Mechanical Vibrations: Theory and Applications [2 ed.] 0205066704, 9780205066704

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ECHANICAL ;

_ VIBRATIONS ~ THEORY AND APPLICATIONS

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_ SECOND EDITION

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Francis S. Tse

Ivan E. Morse | Rolland T. Hinkle

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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 0-205-05940-6 ISBN 0-205-06670-4 (International)

77-20933

Contents

Preface

xi

CHAPTER1 1-1 1-2 1-3 1-4 1-5 -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 2-1 2-2 2-3 2-4 2-5

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

2-6

, 2-7

2-8 2-9

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 3-5

3-6

| WwW ~ WW \o 0

3

51

SYSTEMS WITH ONE DEGREE OF FREEDOM— APPLICATIONS

69

Introduction 69 Undamped Free Vibration 70 Damped-Free 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

4-1

Introduction

4-2

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 Myklestad-Prohl Method 207 Summary 213 Problems 214

CHAPTER6

218

DISCRETE SYSTEMS

6-1 6-2

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 6-13 Damped Forced Vibration—Modal Analysis 243

226

viii

Contents

6-14

Summary Problems

CHAPTER7

245 246

CONTINUOUS

SYSTEMS

253

7-1 7-2

Introduction 253 Continuous Systems— A Simple Exposition 253 7-3 Separation of the Time and Space Variables 256 7-4 Problems Governed by the Wave Equation 258 258 Longitudinal Vibration of Rods 261 Torsional Vibration of Shafts 7-5 Lateral Vibration of Beams 262 265 7-6 Rotary Inertia and Other Effects Shear Deformation and Rotary Inertia Effects 266 Effect of Axial Loading 268 7-7 The Eigenvalue Problem 270 7-8 Orthogonality 94 G2 Boundary Conditions Independent of A 273 Boundary Conditions Dependent on A 275 7-9 Lagrange’s Equations 280 7-10 Undamped Forced Vibration—Modal Analysis 285 7-11 Rayleigh’s Quotient 288 -12 Rayieigh-Ritz Method 290 -13 Summary 294 Problems 295

CHAPTER8 8-1 8-2 8-3 8-4 8-5

-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 Self-excited Oscillations 321 Analytical Methods 323

300

Contents

8-8

8-9

8-10

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 9-1 9-2 -3 -4 5 -6

9-7 9-8

9-9

325

SOLUTIONS BY DIGITAL COMPUTERS

339

Introduction 339 One-degree-of-freedom Systems—Transient Response 341 Program—TRESP1 342 Program—TRESPSUB 346 Program—TRESPF1 349 One-degree-of-freedom Systems—Harmonic Response 352 N-degree-of-freedom Systems—Harmonic Response 356 Transient Response of Undamped Discrete Systems 360

Rayleigh’s Method—Undamped

Multirotor Systems 365 9-10 Myklestad-Prohl Method—Transfer Matrix Technique 369 9-11 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

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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 one-year 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 one-dimensional 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 9-1 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 3-5 can be used as assigned reading. Sections 3-6 through 3-8, Section 4-9, and Sections 5-4 through 5-6 may be omitted without loss of continuity. For a one-semester senior or dual-level 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 one-year 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

1-1

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

1-2

Introduction

ELEMENTS

OF A VIBRATORY

CHAP. 1

SYSTEM

The elements that constitute a vibratory system are illustrated in Fig. 1-1. 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. 1-1, 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. 1-1.

Elements of a vibratory system.

SEC. 1-2

Elements of a Vibratory System

3

2s atPeriod is OA

ee

Displacement x

Period r af

Fic. 1-2.

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. 1-1. 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. 1-2. 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 a-system. 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. 1-1 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. 1-1 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 Jumped-parameter, 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 lumped-parameter 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 1-1. The comparison is

SEC. 1-3

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 2-2 and 2-3. It is apparent from the comparison that the concept of rectilinear systems can be extended easily to rotational systems. 1-3

EXAMPLES

OF VIBRATORY

MOTIONS

To illustrate different types of vibratory motion, let us choose various combinations of the four elements shown in Fig. 1-1 to form simple dynamic systems. The spring-mass system of Fig. 1-3(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 free-body 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

Free-body

~

sketch

tatic

deflection

(b)

Static equilibrium position

Damped free vibration

Fic. 1-3.

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 mass-spring system with damping is shown in Fig. 1-3(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. 1-3

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. 1-3(a) and (b) are illustrated in Fig. 1-4. 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, vibration-measuring 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. 1-3(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 steady-state response. Hence the steady-state response must be at the excitation frequency regardless of the initial

*o

Critically damped Overdamped

°o Displacement x

Undamped

Fic. 1-4.

Underdamped

Free vibration of systems shown in Figs. 1-3(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 steady-state 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 steady-state condition, the net energy-input 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.

1-4

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) steady-state 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. 1-5. If the distance OP is

OP = x(t)= X cos ot

(1-1)

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. (1-1), 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. 1-4

Simple Harmonic Motion

Fic. 1-5.

9

Simple harmonic motion: x(t) = X cos ot.

that is,

2 r=— Bes

Period is

A

=a

(1-2)

QO

1 \Frequency f=-=—— cycle/s, soror

He)

(1-3)

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

(1-4)

x = —wX sin wt = wX cos(wt+ 90°)

(1-5)

Acceleration ¥ = —w*X cos wt = w*X cos(wt + 180°)

(1-6)

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. (1-4) and (1-6). X= -w’x (1-7) 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. (1-7) 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. 4-1.

(b)

kx,

k(x, —x5)

Lm

—X,)

KO

CX

KX

5)

Free-body sketches

A two-degree-of-freedom system.

144

Systems with More Than One Degree of Freedom

CHAP. 4

For conciseness, Eq. (4-1) 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)}

(4-3)

By simple matrix operations, it can be shown that Eqs. (4-1) and (4-2) are

equivalent. The quantities in Eq. (4-3) can be identified by comparing with Eq. (4-2). 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. 4-4 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.

(4-2). 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 two-degree-offreedom 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}

(4-5)

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. (4-5) also describes the motion of an ndegree-of-freedom that is,

system if the matrices M, C, and K are of nth-order,

M=[m,],

C=[e,],

K =[k;]

(4-6)

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}

(4-7) (4-8)

SEC. 4-3

Undamped

Free Vibration: Principal Modes

145

Hence Eq. (4-5) is also the general form of the equations of motion of an n-degree-of-freedom system.

4-3

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. 4-1

reduces to that shown in Fig. 4-2(a). Hence the equations of motion from Eq. (4-2) are

Fie alelilecmvewelle ely|e 0) 0

m,JLxX,

oils

k+k,\LX>

0

The equations are linear and homogeneous and are in the form of Eq. (D-47), App. D. Hence the solutions can be expressed as x,= B,e*

(4-10)

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. 4-2.

(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)

,

(4-11)

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. (4-11) in (4-9), dividing out the factor sin(wt+w), and

rearranging, we have (k =

k, oe w*>m,)A,—

kA,

=

0

(4-12)

—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=

AD-BCz2O

=0

(4-13)

From linear algebra, Eq. (4-12) 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,

(4-14)

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. (4-14) are +w, and +w. Since the solutions in Eq. (4-11) 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 twodegree-of-freedom system. Each of the solutions of Eq. (4-9) has two harmonic components at the frequencies w, and w, respectively. By * Note that the values of s in Eq. (4-10) are +jw, and +jw, in order to have the periodic

solutions assumed in Eq. (4-11). If w* is not real and positive, it can be shown that the solutions by Eq. (4-10) would either diminish to zero or increase to infinity withiincreasing

time.

SEC. 4-3

Undamped Free Vibration: Principal Modes

J-Losmoweon(}

147

superposition, the solutions from Eq. (4-11) are

[

=

A,

Pinlort+ da) w,)++]4"

sin(w,t+

fsin(wat+ + ya)

4-15 (4-18)

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. (4-15) are defined in Eq. (4-12). Substituting @, and w, in Eq. (4-12) 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

(4-16)

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. (4-15) becomes

[2]23 i Aj, Sin(@,t+ p,)+ EB|Ay sin(w.t+W)

(4-17)

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 second-order differential equations. Note that

(1) from the homogeneous equation in Eq. (4-12), 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. 4-2(b). in Eq. (4-17), that is,

Ze ie Aj, sin(w,t+ y,)

For example, the first mode occursifA,,=0

Ory

xy

fed p(t) = {u}, px(t)

(4-18)

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. (4-17) is zero, that is,

[|= [2] Avsinoat+ us or c)=[“"] pS tube 4-19) X2

The modal vector for the second mode is {u},. The harmonic functions of the motions x,(t) and x,(t) in Eq. (4-17) can be expressed as =| 4

X

a

1

1 |

U;

sin(w,t+ ie A ba

UzJLAj;> sin(wzt+ Yr)

.

se|kevel

Ur,

(4-20)

Us2JLp2(t)

Or

{x}=[u]{p}

(4-21)

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,).

(4-22) Note

that in

Eqs. (4-18) through (4-20) only the relative values in a modal vector can

be defined, as shown in Eq. (4-16). A modal matrix [u] in Eq. (4-22) is simply a combination of the modal vectors. The actual motions {x} in Eq. (4-20) 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. (4-18) and (4-19). Principal coordinates will be further discussed in Sec. 4-5 and Chap. 6. The coordinate transformation between the {x} and {p} coordinates is shown in Eq. (4-21). The extension of the concept to n-degree-of-freedom systems is immediate. For example, a system may be described by the {x}= {x; X,°+*x,} coordinates. Analogous to Eq. (4-17), 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;; |

(4-23)

where i,j=1,2,...,n. The transformation between the {x} and the {p}

coordinates is analogous to that in Eq. (4-21).

SEC. 4-3

Undamped Free Vibration: Principal Modes

149

Example 1 Referring

to Fig. 4-2(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. (4-14), the natural frequencies are

wo, =Vkim

and

w=

V3k/m

Substituting w, and w, in Eq. (4-16), we obtain u,=1 and u,=—1. Hence the displacement vector {x} from Eq. (4-20) 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. (4-17), we obtain

| Xone

An == | cos Dal

k [| 4/—t+— COS i

Pali

3k |——'t m

The motions are plotted in Fig. 4-3 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. 4-3.

Example 2.

Superposition of modes of vibration: Example

Natural modes

3k/m

t

1.

~ /\9_ ©. A140

Find the initial conditions that would set a two-degree-of-freedom into its natural modes of vibration, that is, Aj; or Aj,in Eq. (4-17) becomes zero. Solution: From

Eq. (4-20), 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. 4-3

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. 4-4. 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. 3-5.) Because of this large separation.

of_natural frequencies between the wheels and the car body, the problem in Fig. 4-5. 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. 4-4.

Schematic of an automobile.

CHAP. 4

Systems with More Than One Degree of Freedom

152

Fic. 4-5.

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. (4-9). The frequency equation from Eq. (4-13) 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. 4-3

Undamped Free Vibration: Principal Modes

(a)

First mode: f= 1.09 Hz

Fic. 4-6.

(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

50-02, | 0.42 m/rad

The two principal modes of vibration are shown schematically in Fig. 4-6. 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. (4-17), 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 three-degree-of-freedom system

A torsional system with three degrees of freedom is shown in Fig. 4-7. (a) Determine the equations of motion and the frequency equation. (b) If

154

Systems with More Than One Degree of Freedom

Fic. 4-7.

CHAP. 4

A three-degree-of-freedom 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)

(4-24)

Substituting 6,=0, sin(wt+y), for i=1,2, and 3, in Eq. factoring out the sin(wt+) term, and rearranging, we have

(4-24),

(k,1+ k,2— @7J,)O, — k,,O, =0 — k,2@, + (k,2+ k,3— 7 Jz)O2— k,303 =0

(4-25)

— 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 5|0.198 |Hz 0.288

From Eq. (4-25), 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. 4-4

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. 4-8. 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. (4-21).

[6;]=[{6},

{6},

it]

1 1 90 0.445 295-0802)

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. 4-8. _ 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.

4-4

GENERALIZED COORDINATES COORDINATE COUPLING

AND

The general form of the equations of motion of a two-degree-offreedom system is shown in Eq. (4-4). For undamped free vibration, we have [2s M2,

i

alr

Mg2ILX2

Koy

EE Kaz JLX2

(4-26) 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

4-9. Generalized coordinates and coordinate coupling: a twodegree-of-freedom 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. 4-9 by the displacement vectors {x,

9},

(Xe ealids, X5.0m00 } Referring to Fig. 4-9(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

4-27

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. 4-9(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. 4-4

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~

mel|l[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. 4-9(c). It can be shown that the equations of motion are

m

ie: (faa

Selb 3]

(4-29)

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. 4-9. 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. 4-2 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. (4-9) for the same system, we get

parallels emcee lle 4-5

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 two-degree-of-freedom system is uncoupled by the principal coordinates {p}. The corresponding equations of motion from Eq. (4-4) are

|0

“ele

=mMy2ILP2

0

kp2JSLP2

=

0

4-30

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)

(4-31)

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. 4-3. At a given mode, the system resembles an independent one-degree-of-freedom system. Now, assume the same system is described by the generalized coordinates {q} and the equations of motion are coupled. From Eq. (4-17), the

SEC. 4-5

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,,

(4-32)

Up,} are the modal vectors for the frequencies

w, and w., respectively.

Substituting Eq. (4-31) in (4-32) and simplifying, we get | = be q2 Ur,

alles U2 ILP2

(4-33)

= {p}=[u]}‘{q}

(4-34)

where [u]' is the inverse of the modal matrix [uJ], (4-22). The transformation between the {p} and the {q} (4-34) is identical to that shown in Eq. (4-21).* 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. (4-34). This can be done, but the general theory in Sec. 6-4 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. 4-2(a) if m,=m,=m

and k,=k,=k.

Solution: From Example

1, we have u, = 1 and u,=—1. Hence Eq. (4-33) 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. (4-9) 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.

4-6

MODAL ANALYSIS: TRANSIENT VIBRATION OF UNDAMPED SYSTEMS

Consider the steps to solve the equations of motion of an undamped system. From Eq. (4-5), we get

M{q}+ Kiq}={Q(t)}

(4-35)

(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 onedegree-of-freedom system. (3) Applying the coordinate transformation in Eq. (4-34), 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. (4-35), the equations of motion of an unforced system are

M{q} + Kiq}= {0}

(4-36)

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. (4-36) and simplifying, we get [—w*M+ K {q} = {0}

(4-37)

‘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. (4-37) is a set of homogeneous algebraic equations, it possesses a solution only if the characteristic determinant A(w) is zero;

SEC. 4-6

Modal Analysis: Transient Vibration of Undamped Systems

161

that is,

A(w) =|K —w?M|=0

(4-38)

This is the characteristic of the frequency equation, which may be compared with Eq. (4-13). Previously, the frequency equation was solved by hand calculations as shown in Eq. (4-14). 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. (4-16), computers can be used to solve for {q} in Eq. (4-37). 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.

(4-23). Example 8 Find the coefficients of the frequency equation for the system shown in Fig. 4-10(a).

80 (a)

Vibratory system

Fy() 10

(c)

Fic. 4-10.

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) 1-20 —80: -200-1Ex Gs Seo & or ale,

(4-39)

or

M{x} + K{x} = {F(t)} The program COEFF to find the coefficients of the frequency equation is listed in Fig. 4-11(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.

(4-36) 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 print-out is listed in Fig. 4-11(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 self-explanatory. 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. (4-37) 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}

(4-52)

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,

(4-53)

or ie 224

mele | = F| Z22

X>

(4-54)

F,

or

Z(w){X} = {F}

(4-55)

where

Zy = (ky wm; + jac;;)

for

ie eal

lass

(4-56)

170

Systems with More Than One Degree of Freedom

CHAP. 4

and Z(w)=[z,] is the impedance matrix. In other words, Eqs. (4-51) to

(4-55) are different forms of the same equation, all of which can be summarized by Eq. (4-55). The solution {X} gives the amplitude and phase angle of the response relative to the excitation {F}. Premultiplying both sides of Eq. (4-55) by

the inverse Z(w) ' of Z(w) gives

{X}= Z(w)'{F}

(4-57)

For a two-degree-of-freedom system this can be written explicitly as ae: —

Z2F) — 212F 2

xX, _

and

7 2ZaFit Z11F2

411222

_ 2112227 2122217A (us

(4-58)

— 21222 vt sae)

Equations (4-55) to (4-57) are equally applicable to n-degree-offreedom systems. The elements of the impedance matrix Z(w) in Eq. (4-55) become Zij

=(k

ij >

wo mi + joc)

and Z(w) is of order

for

i J=

Le De seeleie

LU

(4-59)

n. Note that each z; is identical in form to the

mechanical impedance in Eq. (2-51) and Z(w) is symmetric by the proper choice of coordinates as discussed in Sec. 4-4. Example 12.

Undamped dynamic absorber

Excessive vibration, due to near resonance

conditions, is encountered

in a

constant speed machine shown in Fig. 4-16(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. 4-16.

(b)

Equivalent system

Undamped dynamic absorber: Example 12.

SEC. 4-8

Forced Vibration—Harmonic Excitation

171

Solution: The equations of motion for the equivalent system in Fig. 4-16(b) are MX, = —k1X,— k(x, — X2)+ Feq Sin wt

M2X_ = —k2(x2— X,) The impedance method can be applied harmonic. From Eq. (4-53), we have nee

directly, since the excitation

Ke

—k,

is

eee

k,— wm,

Xn

0

(a) Following Eq. (4-13), 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

(4-60)

From Eq. (4-58), the phasors of the responses are

ms

1

X,= Alo) (k2- w°m2)F.,

1

and

Xone AG)) k,F.,

(4-61)

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. (4-60) 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

(4-62)

From Eq. (4-61), the responses can be expressed as x

es

F./ky

1 ae r’

r—(2+m,/m,)r+1

Re ay F./k,

1

(4-63)

r*—(2+m,/m,)r°+1

The equations are plotted in Fig. 4-17 for m,/m,=0.3. The plus or minus sign of the amplitude ratio denotes that the response is either in-phase or 180° out-of-phase 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. (4-61) that this condition occurs when the excitation F., sin ot is balanced by the spring force —k2xp.

(c) The frequency equation, Eq. (4-62), is plotted in Fig. 4-18 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,

. 4-17. Typical harmonic response of a two-degree-of-freedom 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. (4-63) 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. 4-18.

Undamped

Example 12.

dynamic

absorber:

effect of mass

ratio m,/m,;

SEC. 4-8

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. 4-19(b). Briefly discuss the problem. Solution:

From Eq. (4-53), the equations of motion in phasor notations are er

—k,—jwc

—k2—jwce

Be

k.— w?m,+ jc

|LX,

0

From Eq. (4-13), 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. (4-58), 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. 4-19(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. 4-19.

(b)

Vibratory system

Dynamic absorber with damping: Example 13.

* J. D. Den Hartog, Mechanical

Vibrations, 4th ed., McGraw-Hill

Book Company, New

York, 1956, pp. 93-102. Note that the “tuned” condition of k,/m,=k,/m, in Exampie 12 is only for undamped absorbers. See Prob. 4-27 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. 4-20.

Example 14.

Vibration isolation: Example

14.

Vibration isolation

A constant speed machine is isolated as shown in Fig. 3-23(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. 4-20, 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. (4-53), 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. (4-58), 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. 4-9

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|

2-4r?

Mei: WHI

x Big2

i owes 1eee 6r+4r

F

eTyi

the given values,

resonance

occurs

1

are FB,eee 1-6r+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 or-r=2, X,=|X,|=0.053F,/k, and 'F;= |F;| = 0.053F.,, Hence the force transmitted is higher than that in proposal 1.

4-9

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. 4-21.

CHAP. 4

|

Method of influence coefficients.

|applied at station j, when this force is the only force applied. Consider the beam shown in Fig. 4-21. 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 reciprocity-theorem. For the general case, we have

\a; = dy |

for

ibalaese eae

(4-64)

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. 4-9

Influence Coefficients

masses. For the undamped system, we have

177

free vibration of a two-degree-of-freedom

AE.

(3

=

palleane |

dz,

é dz2JL—m2q,

4-65

or

{q}= Ld {- md}

(4-66)

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. (4-26), 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}

(4-67)

where [k,,] is the stiffness matrix. Premultiplying Eq. (4-66) 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] ==

(4-68)

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. 4-2(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. (4-66) are

{x} =[d, {-—mx} Xo

kiko+

k(k,+k,)

k

(4-69) k+k,

—M2X2

Note that the stiffness matrix from Eq. (4-9) is

ee

jee

—k

It can be shown readily that [k,;]=[d,,]"’. Moreover, the premultiplication of Eq. (4-70) by [d,]"* will give Eq. (4-9), 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. (4-69) 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,

(4-71)

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

(4-72)

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. (4-14) by Newton’s second law for the same system.

SEC. 4-9

Influence Coefficients

179 mx

(b)

\ Fic. 4-22.

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. 4-22(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. 4-22(a) and the influence coefficients

are defined in Fig. 4-22(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. (4-65) are

Sl-Las gall 6

as

dy»

-~J6

or

S l-se ea

6}

6EIL3L

pa (tied

6

SLJ6

Following the last example, this can be expressed in phasor notations as ; shown in Eq. (4-71).

6EI-20’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?

4-10

and

16.3VEI/mL?

SUMMARY

The chapter introduces the theory of discrete systems from the generalization of a two-degree-of-freedom system shown in Fig. 4-1. The equations of motion in Eq. (4-1) through (4-4) 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. 4-3 for undamped free vibrations. The natural frequencies are obtained from the characteristic equation in Eq. (4-13). 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. (4-16) and illustrated in Fig. 4-2(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. (4-17). A system can be described by more than one set of generalized

coordinates {q}. In Eqs. (4-27) through (4-29), 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. (4-34). A method for finding the modal matrix is shown in Sec. 4-6. The equations of motion can be uncoupled by means of the modal matrix. Thus, each uncoupled equation can be treated as an independent onedegree-of-freedom 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. 4-7. 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. (4-55) and the response as {X}= Z(w) ‘{F} in Eq. (4-57). The method of influence coefficients in Sec. 4-9 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. 4-1 Consider the system in Fig. 4-2(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). 4-2 Repeat Prob. 4-1 if m,;=m,=10kg,

k,=40 N/m,

k,=140 N/m, and k=

60N/m. Are the motions {x(t)} periodic? ON 200-kg uniform bar is supported by springs at the ends as illustrated in (

rig. 4-5. 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. 4-6.

4-4 For the three-degree-of-freedom

system in Fig. 4-7, if J;=2J,,

J,=2Js,

k,,=2k,2, and k,.=2k,3, find the motions 6,(t), 02(t), and 6(t). 4-5 For each of the systems shown in Fig. P4-1, 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

5-6

MYKLESTAD-PROHL

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. 5-10, a structure or a beam can be divided into segments. A typical segment of a beam, as shown in Fig. 5-12, 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 free-body sketch of a uniform beam of length L in span n as shown in Fig. 5-12(a). For equilibrium, we require

VE=VR,

and

ML=M®",-L,VE,

(5-26)

where M and V are the moment and the shear force, respectively. The symbols in this section are defined in Fig. 5-9 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.

153-162. M. A. Prohl, ““A General Method for Calculating Critical Speeds of Flexible Rotors,” Trans. ASME, vol. 66 (1945), p. A-142.

208

Methods for Finding Natural Frequencies

(a)

Free-body sketch of span

Fic. 5-12.

(b)

CHAP. 5

Free-body sketch of mass

Derivation of transfer matrix of a beam.

5-12(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. (5-26) in (5-27) and rearranging, we get L o:-08 ,+(4) ME (ie

ve,

(5-28)

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. (5-26) in (5-29) and rearranging, we obtain

Yi=¥8,+1,08,+ (5) ms,-(=) vR pal § heoae GEL)

(5-30)

The field transfer matrix is obtained by writing Eqs. (5-26), (5-28), and

SEC. 5-6

Myklestad-Prohl Method

209

(5-30) in the matrix form. I,

L2

ue

HSS

M

L3

ra

Sela

L

|e

BY

= 2EI ==

Og.

Aaa |ee US di eed

R

aaelee (5-31)

M

JS all thBAYA

n=1

To derive the point transfer matrix, consider the free-body sketch of m,,

in Fig. 5-12(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=Vi-w?m,Y,

and

M=Mi-@7J,®;

(5-32)

yi= ys

(5-33)

For the rigid body motion of m,, we have

p= DF

and

The point transfer matrix is obtained from Eqs. (5-32) and (5-33). 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. (5-31) in (5-34). 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

(5-35)

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. (5-19). Hence the problem can be solved by its repeated application as indicated in Eq.

(5-20). 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

(6-56)

If one more iteration is performed, it can be seen that the result is {V}.41= c,Aq"{u}, =A, AV}.

(6-57)

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

(6-58)

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}

(6-59)

where S, as defined above is called the sweeping matrix. The matrix H in Eq. (6-52) is modified by S, to give a new matrix H,. H,

=

HS,

(6-60)

Now, H, can be used in Eq. (6-52) for the iteration of the second mode,

since the first mode has been suppressed. Equations (6-55) to (6-57) can be used for the iterations as before. The v’s in Eq. (6-58), however, cannot be determined directly, because only {u}, is found and [u] '=[v] is as yet unknown. From the orthogonal

property in Eq. (6-35), 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

(6-61)

Thus, the sweeping matrix in Eq. (6-59) can be constructed. Example 6 Determine the natural frequencies and the modal vectors of the system in Fig. 6-5.

eae Fic. 6-5.

A three-degree-of-freedom system.

SEC. 6-9

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. (6-55),

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. (6-58). The v’s are found from Eq. (6-61) 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. (6-59) 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. (6-60), 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},= {f-1 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. (6-61) as before. Thus, the sweeping matrix S, from Eq. (6-59) 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. (6-60) 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

6-10

| 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 second-order 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. (6-37) to uncouple the equations of motion in order to describe the

SEC. 6-10

Undamped Forced Vibration—Modal Analysis

239

equations in the principal coordinates {p}, (2) employing the transformation {p(0)}=[u]” {q(0)} in Eq. (6-23) 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. (2-74), and (4) finding the response in the original coordinates {q} by the transformation {q}=[u]{p} as shown in Eq. (6-23). 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)}

(6-62)

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. (6-39) to relate {q} and the normal coordinates {n}. Thus, Eq. (6-62) yields

M[ulki}+ K[w in} = {Q(t} Premultiply this by [uw]. Note that from Eq. (6-43) [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

(6-63)

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. (6-63) can be written as

7H + w7n; = N;(t)

fora =1,2;...,7n

(6-64)

The equation above can be treated as an independent one-degree-offreedom system. From Eq. (2-71), the particular solution due to N,(t) is

ioe [ee irae (0)

where h;(t) is the impulse response of the system in Eq. (6-64). From Eq. (2-69), 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

(6-65)

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. (6-39), we get

{n(O)}}=[u] {q(0)}

= and = {n(0)}=[uT '{4(0)}

(6-66)

Let n;, and 7;, be the initial conditions in the normal coordinates for the ith mode. Hence the complementary solution of Eq. (6-64) due to the initial conditions is

1 n;(t) = Nip COS @t +— io SiN ajt

(6-67)

The complete solution for the ith mode is the sum of the responses due to (1) the excitation in Eq. (6-65) and (2) the initial condition in Eq.

(6-67). As indicated in Eq. (2-74), 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. (6-39). Let ni(t) be the sum of the responses from Eqs. (6-65) and (6-67). Thus,

aO}= Liukin)

or

{a(0}=(un}

(6-68)

Since the transient excitation {Q(t)} is arbitrary, only the formal solution, as shown in Eq. (6-68), 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.

6-11

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}

(6-69)

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. 6-12

Orthogonality of Modes of Damped Systems

generalized coordinates. Proportional damping* matrix D can be expressed as D=aM+

241

occurs if the damping

BK

(6-70)

where a and 8B are constants. To uncouple the equations, we substitute Eq. (6-70) and the transformation {q}=[]{n} from Eg. (6-39) in (6-69).

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. (6-43), we get

{H}+[al+ BAKh}+ A{n}=[p]{O(D}={N(O}

and

(6-71)

OT

H, + (a + Bw7)H; + @77; = N;(t)

[Ota

= oe th

(6-72)

Hence the equations are uncoupled. Equation (6-71) can be treated as a set of one-degree-of-freedom systems.

6-12

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 first-order 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}

(6-73)

The reduced equations are formed by introducing a 2nX1

state vector

o-[4]

{y}.

(6-14)

q

It can be shown that Eq. (6-73) 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.

(6-75)

242

Discrete Systems

CHAP. 6

or

Aty}+ Bty} = {0}

(6-76)

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. (6-73) is reduced to a set of 2n first-order equations shown in Eq. (6-76). It can

be expressed in a more convenient form by premultiplying by A~* and

defining

H=—A~'B. Thus,

(6-77)

{y}— H{y}= {0}

The eigenvalues are obtained by assuming the solution of Eq. (6-77) is of the form

tyre"

(6-78)

where y is a complex number and {WV} a 2n X 1 modal vector with complex elements. Substituting Eq. (6-78) in (6-77) yields

[yIl— H]{¥} = {0}

(6-79)

Hence the characteristic equation is

A(y) =|yI-— H|=0

(6-80)

This may be compared with Eq. (6-21) 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. (6-79) 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 |

(6-81)

Let us show that the modal vectors {VW} are orthogonal relative to the

matrices A and B. Substituting Eq. (6-78) in (6-76) 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. 6-13

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

(6-82)

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

(6-83)

for

EFAS

(6-84)

that

[Vv], B{V}, =0

The last two equations are the orthogonal relations for systems with viscous damping. They may be compared with Eqs. (6-35) and (6-36). It is implicit that Eqs. (6-83) and (6-84) 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]=F-B_]

(6-85)

where [~A~] and [~BY] are diagonal. This may be compared with Eq. (6-37) for undamped systems. If the matrices A and B are diagonalized as shown above, then the equations of motion in Eq. (6-76) are uncoupled correspondingly.

6-13

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. (6-73) becomes

M{q}+ Ciq}+ Kig} ={Q(t)} As shown in Eqs. (6-75) and (6-76), this can be reduced to a set of 2n

CHAP. 6

Discrete Systems

244

simultaneous first-order 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. (6-43).

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. 6-8 for the system shown in Fig. 4-11, that is,

° offs] 2

0

00.

O7F%X:

cP

2ILs.

OR

6-10 Determine shown

the modal

pales

pel Cea Up | p55 lan

F,(t)

lex;

F;(t)

matrix of the eee

in Fig. P5-3(a).

Assume

J,=J,=6,

system with fixed ends as

J,;=9,

k,=6,

and

k,,=k,.=

k,3= 18. 6-11 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.

6-12 Find the natural frequencies and the modal matrices of each of the systems in the following figures by matrix iteration: (a) A two-mass system in Fig. 4-2(a). Assume 20 kN/m, k,=2kN/m, and k=4kN/m. (b) A

double

pendulum

shown

m,=4kg,

in Fig. P4-1(a).

Assume

m.=1kg,

k,=

m,=m,,

and

L,= Ly. (c) The torsional system in Fig. P5-3(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. 4-9. 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

6-13 Referring to the equations of motion in Prob. P6-8, 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.

6-14 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. (6-86). 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:

6-15 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. (6-10), the equations become {g}+ S{g}= {0}, where S is symmetric. The form of [t] is as shown in Eq. (6-13). (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

6-16 Use

the program ITERATE listed in Fig. 9-14(a) to find the natural frequencies and the modal matrices of each of the systems described in Prob. 4-35.

6-17 Rewrite the program ITERATE listed in Fig. 9-14(a) as a subroutine. Repeat Prob. 6-16, using this subroutine.

6-18 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. 6-17 to find the natural frequencies and the modal

matrix, and (c) to express the modal

matrix in the original coordi-

nates.

6-19 Referring to the cantilever shown in Fig. 9-12, write a program (a) to formulate the problem by the method of influence coefficients, and (b) to use the subroutine in Prob. 6-17 to find the natural frequencies and the modal matrix.

6-20 Rewrite the subroutine $CMODL

listed in Fig. C-11 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

6-21 Modify the program in Prob. 6-20 to obtain the print-outs 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. (6-85).

(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 print-out.

6-22 Use the program TRESPDAM in Fig. 9-15(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 print-out 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

6-23 Modify the program TRESPDAM

CHAP. 6

in Fig. 9-15(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. 9-5(a) to plot the results. 6-24 Modify the program TRESPDAM in Fig. 9-15(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. 9-5(a) to plot the results.

7 Continuous Systems

7-1

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 boundary-value 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. One-dimensional continuous systems are treated in some detail.

7-2

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. 7-1.

CHAP. 7

(c)

dx Element

Initial deflection

Lateral vibration of a flexible taut string.

in Fig. 7-1(a). The lateral deflection u along the string is a function of the

space variable x and time t.

u=u(x,t)

(7-1)

The free-body sketch of an element dx of the string is shown

in Fig.

7-1(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

(7-2)

Substituting 6 =0u/dx and simplifying, the equation becomes Faas a a?

(7-3)

where c*= T/m. This is a one-dimensional wave equation. The

solution

of Eq. (7-3) 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

(7-4)

Two of the constants are specified by Eq. (7-4); the other two by the initial conditions u(x,0) = f(x)

and

oe(x,0) = g(x)

(7-5)

For example, if the string is plucked at some intermediate point as shown in Fig. 7-1(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. (7-2). (2) The resulting equation is a partial differential equation, Eq. (7-3). (3) The string problem has four arbitrary constants to be evaluated by the boundary conditions, Eq. (7-4), and the initial conditions, Eq. (7-5). (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. 7-2

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. (7-3) is of the form

u(x,t) = b(x)sin(wt +p)

(7-6)

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. (7-6) in (7-3) and factoring out the sine term, we get

ae tz O(x) =0

(7-7)

a a o(x) | w?

The solution of this equation is o(x)=A sin ae

cos

(7-8)

Combining this with Eq. (7-6) gives the formal solution u(x,t) = (4sin Be B cos =) snot + w) The boundary condition u(0,t)=0 requires that

(7-9)

B=0. The condition

u(¢,t)=0 gives wot

sin a

0

ot

or

—=

i

for

Carle 2 eee

(7-10)

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. (7-10).* 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;,)

(7-11)

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. (7-11) 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)

(7-12)

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. (7-5) yields

u(x,0) = y D, sin _ = f(x) a ou

- (x,0)

—

(6) =

(7-13)

»: 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

|

|

(7-14)

(6;

C, and D, are found by equating the coefficients in Eqs.

(7-13) and (7-14), that is, Cw,;=g, and D,=f, Since C,; and D, are associated with the vibratory motion at w,, in Eq. (7-12) the initial conditions determine the contribution of each principal mode to the vibration of the system.

7-3

SEPARATION OF THE SPACE VARIABLES

TIME

AND

Many methods can be used to solve the wave equation shown in Eq.

(7-3). 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. (7-3) 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. (7-3) be satisfied by the functions of the form

u(x,t) = o(x)q(t)

(7-15)

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. (7-15) in (7-3) and simplifying, we obtain

ldb_1aq ob dx? q dt?

(7-16)

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. 7-3

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

(7-53)

(A) $,M(¢,) da=0

(7-54)

z

—=

-

Since A,#A,, we deduce that |o,M(¢,) da=0

for

r#zS8

(7-55)

for

r#Ss

(7-56)

>3

Similarly, it can be shown that |o,L(¢,) da=0 z

The orthogonal relations are as shown in Eqs. (7-55) and (7--56). 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. 7-8

Orthogonality

Case 1.

Boundary Conditions Independent of

273

(1) Consider the wave equation. Substituting Eq. (7-49) in (7-53) 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

(7-57)

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. (7-25) and (7-30) are of the form.*

ad + bd'=0 where-a

(7-58)

and b may depend on the boundary points.

From Eggs. (7-55) and (7-56), the orthogonal relations of the wave

equation are

£,

|m(x)o,o, dx = 0

for

rz 8

(7-59)

for

#5

(7-60

‘0

| koi; dx—Lkd,bsf=0 €

0

Neglecting the spring load, Eq. (7-60) becomes €

|ko/o, dx =0

for

r#8

(7-61)

‘0

Note that k = T for the vibrating string, k = AE in Eq. (7-23) for the longitudinal vibration of a bar, and k=I,G in Eq. (7-28) for the torsional oscillation of a rod. (2 Ne Consider the beam equation. Substituting Eq. (7-50) in (7-53) 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, (7-62) *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. (7-62) is identically zero if the boundary conditions at each end can be expressed as* ad + b(kh")' =0

(7-63)

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. (7-36) are described by Eq. (7-63). The

orthogonal relations for the beam

equation

from Eqs. (7-55)

and (7-56) are {m(x)d,, dx =0

for

r#S

(7-64)

for

ré#s

(7-65)

‘0

[-kolo! dx +[6,(kb)'—kobby]o=0 0

Neglecting the spring load, Eq. (7-65) becomes [kor: dx =0

for

r#S

(7-66)

where k is the flexural stiffness EI of the beam. Example 7.

Uniform Free-free Beam

Components

of machines

are often tested

under

free-free

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 free-free beam.

Solution: Let the lateral deflection be u(x,t)= (x)q(t). Comparing the beam equation in Eq. (7-34) and the generalized equation in Eq. (7-44), we have

M=m

and L = EId*/dx*. Thus, the generalized equations in Eqs. (7-47)

and (7-48) 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. (7-36) are

(1) (3) * Tong, ibid, p. 269.

¢",-0=0 6" 2,=0"

(2) 4)”

second

SEC. 7-8

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 free-free 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 single-span uniform beam for different end conditions are shown in Fig. 7-9. The corresponding equations are summarized in Table 7-1. The roots B¢ of the frequency equations are listed in Table 7-2.

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.

(7-59) 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. 7-3(b). The boundary conditions are as given in Eq. (7-25). 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. 7-9. Modes of vibration of single-span 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 — @)

(7-67)

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

sg-f=O

s

fos i

[80-6 ae=1

For any arbitrary function f(x), this gives

| 8-Ofl) ar=110 an

Eq. (2-64) we have

SEC. 7-8

Orthogonality

Taste 7-1. Beam.

END

if

277

Summary of Equations for Lateral Vibration of Single-span 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 7-2.

Roots B€ of Frequency Equation for Single-span 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 7-1. 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. (7-59) is

[om +m,6(x — €)]6,d, dx = |mo,o, dx + m.,(€) ,(¢) =0

for

r#zS

(7-68)

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. (7-56) is not affected by m,, the corresponding orthogonal relation remains as shown in Eq. (7-60) or (7-65). Example 8 A long uniform rod with an attached mass m, at x =@ is shown in Fig. 7-3(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. (7-15). From Eq. (7-18), we get 5)

BE) = Cusin

@

X cia, 008%

q(t) = C; sin wt + C, cos wt

SEC. 7-8

Orthogonality where

c’=E/p,

p=mass/volume

279

of the rod, and

C’s are

constants.

From Eq. (7-25), 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.

==

(7-71)

SEC. 7-9

Lagrange’s Equations

281

where é

mij =| mx); dx

(7-72)

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. (7-55), (7-59), and (7-64). If the system has a mass m, attached at x=, from Eq. (7-68), the generalized mass m,; in Eq. (7-71) is mi = |m(x)o7 dx + ma; (€)

(7-73)

Similarly, the operator L describes the elastic properties of the system. Hence the potential energy function U is U=; |uL(u) dx

(7-74)

Substituting u(x,t) from Eq. (7-69) and noting the orthogonal property in Eq. (7-56), we have

bat( ll

a) dx

ai,| $,L(¢;) dx

4M ims

1ay8

qi ki Somes Pedlieds. Vie NIP tes ih

(7-75)

where é k= |:L(¢;) dx

(7-76)

and k;; is the generalized stiffness for the ith mode. This may be compared with the orthogonal relations in Eqs. (7-56), (7-60), and (7-65). Furthermore, the expressions for the kinetic energy T and the potential energy U

in Eqs. (7-71) and (7-75) may be compared with the corresponding expressions in Eq. (6-43) 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;

(7-77)

Substituting Eqs. (7-71) and (7-75) into (7-77), we obtain mig: + kiiq; = 9

(7-78)

qi + w; Gi =)

(7-79)

Or

282

Continuous Systems

CHAP. 7

where w; = k;,/m,;. This was anticipated, since T and U in Eqs. (7-71) and

(7-75) 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. (7-13) and (7-14), to our present discussion. From Eq. (7-69) the initial conditions can be expressed as

(x40) = ¥ di(x)ai(0)= fo)

(7-80)

ti(x,0) = 2,b,(x)4i(0) = g(x)

(7-81)

Using the orthogonal relation in Eq. (7-68) with m, at x=, corresponding initial conditions in the normal coordinates are

the

a0)=——|[ mixiflardar+mfOo(|

(7-82)

ain=—| i mi |[" 4 mise(a)d, S ; dx tin.e(O)d ae ; 6)

(7-83)

1

“@

where m, is defined in Eq. (7-73). 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. (7-76). Except for the zero mode, neglecting the spring load as in Eq. (7-61) for simplicity, for the wave equation, we get

q,(0) =i [eres; dx

HO=— |ke'(x)6y dx il

JJij

€

(7-84)

~O

where k is as defined in Eq. (7-49) for a general problem. Similarly, neglecting the spring load as in Eq. (7-66), the beam equation gives

q;(0) == {EIf"(x)$j dx

i ;

(7-85)

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. 7-9

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. (7-36) 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. (7-72) m, =J m@; dx = a,m, a normalized mass. For a uniform beam, k,;; can be calculated from Eq. (7-66) instead of

(7-76), that is, k;, =f EI(o!)? dx. Note that (1) f (!")¢; dx =J (/)” dx from

Eqs. (7-65) and (7-66), 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 7-1, 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. (7-72) 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. (7-66), 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. (7-82) and (7-83) 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. (7-72) to find m,; and then (7-82) 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. (7-83), C, =0. Thus, 8FE

—

(-1)"

ye UCSDt==a Oia

.

@n+1)ax

ae

(2n+1)act cos ———————

2e

SEC. 7-10 7-10

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. (7-44) is M(ii)+ L(u) = F(x, t)

(7-86)

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

(7-87)

Analogous to the modal analysis of discrete systems in Sec. 6-11, Eq. (7-86) can be uncoupled by substituting Eq. (7-87) in (7-86) and using the orthogonal properties in Eqs. (7-55) and (7-56). It can be shown readily that

miGi + kgs = |F(x,t1)6,(x) dx=Q(t)

(7-88)

where Q,(t) as defined above is the generalized force for the ith mode and

m; and k;; are evaluated as shown in Eqs. (7-73) and (7-76). The complementary and the particular solutions of Eq. (7-88) can be obtained as described in Sec. 2-7. The solutions q;(t) are then substituted in Eq. (7-87) 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. (7-88) is

Q; = |[F(t)5(x — €)]b;(x) dx

(7-89)

QO; = F(t)¢;(€)

(7-90)

Since Eq. (7-88) is of the same form as (6-64), its particular solution can be found by the convolution integral as shown in Eq. (6-65). 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

(7-91)

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. (7-82) and (7-83) or as shown in Example 10. Analogous (6-67), the complementary solution of Eq. (7-88) is 1 q; = G(O)cos wt +— q,(0)sin w;t

to Eq.

(7-92)

@}

The complete solution for the ith mode (7-91) and (7-92).

|

is the superposition

1

of Eqs.

et

4p= Gi O)oos of -§(O)sin ot |

| + ol) [F(r)sin (t=) dr aa a “Ave |

(7-93)

me

This is substituted into Eq. (7-87) 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. (7-92). 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. 7-11(a) is suddenly applied to a simply-supported 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 7-1.

(x) =A; sin y; The generalized mass

and

ae

El(iz)*

for

mt*

il Pass Wea

m, from Eq. (7-72) 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. 7-11.

(b)

Moving load F,

Forced vibration of beam.

and the

SEC. 7-10

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. (7-88) gives

From Eq. (7-88), the equation of motion is Grats @: i = Q/me

where Q; is as defined above. From the convolution integral in Eq. (6-65), the particular solution for the ith mode is

a(t) = (1)

Py

mtitrw

= (-1)""Po

V2

3

TW;

{sin @;(t—7) dr i

(1-08 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. 7-11(b). Find the deflection u(x,t) of the bridge. Assume zero initial conditions. Solution:

Let us approximate the bridge as a simply-supported uniform beam of m mass/length. The concentrated force Fy is applied atx =€=vt for O=vts €. The generalized force Q, from Eq. (7-88) 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. (6-65). 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. (7-87), 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. (7-92).

7-11

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. 6-7. This section prepares the background for the Rayleigh-Ritz 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. (6-48). Let the deflection u(x,t) of a continuous system be approximated by

u(x,t) = b(x)q(t)

(7-94)

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. (7-94) in (7-70) and (7-74), the

Rayleigh’s quotient is

|WL (wh) dx

AglW(x)]=—————_ |WM (us) dx

(7-95) :

0

where L and M are linear differential operators of the variable x as defined in Eq. (7-44). 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. 7-11

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. (7-95), noting from Eq.

(7-34) 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 7-2 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

(7-96)

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

(7-97)

0

Equating the maximum potential and kinetic energies, we get Oss

w? = Te

where

(7-98)

T*,, =3/6 m(x)W? dx is called the reference kinetic energy. The

quotient in Eq. (7-98) is equivalent to Ap in Eq. (7-95).

290 Z, 7-12

Continuous Systems RAYLEIGH-RITZ

CHAP. 7

METHOD

The Rayleigh-Ritz method assumes the deflection curve (7-94) can be approximated by a finite series.

Woo) = ¥ aro)

(x) in Eq.

(7-99)

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 ndegree-of-freedom discrete system. Since (1) the assumed mode in Eq. (7-99) 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 Rayleigh-Ritz method gives a procedure to minimize the estimated frequencies. Substituting Eq. (7-99) into the Rayleigh’s quotient in Eq. (7-95), 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

(7-100)

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

(7-101)

0

Note that k,, and m, are known values for the assumed values of y’s in Eq. (7-99). The problem is selfadjoint and

Ki = kj

and

Mm, = Mj,

(7-102)

The problem is to minimize Ag above with respect to the a’s. Let dAp/da,; =O for minimum. Thus, Eq. (7-100) yields OAR _ Dr(Np/0a;)—

da;

DE

Ne(ODp/0a;) _

7

0

(7-103)

SEC. 7-12

Rayleigh-Ritz Method

291

If w* is the minimum of Ap, from Eq. (7-100) 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

(7-105)

n .

=

0a;

La] [m,y

0a;

an

Ka}

2

%

ij 2 MG;

j

l

for

Substituting Eqs. (7-104) and (7-105) in (7-103) numerator to zero for minimum, we obtain

» (ky—@?m,)a,=0

for

il. De

=

and

f=1,2,...,n

ee

> n

equating

the

(7-106)

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

(7-107)

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. (7-96) 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. (7-105), we obtain

kp =4BIC

kjk, =ORI

kag

2B

CHAP. 7

Continuous Systems

292 The maximum

reference kinetic energy from Eq. (7-97) 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. (7-105) are

m,=mO/5

my =m,=me*/6

My = me’ /7

Substituting k,, and m, in Eq. (7-107) 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 7-2 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. (7-99) 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. (7-98) needs to satisfy only the geometric boundary conditions. Example 16 Find the fundamental frequency of the wedge-shape cantilever beam shown in Fig. 7-12. 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. 7-12

Rayleigh-Ritz Method

Fic. 7-12.

293

Lateral vibration of wedge-shape 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. (7-96), the maximum potential energy U,,.x is EH?

€

Ca — ome |(€—x)*(2a,+6ax)* dx

6

_ EH :

3

REA Applying Eq. (7-105) 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

kerku-pas

kn

EH? 3

es

reference kinetic energy from Eq. (7-97) is H

é

hs oe |(€—x)(a,x°> +ax°)

dx

0

pH < et ee ae (i+ 20105403 = The partial derivatives are evaluated by Eq. (7-105) 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 (7-107). 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’.

7-13

SUMMARY

The one-dimensional 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. 7-2. (1) The motion is a function of the space and time variables. The partial differential equation in Eq. (7-3) 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. 7-3. Examples of problems governed by the wave equation are shown in Sec. 7-4. The Euler-Bernouli beam equation is discussed in Sec. 7-5. 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. 7-6. Again, the general method is the same as described above, but the procedure can be quite involved. The continuous problem is generalized in Sec. 7-7, 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. (7-49) and (7-50). The boundary conditions are generalized in Eqs. (7-51) and (7-52). The proof of orthogonality of the eigenfunctions in Sec. 7-8 essentially follows the procedure for discrete systems. The orthogonal relations for

Problems

295

the wave equation are shown in Eqs. (7-59) to (7-61) and for the beam

equation in Eqs. (7-64) to (7-66). If the system has an inertia load, the orthogonal relation is modified as shown in Eq. (7-68). The modal analysis in Sec. 7-10 closely resembles that for discrete systems. The equation of motion is first uncoupled by means of the eigenfunctions. The uncoupled equations in Eq. (7-88) 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. (7-80)

to (7-85). The

complete solution for each mode is shown in Eq. (7-93). The general solution is the superposition of the modes u(x,t)= > o;(x)q,(t). The Rayleigh’s quotient in Eq. (7-95) is an extension of Eq. (6-47) for discrete systems. The Rayleigh-Ritz method approximates a deflection curve W(x) by a finite series in Eq. (7-99), where the a’s are arbitrary. The estimated natural frequency by the Rayleigh’s method tends to be higher than the actual value. The Rayleigh-Ritz method gives a procedure to minimize the Rayleigh’s quotient and to obtain a frequency equation.

PROBLEMS 7-1 Find the motion u(x,t) for the lateral vibration of a taut string shown in Fig. 7-1 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. 8-6(a) and (b). From Eq. (8-18), we have

LA dz,

=——— AZ;

Gas

or

Zs

Ah Z

which can be integrated directly to yield Ce fad Wares

(8-19)

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. (8-18). 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. 8-3.* 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. 8-6(c). Since the ratio A,/A, is negative, Eq. (8-19) can be expressed as

Zaztt

OF

24232 9= C

(8-20)

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. (8-18) 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. 8-4

Stability of Equilibrium

(a)

Vortex: Aid imaginary

Fic. 8-7.

(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. 1-5. Hence, the resulting motion in the (z,,z,) plane is a combination of two harmonic motions, which is an ellipse as shown in Fig. 8-7(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. (8-18), we get Z,=(a+jB)z,

and

Zo=(a—jB)z,

Zi (zne: en

and

Za=(zene je

V—1 and

(8-21)

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. 8-7(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. (8-24) 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. 8-9, it is necessary to map the z,> axes in the (y,,y,) plane. From Eq. (8-26) we have

SEC. 8-4

Stability of Equilibrium

Fic. 8-9.

315

Stability of equilibrium; Example 4.

Ay2= 1.05 and —0.95. The matrix B and the similarity transformation {y}= B{z} in Eq. (8-15) can be obtained by the method shown in

Example 6, App. A. The lambda matrix [f(A)] of matrix A in Eq. (8-11) is

tay=[*

—Az,

ee]

A-Ay

The adjoint matrix F(A) of [f(A)] is

F)=|

A- ar.

i

az,

A-ay

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

(8-28)

pele SIE( IE] om

Hence the similarity transformation is*

y2

Substituting A,.=1.05

boi

baa JL22

and —0.95

Mi

B2JLZ2

in Eq. (9-28) 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. (4-16) and

(8-28), and Eqs. (4-20) and

(8-29).

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. 8-9.

8-5

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. (8-4) as ; } 2

AE Abn Bok aaFg) dx

The

x

(8-30)

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. 8-10. 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. 8-10.

Tangential line segments

Isocline method; linear system in Example 2.

SEC. 8-5

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 a-a and b-b of the trajectories in Fig. 8-10 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)

(8-31)

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

(8-32)

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. 8-11(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. 8-11(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. (8-32) 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. (8-32) 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, 8-11,

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}..

(9-36)

{V},= G{V}i-1

As shown in Eq. (6-57) 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},

(9-37)

frequency in rad/s and {V}, the corres-

SEC. 9-11

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 1D-7

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 0-0/ 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. C-9.

EQUATIONS *** (3) $CSUBN

X(10)4

YC10)

Complex

Solution of homogeneous algebraic equations

Comparing the subroutine $HOMO and Example C-3, 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. C-9(b), is the same program.

SEC. C-11 C-11

$MODL—Modal Matrix of Undamped Discrete Systems

$MODL—MODAL MATRIX DISCRETE SYSTEMS.

423

OF UNDAMPED

The subroutine $MODL finds the modal matrix of positive-definiteundamped-discrete 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. (4-36) and (4-38) are i

M{q}+ K{q} = {0}

(C-21)

A(w) =|K— w?M|=0

(C-22)

Alternatively, premultiplying Eq. (C-21) by M“* and defining Eq. (C-22) can be expressed as

H = M“'K,

A(A) =|AI- H]|=0 where A = w~. Thus, the subroutine $COEFF

(C-23) 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}

(C-24)

The displacement vector {q} at a principal mode is also a modal vector. Substituting Eq. (C-24) into (C-21) yields

[-w*M + K]{q}= {0}

(C-25)

The equation above is identical to Eq. (4-37). 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. (4-23). The equations above can be traced readily in $MODL, as listed in Fig. C-10. $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. C-10.

$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. (C-23), 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. (C-25) to yield a modal

4. The modal matrix is formed in the last DO loop.

C-12

$CMODL—MODAL MATRIX OF DISCRETE SYSTEMS WITH VISCOUS DAMPING.

The subroutine $CMODL finds the modal matrix of positive-definitediscrete systems with viscous damping, assuming the eigenvalues are complex and distinct. Except for the reduced equations, as described in Sec. 6-12, the technique for this program runs parallel to that of undamped systems. The equations of motion and the corresponding reduced equations from Eqs. (6-73), (6-75) and (6-77) are

where {y}={q

Mig} + Clq}+ K{q}={0}

(C-26)

lw cllal*Lo allele]

ea

(¥}— Hy} = 10}

(C-28)

4}, -H=[M"6 I

Ab 0

I= unit matrix of order n, H1=M 'C,

lala min yy

2 0

(C-29)

and H2=M_'K. The characteris-

tic equation from Eq. (C-28) or (6-80) is

A) where y is a complex The solution of Eq. modal vector. Thus, {q qt=yv{q qh}, that Eq. (C-26) yields

yi

0

(C-30)

root. (C-28) 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}

(C-31)

Note that this may be compared with Eq. (C-25) 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. C-12

$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. C-11. Modal matrix of discrete systems with viscous damping; roots distinct.

Let us compare the equations above with the statements in };CMODL listed in Fig. C-11.

1. the CALL $INVS statement gives M~' (MINVS). The two CALL $MPLY statements give M 'C=H1 and M 'K = H2 in Eq. (C-29). 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. (C-30).

statement

yields

the

complex

roots

y,

4. The coefficient matrix [My?+ Cy+K] in Eq. (C-31) 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

C-13

APP. C

$CRKUT—SOLUTION OF COMPLEX 1STORDER DIFFERENTIAL EQUATION. METHOD: 4TH-ORDER RUNGE-KUTTA.

The subroutine $CRKUT equation

finds the solution of a first-order differential dz pe Gz

(C-32)

Az=At(—Cz+F)

(C-33)

or

where C and F are constants. For the initial condition z,, applying the fourth-order Runge-Kutta method gives K, =At(—CXz,+F

(C-34)

K,;=At[—CX(z,+K,/2)+ F] K,=At(=CX(@,+K,)+F]

Az=(K,+2K,+2K,+K,)/6

(C-35)

Z=2,+Az

(C-36)

Let us compare the equations above with the statements in $CRKUT listed in Fig. C-12. 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. (C-34) to (C-36) can be identified readily in the DO 41 loop. $CRK UT " m0 be

x&#*

SOLUTION OF COMPLEX 1ST-ORDER DIFFERENTIAL EQUATIONS RUNGE-KUTTA METHOD. ASSUME CONSTANT FORCE. SUBROUTINE CALCULATES VARIABLE-Z FOR EVERY DT/4.

SUBROUTINE

$CRKUT

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Fic. C-12.

Solution of 1st-order differential equations.

*xx*

SEC. C-14

C-14

$PLOTF—To Plot Data from a File

$PLOTF—TO

PLOT DATA

427

FROM A FILE.

The subroutine $PLOTEF listed in Fig. C-13 pots data stored in a file. It assumes that (1) the number of variables NVAR

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INDEX

Absorber, dynamic: centrifugal pendulum, 83-86 damped, 129, 173, 186 undamped, 170-172, 185 Acceleration, 33, 34, 69 due to moving support, 98-100 SI units, 17, 59 vectorial representation of, 13 Acceleration ratio in harmonic response, 41 Accelerometer, 102-106 Adjoint matrix, 223, 225, 399-401 (see also Matrix inversion) Amplitude: complex, 14 (see also Phasor) of free vibration, 6-7, 31, 36, 78 (see also Modal vector) at resonance, 8, 40, 81 (see also Forced vibration) Analogy: electro-mechanical, 48 of rectilinear and torsional systems, 58-60 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, 262-264 boundary conditions, 210, 264 discrete system representation, 207,

213 effect of axial load, 268-270 effect of shear deformation and rotary inertia, 266-268 as eigenvalue problem, 271 frequency equation, 265, 277, 291 Lagrange’s equations, 280, 284 modal analysis, 285, 288 modes, 276 orthogonality of modes, 273-275 Rayleigh’s method, 64, 288 Rayleigh-Ritz method, 290 Bearings with elastic support, 91-93 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, 82-86 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, 281-282 Convolution integral, 55-57 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, 155-157 of nonsymmetrical matrices, 158 Coordinate transformation: of continuous systems, 280-282 of discrete systems, 159, 218, 220, 225, 244 of nonlinear systems, 308 Coordinate translation, 307

Coulomb damping, 125-126 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, 91-93 Cunningham, W. J., 388 D’Alembert’s principle, 34, 403 Damping: Coulomb, 125-126 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, 35-37, 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, 24-26 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, 431-434 complementary solution, 55, 240, 245 computer solution by impedance method, 352, 356 Runge-Kutta method, 342, 410, 426 defined, 429 general solution, 34, 431, 437 due to arbitrary excitation, 55, 240, 245 due to harmonic excitation, 42-44 graphical solution, 316-320 impedance method, 45, 86, 169 linear, with constant coefficients, 4, 429 nonlinear autonomous systems, 304 analytical solutions, 323-328 particular integral by convolution integral, 55, 120, 239, 244, 285 method of undetermined coefficients, 435-437 particular solution, defined, 438 simultaneous, linear, 438-439 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, 190-193

Dynamic absorber (see Absorber, dynamic) Dynamic (inertia) coupling, 157 Dynamic matrix, 221, 234, 374 (see also Equations of motion)

Eigenfunction, continuous systems, 255-256, 277 (see also Orthogonality) Eigenvalue, 224, 227, 230, 234, 308,

398-399 Eigenvalue problem, continuous systems, 270-271 Eigenvector (see Modal vector) Elastic (static) coupling, 156 Energy: damping, 8, 40, 123-127, 407 excitation, 3 as initial conditions, 4 kinetic and potential, 27-28

in Lagrange’s equations, 219-221, 280-282, 405-406 in Rayleigh’s method, 31-32,

194-196, 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, 27-28, 33-34, 143, 301 Equilibrium points, nonlinear systems, 309-312 Equivalent damping, 69 (see also Damping)

444

Index

Equivalent mass, 31, 32, 69-71, 218 of continuous systems, 281 Equivalent spring, 69, 72-76, 168, 218 of continuous systems, 281 Euler-Bernouli 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, 328-333 due to support (base) motion, 98, 101, 123 transient (arbitrary), 55, 118, 239, 244, 285 Expansion theorem, 230 Faddeeva, V. N. 389 Faddeev-Leverrier 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, 169-175, 356 one-degree-of-freedom systems, 34, 38, 80, 86-87, 352 impulse response, 53-55 modal analysis continuous systems, 285-288

discrete systems, damped, 243-245, 376 discrete systems, undamped, 160, 164, 238-240, 360 nonlinear systems damped, 331-334 undamped, 328-330 periodic excitation, 110-114 resonance, 8, 40, 81, 172 transient excitation, 55-58, 116-122, 238, 243 Fourier series, 109-112, 228, 256, 327 Fourier spectrum, 112-114 Frazer, R. A., 224, 241, 389 Free vibrations:

of continuous systems assumed mode method, 255 as eigenvalue problem, 270-271 Lagrange’s equations, 280-282 Rayleigh-Ritz method, 290-293 separation of variables method, 256 defined, 3 of discrete systems, damped, 241-243 of discrete systems, undamped » Dunkerley’s equation, 190-193 influence coefficient method, 175-179 Holzer method, 197 Lagrange’s equations, 219, 407 matrix iteration, 234-238 modal analysis, 145-148, 160, 165, 223-225, 232 Myklestad-Prohl method, 207 Rayleigh’s method, 193-195, 231 of nonlinear systems analytical methods, 323-328 graphical methods, 316-320 of one-degree-of-freedom systems energy method, 27-29 Newton’s second law, 33-34, 69 Rayleigh’s method, 31-32 Frequency: beat, 10 circular, 9, 35 defined, 3 fundamental, harmonics, 147 natural (see also Frequency equation)

445

Index

Frequency (continued) defined, 5-6 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

Rayleigh-Ritz method, 291 of discrete systems, 146, 161, 178,

224 of semidefinite systems, 166, 198,

234 Frequency ratio, 38 Frequency response method, 38-43, 45 (see also Forced vibrations) computer programs for, 352-360 of discrete systems, 172-175 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, 45-49, 86-87 computer applications, 352-360 for discrete systems, 169-170, 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, 411-414 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, 175-180 defined, 176

relating to stiffness matrix, 179 Initial conditions: of continuous systems, 254-256 generalized, 282 defined, 3

446

Index

Initial Conditions (continued) of discrete systems, 150-151, 165,

240, 245, 360, 378 in nonlinear analysis, 305, 319, 321, 324 of one-degree-of-freedom 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 two-degree-of-freedom systems, 174 Jacobsen, L. S., 121, 122, 319 OWES IDE Ure 1A Jump Phenomenon, 330-332

Karman, Theodore von, 130, 389 Kelly, R. D., 388 Nettera ike los iuL Kilogram, mass, defined 16 Kinetic energy (see Energy) Lagrange’s equations, 219-221,

402-407 of continuous systems, 280-284

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, 391-401

Linear differential equations, defined, 430 Linear differential operator, 431 of continuous systems, 270-271 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, 38-41, 47, 51 Mass: equivalent, 31, 69-72, 155-157, 219

generalized, of continuous systems, 281 Mass matrix, 144, 155-157, 220 Matrix:

damping, 144, 241, 243, 245 defined, 391 dynamic, 221, 234, 374 mass,

144, 155-157,

nonsymmetrical,

164, 220

157

stiffness, 144, 155-157,

164, 177,

220 types of, 391-393 Matrix algebra, elements of, 391-401 Matrix inversion, 393, 395-397 computer method, 411, 414 Matrix iteration, 234-238 computer method, 371-376 Matrix operations, 393-396 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, 285-288 of discrete systems, damped, 243, 376-380 of discrete systems, undamped, 160-165, 238-240, 360-365 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, one-degree-of-freedom systems, 4, 33, 49, 69 Modes: natural, 3-4

447

Operators (see Linear differential operators) Ordinary point, 306 Orthogonal functions, 228 Orthogonality property: of continuous systems, 272-280 boundary condition dependent

of A, 275 boundary condition independent of A, 273 of discrete systems, damped, 241-243 of discrete systems, undamped, 164, 226, 229

Overdamped systems, 7, 36

orthogonality of (see Orthogonality) principal, 145-155, 223-226 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, 277-278 Moment of inertia, 29 equivalent, 70-71 IMOErOWs Ce Le) hy Moving support, 82, 98, 101, 121, 123 Multi-degree-of-freedom systems (see Discrete systems, Forced vibrations, Free vibrations) Multi-rotor systems, 154, 165, 194, 197, 232 computer application, 365 Myklestad, N. O., 207, 211 Myklestad-Prohl method, 207 computer application, 369-371 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, 82-86 compound, 182 simple, 29-30 spherical, 26, 182 torsional, 60, 74 Period, of vibration, 3, 9 Periodic excitation, 110-114 Periodic motion, defined, 3 (see also Harmonic motion) Perturbation method, 323 Phase angle, 9, 14, 38-39, 112-113

in harmonic excitation, 87 as time delay, 104 Phase plane, 303-306 Phase spectrum, 112-114 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 Rayleigh-Ritz method, 290-294 Rayleigh’s method, 31, 193, 196

computer application, 365-369 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, 330-331 Response (see Forced vibrations, Free vibrations) Richman, G., 388 Rods: in lateral vibration (see Beams) in longitudinal vibration, 258-261 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 Runge-Kutta 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, 101-106

Self-excited oscillation, 321-323 Semidefinite systems, 165-169, 197,

206, 232-234 amplitude ratio of, 166-167, 197, 234 constraint matrix, 233 computer application, 250, 365 as continuous systems, 276-278 Separation of variables, 256-258 Shafts, rotating, 89, 91, 195 Shafts in torsional vibration: continuous systems, 261-262 as eigenvalue problem, 270 orthogonality of modes, 273, 275 discrete systems, 155, 165-168, 197, 206, 233 Shapiro, W., 389 Shock spectrum, 116-122 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, 74-75 in parallel, 73 in series, 72 equivalent mass of, 27-28 force, defined, 2 hard, 325, 328, 331 soft, 325, 330 Stability of oscillation, 301-302 analysis of, 306-316 trajectories about equilibrium, 309, Sil State variables, 202, 304 State vector, 202, 205, 241, 304, 375 Static (elastic) coupling, 156 Steady-state response, 7-8 (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, 253-256 as eigenvalue problem, 271 as nonlinear system, 303 Structural damping, 128 Subharmonic oscillations, 332-334 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, 2-4, 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, 202-207, 209 computer application, 369-371 Transient vibration (see also Forced vibration, Free vibration): computer application, 341-352, 360, 376 free vibration, 7, 36, 43 nonperiodic excitation, 52-58, 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, 13-15 differentiation of, 12 displacement, velocity, acceleration, 13, 46 multiplication of, 15 representation of harmonic functions, 12, 46, 86-87 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., 16-18 Virtual work, 403, 404 Vortex (center), as singular point, 310, 312

Wave equation, 258-262 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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