N N known coefficient matrix with det ðMÞ 6¼ 0; Fðu,
N 1 vector, whose elements are nonlinear functions of the components of
_ and pðt Þ is the vector of the N given load functions and u0 , u_ 0 given conu, u;
stant vectors.
TABLE 14.11.3 Analog equation method (AEM). Nonlinear equations.
A. Data
_ uÞ, u0 , u_ 0 , pðt Þ, ttot
1. Read: M, Fðu,
B. Initial computations
1. Choose: h≔Dt and compute ntot
2. Compute: c1 ≔h 2 =2 c2 ≔h
q0 ≔M1 ½p0 Fðu_ 0 , u0 Þ :
C. Compute solution
for n≔1 to ntot solve for fqn u_ n un gT the system of the nonlinear algebraic equations:
Mqn + Fðu_ n , un Þ ¼ pn
hI I u_ n
0 I
u_ n1
¼
I 0 un
I 0 un1
end
2 c 3
2 c 3
1
1
I
I
6
7
6
7
+ 4 c2 5qn + 4 c2 5qn1
2
2
I
I
2
2
The solution procedure is similar to that for the linear systems. Thus,
Eq. (14.11.2a) for t ¼ 0 gives the initial acceleration vector
€
q0 ¼ M1 ½p0 Fðu_ 0 , u0 Þ, q0 ¼ u
(14.11.3)
Subsequently, we apply Eq. (14.11.2a) for t ¼ tn
Mqn + Fðu_ n , un Þ ¼ pn
(14.11.4)
Apparently, the second and third of Eqs. (14.10.19a), (14.10.19b) are valid
in this case too, and can be written as
2 c 3
2 c 3
1
1
I
I
hI I
u_ n
0 I
u_ n1
6 2 7
6 2 7
¼
+ 4 c 5qn + 4 c 5qn1
2
2
I 0 un
I 0 un1
I
I
2
2
(14.11.5)
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
685
Eqs. (14.11.4), (14.11.5) are combined and solved for qn , u_ n , un with
n ¼ 1, 2, …. Note that Eq. (14.11.5) are linear and can be solved for u_ n , un . Then
substitution into Eq. (14.11.4) results in a nonlinear algebraic equation, which
can be solved to yield qn . In our examples, the function fsolve of MATLAB has
been employed to obtain the numerical results. The steps of the solution procedure are presented in Table 14.11.3. Following these steps, a computer program
called aem_nlin_MDOF.m has been written in MATLAB. The program list is
available on this book’s companion website.
Example 14.11.1 The structure in Fig. E14.11a consists of two extensible
cables with length l ¼ l0 + d0 supporting a concentrated mass m. Study the
dynamic response of the system when the mass undergoes the displacements
u0 ¼ 20 cm, v0 ¼ 15 cm from the initial position. The cables of cross-sectional
area A are assumed massless and are prestressed by the force S0 ¼ EAd 0 =l0 , so
that their stress is tensile during the response. Compute the minimum value of d0
that ensures tension. Assume: A ¼ 3:14 cm2 , l0 ¼ 3:00 m, m ¼ 10 kN m1 s2 ,
g ¼ 9:81 m=s2 , E ¼ 2 107 kN=m2 , and px ¼ py ¼ 0.
(a)
(b)
(c)
FIG. E14.11 Structure in Example 14.11.1 (a); degrees of freedom (b); forces acting on the
mass m (c).
Solution
The system has two degrees of freedom, Fig. E14.11b. The equations of
motion are formulated using the method of equilibrium of forces. The forces
acting on the mass m are shown in Fig. E14.11c. The equilibrium of forces
gives
m u€ + S1 sin q1 + S2 sin q2 ¼ px
(1a)
m v€ S1 cos q1 + S2 cos q2 ¼ py mg
(1b)
The elastic forces S1 and S2 are obtained from the relations
S1 ¼ EA
l 1 l0
l0
(2a)
S2 ¼ EA
l 2 l0
l0
(2b)
686 PART
II Multi-degree-of-freedom systems
Referring to Fig. E14.11b, we have
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
l1 ¼ ðl v Þ2 + u 2 ¼ l0 ð1 + d 0 y Þ2 + x 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
l2 ¼ ðl + v Þ + u ¼ l0 ð1 + d 0 + y Þ2 + x 2
(3a)
(3b)
where
u
x¼ ,
l0
v
y¼ ,
l0
d0 ¼
d0
l0
Substituting Eqs. (3a), (3b) into Eqs. (2a), (2b) gives
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
ð1 + d 0 y Þ + x 1
S1 ¼ EA
(4a)
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
S2 ¼ EA
ð1 + d 0 + y Þ + x 1
(4b)
Moreover, we have
cos q1 ¼
l v
1 + d0 y
¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
l1
ð1 + d 0 y Þ 2 + x 2
u
x
sin q1 ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
l1
ð1 + d 0 y Þ2 + x 2
cos q2 ¼
l v
1 + d0 + y
¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
l2
ð1 + d 0 + y Þ2 + x 2
u
x
sin q2 ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
l2
ð1 + d 0 + y Þ2 + x 2
(5a)
(5b)
(5c)
(5d)
Finally, substituting Eqs. (4a) and (4b), (5a)–(5d) into Eqs. (1a), (1b) yields
the equations of motion
3
2
1
1
7
6
ml 0 x€ + EAx 42 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 ¼ px
2
2
x 2 + ð1 + d 0 y Þ
x 2 + ð1 + d 0 + y Þ
2
3
(6a)
1 + d0 y
1 + d0 + y
6
7
ml 0 y€ + EA42y + qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 ¼ py mg
x 2 + ð1 + d 0 y Þ2
x 2 + ð1 + d0 + y Þ2
(6b)
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
687
Eqs. (6a), (6b) are solved numerically using the AAM and AEM methods.
Fig. E14.12 shows the graphs of u ðt Þ and v ðt Þ while Fig. E14.13 shows
the graphs of S1 ðt Þ and S2 ðt Þ. The results are graphically identical. The minimum value of d0 is obtained iteratively until min ðS1 , S2 Þ ¼ 0. This gives
min d0 ¼ 16:5 cm.
FIG. E14.12 Time history of the displacements u(t) and v(t) in Example 14.11.1.
FIG. E14.13 Time history of the elastic forces S1(t) and S2(t) in Example 14.11.1.
688 PART
II Multi-degree-of-freedom systems
14.12 Problems
FIG. P14.1 Dynamic model of the television tower in problem P14.1.
Problem P14.1 The television tower modeled by the system in Fig. P14.1 is
subjected (i) to the wind pressure of Fig. E14.6 with p0 ¼ 2 kN=m2 ,
t1 ¼ 0:5 s and (ii) to the seismic ground motion ug ðt Þ ¼ 2 sin 5t. The column
is approximated by three constant elements of length l ¼ h=3. Adopting the
lumped assumption, derive the equation of motion of the structure and analyze
its response using the mode superposition method. Give the graphs of the time
history of the displacements, the base shear force Qb and overturning moment
Mb . Compute the peak value of the shear force and the bending moment at the
base cross-section of the column as well as at the cross-sections above and
beneath the body B representing a rotating restaurant. The ground is elastic.
The reaction moment of the elastic ground is represented by the expression
MR ¼ CR f, where CR ¼ KI f ; If is the moment of inertia of the planform of
the fundament and K ¼ E=10h the foundation modulus with E being the modulus of elasticity of the material of the structure. The cross-sections of the flexible column, the planform of the fundament, and the body B are circular with
diameters D, Df ¼ 8D, and DB ¼ 5D, respectively. The specific weight of the
material is g. The fundament and the body B are assumed rigid. Data h ¼ 60 m,
D ¼ 2 m, E ¼ 2:1 107 kN=m2 , g ¼ 24 kN=m3 .
Problem P14.2 The shear frame in Fig. P14.2a is modeled as shown in
Fig P14.2b. Compute the modal contribution factor of (i) the base shear
force Qb , (ii) the overturning moment Mb , and (iii) the top displacement u4
t. Assume:
when the excitation is due to ground motion ug ðt Þ ¼ uo sin w
¼ 2:0 kN m1 s2 =m; a ¼ 3:0 m; columns cross-sections: first and second
m
floor 0:30 0:30 m2 , third and fourth floor 0:25 0:25 m2 ; modulus of elasticity E ¼ 2:1 107 kN=m2 .
Multi-degree-of-freedom systems: Forced vibrations Chapter
(a)
14
689
(b)
FIG. P14.2 Shear frame in problem P14.2 (a); dynamic model (b).
Problem P14.3 For the shear frame of Problem P14.2, derive Ritz vectors using
(i) the method of the natural mode shapes and (ii) the method of load dependent
Ritz vectors, and use them to approximate the eigenfrequencies. Then reduce
the number of equations of motion to two.
Problem P14.4 For the shear frame of Problem P14.2, compute the modal
magnification factors Dn and examine whether the static correction method
is recommended for application.
Problem P14.5 For the shear frame of Problem P14.2, compute the peak modal
values Qbn0 , Mbn0 , and u4n0 of the base shear force, overturning moment, and
top displacement, respectively. Then use the ABSSUM, CQM, and SRSS
methods to compute the peak values of these quantities. Compare them with
those resulting from the RHA (Response History Analysis).
Problem P14.6 The two one-story buildings of Fig. P14.6 are connected by a
beam as shown in Fig. P14.6. Formulate the equations of motion. Then compute
the truncation error as a function of the number of employed eigenmodes.
Use both the natural mode shapes and derived Ritz vectors and compare the
computed error. Data: column cross-sections 0:30 0:30 m2 , beam crosssection 0:20 0:30 m2 , h ¼ 3:5 m, a ¼ 4:0 m, b ¼ 4:0 m, q ¼ 20 kN=m2 ,
b ¼ p=6, and F ðt Þ ¼ 2 cos 5t.
FIG. P14.6 Two one-story buildings in problem P16.6.
690 PART
II Multi-degree-of-freedom systems
Problem P14.7 Consider the two-bar system of Fig. P14.7. While the system is
in static equilibrium under the gravity load W ¼ mg, the mass m is given the
additional displacements u0 , v0 . Compute the time history of the displacements
u ðt Þ,v ðt Þ in the horizontal and vertical directions, respectively, as well as the
axial forces S1 ðt Þ, S2 ðt Þ of the bars and give their graphs. Use the numerical
methods CDF, AAM, and AEM to solve the nonlinear equations of motion.
Assume: A ¼ 3:14 cm2 , l1 ¼ 4:0 m, l2 ¼ 3:0 m, m ¼ 10 kN m1 s2 ,
g ¼ 9:81 m=s2 , and E ¼ 2 108 kN=m2 . The bars are assumed massless.
FIG. P14.7 Two-bar system in problem P14.7.
Problem P14.8 The two-story building of Fig P14.8 is modeled as a space
frame. Reduce the degrees of freedom of the structure by considering the axial
restraints of columns as well the diaphragmatic function of the slabs. Adopt the
lumped mass assumption for the columns.
FIG. P14.8 Two-story building in problem P14.8.
Problem P14.9 The shear frame of Fig. P14.9 is supported on an elastic foundation that is modeled by two nonlinear springs producing
the force
FT ¼ CT ðu + u 2 =4a Þ and the moment MR ¼ CR f + f2 =4 , where
CT ¼ EI =10a 3 and CR ¼ EI =5a. Formulate the equations of motion of the
structure and determine the time history of the shear forces and bending
moments of the columns. Then compute their peak values. Data:
¼ 2 kN m1 s2 =m, a ¼ 10 m, h1 ¼ 5h2 =3, h2 ¼ 3:0 m; g ¼ 9:81 m=s2 ,
m
E ¼ 2 107 kN=m2 ; cross-section area of columns k1 0:30 0:30 m2 and k2
0:25 0:25 m2 ; p1 ðt Þ ¼ 5H ðt Þ, and p2 ðt Þ ¼ 8H ðt Þ.
Multi-degree-of-freedom systems: Forced vibrations Chapter
14
691
FIG. P14.9 Shear frame in problem P14.9.
Problem P14.10 The steel mast of Fig. P14.10a is spherically hinged at point O
and is supported by four massless cables of cross-section A. At the top of the
mast is a steel box of side a and thickness d1 . The mast consists of a circular
steel tube with thickness d2 . Its cross-section varies linearly as shown in the figure. The external diameters at the cross-sections are D1 at the top C, D2 at B, and
D3 at the base O. The planform of the structure is shown in Fig. P14.10b. The
box is subjected to a wind blast of intensity pðt Þ in the direction b with respect to
the x axis, whose time variation is given in Fig. E14.6. The cables are prestressed to undertake compressive forces. Considering small displacements,
determine the response of the structure. Data: L ¼ 60:0 m, a ¼ 5:0 m,
d1 ¼ d2 ¼ 2 cm, D1 ¼ D3 ¼ 1:0 m, D2 ¼ 1:5 m, p0 ¼ 2:0 kN=m2 , b ¼ p=6,
A ¼ 3:14 cm2 , E ¼ 2 108 kN=m2 , and g ¼ 9:81 m=s2 .
(a)
(b)
FIG. P14.10 Steel mast in problem P14.10 (a); planform (b).
692 PART
II Multi-degree-of-freedom systems
References and further reading
[1] A.K. Chopra, Modal analysis of linear dynamic systems: physical interpretation, ASCE J.
Struct. Eng. 122 (1996) 517–527.
[2] J.T. Katsikadelis, A generalized Ritz method for partial differential equations in domains of
arbitrary geometry using global shape functions, Eng. Anal. Bound. Elem. 32 (5) (2008)
353–367, https://doi.org/10.1016/j.enganabound.2007.09.001.
[3] E.L. Wilson, A. Der Kiureghian, E.P. Bayo, A replacement for the SRSS method in seismic
analysis, Int. J. Earthquake Eng. Struct. Dyn. 9 (1981) 187–194.
[4] N.M. Newmark, E. Rosenblueth, Fundamentals of Earthquake Engineering, Prentice-Hall,
Englewood Cliffs, NJ, 1971.
[5] J.T. Katsikadelis, N. Babouskos, Nonlinear flutter instability of thin damped plates. An AEM
solution, J. Mech. Mater. Struct. 4 (7–8) (2009) 1394–1414.
[6] J.T. Katsikadelis, A new direct time integration method for the equations of motion in structural dynamics, ZAMM Z. Angew. Math. Mech. 94 (9) (2014) 757–774, https://doi.org/
10.1002/zamm.20120024.
[7] M.A. Dokainish, K. Subbaraj, A survey of direct time-integration method in computational
structural dynamics—I. Explicit methods, Comput. Struct. 32 (1989) 1371–1386.
[8] K. Subbaraj, M.A. Dokainish, A survey of direct time-integration methods in computational
structural dynamics—II. Implicit methods, Comput. Struct. 32 (1989) 1387–1401.
[9] W. Weaver Jr., P.R. Johnston, Structural Dynamics by Finite Elements, Prentice Hall,
Englewood Cliffs, NJ, 1987.
[10] K.J. Bathe, E.L. Wilson, Numerical Methods in Finite Element Analysis, Prentice-Hall, Inc.,
Englewood Cliffs, NJ, 1976.
[11] J.W. Leonard, Tension Structures, McGraw-Hill, New York, 1988.
Chapter 15
Dynamic analysis of multistory
buildings
Chapter outline
15.1 Introduction
15.2 The multistory building
15.2.1. The concept of the
multistory element
15.2.2. Nodal displacement
matrix, nodal force
matrix, transformation
matrix, and stiffness
matrix of the MSE
15.2.3. Mass matrix of the
MSE and multistory
building
693
695
695
15.2.4. Equation of motion
of the multistory
building
15.3 Dynamic response of
multistory buildings due to
ground motion
15.4 Problems
References and further reading
702
715
723
724
696
701
15.1 Introduction
The load-bearing systems of modern multistory buildings are skeletal structures
made of steel, RC (reinforced concrete), or a combination of them (composite structures). The load transfer path is from slabs (plates) to beams, from beams to columns, and from columns to the foundation, which may consist of individual
footings, strip footings, and raft foundations. The first modeling of a building is
the shear building. It is a simple model that is used to approximate the dynamic
response of a building. According to this model, the beam-reinforced slabs are
encountered as plane rigid bodies. The obtained response is acceptable when the
beam-slab system is very stiff, which is a usual case for buildings of reinforced concrete. However, in the beam-slab system, the flexible slabs are connected with the
beams either by various types of connectors, as in composite structures, or rigidly,
as in concrete structures, and they cooperate in carrying the live and dead loads. The
deformation of the slab within its plane is very small and can be neglected. Thus,
the functioning of the slab in the horizontal motion can be simulated with that of a
plane rigid body, called diaphragm. This functioning constrains the horizontal displacements and the rotations about the vertical axis at the ends of the columns as
well as the axial deformation of the beams. On the other hand, this approximation
enforces the cooperation of slabs and beams in the vertical direction.
Dynamic Analysis of Structures. https://doi.org/10.1016/B978-0-12-818643-5.00015-7
© 2020 Elsevier Inc. All rights reserved.
693
694 PART
II Multi-degree-of-freedom systems
The aforementioned functioning of the load-bearing system of the building
allows the approximation of its dynamical response by the following three models:
(a) The slab together with the beams behaves as a rib-reinforced plate (see
Fig. 15.1.1) and is treated as a substructure connected to the columns.
Though this model represents the actual response of the building, it exhibits
considerable difficulties in the analysis of the beam-plate system [1–3],
which increase when the directions of the beam principal axes are inclined
with respect to the middle plane of the plate [4]. On the other hand, the
employed simplified methods to approximate rib-reinforced plates [5]
introduce a considerable error, which cancels the advantages of this model
and prevents its use in building analysis.
(b) The building is approximated by a space frame in which the plates participate with their effective breadth, that is, a strip of the plate that increases
the stiffness of the beam, and the beams perform as beams with flanges, that
is, T-beams or G-beams (Fig. 15.1.2). The effective breadth assumption is
adopted as a recommendation by several specification codes. Nevertheless,
the resulting error is not controlled because the effective breadth cannot be
estimated theoretically [6].
(c) A third way to approximate the response of a building is the finite element
method. In this method, the columns are modeled as beam elements and the
beams as beam structures interconnected with the plate, which is treated as
a surface structure. Then the degrees of freedom are reduced by introducing
nodal constraints, which ensure the diaphragmatic functioning of the plates
and if allowed, the omitting of the axial deformation of columns. The finite
element method describes the actual response of the structure reliably and
has been very popular recently because of the increased cheap computational power. Therefore, it is used in most available professional codes
for the dynamic analysis of buildings. Nevertheless, the method exhibits
a great shortcoming as the number of the eigenmodes is very large and a
procedure should be developed to determine the least number of eigenmodes that dominate the dynamic response.
In all three models, the slab functions as a diaphragm. In the second model, the
equation of motion of the building can be derived from the equation of motion of
the space frame, as formulated in Section 11.9. Then, the lumped mass assumption
for the columns eliminates the end rotations about the horizontal axes while their
horizontal displacements and rotations about the vertical axis are constrained by
the diaphragm, as described in Section 11.11. If the height of the building is small
compared with its other dimensions, the axial displacements of the columns may
be neglected. The masses of the building are concentrated at the levels of the slabs.
This process reduces the degrees of freedom to N ¼ 3n, where n is the number of
floors. We have illustrated it in Section 14.4.2, where the degrees of freedom of a
three-story building are reduced to 9.
Dynamic analysis of multistory buildings Chapter
15
695
Several professional codes have been developed for the analysis of buildings
and in general of structures, such as ETABS, SAP2000, CSiBridge, STAAD,
MIDAS, etc. These codes can be used by professional engineers to efficiently
analyze structures made from various materials. However, the engineers must
be aware of the assumptions adopted by the code. It is highly essential that
the user of the code can check the correctness of the obtained results. This
presupposes a deep understanding of the static and dynamic response of the
structures. Therefore, it is recommended that these codes are employed only
by experienced engineers who master the static and dynamic structural analysis.
FIG. 15.1.1 Plate reinforced with beams.
Effective breadth
FIG. 15.1.2 Plate modeling with a grid of T- and G-beams (cross-section).
In this chapter, we will study the dynamic response of multistory buildings
on the basis of a simpler building model, which takes into account the framing
of horizontal beams and vertical columns. This functioning is not limited individually to the floors but extends to the entire height of the building, including
the influence of the elastic support on the ground. The basic concept introduced
by this model is that of the multistory element (MSE). This element pertains
along the whole height of the building and it may be an individual column, a
wall, a frame, an elevator core, a staircase core, or elements that result from their
coupling. The slabs function as diaphragms and the axial deformation of the
multistory elements may be neglected. The masses are considered lumped at
the levels of the floors. Thus, the active degrees of freedom are three for each
floor, two translations and one rotation. This implies that in a building with n
floors, the motion is governed by 3n equations. The method is illustrated with
two buildings, a one-story building and a two-story building.
15.2 The multistory building
15.2.1 The concept of the multistory element
We consider the multistory building of Fig. 15.2.1, which consists of n horizontal plates connected to each other by K vertical elastic elements. The plates
are assumed undeformable in their plane, that is, they function as diaphragms.
696 PART
II Multi-degree-of-freedom systems
The vertical elements are rigidly connected with the plates so that the actions
(forces and moments) can be transferred from the plates to the vertical elements
and vice versa.
FIG. 15.2.1 Multistory building.
The vertical elements may connect two or more, or even all, slabs of the
structure. They may be fixed, hinged, or elastically supported on the ground.
These elements will be referred to as multistory elements (MSE). The MSEs
may be columns, frames, walls, isolated or framed with beams, closed sections,
staircase cores, etc. The directions of the principal axes of the MSE may vary
from floor to floor. Nevertheless, their treatment is easier if their principal
directions are the same along the height.
15.2.2 Nodal displacement matrix, nodal force matrix,
transformation matrix, and stiffness matrix of the MSE
We consider an MSE whose axis coincides with the vertical z axis. Its horizontal axes x, y coincide with the directions of the common principal axes
when they do not change from floor to floor or with the predominant principal axes. The z axis of all MSEs is parallel to the z axis of the global system O xyz of the structure and has the same direction with that. The x, y
axes are generally different for each MSE and are rotated with respect to
the global system of axes (see Fig. 15.2.2). The system of axes i xyz of i MSE is its local system of axes. The MSE is idealized with its axis. The
points 1, 2, …,n at which the z axis of the MSE intersects the horizontal
slabs are referred to as the nodes of the i MSE.
Dynamic analysis of multistory buildings Chapter
15
697
Node
Node
Node
Node
Node
FIG. 15.2.2 Displacements and elastic forces at the node j of the i MSE.
The MSE is deformed due to the deformation of the structure. Its deformation can be determined by the nodal displacements, which are three for each
node, two translational in the directions of the x,y axes and one rotational about
the z axis. The displacements and elastic forces at the j node of i MSE will be
ij and
denoted by uji , vji , wij and Xji , Yji , Mji in the local axes and by u ij , v ij , w
i
i
i
in the global axes, respectively.
X j , Y j , M
j
Thus we define the following matrices for the displacements and elastic
forces at the j node of i MSE:
In the local axes
n
oT
Dij ¼ uji vji wij
(15.2.1a)
In global axes
T
FiSj ¼ Xji Yji Mji
(15.2.1b)
n
oT
i
i ¼ u i v i w
D
j
j j j
(15.2.2a)
n
i ¼ X i
F
Sj
j
i
Y j
i
M
j
oT
(15.2.2b)
We assume that the local axes of the MSE are the same on all floors. The
transformation matrix from the global system of axes to the local one is
2
3
cos fi sin fi 0
Rij ¼ Ri ¼ 4 sin fi cos fi 0 5
0
0
1
(15.2.3)
698 PART
II Multi-degree-of-freedom systems
Hence, the known transformation relations hold
i
Dij ¼ Ri D
j
i T i
i
¼ R D
D
j
j
(15.2.4b)
i
FiSj ¼ Ri F
Sj
(15.2.5a)
i
F
Sj
¼ R
i T
(15.2.4a)
FiSj
(15.2.5b)
Subsequently, we formulate the total vector of displacements of the MSE
in the local and global systems of axes. To be consistent with Fig. 15.2.2,
we place the displacements in the corresponding vectors starting with the n
node and ending with the first, namely
8 i9
D1 >
>
>
>
>
=
< Di >
2
,
(15.2.6a)
Di ¼
>
>
>
>
>⋮ >
;
:
Din
8 i9
>
>
D
>
1>
>
>
>
=
< i >
i
D
2
¼
(15.2.6b)
D
>
>
>
>
>⋮ >
>
>
: i ;
Dn
Obviously, these vectors are associated by the relations
^ iD
i
Di ¼ R
i T i
i ¼ R
^ D
D
where
2
Ri
60
^i ¼ 6
R
6
4⋮
0
0
Ri
⋮
0
⋯
⋯
⋱
0
(15.2.7a)
(15.2.7b)
3
0
0 7
7
7
⋮ 5
Ri
(15.2.8)
By the same reasoning, we define the total vectors of the nodal elastic forces
8 i 9
FS1 >
>
>
>
>
=
< Fi >
S2
i
,
FS ¼
>
>
>⋮ >
>
>
: i ;
FSn
(15.2.9a)
Dynamic analysis of multistory buildings Chapter
8 i 9
>
>
F
>
S1 >
>
>
> i >
=
<
i ¼ FS2
F
S
>
>
>⋮ >
>
>
>
;
: i >
F
15
699
(15.2.9b)
Sn
which are associated with the relations
^ iF
i
FiS ¼ R
(15.2.10a)
S
T
i ¼ R
^ i Fi
F
(15.2.10b)
S
S
Among the components of the nodal elastic forces and the nodal displacements of the MSE, a relation of the following form can be established
ð15:2:11Þ
or
FiS ¼ ki Di
(15.2.12)
i
The matrix k with dimensions N N ðN ¼ 3n Þ is the stiffness matrix of the
i MSE with respect to its nodal displacements. In global axes, this matrix is
transformed as
i D
i ¼ k
i
F
(15.2.13)
S
where
i T i i
i ¼ R
^ kR
^
k
(15.2.14)
The plates undergo translational displacements in the direction of the
y axes and rotations about the z axis. We denote the displacement vector
x,
of the j plate with respect to the O xyz with
8 9
< U j =
Uj ¼ V j , j ¼ 1, 2, …,n
(15.2.15)
: ;
Wj
700 PART
II Multi-degree-of-freedom systems
Hence, the displacement vector of all plates is defined as
8 9
>
U
>
>
=
< 1 >
U
2
¼
U
⋮ >
>
>
;
: >
Un
(15.2.16)
The origin Oj of the axes to which the motion of the plate is referred may
i with U
j is
differ from plate to plate. The relation that links the displacements D
j
T
i ¼ Tj U
j , j ¼ 1, 2,…,n
D
(15.2.17)
i
j
where
2
3
1 0 0
Tji ¼ 4 0 1 0 5
y ij xij 1
(15.2.18)
is the transformation matrix, in which xij , yij represent the coordinates of the
point i referred to the system of axes with origin the point Oj (see
Section 11.11).
Eqs. (15.2.17) are combined as
i ¼ TT U
D
i
where
2
T1i
60
Ti ¼ 6
4⋮
0
0
T2i
⋮
0
⋯
⋯
⋱
⋯
(15.2.19)
3
0
0 7
7
⋮ 5
Tni
(15.2.20)
is the transformation matrix of the i MSE due to the transfer.
Similarly, the elastic forces are transformed to point Oj according to
Eq. (15.2.17)
oi ¼ Tj F
i , j ¼ 1, 2, …,n
F
Sj
i
Sj
which are combined to
8 oi 9
2 1
>
>
F
>
S1 >
T
>
>
>
= 6 i
< oi >
0
FS2Þ
¼6
4⋮
>
>
>⋮ >
>
>
>
>
: oi ;
0
F
Sn
0
T2i
⋮
0
⋯
⋯
⋱
⋯
8 i 9
3> F
>
S1 >
0 >
>
>
>
=
<
i >
7
0 7 FS2
⋮ 5>
>
>
>
>⋮ >
;
: i >
Tni >
F
(15.2.21)
Sn
or
oi ¼ Ti F
i
F
S
S
(15.2.22)
Using Eqs. (15.2.13), (15.2.19), we write Eq. (15.2.22) as
oi U
oi ¼ k
F
S
(15.2.23)
Dynamic analysis of multistory buildings Chapter
15
701
where
i TT
oi ¼ Ti k
k
i
(15.2.24)
is the global stiffness matrix of the i MSE with respect to points Oj .
15.2.3 Mass matrix of the MSE and multistory building
In order to formulate the mass matrix of the i MSE, we assume that it consists
of successive single-story elements. Thus, the mass of the MSE is lumped at its
nodes. If mji is the mass at the j node and Iji the moment of inertia about the local
vertical axis, then the inertial force at the level of the j plate is
i
€
,
i ¼ m
ij D
F
j
Ij
where
2
mji 0
6
ij ¼ 4 0
m
0
0
j ¼ 1, 2, …,n
(15.2.25)
3
mji 0 7
5,
0 Iji
j ¼ 1, 2, …,n
(15.2.26)
Eqs. (15.2.25) are combined to
€
i ¼ m
iD
F
I
where
8 i 9
>
F
>
>
>
> Ii 1 >
<
=
i ¼ F
I2 ,
F
I
>
>
>⋮ >
>
;
: i >
F
In
2
mi1
60
6
i ¼6
m
4⋮
0
i
0
mi2
⋮
0
(15.2.27)
⋯
⋯
⋱
⋯
3
0
0 7
7
7
⋮ 5
min
(15.2.28)
Eq. (15.2.27) is transformed to point Oj as
€
oi ¼ m
oi U
F
I
(15.2.29)
oi ¼ Ti m
i TTi
m
(15.2.30)
where
If mj is the mass of the j plate and Ijc its moment of inertia about the vertical
axis with respect to the center of mass Cj , then the inertial force of the j plate is
€
cU
c ¼ M
c
F
j j , j ¼ 1, 2, …,n
Ij
where
3
mj 0 0
7
c ¼6
M
4 0 mj 0 5, j ¼ 1, 2, …,n
j
0 0 Ijc
(15.2.31)
2
(15.2.32)
702 PART
II Multi-degree-of-freedom systems
Eq. (15.2.31) is combined to
€
cU
c ¼ M
c
F
I
(15.2.33)
where
8 c 9
2 c
FI 1 >
M1
>
>
>
>
>
c =
<F
60
I2
c ¼6
c ¼
, M
F
6
I
>
>
4⋮
⋮
>
>
>
>
: c ;
0
FIn
3
0 ⋯ 0
c ⋯ 0
M
2
⋮
⋱ ⋮
c
⋯ M
0
7
7
7,
5
n
8 c 9
U1 >
>
>
>
>
=
<U
c >
2
c
U ¼
>
⋮ >
>
>
>
;
: c>
U
(15.2.34)
n
Eq. (15.2.33) is transformed to point Oj as
€
oU
o ¼ M
F
I
which
2
T1c
6
o ¼ Tc M
c TT , Tc ¼ 6 0
M
c
4⋮
0
0
T2c
⋮
0
⋯
⋯
⋱
⋯
3
2
3
0
1 0 0
0 7
7 , Tj ¼ 4 0 1 0 5
c
⋮ 5
ycj xcj 1
n
Tc
(15.2.35)
(15.2.36)
15.2.4 Equation of motion of the multistory building
A
A
A T
If PA
j ¼ fPjx Pjy Mjz g denotes the external force that is applied at the point
Aj of the j plate, then it is transformed to point Oj as
Poj ¼ TjA PA
j , j ¼ 1, 2, …,n
(15.2.37)
P o ¼ T A PA
(15.2.38)
which are combined to
where
8 o9
2 1
P1 >
TA
>
>
>
>
>
< Po =
60
6
2
, TA ¼ 6
Po ¼
>
>
4⋮
⋮ >
>
>
;
: o>
Pn
0
T2A
⋮
0
3
2
3
1 0 0
7
⋯ 0 7
6
7
j
7, TA ¼ 4 0 1 0 5 (15.2.39)
5
⋱ ⋮
A
yA
j x
j 1
⋯ TnA
⋯ 0
0
The dynamic equilibrium of the j plate yields
o +
F
Ij
K
X
i¼1
oi +
F
Ij
K
X
i¼1
oi ¼ Po , j ¼ 1, 2,…,n
F
Sj
j
(15.2.40)
Dynamic analysis of multistory buildings Chapter
15
703
which are combined to
o +
F
I
K
X
i¼1
oi +
F
I
K
X
i¼1
oi ¼ Po
F
S
(15.2.41)
Further, using Eqs. (15.2.23), (15.2.29), and (15.2.35), we obtain
where
€ +K
U
U
¼P
M
K
X
¼M
o+
oi
m
M
(15.2.42)
(15.2.43a)
i¼1
¼
K
K
X
oi
k
(15.2.43b)
i¼1
¼ Po
P
(15.2.43c)
The modelling of buildings using the concept of the MSE is not recommended
for the engineering praxis where the analysis is performed using commercial
computer codes. However, it can be used for educational purposes to analyze simple buildings aiming at understanding their dynamic response. From the examples
presented below we conclude that method of MSEs provides good results.
Example 15.2.1 Formulate the equation of motion of the one-story building of
Fig. E15.1 with respect to the center of mass O of the slab for the ground motion
ug ðt Þ in the direction b. Assume:
(a) The columns k1 , k2 , k3 , k4 have a square cross-section ða a Þ with
moments of inertia Ix ¼ Iy ¼ I .
(b) The beams b1 , b2 have a rectangular cross-section ða2a Þ, hence a
moment of inertia Ib ¼ 8I . The effective breadth of the beams as well as
their torsional stiffness are neglected.
(c) Shear modulus G ¼ 0:40E and mass of the slab r per square meter.
FIG. E15.1 One-story building in Example 15.2.1.
704 PART
II Multi-degree-of-freedom systems
Remarks and recommendations:
(i) The theoretical height of the columns is L.
(ii) The theoretical span of the frame k1 b1 k4 is 2L and that of the frame
k2 b2 k1 is L.
(iii) The mass of the columns as well as of the beams below the slab is
neglected. Hence the center of mass of the slab coincides with its geometrical center of mass.
(iv) The slab is assumed undeformable in its plane. Therefore, the axial
deformation of the beams is neglected.
(v) The axial deformation of the columns is neglected. When formulating the
stiffness of the structure, the beams and columns function as frames.
(vi) The columns are assumed fixed on the ground.
(vii) Due to the small stiffness of the slab, the connection between the column
k3 and the slab is a hinge about the horizontal axes.
Solution using the concept of the MSE
Step 1
We determine the MSEs of the structure. From Fig. E15.1 we distinguish the
following MSEs, which are actually one-story elements. Below, they are shown
in their local element axes
It contributes to the stiffness of the
structure with its translational
stiffness kx along the local axis x.
3
kx 0 0
k ¼ 40 0 05
0 0 0
It contributes to the stiffness of the
structure with its translational
stiffness kx along the local axis x.
3
kx 0 0
4
k ¼ 0 0 05
0 0 0
It contributes to the stiffness of the
structure only with its torsional
stiffness kw because its translation
stiffnesses along local axes x,y have
been considered in MSE 1 and MSE 2.
It contributes to the stiffness of the
structure with its translational
stiffness kx along the local axis x and
with the torsional stiffness kw . The
translational stiffness along the local
axis y has already been considered in
MSE 2.
2
1
2
2
2
3
0 0 0
k ¼ 40 0 0 5
0 0 kw
3
2
3
kx 0 0
k4 ¼ 4 0 0 0 5
0 0 kw
Dynamic analysis of multistory buildings Chapter
It contributes to the stiffness of the
structure with its translational
stiffnesses kx and ky along the local
axes x and y as well as with the
torsional stiffness kw .
It contributes to the stiffness of the
structure with its translational
stiffnesses ky along the local axes y
and with the torsional stiffness kw .
The translational stiffness along the
local axis x has been considered in
MSE 1.
15
705
2
3
kx 0 0
k5 ¼ 4 0 ky 0 5
0 0 kw
2
3
0 0 0
4
k ¼ 0 ky 0 5
0 0 kw
6
Step 2
Formulation of the element transformation matrices Ri due to rotation of the local
axes of the MSE. From Fig. E15.1, we have f1 ¼ f3 ¼ f4 ¼ f5 ¼ f6 ¼ 0,
f2 ¼ p=2. Hence
2
3
2
3
1 0 0
0 1 0
1
3
4
5
6
2
R ¼ R ¼ R ¼ R ¼ R ¼ 4 0 1 0 5, R ¼ 4 1 0 0 5
0 0 1
0 0 1
Step 3
i of the ith element depend on the
Because the plate is rigid, the displacements D
displacements U of the point O. Hence the relation that connects them results
uj ¼ D
i , that is
from Eq. (11.11.13) by setting uJ ¼ U,
i ¼ TT U
D
i
where
3
1 0 0
Ti ¼ 4 0 1 0 5
y i xi 1
(1)
2
(2)
The transformation matrix is identical to that of the plane frame
(see Table 11.11.1).
Referring to Fig. E15.1 and using Eq. (2), we obtain
2
3
1
0 0
x1 ¼ 0, y1 ¼ L=2,
T1 ¼ 4 0
1 05
0:5L 0 1
2
3
1 0 0
x2 ¼ L, y2 ¼ 0,
T2 ¼ 4 0 1 0 5
0 L 1
706 PART
II Multi-degree-of-freedom systems
2
x3 ¼ L, y3 ¼ L=2,
x4 ¼ L, y4 ¼ L=2,
x5 ¼ L, y5 ¼ L=2,
x6 ¼ L, y6 ¼ L=2,
1
T3 ¼ 4 0
0:5L
2
1
T4 ¼ 4 0
0:5L
2
1
T5 ¼ 4 0
0:5L
2
1
T6 ¼ 4 0
0:5L
3
0 0
1 05
L 1
3
0 0
1 05
L 1
3
0 0
1 05
L 1
3
0 0
1 05
L 1
Step 4
Formulation of the local stiffness matrices ki of the elements and total stiffness
of the building.
matrix K
MSE 1 (frame k1 b1 k4 )
kx is the translational stiffness of the frame of Fig. a. Assuming flexural
vibrations and working as in Example 11.5.2, the stiffness matrix of the frame
with respect to its three degrees of freedom is
2
3
24 6L
6L
EI
k1 ¼ 3 4 6L 20L2 8L2 5
L
6L 8L2 20L2
and after the static condensation of the rotational degrees of freedom u2 ,u3
2
3
21:43 0 0
EI
0 05
k1 ¼ 3 4 0
L
0
0 0
MSE 2 (frame k1 b2 k2 )
Dynamic analysis of multistory buildings Chapter
15
707
kx is the translational stiffness of the frame of Fig. b. The stiffness matrix of the
frame with respect to its three degrees of freedom is
2
3
24 6L
6L
EI
k2 ¼ 3 4 6L 36L2 16L2 5
L
6L 16L2 36L2
and after the static condensation of the rotational degrees of freedom u2 , u3
2
3
22:61 0 0
EI
0 05
k2 ¼ 3 4 0
L
0
0 0
MSE 3, MSE 4, MSE 5, MSE 6
The stiffnesses kx , ky , kw of elements are computed from the relations
kx ¼ 3EI y =L3 , ky ¼ 3EI x =L3 , kw ¼ GI t =L
For a square cross-section a a it is: Ix ¼ Iy ¼ a 4 =12, It ¼ 0:140a 4 ¼ 1:68I .
Hence
2
3
2
3
0 0 0
3 0 0
EI
EI
5, k4 ¼ 4 0 0 0
5
k3 ¼ 3 4 0 0 0
L
L3
2
2
0 0 0:672L
0 0 0:672L
2
3
2
3
3 0 0
0 0 0
EI 4
EI
5
6
5, k ¼ 4 0 3 0
5
k ¼ 3 0 3 0
L
L3
2
2
0 0 0:672L
0 0 0:672L
of the building with respect to point O is computed
The stiffness matrix K
from the relation
o ¼
K
6
X
T
T i Ri k i Ri ð T i Þ T
i¼1
¼
2
EI 6
4
L3
27:43
0
7:71L
3
7
0
28:61 16:61L 5
7:71L 16:61L 38:16L2
Step 5
of the building with respect to point O.
Formulation of the mass matrix M
Because it is assumed that only the mass of the plate is taken into account,
the mass matrix will result from the inertial properties of the plate. Given that
point O coincides with the mass center of the plate, we have
m ¼ 3:36L2 r
Io ¼
ð2:4LÞ ð1:4LÞ3
ð2:4LÞ3 ð1:4LÞ
r ¼ 2:16L4 r
r +
12
12
708 PART
Hence
II Multi-degree-of-freedom systems
2
3
3:36 0
0
o ¼ rL2 4 0
5
M
3:36 0
2
0
0
2:16L
Step 6
The equation of motion reads
€
U
+K
U
¼ Mb
u€g ðt Þ
M
is the vector of the relative displacement with respect to the ground
where U
and b ¼ f cos b sin b 0 gT .
For a ¼ 0:30 m, L ¼ 3 m, E ¼ 2:1 107 kN=m2 , r ¼ 0:4 kN m1 s2 =m2
we obtain
2
3
14400:75
0
12143:25
¼4
K
0
15020:25 26160:75 5
12143:25 26160:75
180306:0
2
3
12:09 0
0
¼4 0
M
12:09 0 5
0
0
69:98
Solution using model (b)
The building is approximated by the space frame of Fig. E15.2. The top
of element 2 is assumed hinged to the slab permitting the relative rotation about
the horizontal axes. The slab behaves as a plane rigid body, which restrains the
horizontal displacements and the rotations about the z axis.
This procedure yields a stiffness matrix, which is practically identical to that
obtained using the concept of the MSE.
Another approach that facilitates the solution is to assume two additional
beam elements (elements 5 and 6) with a very small stiffness.
FIG. E15.2 Space frame modeling of the one-story building in Example 15.2.1.
Dynamic analysis of multistory buildings Chapter
15
709
Table E15.1 shows the eigenfrequencies obtained using the MSE modeling
and SAP2000. In both models the torsional stiffness of the beams has been
neglected.
TABLE E15.1 Eigenfrequencies with MSE modeling and
SAP2000.
wn
MSE modeling
SAP2000 modeling
w1
26.5784
26.5776
w2
34.6436
34.6437
w3
55.6964
55.6911
Example 15.2.2 Formulate the equation of motion of the two-story building shown
in Fig. E15.3 with respect to the origin of the global axes O (center of column 1)
when it is subjected to the ground motion ug ðt Þ in the x direction and evaluate
the eigenfrequencies and eigenmodes. The building is made from reinforced concrete with material constants E ¼ 2:1 107 kN=m2 , n ¼ 0:25. The thickness of the
slabs is 15 cm, the cross-sectional area of the columns is 40 40 cm2 , and of the
beams 30 50 cm2 . The columns are assumed fixed at the ground. The loads of the
slabs are: 5 kN=m2 live load and 1 kN=m2 covering.
Planform of the stories
Section
FIG. E15.3 Two-story building in Example 15.2.2.
Solution using the concept of the MSE
Cross-sectional moment of inertia and torsional constant of the columns:
Ix ¼ Iy ¼
0:404
¼ 21:33 104 m4 , It ¼ 0:141 0:404 ¼ 36:01 104 m4
12
Cross-sectional moment of inertia of the beams:
To simplify the solution and limit the numerical computations, the effective
breadth of the beams is neglected. Thus, we have
710 PART
II Multi-degree-of-freedom systems
Ib ¼
0:30 0:503
¼ 31:25 104 m4
12
a. Equation of motion
Step 1
Determination of the MSEs of the structure. Referring to Fig. E15.3, we distinguish six MSEs of the frame type, three in the direction of the x axis and three in
the direction of the y axis. The columns are treated as additional MSEs contributing with their torsional stiffness because their translational stiffnesses are
included in the frames. The stiffness matrices result using the method presented
in Example 11.5.2. The stiffness matrices of all MSEs are given below after the
static condensation of the rotational degrees of freedom.
2
29237:03
6
0
6
6
0
1
6
k ¼6
6 16737:43
4
0
0
2
44851:12
6
0
6
6
0
2
6
k ¼6
6 26194:78
4
0
0
2
44851:12
6
0
6
6
0
3
6
k ¼6
6 26194:78
4
0
0
2
41960:02
6
0
6
6
0
4
6
k ¼6
23362:92
6
4
0
0
0
0
0
0
0
0
3
0 16737:43 0 0
0
0
0 07
7
0
0
0 07
7
0 13897:72 0 0 7
7
0
0
0 05
0
0
0 0
0
0
0
0
0
0
3
0 26194:78 0 0
0
0
0 07
7
0
0
0 07
7
0 22307:63 0 0 7
7
0
0
0 05
0
0
0 0
0
0
0
0
0
0
3
0 26194:78 0 0
0
0
0 07
7
0
0
0 07
7
0 22307:63 0 0 7
7
0
0
0 05
0
0
0 0
0
0
0
0
0
0
3
0 23362:92 0 0
0
0
0 07
7
0
0
0 07
7
0 18828:40 0 0 7
7
0
0
0 05
0
0
0 0
Dynamic analysis of multistory buildings Chapter
2
41960:02
6
0
6
6
0
5
k ¼6
6 23362:92
6
4
0
0
2
27145:69
6
0
6
6
0
6
6
k ¼6
14736:02
6
4
0
0
2
0
60
6
60
k7, 8, …, 14 ¼ 6
60
6
40
0
15
711
0
0
0
0
0
0
3
0 23362:92 0 0
0
0
0 07
7
0
0
0 07
7
0 18828:40 0 0 7
7
0
0
0 05
0
0
0 0
0
0
0
0
0
0
3
0 14736:02 0 0
0
0
0 07
7
0
0
0 07
7
0 11490:06 0 0 7
7
0
0
0 05
0
0
0 0
3
0
0
0 0
0
0
0
0 0
0 7
7
0 15364:52 0 0 8642:40 7
7
0
0
0 0
0 7
7
0
0
0 0
0 5
0 8642:40 0 0 8642:40
Step 2
Formulation of the transformation matrices Ri due to the rotation of the local
axes. From Fig. E15.3, we have f1 ¼ f2 ¼ f3 ¼ f7 ¼ f8 ¼ ⋯ ¼ f14 ¼ 0,
f4 ¼ f5 ¼ f6 ¼ п=2. Hence
3
3
2
2
1 0 0 0 0 0
0 1 0 0 0 0
60 1 0 0 0 07
6 1 0 0 0 0 0 7
7
7
6
6
7
6
6 0 0 1 0 0 07
0
0
1
0
0
0
1
2
3
4
5
6
^
^
^
^
^
^
7
7
6
6
R ¼R ¼R ¼6
7, R ¼ R ¼ R ¼ 6 0 0 0 0 1 0 7
7
60 0 0 1 0 07
6
40 0 0 0 1 05
4 0 0 0 1 0 0 5
0 0 0 0 0 1
0 0 0 0 0 1
2
1
60
6
6
^7 ¼ R
^8 ¼ ⋯ ¼ R
^ 14 ¼ 6 0
R
60
6
40
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
3
0
07
7
07
7
07
7
05
1
Step 3
Formulation of the transformation matrices Ti due to translation of the local
axes. Apparently, in this case, any point of the local x axis can be taken as point i.
Referring to Fig. E15.3 and applying Eq. (15.2.20), we obtain
712 PART
II Multi-degree-of-freedom systems
3
1 0 0 0 0 0
60 1 0 0 0 07
7
6
60 0 1 0 0 07
1
1
1
1
7
6
MSE 1 : x1 ¼ x2 ¼ 0, y1 ¼ y2 ¼ 0, T1 ¼ 6
7
60 0 0 1 0 07
40 0 0 0 1 05
0 0 0 0 0 1
3
2
1 0 0 0 0 0
6 0 1 0 0 0 07
7
6
6 8 0 1 0 0 0 7
7
MSE 2 : x21 ¼ x22 ¼ 0, y21 ¼ y22 ¼ 8:0, T2 ¼ 6
6 0 0 0 1 0 07
7
6
4 0 0 0 0 1 05
0 0 0 8 0 1
3
2
1 0 0 0 0 0
6 0 1 0 0 0 07
7
6
6 16 0 1 0 0 0 7
3
3
3
3
7
6
MSE 3 : x1 ¼ x2 ¼ 0, y1 ¼ y2 ¼ 16:0, T3 ¼ 6
7
6 0 0 0 1 0 07
4 0 0 0 0 1 05
0 0 0 16 0 1
3
2
1 0 0 0 0 0
60 1 0 0 0 07
7
6
60 0 1 0 0 07
7
6
MSE 4 : x41 ¼ x42 ¼ 0, y41 ¼ y42 ¼ 0, T4 ¼ 6
7
60 0 0 1 0 07
7
6
40 0 0 0 1 05
0 0 0 0 0 1
3
2
1 0 0 0 0 0
60 1 0 0 0 07
7
6
60 5 1 0 0 07
5
5
5
5
7
6
MSE 5 : x1 ¼ x2 ¼ 5:0, y1 ¼ y2 ¼ 0, T5 ¼ 6
7
60 0 0 1 0 07
40 0 0 0 1 05
0 0 0 0 5 1
3
2
1 0 0 0 0 0
60 1 0 0 0 07
7
6
6 0 10 1 0 0 0 7
7
MSE 6 : x61 ¼ x62 ¼ 10:0, y61 ¼ y62 ¼ 0, T6 ¼ 6
60 0 0 1 0 07
7
6
40 0 0 0 1 05
0 0 0 0 10 1
2
The reaming MSEs are columns, which contribute only with their rotational
stiffness. Thus, we have independently from the directions of their local axes
Dynamic analysis of multistory buildings Chapter
2
1
60
6
60
T7 ¼ T8 ¼ ⋯ ¼ T14 ¼ 6
60
6
40
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
15
713
3
0
07
7
07
7
07
7
05
1
Step 4
Formulation of the stiffness matrix of the structure. Applying Eq. (15.2.43b)
yields the stiffness matrix of the structure with respect to the origin O of
the axes.
Ko ¼
14
X
T i
^i k R
^ i TT
Ti R
i
3
2 i¼1
118939:27
0
1076426:90 69126:99
0
628674:79
7
6
0
111065:74
481257:02
0
61461:87
264174:87 7
6
6
7
6 1076426:90
7
481257:02
18238842:00
628674:79 264174:87 10509145:00 7
¼6
6 69126:99
0
628674:79
58512:99
0
535383:28 7
6
7
6
7
4
0
61461:87
264174:87
0
49146:87
209042:62 5
628674:79 264174:87 10509145:00 535383:28
209042:62
8827299:10
Step 5
Formulation of the mass matrix of the structure. The surface of the plates is
120:0 m2 . Neglecting the mass of the MSEs, we have
24
6
+
m1 ¼ m2 ¼ ð8:40 10:40 + 8:0 5:40Þ 0:15 9:81 9:81
¼ 127:76 kN m1 s2
xc1 ¼ xc2 ¼ 4:17, yc1 ¼ yc2 ¼ 9:29, I1c ¼ I2c ¼ 3680:24
Hence the mass matrix with respect to the center of mass is
3
2
127:76
0
0
0
0
0
6 0
127:76
0
0
0
0 7
7
6
6
0
0
3680:24
0
0
0 7
c
7
6
M ¼6
0
0
127:76
0
0 7
7
6 0
4 0
0
0
0
127:76
0 5
0
0
0
0
0
3680:24
Moreover, the transformation matrix from C to O is
714 PART
II Multi-degree-of-freedom systems
2
1
6 0
6
6 9:29
Tc ¼ 6
6 0
6
4 0
0
0
1
4:17
0
0
0
0 0
0 0
1 0
0 1
0 0
0 9:29
0
0
0
0
1
4:17
3
0
07
7
07
7
07
7
05
1
and applying Eq. (15.2.36) yields the mass matrix with respect to point O
Mo ¼ Tc Mc TTc
2
127:76
0
6
0
127:76
6
6 1186:89 532:75
¼6
6
0
0
6
4
0
0
0
0
1186:89
0
0
532:75
0
0
16928:06
0
0
0
127:76
0
0
0
127:76
0
1186:89 532:75
3
0
0 7
7
0 7
7
1186:89 7
7
532:75 5
16928:06
Therefore the equation of motion is
€
+ Ko U
¼ Mo bug ðt Þ, b ¼ f 1 0 0 1 0 0 gT
Mo U
b. Eigenfrequencies and eigenmodes
They are computed using the function [PHI,D]¼eig(K,M) of MATLAB.
Table E15.2 shows the eigenfrequencies obtained using the MSE modeling
and SAP2000. In both models the torsional stiffness of the beams has been
neglected.
TABLE E15.2 Eigenfrequencies MSE modeling and SAP2000.
wn
MSE modeling
SAP2000 modeling
w1
9.390
9.377
w2
10.192
10.085
w3
13.794
13.688
w4
34.129
33.780
w5
35.814
35.299
w6
48.829
48.332
Dynamic analysis of multistory buildings Chapter
2
0:0014
0:0455
0:0836 0:0025
15
0:0684 0:1323
715
3
7
6
6 0:0471 0:0011 0:0363 0:0768 0:0015 0:0557 7
7
6
7
6
6 0:0002 0:0003 0:0088 0:0003 0:0006 0:0139 7
7
6
F¼6
7
6 0:0018 0:0701 0:1315 0:0019 0:0440 0:0844 7
7
6
6 0:0762 0:0014 0:0561 0:0475 0:0013 0:0356 7
5
4
0:0002 0:0004
0:0139
0:0003
0:0005
0:0088
15.3 Dynamic response of multistory buildings
due to ground motion
We consider a multi-story building under uniform support excitation (Fig.
15.3.1). We first formulate the mass and stiffness matrices of the free structure
using any method described in Section 15.1 or 15.2.
The displacement vector of the j slab is written as
j ¼ Uj Iug
U
(15.3.1)
j are the relative displacements of the j slab with respect to the
where U
T
ground, ug ¼ f ug vg wg g the vector of the ground displacements, and I
(a)
(b)
FIG. 15.3.1 Multistory building (a) with support excitation (b).
716 PART
II Multi-degree-of-freedom systems
the 3 3 unit matrix. Usually, it is wg ¼ 0. Then the total displacement vector
of the building is written
+ rug
u¼u
(15.3.2)
where
r ¼ f I I ⋯ I gT
(15.3.3)
Taking into account that the elastic and damping forces are produced only
by the relative displacements, we can write the equation of motion as (see
Section 14.7.2)
€ + Cu
_ + K
Mu
u ¼ Mr€
ug
(15.3.4)
it is
For ground motion in direction x,
r ¼ f 1 0 0 1 0 0 ⋯ 1 0 0 gT
(15.3.5)
Apparently, it is convenient to refer the displacements to the center of mass
c ¼ TTc u
, u
¼ TT
c and
of the plates. Then, taking into account that u
c u
c T
M ¼ Tc M Tc , Eq. (15.3.4) becomes
c
c
€ + Tc Cu
_ + Tc K u
c ¼ Tc Mc TTc r€
Tc M c u
ug
(15.3.6)
Then premultiplying by T1
c gives
c
c
€ + Cc u
_ + Kc u
c ¼ Mc rc u
€g
Mc u
(15.3.7)
T
c
1
T
Cc ¼ T1
c CTc , K ¼ Tc KTc
(15.3.8)
where
are the transformed damping and stiffness matrices with respect to the center of
mass of the plates and
8 T 9
3
2 1
>
0
⋯ 0
T1c >
T1c
>
>
>
>
2 1
< 2 T =
7
6
7
6
Tc
⋯ 0
Tc
(15.3.9)
, T1
rc ¼ T T
¼ 60
7
c r¼>
c
>
5
4
⋮
⋮
⋱
⋮
⋮
>
>
>
>
1
: n T ;
0
0
⋯ Tnc
Tc
To avoid the inversion Tc , we can readily show that (see Section 11.11)
2
3
1
0 0
j 1
1 0 5 j ¼ 1, 2, …,n
Tc
¼ 40
c
yj xcj 1
1 T
We recall the notation TT
(see Section 12.5.3).
c ¼ Tc
(15.3.10)
Dynamic analysis of multistory buildings Chapter
15
717
Example 15.3.1 The building in Example 15.2.2 is subjected to the ground
motion in the x direction due to the 1999 Athens earthquake (see Chapter 6).
Determine:
2 at the top of column 1, the base shear force Qbx , the
2, W
(i) The peak values U
overturning moment Mbx , and the base torsion moment Tb .
(ii) The peak values of the shear force Qx , the bending moment Mx , and the
torsion moment at the base of column 1 (cross-section a).
Assume modal damping x ¼ 0:05 for all mode shapes
Solution
2 , Qbx , Mbx , Tb
2, W
(i) Peak values U
The eigenfrequencies and mode shapes were computed in Example 15.2.2.
Adhering to the steps in Example 14.8.1, we have
Computation of the vectors Rn .
It is R ¼ Mo r, where r ¼ f 1 0 0 1 0 0 gT . Then Eq. (14.3.14) gives
G1 ¼ 0:057, G2 ¼ 15:640, G3 ¼ 0:478
G4 ¼ 0:035, G5 ¼ 3:263, G6 ¼ 0:190
and we obtain from Eq. (14.3.13)
R1
R2
R3
0:001
96:773 0:072
0:338
0:367 0:041
1:360 918:659 15:131
0:002
147:830 0:157
0:549
0:639 0:084
2:207 1399:871 23:322
R4
R5
0:004
30:823
0:341
0:404
1:339 295:177
0:003 20:176
0:210
0:243
0:811 194:117
R6
0:086
0:051
9:115
0:005
0:027
5:854
st , Q st , M st , T st . The
st , W
First, we compute the quantities U
2n
2n
bxn
bxn
bn
n ¼ K1 Rn , which
modal displacements are computed from the relation U
by virtue
of Eq. (14.3.14) and the relation K ¼ w2n M becomes
2
n ¼ Gn =w fn . Hence
U
n
st ¼ Gn f
st ¼ Gn f1n , W
U
2n
2n
w2n
w2n 3n
and
st
st
st
Qbxn
¼ R1n + R4n , Mbxn
¼ h1 R1n + ðh1 + h2 ÞR4n , Tbn
¼ R3n + R6n
Their values are given in Table E15.3. The peak modal values were
computed from Eq. (14.8.4), namely
qn0 ¼ qnst Spa ðx n , Tn Þ
2 , Qbx , Mbx , Tb Þ and Spa ðx , Tn Þ the spectral pseudoacce 2, W
where q U
n
leration, which was computed numerically using the program response_
spectrum_aem.m (see Section 6.2) with x ¼ 0:05. The computed values
718 PART
II Multi-degree-of-freedom systems
TABLE E15.3 Modal quantities.
2 105
W
n
st
U 2n 105
st
Qbxn
st
Mbxn
st
Tbn
1
0.119
0.015
0.003
0.021
3.567
2
1055.796
6.248
244.603
1618.120
2318.530
3
33.054
3.489
0.229
1.578
38.454
4
0.006
0.0008
0.001
0.003
0.528
5
11.196
0.120
10.647
22.705
101.060
6
0.673
0.071
0.036
0.012
3.261
are given in Table E15.4 together with the peak values of the modal
quantities.
Applying Eq. (14.8.8) gives the correlation matrix
2
1000:00 602:02
61:70
4:29
3:89
6 602:02 1000:00
96:66
5:08
4:59
6
6
6 61:70
96:66 1000:00
10:19
9:04
3
r ¼ 10 6
6
4:29
5:08
10:19 1000:00 811:16
6
6
4
3:89
4:59
9:04 811:16 1000:00
2:16
2:51
4:52
70:45
3
2:16
2:51 7
7
7
4:52 7
7
70:45 7
7
7
92:45 5
92:45 1000:00
Obviously, the off-diagonal elements are not negligible. That was anticipated because there are eigenfrequencies close to each other. Therefore, the
CQC method should be used to compute the peak values. The resulting
values are given in Table E15.5 as compared with those obtained by direct
solution of the equation of motion (Response History Analysis), namely
€
+ Ko U
¼ Mo bug ðt Þ, b ¼ f 1 0 0 1 0 0 gT
Mo U
which has been solved using the MATLAB program aem_lin_MDOF.m
(see Section 14.10.3).
Apparently, although the results obtained by the CQC method are close
to those obtained by the RHA for the displacements, the error may be appreciable for the forces.
(ii) Peak values of Qx and Mx at cross-section a of column 1.
The degrees of freedom of the frame are shown in Fig. E15.4. The displacements u1 , u2 result from the displacements of the MSE 1 while the rotation
from the MSE 7. From the relations (15.2.7a), (15.2.19), we have
^ i TT U
Di ¼ R
i
TABLE E15.4 Peak values of the modal quantities.
Spa ðx n , Tn Þ
U 2n0 105
2n0 105
W
Qbxn0
Mbxn0
Tbn0
(m)
(rad)
(kN)
(kNm)
(kNm)
n
Tn
(m/sec )
1
0.669
3.7824
0.451
0.055
2
0.616
4.1419
4373.003
25.866
3
0.455
3.4180
112.98
4
0.184
5.3077
5
0.175
6
0.129
2
0.0123
0.081
13.493
1012.123
6702.093
9603.121
11.924
0.782
5.395
131.435
0.031
0.004
0.007
0.0179
6.0243
67.447
0.723
64.141
136.782
608.816
4.6813
3.151
0.331
0.169
0.055
15.264
2.803
720 PART
II Multi-degree-of-freedom systems
TABLE E15.5 Peak values.
U 2 max
W
2
max
jQbx jmax
jMbx jmax
jTb jmax
(cm)
(rad)
(kN)
(kNm)
(kNm)
CQC
4.392
0.00027
1015.55
6703.40
9605.00
RHA
4.440
0.00032
1028.28
6726.98
9806.23
FIG. E15.4 Degrees of freedom of MSE 1.
Hence
8 9
8 9
0 >
U1>
>
>
>
>
>
>
>
>
>
>
>
> V >
>
>
>
>
0 >
1
>
>
>
>
>
>
>
>
>
>
>
> =
=
<
<
W
0
1
1
1
T
7
7
T
^ T U
^ T U
¼
¼
,
D
D ¼R
¼
R
1
7
>
>
0 >
>
>
>
> U 2 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
V
>
>
>
>
2
>
>
>
;
: >
: ;
W2
0
which yields
1
u1 ¼ u 1 ¼ U 1 , u2 ¼ u 1 ¼ U 2 , w7 ¼ W