Mechanical Vibrations: Theory and Applications (Allyn and Bacon series in mechanical engineering and applied mechanics)
2,788 88 19MB
English Pages 464 [468] Year 1978
ECHANICAL ;
_ VIBRATIONS ~ THEORY AND APPLICATIONS

_ SECOND EDITION
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This book is part of the
ALLYN AND BACON SERIES IN MECHANICAL AND APPLIED MECHANICS Consulting Editor:
FRANK KREITH University of Colorado
ENGINEERING
FRANCIS S. TSE University of Cincinnati
IVAN E. MORSE University of Cincinnati
ROLLAND
T. HINKLE
Michigan State University
Mechanical Vibrations Theory and Applications SECOND
Allyn and Bacon, Inc. Boston / London / Sydney / Toronto
EDITION
Copyright © 1978, 1963 by Allyn and Bacon, Inc. 470 Atlantic Avenue, Boston, Massachusetts 02210.
All rights reserved. Printed in the United States of America. No part of the material protected by this copyright notice may be reproduced or utilized in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the copyright owner.
Printing number and year (last digits) :
10987654
85 84 83 82 81 80
Library of Congress Cataloging in Publication Data Tse, Francis Sing Mechanical vibrations.
(Allyn and Bacon series in Mechanical engineering and applied mechanics) Includes index. 1. Vibrations. I. Morse, Ivan E., joint author. II. Hinkle, Rolland Theodore, joint author. ill. Title.
TA355.T77
1978
620.3
ISBN 0205059406 ISBN 0205066704 (International)
7720933
Contents
Preface
xi
CHAPTER1 11 12 13 14 15 6 7
Primary Objective 1 Elements of a Vibratory System Examples of Vibratory Motions Simple Harmonic Motion 8 Vectorial Representation of Harmonic Motions et Units 16 Summary 19 Problems 20
CHAPTER 21 22 23 24 25
INTRODUCTION
2
2 5
SYSTEMS WITH ONE DEGREE OF FREEDOM—THEORY
Introduction 23 Degrees of Freedom 25 Equation of Motion—Energy Method Equation of Motion—Newton’s Law of Motion 33 General Solution 34 Complementary Function 34 Particular Integral 38 General Solution 42
py]
vi
Contents
26
, 27
28 29
Frequency Response Method 45 Impedance Method 45 Transfer Function 49 Resonance, Damping, and Bandwidth Transient Vibration SZ Impulse Response 53 Convolution Integral 55 Indicial Response 57 Comparison of Rectilinear and Rotational Systems 58 Summary 58 Problems 62
CHAPTER 1 2 3 4 35
36
 WwW ~ WW \o 0
3
51
SYSTEMS WITH ONE DEGREE OF FREEDOM— APPLICATIONS
69
Introduction 69 Undamped Free Vibration 70 DampedFree Vibration 77 Undamped Forced Vibration— Harmonic Excitation 80 Damped Forced Vibration— Harmonic Excitation 86 Rotating and Reciprocating Unbalance 87 Critical Speed of Rotating Shafts 89 Vibration Isolation and Transmissibility 94 Systems Attached to Moving Support 98 Seismic Instruments 101 Elastically Supported Damped Systems 106 Damped Forced Vibration— Periodic Excitation 109 Transient Vibration—Shock Spectrum 116 Equivalent Viscous Damping 122 Summary 129 Problems 37
CHAPTER 4
SYSTEMS WITH MORE THAN ONE DEGREE OF FREEDOM
41
Introduction
42
Equations of Motion:
142
Newton’s Second Law
143
142
Vii
Contents
Undamped Free Vibration: Principal Modes 145 Generalized Coordinates and Coordinate Coupling 155 Principal Coordinates 158 Modal Analysis: Transient Vibration of Undamped Systems 160 Semidefinite Systems 165 Forced Vibration—Harmonic Excitation Influence Coefficients 175 Summary 180 Problems
CHAPTER
5
169
18]
METHODS FOR FINDING NATURAL FREQUENCIES
190
Introduction 190 Dunkerley’s Equation 190 Rayleigh Method 193 Holzer Method 197 Transfer Matrix 202 MyklestadProhl Method 207 Summary 213 Problems 214
CHAPTER6
218
DISCRETE SYSTEMS
61 62
Introduction 218 Equations of Motion—Undamped Systems 219 Undamped Free Vibration—Principal Modes 223 Orthogonality and Principal Coordinates Normal Coordinates 229 Expansion Theorem 230 Rayleigh’s Quotient 231 Semidefinite Systems 232 Matrix Iteration 234 Undamped Forced Vibration—Modal Analysis 238 Systems with Proportional Damping 12 Orthogonality of Modes of Damped Systems 241 613 Damped Forced Vibration—Modal Analysis 243
226
viii
Contents
614
Summary Problems
CHAPTER7
245 246
CONTINUOUS
SYSTEMS
253
71 72
Introduction 253 Continuous Systems— A Simple Exposition 253 73 Separation of the Time and Space Variables 256 74 Problems Governed by the Wave Equation 258 258 Longitudinal Vibration of Rods 261 Torsional Vibration of Shafts 75 Lateral Vibration of Beams 262 265 76 Rotary Inertia and Other Effects Shear Deformation and Rotary Inertia Effects 266 Effect of Axial Loading 268 77 The Eigenvalue Problem 270 78 Orthogonality 94 G2 Boundary Conditions Independent of A 273 Boundary Conditions Dependent on A 275 79 Lagrange’s Equations 280 710 Undamped Forced Vibration—Modal Analysis 285 711 Rayleigh’s Quotient 288 12 RayieighRitz Method 290 13 Summary 294 Problems 295
CHAPTER8 81 82 83 84 85
6 7
NONLINEAR
SYSTEMS
Introduction 300 Stability and Examples of Nonlinear Systems 301 The Phase Plane 303 Stability of Equilibrium 306 Graphical Methods 316 Isocline Method 317 Pell’s Method 319 Selfexcited Oscillations 321 Analytical Methods 323
300
Contents
88
89
810
Free Vibration 323 Perturbation Method 323 Variation of Parameter Method Harmonic Balance 327 Forced Vibration 328 Jump Phenomenon 328 Subharmonic Oscillation 332 Summary 334 Problems 335
CHAPTER9 91 92 3 4 5 6
97 98
99
325
SOLUTIONS BY DIGITAL COMPUTERS
339
Introduction 339 Onedegreeoffreedom Systems—Transient Response 341 Program—TRESP1 342 Program—TRESPSUB 346 Program—TRESPF1 349 Onedegreeoffreedom Systems—Harmonic Response 352 Ndegreeoffreedom Systems—Harmonic Response 356 Transient Response of Undamped Discrete Systems 360
Rayleigh’s Method—Undamped
Multirotor Systems 365 910 MyklestadProhl Method—Transfer Matrix Technique 369 911 Matrix Iteration—Undamped Discrete Systems 371 12 Transient Response of Damped Systems 13 Summary 380 Problems 385
APPENDIX A_ APPENDIX B_ APPENDIX C_ APPENDIX D_
Index
ix
376
Elements of Matrix Algebra Lagrange’s Equations Subroutines Linear Ordinary Differential Equations with Constant Coefficients
391 402 408
429
441
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Preface
Vibration is the study of oscillatory motions. The ultimate goals of this study are to determine the effect of vibration on the performance and safety of systems, and to control its effects. With the advent of high performance machines and environmental control, this study has become a part of most engineering curricula. This text presents the fundamentals and applications of vibration theory. It is intended for students taking either a first course or a oneyear sequence in the subject at the junior or senior level. The student is assumed to have an elementary knowledge of dynamics, strength of materials, and differential equations, although summaries of several topics are included in the appendices for review purposes. The format of its predecessor is retained, but the text material has been substantially rewritten. In view of the widespread adoption of the International System of Units (SI) by the industrial world, SI units are used in the problems. The objectives of the text are first, to establish a sense of engineering reality, second, to provide adequate basic theory, and finally, to generalize
these concepts for wider applications. The primary focus of the text is on the engineering significance of the physical quantities, with the mathematical structure providing a supporting role. Throughout the text, examples of applications are given before the generalization to give the student a frame of reference, and to avoid the pitfall of overgeneralization. To further enhance engineering reality, detailed digital computations for discrete systems are presented so that the student can solve meaningful numerical problems. The first three chapters examine systems with one degree of freedom. General concepts of vibration are described in Chapter 1. The theory of
xi
xii
Preface
time and frequency domain analysis is introduced in Chapter 2 through the study of a generalized model, consisting of the mass, spring, damper, and excitation elements. This provides the basis for modal analyses in subsequent chapters. The applications in Chapter 3 demonstrate that the elements of the model are, in effect, equivalent quantities. Although the same theory is used, the appearance of a system in an engineering problem may differ greatly from that of the model. The emphasis of Chapter 3 is on problem formulation. Through the generalization and classification of problems in the chapter, a new encounter will not appear as a stranger. Discrete systems are introduced in Chapter 4 using systems with two degrees of freedom. Coordinate coupling is treated in detail. Common methods of finding natural frequencies are described in Chapter 5. The material in these chapters is further developed in Chapter 6 using matrix techniques and relating the matrices to energy quantities. Thus, the student would not feel the artificiality in the numerous coordinate transformations in the study. The onedimensional wave equation and beam equation of continuous systems are discussed in Chapter 7. The material is organized to show the similarities between continuous and discrete systems. Chapter 8, on nonlinear systems, explains certain common phenomena that cannot be predicted by linear theory. The chapter consists of two main parts, conforming to the geometric and analytical approaches to nonlinear studies. The digital computation in Chapter 9 is organized to follow the sequence of topics presented in the prior chapters and can be assigned concurrently with the text material. The programs listed in Table 91 are sufficient for the computation and plotting of results for either damped or undamped discrete systems. Detailed explanations are given to aid the student in executing the programs. The programs are almost conversational and only a minimal knowledge of FORTRAN is necessary for their execution. The first five chapters constitute the core of an elementary, onequarter terminal course at the junior level. Depending on the purpose of the particular course, parts of Section 35 can be used as assigned reading. Sections 36 through 38, Section 49, and Sections 54 through 56 may be omitted without loss of continuity. For a onesemester senior or duallevel course, the instructor may wish to use Chapters 1 through 4, Chapter 6, and portions of Chapter 7 or 8. Some topics, such as equivalent viscous damping, may be omitted. Alternatively, the text has sufficient material for a oneyear sequence at the junior or senior level. Generally, the first course in mechanical vibrations is required and the second is an elective. The material covered will give the student a good background for more advanced studies.
We would like to acknowledge our indebtedness to many friends, students, and colleagues for their suggestions, to the numerous writers who
Preface
xiii
contributed to this field of study, and to the authors listed in the references. We are especially grateful to Dr. James L. Klemm for his suggestions in Chapter 9, and to K. G. Mani for his contribution of the subroutine $PLOTF in Appendix C. Francis S. Tse Ivan E. Morse Rolland T. Hinkle

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1 Introduction
11
PRIMARY
OBJECTIVE
The subject of vibration deals with the oscillatory motion of dynamic systems. A dynamic system is a combination of matter which possesses mass and whose parts are capable of relative motion. All bodies possessing mass and elasticity are capable of vibration. The mass is inherent of the body, and the elasticity is due to the relative motion of the parts of the body. The system considered may be very simple or complex. It may be in the form of a structure, a machine or its components, or a group of machines. The oscillatory motion of the system may be objectionable, trivial, or necessary for performing a task. The objective of the designer is to control the vibration when it is objectionable and to enhance the vibration when it is useful, although vibrations in general are undesirable. Objectionable vibrations in a machine may cause the loosening of parts, its malfunctioning, or its eventual failure. On the other hand, shakers in foundries and vibrators in
testing machines require vibration. The performance of many instruments depends on the proper control of the vibrational characteristics of the devices. The primary objective of our study is to analyze the oscillatory motion of dynamic systems and the forces associated with the motion. The ultimate goal in the study of vibration is to determine its effect on the performance and safety of the system under consideration. The analysis of the oscillatory motion is an important step towards this goal. Our study begins with the description of the elements in a vibratory system, the introduction of some terminology and concepts, and the discussion of simple harmonic motion. These will be used throughout the text. Other concepts and terminology will be introduced in the appropriate places as needed.
2
12
Introduction
ELEMENTS
OF A VIBRATORY
CHAP. 1
SYSTEM
The elements that constitute a vibratory system are illustrated in Fig. 11. They are idealized and called (1) the mass, (2) the spring, (3) the damper,
and
(4) the
excitation.
The
first three
elements
describe
the
physical system. For example, it can be said that a given system consists of a mass, a spring, and a damper arranged as shown in the figure. Energy _may be stored in the mass and the spring and dissipated in the damper in the form of heat. Energy enters the system through the application of an _ excitation. As shown in Fig. 11, an excitation force is applied to the mass m of the system. The mass m is assumed to be a rigid body. It executes the vibrations and can gain or lose kinetic energy in accordance with the velocity change
of the body. From Newton’s law of motion, the product of the mass and its acceleration is equal to the force applied to the mass, and the acceleration takes place in the direction in which the force acts. Work is force times displacement in the direction of the force. The work is transformed into the kinetic energy of the mass. The kinetic energy increases if work is positive and decreases if work is negative. The spring k possesses elasticity and is assumed to be of negligible mass. A spring force exists if the spring is deformed, such as the extension or the compression of a coil spring. Therefore the spring force exists only if there is a relative displacement_between the two ends of the spring. The work done in deforming a spring is transformed into potential energy, that
is, the strain energy stored in the spring. A linear spring is one that obeys Hooke’s law, that is, the spring force is proportional to the spring deformation. The constant of proportionality, measured in force per unit deformation, is called the stiffness, or the spring constant k. The damper c has neither mass nor elasticity. Damping force exists only if there is relative motion between the two ends of thedamper. The work or the energy input to a damper is converted into heat. Hence the damping element is nonconservative. Viscous damping, in which the damping force is proportional to the velocity, is called linear damping. Viscous damping, or its equivalent, is generally assumed in engineering.
Spring k
‘a
Static equilibrium position
Damper
Excitation 0 bs
force F(t)
Displacement
x
Fic. 11.
Elements of a vibratory system.
SEC. 12
Elements of a Vibratory System
3
2s atPeriod is OA
ee
Displacement x
Period r af
Fic. 12.
A periodic motion.
The viscous damping coefficient c is measured in force per unit velocity. Many types of nonlinear damping are commonly encountered. For example, the frictional drag of a body moving in a fluid is approximately proportional to the velocity squared, but the exact value of the exponent is dependent on many variables. Energy enters a system through the application of an excitation. An excitation force may be applied to the mass and/or an excitation motion applied to the spring and the damper. An excitation force F(t) applied to the mass m is illustrated in Fig. 11. The excitation varies in accordance with a prescribed function of time. Hence the excitation is always known at a given time. Alternatively, if the system is suspended from a support, excitation may be applied to the system through imparting a prescribed motion to the support. In machinery, excitation often arises from the unbalance of the moving components. The vibrations of dynamic systems under the influence of an excitation is called forced vibrations. Forced vibrations, however,
are often defined
as the vibrations that are caused
and maintained by a periodic excitation. If the vibratory motion is periodic, the system repeats its motion at equal time intervals as shown in Fig. 12. The minimum time required for the system to repeat its motion is called a period 7, which is the time to complete one cycle of motion. Frequency f is the number of times that the motion repeats itself per unit time. A motion that does not repeat itself at equal time intervals is called an aperiodic motion. A dynamic system can be set into motion by some initial conditions, or disturbances at time equal to zero. If no disturbance or excitation is applied after the zero time, the oscillatory motions of the system are called free vibrations. Hence free vibrations describe the natural behavior
or the natural modes of vibration of asystem. The initial condition is an energy input. If a spring is deformed, the inputis potential energy. If a mass is given an initial velocity, the input is kinetic energy. Hence initial conditions are due to the energy initially stored in the system. If the system does not possess damping, there is no energy dissipation. Initial conditions would cause the system to vibrate and the free vibration of an undamped system will not diminish with time. If a system possesses
4
Introduction
CHAP. 1
damping, energy will be dissipated in the damper. Hence the free vibrations will eventually die out and the system then remain at its static equilibrium position. Since the energy stored is due to the initial conditions, free vibrations also describe the natural behavior of the system as it relaxes from the initial state to its static equilibrium. For simplicity, lumped masses, linear springs, and viscous dampers will be assumed unless otherwise stated. Systems possessing these characteristics are called linear systems. An important property of linear systems is that they follow the principle of superposition. For example, the resultant motion of the system due to the simultaneous application of two excitations is a linear combination of the motions due to each of the excitations acting separately. The values of m, c, and k of the elements in Fig. 11 are often referred to as the system parameters. For a given problem, these values are assumed time invariant. Hence the coefficients or the parameters in the equations are constants. The equation of motion of the system becomes a linear ordinary differential equation with constant coefficients,
which can be solved readily. Note that the idealized elements in Fig. 11 form a model of a vibratory system which in reality can be quite complex. For example, a coil spring possesses both mass and elasticity. In order to consider it as an idealized spring, either its mass is assumed negligible or an appropriate portion of its mass is lumped together with the other masses of the system. The resultant model is a Jumpedparameter, or discrete, system. For example, a beam has its mass and elasticity inseparably distributed along its length. The vibrational characteristics of a beam, or more generally of an elastic body or a continuous system, can be studied by this approach if the continuous system is approximated by a finite number of lumped parameters. This method is a practical approach to the study of some very complicated structures, such as an aircraft. In spite of the limitations, the lumpedparameter approach to the study of vibration problems is well justified for the following reasons. (1) Many physical systems are essentially discrete systems. (2) The concepts can be extended to analyze the vibration of continuous systems. (3) Many physical systems are too complex to be investigated analytically as elastic bodies. These are often studied through the use of their equivalent discrete systems. (4) The assumption of lumped parameters is not to replace the basic understanding of a problem, but it simplifies the analytical effort and renders a technique for the computer solution. So far, we have discussed only systems with rectilinear motion. For systems with rotational motions, the elements are (1) the mass moment of inertia of the body J, (2) the torsional spring with spring constant k,, and (3) the torsional damper with torsional damping coefficient c,. An angular displacement @ is analogous to a rectilinear displacement x, and an excitation torque
T(t) is analogous to an excitation force F(t). The two
types of systems are compared as shown in Table 11. The comparison is
SEC. 13
Examples of Vibratory Motions
t"Tanies gs: Comparison Rotational Systems
of
5
Rectilinear
RECTILINEAR
ROTATIONAL
Spring force = kx
Spring torque = k,6
: Damping force =
x c—
dt
d*x
Inertia force=m ag
and
dé Damping torque = c,—
dt

d’*0
Inertia torque = heres
shown in greater detail in Tables 22 and 23. It is apparent from the comparison that the concept of rectilinear systems can be extended easily to rotational systems. 13
EXAMPLES
OF VIBRATORY
MOTIONS
To illustrate different types of vibratory motion, let us choose various combinations of the four elements shown in Fig. 11 to form simple dynamic systems. The springmass system of Fig. 13(a) serves to illustrate the case of undamped free vibration. The mass m is initially at rest at its static equilibrium position. It is acted upon by two equal and opposite forces, namely, the spring force, which is equal to the product of the spring constant k and the static deflection 6,, of the spring, and the gravitational force mg due to the weight of the mass m. Now assume that the mass is displaced from equilibrium by an amount x, and then released with zero initial velocity. As shown in the freebody sketch, at the time the mass is released, the spring force is equal to k(x)+6,,). This is greater than the gravitational force on the mass by the amount kx . Upon being released, the mass will move toward the equilibrium position. Since the spring is initially deformed by x, from equilibrium, the corresponding potential energy is stored in the spring. The system _is conservative because there is no damper to dissipate the energy. When the mass moves upward and passes through equilibrium, the potential energy of the system is zero. Thus, the potential energy is transformed to become the kinetic energy of the mass. As the mass moves above the equilibrium position, the spring is compressed and thereby gaining potential energy from the kinetic energy of the mass. When the mass is at its uppermost position, its velocity is zero. All the kinetic energy of the mass has been transformed to become potential energy. Through the exchange of potential and kinetic energies between the spring and the mass, the system oscillates periodically at its natural frequency about its static
6
Introduction
CHAP. 1
Free length
of spring
Freebody
~
sketch
tatic
deflection
(b)
Static equilibrium position
Damped free vibration
Fic. 13.
k(x, + 6,,)
(c)
Forced vibration
Simple vibratory systems.
equilibrium position. Hence natural frequency describes the rate of energy exchange between two types of energy storage elements, namely, the mass and the spring. It will be shown in Chap. 2 that this periodic motion is sinusoidal or simple harmonic. Since the system is conservative, the maximum displacement of the mass from equilibrium, or the amplitude of vibration, will not
diminish from cycle to cycle. It is implicit in this discussion that the natural frequency is a property of the system, depending on the values of m and k. It is independent of the initial conditions or the amplitude of the oscillation. A massspring system with damping is shown in Fig. 13(b). The mass at rest is under the influence of the spring force and the gravitational force, since the damping force is proportional to velocity. Now, if the mass is displaced by an amount x, from its static equilibrium position and then released with zero initial velocity, the spring force will tend to restore the mass to equilibrium as before. In addition to the spring force, however, the mass is also acted upon by the damping force which opposes its motion. The resultant motion depends on the amount of damping in the system. If the damping is light, the system is said to be underdamped and the motion is oscillatory. The presence of damping will cause (1) the eventual dying out of the oscillation and (2) the system to oscillate more slowly than without damping. In other words, the amplitude decreases
SEC. 13
Examples of Vibratory Motions
1
with each subsequent cycle of oscillation, and the frequency of vibration with viscous damping is lower than the undamped natural frequency. If the damping is heavy, the motion is nonoscillatory, and the system is said to be overdamped. The mass, upon being released, will simply tend to return to its static equilibrium position. The system is said to be critically damped if the amount of damping is such that the resultant motion is on the border line between the two cases enumerated. The free vibrations of the systems shown in Figs. 13(a) and (b) are illustrated in Fig. 14. All physical systems possess damping to a greater or a lesser degree. When there is very little damping in a system, such as a steel structure or a simple pendulum, the damping may be negligibly small. Most mechanical} systems possess little damping and can be approximated as undamped systems. Damping is often built into a system to obtain the desired performance. For example, vibrationmeasuring instruments are often built with damping corresponding to 70 percent of the critically damped value. If an excitation force is applied to the mass of the system as shown in Fig. 13(c), the resultant motion depends on the initial conditions as well
as the excitation. In other words, the motion depends on the manner by which the energy is applied to the system. Let us assume that the excitation is sinusoidal for this discussion. Once the system is set into motion, it will tend to vibrate at its natural frequency as well as to follow the frequency of the excitation. If the system possesses damping, the par of the motion not sustained by the sinusoidal excitation will eventually di out. This is the transient motion, which is at the natural frequency of th system, that is, the oscillation under free vibrations.
The motion sustained by the sinusoidal excitation is called the steadystate vibration or the steadystate response. Hence the steadystate response must be at the excitation frequency regardless of the initial
*o
Critically damped Overdamped
°o Displacement x
Undamped
Fic. 14.
Underdamped
Free vibration of systems shown in Figs. 13(a) and (b). Initial
displacement = x,; initial velocity = 0.
CHAP. 1
Introduction
8
conditions or the natural frequency of the system. It will be shown in Chap. 2 that the steadystate response is described by the particular integral and the transient motion by the complementary function of the differential equation of the system.  Resonance occurs when the excitation frequency is equal to the natural / frequency of the system. No energy input is needed to maintain the vibrations of an undamped system at its natural frequency. Thus, any energy input will be used to build up the amplitude of the vibration, and the amplitude at resonance of an undamped system will increase without limit. In a system with damping, the energy input is dissipated in the damper. Under steadystate condition, the net energyinput per cycle is equal to the energy dissipation per cycle. Hence the amplitude of vibration at resonance for systems with damping is finite, and it is determined by the amount of damping in the system.
14
SIMPLE
HARMONIC
MOTION
Simple harmonic motion is the simplest form of periodic motion. It will be shown in later chapters that (1) harmonic motion is also the basis for more complex analysis using Fourier technique, and (2) steadystate analysis can be greatly simplified using vectors to represent harmonic motions. We shall discuss simple harmonic motions and the manipulation of vectors in some detail in this section. A simple harmonic motion is a_reciprocating motion. It can be represented by the circular functions, sine or cosine. Consider the motion of the point P on the horizontal axis of Fig. 15. If the distance OP is
OP = x(t)= X cos ot
(11)
where t=time, w = constant, and X = constant, the motion of P about the
origin O is sinusoidal or simple harmonic.* Since the circular function repeats itself in 27 radians, a cycle of motion is completed when wt = 27, * A sine, a cosine, or their combination motion. For example, let :
can be used to represent a simple harmonic
Ane
20s
x(t) = X, sin ot +X, cos ot = X xX sin ott. cos wt
= X(sin wi cos a+ cos wt sin a) = X sin(wt+ a)
where X =v X{+X3 and a=tan '(X,/X,). It is apparent that the motion x(t) is sinusoidal and, therefore, simple harmonic. For simplicity, we shall confine our discussion to a cosine function. In Eq. (11), x(t) indicates
that x is a function of time
equation, we shall omit (t) in all subsequent equations.
t Since this is implicit in the
SEC. 14
Simple Harmonic Motion
Fic. 15.
9
Simple harmonic motion: x(t) = X cos ot.
that is,
2 r=— Bes
Period is
A
=a
(12)
QO
1 \Frequency f==—— cycle/s, soror
He)
(13)
et Tie
w is called the circular frequency item lin rd) If x(t) represents the displacement of a mass in a vibratory system, the velocity and the acceleration are the first and the second time derivatives of the displacement, that is,
Displacement x = X cos wt Velocity
(14)
x = —wX sin wt = wX cos(wt+ 90°)
(15)
Acceleration ¥ = —w*X cos wt = w*X cos(wt + 180°)
(16)
These equations indicate that the velocity and acceleration of a harmonic displacement are also harmonic of the same frequency. Each differentiation changes the amplitude of the motion by a factor of » and the phase angle of the circular function by 90°. The phase angle of the velocity is 90° leading the displacement and the acceleration is 180° leading the displacement. Simple harmonic motion can be defined by combining Eqs. (14) and (16). X= w’x (17) where w” is a constant. When the acceleration of a particle with rectilinear motion is always proportional to its displacement from a fixed point on the path and is directed towards the fixed point, the particle is said to have simple harmonic motion. It can be shown that the solution of Eq. (17) has the form
of a sine and
a cosine
function
with
circular
frequency equal to w. *In 1965, the Institute of Electrical and Electronics Engineers, Inc. (IEEE) adopted new standards for symbols and abbreviation (IEEE Standard No. 260). The unit hertz (Hz)
replaces cycles/sec (cps) for frequency. Hz is now commonly used in vibration studies. + The symbols x and x represent the first and second time derivatives of the function x(t),
respectively. This notation is used throughout the text unless ambiguity may arise.
10
Introduction
CHAP. 1
The sum of two harmonic functions of the same frequency but with different phase angles is also a harmonic function of the same frequency. For example, the sum of the harmonic motions x, = X, cos wf and x,= X, cos(wt+a)
is
x=x,+x,=X, cos wt+X, cos(wt+ a) = X, cos wt + X,(cos wt cos a—sin wt sin a) = (X,+ X, cos a)cos wt— X, sin a sin wt
= X(cos 6 cos wt—sin B sin wf) = X cos(wt+ B) where X =V(X,+X, cos a)*+(X, sin a)’ is the amplitude of the resultant harmonic motion and $=tan ‘(X,sin a)/(X,+X,cosa) is its phase angle. The sum of two harmonic motions of different frequencies is not harmonic. A special case of interest is when the frequencies are slightly different. Let the sum of the motions x, and x, be
X=x,+x,=X cos wt+ X cos(w+e)t = X[cos wt+cos(w + €)t} =
E E IX COs =: cosa —
tf
where ¢) c(x,
(a)
System
Fic. 41.
(b)
kx,
k(x, —x5)
Lm
—X,)
KO
CX
KX
5)
Freebody sketches
A twodegreeoffreedom system.
144
Systems with More Than One Degree of Freedom
CHAP. 4
For conciseness, Eq. (41) can be expressed in matrix notations as
K 0
0 fe [ 2 mJLX
=—C.
—c a se FC.IL
a
k
k+k,JjLx,
F,(t)
or
M{x}+ C{x}+ K{x} = {F(t)}
(43)
By simple matrix operations, it can be shown that Eqs. (41) and (42) are
equivalent. The quantities in Eq. (43) can be identified by comparing with Eq. (42). The 2x2 matrices M,C, and K are called the mass matrix, damping matrix, and stiffness matrix, respectively. The 2 x 1 matrix {x} is called the displacement vector. The corresponding velocity vector is {x} and the acceleration vector is {x}. The 2X1 matrix {F(t)} is the force vector. It will be shown in Sec. 44 that if another set of coordinates {q,q,} is
used to describe
the motion
of the same
system,
the values
of the
elements in the matrices M, C, and K will differ from those shown in Eq.
(42). The inherent properties of the system, such as natural frequencies, must be independent of the coordinates used to describe the system. Hence the general form of the equations of motion of a twodegreeoffreedom system is M1, ke
M1211
qi
Ci
ABE:
Cia  ellie
Kay
abe eoe
ee ky SLq>
Q,(t)
)
or
M{q}+Cig}+K{q}={Q(o}
(45)
The 2 x 2 matrices M, C, and K associated with the coordinates {q} can be
identified by comparing the last two equations. The 2X 1 matrix {Q(t)} is the force vector associated with the displacement vector {q}. Generalizing the concept, Eq. (45) also describes the motion of an ndegreeoffreedom that is,
system if the matrices M, C, and K are of nthorder,
M=[m,],
C=[e,],
K =[k;]
(46)
where i, j=1,2,3,...,n. The coefficients m;,,ij» c;,,‘ij and k;, are the elements
of the matrices M, C, and K, respectively. The generalized coordinates {q} and the generalized force vector {Q(t)} are
{Gia tdi.
a.)
{O(t)}} ={Q,(t) Q,(0}
(47) (48)
SEC. 43
Undamped
Free Vibration: Principal Modes
145
Hence Eq. (45) is also the general form of the equations of motion of an ndegreeoffreedom system.
43
UNDAMPED MODES
FREE
VIBRATION:
PRINCIPAL
A dynamic system has as many natural frequencies and modes of_ vibration as the degrees of freedom. The general motion is the superposition of the modes. We shall discuss (1) a method to find the natural frequencies, and (2) the modes of vibration of an undamped system at its natural frequencies. In the absence of damping and excitation, the system in Fig. 41
reduces to that shown in Fig. 42(a). Hence the equations of motion from Eq. (42) are
Fie alelilecmvewelle elye 0) 0
m,JLxX,
oils
k+k,\LX>
0
The equations are linear and homogeneous and are in the form of Eq. (D47), App. D. Hence the solutions can be expressed as x,= B,e*
(410)
x,= Be" Motions at first mode
Ay, Time
25)
ujAy, Time Motions at second mode
aT Ai, Time
225)
U,Aj9 (a)
Vibratory system
Fic. 42.
(b)
Motions at principal modes
Modes of vibration.
146
Systems with More Than One Degree of Freedom
CHAP. 4
where B,, B,, and s are constants. Since the system is undamped, it can be shown that the values of s are imaginary, s=+jw. By Euler’s formula,
e~’'=cos
wt+j sin wt, and recalling that the x’s are real, the
solutions above must be harmonic and the general solution must consist of a number of harmonic components. Assume one of the harmonic components is x, =A, sin(wt+ w)
,
(411)
x, = A, sin(wt+ w)
where A,, A>, and w are constants and w is a natural frequency of the system. If the motions are harmonic,the choice of sine or cosine functions is arbitrary. Substituting Eq. (411) in (49), dividing out the factor sin(wt+w), and
rearranging, we have (k =
k, oe w*>m,)A,—
kA,
=
0
(412)
—kA,+(k+k,—@?m,)A,=0
which are homogeneous linear algebraic equations in A, and A,. The determinant A(w) of the coefficients of A, and A, is called the characteristic determinant. If A(w) is equated to zero, we obtain the characteris
tic or the frequency equation of the system from which the values of w are found, that is,
n=
ADBCz2O
=0
(413)
From linear algebra, Eq. (412) possesses a solution only if the determinant A(w) is zero. Expanding the determinant and rearranging, we get
4
wo’ —
[ Mm,
My
G24
ERR m,M,
(414)
which is quadratic in w*. This leads to two real and positive values* for
w*. Calling them w7 and w3, the values of w from Eq. (414) are +w, and +w. Since the solutions in Eq. (411) are harmonic, the negative signs for w merely change the signs of the arbitrary constants and wouldnot lead to new solutions. Hence the natural frequencies are w, and w». The example shows that there are two natural frequencies in a twodegreeoffreedom system. Each of the solutions of Eq. (49) has two harmonic components at the frequencies w, and w, respectively. By * Note that the values of s in Eq. (410) are +jw, and +jw, in order to have the periodic
solutions assumed in Eq. (411). If w* is not real and positive, it can be shown that the solutions by Eq. (410) would either diminish to zero or increase to infinity withiincreasing
time.
SEC. 43
Undamped Free Vibration: Principal Modes
JLosmoweon(}
147
superposition, the solutions from Eq. (411) are
[
=
A,
Pinlort+ da) w,)++]4"
sin(w,t+
fsin(wat+ + ya)
415 (418)
sin(@>t
where the A’s and w’s are arbitrary constants. The lower frequency term _is_called the fundamental and the others are the harmonics. Double subscripts are assigned to the amplitudes; the first subscript refers to the coordinate and the second to the frequency. For example, Aj, is the _amplitude of x,(t) at the frequency @ = @,. The relative amplitudes of the harmonic components in Eq. (415) are defined in Eq. (412). Substituting @, and w, in Eq. (412) and rearranging, we obtain
aie Aoi
k
_k+k,@jm,
K+k,—wim,
k
k+k,ow3m,_
1
in
An_ Ki +k 3m, A>>
una
k
(416)
ya 1 ee
where the u’s are constants, defining the relative amplitudes of x, and x, at each of the natural frequencies w, and w,. Thus, Eq. (415) becomes
[2]23 i Aj, Sin(@,t+ p,)+ EBAy sin(w.t+W)
(417)
X2
where Ajj, Ajo, &, and ys are the constants of integration, to be deter. mined by the initial conditions. There are four constants because the system is described by two secondorder differential equations. Note that
(1) from the homogeneous equation in Eq. (412), only the ratios 1:u, and 1:u, can be found, and (2) the relative amplitudes at a given natural frequency are invariant, regardless of the initial conditions. A principal or natural mode of vibration occurs when the entire system
executes synchronous harmonic motionat one of the natural frequencies as illustrated in Fig. 42(b). in Eq. (417), that is,
Ze ie Aj, sin(w,t+ y,)
For example, the first mode occursifA,,=0
Ory
xy
fed p(t) = {u}, px(t)
(418)
X2
where {u}, is called a modal vector or eigenvector. Note that ive {u11 U>,;}={1 u,} as shown above. It represents the relative amplitude, or the
mode
shape,
of the motions
x,(t) and
x,(t) at w=@,.
Hence
a
principal mode is specified by the modal vector at the given natural Aj; sin(w;t+) is harmonic. It shows frequency. The quantity p,(t)=
CHAP. 4
Systems with More Than One Degree of Freedom
148
that the entire system executes synchronous harmonic motion at a principal mode. Similarly, the second mode occurs if A,, in Eq. (417) is zero, that is,
[= [2] Avsinoat+ us or c)=[“"] pS tube 419) X2
The modal vector for the second mode is {u},. The harmonic functions of the motions x,(t) and x,(t) in Eq. (417) can be expressed as = 4
X
a
1
1 
U;
sin(w,t+ ie A ba
UzJLAj;> sin(wzt+ Yr)
.
sekevel
Ur,
(420)
Us2JLp2(t)
Or
{x}=[u]{p}
(421)
where the modal matrix [u] is Z
[u} = ——
Or
Uy,
Uj2 Ur2
U4
Cia
and p,(t)=Aj,,sin(w,t+wW,)
=
1 al
I ; U5

te eC
and p,(t)= Aj, sin(w,t+y,).
(422) Note
that in
Eqs. (418) through (420) only the relative values in a modal vector can
be defined, as shown in Eq. (416). A modal matrix [u] in Eq. (422) is simply a combination of the modal vectors. The actual motions {x} in Eq. (420) are specified by the constants A’s and wW’s, which are determined
by the initial conditions. The vector {p} consists of a set of harmonic functions at the frequencies }@, and wz. The vector {p} is called the principal coordinates. Each “principal coordinate p,(t) and its associated modal vector {u}; describe a mode of vibration as shown in Eqs. (418) and (419). Principal coordinates will be further discussed in Sec. 45 and Chap. 6. The coordinate transformation between the {x} and {p} coordinates is shown in Eq. (421). The extension of the concept to ndegreeoffreedom systems is immediate. For example, a system may be described by the {x}= {x; X,°+*x,} coordinates. Analogous to Eq. (417), each of the motions x;(t) has n harmonic components. A principal mode occurs if the entire system executes synchronous harmonic motion at one of the natural frequencies. The corresponding principal coordinates is {p}= {Pp P2*** p,}. The modal matrix [u] consists of n modal vectors {u},. [ujJ=[{u},
{u}>+ + {u},]= Lu;; 
(423)
where i,j=1,2,...,n. The transformation between the {x} and the {p}
coordinates is analogous to that in Eq. (421).
SEC. 43
Undamped Free Vibration: Principal Modes
149
Example 1 Referring
to Fig. 42(a),
let m;=m,=m
and
k,=k,=k.
If the initial
conditions are {x(0)}={1 0} and {x(0)}={0 0}, find the natural frequencies of the system and the displacement vector {x}. Solution: From
Eq. (414), the natural frequencies are
wo, =Vkim
and
w=
V3k/m
Substituting w, and w, in Eq. (416), we obtain u,=1 and u,=—1. Hence the displacement vector {x} from Eq. (420) is {x}= E alk sin(w,t+ ed
Aj2 Sin(wot + Wr) For the initial conditions {x(0)}={1 Ree
iH
il
Ge
0}, we get Aj,

sin
oe Me
‘
ae
f FA

Premultiplying the equation by the inverse [u]’ of [u] ee
[arsed22 aflol=2L:] Ajo sin
Wo
2
1
—1
0
2;
1
or
Au
1
=
d
2 sin Wy,
For the initial conditions {x(0)}={0
1 sin W
0}, we have >
1 lone cos “
Bae
1
0
A 2
ae
[uo
ip
—1JLw2A;. cos py
Azo
+
.
y
—>
Cee
Premultiplying the equation by the inverse [u]* of [u] gives ee @2A
cos
i> cos
eles Wo
2
1 el(ol 1
—1
0
0
Since the A’s and w’s are nonzero, we have cos W,=cos ~,=0. Let f= ma/2 and W.=n/2, where m and n are odd integers. It canbe shown that
the choice_of m_and nother than 1 will not lead to new solutions. Thus, Ain = A=
1/2.
From Eq. (417), we obtain
 Xone
An ==  cos Dal
k [ 4/—t+— COS i
Pali
3k ——'t m
The motions are plotted in Fig. 43 for Vk/m=27. The example can be repeated for different initial conditions to show that the relative amplitudes of the principal modes remain unchanged. This is left as an exercise.
Y,
CHAP. 4
Systems with More Than One Degree of Freedom
150
Displacement x, aA
\
"aa ee
0.5 cosV
e
0.5 cosvV
3k/m t
k/m t
0.5 cosV k/m hoe 7
wn
1 sec
AS
Displacement x,
—0.5 cosV
7
3k/m esi
5 = 0.5 cosV k/m t — 0.5 cosvV
Fic. 43.
Example 2.
Superposition of modes of vibration: Example
Natural modes
3k/m
t
1.
~ /\9_ ©. A140
Find the initial conditions that would set a twodegreeoffreedom into its natural modes of vibration, that is, Aj; or Aj,in Eq. (417) becomes zero. Solution: From
Eq. (420), we have
 4 ib X2
Gl Hie sin(w,t+ vl
U;
Uz
JLAj2 Sin(wot + Wr)
Applying the initial conditions {x(0)}={xio etal. X20
X20}, we get
aber Al
U,
UrJLAj2 Sin YW
Premultiplying the equation by the inverse [u]’ of [u] gives be sin
a
A, sin
1
Uz— UU,
 uz
melee
L—Uy,
1
X20
Or 
UxXi9
Ag= 
—X
pand)
Ae
(u2—u,)sin Wy,
U2X109— X20
=
SSS
SS SS
@ (Uz — U;)COs Wy
—
(uz —u,)sin YW
Similarly, using the initial conditions {x(0)}={Xi0 Ai
UX ee
Xoq
and
Aa =
X20}, we get UR
ee
ee es
@(Uz— U;)cos Wr
SEC. 43
Undamped Free Vibration: Principal Modes
151
The first mode occurs at @, if A,.=0, that is,
X20 = UyX10 In other
words,
the
system
and
X20 = Uy X10
will vibrate
at its first mode
if the
initial
conditions are {X;9 X20}={1 u,} with zero initial velocities. Alternatively, the initial conditions can be {Xo X2o}={1 u,} with zero initial displacements. Any combination of the above conditions would also set the system in its first mode. It is only necessary to set the initial values of {x} and/or {x} _to conform to their relative values for the first mode, as indicated by the
corresponding modal vector {u},={1
u,}.
Similarly, the second mode occurs when A,,=0, that is, X59 = UX 9 and
X20 = U2Xyo.Any combination of these conditions will give the second mode.
Example 3.
Vehicle suspension
An automobile is shown schematically in Fig. 44. Find the natural frequencies of the car body. Solution:
An automobile the car moves vertical motion body about its
has many degrees of freedom. Simplifying, we assume that in the plane of the paper and the motion consists of (1) the of the car body, (2) the rotational pitching motion of the mass center, and (3) the vertical motion of the wheels. Even
then, the system has more
than two degrees of freedom.
When the excitation frequency due to the road roughness is high, the wheels move up and down with great rapidity but little of this motion is transmitted to the car body. In other words, the natural frequency of the car body is low and only the low frequency portion of the road roughness is _being transmitted. (See Case 4 in Sec. 35.) Because of this large separation.
of_natural frequencies between the wheels and the car body, the problem in Fig. 45. Assuming small oscillations, the equations of motion in the x(t) and 6(t) coordinates are
can be further simplified by neglecting the wheels as shown
6
Car body
Fic. 44.
Schematic of an automobile.
CHAP. 4
Systems with More Than One Degree of Freedom
152
Fic. 45.
Simplified representation of an automobile body.
and
IO = ys(moments),
Jo6 = k,(x—L,6)L,— k(x + L26)L, Rearranging, we obtain
i él+ k+k Sr yentce 0 JIojLé ~—(k,lL,—k LL.) k,L2+k12
He je 0
which is of the same form as Eq. (49). The frequency equation from Eq. (413) is A
= ()
k,+k,—o'm kote
kel
kL
k,Li+
— kk
=0
k,L3—w’J,
Expanding the determinant and solving the equation, we get
0? +4 Jae 12
2
ee Ty
m
tes ke ee
Jo _ m
ee Jo
rd mJ
The natural frequencies are w,/27 and w,/27 Hz.
Example 4 A vehicle has a mass of 1,800 kg (4,000 lb,,) and a wheelbase of 3.6m (140 in.). The mass center cg is 1.6 m (63 in.) from the front axle. The radius
of gyration of the vehicle about cg is 1.4 m (55 in.). The spring constants of the front and the rear springs are 42 kN/m (240 Ib,/in.) and 48 kN/m (275 lb,/in.), respectively. Determine (a) the natural frequencies, (b) the principal modes of vibration, and (¢c) the motion x(t) and 6(t) of the vehicle.
SEC. 43
Undamped Free Vibration: Principal Modes
(a)
First mode: f= 1.09 Hz
Fic. 46.
(b)
153
Second mode: f= 1.50 Hz
Principal modes of vibration of a car body (not to scale).
Solution:
(a) From the given data and the equations in Example 3, we have kit
ke
50
kiL,—k2L2
m kk,atLk2
2
_
m RAO
4k,k.(L,+L,) al o( 1 2)
lf
=
—16.0 16457
mJo
46. (oe
w?=4[50+84.9 V(50+ 84.9) — 16 a= (nor =
eee rad/s = 1.09 Hz 9.40 rad/s = 1.50 Hz
(b) The amplitude ratios for the two modes of vibration are
x_ (k,L,—k,L2)/m
~(k;+k)/im—o2,
—_
—16
leaps m/rad
5002,  0.42 m/rad
The two principal modes of vibration are shown schematically in Fig. 46. The mode shape at 1.09 Hz is {X @}={1 —1/4.69}. Thus, when x(t) is positive, @(t) is negative from the assumed direction of rotation. When x(t)=1m, 6(t)=—1/4.69 rad, that is, the node is 4.69m from the cg of the car body. Similarly, at 1.50 Hz, the mode shape is
{X
@}={1
1/0.42}.
(c) From Eg. (417), the x(t) and @(t) motions are
;=  : 6
where
Example 5.
—1/4.69
A, sin(6.83t+y,)+
1
pres
A,
12 sin(9.40t + w) i
.
+
Aj;, Aj2, Wi, and yw, are the constants of integration.
A threedegreeoffreedom system
A torsional system with three degrees of freedom is shown in Fig. 47. (a) Determine the equations of motion and the frequency equation. (b) If
154
Systems with More Than One Degree of Freedom
Fic. 47.
CHAP. 4
A threedegreeoffreedom torsional system; Example 5.
J,=J,=J,=J and k,,=k,.=k,,;=k,, find the natural frequencies and the equation for the displacement {6}. Solution:
(a) From Newton’s second law, the equations of motion are
J,6,= —ky101— ki2(01 — 92) Jo = —k,2(02— 01) — k,3(02— 93) J03 = —k,3(83— 62)
(424)
Substituting 6,=0, sin(wt+y), for i=1,2, and 3, in Eq. factoring out the sin(wt+) term, and rearranging, we have
(424),
(k,1+ k,2— @7J,)O, — k,,O, =0 — k,2@, + (k,2+ k,3— 7 Jz)O2— k,303 =0
(425)
— k,30, + (k,3— w7J3)03 = 0 The
frequency
equation
is obtained
by equating
the determinant
A(q@) of the coefficientsof @,,@,, and ©, to zero. kitke—w7J,
—k,2
—k,2
ki2+ kis — w’J,
0
= Kes
A(w)=
(b) If J;
Beet 3 e
‘a
pay
The
=J,=J3=J and k,,=k,.=k,3=k,,
act
roots
5
w°—5
of
the
k,
(5)w*+6
equation
0
=6 Js
the frequency equation is
k,\?
k,\3
(7) o>
are
—k,; k,3—
(7) =0
w*=0.198(k/J),
1.55(k/J),
3.25(k,/J). The corresponding frequency vector {f} is
AE: V0.198 {flea ipl) ESE: = V3.25
0.071 50.198 Hz 0.288
From Eq. (425), the amplitude ratios are 0,
— 2st 0,
ki2
eee
kit k.2@
2
and
Ji
Seby
=
0,
ki = oJ,
0,
ki3
tee
se, ee
/
\
=SF
rae)
va
and
SEC. 44
Generalized Coordinates and Coordinate Coupling
155
f= 0.071 VK,
kK,t
™s ae
f= 0.288VK,/J ‘
Ss
ee
f= 0.198VK,J
VE
Fic. 48. Principal modes of vibration; amplitudes plotted normal to axis of rotation; Example 5S. Thus, a modal vector {O9, ©, @©3}; can be calculated for each of the natural frequencies w;. The modal matrix [6,], for i,j=1,2, and 3, is formed from a combination of the modal vectors {6}; as in Eq. (421).
[6;]=[{6},
{6},
it]
1 1 90 0.445 2950802)
1 “1201 0.555
where {6},, {6}, and {6}, are the modal vectors for the frequencies @,=V0.198k/J, w.=V1.55k/J, and w3=V3.25k,/J, respectively. The principal modes of vibration are illustrated in Fig. 48. _ By superposition of the principal modes, the motions {6} of the rotating disks are
1 1 {0}= Ea ©,; sin(w,t+ yy) ‘ 0.445  ©, sin(wot+Wp) Deas
—0.802
1 aia
201] 0,3 sin(w3t 45 Ws)
0.555
The ©’s and w&’s are to be determined by the initial conditions.
44
GENERALIZED COORDINATES COORDINATE COUPLING
AND
The general form of the equations of motion of a twodegreeoffreedom system is shown in Eq. (44). For undamped free vibration, we have [2s M2,
i
alr
Mg2ILX2
Koy
EE Kaz JLX2
(426) 0
The system is described by the coordinates x, and x2, which are the elements of the displacement vector {x}. The coupling terms in the
equations are M12, M1, ky, and k,. We shall show that the valuesof the.
156
Systems with More Than One Degree of Freedom
(a)
Fic.
Static coupling
(b)
Dynamic coupling
(c)
CHAP. 4
Static and dynamic coupling
49. Generalized coordinates and coordinate coupling: a twodegreeoffreedom system described by the (x,,@), (x2,0), and (x3,0) coordinates.
elements in the matrices M and K are dependent on the coordinates selected for the system description.
A vibratory system can be described by more than one set of independent spatial coordinates, each of which can be called a set of generalized coordinates. We often use the displacements from the static equilibrium positions of the masses and the rotations about the mass centers for the coordinates.
This choice
is convenient,
but, nonetheless,
arbitrary. We
shall describe the system in Fig. 49 by the displacement vectors {x,
9},
(Xe ealids, X5.0m00 } Referring to Fig. 49(a) and assuming small oscillations, the equations of motion in the (x,,@) coordinates are mx, =—k,(x,—L,0)—k,(x,+ L,6) Te =k,(x,;—L,0)L,—k,(x,+L,6)L,
Rearranging, we obtain m
0
4
=»
li Mle
{+
(k,
+k.)
—(k,L,—
kL)
Raver
le
(k,L{+kL5)
6
=
a
427
0
and the system is said to be statically or elastically coupled (see Examples 3 and 4). If a static force is applied through the cg at point 1, the body will rotate as well as translate in the x, direction. Conversely, if a torque is applied at point 1, the body will translate as well as rotate in the 6 direction. The same system is described by the (x2,@) coordinates in Fig. 49(b). The distance e is selected to give k,L3=k L,. If a static force is applied at point 2 to cause a displacement x2, the body will not rotate. Hence no static coupling is anticipated in the equations of motion. During vibration, however, the inertia force mx, through cg will create a moment mx,e
SEC. 44
Generalized Coordinates and Coordinate Coupling
157
about point 2, tending to rotate the body in the @ direction. Conversely, a rotation 6 about point 2 will give a displacement e@ at cg and therefore a force me@ in the x, direction. Hence. dynamic coupling is anticipated in the equations. The equations of motion in the (x5,0) coordinates are
mx, = —k,(x,— L30)— k(x,+L,0)— med J26 = k3(x.— L30)L,— k(x, + L,0)L,—mex, Or
m~
mell[x,
cas
ki, +k,
0
\2]
=
P
:
The coupling terms are associated with the inertia forces and the system is said to be dynamically, or inertia, coupled.  Lastly, let the same system be described by the (x3,0) coordinates as— shown in Fig. 49(c). It can be shown that the equations of motion are
m
ie: (faa
Selb 3]
(429)
For the (x3,6) description, the equations are statically and dynamically coupled. Note that (1) the choice of coordinates for the system description is a mere convenience, (2) the system will vibrate in its own natural way regardless of the coordinate description, (3) the equations for one coordinate description can be obtained from those for other descriptions, and (4) coupling in the equations is not an inherent property of the system, such as natural frequencies. The examples above show that the matrices M and K are symmetric, that is, m,>= mp», and k,,=k,,. Symmetry is assuredif the deflections are measured from a fixed position in space. This can be deduced from Maxwell’s reciprocity theorem in Sec. 49. Let us select a set of generalized coordinates based on relative deflections to illustrate the nonsymmetric matrices in the equations of motion. Example 6 Consider the system shown in Fig. 42 and assume the generalized coordinates q, =X, and q,=X,—%,, that is, q, is proportional to the spring force due to k. Find the equations of motion of the system. Solution:
The coordinates {q} and {x} are related by
158
Systems with More Than One Degree of Freedom
CHAP. 4
or
ele alle. 1%)
it
A
ahtae
Replacing the displacement and acceleration vectors {x} and {x} by {q} and {q} in Eq. (49) for the same system, we get
parallels emcee lle 45
PRINCIPAL
COORDINATES
It was shown in the last section that the elements of the matrices M and K depend on the coordinates selected for the system description. It is possible to select a particular set of coordinates, called the principal
coordinates, such that there is no coupling terms in the equations of motion, that is, the matrices M and K become diagonal matrices. Hence each of the uncoupled equations can be solved independently. In other words, when the system is described in terms of the principal coordinates, the equations of motion are uncoupled, and the modes of vibration are mathematically separated. Thus, each of the uncoupled equations can be solved independently, as if for systems with one degree of freedom. Assume an undamped twodegreeoffreedom system is uncoupled by the principal coordinates {p}. The corresponding equations of motion from Eq. (44) are
0
“ele
=mMy2ILP2
0
kp2JSLP2
=
0
430
Expanding the equations gives mM,,pi+ kiiPi ==)
M2P2+ kop> = 0
The solutions of the equations are Pi= Aj, sin(@,t+ W,) P2 = Ajp Sin(@zt+ Wo)
(431)
where {= k,,/m,,, @3=k /m 2, and the A’s and w’s are constants. Evidently, each of the solutions above represents a mode of vibration as discussed in Sec. 43. At a given mode, the system resembles an independent onedegreeoffreedom system. Now, assume the same system is described by the generalized coordinates {q} and the equations of motion are coupled. From Eq. (417), the
SEC. 45
Principal Coordinates
159
motions in the {q} coordinates are “ a ea Aj, Sin(w,t+ w,)+ [se A>, Sin(wst+W) q2 Ur 22
where {u,;
U;} and {u,,
(432)
Up,} are the modal vectors for the frequencies
w, and w., respectively.
Substituting Eq. (431) in (432) and simplifying, we get  = be q2 Ur,
alles U2 ILP2
(433)
= {p}=[u]}‘{q}
(434)
where [u]' is the inverse of the modal matrix [uJ], (422). The transformation between the {p} and the {q} (434) is identical to that shown in Eq. (421).* The discussion implies that the equations of motion by means of a coordinate transformation. In other coupled equations in the {q} coordinates, the equations
as defined in Eq. coordinates in Eq.
or
{q}=[ul{p}
and
can be uncoupled words, given the can be uncoupled
by substituting {p} for {q} as shown in Eq. (434). This can be done, but the general theory in Sec. 64 will be needed. In the mean time, we shall illustrate with another example and then show the general technique in the next section. Example 7 Determine the principal coordinates for the system shown in Fig. 42(a) if m,=m,=m
and k,=k,=k.
Solution: From Example
1, we have u, = 1 and u,=—1. Hence Eq. (433) becomes
[Ab alk =
Xo
The
1
1
fFeh Sle
and
=>
P2
P2
Dy,
1
—1
X2
transformation above indicates that, if p, = 3(x,+X2) and p= 3(xX;— Xo),
the equations of motion in the {p} coordinates are uncoupled. Let us further examine this statement. The equations of motion for the same system from Eq. (49) are mx, te 2kx,
mxX>+
—
kx> a
0)
2KX>—
kx, =
0
* We assume that the matrices M and K in the generalized coordinates are symmetric. An inherent symmetry in the system can be assumed. We shall not discuss nonsymmetric matrices as illustrated in Example 6.
160
Systems with More Than One Degree of Freedom
CHAP. 4
Adding and subtracting the equations, we obtain m(X, + X2)+ k(x,+ x2)
=0
m(X, — X2)+3k(x,— x2) =0
mp,+ kp, =0 Or
%
mp, + 3kp,=0
Again, the equations are uncoupled if we define Pi =(xX,+%X) and p2= (x,;—Xx). Since the amplitudes of oscillation are arbitrary, the factor (5) between the two definitioris of p, and p, in the problem is secondary.
46
MODAL ANALYSIS: TRANSIENT VIBRATION OF UNDAMPED SYSTEMS
Consider the steps to solve the equations of motion of an undamped system. From Eq. (45), we get
M{q}+ Kiq}={Q(t)}
(435)
(1) The equations can be uncoupled by means of the modal matrix [u] and expressed in the principal coordinates {p} as shown in Example 7. (2) Each of the uncoupled equations can be solved as an independent onedegreeoffreedom system. (3) Applying the coordinate transformation in Eq. (434), the solution can be expressed in the {p} or {q} coordinates as desired. The steps enumerated are conceptually simple. Except for the formula to uncouple the equations of motion, we have the necessary information for the modal analysis of transient vibration of undamped systems. For systems with more than two degrees of freedom, however, computer solutions, as illustrated in Chap. 9, are manditory to alleviate the numerical computations. The modal matrix of undamped systems can be found by the method described in the previous sections. From Eq. (435), the equations of motion of an unforced system are
M{q} + Kiq}= {0}
(436)
A principal mode occurs if the entire system executes synchronous harmonic motion at a natural frequency w. Thus, the acceleration of q; is
G; =—wq, or {G}={w*q} = —w*{q}. Substituting —w7{q} for {g} in Eq. (436) and simplifying, we get [—w*M+ K {q} = {0}
(437)
‘Since the system is at a principal mode, the displacement vector {q} is also
a modal vector at the natural frequency w. In other words, {q} in Eq. (4 37) gives the relative amplitude of vibration of the masses of the / system at the given natural frequency. Since Eq. (437) is a set of homogeneous algebraic equations, it possesses a solution only if the characteristic determinant A(w) is zero;
SEC. 46
Modal Analysis: Transient Vibration of Undamped Systems
161
that is,
A(w) =K —w?M=0
(438)
This is the characteristic of the frequency equation, which may be compared with Eq. (413). Previously, the frequency equation was solved by hand calculations as shown in Eq. (414). Computer solutions, however, are necessary for systems with more than two degrees of freedom. Likewise, instead of solving for the modal vector by hand calculations as
shown in Eq. (416), computers can be used to solve for {q} in Eq. (437). A modal vector is found for each natural frequency. The modal matrix {u] is formed from a combination of the modal vectors as shown in Eq.
(423). Example 8 Find the coefficients of the frequency equation for the system shown in Fig. 410(a).
80 (a)
Vibratory system
Fy() 10
(c)
Fic. 410.
Transient response
Transient vibration of undamped system: Examples 8 to 10.
162
Systems with More Than One Degree of Freedom
CHAP. 4
Solution: Applying Newton’s second law to each of the masses of the system, we have
3X, = —60x, — 60(x, — x2) — 20(x, — x3) + F,(t) 1X, = —80x,—60(x2— x,) — 80(x,—x3)+ F(t) 2%, =
—100x3—20(x3—x,) — 80(x3— x2) + F3(t)
PRO [140 60 20]fx] 017%] o ¢+]60 220 80]] x. = F(t) F;(t) 120 —80: 2001Ex Gs Seo & or ale,
(439)
or
M{x} + K{x} = {F(t)} The program COEFF to find the coefficients of the frequency equation is listed in Fig. 411(a). The values of the matrices M and K are entered (READ) and verified (WRITE) in Par. #1. The computations in Par. #II first change Eq.
(436) into the form
{q}+M 'K{q}= {0} The subroutine $INVS is called to find the inverse M_' (MINVS) of M. The
subroutine $MPLY performs the matrix multiplication MINVS*K =H, where H = M 'K is the dynamic matrix. The subroutine $COEFF is called to give the coefficients of the frequency equation. The printout is listed in Fig. 411(b). We first give the command
MERGE
COEFF, $COEFF, $INVS, $MPLY, $SUBN
to merge the main program COEFF with the necessary subroutines. The subroutine $SUBN is for the matrix substitutions in the calculations. When the computer is READY, the command RUN is given to start the program. The data entries are selfexplanatory. The frequency equation is
1—45.685 X 10 7w?+ 515.94
10 °w*— 1.4071 x 10 °w° =0
Example 9 Assume the roots of the frequency equation in Example 8 are w*= 33.23, 86.67, and 246.8. Find the modal matrix of the system. Solution:
For the given values of M and K and w*= 33.23, the direct application of Eq. (437) gives
140 — 33.23 x3 
—60 =20)
—60 PAYS SNS I2S) > Il —80
el) —80
xX, e
200—3323
a
or
[K — w?M + joCX} ={F}
(452)
where the matrices M, C, and K can be identified readily.
The equations above can be alternatively expressed as ie—@7My,+ jac,
ky1—@7Mg1+ jaca,
ki. Oe
ze
ky2— @? M2 + jocy2

= 4
ILX2
F,
(453)
or ie 224
mele  = F Z22
X>
(454)
F,
or
Z(w){X} = {F}
(455)
where
Zy = (ky wm; + jac;;)
for
ie eal
lass
(456)
170
Systems with More Than One Degree of Freedom
CHAP. 4
and Z(w)=[z,] is the impedance matrix. In other words, Eqs. (451) to
(455) are different forms of the same equation, all of which can be summarized by Eq. (455). The solution {X} gives the amplitude and phase angle of the response relative to the excitation {F}. Premultiplying both sides of Eq. (455) by
the inverse Z(w) ' of Z(w) gives
{X}= Z(w)'{F}
(457)
For a twodegreeoffreedom system this can be written explicitly as ae: —
Z2F) — 212F 2
xX, _
and
7 2ZaFit Z11F2
411222
_ 2112227 2122217A (us
(458)
— 21222 vt sae)
Equations (455) to (457) are equally applicable to ndegreeoffreedom systems. The elements of the impedance matrix Z(w) in Eq. (455) become Zij
=(k
ij >
wo mi + joc)
and Z(w) is of order
for
i J=
Le De seeleie
LU
(459)
n. Note that each z; is identical in form to the
mechanical impedance in Eq. (251) and Z(w) is symmetric by the proper choice of coordinates as discussed in Sec. 44. Example 12.
Undamped dynamic absorber
Excessive vibration, due to near resonance
conditions, is encountered
in a
constant speed machine shown in Fig. 416(a). The original system consists of m, and k,. It is not feasible to change m, and k,. (a) Show that a dynamic absorber, consisting of m, and k,, will remedy the problem. (b) Plot the response curves of the system, assuming gate the effect of the mass ratio m,/m,.
m,/m,=0.3.
(c) Investi
ee sin wt
(a)
Vibratory system
Fic. 416.
(b)
Equivalent system
Undamped dynamic absorber: Example 12.
SEC. 48
Forced Vibration—Harmonic Excitation
171
Solution: The equations of motion for the equivalent system in Fig. 416(b) are MX, = —k1X,— k(x, — X2)+ Feq Sin wt
M2X_ = —k2(x2— X,) The impedance method can be applied harmonic. From Eq. (453), we have nee
directly, since the excitation
Ke
—k,
is
eee
k,— wm,
Xn
0
(a) Following Eq. (413), the frequency equation is obtained by equating the characteristic determinant A(w)of the coefficient matrixof {X} to zero, that is, A(@)
=
(k,
+ k2—
w*m,)(ky—
@*m)—
k3 =0
(460)
From Eq. (458), the phasors of the responses are
ms
1
X,= Alo) (k2 w°m2)F.,
1
and
Xone AG)) k,F.,
(461)
where A(w) is the characteristic determinant. Note that the amplitude X, becomes zero at the excitation frequency w = Vk,/m,. An undamped dynamic absorber is “tuned” for k,/m,=k,/mp, tach that X, approaches zero at the resonance frequency of the original system.
(b
ae
The frequency equation in Eq. (460) can be expressed as
mm.
ae
,
k,2)mM
ms] BP
ae =1+—)—+— Shean
Since
k,/k,=m,/m,,

e
oti
w?+1=0
a frequency ratio r is defined as r=a@/Vk,/m,=
w/Vk,/mz. The frequency equation reduces to r?—(2+m,/m,)r°>+1=0
(462)
From Eq. (461), the responses can be expressed as x
es
F./ky
1 ae r’
r—(2+m,/m,)r+1
Re ay F./k,
1
(463)
r*—(2+m,/m,)r°+1
The equations are plotted in Fig. 417 for m,/m,=0.3. The plus or minus sign of the amplitude ratio denotes that the response is either inphase or 180° outofphase with the excitation. Resonances occur at
r=(0.762 and 1.311. Note that x,(t)=0 when k,—w’m,=0. It can be shown from Eq. (461) that this condition occurs when the excitation F., sin ot is balanced by the spring force —k2xp.
(c) The frequency equation, Eq. (462), is plotted in Fig. 418 to show the effect of the mass
ratio
m,/m,.
When
m,/m,
is small, the resonant
172
Amplitude ratio
SIG,
ee /k, )
m,/m, = 0.3
Frequency ratio
r= w/Vk,/m,
. 417. Typical harmonic response of a twodegreeoffreedom system: Example 12. frequencies are close together about the resonance frequency of the original system. This means that there is little tolerance for variations in the excitation frequency, although x,(t)=0 when r= 1. Furthermore, it is observed in Eq. (463) that the amplitude X, of the absorber at r= 1 can be large for small values of m,/m,. When m,/m, is appreciable, the resonant frequencies are separated. For example, when m,/m,=0.4, 1.6
k,/m,
1.4
Nae
1.0
0.8 w/ ratio Frequency r= 0.6
Mass ratio m,/m, Fic. 418.
Undamped
Example 12.
dynamic
absorber:
effect of mass
ratio m,/m,;
SEC. 48
Forced Vibration—Harmonic excitation
173
resonances occur at r equal to 0.73 and 1.37 times that of the original system. The amplitude X, of the absorber mass is correspondingly reduced at r=1 for larger mass ratios.
Example 13.
Dynamic absorber with damping
Consider the dynamic absorber in Example 12 in which a viscous damper c is installed in parallel with the spring k, as shown in Fig. 419(b). Briefly discuss the problem. Solution:
From Eq. (453), the equations of motion in phasor notations are er
—k,—jwc
—k2—jwce
Be
k.— w?m,+ jc
LX,
0
From Eq. (413), the corresponding frequency equation 1s A(w)
=
k,+k,—w?m,+j 1 2 @ ‘7 Jac —k,—jwc
—k,j Z es ky— w* m+
=
joc
From Eq. (458), the phasors of the responses are X=
(ho w?ma+ jwe)Fa
and
X=
57 (kot joc)
where A(w) is the characteristic determinant. The values of X, and X, can be calculated readily using the programs in Chap. 9. The response curve of a properly tuned* dynamic absorber with appropriate damping is shown in Fig. 419(a). Curve 1 is that of an undamped system and curve 2 corresponding to c=. Curve 3 of intermediate damping must pass through the intersections of these curves.
Amplitude X,
wh ‘\ AES
ENP x Se
Curve 3_/\
/ \/
Ss Curve
~
2/0
Excitation frequency w (a)
Dynamic absorber with damping
Fic. 419.
(b)
Vibratory system
Dynamic absorber with damping: Example 13.
* J. D. Den Hartog, Mechanical
Vibrations, 4th ed., McGrawHill
Book Company, New
York, 1956, pp. 93102. Note that the “tuned” condition of k,/m,=k,/m, in Exampie 12 is only for undamped absorbers. See Prob. 427 for discussion of dynamic absorbers with viscous damping.
174
Systems with More Than One Degree of Freedom
CHAP. 4
ys sin wl
0
oundation
Fic. 420.
Example 14.
Vibration isolation: Example
14.
Vibration isolation
A constant speed machine is isolated as shown in Fig. 323(a) and the girls in the office complain of the annoying vibration transmitted from the machine. It is proposed (1) to mount the machine m, on a cement block m, as shown in Fig. 420, or (2) to bolt m, rigidly to mz. Assume m,/m, = 4, k,=k,, and the excitation frequency »=2,,;=2Vk,/m,. (a) Find the
magnitudes of x,(t) and x,(t) and the force F; transmitted to the rigid foundation. (b) Neglecting the damping in the system for the estimation, would you approve proposal 1 or 2? Solution: (a) From
Eq. (453), the equations of motion Bier
seaciat
—k,—
—k,—joc,
in phasor notations are
Joc;
ki +k.w*m,+
jo(c,
lela +c)
x
0
The characteristic determinant A(w) can be identified from the equation above. Equating A(w) to zero gives the frequency equation. Thus, k,—w’m,+ A(@)= 
jac,
—k,—joc,
—k,—joc, ki +k,w?m,—
z
jw(c;
+c)
From Eq. (458), the response of the system is
=
1
x =
2
a) [k,+k,— wm
+ jwo(cy + C2) Fog
1
X,= AC@) (k, + jwc,) Fog
“ The
force F, is transmitted
to the foundation
through the spring
'k, and the damper cy). Thus,
=
os
[Fixe coe X(k2+
1 jwc2) =
A(@)
; (k, + joc,)(k.+
; j@C2) Feq
SEC. 49
Influence Coefficients
175
The solution {X} and the force transmitted F; can be calculated from the equations above.
(b) Proposal 1: The ee
A(w)
=
equation of the undamped system is —@® =m,
ie ie
=()
—k,
—k,
k,+k.—w’*m,
Substituting the given conditions m,/m, = 4, etc., the frequency equation can be expressed as A(@)
=
k,k,1
—6r?+4r*)
=s 0
where r= @/@,; = @/Vk,/m,. Correspondingly, we get
=
X,=—
x
(k, +k,—w’m,)F
=
1
:
1
Ya *weilayre ik 3 ae PAG
S For
S
24r?
Mei: WHI
x Big2
i owes 1eee 6r+4r
F
eTyi
the given values,
resonance
occurs
1
are FB,eee 16r+47
RH
at r=0.437
and
1.14. At
@=20,; or r=2, we have X, =X,=0.341F,/k,, Xo = 0.024K,/k,, and Fy =F;=0.024F.,,. Proposal 2: If m, is attached to m2, the system has one degree of freedom. The equation of motion in phasor notations is [k. w*(m, ar m)X> = rae
\
Using the given data and normalizing the results by k,, we get
>
rs
Fike
— Si
Fr =k,
ie
ial
Gaeibs d
——=
SEeaalea Sr: eq
Resonance occurs at r=0.45. At the excitation frequency w= Qa orr=2, X,=X,=0.053F,/k, and 'F;= F; = 0.053F.,, Hence the force transmitted is higher than that in proposal 1.
49
INFLUENCE
The method formulate the in the analysis by its stiffness influence theorem,
COEFFICIENTS
of influence coefficients gives an alternative procedure to equations of motion of a dynamic system. It is widely used of structures, such as an aircraft. A spring can be described or its compliance, which is synonymous to the flexibility
coefficient. We shall first (1) show Maxwell’s reciprocity (2) relate the stiffness and flexibility matrices, and then (3)
illustrate the method of influence coefficients.
176
Systems with More Than One Degree of Freedom
Station
Fic. 421.
CHAP. 4

Method of influence coefficients.
applied at station j, when this force is the only force applied. Consider the beam shown in Fig. 421. The vertical force F; is applied at station 1 and F, at station 2. First, let F, be applied to station 1 and then F, to station 2. When F, is applied alone, the deflection at station 1 is F,d,,. The potential energy in
the beam, by virtue of its deflection, is }F{d,,. Now, when F, is applied, the additional deflection at station 1 due to F, is F,d,,. The work by F, corresponding to this deflection is F,(F,d,,). Thus, the total potential energy U of the system due to F, and F, is U= 3F id; + F,(F2d,2) +3F3d2.
Secondly, let F, be applied to station 2 and then F, to station 1. It can be shown that the potential energy U due to F, and F, is U =$Fsd,,.+
F,(Fyd>,)
+3F id,
The potential energies for the two methods of loading must be the same, since the final states of the system are identical. Comparing the expressions for U, we deduce that d,,= d,, for the system with two loads. This is called Maxwell’s reciprocitytheorem. For the general case, we have
\a; = dy 
for
ibalaese eae
(464)
which holds for all linear systems. When the force F is generalized to represent a force or a moment, the influence coefficient d;,; correspondingly represents a rectilinear or an angular displacement. Furthermore, when
the deflections due to the inertia forces are considered, we obtain
the equations of motion of the system. During vibration, the inertia force associated with each mass is transmitted throughout the system to cause a motion at each of the other
SEC. 49
Influence Coefficients
masses. For the undamped system, we have
177
free vibration of a twodegreeoffreedom
AE.
(3
=
palleane 
dz,
é dz2JL—m2q,
465
or
{q}= Ld { md}
(466)
where {q} is a generalized displacement vector, {mg} the generalized force vector of the inertia forces, and [d,,] is the flexibility (influence coefficient) matrix. For example, the deflection q, at station 1 is due to the combined
effect of the inertia forces —m,g,
and —m,q). The total
deflection q, is [d,,(—m,4,) + dy.(—m2q,)].
From Eq. (426), in the absence of dynamic coupling, the equations of motion for the free vibration of an undamped system can be expressed as {mq}= — Lk: Kq}
(467)
where [k,,] is the stiffness matrix. Premultiplying Eq. (466) by the inverse [d,,]* of [d,;] and rearranging, we get {mq} = —[d,}“{q}
Comparing the last two equations, it is evident that [k;;] = [d,1*
or
[ki Idi] ==
(468)
where I is a unit matrix. This is to say that [k,,] is the inverse of [d,,] and
Vice versa. Example 15 Write the equations of motion for the system shown in Fig. 42(a) by the method of influence coefficients and find the frequency equation. Solution:
To find the influence coefficients, let a unit static force be applied to m,. The springs k and k, are in series and their combination is in parallel with k,. Thus, Keg = ky +
kk k+k,
The corresponding deflection of m, is dj.
ee
1
as eee oe k+k, ee
1 Kea Kika + k (ky + ka)
Since the deflection of a spring is inversely proportional to its stiffness and the deflection of m, is d,,, it can be shown that the corresponding deflection
CHAP. 4
Systems with More Than One Degree of Freedom
178
d,, of m, is
k
k
Gis at
py Pk
d1>
Similarly, considering a unit static force at m2, we get
k+k; Ce ee REI) Combining the influence coefficients yields
tayl=2" ae tlds
1
tds
bee
leekglnEka
Be)
k
ki
Oe Be
The equations of motion from Eq. (466) are
{x} =[d, {—mx} Xo
kiko+
k(k,+k,)
k
(469) k+k,
—M2X2
Note that the stiffness matrix from Eq. (49) is
ee
jee
—k
It can be shown readily that [k,;]=[d,,]"’. Moreover, the premultiplication of Eq. (470) by [d,]"* will give Eq. (49), which are the equations of motion of the same system by Newton’s second law. To find the frequency equation, we substitute (jw)? for the second time derivative and express Eq. (469) in phasor notation as 
_ ie
x
hl
dx,
da
ee
w?mX>
Or
een —d>,@
mM,
Bac oike 2] ~ B 1—d,.0
M2
X,
(471)
0
The frequency equation isobtained by equating the determinant A(w) of the coefficient matrix of {X} to zero. 2
1—d,,0
A(@) = 
—a51@
2
my,
mM,
=o)
t—id@
2
mM,
2
=0
(472)
Ms
Substituting the values for d;, expanding the determinant A(w), and simplifying, the frequency equation becomes m,m,w" —[(k +k,)m,+(k+
kz)m, Jo? +[k,k,+ k(k,+ k>)] =0
This is identical to the frequency equation in Eq. (414) by Newton’s second law for the same system.
SEC. 49
Influence Coefficients
179 mx
(b)
\ Fic. 422.
Determination of influence coefficients
Influence coefficients due to force and moment; Example 16.
Example 16
Determine the natural frequency of the system shown in Fig. 422(a). Assume that (1) the flexural stiffness of the shaft is EI, (2) the inertia effect of the shaft is negligible, (3) the shaft is horizontal in its static equilibrium \
position, and (4) the mass moment of inertia of the disk is J = mR*/4 where R=L/4.
Solution: The inertia forces are as shown in Fig. 422(a) and the influence coefficients
are defined in Fig. 422(b). From elementary beam theory, it can be shown that the influence coefficients are Gu SEL,
dy, = L/EI,
dp =d,,=L7/2EI
The equations of motion from Eq. (465) are
SlLas gall 6
as
dy»
~J6
or
S lse ea
6}
6EIL3L
pa (tied
6
SLJ6
Following the last example, this can be expressed in phasor notations as ; shown in Eq. (471).
6EI20’mL? 30’mL?
—3° JL’ IEEE 6EI—6w7JLILO} [0
180
Systems with More Than One Degree of Freedom
CHAP. 4
The frequencyis obtained by equating the determinant A(w) of the coefficients of {X ©} to zero. 6EI—2w*mL?
—3w° JL”
May= ONare
a
GET er IE ©
Substituting J= mR7/4 and R=L/4, and expanding A(w), we get
w*—268(El/mL’)w? +768(El/mL’)* = 0 Hence
w=1.70VEI/mL?
410
and
16.3VEI/mL?
SUMMARY
The chapter introduces the theory of discrete systems from the generalization of a twodegreeoffreedom system shown in Fig. 41. The equations of motion in Eq. (41) through (44) are coupled, because the equation for one mass is influenced by the motion of the other mass of the system. The modes of vibration are examined in Sec. 43 for undamped free vibrations. The natural frequencies are obtained from the characteristic equation in Eq. (413). A mode of vibration, called the principal mode, is associated with each natural frequency. At a principal mode, (1) the entire system executes synchronous harmonic motion at a natural frequency and (2) the relative amplitudes of the masses are constant, as shown in Eq. (416) and illustrated in Fig. 42(b). The relative amplitudes define the modal vector for the given mode. The general motion is the superposition of the modes, ‘as shown in Eq. (417). A system can be described by more than one set of generalized
coordinates {q}. In Eqs. (427) through (429), it is shown that the elements of the mass matrix and the stiffness matrix as well as the type of coupling in the equations of motion are dependent on the coordinates selected for the system description. Hence coordinate coupling is not an inherent property of the system. The coordinates that uncouple the equations are called the principal coordinates {p}. The coordinates {p}
and {q} are related by the modal matrix [u] as shown in Eq. (434). A method for finding the modal matrix is shown in Sec. 46. The equations of motion can be uncoupled by means of the modal matrix. Thus, each uncoupled equation can be treated as an independent onedegreeoffreedom system. The results can be expressed in the {p} or {q} coordinates as desired. The technique is conceptually simple, but computers are necessary for the numerical solutions. Many practical problems can be represented as semidefinite systems as discussed in Sec. 47. A system is semidefinite if it can move as a rigid body. Correspondingly, at least one of its natural frequencies is zero.
Problems
181
The harmonic response of discrete systems can be found readily by the mechanical impedance method. Using phasor notations, the equations of motion can be expressed as Z(w){X} ={F} in Eq. (455) and the response as {X}= Z(w) ‘{F} in Eq. (457). The method of influence coefficients in Sec. 49 gives an alternative procedure to formulate the equations of motion. From Maxwell’s reciprocity theorem, the flexibility matrix [d;,] is symmetric. The inverse [d, ]”* of the flexibility matrix is the stiffness matrix [k,]. Thus, except for the technique in obtaining the equations of motion and certain advantages in its application, the concepts of vibration in the previous sections can be applied readily in this method. PROBLEMS Assume all the systems in the figures to follow are shown in their static equilibrium positions. 41 Consider the system in Fig. 42(a). Let m;=m,=10kg, kj =k,=40 N/m, and k=60N/m. (a) Write the equations of motion and the frequency equation. (b) Find the natural frequencies, the principal modes, and the modal matrix. (ec) Assume {x(0)}={1 0} and {x(0)}={0 1}. Plot x,(t) and x(t) and their harmonic components. (d) Assume {x(0)}={0 0} and {x(0)}={1 —1}. Find x,(t) and x,(t). 42 Repeat Prob. 41 if m,;=m,=10kg,
k,=40 N/m,
k,=140 N/m, and k=
60N/m. Are the motions {x(t)} periodic? ON 200kg uniform bar is supported by springs at the ends as illustrated in (
rig. 45. The total length is L=1.5 m, k; =18 kN/m, and k,=22 kN/m. (a) Write the equations of motion and the frequency equation. (b) Find the [ natural frequencies, the principal modes, and the modal matrix. (ec) If x(0)=1, x(0) = 6(0) = 6(0) =0, find the motions x(t) and @(t). (d) Illustrate the principal modes, such as shown in Fig. 46.
44 For the threedegreeoffreedom
system in Fig. 47, if J;=2J,,
J,=2Js,
k,,=2k,2, and k,.=2k,3, find the motions 6,(t), 02(t), and 6(t). 45 For each of the systems shown in Fig. P41, specify the coordinates to describe the system, write the equations of motion, and find the frequency equation.
ay A double pendulum. Aby The arm is horizontal in its static equilibrium position.
Ac Three identical pendulums. Ad) A double compound pendulum. A) A schematic representation of an overhead crane.
“ty The system is constrained to move in the plane of the paper. (g) The bar and the shaft are initially horizontal. The shaft deflects vertically. The bar moves vertically as well as rotates in a vertical plane.
ae 9
N
ae
182 ~r fear
—>
tee
>) Pe.
mgt V rr a
ali, tae td
big aie CNLoO
.
>
ae
QO
20
Systems with More Than One Degree of Freedom Z,+9,,2,) Ke) ry) ;
mw
_
CHAP. 4
\ > 92a Laun9 >Om STAp=
Pod
.
bale IEA
2
1
Wks
—w’ J;
wo J5/k,
(oie
56
MYKLESTADPROHL
Bab JUN
1
tee
METHOD*
Myklestad and Prohl developed a tabular method to find the modes and natural frequencies of structures, such as an airplane wing. It is generally known as the Myklestad method. We shall use the transfer matrix technique for this discussion. Following the finite element approach illustrated in Fig. 510, a structure or a beam can be divided into segments. A typical segment of a beam, as shown in Fig. 512, consists of a massless span and a point mass. The flexural properties of the segment is described by the field transfer matrix of the span; the inertial effect of the segment is described by the point transfer matrix of the mass. Hence the procedure is identical to that described in the last section, except for the state variables associated with the problem. To describe the field transfer matrix, consider the freebody sketch of a uniform beam of length L in span n as shown in Fig. 512(a). For equilibrium, we require
VE=VR,
and
ML=M®",L,VE,
(526)
where M and V are the moment and the shear force, respectively. The symbols in this section are defined in Fig. 59 and the subscripts and superscript notations are the same as the last section. Referring to Fig. *N. O. Myklestad, ‘““A New Method of Calculating Natural Modes of Uncoupled Bending Vibrations of Airplane Wings and Other Types of Beams,” J. Aeron. Sci., (April, 1944), pp.
153162. M. A. Prohl, ““A General Method for Calculating Critical Speeds of Flexible Rotors,” Trans. ASME, vol. 66 (1945), p. A142.
208
Methods for Finding Natural Frequencies
(a)
Freebody sketch of span
Fic. 512.
(b)
CHAP. 5
Freebody sketch of mass
Derivation of transfer matrix of a beam.
512(a), the change in the slope ® of the span is due to moment M/, and the shear Vi E i '—@R .=(—] Mit (=) \ee 27
seers
bale PREM
=
GER
Substituting M‘ and V/; from Eq. (526) in (527) and rearranging, we get L o:08 ,+(4) ME (ie
ve,
(528)
The change in the deflection Y of the span is L VE)
The first term on span, the second force. The shear stituting My and
R R L? Cae +(2EI ) 1 = 1,08, ot ee+
i 3EI
) ae=

(5 29)
the right is the deflection due to the initial slope of the term due to the moment and the third term the shear deformation of the beam is assumed negligible. SubV‘; from Eq. (526) in (529) and rearranging, we obtain
Yi=¥8,+1,08,+ (5) ms,(=) vR pal § heoae GEL)
(530)
The field transfer matrix is obtained by writing Eqs. (526), (528), and
SEC. 56
MyklestadProhl Method
209
(530) in the matrix form. I,
L2
ue
HSS
M
L3
ra
Sela
L
e
BY
= 2EI ==
Og.
Aaa ee US di eed
R
aaelee (531)
M
JS all thBAYA
n=1
To derive the point transfer matrix, consider the freebody sketch of m,,
in Fig. 512(b). The D’Alembert’s inertia loads are —w?m,Y"% and —w’J,®*, where J, is the mass moment of inertia of m, about its axis normal to the (x,y) plane. Neglecting the applied force P and the torque T, the equations for the shear and moment are
VR=Viw?m,Y,
and
M=Mi@7J,®;
(532)
yi= ys
(533)
For the rigid body motion of m,, we have
p= DF
and
The point transfer matrix is obtained from Eqs. (532) and (533). ys
1
0
OO
®
0
1
0
O}]
®
0
w7J
1
0
M
0
0
1
=
M V
n
—w*m
Yale
n
34
Cae
V
n
The transfer matrix for the segment n is obtained by substituting the
state vector {Z}* from Eq. (531) in (534). R
IEZ
Y
=
®D M Var,
il
0
ree)
1
It
0
1
Ome
TAO
ad
0)
Ord
MAP
—w>m
0
We
1b,
ee EI
®
ENOINOSO
1
=I hy
M
tObRe
0
Org
=a EI
il
L?2
is 
R
= 6EI
R
D
Lb
EI
:
.
2EI
0
1
L — EI
HV
Net
jes
=
=
R
4
6EI L?
®
2EI
(535)
IL?
M
0
V
—@?m
mes
w~J
Le 1—@
RON pee
ay,
aay
er L+q@ Se
2
[
—w?mL
w*m
2EI
ee
1+@?m
6EiSIeL
M
V
2,
210
Methods for Finding Natural Frequencies
CHAP. 5
This is the recurrence formula analogous to Eq. (519). Hence the problem can be solved by its repeated application as indicated in Eq.
(520). The common boundary conditions for the beam problem are
Simple support Free Fixed
Y
@
Mav
0 Y 0
@® A,>:>A,. sufficiently large, Aj > Aj, for i=2,...,n. Thus, we obtain {V}, = H*{V}o=c,Ait{u}i
If s is
(656)
If one more iteration is performed, it can be seen that the result is {V}.41= c,Aq"{u}, =A, AV}.
(657)
Thus, the first eigenvalue A, and the corresponding modal vector {u}, can
be obtained. The second mode is found by suppressing the first mode p,, that is, by introducing the constraint p;=0. Then the equations of motion are modified accordingly and the iteration procedure repeated. To suppress the first mode, consider the transformation
{q}=[uKp}
and = {p}=[u} {q}=[vl{q}
Using the constraint equation p, =0 gives
Pu= 01191 + Vingot + * + + 01,9, =9
(658)
The rest of the coordinates remain unchanged, that is, Gi = Gi
for
Peano ee coetent ()
The last two equations can be conveniently expressed in matrix notations.
236
Discrete Systems
CHAP. 6
Using a 3X3 matrix to illustrate the last two equations, we have U1,
«Ui2)
—U437
11
0
0
0
0
1
ae
0
liLq,
qi
Vi
Vin
0137
0
So
q3
)
)
1
0)
So 1S ©=
0474;
qi
1jLq;
or
{q} = S,{q}
(659)
where S, as defined above is called the sweeping matrix. The matrix H in Eq. (652) is modified by S, to give a new matrix H,. H,
=
HS,
(660)
Now, H, can be used in Eq. (652) for the iteration of the second mode,
since the first mode has been suppressed. Equations (655) to (657) can be used for the iterations as before. The v’s in Eq. (658), however, cannot be determined directly, because only {u}, is found and [u] '=[v] is as yet unknown. From the orthogonal
property in Eq. (635), we have
[u}"M[u]=[M.] where [~ML] is diagonal. Premultiplying both sides of the equation by
[~“M.]"* and postmultiplying by [u]~’, we get
[ML] '[u}"M = [uy '= [v] Since [~M.] is diagonal, the first row of this equation gives [011
012°** U1,)]= (constant) u ,M
(661)
Thus, the sweeping matrix in Eq. (659) can be constructed. Example 6 Determine the natural frequencies and the modal vectors of the system in Fig. 65.
eae Fic. 65.
A threedegreeoffreedom system.
SEC. 69
Matrix Iteration
237
Solution:
The equations of motion are
men
Ome
:m
0
0
Ost
tex:
Ski
=k
) O16
x;
offs: [> 2k
9 1
1
‘
fir Fet
where Mar
Spe)
or
Glu} =u) Let {V},={1
1
1} arbitrarily to start the iteration. From Eq. (655),
we have

3 
{V}, = G{V}= > 9 ie
sypil
9
1.0
 = [5= .
oe
oll  al
9
1.0
The constant 9 is discarded. Continuing the iteration gives
Dios 39 iS}
man
t1f 1.0 AL 1.0 >frs7][as] ufos] Saka 11 1.0
Again, the constant 11 is discarded. The fourth and fifth iterations give
1.0 19420
ine
and
ins
1.0 129)20
1.0
Hence, we choose {u},={1
1.0
2
1} and 1/A*=12=12k/mA,,
,
12k
Oa
or
k
= aes a ee
12m
m
The first mode is suppressed by the constraint p;=0 as shown in Eq. (658). The v’s are found from Eq. (661) as wm 2
lou
VisJ=[1
2
WO
@ 
nt m OO
The
common
constant
terms
in the vector
i
v] can
be discarded.
Ota
at
sweeping matrix from Eq. (659) is borzinin
spo
m0m0
sf 1  2 1 o[o Oreoret i Od Omer
2
1  Oars
The
238
Discrete Systems
CHAP. 6
From Eg. (660), the new matrix G, is
53
lire
G,=GS,= > 9 Lod
=2n
fo SIL
8
Oi
=i
1 = ° 3 Ops 2 en
 Same
The second mode is obtained by assuming an arbitrary vector {V},)= {1 1 1} and iterating on G,. After 14 iterations we have {u},= {f1 0 1} and 1/A}=4=12k/mdA,. Hence
The first and the second modes are suppressed by the constraints p;= p2=0. The vectors v], and v], can be found from Eq. (661) as before. Thus, the sweeping matrix S, from Eq. (659) is 1, S,=1
2
a0.
0 0
1
0.
0
Unet
0
0
OVOl—I8>
Urea
0
I
0.1 On
Oana!
The new matrix G, from Eq. (660) is
Ou
Rh
G,=G,S,= ° 3 C1. {1 {J
*4nrOhe of 0 “410.0
4
64
Dia
= = °0 3 oO
The third mode is obtained by assuming an arbitrary vector {V})= 1 1} and iterating with G,. After two iterations we obtain {u};= 1 1} and 1/A3=3=12k/mdA3. Hence
3m
m
Summarizing the calculations, the natural frequencies w= V3k/m, o,=~V4k/m. The modal matrix is
1 wa
iby)
2 0 ieee
610
 238
UNDAMPED FORCED MODAL ANALYSIS
are
w,;=Vk/m,
Pal “ 1
VIBRATION—
The method of modal analysis transforms the simultaneous equations of motion of a discrete system into a set of independent secondorder differential equations. Each of the uncoupled equations in the principal coordinate
p,(t) describes a mode
of vibration. The resultant motion is
obtained from the superposition of the motions of the modes. The procedure consists of (1) using the orthogonal relation in Eq. (637) to uncouple the equations of motion in order to describe the
SEC. 610
Undamped Forced Vibration—Modal Analysis
239
equations in the principal coordinates {p}, (2) employing the transformation {p(0)}=[u]” {q(0)} in Eq. (623) to find the contribution of the initial conditions towards the excitation of each mode p,, (3) superposing the solutions due to the initial conditions and the excitation for each mode p,
as shown in Eq. (274), and (4) finding the response in the original coordinates {q} by the transformation {q}=[u]{p} as shown in Eq. (623). The method is conceptually simple. Computers, however, are necessary for the numerical solution of the problem. We shall first discuss the particular solution due to the transient excitation and then the complementary solution due to the initial conditions, since the two can be handled separately. Consider the equations of motion of a conservative system
Mi4}+ K{q}={Q(t)}
(662)
where M is the mass matrix, K the stiffness matrix, {q} the generalized coordinates, and {Q(t)} the generalized excitation force. For convenience, we use the transformation {q}=[u]{n} from Eq. (639) to relate {q} and the normal coordinates {n}. Thus, Eq. (662) yields
M[ulki}+ K[w in} = {Q(t} Premultiply this by [uw]. Note that from Eq. (643) [w]’M[w]=f, a unit matrix, and [w]’K[u]=A, motion become
a diagonal matrix. Hence
{a}+ A{n} =[u]7{Q(0)} ={N(o)}
the equations of
(663)
where {N(t)} is a column vector representing the normalized excitation force. In other words, {N(t)} represents the contribution of the transient force {Q(t)} toward the excitation of the normal modes. Since A is diagonal and A; = w7, each of the uncoupled equations in Eq. (663) can be written as
7H + w7n; = N;(t)
fora =1,2;...,7n
(664)
The equation above can be treated as an independent onedegreeoffreedom system. From Eq. (271), the particular solution due to N,(t) is
ioe [ee irae (0)
where h;(t) is the impulse response of the system in Eq. (664). From Eq. (269), the impulse response of an undamped system is h;(t) = a sin w;t Combining the last two equations gives the transient response for the ith
240
Discrete Systems
mode
CHAP. 6
for zero initial conditions; that is,
Lea ni (t)=— N;,(7) sin @,(t—7) dt @;
fori=1,2,...,n
(665)
Jo
For the complementary solution, we first find the contribution of the initial conditions towards the excitation of the normal modes. The original initial conditions are {q(0)} and {q(0)}. By the transformation {n}= [u] {q} in Eq. (639), we get
{n(O)}}=[u] {q(0)}
= and = {n(0)}=[uT '{4(0)}
(666)
Let n;, and 7;, be the initial conditions in the normal coordinates for the ith mode. Hence the complementary solution of Eq. (664) due to the initial conditions is
1 n;(t) = Nip COS @t +— io SiN ajt
(667)
The complete solution for the ith mode is the sum of the responses due to (1) the excitation in Eq. (665) and (2) the initial condition in Eq.
(667). As indicated in Eq. (274), these responses are added directly to give the complete response. The complete solution in the generalized coordinates {q} is the superposition of the responses of the modes as indicated in Eq. (639). Let ni(t) be the sum of the responses from Eqs. (665) and (667). Thus,
aO}= Liukin)
or
{a(0}=(un}
(668)
Since the transient excitation {Q(t)} is arbitrary, only the formal solution, as shown in Eq. (668), can be presented. Computer methods
for the numerical solutions will be discussed in Chap. 9. The equations above can be expressed in alternative forms. Since these do not introduce additional concepts, we shall not further reduce the equations.
611
SYSTEMS
WITH
PROPORTIONAL
DAMPING
If a system possesses damping, the modal analysis described in the last section generally does not apply. The equations of motion cannot be uncoupled by the modal matrix of undamped systems except for systems with proportional damping. Consider the equations of motion of a system with damping.
M{q}+Dig}+ K{q} = {Q(x}
(669)
where the mass matrix M, the damping matrix D, and the stiffness matrix
K
are
symmetric.
{Q(t)} denotes
the generalized
force
and
{q} the
SEC. 612
Orthogonality of Modes of Damped Systems
generalized coordinates. Proportional damping* matrix D can be expressed as D=aM+
241
occurs if the damping
BK
(670)
where a and 8B are constants. To uncouple the equations, we substitute Eq. (670) and the transformation {q}=[]{n} from Eg. (639) in (669).
M[uKn}+laM+ BK][w]in}+ K[w Hn} = {Q()} where {n} are the normal coordinates of the corresponding conservative
system. Premultiplying this by [u]* and recalling that [w]7’M[]=I [u]’K[w]=A from Eq. (643), we get
{H}+[al+ BAKh}+ A{n}=[p]{O(D}={N(O}
and
(671)
OT
H, + (a + Bw7)H; + @77; = N;(t)
[Ota
= oe th
(672)
Hence the equations are uncoupled. Equation (671) can be treated as a set of onedegreeoffreedom systems.
612
ORTHOGONALITY SYSTEMS
OF MODES
OF DAMPED
The equations of motion of systems with viscous damping can be uncoupled by reducing the equations to a set of firstorder differential equations.+ Operating on the reduced equations, the analyses are essentially the same as those for undamped systems. For example, the same techniques are used to find the characteristic equation and to derive the orthogonality of the modes. The eigenvalues and the modal vectors, however, are complex quantities. Computer solutions, as illustrated in Chap. 9, are necessary for the numerical solution of the problem. Consider the equations of motion for the free vibration of systems with viscous damping, in which the matrices M, C, and K are symmetric of
babes
M{q}+ C{4}+ K{q} = {0}
(673)
The reduced equations are formed by introducing a 2nX1
state vector
o[4]
{y}.
(614)
q
It can be shown that Eq. (673) can be expressed as
me ech)  epee Loo Tepeedero eee 2nX2neenx 
Die aeeee
Tew
*Lord Rayleigh, The Theory of Sound, vol. 1, Dover, New York, 1945, p. 130. T Frazer, Duncan, and Collar, op. cit., p. 289.
(675)
242
Discrete Systems
CHAP. 6
or
Aty}+ Bty} = {0}
(676)
where A and B are defined by comparing the last two equations. The matrices M, C, and K become submatrices of A and B. Hence Eq. (673) is reduced to a set of 2n firstorder equations shown in Eq. (676). It can
be expressed in a more convenient form by premultiplying by A~* and
defining
H=—A~'B. Thus,
(677)
{y}— H{y}= {0}
The eigenvalues are obtained by assuming the solution of Eq. (677) is of the form
tyre"
(678)
where y is a complex number and {WV} a 2n X 1 modal vector with complex elements. Substituting Eq. (678) in (677) yields
[yIl— H]{¥} = {0}
(679)
Hence the characteristic equation is
A(y) =yI— H=0
(680)
This may be compared with Eq. (621) for undamped systems. Since H is a square matrix of order 2n, there are 2n eigenvalues, which are necessarily complex conjugates. We assume the eigenvalues are distinct. A modal vector {¥} is found by substituting an eigenvalue y in Eq. (679) and solving the corresponding homogeneous algebraic equations for {¥}. The process was illustrated in Example 9, Chap. 4, and it is conceptually simple. The subroutine $CHOMO, listed in App. C, can be used to solve the complex homogeneous algebraic equations. Since the matrix H is of order 2n, there are 2n modal vectors, which are necessarily complex conjugates. The modal matrix [VY] is a linear combination of the eigenvectors and is of order 2n. [v]= LW}, {V},
Be
{Vhon 
(681)
Let us show that the modal vectors {VW} are orthogonal relative to the
matrices A and B. Substituting Eq. (678) in (676) and factoring out the e”' term, we have
yA{V¥} + BLY} = {0} Substituting y, and y, for the rth and sth modes in the equation above gives
VAT a BAe
{UF
VAM} + Biv ={0}
SEC. 613
Damped Forced Vibration—Modal Analysis
243
Premultiplying the first of these equations by the transpose W], of {V}, and transposing the resulting equations, and premultiplying the second equation by V],, we get
yr, LW] ALY}, +P] Biv}, =0
TAMAS UNG Garo Bene eed
(682)
The difference of these equations gives
(y,— ys) LV] ALY}, = 0 Since y,# y,, we deduce that
[v A{v},=0 Similarly, it can be shown
for
rs
(683)
for
EFAS
(684)
that
[Vv], B{V}, =0
The last two equations are the orthogonal relations for systems with viscous damping. They may be compared with Eqs. (635) and (636). It is implicit that Eqs. (683) and (684) are not zero for r=s. By virtue of the orthogonal property, using the modal matrix [WV] to include all the modal vectors we obtain
[¥]"A[¥J=[A]
and
= [¥]*B[W]=FB_]
(685)
where [~A~] and [~BY] are diagonal. This may be compared with Eq. (637) for undamped systems. If the matrices A and B are diagonalized as shown above, then the equations of motion in Eq. (676) are uncoupled correspondingly.
613
DAMPED FORCED VIBRATION— MODAL ANALYSIS
The technique for the modal analysis of systems with viscous damping is similar to that for undamped systems, except that the complex modal
matrix [WV] is applied to the reduced equations. We shall first uncouple the equations of motion, find the particular solution due to the excitation, and then the complementary solution due to the initial conditions. When the excitation {Q(t)} is applied to the system, the equations of motion in Eq. (673) becomes
M{q}+ Ciq}+ Kig} ={Q(t)} As shown in Eqs. (675) and (676), this can be reduced to a set of 2n
CHAP. 6
Discrete Systems
244
simultaneous firstorder equations as
Ie cla +00 ella]  Low. 2nx2n
Diis 6\
Problems
(b) Evaluate the modalamet [ul]. \UA

L
wri aN ues RA
Lal K Kul=LK
J
(c) Find the atelier’ cheegter erpremed in the principal coordinates.
A (4) >) (d) Verify that [uJ ‘H[u]=A, wieiee A is a diagonal matrix with oe ’s as the diagonal elements. Yat OF ob] (e) Normalize [u] such that the mas M becomes a unit matrix as shown in Eq. (643).
a a = My VA Te
Ay Gar
My
HMM
wae
t
Gay 4 mar
Mir G Gut
(f) Express the uncoupled equations in the normal coordinates.
Wa beruytm, Ay
oy Mee
Vine 23 e
Repeat Prob. 68 for the system shown in Fig. 411, that is,
° offs] 2
0
00.
O7F%X:
cP
2ILs.
OR
610 Determine shown
the modal
pales
pel Cea Up  p55 lan
F,(t)
lex;
F;(t)
matrix of the eee
in Fig. P53(a).
Assume
J,=J,=6,
system with fixed ends as
J,;=9,
k,=6,
and
k,,=k,.=
k,3= 18. 611 A semidefinite system can be made positive definite by suppressing the zero mode as illustrated in Example 5. Consider the free vibration of each of the following semidefinite systems: System
J,
(1) (2) (3)
JS
2 5 pelle 8 2
J;
Kea
ke
8 544 wii es cap 8 8
36 tate) 8
(a) Find the modal matrix [u]. Show that [u] 'H[u]=A. (b) Write the uncoupled equations. (c) Normalize [u] such that the mass matrix becomes a unit matrix.
(d) Express the uncoupled equations in the normal coordinates.
612 Find the natural frequencies and the modal matrices of each of the systems in the following figures by matrix iteration: (a) A twomass system in Fig. 42(a). Assume 20 kN/m, k,=2kN/m, and k=4kN/m. (b) A
double
pendulum
shown
m,=4kg,
in Fig. P41(a).
Assume
m.=1kg,
k,=
m,=m,,
and
L,= Ly. (c) The torsional system in Fig. P53(a). Assume J, = J,=6, J3=9, ko = 6, and
ki=k2=
kz =
18.
:
(d) Each of the configurations of the system shown in Fig. 49. Assume m=1.0kg, Jog=0.6 m* kg, k;=60 N/m, k,=40 N/m, L,=1.5 m, and 1
=
1.0m.
mi
Gi
250
Discrete Systems
CHAP. 6
613 Referring to the equations of motion in Prob. P68, find the transient response for each of the following initial conditions:
(a) {x(0)}= {0}, {x(0)} = {0}; (b) {x(O)}={1
O}, {xC)}={2
1.
Check the solutions by the classical method.
614 Consider each of the following systems:
fo sills EE SIFtal [5 cllelLo2 calle Ls aleILo Silla
=1
0
A
2)Lx,
PANS,
6,
eA
es
8
JLx2
4
0
4)Lx,
0
Find the characteristic equation by means of the reduced equation as shown in Eq. (686). Check the characteristic equation by substituting D(= d/drt) in the system equations above and expanding the determinant of the coefficient matrix of {x}. Computer problems:
615 Given the equations of motion
M{q}+K{q}={0}, where M and K are symmetric. By means of the transformation {g}=[t]{q} in Eq. (610), the equations become {g}+ S{g}= {0}, where S is symmetric. The form of [t] is as shown in Eq. (613). (a) Write a subroutine for the transformation matrix [tf].
(b) Write a program to use this subroutine and to show that [t]’[t]=M,
[t]’K[t]=S, and [[¢]’?'M[t)'=1 (c) Use the following data to verify your program:
6. 5. 397d.
reese =)
—lirg,)
373
Lie
11
Sg,
2)
one
alee
616 Use
the program ITERATE listed in Fig. 914(a) to find the natural frequencies and the modal matrices of each of the systems described in Prob. 435.
617 Rewrite the program ITERATE listed in Fig. 914(a) as a subroutine. Repeat Prob. 616, using this subroutine.
618 Given the semidefinite system (see Example 5, Chap. 5) 4
0
0
07174
0.1 0 0))@) 0
0
4
0}
6
el
oli
0
2 ae A
O77
4
0
2. OlPer_0 a
0
Problems
251
Write a program (a) to transform the system to become positive definite, (b)
to use the subroutine in Prob. 617 to find the natural frequencies and the modal
matrix, and (c) to express the modal
matrix in the original coordi
nates.
619 Referring to the cantilever shown in Fig. 912, write a program (a) to formulate the problem by the method of influence coefficients, and (b) to use the subroutine in Prob. 617 to find the natural frequencies and the modal matrix.
620 Rewrite the subroutine $CMODL
listed in Fig. C11 for the modal matrix of discrete systems with viscous damping as a main program. Find the modal matrix of the system shown below but do not list the complex conjugates.
3 0 OTF%, OTL ROO
10
3
—27/7%,
140
Opie 3. eb xe a
8 ot,
= 45 esi ghAlf ee
—60) 220) =80)) x54 = 20. 5=360) 200/)Ix.
60
=207[x:
0 0 0
621 Modify the program in Prob. 620 to obtain the printouts as follows: (a) The complex conjugate roots y of the characteristic equation. (b) The modal matrix [V].
(ce) The
product [¥]’A[W]=[A~]
and [¥]’B[¥]=[B_]
to show
the
[yA]
is a
orthogonal relations as shown in Eq. (685).
(d) The
product
[W] ’H[¥]=Fy—],
where
H=A™'B
and
diagonal matrix with y’s as the diagonal elements.
Note that the modal matrix [VW] of order 2n can be partitioned into two submatrices.
Vv [v] = Be where
anxn [v.] nX2n
=
L [¥%]
Cw
nX2n2nX2n
Verify the equations above from the printout.
622 Use the program TRESPDAM in Fig. 915(a) to find the transient response of a discrete system with viscous damping. Choose the appropriate initial conditions, {x(0)} and {x(0)}, and excitations {F(t)}. Consider the problem in three parts as follows:
(a) {F(t)}#{O}, {x(O)} # {0}, and {x(0)} # {0}.
(b) {F(t)} #{0}, {x(0)} ={0}, and {x(0)} ={0}. (c) {F(t)} ={0}, {x(0)} ¥{0}, and {x(0)} #{0}. Verify from the computer printout that the values of {x(t)} and {x(t)} from part a is the sum of that of parts b and c.
252
Discrete Systems
623 Modify the program TRESPDAM
CHAP. 6
in Fig. 915(a) such that the values of
{x(t)} are stored in one file and that of {x(t)} in another.
Choose
the
appropriate initial conditions and/or excitations. Execute the program. Use the program PLOTFILE in Fig. 95(a) to plot the results. 624 Modify the program TRESPDAM in Fig. 915(a) such that only every nth computed value of {x(t)} is entered into a data file and every nth value of {x(t)} into another. Choose the appropriate initial conditions and/or excitations. Execute the program. Use the program PLOTFILE in Fig. 95(a) to plot the results.
7 Continuous Systems
71
INTRODUCTION
A beam is an example of a continuous system. Its mass is distributed, inseparable from the elasticity of the beam. A continuous system has its mass, elasticity, and damping distributed. By contrast, a discrete system consists of distinct, separate, and idealized masses, springs, and dampers. Although complex continuous systems are often modeled as discrete systems, a basic understanding of the continuous system is helpful in formulating the equivalent discrete system. A continuous model, however, may be justified for many applications. The analysis of a continuous system is more involved. Its vibratory motion is described in terms of the space and time coordinates, and the equation of motion is a partial differential equation. Furthermore, the elasticity of a beam and the manner of its support must be considered. Thus, two additional aspects are introduced in the analysis, namely, elasticity and boundaryvalue problem. These two topics are separate studies in themselves. Fortunately, the conceptual aspect of its vibration theory closely resembles that of discrete systems. This chapter is an introductory study of continuous systems. The continuous model is assumed linear and the material in the system homogeneous and isotropic. The topics selected are similar to those for discrete systems. Onedimensional continuous systems are treated in some detail.
72
CONTINUOUS
SYSTEMS—A
SIMPLE
EXPOSITION
To illustrate a continuous system, we shall first derive the equation of motion for the lateral vibration of a taut string and then describe the solution in terms of the principal modes as for discrete systems. A flexible string of mass m per unit length under the tension T is shown 253
254
Continuous Systems
(a)
(b)
Lateral vibration
Fic. 71.
CHAP. 7
(c)
dx Element
Initial deflection
Lateral vibration of a flexible taut string.
in Fig. 71(a). The lateral deflection u along the string is a function of the
space variable x and time t.
u=u(x,t)
(71)
The freebody sketch of an element dx of the string is shown
in Fig.
71(b). If the lateral deflection is small, the change in T due to the deflection is negligible. Applying Newton’s second law and assuming small deflections u and @, the equation of motion is 2
mdx
Ou
at?
00
=T Cieee ar)—T0
(72)
Substituting 6 =0u/dx and simplifying, the equation becomes Faas a a?
(73)
where c*= T/m. This is a onedimensional wave equation. The
solution
of Eq. (73) has four arbitrary constants.
The problem
must be compatible with the boundary conditions at the fixed ends, which are u(0,t)=0
and
Uo
0
(74)
Two of the constants are specified by Eq. (74); the other two by the initial conditions u(x,0) = f(x)
and
oe(x,0) = g(x)
(75)
For example, if the string is plucked at some intermediate point as shown in Fig. 71(c) and released with zero initial velocity, then f(x) defines the initial deflection and g(x)=0.
Let us pause for a moment to consider the problem formulation. (1) The equation of motion is derived from Newton’s second law, Eq. (72). (2) The resulting equation is a partial differential equation, Eq. (73). (3) The string problem has four arbitrary constants to be evaluated by the boundary conditions, Eq. (74), and the initial conditions, Eq. (75). (4) It may be anticipated that the boundary conditions specify the type or the
mode of vibration, that is, the possible ways that the system will oscillate.
SEC. 72
Continuous Systems—A Simple Exposition
255
(5) The initial conditions give the actual motion, that is, the degree of participation of each mode. Since the system is undamped, we assume that a mode of vibration is harmonic as for discrete systems. Thus, the solution of Eq. (73) is of the form
u(x,t) = b(x)sin(wt +p)
(76)
where which
is a natural frequency, w a constant, and #(x) eigenfunction, an describes _the mode shape of the string atthe frequency—w. Substituting Eq. (76) in (73) and factoring out the sine term, we get
ae tz O(x) =0
(77)
a a o(x)  w?
The solution of this equation is o(x)=A sin ae
cos
(78)
Combining this with Eq. (76) gives the formal solution u(x,t) = (4sin Be B cos =) snot + w) The boundary condition u(0,t)=0 requires that
(79)
B=0. The condition
u(¢,t)=0 gives wot
sin a
0
ot
or
—=
i
for
Carle 2 eee
(710)
This is called the frequency or the characteristic equation. For example, the frequency for the ith mode is w; = imc/¢ rad/s. A continuous system has an infinite number of natural frequencies as indicated in Eq. (710).* Since the system is linear, the general solution is the superposition of the principal modes, that is,
.
Seige
u(x,t) = y A; sin 5 sin(@,t+ W;,)
(711)
i=1
where @; = imc/€. Since $(x) =sin imx/€ was determined by the boundary conditions, the modes are stipulated by the boundary conditions. The constants A; and w; in Eq. (711) are evaluated by the initial conditions. Let the equation be expressed in a more convenient form u(x,t) = »; (sinZC
sin w,t + D, cos «;t)
(712)
i=1
* A continuous system is said to have an infinite number of degrees of freedom. Then it is
necessary to explain that the natural frequencies are countable instead of being ‘“‘distributed.” An alternative viewpoint is that the mass and elasticity of continuous systems are distributed. If principal modes exist, then the natural frequencies are countable, but there are an infinite number of them.
CHAP. 7
Continuous Systems
256
Applying the initial conditions from Eq. (75) yields
u(x,0) = y D, sin _ = f(x) a ou
 (x,0)
—
(6) =
(713)
»: C.@; sin on = g(x) i=1
Consider the Fourier series expansion of f(x) and g(x). 
fay sit
hake
g(x)= ) g, sin t—
The constants


(714)
(6;
C, and D, are found by equating the coefficients in Eqs.
(713) and (714), that is, Cw,;=g, and D,=f, Since C,; and D, are associated with the vibratory motion at w,, in Eq. (712) the initial conditions determine the contribution of each principal mode to the vibration of the system.
73
SEPARATION OF THE SPACE VARIABLES
TIME
AND
Many methods can be used to solve the wave equation shown in Eq.
(73). The D’Alembert’s solution expresses the wave motion as the _ superposition of two travelling waves in opposite directions, that is, u(x,t) = Fe
ct) Foo + ot)
where F;, and F, are arbitrary functions. Laplace transform can be used to transform Eq. (73) with respect to either x or t. We shall discuss the method of the separation of variables, which has general applications. Let the partial differential equation in Eq. (73) be satisfied by the functions of the form
u(x,t) = o(x)q(t)
(715)
where (x) is a function of the space variable x alone and q(t) a function
of t alone. Hence, they are written as ¢ and q in subsequent equations; their differentiations give dd/dx and dq/dt instead of partial derivatives. Substituting Eq. (715) in (73) and simplifying, we obtain
ldb_1aq ob dx? q dt?
(716)
Since the left side of the equation is independent of t and the right side is independent of x and the equality holds for all values of t and x, it follows
SEC. 73
Separation of the Time and Space Variables
that each side must
be a constant.
257
Let the constant be —w*. Thus, we
obtain two ordinary differential equations ee
2
s
>
By virtue of the selfadjoint property, we get
I[AiG ene do=0
(753)
(A) $,M(¢,) da=0
(754)
z
—=

Since A,#A,, we deduce that o,M(¢,) da=0
for
r#zS8
(755)
for
r#Ss
(756)
>3
Similarly, it can be shown that o,L(¢,) da=0 z
The orthogonal relations are as shown in Eqs. (755) and (756). We shall apply these ‘relations to the wave and the beam«equation. * The eigenfunctions are selfadjoint if
6,L(¢,) do= $,L(¢,) do =
I$,M(¢,) do=
=
:
$,.M(¢,) dao
This is analogous to the symmetry of the stiffness matrix K and the mass matrix M of discrete systems, that is,Mik,, ==k; and m, ==m,
SEC. 78
Orthogonality
Case 1.
Boundary Conditions Independent of
273
(1) Consider the wave equation. Substituting Eq. (749) in (753) and then integrating by parts, we obtain
[[4.(k60) —4,(kay ax
=[4.ko5 i [ko6; dx—[4,kor6— [kbps dr €
rg
0)
=[4,kb.— bkbt5
(757)
The problem is selfadjoint and the eigenfunctions orthogonal if this equation is equal to zero for r#s. The equation is satisfied if the boundary conditions in Eqs. (725) and (730) are of the form.*
ad + bd'=0 wherea
(758)
and b may depend on the boundary points.
From Eggs. (755) and (756), the orthogonal relations of the wave
equation are
£,
m(x)o,o, dx = 0
for
rz 8
(759)
for
#5
(760
‘0
 koi; dx—Lkd,bsf=0 €
0
Neglecting the spring load, Eq. (760) becomes €
ko/o, dx =0
for
r#8
(761)
‘0
Note that k = T for the vibrating string, k = AE in Eq. (723) for the longitudinal vibration of a bar, and k=I,G in Eq. (728) for the torsional oscillation of a rod. (2 Ne Consider the beam equation. Substituting Eq. (750) in (753) and then integrating by parts, we obtain
[£4.
k60" 6.(k0" a
= {4,(kon 41(ko4y ax] {Lak [oxaary ax}
= Fo.(koy’— ko] + [Kove.ax}
{ta.c4ony—axaan+ kocos ax} = [,(ko")'— 4,(kb")' —61kb!+ 8kONI, (762) *K.N.
Tong,
1960, p. 264.
Theory of Mechanical
Vibrations, John Wiley & Sons, Inc., New
York,
274
Continuous Systems
CHAP. 7
The problem is selfadjoint and Eq. (762) is identically zero if the boundary conditions at each end can be expressed as* ad + b(kh")' =0
(763)
co'+d(ko")=0 The
constants
a, b, c, and
d depend
on the boundary
conditions.
Except for the inertia load, the two boundary conditions at each end of a beam in Eq. (736) are described by Eq. (763). The
orthogonal relations for the beam
equation
from Eqs. (755)
and (756) are {m(x)d,, dx =0
for
r#S
(764)
for
ré#s
(765)
‘0
[kolo! dx +[6,(kb)'—kobby]o=0 0
Neglecting the spring load, Eq. (765) becomes [kor: dx =0
for
r#S
(766)
where k is the flexural stiffness EI of the beam. Example 7.
Uniform Freefree Beam
Components
of machines
are often tested
under
freefree
conditions,
be
cause boundary conditions, such as fixed or hinged, are difficult to achieve.
Determine
the frequency equation and the eigenfunctions for the lateral
vibration of a uniform freefree beam.
Solution: Let the lateral deflection be u(x,t)= (x)q(t). Comparing the beam equation in Eq. (734) and the generalized equation in Eq. (744), we have
M=m
and L = EId*/dx*. Thus, the generalized equations in Eqs. (747)
and (748) reduce to
d*q
qe te
where w =A
and
4=0
d*h
rieaaa B*>
and
B*=@’/a*>=w’m/EI.
The
=0
solution
of the
equation 1s
(x) = C, sin Bx + C, cos Bx + C; sinh Bx + C, cosh Bx
The boundary conditions from Eq. (736) are
(1) (3) * Tong, ibid, p. 269.
¢",0=0 6" 2,=0"
(2) 4)”
second
SEC. 78
Orthogonality
275
Substituting conditions (1) and (2) in (x), we get arta)
and
(Ora teas0)
Conditions (3) and (4) require that
(sin Bf
—sinh B@)C;+(cos B—cosh BC) C,=0
(cos B€—cosh B¢)C;—(sin
BE
+sinh Bf) C,=
A nontrivial solution for C,; and C, requires that the determinant of the coefficients of C; and C, be zero. Thus, the frequency equation is
cos Bf cosh Bf = 1 When 6 =0, the solutions of the differential equations are q(t)=A,+Adzt
(x) = Bo+ Bix + B.x°+B3x° The function q(t) indicates that the freefree beam is semidefinite. The boundary conditions (3) and (4) require that B,=B,=0. Let the corresponding eigenfunctions be
@(x)=B,
and
,(x)=B,(x—¢/2)
where B, and B, are arbitrary. This shows that, corresponding to 6 =0 or @ =0, the rigid body motion can be a translation and/or a rotation about the mass center of the system. When 8#0, we obtain from the equations above C3 ae,
(cos B&—cosh Bf) _ (sin Bf
(Gn
(sin B€—sinh B@)
+sinh Bf)
(cos B&—cosh BE)
é,(x) = C [(sin B,¢ + sinh B,¢)(sin B,x + sinh B,x) +(cos B,f—cosh B,€)(cos B,x + cosh B,x)] where
r=2,3,...
The modes of singlespan uniform beam for different end conditions are shown in Fig. 79. The corresponding equations are summarized in Table 71. The roots B¢ of the frequency equations are listed in Table 72.
Case 2.
Boundary Conditions Dependent on A
_An eigenvalue A will appear in the boundary condition if the system has an
inertia load. Under
such conditions,
the eigenfunctions
in Eq.
(759) will not be orthogonal. We shall derive the orthogonal relation by redefining the mass distribution in the system. Consider the longitudinal vibration of a long uniform fod of length ¢ and mass/length m. Let a mass m, be attached at the free end as shown in Fig. 73(b). The boundary conditions are as given in Eq. (725). Let m,
276
Continuous Systems Fixed
Beam
Fixed
CHAP. 7
Free
j
Serres
Ve.
ak
Fee
7
SS >
Free
eee
Modes
ye.
Beam
ee
Modes
Pa
Ge
“
>
Wa
a," 5a Be
\ FZ
\
eK,
Fixed
Beam
Free
Hinged
Beam
Modes
Modes
Ce SIN
Hinged
ea
Ze
>, —
Zero mode ———_—_——_ 5 2)
Hictucdes
—_
—_ Second mode
SSeS
Third mode
Fic. 79. Modes of vibration of singlespan uniform beam for different end conditions.
be an integral part of the rod. Hence
Total mass = m@+ m,
or Mass/length = m+ m,6(x — @)
(767)
Since_m, is a concentrated mass, its mass/length becomes a delta function* m,6(x—@). If m, is part of the rod, then the boundary condition beyond m, is that of a free end. * Dirac’s delta function was described
in Chap. 2. From
sgf=O
s
fos i
[806 ae=1
For any arbitrary function f(x), this gives
 8Ofl) ar=110 an
Eq. (264) we have
SEC. 78
Orthogonality
Taste 71. Beam.
END
if
277
Summary of Equations for Lateral Vibration of Singlespan Uniform
CONDITIONS
EQUATIONS
(1)
u(0,t)=u'(0,t)=u(én
(2)
cos Bé cosh B6=1
=u'(ét)
=0
(3)
(x)=A(cos Bx —cosh Bx) +(sin Bx —sinh Bx) ot sin B€—sinh BE _ cos Bf cos B€—cosh BE
0
Q *
—cosh BE
sin B€ + sinh B¢
(1)
u"(0,t)=u'"(0,t) = u"(€,t) = u'"(€,t) =0
(2) (3)
cos B€ cosh Bé = 1* (x)=A(cos Bx + cosh Bx) +(sin Bx +sinh Bx) Ore sin B€—sinh Bf _ cos Bf cos B€—cosh Bf
—cosh BE
sin B+ sinh BC
(1)
u(O0,1)=u'(0,1)=u(G1) =u'(¢t) =0
(2) (3)
tan B&=tanh BE (x)=A(cos Bx —cosh Bx) +(sin Bx —sinh Bx) ae
(1) (2) (3)
sin Bf—sinh BE _—
sin Bf + sinh BE
cos Bf —cosh BE
cos B+ cosh Bé
u(0,t)=u"(0,t) =u"(61) = u'"(¢,t) = 0 tan B€ =tanh Be* (x)=A sin Bx +sinh Bx _ sinh B€_ cosh B¢ ~ sin BE ~ cos Be
(1) (2)
u(0,t)=u"(0,t) =u(G6t) =u'(é,t) =0 sin B€=0
(3)
o(x)=A sin Bx
(1) (2)
u(t) =u'(0,t) =u"(60 =u"(€b =0 cos B€ cosh BE =—1 (x)= A(cos Bx —cosh Bx) +(sin Bx —sinh Bx)
(3)
sin Bf+sinh BE _ cos B&+cosh BE cos Bf +cosh BE Lateral deflection=¥ $(x)q(t) .
Beam equation
d>u atuce iy ease 0
eee
m = mass/length
d* dnt
4
Bd
=)
o= freq., rad/s
d*q 27 eee
+
= 0
B*=w?/a? = mw?/EI
(1)
End condition
* Semidefinite systems
(2)
Frequency equation
(3)
Ejigenfunction
sin B€
—sinh BE
CHAP. 7
Continuous Systems
278 Taste 72.
Roots B€ of Frequency Equation for Singlespan Uniform Beam*
END CONDITIONS
oe
*
B,¢
Bx€
B3¢
Bue
4.730041
7.853205
10.995608
14.137166
4.730041
7.853205
10.995608
14.137166
3.926602
7.068583
10.210176
13.351768
3.926602
7.068583
10.210176
13.351768
3.141593
6.283185
9.424778
12.566370
1.875104
4.694091
7.854757
10.995541
* Semidefinite system. Zero modes not included.
+ The roots B€ are computed from the frequency equations in Table 71. Similar tables are found in the literature. See, for example, R. Bishop and D. Johnson, Vibration Analysis Tables, Cambridge
University Press, New York,
1955.
Using m(x)=m+m,6(x— ¢), the orthogonal relation in Eq. (759) is
[om +m,6(x — €)]6,d, dx = mo,o, dx + m.,(€) ,(¢) =0
for
r#zS
(768)
The same technique can be used to find the orthogonal relation of a torsional shaft or a vibrating beam with an inertia load. The proof of this statement is left as an exercise. Since the operator L in Eq. (756) is not affected by m,, the corresponding orthogonal relation remains as shown in Eq. (760) or (765). Example 8 A long uniform rod with an attached mass m, at x =@ is shown in Fig. 73(b). For the longitudinal vibration, find (a) the frequency equation, and (b) the frequencies of the first four modes. (e) Illustrate that the eigenfunctions are orthogonal and sketch the mode shapes of the first four modes. Solution:
(a) Let the longitudinal displacement be u(x,t) = }(x)q(t) as defined in Eq. (715). From Eq. (718), we get 5)
BE) = Cusin
@
X cia, 008%
q(t) = C; sin wt + C, cos wt
SEC. 78
Orthogonality where
c’=E/p,
p=mass/volume
279
of the rod, and
C’s are
constants.
From Eq. (725), the boundary conditions are (1)
u(0,t)=0
(2)
m,u(€,t) =— EAu'(€,t)
Condition (1) gives C,=0. Condition (2) yields
mw sin wf/c = EAw/c cos w/c This can be rearranged to give the frequency equation wl wl pA€_ — tan — Cc
Cc
m,
mass of rod attached mass
This may be compared with the frequency equation in Example 3 for a torsional shaft with an inertia load. (b) The values of w€/c in the frequency equation are obtained by numerical solution or from tables.* For numerical
Rod: mass/length,
m=20kg/m,
€=0.25m,
E=200GPa, attached mass: m, =5 kg, mass the first four modes are: Mode Frequency,Hz
(ce) Let
us
use
¢,(x)
and
p=8000kg/m°*
c=VE/p=5 km/s ratio = 20
NI NIe NIRF
Q.
==
(771)
SEC. 79
Lagrange’s Equations
281
where é
mij = mx); dx
(772)
lo
and m, is the generalized mass of the continuous system for the ith mode. This may be compared with the orthogonal relations in Eqs. (755), (759), and (764). If the system has a mass m, attached at x=, from Eq. (768), the generalized mass m,; in Eq. (771) is mi = m(x)o7 dx + ma; (€)
(773)
Similarly, the operator L describes the elastic properties of the system. Hence the potential energy function U is U=; uL(u) dx
(774)
Substituting u(x,t) from Eq. (769) and noting the orthogonal property in Eq. (756), we have
bat( ll
a) dx
ai, $,L(¢;) dx
4M ims
1ay8
qi ki Somes Pedlieds. Vie NIP tes ih
(775)
where é k= :L(¢;) dx
(776)
and k;; is the generalized stiffness for the ith mode. This may be compared with the orthogonal relations in Eqs. (756), (760), and (765). Furthermore, the expressions for the kinetic energy T and the potential energy U
in Eqs. (771) and (775) may be compared with the corresponding expressions in Eq. (643) for discrete systems. T is always positive definite. U can be positive definite or semidefinite. The Lagrange’s equations of motion for the small oscillations of a conservative system are £ (22) SO dt \dq;/ 9q;
(777)
Substituting Eqs. (771) and (775) into (777), we obtain mig: + kiiq; = 9
(778)
qi + w; Gi =)
(779)
Or
282
Continuous Systems
CHAP. 7
where w; = k;,/m,;. This was anticipated, since T and U in Eqs. (771) and
(775) are expressed in the eigenfunctions, or normal modes. There is no coupling in the equations of motion. Let us relate the initial conditions, as shown in Eqs. (713) and (714), to our present discussion. From Eq. (769) the initial conditions can be expressed as
(x40) = ¥ di(x)ai(0)= fo)
(780)
ti(x,0) = 2,b,(x)4i(0) = g(x)
(781)
Using the orthogonal relation in Eq. (768) with m, at x=, corresponding initial conditions in the normal coordinates are
the
a0)=——[ mixiflardar+mfOo(
(782)
ain=— i mi [" 4 mise(a)d, S ; dx tin.e(O)d ae ; 6)
(783)
1
“@
where m, is defined in Eq. (773). The terms due to m, are neglected if no mass is attached at x = @. Similarly, q;(0) and q,(0) can be obtained from the generalized stiffness
in Eq. (776). Except for the zero mode, neglecting the spring load as in Eq. (761) for simplicity, for the wave equation, we get
q,(0) =i [eres; dx
HO=— ke'(x)6y dx il
JJij
€
(784)
~O
where k is as defined in Eq. (749) for a general problem. Similarly, neglecting the spring load as in Eq. (766), the beam equation gives
q;(0) == {EIf"(x)$j dx
i ;
(785)
qi(0) =F EIg"(x)o; dx Example 9 Show that k;/m, =«; for the first three modes of vibration of a uniform cantilever beam of m mass/length.
SEC. 79
Lagrange’s Equations
283
Solution: Let the lateral deflection of the beam be u(x,t)=$(x)q(t). From Example 4, the formal solutions of the beam equation are (x)= C, sin Bx + C, cos Bx + C; sinh Bx + C, cosh Bx
q(t) = Cs sin w@t +Cg cos wt
where B* =@*/a*= mw /EI and m=mass/length. The boundary conditions for a cantilever beam from Eq. (736) are
(1)
u(0,t) =0
(3) Sete (CEO 8
(2)
u'(0,t) =0
bes
(Gt) =0
Conditions (1) and (2) give C,+C,=0 and C,+C,=0. Conditions (3) and (4) yield (sin B€+ sinh B€)C, +(cos B+ cosh B’)C, = 0 (cos B€+ cosh B€)C, —(sin B&—sinh B¢)C, =0
By letting the determinant of the coefficients of C, and C, equal zero, we obtain the frequency equation cos Bf cosh Bf =—1
Let us normalize m, in order to show that w7 =k;,/m,. Let J 67 dx =a, a constant. Hence, from Eq. (772) m, =J m@; dx = a,m, a normalized mass. For a uniform beam, k,;; can be calculated from Eq. (766) instead of
(776), that is, k;, =f EI(o!)? dx. Note that (1) f (!")¢; dx =J (/)” dx from
Eqs. (765) and (766), and (2) d*¢/dx*= B*¢. Thus, we have €
Le
€
k= Er {(¢,)° dx = EI {(¢,)o,dx = EIB! ? dx (0)
(0)
= EIB;a; k,; EI (“:) (mw? =; Dy aainaee EIBta; Bia =—EI, p*=— M;;
ma;
m
forall modes
m
Alternatively, let us calculate m, and k,, in order to find w;. From Table 71, the expression for (x) is
¢; = A; (cos B;x — cosh B,x) + (sin B;x — sinh Bx) where A; =(cos B€ +cosh B,€)/(sin Bf —sinh Bf). The calculated values of A, for the first three modes are —1.3622, —0.9818, and —1.0008. Thus, o;(x) can be substituted into Eq. (772) to evaluate mj. For i=1, we get é
m
M4, = mo; OB
B,¢
Fall
$1 dB:x
= pail [A,(cos,x —cosh B,x) + (sin B,x—sinh B,x)PdB,x B,e
: (7).Ai4.8155) +2.4,(2.4865)+(1.3181)] = 3.4795 m/B,
284
Continuous Systems
CHAP. 7
Obtaining ¢(x) from #,(x) and substituting ¢ (x) into Eq. (766), we get
k,, = EIB3(17.2120 — 23.3269 + 9.5944) = 3.4795 EIB} 2”. ky,/m,, = Elpt/m = (Ell/m)(mo7/ED = o{ The results follows:
for the first three modes
i= Siu 1 Koy 3.4795m/B, 3.4795 EIB;
Bit A; Mix ki
to show
that w;=k,,/m,
i=2 4.6941 —0.9819 4.5254m/B, 4.5254EIB3
are as
i=3 7.8548 —1.0008 7.8418m/B; 7.8418EIB3
Example 10 Repeat Example 2, using Eqs. (782) and (783) to evaluate the constants of integration from the initial conditions.
Solution:
From Example
2, the formal solution in the form of u(x,t)=) 6(x)q(t) is —
u(x,t) =
lx
(1) ee
where c*=E/p, conditions are
imct
Sin——= CSS
20 ( Ol
E=Young’s
modulus
u(x,0) = = = f(x)
and
and
imct
FD?
a)
aa ee
p=mass/volume.
The
initial
u(x,0) = g(x) =0
To evaluate q,(0), we use Eq. (772) to find m,; and then (782) to find q(0). For 6;(x) =sin(imx/2¢), we have
m
4
ii =?
{pAd; A
ier.
a
k
pAt
for i=1,3,5,....
0
dx
x
=
pAA sin i
lq
50%
—
Fx it A —— sin— x dx =
AE»
jy
2¢
d
x
SFE
pA€ 5
=
(H1) ee
WAR
—
Defining i=(2n +1), we get
SFO n
qn.(0)
aa
=
(e1)°
for
Aan
aw? AE
(2n+1)*
ri — Sle 23) eee
This gives D, in u(x,t). From Eq. (783), C, =0. Thus, 8FE
—
(1)"
ye UCSDt==a Oia
.
@n+1)ax
ae
(2n+1)act cos ———————
2e
SEC. 710 710
Undamped Forced Vibration—Modal Analysis
UNDAMPED FORCED MODAL ANALYSIS
285
VIBRATION—
The partial differential equation for the undamped forced vibration of a continuous system from Eq. (744) is M(ii)+ L(u) = F(x, t)
(786)
where M and L are linear differential operators and F(x,t) is an excitation force. Let the deflection u(x,t) be
ue) = Ldexdalo
(787)
Analogous to the modal analysis of discrete systems in Sec. 611, Eq. (786) can be uncoupled by substituting Eq. (787) in (786) and using the orthogonal properties in Eqs. (755) and (756). It can be shown readily that
miGi + kgs = F(x,t1)6,(x) dx=Q(t)
(788)
where Q,(t) as defined above is the generalized force for the ith mode and
m; and k;; are evaluated as shown in Eqs. (773) and (776). The complementary and the particular solutions of Eq. (788) can be obtained as described in Sec. 27. The solutions q;(t) are then substituted in Eq. (787) to obtain the deflection u(x,t). Example 11.
Concentrated Force
Find the deflection u(x,t) of a continuous force F(t) applied at x =&.
system
due to a concentrated
Solution:
The concentrated force at x =€ can be expressed as F(t)5(x —€&), where 5(x—€) isa delta function occurring at x = & The generalized force Q,(t) for the ith mode from Eq. (788) is
Q; = [F(t)5(x — €)]b;(x) dx
(789)
QO; = F(t)¢;(€)
(790)
Since Eq. (788) is of the same form as (664), its particular solution can be found by the convolution integral as shown in Eq. (665). Defining o;= k,/m, and the force Q,(t) as shown above, the particular solution for the ith mode is
_b8) MQ;
IF(r)sin w;(t— 7) dr
(791)
286
Continuous Systems
CHAP. 7
The initial conditions q,(0) and q,(0) are found from u(x,0) and u(x,0) by
Eqs. (782) and (783) or as shown in Example 10. Analogous (667), the complementary solution of Eq. (788) is 1 q; = G(O)cos wt +— q,(0)sin w;t
to Eq.
(792)
@}
The complete solution for the ith mode (791) and (792).

is the superposition
1
of Eqs.
et
4p= Gi O)oos of §(O)sin ot 
 + ol) [F(r)sin (t=) dr aa a “Ave 
(793)
me
This is substituted into Eq. (787) to obtain u(x,t)= > (x)q,(t). Note that if the system is positive definite, the complementary solution due to the initial conditions is always in the form shown in Eq. (792). Hence, we shall examine only the particular solutions due to typical excitations in the examples to follow. Example 12.
Distributed Force
A distributed force F(x,t) shown in Fig. 711(a) is suddenly applied to a simplysupported uniform beam. Find the generalized force and the motion of the beam. Assume zero initial conditions. Solution:
The eigenfunction ¢,(x) and the frequency equation to obtain w; are given in Table 71.
(x) =A; sin y; The generalized mass
and
ae
El(iz)*
for
mt*
il Pass Wea
m, from Eq. (772) is
3
€
lo
2
my =m o; dx =mA? = where
m
is mass/length.
Let A,= V2.
Thus,
the normal
mode
Velocity
—_
(a)
Distributed force P(x) = Po x/2k
Fic. 711.
(b)
Moving load F,
Forced vibration of beam.
and the
SEC. 710
Undamped Forced Vibration—Modal Analysis
287 a
ren Cea
g An
normalized m, become
ea \, Px, &)
vrUm)MO
Ar
;(x) =V2 sin ze
and
m, = me
Yi (ne)Me
PIM, *)= frr
LRAre
Ya (a) = D (a)
The distributed force F(x,t) is
F (x,t) = p(x) = Pox/€ which is suddenly applied at t=0. Substituting this in Eq. (788) gives
From Eq. (788), the equation of motion is Grats @: i = Q/me
where Q; is as defined above. From the convolution integral in Eq. (665), the particular solution for the ith mode is
a(t) = (1)
Py
mtitrw
= (1)""Po
V2
3
TW;
{sin @;(t—7) dr i
(108 wt)
for
i
Wee es eee
For zero initial conditions, the complementary solution becomes zero. Thus, the motion of the beam is
ulxt)= Lda = Lv2 sin qo Example 13.
Concentrated Moving Force
A moving load on a bridge with constant velocity v is shown in Fig. 711(b). Find the deflection u(x,t) of the bridge. Assume zero initial conditions. Solution:
Let us approximate the bridge as a simplysupported uniform beam of m mass/length. The concentrated force Fy is applied atx =€=vt for O=vts €. The generalized force Q, from Eq. (788) is
Q;= [F8(x — vt) ]o;(vt) dx
From
x (eee
for
Ostsd/v
Pad
for
t>v
the last example,
we
have
$;(vt) =V2 sin(imvt/@),
m,;=mé,
w? = El(ir)*/(m€*). The particular solution is obtained by means
and
of the
288
Continuous Systems
CHAP. 7
convolution integral as shown in Eq. (665). Thus, for 0=t=//v, the ith mode is 1 ; i Gqit)=——  (F.v2 sin aetsin @;(t—7) dr MO}
=
q;(t) for
‘0
V2F,
1
>
(4; sin w;t— w; sin a;t)
mtw; a; — o;
where a;=imv/€. The complementary solution is zero if the bridge was originally at rest. Substituting q,(t) in u(x,t) in Eq. (787), we obtain u(x,t) = 3 See ea i=1
For
t>€/v,
@;,
the bridge
A;—W;
:
.
,
ax
(a; sin w;t—w; sin ait)(sin=) ee
is in free vibration.
The
initial conditions
are
q)(€/v) and q,(€/v), which are the end conditions for the interval 0= t= ¢/v. The free vibration can be obtained from Eq. (792).
711
RAYLEIGH’S
QUOTIENT
The Rayleigh method is perhaps the basis of a majority of approximate methods for vibration analysis. The technique used for continuous systems is similar to that for discrete systems in Sec. 67. This section prepares the background for the RayleighRitz method in the next section. The Rayleigh’s quotient is formed from the kinetic and the potential energy functions as for the discrete system shown in Eq. (648). Let the deflection u(x,t) of a continuous system be approximated by
u(x,t) = b(x)q(t)
(794)
where {x)~1s~an_assumed mode shape. Note that we defined u(x,t)= o(x)q(t) in the previous sections, where $(x) is an eigenfunction or the exact mode shape. Substituting Eq. (794) in (770) and (774), the
Rayleigh’s quotient is
WL (wh) dx
AglW(x)]=—————_ WM (us) dx
(795) :
0
where L and M are linear differential operators of the variable x as defined in Eq. (744). The quotient is a function of the function W(x) and is called a functional. There are restrictions, however, on the assumed mode w(x). Essentially, the restrictions are that (1) the functions L(w) and M(ys) must exist and the integration possible, and (2) w(x) must
SEC. 711
Rayleigh’s Quotient
289
satisfy all the boundary conditions. It will be shown in the next section that the boundary conditions can be relaxed. Example 14
Find the fundamental frequency of a uniform cantilever beam. Assume the mode shape (x) is that of a static curve, when the beam is deflected by its own weight. Solution:
From elementary strength of materials, the static curve of a cantilever from the fixed end is mg
WO) = Ta ET where
m=mass/length.
(x*46x°+6€°x”)
Substituting w(x) in Eq. (795), noting from Eq.
(734) that L = EIo*/ax* and M=~m for a uniform beam, and performing the integration, we obtain Ap =12.48EI/mé* or w =3.530VEI/mé*. The exact value from Table 72 is w =(B¢)’V El/m€* = 3.516VEI/mé’.
Let us write Rayleigh’s quotient in terms of the maximum potential and kinetic energies. Let the beam deflection be u(x,t) = W(x)q(t), where q(t) is harmonic. It can be shown from strength of materials that the potential energy U of a beam due to a bending moment is
mele en d*p\? i U=54  (” EI(x)(S5) Assuming q(t) max = 1, the maximum potential energy is il
(ee
€
dz
2
{E10) 2 dx
dx
(796)
The kinetic energy of the beam can be expressed as
Tope
du)?
1
F
te, m(x)(2) dx => ro m(x)b? dx
Assuming q(t)max= 1, the maximum kinetic energy is 1
€
tbe. as4 o m(x) pb? dx 4 rod Rear
(797)
0
Equating the maximum potential and kinetic energies, we get Oss
w? = Te
where
(798)
T*,, =3/6 m(x)W? dx is called the reference kinetic energy. The
quotient in Eq. (798) is equivalent to Ap in Eq. (795).
290 Z, 712
Continuous Systems RAYLEIGHRITZ
CHAP. 7
METHOD
The RayleighRitz method assumes the deflection curve (794) can be approximated by a finite series.
Woo) = ¥ aro)
(x) in Eq.
(799)
where the a’s are parameters and y,(x) form a generating set, consisting of linearly independent functions satisfying all the boundary conditions. In limiting the generating set to a finite number of functions, the analysis can be interpreted as approximating a continuous system by means of a ndegreeoffreedom discrete system. Since (1) the assumed mode in Eq. (799) is a finite series instead of infinite, and (2) the assumed mode is not
exact, the results can be interpreted as geometric constraints, both tending to raise the estimated natural frequency. The RayleighRitz method gives a procedure to minimize the estimated frequencies. Substituting Eq. (799) into the Rayleigh’s quotient in Eq. (795), we obtain é Lg
 Ee _ Jo
ayLO
i4 a;y;) dx
 yah a;y,M(y7=1 a;Y;) dx 0 €
gies
aay] y,L(y;) ax
:es ey = Ve é at ;=1 42,0;M,;, i=
seas,
1d
y¥M(y;) dx fat Ln a,  0
which can be expressed in matrix notations as
_ laltk,Ha} s Ne AR (4,4,
me
tsa a,)
la] [m,
a}
Dz
(7100)
where {a} is a vector of the parameter a’s and its transpose is a]. é
€
k,, = yiL(y;) dx
and
m,; = yiM(y,) dx
(7101)
0
Note that k,, and m, are known values for the assumed values of y’s in Eq. (799). The problem is selfadjoint and
Ki = kj
and
Mm, = Mj,
(7102)
The problem is to minimize Ag above with respect to the a’s. Let dAp/da,; =O for minimum. Thus, Eq. (7100) yields OAR _ Dr(Np/0a;)—
da;
DE
Ne(ODp/0a;) _
7
0
(7103)
SEC. 712
RayleighRitz Method
291
If w* is the minimum of Ap, from Eq. (7100) we get
ae “ Z
=(%
min
ie
a
Taking the partial derivatives with respect to a; and recalling k,, = k;, and m,; =m, ji? we have ON,
ra)
oD
r)
n
=2) kya, la[kjHa}=i ac 1 he i
i=1,2,...,n
for
(7105)
n .
=
0a;
La] [m,y
0a;
an
Ka}
2
%
ij 2 MG;
j
l
for
Substituting Eqs. (7104) and (7105) in (7103) numerator to zero for minimum, we obtain
» (ky—@?m,)a,=0
for
il. De
=
and
f=1,2,...,n
ee
> n
equating
the
(7106)
j=l
This is a set of n homogeneous equations with a’s as the unknowns. For a nontrivial solution, we set the determinant of the coefficients to zero.
k,; —w7m,,=0
(7107)
This gives the frequency equation, resembling that of a discrete system. Example 15 Find the fundamental frequency of a uniform cantilever beam and compare the answer with that given in Example 14. Assume h(x) = a,x7+ ax”. Solution:
The maximum
potential energy from Eq. (796) is 1
¢:
Usnax = 5 e1 (2a,+6ax)* dx 0)
1 = 5 El(4aj€+12a,a,€7 + 12a3¢°)
aU — BI(4a,¢+ om 6a,¢2)* Y; kya, 0a, if)
aU —2* 0a,
From
; koja, — FI(6a,€7 + 12a,€°) = ), a=
the derivatives above and Eq. (7105), we obtain
kp =4BIC
kjk, =ORI
kag
2B
CHAP. 7
Continuous Systems
292 The maximum
reference kinetic energy from Eq. (797) is
Thx
m
(*
m
2e5
m oe
{°
+a x°*) dx {We? dx = — (a,x?
Dae
ie (a,Far
eS
t’
as ‘Ga
Similarly, m, from Eq. (7105) are
m,=mO/5
my =m,=me*/6
My = me’ /7
Substituting k,, and m, in Eq. (7107) gives the frequency equation
4EI€ —w° mé?/5 eeeeeryat #2
6EIl* — w me°/6 ae Oy
or
w*—1,224(EI/mé*)w? + 15,120(El/mé*)” = 0 The estimated frequencies are
WaSS3GEL P ye
o=
and
_ 34.81
a
/EI a
The exact frequencies for the first two modes from Table 72 are
=
3.516
/EI == m
ie
and
a=
22.03 = La
/EI = m
Using a simple power series for the problem above, the method gives good results for the fundamental frequency, but the correlation for the second mode is poor. Greater accuracy in the second mode can be achieved if the mode w(x) is approximated by three or more terms. It may be difficult to form a function w(x) in Eq. (799) to meet all the boundary conditions. In fact, the w(x) in Example 15 meets only the boundary conditions at the fixed end of the cantilever. The types of boundary conditions are (1) geometric, such as deflection and slope, and (2) natural, such as moment
and shear. It can be shown*
the boundary
conditions can be relaxed and the quotient in Eq. (798) needs to satisfy only the geometric boundary conditions. Example 16 Find the fundamental frequency of the wedgeshape cantilever beam shown in Fig. 712. Assume the beam is of constant unit width. *See, for example, L. Meirovitch, Company, New York, 1967, p. 227.
Analytical
Methods
in Vibrations,
The
Macmillan
SEC. 712
RayleighRitz Method
Fic. 712.
293
Lateral vibration of wedgeshape cantilever.
Solution: Since the height H is linear with x, we have m(x)= pA(x) = pH(1—x/€)
El(x) = E[H(1—x/@)}/12 where
m(x)=mass/length,
p =mass/volume,
and A =area of section. Let
Ww =a,x*>+ax°. From Eq. (796), the maximum potential energy U,,.x is EH?
€
Ca — ome (€—x)*(2a,+6ax)* dx
6
_ EH :
3
REA Applying Eq. (7105) yields
dU. au
—
Sten, da
EH?
3
128
(ae++3 ast?)=
EE =
. Y kas
TS
128
2
(: a,t
eS
ast’)=
2
hasty
Thus,
os Fe
EH 3
kn=Tpe® The maximum
kerkupas
kn
EH? 3
es
reference kinetic energy from Eq. (797) is H
é
hs oe (€—x)(a,x°> +ax°)
dx
0
pH < et ee ae (i+ 20105403 = The partial derivatives are evaluated by Eq. (7105) to give my,= ae 11 30
?
Myy=m 12
21
(sts 42
>
and
My. De, = gee
56
294
Continuous Systems The frequency equation is obtained (7107). Simplifying, we get
CHAP. 7
by substituting k, and m, in Eq.
(a Eee a ee of" 20 12 30/\60 56/ \60 42 where b* = EH’/p¢*. The frequency equation can be reduced to 10w*—273b*w* + 588b* =0
Hence the fundamental frequency is w = 1.535HVE/pe’.
713
SUMMARY
The onedimensional wave equation and beam equation are treated in this chapter. The vibration of continuous systems involves (1) the elasticity of the material, (2) the boundary value problem, and (3) vibration theory. The chapter shows the similarity of the vibrational aspect of continuous and discrete systems. The methods of problem formulation and solution of continuous systems are illustrated in Sec. 72. (1) The motion is a function of the space and time variables. The partial differential equation in Eq. (73) is derived from Newton’s law of motion. (2) The problem is governed by the boundary conditions and the initial conditions. The boundary conditions stipulate the frequency equation and the modes of vibration. The initial conditions define the degree of participation of each mode. (3) Similar to discrete systems, the motion of continuous systems is obtained from the superposition of the modes. The procedure above is formalized by the classical method of separation of variables in Sec. 73. Examples of problems governed by the wave equation are shown in Sec. 74. The EulerBernouli beam equation is discussed in Sec. 75. The methods of problem formulation and solution are essentially as outlined
above.
;
Rotary inertia and other effects will influence the effective inertia and stiffness of a beam in vibration. A decrease in stiffness and/or an increase in inertia will decrease the natural frequency and vice versa. Some of the effects on beam vibration are discussed in Sec. 76. Again, the general method is the same as described above, but the procedure can be quite involved. The continuous problem is generalized in Sec. 77, using the linear differential operators L and M. These may be compared with the matrices K and M of discrete systems. The forms of the operator L are as shown in Eqs. (749) and (750). The boundary conditions are generalized in Eqs. (751) and (752). The proof of orthogonality of the eigenfunctions in Sec. 78 essentially follows the procedure for discrete systems. The orthogonal relations for
Problems
295
the wave equation are shown in Eqs. (759) to (761) and for the beam
equation in Eqs. (764) to (766). If the system has an inertia load, the orthogonal relation is modified as shown in Eq. (768). The modal analysis in Sec. 710 closely resembles that for discrete systems. The equation of motion is first uncoupled by means of the eigenfunctions. The uncoupled equations in Eq. (788) are expressed in terms of the generalized mass m,,, stiffness k,;,, and force Q,(t). The initial conditions
for the modes
are
found
from
Eqs. (780)
to (785). The
complete solution for each mode is shown in Eq. (793). The general solution is the superposition of the modes u(x,t)= > o;(x)q,(t). The Rayleigh’s quotient in Eq. (795) is an extension of Eq. (647) for discrete systems. The RayleighRitz method approximates a deflection curve W(x) by a finite series in Eq. (799), where the a’s are arbitrary. The estimated natural frequency by the Rayleigh’s method tends to be higher than the actual value. The RayleighRitz method gives a procedure to minimize the Rayleigh’s quotient and to obtain a frequency equation.
PROBLEMS 71 Find the motion u(x,t) for the lateral vibration of a taut string shown in Fig. 71 for each of the following sets of initial conditions: and
(x,0) =
—
(a)
u(x,0) =0
(b)
u(x,0) = f(x) = Hx(€—x)/€?
where H and V
2 Vx/€
DV
and
for
s9K2)) OU_
a
for
for
Osx=2
U2
ee,
O) and (y,,y) phase planes are illustrated in Figs. 86(a) and (b). From Eq. (818), we have
LA dz,
=——— AZ;
Gas
or
Zs
Ah Z
which can be integrated directly to yield Ce fad Wares
(819)
where C =constant. The exact plot depends on the values of C and the
ratio A,/A,. For example, if A,/A, =2 and C=1, we get z,= 27. If A,=A,, the trajectories are simply radial lines from the origin. The directions of the trajectories can be deduced from Eq. (818). If both A, and A, are negative, the system is stable and the trajectories converge towards the equilibrium. Conversely, if A, are positive, the system is unstable and the trajectories point away from equilibrium. Note that the trajectories in the (z,,z,) plane are symmetrical, but those in the (y,,y>) plane are governed by the characteristics discussed in
Sec. 83.* Case 2.
Saddle Point: A, Real and Opposite Sign
A saddle point occurs if A, are real and of opposite sign. Since one of the A’s is positive, the system in the neighborhood of a saddle point is always unstable as shown in Fig. 86(c). Since the ratio A,/A, is negative, Eq. (819) can be expressed as
Zaztt
OF
24232 9= C
(820)
This indicates that the trajectory is a hyperbola. Case 3.
Vortex: A,, Imaginary
A vortex, or center, occurs if A, »= +jB are imaginary, where j= /—1 and 6 =constant. From Eq. (818) we get Z1 = JBzZ,
and
7 Naor
 oS
Z4=Z ze".
and
7 ee SN
OT
* Since (y,,y>) are obtained from (x,,x,) by a coordinate translation, the x’s and y’s have the same physical interpretation. For example, if x, denotes a displacement, and so does y,. Similarly, if x, denotes a velocity, and so does y,. Analogous to principal coordinates, (2,2) do not necessarily have such physical interpretations. Thus, the characteristics of phase trajectories, as discussed in the last section, do not apply to the plots in the (z,,z,) phase plane.
SEC. 84
Stability of Equilibrium
(a)
Vortex: Aid imaginary
Fic. 87.
(b)
311
Unstable focus: A, » complex
Phase trajectories for A, imaginary and complex.
where (219,229) = constants. The factor e*'* represents a harmonic motion of unit magnitude with circular frequency B as discussed in Sec. 15. Hence, the resulting motion in the (z,,z,) plane is a combination of two harmonic motions, which is an ellipse as shown in Fig. 87(a). The motion in the neighborhood of equilibrium in the (z,,z,) or (y,,y2) plane forms a closed curve. By definition, the system is stable. Case 4.
Focus: \,, Complex Conjugates
A focus occurs if A, ,= a+jB are complex conjugates, where j= a and @ are constants. From Eq. (818), we get Z,=(a+jB)z,
and
Zo=(a—jB)z,
Zi (zne: en
and
Za=(zene je
V—1 and
(821)
Or
where (Z,9,Z9) = constants. The factor e*'* represents a harmonic motion of unit magnitude with circular frequency B as before. If a>0, e™ increases exponentially with time t and the trajectory in the (z,,z,) plane is a divergent logarithmic spiral as shown in Fig. 87(b). Hence for a>0, the equilibrium is unstable. If a4v and v>0, with u*=4v as a limiting case. The node is stable if u0.
Case 2. Saddle point: A, real and opposite sign. This requires u*>4v and v imaginary. This requires definition, the system is stable.
u*, = 0.1. From Eq. (824) we get u=0.1
and
yit3y3
are a,,;=0,
a,;.=1, a,,=1,
v=l
It can be shown that the equilibrium at (—1,0) is a saddle point, which is unstable, as shown in Case 2. (ec) To plot the trajectories in the,(y;,y2) plane in Fig. 89, it is necessary to map the z,> axes in the (y,,y,) plane. From Eq. (826) we have
SEC. 84
Stability of Equilibrium
Fic. 89.
315
Stability of equilibrium; Example 4.
Ay2= 1.05 and —0.95. The matrix B and the similarity transformation {y}= B{z} in Eq. (815) can be obtained by the method shown in
Example 6, App. A. The lambda matrix [f(A)] of matrix A in Eq. (811) is
tay=[*
—Az,
ee]
AAy
The adjoint matrix F(A) of [f(A)] is
F)=
A ar.
i
az,
Aay
Any nonzero column of F(A;) can be used for a column {b,;
_b,;} in the
matrix B, that is,
og
ee SE
i
for i=1,2
(828)
pele SIE( IE] om
Hence the similarity transformation is*
y2
Substituting A,.=1.05
boi
baa JL22
and —0.95
Mi
B2JLZ2
in Eq. (928) and for the values
4,,=0, a;2=1, a,,=1, and a,.=0.1, we get
a % Fe Seale The equation of the z, axis in the (y,,y2) plane is z,=0. Thus, from the equation above we get
Y2 = p= 1.05 yi *Note
the similarity between
Eqs. (416) and
(828), and Eqs. (420) and
(829).
316
Nonlinear Systems
CHAP. 8
Similarly, the equation of the z, axis in the (y;,y2) plane is
y2_
pn = 0.95
yi
The z,, axes and the trajectories about the equilibrium (—1,0) are as
shown in Fig. 89.
85
GRAPHICAL
METHODS
Graphical methods may serve (1) as a supplementary means for solving nonlinear differential equations, and (2) as an exploratory tool to obtain the perspective of a problem. We shall discuss the isocline and the Pell’s method. Let us first describe the basis of the graphical methods by plotting the trajectories from Example 2. The equation of motion ¥+2lw,% + w2x =0
can be expressed in the form of Eq. (84) as ; } 2
AE Abn Bok aaFg) dx
The
x
(830)
slope of the trajectory in the (x,x) phase plane is F(x,x), that is,
F(x,x) is tangential to the trajectory at the point (x,x). Substituting the values of (X9,X,) in F(x,x) gives the slope at (x,,X,). A small tangential line segment through (x 9,X)) can be drawn as shown in Fig. 810. In other words, the trajectory is extrapolated from (x9,Xo) to (x,x) by following this
line segment. Using the end point of the previous step as the beginning point of the next step, a new slope can be calculated for additional
—3
Fic. 810.
Tangential line segments
Isocline method; linear system in Example 2.
SEC. 85
Graphical Methods
317
extrapolations. The process is repeated to obtain a trajectory. The phase plane can be filled with a family of trajectories to show the pattern of the solution of the differential equation. Isocline Method
An isocline in a phase plane is the locus joining the points of trajectories having the same slope. For example, the tangential line segments aa and bb of the trajectories in Fig. 810 have the same slope. A locus joining these segments is called an isocline. All the line segments drawn on the isocline have the same slope as shown in the figure. Consider an autonomous system
&dx _ F(x,3)
(831)
where F(x,x) is the slope of the phase trajectory at (x,x). An isocline is defined by the equation F(x,x)=C
(832)
where C=constant. This can be plotted in the (x,x) phase plane to obtain one isocline. A different value of C gives another isocline. Thus, a family of isoclines can be plotted. The procedure is (1) to plot the isoclines in the phase plane to cover the area of interest, (2) to draw the line segments on each isocline to indicate the respective slopes, and (3) to use the slopes provided to guide the extrapolation in order to sketch the trajectories. Example 5 A piecewise linear system consisting of a mass m shunting between two stoppers is shown in Fig. 811(a). The springs k and dampers c are linear, the clearance between the stoppers is 2A, and A=1. When m is in contact with the right stopper, the equation of motion is X+0.2x+(x—A)=0 Plot the trajectory for the initial conditions (x9,Xo) = (2.5,3.2). Solution:
Let us divide the phase plane into three regions as shown in Fig. 811(b). Due to symmetry, we need to calculate only the isoclines of regions (1) and
(2). Since the system is linear within each region, Eq. (832) must be a linear algebraic equation. In other words, the isoclines im a linear region must be straight lines*. *Tt should be noted that if Eq. (832) is a nonlinear algebraic equation, the corresponding isocline is not a straight line.
CHAP. 8
Nonlinear Systems
318
Region
4
S topper; k
Stopper oppe k
ribs
—sl Ae

AK
O—t x
(a4)
Mechanical
C=
system
())
Fig, 811,
In region (1) for
0.2\c=
0
Phase trajectory
Isocline method; Example 5.
A/28X,"C4) YO = INITIAL VELOCITY',/,8X,'(5) DT = TIME INCRE’ % "MENT ',»/s8X.'(6) NDATA = NO DATA POINTS FOR FORCECT)/M (
G{V},.
{V};=
G{V}..
(936)
{V},= G{V}i1
As shown in Eq. (657) and in Example 6, Chap. 6, a constant can be factored from {V}; after each iteration. For the (s+1)th iteration and for s sufficiently large, we have {V},41 =A,
where VA, is the fundamental ponding modal vector.
V},
(937)
frequency in rad/s and {V}, the corres
SEC. 911
Matrix Iteration—Undamped Discrete Systems
T™X CANT!
APPROXIMATE NATURAL FREQUENCIES OF UNIFORM CANTILEVER. TRANSFER MATRIX TECHNIQUE. MYKLESTAD METHOD. ENTER:
25> N=
xxx
N
=
NO
NITER
=
MAX.
DW1
=
INITIAL
INCREMENT
DEL 400
=
DW FOR 1D7
APPROXIMATE
Ss
IS
1
NITER
THIS
=
STATIONS NO
400
CORRECT?
xk
N
ey
eee where
Z ={z}={xi/x,
X2/X4
 en
—2
%s/Xat={x1
jo
y3
X2
Xs}. Thus,
ie A a
Z=B.Yo= 2) 54
1 WeSal Recalling x,=1, the solution is X={2
T=
2 —3
1
2
x,
Weed =
X5:
1
X3
1}.
* A nontrivial solution for X can be found if the determinant of the matrix A is zero.
+ Any (n—1) of the (n) simultaneous equations can be used to find the solution.
APP. C
Subroutines
422 $HOMO " ne
#**
40
10
SOLUTION OF SUBROUTINES
REAL ALGEBRAIC HOMOGENEOUS EQUATIONS *** REG@D: (1) $INVS (2) $MPLY (3) $SUBN
SUBROUTINE $HOMO (As Xs N) REAL*8 AC10,10), BC10,10), BINVSC10,10), X(N) = 1 NM! = N  1 DO 40 I=1,NM1 YCI) = ACI.N) DO 40 J=1,NM1 BCIls,J) = ACIsJ) IF (NMI = 1) BINVS(1,1) = 1/BC1,1) IF (NM] = 1) GOTO 10 CALL $INVS (Bs BINVS,s NM1) DO 41 I=1,NM1
X¢1) = 0 DO 41 J=1,NM1 4! XCL) = XCI) RETURN END
X¢€10)4
YC10)
+ BINVS(IsJd)*YCU)
(a) Real
$CHOMO "eee MH
SOLUTION OF SUBROUTINES
COMPLEX ALGEBRAIC REQD: (1) $CINVS
HOMOGENEOUS (2) $CMPLY
SUBROUTINE $CHOMO (A, X, N) COMPLEX*16 AC10,10), BC10,10). BINVS(10.10). REAL*8 U, Z DATA Us Z /1.02 00/ X(N) = DCMPLX(U, Z) NMI = N  1 DO 40 I=1,NM1 YCL) = ACILN) DO 40 J=1,NMI 40 B(I,J) = ACIsJ) IF (NMI = 1) BINVSC1,1) = 1e/BC1s1> IF (NMI = 1) GOTO 10 CALL $CINVS (B, BINVS, NM1) 10 DO 41 I=1,NM1 X(1) = DCMPLX(Z, Z) DO 41 J=1,NM1 4l XCI) = XCI) + BINVSCI»J)*YCU) RETURN END (b) Fic. C9.
EQUATIONS *** (3) $CSUBN
X(10)4
YC10)
Complex
Solution of homogeneous algebraic equations
Comparing the subroutine $HOMO and Example C3, for which n = 4, first we define X(N) =1, that is, x,=1. Then matrix B and vector Y are
formed in the DO 40 loop. The CALL $INVS statement gives B™! (BINVS). Then the product of B~' and Y is obtained in the DO 41 loop. The solution vector X follows. Except for the complex mode, the subroutine $CHOMO, listed in Fig. C9(b), is the same program.
SEC. C11 C11
$MODL—Modal Matrix of Undamped Discrete Systems
$MODL—MODAL MATRIX DISCRETE SYSTEMS.
423
OF UNDAMPED
The subroutine $MODL finds the modal matrix of positivedefiniteundampeddiscrete systems with distinct roots. We shall follow the presentation in Example 9, Chap. 4, to derive the modal matrix. The equations of motion and the frequency equation from Eqs. (436) and (438) are i
M{q}+ K{q} = {0}
(C21)
A(w) =K— w?M=0
(C22)
Alternatively, premultiplying Eq. (C21) by M“* and defining Eq. (C22) can be expressed as
H = M“'K,
A(A) =AI H]=0 where A = w~. Thus, the subroutine $COEFF
(C23) can be used to find the A’s
and therefore the natural frequencies w’s At a principal mode, the entire system executes synchronous harmonic notion at a natural frequency w. Hence
{g} ={w*q} = —w*{q}
(C24)
The displacement vector {q} at a principal mode is also a modal vector. Substituting Eq. (C24) into (C21) yields
[w*M + K]{q}= {0}
(C25)
The equation above is identical to Eq. (437). A modal vector is obtained from the solution of the homogeneous equations as described in the last section. Considering all modes of vibration of the system, the modal matrix is formed from a combination of all the modal vectors as shown in Eq. (423). The equations above can be traced readily in $MODL, as listed in Fig. C10. $40 DL x%**x
41
40
CALCULATION OF THE MODAL MATRIX OF *** UNDAMPED POSITIVE DEFINITE DISCRETE SYSTEMS. ¢3) $MPLY (2) $INVS C1) REQD: SUBROUTINES
SUBROUTINE $MODL (Ms Ks Us ROOT, ERROR» NITER, K(10,10)4 H(10,10), ERRORs DUM(10,10), REAL*8 MINVS(10,10), ROOTC10), UC10,10)4 X¢10) CALL $INVS (M, MINVS, N) CALL $MPLY (MINVSs Ks Hs N) CALL $h00T (Hs ROOT, ENKKOR, WATLR ND DO 40 L=1,4N DO 4! I=1,4N DO 41 J=14N + K(Isd) =  MCIsJ)*ROOT(L) DUM(1s.J) CALL $HOMO (DUM, Xs N) DO 40 11=1,4N UCIisLy = exc hI) RETURN END
Fic. C10.
$ROOT
N) MC10,10)5
¢4)
$HOMO
&%
Modal matrix of undamped discrete systems; roots distinct.
424
Subroutines
1. The CALL The CALL r=1 ton.
APP. C
$INVS and CALL
$MPLY
$ROOT
finds the roots A, of Eq. (C23), for
statement
statements give
H=M °K.
2. For a given frequency w, the coefficient matrix [w*M+ K] is obtained by means of the DO 41 loop, where ROOT= —w’. 3. The CALL vector.
$HOMO
statement solves Eq. (C25) to yield a modal
4. The modal matrix is formed in the last DO loop.
C12
$CMODL—MODAL MATRIX OF DISCRETE SYSTEMS WITH VISCOUS DAMPING.
The subroutine $CMODL finds the modal matrix of positivedefinitediscrete systems with viscous damping, assuming the eigenvalues are complex and distinct. Except for the reduced equations, as described in Sec. 612, the technique for this program runs parallel to that of undamped systems. The equations of motion and the corresponding reduced equations from Eqs. (673), (675) and (677) are
where {y}={q
Mig} + Clq}+ K{q}={0}
(C26)
lw cllal*Lo allele]
ea
(¥}— Hy} = 10}
(C28)
4}, H=[M"6 I
Ab 0
I= unit matrix of order n, H1=M 'C,
lala min yy
2 0
(C29)
and H2=M_'K. The characteris
tic equation from Eq. (C28) or (680) is
A) where y is a complex The solution of Eq. modal vector. Thus, {q qt=yv{q qh}, that Eq. (C26) yields
yi
0
(C30)
root. (C28) is of the form {y}={WV}e”, where {WV} is a {y}=y{y}. Since {y}={q q}, we deduce that is, {q}= y{q} and {q}= y7{q}. Substituting these in
[My* + Cy + K]{q} = {0}
(C31)
Note that this may be compared with Eq. (C25) for undamped systems in the last section. In other words, except for the reduced equations and complex numbers, the same technique is applicable to both programs.
SEC. C12
$CMODL—Modal
Matrix
425
$CMODL "ee
CALCULATION OF THE MODAL MATRIX POSITIVE DEFINITE SYSTEMS WITH SUBROUTINES REQD: (1) $COEFF €4) $CINVS (7) $INVS
C4 7 oa
SUBROUTINE REAL*8&
K(10,10)5 COMPLEX*!16 DATA UNIT, NT2 = Nxe DO
40
43
4e
(Ms
Cs
ERROR,
Ks
ERROR,
NITER,
N)
H1¢10,10),
H2¢10510),
%
UNITC 10,10), ZEROC10,10) UC10,10)5 X10)
I=14N
N)
CALL
(MINVSs
Cs
Hl»
N)
CALL $MPLY (MINVS, DO 4! I=lsw
Ks
He,
N)
41
Us
H(10510)5
MC€10,10), MINVS( 10,10). DUM(10,10), ROOTC2,5), ZERO /200*0.0/
UNE TGa ED = Le CALL $INVS (Ms MINVS,
DO
4
40
$CMODL
C¢10,10),
OF *x* VISCOUS DAMPING. ¢2) $CHOMO (3) $CROOT (¢5) $CMPLY (6) $CSUBN (8) $MPLY (9) $SUBIN
$MPLY
J=I5N
H€loeJd) = =HICIsJ) HCI,NtJ) = HeCIsdJ) H(N+I4Jj) = UNITCIsJ) H(N+I,NtJ) = ZEROCIsJ) CALL $CROOT (Hs ROOT, ERROR, NITERs NT2) DO 42 JJ=1,5N DO 42 I11=1,2 DO 43 I=1,4N DO 43 J=1,N DUMCIsJd) = MCI,d)*ROOT(IIs,dJ) te + CCILs,J)*ROOTC II sud) CALL $CHOMO (DUM, X, N) JIN = ek CSI. eT DO 42 L=1,N UCN+LsJI) = XC(L) UCL»JI) = UCN+L,JI)*ROOTCIIsJdJ) RETURN END
+
KOTsJ)
Fic. C11. Modal matrix of discrete systems with viscous damping; roots distinct.
Let us compare the equations above with the statements in };CMODL listed in Fig. C11.
1. the CALL $INVS statement gives M~' (MINVS). The two CALL $MPLY statements give M 'C=H1 and M 'K = H2 in Eq. (C29). 2. The matrix H is formed in the DO 41 loop. The rest of the program is almost identical to $MODL. 3. The CALL $CROOT in Eq. (C30).
statement
yields
the
complex
roots
y,
4. The coefficient matrix [My?+ Cy+K] in Eq. (C31) is obtained by means of the DO 43 loop, to be followed by the CALL $CHOMO statement for finding the complex modal vectors.
5. The modal matrix is determined by means of the last DO loop in the program.
426
Subroutines
C13
APP. C
$CRKUT—SOLUTION OF COMPLEX 1STORDER DIFFERENTIAL EQUATION. METHOD: 4THORDER RUNGEKUTTA.
The subroutine $CRKUT equation
finds the solution of a firstorder differential dz pe Gz
(C32)
Az=At(—Cz+F)
(C33)
or
where C and F are constants. For the initial condition z,, applying the fourthorder RungeKutta method gives K, =At(—CXz,+F
(C34)
K,;=At[—CX(z,+K,/2)+ F] K,=At(=CX(@,+K,)+F]
Az=(K,+2K,+2K,+K,)/6
(C35)
Z=2,+Az
(C36)
Let us compare the equations above with the statements in $CRKUT listed in Fig. C12. 1. Since A= DT/4, the subroutine computes a value of Z for every DT/4 and returns Z for the given DT.
2. The DO
40 loop accounts for the number of equations N to be solved
by the subroutine. 3. The Egs. (C34) to (C36) can be identified readily in the DO 41 loop. $CRK UT " m0 be
x&#*
SOLUTION OF COMPLEX 1STORDER DIFFERENTIAL EQUATIONS RUNGEKUTTA METHOD. ASSUME CONSTANT FORCE. SUBROUTINE CALCULATES VARIABLEZ FOR EVERY DT/4.
SUBROUTINE
$CRKUT
REAL*8 A, DT COMPLEX*16 Kl, A = DT/4. DO 40 I=1,N DO 41 J=1,4
4l 40
(F,
K2s
Zl,
K3,
Cs
K4,
=
A¥CCCI)*Z1¢1)
K2 K3 K4
= = =
Ax€CCI)D*(Z1C1)+K1/26) A*x(CCI)*(Z1C1)+K2/2.) A*C=CCOI)*(Z1C1)+K3) +
Z1CI)
=
ZCI)
+
CKL
+
Zs
CC10)5
KL
OCI).
+
DT,
N)
FC10)s
2610),
Z1¢10)
FCI))
2.*K2
+ FCI)) + FCI)) FCI)D)
+
26*K3
+
K4)/6.
S211)
RETURN END
Fic. C12.
Solution of 1storder differential equations.
*xx*
SEC. C14
C14
$PLOTF—To Plot Data from a File
$PLOTF—TO
PLOT DATA
427
FROM A FILE.
The subroutine $PLOTEF listed in Fig. C13 pots data stored in a file. It assumes that (1) the number of variables NVAR
to be plotted is of Eq. (D19) and the general solution of (D18) are
Si2=5

(et ve*—4mk)
x, = Ce" + Cye’"
(D20)
(D21)
Note that c?>4mk in order that the roots be real. Since Vc?—4mk
;
dak 4 é
7 al
ami
Ditéeckrshiy eT
ei)
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aa 4
Di rommtanyrs oi) Wy «
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7
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we on ityone
ro,
ttly AAU
:
Oiemndt apd > See
“Ay
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iv iw Ye Wty eis y ue a ty au Ake ‘(ALras.
Th
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=
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9a

soe
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7
INDEX
Absorber, dynamic: centrifugal pendulum, 8386 damped, 129, 173, 186 undamped, 170172, 185 Acceleration, 33, 34, 69 due to moving support, 98100 SI units, 17, 59 vectorial representation of, 13 Acceleration ratio in harmonic response, 41 Accelerometer, 102106 Adjoint matrix, 223, 225, 399401 (see also Matrix inversion) Amplitude: complex, 14 (see also Phasor) of free vibration, 67, 31, 36, 78 (see also Modal vector) at resonance, 8, 40, 81 (see also Forced vibration) Analogy: electromechanical, 48 of rectilinear and torsional systems, 5860 Anderson, R. A., 388 Andronow, A., 388 Aperiodic motion, 3, 35 (see also Forced vibration, Free vibration, Transient vibration) Assumed mode method, continuous
systems, 255, 288, 290 (see also
Separation of variables) Asymptotic stability, defined, 302 Autonomous systems, defined, 302 Auxiliary equation (see Characteristic equation) J iHies Wo, Wos LUG Wil, Bandwidth, near resonance, 51 Bars (see Beams, Rods, Shafts) Beams, lateral vibration, 262264 boundary conditions, 210, 264 discrete system representation, 207,
213 effect of axial load, 268270 effect of shear deformation and rotary inertia, 266268 as eigenvalue problem, 271 frequency equation, 265, 277, 291 Lagrange’s equations, 280, 284 modal analysis, 285, 288 modes, 276 orthogonality of modes, 273275 Rayleigh’s method, 64, 288 RayleighRitz method, 290 Bearings with elastic support, 9193 Beat frequency, 10 Bendixson, T., 323 Bendixson’s second theorem, 323
442
Index
Biot, M. A., 30, 389 Bishop, R. D., 203 Blake, M. P., 45 Borel’s theorem, 55 Boundary conditions, 253, 254 beams in lateral vibration, 210, 264, 277, 292 effect on orthogonality, 273, 275 rods in longitudinal vibration, 259 shafts in torsional vibration, 262 Center, as singular point, 310, 312 Centrifugal (bifilar) pendulum, 8286 Chaiken, S., 388 Characteristic determinant (function), 399 of damped systems, 242 of undamped systems, 146, 160, 166, 178, 224 Characteristic equation, 35, 146, 224 (see also Frequency equation) computer solution of coefficients, 163, 416 computer solution of roots, 417, 419 of damped systems, 242 of a square matrix, 308, 398 Circular frequency, 9 Cofactor, 393 Collar, A. R., 224, 234, 241, 389 Complementary function, 34 (see also Differential equation) Complementary solution, 55 (see a/so Impulse response) Complex numbers, products of, 15 (see also Vectors) Compliance (see Influence coefficients) Computer programs: listing, 340, 409 using data file for plotting, 349 using subroutines, 346 Conservative systems, 27, 31, 407 Constraint equation, 25, 26, 403 Constraint matrix, 233 Continuous systems: defined, 253 generalized mass, stiffness, initial condition, 281282 Convolution integral, 5557 for continuous systems, 285 for discrete systems, 118, 239, 244
Coordinates: Cartesian, 26, 403 generalized, 144, 155, 225, 360 normal, 229 principal, 148, 158, 165, 225, 227, 360 Coordinate coupling, 155157 of nonsymmetrical matrices, 158 Coordinate transformation: of continuous systems, 280282 of discrete systems, 159, 218, 220, 225, 244 of nonlinear systems, 308 Coordinate translation, 307
Coulomb damping, 125126 Coupling (see Coordinate coupling) Cramer’s rule, 107, 396 Crandal, S. H., 338 Crede, C. F., 117, 120, 121, 388 Cress, P., 389 Critical speed of shafts, 89, 195 effect of elastic bearing supports, 9193 Cunningham, W. J., 388 D’Alembert’s principle, 34, 403 Damping: Coulomb, 125126 critical, 36 energy dissipation in, 123 equivalent viscous, 125 hysteretic (structure, solid), 127 nonlinear, 304, 319, 322 proportional, 241 velocity squared (quadratic), 126 Damping, coefficient, 3, 5, 59, 60, 69 factor, 3537, 51 force (viscous), 2 function, 407 matrix, 144, 241, 243 Damping, linear, 2, 4, 59, 60 Dashpot (see Damping) Decoupled equations (see Modal analysis) Degrees of freedom: defined, 25 examples of, 2426 Delta (Dirac) function (see also Impulse response): as concentrated force, 285
443
Index
Delta (Dirac) Function (continued) as concentrated mass, 276 defined, 53 Delta method, 319 Den Hartog, J. P., 193, 197, 388 Derkley, T. F.; 123 Determinant, 392 (see also Characteristic determinant) Differential equation: complementary function, 34, 70, 77, 431434 complementary solution, 55, 240, 245 computer solution by impedance method, 352, 356 RungeKutta method, 342, 410, 426 defined, 429 general solution, 34, 431, 437 due to arbitrary excitation, 55, 240, 245 due to harmonic excitation, 4244 graphical solution, 316320 impedance method, 45, 86, 169 linear, with constant coefficients, 4, 429 nonlinear autonomous systems, 304 analytical solutions, 323328 particular integral by convolution integral, 55, 120, 239, 244, 285 method of undetermined coefficients, 435437 particular solution, defined, 438 simultaneous, linear, 438439 solution, defined, 430 Differential operator (see Linear differential operator) Differentiation of matrices, 395 Digital computers (see Computer programs) Dirac delta function (see Delta function) Dirksen, P., 389 Discrete systems: defined, 4 discrete and continuous systems compared, 253 Drop test, 56 Duffing’s equation, 328, 332
Duhamel’s equation, 58 Duncan, W. J., 224, 234, 241, 389 Dunkerley, S., 190 Dunkerley’s equation, 190193
Dynamic absorber (see Absorber, dynamic) Dynamic (inertia) coupling, 157 Dynamic matrix, 221, 234, 374 (see also Equations of motion)
Eigenfunction, continuous systems, 255256, 277 (see also Orthogonality) Eigenvalue, 224, 227, 230, 234, 308,
398399 Eigenvalue problem, continuous systems, 270271 Eigenvector (see Modal vector) Elastic (static) coupling, 156 Energy: damping, 8, 40, 123127, 407 excitation, 3 as initial conditions, 4 kinetic and potential, 2728
in Lagrange’s equations, 219221, 280282, 405406 in Rayleigh’s method, 3132,
194196, 231, 288, 290 reference kinetic, 289, 292 Energy method, 27 (see also Lagrange’s equations, Rayleigh’s method) Equations of motion, 27, 34, 69, 112 of continuous systems, 259, 263,
270, 285 of discrete systems, damped,
143, 144, 243, 377 of discrete systems, undamped, 145, 160, 220, 239, 360, 374 in influence coefficients, 177 of Lagrange, 219, 281, 407
Equilibrium: dynamic, 46
as singular point, 306 static, 4, 2728, 3334, 143, 301 Equilibrium points, nonlinear systems, 309312 Equivalent damping, 69 (see also Damping)
444
Index
Equivalent mass, 31, 32, 6971, 218 of continuous systems, 281 Equivalent spring, 69, 7276, 168, 218 of continuous systems, 281 EulerBernouli beam, 262 Euler’s formula, 12, 36 Excitation: defined, 3 equivalent, 69 harmonic, 34, 80, 112, 123 of continuous systems, 297 by impedance method, 45, 86, 169 of nonlinear systems, 328333 due to support (base) motion, 98, 101, 123 transient (arbitrary), 55, 118, 239, 244, 285 Expansion theorem, 230 Faddeeva, V. N. 389 FaddeevLeverrier method, 414, 416 Field transfer matrix, 203, 207 Finite element method, 203 Flexibility matrix, 177, 220 Focus, as singular point, 311, 312 Force: damping, 2, 123, 407 dynamic, 34, 143 equivalent, 69 generalized, 144, 239, 244, 285, 406 impulse, 53 inertia, 34, 177, 404 spring, 2 static, 34 Force polygon, 47 (see also Vectors) Force reduction, isolation mount, 97 Force transmission (see Trans
missibility) Forced vibrations: defined, 3 harmonic excitation (see also Impedance method) continous systems, 297 discrete systems, 169175, 356 onedegreeoffreedom systems, 34, 38, 80, 8687, 352 impulse response, 5355 modal analysis continuous systems, 285288
discrete systems, damped, 243245, 376 discrete systems, undamped, 160, 164, 238240, 360 nonlinear systems damped, 331334 undamped, 328330 periodic excitation, 110114 resonance, 8, 40, 81, 172 transient excitation, 5558, 116122, 238, 243 Fourier series, 109112, 228, 256, 327 Fourier spectrum, 112114 Frazer, R. A., 224, 241, 389 Free vibrations:
of continuous systems assumed mode method, 255 as eigenvalue problem, 270271 Lagrange’s equations, 280282 RayleighRitz method, 290293 separation of variables method, 256 defined, 3 of discrete systems, damped, 241243 of discrete systems, undamped » Dunkerley’s equation, 190193 influence coefficient method, 175179 Holzer method, 197 Lagrange’s equations, 219, 407 matrix iteration, 234238 modal analysis, 145148, 160, 165, 223225, 232 MyklestadProhl method, 207 Rayleigh’s method, 193195, 231 of nonlinear systems analytical methods, 323328 graphical methods, 316320 of onedegreeoffreedom systems energy method, 2729 Newton’s second law, 3334, 69 Rayleigh’s method, 3132 Frequency: beat, 10 circular, 9, 35 defined, 3 fundamental, harmonics, 147 natural (see also Frequency equation)
445
Index
Frequency (continued) defined, 56 determination of, 29, 31, 81, 190,
1932319305 with elastic support, 93 matrix iteration, 234, 371 with viscous damping, 36 resonance, of elastic absorbers, 129 Frequency, effect of nonlinear spring, 325 Frequency. equation (see also Characteristic equation): coefficients of, 163, 416 computer solutions of, 417, 419 of continuous systems, 255, 277, 278
RayleighRitz method, 291 of discrete systems, 146, 161, 178,
224 of semidefinite systems, 166, 198,
234 Frequency ratio, 38 Frequency response method, 3843, 45 (see also Forced vibrations) computer programs for, 352360 of discrete systems, 172175 Frequency spectrum, 112 Functional, 288 Functions:
Goldstein, H., 25 Graham, J. W., 389 Gravitational constant g, 20
Half power point, 51
Hamming, R. W., Harmonic balance Harmonic motion, Vectors) addition of, 10,
341, 389 method, 327 simple (see also
13, 14 defined, 8 differentiation of, 9, 12
vectorial representation of, 11, 13 Harmonic response (see Forced vibrations) Harris, C. M., 117, 388 Hertz (Hz), 9
Hildebrand, F. B., 224, 230, 398, 432 Hinkle, R. T., 89
Holzer method, 197 Homogeneous algebraic equations, computer solution, 421, 422 Hooke’s law, 2 Horner, G. C., 389 Hysteresis loop, 123, 124 (see also Damping, equivalent)
even and odd, 111
Impedance, mechanical, defined, 48
positive definite quadratic, 219 semidefinite, 219
Impedance matrix, 170, 356 Impedance method, 4549, 8687 computer applications, 352360 for discrete systems, 169170, 356 for periodic excitation, 112 Impulse, defined, 53 (see also Delta function) Impulse response, 53, 118, 239, 244, 285 (see also Forced vibrations) Indicial response, 57
Gaussian elimination method, matrix inversion, 397, 411414 Geared systems, 166 branched, 206 Generalized: coordinates, 144, 155, 224 relating principal coordinates, 239, 244, 360, 378 force, 144, 239, 243, 285, 406 initial conditions, 240, 245 of continuous systems, 282 mass and stiffness, 144, 155, 158, 219, 226 of continuous systems, 270, 281 Generating solution, nonlinear systems, 323 Gladwell, G. L., 203
Inertia (dynamic) coupling, 157 Inertia force, 34, 177, 404 Inertia torque (see Force, equivalent) Influence coefficients, 175180 defined, 176
relating to stiffness matrix, 179 Initial conditions: of continuous systems, 254256 generalized, 282 defined, 3
446
Index
Initial Conditions (continued) of discrete systems, 150151, 165,
240, 245, 360, 378 in nonlinear analysis, 305, 319, 321, 324 of onedegreeoffreedom systems, 7, 28, 36, 42, 341 due to unit impulse, 54 Initial value problem, 341, 360, 378 Instrument mounting, 99 Isocline method, 317 Isolation, vibration, 94, 99, 108, 123 force reduction in, 97 of rotating unbalance, 94, 96 in twodegreeoffreedom systems, 174 Jacobsen, L. S., 121, 122, 319 OWES IDE Ure 1A Jump Phenomenon, 330332
Karman, Theodore von, 130, 389 Kelly, R. D., 388 Nettera ike los iuL Kilogram, mass, defined 16 Kinetic energy (see Energy) Lagrange’s equations, 219221,
402407 of continuous systems, 280284
Lagrangian, 407 Lambda matrix, 225, 399 LaSalle, J., 301 Lateral vibration of rods (see Beams in lateral vibration) Wazany Bele 122 Lefschetz, S., 301
is 1, 1, Tey Lienard, A., 319 Lienard method, 319 Limit cycle, 321, 322 Linear algebra, elements of, 391401
Linear differential equations, defined, 430 Linear differential operator, 431 of continuous systems, 270271 Linear systems, defined, 4 (see also Forced vibrations, Free vibrations) Linearization of nonlinear systems, 3023307525
Linearly independent set, 224, 230 Logarithmic decrement, 78 (see also Free vibration) Loss coefficient, damping, 128
Magnification factor, 3841, 47, 51 Mass: equivalent, 31, 6972, 155157, 219
generalized, of continuous systems, 281 Mass matrix, 144, 155157, 220 Matrix:
damping, 144, 241, 243, 245 defined, 391 dynamic, 221, 234, 374 mass,
144, 155157,
nonsymmetrical,
164, 220
157
stiffness, 144, 155157,
164, 177,
220 types of, 391393 Matrix algebra, elements of, 391401 Matrix inversion, 393, 395397 computer method, 411, 414 Matrix iteration, 234238 computer method, 371376 Matrix operations, 393396 Maxwell’s reciprocity theorem, 157, 176 Mechanical impedance, defined, 48 (see also Impedance method, Forced vibrations) Meirovitch, L., 292, 388 Meter, length, defined, 16 Michealson, S., 203 Mindlin, R. D., 56 Minorisky, N., 388 Mitchell, Wm. S., 45 Modal analysis: of continuous systems, 285288 of discrete systems, damped, 243, 376380 of discrete systems, undamped, 160165, 238240, 360365 Modal matrix: computer method, 423, 424 of discrete systems, damped, 242, 244 of discrete systems, undamped, 148, 159, 163, 224, 226 normalization of, 229, 399
Index
Modal vectors: of discrete systems, damped, 242 of discrete systems, undamped, 147, 160, 223, 225, 398 by matrix iteration, 235, 374 orthogonality of, 164, 228, 243 Mode shape: of continuous systems, 276, 280, 288 (see also Eigenfunction) of discrete systems, 147, 193, 231 (see also Modal vectors) Model, onedegreeoffreedom systems, 4, 33, 49, 69 Modes: natural, 34
447
Operators (see Linear differential operators) Ordinary point, 306 Orthogonal functions, 228 Orthogonality property: of continuous systems, 272280 boundary condition dependent
of A, 275 boundary condition independent of A, 273 of discrete systems, damped, 241243 of discrete systems, undamped, 164, 226, 229
Overdamped systems, 7, 36
orthogonality of (see Orthogonality) principal, 145155, 223226 amplitude ratio in, 147, 155 defined, 147 harmonic components of, 146 initial conditions for, 150 of semidefinite systems, 167, 200, 233 of single span uniform beams, 276, 277278 Moment of inertia, 29 equivalent, 7071 IMOErOWs Ce Le) hy Moving support, 82, 98, 101, 121, 123 Multidegreeoffreedom systems (see Discrete systems, Forced vibrations, Free vibrations) Multirotor systems, 154, 165, 194, 197, 232 computer application, 365 Myklestad, N. O., 207, 211 MyklestadProhl method, 207 computer application, 369371 Natural frequency (see Frequency) Natural modes (see Modes) Newton’s second law (see Equations of motion) Nodes, as singular point, 310, 312 Nonlinear systems, 300, 341, 430 stability of, 301 Nonperiodic excitation (see Forced vibrations) Nonsymmetrical matrices, 157 Normal coordinates, 229
Page, C. H., 16, 17, 18 Parameters (see Systems, elements of) Particular solution (see Differential equations) Pell, W. H., 319 Pell’s method, 319 Pendulum: centrifugal (bifilar, dynamic absorber), tuned, 8286 compound, 182 simple, 2930 spherical, 26, 182 torsional, 60, 74 Period, of vibration, 3, 9 Periodic excitation, 110114 Periodic motion, defined, 3 (see also Harmonic motion) Perturbation method, 323 Phase angle, 9, 14, 3839, 112113
in harmonic excitation, 87 as time delay, 104 Phase plane, 303306 Phase spectrum, 112114 Phase trajectory, 304 characteristics of, 306 Phasor, 14, 46, 86 Pilkey, W. D., 389 Pipe Ae224: Positive definite quadratic function, 219 Potential energy (see Energy) Prawel, S. D., 371 Principal coordinates (see Coordinates) Principal modes (see Modes)
448
Index
Principle of superposition (see Superposition) Programs (see Computer programs) Prohl, M. A., 207 Proportional damping, 241
Quadratic damping,
126
Quadratic function, positive definite,
219 Quality factor, sharpness of resonance, 52
Rayleigh, Lord, 241 RayleighRitz method, 290294 Rayleigh’s method, 31, 193, 196
computer application, 365369 Rayleigh’s quotient, 231, 288 Reciprocal unbalance, 87 (see also
Forced vibrations) Reciprocity theorem, Maxwell’s, 157, 176 Redheffer, R. M., 109
Regular point, 306 Relative amplitude (see Modes) Resonance, 8, 40, 51, 81, 129 of discrete systems, 172 of nonlinear systems, 330331 Response (see Forced vibrations, Free vibrations) Richman, G., 388 Rods: in lateral vibration (see Beams) in longitudinal vibration, 258261 as eigenvalue problem, 271 orthogonality of, 273, 275 Rayleigh method, 196 in torsional vibration (see Shafts) Rotating unbalance, 87, 90, 92 (see also Forced vibrations) Rubin, S., 117 RungeKutta method, 342, 410, 426 Ruzicka, J. E., 123 Saddle point, as singular point, 310,
B12 Scarborough, J. B., 342, 389
Second, time, defined, 17 Secular term, 325 Seismic instruments, 101106
Selfexcited oscillation, 321323 Semidefinite systems, 165169, 197,
206, 232234 amplitude ratio of, 166167, 197, 234 constraint matrix, 233 computer application, 250, 365 as continuous systems, 276278 Separation of variables, 256258 Shafts, rotating, 89, 91, 195 Shafts in torsional vibration: continuous systems, 261262 as eigenvalue problem, 270 orthogonality of modes, 273, 275 discrete systems, 155, 165168, 197, 206, 233 Shapiro, W., 389 Shock spectrum, 116122 initial, 120 residual, 121 SI units (see Units) Similarity transformation, 308 Simultaneous differential equations, 438 Singular points, defined, 306 Snowdon, J. C., 108, 129 Sokolnikoff, I. S., 109 Solid damping, 128 Spring: constant (stiffness), 2, 69 equivalent, 72, 168, 218 in centrifugal field, 76 in gravitational field, 75 due to orientation, 7475 in parallel, 73 in series, 72 equivalent mass of, 2728 force, defined, 2 hard, 325, 328, 331 soft, 325, 330 Stability of oscillation, 301302 analysis of, 306316 trajectories about equilibrium, 309, Sil State variables, 202, 304 State vector, 202, 205, 241, 304, 375 Static (elastic) coupling, 156 Steadystate response, 78 (see also Forced vibrations)
Index Suffness (see a/so Spring): complex, 128 (see also Damping, hysteretic) equivalent, 69 generalized, of continuous systems, 281 Stiffness matrix, 144, 155, 219, 241
relating to flexibility matrix, 178 Stoker, J. J., 388
String in lateral vibration, 253256 as eigenvalue problem, 271 as nonlinear system, 303 Structural damping, 128 Subharmonic oscillations, 332334 Subroutines, listing of, 409 Superposition, 4, 52, 55, 147, 255, 300, 432 Support (base) motion, 98, 101, 121, 123 Sweeping matrix (see Matrix iteration) System function (see Transfer function) Systems, dynamic: defined, 1 elements (parameters) of, 24, 69 torsional and rectilinear, compared,
4, 59, 60
Timoshenko, S., 33, 388 Timoshenko beam, 266
Tong, K. N., 273, 274, 388 Torsional systems (see Shafts in torsional vibration, Systems, dynamic) Trajectories in phase plane, 304, 306 about equilibrium, 309, 311 mapping of, 314 Transfer function, 49 sinusoidal, 50 Transfer matrix, 202207, 209 computer application, 369371 Transient vibration (see also Forced vibration, Free vibration): computer application, 341352, 360, 376 free vibration, 7, 36, 43 nonperiodic excitation, 5258, 161 shock spectrum, 117
449 Transmissibility (see Isolation, vibration) Underdamped systems, 6, 36 (see also Forced vibration, Free vibration)
Unit impulse, defined 53 Unit step function, defined, 57 Units: English engineering, 16 English to SI conversion, 18 international systems SI, 16 prefixes, 18
symbols, 18 for rectilinear and torsional systems, 59 Van der Pol equation, 321, 327 Variation of parameter method, 325 Vectors: addition of, 10, 1315 differentiation of, 12 displacement, velocity, acceleration, 13, 46 multiplication of, 15 representation of harmonic functions, 12, 46, 8687 Vehicle suspension, 99, 151 Velocity ratio in harmonic response, 41 Velocity squared damping, 126 Vernon, J. B., 118, 388 Vibration absorber (see Absorber, dynamic) Vibrometer, 102 Vigoreux, P., 1618 Virtual work, 403, 404 Vortex (center), as singular point, 310, 312
Wave equation, 258262 of bars in longitudinal vibration,
259 of shafts in torsional vibration, 261 of strings in lateral vibration, 254 Weaver, W., 388 ; Whirling of shafts (see Critical speed of shafts) Wylie, C. R. Jr., 389 Young, D. H., 33, 388
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