Vibrations [3 ed.] 1108427316, 9781108427319
This new edition explains how vibrations can be used in a broad spectrum of applications and how to meet the challenges
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Table of contents :
Contents
List of Examples
List of Interactive Graphics
List of Symbols
Preface to the Third Edition
1 Introduction
1.1 Introduction
1.2 A Brief History of Vibrations
1.3 About This Book
2 Modeling of Vibratory Systems
2.1 Introduction
2.2 Inertia Elements
2.3 Stiffness Elements
2.3.1 Introduction
2.3.2 Linear Springs
2.3.3 Nonlinear Springs
2.3.4 Other Forms of Potential Energy Elements
2.3.5 Summary of Equivalent Spring Constants
2.4 Dissipation Elements
2.4.1 Viscous Damping
2.4.2 Combinations of Viscous Dampers and Linear Springs
2.4.3 Other Forms of Dissipation
2.5 Model Construction
2.5.1 Introduction
2.5.2 A Few Simple Models
2.5.3 A Microelectromechanical System
2.5.4 The Human Body
2.5.5 A Ski
2.5.6 Cutting Process
2.6 Design for Vibration
2.7 Summary
Exercises
3 Single Degree-of-Freedom Systems: Governing Equations
3.1 Introduction
3.2 Force-Balance and Moment-Balance Methods
3.2.1 Force-Balance Methods
3.2.2 Moment-Balance Methods
3.3 Natural Frequency and Damping Factor
3.3.1 Natural Frequency
3.3.2 Damping Factor
3.4 Governing Equations for Different Types of Damping
3.5 Governing Equations for Different Types of Applied Forces
3.5.1 System with Base Excitation
3.5.2 System with Unbalanced Rotating Mass
3.5.3 System with Added Mass Due to a Fluid
3.6 Lagrange’s Equations
3.7 Summary of Natural Frequency Equations for Single Degree-of-Freedom Systems
3.8 Summary
Exercises
4 Single Degree-of-Freedom Systems: Free-Response Characteristics
4.1 Introduction
4.2 Free Responses of Undamped and Damped Systems
4.2.1 Introduction: Damping Cases
4.2.2 Free Response of Underdamped Systems: Kelvin–Voigt Model
4.2.3 Free Response of Underdamped Systems: Maxwell Model
4.3 Stability of a Single Degree-of-Freedom System
4.4 Single Degree-of-Freedom Systems with Nonlinear Elements
4.4.1 Nonlinear Stiffness
4.4.2 Nonlinear Damping
4.5 Summary
Exercises
5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations
5.1 Introduction
5.2 Response to Harmonic Excitation
5.2.1 Excitation Applied from t = 0
5.2.2 Excitation Present for All Time
5.2.3 Response of Undamped System and Resonance
5.2.4 Magnitude and Phase Information: Mass Excitation
5.2.5 Magnitude and Phase Information: Rotating Unbalanced Mass
5.2.6 Magnitude and Phase Information: Base Excitation
5.2.7 Summary of Results of Sections 5.2.4, 5.2.5, and 5.2.6
5.2.8 Harmonic Excitation of a System with a Maxwell Model
5.3 Response to Excitation with Harmonic Components
5.4 Frequency-Response Function
5.4.1 Introduction
5.4.2 Curve Fitting and Parameter Estimation
5.4.3 Amplitude Response Function and Filter Characteristics
5.4.4 Relationship of the Frequency-Response Function to the Transfer Function
5.4.5 Alternative Forms of the Frequency-Response Function
5.5 Acceleration Measurement: Accelerometer
5.6 Vibration Isolation
5.7 Energy Dissipation and Equivalent Damping
5.8 Influence of Nonlinear Stiffness on Forced Response
5.9 Summary
Exercises
6 Single Degree-of-Freedom Systems: Subjected to Transient Excitations
6.1 Introduction
6.2 Response to Impulse Excitation
6.3 Response to Step Input Excitation
6.4 Response to Rectangular Pulse Excitation
6.5 Response to Other Excitation Waveforms
6.5.1 Significance of the Spectral Content of the Applied Force: An Example
6.6 Impact Testing
6.7 Summary
Exercises
7 Multiple Degree-of-Freedom Systems: Governing Equations, Natural Frequencies, and Mode Shapes
7.1 Introduction
7.2 Governing Equations
7.2.1 Force-Balance and Moment-Balance Methods
7.2.2 General Form of Equations for a Linear Multi-Degree-of-Freedom System
7.2.3 Lagrange’s Equations of Motion
7.3 Free Response Characteristics
7.3.1 Undamped Systems: Natural Frequencies and Mode Shapes
7.3.2 Natural Frequencies and Mode Shapes: A Summary
7.3.3 Undamped Systems: Properties of Mode Shapes
7.3.4 Characteristics of Damped Systems
7.3.5 Conservation of Energy
7.4 Rotating Shafts on Flexible Supports
7.5 Stability
7.6 Summary
Exercises
8 Multiple Degree-of-Freedom Systems: General Solution for Response and Forced Oscillations
8.1 Introduction
8.2 Normal-Mode Approach
8.2.1 General Solution
8.2.2 Response to Initial Conditions
8.3 Response to Arbitrary Forcing and Initial Conditions: Direct Numerical Approach
8.4 Response to Harmonic Forcing and the Frequency-Response Function
8.4.1 Frequency-Response Function
8.5 Vibration Absorbers
8.5.1 Undamped Vibration Absorber
8.5.2 Damped Linear Vibration Absorber
8.5.3 Centrifugal Pendulum Vibration Absorber
8.5.4 Bar Slider System
8.5.5 Pendulum Absorber
8.5.6 Particle Impact Damper
8.5.7 Vibration Absorbers: A Summary
8.6 Vibration Isolation: Transmissibility Ratio
8.7 Systems with Moving Base
8.8 Summary
Exercises
9 Vibrations of Beams
9.1 Introduction
9.2 Governing Equations of Motion
9.2.1 Preliminaries from Solid Mechanics
9.2.2 Potential Energy, Kinetic Energy, and Work
9.2.3 Derivation of the Equations of Motion
9.2.4 Beam Equations for a General Case
9.3 Free Oscillations: Natural Frequencies and Mode Shapes
9.3.1 Introduction
9.3.2 General Solution for Natural Frequencies and Mode Shapes for Beams with Constant Cross-Section
9.3.3 Orthogonality of the Mode Shapes
9.3.4 Natural Frequencies and Mode Shapes of Constant Cross-Section Beams Without In-Span Attachments: Effects of Boundary Conditions
9.3.5 Effects of Stiffness and Inertial Elements Attached at an Interior Location
9.3.6 Effects of an Axial Force and an Elastic Foundation on the Natural Frequency
9.3.7 Tapered Beams
9.4 Forced Oscillations
9.5 Summary
Exercises
Appendices
A Preliminaries from Dynamics
B Laplace Transform Pairs
C Solutions to Ordinary Differential Equations
D Matrices
E Complex Numbers and Variables
F State-Space Formulation
G Natural Frequencies and Mode Shapes of Bars, Shafts, and Strings
H Evaluation of Eq. (9.120)
Answers to Selected Exercises
Glossary
Index
Citation preview
Vibrations
This new edition explains how vibrations can be used in a broad spectrum of applications and how to meet the challenges faced by engineers and system designers.
The
text
integrates
linear
and
nonlinear
systems
and
covers
the
time
domain
and
the
frequency domain, responses to harmonic and transient excitations, and discrete and continuous
system
models.
It
focuses
on
modeling,
analysis,
prediction,
and
measurement
to
provide a complete understanding of the underlying physical vibratory phenomena and their relevance for engineering design.
Knowledge is put into practice through numerous examples with real-world applications in a range of disciplines, detailed design guidelines applicable to various vibratory systems, and over 40 online interactive graphics which provide a visual summary of system behaviors and enable students to carry out their own parametric studies. Thirteen new tables act as a quick reference for self-study, detailing key characteristics of physical systems and summarizing important results.
This is an essential text for undergraduate and graduate courses in vibration analysis, and a valuable reference for practicing engineers.
Balakumar Balachandran is a Minta Martin Professor of Engineering at the University of
Maryland. He has authored and co-authored many books, chapters, and journal articles related to dynamics and vibrations, and he has several patents to his credit. He is a fellow of the American Society of Mechanical Engineers and the American Institute of Aeronautics and Astronautics.
Edward B. Magrab is Emeritus Professor in the Department of Mechanical Engineering at
the University of Maryland. He has extensive experience in analytical and experimental studies of vibrations and acoustics, serving as an engineering consultant to over 20 companies and authoring or co-authoring a number of books on vibrations, noise control, instrumentation, integrated product design,
Matlab,
and Mathematica. He is a Life Fellow of the
American Society of Mechanical Engineers.
Vibrations Third Edition
BALAKUMAR BALACHANDRAN
University of Maryland, College Park EDWARD B. MAGRAB
University of Maryland, College Park
University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314
–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi –
110025, India
–
79 Anson Road, #06 04/06, Singapore 079906
Cambridge University Press is part of the University of Cambridge. It furthers the University’ s mission by disseminating knowledge in the pursuit of education, learning, and research at the highest international levels of excellence.
www.cambridge.org Information on this title: www.cambridge.org/9781108427319 DOI: 10.1017/9781108615839
©
Balakumar Balachandran and Edward B. Magrab 2019
This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. This book was previously published by Cengage Learning 1980, 2008 Third edition published by Cambridge University Press 2019 Printed in the United Kingdom by TJ International Ltd. Padstow Cornwall
A catalogue record for this publication is available from the British Library. Library of Congress Cataloging-in-Publication Data Names: Balachandran, Balakumar, author.
|
Magrab, Edward B., author.
Title: Vibrations / Balakumar Balachandran (University of Maryland, College Park), Edward Magrab (University of Maryland, College Park). Description: 3rd edition. | Cambridge, United Kingdom ; New York, NY : Cambridge University Press, 2019.
|
Includes bibliographical references
and index. Identi
fiers: LCCN 2018034163 | ISBN 9781108427319 (hardback : alk.
paper) Subjects: LCSH: Vibration. Classi
|
Vibration—Mathematical models.
fication: LCC TA355 .B28 2019 | DDC 620.3—dc23
LC record available at https://lccn.loc.gov/2018034163 ISBN 978-1-108-42731-9 Hardback Additional resources for this publication at www.cambridge.org/vibrations. Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.
To Malini, Ragini, and Nitin and In memory of T. R. Balachandran (1933–2017) For June Coleman Magrab Still my muse after all these years
CONTENTS
List of Examples List of Interactive Graphics List of Symbols Preface to the Third Edition
1
2
page xii xv xvii xxiii
Introduction
1
1.1 Introduction
1
1.2 A Brief History of Vibrations
6
1.3 About This Book
8
Modeling of Vibratory Systems
11
2.1 Introduction
11
2.2 Inertia Elements
13
2.3 Stiffness Elements
18
2.3.1 Introduction
18
2.3.2 Linear Springs
20
2.3.3 Nonlinear Springs
32
2.3.4 Other Forms of Potential Energy Elements
38
2.3.5 Summary of Equivalent Spring Constants
44
2.4 Dissipation Elements
44
2.4.1 Viscous Damping
44
2.4.2 Combinations of Viscous Dampers and Linear Springs
49
2.4.3 Other Forms of Dissipation
52
2.5 Model Construction
55
2.5.1 Introduction
55
2.5.2 A Few Simple Models
55
2.5.3 A Microelectromechanical System
60
2.5.4 The Human Body
62
2.5.5 A Ski
63
2.5.6 Cutting Process
64
2.6 Design for Vibration
65
2.7 Summary
66
Exercises
67
Contents
viii
3
Single Degree-of-Freedom Systems: Governing Equations
76
3.1 Introduction
76
3.2 Force-Balance and Moment-Balance Methods
77
3.2.1 Force-Balance Methods
77
3.2.2 Moment-Balance Methods
83
3.3 Natural Frequency and Damping Factor
87
3.3.1 Natural Frequency
87
3.3.2 Damping Factor
92
3.4 Governing Equations for Different Types of Damping
96
3.5 Governing Equations for Different Types of Applied Forces
97
3.5.1 System with Base Excitation
97
3.5.2 System with Unbalanced Rotating Mass
99
3.5.3 System with Added Mass Due to a Fluid
100
3.6 Lagrange’s Equations
102
3.7 Summary of Natural Frequency Equations for Single Degree-of-Freedom Systems
4
129
3.8 Summary
135
Exercises
136
Single Degree-of-Freedom Systems: Free-Response Characteristics
148
4.1 Introduction
148
4.2 Free Responses of Undamped and Damped Systems
150
4.2.1 Introduction: Damping Cases
150
–
5
4.2.2 Free Response of Underdamped Systems: Kelvin Voigt Model
157
4.2.3 Free Response of Underdamped Systems: Maxwell Model
176
4.3 Stability of a Single Degree-of-Freedom System
178
4.4 Single Degree-of-Freedom Systems with Nonlinear Elements
182
4.4.1 Nonlinear Stiffness
182
4.4.2 Nonlinear Damping
186
4.5 Summary
187
Exercises
189
Single Degree-of-Freedom Systems Subjected to Periodic Excitations
197
5.1 Introduction
197
5.2 Response to Harmonic Excitation
200
5.2.1 Excitation Applied from
t
= 0
200
5.2.2 Excitation Present for All Time
210
5.2.3 Response of Undamped System and Resonance
214
5.2.4 Magnitude and Phase Information: Mass Excitation
217
5.2.5 Magnitude and Phase Information: Rotating Unbalanced Mass
221
Contents
ix
5.2.6 Magnitude and Phase Information: Base Excitation
226
5.2.7 Summary of Results of Sections 5.2.4, 5.2.5, and 5.2.6
230
5.2.8 Harmonic Excitation of a System with a Maxwell Model
233
5.3 Response to Excitation with Harmonic Components
238
5.4 Frequency-Response Function
248
5.4.1 Introduction
248
5.4.2 Curve Fitting and Parameter Estimation
249
5.4.3 Amplitude Response Function and Filter Characteristics
250
5.4.4 Relationship of the Frequency-Response Function to the Transfer Function 5.4.5 Alternative Forms of the Frequency-Response Function
261
5.6 Vibration Isolation
263
5.7 Energy Dissipation and Equivalent Damping
270
5.8 In
282
5.9 Summary
289
Exercises
290
Single Degree-of-Freedom Systems: Subjected to Transient Excitations
297
6.1 Introduction
297
6.2 Response to Impulse Excitation
300
6.3 Response to Step Input Excitation
310
6.4 Response to Rectangular Pulse Excitation
316
6.5 Response to Other Excitation Waveforms
322
6.5.1 Signi
ficance of the Spectral Content of the Applied Force:
An Example
7
259
5.5 Acceleration Measurement: Accelerometer
fluence of Nonlinear Stiffness on Forced Response
6
255
334
6.6 Impact Testing
338
6.7 Summary
340
Exercises
341
Multiple Degree-of-Freedom Systems: Governing Equations, Natural Frequencies, and Mode Shapes
344
7.1 Introduction
344
7.2 Governing Equations
346
7.2.1 Force-Balance and Moment-Balance Methods
346
7.2.2 General Form of Equations for a Linear Multi-Degree-of-Freedom System 7.2.3 Lagrange’s Equations of Motion 7.3 Free Response Characteristics
356 359 378
7.3.1 Undamped Systems: Natural Frequencies and Mode Shapes
378
7.3.2 Natural Frequencies and Mode Shapes: A Summary
402
Contents
x
8
7.3.3 Undamped Systems: Properties of Mode Shapes
402
7.3.4 Characteristics of Damped Systems
411
7.3.5 Conservation of Energy
421
7.4 Rotating Shafts on Flexible Supports
423
7.5 Stability
434
7.6 Summary
441
Exercises
442
Multiple Degree-of-Freedom Systems: General Solution for Response and Forced Oscillations
458
8.1 Introduction
458
8.2 Normal-Mode Approach
460
8.2.1 General Solution
460
8.2.2 Response to Initial Conditions
465
8.3 Response to Arbitrary Forcing and Initial Conditions: Direct Numerical Approach 8.4 Response to Harmonic Forcing and the Frequency-Response Function 8.4.1 Frequency-Response Function 8.5 Vibration Absorbers
9
472 475 475 489
8.5.1 Undamped Vibration Absorber
489
8.5.2 Damped Linear Vibration Absorber
492
8.5.3 Centrifugal Pendulum Vibration Absorber
504
8.5.4 Bar Slider System
508
8.5.5 Pendulum Absorber
511
8.5.6 Particle Impact Damper
515
8.5.7 Vibration Absorbers: A Summary
525
8.6 Vibration Isolation: Transmissibility Ratio
525
8.7 Systems with Moving Base
536
8.8 Summary
538
Exercises
539
Vibrations of Beams
545
9.1 Introduction
545
9.2 Governing Equations of Motion
547
9.2.1 Preliminaries from Solid Mechanics
548
9.2.2 Potential Energy, Kinetic Energy, and Work
550
9.2.3 Derivation of the Equations of Motion
557
9.2.4 Beam Equations for a General Case
559
9.3 Free Oscillations: Natural Frequencies and Mode Shapes 9.3.1 Introduction
566 566
Contents
xi
9.3.2 General Solution for Natural Frequencies and Mode Shapes for Beams with Constant Cross-Section 9.3.3 Orthogonality of the Mode Shapes
570 580
9.3.4 Natural Frequencies and Mode Shapes of Constant Cross-Section Beams Without In-Span Attachments: Effects of Boundary Conditions
583
9.3.5 Effects of Stiffness and Inertial Elements Attached at an Interior Location
599
9.3.6 Effects of an Axial Force and an Elastic Foundation on the Natural Frequency 9.3.7 Tapered Beams
616 617
9.4 Forced Oscillations
622
9.5 Summary
639
Exercises
640
Appendices A
Preliminaries from Dynamics
647
B
Laplace Transform Pairs
661
C
Solutions to Ordinary Differential Equations
669
D
Matrices
679
E
Complex Numbers and Variables
683
F
State-Space Formulation
688
G
Natural Frequencies and Mode Shapes of Bars, Shafts, and Strings
695
H
Evaluation of Eq. (9.120)
705
Answers to Selected Exercises
708
Glossary
714
Index
719
EXAMPLES
page
2.1
Determination of mass moments of inertia
2.2
Slider mechanism: system with varying inertia property
17
2.3
Equivalent stiffness of a beam-spring combination
28
2.4
Equivalent stiffness of a cantilever beam with a transverse end load
29
2.5
Equivalent stiffness of a beam with a
fixed end and a translating support at
the other end 2.6
16
30
Equivalent stiffness of a microelectromechanical system (MEMS)
flexure
fixed-fixed 31
2.7
Equivalent stiffness due to gravity loading
2.8
Equivalent damping coef
43
2.9
ficient and equivalent stiffness of a vibratory system Equivalent linear damping coeffi cient of a nonlinear damper
3.1
Wind-driven oscillations about a system’s static-equilibrium position
81
3.2
Eardrum oscillations: nonlinear oscillator and linearized systems
82
3.3
Hand biomechanics
85
3.4
Natural frequency from static de ection of a machine system
3.5
Static de ection and natural frequency of the tibia bone in a human leg
89
3.6
System with a constant natural frequency
90
3.7
Effect of mass on the damping factor
94
3.8
Effects of system parameters on the damping factor
94
3.9
fl
fl
50 51
88
Equation of motion for a linear single degree-of-freedom system
105
3.10
Equation of motion for a system that translates and rotates
106
3.11
Governing equation for an inverted pendulum
108
3.12
Governing equation for motion of a disc segment
111
3.13
Governing equation for a translating system with a pretensioned or precompressed spring
114
3.14
Using negative stiffness to isolate a car seat
115
3.15
Equation of motion for a disc with an extended mass
117
3.16
Lagrange formulation for a microelectromechanical system (MEMS) device
119
3.17
Equation of motion of a slider mechanism
122
3.18
Oscillations of a crankshaft
124
3.19
Vibration of a centrifugal governor
127
3.20
Oscillations of a rotating system
128
4.1
Free response of a microelectromechanical system
153
4.2
Free response of a car tire
154
Examples
xiii
4.3
Free response of a door
155
4.4
Impact of a vehicle bumper
167
4.5
Impact of a container housing a single degree-of-freedom system
169
4.6
Collision of two viscoelastic bodies
172
4.7
Estimate of damping ratio using the logarithmic decrement
173
4.8
Inverse problem: information from a phase-plane plot
174
4.9
Instability of an inverted pendulum
181
5.1
Estimation of system damping ratio to tailor transient response
207
5.2
Start-up response of a
5.3
Forced response of a damped system
212
5.4
Forced response of an undamped system
215
5.5
Fraction of applied force that is transmitted to the base
224
5.6
Response of an instrument subjected to base excitation
229
5.7
Single degree-of-freedom system subjected to a periodic pulse train
243
5.8
Damping ratio and bandwidth to obtain a desired
5.9
flexibly supported rotating machine
Q
208
254
Frequency-response function of a tire for pavement design analysis
256
5.10
Design of an accelerometer
263
5.11
Design of a vibration isolation mount
267
5.12
Transmissibility of a
5.13
Vibratory system with structural damping
5.14
Estimate for response amplitude of a system subjected to
floating vibration isolation system
268 281
fluid damping
282
6.1
Response of a linear vibratory system to multiple impacts
306
6.2
Use of an additional impact to suppress the transient response
307
6.3
Stress level under impulse loading
309
6.4
Vehicle response to a step change in the road pro
6.5
Response of a slab
6.6
Response to half sine pulse base excitation
329
6.7
Single degree-of-freedom system with moving base and nonlinear spring
331
7.1
Modeling of a milling machine on a
7.2
Conservation of linear momentum in a multiple degree-of-freedom system
350
7.3
System with bounce and pitch motions
351
7.4
Governing equations of a rate gyroscope
354
7.5
System with a translating mass attached to an oscillating disc
363
7.6
System with bounce and pitch motions revisited
364
7.7
Pendulum absorber
366
7.8
Bell and clapper
368
7.9
floor to transient loading
file
flexible floor
314 327
348
Three coupled nonlinear oscillators: Lavrov’s device
370
7.10
Governing equations of a rate gyroscope revisited
372
7.11
Governing equations of hand-arm vibrations
373
7.12
Natural frequencies and mode shapes of a two degree-of-freedom system
382
7.13
Rigid-body mode of a railway car system
389
xiv
Examples 7.14
Natural frequencies and mode shapes of a two-mass-three-spring system
7.15
Natural frequencies and mode shapes of a pendulum attached to a translating mass
390
392
7.16
Natural frequencies and mode shapes of a system with bounce and pitch motions
395
7.17
Inverse problem: determination of system parameters
399
7.18
System of oscillators attached to a single degree-of-freedom system
400
7.19
Orthogonality of modes, modal masses, and modal stiffness of a spring-mass system
410
7.20
Nature of the damping matrix
417
7.21
Free oscillation characteristics of a proportionally damped system
418
7.22
Free oscillation characteristics of a system with gyroscopic forces
420
7.23
Conservation of energy in a three degree-of-freedom system
423
7.24
Stability of an undamped system with gyroscopic forces
434
7.25
Wind-induced vibrations of a suspension bridge deck: stability analysis
435
7.26
Disc brake squeal
437
8.1
Undamped free oscillations of a two degree-of-freedom system
467
8.2
Damped and undamped free oscillations of a two degree-of-freedom system
468
8.3
Interaction of a rigid structure with its soil foundation
485
8.4
Absorber for a diesel engine
491
8.5
Absorber design for a rotating system with mass unbalance
502
8.6
Absorber design for a machine system
504
8.7
Design of a centrifugal pendulum vibration absorber for an internal combustion engine
508
8.8
Design of machinery mounting to meet a transmissibility ratio requirement
535
9.1
Boundary conditions for a cantilever beam with an extended mass
564
9.2
Beams with attachments: maintaining a constant
9.3
Determination of the properties of a beam supporting rotating machinery
614
9.4
Natural frequencies of a tapered cantilever beam
619
9.5
Mode shape of a baseball bat
620
9.6
Impulse response of a cantilever beam
629
9.7
Frequency-response functions of a beam
631
9.8
Use of a second impact to suppress the transient response
633
9.9
Moving load on a beam
635
first natural frequency
597
INTERACTIVE GRAPHICS
Interactive graphics are available online www.cambridge.org/vibrations
1.1
Normalized response of a spring-mass-damper system to various excitations and damping
flection of two springs in series flection of two springs in parallel and their equivalent stiffness
page
10
2.1
De
23
2.2
De
25
2.3
Geometric nonlinear stiffness of a combination of a vertical spring and an inclined spring
2.4
36
figurations for zero
Geometric nonlinear stiffness of three springs and con stiffness
38
3.1
Animation of the system shown in Figure 3.7
100
3.2
Animation of the system shown in Figure 3.9
106
3.3
Animation of the system shown in Figure 3.11
111
3.4
Animation of the system shown in Figure 3.13
116
3.5
Animation of the system shown in Figure 3.14
117
3.6
Animation of the system shown in Figure 3.16
122
3.7
Animation of the system shown in Figure 3.17
125
4.1
Response to an initial velocity: damping factor comparisons
153
4.2
Displacement, velocity, and acceleration response variations with respect to initial conditions
161
4.3
Logarithmic decrement and the determination of the damping factor
163
4.4
Phase-plane plot, initial conditions, and the extrema values indicated in Table 4.1
4.5
165
–
Maxwell and Kelvin Voigt models: comparison of force transmitted to the base
4.6
Comparison of responses of systems with linear springs, linear elastic stops, and nonlinear springs
5.1
178
184
Amplitude and phase response functions and response to suddenly applied sine wave or cosine wave
207
5.2
Harmonic excitation with time-dependent frequency
209
5.3
Amplitude and phase response functions for mass excitation, unbalanced mass excitation, and moving base excitation
5.4
232
Phase relations amongst the excitation, displacement, and velocity for three types of excitation
232
xvi
Interactive Graphics 5.5
Comparison of the amplitude response and phase response for systems with
–
Kelvin Voigt and Maxwell models 5.6
237
Responses to different periodic waveforms in the time domain and the frequency domain
247
filter
5.7
Amplitude response function as a
5.8
Transmissibility ratio for a system with a Kelvin Voigt model and a
254
–
system with a Maxwell model
266
Hysteresis loops for different dissipation models
278
Amplitude response of a system with cubic hardening spring
285
6.1
Response to a rectangular pulse: convolution integral based approach
305
6.2
Spectral content of aperiodic and periodic waveforms
320
6.3
Response to transient excitation: time and frequency representations
327
6.4
Response of system with a nonlinear spring subjected to transient base
5.9 5.10
excitation
334
6.5
Effect of natural frequency and input force spectrum on system response
338
7.1
Natural frequencies of translating two degree-of-freedom systems
387
7.2
Two degree-of-freedom systems with translation: natural frequencies and mode shapes
7.3
frequencies and mode shapes 8.1
398
Two degree-of-freedom system: displacement responses of masses to initial conditions
8.2
391
Two degree-of-freedom systems with translation and rotation: natural
473
Two degree-of-freedom system: displacement responses of masses to transient forcing
474
8.3
Two degree-of-freedom system: frequency-response functions
483
8.4
Frequency-response function for an optimal choice of parameters
500
8.5
Particle impact damper
525
8.6
Transmissibility ratio (TR) variation with respect to system parameters
533
9.1
Natural frequencies, mode shapes, and node points of beams with various boundary conditions
599
9.2
Natural frequencies and mode shapes of beams with in-span attachments
610
9.3
First mode shape of a double-tapered free-free beam
622
9.4
Response of a cantilever beam to an impulse force
631
SYMBOLS
ai, bi a n, a n, b a aG a(t), a(τ ) ass(t) c, cj cb cc cd ceq cjn cr ct c ci(Ωi) 1
2
Fourier series coef
n,
1
32
b
n,
2
ficients
boundary condition parameters for beam acceleration vector absolute acceleration of center of mass of a system acceleration steady-state portion of
a (t)
ficient, translation motion
damping coef
longitudinal speed in beam
ficient fluid damping coefficient critical damping coef
equivalent viscous damping
ficient
damping coef
pulse duration-bandwidth product
ficient, rotational motion ratio of damping coeffi cients, c / c coeffi cients in response of a single degree-of-freedom damping coef
3
2
system to periodic forcing
e,e f(t), f (τ) f(x,t), f (η, τ) fn fnc g h i, j, k k, kj , ks ke kf kjn kt, ktj k m, mi , Ms me mo 1
32
2
fixed unit vectors
body-
external forcing external forcing on beam natural frequency, Hz non-conservative force per unit length on a beam gravitational constant elevation from ground or thickness of a beam unit vectors
fixed in inertial reference plane
translation spring constant, spring stiffness equivalent spring constant stiffness of elastic foundation
ficient
stiffness coef
torsion spring constant stiffness ratio,
k /k 3
2
mass of a particle or a rigid body equivalent mass mass of a beam
xviii
Symbols
mr p( x, t) p, pi qj q_ j r
mass ratio,
m /m 2
1
axial force on beam linear momentum vector generalized coordinate generalized velocity radius of gyration of beam cross-section
r, ri u( t) s, sj
position vector unit step function Laplace transform parameter, roots of a polynomial in the parameter
t to
s
time initial time, time delineating a characteristic of a waveform, characteristic time of a beam
tr v v( t), v(τ ) vss( t) w(x,t), w(η , τ) xð0Þ; x_ ð0Þ; xj ð0Þ; x_ j ð0Þ x(t), x(τ ), xj(t) x xo xss(t) x_ ðtÞ; x_ ðτÞ; x_ j ðtÞ x€ðtÞ; x€ðτÞ; € xj ðtÞ y( t) €yðtÞ z( t) A Ao [ A] Bj Bw Cn [ C] D D (Ω) E Ed, Ediss ET E(ω) max
rise time velocity vector velocity steady-state portion of
v(t)
transverse displacement of beam initial displacement, initial velocity displacement magnitude of
x(τ m)
static-equilibrium position steady-state portion of
x(t)
velocity acceleration displacement of base or nondimensional displacement acceleration of base relative displacement between mass and base cross-sectional area of beam magnitude of displacement due to initial conditions state matrix nondimensional torsion stiffness bandwidth of a
filter
beam eigenfunction parameter damping matrix Rayleigh dissipation function denominator of
H(Ω)
Young’s modulus dissipation energy total energy in a signal signal energy as a function of frequency
Symbols
F, F i Fs FT ( t) Fo F(x) F(s) G GB G , GL Gij (jΩ) 0
force vector internal force vector force transmitted to the base magnitude of
f(t)
spring force Laplace transform of forcing shear modulus beam energy function beam energy function at boundaries frequency-response function of inertial element applied to inertial element
G(Ω) H(Ω) Hij ( Ω)
i to force
j
frequency-response function amplitude response function, single degree-of-freedom system amplitude response function of inertial element applied to inertial element
Hst( Ω)
f(t) or f(τ)
i to force
j
amplitude response function, system with structural damping
Hmb(Ω) Hub(Ω)
amplitude response function, system with base excitation amplitude response function, system with rotating unbalanced mass
H I J JG JO Jj Jo Kf Kj Ks [K ] L LT M M M(t) Mjn Mj Mo Mso [M]
angular momentum moment of inertia of beam cross-section about bending axis mass moment of inertia about axis of rotation mass moment of inertia about center of mass mass moment of inertia about point mass moment of inertia of mass moment of inertia of
“O”
mj mo
nondimensional stiffness of elastic foundation nondimensional spring stiffness stiffness ratio,
k sL /(EI) 3
stiffness matrix length of an element Lagrangian for a beam net moment about a magnitude of
fixed point or the center of mass
M
external moment
ficient
inertia coef
mass attached to beam boundaries mass attached to beam mass ratio,
Ms /m o
inertia matrix, mass matrix
xix
xx
Symbols
Nn Nd ^ P Po Q Q (ϕ) Qi R R (ϕ) S , Sj S (ϕ) T T(ϕ) Td TR To, T U V Vo W W(x), W(η ) W(0), W′ (0), W″ (0), W″′(0) Wn(η ) X (s) Xj X ij Xo { X} { X} j 1
α
square of the norm of the beam eigenfunction number of periods to reach
τd
nondimensional axial force on beam percentage overshoot quality factor spatial beam function generalized force reduction in transmissibility spatial beam function sensitivity spatial beam function period of oscillation at frequency
ω,
kinetic energy
spatial beam function period of oscillation at frequency
ωd
transmissibility ratio spring tension potential energy of beam potential energy, shear force on beam initial velocity work transverse displacement of a beam
beam shape function and its derivatives evaluated at
η
= 0
beam mode shape Laplace transform of
x(t) or x(τ )
magnitude of displacement response to harmonic force elements of {
X}
initial displacement displacement column vector
jth mode shape corresponding to Ωj
fi
fi
nonlinear spring stiffness coef cient, coef cient in proportional damping matrix
α β
angular acceleration vector
fi
structural damping constant, coef cient in proportional damping matrix
δ δ nm
logarithmic decrement Kronecker delta
δ (t) δ st
static displacement of a spring
ε
coef cient of restitution, percentage error
delta function
fi
Symbols
ζ ζj η
fηðtÞg; fη_ ð tÞ g; f€ ηðtÞ g θ; θ_ ; € θ
damping ratio, damping factor modal damping factor nondimensional beam coordinate,
x/L
modal amplitude or displacement, modal velocity, modal acceleration column matrices angular displacement, angular velocity, angular acceleration
θ( Ω) θt(Ω)
phase angle response to harmonic excitation, transient
θst (Ω)
phase angle response to harmonic excitation, system with
phase angle response to harmonic excitation, steady state
structural damping
κ λj μ
curvature roots of a polynomial in the parameter coef
λ
ficient of friction, kinematic viscosity, overlap factor in
turning
ρ σ τ τd
mass density bending stress in a beam nondimensional time,
ωnt
or
ωn1t
nondimensional time it takes for a system to decay to a
fied level
speci
τm τr
_ ;ϕ € ϕ; ϕ
nondimensional time at which
x(τ ) is a maximum
nondimensional rise time angular displacement, angular velocity, angular acceleration
φ
phase angle associated with
φd ψn(τ )
temporal separation of variables function
ψ(Ω) ω ωc ωd ωdj ωn ωnj ωr ω
Φ] Ω, Ωi , Ωo Ω Ωc Ωcl [
ζ
phase angle associated with initial conditions
phase response, system with harmonic base excitation excitation frequency, rad/s cutoff frequency, rad/s damped natural frequency, rad/s damped natural frequency of
jth mode
natural frequency of single degree-of-freedom system, rad/s uncoupled natural frequency of frequency ratio,
jth spring-mass system, rad/s
ωn2/ωn1
angular velocity vector modal matrix nondimensional frequency ratio,
fi
ω/ ωn or ω/ωn1, ωi / ωn, ωo/ ωn
nondimensional frequency coef cient for a beam center frequency ratio of a
filter
lower cutoff frequency ratio of a
filter
xxi
xxii
Symbols
Ωcu Ω Ωn
max
upper cutoff frequency ratio of a frequency at which
filter
H( Ω) is a maximum
fi
nondimensional natural frequency coef cient for a beam at the
nth natural frequency
PREFACE TO THE THIRD EDITION
Vibration is a classical subject whose principles have been known and studied for many centuries and presented in many books. Over the years, the use of these principles to understand and design systems has seen considerable growth in the diversity of systems that are designed with vibrations in mind: mechanical, aerospace, electromechanical and microelectromechanical devices and systems, biomechanical and biomedical systems, ships and submarines, and civil structures. As the performance envelope of an engineered system is pushed to higher limits, nonlinear effects also have to be taken into account.
AIMS OF THE BOOK
This book has been written to enable the use of vibration principles in a broad spectrum of applications and to meet the wide range of challenges faced by system analysts and designers. To this end, the authors have the following goals.
•
To provide an introduction to the subject of vibrations for undergraduate students in engineering and the physical sciences.
•
To present vibration principles in a general context and to illustrate the use of these principles through carefully chosen examples from different disciplines.
•
To use a balanced approach that integrates principles of linear and nonlinear vibrations with modeling, analysis, prediction, and measurement so that physical understanding of the vibratory phenomena and their relevance for engineering design can be emphasized.
•
To deduce design guidelines that are applicable to a wide range of vibratory systems.
In writing this book, the authors have used the following guidelines. The material presented should have, to the extent possible, a physical relevance to justify its introduction and development. The examples should be relevant and wide ranging. There should be a natural integration and progression between linear and nonlinear systems, between the time domain and the frequency domain, among the responses of systems to harmonic and transient excitations, and between discrete and continuous system models. There should be a minimum emphasis placed on the discussion of numerical methods and procedures, per se,
®
and instead, advantage should be taken of tools such as Matlab
and Mathematica
®
for
generating the numerical solutions to complement the analytical solutions. In addition,
xxiv
Preface to the Third Edition numerical tools should be used in concert with analysis to extend studies on linear systems to include nonlinear elements. Finally, there should be a natural and integrated interplay and presentation between analysis, modeling, measurement, prediction, and design so that a reader does not develop arti
ficial distinctions among them.
NEW MATERIAL
In this third edition, the authors have signi
ficantly
enhanced the previous editions by
creating the following new materials:
•
Added numerous examples of model construction
•
Introduced two ways in which a linear spring can be used to create zero stiffness
•
Expanded the treatment of the Maxwell model
•
Increased the number of waveforms analyzed for both periodic and transient excitations
•
Doubled the number of vibration absorbers considered
•
Presented a general way to determine the natural frequencies and mode shapes of 19 different undamped linear two degree-of-freedom systems and subsequently showed how to use these results to obtain the frequency-response functions for damped systems
•
Added several new examples that illustrate novel applications of vibration analysis and design.
REARRANGEMENT OF MATERIAL
We have rearranged some of the material to make it more cohesive and better focused, to allow special cases to be easily considered, and to allow for the results to be more easily generalized. This is especially true for the material covering single degree-offreedom systems subject to harmonic excitation and to transient excitation, to systems with two degrees of freedom, and to beams.
TABLES
We have created 13 new tables in which we have collected, summarized, and in many cases extended, the results appearing in the main body of the text. The material presented in the tables has been organized so that the similarity in the vibration model features, system response, or other characteristics of seemingly different physical systems can be emphasized. These tables have many purposes: they are used to summarize the important results, they serve as a reference source and as a study guide, they extend basic results, and, in some cases, they can be used to create exercise problems.
Preface to the Third Edition
xxv
INTERACTIVE GRAPHICS
We have created over 40 real-time interactive graphics that are keyed to the book’s text and
figures. The interactive graphics require no programming experience, only the use of
a mouse or touch pad. (It does require one to download a free program from Wolfram.) Additionally, appearing in appropriate places in the text are guidelines that direct the reader on what to note in each and the major conclusions that can be reached through the use of an interactive graphic. The interactive graphics materials are intuitive to use and self-explanatory. All
figures are enhanced using different colors, line types and labels
and in many of them the authors display numerical values of special quantities of interest such as a maximum/minimum value or an optimum value. There are numerous
fi
advantages and bene ts of these interactive graphics materials, including the following: they are easy to use since no knowledge of a programming language is required; realtime parametric investigations can be conducted and “what-if” scenarios can be explored while making comparisons with special cases; they complement textbook material where
fi
graphics can only provide speci c instances of the results; and they can be used to enhance homework assignments by asking questions the answers to which can help further the understanding of the material. With the inclusion of the interactive graphics, the analytical and numerical results are now extensively complemented with the ability to easily visualize and explore them. These interactive graphics also allow the student to verify the design guidelines that appear throughout the book and to realize that the interactive graphics can be used as a design aid for applications beyond the classroom. By conducting parametric studies with the aid of the interactive graphics, a reader can also get a better appreciation for the range of vibratory behaviors possible. Many of these interactive graphics can be used to reveal interesting phenomena, which the authors believe will help further a reader’s understanding of vibrations.
FEATURES RETAINED FROM THE PREVIOUS EDITION
In addition to the new enhancements mentioned above, this book retains the following features.
•
Newton’ s laws and Lagrange’s equations
are used to develop models of systems. Since
an important part of this development requires kinematics, kinematics is reviewed in Appendix A.
•
Laplace transforms are used to develop analytical solutions for linear vibratory systems and, from the Laplace domain, extend these results to the frequency domain. The responses of these systems are discussed in both the time and frequency domains to emphasize their duality.
xxvi
Preface to the Third Edition •
Notions of transfer functions and frequency-response functions also
are used throughout
the book to help the reader develop a comprehensive picture of vibratory systems.
•
Design for vibration (DFV) guidelines are introduced and are based on vibration principles developed throughout the book. The guidelines appear at the appropriate places in each chapter. These design guidelines serve the additional function of summarizing the preceding material by encapsulating the most important elements as they relate to some aspect of vibration design.
•
Introduction
fically
to each chapter provides a discussion on what speci
will be covered
in that chapter.
•
Examples
have been chosen so that they are of different levels of complexity, cover a
wide range of vibration topics and, in most cases, have practical applications to realworld problems.
•
Exercises
have been organized to correlate with the most appropriate section of the
text.
•
Appendices are included on the following: ○ Preliminaries from dynamics ○ Laplace transform pairs ○ Solution methods to second-order ordinary differential equations ○ Matrices ○ Complex numbers and variables ○ State-space formulation ○ Natural frequencies and mode shapes of bar, shafts, and strings ○ Derivation details related to beam vibrations.
•
A glossary is
included to list in one place the de
finitions of the major terms used in the
book.
CONTENTS AND ORGANIZATION
The book is organized into nine chapters, with the topics covered ranging from pendulum systems and spring-mass-damper prototypes to beams. In the
first
chapter, a brief
introduction to the subject of vibrations is provided, related history is reviewed, and examples of scenarios where this subject is relevant are provided. In the second chapter, the inertia, stiffness, and damping elements that are used to construct a vibratory system model are introduced, the notion of equivalent spring stiffness is presented in different physical contexts, the modeling of nonlinear springs is addressed, damping models are discussed, and many examples of modeling physical systems are shown. In Chapter 3, the equation governing a single degree-of-freedom vibratory system is derived by using the principles of linear momentum balance and angular momentum balance and the Lagrange equations. The notions of natural frequency and
Preface to the Third Edition
xxvii
damping factor are introduced and mass excitation, base excitation, and unbalanced mass excitation are examined. The linearization of governing equations for nonlinear systems is also discussed. In Chapter 4, the responses of linear single degree-of-freedom
–
systems to initial conditions are examined for the Kelvin Voigt material and for a Maxwell material, and the effects of nonlinear springs and damping are determined. In addition, response stability, nonlinear springs, and nonlinear dampers are discussed. In Chapter 5, the responses of single degree-of-freedom systems subjected to periodic excitations are considered and the notions of resonance, frequency-response functions, and transfer functions are introduced. The relation between the information in the time domain and the frequency domain is examined in detail. The concepts used for vibration isolation and accelerometers are presented and the notion of equivalent damping is introduced. Alternative forms of the frequency-response function are discussed. The forced response of a nonlinear oscillator is also treated. In Chapter 6, the responses of single degree-of-freedom systems to different types of transient excitations are analyzed in terms of their frequency spectra relative to the amplitude response function of the system. The notion of rise time, overshoot, and settling time are presented. The transient response of a nonlinear oscillator is also examined. Multiple degree-of-freedom systems are treated in Chapters 7 and 8 leading up to continuous systems in Chapter 9. In Chapter 7, the derivation of governing equations of motion of a system with multiple degrees of freedom is addressed by using the principles of linear momentum balance and angular momentum balance and Lagrange’s equations. The natural frequencies and mode shapes of undamped systems are studied and the notion of a vibratory mode is explained. The linearization of governing system for nonlinear systems is treated and the stability and vibrations of rotating shafts on
flexible
mounts is presented in detail. In Chapter 8, the general solution for the responses of systems with two degrees of freedom subjected to initial conditions and arbitrary forcing is presented by using the normalmode approach. The limitation of this approach with regard to the type of damping that can be considered is addressed. The notions of resonance, frequency-response functions, and transfer functions for a multiple degree-of-freedom system are discussed with respect
fi
to their application for system identi cation and for the design of vibration absorbers and for vibration isolation. The vibration-absorber material includes the traditional treatment of linear vibration absorbers and a brief introduction to the design of several distinctly different types of nonlinear vibration absorbers, which include bar-slider systems, pendulum absorbers, and particle-impact dampers. Techniques that can be used to determine optimal choice of absorber and isolator parameters are also presented. In Chapter 9, the free and forced oscillations of thin elastic beams are treated for a large number of boundary conditions, in-span attachments, and beam geometry. Considerable attention is paid to the determination of natural frequencies and mode shapes for these
fi
con gurations, which include effects of axial forces and an elastic foundation. In this
xxviii
Preface to the Third Edition chapter, the power of the Laplace transform approach to solve the beam response for complex boundary conditions and in-span attachments becomes apparent. In addition, an appendix on the natural frequencies and mode shapes associated with the free oscillations of strings, bars, and shafts, each for various combinations of boundary conditions including an attached mass and an attached spring is included.
ACKNOWLEDGMENTS
We express our sincere thanks to our former students for their spirited participation with regard to earlier versions of this book and for providing feedback, to the reviewers of this manuscript for their constructive suggestions, and to our colleagues worldwide for their valuable feedback on previous editions of this book.
B. Balachandran E. B. Magrab College Park, MD
1
Introduction
page
1.1 Introduction
1.1
1
1.2 A Brief History of Vibrations
6
1.3 About this Book
8
INTRODUCTION
Vibrations are oscillations about an equilibrium position. They occur in many aspects of our everyday experiences. They are an integral part of human life: low-frequency oscillations of the lungs and the heart; high-frequency oscillations of the ear; oscillations of the larynx to create speech; and oscillations induced by rhythmical body motions such as walking, jumping, and dancing.
fit
Sometimes vibrations are used for our bene
in such devices as loudspeakers, vibra-
tory feeders, paint mixers, electrodynamic shakers, massage chairs and beds, plate compactors,
ultrasonic
devices
used
in
non-invasive
diagnostics,
sirens
and
alarms
for
warnings, and stimulation of bone growth. Illustrations of these types of devices are
ficial
shown in Figure 1.1. Other bene
uses of vibrations include atomic clocks that are
based on atomic vibrations and ultrasonic instrumentation used in eye and other types of surgeries. There are situations in which vibrations are unwanted and can be due to manmade or
fi
natural sources. Examples of unwanted manmade disturbances are vehicular traf c, construction site machinery, factory metal forming machines, aircraft and helicopter
fly-
overs, railroads, rotating machinery, wind farms, and electric transformers. Examples of these types of sources are shown in Figure 1.2. Examples of disturbances from natural phenomena are earthquakes, thunder, ocean waves, and wind. Vibrations are also undesirable when performing measurements with precision instruments such as an electron microscope and when fabricating microelectromechanical systems. In vehicle design, noise due to vibrating panels needs to be reduced. However, vibrations that can be responsible for unpleasant sounds, which is called noise, are also responsible for the music that we hear. Frequently, systems must be designed to withstand vibration environments. Examples include buildings designed to withstand earthquakes and wind; offshore drilling platforms
2
Introduction
(a)
(b)
(c)
(d) Figure 1.1.
(e)
ficial uses of vibrations: (a) paint mixer, (b) electrodynamic
Examples of the bene
shaker, (c) massage chair, (d) loudspeaker, and (e) parts feeder.
Source : Photo of paint mixer courtesy of Radia, Plymouth
MN www.radiaproducts.com/
products/paint-mixers-shakers/; Photo of electrodynamic shaker courtesy of Bruel & Kjaer Sound and Vibration Measurement A/S; Photo of massage chair courtesy of Titan World LLC/Titan World LLC; Photo of loudspeaker courtesy of Mitek Audio, Mitek USA; Photo of parts feeder by Richdsu https://commons.wikimedia.org/wiki/File:Bowl_Feeder.jpg and reprinted under Creative Commons Attribution 3.0.
to withstand ocean waves; chimneys and bridges to withstand wind; aircraft, helicopters, ships, and railroad cars to withstand their internally generated disturbances; machine tools; washing machines; and fragile packages sent from one destination to another. Examples of these types of systems are shown in Figure 1.3.
fically to reduce or minimize the effects of a particular
Some systems are designed speci
type of vibratory disturbance such as automobile suspension systems, helicopter rotors, submarine stealth, vibration isolation tables, and disc drives. Examples of these types of systems are shown in Figure 1.4.
1.1 Introduction
(a)
(b)
(d)
3
(c)
(e)
Figure 1.2. Examples of unwanted manmade environmental vibratory disturbances: (a) hydraulic
breaker (jackhammer), (b) pile driver, (c) airplane, (d) train, and (e) vehicular traf
fic.
Source: Photo of hydraulic breaker courtesy of The Toro Company; Photo of pile driver courtesy of Hammer & Steel Inc. Hazelwood, MO; Photo of Boeing 777X-9 ascending copyright
©
Boeing and reprinted with permission of Boeing/Boeing Images; Photo of train by David
Gubler from https://commons.wikimedia.org/wiki/File:UP_EMD_SD9043AC_ Joso_Bridge,_USA.jpg and reprinted under Creative Commons Attribution 3.0.
Vibrations manifest themselves in many different ways. Temporally, they can occur intermittently, continuously, or randomly, and physically they can manifest themselves as a displacement, an unbalanced force or moment, an acoustic wave, or a pressure wave. Based on the discussion so far, one can roughly place the design of devices and systems in cases in which some aspect of vibrations is involved into one of three categories. These categories are: (i) cases in which one seeks to control vibrations and puts them to
fi
bene cial use; (ii) systems whose oscillatory motions must be prevented or minimized from impacting itself, another system, or its environment; and (iii) systems that must be able to withstand a vibratory environment and perform as intended. When it is known that a system will have to operate in an environment that will subject the system to vibrations or that the system itself will be a source of vibrations, one must consider these vibrations in the system’s design stage. If this is not done, one may end up with systems that
•
Experience catastrophic failure
•
Undergo excessive wear
4
Introduction
Figure 1.3.
(a)
(b)
(c)
(d)
Examples of systems designed to withstand vibration environments: (a) isolation
table, (b) ship, (c) earthquake resistant building, and (d) suspension bridge.
Source : Photo of isolation
table, permission to use granted by Newport Corporation, all rights
reserved; Photo of ship used under the UK Open Government License v3.0; Photo of earthquake resistant building by Alastair McLean, courtesy of Te Ara, The Encyclopedia of New Zealand; Photo of suspension bridge by Octagon in https://commons.wikimedia.org/wiki/File:Golden_ Gate_Bridge_.JPG and reprinted under ShareAlike 3.0 Unported.
•
Experience excessive displacements and stresses
•
Are hard to control
•
Produce unacceptable disturbances to the surroundings, both vibratory and acoustic
•
Waste energy.
On the other hand, systems that are designed with vibrations in mind tend to
•
Operate as intended
•
Undergo minimum wear
•
Have long life
•
Are controllable
•
Interfere minimally with their environment
•
Perform useful tasks
•
Operate with high mechanical ef ciency.
fi
1.1 Introduction
(a)
(b)
(c)
(d)
Figure 1.4. Examples of systems designed speci
5
fically to reduce the effects of a particular type
of vibratory disturbance: (a) automobile suspension system, (b) helicopter rotors, (c) disc drive, and (d) submarine silencing.
Source: Photo of racing car courtesy of Terps Racing, University of Maryland, College Park, MD; Photo of helicopter rotors, US Coast Guard photo by Adam Eggers; Photo of disc drive courtesy of Western Digital; Photo of submarine US Navy photo.
Additionally, when undesirable vibratory motions are reduced
•
Human comfort is increased
•
Humans and machinery operate more ef
•
Systems tend to be more reliable and predictable
•
There is a decrease in material fatigue and in human fatigue.
ficiently and safely
Vibrations, a Subset of Dynamics As mentioned earlier, vibrations involve oscillatory motions that occur about an equilibrium position of a system. As such, vibrations are considered a subset of the subject area called dynamics, which covers all types of motions. Preliminaries of dynamics are covered in Appendix A. These preliminaries are needed for the determination of the displacement,
6
Introduction velocity, and acceleration of a mass element; the number of degrees of freedom of a system; the kinetic energy and work of a system; and the governing equations of motion.
1.2
A BRIEF HISTORY OF VIBRATIONS
It is likely that the early interest in vibrations was due to the development of musical instruments such as whistles and drums. As early as 4000
BCE,
it is believed that in India
and China there was an interest in understanding music, which is described as a pulsating effect due to rapid change in pitch. The origin of the harmonica can be traced back to 3000
BCE,
when in China a bamboo reed instrument called a
sheng
was introduced.
From archeological studies of the royal tombs in Egypt, it is known that stringed instruments have also been around from about 3000
BCE.
A
first
fic
scienti
study into such
instruments is attributed to the Greek philosopher and mathematician Pythagoras (582 507
BCE).
–
He showed that if two like strings are subjected to equal tension, and if one is
half the length of the other, the tones they produce are an octave (a factor of two) apart. It is interesting to note that although music is considered a highly subjective and personal
art,
it
is
closely
governed
by
vibration
principles
such
as
those
determined
by
Pythagoras and others who followed him.
–
The vibrating string was also studied by Galileo Galilei (1564 1642), who was the
first to
show that pitch is related to the frequency of vibration. Galileo also laid the foundations for studies of vibrating systems through his observations made in 1583 regarding the motions of a lamp hanging from a cathedral in Pisa, Italy. He found that the period of motion was independent of the amplitude of the swing of the lamp. This property holds for all vibratory systems that can be described by linear models. The pendulum system studied by Galileo has been used as a paradigm to illustrate the principles of vibrations for many centuries. Galileo and many others who followed him laid the foundations for vibrations, which is a discipline that is generally grouped under the umbrella of mechanics. A brief summary of some of the major contributors and their contributions is provided in Table 1.1. The biographies of many of the individuals listed in this table can be found in the
Scientific Biography.
1
Dictionary of
It is interesting to note from Table 1.1 that the early interest of the
investigators was in pendulum and string vibrations, followed by a phase where the focus was on membrane, plate, and shell vibrations, and a subsequent phase in which vibrations in practical problems and nonlinear oscillations received considerable attention. Lord Rayleigh’s book
The Theory of Sound,
2
which was
first published in 1877, is one
of the early comprehensive publications on vibrations. In fact, many of the mathematical developments that are commonly taught in a vibrations course can be traced back to the 1800s and before. However, since then, the use of these principles to understand and
1 2
C. C. Gillispie, editor,
Dictionary of Scientifi c Biography, 18 Volumes, Scribner, New York, 1970 –1990. The Theory of Sound, 2 Volumes, Macmillan, London, 1877, 1878.
Rayleigh, J.W. Strutt, Lord,
1.2 A Brief History of Vibrations
Table 1.1.
Major contributors to the
Contributor
Galileo Galilei (1564–1642) Marin Mersenne (1588–1648) John Wallis (1616–1703)
field of vibrations and their contributions Areas of contributions
Pendulum frequency measurement, vibrating string Vibrating string String vibration: observations of modes and harmonics
Christian Huygens (1629–1695) Robert Hooke (1635–1703)
Nonlinear oscillations of pendulum
–
Pitch frequency relationship; Hooke’s law of elasticity
Isaac Newton (1642–1727) Gottfried Leibnitz (1646–1716) Joseph Sauveur (1653–1716)
Laws of motion; calculus Calculus String vibration: coined the name “fundamental harmonic” for lowest frequency and “harmonics” for higher frequency components
Brook Taylor (1685–1731)
Vibrating string frequency computation; Taylor ’s
Daniel Bernoulli (1700–1782)
Principle of liner superposition of harmonics; string
Leonhard Euler (1707–1783)
Angular momentum principle; complex numbers;
Jean d’Alembert (1717–1783)
d’Alembert’s principle; equations of motion;
theorem
and beam vibrations
Euler’s equations; beam, plate, and shell vibrations
wave equation
Charles Coulomb (1736–1806) Joseph Lagrange (1736–1813)
Torsional vibrations; friction Lagrange’s equations; frequencies of open and closed organ pipes
E. F. F. Chladni (1756–1824) Jacob Bernoulli (1759–1789) J. B. J. Fourier (1768–1830) Sophie Germain (1776–1831) Simeon Poisson (1781–1840) G. R. Kirchhoff (1824–1887) R. F. A. Clebsh (1833–1872) Lord Rayleigh (1842–1919)
Plate vibrations: nodal lines Beam, plate, and shell vibrations Fourier series Equations governing plate vibrations Plate, membrane, and rod vibrations; Poisson’s effect Plate and membrane vibrations Vibrations of elastic media Energy methods: Rayleigh’s method; Strutt diagram; vibration treatise
C. G. P. De Laval (1845–1913)
Vibrations of unbalanced rotating disc: practical solutions
Gaston Floquet (1847–1920) Henri Poincaré (1854–1912) A. M. Liapunov (1857–1918) Aurel Stodola (1859–1943) Balthasar van der Pol (1889–1959) Jacob Pieter Den Hartog (1901–1989)
Stability of periodic oscillations: Floquet theory Nonlinear oscillations; Poincaré map; stability; chaos Stability of equilibrium Beam, plate, and membrane vibrations; turbine blades Nonlinear oscillations: van der Pol oscillator Nonlinear systems with Coulomb damping; vibration of rotating and reciprocating machinery; vibration textbook
7
8
Introduction design systems has seen considerable growth in the diversity of systems that are designed with
vibrations
in
mind:
mechanical,
electromechanical
and
microelectromechanical
devices and systems, biomechanical and biomedical systems, ships and submarines, and civil structures.
1.3
ABOUT THIS BOOK
Vibration analysts and designers face a broad spectrum of applications and challenges. To meet these challenges, we have done the following with the material in this book: (1) we present vibration principles in a general context and illustrate their use through carefully chosen examples from different disciplines; (2) we use an approach that integrates principles of linear and nonlinear vibrations with modeling, analysis, prediction, and measurement so that physical understanding of the vibratory phenomena and their relevance for engineering design can be emphasized; (3) we deduce design guidelines that are applicable to a wide range of vibratory systems; (4) we present tables summarizing major results and include additional results for reference; and (5) we make available interactive graphics that provide an environment from which the reader can explore and analyze in real time all major results.
Topics The major topics covered in the remaining eight chapters are as follows
•
In Chapter 2, the inertia, stiffness, and damping elements that are used to construct a vibratory system model are introduced, the notion of equivalent spring stiffness is presented in different physical contexts, the modeling of nonlinear springs is addressed, and many examples of modeling physical systems are shown.
•
In Chapter 3, the equation governing a single degree-of-freedom vibratory system is derived using the principles of linear momentum balance and angular momentum balance and the Lagrange equations. The notions of natural frequency and damping factor
are
introduced
and
mass
excitation,
base
excitation,
and
unbalanced
mass
excitation are examined. The linearization of governing equations for nonlinear systems is also discussed.
•
In Chapter 4, the responses of linear single degree-of-freedom systems to initial condi-
–
tions are examined for the Kelvin Voigt material and for a Maxwell material and the effects of nonlinear springs and damping are determined. In addition, stability, nonlinear springs, and nonlinear dampers are covered.
•
In Chapter 5, the responses of single degree-of-freedom systems subjected to periodic excitations are considered and the notions of resonance, frequency-response functions, and transfer functions are introduced. The relation between the information in the time domain and the frequency domain is examined in detail. The concepts used for
1.3 About this Book
9
vibration isolation and accelerometers are presented and the notion of equivalent damping is introduced. The forced response of a nonlinear oscillator is also treated.
•
In Chapter 6, the responses of single degree-of-freedom systems to different types of transient excitations are analyzed in terms of their frequency spectra relative to the amplitude response function of the system. The notions of rise time, overshoot, and settling time are presented. The transient response of a nonlinear oscillator is examined.
•
In Chapter 7, the derivation of governing equations of motion of a system with multiple degrees of freedom is addressed by using the principles of linear momentum balance
and
angular
momentum
balance
and
Lagrange’s
equations.
The
natural
frequencies and mode shapes of undamped systems are studied and the notion of a vibratory mode is explained. The linearization of governing system for nonlinear systems is treated and stability is addressed.
•
In Chapter 8, the general solution to a two degree-of-freedom system subjected to initial conditions and arbitrary forcing is presented using the normal-mode approach. The limitation of this approach with regard to the type of damping that can be considered
is
addressed.
The
frequency-response
function
for
a
general
two
degree-of-
freedom system is discussed in detail. The notion of a vibration absorber is presented for several distinctly different types of systems. The responses of nonlinear systems are also considered.
•
In Chapter 9, the free and forced oscillations of thin elastic beams are treated for a large
number
of
boundary
conditions,
in-span
attachments,
and
beam
geometry.
Considerable attention is paid to the determination of natural frequencies and mode shapes for these con
figurations.
Interactive Graphics In subsequent chapters all major aspects of the material are supplemented with real-time interactive graphics, which are available on the publisher’s website. The interactive graphics require no programming experience, only that the reader download a free program from Wolfram. Appearing in appropriate places in the text are guidelines that direct the reader on what to note in each interactive graphic and the major conclusions that can be reached from the use of that interactive graphic. These interactive graphics are introduced at the appropriate places in the respective chapters and have been created with the following aims.
•
They complement and augment the material in the text and enhance understanding of the topics.
•
They
provide
real-time
ability
to
perform
parametric
investigations
“what-if” scenarios: typically, a wide range of parameters and con
and
figuration
explore
combina-
tions can be explored, with comparisons made for special cases.
•
They are intuitive to use and self-explanatory in the context of the material that they are illustrating.
10
Introduction •
When appropriate, they provide numerical values for important quantities of interest such as the maximum or minimum or optimum.
With regard to the interactive graphics, the reader is encouraged to explore Interactive Graphic 1.1. This interactive graphic is used to illustrate the fundamental aspects of the response of a single degree-of-freedom vibratory system for different amounts of damping and for different types of excitation to the mass: initial displacement, impact force, and harmonically varying force.
INTERACTIVE GRAPHIC 1.1: NORMALIZED RESPONSE OF A SPRINGMASS-DAMPER SYSTEM TO VARIOUS EXCITATIONS AND DAMPING
Summary Tables Numerous summary tables have been created to
•
Summarize the important results
•
Serve as a reference source
•
Serve as a study guide
•
Extend basic results
•
Serve, in some cases, as exercises
•
Organize certain material to illustrate and emphasize the similarity in the vibration model’s features, the system response, or other characteristic of different physical systems.
2
Modeling of Vibratory Systems
2.1 Introduction
11
2.2 Inertia Elements
13
2.3 Stiffness Elements
18
2.3.1 Introduction
18
2.3.2 Linear Springs
20
2.3.3 Nonlinear Springs
32
2.3.4 Other Forms of Potential Energy Elements
38
2.3.5 Summary of Equivalent Spring Constants
44
2.4 Dissipation Elements
44
2.4.1 Viscous Damping
44
2.4.2 Combinations of Viscous Dampers and Linear Springs
49
2.4.3 Other Forms of Dissipation
52
2.5 Model Construction
55
2.5.1 Introduction
55
2.5.2 A Few Simple Models
55
2.5.3 A Microelectromechanical System
60
2.5.4 The Human Body
62
2.5.5 A Ski
63
2.5.6 Cutting Process
64
2.6 Design for Vibration
2.1
page
65
2.7 Summary
66
Exercises
67
INTRODUCTION
In this chapter, the elements that comprise a vibratory system model are described and the use of these elements to construct models is illustrated with examples. There are, in general, three elements that comprise a vibrating system: (i) inertia elements, (ii) stiffness elements, and (iii) dissipation elements. In addition to these elements, one must also consider externally applied forces and moments and external disturbances from prescribed initial displacements and/or initial velocities.
12
Modeling of Vibratory Systems
Table 2.1.
Units of components comprising a vibrating mechanical
system and their customary symbols Quantity
Units
Translational motion Mass,
m
kg
k Damping, c
Stiffness,
N/m
⋅
N s/m
External force,
F
N
Rotational motion Mass moment of inertia, Stiffness,
kt ct
J
2
kg m
⋅ ⋅ ⋅ ⋅
N m/rad
Damping,
N m s/rad
External force,
The
inertia element
M
N m
stiffness element damping element is used
stores and releases kinetic energy, the
stores and
dissipation
to express
releases potential energy, and the
or
energy loss in a system. Each of these elements has different excitation-response characteristics and the excitation is in the form of either a force or a moment and the corresponding
response
of
the
element
is
in
the
form
of
a
displacement,
velocity,
or
acceleration. The inertia elements are characterized by a relationship between an applied force (or moment) and the corresponding acceleration response. The stiffness elements are characterized by a relationship between an applied force (or moment) and the corresponding displacement (or rotation) response. The dissipation elements are characterized by a relationship between an applied force (or moment) and the corresponding velocity response. The natures of these relationships, which can be linear or nonlinear, are presented in this chapter. The units associated with these elements and the commonly used symbols for the different elements are shown in Table 2.1. In this chapter, we shall show how to do the following.
•
Compute the mass moment of inertia of rotational systems.
•
Determine the stiffness of various linear and nonlinear elastic components in translation and torsion and the equivalent stiffness when individual linear components are combined.
fluid, gas, and pendulum elements.
•
Determine the stiffness of
•
Determine the potential energy of stiffness elements.
•
Determine the damping for systems that have different sources of dissipation: viscosity, dry friction,
•
fluid, and material.
Construct models of vibratory systems. In this and subsequent chapters, many aspects of the material are supplemented with
interactive graphics that are available on the publisher’s website and have been created to better visualize various aspects of the material. They are introduced and discussed
2.2 Inertia Elements
13
at the appropriate places in the respective chapters and have been created with the following aims.
•
They are intuitive to use and self-explanatory in the context of the material that they are meant to illustrate.
•
They complement, enhance, and augment the material.
•
They
provide
“what-if” •
real-time
ability
to
perform
parametric
investigations
and
explore
scenarios.
When appropriate, they provide numerical values for important quantities of interest such as the maximum or minimum or optimum. For this chapter, the following interactive graphics have been created.
fl Defl ection of two springs in parallel and their equivalent stiffness
Interactive Graphic 2.1: De ection of two springs in series Interactive Graphic 2.2:
Interactive Graphic 2.3: Geometric nonlinear stiffness of a combination of a vertical spring and an inclined spring
fi
Interactive Graphic 2.4: Geometric nonlinear stiffness of three springs and con gurations for zero stiffness
2.2
INERTIA ELEMENTS
Translational motion of a mass is described as motion along the path followed by the center of mass. The associated inertia property depends only on the total mass of the system and is independent of the geometry of the mass distribution of the system. The inertia property of a mass undergoing rotational motion, however, is a function of
fi
fined fixed point O. When the mass oscillates about a fixed point
the mass distribution, speci cally the mass moment of inertia, which is usually de about its center of mass or a
O or a pivot point O , the rotary inertia JO is given by JO = JG + md
2
(2.1)
m is the mass of the element, JG is the mass moment of inertia about the center of d is the distance from the center of gravity to the point O. In Eq. (2.1), the mass moments of inertia JG and JO are both defi ned with respect to axes normal to the plane of
where
mass, and
the mass. This relationship between the mass moment of inertia about an axis through the
G and a parallel axis through another point O follows from the parallel-axes theorem. The mass moments of inertia of some common shapes are given in Table 2.2.
center of mass
The questions of how the inertia properties are related to forces and moments and how
these
properties
Figure 2.1, a mass
m
affect
the
kinetic
energy
of
a
system
are
translating with a velocity of magnitude
shown. The direction of the velocity vector is also shown in the direction of the force acting on this mass.
x_
examined in the
figure,
X–Y
next.
In
plane is
along with the
14
Modeling of Vibratory Systems
Table 2.2.
–
Mass moments of inertia about the z-axis normal to the x y plane and
passing through the center of mass
L
Slender bar
JG = L/2
z Circular disc
1 12
mL
R G
JG = mR
2
R
JG = mR
2
1
2
2
z Sphere
2 5
z y
Circular cylinder
Jx = J y =
h G
R
Jz = mR 1
x
.
±
m 3R
2
+h
2
²
2
2
z
Figure 2.1. Mass in translation.
xi
Y
1 12
Fi
m j
k
O
X i
Z
From the principle of linear momentum (see Eqs. (A.21) and (A.23) of Appendix A), the equation governing the motion of the mass is
Fi= which, when
d _ Þ ð mxi dt
m and i are independent of time, simplifies to F = m€x
(2.2)
On examining Eq. (2.2), it is evident that
for translational motion, the inertia property m
is the ratio of the force to the acceleration.
The units for mass shown in Table 2.1 should
also be evident from Eq. (2.2). From Appendix A, it follows that the kinetic energy of mass
m is given by T=
1 2
mðxi _ ⋅ xi _ Þ=
1 2
mx_
2
(2.3)
2.2 Inertia Elements
15
i ⋅ i = 1. From the definition given by Eq. (2.3), it is clear the kinetic energy of translational motion is linearly proportional to the mass.
and we have used the identity that
Furthermore, the kinetic energy is proportional to the second power of the velocity magnitude. To arrive at Eq. (2.3) in a different manner, let us consider the work-energy theorem given by Eq. (A.33) of Appendix A. We assume that the mass shown in Figure 2.1 is translated from an initial rest state, where the velocity is zero at time state at time
to,
to the
final
tf. Then the work W done under the action of a force Fi is
ð
x
ð
ð
0
0
x
x
W = F i ⋅ dxi = m€ xi ⋅ dxi = m€ x dx 0
ð
ð
tf
=
tf
m€ xx_ dt = mxd _ x_ = m x_
to where we have used the relation
dx
1
to
= x_ dt
W = T jt=tf
2
(2.4)
2
. Hence, the kinetic energy is
³
− T ³³
=
0
t= to
=
1 2
mx_
2
(2.5)
which is identical to Eq. (2.3). For a rigid body undergoing only rotation in the plane with an angular speed
θ_ ,
one
can show from the principle of angular momentum given by Eqs. (A.25a) and (A.28) of Appendix A that
M = J €θ where
M
is the moment acting about the center of mass
(2.6)
G
or a
fixed
point
O (as
shown
J is the associated for rotational motion, the inertia property J is the ratio of the moment to the angular acceleration. Again, one can verify that the units of J shown in Table 2.1 are consistent with Eq. (2.6). This inertia property is also referred to as rotary inertia. Furthermore, to determine how the inertia property
in Figure 2.2) along the direction normal to the plane of motion and mass moment of inertia. From Eq. (2.6), it follows that
O G m, JG
(a)
Figure 2.2. (a) Uniform disc hinged at a point on its perimeter
O
R
L/2 c.g.
and (b) bar of uniform mass hinged at one end.
m G
(b)
L
16
Modeling of Vibratory Systems
J
affects the kinetic energy, we use Eq. (A.32) of Appendix A to show that the kinetic
energy of the system is
T=
1 2
J θ_
2
(2.7)
Hence, the kinetic energy of rotational motion only is linearly proportional to the inertia property
J,
the mass moment of inertia. Furthermore, the kinetic energy is proportional
to the second power of the angular velocity magnitude. In the discussions of the inertia properties of vibratory systems provided thus far, the inertia properties are assumed to be independent of the displacement of the motion. This assumption is not valid for all physical systems. For a slider mechanism discussed in Example 2.2, the inertia property is a function of the angular displacement. Other examples can also be found in the literature.
EXAMPLE 2.1
1
Determination of mass moments of inertia
We shall illustrate how the mass moments of inertia of several different rigid body distributions are determined.
Uniform Disc Consider the uniform disc shown in Figure 2.2a. If
JG
is the mass moment of inertia
about the disc’s center, then from Table 2.2
JG =
1 2
mR
2
Therefore, the mass moment of inertia about the point from point
O,
which is located a distance
R
G , is JO = JG + mR
2
=
1
mR
2
+ mR =
2
3
2
2
mR
2
(a)
Uniform Bar A bar of length
L
is suspended as shown in Figure 2.2b. The bar’s mass is uniformly
distributed along its length. Then the center of gravity of the bar is located at
L/2. From
Table 2.2, we have that
JG =
1 12
mL
2
Therefore, after making use of the parallel axis theorem, the mass moment of inertia about the point
O is
´L µ
JO = JG + m 1
J. P. Den Hartog,
2
2
=
1 12
mL
2
+
1 4
mL
2
Mechanical Vibrations, Dover, New York, 1985, p. 352.
=
1 3
mL
2
(b)
2.2 Inertia Elements EXAMPLE 2.2
17
Slider mechanism: system with varying inertia property
In Figure 2.3, a slider mechanism with a pivot at point
O
is shown. A slider of mass
ms
ml . Another bar, which is pivoted at point O′, has a m b and another portion of length e that has a mass inertia JO of this system and show its dependence on
slides along a uniform bar of mass portion of length
me.
b
that has a mass
We shall determine the rotary
φ.
the angular displacement coordinate
Figure 2.3. Slider mechanism.
O l
r a
ms b mb
ml
If
ae
O'
me
e
is the distance from the midpoint of bar of mass
from the midpoint of bar of mass
me
to
O
and
ab
is the distance
m b to O, then from geometry we find that
r ð φÞ = a
+ b − ab abð φÞ = ðb Þ + a − ab ae ð φÞ = ðe Þ + a − ae 2
2
2
2
=2
2
=2
2
cos
2
2
2
2
φ φ
cos
(a)
π − φÞ
cos ð
Hence, all motions of the system can be described in terms of the angular coordinate The rotary inertia
φ.
JO of this system is given by JO = Jml
+ Jm ðφÞ + Jm ð φÞ + Jm ð φÞ s
b
(b)
e
where
Jml
=
1 3
ml l
Jmb ðφÞ = m b
b
Jme ðφÞ = m e
e
2
2
12 2
12
;
Jms ðφÞ = ms r ðφÞ 2
+ mbab = mb
"
2
+ me ae = me 2
"
b
2
3
e
2
3
+ a − ab 2
+ a − ae 2
# φ
cos
#
(c)
π − φÞ
cos ð
In arriving at Eqs. (b) and (c), the parallel-axes theorem has been used in determining
J ml ; Jme ; and Jmb : From Eqs. (b) and (c), it is clear JO of this system is a function of the angular displacement φ.
the bar inertias
that the rotary inertia
18
Modeling of Vibratory Systems
2.3
STIFFNESS ELEMENTS
2.3.1
Introduction Stiffness elements are manufactured from different materials and they have many different shapes. One chooses the type of element depending on the requirements, for example, to minimize vibration transmission from machinery to the supporting structure, to isolate a building from earthquakes, or to absorb energy from systems subjected to impacts. Some representative types of stiffness elements that are commercially available are shown in Figure 2.4 along with their typical application. The stiffness elements store and release the potential energy of a system. To examine how the potential energy is de in which a spring is held
fixed
fined,
let us consider the illustration shown in Figure 2.5,
at end
O,
and at the other end a force of magnitude
F
is
Motion Motion
Rubber mounts
(b)
(a)
Motion
Motion
(c) Figure 2.4.
Motion
(d)
(e)
(a) Highway base isolation for lateral motion using cylindrical rubber bearings; (b)
wire rope isolators to isolate vertical motions of machinery; (c) air springs used in suspension systems to isolate vertical motions; (d) typical steel coil springs for isolation of vertical motions; and (e) steel rope isolators used in a chimney tuned mass damper to suppress lateral motions
Source : Photo of highway isolation mounts courtesy of Dynamic Isolation Systems Inc.; Photo of wire rope isolators courtesy of ITT Connect & Control Technologies, Enidine Inc., www.enidine. com/en-US/Products/WireRopeIsolator/; Photo of air springs courtesy of ITT Connect & Control Technologies, Enidine Inc., www.enidine.com/en-US/Products/AirSprings/; Photo of steel coil springs courtesy of Isolation Technology Inc.; Photo of tuned mass damper courtesy of Industrial Environmental Systems Inc. www.iesysinc.com/solutions/tuned-mass-dampers.
2.3 Stiffness Elements
19
Figure 2.5. (a) Stiffness element with a force acting on it and
(b) its free-body diagram.
Stiffness element
j
Fs
F F (a)
(b)
directed along the direction of the unit vector
j.
Under the action of this force, let the
element stretch from an initial or unstretched length direction of
j.
Lo
Lo + x along the between F and x can be
to a length
In undergoing this deformation, the relationship
linear or nonlinear as discussed subsequently. If
Fs
represents the internal force acting within the stiffness element, as shown in the
free-body diagram in Figure 2.5b, then in the lower spring portion this force is equal and opposite to the external force
F, that is, Fs = − F j
Since the force
Fs
figuration, it
tries to restore the stiffness element to its undeformed con
restoring force. As the stiffness element is deformed, energy is stored in The potential energy V is defined as the work done to take the stiffness element from the deformed position to the undeformed position, that is, the work needed to undeform the element to its original shape. For the element shown in Figure 2.5, this is given by
is referred to as a
this element, and as the stiffness element is undeformed, energy is released. 2
ð 0
V ðxÞ = F s ⋅ dx x
ð
ð
x
0
x
0
= − Fj ⋅ dxj = where we have used the identity potential energy
2
A general de
j
⋅j=
1 and
V is a scalar-valued function.
Fs
F dx
= −Fj
. Like the kinetic energy
(2.8)
T,
the
finition of potential energy V takes the form V ðxÞ ¼
where the force
F s is
ð
initial or reference position deformed position
F s ⋅ dx
a conservative force. The work done by a conservative force is independent of the path fol-
lowed between the initial and
final positions.
20
Modeling of Vibratory Systems The relationship between the deformation experienced by a spring and an externally applied force may be linear as discussed in Section 2.3.2 or nonlinear as discussed in Section 2.3.3. The notion of an equivalent spring element is also introduced in Section 2.3.2.
2.3.2
Linear Springs
Translation Spring F is applied de fl ection x such that
If a force
to a linear spring as shown in Figure 2.6a, this force produces a
F ðx Þ = kx where the coef
ficient k
is called the
(2.9)
spring constant
and there is a linear relationship
between the force and the displacement. Based on Eqs. (2.8) and (2.9), the potential energy
V stored in the spring is given by
ð x
V ðxÞ =
ð
ð
x
F ðxÞdx =
0
x
kx dx = k x dx =
0
1 2
kx
2
(2.10)
0
Hence, for a linear spring, the associated potential energy is linearly proportional to the spring stiffness
k and proportional to the second power of the displacement magnitude.
Torsion Spring If a linear torsion spring is considered and if a moment end while the other end of the spring is held
fixed, then
τ ðθÞ = k tθ
k
x
is applied to the spring at one
(2.11)
k2
k1
τ
k1
x F
k2
F x
F (a) Figure 2.6.
(b)
(c)
fi
Various spring con gurations: (a) single spring, (b) two springs in parallel, and (c)
two springs in series.
2.3 Stiffness Elements where
kt
θ
is the spring constant and
21
is the deformation of the spring. The potential
energy stored in this spring is
ð
ð
θ
V ð θÞ =
θ
τ ðθÞd θ =
0
kt θd θ =
1 2
ktθ
2
(2.12)
0
Combinations of Linear Springs Different combinations of linear spring elements are now considered and the equivalent stiffness of each of these combinations is determined. First, combinations of translation springs shown in Figures 2.6b and 2.6c are considered and following that, combinations of torsion springs shown in Figures 2.7a and 2.7b are considered. When there are two springs in parallel as shown in Figure 2.6b and the bar on which the force
F
acts remains parallel to its original position, then the displacements of both
springs are equal and, therefore, the total force is
F ðxÞ = F ðxÞ + F ðxÞ = k x + k x = ðk 1
2
1
where
F j(x)
2
is the resulting force in spring
+k
1
k j, j =
2
Þx
= k ex
1,2, and
ke is
(2.13)
the equivalent spring con-
stant for two springs in parallel given by
ke = k
+k
1
(2.14)
2
When there are two springs in series, as shown in Figure 2.6c, the force on each spring is the same and the total displacement is
x= x
1
=
F k
1
+x +
2
F k
=
2
1
k
+
1
1
k
2
!
F=
F ke
t
τ
kt1
kt2
(a)
kt2
kt1
(b)
Figure 2.7. Two torsion springs: (a) parallel combination and (b) series combination.
(2.15)
22
Modeling of Vibratory Systems where the equivalent spring constant
ke = In general, for
´
ke is
+
1
k
µ−
1
1
1
kk +k 1
k
2
2
1
(2.16) 2
N springs in parallel, we have
X N
ke = and for
=
k
N springs in series, we have ke =
i=1
ki
(2.17)
"X #− N
1
1
(2.18)
ki
i=1
The potential energy for the spring combination shown in Figure 2.6b is given by
V ðxÞ = V ðxÞ + V ðxÞ 1
where
V (x) 1
2
is the potential energy associated with the spring of stiffness
the potential energy associated with the spring of stiffness to determine
V (x) and V ( x), we find that 1
k
2.
k
1
and
V (x) 2
is
Making use of Eq. (2.10)
2
V ðx Þ =
1 2
k x
2
1
+
1 2
k x
2
2
=
1 2
ðk1
+ k Þx
2
2
For the spring combination shown in Figure 2.6c, the potential energy of this system is given by
V ðx ; x 1
2
Þ
= V ðx Þ + V ð x Þ 1
=
1 2
1
kx 1
2
2 1
+
1 2
2
kx 2
2 2
where again Eq. (2.10) has been used. Expressions constructed from the potential energy of systems are useful for determining the equations of motion of a system, as discussed in Sections 3.6 and 7.2. For two torsion springs in series and parallel combinations, we refer to Figure 2.7. From Figure 2.7a, the rotation
θ
of each spring is the same and, therefore,
τ ðθÞ = τ 1ðθÞ + τ2 ðθ Þ
= kt θ + k t θ = ðkt + k t Þθ = kte θ 1
where
τj
2
is the resulting moment in spring
1
ktj, j =
2
1,2, and
kte
(2.19)
is the equivalent torsional
stiffness given by
kte = kt
1
+ kt
2
(2.20)
2.3 Stiffness Elements
23
For torsion springs in series, as shown in Figure 2.7b, the torque on each spring is the same, but the rotations are unequal. Thus,
θ = θ1 + θ2 = where the equivalent stiffness
kte
is
kte = The
potential
energy
for
the
τ
kt
´
+
1
1
kt
1
+
τ
kt
=
´
kt
2
µ−
1
kt
1
2
torsion-spring
+
1
=
1
µ
1
kt
τ
τ=
(2.21)
k te
2
kt kt kt + k t 1
2
1
(2.22)
2
combination
shown
in
Figure
2.7a
is
given by
V ðθÞ = V ðθÞ + V ðθÞ 1
= where
we
have
used
Eq.
1 2
2
kt θ
2
1
(2.12).
+
1 2
For
kt θ
2
2
the
=
1 2
ð kt
1
+ k t Þθ
2
2
torsion-spring
combination
shown
in
Figure 2.7b, the system potential energy is given by
V ðθ ; θ 1
2
Þ
= V ðθ Þ + V ðθ 1
=
1 2
1
kt θ 1
2
2 1
+
1 2
2
Þ
kt θ 2
2 2
INTERACTIVE GRAPHIC 2.1: DEFLECTION OF TWO SPRINGS IN SERIES This interactive graphic is used to visualize the total deflection of two springs in series and the contribution of each spring to the total deflection as a function of the stiffness of each spring. In this interactive graphic, the following should be noted. • • •
When k2 remains fixed, the top spring moves as a rigid entity as the stiffness k1 decreases. When k1 remains fixed, the bottom spring does not compress further as the stiffness of k2 decreases. The displacement of each spring is independent of the displacement of the other spring; however, the total displacement of the spring combination is the sum of the displacements of the individual springs.
Equivalent Stiffness of Two Springs in Parallel: Removal of a Restriction Let us reexamine the pair of springs in parallel shown in Figure 2.6b. Now, however, we remove the restriction that the bar to which the force is applied has to remain parallel to its original position. Then, we have the con
fi
figuration
shown in Figure 2.8. The equiva-
lent spring constant for this con guration will be determined.
24
Modeling of Vibratory Systems F b
a
b
a
x2 k2
x
x1
k1
Figure 2.8.
Parallel springs subjected to unequal forces.
From similar triangles, we see that
x= x
F
1
is the force acting on
k
1
x
a+ b
F
and
ðx1
a+ b
b
= If
b
+
2
1
−x Þ
a
+
a +b
2
x
is the force acting on
2
(2.23)
2
k
, then from the summation
2
of forces and moments on the bar we obtain, respectively,
F = F +F bF = aF 1
2
2
(2.24)
1
Thus, Eqs. (2.24) lead to
F
1
F
2
=
bF a+b
=
aF a+b
(2.25)
Therefore,
x
1
x
2
=
F k
=
bF k ða + bÞ
=
F k
=
aF k ða + bÞ
1 1
2
1
2
2
"
#
From Eqs. (2.23) and (2.26), we obtain
x=
=
b
bF a + b k ða + bÞ 1
"
F
ða
+bÞ
2
(2.26)
a
aF a + b k ða + bÞ
+
k b +k a k k 2
2
2
"
#
#
2
1
1
2
(2.27)
2.3 Stiffness Elements
25
Comparing Eq. (2.27) to the form
F ke
x= we
(2.28)
find that the equivalent spring constant ke is given by ke =
k k ða +bÞ k b +k a
2
1
2
2
2
1
(2.29)
2
It is noted that Eq. (2.29) resembles Eq. (2.16), which was obtained for two springs in series. This similarity is due to the fact that Eq. (2.23) resembles Eq. (2.15), which takes
fl
into account that the spring de ections are unequal.
INTERACTIVE GRAPHIC 2.2: DEFLECTION OF TWO SPRINGS IN PARALLEL AND THEIR EQUIVALENT STIFFNESS This interactive graphic is used to do the following: (1) visualize the deflections of two parallel springs when the bar connecting them is no longer required to remain parallel to its undeformed state; and (2) display the values of the equivalent spring constant with respect to variations in the stiffness of each spring and the position of the force. In this interactive graphic, the following should be noted. • • •
When the position of F is constant (β is constant), a change in relative stiffness causes one spring to displace more or less than the other spring. When the force is placed over either spring, the displacement of the other spring is zero. In general, the displacements of each spring are equal when α = β/(1 −β). At this value of α, the maximum value of ke/k1 occurs. Conversely, the maximum value of ke/k1 occurs when β = α/(1 + α). Use the interactive graphic to verify these relations.
Equivalent Spring Constants for Common Structural Elements Used in Vibration Models To determine the spring constant for numerous elastic structural elements one can make use of known relationships between force and displacement. Many such spring constants that have been determined for different geometry and loading conditions are presented in Table 2.3. For modeling purposes, the inertias of the structural elements such as the beams of Cases 4 to 6 in Table 2.3 are usually ignored. In Chapter 9, it is shown under what conditions it is reasonable to make such an assumption.
3
Since it may not always be possible to obtain a spring constant for a given system through analysis, often one has to experimentally determine this constant. As a representative
3
See Eq. (9.167).
26
Modeling of Vibratory Systems
Table 2.3. 1
Spring constants for some common elastic elements
Axially loaded rod or cable
L
k=
AE L
k=
π Ed 1 d2
F, x 2
d1
Axially loaded tapered rod
L
4
L d2 F, x 3
L
5
Hinged-hinged beam
F, x
a
b
k=
3
b
k=
3
(simply supported)
6
7
Clamped-clamped beam
F, x
a
Two circular rods in torsion
8
Two circular rods in torsion
,
L2
kt1
L2
D
Coil spring
d n turns F, x
10
Clamped rectangular plate, constant thickness, force at center
a
F
h
a/2 b /2 b
2
EI ð a +bÞ a b 3
ktj = 9
EI ð a + bÞ ab 2
kte =
kt2
L1
F ð xÞ x < = > kL L:
or
v
The spring coef
ficient
1
2
h
δo
Fs x pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi L +x pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2
+
L
2
2
i
+x − L 2
ð ffiÞ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ko x δo − L L +x 2
(2.34)
2
2 39> k ðδ =L − Þ = + 64 + qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi75> k ; o
1
o
1
1
+ ðx=LÞ
(2.35)
2
in the vertical direction is obtained by differentiating Eq. (2.35)
x, which results in
ð Þ = dFdxv ðxÞ = k + ko + ¶koðδo=L −· Þ= + ðx=LÞ 1
kv x
1
x, F
2
3 2
(2.36)
Figure 2.15. Nonlinear stiffness due to
ko L,To k
(2.33)
x-direction is obtained from Eq. (2.33) and Figure 2.15 as Fv x
with respect to
0, the inclined spring is under an
and the spring, therefore, is initially under a tension force
When the spring is moved up or down an amount
The force in the
=
γ
geometry: spring under an initial tension, one
Fs ko To
g
end of which is constrained to move in the vertical direction.
2.3 Stiffness Elements or
kv ðxÞ k
=
1
35
2 3 k ð δ =L − Þ 7 + 64 + ¶ · 5 k o
o
1
1
1
(2.37)
=
+ðx=LÞ
3 2
2
From Eq. (2.34), it is clear that the spring force opposing the motion is a nonlinear function of the displacement
x.
Furthermore, from Eq. (2.36) one can see that the spring
constant is also a nonlinear function of
x. Hence, a vibratory model of the system shown
in Figure 2.15 will have nonlinear stiffness. When
−δo
3 2 k ðδ =L + Þ 7 + 64 − ¶ · 5 k
is the amount that the spring is intially compressed, then Eq. (2.37) becomes
kv k
=
1
o
o
1
1
+ ðx=LÞ
2
1
(2.38)
=
3 2
fi
For this case, it is possible for this spring con guration to have negative stiffness. After differentiating Eq. (2.38) with respect to mine the value of
x
this operation gives when
at which
x
=
k v will
x
and setting the result to zero, one can deter-
have either a minimum or a maximum. Performing
0. From Eq. (2.38), it is found that
kv
will equal zero at
k = koδo/ L. We shall show the benefits of this result in Example 3.13. x/L is small, we can approximate Eqs. (2.34) and (2.36), respectively, by
When
Fv ðxÞ ≃ ðkL + ko Lðδo=LÞÞ
x L
+
k oL 2
ð1
and
k vðxÞ ≃ ðk + k oδo =LÞ +
ko
3
2
ð1
¶x ·
− δo=LÞ
3
L
¶x ·
− δ o=LÞ
2
L
+⋯
x
=
0
(2.39)
+⋯
(2.40)
When the quadratic term in Eq. (2.40) is negligible, Eq. (2.40) becomes
kv Thus,
kv
≃ ðk + ko δo=LÞ
(2.41)
is a constant. As discussed previously, when the spring with constant
initially under compression,
δo
→ −δ
o, and there is a set of values for which
kv
ko
=
is 0,
that is,
k = k oδo =L
(2.42)
which, because of the approximation made in arriving at Eq. (2.42), occurs at
x/ L
This result agrees with that obtained from Eq. (2.38). Equations (2.35) and (2.37) are plotted in Figure 2.16 for the case when and
ko/k =
1.43 where for these parameters
k v/k ≈
0.
δo/ L
≈
=−
0.
0.7
36
Modeling of Vibratory Systems 2 F/(kL) k /k n
1 edutilpmA 0
–1 –1.0
Figure 2.16.
–0.5
0.0 x/L
1.0
0.5
Force in the vertical direction and the corresponding spring constant for the spring
combination in Figure 2.15 when
δo /L
=−
0.7 and
k o/k
=
1.43.
INTERACTIVE GRAPHIC 2.3: GEOMETRIC NONLINEAR STIFFNESS OF A COMBINATION OF A VERTICAL SPRING AND AN INCLINED SPRING This interactive graphic is used to visualize the force and the spring constant of the two orthogonal spring system shown in Figure 2.15 and the combination of parameters for which the spring constant in the vertical direction is zero. In this interactive graphic, the following should be noted. • • •
The equivalent spring constant in the vertical direction is symmetrical about x/L = 0 for all combinations of δo/L and ko/k whereas the force-deflection relation is asymmetrical about x/L = 0. The equivalent spring constant in the vertical direction can never be zero when the horizontal spring is in tension. The value at which the equivalent spring constant in the vertical direction is zero in the vicinity of x/L = 0 occurs when ko/k = −1/(δ o/L).
System 2 The second system is shown in Figure 2.17a. The symbols used to describe the extension of
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
one of the non-vertical springs are given in Figure 2.17b. From this
Lo =
a
2
+ ho 2
;
sin
θo =
ho ; Lo
L=
ðho
−x Þ + a 2
2
;
figure, it is seen that
sin
θ=
ho − x L
(2.43)
2.3 Stiffness Elements
x x, F
37
Lo
ho
L
Lo
Lo ko
qo
ko
a
k
qo
(a) Figure 2.17.
q
(b)
fi
(a) Three-spring con guration and (b) the symbols used to describe the deformation
of one of the non-vertical springs.
The forces in the inclined springs add up to
F c = 2koðLo − LÞ
(2.44)
x-direction is
and the total force acting in the downward
Fx ðxÞ = kx + F c sin θ = kx + 2ko ðLo − LÞsin θ
0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 a +h − C = kx + k ðh − xÞ B @qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A ðh −xÞ + a 8 0 < = kL : ^x + αð θ − ^xÞ@pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi − 2
2
o
2
o
o
o
o
2
2
1
o
sin
1
2
1
−
^x sin θo + x^
2
2
19= A;
1
(2.45)
where we have used Eq. (2.43) and
^x =
fi
The corresponding spring coef cient respect to
x, which results in kx ðxÞ k
x ; Lo
kx
= dF x ðxÞ = + dx
1
ko k
α=
(2.46)
is determined by differentiating Eq. (2.38) with
0 α@ − ±
2
1
cos
−
1
^x sin
2
1 θ ² A θ +^ x
2
o
2
o
3=2
(2.47)
fl
For this spring system, it is possible to have zero stiffness at a stationary point of in ection, which is determined by setting the derivative of
k x with respect to x^ to 0. When this
fl
operation is performed with Eq. (2.47), it is found that this in exion point occurs when
^x = sinθo. At this value of x^, we find from Eq. (2.47) that for zero stiffness α=
cos 2ð1
−
θo
cos
θoÞ
(2.48)
38
Modeling of Vibratory Systems Thus,
α
and
θo
fl
are not independent at this in ection point.
figuration
Notice that this spring con
does not need an initial compression or tension
of the springs to be able to create a spring system with zero stiffness and, therefore, differs from the spring con
figuration
examined previously. Equations (2.45) and (2.48) are
plotted in Figure 2.18 for the case when 0 at
x/L =
θo
= ° 70
and
k o/k
=
0.26. At these values,
kx
=
0.94.
INTERACTIVE GRAPHIC 2.4: GEOMETRIC NONLINEAR STIFFNESS OF THREE SPRINGS AND CONFIGURATIONS FOR ZERO STIFFNESS This interactive graphic is used to visualize the vertical force and the spring constant of the three-spring system shown in Figure 2.17 and the combination of parameters for which the spring coefficient in the vertical direction is zero. In this interactive graphic, the following should be noted. •
•
•
2.3.4
The strong interaction that the orientations of the two springs to the vertical spring and their relative stiffness have on the magnitude and linearity of the equivalent spring constant in the vertical direction. The value of x/L where the vertical equivalent spring constant equals zero varies with respect to the values of θo and k/ko, which are interdependent according to the relation k/ko = 0.5 cos( θo)/(1 − cos(θo)). Additionally, the vertical equivalent spring constant is symmetric about this value. There is always a set of values of x/L, θo, and k/ko for which the vertical equivalent spring constant equals zero.
Other Forms of Potential Energy Elements In the previous two subsections, we discussed stiffness elements in which the source of the restoring force is a structural element. Here, we consider other stiffness elements in which there is a mechanism for storing and releasing potential energy. The source of the restoring force is a
fluid element or a gravitational loading.
Fluid Element
fluid element, consider the manometer shown in Figure 2.19 in which the fluid is displaced by an amount x in one leg of the manometer. Consequently, the fluid has been displaced a total of 2 x. If the fluid has a density ρ and the manometer has an area A o, then the magnitude of the total force of the displaced fluid acting on the rest of the fluid is As an example of a
8
Fm ðxÞ = 2ρgAo x 8
For a proposed practical application of this type of system see: S. D. Xue,
(2.49)
“Optimum parameters of tuned liquid ” Journal of Sound and Vibration,
column damper for suppressing pitching vibration of an undamped structure,
–
235(2) (2000) 639 653.
2.3 Stiffness Elements
39
2.0 F/(kLo ) kx/k 1.5
edutilpmA
1.0
0.5
0.0 0.0
0.5
Figure 2.18.
1.0 x/L
1.5
2.0
fi
Force in the vertical direction and the corresponding spring coef cient for the three-
figuration of Figure 2.17a when θo =
spring con
x/L = 0.94. x
Ao
°
70
and
k o/k
=
0.26; for these values
k x/k
=
0 at
Figure 2.19. Manometer.
g x
r
Consequently, the equivalent spring constant of this
ke = from which it is clear that the ometer cross-sectional area
Ao,
dF m dx
=
fluid-element
fluid system is
ρgAo
2
(2.50)
stiffness depends on the density
and the acceleration due to gravity
g.
ρ,
man-
The corresponding
potential energy is
V ðx Þ =
1 2
k ex
2
= ρgAox
2
(2.51)
Alternatively, the potential energy can also be obtained directly from the work done
ð
x
V ðxÞ = Fm ðxÞdx 0
ð
x
=
2
ρgA 0 x dx = ρgA0 x2 0
(2.52)
40
Modeling of Vibratory Systems
Compressed Gas
Po and Lo is the origiamount x along
Consider the piston shown in Figure 2.20 in which gas is stored at a pressure entrained in the volume
Vo
= Ao Lo
Ao
, where
is the area of the piston and
nal length of the cylindrical cavity. Thus, when the piston moves by an
Vo
the axis of the piston,
decreases to a volume
Vc ðxÞ = V
0
−A x
=A L 0
Vc, where
!
0
0
1
−
x L
(2.53)
0
The equation of state for the gas is
PV nc = PoV on = co = constant
(2.54)
When a gas is compressed slowly, the compression is isothermal and compressed rapidly, the compression is adiabatic and
n
= cp cv /
n
=
1. When it is
fic heats
, the ratio of speci
of the gas, which for air is 1.4. To determine the spring constant, we note that the magnitude of the force on the piston is
F = A oP = A ocoV c−n Upon substitution of
(2.55)
Vc given by Eq. (2.54), we obtain F ðxÞ = A oco Vo−nð1 − x=LoÞ−n = A oPoð1 − x=LoÞ−n
Thus, the pressure-
filled
(2.56)
gas element provides the stiffness. Equation (2.56) describes a
nonlinear force versus displacement relationship. As discussed earlier, this relationship may be replaced in the vicinity of
x
= xl
by a straight line with a slope
ke ,
where
ke
is
the stiffness of an equivalent linear stiffness element. This equivalent stiffness is given by
dF ke = dx
=
Gas at P o Lo
Ao
Vo
x
Gas at P V Lo
Ao
x
³³ ³³ ³=
x xl
nAoPo −n− ð1 −xl =Lo Þ Lo
Figure 2.20.
1
Gas compression with a piston.
(2.57a)
2.3 Stiffness Elements For
|xl Lo| ≪ /
41
fi
1, Eq. (2.57a) is simpli ed to
nAoPo Lo
ke = For arbitrary
x/Lo,
the potential energy
Eq. (2.56) as
V(x)
ð x
0
ð
ð1
−x=Lo Þ−n dx
0
= −A o P o L o =
is determined by using Eq. (2.10) and
x
V ðx Þ = F ðxÞdx = A oPo
(2.57b)
A oPoLo n−1
h
lnð1
− x=LoÞ
− x=Lo Þ −n − 1
ð1
Pendulum Systems
m
Consider the bar of uniformly distributed mass
i
n =1 n ≠1
1
(2.58)
shown in Figure 2.21a that is pivoted
L/2
at its top. Let the reference position be located at a distance where the center of mass is located when the pendulum is at
θ
=
below the pivot point,
0. When the bar rotates
either clockwise or counterclockwise, the vertical distance through which the center of gravity of the bar moves up from the reference position is
x= Since
L 2
−
L 2
cos
θ=
L 2
ð1
−
cos
θÞ
(2.59)
F = mg, the increase in the potential energy is
ð
x
V ðxÞ =
ð x
F ðxÞdx =
0
mg dx = mgx
(2.60a)
0
m1 L/2
L/2 L
x L
L c.g.
x
m1 g L
L
x
mg m1
m1g (a) Figure 2.21.
(b)
(c)
Pendulum systems: (a) bar with uniformly distributed mass; (b) mass on a weightless
rod; and (c) inverted mass on a weightless rod.
42
Modeling of Vibratory Systems or, from Eq. (2.59),
V ðθ Þ = When the angle of rotation
θ
mgL 2
−
ð1
cos
θÞ
(2.60b)
about the upright position
the Taylor series approximation
θ
=
0 is
“small,”
we can use
9
θ2
θ = 1−
cos
+⋯
2
(2.61)
and substitute this expression into Eq. (2.60b) to obtain
V ð θÞ ≈
1
´mgLµ
2
2
θ2 =
1 2
ke θ
2
(2.62)
where the equivalent spring constant is
ke = In a similar manner, for
mgL
(2.63)
2
“small” rotations about the upright position θ = 0 in Figure 2.21b,
we obtain the increase in potential energy for the system. Here, it is assumed that a weightless bar supports the mass
m
1.
Choosing the reference position as the bottom position, we
obtain
V ðθÞ ≈
1 2
m gLθ
2
1
=
1 2
ke θ
2
(2.64)
where the equivalent spring constant is
k e = m gL
(2.65)
1
figuration shown in Figure 2.21b, if the weightless bar is replaced by one that has
In the con
a uniformly distributed mass
V ðθ Þ ≈
1 4
m, then the total potential energy of the bar and the mass is
mgLθ
2
+
1 2
m gLθ
2
1
where the equivalent spring constant is
ke =
¶m 2
=
1
¶m
2
+m
1
2
+m
·
1
gLθ
2
=
1 2
k eθ
2
·
gL
(2.66)
(2.67)
When the pendulum is inverted as shown in Figure 2.21c, then there is a decrease in potential energy, that is,
V ðθÞ ≈ − m gLθ 1 2
9
T. B. Hildebrand,
2
1
Advanced Calculus for Applications, Prentice Hall, Englewood Cliffs, NJ, 1976.
(2.68)
2.3 Stiffness Elements
EXAMPLE 2.7
43
Equivalent stiffness due to gravity loading
Consider the pivoted bar shown in Figure 2.22. The lower portion of the bar has a mass
m
1
portion portion
θ
=
m . The distance of the center of gravity of the upper of the bar to the pivot is a and the distance of the center of gravity of the lower of the bar to the pivot is b. For “small” rotations about the upright position
and the upper portion a mass
2
0, the potential energy is
V ðθÞ =
1 2
m gbθ
2
1
−
1 2
m gaθ
=
2
2
1 2
ðm 1 b
−m
aÞg θ
2
2
There is a gain or loss in potential energy depending on whether
(a)
mb 1
.
>m
a
2
or vice
versa
A special case of Eq. (a) is when the bar has a uniformly distributed mass
m
=
m
=
1
2
where
L=L
1
+L
. Then, since
2
b=L
= where the equivalent stiffness
1
L
1ð
1
2
−L Þ 2
2
length.
L2 a
b m1
L1
(b)
2
−
a
=L
/2, Eq. (a) becomes
L 4 ðL + L
mg θ
!
2
2
2
1
2
=
1 2
2Þ
ke θ
mg θ
2
2
(c)
ke is
Figure 2.22.
g
2Þ
1
ke =
m2
1
/2 and 2
V ð θÞ =
mL L mL L
1
L 4ð L + L
m so that
ðL1
−L Þ 2
2
mg
Pivoted bar with unequally distributed mass along its
(d)
44
Modeling of Vibratory Systems Hence, the stiffness is due to gravity loading. When
L
1
=L
,
2
V=
0. It is mentioned that
if the systems were to lie in the plane of the page, that is, normal to the direction of gravity, then there is no restoring force when the bar is rotated an amount relationships
are
Eq. (2.63) with
2.3.5
L
not
applicable.
replaced by
L
Furthermore,
when
L
2
=
0,
Eq.
(d)
θ
and these
reduces
to
.
1
Summary of Equivalent Spring Constants The results of Sections 2.3.2 to 2.3.4 are summarized in Table 2.4 along with other spring combinations.
2.4
DISSIPATION ELEMENTS
Damping elements are assumed to have neither inertia nor the means to store or release potential energy. The mechanical motion imparted to these elements is converted to heat or sound and, hence, they are called
nonconservative
or
dissipative
because this
energy is not recoverable by the mechanical system. There are four common types of damping mechanisms used to model vibratory systems. They are (i) viscous damping, (ii) Coulomb or dry friction damping, (iii) material or solid or hysteretic damping, and (iv)
fluid
damping. In all these cases, the damping force is expressed as a function
of velocity.
2.4.1
Viscous Damping When a viscous
fluid flows through a slot
or around a piston in a cylinder, the damping
force generated is proportional to the relative velocity between the two boundaries con-
fining the fluid. A common representation of a viscous damper is a cylinder with a piston head, as shown in Figure 2.23. In this case, the piston head moves with a speed to the cylinder housing, which is
fixed.
The damper force magnitude
F
x_ relative
always acts in a
direction opposite to that of velocity. Depending on the damper construction and the velocity range, the magnitude of the damper force
F ðx_ Þ
is a nonlinear function of velo-
city which, in some cases, can be approximated as a linear function of velocity. In the linear case, the relationship is expressed as
F ðx_ Þ = cx_ where the constant of proportionality denoted by damping coef
ficient c
Eq. (2.69) is also called
(2.69)
c is
called the
damping coefficient.
The
has units of N/(m/s). Viscous damping of the form given by
slow-fluid damping.
^xo θ nis oθ
2
soc
2
o
ok 6
nis soc
− oθ
^x2− 1
+k
nehW
ok 2
≪ x^
nis
2
;1
1
1
/
o
¶+
≪Lx = oT
oδ k
,1
+k
+k
1
ek
k+ k
ok
erehw
nehW
2
± − B@ k 0
L = oh = o θ nis dna oL =x = x ^ erehw
:o
2=3
o
2
x ^+ θ CA ² θ 1 L= oδ o k + k
+
δ ðo k
Þ L=x ð
2
=o
−L
·
.ecrof elisnet a si
2=3
Þ1
2
)m/N(
sgnirps fo
ek
oL
ok
k
F ,x
oh
2k
oq
k
ok
T,L o
ok
oL
F ,x
F
1k
lellaraP
stnatsnoc gnirps tnelaviuqE
−
k 2
1
+
´ ek
k 1
1
)m/N(
2k
1k
F
3
2
1
.4.2 elbaT
seireS
stnatsnoc gnirps tnelaviuqE
µ ek
1
smaeb dna sgnirps fo ek stnatsnoc gnirps tnelaviuqE
secived egarots ygrene laitnetop rehto dna ,smaeb ,sgnirps fo
45
Þb
2
ba
2
3
3
hh k
3
3
= hhk
= cck +k
k
e
+k
= hhk
cc k
3
a = fck IE 3 fc k + k
+ fck
L = fck IE 3
+ að IE
3
ba
3
+að IE
2
ba
2
+ að IE
Þb
3
Þb
hh k
)m/N(
sgnirps fo
b
b
b
F
F
F
F
k
L
k
k
a
a
a
L a
Þb 3
Þb
1
3
fck 1
3
cck 1
+
+
k 1
k 1
@ 0
= cck
@ 0
a = fck IE 3
A −1 3
ba
3
A 1
+a ð IE
− 1
2
A 1
3
hhk 1
k 1
@ 0
= hhk +
)m/N(
k
e
))dar/m·N( ek :01 esaC(
−
ba
2
1
+ að IE
sgnirps lellarap tnelaviuqE
lellaraP
k stnatsnoc gnirps tnelaviuqE
e
b
b
F k
k
F
L
F
k
a
a
a
seireS
7
6
5
4
).tnoc( .4.2 elbaT
46
2
2
2
1
1
b k+ a k + að k k
3
3
fc k ′
= cck
+ fck
L = fck IE 3
3
a = fc′k IE 3
Þb
2
2
3
+að IE
ba
3
θ 2nat Þ 2k + 1k ð
Þb 3
3
L = fck IE 3 cc k + fc k
2k
F
F
F
L q
L
L F
L
1k
a
L a
L a
⋅
m N
2
2
Lgm
1k
)derongi
1
k k
Þ 2D= 1D ð2 k
+
1
k
sraeg fo aitreni( 2
2D
1k
2/L
gm
q
m
1D
2k
q
secived egarots ygrene laitnetop rehto fo ek stnatsnoc gnirps tnelaviuqE
2k
b
b
b
21
11
01
9
8
47
)m/N(
k
e
sgnirps fo
ria rof 4: 1
=n ;1
o
L ≪ ³³ ³³ ³³ x ³³
⋅
)lellarap ni 31 dna 21 sesaC( m N
lellaraP
k stnatsnoc gnirps tnelaviuqE
e
m=N
1
oL o Po An
2
m¶
1
k
e
Lg m
Lg m +
·
⋅
m N
)m/N(
oA
x
L
q
g1m
m
q
gm
g1m
oP
1m
1m
2/L
L
oL
seireS
51
41
31
).tnoc( .4.2 elbaT
48
2.4 Dissipation Elements
.
Figure 2.23.
x c
Representation of a viscous damper
49
c.
m
c m
F
F
F
Figure 2.24.
x
ke x
ce x
ke
ce
–
(a) Nomenclature for the Kelvin Voigt
model; and (b) free-body diagram.
x
ke
ce
ke x
(a)
ce x
(b)
In the case of a nonlinear viscous damper described by a function
³ dF ðx_ Þ³³ c = d x_ ³ =
linear viscous damping around an operating speed
e
x_ = x_ l
F ðx_ Þ, the equivalent
is determined as follows:
(2.70)
x_ x_ l
Linear viscous damping elements can be combined in the same way that linear springs are, except that the forces are proportional to velocity instead of displacement.
Energy Dissipation The energy dissipated by a linear viscous damper is given by
Ed =
2.4.2
ð
F dx =
ð
F x_ dt =
ð
ð
cx_ dt = c x_ dt 2
2
(2.71)
Combinations of Viscous Dampers and Linear Springs In modeling systems for vibration analysis, there are two models that are frequently employed. The
first, and most common, is the placement of an equivalent spring in
par-
allel with an equivalent viscous damper as shown in Figure 2.24. The magnitude of the force
dx/ dt
F
required to displace one end of these components an amount
= x_
x
with a speed
is the sum of the spring force and the viscous force, that is,
F = k ex + ce x_
N
(2.72)
50
Modeling of Vibratory Systems F
Figure 2.25. (a) Nomenclature for the Maxwell
model; and (b) free-body diagram.
kx
c x – xd
k
c c x – xd
kx
F
k1xd
x
k1
xd
c k k1
kx
k1xd
(a)
(b)
–
This combination is often referred to as a Kelvin Voigt model and can be thought of as a representation of a material that has elasticity and viscosity, the latter being the mechanism whereby the material dissipates energy. Notice also that the magnitude given by Eq. (2.72) is also the magnitude of the force acting on the Another
model
that
has
proven
useful
in
certain
fixed boundary.
applications
is
that
shown
F in
Figure 2.25a. In this model, an additional spring is placed in series with the viscous damper.
This
combination
of
elements
is
called
a
Maxwell
model.
After
using
the
F
required
to displace the common end of these components by an amount shown in the
is
free-body model in Figure 2.25b, it is found that the magnitude of the force
given by the following two coupled
first-order ordinary differential equations
F = kx + cðx_ − x_ d Þ k xd = cðx_ − x_ d Þ 1
figure
N (2.73) N
Notice from the free-body diagram of Figure 2.25b that the magnitude of the force acting on the
fixed boundary FB is given by FB = kx + k xd 1
EXAMPLE 2.8
Equivalent damping coef
N
(2.74)
ficient and equivalent stiffness of a
vibratory system
Consider the vibratory system shown in Figure 2.26a in which the motion of mass
m
is
restrained by a set of linear springs and linear viscous dampers. A free-body diagram of this system is shown in Figure 2.26b. In determining the spring force generated by the springs and
k
3,
we have made use of the fact that these springs are in series and used Eq. (2.18).
k
2
2.4 Dissipation Elements
51
x k1
k2
k3
k1x
m
m
.
c 1x
c2
c1
k2k3/(k2
(a)
k3 )x
.
c2 x
(b)
x ke m ce
(c) Figure 2.26.
(a) Linear vibratory system; (b) free-body diagram of mass
m; and (c) equivalent
system.
In Figure 2.26c, the springs and dampers shown in Figure 2.26a have been collected and expressed as an equivalent spring and equivalent damper combination. Thus, we have
kk
ke = k
+ k +k ce = c + c 2
3
1
2
1
EXAMPLE 2.9
3
2
ficient of a nonlinear damper
Equivalent linear damping coef
It has been experimentally determined that the damper force-velocity relationship is given by the function
F ðx_ Þ = ð4 N ⋅ s=mÞ x_ +
±
0:3 N
⋅
3
3
s =m
²x_
3
N
(a)
fi
We shall determine the equivalent linear damping coef cient around an operating speed of
³ dF ð x_ Þ³³ = ³ d x_ ³ =
fi
3 m/s. To determine this damping coef cient, we use Eq. (2.70) and Eq. (a) and arrive at
ce
= =
4 N
⋅
x_
⋅
4 N
⋅
s=m
+
3 m=s
s=m
12:1 N
±
=
+
s=m
0:9 N
⋅
3
s =m
3
±
0:9 N
²×±
2
3
⋅
3
3
s =m
2
2
m =s
²
²x_ ³³ 2
x_ =3
m=s
(b)
52
Modeling of Vibratory Systems
2.4.3
Other Forms of Dissipation
Coulomb Damping or Dry Friction This type of damping is due to the force caused by friction between two solid surfaces. The force acting on the system must oppose the motion; therefore, the sign of this force must have the opposite sense (direction) of velocity as shown in Figure 2.27. If the kinetic coef
ficient of friction is μ and the force compressing the surfaces is N , then F ð x_ Þ = μN sgnðx_ Þ
(2.75)
where sgn is the signum function, which takes on the value of the argument (
x_
in this case),
–1
+
1 for positive values of
for negative values of the argument, and 0 when the
argument is zero. If the normal force is due to the system weight, then
N
= mg
and we
have
F ðx_ Þ = μmg sgnð x_ Þ
ð
The energy dissipated in this case is given by
Ed =
F dx =
= μmg
ð
ð
F x_ dt x_ Þx_ dt
(2.77)
ficiency
of internal combustion engines,
sgnð
Although dry friction can result in loss of ef
(2.76)
wear on contacting parts, and the loss of position accuracy in servomechanisms, it has been used to enhance the performance of turbomachinery blades, certain built-up structures, and earthquake isolation.
10
Fluid Damping This type of damping is associated with a system whose mass is vibrating in a ium. It is often referred to as
velocity-squared damping.
fluid med-
This force always acts in a direc-
tion opposite to that of the velocity of the mass. The magnitude of the damping force is given by
F ðx_ Þ = cd x_
m
. x
2
x_ Þ = cd ∣x_ ∣x_
sgnð
(2.78)
m . F(x)
Figure 2.27.
10
Dry friction.
“Friction damping and isolation systems,” ASME Journal of Vibrations and Acoustics, Special 50th Anniversary Design Issue, 117 (1995) 196–206.
A. A. Ferri,
2.4 Dissipation Elements
53
where
cd = C is a drag coefficient, A is the x_ , and ρ is the density of the fluid.
and
1 2
C ρA
(2.79)
projected area of the mass in a direction normal to
The energy dissipated is
Ed =
ð
= cd
F dx =
ð
ð
F x_ dt
x_ Þx_ dt 3
sgnð
Fluid damping of the form given by Eq. (2.78) is often referred to as
(2.80)
fast-fluid damping.
Structural or Solid or Hysteretic Damping This type of damping describes the losses in materials due to internal friction. The damping force is a function of displacement and velocity and is of the form
F = k πβh sgnð x_ Þ∣ x_ ∣ where
βh is
ð
ð
(2.81)
an empirically determined constant. The energy dissipated is
Ed =
F dx =
= k πβh
ð
F x_ dt x_ Þ∣x∣x_ dt
sgnð
(2.82)
From the discussions in Sections 2.4.1 and 2.4.2, it is clear that the damping force is linearly proportional to velocity for linear viscous damping, and the damping force is a nonlinear function of velocity for nonlinear viscous damping, Coulomb damping, and
fluid
damping. The structural damping model is used only in the presence of harmonic excitation, as discussed further in Section 5.7. Although damping models were presented only for translational motions, they are equally valid for rotational motions.
Damping in Microelectromechanical Systems (MEMS) There are several damping mechanisms that apply only to vibrating systems that are small in size, on the order of fractions of a millimeter. Two of the most commonly
fl
encountered ones are brie y described below.
Squeeze Film Air Damping Squeeze
film
11
air damping is caused by the entrapment of air between two parallel sur-
faces that are moving relative to each other in the normal direction. The reactive force
11
E. B. Magrab,
Vibrations of Elastic Systems: with Application to MEMS and NEMS,
Applications 184, Springer Science+Business Media B.V., New York, 2012, Section 2.3.
Solid Mechanics and Its
54
Modeling of Vibratory Systems brought about by the relative movement of the surfaces consists of two components.
flow
One component is due to the viscous
of air that is squeezed out of the volume
between the two surfaces and the second component is due to compression of the air between
the
damping
csf
two
surfaces.
The
former
component
results
in
a
frequency-dependent
and the latter component results in a frequency-dependent stiffness
nondimensional parameter that determines whether the squeeze
ksf .
The
film damping will be sig-
ficant is called the squeeze number, which is defined for a circular surface as
ni
μ aωb2o
σ= In Eq. (2.83),
ω
(rad/s) is the frequency,
radius of the circular surface,
Pa
12
(2.83)
Pa ho 2
μa
⋅
2
(N s/m ) is the viscosity of air,
2
(N/m ) is the atmospheric pressure, and
distance between the vibrating surface and the
fixed
surface
–
bo ho
(m) is the (m) is the
the air gap. The stiffness
and damping are determined from
ksf
= Pa bo Sk ðσ Þ
csf
Pa bo Sc ðσ Þ ωho
=
N m
ho
=
N
⋅
(2.84)
=
s m
S (σ) are expressions involving Kelvin functions. ksfho/( Pbo) is less than 0.05 when σ < 1.6 and is greater than 0.8 when σ > 50.3. The quantity csf hoω/(Pbo) has a maximum value of 0.377 when σ = 6.233. It is less than 0.1 when σ < 0.82 and when σ > 179. For air at standard conditions, Pa = 10 N/m − N⋅s/m . If we assume that h = 2 μm, b = 0.5 mm, and f = 2500 Hz, and μa = 1.87 × 10 o o then σ = 2.2. On the other hand, if ho = 0.2 mm, bo = 0.05 m, and f = 25 Hz, then σ = 0.022 where
The quantity
5
5
and the squeeze
2
2
film effects are virtually nonexistent.
Viscous Fluid Damping
12
The motion of an oscillating rigid cylindrical body of radius merged in a viscous
fluid
is resisted by the
fluid
b
and length
L
that is sub-
pressure distribution on the surface of
the body. When the pressure distribution is converted to a force in the direction of the motion of the rigid body, it is found that the resisting force consists of two components, one is a frequency-dependent damping mass
m a. These quantities are given by
ca
and the other is a frequency-dependent added
ma = ρf ALΓ mð ReÞ
ca = ρf ALωΓ cð ReÞ
12
E. B. Magrab,
ibid., Section 2.4.
kg
=
kg s
ðN
⋅
=
s mÞ
(2.85)
2.5 Model Construction
55
where
Re
is the Reynolds number,
A = πb
2
ω
=
ρf ωb2
(2.86)
μf
4
(rad/s) is the frequency,
ρf
/4 (m ) is the cross-sectional area of the cylinder, and
viscosity of the
fluid.
In Eq. (2.85),
Γ =Γ m+fΓc
pffiffiffiffiffiffi
placed by the cylinder. When Re
Γc ≈
2.83/
Re (within 5%) and
=
For water, Re
= b=
>
1.57
×
fb
6
10
ca
≈
, where
ρfALω/ /2 π. For
2.83
=ω
f
0.025. At this Reynolds number,
hand, when
0.3 m and
ρfAL.
0.024
=
10 Hz, Re
=
(N
⋅
fluid,
2
s/m ) is the dynamic
1.4
p ffiffiffiffiffiffi
×
Γm
1
2.83/
a cylinder with 10 and
Γc
6
10 . In this case,
>
dis-
Re: When Re
Re:
>
fluid
is the mass of the
pΓffiffiffiffiffimffi ≈ +
1000 Hz, Re
f
ρfAL
4, to within 1%,
2
μf
is called a hydrodynamic function, which
is a function of the Reynolds number. The quantity
200,
3
(kg/m ) is the density of the
2
b
ma
=
4
μm
and
f
>
=
20. On the other
≈ ρfAL
and
ca
≈
This high Reynolds number case is examined in Section 3.5.3.
2.5
MODEL CONSTRUCTION
2.5.1
Introduction As discussed in the subsequent chapters, the mass, stiffness, and damping of a system appear as parameters in the governing equations of the system. When only discrete elements
are
used
to
model
equations is referred to as a
a
physical
system,
the
associated
system
discrete system or a lumped-parameter system.
as will become evident in later chapters, since a
fi
finite
of
governing
In these cases,
number of independent displace-
ment or rotation coordinates suf ces to describe the position of a physical system, discrete systems are also referred to as element
is
used
to
model
equations is referred to as a
a
finite degree-of-freedom systems. When a
physical
system,
the
associated
distributed-parameter system
or a
system
of
distributed governing
continuous system.
In this
case, one or more displacement functions are needed to describe the position of a physical system. Since a function is equivalent, in a sense, to specifying the displacement at
fi
every point of the physical system or the displacements at an in nite number of points, distributed-parameter systems are also referred to as
2.5.2
in finite degree-of-freedom systems.
A Few Simple Models The representation of a physical system as a model composed of discrete masses, springs, and dampers is a bit of an art. As a rule, one should start with the simplest model that will preserve the important characteristics of the system and then improve the model as necessary. Before creating the model, one should determine how it is or will be excited: forces
56
Modeling of Vibratory Systems on the mass, displacements of its base, or self-excited. Examples of these types of excitations are: Forces on the mass or inertia element Flow of air over an aircraft wing Wind on tall buildings Wave forces on ships and offshore structures Magnetic forces on loudspeakers and electrodynamic shaker tables Electrostatic actuation of microelectromechanical systems Unbalance in all types of rotating machinery: motors, fans, compressors, gearboxes, internal combustion engines, wind turbines, washing machines, etc. Impact machines: pile drivers, drop forges, jack hammers, etc. Base excitation Buoyancy of ship hulls Road surfaces on vehicle tires Earthquakes
Forces on the Mass and Displacement of the Base of a Civil Structure Civil structures are typically subject to wind and, in certain parts of the world, earthquakes. The wind is usually represented as a force on the mass and an earthquake as a displacement
of
the
base.
Seattle’s space needle. As a
A
representative
civil
structure
is
shown
in
Figure
2.28:
first approximation, the vertical support structure is modeled
as a uniform cantilever beam whose stiffness is estimated from Case 4 of Table 2.3. If the structures are subjected to wind, then the model is that denoted as Model 1 in Figure 2.28. This type of system is discussed in Chapters 4 through 6. If the excitation is at the base, then the model is denoted as Model 2, where in this model the rotational inertia
JO
of the structure is restrained by the elasticity of the structure’s foundation
with the soil. This elasticity is captured as a torsion spring. In this case, the base is attached to the torsion spring that is subjected to a rotation
θquake.
This type of system is
discussed in Sections 3.5.1, 5.2.6, and Examples 6.5 and 6.7. An improved approximation would be to consider the vertical support structure as a cantilever beam with a rigid mass attached to its free end. In this approximation, one should also take into account the compressive force created by the weight of the observation deck of the space needle. This model is shown as Model 3 in Figure 2.28 and is discussed in Sections 9.3.4 and 9.3.6.
Magnetic Forces on the Mass: Loudspeaker A common device for which vibrations are desirable is a loudspeaker, a cutaway of which is shown in Figure 2.29. This device has a permanent magnet that surrounds a cylinder that is wrapped with copper wire. One end of the cylinder is attached to a conical diaphragm. When a current
flows in the coil, the cylinder moves in its axial direction
2.5 Model Construction fwind
57
fwind
mdeck
mdeck
kcant
kcant
JO
Model #1
Model #2 ksoil O quake
Cantilever beam Model #3 fwind Figure 2.28.
mdeck
Tall structure modeled several ways depending on the type of excitation and the
sophistication of the model.
Source: Photo by Cacophony at https://commons.wikimedia.org/wiki/File:SpaceNeedleQAClose. jpg and reprinted under Creative Commons Attribution 3.0.
Model #4
kdiap mcoil
fcoil
fcoil mcoil kdiap
Magnet
Figure 2.29.
Loudspeaker modeled as a spring-mass system.
Source: Courtesy of Audiotechnology ApS Denmark www.audiotechnology.dk/iz.asp?id=4|a | 119
|||
.
and the diaphragm acts as both a spring and an acoustic radiator. The loudspeaker can be modeled as the spring-mass system denoted as Model 4. Notice that Model 1 for the civil structure and Model 4 for the loudspeaker are the same.
Forces on an Aircraft Wing Consider the two-engine jet aircraft shown in Figure 2.30. One way to model only the wing is to use the spring-mass model denoted as Model 5. In this model, the mass is the
58
Modeling of Vibratory Systems
mengine
Tapered cantilever beam
kfuselage
kwing
flift flift
mengine Model #6
flift mengine
Model #5
kwing
Figure 2.30.
Two models for an aircraft wing.
Source : Photo of Boeing 777–300ER in flight copyright
©
Boeing and reprinted with permission
of Boeing/Boeing Images.
mass of the engine and the spring stiffness is approximated by the bending stiffness of a cantilever beam that extends from the fuselage to the engine. The force is the total force on the wing and is assumed to be concentrated on the mass. It is noticed that Models 1, 4, and 5 discussed for the three different physical systems are qualitatively the same. In an improved model, the entire wing can be considered as a tapered beam carrying a rigid mass (the engine) at an in-span location. The elasticity of the joint where the wing meets the fuselage is represented by a torsion spring. In this model (Model 6), the lift force is assumed to be distributed along the length of the beam. This type of system is discussed in Sections 9.3.5 and 9.4.
Flexible System: Pendulum Pendulum
systems
occur
quite
frequently
with
many
types
of
construction
cranes.
Shown in Figure 2.31 is a tower crane with its payload. This system can be modeled as a simple pendulum, which is denoted as Model 7 in Figure 2.31. This type of system is examined in Example 3.11 and Table 3.2.
2.5 Model Construction
L
59
Model #7 L
mobject mobject
Figure 2.31.
Tower crane payload modeled as a pendulum.
Model #8 c.g. ktire
Figure 2.32.
ktire
Spring-mass model of a motorcycle, which can both translate and rotate.
Source: Courtesy Vincent Nguyen.
System that Translates and Rotates: Motorcycle Many systems can both translate and rotate. One such common system is a motorcycle, as shown in Figure 2.32 along with its model, which is denoted as Model 8. The model allows its center of gravity to move up and down vertically and to rotate about its center of gravity. This model is known as a two degree-of-freedom model and is discussed in Examples 7.3 and 7.16 and Table 8.1.
Impact Machines: Drop Forge One aspect of vibration design requires the minimization of the transmission of impact forces
from
devices
to
the
surrounding
environment.
For
the
drop
forge
shown
in
Figure 2.33, an effective way to achieve this goal is to mount the system on elastic supports. Such a system can be modeled as a spring-mass system and is denoted as Model 9 in the
figure.
Notice that Model 9 is the same as Model 1. This method of isolation can
be improved by mounting the elastic supports beneath the forge on a massive mass and
fi
then placing this mass on its own elastic supports. This modi ed system is denoted Model 10 in the
figure. This type of system is discussed in Section 8.5.
System with an Unbalanced Mass: Axial Fan Consider the axial fan shown in Figure 2.34. The motor that drives the fan blades is supported by elastic mounts that are attached to the fan’s shroud. If one or more of the fan
60
Modeling of Vibratory Systems Ground Impact Force Reduction Better
Good
fimpact
fimpact mforge
mforge
mforge
kisolation
kisolation
misolation Model #9
k’isolation Model #10
Figure 2.33.
Isolation of impact forces from a drop forge.
Source : Photo by F. Broer https://commons.wikimedia.org/wiki/File:Drop_forging_Gesenkschmieden. jpg and reprinted under Creative Commons Attribution 3.0.
ksupport mmotor + fan mfan unbalance
4ksupport Motor
Model #11
Figure 2.34.
Axial fan and its spring-mass model to represent a rotational out-of-balance system.
Source : Copyright © 2017 Nuaire: A Trading Division of Polypipe Limited. www.nuaire.co.uk/ our-products/catalogue/oem-products/oem-ambient/. Reprinted with permission of Nuaire. Note: The image of the fan is a representation of a typical axial fan and is not meant to imply that this manufacturer ’s fan is out of balance.
blades is out of balance, then this system can be represented by Model 11. This type of system is discussed in Sections 3.5.2 and 5.2.5.
2.5.3
A Microelectromechanical System In Figure 2.35, a microelectromechanical accelerometer
13
is shown along with the vibra-
tory model of this sensor. In this sensor, the dimensions of the cantilevered structure are
13
K. E. Petersen, A. Shartel and N. F. Raley, “ Micromechanical accelerometer integrated with MOS detection circuitry,”
IEEE Transactions on Electronic Devices , 29(1) (1982) 23–24.
2.5 Model Construction
61
End mass m
Ground Output Fixed end
Free end
Physical system
Electrostatic force Fe End mass m
x
Equivalent structure stiffness k
..
Base acceleration y
Single degree-of-freedom system Figure 2.35.
Microelectromechanical accelerometer and a vibratory model of this sensor.
Source: K. E. Petersen, A. Sharted and N. F. Raley, “Micromechanical accelerometer integrated with MOS detection circuity, ” IEEE Transactions on Electronic Devices , 29(1) (1992) 23 –27. Copyright
©
IEEE. Reprinted with permission.
of the order of micrometers and the weight of the end mass is of the order of micrograms. A coating on top of the structure serves as one of the faces of a capacitor and another layer below the structure serves as another face of the capacitor. The gap between the capacitor faces changes in response to the accelerations experienced by the sensor, and the change in voltage across this capacitor is sensed to determine the acceleration. In constructing the vibratory model, the inertia of the cantilevered structure is ignored and this structure is represented by an equivalent spring with stiffness
k. The mass of the
cantilevered structure is assumed to be negligible and the end mass is modeled as a point mass of mass
m.
Consequently, the motion of this inertial element is described by a sin-
gle generalized coordinate
x,
and the model is an example of a single degree-of-freedom
system. The electrostatic force due to the capacitor acts directly on the mass, while the acceleration to be measured
€y
acts at the base of the vibratory model. The electrostatic
force that acts directly on the inertial element is an example of a direct excitation, while
62
Modeling of Vibratory Systems z
Head Eyeballs Upper torso Arms and shoulders Spinal column (stiff elasticity)
Thorax-abdomen Hips
Legs y
Feet
Applied force
x Figure 2.36.
Vibrating platform Human body and a vibratory model.
Source : M. P. Norton, Fundamentals of Noise and Vibration Analysis for Engineers, Cambridge University Press, New York, 1989. Copyright
©
1989 Cambridge University Press. Reprinted
with the permission of Cambridge University Press.
the acceleration acting at the base is an example of a base excitation. In a re
fined model
of the system, the mass of the cantilevered structure can also be lumped together with that of the end mass to obtain an effective point mass. No damping elements are used in constructing the vibratory model because the physical system has
“very
low
”
damping
levels. This model is similar to Model 1 when subjected to mass excitation; the case of base excitation is discussed in Section 5.5.
2.5.4
The Human Body In Figure 2.36, the human body and a vibratory model used to study the response of this physical system when subjected to vertical excitations are shown. While the vibratory model used in the previous section has only one discrete inertia element and one discrete spring element, the model
14
shown in Figure 2.36 has many inertial, spring, and
damper elements. Since many independent displacement variables are needed to describe the motion of this physical system, this vibratory model is an example of a system with multiple degrees of freedom treated in Chapters 7 and 8.
14
M.
P.
Norton,
Fundamentals of Noise and Vibration Analysis for Engineers,
New York, 1989.
Cambridge
University
Press,
2.5 Model Construction
63
The human body is highly sensitive to vibration levels. While the body may sense displacements with amplitudes in the range of a hundredth of a millimeter, some of the components of the ear can sense even smaller displacements. In the low-frequency range from 1 Hz to 10 Hz, the perception of motion is said to be proportional to acceleration, and in the mid-frequency range from 10 Hz to 100 Hz, the perception of motion is said to be proportional to velocity. In addition, the level of stimulation also needs to be considered. The responses of different parts of a human body are also dependent upon the frequency content of the excitation. For example, the thoraxabdomen system is highly responsive to vibrations in the range of 3 Hz to 6 Hz, the head-neck-shoulder system to vibrations in the range of 20 Hz to 30 Hz, and the eyeball to vibrations in the range of 60 Hz to 90 Hz. In terms of modeling, the detailed model shown in Figure 2.36 can be further simpli
fied
based on the frequency content
of the excitation. If, for example, the excitation has no frequency content below 20 Hz, then it is not necessary to have a detailed spring-mass-damper model for the thoraxabdomen system. In biomechanics and biomedical engineering, there is an area called
tion
whole-body vibra-
where one is concerned with the response of a human body to different types of
vibratory excitations and medical aspects of occupational exposure to these excitations. 15
There are detailed international standards
that provide acceptable vibration levels in
terms of acceleration magnitudes for horizontal and vertical vibrations based on exposure times and frequency content. In another area, called
hand-arm vibration, one is
con-
cerned with the response of the hand-arm system to vibratory excitations and medical aspects of occupational exposure to such excitations. These are also covered by interna16
tional standards.
2.5.5
A Ski A cross-country ski is shown in Figure 2.37. The corresponding vibratory model is
fi
usually a system with an in nite number of degrees of freedom. In the model shown here, the ski is modeled as a collection of discrete inertial elements. In order to take into account that each end or boundary of the ski is free, the inertial elements at the boundaries are only constrained by a spring element on one side in the
X-direction.
Furthermore, each of these inertial elements is allowed two translational degrees of
15
ISO 2631/1,
“Evaluation
of human exposure to whole-body vibration: General requirements,
International Standards Organization, Geneva, 1985 and ISO 2631/2, body
vibration:
Continuous
and
shock-induced
vibration
in
“ Evaluation
buildings
(1
to
”
ISO 2631, Part 1,
of human exposure to whole-
80
”
Hz),
ISO
2631,
Part
2,
International Standards Organization, Geneva, 1985. 16
ISO 5349/1, ments,
”
“Measurement
and evaluation of human exposure to hand-transmitted vibration: General require-
ISO 5349, Part 1, International Standards Organization, Geneva, 1999 and ISO 5349/2,
“ Measurement
and evaluation of human exposure to hand-transmitted vibration: Practical guidance for measurement at the
”
workplace,
ISO 5349, Part 2, International Standards Organization, Geneva, 2001.
64
Modeling of Vibratory Systems
Physical system
Y
Vibratory model
X Z
Figure 2.37.
Cross-country ski, which is a physical system with distributed stiffness and inertia
properties, and its vibratory model.
freedom, one along the ments are
X -direction
considered as point
and the other along the
masses
only, then they
Y-direction.
do not have
If these ele-
any
rotational
degrees of freedom. However, if the inertial elements are treated as rigid bodies, then rotational degrees of freedom about the axis oriented along the
Z -direction
also need
to be taken into account. In the limit, as the ski is broken up into a collection of in
finitesimal
segments, where
each segment is modeled as either a point mass or a rigid body, one ends up with a vibratory model with an in
finite number of degrees of freedom. The process of discretiz-
ing a spatially distributed system into a collection of inertial elements can be realized through various means. This aspect is not addressed in this book; however, if the model of the ski is a beam, then the results of Chapter 9 may be applicable.
2.5.6
Cutting Process Consider the cutting process model
17
shown in Figure 2.38. The physical system consists
of different components of the turret, lathe bed, and the tool used for cutting the workpiece. Unlike the previous vibratory models, a more complex model is constructed. While previously only lumped masses or discrete inertia elements were used, here, a distributed inertia element, a beam, is also used in the model. In the vibratory model, the bed is modeled as an elastic beam with a length
m b, area moment of inertia Ib about elasticity E b. The turret is modeled as a
mass per unit length modulus
of
degree of freedom
17
θ
Lb,
the bending axis, and Young’s rigid
body
with
a
rotational
and a translational degree of freedom along the vertical direction.
M. U. Jen and E. B. Magrab, “The dynamic interaction of the cutting process, work piece, and lathe’ s structure in facing,”
ASME Journal of Manufacturing Science and Engineering , 118 (1996) 348–358.
2.6 Design for Vibration
Lw, mw, Ew, Iw
Turret
65
Cantilever beam (work piece)
Tool slide
Tool
Tool
Cutting force Turret J , Mm
Turret and slide stiffness and damping
c
Work piece Cutting force
Machine bed Beam fixed at each end (machine bed)
Physical system Figure 2.38.
Lb , mb , Eb , Ib
Vibratory model
Workpiece-tool turning system and vibratory model of this system.
The mass moment of inertia of the turret is represented by and the tool together is represented by
Mm .
Jθ
and the mass of the turret
Spring elements are introduced between the
turret and the bed, and a damper element is introduced to model damping in the turret. A viscous-damping element
c
is used to model the damping experienced by the turret
and the damping experienced in the tool-slide system. The workpiece is also modeled as
mw , area moment of inertia Iw about the bending axis, and Young’s modulus of elasticity E w. Since the model a distributed inertia element with a length
Lw,
mass per unit length
fi
shown in Figure 2.38 consists of distributed inertial elements, this model has in nite number of degrees of freedom. In the modeling, the beam model used for the turret is
fixed at both ends, while the beam model used for the workpiece fixed at one end and hinged or free at the other end where it is being cut.
considered ered
is consid-
The stability of a machine tool is determined from a vibratory model to avoid undesirable cutting conditions called chatter. This type of analysis can be used to choose parameters such as width of cut, spindle rpm, and so on. In Chapter 9, vibrations of beams used in the model in Figure 2.38 are discussed at length.
2.6
DESIGN FOR VIBRATION
Principles that govern single degree-of-freedom systems, multiple degree-of-freedom systems, and continuous vibratory systems are covered in this book and are presented along with information needed to experimentally, numerically, and analytically investigate a vibratory system. In Figure 2.39, we show how these different aspects are used to design
fi
a system to have speci c vibratory characteristics.
66
Modeling of Vibratory Systems Design for Vibration Objectives Isolation Absorption Natural frequency Damping ratio Filter Mode shape
Model Assumptions 1 degree-of-freedom 2 degree-of-freedom
N degree-of-freedom Beam Excitation Force Base
System
Experiments and Measurements Data Collection and Interpretation Parameter Identification Mass Stiffness Linear Nonlinear Damping Viscous Fluid Material Dry friction System Characterization Natural frequency Impulse response Transfer function Excitation Characterization Harmonic Transient Random
Velocity Displacement
Figure 2.39.
2.7
Numerical Evaluation and Analysis Response Characteristics Time Domain Frequency Domain Displacement Velocity Acceleration Force at base
Design for vibration.
SUMMARY
In this chapter, the inertia elements, stiffness elements, and dissipation elements that are used to construct a vibratory model were discussed. An ideal inertia element, which does not have any stiffness or dissipation characteristics, can store or release kinetic energy. An ideal stiffness element, which does not have any inertia or dissipation characteristics, can store or release potential energy. The stiffness element can be due to a structure, a
fluid element, or gravity loading. The notion of equivalent stiffness has been introduced. An ideal dissipation element, which does not have any stiffness or inertia characteristics, is used to represent energy losses of the system. The notion of equivalent damping has also been introduced. Numerous examples of modeling physical systems with springmass models and beam models have been provided. The use of interactive graphics material to carry out parametric studies in modeling efforts has also been provided.
Exercises
67
Exercises
Section 2.1 2.1
Examine Eqs. (2.1) and (2.5) and verify that the units (dimensions) of the different terms in the respective equations are consistent.
Section 2.2 2.2
Consider the slider mechanism of Example 2.2 and show that the rotary inertia about the pivot point
2.3
O′
is also a function of the angular displacement
JO′
φ.
Consider the crank-mechanism system shown in Figure E2.3. Determine the rotary inertia of this system about the point
O
and express it as a function of the angular
Jd about the point O. The crank has a mass mG and rotary inertia JG about the point G at the center of mass of the crank. The mass of the slider is mp. θ.
displacement
The disc has a rotary inertia
Figure E2.3.
l
a
b
G
r
d O
mG , JG
Jd
mp
Section 2.3.2 2.4
Find the equivalent length
Le
that
constant
has
the
same
spring
of a spring of constant cross-section of diameter as
the
tapered
spring
Table 2.3. Both springs have the same Young’s modulus 2.5
shown
in
Case
2
d
2
of
E.
Extend the spring combinations shown in Figures 2.6b and 2.6c to cases with three springs. Verify that the equivalent stiffness of these spring combinations is consistent with Eqs. (2.14) and (2.16), respectively.
2.6
Consider the mechanical spring system shown in Figure E2.6. Assume that the bars are rigid and that
|x
a
/( /2)
|≪
2
1 and determine the equivalent spring constant
ke, which we can use in the relation F = kex.
68
Modeling of Vibratory Systems a
Figure E2.6.
k
L
L
F, x Consider the three beams connected as shown in Figure E2.7. The beam that is
2.7
attached to the ends of the two cantilever beams is pinned so that its ends can rotate unimpeded. Determine the system’s equivalent spring constant for the transverse loading shown. Figure E2.7.
E1 , I1 , L1
F L/2
E1, I1, L 1
L
E, I
For the weightless pulley system shown in Figure E2.8, determine the equivalent
2.8
spring constant. Recall that when the center of the pulley moves by an amount the rope moves an amount 2
x,
x.
Figure E2.8.
k1
F, x
2.9
k2
Determine the equivalent stiffness for each of the systems shown in Figure E2.9. Each system consists of three linear springs with stiffness Figure E2.9.
k1
k2
k3
k1 k3 k2
k,k 1
, and
2
k
.
3
Exercises
69
For the two cantilever beams whose free ends are connected to springs shown in
2.10
Figure E2.10, give the expressions for the spring constants mine the equivalent spring constant
L1
k3
L2 k2
E2, I2
k
1
and
k
2
and deter-
k e for the system.
Figure E2.10.
k1
E1 , I1
k4 F
For the system of translation and torsion springs shown in Figure E2.11, deter-
2.11
mine the equivalent spring constant for torsional oscillations. The disc has a radius
b, and the translation springs are tangential to the disc at the point of attachment. kt1
k1
Figure E2.11.
k2
k3
kt2 kt3
2.12
For
the
system
of
translation
springs
shown
in
Figure
equivalent spring constant for motion in the horizontal ( that
|Δx| ≪
1 so that the
θj
E2.12,
determine
the
x) direction only. Assume
remain constant.
Figure E2.12.
k6 k5
k4
1
2
x
3
k3
k2 k1
2.13
Consider the system shown Figure E2.13, which lies in the tem, which is called a crab-leg A load along the
X-direction
flexure,
X–Y
plane. This sys-
is used in microelectromechanical sensors.
is applied to the mass to which all of the four
70
Modeling of Vibratory Systems
flexures
flexure has a shin of length L along the Y -direction, and thickness h along the Z-direction. The thigh of each fl exure has a length L t along the Y -direction, a width bt along the X -direction, and a thickness h along the Z -direction. The equivalent stiffness of each of these fl exures in the X-direction can be shown
crab-leg
X-direction,
are attached. Each
width
b
along the
to be
Ehb
¶ · L + L ðb=b Þ ¶ ·
3
k flexure =
t
4
L L + Ltðb=bt Þ 3
3
where the Young’s modulus of elasticity is the polysilicon material from which the each crab leg are as follows: length
Lt
L
=
100
b
E,
which has a value of 160 GPa for
flexure
μm
and
spanning the range from 10 to 75
stiffness of the system versus
3
t
b
μm,
is fabricated. The dimensions of
= bt = h =
2
μ m.
For the thigh
plot the graph of the equivalent
L t. Figure E2.13.
L
Source: See Figure 2.12.
Lt Y Mass X
Thigh
Crab-leg flexure
2.14
Shin
Based on the expression for equivalent stiffness of each thigh width
dkflexure/ dbt,
bt
kflexure provided in
flexure
Exercise 2.13, the sensitivity of the
with respect to the
flexure
width
can be assessed by determining the derivatives
b
and the
dk flexure/db
and
respectively. Carry out these operations and discuss the expressions
obtained. 2.15
Find the torsion-spring constant
k te
of the stepped shaft shown in Figure E2.15,
where each shaft has the same shear modulus length of a shaft of constant diameter
De
G.
Determine the equivalent spring
and length
Le
that has the same spring
constant as the torsion spring shown in Figure E2.15.
L1 , D1 L , D 2 2
2.16
L3 , D3
Figure E2.15.
Verify the expressions for the equivalent spring constants for Case 11 of Table 2.4.
Exercises
71
Section 2.3.3 2.17
Consider the two nonlinear springs in parallel that are shown in Figure E2.17. The force-displacement relations for each spring are, respectively,
Fj ðxÞ = kj x + k j αx
3
j = 1; 2
(a) Obtain the expressions from which the equivalent spring constant can be determined at a displacement (b) If
F
=
1000 N,
k
1
=k = 2
xo.
50 000 N/m, and
xo and the equivalent spring constant.
α
=
2 m
–2
, determine the value of
Figure E2.17.
F1 (x)
F2 (x)
x
F
2.18
Consider the two nonlinear springs in series shown in Figure E2.18. The forcedisplacement relations for each spring are, respectively,
Fj ðxÞ = kj x + k j αx
3
j = 1; 2
(a) Obtain the expressions from which the equivalent spring constant can be determined at a displacement (b) If
F
=
1000 N,
k
1
=
xo.
50 000 N/m,
the equivalent spring constant.
Figure E2.18.
F1 (x)
F2 (x)
x F
k
2
=
25 000 N/m, and
α
=
2 m
–2
, determine
72
Modeling of Vibratory Systems 2.19
Consider the data in Table E2.19 in which the experimentally determined tire loads
fl
versus tire de ections have been recorded. These data are for a set of dual tires and a single wide-base tire.
18
fl
2
The in ation pressure for all tires is 724 kN/m . Examine
the stiffness characteristics of the two different tire systems and discuss them.
Table E2.19.
flection data
Tire load versus de
flection (mm)
Tire de
Tire load (N)
0
Dual tire
Single wide-base tire
0
8896.4
0
7.62
10.2
17793
14
19
26689
19
27.9
35586
24.1
35.6
44482
27.9
41.9
Section 2.3.4 2.20
Consider the manometer shown in Figure 2.19 and seal the ends. Assume that the initial gas pressure of the sealed system is
Po
and that
Lo
is the initial length of
the cavity. Determine the equivalent spring constant of the system when the column of liquid undergoes “small” displacements. 2.21
Consider
“small”
amplitude
angular
oscillations
of
the
pendulum
shown
Figure E2.21. Considering the gravitational loading and the torsion spring
kt
in at
the pivot point, determine the expression for the system’s equivalent spring constant. The masses are held with rigid, weightless rods for the loading shown.
Figure E2.21.
m2 g
b
kt a
m1
18
J. C. Tielking, “Conventional and wide base radial tyres,” in
on Heavy Vehicle Weights and Dimensions,
Proceedings of the Third International Symposium –
Cambridge, UK, 28 June 2 July, D. Cebon and C. G. B. Mitchell,
–
editors, Thomas Telford, London, 1992, pp. 182 190.
Exercises
73
Section 2.4.1 2.22
Determine the equivalent damping of the system shown in Figure E2.22.
c1
Figure E2.22.
c3 c2
2.23
Determine the equivalent damping for the system shown in Figure E2.23.
c1
2.24
c2
Figure E2.23.
c3
Representative
damping-force
magnitudes
versus
Table E2.24 for a racecar damper in compression.
19
speed
data
are
given
in
Examine these data and dis-
cuss the type of damping model that can be used to represent them.
Table E2.24.
Racecar damper
Damper force magnitude (N)
0
0
57.83
2.54
115.7
5.08
177.9
7.62
266.9
19
10.2
355.9
12.7
444.8
15.2
489.3
17.8
533.8
20.3
578.3
22.9
622.8
25.4
667.2
27.9
711.7
30.5
756.2
33
800.7
35.6
845.2
38.1
889.6
40.6
Racecar dampers are called pression (called
Damper rod speed (mm/s)
“bump”)
“shocks”
in the USA, and they have different damping characteristics during com-
“rebound”) phases. For material on racecar damping, Inside Racing Technology , TV Motorsports, Springfi eld, IL, 1999.
and expansion (called
example, P. Haney and J. Braun,
see, for
74
Modeling of Vibratory Systems 2.25
ficient for the following nonlinear damper:
Determine the equivalent damping coef
F ðx_ Þ = c x_ + c x_
3
1
where
c
1
=
⋅
5 N s/m and
c
3
=
⋅
3
3
3
0.6 N s /m . The damper is to be operated around a
speed of 5 m/s. 2.26
Represent the vibratory system given in Figure E2.26 as an equivalent vibratory system with mass
m, equivalent stiffness ke , and equivalent damping coefficient ce. m
x
2.27
k1
k2
c1
c2
Represent the vibratory system given in Figure E2.27 as an equivalent vibratory system with mass
m, equivalent stiffness ke , and equivalent damping coefficient ce.
x
k2
k1 m
c3
The vibration of a system with stiffness
x(t)
Figure E2.27.
k3
c2
c1
2.28
Figure E2.26.
= Xo
sin
ωt.
k and damping coefficient c
is of the form
This type of response, which is called a harmonic response, is
possible when a vibratory system is excited by a harmonic force. Evaluate the work done by the spring force and the work done by the damper force, which are given by the integrals
ð
xð 2π = ω Þ
Ek =
kxdx =
xð 0Þ xð2 π =ω Þ
Ed =
ð
x ð0Þ
2
cxdx _ =
ð
π=ω
kxx_ dt
0
2
ð
π =ω
0
cx_ dt 2
Exercises 2.29
For the system of Exercise 2.28, assume 0.1 m, and
ω
=
k
=
1000 N/m,
c
=
2500 N/(m/s),
75
Xo
=
9 rad/s. Plot the graph of the sum of the spring force and damper
force versus the displacement, that is,
kx + cx_ versus x.
Section 2.4.2 2.30
Consider the viscous-damping model given by Eq. (2.69) and the dry-friction model given by Eq. (2.75). Sketch the force versus velocity graphs in each case for the following parameter values:
c=
100 N/(m/s),
m = 100 kg, and μ = 0.1.
Single Degree-of-Freedom Systems:
3
Governing Equations
page
3.1 Introduction 3.2 Force-Balance and Moment-Balance Methods
76 77
3.2.1 Force-Balance Methods
77
3.2.2 Moment-Balance Methods
83
3.3 Natural Frequency and Damping Factor
87
3.3.1 Natural Frequency
87
3.3.2 Damping Factor
92
3.4 Governing Equations for Different Types of Damping
96
3.5 Governing Equations for Different Types of Applied Forces
97
3.5.1 System with Base Excitation
97
3.5.2 System with Unbalanced Rotating Mass
99
3.5.3 System with Added Mass Due to a Fluid
100
3.6 Lagrange’s Equations
102
3.7 Summary of Natural Frequency Equations for Single Degree-of-Freedom Systems
3.1
129
3.8 Summary
135
Exercises
136
INTRODUCTION
In this chapter, two approaches are illustrated for deriving the equation governing the motion
of
a
single
degree-of-freedom
system.
In
one
approach
force-balance
and
moment-balance methods are used. The other approach is based on Lagrange’s equations,
first
introduced in this chapter and extensively used throughout the remaining
chapters. Based on the parameters that appear in the governing equation, the expressions for the natural frequency and damping factor are de
fined. The physical significance
of the natural frequency and damping factor is brought forth in Section 4.2, when free oscillations of single degree-of-freedom systems are considered. Furthermore, system stability can also be assessed from the system parameters, as discussed in Section 4.3. When
either
an
analytical
or
a
numerical
solution
of
the
governing
equation
is
obtained, one can determine the relative effects that the system’s components have on the system’s response to various externally applied forces and initial conditions. One can
3.2 Force-Balance and Moment-Balance Methods also
study
vibratory
systems
by
determining
the
transfer
function
of
the
77
system
(Section 5.4.4) or the system’s impulse response (Section 6.2). In this chapter, we will only derive the governing equations for various systems. We shall then obtain the general solutions in Chapter 4 and Appendix C and illustrate numerous applications and interpretations of the solutions in Chapters 4, 5, and 6. In this chapter, we shall show how to do the following.
•
Obtain the governing equation of motion for single degree-of-freedom translating and rotating systems by using force balance and moment balance methods.
•
Obtain the governing equation of motion for single degree-of-freedom translating and rotating systems by using Lagrange’s equations.
•
Determine the equivalent mass, equivalent stiffness, and equivalent damping of a single degree-of-freedom system.
•
Determine the natural frequency and damping factor of a system. In addition, the following interactive animations have been created to better visualize
the motions of several translating and rotating single degree-of-freedom systems. Interactive Graphic 3.1: Animation of the system shown in Figure 3.7 Interactive Graphic 3.2: Animation of the system shown in Figure 3.9 Interactive Graphic 3.3: Animation of the system shown in Figure 3.11 Interactive Graphic 3.4: Animation of the system shown in Figure 3.13 Interactive Graphic 3.5: Animation of the system shown in Figure 3.14 Interactive Graphic 3.6: Animation of the system shown in Figure 3.16 Interactive Graphic 3.7: Animation of the system shown in Figure 3.17
3.2
FORCE-BALANCE AND MOMENT-BALANCE METHODS
In this section, we illustrate the use of force-balance and moment-balance methods for deriving governing equations of motion of single degree-of-freedom systems, show how the staticequilibrium position of a vibratory system can be determined, and carry out linearization of a nonlinear system for “small” amplitude oscillations about a system’s equilibrium position.
3.2.1
Force-Balance Methods Consider the principle of linear momentum, which is Newton’s second law of motion. The statement of dynamic equilibrium given by Eq. (A.21) of Appendix A is recast in the form
F − p_ = 0 where
F
(3.1a)
is the net external force vector acting on the system,
momentum
of
the
considered
system
and
the
overdot
p
is the absolute linear
indicates
the
derivative
with
78
Single Degree-of-Freedom Systems: Governing Equations
m
respect to time. For a system of constant mass absolute acceleration
a,
whose center of mass is moving with
the rate of change of linear momentum
p_ = ma
and Eq. (3.1a)
lead to
F − ma = 0
−ma
The term
is referred to as the
inertia force.
(3.1b)
The interpretation of Eq. (3.1b) is that
the sum of the external forces and inertial forces acting on the system is zero; that is, the system is in equilibrium under the action of external and inertial forces.
1
Vertical Vibrations of a Spring-Mass-Damper System In Figure 3.1, a spring-mass-damper model is shown. A linear spring with stiffness
fi
a viscous damper with damping coef cient ment
m.
c
k and
are connected in parallel to the inertia ele-
In addition to the three system elements that were discussed in Chapter 2, an
external forcing is also considered. We would like to obtain an equation to describe the motions of the system in the vertical direction. In order to derive such an equation for translation motions, the force-balance method is used. Prior to obtaining the governing equation of motion for the system of Figure 3.1, we choose a set of orthogonal unit vectors
i
and
j fixed
in an inertial reference frame and a coordinate system with
axes and an origin
O that is fixed.
Since the mass
m translates only along the j
X
and
Y
direction,
the force balance is considered only in this direction. Let the unstretched length of the spring shown in Figure 3.1 be
L + δst + xÞj
located at the position ð
from the
fixed
determined shortly and explained. After determining
L.
Then the mass is
surface, where the term
δst,
δ st
will be
the equation of motion will be
i
O
Y
j
L
k
c
k
X
c
k(x + dst )
. cx
k(x + dst )
. cx
xst
dst
Equilibrium position (x = 0)
m
x
mg mg Figure 3.1.
1
.. mx
m f(t)
Vertical vibrations of a spring-mass-damper system.
d’Alembert’s principle. According to the generalized form of this principle, virtual displacements are imposed on the system of interest, the net work done by the exterforces is zero. See, for example, D. T. Greenwood, Principles of Dynamics, Prentice Hall,
This statement is also referred to as when a set of so-called nal forces and inertial
Upper Saddle River, NJ, 1988, Chapter 8.
3.2 Force-Balance and Moment-Balance Methods
x.
developed in terms of the displacement variable the
fixed point O is given by
79
The position vector of the mass from
r = rj = ðL + δst + xÞ j
(3.2)
The directions of different forces along with their magnitudes are shown in Figure 3.1. Note that the inertia force
−m€xj
is also shown along with the free-body diagram of the
inertia element. Since the spring force is a restoring force and the damper force is a resistive force, they oppose the motion as shown in Figure 3.1. Based on Eq. (3.1b), we can carry out a force balance along the
f ðtÞj + mgj
−
|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}
j direction and obtain the following equation
kx + k δst Þj
−
|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} ð
External forces acting
Spring force acting
on mass
on mass
c
dr j dt
−
|ffl{zffl} Damping force
m
d r j dt 2
=
|fflfflffl{zfflffl} 2
0
(3.3)
Inertia force
acting on mass
After making use of Eq. (3.2), noting that
L
and
δst
are constants, and rearranging
terms, Eq. (3.3) reduces to the following scalar differential equation
m
d x dt 2
2
+c
dx dt
+ k x + δst = f ð
Þ
t
ð Þ
+ mg
(3.4)
Static-Equilibrium Position The
static-equilibrium position
of a system is the position that corresponds to the sys-
tem’s rest state, that is, a position with zero velocity and zero acceleration. Dropping the time-dependent
forcing
Eq. (3.4) to zero, we
term
f (t)
and
setting
the
velocity
and
acceleration
terms
in
find that the static-equilibrium position is the solution of k ðx + δst Þ = mg
(3.5)
If, in Eq. (3.5), we choose
δ st = we
find that x
=
mg k
0 is the static-equilibrium position of the system. Equation (3.6) is inter-
preted as follows. Due to the weight of the mass so that the spring force balances the weight
displacement.
(3.6)
mg.
m,
the spring is stretched an amount
For this reason,
Recalling that the spring has an unstretched length
position measured from the origin
L,
δst
is called the
δ st
static
the static-equilibrium
O is given by xst = xst j = ðL + δst Þj
(3.7)
which is the rest position of the system. For the vibratory system of Figure 3.1, it is clear from Eq. (3.6) that the static-equilibrium position is determined by the spring force and gravity loading. An example of another type of static loading is provided in Example 3.1.
80
Single Degree-of-Freedom Systems: Governing Equations
Equation of Motion for Oscillations about the Static-Equilibrium Position Upon substituting Eq. (3.6) into Eq. (3.4), we obtain
m
d x dt 2
2
+c
dx dt
+ kx = f
t
ð Þ
(3.8)
Equation (3.8) is the governing equation of motion of a single degree-of-freedom system for oscillations about the static-equilibrium position given by Eq. (3.7). Note that the gravity loading does not appear explicitly in Eq. (3.8). For this reason, in development of models of linear vibratory systems, the measurement of displacement from the staticequilibrium position turns out to be a convenient choice, since one does not have to explicitly take the static loading into account. The left-hand side of Eq. (3.8) describes the forces from the components that comprise a single degree-of-freedom system. The right-hand side represents the external force acting on the mass. Examples of external forces acting on a mass are sure loading such as that on the wing of an aircraft,
fluctuating
fluctuating
air pres-
electromagnetic forces
such as in a loudspeaker coil, electrostatic forces that appear in some microelectromechanical
devices,
forces
caused
by
an
Section 3.5.2), and buoyancy forces on
unbalanced
floating
mass
in
rotating
machinery
(see
systems. The system represented by
Eq. (3.8) is a linear, ordinary differential equation with constant coef
ficients m, c, and k.
As mentioned in Sections 2.2, 2.3, and 2.4, these quantities are also referred to as
system
parameters.
Horizontal Vibrations of a Spring-Mass-Damper System In Figure 3.2, a mass moving in a direction normal to the direction of gravity is shown. It is assumed that the mass moves without friction. The unstretched length of the spring is
L,
and a
fixed
point
O
c
Figure 3.2.
Y k
j
X
O i
c m
g
k . cx
x
. cx
.. mx m
kx
is located at the unstretched position of the spring, as shown in
kx
f (t)
Horizontal vibrations of a spring-mass-damper system.
3.2 Force-Balance and Moment-Balance Methods the
figure. Noting that the spring does not undergo any static deflection
a force balance along the position
i
81
and carrying out
direction gives Eq. (3.8) directly. Here, the static-equilibrium
x = 0 coincides with the position corresponding to the unstretched spring.
Force Transmitted to a Fixed Surface From Figure 3.1, we see that the total reaction force due to the spring and the damper on the
fixed surface is the sum of the static and dynamic forces. Thus, FR =
+ kx + c
kδ |{z}
dx dt
|fflfflfflfflffl{zfflfflfflfflffl}
st
Static
(3.9)
Dynamic
component
component
– that is, only those forces created by the motion x (t) from its static-equilibrium position – then Eq. (3.9) leads to
If we consider only the dynamic part of the reaction force
FRd = c where
x(t)
dx dt
+ kx
(3.10)
is the solution of Eq. (3.8). This result agrees with Eq. (2.72). Equation (3.10)
is used in later chapters to determine the force transmitted to the ground (Section 5.6) or the
force
transmitted
to
the
mass
when
the
base
is
in
motion
(Section
5.2.6
and
Examples 5.5 and 6.7). In the following two examples, we show how to obtain the governing equation for a system subjected to an external force that has a static component and how to linearize a system that has a spring with a quadratic nonlinearity.
EXAMPLE 3.1
Wind-driven oscillations about a system’s static-equilibrium position
In examining wind
flow
across civil structures such as buildings, water towers, and 2
lampposts, it has been found
that the wind typically generates a force on the structure
that consists of a steady-state part and a force
fluctuating
part. In such cases, the excitation
f( t) is represented as f ðtÞ = fss + fd ðtÞ
where
fss
(a)
is the time-independent steady-state force and
fd (t)
is the
fluctuating
time-
dependent portion of the force. A single degree-of-freedom model of the vibrating structure has the form of Eq. (3.8), where
x(t) is the displacement of the structure in the direction of
the wind loading and the forcing is given by Eq. (a), that is,
m
2
E. Naudascher and D. Rockwell, 1994, p. 103.
d x dt 2
2
+c
dx dt
+ kx = f ss + fd
t
ð Þ
Flow-Induced Vibrations: An Engineering Guide,
(b)
A. A. Balkema, Rotterdam,
82
Single Degree-of-Freedom Systems: Governing Equations We
shall
determine
the
static-equilibrium
position
of
this
system
and
obtain
the
equation of motion governing oscillations about this static-equilibrium position. To this end, we assume a solution of the following form for Eq. (b)
xð tÞ = xo + xd ðtÞ where
xo
is determined by the static loading and
(c)
xd (t)
determines motions about the sta-
tic position. Thus, after substituting Eq. (c) into Eq. (b) and noting that dent of time, we
find that
xo
is indepen-
xo = fss =k and that
(d)
xd ( t) is the solution to the equation m
d xd dt 2
2
+c
dxd dt
+ kxd = fd
t
ð Þ
(e)
which is the governing equation of motion for oscillations about the static-equilibrium position
xo
= f ss k
/ .
EXAMPLE 3.2
3
Eardrum oscillations: nonlinear oscillator
and linearized systems
In this example, we consider a nonlinear oscillator used to study eardrum oscillations. We shall determine the static-equilibrium positions of this system and illustrate how the governing
nonlinear
equation
can
be
linearized
to
study
oscillations
local
to
an
equilibrium position. The governing nonlinear equation has the form
m
d x dt 2
2
+ kx +
k αx |ffl{zffl} 2
=
0
(a)
Quadratic nonlinearity
where
α
(m
−
1
) is a constant. The restoring force of the eardrum has a component with a
quadratic nonlinearity.
Static-Equilibrium Positions Noting that there are no external time-dependent forcing terms in this problem and setting the acceleration term to zero, we solutions of the algebraic equation
find
±
that the equilibrium positions
k xo + αxo 2
²=
0
x
= xo
are
(b)
The solutions of Eq. (b) provide us with two static-equilibrium positions for the eardrum, namely,
xo
1
xo
2
3
R. E. Mickens,
= =− 0
=α
1
Oscillations in Planar Dynamic Systems, World Scientific, Singapore, 1996.
(c)
3.2 Force-Balance and Moment-Balance Methods
83
Linearization Next, we substitute
xð tÞ = xo + ^xðtÞ
(d)
into Eq. (a) and linearize the nonlinear stiffness term for oscillations about one of the
x^ðtÞ: To this end, we note that
equilibrium positions in terms of the variable
x ð tÞ = ðxo +x^ðtÞÞ
2
2
≈ xo +
xo x^ðtÞ + ⋯
2
2
(e)
In addition,
dx d ðxo + ^xÞ = dt dt
=
d x d ðxo + ^xÞ = dt dt 2
2
2
2
d x^ dt
=
(f)
d ^x dt 2
2
Making use of Eqs. (e) and (f) in Eq. (a), we obtain
m
d ^x dt 2
2
+k + ð1
αxo Þ^ x + kxo ð1 + αxoÞ = 0
2
(g)
Linearized System for “Small” Amplitude Oscillations Around xo1 = 0 Setting x o = x o = 0 in Eq. (g), we arrive at the linear equation 1
m
d ^x dt 2
2
+ kx^ =
0
(h)
Linearized System for “Small” Amplitude Oscillations Around xo2 = - 1=α Setting x o = x o = –1/α , we arrive at the linear equation 2
m
d ^x dt 2
2
− kx^ =
0
(i)
Comparing Eqs. (h) and (i), it is clear that the equations have different stiffness terms.
3.2.2
Moment-Balance Methods For single degree-of-freedom systems that undergo rotational motion, such as the system shown in Figure 3.3, the moment-balance method is useful in deriving the governing equation. A shaft with torsional stiffness
kt
is attached to a disc with rotary inertia
about the axis of rotation, which is directed along the
M(t)
k-direction.
JG
An external moment
fi
acts on the disc, which is immersed in an oil- lled housing. Let the variable
θ
describe the rotation of the disc, and let the rotary inertia of the shaft be negligible in comparison to that of the disc. Note that a positive shown in the
θ
rotation is along the
k-direction
figure; therefore, this is the positive direction for the external moment.
84
Single Degree-of-Freedom Systems: Governing Equations
q
M(t)
(t)
ct
kt
k
.
Axis of rotation Disc with rotary inertia JG about rotation axis
Shaft with equivalent torsional stiffness k t
JG
G
..
M(t)
Housing filled with oil (a) Figure 3.3.
(b)
(a) A disc undergoing rotational motions and (b) free-body diagram of this disc in the
plane normal to the axis of rotation.
The principle of angular momentum given by Eq. (A.25a) of Appendix A is applied to obtain the equation governing the disc ’s motion. First the angular momentum
H
of
the disc is determined. Since the disc is a rigid body undergoing rotation in the plane, Eq. (A.27) of Appendix A is used to write the angular momentum about the center of mass of the disc as
H = JG θ_ k Since the rotary inertia
JG and the unit vector k
do not change with time, Eq. (A.25a) of
Appendix A leads to
M − JG €θk = 0 where
M
(3.11)
is the total external moment acting on the free disc. Based on the free-body
diagram shown in Figure 3.3, which also includes the inertial moment
− JG €θk
; the gov-
erning equation of motion is
|fflfflffl{zfflfflffl} M ðtÞk
−
−
|ffl{zffl} k tθk
External moment
Restoring moment due
acting on disk
to shaft stiffness
ct
dθ k dt
−
|fflfflffl{zfflfflffl} Damping moment due
JG
d θ k dt 2
|fflfflfflffl{zfflfflfflffl} 2
=
0
(3.12)
Inertial moment
to oil in housing
After collecting the scalar coef
ficients
of the different vector terms in Eq. (3.12) and set-
ting them to zero, we arrive at the following scalar equation
JG
d θ dt 2
2
+ ct
dθ dt
+ ktθ = M
t
ð Þ
(3.13)
The form of Eq. (3.13) is identical to Eq. (3.8), which was obtained for a translating vibratory system; that is, the element
JG,
first
term on the left-hand side is determined by the inertia
the second term on the left-hand side is determined by the damping element
3.2 Force-Balance and Moment-Balance Methods
ct,
85
kt, and the M(t). In general, all linear single degree-of-freedom vibratory systems are governed by a linear second-order ordinary differential equation with an inertia term, a stiffness term, a damping term, and a term related to the external forcing imposed on the system. the third term on the left-hand side is determined by the stiffness element
right-hand side contains the external forcing, the moment
EXAMPLE 3.3
Hand biomechanics
4
X–Y
Consider the rotational motion of the hand in the motion is described by the angle forearm
has
a
mass
m
and
a
θ.
plane shown in Figure 3.4. This
M
An object of mass
length
l.
If
we
use
is held in the hand, and the
fi
simpli ed
models
for
the
forces
Fb = − kb θ, where kb is a constant, and the triceps provide a force of magnitude F t = Kt v , where Kt is a constant and v is the magnitude of the velocity with which the triceps are stretched. It generated by the muscles, then the biceps provide a force of magnitude
is
further
assumed
that
the
forearm
can
be
treated
as
a
uniform
rigid
beam.
The
governing nonlinear equation of motion is obtained, and following along the lines of Example 3.2, the nonlinear system is linearized for
“small”
oscillations about the system
equilibrium position. The equation of motion of the system of Figure 3.4 is derived from a moment balance about either the center of the forearm or the pivot point
O (the
elbow) if the pivot point
Upper arm
Biceps Fb
Forearm
X
Ft Triceps
k
Z Y
Pivot point O
mg
Mg
a
a
l Figure 3.4. Hand motion.
Source: P. Maróti,
L. Berkes and F. Tölgyesi,
Akademiai Kiado, Budapest, 1998. Copyright
Biophysics Problems: A Textbook with Answers,
©
1998 P. Maróti, L. Berkes and F. Tölgyesi.
Reprinted with permission.
4
P.
Maróti,
L.
Budapest, 1998.
Berkes
and
F.
Tölgyesi,
Biophysics Problems: A Textbook with Answers,
Akademiai
Kiado,
86
Single Degree-of-Freedom Systems: Governing Equations is a
fixed
point. Assuming that the pivot point is a
carried out about point
fixed
point, a moment balance is
O . This leads to M − Jo€θ k = 0
where
Jo
moment
(a)
is the rotary inertia of the forearm and the object held in the hand. The net
M
O
acting about the point
due to gravity loading and the forces due to the
biceps and triceps is given by
M = − Mgl cos θk − mg Therefore, from Eqs. (a) and (b), we
− Mgl
2
cos
θk + F bak − Ft ak
(b)
find that the governing equation takes the form
θk − mg
cos
l
l 2
cos
θk + Fb ak − F tak − Jo € θk = 0
(c)
Recognizing that
Fb = −k bθ
Ft = Kt v = Kt aθ_
Jo =
1 3
ml
2
+ Ml
(d) 2
ficients of all the vector terms in Eq. (c) and setting them to
and collecting the scalar coef
³
´
³
zero, we obtain the scalar equation
M+
m 3
´
m l €θ + K ta θ_ + kb a θ + M + gl cos θ = 0 2
2
(e)
2
In this equation, the inertia term is due to the rotary inertia of the forearm and rotary inertia of the end mass, the damping term is due to the triceps, the stiffness term is due to the biceps, and there is an additional term due to gravity that makes the equation nonlinear due to the presence of the cos
θ
fl fluence depends on the magnitude of θ.
term. This latter term in uences the static-equilibrium
position and the stiffness of the system, and this in
Static-Equilibrium Position Noting that time-dependent external moment terms are absent in Eq. (e) and setting the velocity term and the acceleration term to zero, we
θ
= θo
³
´
find
that the equilibrium position
is a solution of the transcendental equation
k baθo + M +
m 2
gl cos θo = 0
(f)
Linear System Governing “Small” Oscillations about the Static-Equilibrium Position We
now
consider
angular variable
oscillations
θ( t)
about
the
static-equilibrium
position
and
expand
the
in the form
θðtÞ = θo + ^ θð tÞ
(g)
3.3 Natural Frequency and Damping Factor and linearize the nonlinear term cos 5
series expansions
±θ
θ
in Eq. (e). To do this, we carry out the Taylor-
² + ^θ ≈
of this term and retain only linear terms in cos
θ = cos
Evaluating the time derivatives of
o
θ( t),
θ_ ðtÞ =
87
we
cos
θo − ^ θ sin θo +⋯
(h)
find that
±
² = ^θ_ t
±
² = €^θ t
d θo + ^ θ dt
ð Þ
€θðtÞ = d θo + ^θ dt 2
^θ. This leads to
2
(i)
ð Þ
On substituting Eqs. (i) and (h) into Eq. (e) and making use of Eq. (f), we arrive at the following linear equation of motion governing “small” oscillations of the forearm about
³
´
the system-equilibrium position:
M+
m 3
l ^€θ + K ta ^θ_ + ke ^θ = 0 2
³
where
2
ke = k ba − M +
(j)
´
m 2
gl sin θo
(k)
It is noted out that the “linear” stiffness of the linearized system is in
flected by the second term in k e.
fluenced
by the
gravity loading as re
3.3
NATURAL FREQUENCY AND DAMPING FACTOR
fi
In this section, we de ne the natural frequency and damping factor for a vibratory system and explore how these quantities depend on various system properties. These quantities are discussed without explicitly determining the solution of the system. They are only a function of the system’s inertia, stiffness, and damping parameters and independent of the external time-dependent forcing imposed on a system. The solutions for the responses of vibratory systems represented by Eqs. (3.8) and (3.13), which are discussed in Chapters 4 to 6, can be characterized in terms of the system’s damping factor.
3.3.1
Natural Frequency
Translation Vibrations: Natural Frequency For translation oscillations of a single degree-of-freedom system, the
ωn of
the system is de
fined as ωn = 2π fn =
5
T. B. Hildebrand,
ibid.
natural frequency
rffiffiffiffi
k m
=
rad s
(3.14)
88
Single Degree-of-Freedom Systems: Governing Equations
k is
where
the stiffness of the system and
m is
the system mass. The quantity
fn,
which is
also referred to as the natural frequency, has the units of Hz.
fi
For the con guration shown in Figure 3.1, the vibratory system exhibits vertical oscillations. For such oscillations, we make use of Eq. (3.6) and Eq. (3.14) and obtain
ωn = 2π fn = where
δ st
rffiffiffiffigffi
rad=s
δst
(3.15)
fl
is the static de ection of the system.
Rotational Vibrations: Natural Frequency
finition of natural frequency of translation motions of a single degree-of-freedom system, the natural frequency for rotational motion is de fi ned as
Drawing a parallel to the de
ωn = 2π fn =
kt
where
rffiffiffiffi kt J
rad=s
is the torsion stiffness of the system and
J
(3.16)
is the mass moment of inertia of the
system.
Design Guideline:
For single degree-of-freedom systems, an increase in the stiffness or
a decrease in the mass or mass moment of inertia increases the natural frequency, whereas a decrease in the stiffness and/or an increase in the mass or mass moment of inertia decreases the natural frequency. Equivalently, when applicable, the greater the static
displacement
the
lower
the
natural
frequency;
however,
from
practical
considerations, too large a static displacement may be undesirable.
Period of Undamped Free Oscillations For an unforced and undamped system, the
period of free oscillation
of the system is
given by
T=
1
fn
=
π
2
(3.17)
ωn
Thus, increasing the natural frequency decreases the period and vice versa. In the following examples, we show how the static displacement can be used to determine the natural frequency of a single degree-of-freedom system.
EXAMPLE 3.4
For
a
Natural frequency from static de
particular
choice
machinery is found to be this de
of
machinery
δ st1 = 0:1 mm.
flection is found to be δ st
2
=
flection of a machine system
mounting,
the
static
de
flection
of
a
piece
of
For two other choices of machinery mounting,
1 mm and
δ st3 = 10 mm.
fl
Based on the static de ection,
we will determine the natural frequency for vertical vibrations for each of the three
3.3 Natural Frequency and Damping Factor machinery
mountings.
To
acceleration due to gravity
this
end,
g
9.81 m/s , we arrive at the following for the natural
=
we
make
use
of
Eq.
(3.15).
Noting
89
that
the
2
frequencies of the three machinery mountings:
fn
EXAMPLE 3.5
1
=
fn
2
fn
3
Static de
sffiffiffiffiffiffi g
1
π
δst1
2
=
=
2
9:81 m=s
2
0:1
×
3
10
49:85 Hz
m
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u = t = π δ π × − v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffi u g u = t − = π δ π sffiffiffiffiffiffi g
1
2
1
st2
2
=
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u t − = π
1
9:81 m=s
2
1
1
9:81 m=s
2
st3
15:76 Hz
m
2
1
2
3
10
10
×
3
10
4:98 Hz
m
flection and natural frequency of the tibia bone in a human leg
Consider a person of 100
kg mass standing upright. We shall determine the
static
fl
de ection in the tibia bone and an estimate of the natural frequency of axial vibrations. The tibia has a length of 40 cm, and it is modeled as a hollow tube with an inner diameter of 2.4 cm and an outer diameter of 3.4 cm. The Young’s modulus of elasticity of the bone material is 2
fl
×
10
10
2
N=m .
The static de ection will be determined by using Eq. (3.6), and Eq. (3.15) will be used to determine the natural frequency. We assume that both legs support the weight of the
µ
¶ ± ×
person equally, so that the weight supported by the tibia is
mg =
k=
=
EA = L 22:78
×
2
kg
2
2
9:81 m=s
²=
490:5 N
(a)
k of the tibia, we use Case 1 of Table 2.3 to obtain
To determine the stiffness
±
100
×
10
10
2
N=m
² × π h±
3:4
4
40 6
10
×
−
×
10
−
10
2
2
² −± 2
2:4
×
−
10
2
²i 2
2
m
m
N=m
(b)
flection is given by
Hence, from Eqs. (3.6), (a), and (b), the static de
δ st =
mg k
=
=
490:5 N 22:78
×
6
10
N=m
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
21:53
μm
(c)
and, from Eqs. (3.15) and (c), the natural frequency is
fn =
1
π
2
rffiffiffiffigffi δst
=
1
π
2
2
9:81 m=s 21:53
μm
=
107:4 Hz
(d)
90
Single Degree-of-Freedom Systems: Governing Equations
EXAMPLE 3.6
System with a constant natural frequency
In many practical situations, different pieces of machinery are used with a single springmounting system. Under these conditions, one would like the system natural frequency to be constant for the different machinery-mounting-system combinations; that is, we are looking for a system whose natural frequency does not change as the system mass is changed. Through this example, we examine how the spring-mounting system can be designed and discuss a realization of this spring in practice. From Eq. (3.14), it is clear that the natural frequency depends on the mass of the system. In order to realize the desired objective of constant natural frequency regardless of the system weight, we need a nonlinear spring whose equivalent spring constant is given by
k = AW where
A
is a constant, the weight
W
rffiffiffiffi
= mg
fn =
k m
π
2
g
, and
rffiffiffiffiffi
from Eqs. (3.15) and (a), we arrive at 1
(a)
=
1
is the gravitational constant. Then,
kg 1 = W 2π
π
2
pffiffiffiffiffiffi Ag
Hz
(b)
from which it is clear that the natural frequency is constant irrespective of the weight of the mass.
Nonlinear Spring Mounting When the sidewalls of a 6
region,
rubber cylindrical tube
are compressed into the
nonlinear
the equivalent spring stiffness of this system approximates the characteristic
given by Eq. (a). For illustrative purposes, consider a spring that has the general force-
³x ´
displacement relationship
F ðxÞ = a where weight
µW ¶
“small”
·· ·· ·= !
=c
dF ðxÞ keq = dx
the linear equivalent stiffness of this spring
=
E. I. Riven,
=
x xo
ac W b a
(d)
a
xo,
amplitude vibrations about
for a machinery of
1
is determined from Eqs. (c) and (d) to be
6
(c)
a and b are scale factors and c is a shape factor. Noting that W, the static deflection xo is determined by using Eq. (c) as xo = b
For
c
b
ð
ac xo b b
!− c
1
c− 1 Þ=c
Stiffness and Damping in Mechanical Design , Marcel Dekker, New York, 1999, pp. 58–61.
(e)
3.3 Natural Frequency and Damping Factor
91
Then, from Eqs. (3.14) and (e), we determine the natural frequency of this system as
fn =
=
1
π
2
1
sffiffiffiffiffiffiffiffiffiffi k eq
W=g
sffiffiffiffigcffi b
π
2
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !− ffi u u gc W t 1
=
W a
1
π
2
!−
b
=c
a
= c
1 ð2 Þ
Hz
(f)
3500 3000 2500
)N( W
2000 1500 1000 500 0 0.004
0.006
0.008
(a)
0.01
0.012
x (m) 13 12 11
)zH( nf
10 9 8 7
(b)
0
500
1000
1500 2000 W (N)
2500
3000
3500
Figure 3.5. (a) Curve fit of nonlinear spring data: squares, experimental data values; solid line, fitted curve; (b) natural frequency for data values in (a) above; horizontal broken lines are within
10% from the solid horizontal line.
92
Single Degree-of-Freedom Systems: Governing Equations
Representative Spring Data We now consider the representative data of the nonlinear spring shown in Figure 3.5a.
fitting procedures,
7
By using standard curve-
c
=
we
find that a = 135, 203 N, b = 0.0448 m, and
2.839. After substituting these values into Eq. (f), we arrive at the natural frequency
values shown in Figure 3.5b. It is seen that over a sizable portion of the load range, the natural frequency of the system varies within the range of
±
10% about a frequency
³ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi =
´
of 8.2 Hz. The natural frequency of a system with a linear spring whose static displacement
³ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi =
´
ranges from 12 mm to 7 mm varies approximately from 4.5 Hz
6.0 Hz
3.3.2
π
9:8=0:007 =2
Hz
or approximately
±
π Hz
9:8=0:012 =2
to
14% about a frequency of 5.25 Hz.
Damping Factor
Translation Vibrations: Damping Factor For translating single degree-of-freedom systems, such as those described by Eq. (3.8), the
damping factor or damping ratio ζ ζ=
where
c
fi
is de ned as
c 2mωn
=
ffiffiffiffiffiffi = cωk
c km
n
p 2
ficient, k
is the system damping coef
(3.18)
2
is the system stiffness, and
m
is the system
mass. The damping factor is a nondimensional quantity.
Critical Damping, Underdamping, and Overdamping De
fining the quantity cc , called the critical damping, as p
ffiffiffiffiffiffi
cc = 2mωn = 2 km
(3.19)
the damping ratio is rewritten in the form
ζ= When
c = cc ; ζ = 1.
fi
The signi cance of
cc is
c cc
(3.20)
discussed in Section 4.2.1, where free oscilla-
tions of vibratory systems are considered. A system for which 0
underdamped
system and a system for which
tem for which
ζ=1
is called a
ζ>1
is called an
0, β > 0, and γ > 0 Parameter Case
changed
Natural frequency
c = α co m = mo k = ko
1
ωn =
c = co m = βmo k = ko
2
ωn =
c = co m = mo k = γ ko
3
ωn =
rffiffiffiffiffi ffi k
Damping factor
ζ=
= ωno
o
mo
rffiffiffiffiffiffiffiffi k o
β mo
rγffiffiffiffiffiffik ffi o
mo
=
ffiffiβffi
ωno
=
p
p
ffiffiγω
ζ=
no
ζ=
ffiffiffiffiffiffiffiffiffiffi = αζ km
αco
p
δstatic =
mo g ko
c ffiffiffiffiffiffiffiffiffiffiffiffi ffi = ζ ffiffiβffi k βm
δstatic =
β mo g
c ζ ffiffiffiffiffiffiffiffiffiffiffiffi = ffiffi γk m γ
δstatic =
mo g γ ko
o
2
p
o
o
p 2
o
o
p
o
2
Static displacement
o
o
o
o
o
p
ko
= δo static ;
= βδ o static ;
=
δo;static γ
From Eqs. (3.14) and (3.16), we see that the stiffness and inertia properties affect the natural frequency. From Eqs. (3.18) and (3.21), we see that the damping ratio is affected by any change in the stiffness, inertia, or damping property. However, one can change more
than
one
system
parameter
in
such
a
way
that
the
net
effect
on
ζ
remains
unchanged. These interactions are summarized in Table 3.1 and further examined in Example 3.8.
Governing Equation of Motion in Terms of Natural Frequency and Damping Factor fi
Introducing the de nitions given by Eqs. (3.14) and (3.18) into Eq. (3.8), we obtain
d x dt 2
2
fi
The signi cance of the quantities
+
ζωn
dx dt
and
ζ
2
ωn
+ ωn x = f 2
ðtÞ m
(3.22)
will become apparent when the solution to
Eq. (3.22) is discussed in detail in the subsequent chapters. If we introduce the dimensionless time
τ = ωnt,
then Eq. (3.22) can be written as
d x dτ 2
2
+
ζ
2
dx dτ
+x =
f ðτÞ k
(3.23)
94
Single Degree-of-Freedom Systems: Governing Equations It is seen from Eq. (3.23) that the damping factor
ζ
is the only system parameter that
appears explicitly on the left-hand side of the equation. We shall use both forms of Eqs. (3.22) and (3.23) in subsequent chapters.
f (τ )
In the absence of forcing, that is, when
=
0, the motion of a vibratory system
expressed in terms of nondimensional quantities can be described by just one system parameter.
This
fact
is
further
elucidated
in
Section
4.2,
where
free
oscillations
are considered and it is shown that the qualitative nature of these oscillations can be
completely
that is,
f( τ)
≠
characterized
by
the
damping
ζ
0, both the damping factor
tant for characterizing
the
nature
of the
factor.
In
the
presence
and the natural frequency
response.
This
is
further
ωn
of
forcing,
are impor-
addressed
when
the forced responses of single degree-of-freedom systems are considered in Chapters 5 and 6.
fi
Since the damping coef cient is one of the most important descriptors of a vibratory system, it is important to understand its interrelationships with the component’s parameters
m
(or
J ), c
(or
ct),
and
k
(or
kt).
We shall illustrate some of these relationships
shown in Table 3.1 with the following examples.
EXAMPLE 3.7
Effect of mass on the damping factor
A system is initially designed to be critically damped, that is, with a damping factor of
ζ = 1.
Due to a design change, the mass of the system is increased 20%, that is, from
to 1.2
m.
m
Will the system still be critically damped if the stiffness and the damping
ficient of the system are kept the same? The de fi nition of the damping factor is given
coef
by Eq. (3.18) and that for the critical
damping factor is given by Eq. (3.19). Then, the damping factor of the system after the design change is given by
ζnew =
c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi = k m
2
ð1:2
0:91
Þ
ffiffiffiffiffiffi =
c km
p 2
0 :91
c cc
=
0:91
× = 1
0:91
Therefore, the system with the increased mass is no longer critically damped; rather, it is now underdamped.
EXAMPLE 3.8
An engineer
finds
spring constant
δ st
Effects of system parameters on the damping factor
that a single degree-of-freedom system with mass
k has
too much static de
flection δst .
m,
damping
c,
and
The engineer would like to decrease
by a factor of 2, while keeping the damping factor constant. We shall determine the
different options.
3.3 Natural Frequency and Damping Factor
95
Noting that this is a problem involving vertical vibrations, it is seen from Eqs. (3.6), (3.15), and (3.18) that
mg k
δst =
sffiffiffiffiffi sffiffiffiffiffiffiffiffi c δ δ
ζ=
2
st
m
g
=c
st
gm
2
v ffiffiffiffiffiffiffiffiffi u u = tc δ m g 2
1
(a)
st
From Eqs. (a), we see that there are three ways that one can achieve the goal.
First Choice For the
first choice, let c remain constant. Then, when δst is reduced by one-half, ζ=c
sffiffiffiffiffiffiffiffi
→c gm δst
2
sffiffiffiffiffiffiffiffiffiffi δst
gm
2
2
(b)
2
p
ffiffi
and, therefore, the mass has to be reduced by a factor of p increased by a factor of
2, that is,
m
→ mffiffi p
k
and
2
→k
ffiffi
2 and the stiffness has to be
ffiffi
p
2
(c)
since,
mg k
δst = Thus, for
δ st
→δ = st
2 and
→ mffiffi k g ffiffi = δ ffiffi k → k ffiffi m → m= p
st
p
2
2
p
c held constant,
(d)
2
p
2 and
2.
Second Choice For the second choice, let
m remain constant. Then, when δ st is reduced by one-half, ζ=
2
sffiffiffiffiffiffiffiffiffi c δ st g 2
1
m
ficient
and, therefore, the damping coef
→m
sffiffiffiffiffiffiffiffiffi c δ st 2g 2
1
(e) p
has to be increased by a factor of
ffiffi
2 and the
stiffness has to be increased by a factor of 2, that is,
c
→c
ffiffi
p
2
and
k
→
k
2
(f)
since,
δst = Thus, for
δ st
→δ = st
2 and
mg k
m held constant,
→ mgk = δ ffiffi k→ k c→ c st
2
2
p
2
and
2.
(g)
96
Single Degree-of-Freedom Systems: Governing Equations
Third Choice For the last choice, let
k remain constant. Then, when δst is reduced by one-half, δ st =
→m
mg k
g k
2
=
δ st
(h)
2
Thus, the mass has to be reduced by a factor of 2, that is,
m
→m
(i)
2
Furthermore, since
ζ=
2
the damping coef
ficient has to be reduced by a factor of c
Thus, for
δst
c ffiffiffiffiffiffi km
p
→δ = st
→ cffiffi m → m=
p
p
2
2 and
k held constant,
2 and
c
(j)
ffiffi
2, that is,
→ c= ffiffi p
2.
Notice that in all three cases the natural frequency increases by a factor of
(k)
ffiffi
p
2. The results
of this example can be generalized to a design guideline (see Exercises 3.19 and 3.20).
In the next two sections, the governing equations for different types of damping models and forcing conditions are presented. For all of these cases, translational motions are considered for illustrative purposes, and the equations are obtained by carrying out a force balance along the direction of motion. The form of the governing equations will be similar for systems involving rotational motions.
3.4
GOVERNING EQUATIONS FOR DIFFERENT TYPES OF DAMPING
The governing equations of motion for systems with different types of damping are obtained by replacing the term corresponding to the force due to viscous damping with the force due to either
fluid, structural, or dry friction type damping. Solutions for differ-
ent periodically forced systems are given in Section 5.7, where equivalent viscous damping coef
ficients for different damping models are obtained.
Coulomb or Dry Friction Damping After using Eq. (2.76) to replace the
cx_
term in Eq. (3.8), the governing equation of
motion takes the form
m
d x dt 2
2
+ kx +
x_ |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}
μmg
sgnð Þ
Dry friction force
=f
t
ð Þ
(3.24)
3.5 Governing Equations for Different Applied Forces
97
which is a nonlinear equation because the damping characteristic is piecewise linear. This piecewise linear property can be used to
Fluid Damping After using Eq. (2.78) to replace the
cx_
find the solution of this system.
term in Eq. (3.8), the governing equation takes
the form
m
d x dt 2
2
+
cd ∣x_ ∣ x_
|fflffl{zfflffl}
+ kx = f
t
ð Þ
(3.25)
Fluid damping force
which is a nonlinear equation due to the nature of the damping.
Structural Damping After using Eq. (2.81) to replace the
m
d x dt 2
2
+
cx_ term in Eq. (3.8), we arrive at the governing equation k βπ sgnð x_ Þ ∣ x∣
|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}
+ kx = f
t
ð Þ
(3.26)
Structural damping force
Equation (3.26) is further addressed in Section 5.7.
3.5
GOVERNING EQUATIONS FOR DIFFERENT TYPES OF APPLIED FORCES
In Section 3.2, we addressed governing equations of single degree-of-freedom systems whose inertial elements were subjected to direct excitations. Here, we address governing equations of single degree-of-freedom systems subjected to base excitations, systems excited by rotating unbalance, and systems immersed in a
3.5.1
fluid.
System with Base Excitation The base-excitation model is a prototype that is useful for studying buildings subjected to earthquakes, packaging during transportation, vehicle response, and for designing accelerometers (see Section 5.5). Here, the physical system of interest is represented by a single degree-of-freedom system whose base is subjected to a displacement disturbance
y(t),
and an equation governing the motion of this system is sought to determine the
response of the system
x(t).
If the system of interest is an automobile, then the road surface on which it is traveling can be a source of the disturbance
y(t) and the vehicle response x(t) is to be determined. To
avoid failure of electronic components during transportation, a base-excitation model is used to predict the vibration response of the electronic components. For buildings located
98
Single Degree-of-Freedom Systems: Governing Equations Package
x(t)
m
Y
k
y(t) X
O Figure 3.6.
. k(x – y) c(x – y)
c
j
.. mx
m
Base
Base excitation and the free-body diagram of the mass.
above or adjacent to subways or above ground railroad tracks, the passage of trains can act as a source of excitation to the base of the building. In designing accelerometers, the accelerometer responses to different base excitations are studied to determine the appropriate accelerometer system parameters, such as the natural frequency and damping factor. A prototype of a single degree-of-freedom system subjected to a base excitation is illustrated in Figure 3.6. The system represents an instrumentation package being transported in a vehicle. The vehicle provides the base excitation
y(t)
to the instrumentation package
modeled as a single degree-of-freedom system. The displacement response
x(t) is measured
from the system’s static-equilibrium position. In the system shown in Figure 3.6, it is assumed that no external force is applied directly to the mass; that is,
f(t)
=
0. Based on
the free-body diagram shown in Figure 3.6, we use Eq. (3.1b) to obtain the following governing equation of motion
d 2x dt2
m
+c
dx dt
+ kx = c
dy dt
+ ky
(3.27)
which, on using Eqs. (3.14) and (3.18), takes the form
d x dt 2
2
The displacements
y(t)
reference frame and a
and
+
x(t)
fixed
ζω n
2
dx dt
+ ωn x = 2
ζω n
2
are measured from a
dy dt
+ ωn y 2
(3.28)
fixed point O located in an inertial
point located at the system’s static-equilibrium position,
respectively. If the relative displacement is desired, then we let
zðtÞ = xð tÞ − yðtÞ
(3.29)
and Eq. (3.27) is written as
m
d 2z dt2
+c
dz dt
+ kz = −m
d2 y dt2
(3.30)
while Eq. (3.28) becomes
d z dt 2
2
where
+
ζω n
2
yðtÞ is the acceleration of the base. €
dz dt
+ ωn z = − 2
d y dt 2
2
(3.31)
3.5 Governing Equations for Different Applied Forces 3.5.2
99
System with Unbalanced Rotating Mass As discussed in Chapter 1, many rotating machines such as fans, clothes dryers, internal combustion engines, and electric motors, have a certain degree of unbalance. In modeling such systems as single degree-of-freedom systems, it is assumed that the unbalance generates a force that acts on the system’s mass. This force, in turn, is transmitted through the spring and damper to the
mo
that rotates with an angular speed
fixed
ω,
base. The unbalance is modeled as a mass
and this mass is located a
fixed distance ε from
the center of rotation as shown in Figure 3.7. Note that in Figure 3.7, include the unbalance
M
does not
m o.
For deriving the governing equation, only motions along the vertical direction are considered, since the presence of the lateral supports restricts motion in the displacement position. The
of
the
fixed
system
point
O is
x(t)
is
measured
from
the
system’s
j-direction.
The
static-equilibrium
chosen to coincide with the vertical position of the static-
equilibrium position. Based on the discussion in Section 3.2, gravity loading is not explicitly taken into account. From the free-body diagram of the unbalanced mass the point
O′ are given by
±
Nx = − mo x€ − εω Ny = mo εω
2
cos
2
sin
ωt
mo,
X
t
i
k
Nx O' M kx
2
Nx Ny
O'
j
c
.. mo (x .. Mx
Y
O
M
sin t)
mo
2 cos
that the reactions at
(3.32)
ωt
O'
Clothes dryer
find
²
mo Fan
we
t
Ny
. cx
Figure 3.7. System with unbalanced rotating mass and free-body diagrams.
100
Single Degree-of-Freedom Systems: Governing Equations and from the free-body diagram of mass
M Then, substituting for
Nx
d x dt 2
M we find that
+c
2
dx dt
+ kx = Nx
(3.33)
from Eqs. (3.32) into Eq. (3.33), we arrive at the equation of
motion
M + m oÞ
ð
d x dt 2
2
+c
dx dt
dx dt
+ ωn x =
+ kx = mo εω
2
sin
ωt
(3.34)
which is rewritten as
d x dt 2
2
+
ζωn
2
F ðωÞ sin ωt m
2
(3.35)
where
m = M + mo ωn =
sffiffiffikffi
(3.36)
m
F ðωÞ = m o εω
2
F(ω) is the magnitude of the unbalanced force. This magnitude depends on the unbalanced mass mo and it is proportional to the square of the excitation
In Eqs. (3.35) and (3.36),
frequency. From Eq. (3.6), it follows that the static displacement of the spring is
δst =
INTERACTIVE GRAPHIC 3.1: IN FIGURE 3.7
M + mo Þg k
ð
=
mg k
(3.37)
ANIMATION OF THE SYSTEM SHOWN
With this interactive graphic, one can animate the system shown in Figure 3.7.
3.5.3
System with Added Mass Due to a Fluid Consider a rigid body that is connected to a spring as shown in Figure 3.8. The entire system is immersed in a
fluid.
From Eq. (3.8) and Figure 3.8, and noting that
c
=
0
because there is no damper, the equation of motion of the system is
m
d x dt 2
2
+ kx = f
t
ð Þ
+f
tÞ
1ð
(3.38)
3.5 Governing Equations for Different Applied Forces Figure 3.8. Vibrations of a system immersed in a
x(t)
fluid.
Container
f(t)
k
101
m f1 (t)
Fluid
where
x(t )
is measured from the unstretched position of the spring,
f (t)
applied force, and
1
fluid on the mass due to the motion of fluid on the rigid body is 8
f ðtÞ = −Ko M 1
M
is the externally
is the force exerted by the
the mass. The force generated by the
where
f(t)
is the mass of the
fluid
d x dt 2
2
− Cf
dx dt
displaced by the body,
(3.39)
Ko
fi
is an added mass coef cient
that is a function of the shape of the rigid body and the shape and size of the container holding the
fluid,
and
Cf
is a positive
fluid
fi fluid, and the frequency of oscilla-
damping coef cient that is a function of the
shape of the rigid body, the kinematic viscosity of the
tion of the rigid body. If the mass has the shape of a cylinder and the direction of oscillation of the cylinder is perpendicular to the cylinder’s axis, then one can place
ma
and
Cf
with
ca,
is assumed that Re
where
>
ma
and
ca
KoM
with
are given by Eq. (2.85). In using this assumption, it
200.
After substituting Eq. (3.39) into Eq. (3.38), we obtain the governing equation of motion
ð
where
m + KoM Þ
d x dt 2
2
+ Cf
dx dt
+ kx = f
t
ð Þ
(3.40)
K oM is the added mass due to the fluid. From Eq. (3.40), we see that placing a sin-
gle degree-of-freedom system in a 9
viscous damping to the system.
fluid
increases the total mass of the system and adds
A practical application of the
fluid
mass loading is in
10
modeling offshore structures.
In this section, the use of force-balance and moment-balance methods for deriving the governing equation of a single degree-of-freedom system was illustrated. In the next section, a different method to obtain the governing equation of a single degree-of-freedom
8
K. G. McConnell and D. F. Young, “Added mass of a sphere in a bounded viscous
Engineering Mechanics Division , 91(4) (1965) 147–164. 9
A similar result is obtained when one considers the acoustic radiation loading on the mass, when the surface of the mass is used as an acoustically radiating surface. See L. E. Kinsler and A. R. Frey, 2nd edn., John Wiley and Sons, New York, 1962, pp. 180
10
fluid,” Journal of the
Fundamentals of Acoustics ,
–183.
A. U´ scilowska and J. A. Kolodzeij, “Free vibration of immersed column carrying a tip mass,”
and Vibration, 216(1) (1998) 147–157.
Journal of Sound
102
Single Degree-of-Freedom Systems: Governing Equations system is presented. This method is based on Lagrange’s equations, where one makes use of scalar quantities such as kinetic energy and potential energy for obtaining the equation of motion.
3.6
LAGRANGE ’S EQUATIONS
11
Lagrange’s equations
can be derived from differential principles such as the principle of
virtual work and integral principles such as those discussed in Chapter 9. We will not derive
Lagrange’s
equations
equations of holonomic
12
here,
but
systems. We
use
first
these
equations
to
obtain
governing
present Lagrange’s equations for a system
with multiple degrees of freedom and then apply them to vibratory systems modeled with a single degree of freedom. In Chapter 7, we illustrate how the governing equations of systems with multiple degrees of freedom are determined by using Lagrange’s equations.
N degrees of freedom that is described by a set of N gen2,…, N. These coordinates are unconstrained, independent
Let us consider a system with eralized coordinates
qi, I
=
1,
coordinates; that is, they are not related to each other by geometrical or kinematical conditions. Then, in terms of the chosen generalized coordinates, Lagrange’s equations have the form
!
d ∂T dt ∂ q_ j where
q_ j
−
∂T ∂ qj
+
∂D ∂q_ j
∂V ∂qj
+
are the generalized velocities,
T
= Qj
j = 1; 2; …; N
is the kinetic energy of the system,
(3.41)
V
is the
D is the Rayleigh dissipation function, and Qj is the genj th equation. Let us suppose that the system is coml bodies. Then, the generalized forces Qj are given by
potential energy of the system,
eralized force that appears in the posed of
Qj = Fl
X l
Fl ⋅
∂rl ∂qj
+
X l
Ml ⋅
∂ωl ∂q_ j
(3.42)
Ml are the vector representations of the externally applied forces and lth body, respectively, rl is the position vector to the location where the force F l is applied, and ω l is the l th body’s angular velocity about the axis along which the where
and
moments on the
⋅
considered moment is applied. The “ ” symbol in Eq. (3.42) indicates the scalar dot product of two vectors. Note that the generalized force may be a function of the generalized coordinates depending on the problem.
11 12
For a derivation of the Lagrange equations see D. T. Greenwood,
ibid., Section 6– 6.
As discussed in Appendix A, holonomic systems are systems subjected to holonomic constraints, which are integrable constraints. Geometric constraints discussed in Appendix A are in this category.
3.6 Lagrange’ s Equations
103
Linear Vibratory Systems For vibratory systems with linear inertial characteristics, linear stiffness characteristics, and linear viscous damping characteristics, the quantities
T, V , and D take the
following
13
form:
T=
1
V=
1
D=
1
XX N
2
N
j =1 n =1
mjn q_ j q_ n
XX N
2
N
j =1 n =1
kjn qj qn
(3.43)
XX
2
N
N
j=1 n =1
cjn q_ j q_ n
In Eqs. (3.43), each of the summations is carried out over the number of degrees of freedom
N,
fined
which are de
fi
coef cients
fi
in Appendix A. Also, the inertia coef cients
kjn , and the damping coefficients cjn are positive quantities.
Single Degree-of-Freedom Systems In this chapter, we shall limit the discussion to the case when
µ ¶
N
=
mjn ,
the stiffness
1, which is the case of
a single degree-of-freedom system. Then Eqs. (3.41) reduce to
d ∂T dt ∂q_
−
1
∂T ∂ q1
+
∂D ∂q_ 1
∂V ∂ q1
+
=Q
1
(3.44)
where the generalized force is obtained from Eq. (3.42), that is,
Q
1
=
X l
Fl ⋅
∂ rl ∂ q1
Q
In general, the resulting generalized force nates
q
1.
1
+
X l
Ml ⋅
∂ ωl ∂ q_ 1
(3.45)
is not a function of the generalized coordi-
However, in certain instances, they can be, as shown in Example 3.20.
Linear Single Degree-of-Freedom Systems For linear single degree-of-freedom systems, the expressions for the system kinetic energy, the system potential energy, and the system dissipation function given by Eqs. (3.43) reduce to
T=
13
The quadratic forms shown for In addition, the kinetic energy
1 2
XX 1
1
j =1 n =1
mjn q_ j q_ n =
1 2
m q_
2
11
1
T, V, and D in Eqs. (3.43) are strictly valid for systems with linear characteristics. T is not always a function of only velocities as shown here. Systems in which the
kinetic energy has the quadratic form shown in Eqs. (3.43) are called natural systems. For a general system with holonomic constraints, Eqs. (3.41) will be used directly in this book to obtain the governing equations.
104
Single Degree-of-Freedom Systems: Governing Equations
V= D=
1 2
1 2
XX 1
1
kjn qj qn =
j =1 n =1
XX 1
1
j=1 n= 1
T
Comparing the forms of the kinetic energy with Eqs. (2.3) and (2.10), we
1
cjn q_ j q_ n
find that T=
1
V=
1
2
2
1
k q
2
11
2
1
(3.46a)
c q_
2
11
2
1
and the potential energy
V,
respectively,
me q_
2 1
(3.46b)
ke q
2 1
For the Rayleigh dissipation function, we set
D= In these equations,
me
1
ce q_
2
(3.46c)
1
2
is the equivalent mass,
ke
is the equivalent stiffness, and
ce
is the
equivalent viscous damping; they are given by
me = m ke = k ce = c
11
(3.46d)
11
11
!!
!
On substituting Eqs. (3.46) into Eq. (3.44), the result is
d ∂ dt ∂ q_
1
1 2
me q_
2 1
−
∂
1
∂ q1
2
me q_
2 1
+
∂ ∂q_ 1
1 2
!
ceq_
2 1
d ðm q _ dt e
1
Þ
+
∂ ∂q1
1 2
! ke q
2
=Q
1
1
− + ce q_ + k eq = Q 0
m e€q
1
1
1
(3.47)
1
+ ce q_ + k eq = Q 1
1
1
Thus, to obtain the governing equation of motion of a linear vibrating system with viscous damping, one
first
obtains expressions for the system kinetic energy, system poten-
tial energy, and system dissipation function. If these quantities can be grouped so that an equivalent mass, equivalent stiffness, and equivalent damping can be identi
fied,
then,
after the determination of the generalized force, the governing equation is given by the last of Eqs. (3.47). We see further from the de
ωn = ζ=
sffiffiffiffiffi k
finitions Eqs. (3.14) and (3.18) that
e
me ce 2m eωn
=
ffiffiffiffiffiffiffiffiffiffi
ce ke me
p 2
(3.48)
3.6 Lagrange’ s Equations
105
It is noted that depending on the choice of the generalized coordinate, the determined equivalent inertia, equivalent stiffness, and equivalent damping properties of a system will be different; however, the numerical values of the natural frequency and the damping factor will remain the same. In the rest of this section, the use of the Lagrange equations is illustrated with twelve examples. In these examples, we show how the Lagrange equations can be used to derive the governing equations for a wide range of single degree-of-freedom systems. As illustrated in these examples, we use the last of Eqs. (3.47) to obtain the governing equations of motion if the system kinetic energy, system potential energy, and dissipation function are in the form of Eqs. (3.46b); otherwise, we use Eq. (3.44) directly to obtain the governing equation of motion. It is noted that only the system displacements and velocities are needed from the kinematics to use Lagrange’s method whereas, to use the force-balance and moment-balance methods, one also needs system accelerations and one has to deal with internal forces. In addition, with the increasing use of symbolic manipulation programs, it has become more common to have these programs derive the governing equations directly from the Lagrange equations.
EXAMPLE 3.9
Equation of motion for a linear single degree-of-freedom system
For
system
the
linear
of
Figure
3.1,
the
equation
of
motion
is
derived
Lagrange’s equations. After choosing the generalized coordinate to be
x,
by
using
we determine
the system kinetic energy, the system potential energy, and the dissipation function for the system. From these quantities and the determined generalized force, the governing equation of motion of the system is established for motions about the static-equilibrium position. First, we identify the following
q
1
where
j is
=x
F l = f ðtÞj ;
;
rl = xj ;
and
Ml =0
(a)
the unit vector along the vertical direction. Making use of Eqs. (3.45) and (a),
we determine the generalized force as
Q
1
=
X
From Eqs. (2.3) and (2.10), we
l
Fl ⋅
find
∂ rl ∂ qj
+ =f 0
t j⋅
ð Þ
∂xj ∂x
=f
t
ð Þ
(b)
that the system kinetic energy and potential energy
are, respectively,
T=
1
V=
1
2
2
mx_
2
(c)
kx
2
106
Single Degree-of-Freedom Systems: Governing Equations and, from Eqs. (3.46), the dissipation function is
D=
1 2
cx_
2
(d)
Comparing Eqs. (c) and (d) to Eqs. (3.46), we recognize that
me = m;
ke = k;
ce = c
and
(e)
Hence, from Eqs. (e) and the last of Eqs. (3.47), the governing equation of motion has the form
m
d x dt 2
2
+c
dx dt
+ kx = f
t
ð Þ
(f)
which is identical to Eq. (3.8).
INTERACTIVE GRAPHIC 3.2: IN FIGURE 3.9
ANIMATION OF THE SYSTEM SHOWN
With this interactive graphic, one can animate the system shown in Figure 3.9 for the undamped case.
EXAMPLE 3.10
Equation of motion for a system that translates and rotates
In Figure 3.9, a system that translates and rotates is illustrated. After choosing a generalized coordinate, we construct the system kinetic energy, the system potential energy, and the dissipation function, and then noting that they are in the form of Eqs. (3.46), we determine the equivalent inertia, equivalent stiffness, and equivalent damping coef
ficient.
Based on
these equivalent system properties and the last of Eqs. (3.47), we obtain the governing equation of motion of this system. We also determine the expressions for the natural frequency and the damping factor.
Y
Figure 3.9. Disc rolling and translating.
j k
O
Z k M(t)
r G m, JG
X i
x
c
3.6 Lagrange’ s Equations
m and a mass moment of
As shown in Figure 3.9, the disc has a mass its center
G.
inertia
The disc rolls without slipping. The horizontal location of the
107
JG
about
fixed point O
is chosen to coincide with the unstretched length of the spring, and the horizontal translations of the center of mass of the disc are measured from this point center of the disc translates an amount where
θ
x
along the horizontal direction
i,
O.
then
is the corresponding rotation of the disc about an axis parallel to
choose either
x
or
θ
When the
k.
x
= rθ
,
We can
as the generalized coordinate and express the one that is not chosen
θ
in terms of the other. Here, we choose
as the generalized coordinate. Furthermore, we
recognize that
q
1
=θ
F l = 0;
;
M l = M ðtÞk;
ωl = θ_ k
and
(a)
Then, making use of Eqs. (3.45) and (a), we determine the generalized force to be
Q
1
=
X l
∂ωl ∂ q_ 1
Ml ⋅
=M
∂θ_
t k⋅
ð Þ
∂θ_
k = M ðtÞ
(b)
To determine the potential energy, we note that we have a linear spring. Therefore, we make use of Eq. (2.10) and arrive at
V=
1 2
kx
2
=
1
k ðrθÞ
2
2
=
1 2
kr θ 2
2
(c)
From Eq. (c), we recognize the equivalent stiffness of the system to be
ke = kr
2
(d)
To determine the kinetic energy of the system, we make use of Eq. (A.30) of Appendix A; that is, the kinetic energy of the disc is the sum of the kinetic energy due to translation of the center of mass of the disc and the kinetic energy due to rotation about the center of mass. Thus, we
find that T=
=
1
mx_
+
2
|fflffl{zfflffl} 2
1
JG θ_
2
|fflffl{zfflffl} 2
Translation
Rotational
kinetic energy
kinetic energy
1 2
¸mr
where from Table 2.2, we have used
2
+ JG JG
¹θ_
=
2
=
mr
1 2
"
3 2
#
mr θ_ 2
2
(e)
2
/2. From Eq. (e), we recognize that the
equivalent mass of the system is
me =
3 2
mr
2
(f)
108
Single Degree-of-Freedom Systems: Governing Equations The dissipation function takes the form
D=
1 2
cx_
2
=
1 2
± ² = ±cr ²θ_
c r θ_
1
2
2
2
2
from which we identify the equivalent damping coef
(g)
ficient to be
ce = cr
2
(h)
Hence, from the determined generalized force and the equivalent inertia, equivalent stiffness, equivalent damping properties, and the last of Eqs. (3.47), we obtain the governing equation of motion
3 2
mr €θ + cr θ_ + kr θ = M ðtÞ 2
2
2
(i)
Natural Frequency and Damping Factor To determine the natural frequency and the damping factor, we make use of Eqs. (3.48) and
find that
v ffi sffiffiffiffiffiffi sffiffiffiffiffi uffiffiffiffiffiffiffiffiffiffiffiffiffiffi k u kr k = ω = =t m mr = m n
ζ=
EXAMPLE 3.11
2
e
e
ce 2me ωn
3
=
2
2
2
3
pffiffiffiffiffiffiffiffiffiffiffiffiffi ffi = ffiffiffiffiffiffiffiffiffi k= m
cr 2ð3mr =2Þ
2
2
2
3
c p 6km
(j)
Governing equation for an inverted pendulum
For the inverted pendulum shown in Figure 3.10, we obtain the governing equation of motion for
“small”
oscillations about the upright position. The natural frequency of the
inverted pendulum is also determined and the natural frequency of a related pendulum
x1
Figure 3.10.
Inverted planar pendulum restrained by a spring
and a viscous damper.
2r c
k
m1 m2
L2
L1 c.g.
L 2 /2 O
L2
r
3.6 Lagrange’ s Equations
109
system is examined. In the system of Figure 3.10, the bar, which carries the sphere of mass
m , has a mass m that is uniformly distributed along its length. A linear spring of stiffness k and a linear viscous damper with a damping coefficient c are attached to the sphere. 1
2
Before constructing the system kinetic energy, we determine the mass moments of inertia of the sphere of mass
m
m
and the bar of mass
1
about the point
2
O.
The total
rotary inertia of the system is given by
JO = JO where
JO
1
axes theorem, we
O.
find that JO
=
JO
=
1
2
After choosing
1
q
1
=θ
2 5
+m L
m r
2
1
1 12
1
m L 2
2 2
(a)
2
about point
1
O
and
JO
2
is the mass
Making use of Table 2.2 and the parallel-
2 1
L
+m
2
!
(b)
2
=
2
2
1
m L 2
3
2 2
as the generalized coordinate, and making use of Eqs. (a) and (b),
find that the system kinetic energy takes the form
For
m
m
is the mass moment of inertia of
moment of inertia of the bar about point
we
+ JO
1
“small”
T=
1
=
1
2
JO θ_
=
2
"
2
2
5
1 2
m r
2
1
½
JO
+m
1
1
+ JO
L
2 1
θ_
2±
+
1 3
2
#
m L θ_ 2
2
2
2
(c)
rotations about the upright position, we can express the translation of mass
as
x
1
≈L
1
θ
(d)
Then, making use of Eqs. (2.10), (2.62), (2.68), and (d), the system potential energy is constructed as
V=
1
=
1
2
2
kx
2
"
1
−
kL
1 2
2
1
m gL θ 1
1
−m
1
gL
1
2
−
1 2
−m
2
mg
L
2
g
L
2
2
2
#
2
θ2
θ2
(e)
The dissipation function takes the form
D=
1 2
cx_
2 1
=
1 2
cL θ_ 2
1
2
(f)
110
Single Degree-of-Freedom Systems: Governing Equations
find
Comparing Eqs. (c), (e), and (f) to Eqs. (3.46), we
that the equivalent inertia, the
equivalent stiffness, and the equivalent damping properties of the system are given by, respectively,
me =
2 5
m r
2
1
ke = kL
2
ce = cL
2
1
+m L +
1
2
1
−m
1
gL
1
1
3
−m
m L 2
L
g
2
2 2
(g)
2
2
1
Noting that the only external force acting on the system is gravity loading, and that this has already been taken into account, the governing equation of motion is obtained from the last of Eqs. (3.47) as
m e€θ + ceθ_ + k e θ = 0 Then, from the
(h)
first of Eqs. (3.48) and (g), we find that
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi u sffiffiffiffiffi v L u k tkL − m gL − m g ω = = 2
2
n
It is pointed out that
k e can
1
e
1
JO
me
2
1
1
+ JO
2
(i)
2
be negative, which affects the stability of the system as dis-
cussed in Section 4.3. The equivalent stiffness
kL
2 1
>m
1
gL
1
k e is positive when
+m
2
g
L
2
(j)
2
that is, when the net moment created by the gravity loading is less than the restoring moment of the spring.
Natural Frequency of Pendulum System In this case, we locate the pivot point O in
Figure 3.10 on the top, so that the sphere is
now at the bottom. The spring combination is still attached to the sphere. Then this pendulum system resembles the combination of the systems shown in Figures 2.21a and 2.21b. The equivalent stiffness of this system takes the form
k e = kL
1
2
+m
1
gL
1
+m
2
g
L
2
(k)
2
Noting that the equivalent inertia of the system is the same as in the inverted-pendulum case, we
find the natural frequency of this system is
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u sffiffiffiffiffi v L u k tkL + m gL + m g 2
2
ωn =
e
me
=
1
1
JO
1
2
1
+ JO
2
2
(l)
3.6 Lagrange’ s Equations If
≪m
m
2
,
1
r
≪L
, and
1
k=
111
0, then from Eqs. (b) and (l), we arrive at
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v µ mL¶ u u rffiffiffiffiffig m gL + u u m L =u tm L µ + r ¶ → L 1
ωn
1
2
1
2
1
2
1
1
1
2
2
1
L
5
(m)
1
2
1
which is the natural frequency of a pendulum composed of a rigid, weightless rod carrying a mass a distance
L
1
from its pivot. We see that the natural frequency is independent
of the mass and inversely proportional to the length
EXAMPLE 3.12
L
.
1
Governing equation for motion of a disc segment
For the half-disc shown in Figure 3.11, we will choose the coordinate
θ
as the generalized
coordinate and establish the governing equation for the disc. Through this example, we illustrate
how
the
approximated for
system
“small”
kinetic
energy
and
the
system
potential
amplitude angular oscillations, so that the
energy
final
can
be
form of the
governing equation is linear. During the course of obtaining the governing equation, we determine the equivalent mass and equivalent stiffness of this system. The natural frequency of the disc is determined and it is shown that the disc can be treated as a pendulum with a certain effective length. After determining the equivalent system properties, we determine the governing equation of motion based on the last of Eqs. (3.47).
INTERACTIVE GRAPHIC 3.3: ANIMATION OF THE SYSTEM SHOWN IN FIGURE 3.11 With this interactive graphic, one can animate the system shown in Figure 3.11. As shown in Figure 3.11, the half-disc has a mass about
the
center
of
mass
G.
The
system
Figure 3.11. j
R b G m, JG r
O
i k
C
is
m and
assumed
a mass moment of inertia
to
oscillate
Half-disc rocking on a surface.
without
JG
slipping.
112
Single Degree-of-Freedom Systems: Governing Equations The point
O
is a
fixed
R θ from the fixed i, j, and k are fixed in an fixed point O to the center of mass
point, and the point of contact
θ.
point for an angular motion
inertial reference frame. The position vector from the
G
C
is a distance
The orthogonal unit vectors
is given by
r = ð − Rθ + b sin θÞi + ðR − b cos θÞj
(a)
and the absolute velocity of the center of mass is determined from Eq. (a) to be
r_ = − ðR − b cos θÞθ_ i + b sin θθ_ j
(b)
q
Then, using Eq. (A.31) of Appendix A and selecting the generalized coordinate
1
=θ
,
the system kinetic energy takes the form
Choosing the
T=
1
=
1
=
1
2
2
2
JG θ_
2
JG θ_
2
JG θ_
2
+
1
+
1
+
1
2
2
2
mðr_ ⋅ r_ Þ
h
m ð R −b cos θÞ
+b
2
¸
mR
2
+b − 2
2
2
sin
¹
bR cos θ θ_
2
i
2 θ θ_
2
(c)
fixed ground as the datum, the system potential energy takes the form V = mgð R − b cos θÞ
(d)
Note that the system kinetic energy and the system potential energy, which are given by Eqs. (c) and (d), respectively, are not in the form of Eqs. (3.46a).
Small Oscillations about the Upright Position If we use the expressions for the system kinetic energy and system potential energy in Eq. (3.44), then the resulting equation of motion will be a nonlinear equation. Since our
final
objective
is
to
have
a
linear
equation
that
can
be
used
to
describe
amplitude angular oscillations about the upright position of the disc (i.e.,
θo
“small”
=
0), we
express the angular displacement as
θðtÞ = θo + ^ θð tÞ and
expand
the
trigonometric
expansions:
terms
±
²≈
cos
θ = cos θo + ^ θ
sin
θ = sin θo + ^ θ
±
²≈
sin
cos
sin
θ
and
cos
θo − ^ θ sin θo
(e)
θ
−
as
the
^θ
cos
1 2
2
1 2 θo + ^ θ cos θo − ^ θ 2
sin
following
Taylor-series
θo + ⋯
θo + ⋯
(f)
3.6 Lagrange’ s Equations Since
θo =
113
0, Eqs. (f) become
cos
sin
θ ≈1 −
1 2
^θ
2
(g)
θ ≈^ θ
In the expansions given by Eqs. (g), we have kept up to the quadratic term sin
^θ,
since
quadratic terms in the expressions for kinetic energy and potential energy lead to linear terms in the governing equations. On substituting Eqs. (e) and (f) into the expressions (c) and (d), noting that
_
θ_ = ^ θ,
and retaining up to quadratic terms in
sions, we arrive at
T≈
1 2
h
i
^θ
and
^θ_ in these expres-
JG + mðR −bÞ ^θ_ 2
V ≈ mgð R − bÞ +
1 2
2
(h)
mgb ^θ
2
On comparing Eqs. (h) with Eqs. (3.46), we see that while the kinetic energy is in the standard form, the potential energy is not in the standard form because of the constant term. However, since the datum for the potential energy is not unique, we can shift the datum for the potential energy from the
fixed
R−b)
ground of Figure 3.11 to a distance (
above the ground. When this is done, this constant term does not appear and, from Eqs. (h) and (3.46), we identify that the equivalent inertia and stiffness properties of the system are
m e = JG + mðR −bÞ k e = mgb
2
(i)
Since the gravity loading has already been taken into account, the generalized force is
fi
zero. Furthermore, since there is no damping, the equivalent damping coef cient
ce
is zero. Hence, from Eqs. (3.47) and (i), we arrive at the governing equation
h
i
JG + mðR− bÞ €^θ + mgb ^θ = 0 2
(j)
Natural Frequency From Eq. (j), we
find that the natural frequency is
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u mgb u ω =t J + m R −b v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi u g u i = th n
G
ð
2
Þ
JG + mðR− bÞ =mb 2
(k)
On comparing the form of Eq. (k) to the form of the equation for the natural frequency of a planar pendulum of length
L
1
given by Eq. (m) of Example 3.11, we note that
114
Single Degree-of-Freedom Systems: Governing Equations Eq. (k) is similar in form to the natural frequency of a pendulum with an effective length
JG + mðR− bÞ mb
2
Le = EXAMPLE 3.13
(l)
Governing equation for a translating system with a pretensioned or
precompressed spring
We revisit the pretensioned spring system shown in Figure 2.15 and add to the spring combination a mass
m
as shown in Figure 3.12. We shall use Lagrange’s equations to
derive the governing equation of motion for vertical translations
x
of the mass about the
static-equilibrium position of the system and examine how the initially horizontal spring with linear stiffness
k
1
affects the natural frequency of this system. The equation of
motion will be derived for “small” amplitude vertical oscillations, that is,
| x L| ≪
From Eq. (2.41) it is found that when
/
1, the total stiffness
|x L| ≪
kv
/
1.
in the vertical
direction, in the notation of Figure 3.12, is given by
kv = k where
δo
=
T /k 1
1
+ δo k =L
2
(a)
1
when the spring is initially in tension. Then the expression for the
potential energy is
V ðxÞ =
1 2
k vx
2
=
1 2
k
ð
2
+ δo k =L x = 2
Þ
1
1 2
k
ð
2
+ T =L x 1
Þ
2
(b)
The expression for the kinetic energy is
T ðxÞ = Noting that the dissipation function
D
=
1 2
mx_
2
(c)
0 and that the generalized force
Q
1
=
0, we
substitute Eqs. (b) and (g) into Eq. (3.44) to obtain the following governing equation of
µ
motion
m€ x+ k
2
L
Figure 3.12.
+
T L
1
¶
x =0
Single degree-of-freedom system with the horizontal
spring under an initial tension
k1, T1
x m
k2
(d)
T
1
.
3.6 Lagrange’ s Equations
115
From Eq. (d), we recognize the natural frequency to be
ωn =
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k
2
+ T =L 1
(e)
m
It is seen that the effect of a spring under tension, which is initially normal to the direction of motion, is to increase the natural frequency of the system. If the spring of constant can replace
T
1
by
−T
1
k
1
is initially compressed instead of being in tension, then we
and Eq. (e) becomes
ωn =
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k
2
− T =L 1
(f)
m
From Eq. (f), it is seen that the natural frequency can be made very low by adjusting the compression of the spring with stiffness
k
. At the same time, the spring with stiffness
1
k
2
can be made stiff enough so that the static displacement of the system is not excessive. 14
This type of system is the basis of at least one commercial product.
EXAMPLE 3.14
15
Using negative stiffness to isolate a car seat
Consider the system shown in Figure 3.13a in which the mass that the horizontal springs with constant
kh
m
fi
is suf ciently heavy so
are in compression when the system is in
static equilibrium. We shall use the results of Section 2.3.3 to derive the governing equation of motion using the Lagrange equation. The displacement of the mass with
x
b m
y
a
x
z kh
kh
c
kv
y
z
a a
L
Δ
b –L (a)
Figure 3.13.
(b)
(a) Seat suspension model and (b) geometry to determine horizontal spring
displacement.
Source: T. D. Le and K. K. Ahn, “A vibration isolation system in low frequency excitation region using negative stiffness structure for vehicle seat,” Journal of Sound and Vibration, 330 (2011) 6311– 6335. Copyright © 2011, with permission from Elsevier Science.
14 15
Minus K Technology, 420 S. Hindry Avenue, Unit E, Inglewood, CA, USA (www.minusk.com).
“A vibration isolation system in low frequency excitation region using negative stiffness ” Journal of Sound and Vibration , 330 (2011) 6311–6335.
T. D. Le and K. K. Ahn, structure for vehicle seat,
116
Single Degree-of-Freedom Systems: Governing Equations
INTERACTIVE GRAPHIC 3.4: IN FIGURE 3.13
ANIMATION OF THE SYSTEM SHOWN
With this interactive graphic, one can animate the system shown in Figure 3.13 for the undamped case. respect to the inertial frame is inertial frame is
=
z
−
x
y.
y, and the
x,
the displacement of the base with respect to the
displacement of the mass with respect to the base is
y
Therefore, when
equilibrium position. When
=
0,
z
=
x.
z; that is,
All these quantities are about the static-
y is different from zero, we see from Figure 3.13b that ð
b− L+ ΔÞ
+ z =a
2
2
2
(a)
where Δ is the displacement of the horizontal springs. Therefore,
ffiffiffiffiffiffiffiffiffiffiffiffiffi
p
Δ= L− b+
a
2
−z
2
(b)
The kinetic energy of the system is
T = m x_ 1
2
2
=
1 2
m ðz_ + y_ Þ
2
the potential energy is
V=
1 2
kv z
2
+ k hΔ = 2
1 2
k vz
2
+ kh
(c)
³
ffiffiffiffiffiffiffiffiffiffiffiffi´
p
L −b + a
2
−z
2
2
(d)
and the dissipation function is
D=
1 2
c_z
2
(e)
After substituting Eqs. (d), (e), and (f) into Lagrange’s equation given by Eq. (3.44) with
q
1
=z
, we obtain
L −b z = m€y m€ z + cz_ + ð kv − 2khÞ z − 2kh p a −z
ffiffiffiffiffiffiffiffiffiffiffiffiffi 2
which is a nonlinear equation. To linearize the equation, we assume that use the approximation
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ≈ a − z =a a 1
1
Then, Eq. (f) becomes
1
2
µ
µ 1
2
m€z + c_z + kv − 2kh
+
º 1
+
z 2a
2 2
(f)
2
¶
+⋯ ≈
1
a
|z a| ≪ /
1 and
(g)
Ȧ
L−b a
z = m€y
(h)
3.6 Lagrange’ s Equations
117
The natural frequency of this system is
ωn =
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi µ º L − b»¶ 1
m
kv − 2k h
which equals zero when
k v = 2kh
µ 1
+
1
+
(i)
a
¶
L− b a
(j)
INTERACTIVE GRAPHIC 3.5: ANIMATION OF THE SYSTEM SHOWN IN FIGURE 3.14 With this interactive graphic, one can animate the system shown in Figure 3.14 for the undamped case.
EXAMPLE 3.15
Equation of motion for a disc with an extended mass
We shall determine the governing equation of motion and the natural frequency for the system shown in Figure 3.14, for
“small”
angular motions of the pendulum. The system
shown in Figure 3.14 is similar to the system shown in Figure 3.9, except that there is an additional pendulum of length
L
and rigid mass
rolls without slipping. The position of unstretched
length
of
the
spring,
coordinate, and the translation
Y
x = − R θ.
Figure 3.14.
fixed
j
O
Z
X i
2R x L
c
G mD, JG
m
that is attached to the disc. The disc
point
coordinate
O θ
is chosen to coincide with the is
chosen
as
the
generalized
Disc that is rolling and translating and has a rigidly
attached extended mass.
k
k
the
m
118
Single Degree-of-Freedom Systems: Governing Equations The kinetic energy of the system is given by
T = Tdisk + Tpendulum
(a)
where the kinetic energy of the disc is given by Eq. (e) of Example 3.10. The kinetic energy of the pendulum mass
m is given by Tpendulum =
m ðV m ⋅ V mÞ
1 2
(b)
where
Vm =
drm dt
±
d ½ðx + L sin θ Þi + ð L − L cos θ Þ j± dt
=
= −Rθ_ + Lθ_ and
rm is
the position vector from
cos
O to
²
θ i + Lθ_
the mass
sin
m.
θj
(c)
On substituting for the velocity vec-
tor from Eq. (c) into Eq. (b) and executing the scalar dot product, we obtain
Tpendulum =
1
=
1
=
1
2
2
2
m
h± _ _ −R θ +L θ h
m R θ_ 2
¸
mR
2
2
+ L θ_ − 2
2
+L − 2
θ
cos
²
2
+ L θ_ 2
LRθ_
2
2
²
this end, we expand the cos
θ
θ = 0,
θ
i
i
2
(d)
Since the objective is to obtain the governing equation for of the pendulum about the position
2
sin
θ
cos
LR cos θ ± θ_
2
2
“small”
angular oscillations
we retain up to quadratic terms in Eq. (d). To
term as
cos
θ ≈ 1−
1 2
θ2 + ⋯
(e)
substitute Eq. (e) into Eq. (d), and retain up to quadratic terms to obtain
Tpendulum =
1 2
¸
mL
2
+R − 2
¹
LR θ_
2
2
=
1 2
m ðL − RÞ θ_ 2
2
(f)
Making use of Eq. (e) of Example 3.10 and Eq. (f), we construct the system kinetic energy from Eq. (a) as
± =²
T θ_
=
1 2 1 2
mðL− RÞ θ_ 2
h
mðL −R Þ
2
2
+
1 2
mD x_
2
+
+ mDR + JG 2
1 2
JG θ_
i_
θ
2
2
(g)
3.6 Lagrange’ s Equations
119
The potential energy of the system is
V ð θÞ =
1 2
kx
2
+ mg L − L ð
cos
θÞ =
1 2
kR θ 2
+ mgL −
2
ð1
θÞ
cos
(h)
where the datum for the potential energy of the pendulum is located at the bottom position and we have used Eq. (2.59) with
L/2
replaced by
θ
of the pendulum, we use the expansion for the cos
V ðθ Þ =
1
=
1
2
2
kR θ 2
±kR
2
2
+
1 2
L.
To describe small oscillations
term given in Eq. (e) to obtain
mgLθ
²θ
+ mgL
2
2
(i)
The dissipation function is given by
± =²
D θ_
1 2
cx_
=
2
1 2
cR θ_ 2
2
Comparing Eqs. (g), (i), and (j) to Eqs. (3.46), we
(j)
find
that the equivalent system
properties are given by
me = mðL − R Þ
2
ke = kR ce = cR
2
+ mDR + JG 2
+ mgL
(k)
2
Thus, making use of the last of Eqs. (3.47) and Eqs. (k) and noting that the generalized force
Q
=
1
0, we arrive at the governing equation
me €θ + ce θ_ + ke θ = 0 From Eqs. (k) and the
first of Eqs. (3.48), we determine that the system natural frequency is ωn =
EXAMPLE 3.16
(l)
sffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ke = me
kR
2
mðL − RÞ
2
+ mgL + mD R + JG
(m)
2
Lagrange formulation for a microelectromechanical system (MEMS)
device
We shall determine the governing equation of motion and the natural frequency for the 16
microelectromechanical system
shown in Figure 3.15. The mass
micro mirror whose typical dimensions are 300
μm
rigid bar. The torsion springs are rods that are 50
16
400
μm
2
is the scanning
μ m. This mass is modeled as a μm in area and
in length and 4
2
“Electrostatic comb-drive-actuated micromirrors ” Journal of Microelectromechanical Systems , 7(1) (1998) 27 –37.
M.-H. Kiang, O. Solgaard, K. Y. Lau and R. S. Muller, laser-beam scanning and positioning,
×
m
for
120
Single Degree-of-Freedom Systems: Governing Equations
x2
c Jo , m2
L1
O kt k MEMs device
m1 x1
xo (t)
(a) Figure 3.15.
L2
(b)
(a) MEMS device and (b) single degree-of-freedom model.
Source : M. H. Kiang, O. Solgaard, K. Y. Lau and R. S. Muller,
“Electrostatic comb-
Journal of Microelectromechanical Systems , 7(1) (1998) 27– 37. Copyright © 1998 IEEE. Reprinted with drive-actuated micromirrors for laser-beam scanning and positioning”,
permission.
are collectively modeled by an equivalent torsion spring of stiffness
m
1
kt in the figure. Mass
fingers.”
is the mass of the electrostatic comb drive, which comprises 100 interlaced “
The comb
fingers
are 2
μm
μm
wide and 40
long. The comb drive is connected to the
displacement drive through an elastic member that has a spring constant is connected to the bar
m
2
We shall use the angular coordinate
ϕ
x
1
1
as the generalized coordinate, and derive the
xo( t)
equation of motion for “small” angular oscillations. The translation and the translations
k. The mass m
by a rigid, weightless rod.
and
x
2
is prescribed,
are approximated as
x x
1
2
=L ϕ =L ϕ 2
(a)
1
The system potential energy is constructed as
V =V
1
where
V
1
+V +V 2
(b)
3
is the potential energy of the torsion spring,
translation spring, and
V
3
V
2
is the potential energy of the
is the gravitational potential energy of the bar. For “small”
angular oscillations of the bar, Eq. (c) of Example 2.7 is used to describe the bar ’s potential energy. Thus, we arrive at
V ðxo; ϕÞ =
1
=
1
2
2
kt ϕ
+
1
kt ϕ
+
1
2
2
2
2
kð xo ðtÞ−x
2
1Þ
+
1 4
kð xo ðtÞ−L ϕÞ
+
=
L
where we have made use of Eqs. (a). When
2
2
L
2
m gðL 2
1 4
2
−L
m gðL 2
ϕ2
1Þ
2
−L
ϕ2
1Þ
(c)
, the effects of the increase and
1
decrease in the potential energy of each portion of the bar of mass
m
2
cancel.
3.6 Lagrange’ s Equations
121
The system’s kinetic energy is determined as
± =²
_ T ϕ
1 2
=
1 2
Jo ϕ_
2
±J
o
+
1 2
+m
m x_ 1
²
1 2
_ Jo ϕ
2
+
1 2
m L ϕ_
2
2
1
2
L ϕ_ 2
1
=
2 1
2
(d)
2
and the system dissipation function is given by
± =²
D ϕ_
1 2
cx_
=
2 2
1 2
cL ϕ_
2
2
(e)
1
where we have again used Eqs. (a). Comparing the forms of Eqs. (c), (d), and (e) to Eqs. (3.46), we
find
that the potential energy is not in the standard form. Thus, we will
make use of Eq. (3.44) to determine the governing equation of motion. To this end, we
find from Eq. (c) that ∂V ∂ = ∂ϕ ∂ϕ
"
1 2
kt ϕ
2
= k tϕ − kL =
2ð
"
k t + kL
2 2
+
1 2
kð xo ðtÞ−L ϕÞ
2
2
xo ðtÞ − L ϕÞ +
+
1
2
1 2
m g ðL 2
2
2
−L
+
1
m g ðL 2
#
1Þ
m gðL 2
4
2
2
−L
−L
# ϕ
2
1Þ
ϕ
1Þ
ϕ − kL2 xo ðtÞ
To obtain the governing equation of motion, we recognize that
(f)
q
1
=ϕ
, substitute for the
system kinetic energy and the dissipation function from Eqs. (d) and (e), respectively, into Eq. (3.44), make use of Eq. (f), and note that the generalized force
Q
1
=
0. Thus,
we obtain
€ + cL ϕ_ + ke ϕ = kL xoðtÞ me ϕ 2
(g)
2
1
where
me = Jo + m L 1
ke = k t + k L
2 2
2 2
+m
2
gðL
2
From the inertia and stiffness terms of Eq. (g), we is given by
ωn =
−L
=2
1Þ
find that the system natural frequency
sffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ke = me
kt + kL
2 2
+m g L −L Jo + m L 2
ð
2
2
1
=2
1Þ
(h)
2
An experimentally determined value for the natural frequency of a typical system is 2400 Hz. If the rod connecting the mass
m
1
to the bar were not assumed rigid and
weightless, one would need to consider additional coordinates to describe the system of Figure 3.14. Systems with more than one degree of freedom are treated in Chapters 7 to 9.
122
Single Degree-of-Freedom Systems: Governing Equations
INTERACTIVE GRAPHIC 3.6: IN FIGURE 3.16
ANIMATION OF THE SYSTEM SHOWN
With this interactive graphic, one can animate the system shown in Figure 3.16 without the spring kd.
EXAMPLE 3.17
Equation of motion of a slider mechanism
17
fi
We revisit a modi ed revision of the slider mechanism system of Example 2.2 and obtain the equation of motion of this system by using Lagrange’s equations. Gravity loading is
ms slides m l that is pivoted at the point O . A linear spring of stiffness k restrains the motions of the mass ms . Another uniform bar of mass (mb + me) is pivoted at O′, which is attached to a linear spring of stiffness kd at one end and attached to the mass m s at the other end. An excitation d(t) is imposed at one end of the spring with stiffness k d. assumed to act normal to the plane of the system shown in Figure 3.16. The mass
along a uniform bar of mass
We choose the angular coordinate
φ
as the generalized coordinate, and we shall
determine the equation of motion in terms of this coordinate. The geometry imposes the following constraints on the motion of the system:
r ðφÞ = a + b − 2ab cos φ r ðφÞsin β = b sin φ a = rð φÞcos β + b cos φ 2
At a
first
2
2
(a)
glance, the slider mechanism appears to be a system that would need more
than one coordinate for its description; however, due to the constraints given by Eqs. (a)
O l
ml
Slider mechanism.
r a
ms k
Figure 3.16.
b mb
Spring end fixed to m l
e
O' kd
me
d(t)
17
J. Pedurach and B. H. Tongue, “Chaotic response of a slider crank mechanism,”
Acoustics, 113 (1991) 69–73.
Journal of Vibration and
3.6 Lagrange’ s Equations
123
the system is a single degree-of-freedom system that can be described by the independent coordinate
φ.
System Kinetic Energy The total kinetic energy of the system is
±
T r; r_ ; β_ ; φ_ where
Jmb and Jme
²=
1 2
Jmb + Jme ±φ_
½
2
+
1 2
Jml β_
2
+
1 2
ms r_ ðφÞ +
1 2
ms r β_
are the mass moments of inertia of bar of mass
2
2
(b)
mb
and bar of mass
ml l
(c)
me about the fixed point O′, respectively, and Jml is the mass moment of inertia of bar ml about the fixed point O. Then, making use of Eq. (c) of Example 2.2 and Figure 3.16, we
find that
Jmb = Since the angle
β
1
mbb
2
3
Jme =
;
r(φ)
and the length
obtain expressions for
β_
and
r_ ðφÞ
1 3
me e
2
;
Jml =
and
are each related to
in terms of
φ_ .
φ
1 3
2
by Eqs. (a), we proceed to
We differentiate the
first of Eqs. (a) with
respect to time to obtain
r φÞ r_ ðφÞ = 2ab φ_ sin φ
2 ð
which leads to
r_ ðφÞ =
ab φ_ sin φ rðφÞ
(d)
Upon differentiating the third of Eqs. (a) with respect to time, we arrive at
r_ ðφÞ cos β − rðφÞ β_
sin
β − b φ_ sin φ = 0
which results in
β_ =
r_ ðφÞcos β − b φ_ sin φ rðφÞsin β
We now use Eqs. (a) and Eq. (d) in Eq. (e) to obtain
β_ =
= =
(
$
ab a − b cos φ φ_ sin φ b sin φ rð φÞ rðφÞ 1
φ_
r ð φÞ 2
φ_
r ð φÞ 2
¸a
− ab
¸ab
cos
2
cos
φ − r2ð φÞ
φ − b2
¹
(e)
%
− bφ_
) sin
φ
¹
(f)
124
Single Degree-of-Freedom Systems: Governing Equations After substituting Eqs. (d) and (f) into Eq. (b), we arrive at the following expression
φ_
for the total kinetic energy in terms of
T ðφ; φ_ Þ = where
mð φÞ = Jmb + Jme
± + J
ml
1
mð φÞ φ_
2
2
µab ² +m r
cos
2
s
(g)
φ− b2
¶
2
r ðφÞ 2
+ ms
µ ab rð φÞ
sin
φ
¶
2
(h)
System Potential Energy The system potential energy is given by
V=
1 2
kr ðφÞ + k d ½d ðt Þ−eφ± 1
2
2
(i)
2
Equation of Motion Since the expressions for kinetic energy and potential energy are not in the standard form of Eqs. (3.46), we make use of the Lagrange equation given by Eq. (3.44) to obtain the equation of motion, that is,
µ ¶
d ∂T dt ∂φ_
−
∂T ∂φ
+
∂D ∂ φ_
+
∂V ∂φ
=
0
(j)
where we have used the fact that the generalized force is zero. Noting that there is no dissipation in the system (that is,
D
=
0) we substitute for the kinetic energy and poten-
tial energy from Eqs. (g) and (i), respectively, into Eq. (j), and carry out the differentiation operations to obtain the following nonlinear equation
mðφÞφ €+
1 2
m′ð φÞ φ_
2
+ kr φ ð
Þ
r′ ðφÞ + k d e φ = k d ed ðtÞ 2
where the prime denotes the derivative with respect to
EXAMPLE 3.18
Oscillations of a crankshaft
(k)
φ.
18
k JG about its center of mass
Consider the model of a crankshaft shown in Figure 3.17 where gravity is acting in the direction. The crank of mass
mG
is connected to a slider of mass
Jd
about the
fixed point O .
and mass moment of inertia
mp
at one end and to a disc of mass moment of inertia
Choosing the angle
θ
as the generalized coordinate, we shall
first derive the governing equation of motion of the system, and then from this equation, determine the equation governing oscillations about a steady rotation rate.
18
Vibration of Structures and Machines: Practical Aspects, 2nd edn., Springer-Verlag, New York, 1995, “Torsional vibration of crankshafts: effects of nonconstant moments of inertia,” Journal of Sound and Vibration, 205(2) (1997) 135 –150.
G. Genta,
–
pp. 338 341; and E. Brusa, C. Delprete and G. Genta,
3.6 Lagrange’ s Equations
M(t)
l
a G
r
j
125
b d
O
mG , JG
i
mp
Jd
Figure 3.17.
Crankshaft model.
INTERACTIVE GRAPHIC 3.7: ANIMATION OF THE SYSTEM SHOWN IN FIGURE 3.17 With this interactive graphic, one can animate the system shown in Figure 3.17.
Kinematics From Figure 3.17, we see that the position vector of the slider mass point
mp
with respect to
O is rp = ðr cos θ + l cos γ Þi + dj
and that the position vector of the center of mass
G
(a)
of the crank with respect to point
O is rG = ðr cos θ + a cos γ Þi + ðr sin θ − a sin γ Þ j Furthermore, from geometry, the angle
γ
and the angle
θ
(b)
are related by the relation
r sin θ = d + l sin γ
(c)
To determine the slider velocity, we differentiate the position vector time and obtain
vp =
±− r θ_
sin
θ − l γ_
sin
rp
²
γ i
with respect to
(d)
By differentiating Eq. (c) with respect to time, we obtain the following relationship between
γ_
and
θ_ :
γ_ =
r l
θ_ θ cos γ cos
(e)
After substituting Eq. (e) into Eq. (d), we obtain the slider velocity to be
vp = −rθ_ ðsin θ + tan γ cos θÞi
(f)
126
Single Degree-of-Freedom Systems: Governing Equations The velocity of the center of mass
±− rθ_
G
of the crank is obtained in a similar manner. We
² ±
²
differentiate Eq. (b) with respect to time to obtain
vG =
sin
θ − aγ_
sin
γ i + rθ_
cos
³
vG = −
θ+
sin
´
cos
a
+b=l
µ
¶
After substituting Eq. (e) into Eq. (g) and noting that of the crank’s center of mass to be
θ − a_γ
γ j
(g)
, we obtain the velocity
a _ i + b cos θ rθ_ j tan γ cos θ rθ l l
(h)
System Kinetic Energy The total kinetic energy of the system is given by
T=
1 2
Jd θ_
2
+
1 2
mG ðvG ⋅ vG Þ +
1 2
JG γ_
+
2
1 2
±
mp vp ⋅ vp
²
(i)
We now substitute Eqs. (e), (f), and (h) into Eq. (i) to obtain
T= where
J ðθ Þ = Jd + r mG 2
+ JG
8< :
r cos θ l cos γ
1 2
!
+r
2
&r
q
Noting that the generalized coordinate zero,
d dt
!
2
+
b cos l
mpð sin θ+ tan γ cos θÞ
l
Equation of Motion function is
(j)
! 9= θ ; 2
2
γ = sin−1
system dissipation
2
a tan γ cos θ sin θ + l
and from Eq. (c)
Eq. (3.44) takes the form
J ðθ Þ θ_
1
and
µ∂T¶ ∂ θ_
θ−
sin
=φ
(k)
' (l)
, the system potential energy is zero, the
that
−
d l
2
∂T ∂θ
the
generalized
= −M
t
ð Þ
moment
Q
1
= −M t
( ),
(m)
Upon substituting Eq. (j) into Eq. (m) and performing the differentiation operations, we obtain
J ð θÞ €θ +
1 2
J ′ ð θÞ θ_
2
= −M
where the prime denotes the derivative with respect to
t
ð Þ
θ.
(n)
3.6 Lagrange’ s Equations The angle
θ
127
can be expressed as the superposition of a rigid-body motion at a constant
angular velocity
ω
and an oscillatory rotation
ϕ,
that is,
θðtÞ = ωt + ϕðtÞ Then, from Eqs. (n) and (o), we arrive at
€+ J ðθ Þϕ EXAMPLE 3.19
1 2
±
J ′ ðθÞ ω +ϕ_
²
2
(o)
= −M
t
ð Þ
(p)
19
Vibration of a centrifugal governor
A centrifugal governor is a device that automatically controls the speed of an engine and prevents engines from exceeding certain speeds or prevents damage from sudden changes in torque loading. We shall derive the equation of motion of such a governor by using Lagrange’s equations. A model of this device is shown in Figure 3.18. The velocity vector relative to point
o
of the left-hand mass is given by
V m = − L φ_ cos φ i + L φ_ sin φ j + ðr + L sin φÞωk
(a)
From Eq. (A.29) of Appendix A and Eq. (a), the kinetic energy is
"
T ðφ; φ_ Þ = 2
1
h
2
mðV m ⋅ V mÞ
= m − Lφ_ = mω r + L ð
2
cos
ð
sin
#
φÞ2 + ðLφ _ sin φÞ2
+ r+L ðð
sin
φÞωÞ2
i
φÞ2 + mφ_ 2L2
(b)
Figure 3.18. Centrifugal governor.
j
i
r o
Lϕ
L L k
m
(r Lsin ) [ to page]
m
Controlled arm
19
J. P. Den Hartog,
ibid.,
p. 309; and Z.-M. Ge and C.-I. Lee, “Nonlinear dynamics and control of chaos for a
rotational machine with a hexagonal centrifugal governor with a spring,”
–
(2003) 845 864.
Journal of Sound and Vibration,
262
128
Single Degree-of-Freedom Systems: Governing Equations The potential energy with respect to the static equilibrium position is
V ðφÞ =
1 2
k ð2Lð1 −cos φÞÞ
2
−
mgL cos φ
2
(c)
where the factor of 2 inside the parenthesis is because each pair of linkages compresses the spring from both the top and the bottom. Upon
using
Eq.
(3.44)
with
q
=
1
φ,
noting
that
Q
1
=D=
0,
and
performing
the
required operations, we obtain the following governing equation
mL € φ − mrLω 2
2
cos
±
²
φ − mω2 + 2k L2
sin
φ cos φ + L ðmg + 2kLÞsin φ = 0
(d)
Introducing the quantities
γ=
r ; L
cos
φ − ω2 + ω2n
Eq. (d) is rewritten as
φ € − γω2
ω2p =
±
If we assume that the oscillations
fi
and Eq. (f) simpli es to
φ
³
g ; L
²
sin
ω2n =
and
³
k m
2
φ =0
(f)
0 are small, then cos
ϕ ≈ 1; sin ϕ ≈ ϕ,
φ cos φ + ω2p + ω2n
about
φ
=
´
(e)
sin
´
φ € + ω2p − ω2 φ = γω2
ficient ficient is negative.
From the stiffness coef ness coef
EXAMPLE 3.20
(g)
in the equation of motion, we see that for
ω > ωp
the stiff-
Oscillations of a rotating system
A cylindrical wheel is placed on a platform that is rotating about its axis with an angular speed constant
k,
Ω.
The center of the wheel is attached to the platform by a spring with
as shown in Figure 3.19. We shall determine the change in the equilibrium
position of the wheel and the natural frequency of the system about this equilibrium position. When
Ω = 0,
the center of the wheel is at a distance
R
from the axis of
rotation, which is the length of the unstretched spring.
Ω
Figure 3.19. Elastically restrained wheel on a rotating platform.
R k
m r x
3.7 Summary of Natural Frequency Equations If we denote the change in the equilibrium of the spring due to the rotation
129
Ω
as
δ,
then at equilibrium, the spring force is equal to the centrifugal force, which can be represented as
k δ = mðR + δÞΩ Upon solving for
δ;
2
we obtain
R
δ=
ω21n =Ω2 − 1
where
ω21n =
k m
For small angles of rotation, the kinetic energy is
T=
1 2
µ
1 2
¶µx_ ¶ 2
mr
2
r
+
1 2
m x_
2
=
1
µ ¶
2
3 2
m x_
2
The potential energy for oscillations about the equilibrium position is
V=
µ ¶
1 2
kx
2
The Lagrange equation for this undamped system is
d ∂T dt ∂ x_ where the centrifugal force
mxΩ
2
−
∂T ∂x
+
∂V ∂x
= Qx = mxΩ
2
is treated as an external force. In this problem, the
generalized force is a function of the generalized coordinate
µ ¶
x, and as seen next it
affects
the system natural frequency. Thus, the governing equation is 3 2
±
²
m €x + k − mΩ x = 0 2
fi
Examining the equation of motion, we see that the effective stiffness is modi ed by the rotation. From the stiffness and inertia terms in the governing equations of motion, the natural frequency is
ωn
3.7
ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ± ² = ω −Ω 2 3
2 1
n
2
rad=s
SUMMARY OF NATURAL FREQUENCY EQUATIONS FOR SINGLE DEGREE-OF-FREEDOM SYSTEMS
The equations for the natural frequencies of many of the systems analyzed in this chapter are summarized in Table 3.2 along with those for many other rotating, translating, and pendulum type single degree-of-freedom systems.
130
Single Degree-of-Freedom Systems: Governing Equations
Table 3.2.
Single degree-of-freedom systems and expressions for their natural
frequencies Case
System
Natural frequency (Hz)
Translation 1
fn =
m
1
π
2
rffiffiffiffi k m
k
L
2
fn = k1 T1
+T −T
m
1
π
2
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k
2
±T
1
=L
m
1
for tension
1
for compression
k2
a
3
b
fn =
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1
π
2
ka
2
mða +bÞ
2
m k
a
4
b
k1
fn =
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u t ³ kka ´ π 2
1 2
1
m ka 1
2
2
+ k a +b 2ð
Þ
k2
m
L
5
fn =
a
m
m1 k
1
π
2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k
m + m ðL=a − 1Þ 1
2
2
3.7 Summary of Natural Frequency Equations
Table 3.2. (cont.) Case
System
Natural frequency (Hz)
6
fn =
a
1
π
2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k
m + m ðL=aÞ
2
1
m1 m k
Rotation 7
fn =
m, J O kt
sffiffiffiffiffi 1
π
2
kt 1 = JO 2π
rffiffiffiffiffiffiffiffiffi kt mR 2
2
O R
sffiffiffiffiffiffiffi
L
8
fn =
a
1
π
2
ka JO
2
sffiffiffiffiffiffiffiffiffi =
1
ka mL
3
π
2
2
2
m, JO O k
Translation and rotation 9
k
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
m, JO
fn =
O
1
π
2
kR 1 = 2 π mR + J O 2
2
rffiffiffiffiffiffi
k m
2
3
R
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
10
k
R
O m, J O
a
fn =
1
π
2
k ðR +a Þ 1 = mR + J O 2 π 2
2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2
k ðR+ aÞ 3mR 2
2
131
132
Single Degree-of-Freedom Systems: Governing Equations
Table 3.2. (cont.) Case
System
Natural frequency (Hz)
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
11
fn =
m, JO a g
k ðR +aÞ mR + JO 2
1
π
2
2
k
O
R a
12
fn = m, JO
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kR = m R + JO
k m=2 + m
2
1
π
2
2
1
R
O k
m1
13
k
R
fn =
O
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi + mD gL L −R + mR + JO kR
1
π
2
2
mD ð
Þ
2
2
m, JO L
mD
14
fn =
k
m2 , JO R
m1
=
1
π
2
1
π
2
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u kR t 2
m R 2
2
+
4
m R
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k
m
3
2
+
m
4
1
1
2
+
JO
4
1
3.7 Summary of Natural Frequency Equations
Table 3.2. (cont.) Case
System
Natural frequency (Hz)
k
15
L2
fn =
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
k m + ð2 JO =r
β
tan
2
1
L2
L1 r
L πL 1
Þtan
2
L1 m
b
r O
JO O
JO
Pendulum (acceleration due to gravity, g 16
= 9.8 m/s2) fn =
m1 , JO1 R L 2/2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1
kL
2 1
π
2
− m gL − m gL = JO + JO 1
2
1
1
2
L1 m2 , JO2
L 2/2 O 17
fn =
R G
1
π
2
b
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mgb
JG + m ðR − bÞ
m, JG
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
18
a
fn =
k
1
π
2
g L
+
ka mL
2
2
L
m
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
19
fn =
m k L a
1
π
2
ka
2
mL
2
−
g L
2
2
2
β
133
134
Single Degree-of-Freedom Systems: Governing Equations
Table 3.2. (cont.) Case
System
Natural frequency (Hz)
20
fn =
kt
1
π
2
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g + kt L mL
2
L m
21
fn =
O L
1
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mgL γ
JO =
cos
JO
π
2
2
mL
2
3
L g
g
m, JO
fn =
22
R
π
2
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mg R − rÞðm + JO =r
ð
2
J O = mr 1
Þ
2
2
m, JO
O 2r 23
kt
fn =
L1
O L2
m 2, JO k
1
m1
L3
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k + kL + m g L − L = + m gL ± ² π m L +L = +m L
1 2
t
2
2
3
2
ð
1Þ
2
2
2
2
1
3
2 1
1
2 3
3
3.8 Summary 3.8
135
SUMMARY
In this chapter, the use of two different methods to derive the governing equation of motion of a single degree-of-freedom system was illustrated. One of these methods is based on applying force and/or moment balance and the other method is based on Lagrange’s equations. The underlying approach for each of these methods will be used again for deriving governing equations of systems with multiple degrees of freedom in
fi
Chapter 7. De nitions of natural frequency and damping factor were also introduced. It was shown how the static-equilibrium position of a vibratory system can be determined and how nonlinear systems can be linearized to describe small oscillations about a system’s equilibrium position. In addition, interactive graphics material has been included to facilitate visualization of motions of many single degree-of-freedom systems.
136
Single Degree-of-Freedom Systems: Governing Equations
Exercises
Section 3.2.1 3.1
A vibratory system with a hardening nonlinear spring is governed by the following equation
±
m€x + cx_ + k x + αx
3
²=
0
Determine the static-equilibrium position of this system for
α
=
1 m
−
2
and linearize
the system for “small” oscillations about the system’s static-equilibrium position. 3.2
A vibratory system with a softening nonlinear spring is governed by the following equation
±
m€x + cx_ + k x − αx
3
²=
0
Determine the static-equilibrium positions of this system for
α
=
−
1 m
2
and linear-
ize the system for “ small” oscillations about each of the system’s static-equilibrium positions. 3.3
Determine the equation governing the system studied in Example 3.13 by carrying out a force balance.
3.4
m
A mass
is attached to the free end of a thin cantilever beam of length
shown in Figure E3.4. The
that is rotating about its axis at a speed of beam
is
negligible
and
L,
its
equivalent
Ω
rad/s. Assume that the mass of the
stiffness
is
k b.
Derive
the
governing
equation of motion for transverse vibrations of the beam in terms of the variable
2a
as
fixed end of this beam is attached to a shaft of radius a x.
Figure E3.4.
L m
kb
x
Section 3.2.2 3.5
Determine the equation governing the system studied in Example 3.16 by carrying out a moment balance.
Section 3.3.1 3.6
A cylindrical buoy with a radius of 1.5 m and a mass of 1000 kg
ρ
(
=
3
floats in salt water
1026 kg/m ). Determine the natural frequency of this system.
Exercises 3.7
137
A 10 kg instrument is to be mounted at the end of a cantilever arm of annular cross-section. The arm has a Young’s modulus of elasticity a mass density
ρ
=
E
=
72
×
9
10
N/m
2
and
3
2800 kg/m . If this arm is 500 mm long, determine the cross-
sectional dimensions of the arm so that the
first
natural frequency of the system is
above 50 Hz. 3.8
The static displacement of a system with a motor weight of 385.6 kg is found to be 0.0254 mm. Determine the natural frequency of vertical vibrations of this system.
3.9
A rotor is attached to one end of a shaft that is rotary inertia of the rotor be
JG,
fixed
at the other end. Let the
and assume that the rotary inertia of the shaft is
negligible compared to that of the rotor. The shaft has a diameter and it is made from material with a shear modulus
G.
d,
a length
L,
Determine an expression for
the natural frequency of torsional oscillations. Obtain
3.10
an
expression
for
the
natural
frequency
for
the
system
shown
in
Figure E3.4. Consider the hand motion discussed in Example 3.3 and let the hand move in the
3.11
horizontal plane; that is, the gravity force acts normal to this plane. Assume that the length of the forearm
l
is 25 cm, the mass of the forearm
object being carried in the hand has a mass
M
with the restoring force of the biceps is 2
10
with the triceps is 2
×
3
10
×
=
3
m
5 kg, the constant
N/rad, the constant
N/rad, and the spacing
a
=
is 1.5 kg, the
kb Kt
associated associated
4 cm. Determine the
equation of motion of this system, and from this governing equation,
find the nat-
ural frequency and damping factor of the system. A spring elongates 2.5 mm when stretched by a force of 5 N. Determine the static
3.12
de
flection and the period of vibration if a mass of 8 kg is attached to the spring.
Determine the natural frequency in Hz of the steel disc with torsion spring shown
3.13
in Figure E3.13 when
h
=
kt = 0:488 N ⋅ m=rad, d
=
50 mm,
ρdisc
=
3
7850 kg/m , and
2 mm.
h
Figure E3.13.
kt d
3.14
Consider
a
relationship
nonlinear
spring
that
is
governed
³x ´
F ðxÞ = a
c
b
by
the
force-displacement
138
Single Degree-of-Freedom Systems: Governing Equations where
a
=
3000 N,
b
=
0.015 m, and
c
=
2.80. If this spring is to be used as a
mounting for different machinery systems, obtain a graph similar to that shown in Figure 3.5b and discuss how the natural frequency of this system changes with the weight of the machinery. 3.15
The is
fl
static
found
de ection in the
to
be
25
μm.
tibia bone
Determine
the
of
a
120
associated
kg
person standing
natural
frequency
upright of
axial
vibrations. 3.16
A solid wooden cylinder of radius
r, height h, and speci fic gravity sw is placed in a
container of tap water such that the axis of the cylinder is perpendicular to the surface of the water. Assume that the density of the water is
ρH
2O
. It is assumed
the wooden cylinder stays upright under small oscillations. (a) If the cylinder is displaced a small amount, then determine an expression for its natural frequency.
fi
(b) If the tap water is replaced by salt water with speci c gravity of 1.2, then determine whether the natural frequency of the wooden cylinder increases or decreases and by what percentage. 3.17
Consider the pulley system shown in Figure E3.17. The mass of each pulley is small compared with the mass
m and, therefore, can be ignored. Furthermore, the
cord holding the mass is inextensible and has negligible mass. Obtain an expression for the natural frequency of the system.
Figure E3.17.
k
m k
3.18
A rectangular block of mass
m
rests on a stationary half-cylinder, as shown in
Figure E3.18. Find the natural frequency of the block when it undergoes small oscillations about the point of contact with the cylinder.
L h g
r
Figure E3.18.
Exercises
139
Section 3.3.2
Formulate a design guideline for Example 3.8 that would enable a vibratory sys-
3.19
flection by ficient constant.
tem designer to decrease the static de
n
a factor
while holding the
damping ratio and damping coef
Formulate a design guideline for Example 3.8 that would enable a vibratory sys-
3.20
tem designer to decrease the static de damping ratio and mass
m constant.
flection
An instrument’s needle indicator has a rotary inertia of 1:4
3.21
to a torsion spring whose stiffness is
fi
coef cient
− 1 :1 × 10
5
n
by a factor
N
⋅
×
10
6
while keeping the
kg
⋅
2
m . It is attached
m=rad and a viscous damper of
c. What is the value of c needed so that the needle is critically damped?
Determine the natural frequency and damping factor for the system shown in
3.22
Figure E2.26. Determine the natural frequency and damping factor for the system shown in
3.23
Figure E2.27.
Section 3.6
Derive
3.24
the
governing
equation
of
motion
for
the
rocker-arm
valve
assembly
Jo is the mass O of the rocker arm of length (a + b), k is the stiffis fi xed at one end, and M is the external moment
shown in Figure E3.24. Assume “small” motions. The quantity moment of inertia about point ness of the linear spring that
imposed by the cam on the system. This moment is produced by the contact force generated by the cam at one end of the rocker arm.
a
b O
Figure E3.24.
Jo k
m 3.25
Derive the governing equation of motion for the system shown in Figure E3.25. The
O is JO, and the torsion stiffness kt. Assume that there is gravity loading.
mass moment of inertia of the bar about the point of the spring attached to the pivot point is
Figure E3.25.
c.g.
L/2 O
L kt
140
Single Degree-of-Freedom Systems: Governing Equations 3.26
For the base-excitation prototype shown in Figure 3.6, assume that the base displacement
y(t)
is known, choose
x(t )
as the generalized coordinate, and derive the
equation of motion by using Lagrange’s equations. 3.27
Obtain the equation of motion for the system with rotating unbalance shown in Figure 3.7 by using Lagrange’s equations.
3.28
Obtain the equation of motion for the system shown in Figure 3.10 by using moment balance and compare it to the results obtained by using Lagrange’s equations.
3.29
fluid-float
For the
tia about point
system shown in Figure E3.29,
JO
is the mass moment of iner-
O. Assume that the mass of the bar is mb. Answer the following.
(a) For “small” angular oscillations, derive the governing equation of motion for the
fluid float system.
(b) What is the value of the damping coef
ficient c
for which the system is criti-
cally damped?
Figure E3.29.
c
L
O
c.g.
Rigid, weightless connection
Jo
a m
L 2
3.30
d
Determine the nonlinear governing equation of motion for the kinematically constrained system shown in Figure E3.30. Consider only vertical motions of
Figure E3.30.
k2 m1 a
m2
b
Rigid, weightless rod
m3
k3
m
.
1
Exercises 3.31
141
Obtain the linearized equation of motion and an expression for the natural frequency for the following cases in Table 3.2: (a) Case 4 (b) Case 5 (c) Case 6 (d) Case 8 (e) Case 10 (f) Case 12 (g) Case 14 (h) Case 18 (i) Case 19 (j) Case 20 (k) Case 21 (l) Case 22
3.32
Determine the natural frequency for the vertical oscillations of the system shown in Figure E3.32. Let and 0
0 that is, while the spring-damper combination is being compressed. At the instant when they are no longer in compression the acceleration is zero, that is, the time at which the
first time instance at which the acceleration
sum of these forces on the mass is zero. The is zero is given by Eq. (T.15) for
=
p
0 and the corresponding velocity is given by
Eq. (T.14) of Table 4.1.
ficient of restitution ε is defined as
Based on Newton’s law of impact, the coef
ε=
− vvehicle − vbarrier after impact − vvehicle after impact = vvehicle − vbarrier before impact vvehicle before impact ð
Þ
ð
ð
Þ
Þ
ð
(a)
Þ
vvehicle is the vehicle velocity, v barrier is the velocity of the barrier, and the assumption that vbarrier is zero has been used; that is, the barrier is fi xed. Then, making use of Eq. (a) and Eq. (T.14) of Table 4.1 with p = 0, we fi nd that where
ε=
4
See
also,
V.
Appendix I.
I.
Babitsky,
−x_ tv ð
x_ ð0Þ
Þ
=
V oe−
2
φv = tan φv
Vo
= e− φ = 2
v
tan
φv
Theory of Vibro-Impact Systems and Applications,
(b)
Springer-Verlag,
Berlin,
1998,
168
Single Degree-of-Freedom Systems: Free Responses Barrier
Vo
c
v(t) c
v(tvm)
c
m
m
k
m
k
Bumper makes contact with barrier at t to
k
Bumper in contact with barrier for to t tvm to
Bumper no longer in contact with barrier for t tvm to
(a) Vo Rigid system v(t) to
to
0
tvm
k
t Oscillator
m
Rigid system
v(tvm)
c
Vo (c)
(b) Figure 4.8.
(a) Model of a car bumper colliding with a stationary barrier, (b) time history of
velocity of mass, and (c) equivalent impact con
figuration. In this equivalent configuration,
V o impacts a barrier, which is represented by a spring
a mass moving with a velocity and damper combination.
We now make use of Eq. (b) to examine how the coef the damping factor Table 4.1 that
ζ.
Considering
φv/tan φv
→
0 as
ζ
→
first
ficient
of restitution
ε
depends on
the undamped case, we note from Eq. (T.6) of
0, and, therefore,
ε
→
1. In other words, there are no
losses and the system leaves with the same velocity with which it arrived. This is consistent with the fact that this is an elastic collision. When cally damped and
φv/tan φv
→
ζ
→
1, the system becomes criti-
1 and, therefore, from Eq. (b), we
is, the mass leaves the barrier with a velocity of 0.135
V o.
find
that
The amount of energy that the system dissipates during the interval 0
ε
→ e−
2
≤ t ≤ tv
; that
is the
difference between the initial kinetic energy and the kinetic energy at separation. Note that the vehicle does not have any potential energy when it is not in contact with the barrier. Thus,
Ediss =
1 2
mV o − m½x_ ðtv Þ± 2
1 2
2
=
1 2
mV o − mV o e− 1
2
2
or
Ediss Einit
= − e− φ = 1
4
v
tan
φv
2
4
φv =tan φv
4.2 Free Responses of Undamped and Damped Systems
169
1.0 0.8
Ediss /Einit
tini E/ ssidE
0.6
,e
0.4
e
0.2 0
0
0.2
0.4
0.6
0.8
1.0
ficient of restitution and fraction of energy dissipation for impacting single
Figure 4.9. Coef
degree-of-freedom system.
where
Einit
is given by Eq. (4.35). These results are summarized in Figure 4.9. Only colli-
sion with a single impact has been considered.
EXAMPLE 4.5
Impact of a container housing a single degree-of-freedom system
floor a system that resides inside respect to the floor. The system is
We shall now consider the effects of dropping onto the
fi
ε
a container that has a coef cient of restitution
with
shown in Figure 4.10. If the container falls from a height velocity at the time of impact with the
floor is Vo =
h,
then the magnitude of the
pffiffiffiffiffiffiffi gh
2
t = 0+ after impact, the container bounces upwards with a velocity whose εVo. Then at t = 0+, the container and the single degree-of-freedom system
At the instant magnitude is
can be modeled as a single degree-of-freedom system with a moving base as discussed in
fi
Section 3.5.1. Thus, if we de ne the relative displacement
zðtÞ = xðtÞ − yðtÞ
(a)
then, from Eq. (3.30) we have
m However,
€ y = −g,
d z dt 2
2
+c
dz dt
+ kz = − m
d y dt 2
2
(b)
since the container is decelerating during the rebound upwards. Then
Eq. (b) becomes
m
d z dt 2
2
+c
dz dt
+ kz = mgu t
ð Þ
(c)
170
Single Degree-of-Freedom Systems: Free Responses
m
x
Vo
g c
k
y
Vo
m c
k
h
(a) Figure 4.10.
(b)
Single degree-of-freedom system inside a container: (a) dropped from a height
(b) on rebound immediately after impact with the
where
floor.
h and
u(t ) is the unit step function. The initial conditions are zð0Þ = xð0Þ − yð 0Þ = 0 z_ ð0Þ = x_ ð0Þ − y_ ð 0Þ =−Vo − εV o
= − + ε Vo = − + ε 0
value. This is a boundedness condition, which requires the system
be bounded for bounded system inputs. If this is not the case, then the sys-
tem is considered
unstable.
8
For the systems that are dealt with in this book, the unstable
responses grow either linearly with time or exponentially with time.
Instability of Unforced System We consider an unforced vibratory system subjected to
finite
initial conditions and study
when this system can be unstable. To this end, we start from the solution for the response given in the Laplace domain by Eq. (C.2) of Appendix C and set
X ðsÞ =
8
sxð0Þ D ðs Þ
+
F(s)
ζωn xð0Þ + x_ ð0Þ
2
DðsÞ
=
0 to obtain
(4.40)
Applied Nonlinear Dynamics: Analytical, Computational and Experimental Methods , John Wiley & Sons, New York, 1995.
Other notions of stability can be found in A. H. Nayfeh and B. Balachandran,
4.3 Stability of a Single Degree-of-Freedom System
D (s),
Now let us examine the denominator
179
which is given by Eq. (C.3) of Appendix C,
that is,
DðsÞ = s
+
2
ζω ns + ω2n = s 2 + ðce =me Þs + k e=m e
2
= s− s ð
where we have used
1 Þð
ζ = ce =ð2me ωnÞ;
s
1;2
=
s− s
2Þ
pffiffiffiffiffiffiffiffiffiffiffiffi ω = k m ¹ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiº e=
n
e;
−ce ±
1
me
2
(4.41)
and
ce − 4m ek e 2
(4.42)
and we have switched to the more general equivalent forms for the inertia, stiffness, and damping. From Eq. (4.40), we see that there are two terms on the right-hand side that involve, respectively, the polynomial ratios 1
DðsÞ
=
s −s
ð
1
s −s
1 Þð
2Þ
=
¹
1
−s
s
ð
2Þ
1
and
s DðsÞ
=
s −s
ð
s
s −s
1 Þð
2Þ
=
ð
−s
s
1Þ
2
s− s
1Þ
¹
1
ð
1
s
2
ð
s− s
2Þ
−
º
1
s −s
ð
º
s
−
(4.43)
2Þ
1
s −s
ð
(4.44)
1Þ
Since
xðtÞ = L− ½X ðsÞ ± 1
where
L–
1
indicates the inverse Laplace transform, it is seen that in order to obtain
x(t)
one needs the inverse Laplace transforms of Eqs. (4.43) and (4.44), which are
L−
1
¹
º
1
s− s
1; 2
= es
1;2
t
where we have used Laplace transform pair 7 in Table B.1 of Appendix B. The condition under which
es
t
1; 2
remains
finite for t > 0 is
µs ¶ ≤
Re
1;2
0
(4.45)
that is, the real parts of the roots have to be less than or equal to zero. From Eq. (4.42),
fi
it is seen that this condition is satis ed when
ke
0, and
k e = kL
2 1
−m
1
gL
1
−m
2
g
L
2
2
ke > 0, that is, when
Thus, the system is stable as long as
kL
2 1
≥m
1
gL
1
+m gL
2
2
2
me
>
0,
182
Single Degree-of-Freedom Systems: Free Responses
4.4
SINGLE DEGREE-OF-FREEDOM SYSTEMS WITH NONLINEAR ELEMENTS
4.4.1
Nonlinear Stiffness We illustrate the effects that two different types of nonlinear springs can have on the free response of a system when subjected to either an initial displacement or an initial velocity.
System with Hardening Cubic Spring First, we consider a system that has a spring whose stiffness includes a component that varies as the cube of the displacement. After using Eq. (2.30) for the nonlinear spring force, the governing equation is
d x dτ 2
2
+
ζ
2
where the nondimensional time variable
dx dτ
+ x + αx = 3
= ωnt
τ
0
10
. We solve Eq. (4.52) numerically
it does not have an analytical solution. We assume that the initial conditions are
Xo
=
1 cm and
Vo
=
(4.52)
α
=
3 cm
–2
,
ζ
=
since
0.15, and that
0. The results are shown in Figure 4.17,
along with the solution for the system with a linear spring, that is, when
α
=
0.
We see from these results that the response of the system with the nonlinear spring is distinctly different from that with the linear spring. First, the response of the system with the nonlinear spring does not decay exponentially with time. Second, the displacement response does not have a constant period of damped oscillation. These differences provide one with a means of distinguishing one type of nonlinear system from a linear system based on an examination of the response to an initial displacement. In practice, the nonuniformity of the period is easier to detect, since the dependence of frequency (or period) on the amplitude of free oscillation is a characteristic of a nonlinear system.
System with Piecewise Linear Springs We now consider a second nonlinear system shown in Figure 4.18. In this case, the springs are linear; however, the mass is straddled by two additional linear elastic spring-stops that are not contacted until the mass has been displaced by an amount
d
in either direction. The
stiffness of the springs is proportional to the attached spring by a constant of proportionality
μ (μ
≥
and when
0). When
μ
>
μ
=
0, we have the standard linear single degree-of-freedom system,
1, the elastic spring-stops are stiffer than the spring that is permanently
attached to the mass. The governing equation describing the motion of the system is
d y dτ 2
2
10 11
Using Matlab,
+
ζ
2
dy dτ
+ y + μh y =
0
ode45; using Mathematica, NDSolveValue.
“ Primary resonance of a harmonically forced Journal of Sound and Vibration, 207(3) (1997) 393–401.
H. Y. Hu,
ð Þ
11
oscillator with a pair of symmetric set-up elastic stops,
”
4.4 Single Degree-of-Freedom Systems with Nonlinear Elements
183
1.0 Nonlinear Linear ±e–
0.5
zt
oX /)t( raenilx ,) t( y
0.0
–0.5
–1.0
0
10
20
(a)
Linear zero– crossing intervals
Nonlinear zero– crossing intervals
3.178 3.178 3.178 3.178 3.178 3.178 3.178 3.178 3.178 3.178
2.034 2.332 2.632 2.879 3.038 3.118 3.154 3.168 3.174 3.176
30 t
1.0
Nonlinear Linear
0.5 t
d/raenilxd ,td/yd
0.0
–0.5
–1.0
–1.5
–1.0
–0.5
(b) Figure 4.17.
0.0 y(t ), xlinear(t )/Xo
0.5
1.0
Comparison of the responses of linear (dashed lines) and nonlinear (solid lines)
systems with prescribed initial displacement: (a) displacement and (b) phase-plane plot.
d m
k
d x
Figure 4.18.
m
m k
Single degree-of-freedom system with additional
springs that are not contacted until the mass displaces a
c
k
distance d in either direction. Source : H. Y. Hu, “ Primary resonance of a harmonically forced oscillator with a pair of symmetric set-up elastic stops, ” Journal of Sound and Vibration , 207(3) (1997) 393– 401. Reprinted by permission of Federation of the European Biochemical Societies. Copyright from Elsevier Science.
©
1997, with permission
184
Single Degree-of-Freedom Systems: Free Responses where
hðyÞ = 0 = y − sgnðyÞ y
and the signum function sgn( ) is
+
1 when
we have employed the following de
τ = ωnt; Although it is possible to 12
obtain a numerical
finitions:
ωn =
find
rffiffiffiffi
k ; m
y
y=
>
∣y∣ ≤ ∣y∣ > −
0 and is
x ; d
and
1 1 1 when
ζ=
2
ζ
4.21
0 and linear in the region are
determined
x_ (τ )
numerically.
0
(5.20)
5.2 Response to Harmonic Excitation where
finite
A is a positive finite number. Hence, for Ω ≠
215
1, an undamped system excited by a
harmonic excitation is stable in the sense of boundedness introduced in Section
=
Ω
4.3. However, this is not true when
1. When
Ω
=
1, the term
τ
τ
cos( ) in Eq. (5.19)
grows linearly in amplitude with time and, hence, it becomes unbounded after a long time. This special ratio
Ω
=
1 (
= ωn
ω
resonance relation; that is, the linear resonance when the excitation frequency is
) is called a
system given by Eq. (5.16) is said to be in equal to the natural frequency.
From a practical standpoint, the question of boundedness is important. Since there is always some amount of damping in a system, it is clear from Eqs. (5.7) to (5.9) that the
Ω=
response remains bounded when excited at the natural frequency, that is, for
→∞
lim
τ
xð τÞ =
1
Fo sin ð τ − π=2Þ 2ζk
(5.21)
fi
The response given by Eq. (5.21) satis es the boundedness condition given by Eq. (5.20), even though as the damping decreases in magnitude the response increases in magnitude.
EXAMPLE 5.4
Forced response of an undamped system
Consider translational motions of a vibratory system with a mass of 100 kg and a stiffness of 100 N/m. When a harmonic forcing of the form the system, where
Fo
=
them for the following cases: (i) Let
the
variable
x
Fosin( ωt ) acts on the mass of
1.0 N, we shall determine the responses of the system and plot
be
ω
used
=
to
0.2 rad/s, (ii)
describe
the
ω=
1.0 rad/s, and (iii)
translation
of
the
ω=
mass.
2.0 rad/s.
Then,
from
Eq. (5.16),
€x + ωn x = 2
Fo m
sin ð
ωt Þ
(a)
The natural frequency of the system is
ωn =
rffiffiffikffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi = m
=
100 N m 100 kg
=
=
1 rad s
(b)
Hence, for cases (i) and (iii), the excitation frequency is different from the natural frequency, and for case (ii), the excitation frequency is equal to the natural frequency of the system. In cases (i) and (iii), the solution for the displacement response is given by Eq. (5.17), that is,
xðtÞ =
Fo k ð1 − Ω
2
ωtÞ − Ω sin ðωntÞg
fsin ð Þ
(c)
On the other hand, in case (ii), the response is given by Eq. (5.19), that is,
xðtÞ =
Fo fsin ð ωn tÞ − ωn t cos ðωn t Þg 2k
(d)
216
Single Degree-of-Freedom Systems: Periodic Excitations
Figure 5.8.
Displacement response of an undamped system subjected to harmonic forcing at three
frequencies: (i)
ω
=
0.2 rad/s; (ii)
ω
=
1.0 rad/s; and (iii)
ω
=
2.0 rad/s.
For the given values, the responses are as follows.
Case (i) xðtÞ =
1 N
100 N=m
=
±
×
1
t
0:01f sin ð0:2 Þ
−
ð0:2=1:0Þ
−
0:2sin g
2
t
² m
t
fsin ð0:2 Þ
−
ð0:2=1:0Þsin ð1
×t
Þg
(e)
5.2 Response to Harmonic Excitation
217
Case (ii) xð tÞ =
=
2
×
1 N 100 N=m
0:005 fsin
t − tcostg
fsin
t − tcos tg
(f)
m
Case (iii) xðt Þ =
±
1 N
100 N=m
=−
×
1
−
ð0:2=1:0Þ
t
−
0:003 f sin ð2 Þ
2
²
t
0:2 sin g
t
f sin ð2 Þ
−
t
ð0:2=1:0Þ sin g
(g)
m
The graphs of Eqs. (e), (f), and (g) are provided in Figure 5.8. From the graphs, it is clear that the response of the undamped system remains bounded when the excitation frequency is away from the natural frequency, as in cases (i) and (iii). However, in case (ii), the response becomes unbounded after a long period of time since the amplitude increases with time.
5.2.4
Magnitude and Phase Information: Mass Excitation
°
In the case of an undamped linear vibratory system, there is a phase shift of 180
in the
response as one goes from an excitation frequency that is less than the natural frequency to
an
excitation
frequency
that
is
greater
than
the
natural
frequency.
This
can
be
discerned from Eq. (5.17) by noting that the change in sign is brought about by the term 1
−Ω
2
.
For a linear damped vibratory system excited at the resonance frequency
°
response lags the excitation by 90
Ω
=
1, the
as shown by Eq. (5.21) and in Figure 5.2b. This
observation is used in experiments to determine whether the excitation frequency is equal to the undamped natural frequency of the system, that is, response is also large at
ω = ωn ( Ω =
1) when
ζ
ω
= ωn
. The magnitude of the
is small.
Response Characteristics in Different Excitation Frequency Ranges Additional characteristics of the response of an underdamped linear vibratory system can be determined by examining the steady-state response
Ω
≪
1 and
Ω
≫
1 and at the frequency location
Ω
=
xss(τ)
in the frequency ranges
1; that is, in a region considerably
below the natural frequency, in a region well above the natural frequency, and at the natural frequency, respectively. The examination is performed by studying the values of
H(Ω) and θ( Ω) in these three ranges.
218
Single Degree-of-Freedom Systems: Periodic Excitations
Ω≪1 In this region, Eqs. (5.14) lead to
→ →
H ðΩÞ θðΩÞ
1 (5.22)
0
and, therefore, from Eq. (5.13), we obtain
xð τÞ ≅
Fo 1 sin ðΩτÞ = f ðτ Þ k k
(5.23)
where, for convenience, we have omitted the subscript from Eq. (5.13). Since the only system parameter that determines the displacement response is the stiffness this region the
stiffness-dominated region.
k,
we call
In this region, the displacement is in phase
with the force. The velocity response is determined from Eq. (5.23) to be
vð τÞ =
dxðτ Þ dt
= dx τ
ð Þ dτ d τ dt
≅ ΩωnFo k
Ωτ Þ =
cos ð
Ωωn Fo sinðΩτ + π =2Þ k
(5.24)
π/2;
Therefore, as expected, the velocity response leads the displacement response by
that is, the maximum value of the velocity occurs before the maximum value of the displacement.
Ω=1 At this value of
Ω, Eqs. (5.14) simplify to H ð1Þ = θð1Þ =
1
ζ π 2
(5.25)
2
Therefore, the displacement response given by Eq. (5.13) takes the form
xðτ Þ =
Fo Fo sin ðτ − π =2Þ = 2k ζ cωn
Thus, for a given excitation amplitude
Fo
and natural frequency
displacement is determined by the damping coef the
damping-dominated region. Ω
=
ωn,
(5.26)
the amplitude of the
ficient c. This region, therefore, is called
We also see that the displacement lags the force by
and that this phase lag is independent of maximum when
τ − π =2Þ
sinð
ζ.
For
ζ
> 0, it is mentioned that
H(Ω)
π/2
is not a
1. We shall determine this maximum value and the value of
Ω
at
which it occurs subsequently. The amplitude response is characterized by a peak close to the natural frequency for values of Figure 5.2a.
ζ
in the range 0
1=
Ω≥
H ðΩÞ < 1
2;
1 for
ω ≥ ωn
for
Ω
>
p
0,
and
when
ζ < 1=
ffiffi
2;
ffi ffiffiqffiffiffiffiffiffiffiffiffiffiffiffiffi − ζ
p
or
π/2.
(5.30)
2
1
2
2
ffi ffiffiqffiffiffiffiffiffiffiffiffiffiffiffiffi − ζ
p
2
1
2
2
These response characteristics can also be seen in Figure 5.2a. The dependence of the system response on the different system parameters is summarized as a function of the excitation frequency as shown in Figure 5.9. Further discussion about the response magnitude and the response phase as a function of the excitation frequency is provided in Section 5.3.
220
Single Degree-of-Freedom Systems: Periodic Excitations 6 Damping dominated
5
x(t )~1/c
4
) Ω (H
3 Stiffness dominated
2
Inertia dominated
x(t )~1/k
x(t ) ~1/m
1 0
0
0.5
1.0
Ω
1.5
2.0
2.5
(a) 180
Inertia dominated x(t )~1/m
150
120 )°( ) Ω( q
Damping dominated x(t )~1/c
90
60 Stiffness dominated x(t )~1/k
30
0
0
0.5
1.0
Ω
1.5
2.0
2.5
(b) Figure 5.9.
Three regions of a single degree-of-freedom system: (a) amplitude response and
(b) phase response.
5.2 Response to Harmonic Excitation
Observations: Stiffness-dominated region .
When
the
amplitude
of
the
221
harmonic
exciting force is constant and the excitation frequency is much less than the natural frequency of the system, the magnitude of the displacement is determined by the system’s stiffness. The displacement response is in phase with the excitation force.
Damping-dominated region.
When the amplitude of the harmonic exciting force is
constant and the excitation frequency equals the natural frequency of the system, the
fi
magnitude of the displacement response is magni ed for 0
fi
1, θ(Ω) decreases as ζ increases. • For ζ < 0.15, Hmax( Ω) ≈ 1/(2ζ). Unbalanced mass excitation • At Ω = 0, the value of Hub( Ω) is zero. • The phase response is identical to that for excitation of the mass. • For ζ < 0.15, Hmax( Ω) ≈ 1/(2ζ). • For 0 < ζ < 0.7, the maximum value of H(Ω) always occurs at a frequency Ω ≥ 1. • When ζ ≥ 0.7, the maximum value of Hub (Ω) approaches 1. • As Ω increases, Hub (Ω) approaches 1. Base excitation • For ζ < 0.1, Hmax( Ω) ≈ 1/(2ζ). p • For Ω > 2; Hmb( Ω) < 1 for all values of the damping factor; however, the values of Hmb(Ω) in this region approach 1 as ζ increases. • There is no stationary value of θmb (Ω) as ζ changes. • The maximum values of Hmb(Ω) occur at Ω ≤ 1.
ffiffi
INTERACTIVE GRAPHIC 5.4: PHASE RELATIONS AMONGST THE EXCITATION, DISPLACEMENT, AND VELOCITY FOR THREE TYPES OF EXCITATION This interactive graphic is used to visualize the phase relationships amongst the displacement and velocity of the mass and the applied input for three cases of harmonically excited single degree-of-freedom systems: direct mass excitation, base excitation, and excitation by a rotating unbalanced mass.
5.2 Response to Harmonic Excitation
233
In this interactive graphic, one should note the following. • •
•
•
•
5.2.8
The normalized phase relations for the mass excited systems and the unbalanced mass systems are the same; however, the normalizing factors are different. The phase relations are distinctly different when the excitation frequency is less than, equal to, or greater than the natural frequency of the system and are also greatly affected by the damping factor. The phase of the velocity with respect to the force for a mass excited system and for an unbalanced mass is zero at resonance irrespective of the damping factor; however, for a base excited system, in order for the velocity and the force to be in phase the damping factor must be zero. For excitation frequencies well below the resonance frequency and for a ζ < 0.12: (1) for the mass excitation and unbalanced excitation cases, the phase difference between the applied force and the displacement is virtually equal; and (2) for the base excitation case the phase difference between the applied base displacement and the displacement of the mass is relatively small, indicating that the magnitude of the displacement is approximately equal to the static displacement due the magnitude of the applied force. For Ω ≫ 1 and for small damping factor, the phase relations for all three types of excitation are such that the force and the displacement are approximately 180 ° out of phase, a fact which helps explain transmission isolation: the dependency on the damping factor is greatest for base excitation.
Harmonic Excitation of a System with a Maxwell Model In this section, we shall examine the single degree-of-freedom system whose springs are
figured
dampers, con
in the form of the Maxwell model shown in Figure 4.14. For this
model, we shall consider the cases when the mass is harmonically excited, the base is subject to a harmonic excitation, and the mass contains a rotating unbalanced mass. We start with Eq. (4.36), which is repeated here as
d x dτ 2
2
dx + x + 2ζ dτ dx 2ζ dt
where
τ
= ωnt γ = k ,
k Ω
/ ,
1
= ω ωn
the type of forcing as follows.
/
,
ω
−
dxd dτ
−
dxd dt
! !
=
F ðΩÞ k (5.64)
= γ xd
is the excitation frequency, and
F(Ω)/ k depends
on
234
Single Degree-of-Freedom Systems: Periodic Excitations
Mass Excitation F ðΩÞ k
=g
ΩÞejΩτ
1ð
(5.65)
where
Fo k
g ð ΩÞ = 1
(5.66)
Base Excitation F ð ΩÞ k
= yo + j ð1
where
g ðΩÞ = yo
ζ ΩÞejΩτ
2
=g
2ð
ΩÞ ejΩτ
(5.67)
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi +
1
2
ζ ΩÞ ejφb 2
ð2
(5.68)
φb = tan− 2ζ Ω 1
Rotating Unbalanced Mass F ðΩÞ k
=g
ΩÞejΩτ
3ð
(5.69)
where
g ð ΩÞ = 3
and for this case
ωn =
m oε Ω M + mo
2
(5.70)
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k M + mo
(5.71)
In Eqs. (5.65), (5.67), and (5.69), we have assumed for mathematical convenience the complex exponential representation of a harmonic forcing. To solve Eq. (5.64), we assume solutions of the form
xðτ Þ = X pðΩÞejΩτ
xd ðτ Þ = Xp d ðΩÞejΩτ ;
where
p
p = 1; 2; 3
(5.72)
indicates the single degree-of-freedom system being considered as described above.
After substituting Eq. (5.72) into Eq. (5.64) and solving for
X p and X p,d, we obtain
5.2 Response to Harmonic Excitation
γ + j 2ζ Ω
X pðΩÞ =
ao + jbo
Xp d ðΩÞ = ;
235
gp ðΩÞ (5.73)
j 2ζ Ω gp ðΩÞ p = 1; 2; 3 ao + jbo
where
ao = γ ð1 − Ω
2
Þ
bo = 2ζ Ωð1 + γ − Ω Xp ðΩÞ = HM ðΩÞgpð ΩÞ ejθ M
ð
Xp d ðΩÞ = H dM ðΩÞgpð ΩÞ ejθ dM
ð
ΩÞ (5.75)
p = 1; 2; 3
ΩÞ
;
HM ðΩÞ
Þ
find that
Expressing Eq. (5.73) in exponential form, we
where
(5.74) 2
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u u u γ + ζ Ω γ + ζΩ =t =t
HdM ðΩÞ =
2
ð2
ao + bo
Þ
2
2
Þ
γ 2 ð1− Ω2Þ2 + ð2ζ ΩÞ2 ð1 +γ −Ω2 Þ2
2
2
2
ð2
ζΩ ζΩ pffiffiffiffiffiffiffiffiffiffiffiffiffi ffi = qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a +b 2
2
2
γ 2ð1 − Ω2 Þ
2
o
2
o
θM ðΩÞ = tan−1 θdM ðΩÞ = tan−1
ζ Ωao − γ bo
2
γ ao + 2ζ Ωbo
ao bo
=
tan
−
=
tan
+
1
γ
2
ð1
−γ − Ω ζ Ω + γ −Ω
2
2
−
ð1
1
ζ ΩÞ
ð2
2
−Ω
ð1
2
Þ
+ γ− Ω
− +
2
2
Þ
(5.76)
ζ Ωγ
2
ζ ΩÞ
2
2
ð2
ð1
+ γ −Ω
2
Þ
Þ
ð1
2
Þ
–
The Maxwell model reduces to the Kelvin Voigt model when
γ
→∞
as can be seen
from the following limits:
→ ∞ HM ðΩÞ → qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1
lim
γ
→ ∞ HdM ðΩÞ → 0
ð1
−Ω
2
2
Þ
+
ζ ΩÞ
ð2
2
(5.77)
lim
γ The
results
Graphic
5.5
of
this
can
be
section used
to
are
summarized
compare
the
in
Table
amplitude
5.3.
In
response
addition, functions
Interactive and
phase
–
response functions for systems with the Kelvin Voigt and Maxwell models when the excitation is one of the following: mass excited, base excitation, and rotating unbalanced mass.
Single Degree-of-Freedom Systems: Periodic Excitations
236
Table 5.3.
Steady-state displacement response of a single degree-of-freedom system to
differently applied harmonic forces for a Maxwell model
→∞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v u u =u ² γ + ζω ω ± ² t ± γ −ωω + ζω ω +γ − ω ω γ = k 1= k :
Maxwell amplitude and phase functions where Voigt model.
2
HM
2
θM
=
−
tan
1
1
γ2
ð
±
1
−
Mass excitation
ð2
ω=ωn Þ
2
d x dt
m
Fosin w t
c
e w
x
dt
xd
Base excitation x
ð
2
ω=ωn Þ
2
d
o
²
sin
n
m
d
n
o
2
o
d
n d
dt
sin
ωt
M + mo
n
mo ε ω HM sinðωt − θM Þ ðM + m oÞ ω n 2
2
0 1 dx dx + ω x + ζω @ − A = ζy ωω dt dt 0 1 sffiffiffikffi dx dx @ A ζ − = γω x ω = dt dt m qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
d x dt 2
2
m
+γ −
n d
2
n
xðtÞ =
xd
1
2
Fo HM sinðωt − θ M Þ k
dt
k1
yosin wt
±
nÞ
=
0 1 d x dx dx m ε + ω x + ζω @ − A = ω dt dt M +m dt 0 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi dx dx k − A = γω x ω = ζ@
2
c
2
ð
n
dt
2
k
k1
ζω=ωn Þ
ð2
2
n
2
k
+
2
t c
1
ζγ 2ω= ωn
2
xðtÞ =
Rotating unbalanced mass mo
2
nÞ
2
d
2
xd
M
²
=
2
nÞ
2
2
k1
=
–
, this model reduces to the Kelvin
0 1 dx dx + ω x + ζω @ − A = F ωt dt dt m 0 1 sffiffiffikffi dx dx @ A − ζ = γω x ω =
x
k
ð
ð2
2
2
nÞ
=
γ
When
2
2
n
d
2
xðtÞ = y o
1
+
d
n
n d
2
o
2
ωt + yo ω2n sin ωt
+
− 2ζω=ω Þ n
n
ζω=ωn Þ HM sinð ωt − θM
ð2
n cos
tan
1
5.2 Response to Harmonic Excitation
237
INTERACTIVE GRAPHIC 5.5: COMPARISON OF THE AMPLITUDE RESPONSE AND PHASE RESPONSE FOR SYSTEMS WITH KELVIN – VOIGT AND MAXWELL MODELS This interactive graphic is used to visualize the amplitude and phase response functions for three harmonically excited single degree-of-freedom systems: mass excitation, base excitation, and excitation by a rotating unbalanced mass. In addition, these response functions for two different types of spring-damper models are compared: Kelvin –Voigt and Maxwell models. In this interactive graphic, the following should be noted. • •
•
•
•
•
•
•
At Ω = 0, the value of H ub(Ω) is zero. For the system with the Kelvin–Voigt model and for the unbalanced mass in the vicinity of Ω = 1 and ζ > 0, the maximum value of H ub(Ω) occurs at Ω > 1, whereas for the mass excited and unbalanced excitation they respectively occur at Ω < 1; for the system with the Maxwell model and these conditions, all excitations have the maximum values of H(Ω) occuring at Ω > 1. For the system with the Maxwell model and for all excitations, the maximum values of H(Ω) are always greater than those for the corresponding cases of the system with the Kelvin – Voigt model when γ < 1, irrespective of the value of ζ ; however, the magnitudes of these peaks decrease and then increase as ζ increases. The system with the Maxwell model very closely assumes the amplitude and phase characteristics of the system with the Kelvin–Voigt model when γ ≫ 1 and/or ζ ≪ 1 for all three excitation cases. For the system with the Kelvin –Voigt model, the phase response of the mass excited and unbalanced mass systems is always 90 ° at Ω = 1 irrespective of the value of the damping factor; this is not true for the base excited system nor is it true for the system with the Maxwell model for any excitation. For the case of an unbalanced mass excitation and for both models, the value of Hub( Ω) for Ω ≫ 1 approaches 1; for the moving base and the mass excited systems H(Ω) approaches zero. For the case of an unbalanced mass excitation and for the system with a Kelvin –Voigt model, p the value of Hub( Ω) < 1 for Ω > 2 for all values of the damping factor; there is no such fixed value for Ω for the system with the Maxwell model. For the case of direct mass excitation and unbalanced mass excitation, the value of Ω for which H( Ω) < 1 varies with the value of the damping factor for the system with the Kelvin– Voigt model, and in the case of the system with the Maxwell model, it is additionally a function of γ.
ffiffi
238
Single Degree-of-Freedom Systems: Periodic Excitations
5.3
RESPONSE TO EXCITATION WITH HARMONIC COMPONENTS
Responses of vibratory systems subjected to periodic excitations include such widely diverse
applications
as
internal
combustion
engines,
propeller-driven
aircraft,
and
weaving machinery. As explained later in this section, a periodic excitation can be considered as a sum of related harmonic components. Here, we consider the steadystate response of a single degree-of-freedom system subjected to a forcing function that is composed of a collection of harmonic components, each at a different amplitude and frequency.
Excitation with Two Harmonic Components Consider
harmonic
excitation
acting
on
the
mass
of
a
linear
vibratory
system
of
the form
f ðtÞ = B sinðωtÞ
(5.78a)
f ð τÞ = B sinðΩτ Þ
(5.78b)
1
or, in the equivalent form,
1
where the nondimensional time
τ
= ωnt
and the nondimensional frequency
Ω
= ω ωn /
.
Then, from Section 5.2, the steady-state displacement response is given by
x ð τÞ = 1
where the amplitude response
±
²
B H ðΩÞ sin Ωτ − θðΩÞ k
(5.79)
H( Ω) and the phase response θ(Ω) are given by Eqs. (5.8).
When the forcing function is of the form
f ðtÞ = A cosð ωt Þ
(5.80a)
f ðτ Þ = A cosðΩτ Þ
(5.80b)
2
or, in the equivalent form,
2
the corresponding steady-state response is
x ðτ Þ = 2
±
²
A H ðΩÞ cos Ωτ − θðΩÞ k
(5.81)
We now consider the linear combination of forces given by Eqs. (5.78b) and (5.80b), that is,
f ðτ Þ = A cosð Ωτ Þ + B sinðΩτ Þ
(5.82)
5.3 Response to Excitation with Harmonic Components
239
Since the considered system is linear, the solution for the linear combination of forces 8
given by Eq. (5.82) is the linear combination of the individual solutions
given by
Eqs. (5.79) and (5.81). Thus,
xðτ Þ = x ð τÞ + x ðτ Þ 1
where
ψ
=
tan
2
h
±
²
=
H ð ΩÞ A cos Ωτ − θðΩÞ k
=
H ð ΩÞ p A k
− A /B and 1
+B
sin
±
²i
Ωτ − θðΩÞ
± ² ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi +B Ωτ − θ Ω + ψ 2
2
sin
ð
Þ
(5.83)
we have used Eq. (C.12) of Appendix C.
Excitation with Multiple Harmonic Components
N harmonic components and the forcing is of the form
In general, when there are
f ðtÞ =
X N
A i cos ð ωi t Þ + Bi sin ðωi t Þ±
(5.84a)
Ai cos ðΩi τÞ + Bi sin ðΩi τ Þ±
(5.84b)
½
i=1
or, equivalently,
f ðτ Þ = where the
ωi
X N
i=1
½
are distinct, the corresponding displacement response is given by
xðτ Þ =
= =
1
Xn N
k i= 1
N
²
+ Bi
2
+ Bi
2
sin
±
²io
Ωi τ − θðΩi Þ
Ωi τ − θðΩi Þ + ψ i
²
1
N
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ±
H ðΩi Þ A i
2
+ Bi
2
cos
Ωi τ − θðΩi Þ − φi
²
(5.85)
1
In Eqs. (5.84b) and (5.85), the nondimensional time
Ωi = and from trigonometry
8
sin
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ±
H ðΩi Þ A i
X
k i=
±
1
X
k i= 1
h
H ðΩi Þ A i cos Ωiτ − θðΩi Þ
φi
ωi ωn
ψi =
;
tan
− Ai ; 1
Bi
τ
= ωnt φi =
+ ψi = π
/2.
This superposition principle is an important property of linear systems.
and
tan
− Bi 1
Ai
(5.86)
240
Single Degree-of-Freedom Systems: Periodic Excitations 4
4
2
2 )t( xk
)t(f
0
–2
0
–2
–4
–4 0
5
15
10
0
5
10
t
(a) ci
15
t
(b)
c1
c2
Ω1
Ω2
c3
c 1H(Ω 1) c 2H(Ω 2) c3 H(Ω3 )
c iH(Ω i)
Ω1
Ω Ω3
(c)
Ω2
Ω Ω3
(d)
Time and spectral representation of the response of a system subjected to force input
Figure 5.12.
with three harmonic components with different amplitudes: (a) force input history; (b) displacement output history; (c) spectrum of input and time history of individual components;
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
and (d) spectrum of displacement output and time history of individual components. In (c) and (d),
ci =
ai
2
+ bi
2
:
We see that when the input to a linear single degree-of-freedom system is a collection of harmonically varying force components, each at a different frequency and amplitude, the associated displacement response is the weighted combination of these components comprising the input force. In the corresponding response, the amplitude of the component is modi
fied
by
H(Ωi)
and it is delayed an amount
for these results is provided in Figure 5.12. As seen from this sists
ci =
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi of
ai
2
components
+ bi
2
;
i
=
nents comprising
at
the
nondimensional
frequency
θ( Ωi ).
ith
input
The interpretation
figure, the force input con-
ratios
Ωi
with
amplitudes
1, 2, and 3, respectively. The time histories of the individual compo-
f(τ )
are shown along with their spectral information in Figure 5.12c,
that is, information in the frequency domain. In Figure 5.12b, the displacement response
5.3 Response to Excitation with Harmonic Components
241
of the system is shown. The individual components comprising this response are shown along with the corresponding spectral information in Figure 5.12d. These individual components are scaled and phase-shifted versions of the corresponding individual components
shown in
response
the
Figure
5.12c
components
for the
force input.
x(τ)
comprising
are
Although
periodic,
their
in
the
sum
is
corresponding not
necessarily
periodic. For the sum of the harmonic components also to be a periodic function, the different frequencies
ωi
= pωj q /
, where
p
ωi
should be commensurate with respect to each other; that is,
q
are integers. In other words, the ratio of one frequency to
and
another should be a rational number. Next, we examine how a given periodic excitation can be broken up into harmonic components by using Fourier series, before determining the response of a vibratory system.
Fourier Series As a special case of Eq. (5.84a), consider the periodic function
f ðt + nT Þ = f ðtÞ
0
≤t ≤ T
n = 1; 2; 3:::
which can be expressed as
f ðtÞ = where
u(t)
∞ X n= 0
f ðt − nT Þ½uð t − nT Þ − uð t − ðn + 1Þ T Þ±
is the unit step function,
Then
a
f ðt Þ = where the quantities
0
2
+
=
T
∞ X
π ωo
2 /
and
ωo
0
1 are the
=
of
the
periodic
signal
f(t ).
called the higher harmonics
2 we have the second harmonic, when
i
=
3 we have the
third harmonic, and so on.
fi
We notice that in the de nitions of Consequently, the
ai
and
bi
ai
and
bi
we have integrated over the period
T.
are independent of time and these amplitudes are only a
242
Single Degree-of-Freedom Systems: Periodic Excitations
iωo. In other words, we have transformed f(t) into the frequency domain so bi represent the amplitude contributions of the sine and cosine components harmonic iωo comprising f(t). Hence, a plot of these amplitudes as a function
function of that
ai
of the of
iωo
and
i th
would be a plot of discrete values, since the amplitudes are zero everywhere except
at the corresponding frequencies By using Eq. (5.84), we
xð τÞ =
= =
a 2k
+
0
1
k 1
∞n X
k i=
h
±
H ðΩi Þ ai cos Ωi τ − θð Ωi Þ
²
+ bi
sin
±
1
"
a
0
2
"
k
1
iωo.
find that the displacement response is given by
a
0
2
+ +
∞ X i =−1
²#
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi ±
²#
H ðΩi Þ ai
∞ X i =−1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi ± 2
H ðΩi Þ ai
2
+ bi
2
+ bi
2
Ωi τ − θðΩi Þ + ψ i
sin
cos
²io
Ωi τ − θðΩi Þ
Ωi τ − θðΩi Þ − φi
(5.89)
where
Ωi =
iωo
ψi =
;
ωn
tan
− ai ; 1
φi =
bi
tan
− bi 1
ai
(5.90)
Complex Form of the Fourier Series Another form of the Fourier series when
f ðtÞ = jc
0j
where
j=
p
ffiffiffiffiffiffiffi −
+
f(t) is real is
∞ X
2
p =1
j
cpjcosðpωo t − φpÞ
(5.91)
1,
cp =
=
ωo π
2
ωo π
2
ð
π =ω o
−π
f ðtÞe−jpωo tdt
ωo
=
ð
π =ω o
−π
f ðtÞcosðpωot Þdt − j
ωo
=
ωo π
2
ð
π = ωo
−π
=
f ð tÞsinðpωotÞ dt ωo
= Rp − jXp =
cpj e−jφp
j
p = 0;
1; 2 ; :::
(5.92)
5.3 Response to Excitation with Harmonic Components
243
and
ωo
Rp =
π
2
ωo
Xp =
π
ð
π =ω o
−π
f ðtÞcosðpωot Þdt
ωo
ð =
π =ω o
f ðtÞsinð pωo tÞdt
2
cpj =
j
− qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi π=ωo
Rp + Xp ; 2
φp = tan−1
2
It is straightforward to show that
Xp Rp
p = 0;
1; 2; :::
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ci j = 2 Rp + Xp 2
2j
2
=
ai
2
+ bi
2
Fourier series expansions obtained by making use of Eqs. (5.91) and (5.92) for several common periodic waveforms are given in Table 5.4. The responses to excitation inputs in the form of the periodic waveforms shown in this table can be explored by using Interactive Graphic 5.6.
Design Guideline:
When an excitation force to a linear vibratory system is composed
of more than one harmonic waveform, the displacement response of the system is composed of scaled and phase shifted versions of each of these input waveforms. The
fied
displacement response of each force input waveform is modi
by the amplitude
response of the system at that waveform’s frequency and it is delayed by the system’s phase response at that frequency. If the system’s damping ratio is small, then in order to avoid amplifying these input waveforms, the constituent frequency components of the forcing should not be in the system’s damping-dominated region. If the input waveform is composed of commensurate frequency components, then one must select the
period
such
that
neither
the
fundamental
frequency
nor
any
of
the
higher
harmonics falls in the system’s damping-dominated region.
EXAMPLE 5.7
We
shall
Single degree-of-freedom system subjected to a periodic pulse train
consider
the
response
of
a
single
degree-of-freedom
input that is described by Case 1 of Table 5.4. If we let and
Ωp = pΩo, then from Table 5.4 we obtain xðτ Þ =
αFo
k
(
1
+
∞ X
2
p= 1
τ
= ωnt
,
system
τd
to
= ωntd
,
a
Ωo
periodic
= ωo/ ωn
,
)
gpcosðΩp τ − θp − φpÞ
(a)
244
Single Degree-of-Freedom Systems: Periodic Excitations
Table 5.4. Response of a single degree-of-freedom system to periodic excitation of its mass with period T
Fo k
xðtÞ = where
Hp =
³± 1
−
ð
pffiffiffiffiffiffiffiffi ffi k=m ( X ∞
ωn =
Displacement response where
pωo =ωn Þ
2
and
c
²
2
+
0
+
2j
p=1
ωo
π =T
2
cp jHp cosðpωo t − θ p − φp Þ
´−
=
1 2
ð2
ζ pωo =ωn Þ
2
td ωo = 2π td =T
In the relations below, note that
=
θ p = tan−1
)
2 1
−
ζp ωo= ωn
p ωo = ωn Þ
ð
j
c pj
1 j
cp j =
j
c
2
0
j
2
cp j =
j
c
3 j
cp j
=
4 j
c
0j
td ωo 2
0j
=
=
5 j
cp j =
j
c
0
j
π
·· ·· ··
2
·· ·· ··
2 2
2
0 sinð
ptd ωo = 2Þ
=π
/2
φp = tan−1
ptd ωo =2Þ − p td ωo
t d ωo =
p
φp
ptd ωo= 4Þ ptd ωo =4
sin ð
1 4
φp = tan−1
·· ·· ··
cosð
o
φp
·· ·· ··
2
0
0
ptd ωo =2Þ=ðπ
cosð
2
− p t d ωo 2
2
2
Þ
π
p
td ωo π2
·· · t ω ·· ·
cp j = 2
c
π
d
= j
π
ptd ωo =2Þ ptd ωo =2
sin ð
·· · = t ω ·· ·π =
j
2
= td ωo
j
0j
td ωo
·· ·· ··
x(t) above the c p|, p = 1, 2, … .
and that from the expression for
amplitude contribution of the fundamental frequency and each harmonic is 2|
Waveform
2
d
o
1 2
p
0
td ωo 4
π
= td ωo 4
π
sinð
π2
4
·· pt ω = ·· − p t ω ·· d
t d ωo =
0 @
o
2 2
d
φp = tan−1
2Þ
ptd ωo = 2Þ= ð4 π
2
0
2
o
π
2
p
1 A
2
ptd ωo =4Þ ptd ωo =4
sin ð
sinð
φp = tan−1
−
ptd ωo =2Þ cosð ptd ωo =2Þ sinð
− p t d ωo 2
2
2
Þ
5.3 Response to Excitation with Harmonic Components
= ωotd
α
where
π
/(2 )
= td T /
gp =
1
and
·· · H Ω = qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi · − Ω + ζΩ ·
cp j
j
α
φp = tan−1
2
ωnT
2
this value of
ζ.)
overshoots the
Td
=
pπα
2
ð2
pÞ
ζΩp
2 1
(b)
− Ωp 2
p = 1; 2; :::
0 sinð
pπα Þ
=
π Ωo
2 /
=
Ωo
=
0.0424,
148.2 and
π ωd,
2 /
where
ωd
=
α
ωn/ωo =
that the 23rd and 24th harmonics fall in the vicinity of and the damped period
·· ·· ·
pπα Þ
sinð
pÞ
in Figure 5.13 for
Equation (a) is plotted
nondimensional period is
Þ
ð1
θp = tan−1
9
ð
1
245
ωn.
0.4, and
=
Ωo
1/
ζ
=
0.1. Thus, the
23.57, which indicates
(It is noted that the period
T
is given by Eq. (4.5), are virtually equal for
In Figure 5.13a, we see that the normalized output displacement response amplitude
of
the
input
pulse and then exhibits a decaying oscillation
about the pulse’s normalized height. Since
ωn/ωo
=
23.57, the period of the pulse train is
23.57 times longer than the period of the natural frequency of the system; therefore, the nondimensional period of the decaying oscillation is 148.2/23.57
=
π
2 . We will show in
Section 6.3 that this response is equivalent to the response of a single degree-of-freedom system
subjected
to
a
suddenly
applied
constant
force.
When
damping
increases
substantially, these oscillations are almost eliminated as shown in Figure 5.13e. The results plotted in Figure 5.13b are the amplitude spectrum of the pulse train before it is applied to the mass and the amplitude spectrum of the displacement of the mass in response to this force. We have also plotted the system’s amplitude response function
H( Ω)
for reference. We see that the system’s amplitude response function can
greatly magnify the amplitude of those components of the force that have frequency components in the vicinity of the system’s natural frequency
fication
region. It is this magni
–
the damping-dominated
that produces in the time domain the overshoot and
oscillations at a frequency equal to the system’s damped natural frequency. We now see why
large
damping
eliminates
these
oscillations.
As
shown
in
Figure
5.13f,
the
amplitude components in the neighborhood of the system’s natural frequency are all slightly attenuated; thus, the output response more closely follows the input. The next case we consider is where system
is
three
times
the
nondimensional period is Figure
5.13d that
since
fundamental
ωnT the
Ωo
=
π Ωo
2 /
pulse
=
train’s
=
1/3; that is, the natural frequency of the
frequency 18.9 and third
of
the
ωn/ ωo
harmonic
=
pulse
train.
Ωo
3.0. We see from
1/
=
coincides
with
Thus,
the
the
natural
frequency of the system, the amplitude of the third harmonic undergoes the maximum
fication.
magni
9
Thus, the time domain response, which is shown in Figure 5.13c, is
200 terms were used in the summation.
246
Single Degree-of-Freedom Systems: Periodic Excitations
Figure 5.13.
(a)
(b)
(c)
(d)
(e)
(f)
Comparison of responses to pulse train forcing function in the time domain and
frequency domain for two different values of the system damping ratio and two different fundamental excitation frequencies: (a) and (b) (c) and (d)
Ωo
=
ζ
=
=
0.1,
α
=td
/
T=
0.4, and
= α = td Ωo
0.0424;
T = 0.4, and Ωo = 0.333; and (e) and (f) ζ = 0.7, /T = 0.4, and ωnT = 2π /Ωo = 148.2, τd = (2π /Ωo)(td /T) = 59.3, and the location H(Ω) occurs at those integers that are adjacent to n = 1/Ωo. Note: only the shape 0.1,
α
= td
/
0.0424. For these values,
of the peak of of
ζ
H(Ω) is displayed and is unrelated to the vertical axis values.
5.3 Response to Excitation with Harmonic Components
247
dominated by the component whose frequency is coincident with the system’s natural frequency. Consequently, the displacement response does not bear any resemblance to the input forcing function, except that it has the same period. It is noted from Eq. (b) that as train of force impulses of period is
finite.
α
= td T → /
0, the results approach that of a periodic
T. In this case, td
→
0 in such a manner that
If Eq. (a) is solved numerically, it will be found that when
td/ T
td F o
1=
2, the amplitude response does not have an extremum, as is seen in
H(Ω) occurs at Ω =
Figure 5.2a. In this case, the maximum value of
H
max
=H
Ω
ð
max Þ
=
pffiffiffiffiffiffiffiffiffiffiffiffi
ζ
1
−ζ
ffiffi
p
ζ ≤ 1=
1
2
0. Therefore,
2
2
ffiffi
(5.98)
p
=
ζ > 1=
1
2
Cutoff Frequencies of a Filter From Eq. (5.14), the location of the
cutoff frequencies
shown in Figure 5.17 is deter-
mined by solving
H
ffiffi
max
p
2
=
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi = qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1
ζ
2
1
2ð1
−ζ
2
Þ
ð1
−Ω
2
Þ
+
2
(5.99)
ζ ΩÞ
ð2
2
Thus, the upper and lower cutoff frequency ratios are given, respectively, by
Ωcu
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi = − ζ + ζ −ζ q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi 1
Ωcl =
1
−
2
2
ζ
2
2
2
−
2
1
ζ
2
1
(5.100)
−ζ
2
where
Ωcu = are
the
ωcu
Ωcl =
and
ωn
nondimensional cutoff frequencies
and
ωcu
ωcl ωn
and
ωcl
are
the
respective cutoff
frequencies in rad/s. The lower cutoff frequency exists only for those values of
1
Upon solving for
−
ζ, we find that ζ
1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:5
−
p
0:25
2
2
=
0
0:3827:
ζ
for which
5.4 Frequency-Response Function
253
Filter Bandwidth The bandwidth of the system is
Bw =
BWω
=
ωn
Bw = Ωcu
ωcu − ωcl ωn
BWω is the bandwidth in BWf = BWω/2π, and, therefore, ωn=2π
Bw = Ωcu since
Ωcu = ωcu /ωn
=
=
ζ < 0:3827
ζ ≥ 0:3827
for
BWf
for
(5.101a)
where
Bw =
= Ωcu − Ωcl
fcu − fcl fn
rad/s. To determine the bandwidth in Hz, we have
= Ωcu − Ωcl
ζ < 0:3827
for
(5.101b)
ζ ≥ 0:3827
for
πωcu)/(2πωn)
(2
= fcu fn /
Quality Factor
and
Ωcl = ωcl/ωn =
πωcl )/(2πωn)
(2
fi
Another quantity that is often used to de ne the band pass portion of small is the
H(Ω)
/
.
when
ζ
is
quality factor Q, which is given by Q=
where
= fcl fn
Ωc Bw
(5.102)
Ωc is the center frequency ratio defined as the geometric mean of a band pass filter Ωcl and Ωcu as
with the nondimensional cutoff frequencies
Ωc =
pffiffiffiffiffiffiffiffiffiffiffiffiffi Ω Ω cu
(5.103)
cl
fi
The center frequency is not de ned for either a low pass or a high pass
ζ
b
(5.189)
The governing equation for the system shown in Figure 5.28 is, therefore,
me€z + c_z + k ðtÞhð zÞ = f ðtÞ
(5.190)
where we assume that the applied force on the teeth of the pinion is
f ðtÞ =
To r 1
±
1
+α
ωM tÞ
cos ð
²
(5.191)
5.9 Summary and
To
is the steady-state torque applied by the pinion and
α
289
is a parameter used to
select the magnitude of the time-varying portion of the torque. To convert Eq. (5.190) to
fi
a nondimensional form, we introduce the following de nitions
ω2n =
ko me
ζ=
2
τ = ωn t Then, Eq. (5.190) becomes
p + 2ζ p_ + €
±
1
−ε
fo =
c
ΩM
me ωn To r bko
p=
1
²
ΩM τ Þ hðpÞ = f o
cos ð
±
1
ωn (5.192)
z b
+α
where the overdot indicates the derivative with respect to
hðpÞ = 0 = p − sgnð pÞ
ωM
=
τ
²
ΩM τ Þ
cos ð
(5.193)
and
pj ≤ 1 jpj > 1
j
(5.194)
Numerical Results 32
The steady-state responses of the numerical evaluation Figure 5.31. It is seen that for response. As the magnitude of
ε = 0 the steady-state time ε increases, the steady-state
of Eq. (5.193) are shown in histories exhibit a harmonic response is transformed into
a nonharmonic periodic solution with an increasing peak-to-peak value.
5.9
SUMMARY
In this chapter, responses of single degree-of-freedom systems subjected to harmonic and other periodic excitations were studied. Different sources of forcing such as rotating unbalance and base excitation have been considered. The notions of system resonance and frequency-response functions were also introduced and explained. The topic of vibration isolation was discussed at length. The Maxwell model was introduced. The amplitude and phase response functions of a system with this spring-damper combination model were obtained along with the transmissibility and were compared to the corresponding quanti-
–
ties of a system with the Kelvin Voigt model. The underlying principles of an accelerometer were also explained. For the different damping models, the notion of an equivalent viscous damping was introduced and explained. Free-displacement curves for linear and nonlinear damping models were examined in the context of energy dissipation. The forced harmonic oscillations of a single degree-of-freedom system with a cubic spring and a single degree-of-freedom model of two meshing gear teeth with backlash were examined in detail. Interactive graphics material has been included in this chapter to carry out parametric studies and develop further understanding of system behavior.
32
Using Matlab,
ode45; using Mathematica, NDSolveValue.
290
Single Degree-of-Freedom Systems: Periodic Excitations
Exercises
Section 5.2.1 5.1
A vibratory system with a natural frequency of 10 Hz is suddenly excited by a harmonic excitation at 6 Hz. What should the damping factor of the system be so that the system settles down to within 5% of the steady-state amplitude in 200 ms?
5.2
Consider the two independent single degree-of-freedom systems in Figure E5.2 that
are
each
being
forced
to
vibrate
The excitation on system 1 starts at
t
= to
=
t
harmonically
at
the
same
frequency
ω.
0, and the excitation on system 2 starts at
, that is,
f ðtÞ = F 1
f ðtÞ = F 2
ωtÞu ðtÞ
1 sinð
ω½t − to±Þ uðt − t oÞ
2 sinð
Use Eqs. (5.1) to (5.9) to show that the steady-state responses of the two systems are
x
1
x
2
ss ðτ Þ
ss ðτ Þ
±
=
F H ðΩ; ζ k
=
F H ðΩ=ωr ; γζ k
1
1 Þsin
1
2
Ωτ − θð Ω; ζ
±
Þsin 1
1Þ
²
Ω=ωr ðτ − τ o Þ − θðΩ=ωr ; γζ
²
Þ 1
2
where
Ω = ω=ωn
1;
ωr = ωn2=ωn1; γ = ζ 2=ζ 1;
H ða; bÞ =
and
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1
ð1
−a +
θða; bÞ = tan−1
2
2
Þ
baÞ
ð2
2
ba 1− a 2
2
If both systems are operating in their respective mass-dominated regions, then what is the ratio of the magnitudes of the amplitudes of system 2 to that of system 1 and their relative phase?
f1(t)
f2 (t) x1
m1 k1
c1
Figure E5.2.
x2
m2 k2
c2
Exercises
291
Section 5.2.2 5.3
A 150 kg mass is suspended by a spring-damper combination with a stiffness of
×
30
3
10
N/m
and
a
viscous-damping
constant
of
1500
⋅
N s/m.
Calculate
the
steady-state displacement amplitude and phase if the mass is excited by a harmonic force amplitude of 70 N at 3 Hz. 5.4
fication
The dynamic ampli
fi
or attenuation of a single degree-of-freedom system is
de ned as the ratio of the steady-state magnitude of the displacement response to its static displacement magnitude
Fo.
F o/k,
when the mass is being driven by a harmonic force of
Find the dynamic ampli
fication or attenuation of a single degree-of-
freedom system that is being excited at 100 rad/s and has the following system parameters:
m=
100 kg,
k = 20 kN/m, and c
=
⋅
6000 N s/m.
Section 5.2.4 5.5
The control tab of an airplane elevator is shown schematically in Figure E5.5. The mass moment of inertia
JO
of the control tab about the hinge point
known, but the torsional spring constant
ficult
dif
k
1
pkffiffiffiffiffiffiffiffiffiffiffi =J
O
is
associated with the control linkage is
to evaluate, and hence, the natural frequency
ωn =
t
O
fi
is dif cult to
determine. An experiment is designed to determine this natural frequency of the system. In this experiment, the elevator is rigidly mounted, springs with stiffness and stiffness
k
2
e,
as illustrated in the
until resonance occurs at
ω
=
ωr,
figure.
The excitation frequency
k , k , Jo, and L. 2
Figure E5.5.
Control tab
k1
0 Jo L k2 0 e sin t
ω
is varied
and this value is noted. Assuming that the
damping in the system is negligible, determine an expression for
kt
1
are attached to the control tab, and the tab is harmonically excited
at an amplitude
1
k
ωn
in terms of
ωr ,
292
Single Degree-of-Freedom Systems: Periodic Excitations Section 5.2.5 5.6
An air compressor with a total mass of 100 kg is operated at a constant speed of 2000 rpm. The unbalanced mass is 4 kg and the eccentricity is 0.12 m. The properties of the mounting are such that the damping factor
ζ
=
0.15. Determine the
following: (a) the spring stiffness that the mounting should have so that only 20% of the unbalance force is transmitted to the foundation and (b) the amplitude of the transmitted force. 5.7
A motor of mass
flects
m
the beam de
is mounted at the end of a cantilever beam and it is found that
10 mm. When the motor is running at 1800 rpm, an unbalanced
force of 100 N is measured. If the beam damping is negligible and its mass can be neglected, then what speed should the motor operate at so that the amplitude of the dynamic response is less than
ao meters.
Section 5.2.6 5.8
y(t) is the base displacement and x( t) is the dism. Consider the base velocity y_ as the input, the
In the system shown in Figure E5.8, placement response of the mass acceleration response
€xðt Þ
as the output, and determine the frequency-response
function of this system.
Figure E5.8.
m
k
x(t)
2c
c
Base
5.9
y(t)
The damped single degree-of-freedom mass-spring system shown in Figure E5.9 has a mass
m = 20 kg and a spring stiffness coefficient k = 2400 N/m.
(a) Determine the damping coef
ficient
of the system when the mass exhibits a
response with an amplitude of 0.02 m and the support is harmonically excited at the natural frequency of the system with an amplitude
Yo
=
0.007 m.
(b) Determine the amplitude of the dynamic force transmitted to the support.
Exercises
293
Figure E5.9.
m
x(t)
c
k
y(t)
Section 5.3 5.10
Torsional oscillations of a vibratory system is governed by the following equation
€ + ct φ_ + kt φ = M Joφ
1
Jo
where
M
2
=
=
⋅
⋅
20 N m,
ω1
=
=
kt
2
20 kg m ,
ω1tÞ + M2
cos ð
⋅
20 N m/rad,
1.0 rad/s, and
ω2
=
ct
=
cos ð
ω2 tÞ
⋅
20 N m/(rad/s),
M
1
=
⋅
10 N m,
2.0 rad/s. Determine the steady-state
response of the system. 5.11
Determine an expression for the output of an accelerometer with the damping factor
ζ
and natural frequency
ωn,
when it is mounted on a system executing peri-
odic displacement motions of the form
y =A 5.12
A
single
degree-of-freedom
1
sin
ω1t + A2
system
is
sin
ω2 t
driven
by
the
periodic
excitation as shown in Case 5 of Table 5.4. If the period tude the
k=
5.13
fo
=
first
10 N, then
find
T
=
triangular
wave
2 s and the ampli-
the steady-state response of the system by considering
three harmonics of the forcing. Assume that the system parameters are
10 kN/m,
Consider
the
c=
⋅
10 N s/m and
base
excitation
m=
of
a
1 kg. single
degree-of-freedom
system
shown
in
Figure 3.6 and whose motion is described by Eq. (5.45). If the displacement of the base is the periodic waveform described by Case 3 of Table 5.4, then obtain an expression for the displacement of the system’s mass. 5.14
Consider a sine wave forcing of magnitude tion of
N periods tp,
where
tp
=
π ωb.
2 /
Fo
and frequency
ωb
that has a dura-
This sine wave is repeated every period
T,
294
Single Degree-of-Freedom Systems: Periodic Excitations where
T
= Mtp =
π ωo, M
2 /
≥N
N and M
, and
are integers. If this periodic wave-
form is applied to the mass of a single degree-of-freedom system, then
f ðt Þ = Fo where
tb
= Ntp
ωbtÞ ½uð tÞ − uðt − tbÞ±
sinð
f( t)
and
0
≤t ≤ T
;
0
< tb < T
(a)
=ft+T (
). See Figure E5.14.
(a) Expand Eq. (a) in a Fourier series and show that
a
0
=
ai = bi = bi =
0 1
1
πM
−
−
π iN =M Þ
1
−
i MÞ
2
ð =
πiN =M Þ
1 sinð2
πM
1
N M
i ≠M
cosð2
−
i MÞ
ð =
2
i ≠M
i=M
These results were obtained by noting that
ωbT
=
(b)
ωbtb
πM.
=
πN, ωo/ ωb
2
=
M,
and
(b)
and
1/
2
(b) Determine
the
displacement
Eqs. (5.91) and (5.92). (c) Assume that Generate
N
the
=
4,
M
=
equivalent
response
10,
Ωo
figures
=
of
the
mass
by
using
0.0424 or 0.3333, and
for
this
excitation
as
ζ
Eq.
=
those
0.1 or 0.7. shown
in
Figure 5.13. Explain these results. Figure E5.14.
Section 5.4.3 5.15
A micromechanical resonator is to be designed to have a
Q factor of 1000 and a natural
frequency of 2 kHz. Determine the system damping factor and the system bandwidth.
Exercises
295
Section 5.4.4 5.16
Consider the machine of 25 kg mass that is mounted on springs and dampers as shown in Figure E5.16. The equivalent stiffness of the spring combination is
⋅
9 kN/m, and the equivalent damping of the damper combination is 150 N s/m. An excitation force the
figure.
F (t)
is directly applied to the mass of the system, as shown in
Consider the displacement
x(t )
as the output, the forcing
F(t)
as the
input, and determine the frequency response of this system.
F(t)
Figure E5.16.
x(t) m
k 2
c 2
k 2
c 2
Section 5.5 5.17
An
accelerometer
response of
±
is
being
1 % up to
fa
=
designed
if the phase angle is to be less than 2 5.18
to
have
a
uniform
amplitude
frequency
8 kHz. What is the maximum damping ratio allowed
°
over this frequency range?
Determine the phase shift and amplitude error in the output of an accelerometer that has a natural frequency of 25 kHz and a damping ratio of 0.1, when it is measuring vibrations at 1 kHz.
Section 5.6 5.19
A rotating machine runs intermittently. When it is operating, it is rotating at 4200 rpm. The mass of the machine is 100 kg, and it is supported by an elastic structure with an equivalent spring constant of 160 kN/m and a viscous damper of damping coef
ficient of
⋅
2400 N s/m that is in parallel with the spring. The engi-
neer would like to keep the maximum transmission ratio less than 2.3 during the period that it takes the machine to reach the operating speed. If a spring is inserted in series with the damper, then what is the best value of the stiffness of this spring in order to have 5.20
TR < 0.08 when the machine is at its operating speed?
A compressor weighing 1000 kg operates at 1500 rpm. The compressor was originally attached to the
floor
of a building, but it produced undesirable vibrations
296
Single Degree-of-Freedom Systems: Periodic Excitations to the building. To reduce these unstable vibrations, it is proposed that a concrete block be poured that is separated from the building and that the compressor then be mounted to this block. The location of the compressor will permit the block to be 1.8 m by 2.2 m. The soil on which the concrete block will rest has a compres-
fi
sion coef cient
kc =
20
×
6
10
3
N/m . If the density of the concrete is 2.3
×
10
3
3
kg/m ,
then determine the height of the concrete block so that there is an 80% reduction in the force transmitted to the soil.
Section 5.7 5.21
Show that the work done per cycle by a harmonic force acting directly on the mass of a linear spring-mass-damper system is equal to the energy dissipated by the system per forcing cycle.
5.22
A spring-mass system with
fi
m=
20 kg and
a surface with coef cient of friction
μ
=
k=
8000 N/m vibrates horizontally on
0.2. When excited harmonically at 5 Hz,
the steady-state displacement of the mass is 10 cm. Determine the equivalent viscous damping. 5.23
The area of the hysteresis loop of a cyclically loaded system, which is the energy
⋅
dissipated per forcing cycle, is measured to be 10 N m, and the measured maximum response
Xo
fl
of the de ection is 2 cm. Calculate the equivalent viscous
ficient of this system if the driving force has a frequency of 30 Hz.
damping coef
Section 5.8 5.24
Consider the following nonlinear single degree-of-freedom system subjected to a harmonic excitation, where
k
=
2 kN/m, and
m and
v(0)
=
α
=
fo
=
300 N,
ω
= ωn
/3,
m
3
=
100 kg,
c
=
10 kN/m . Assume that the initial conditions are
170 N/m/s,
x(0)
=
0.01
0.1 m/s.
m€x + cx_ + kx − αx
3
= fo
cos
ωt
Compute the time response of the system and associated amplitude spectrum of steady-state motions and compare it with the corresponding quantities of the linear system, that is, when 5.25
α
=
0.
Consider the following nonlinear single degree-of-freedom system subjected to a harmonic excitation, with the same values of parameters as in Exercise 5.24, except that now the excitation frequency is at
m€x + cx_ + kx + αx
3
= fo
ω = ωn. cos
ωt
Compute the response of the system and compare it with that of the corresponding linear system, that is, when
α
=
0. In addition, for this harmonic excitation,
compare the differences between the responses of the system of softening stiffness discussed in Exercise 5.24 and the system of this exercise with hardening stiffness.
Single Degree-of-Freedom Systems:
6
Subjected to Transient Excitations
page
6.1 Introduction 6.2 Response to Impulse Excitation
300
6.3 Response to Step Input Excitation
310
6.4 Response to Rectangular Pulse Excitation
316
6.5 Response to Other Excitation Waveforms
322
6.5.1 Signi
ficance of the Spectral Content of the Applied Force: An Example
6.6 Impact Testing
6.1
297
334 338
6.7 Summary
340
Exercises
341
INTRODUCTION
In Chapter 4, free responses were discussed, and in Chapter 5, responses to harmonic and other periodic excitations were discussed. As illustrated in the last two chapters, a
“sudden”
change in the state of a system brought about by an initial condition or by a
fi
change in the pro le of the forcing function results in transients in the response of the system.
Here,
the
initial
conditions
are
assumed
to
be
zero,
and
the
responses
to
various types of excitations such as impulse excitations, step inputs, ramp inputs, and pulse excitations are considered at length. All of these excitations are characterized by
files
sudden changes in their respective pro
of amplitude with time. When transformed
to the frequency domain, the responses to such transient excitations can also provide a basis
for
determining
the
characteristics
of
a
system.
The
resulting
displacement
response and several design criteria are established based on this information. In the systems considered in this chapter, the inertia element or the base of the system is subjected to a transient forcing.
Solution for Response to Transient Excitation There are two approaches that can be used to obtain the displacement response of the mass. The
first
approach is to use the convolution integral directly and the second
approach is the use the Laplace transform. We shall use both approaches in this chapter.
298
Single Degree-of-Freedom Systems: Transient Excitations
Convolution Integral Based Approach When the initial displacement Xo =
Vo
0 and the initial velocity
equation for a linear underdamped system (0
0 that
ωd t þ ψ Þ
sinð
(6.16)
pffiffiffiffiffiffiffiffiffiffiffi
1
If the maximum force is transmitted to the
dFb/dt
ζ
2
(6.15)
dt
1
−
−ζ
2
ζ2
2
fixed
boundary at the time
tm,
at this time
0. Thus, Eq. (6.16) leads to
ωd tm
¼
φ− ψ
(6.17)
6.2 Response to Impulse Excitation
303
1.3 1.2 )n w of (/ xam,bF
1.1 1.0 0.9 0.8
0.0
0.1
0.2
0.3
0.4
0.5
0.6
z
Figure 6.3. Maximum magnitude of impulse force transmitted to a
where
φ
fixed boundary. 3
is given by Eq. (6.14). Thus, the maximum normalized force
transmitted to the
fixed boundary is given by Fb foωn
;max
¼
e− φ−ψ ð
Þ =tan
φ
(6.18)
This result is plotted in Figure 6.3, where it is seen that
ζ ≈ 0:25; ζ ≈ 0:25;
where
Fb
fo ωnÞ ¼
;max =ð
Fb
;max =ð
f o ωn Þ
is a minimum at
0:81. It is noted that as the damping increases beyond
the magnitude of the transmitted force increases. From Eqs. (6.16) and (6.18)
and Figure 6.3, one can deduce the following design guideline.
Design Guideline:
To
obtain
the
maximum
decrease
in
the
amount
of
force
transmitted to the base of a linear single degree-of-freedom system due to an impulse type loading, one can choose the damping ratio to be between 0.2 and 0.35. For this range of damping values, the maximum attenuation of the force transmitted to the
fixed boundary will be around 18%. For a given damping ratio and magnitude of the impulse force, decreasing the natural frequency decreases the magnitude of the force transmitted to the
The
frequency
Transform of
f (t)
fixed boundary.
domain
information
is
obtained
from
Eq.
(6.5).
The
Laplace
given by Eq. (6.7) is obtained from Laplace transform pair 5 of
Table B.1 of Appendix B as
F ðsÞ ¼ fo 3
Strictly speaking, the force transmitted has an extremum at mum here, we use numerical means.
t
(6.19)
= tm
. In order to determine that this force is a maxi-
304
Single Degree-of-Freedom Systems: Transient Excitations Then, in the frequency domain,
F ð j ωÞ ¼ fo Thus, an impulse of magnitude
ω
≤∞
. Using Eq. (6.5), we
fo
has a constant frequency spectrum over the range 0
find that
X ð j ωÞ ¼
fo f H ð jωÞ = o H ðω=ωnÞe−j θ ω=ωn k k ð
Equation (6.21) is the frequency domain counterpart of
fo
=
(6.20)
x(t)
Þ
≤
(6.21)
given by Eq. (6.9a). When
1, we see from Eq. (5.113) that Eqs. (6.9a) and (6.21) are related by the Fourier
transform.
Impulse Testing From a practical standpoint, an impulse function is hard to realize, since an ideal impulse has an “extremely large” magnitude that lasts an in
finitesimally
short time.
However, a device called an impact hammer can be used in an experiment to apply a pulse of
finite
time duration to the mass of a system. An impact hammer usually has a
built-in transducer to measure the force amplitude-time pro
file
created by the hammer.
The corresponding Fourier transform of the hammer’s pulse has a fairly constant magnitude over a
finite
frequency bandwidth. As discussed in Section 6.4, this bandwidth
becomes larger as the pulse duration is reduced. The notion of an impulse response leads to an experimental procedure to determine the frequency-response function. This procedure is an alternative to that discussed in Chapter 5 where a harmonic excitation was used to construct this function. Thus, one would use an impact hammer or another source to apply a force (or a moment) of very short time duration to the inertia element of a system, measure and digitize the forcing and displacement response signals, use the discrete Fourier transform on these signals to convert the time domain information to the frequency domain, and then determine the
G( jω) based G ( j ω) = H( jω)/k .
frequency-response function as Eq. (6.5) since
on Eq. (5.112c), which appears in this section
Response of a Linear System to Arbitrary Inputs Based on Impulse Response Examining Eq. (6.2), which is in the Laplace domain, one can state that for a linear vibratory system the displacement response is the product of the system’s transfer function and the Laplace transform of the system input. Similarly, from Eq. (6.5), which is in the frequency domain, it can be stated that the displacement response is the product of the system ’s frequency-response function and the Fourier transform of the system input. This input-output relationship, which is characteristic of all linear vibratory systems, is schematically illustrated in Figure 6.4.
6.2 Response to Impulse Excitation
305
Laplace domain F(s) Time domain f(t)
1 h(t) = m
d
e–
nt
sin(
d t)
x(t) =
t
∫0 h(
)f(t – )d
X (s) = F(s)G(s) G(s)
G (s): Transfer function Frequency domain
h(t): Impulse response F(jw)
G( j )
X( j ) = F( j )G(j )
G( j ): Frequency-response function Figure 6.4. System input-output relationships in time and transformed domains.
From the transform pair 4 in Table B.1 of Appendix B, we arrive at the time domain counterpart of Eq. (6.2), that is,
ð t
xðtÞ ¼ hðηÞf ðt − η Þd η
(6.22)
0
where
f (t )
is the forcing applied to the inertia elements and the function
h(t)
is the
impulse response of the system by Eq. (6.10). As noted in Appendix C for Eq. (C.11), Eq. (6.22) is the convolution integral. Thus, the applied forcing is convolved with the system impulse response function over the time interval of interest to determine the system response as a function of time. Interactive Graphic 6.1 has been created so that the reader can explore the functions comprising the convolution integral and see how the impulse response function and the applied force interact.
INTERACTIVE GRAPHIC 6.1: RESPONSE TO A RECTANGULAR PULSE: CONVOLUTION INTEGRAL BASED APPROACH This interactive graphic is used to illustrate the role that the various components comprising the convolution integral given by Eq. (6.1) play in arriving at x(t) when the forcing is a rectangular pulse of duration to. In this interactive graphic, the following should be noted. The time-shifted impulse response h( τ − ξ) is h(τ ) flipped about the y-axis at τ = 0. • The portion of h( τ − ξ ) that contributes to the response of y(τ ) at time τ is that portion of h(τ − ξ) that resides within the shaded region; the time indicated with a red dot is the time that the time-inverted response h( τ − ξ ) has advanced and may or may not reside within the shaded region.
•
306
Single Degree-of-Freedom Systems: Transient Excitations
•
When all meaningful oscillations of h(τ − ξ) reside entirely within f(τ), that is, τ < τ o, then the value of y(τ ) does not change until τ > τo; for example, see the case ζ = 0.25 and τ o = 36.
EXAMPLE 6.1
4
Response of a linear vibratory system to multiple impacts
An impact hammer is used to manually apply an impulse to a model of a single degreeof-freedom
system
as
illustrated
preferred; however, it is dif
ficult
in
Figure
6.5.
In
experiments,
a
single
impact
is
to realize only a single impact and multiple impacts
may occur. Assume that for a system with a mass of 2 kg, stiffness of 8 N/m, and
ficient of 2 N ⋅ s/m, a double impact of the form
damping coef
5
f ðtÞ = δðtÞ + 0:5δð t − 1Þ occurs. We shall determine the system response and compare it to that obtained when the second impact at
t
=
1 s is absent.
For the given parameter values, the natural frequency
ζ
to
Hence, for small values of
ζ, 6
Equation (c) is solved numerically by using a standard optimization procedure determine the earliest time in Figure 6.7 for
ζ
=
to
that gives the smallest value of
0.01 and
ζ
=
0.15. It is seen from the
xrms.
figure
One impulse
Two impulses
One impulse
0.8
to
The results are shown that for a very lightly
damped system, the application of an impulse at the nondimensional time
1.0
we
.
ωnto
=
3.142
Two impulses
0.6 0.5 )n m/ oF(/)t( x
)n m/ oF(/)t( x
0.0
0.4 0.2 0.0
–0.2
–0.5
–0.4 –1.0
0
5
10
15
(a)
20 t
Figure 6.7.
25
30
–0.6
35 (b)
0
5
10
15
20
25
30
35
t
Response of the inertia element subjected to two impulses, where the second impulse
has been applied so that the subsequent response is minimized over the duration of interest: (a)
6
ζ
= 0.01 and (b)
Using Matlab,
ζ=
0.15.
fminsearch from the Optimization Toolbox; using Mathematica, FindMinimum.
6.2 Response to Impulse Excitation
309
virtually eliminates the magnitude of the oscillations immediately after the application of the second impulse. However, the application of the second impact has less effect as
pffiffiffiffiffiffiffiffiffiffiffiffi
the damping is increased. It is noted that the nondimensional half period of damped oscillations is and for
ζ
=
ωnT1=2
¼
π=
1
−ζ
2
; thus, for
ωnT1=2 ¼
0.15, we have that
=
ζ
0.01, we have that
ωnT1=2
¼ 3:1418
3:178. These are the same values as determined
from the minimization of Eq. (c). Therefore, for lightly damped systems, it is possible to greatly reduce the magnitude of oscillations after the application of another impulse if the second impulse is applied at a time approximately equal to the half period of its damped oscillation.
EXAMPLE 6.3
Stress level under impulse loading
Consider the spring-mass-damper system shown in Example 6.1, and assume that the top of the spring is welded to the mass. The welding is over an area the allowed maximum stress for the weld material is
⋅
σw,max
=
Aw
=
2
4 mm , and 2
150 MN/m . For an
impulse of magnitude 100 N s, we shall determine whether the stress level in the weld material will be below the maximum allowed stress. It is assumed that
k=
ωn = 2
800 N/m, and therefore,
rad/s.
The impulse is given by Eq. (6.7) with
fo
=
ζ=
0.25,
m = 200 kg,
⋅
100 N s, and the corresponding displacement
response of the system is determined from Eq. (6.9a). Then, the force acting on the weld material is determined, and from the maximum value of this force the maximum stress experienced
by
the
weld
material
during
the
motions
compared to the maximum allowed stress level,
σw,max.
is
determined.
This
value
is
The force acting on the weld
material is given by
fweld ðtÞ ¼ kxðtÞ
(a)
The maximum force on the weld is given by Eq. (a) when given by Eq. (6.12). From Eq. (6.14), we
φ¼
pffiffiffiffiffiffiffiffiffiffiffiffi¸
− tan
1
1
−ζ
2
¼
ζ
find that − tan
x( t)
=x
max,
where
x
max
is
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi¸ 1
1
−
2
:25
:25
¼ 1:318 rad
(b)
and, therefore,
tan
Then, from Eq. (6.12),
x
max
x
φ¼
tan 1:318 ¼ 3:873
(c)
is
max
¼
100 200
×
2
e−
1:318=3:873
¼ 0:178 m
(d)
By using Eqs. (a) and (d), the maximum force on the weld is
f weld
;max
¼
kx
max
¼ 800
×
0:178 ¼ 142:4 N
(e)
310
Single Degree-of-Freedom Systems: Transient Excitations
σw,max
Hence, the maximum stress
σ w;max
¼
f weld Aw
experienced by the weld material is given by
;max
142:4 N
¼ 4
×
−
6
10
2
m
2
¼ 35:6 MN =m
(f)
This value is well below the maximum allowed stress level in the weld material of 2
150 MN/m . To exceed the given value of maximum allowed stress, the impulse magnitude will have to be at least 4.2 times stronger than the present impulse magnitude.
6.3
RESPONSE TO STEP INPUT EXCITATION
The step input to a vibratory system shown in Figure 6.1 is expressed as
f ðtÞ ¼ Fo uðtÞ where
u(t)
(6.23)
is the unit step function. After substituting Eq. (6.23) into Eq. (6.1a) to deter-
mine the displacement response, we obtain
ð t
F oe−ζωnη ζωn η xðtÞ ¼ e sinð ωd ½t − η ±Þd η m ωd
2 F 4 e− − pffiffiffiffiffiffiffiffiffiffiffi k
3 φ 5u t
0
¼
ζωnt
o
1
1
"
−ζ
ωd t þ
sinð
2
which is rewritten in terms of the nondimensional time
xðτÞ ¼ φ
where the phase angle
Fo k
1
e−ζτ
− pffiffiffiffiffiffiffiffiffiffiffiffi −ζ
=
² qffiffiffiffiffiffiffiffiffiffiffiffi τ
sin
2
1
τ
Þ
1
−ζ
2
(6.24)
ð Þ
ωnt as
þ
φ
³#
uð τÞ
(6.25)
is given by Eq. (6.14) and it has been assumed that the system
is underdamped. The maximum value
x
time derivative of
dxðτ Þ ¼ dτ ¼
is determined from the response given by Eq. (6.25) as fol-
max
lows. The extremum values
x
max
x(τ ) given by Eq. (6.25) is zero, that is,
2 ² qffiffiffiffiffiffiffiffiffiffiffiffi ζ 4pffiffiffiffiffiffiffiffiffiffiffi τ −ζ k −ζ ² qffiffiffiffiffiffiffiffiffiffiffiffi³ F e− pffiffiffiffiffiffiffiffiffiffiffiffi τ − ζ
Fo e−ζτ
1
o
k
are obtained by determining the times
1
2
sin
1
2
³
þ
φ
−
² qffiffiffiffiffiffiffiffiffiffiffiffi
cos
τ
1
−ζ
2
þ
τm
at which the
³3 φ 5
ζτ
−ζ
sin
1
2
(6.26)
2
where we have made use of Eq. (C.12) of Appendix C. Then, setting
find that for τ > 0, the first extremum occurs at τm
¼
π pffiffiffiffiffiffiffiffiffiffiffi ffi 1
−ζ
2
dx( τ)/ dτ
=
0, we
(6.27)
6.3 Response to Step Input Excitation For the value of
τm
x
7
given by Eq. (6.27), Eq. (6.25) evaluates to
max
311
¼
¼
2 ² qffiffiffiffiffiffiffiffiffiffiffiffi e 4 − pffiffiffiffiffiffiffiffiffiffiffi τ xτ −ζ −ζ 2 3 ffiffiffiffiffiffiffi − − F 4 e − pffiffiffiffiffiffiffiffiffiffiffiffi π φ 5 k −ζ i F h − ð
o
mÞ
ζτm
Fo k
¼
2
1
ζπ=
p
1
o
k
1 þ
e
sin
m
1
2
þ
ζ2
1
sinð
2
1
¼
2
1
³3 φ 5
þ
Þ
π = tan φ
(6.28)
where we have made use of Eq. (4.14), that is,
sin
φ
qffiffiffiffiffiffiffiffiffiffiffi
¼
1
−ζ
2
The response given by Eq. (6.25) consists of a constant term decaying sinusoid for 0
ζ.
to
and
ζ
=
0; that is, the system is undamped.
In this case, Eq. (6.41) becomes
xðtÞ ¼ sin ωn t þ sinðωn ð t − to Þ Þ F o=k ωnt − ωn to=2Þcosðωnt o=2Þ
¼ 2 sin ð
Thus, we see that when
ωnto/2
= pπ
/2,
p
=
1, 3,
t > to
…, x(t) = 0. Thus, there is a combination
of parameters for which the displacement response can be minimized after the input force is terminated. This result is similar to that presented in Example 6.2 and is sometimes referred to as pulse shaping. As a
final note, it is pointed out that the displacement response of a single period of a
periodic waveform of rectangular pulses is identical to that of the single rectangular pulse, which can be seen by comparing the results shown in Figure 5.13 to those in Figure 6.15 for the same value of
ζ.
This is not a coincidence. An examination of the
analytic expressions for the discrete amplitudes of a periodic rectangular waveform with those for a rectangular pulse will show that their forms are similar. These analytical expressions are shown in Table 6.1 for the rectangular shape and several other pulse shapes. In addition, Interactive Graphic 6.2 has been created to compare the frequency content of periodic waveforms to a corresponding transient waveform composed of the waveform within one period of the periodic waveform.
INTERACTIVE GRAPHIC 6.2: SPECTRAL CONTENT OF APERIODIC AND PERIODIC WAVEFORMS The normalized frequency contents of different periodic waveforms are compared with the corresponding frequency content of transient waveforms. Each transient waveform is composed of a waveform that lasts for one period of the periodic waveform. In this interactive graphic, the following should be noted. The spectral magnitudes of the waveforms after the first minimum are strongest for the rectangular pair and weakest for the half sine, full sine, and triangular waveforms. • As the number of harmonics for periodic waveforms increases within the region defined by its first minimum, the spacing between the adjacent harmonics cp becomes very small such that the curve connecting their magnitudes approaches the curve for the aperiodic waveform.
•
6.4 Response to Rectangular Pulse Excitation
•
321
The normalized magnitudes of the aperiodic waveforms and the envelope of the magnitudes of the periodic waveforms are the same; however, their respective dimensional magnitudes are very different.
Table 6.1.
Comparison of the normalized frequency content of transient and
periodic waveforms (results taken from Tables 5.4 and 6.2) Waveform Case
πω
(T = 2
/
o
1
)
j
td j
td
2
j
td /2 j
3
j
td j
j
4
td
j
td /2 j
j
5
td /2 td
ω
Transient: |F(
j
j
F ðωÞj ¼ td
ºº ºº ºº
)|
td ω=2Þ td ω=2
sinð
F ðωÞj ¼ 1 td
ω
ω
p
0
td =T jc j td =T 0
2
ω¼
ºº ºº ºº
ω
≠
ºº ºº c t T ººc ºº p
0
d=
p
0
F ðωÞj cos ðt d ω=2Þ ¼ 2π td π − ω td 2 2
2
td =T
ºº ºº ºº
ºº ºº c p
td =T
ω ≠ π= td
F ðωÞj 2 ¼ td π
ω
¼ 0
F ðωÞj 1 ¼ td 2
ω
¼
ºº ºº ºº
≠
ºº ºº c
¼ 0
F ðωÞj sin ðωtd =4 Þ ¼ td ωtd =4 F ðωÞj ¼ 0 td
j
ω
F ðωÞj 1 ¼ td 2
ω
F ðωÞj 1 ¼ td 2 F ðωÞj 1 ¼ td 2
0 @
2
ω
p
td =T
ºº ºº ºº
ºº ºº c
≠
2
p
td =T
π td
2 =
j
¼ 0
0j
ººc ºº p
π td
¼ 2 =
td ω=4Þ
sinð
td ω=4 ¼ 0
c
td =T
1 A ω≠
td =T
2
ω
0j
ººc ºº
F ðωÞj sinð t d ω=2 Þ ¼ 4π td 4π − ω t d F ðωÞj ¼ 0 td
c
td =T
π= td
2
=
Periodic: |cp| (p
ºº ºº ºº
0
ºº ºº c p
td =T jc j td =T 0
¼
ºº ºº ºº
0, 1, 2,
ptd ωo =2Þ ptd ωo =2
sin ð
ºº ºº ºº
…
)
p≠0
¼ 1
pt d ωo =4Þ p≠0 ptd ωo =4 2
¼
sin ð
¼ 0
ºº º π ºº ºπ
¼ 2
¼
¼
t d ωo
≠π
1 2
td ωo
¼
2
p
p≠0
= ;
ptd ωo =2Þ 4π − p t ω d o 2
1
1 2 1 2
2
2
2
π =p;
ºº ºº ºº
p≠0
p¼0
¼ 0
2
≠
2
p≠0
π =p;
sinð
td ωo
¼
2 2
π
¼ 4
¼
2
2
ºº º π ºº º
¼
ºº ºº ºº
ptd ωo =2 Þ − p td ωo
cosð
td ωo
0 @
1 A
ptd ωo =4Þ ptd ωo =4
sinð
p≠0
π p
¼ 2 = ;
2
p≠0
322
Single Degree-of-Freedom Systems: Transient Excitations
6.5
RESPONSE TO OTHER EXCITATION WAVEFORMS
The step forcing and the rectangular forcing are just two of the waveform shapes that a transient forcing may take. They were examined in detail primarily to introduce the major features of the displacement response and how this response relates to the frequency content in the forcing. In this section, we introduce the important characteristics of several other waveforms. These waveforms are given in Table 6.2, wherein we provide expressions for the displacement response, the forcing as a function of time, the frequency content of the waveform, and any special characteristics that the solutions may possess. Representative time responses of Cases 4 to 9 have been plotted in Figure 6.16
τd
for
= ωntd =
=
ζ
20 and
0.15. In addition to these
figures,
all the time and frequency
domain information given in Table 6.2 can be displayed for a wide range of the values
Table 6.2.
Displacement response x(t) of a single-degree-of-freedom system to various
transient forces f (t) applied to the mass and the spectral content of these forces
Notation If s is the Laplace transform parameter, X (s) the Laplace transform of x(t), F( s) the Laplace transform of f (t), and
Hð sÞ ¼ ½ ðs − s
−s
s
1 Þð
−
1
2 Þ±
s
where
= − ζωn ± jωn
1;2
pffiffiffiffiffiffiffiffiffiffiffiffi −ζ
1
2
then
X ð sÞ ¼
1
m
Hð sÞF ðsÞ
and
j
X ð j ωÞj ¼
1
k
j
H ð j ωÞjj F ð j ωÞj
where
H ð j ωÞj ¼
j
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ´ µ 1
1
−ω ð
=
2
ωn Þ
2
ζω =ωn Þ
2
þ ð2
Displacement response functions (p h ðt Þ ¼
−
2
ζ þ ωn t −
−ζωn t
peffiffiffiffiffiffiffiffiffiffiffiffi
h rect ðtÞ ¼ 1 − g ðt; pÞ ¼ 1
g ðt; pÞ ¼ 2
1
−ζ
2
= 0 or 2 in the third and fourth cases below):
² qffiffiffiffiffiffiffiffiffiffiffiffi ° qffiffiffiffiffiffiffiffiffiffiffiffi. − peffiffiffiffiffiffiffiffiffiffiffiffi ω t − ζ − − ζ − ζ ± −ζ ² qffiffiffiffiffiffiffiffiffiffiffiffi °qffiffiffiffiffiffiffiffiffiffiffiffi. !³ ζωn t
2
1
sin
sin
ωn t
1
−ζ
2
2
1
n
þ tan
ω t qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ± ¹
−
1
n d
ω2n t2d
−
p
2
π2
2
p þ2
þ 2
π 2 ω2n t2d ζ
1
´
2 2
sin
1
tan
2
p=2
−ζ
2
1
×
2 2
n d
e−ζωn td sin ωn t
Note 1: When
ζ
is small and
p
2
1
2
π t =td
2
1
1
−
ζ2
2
¹!³
ζ
þ tan
−
1
´
p =2þ1
2
±
πζωn t d =
p
2
π
2
2
− ωntd 2 2
¹µ µ
p =2
πω t pffiffiffiffiffiffiffiffiffiffiffiffiqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ± ¹ −ζ ω t − π π ω t ζ ² qffiffiffiffiffiffiffiffiffiffiffiffi ² qffiffiffiffiffiffiffiffiffiffiffiffi n d2
2
2
−ζ
2
2
p þ2
þ 2
þ tan
ωntd =
2
p/2
−
π,
1
2
2
2
n d
ζ
2
1
2
−ζ
2
»± ζ
ω2n t2d =
2
the denominators of
equal zero at these values. Note 2: When plotted, the following notation is used:
τ
= ω nt
,
g
1
Ω
2
−
¹ω t
2 2
1
and
n d
g
2
= ω ωn /
,
p
þ 2
π2
¼³³
become very small and when
τ d = ωnt d, τ1
=
ωnt1,
and
ζ=
0 they
τ 2 = ωnt 2.
6.5 Response to Other Excitation Waveforms
Table 6.2. (cont.) Case 1 Step
f ðt Þ ¼ Fo u ðt Þ
F ðsÞ ¼
j
F ðωÞj ¼ xðtÞ ¼
Fo s
⋅
s
N
⋅
N
ºº ºº ººF ºº ºº ω ºº o
s
Fo hrect ðtÞ uð tÞ k
´
m
µx ffiffiffiffiffiffiffi −
The maximum (extremum) value p
x
max ð
t
max
Fo Þ ¼ k
−ζπ= 1 þe
1
ζ2
is
max
which occurs at
t
¼
max
Case 2 Step + ramp
F ðsÞ ¼
j
F ðωÞj ¼
1
o
Fo t
2
Fo xðtÞ ¼ k
pπffiffiffiffiffiffiffiffiffiffiffiffi 1
þ
t
2
p
−t
ω2
3Þ
8 < :h
− td
ffiffi 2 4 2
e−st
2
1
9
− e− st = d
s
2
;
−
1
− e−st
−
t
cos ð½
2
−t −t d
−t
N
⋅
ωÞ þ
3±
1
d
o d
s
N
⋅
2
2Þ þ
1 2
ωf ðt2
3Þ
2
h ðt − t Þu ðt − t
2Þ
2
2
−t
t
ð
ºº ºº ºº
N
⋅
þ
1
td − t
2
−t
ω
3Þ
−
0 and
m
t > td
2 Fo cos ðωn ½t − td =2 ±Þsin ðωn td =2 Þ kωn td
Therefore,
2
t ωÞ þ 2 sinðt ωÞg
2 sin ð
2
9 = ;
h ðt − td Þu ðt − td Þ
s
Fo fh rect ðtÞuð tÞ − hrect ðt − td Þuð t − td Þg k ζ=
t − td u ðt − td Þ td − t
s
s
t d ω= 2Þ td ω=2
sinð
'
t−t u ðt − t td − t
3 5
=
1 2
1
rect ðtÞuð tÞ
ºº º F t ºº º
Special case:
xðtÞ = −
1
3
m
f ðt Þ ¼ Fo f 1 − u ðt − td Þg
F ðsÞ ¼ Fo
xðtÞ ¼
&
2
td
F ðωÞj ¼
2
td
Case 3 Rectangle
j
−ζ
f ðt Þ ¼ Fo
8 < F :s ð
ωn
x(t ) = 0 for t > td
when
t
> td
t d = 2n π/ ωn, n 1
1
=
1, 2,
…
N
⋅
s
323
Single Degree-of-Freedom Systems: Transient Excitations
324
Table 6.2. (cont.) Case 4 Double rectangle
F ðsÞ ¼
Fo s
´
f ð tÞ ¼ Fo f½ 1 − u ðt − td = 2Þ± − ½u ðt − td = 2Þ − u ðt − td Þ± g
− e−st = d
1
xðtÞ ¼
N
⋅
s
N
⋅
s
2
sin ð
ωt d =4
Fo f hrect ðtÞu ðt Þ − 2 hrect ðt − td =2 Þuð t − t d =2 Þ þ h rect ðt − td Þu ðt − td Þg k ζ
Special case:
xðtÞ ¼
µ
ωtd =4 Þ
2
F ðωÞj ¼ Fo td
j
2
=
0 and
t > td
Fo sinðωn ½ t − td = 2± Þsin ðωn td =4 Þ k
4
x (t) = 0 for t > td
when
=
td
>
t
2
Therefore,
4
1
F ðωÞj ¼
j
xðtÞ ¼
ð1
Fo
sin ð
ºº ºº ω ºº
xðtÞ ¼
− e−st
d
Þ
N
ºº ºº ºº
td ω=2 Þ td ω=2
⋅
ζ
=
0 and
N
⋅
s
Note: j
&t td
8 < e− − e− F :t − t s ºº ffiffi ºº ºh ºº F ººt t − t ω ººº t − st
td
std
2
þ
2
d
2
→
as
ω
=
1, 2,
0
=
2
t > td
n π /ωn, n 1
1
f ð tÞ ¼ Fo
o
'
t − td u ðt − td Þ td
t > td
x (t) = Fo/k for t > t d when td
1
−
m
Case 6 Double ramp
F ðsÞ ¼
→∞
F ðωÞj
Fo fωn td − sin ωn t þ sin ðωn ðt − td ÞÞg kωn td
Therefore,
…
1, 2,
s
Fo fh ðt Þ − hð t − td Þuð t − td Þg kωn td
Special case:
=
1
f ðt Þ ¼ Fo
Fo td s
2
td
n π/ ωn, n
Case 5 Ramp + constant
F ðsÞ ¼
m
0 @
1
1
t
19 ;
− e− st A= 1
s
1
2
&t
N
t
⋅
−
1
… t−t u ðt − t t 1
1Þ
1
−
t−t u ðt − t td − t 2
'
2Þ þ
2
t − td u ðt − td Þ td − t 2
s
p
F ðωÞj ¼
j
2
1ð 2
2
2
1
−t t 8 F < ht ut k :ω t td
þ ð
xðtÞ ¼
o
dÞ
o
2Þ
1
n
ð1
ð Þ ð Þ
1
t
1 fcos ðð 1
−
t
cos ð½
−t
2
− td ω
ωÞ −
2Þ
1
ωn t1
±
t
ÞÞ þ ð
−
t ωÞ
cos ð
2
h ðt − t Þu ðt − t 1
1Þ
2
−t d
2
Þ ð1
t
cos ðð
−
1
−
−td 1
t ωÞÞ
cos ð
1
ωÞ þ cos ðtd ωÞg±
Þ
ωn ðtd − t2 Þ
hð t − t Þuð t − t 2
=
1 2
2Þ
þ
N
⋅
s
1
ωn ðtd − t2 Þ
9 = ;
hð t − td Þuð t − td Þ
m
6.5 Response to Other Excitation Waveforms
Table 6.2. (cont.) ζ=
Special case:
xðtÞ ¼
2 Fo kωn t ðtd − t
2Þ
1
td − t
þ ð
t > td
0 and
f
ωn ½t − ðt2 þ t3 Þ =2± Þsinðωn ðt 2 − t3 Þ=2 ÞÞ
t
1 ðcos ð
ωn ðt − t1 =2ÞÞsinðωn t1 =2ÞÞg
Therefore,
x(t) =
t > t d when td
0 for
−t = 2
Case 7 Triangle
F ðsÞ ¼
j
F ðωÞj ¼ xðtÞ ¼
n π/ ωn and t
2
1
f ðt Þ ¼ Fo t ds 2
´
2
1
− e−st
d =2
0 @
t d Fo
2
N
1 A
t d ω=4Þ
td ω=4
⋅
1
=
n π/ ωn, n , n
2
2
Fo kωn td 8
Therefore,
ζ=
N
⋅
s
¼
ωn ½t − td ±Þsin ðωn td =4Þ 2
t > t d when td
0 for
1, 2,
2
sin ð
x(t) =
=
s
t > td
0 and
2
=
4
1
Case 8 Half sine wave
m
t > td
n π /ωn, n
1
=
1, 2,
…
f ðt Þ ¼ Fo sin ðπ t=t d Þ½1 − uð t − td Þ± td
F ðsÞ ¼ Fo π t d
j
F ðωÞj ¼ 2 xðtÞ ¼
2 2
ºº º F π t ºº ºπ o
d
2
ºº ºº ºº
td ω=2 Þ − ω td
cos ð
2
2
2
⋅
s
N
⋅
s
ζ=
1
Therefore,
2
2
x(t) =
2
1
2
ωn ðt − td =2ÞÞcos ðωn td =2Þ
t > td
t > t d when td
=
sin ð
0 for
m
t > td
0 and
Fo 2πωn td k td ωn − π
Case 9 Sine wave
N
2
Fo f ½ g ðt; 0 Þ þ g ðt; 0 Þ±u ðt Þ þ ½g ðt − td ; 0 Þ þ g ðt − t d ; 0 Þ±u ðt − td Þg k
Special case:
xðtÞ ¼
e−std þ 1 s td þ π
td
=n
1
π/ ωn, n1
…
Fo ft½ 1 − u ðt − td =2Þ± þ ðtd − tÞ½ uð t − t d =2 Þ − uð t − td Þ±g td
2 Fo fh ðt Þu ðt Þ − 2h ðt − td =2Þu ðt − td =2Þ þ h ðt − td Þuð t − td Þg k ωn td
»
1
2
td
µ
sinð
2
Special case:
xðtÞ ¼
t > td
2 Þðcosð
1, 3, 5,
…
f ðt Þ ¼ Fo sin ð2π t= td Þ½ 1 − u ðt − td Þ±
325
Single Degree-of-Freedom Systems: Transient Excitations
326
Table 6.2. (cont.) F ðsÞ ¼ 2Fo π t d
F ðωÞj ¼
j
xðtÞ ¼
− e−st
d
s td
π2
þ 4
ºº ºº º ºº t ω F π t ºº ºº π − ω t º
4
o
sinð
d
4
d
=2Þ
2
2
2
d
⋅
s
N
⋅
N
s
Fo f ½g ðt; 2Þ − g ðt; 2 Þ±u ðt Þ þ ½g ðt; 2Þ þ g ðt − td ; 2Þ ±u ðt − td Þg k 2
Special case:
xðtÞ ¼
1
2 2
ζ
=
1
0 and
Fo 2 πωn td k td ωn − 2 π 2
Therefore,
2
2
2
ωn ðt − td =2 ÞÞcosðωn t d =2 Þ
m
t > td
sinð
Figure 6.16.
ζ=
2
t > td
x (t) = 0 for t > td
and
1
when
td
=n
1
π/ ωn, n 1
=
1, 3, 5,
…
Representative displacement time responses of Cases 4 to 9 of Table 6.2 for
0.15. Dashed lines represent the shape of the transient forcing.
τd =
20
6.5 Response to Other Excitation Waveforms
327
of each parameter by using Interactive Graphic 6.3. The reader is encouraged to use this graphic material to gain further insights into the behavior of a single degree-of-freedom system when it is subjected to different types of transient excitations.
INTERACTIVE GRAPHIC 6.3: RESPONSE TO TRANSIENT EXCITATION: TIME AND FREQUENCY REPRESENTATIONS The displacement response of the mass of a single degree-of-freedom system is displayed for nine different waveforms of the forcing, along with the corresponding frequency content of the forcing and the amplitude response function of the system and their variations with respect to the pulse duration and the damping factor. In this interactive graphic, the following should be noted. • •
• •
•
Very small τd and very small damping factors ζ produce the same long-time response irrespective of the shape of input, except for the half sine wave and full sine wave inputs. Irrespective of the input waveform, the period Td of the damped oscillation after the cessation of the input pulse is the same, ωn Td ¼ 2π= 1 − ζ 2 ; that is, it is inversely proportional to the natural frequency of the system. The oscillatory nature of the response of the system is caused by the input spectral content that is near the system’s resonance, that is, around ω/ωn = 1. The values of τ d for which there is virtually no oscillatory response after the cessation of the pulse also have zero spectral content at ω/ωn = 1 with the exception of the half sine wave at τ d = π and the full sine wave. Note, however, that the resulting waveform often bears little resemblance to the input waveform. For the half sine wave, the system experiences a resonance at τd = π. For a full sine wave, the system experiences a resonance at τ d = 2π.
pffiffiffiffiffiffiffiffiffiffiffi
EXAMPLE 6.5
Response of a slab
floor to transient loading
Consider the vibratory model of a slab
floor
shown in Figure 6.17. The slab
floor
has a
mass of 1000 kg and the equivalent stiffness of each column supporting the system is 2
×
5
10
N/m. The transient loading
f (t ) acting on the floor has the amplitude versus time x(t)
f(t)
f(t)
1000 N 1
t (s)
Floor slab
m
k
Figure 6.17. Floor slab subjected to ramp forcing.
k
328
Single Degree-of-Freedom Systems: Transient Excitations
fi find the earliest time at which the maximum response occurs.
floor
and
0, and from Figure 6.17 and the parameters given above we
find
x(t)
pro le shown in Figure 6.19. We shall determine the response
We assume that
ζ
=
that
ωn
¼
ffi rffiffiffiffiffik sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi × × 2
m
¼
2
xðtÞ ¼
5
2
10
1000
Making use of Case 5 of Table 6.2 with
ζ
=
of the slab
¼ 20
rad=s
(a)
0 yields
Fo fωn t − sin ωn t − ½ωn ðt − td Þ − sinðωn ðt − td ÞÞ± uðt − td Þg k ωn td
Upon using the parameters given above and Eq. (a), we obtain the displacement of the slab as
xðtÞ ¼ 2:5½ft − 0:05 sinð20tÞguðtÞ − fðt − 1Þ − where we have used the fact that
F o/ k
=
1000/(4
t − 1± Þguðt − 1Þ± mm
0:05 sinð20 ½
×
5
10 ) m
=
2.5
×
–3
10
m
=
2.5 mm.
A graph of Eq. (b) is shown in Figure 6.18. The earliest time for the maximum value is found numerically
Figure 6.18.
9
9
to be
Response of the
Using Matlab,
x
max
=
2.64 mm at
t
max
=
1.13 s.
floor slab shown in Figure 6.17.
fminbnd from the Optimization Toolbox; using Mathematica, FindMaximum.
(b)
x(t )
6.5 Response to Other Excitation Waveforms
EXAMPLE 6.6
The
governing
Response to half sine pulse base excitation
equation
describing
the
motion
of
a
329
10
linear
single
degree-of-freedom
system with a moving base is given by Eq. (3.28). Rewriting this equation in terms of the nondimensional time
τ
= ωnt
, we obtain
d x dx dy þ 2ζ þ x ¼ 2ζ þ y dτ dτ dτ 2
(a)
2
We also recall from Eq. (3.29) that the relative motion of the mass with respect to the base is
z(τ) =x(τ ) – y(τ).
frequency
ωo
If we assume that the motion to the base is a half sine pulse of
and magnitude
yo, then we can describe the base motion as
yðτÞ ¼ yo sinðΩoτ Þ½uðτÞ − uðτ − τ oÞ± where
τo
= ωnto = πωn ωo = π Ωo /
/
,
Ωo
= ωo ωn = T n /
to), ωoto
/(2
(b)
=π
, and
Tn
=
π ωn
2 /
is the
period of the natural frequency of the single degree-of-freedom system. We shall solve Eqs. (a) and (b) using the Laplace transform. If transform of
x(t), Y (s)
is the Laplace transform of
y(t),
X(s)
is the Laplace
and we assume that all initial
conditions are zero, then the Laplace transforms of Eqs. (a) and (b), respectively, are
ζ Y ðsÞ
ð1 þ 2 Þ
X ðs Þ ¼
s
ζs þ 1
2
þ 2
¼
ζ Y ðsÞ
ð1 þ 2 Þ
s −s
ð
s −s
1 Þð
2Þ
(c)
where
Y ðsÞ ¼
yoΩoð1 þ e−sto Þ Ωo þ s
±
2
and
s
1;2
= −ζ ± j
2
¹
(d)
qffiffiffiffiffiffiffiffiffiffiffiffi 1
−ζ
2
To obtain the solution, we take the inverse Laplace transform of
X ðsÞ ¼
e−sto Þ Þðs − s Þ
(e)
gðτÞ þ gðτ − τ oÞ uðτ − τ oÞ
(f)
±yΩΩ o
2
o
ζ ¹ s s −s
oð 1 þ þ
2
2 Þð1 þ
ð
1
2
to arrive at
xðτÞ yo
¼
±
where
¹
Ωo αa ðτÞ − Ωo e−ζτ b ðτÞ gðτÞ ¼ Do 10
1
1
(g)
Extensions of these types of models to include nonlinear springs and dissipative elements can be found in the fol-
“Response of nonlinear dissipative shock isolators,” Journal of Sound and – 603; and N. C. Shekhar et al., “Performance of nonlinear isolators and absorbers to shock excitation,” Journal of Sound and Vibration, 227(2) (1999) 293 –307.
lowing studies: N. C. Shekhar et al.,
Vibration,
214(4) (1998) 589
330
Single Degree-of-Freedom Systems: Transient Excitations and
± ¹ Ω τ ζΩ ±α ζ Ω ζ −Ω ±α ± ζ ¹ ± ζ − ¹±ζ Ω ¹¹ ατ ± ± ζ − ¹Ω ¹ αΩ Ω q ffiffiffiffiffiffiffiffiffiffiffiffi
a ðτÞ ¼ α
2
1
b ðτÞ ¼ 1
Do
¼
α
¼
þ 4
2
o
1
2
2
4
o
2
2
o
þ
þ 1 þ
4
þ 2
2
2
o
2
2
2
sinð
1
2
o
2
Þ þ2
o
þ
2
o
o
sinð
2
þ
¹
ζ 2 − Ω2o − 1
± αζ α
Þ þ 2
2
þ
ζ
2
Ωo τ Þ
cosð
− Ωo − 2
¹
ατÞ
1 cos ð
(h)
þ1
−ζ
2
x(τ) and the when Ωo ≪ 1, the
The solution given by Eq. (f) is used to plot the absolute displacement relative displacement
(a) Figure 6.19.
z(τ )
in Figure 6.19. From this
figure, we see that
(b) (a) Absolute displacement and (b) relative displacement of the mass of a single
degree-of-freedom system subjected to half sine wave displacement to the base with
ζ
=
0.1.
6.5 Response to Other Excitation Waveforms
331
mass follows the movement of the base as if it were rigidly connected to it. As
fi
Ωo
approaches 1, the mass ampli es the base motions and has large excursions relative to the base. This example leads to the following design guideline.
Design Guideline: In order to minimize the amplification of the motion of the base of a single degree-of-freedom system, one should keep the pulse duration of the displacement applied to the base long compared to the period of the natural frequency of the system. Stated differently, for a given pulse loading it is preferable to have a system with as high
fication of the base motion.
a natural frequency as possible to minimize the ampli
EXAMPLE 6.7
Single degree-of-freedom system with moving 11
base and nonlinear spring
Consider
the
displacement
single
y( t)
as
degree-of-freedom shown
in
Figure
system 6.20.
whose The
base
spring
is
is
subjected
nonlinear
to
with
a
known
a
force-
displacement relationship given by Eq. (2.30), that is,
F ðx Þ ¼ kx þ αkx
3
Then, carrying out the force balance for vertical motions of mass
m,
as discussed in
Section 3.2.1, we arrive at the following equation of motion of the system
m€ x þ cðx_ − y_ Þ þ kð x − δ st − yÞ þ k αðx − δst where
− y = −mg 3
Þ
(a)
x is measured from the static equilibrium position of mass m , y is the absolute dis-
placement of the base, and
δst
fl
is the static de ection. From the static equilibrium of the
system, we have that
kδ st þ k αδst 3
Figure 6.20.
m
x(t)g
¼
mg
(b)
Base excitation of a single degree-of-freedom system
with a nonlinear spring. The quantity
L is the unstretched length
of the spring.
L
st
c y(t)
11
N. C.
Shekhar, H. Hatwal and A. K. Mallik,
Sound and Vibration, 214(4) (1998) 589–603.
“Response
of non-linear dissipative shock isolators,
” Journal of
332
Single Degree-of-Freedom Systems: Transient Excitations Upon substituting Eq. (b) into Eq. (a), we obtain
m€ x þ cðx_ − y_ Þ þ k ðx − yÞ þ k αðx − δst
− y = −k αδst 3
3
Þ
(c)
The relative displacement is
zðtÞ ¼ xðtÞ − yðtÞ
(d)
and, after substituting Eq. (d) into Eq. (c), we arrive at
− δst = −k αδst − m€y
m€ z þ c_z þ kz þ k αðz If we set
τ
3
3
Þ
(e)
= ωnt
, then Eq. (e) can be written as
d z dz þ 2ζ þ z þ α ðz dτ dτ 2
2
− δst = − αδst − 3
3
Þ
d y dτ 2
(f)
2
For the displacement to the base of the system, we assume a step-like disturbance that has variable rise time and a rounded shape given by
y ðτÞ ¼ y
max ½1
−
ð1 þ
γτÞe−γτ±
(g)
This waveform was selected instead of a unit step function, in part, because its higherorder
derivatives
are
continuous.
The
normalized
Figure 6.21 for several values of the parameter with respect to
τ,
waveform
y(τ )/y
max
is
shown
in
Taking the second derivative of Eq. (g)
we arrive at
€yðτ Þ ¼ y
max
Figure 6.21.
γ.
γ 2ð1 − γτÞ e−γτ
Base excitation waveform for several values of
γ.
(h)
6.5 Response to Other Excitation Waveforms
333
We now substitute Eq. (h) into Eq. (f) and introduce the nondimensional variable
z(τ )/y
max
d zn dzn þ 2ζ þ zn þ α o ðzn dτ dτ ; αo ¼ α y ; and 2
2
where
δo
¼
δst=y max
max
2
ð1
þ
gðτÞ
(i)
− γτ e−γτ
zn ðτÞ þ y
max ½ 1
αo
=
12
Þ
(j)
−
ð1 þ
γτÞe−γτ ±
(k)
of Eq. (i) yields the results shown in Figure 6.22 for
(b)
(c)
(d)
Figure 6.22.
=
30
Normalized absolute displacement of the mass of a single degree-of-freedom system
with a nonlinear spring with
=
αo
0; that is, when the nonlinear spring is replaced by a linear spring. In
(a)
12
3
zn(τ) is a solution of Eq. (i).
The numerical evaluation
δo
3
Þ
x(τ ) is obtained from Eq. (d) and Eq. (g) that is,
xðτÞ ¼ y
(a)
− δo = −αoδ o
max
The absolute displacement
and for
=
2
gð τÞ = − γ
where
zn(τ)
to obtain
0 and
γ
Using Matlab,
=
1, (b)
δo
=
ζ
=
0.15 and subject to the base excitation shown in Figure 6.21:
0 and
γ
=
4, (c)
δo
=
0.3 and
ode45; using Mathematica, NDSolveValue.
γ
=
1, and (d)
δo =
0.3 and
γ
=
4.
334
Single Degree-of-Freedom Systems: Transient Excitations addition, we have selected the nondimensional static displacements of The value
δo
=
δo =
0.3 and
δo
=
0.
0 corresponds to the case of no static displacement; that is, when the
system is positioned as shown in Figure 6.1. We see that combination of the linear spring and the nonlinear spring is stiffer than the linear spring by itself and, therefore, the period of the oscillation is shorter than that of the linear spring by itself, especially as the rise time
decreases
(
γ
increases).
In
addition,
as
the
rise
time
of
the
input
displacement
decreases, the overshoot increases, which is similar to what was found for the response to a ramp input applied to a linear single degree-of-freedom system.
INTERACTIVE GRAPHIC 6.4: RESPONSE OF SYSTEM WITH A NONLINEAR SPRING SUBJECTED TO TRANSIENT BASE EXCITATION The displacement response of the mass of a base excited single degree-of-freedom system is shown. The variation of this response with respect to the damping factor, static displacement of the spring, relative stiffness of the nonlinear spring, and the rise time of the displacement of the base can be explored by using this interactive graphic. In this interactive graphic, the following should be noted. For a system with a linear spring, the response is very similar to that of a single degree-offreedom system subjected to a ramp input. • When the rise time decreases (γ increases), the response closely resembles that of a system subjected to a step change in force applied to the system mass. • The static displacement of the nonlinear spring and/or the relative stiffness of the nonlinear spring have discernible influence on the nature of the oscillatory portion of the response.
•
6.5.1
ficance of the Spectral Content of the Applied Force: An Example
Signi
There are situations when the time waveform of a force applied to the mass is random; that is, its magnitude as a function of time is not predictable. In such cases, for linear systems one usually represents the force in the frequency domain and conducts the analysis in the frequency domain. For this analysis, one can make direct use of the system’s amplitude response function. We shall illustrate this type of analysis by considering the design of a structure subjected to sustained winds. In particular, we consider the determination of an estimate for the outer diameter
do
of a 10 m high steel lamppost of constant cross-section that is sub-
jected to sustained winds so that the maximum transverse displacement of the lamps on
6.5 Response to Other Excitation Waveforms
335
top of the lamppost does not exceed 5 cm. The mass of the lights on top of the lamppost is 75 kg. We assume that the lamppost is a cylindrical tube whose inner diameter is 95% of
do,
that the system acts as a beam in bending, and that it can be modeled as a single
degree-of-freedom system. In addition, the damping ratio of the system is assumed to be
ζ
=
0.04. The magnitude of the turbulence-induced wind force spectrum has been experi-
mentally determined to be
F ð jf Þ j ¼ 400 fe−
j where
0:667
f
N
(6.43)
f is frequency in Hz. The spectrum given by Eq. (a) is shown in Figure 6.23.
Response in Frequency Domain The solution is obtained by
first
determining from the displacement response in the fre-
k for which the maximum amplitude ∣F ð jf Þ∣. After the magnivalue of d o is determined from the relation-
quency domain the value of the equivalent stiffness
∣X
ð
jf Þ∣
is less than 0.05 m over the entire frequency range of
tude of the equivalent stiffness is known, the
ship for the stiffness of a cantilever beam. The displacement response in the frequency domain is given by Eq. (6.5). This response is converted from the radian frequency notation
ω
to frequency
f
in Hz resulting in
∣X
ð
jf Þ∣
¼
where the natural frequency
∣F
jf Þ∣ k
ð
2 4
2
1
¼
2
þ
n
f n in Hz is fn
Figure 6.23.
² f ³ ¸ ² ζf³ 3− 5 − f f 1
π
2
1=2
2
2
(6.44)
n
rffiffiffiffi
k m
Assumed wind-induced force spectrum acting on the lamppost.
(6.45)
336
Single Degree-of-Freedom Systems: Transient Excitations Upon substituting Eqs. (6.45) into Eq. (6.44) and noting that we obtain
− 400 fe
0:667
X ð jf Þj ¼
j
− 400 fe
¼
f
400
fe− k
0:667
f
75 kg and
ζ
=
0.04,
1=2
2
2
2
þ
2
4
20 0 sffiffiffiffiffi1 1 0 64@ −@ π f A A @ π 2
2
1
k
=
20 0 sffiffiffiffi1 1 0 sffiffiffiffi1 3− 64@ −@ π f mA A @ πζf mA 75 k k 1
k
0:667
¼
f
m
2 4
−
1
75
2
þ
k
f k
2
2960:9
We now plot Eq. (6.46) for three values of
¸
2
4
f k
2
þ 18:95
sffiffiffiffiffi1 3− A 75 f 2
ð0:04Þ
3− 5
1=2
75
k
1=2
(6.46)
k: 20 000, 30 000, and 40 000 N/m. The results
are shown in Figure 6.24. It is seen that as the stiffness increases, the natural frequency
X( jf )| decreases. This decrease is brought about by the fact that the spectral amplitude of |F( jf )| diminishes when f is greater than about 1.6 Hz. increases and the maximum magnitude of |
Since this magnitude of the forcing is decreasing in this frequency region, the magni
fication by
H( jf )| in the damping dominated region is magnifying a smaller quantity; hence, the peak magnitude of |X ( jf )| becomes smaller. the system’s frequency-response function |
Another way to determine approximately the peak responses is to recall the response of a system in the damping dominated region given by Eq. (5.25), that is,
H ð1Þ ¼
Figure 6.24.
1
ζ
2
Displacement response spectrum of lamppost for three values of stiffness.
(6.47)
6.5 Response to Other Excitation Waveforms Then, for
ζ
=
H(1) =
0.04,
337
F( jf )| at the three natural frequencies cork j are, respectively,
12.5. The values of |
responding to the three stiffness values
fn fn fn
2
3
1
¼
¼
¼
1
π
2
1
sffiffiffiffi k 1
m
sffiffiffiffi k 2
π
2
1
¼
m
sffiffiffiffi k 3
π
2
¼
¼
m
1
sffiffiffiffiffiffiffiffiffiffiffiffi
20000
π
2
1
π
2
1
75
¼ 2:6 Hz
sffiffiffiffiffiffiffiffiffiffiffiffi
30000 75
¼ 3:18 Hz
(6.48)
sffiffiffiffiffiffiffiffiffiffiffiffi
40000
π
2
75
¼ 3:68 Hz
Then, from Eqs. (6.43) and (6.48), we have that
Þj ¼ 400
f n e−
0:667
fn
Þj ¼ 400
f n e−
0:667
fn
f n e−
0:667
fn
j
F ð jf n
1
j
F ð jf n
2
j
F ð jf n
3 Þj
¼ 400
1
2
3
1
¼ 400
2
¼ 400
3
¼ 400
× × ×
e−
2:6
0:667
×
2:6
e−
0: 667
e−
0: 667
3:18
3:68
¼ 183:6 N
×
3:18
×
3:68
¼ 152:5 N
(6.49)
¼ 12:4 N
These values could also have been obtained directly from Figure 6.24, but with less precision. Consequently, the peak magnitudes of the displacements at these three frequencies are, respectively,
F ð jf n
j
1
Þj ¼
F ð jf n
Þj ¼
F ð jf n
Þj ¼
j
2
j
3
1
k
F ð jf n
2
F ð jf n
3
j
2
1
k
1
H ð1Þ ¼
183:6
H ð1Þ ¼
152:5
H ð1Þ ¼
126:4
Þj
1
1
k
F ð jf n
j
j
Þj
Þj
3
×
12:5
20000
×
12:5
30000
×
12:5
40000
¼ 11:5 cm
¼ 6:35 cm
(6.50)
¼ 3:95 cm
It is seen that these values correspond to their respective counterparts in Figure 6.24.
Lamppost Parameters The value of
k
that produces a maximum magnitude of | 13
mined numerically
X( jf )|
equal to 5 cm is deter-
to be 34 909 N/m. To determine the diameter of the cylindrical
support, we note from entry 4 in Table 2.3 that
k¼
13
Using Matlab,
NMaxValue.
fzero
and
fminbnd
EI L
3
3
from the Optimization Toolbox; using Mathematica,
(6.51)
FindRoot
and
338
Single Degree-of-Freedom Systems: Transient Excitations where
I
¼
π 32
±d
4
o
− di
4
¹
(6.52)
Upon combining Eqs. (6.51) and (6.52), and noting that
L = 10 m, and E = 2.1 × 10
11
do
k
=
34 909 N/m,
di
=
do,
0.95
2
N/m , we obtain
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v u u t ´ kL µ Eπ − v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u × u t ´× µ 32
4
3
¼
4
1
32
3
4
ð0:95Þ
π × 2:1 × 1011 ×
3
3
34 909
10
1
−
4
ð0:95Þ
¼ 0:235 m
(6.53)
Thus, the outer diameter of the lamppost is 23.5 cm and its inner diameter 22.3 cm, resulting in a wall thickness of 1.2 cm. Interactive Graphic 6.5 has been created for this example so that different aspects and parameter combinations can be explored.
INTERACTIVE GRAPHIC 6.5: EFFECT OF NATURAL FREQUENCY AND INPUT FORCE SPECTRUM ON SYSTEM RESPONSE This interactive graphic is provided as a means to explore how the normalized displacement response of the mass can be ensured to remain less then a specified value for all input frequencies. In this interactive graphic, the following can be noted. For a set of values for the damping factor and the input spectrum shape parameters, the value of f n must be chosen so that the magnitude of the response of the mass for all frequencies is less than the design value. • For a set of values for the input spectrum shape parameters and a given f n, a value of the damping factor must be selected so that the magnitude of the response of the mass for all frequencies is less than the design value. •
6.6
IMPACT TESTING
Impact testing is performed with
standard shock test machines where a test object
impacts a stationary platform. Placing different materials of different geometry between the object and its rigid terminating platform alters the shape of the impact pro
14
C. LaLanne,
Mechanical Shock, Hermes Penton Ltd., 2002, Chapters 6 and 7.
file.
14
6.6 Impact Testing
Test object
Programmer
g
339
Programmer
Guide
Test object
(b)
(a)
Test object Elastomer (programmer)
Programmer Sand (programmer)
Lead (programmer)
Air in
Piston
Air out (d)
(c) Figure 6.25.
Different impact testing methods and the locations of their programmers: (a) free-
fall apparatus, (b) pendulum, (c) different programmers, and (d) pneumatic machine.
These inserts are called
programmers.
These programmers are typically elastomer discs,
lead pellets, and sand. The machines fall into one of three categories, schematic drawings of which are given in Figure 6.25: (1) free-fall machines, which include pendulum type machines; (2) pneumatic machines; and (3) electrodynamic machines. In free-fall machines, the object’s velocity is a function of the release height from the impact platform. In pneumatic machines, the pneumatic driver produces the object’s velocity. In electrodynamic shakers, the shock spectrum is attained by tailoring the time-varying shape, magnitude, and duration of the signal sent to the shaker or by specifying the
ficacy
input spectrum. One application of impact testing is determining the ef
of motor-
cycle helmets, where the acceleration response to impact is required to determine the 15
head injury criteria (HIC) for these devices.
16
Another application
of impact analysis
is to reduce the amount of force that is transferred to the ground (soil) in manufacturing facilities that use drop forges and other heavy machinery that create intermittent loads.
15
R. Willinger, et al., “Dynamic characterization of motorcycle helmets: modeling and coupling with the human head,”
16
Journal of Sound and Vibration, 235(4) (2000) 611–615.
See, for example, A. G. Chehab and M. El Naggar, “Response of block foundations to impact loads,”
Sound and Vibration, 276 (2004) 293–310.
Journal of
340
Single Degree-of-Freedom Systems: Transient Excitations
6.7
SUMMARY
In this chapter, responses of single degree-of-freedom systems subjected to various types of transient excitations have been studied. A common characteristic of the systems studied is that the excitation provides a
“sudden”
change to the state of the system. The
notion of the impulse response, which is an important component for determining the response of linear systems, has been introduced, along with such notions as settling time and rise time of the response. Also illustrated was how the response to an input excitation with an arbitrary frequency spectrum can be determined. Interactive graphics materials have been included for a reader to explore the spectral content of different types of wave forms as well as the responses of single degree-of-freedom systems subjected to different forms of transient excitations.
Exercises
341
Exercises
Section 6.2 6.1
Determine the response of a vibratory system governed by the following equation:
xðtÞ þ 0:2x_ ðtÞ þ 3:5xðtÞ ¼ 1:5 sinðωtÞuðtÞ þ 4δðt − 3Þ € Assume that
m
=
1 kg, the initial conditions are
and the excitation frequency 6.2
ω
=
x(0)
=
0.1 m and
x_ ð0Þ ¼ 0
m/s,
1.4 rad/s. Plot the response.
Consider the following single degree-of-freedom system excited by two impulses when the system is initially at rest
x€ðtÞ þ 2x_ ðtÞ þ 4xð tÞ ¼ δð tÞ þ δ ðt − 5Þ Determine the displacement response of this vibratory system and plot the response. 6.3
From extensive biomechanical tests, the spinal stiffness
k
of a person is estimated
to be 50 000 N/m. Assume that the body mass is 80 kg. Let us assume that this person is driving an automobile without wearing a seat belt. On hitting an obstacle, the driver is thrown upwards, and drops in free fall onto an unpadded seat and experiences an impulse with a magnitude of 100 N
⋅
s: Determine the resulting
motions if an undamped single degree-of-freedom model is used to model the vertical vibrations of this person.
Section 6.3 6.4
Determine the response of an underdamped single degree-of-freedom system that is subjected to the force
f ðtÞ ¼ Fo e−αt uðtÞ:
Assume that the system is initially at
rest. 6.5
Repeat Example 6.4 when the step change in road elevation
a
is 4 cm and the
vehicle speed is 100 km/h. Plot the results. 6.6
–
–
Refer to the Kelvin Voigt Maxwell combination shown in Figure E6.7. Obtain an expression for the displacement response in the Laplace transform domain of the mass when the mass is subjected to a step-function force of magnitude
Figure E6.7.
f(t)
m
k1 k
c
c1
Fo.
342
Single Degree-of-Freedom Systems: Transient Excitations Section 6.4 6.7
A machine system of mass 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During the operations, the machine is subjected to a force of the form shown in Figure E6.8, where the horizontal axis time
t
is in seconds and the
vertical axis is the amplitude of the force in newtons. Assume that the machine system is initially at rest and determine the displacement response of the system.
f(t)
Figure E6.8.
2500 N
0.5 s
6.8
2.0 s
t
In Example 6.5, assume that the motions of the slab
floor
are damped with the
damping factor being 0.2. Determine the response of this damped system for the forcing and system parameters given in Example 6.5. Also,
find
the earliest time at
which the maximum displacement response occurs. Plot the results. 6.9
Consider the Boltzmann sigmoidal function, whose general form is given by
´
S ðτ ; a; b; α; τ oÞ ¼ a þ b where
a, b, α,
and
τo
are constants and
1 þ
eα τo −τ ð
Þ
µ−
1
S (τ o, a, b, α, τo)
=a
+
b/2.
can be used to create a step-like function as shown in Figure E6.9 for 0.1). Determine the
f(τ )
= FoS τ
response of
a
system for
f (τ)
=
F oS( τ,
0, 1,
This function
S (τo, 0, 1, 60, 60, 0.1) and
( , 0, 1, 6, 1) and graphically compare them to the response of the system
to the step input given by Eq. (6.23). The solutions have to be obtained numerically. Assume that
ζ
=
0.15.
Figure E6.9.
Exercises
343
Section 6.5 6.10
Determine the response of the vibratory system discussed in Section 6.5.1 when the forcing due to the wind spectrum is of the following form
∣F where
f
ð
jf Þ∣
fe−
¼ 200
0:5
f
N
is the frequency in Hz. Plot the result for the values of
Example 6.4.
k
3
given in
Multiple Degree-of-Freedom Systems:
7
Governing Equations, Natural Frequencies, and Mode Shapes
7.1 Introduction 7.2 Governing Equations 7.2.1 Force-Balance and Moment-Balance Methods
page
344 346 346
7.2.2 General Form of Equations for a Linear Multi-Degreeof-Freedom System 7.2.3 Lagrange’s Equations of Motion 7.3 Free Response Characteristics
7.1
356 359 378
7.3.1 Undamped Systems: Natural Frequencies and Mode Shapes
378
7.3.2 Natural Frequencies and Mode Shapes: A Summary
402
7.3.3 Undamped Systems: Properties of Mode Shapes
402
7.3.4 Characteristics of Damped Systems
411
7.3.5 Conservation of Energy
421
7.4 Rotating Shafts on Flexible Supports
423
7.5 Stability
434
7.6 Summary
441
Exercises
442
INTRODUCTION
In Chapters 4 through 6, single degree-of-freedom systems and vibratory responses of these systems were studied. Systems with multiple degrees of freedom and their responses are studied in this chapter and the next. Systems that need to be described by more than one independent coordinate have multiple degrees of freedom. The number of degrees of freedom is determined after taking into account the constraints imposed on the system. The governing equations of motion of vibratory systems are determined by using either force-balance and moment-balance methods or Lagrange’s equations. In this chapter, both of these methods will be used to develop the system equations. Furthermore, viscous damping models will be used to model dissipation in the systems. After developing the governing equations of motion, the natural frequencies and mode shapes of multi-degree-of-freedom systems are examined at length. Forced oscillations
7.1 Introduction
345
are considered in the next chapter. To describe the responses of single degree-of-freedom systems, only time information is needed. For systems with multiple degrees of freedom, one also needs spatial information to describe the responses of systems. This spatial information is expressed in terms of mode shapes, which are determined from the freevibration solution. Each mode shape is associated with a natural frequency of the system. This shape provides information about the relative spatial positions of the inertial elements in terms of the chosen generalized coordinates. Therefore, the determination of mode shapes and natural frequencies is discussed in detail in this chapter. As illustrated in the next chapter, the spatial information obtained from the free-vibration problem can also provide a basis for determining the forced response of a system with multiple degrees of freedom. It is also shown there that the properties of mode shapes can be the
used
to
construct
responses
of
the
response
equivalent
of
single
a
multi-degree-of-freedom
degree-of-freedom
systems.
system
This
in
allows
terms the
of use
of the material presented in the preceding chapters for determining the response of a system with multiple degrees of freedom. The notions of stability introduced in Chapter 4 for single degree-of-freedom systems are extended in this chapter to multi-degree-of-freedom systems. In this chapter, we shall show how to do the following.
•
Derive the governing equations for systems with multiple degrees of freedom by using force-balance and moment-balance methods.
•
Derive the governing equations for systems with multiple degrees of freedom by using Lagrange’s equations.
•
Obtain the natural frequencies and mode shapes associated with vibrations of systems with multiple degrees of freedom.
•
Obtain the conditions under which the mode shapes are orthogonal.
•
Interpret the characteristics of damped systems.
•
Determine the vibration characteristics of rotating shafts.
•
Examine the stability of multiple degree-of-freedom systems.
In addition, the following interactive graphics have been created to better visualize the mode shapes of two degree-of-freedom systems.
Interactive Graphic 7.1: Natural
frequencies
of
translating
two
degree-of-freedom
systems Interactive Graphic 7.2: Two degree-of-freedom systems with translation: natural frequencies and mode shapes Interactive Graphic 7.3: Two degree-of-freedom systems with translation and rotation: natural frequencies and mode shapes
346
Multiple Degree-of-Freedom Systems: Governing Equations
7.2
GOVERNING EQUATIONS
In this section, two approaches are presented for determining the governing equations of
first
motion. The
approach is based on force-balance and moment-balance methods
and the second approach is based on Lagrange’s equations. For algebraic ease, the number of degrees of freedom for the physical systems chosen in this chapter is less than or equal to
7.2.1
five.
Force-Balance and Moment-Balance Methods The
underlying
principles
of
the
force-balance
and
moment-balance
methods
are
expressed by Eqs. (A.21) and (A.25a) of Appendix A. Through these equations, the forces and moments imposed on a system are related to the rate of change of linear momentum and the rate of change of angular momentum, respectively.
Force-Balance Method To illustrate the use of force-balance methods, consider the system shown in Figure 7.1. This system consists of linear springs, linear dampers, and translating inertia elements.
m
The free-body diagrams for the point masses
and
1
m
2
are shown, along with the
x
respective inertial forces in Figure 7.2. The generalized coordinates specify the positions of the two masses
m
1
and
m
x
and
2
1
− m x€ |fflffl{zfflffl} − k x |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} + k x −x −c x_ |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} + c x_ − x_ |fflffl +ffl{zfflffl f t |fflfflffl{zfflfflffl} |fflffl{zfflffl} ffl} 1
1
1
2ð
1
1Þ
2
1
2ð
1
1Þ
2
1ð
Þ
Inertia
Force associated
Force associated
Force associated
Force associated
External force
force
with spring of
with spring of
with damper of
with damper of
acting on
k
stiffness
stiffness
1
k
This equation has been rewritten as the diagram of inertial element
m x€ m x€ 1
1
2
2
+ +
c ðc ð
x1 , f1 (t) k1
Figure 7.1.
1
2
end on
c
coefficient
1
c
mass
2
m
=
the
0
1
first of Eqs. (7.1a). Similarly, from the free-body
, the second of Eqs. (7.1a) is obtained:
2
+c +c
x_ Þx _
2Þ
1
3
2
− c x_ + − c x_ + 2
2
2
1
k ðk ð
x2 , f2 (t) k3
k2 m1
c1
m
coefficient
2
are used to
fixed
m and carrying out i, one obtains the following equation
the left side. Based on the free-body diagram of inertial element force balance along the horizontal direction
1
, respectively, from the
2
m2 c2
System with two degrees of freedom.
c3
1
2
+k +k
x Þx
2Þ
1
3
2
−k x =f −k x =f
tÞ ðtÞ
2
2
1ð
2
1
2
(7.1a)
7.2 Governing Equations
..
y
m1x1 k1x1
j
o
x
..
m2 x2
x1, f1
k2(x2
m1
.
.
c2(x2
Figure 7.2. Free-body diagrams for masses
k2(x2
x1)
.
c1x1
i
347
.
m
1
m
and
2
k3 x2
m2
.
c2(x2
x1)
x2, f2
x1)
.
c3 x2
x1)
along with the respective inertial forces
illustrated by broken lines. The origin of the coordinate system is located on the
fixed boundary
at the left end of the left-most spring.
1
These linear differential equations are written in matrix form
°m
1
0
0
m
!& €x ' ° c x €
2
+
1 2
+c −c
1
−c c +c
2
2
2
2
!& x_ ' ° k x_
3
+
1
2
+k −k
1
2
2
as
−k k +k
!& x ' & f '
2
2
3
x
=
1 2
1
f
2
(7.1b)
The off-diagonal terms of the inertia matrix are zero, while the off-diagonal terms of the stiffness and damping matrices are non-zero. In addition, all of these matrices are
metric matrices.
2
off-diagonal
sym-
The equations governing the system are coupled due to these non-zero
terms
in
the
stiffness
and
damping
matrices.
Physically,
the
system
uncoupled when the damper c and the spring k are absent. The excitations f (t) f (t) are directly applied to the inertial elements of the system as shown in the figure. 2
2
1
is
and
2
Moment-Balance Method
flywheels with rotary rotor is treated as a fi xed
We now consider the system shown in Figure 7.3, which has two inertias
Jo
1
and
Jo
2.
The end of the shaft attached to the
end. The drive torque to the and
ϕ2
first flywheel
is
Mo (t ).
are used to describe the rotations of the
The generalized coordinates
flywheels about the axis k
ϕ1
through the
respective centers. The inertias of the shafts are neglected, the torsional stiffness of the shaft in between the
fixed
end and the
flywheel
closest to it is represented by
and the torsional stiffness of the other shaft is represented by
kt
2.
kt
1
,
It is assumed that
flywheels are immersed in housings filled with oil and that the corresponding dissipative effect is modeled by using the viscous damping coeffi cients c t and ct . In the the
1
1 2
See Appendix D for a brief introduction to matrix notation. As discussed in Appendix D, a matrix [
A] is called a symmetric matrix if the elements of the matrix aij
2
= aji
.
348
Multiple Degree-of-Freedom Systems: Governing Equations
Oil housing Jo1
kt1
Jo2
kt1
kt2
Mo(t)
ct1
.
1
Jo1
ct2
..
.
2
Jo2
..
2
1
1
kt2 (
k
1
2)
k
Mo (t)
1 2
(a)
(b)
(a) System of two
Figure 7.3.
flywheels driven by a rotor and (b) free-body diagrams along
with the respective inertial moments illustrated by broken lines.
free-body diagrams of Figure 7.3, the inertial moments
− Jo ϕ€ k 1
1
and
−Jo ϕ€ k 2
2
are
also shown. Based on the free-body diagrams shown in Figure 7.3, we apply the principle of angular momentum balance to each of the
€ Jo ϕ 1
1
+ ct ϕ_ + kt ϕ + kt € + ct ϕ_ + kt Jo ϕ 1
2
"
#( € ) "
flywheels and obtain the governing equations
1
2
ϕ1 − ϕ2Þ = Mo ðtÞ
1
1
2ð
2
2
2ð
#( _ ) "
#( ) (
which are written in matrix form as
Jo
0
1
0
Jo
ϕ1
€2 ϕ
2
+
ct
0
0
ct
1
2
ϕ1 ϕ_ 2
+
kt
(7.2a)
ϕ2 − ϕ1Þ = 0
+ kt −kt
1
2
2
− kt kt
ϕ1
2
ϕ2
2
=
MoðtÞ
) (7.2b)
0
In this case, the equations are coupled because of the non-zero off-diagonal terms in the stiffness matrix, which are due to the shaft with stiffness
kt
.
2
Both of the physical systems chosen for illustration of the force-balance and momentbalance methods are described by linear models and the associated governing system of equations is written in matrix form. This is possible for any linear multi-degree-offreedom system, as illustrated in Example 7.1. For a nonlinear multi-degree-of-freedom system, the governing nonlinear equations of motion must be linearized to obtain a set of linear equations; the resulting linear equations are amenable to matrix form. This is illustrated in Example 7.3.
EXAMPLE 7.1
Modeling of a milling machine on a
flexible floor
A milling machine and a vibratory model of this system are shown in Figure 7.4. We shall derive the governing equations of motion for this system by using the force-balance method. As shown in Figure 7.4b, the milling machine is described by using the three inertial elements
m,m 1
2,
and
m
3
along with discrete spring elements and damper elements. All three
inertial elements translate only along the direction shown in the
i
direction. The external force
figure is a representative disturbance acting on m
. 1
f (t) 1
in the
i
7.2 Governing Equations Flexible support, k1
f1 (t)
f1(t)
Machine tool head, m 1
m1
Machine tool base, m 2
Elastic mount, k2
349
k1
x1 (t) c1
m2
x2(t)
g
c2
k2 m3 k3
x3(t) x c3
Floor, m 3 and k 3
i
y j
Rigid floor support (a)
(b)
f1 (t) .. m1 x1
m1
. . c 1(x1 – x2)
k1 (x1 – x2)
.. m2 x2
m2
. . c 2(x2 – x3)
k2 (x2 – x3)
m3 .. x3
m3 c3 x. 3
k3x3 (c)
Figure 7.4. (a) Milling machine, (b) vibratory model for study of vertical motions, and
(c) free-body diagrams of inertial elements
m,m 1
, and
2
m
3
shown in (b) along with the inertial
forces illustrated by broken lines.
To obtain the governing equations of motion, we use the generalized coordinates
x
,
1
x
, and
2
x
, each measured from the system’s static equilibrium position. Since the
3
coordinates are measured from the static equilibrium position, gravity forces are not considered
below.
In
order
to
apply
the
force-balance
method
to
each
inertial
element, the free-body diagrams shown in Figure 7.4c are used. Applying the forcebalance method along the
i
direction to each of the masses, we obtain the following
equations:
m x€ m x€ + ð k + k Þx − k x − k x m € x + ðk + k Þx 1
2
2
1
2
2
3
3
1
1
2
2
3
1
3
3
+ k x − x + c x_ − x_ = − f + c + c x_ − c x_ − c x_ = − k x + c + c x_ − c x_ = 1ð
ð
2Þ
1
2
2Þ
1
2
ð
2
2
1ð
2Þ
1
tÞ
1ð
1
1
2
3
0
3Þ
3
2
2
0
(a)
350
Multiple Degree-of-Freedom Systems: Governing Equations Equations (a) are arranged in the following matrix form:
2m 64
0
0
m
0
1
0
2
1
2
x€
m
2 k + 64 − k 0
38> x€ 75< x€ >:
0
3
−k k +k −k
1
3
1
0
1
2
1
1
2
0
2
2
2
3
1ð
1
2
2
0
1
0
1
1
9> 2 c − c 38> x_ = 6 < + 4 − c c + c − c 75 x_ > >; − c c + c : x_ 38> x 9> 8> − f t 9> = < = < − k 75 x = >; >: x >; >: k +k 2
3
2
0
3
0
1
2
3
9> = >;
Þ
(b)
We see that the inertia, the stiffness, and the damping matrices are symmetric matrices.
EXAMPLE 7.2
Conservation of linear momentum in a multiple
degree-of-freedom system
We revisit Example 7.1 and discuss when the linear momentum of this multiple degree-
i direction. From Eq. (A.21) of Appendix A, it forces fi (t), the linear momentum of the system is
of-freedom system is conserved along the is clear that in the absence of external conserved, that is,
dp dt Even in the absence of the forcing
=
→p=
0
f (t)
constant
(a)
in Example 7.1, the linear momentum of this
1
three degree-of-freedom system is not conserved because of the forces acting at the base of the system. To examine this, we set
f (t ) 1
=
0 in Eq. (a) of Example 7.1 and arrive at
the following equations:
m €x m x€ + ðk + k Þ x − k x − k x m €x + ðk + k Þx 1
2
2
1
2
2
3
1
3
1
2
2
3
1
3
3
+ k x − x + c x_ − x_ = + c + c x_ − c x_ − c x_ = − k x + c + c x_ − c x_ = 1ð
ð
2Þ
1
2Þ
1
2
ð
2
1ð
2Þ
0
1
1
2
3
0
3Þ
3
2
2
0
2
2
1
(b)
Each of Eqs. (b) was obtained by performing a linear-momentum balance individually for each of the three inertial elements of the system. Adding all three equations of Eqs. (b), we obtain
m x€ 1
1
+ m x€ + m x€ = − 2
2
3
3
k x
ð
3
3
+ c x_ 3
3Þ
(c)
Integrating Eq. (c) with respect to time, that is,
ð t
ð t
m € x
ð
1
1
+m
2
€x
2
+m
3
€x Þdt = −
ð
3
0
kx 3
3
+c
3
x_ Þdt 3
(d)
0
leads to
m x_ 1
1
+m
2
x_
2
+m
3
x_
3
≠
constant
(e)
7.2 Governing Equations
351
From Eq. (e), it follows that due to the presence of the spring and damper forces at the base, the total linear momentum of the system is not conserved. If the spring with stiffness
fi
k
and the damper with damping coef cient
3
c
3
are absent, then the total linear
momentum of the resulting free-free system is conserved.
EXAMPLE 7.3
System with bounce and pitch motions
Consider the rigid bar shown in Figure 7.5a, which can rotate (pitch) about the direction
and
coordinates
y
translate and
θ
(bounce)
along
the
j
direction.
We
locate
the
k
generalized
at the center of gravity of the bar. This model provides a good
representation for describing certain types of motions of motorcycles, automobiles, and other vehicles. This particular example has been chosen to illustrate that both the force-balance and moment-balance methods are needed to obtain the governing equations. In addition, we also illustrate how the equilibrium positions are determined and how linearization of a nonlinear system is carried out.
Governing Equations of Motion The free-body diagram shown in Figure 7.5b will be used to obtain the governing equations of motion. The inertial force and the inertial moment are also shown in Figure 7.5b. Considering force balance and moment balance with respect to the center of mass
G of the rigid bar, we obtain, respectively,
m€ y + ðc
1
+c
y_ − ðc L
2Þ
1
1
−c L 2
L1 m, JG
y k1
j
c1
2Þ
y
θ_
cos
L2
z
i
(b) free-body diagram of the system along
G
with the inertial force and the inertial
k2
(a)
JG
.. .. my
G
k1 (y . c 1( y
L1 sin . ) L1 cos )
mg
(b)
k2 (y . c 2(y
Figure 7.5. (a) Rigid body in the plane
constrained by springs and dampers and
x
k
θ + ðk1 + k2 Þy − ðk1 L1 − k2 L2 Þ sin θ = − mg
L2 sin . ) L2 cos )
c2
moment.
(a)
352
Multiple Degree-of-Freedom Systems: Governing Equations and
θ − ðc L JG € 1
−
k L
ð
1
−c L −k L
1
1
2
2Þ
2Þ
2
± + c L ² θ_ ± ² θ θ+ k L + k L
y_ cos θ + c L 1
y cos
2
2
1
2
2
1
2
2
1
sin
2
θ
Equations (a) and (b) are nonlinear because of the sin
2
cos
2
θ
cos
θ
and cos
(b)
θ =0
terms.
Static-Equilibrium Positions The
equilibrium
positions
yo
θo
and
of
the
system
are
accelerations and velocities to zero in Eqs. (a) and (b). Thus,
³− k L −k L ð
1
1
2
2Þ
1
ð
1
2
2Þ
From the second of Eqs. (c), we
cos
+ k yo − k ± yo + k L + k
k
ð
1
1
−k L ² ´ L θ L
1
2
2Þ
2
2
sin
2
o
obtained
yo and θo
sin
cos
θo =
by
setting
the
are solutions of
− mg
(c)
θo = 0
find that
θo = 0
or
sin
kL kL
θo = yo
1 1
Making use of the second of Eqs. (d) in the
yo = −
±
Note that the equilibrium position
θo
2 1
−k L +k L 2
2
2
(d)
2 2
first of Eqs. (c), we arrive at
+k L L +L
mg k L
2
1
k k 1
1
2ð
1
2
2
²
2
(e)
2
2Þ
1
=π
/2 corresponding to cos
θo =
0 in Eqs. (d) is not
considered because it is not physically meaningful. Examining Eq. (e),
yo
represents a
sag in the position of the bar due to the weight of the bar. From the second of Eqs. (d),
θo
represents a rotation due to the combination of the unequal stiffness at each end and
the unequal mass distribution of the bar. When
k L 1
1
≠kL θo = 2
hence,
k L 1
1
=kL 2
2,
θo
=
0, but
, in the absence of gravity loading or other constant loading,
2
yo yo
≠ =
0. For 0, and
0.
Linearization and Linear System Governing “Small” Oscillations about an Equilibrium Position We now consider “ small” oscillations of the system shown in Figure 7.5 about equilibrium position (y o, θ o). In order to obtain the governing equations,
the we
substitute
yðtÞ = yo + ^yðtÞ θðtÞ = θo + ^ θð tÞ
(f)
7.2 Governing Equations 3
into Eqs. (a) and (b) and carry out Taylor-series expansions
d ðyo + ^ yÞ = €^yðtÞ; dt 2
yðtÞ = €
and cos
θ
and
2
d ðy + ^ yÞ = ^y_ ðtÞ dt o
y_ ðtÞ =
2
d € θðtÞ = dt
±θ + ^θ² = €^θ t d± θ θ_ t = dt ±θ + ^θ² ≈ θ + ^θ θ + ⋯ ±θ + ^θ² ≈ θ − ^θ θ +⋯ ð Þ;
o
2
θ = sin
sin
o
θ = cos
cos
θ
y and θ. To this end, we find that
retain the linear terms in
sin
of sin
353
o
ð Þ
o
cos
o
sin
o
o
cos
²
+ ^θ = ^θ_ t
o
ð Þ
(g)
On substituting Eqs. (g) into Eqs. (a) and (b), making use of Eqs. (c), and retaining only
^ y
the linear terms in
^ θ, we arrive at
and
m€ ^y + ðc + c Þ^ y_ − ð c L − c L Þ^θ_ cos θo + ðk + k Þ^y − ðk L − k L Þ^θ cos θo = 0 ^θ − ðc L − c L Þ^ JG € y_ cos θo + c L + c L ^θ_ cos θo − ðk L − k L Þ^y cos θo + k L + k L ^θ cos θo − sin θo = 0 1
±
2
1
1
1
2
2
1
²±
±
2
2
² 1
2
²
2
2
2
1
1
1
1
2
2
2
2
2
1
1
2
2
2
2
1
1
2
2
2
which in matrix form reads as
"
m
0
0
JG
+
#( €^y ) " +
€ ^θ
"
−k ð
k 1
L
(h)
1
1
c
1
−
cL
ð
1
1
+c
−c
+k 2
L
2Þ
cos
θo
±c L ð
2
L
2Þ
cos
θo
1
1
−c L ² +c L
L
1
2
1
2
1
1
2
2
1
2
2
1
2Þ
2
cos
2
2Þ
2
±k L −+ kk LL ²−± k L θ ð
2
−k
−c
2
2
2
cos
−
o
cos
θo 2
sin
“small”
Equation (i) represents the linear system governing
cos 2
² θ
θo
#( ^y_ )
θo
^_
#( ^y )θ ( ) ^θ
o
=
0
(i)
0
oscillations of the system
shown in Figure 7.5 about the static-equilibrium position given by Eqs. (d) and (e). The matrices appearing in Eq. (i) are symmetric. The static-equilibrium position is determined by the gravity loading and is accounted for in the expressions for loading does not appear explicitly in Eqs. (i). From the second of Eqs. (d) it is seen that if the form
"m 0
0
JG
#( €^y ) " €^θ
+c + −c L + c L "k +k + c
1
1
1
T. B. Hildebrand,
ibid.
1
1
=kL 2
1
L
2
c L
kL
2
1
1
+k L 2
2 2
=
θo
+c L +c L 2
2
y ^θ
θo.
This
#( ^y_ )
2
2
and
0, and Eq. (i) takes
^_
#( ^ ) ( θ)
1
0
,
2
1
2
2
1
1
−c
2
2
0
3
kL
yo
2
=
0 0
(j)
354
Multiple Degree-of-Freedom Systems: Governing Equations If
≠k
k L 1
1
L
"
2
and
2
θo ≈
0, then Eq. (i) takes the form
#( €^y ) "
m
0
0
JG
+c + −c L + c L " k +k + −k L + k L
€ ^ θ
c
1
1
1
2
1
1
When
cL 1
1
=cL 2
2
kL
and
1
1
=
kL 2
−c
+c L c L +c L −k L +k L k L +k L
2
1
2
2
1
2
2
2
1
2
1
L
2
#(θ ) ( )
2
1
2
y^ ^θ
2
2
1
^_
2
2
1
1
2
#( ^y_ )
2
2
1
2
0
=
(k) 0
the equations uncouple: that is, the rotation and
translation motions of the bar are independent of each other. It is noted that Eqs. (k) could have been obtained directly if it had been initially assumed
“small”
that
oscillations
about
the
static-equilibrium
considered and that the static-equilibrium position is In this case, cos
θ
would have been set to 1 and sin
“close”
θ
position
were
being
to the horizontal position.
would have been replaced by
θ
in
Eqs. (a) and (b).
4
EXAMPLE 7.4
Governing equations of a rate gyroscope
We shall obtain the governing equations of motion of a rate gyroscope, also known as a gyro-sensor.
The
physical
system,
along
with
the
vibratory
model,
is
shown
in
Figure 7.6. In the vibratory model shown in Figure 7.6b, the sensor is shown as a mass
m
with two degrees of freedom and its motion is described by the coordinates
the
horizontal
plane.
The
generalized
coordinates
are
both
located
in
reference frame. The sensor is designed to measure the rotational speed
kx
assumed constant. For modeling purposes, a spring with stiffness damper
with
a
damping
coef
ficient cx
are
used
to
constrain
motions
x
and
y
rotating
ωz,
which is
and a viscous along
the
direction. Another spring-damper combination is used to constrain motions along the direction. An external force
fx(t) in the n
1
in
a
n n
1 2
direction is imposed on the system.
n
To obtain the governing equations, force balance is considered along the
1
and
n
2
directions. Making use of the free-body diagram shown in Figure 7.6c, we obtain the following relations
max = − k x x − cxx_ + fx ma y = − k yy − cy y_ where
x
and
y
are the respective displacements along the
rotating reference frame;
x_
and
y_
4
acceleration of
the
mass
m
ax the n
vibrating
rate
gyroscope
1
and
n
2
directions in the
along
ay are the components of the absoand n directions, respectively. From
and
1
O. Degani, D. J. Seter, E. Socher, S. Kaldor and Y. Nemirovsky, micromachined
n
are the respective velocities along these directions in
the rotating reference frame. The quantities lute
(a)
with
modulated
Microelectromechanical Systems, 7(3) (1998) 329 –338.
2
“Optimal
integrative
design and noise consideration of
differential
optical
” Journal of
sensing,
7.2 Governing Equations
355
z z
(2)
(3)
(1)
(4)
y cx
n2
k
m kx cy
(5)
fx
ky
x
n1
(b)
max
kxx cxx
.
cy y
fx
m
.
may ky y
(a)
(c)
Figure 7.6. (a) Micromachined rate gyroscope, (b) vibratory model, and (c) free-body diagram
along with the inertial forces. (1) Suspended proof mass; (2) frame; (3) CMOS chip; (4) photodiodes; and (5) electronic circuitry.
Source: Figure 7.6a from O. Degani, D. J. Seter, E. Socher, S. Kaldor and Y. Nemirovsky, “Optimal design and noise consideration of micromachined vibrating rate gyroscope with modulated integrative differential optical sensing, ” Journal of Microelectromechanical Systems , 7(3) (1998) 329– 338. Copyright © 1998 IEEE. Reprinted with permission. Eqs. (A.6), (A.10), and (A.16) of Appendix A regarding a particle located in a rotating reference frame, it is found that
ax = x€ − 2ωz y_ − ωz x 2
(b)
ay = €y + 2ωz x_ − ωz y 2
In arriving at Eqs. (b), the fact that the rotation
ωz
is constant and that the correspond-
ing angular acceleration is zero has been taken into account. From Eqs. (a) and (b), we arrive at the following set of equations
± ± m €y +
² = −k x − c x_ + f ² ω x_ − ω y = −k y − c y_
m x€ − 2ωz y_ − ωz x 2
2
2
z
z
x
x
y
y
x
(c)
Equations (c) are written in the following matrix form:
M±
½
& x€ ' y €
+
C±
½
& x_ ' y_
+
G±
½
& x_ ' y_
+
& x ' &f '
K±
½
y
=
x
0
(d)
356
Multiple Degree-of-Freedom Systems: Governing Equations where the different square matrices are
M± =
½
½
K± =
The matrix [
G]
°m
"
0
!
;
m
0
kx − mωz
C± =
½
2
0
is called the
°c
x
0
#
cy
0
0
ky − mωz
! ;
½
G± =
°
−
0
m ωz
2
mωz
2
0 (e)
2
gyroscopic matrix.
5
The choice of coordinates in a rotating
reference frame leads to this matrix. The gyroscopic matrix, which is a
matrix,
6
!
leads to coupling between the motions along the
n
and
1
n
2
skew-symmetric
directions. From the
form of the stiffness matrix in Eq. (e), it is clear that the effective stiffness associated with each direction of motion is reduced by the rotation.
7.2.2
General Form of Equations for a Linear Multi-Degree-of-Freedom System Based on
the structure
of
Eqs.
(7.1b) and
(7.2b)
and
the
linear systems treated in
Examples 7.1, 7.3, and 7.4, the general form of the governing equations of motion for a linear
N
degree-of-freedom system described by the generalized coordinates
q,q 1
2
,
…,
qN , are put in the form M ±f € qg + ½½C ± + ½ G ±±fq_ g + ½ ½K ± + ½ H ±± fqg = fQg
½
(7.3)
where the different matrices and vectors in Eq. (7.3) have the following general form:
2m 66 m M =6 64 ⋮ m 2k 66 k K =6 64 ⋮ k 2h 66 h H =6 64 ⋮
11
21
½
±
11
21
11
21
±
hN 5
L. Meirovitch,
1
22
2
½
2
12
1
22
2
½
2c 66 c C =6 64 ⋮ c 2g 66 g G =6 64 ⋮
1
22
2
The gyroscopic matrix [
c
21
c
12
22
⋮
N1
cN
11
g
21
g
±
2
12
11
±
gN
2
12
22
⋮ 1
gN
2
⋯ ⋯ ⋱ ⋯ ⋯ ⋯ ⋱ ⋯
c
N
c
N
1
2
⋮ cNN g
N
g
N
1
2
⋮
3 77 77 5 3 77 77 5
(7.4a)
gNN
2
Computational Methods in Structural Dynamics,
1980, Chapter 2. 6
1
;
±
N1
½
12
;
N1
½
⋯ mN3 m ⋯ m N 777 ⋮ ⋱ ⋮ 75 mN ⋯ mNN k ⋯ kN3 k ⋯ k N 777 ⋮ ⋱ ⋮ 75 kN ⋯ kNN h ⋯ hN3 h ⋯ h N 777 ⋮ ⋱ ⋮ 75 hN ⋯ hNN m
Sijthoff and Noordhoff, Alphen aan den Rijn,
G] is called a skew-symmetric matrix since its elements gij = –gji.
7.2 Governing Equations and
8> q 9> >< q >= => > >: q⋮ >; 1
q
f g
2
8> q_ 9> >< q_ >= => > >: q_⋮ >;
8> €q 9> >< €q >= => > >: €q⋮ >;
1
q_
;
f g
N
2
€q
;
2
f g
N
M],
The inertia matrix [
;
8> Q >< Q => >: Q⋮
1
1
Qg
and f
N
2
N
9> >= >>;
357
(7.4b)
K], and the damping matrix [C ] are each an an N × 1 vector. The N × N matrices [G] and
the stiffness matrix [
N × N matrix, and the force vector {Q } is [H], which are skew-symmetric matrices, are called the gyroscopic matrix and the circulatory matrix, respectively. The N × 1 vector {q} is called the displacement vector, the N × 1 vector fq_ g is called the velocity vector, and the N × 1 vector f€qg is called the acceleration vector. 7
Linear Systems with N Inertial Elements, (N + 1) Linear Spring Elements, and (N + 1) Linear Damper Elements As a special case of linear multi-degree-of-freedom systems, we consider the system shown in Figure 7.7. This system is an extension of the two degree-of-freedom system shown in Figure 7.1. In the
figure, m i
ith inertial element whose qi(t ), which is measured from the point the i direction. The force acting on the ith inertial is the mass of the
motion is described by the generalized coordinate
o
located on the
k1
q1 , Q1 (t)
k2
fixed boundary along q2 , Q2 (t)
m1
k3
kN
...
m2 c2
c1
qN , QN (t)
kN+1
mN
c3
cN
cN+1
(a)
..
mi qi y
ki (qi – qi
o
x
.
.
ci (qi – qi
Qi , qi
1)
ki
1 (qi
mi 1)
ci
.
– qi
1 (qi
1)
.
– qi
1)
i
(b) Figure 7.7. (a) Linear system with
N
N + 1) spring elements, and (N + 1) ith inertial element shown along with the
inertial elements, (
damper elements and (b) free-body diagram of inertial force.
7
Gyroscopic forces and circulatory forces occur in rotating systems such as shafts; see Section 7.4.
358
Multiple Degree-of-Freedom Systems: Governing Equations
Qi( t).
element is represented by
Carrying out a force balance based on the free-body
diagram shown in Figure 7.7b, we obtain the following equation that governs the
ith
inertial element
mi € qi + ð ki + ki + Þ qi − ki qi− 1
1
− k i+ qi+ + 1
1
ci + ci+ Þq_ i − ciq_ i−
ð
1
1
− ci+ q_ i+ = Qi t i= … N 1
ð Þ
1
1; 2;
N
Assembling all of the
;
(7.5a)
equations given by Eqs. (7.5a) into matrix form, we obtain
Eq. (7.3) with the circulatory matrix [
H] =
0 and the gyroscopic matrix [
G] =
0 and with
the following inertia, stiffness, and damping matrices, respectively,
2m 3 ⋯ 66 m 77 6 7 M =6 ⋱ ⋮ 75 4⋮ 2 k + k⋯ −k m 66 − k k + k − k 66 −k K =6 66 66 4 ⋮ ⋮ ⋯ 2 c + c −c 66 −c c + c −c 66 −c 6 C =6 66 64 ⋮ ⋮ 0
1
0
½
0 0
2
±
N
0 1
2
2
2
3
0
½
0
3
⋱
0
0
2
2
0
2
2
3
0
0 0 0
3
0
⋮
−k N − kN − + kN − kN ⋯ ⋯
0
1
−k N k N + kN +
1
⋱
0
⋯
0
− cN− 0
− cN− cN− + cN −cN 1
(7.5b)
1
0
⋮ 0
1
1
3 77 77 77 77 75
3 77 77 77 77 75
0
3
±
0
0
1
−k N −
0
1
0
⋯ ⋯
0
3
±
0
½
0
2
− cN cN + cN+
1
The inertia matrix given by Eq. (7.5b) is a diagonal matrix, while the stiffness matrix and the damping matrix given by Eq. (7.5b) are not diagonal matrices because of the presence of the off-diagonal elements. However, the stiffness and damping matrices are
banded matrices,
with each banded matrix having non-zero elements along three diago-
nals. All the other elements of these matrices are zero.
Conservation of Linear Momentum and Angular Momentum In the absence of external forces, that is,
Q
1
= Q =⋯ = QN = 2
0
(7.6a)
the system linear momentum is conserved. This means that
m q_ 1
1
+ m q_ + ⋯ + mN q_ N = 2
2
constant
(7.6b)
7.2 Governing Equations
359
Equation (7.6b) is obtained directly by making use of Eq. (A.21) of Appendix A and noting Eq. (7.6a). For systems that experience rotational motions, Eq. (A.25a) of Appendix A can be used to examine whether a system’s angular momentum is conserved. For example, in the two degree-of-freedom system shown in Figure 7.3, even in the absence of the exter-
Mo
nal moment, that is,
=
Jo φ_
0, the angular momentum
1
1
+ Jo φ_ ≠
constant along the
N
is the number of
2
2
k direction. This is to be expected because of the fixed boundary at the left end.
7.2.3
Lagrange’s Equations of Motion To use Lagrange’s equations, we start from Eqs. (3.41), where
N
degrees of freedom and the the motion are
q
1,
q
2,
…, qN .
tions is given by
¶
d ∂T dt ∂q_ j where
T
independent generalized coordinates used for describing Repeating Eqs. (3.41), the system of
− ∂T + ∂D + ∂V = Qj ∂qj
∂ q_ j
∂ qj
is the kinetic energy of the system,
the Rayleigh dissipation function, and
Qj
V
for
N
governing equa-
j = 1; 2; …; N
(7.7)
is the potential energy of the system,
is the generalized force acting on the
jth
D
is
iner-
tial element given by Eq (3.42).
Linear Vibratory Systems For vibratory systems with linear inertial characteristics, linear stiffness characteristics, and
linear
viscous
damping
Eqs. (3.43); that is, we have
8
The quadratic forms shown for However, the kinetic energy
T
characteristics,
T, V,
and
D
take
the
form
given
by
8
T=
1
V=
1
D=
1
T, V,
2
2
2
q_
T
q
T
q_
T
f g
f g
f g
and
M ±fq_ g =
½
2
K ±fqg =
1
C ± fq_ g =
1
½
½
D in
1
2
2
XX N
N
j =1 n =1
mjn q_ j q_ n
XX N
N
j= 1 n= 1
kjnqj qn
(7.8)
XX N
N
j =1 n =1
cjn q_ j q_ n
Eqs. (7.8) are strictly valid for systems with linear characteristics.
is not always a function of only velocities as shown here. Systems in which the
kinetic energy has the quadratic form shown in Eqs. (7.8) are called
natural systems.
As discussed later in this
chapter, for systems such as that given in Example 7.4, where rotation effects are present, the form of the kinetic energy
T
is different from that shown in Eqs. (7.8). As noted in Chapter 2, gravity forces can also contribute to
the potential energy of the system. For a general system with holonomic constraints, Eqs. (7.7) will be used directly in this book to obtain the governing equations.
360
Multiple Degree-of-Freedom Systems: Governing Equations For symmetric [
M], [K], and [C] matrices, we have ∂2T ∂2T = ∂ q_ j ∂ q_ k ∂ q_ k ∂q_ j ∂2 V ∂2V = ∂ qj ∂ qk ∂ qk ∂qj
(7.9a)
∂2D ∂2D = ∂ q_ j ∂ q_ k ∂ q_ k ∂q_ j or equivalently
9
mjk = mkj kjk = kkj cjk = ckj
(7.9b)
In light of Eqs. (7.9b), for a two degree-of-freedom system with linear characteristics, Eq. (7.8) is written in expanded form as
T=
1
V=
1
D=
1
2
2
2
XX 2
2
j =1 n =1
m jnq_ j q_ n =
XX 2
2
j =1 n =1
k jnqj qn =
1
m q_
1
k q
2
2
1
11
2
XX N
N
j =1 n =1
2
11
cjn q_ j q_ ncjn q_ j q_ n =
1
1 2
+m
+k
12
c q_ 11
12
q_ q_ 1
2
qq 1
2
1
2
+c
+
12
+
1
1
22
k q
2
q_ q_ 1
m q_
2
2
2
22
2
+
1 2
2
2
(7.10)
c q_ 22
2
2
On substituting Eqs. (7.10) into Eqs. (7.7) and performing the indicated operations, we obtain the following system of two coupled ordinary differential equations
m € q m € q 11
1
12
1
+m +m
12
22
€q €q
2
2
+c +c
11
12
q_ q_
1
1
+c +c
12
22
q_ q_
2
2
+k +k
11
12
q q
1
1
+k +k
12
22
q q
2
2
=Q =Q
1
(7.11)
2
which are written in the matrix form
M ±f€qg + ½ C ±fq_ g + ½K ±fqg = fQ g
½
(7.12)
M] is the inertia matrix, the square matrix [K ] is the stiffness matrix, and square matrix [C] is the damping matrix. They are given by, respectively,
where the square matrix [
½
9
M± =
L. Meirovitch,
°m
11
m
12
m m
ibid., pp. 260–261.
12 22
!
;
K± =
½
°k
11
k
12
k k
12 22
!
;
and
C± =
½
°c
11
c
12
c c
12 22
!
(7.13)
7.2 Governing Equations
q
q_ Q} are given by the column vectors
361
€q
The displacement vector { }, the velocity vector { }, the acceleration vector { }, and the vector of generalized forces {
q €
f g
=
& €q ' 1
q_
;
€ q
f g
=
& q_ ' 1
q_
2
q
;
f g
=
&q ' 1
q
;
Qg =
and
f
& Q' 1
(7.14)
Q
2
2
2
Illustration of Derivation of Governing Equations for a Two Degree-of-Freedom System Consider
again the
two
degree-of-freedom
system
shown
in
Figure
7.1,
with
linear
springs and linear dampers. The kinetic energy and potential energy are given by
T=
1
V=
1
=
1
the dissipation function
2
2
2
m x_ 1
kx 1
k
ð
1
1
= =f
1
and
Q
2
=f
+
1 2 1
m x_
2
2
2
−x +
1
−k x x +
1
−x_ +
1
k ðx 2
2
x
2Þ
2 1
2
1
2Þ
2
1
2
k x
2
3
2
k
ð
2
(7.15)
2
+k
x
3Þ
2
2 2
D takes the form 1
2
1
+
+k
1
Q
1
2
D = c x_
and
2
1 2
ð
2 1
+
1 2
+c
c
c ðx_ 2
x_
2
2Þ
1
1
2
1
2Þ
− c x_ x_ + 2
1
2 1
2
2
c x_
2
3
2
c
ð
2
+c
x_
3Þ
2
(7.16)
2
.
2
fi
Comparing Eqs. (7.15) and (7.16) with Eqs. (7.10), the corresponding inertia coef cients
mjk,
fi
k jk,
stiffness coef cients
and damping coef
ficients cjk
are identi
fied.
Upon
using Eqs. (7.4a), this leads to the governing system
½
M ±f€xg + ½C ±fx_ g + ½K ±fxg = fF g
where the different square matrices are
M± =
½
½
C± =
°m
°
!
; ½K ± = m c +c −c −c c + c 0
1
0
!
2
1
2
2
2
2
°k
& x€ ' 1
x€
2
;
_g= fx
& x_ ' 1
x_
2
−k k +k
2
!
2
2
2
;
and
3
(7.18)
3
and the different column vectors are f€ xg =
+k −k
1
(7.17)
;
xg =
f
&x ' x
1 2
;
and
Fg =
f
&f ' 1
f
2
(7.19)
362
Multiple Degree-of-Freedom Systems: Governing Equations In certain situations, it is necessary to use Eqs. (7.7) directly. To illustrate this procedure, we evaluate the following derivatives based on Eqs. (7.15) and (7.16):
¶
d ∂T dt ∂ x_
¶
d ∂T dt ∂ x_
= m x€
1;
1
1
10
= m x€ 2
2
2
∂T = 0; ∂x1
∂T ∂ x2
=
0
∂V = k1x1 + k2ðx1 − x2 Þ = ðk 1 + k2 Þx1 − k 2x2 ∂x1 ∂V = − k2 ðx1 − x2Þ + k 3x2 = ðk 2 + k 3Þx2 − k 2x1 ∂x2
(7.20)
∂D = c1x_ 1 + c2 ðx_ 1 − x_ 2Þ = ð c1 + c2Þ x_ 1 − c2x_ 2 ∂x_ 1 ∂D = − c2ðx_ 1 − x_ 2 Þ + c3x_ 2 = ðc2 + c3 Þx_ 2 − c2x_ 1 ∂x_ 2 On substituting Eqs. (7.20) into Eqs. (7.7), that is,
¶
d ∂T dt ∂ x_
−
d ∂T dt ∂ x_
∂T − ∂ x2
¶ 1
∂T ∂ x1
+
∂D ∂ x_ 1
+
∂D ∂ x_ 2
+
∂V = Q1 ∂x1
+
∂V = Q2 ∂x2
(7.21a)
2
we obtain
m x€ 1
1
m x€ 2
2
+ +
ð
c
ð
c
1
2
+c +c
x_
2Þ
1
x_
3Þ
2
− c x_ + − c x_ +
+k +k
k
2
2
ð
2
1
ð
x
1
x
2
2Þ
1
k
3Þ
2
−k x =f −k x =f 2
1ð
2
1
2ð
which yields the same matrix form as Eqs. (7.1b).
10
It is mentioned that if
V in Eq. (7.15) is expressed as V
¼
1 2
k x
2
1
1
þ
1 2
k ðx 2
2
−x
2
1Þ
þ
1 2
k x
2
3
2
then
∂V ∂x1
¼
kx
∂V ∂x2
¼
k ðx
1
2
which are the same as given by Eq. (7.20).
1
−k
2
2ð
−x
x
1Þ
2
−x
þ
1Þ
kx 3
2
k
¼ ð
¼ ð
1
k
2
þ
k Þx
þ
2
1
k Þx 3
−k
2
2
tÞ
2
x
2
−k x 2
1
tÞ
(7.21b)
7.2 Governing Equations In
light
of
Eqs.
(7.9a)
and
(7.9b),
the
equations
of
motion
obtained
by
363
using
Lagrange’s equations for linear systems always have symmetric inertia and stiffness matrices. For systems in which the dissipation is modeled by the Rayleigh dissipation function, the damping matrix is also symmetric. These symmetry properties are not necessarily explicit when the governing equations are obtained by using force-balance and moment-balance methods. However, the form of the governing equations obtained by using force-balance and moment-balance methods can be manipulated to obtain the same form as that obtained through Lagrange’s equations. Next, different examples are provided to illustrate the use of Lagrange’s equations for deriving
the
equations
of
motion
of
multi-degree-of-freedom
systems.
When
using
Lagrange’s equations, in cases where the expressions for kinetic energy, potential energy, and dissipation function are in the form of Eqs. (7.8), these equations are directly used
fi
to identify the coef cients in the inertia, stiffness, and damping matrices in the governing equations.
EXAMPLE 7.5
System with a translating mass attached to an oscillating disc
Consider the system shown in Figure 7.8 with linear springs and linear dampers and a rotating element. A massless rigid bar connects the rotating element to the base of the combination of the spring
k
2
and the damper
c
. The inertial element
2
m
1
is treated as a
translating mass and the other inertial element is treated as a rigid body rotating about the point
o.
The position
x
of mass
m
and the other generalized coordinate
is the displacement measured from the
1
θ
fixed
end
is the angular position of the disc measured from
the vertical. We shall use Lagrange’s equations to determine the governing equations of motion of the system. The kinetic energy of the system is given by
T=
1
m x_
2
|fflfflffl{zfflfflffl} 2
1
+
Kinetic energy of mass
m
1
1
Joθ_
2
|fflffl{zfflffl} 2
(a)
Kinetic energy of rigid body
Jo
We assume that the mass center of the disc coincides with the point angular
oscillations
in
y
are
“small.”
The
“small”
angle
x, f(t)
i
A
m1
x
k
c1
assumption
, Mo (t)
k2
k1
j
z
θ
c2
o
o
r
m2, Jo Figure 7.8. Translating and rotating system with two degrees of freedom.
and that the allows
us
to
364
Multiple Degree-of-Freedom Systems: Governing Equations
A
express the horizontal translation of point
as being equal to
rθ.
The potential energy
is then given by
V=
1
+
k x
2
|fflffl{zfflffl} 2
1
1
k ðx+ rθÞ
2
|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}
2
2
Potential energy
Potential energy
associated with
associated with
k
spring
=
1
=
1
kx
2
1
2
k
ð
2
1
+
1
k rθ 2
2
+k
2Þ
x
2
1
2
|fflffl{zfflffl} 2
1
2
2
Dissipation
=
1
=
1
2
2
c x_
2
1
+
2
k x
2
2
kr θ 2
2
(b)
2
2
2
Dissipation associated with
c
damper
1 2
c r θ_
x_
2Þ
1
2
c x_ +rθ_
1
+c
c
ð
c
1
1
± ² |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}
1
associated with damper
2
+ k r θx +
+
c x_
2
+ k rθx +
2
2
The dissipation function takes the form
D=
k
spring
1
2
2
2
2
2
+ c r θ_ x_ + 2
+c
2
rθ_ x_ +
1 2
1 2
c x_
c r θ_ 2
2
2
2
(c)
2
Since Eqs. (a) through (c) are in the standard form of Eqs. (7.8), we identify the inertial, stiffness,
and
damping
coef
x with q
ficients
by
comparing
Eqs.
(a)
through
(c)
to
Eqs.
(7.10)
q . In addition, we make use of Eq. (3.42) to recognize the generalized forces as Q = f(t) and Q = Mo(t). Then, the governing equations become and associating
°m
0
0
Jo
1
and
!& €x ' ° c
θ
with
2
1
€ θ
+
!& x_ ' ° k 2
1
+c
2
rc
2
rc rc
2
2
2
θ_
+
1
+k
rk
2
rk rk
2
!& x ' & θ
2
2
2
=
f ðtÞ Mo ðtÞ
'
(d)
Since the damping and stiffness matrices have non-zero off-diagonal terms, Eqs. (d) are coupled.
EXAMPLE 7.6
System with bounce and pitch motions revisited
Consider the rigid bar shown in Figure 7.5a. The generalized coordinates are which
are
located
at
the
center
of
gravity
of
the
beam.
We
shall
use
y
and
θ,
Lagrange’s
equations to obtain the governing equations of motion of this two degree-of-freedom system.
7.2 Governing Equations For a rigid bar undergoing planar motions, we use Eq. (A.31) of Appendix A to
365
find
that the kinetic energy is given by
T=
1
+
m y_
2
|ffl{zffl} 2
1
J G θ_
2
|fflffl{zfflffl} 2
(a)
Kinetic energy
Kinetic energy
associated with
associated with
translation of center
rotation about
m
of mass
center of mass
Taking into account the potential energy due to the spring displacement and the work done by the gravity loading, the potential energy is
V=
1
k ðy −L
+
2
θÞ
|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} 2
1
sin
1
Potential energy associated
1 2
+k
k
ð
y
2Þ
1
2
k ðy + L
2
sin
k
with spring
ð
kL 1
mgy |ffl{zffl} Potential energy
Potential energy associated
1
−
+
2
θÞ
|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}
2
2
k
with spring
=
1
1
−k
2
2
L Þy sin θ +
±k L
1
2
associated with
1
2
²
gravity loading
+k
2 1
2
L
2 2
θ + mgy
2
sin
(b)
The dissipation function is of the form
± ² + c ±y_ +L θ_ θ² |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
1 D = c y_ − L θ_ cos θ 1
2
1
2
1
2
Dissipation associated with damper
=
1 2
c
ð
1
+c
y_
2Þ
2
c
−
2
Dissipation associated with damper
1
cL
ð
1
2
cos
2
1
−c L 2
c
2
θ_ y_ cos θ +
2Þ
1 2
±c L 1
2
1
+c
2
L
2
2
²θ_
2
2
cos
θ
(c)
Since Eqs. (b) and (c) are not in the standard form of Eqs. (7.8), the equations of motion are obtained directly by using Eqs. (7.7). Thus, recognizing that
q
1
=y
, the
first
equation of Eqs. (7.7) takes the form
· ¸
d ∂T dt ∂y_ Noting that
Qy
=
−
∂T ∂y
+
∂D ∂ y_
+
∂V ∂y
= Qy
(d)
0 and using Eqs. (a) through (d), we obtain for the
first
equation of
motion
m€ y + ðc
1
+c
2Þ
Recognizing that
y_ − ðc L 1
q
2
=θ
1
−c
2
L Þθ_ cos θ + ðk 2
1
+k
y − ðk L
2Þ
1
1
−k
2
L
2 Þsin
θ = − mg
(e)
, the second equation of Eqs. (7.7) assumes the form
· ¸
d ∂T dt ∂θ_
−
∂T ∂θ
+
∂D ∂ θ_
+
∂V ∂θ
= Qθ
(f)
366
Multiple Degree-of-Freedom Systems: Governing Equations Recognizing that
Qθ
=
0 and using Eqs. (a), (b), (c), and (f), we obtain for the second
equation of motion
± ±
² ²
JG € θ − ðc L − c L Þy_ cos θ + c L + c L θ_ cos θ − ðk L − k L Þy cos θ + k L + k L sin θ cos θ = 0 1
1
1
2
1
2
2
2
2
1
1
2
2
1 2
2
1
2
2 2
(g)
2
Equations (e) and (g) are identical to Eqs. (a) and (b) of Example 7.3, which were obtained by using force-balance and moment-balance methods.
EXAMPLE 7.7
Consider
the
generalized coordinate
Pendulum absorber
two
degree-of-freedom
coordinate
θ
x
is
used
to
system locate
shown
the
mass
in
m
Figure 1
and
7.9,
the
in
other
which
the
generalized
is used to specify the angular position of the pendulum. This type of system 11
can model a pendulum absorber, rod of length
L
which is used in many applications. The mass of the
is assumed to be negligible. In this example, we obtain the nonlinear
equations of motion and then linearize the governing nonlinear equations about an equilibrium position of the system.
Governing Equations We see from Figure 7.9 that the position of mass
m
2
is
rm = ðx + L sin θÞi − L cos θj m
Therefore, the velocity of mass
Vm =
2
drm dt
y
±
= x_ + Lθ_
x
o
k
cos
²
θ i + Lθ_
sin
θj
(a)
Figure 7.9. Pendulum absorber.
x
j k
is given by
m1
f(t)
i
z
L m2
11
J. J. Hollkamp, R. L. Bagley and R. W. Gordon,
” Journal of Sound and Vibration,
blades,
“A
centrifugal pendulum absorber for rotating, hollow engine
–
“ Regular and chaotic ” Journal of Sound and
219(3) (1999) 539 549; Z.-M. Ge and T.-N. Lin,
dynamic analysis and control of chaos of an elliptical pendulum on a vibrating basement,
Vibration,
– “Performance of pendulum ” Journal of Sound and Vibration, 229(4) (2000) 913–933.
230(5) (2000) 1045 1068; and A. Ertas et al.,
tem of varying orientation,
absorber for a nonlinear sys-
7.2 Governing Equations
367
Making use of Eq. (A.29) of Appendix A and Eq. (a), the kinetic energy for the system is
T=
1
+
m x_
2
|fflfflffl{zfflffl} 2
1
1 2
Kinetic energy of mass
=
1
=
1
m x_
2
1
2
ð
2
m
1
m
2
Kinetic energy of
h± m x_ +Lθ_
m
² ± θ + Lθ_
pendulum of mass
1
+
m ðV m ⋅ V mÞ
|fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}
1
2
2
+m
x_
2Þ
2
+m L 2
cos
cos
2
2
θx_ θ_ +
1 2
θ
sin
m L θ_ 2
²i 2
2
(b)
2
The potential energy of the system is
+ m gL − θ |ffl{zffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
V=
1 2
kx
2
the
datum
has
been
chosen
(c)
gravity loading
k
at
Þ
associated with
associated with
where
cos
Potential energy
Potential energy
spring
ð1
2
the
bottom
position
of
the
pendulum.
Since
Eqs. (b) and (c) are not in the form of Eqs. (7.8), we use Eqs. (7.7) directly to obtain the
q
equations of motion. On using Eqs. (7.7) with the generalized forces are
Q
1
=ft
¶
( ) and
Q
d ∂T dt ∂x_
2
−
=
∂T ∂x
+
= x, q = θ, and recognizing D = 0, Eqs. (7.7) become
1
0 and
2
∂V = f ðtÞ ∂x
0 1 d @∂ T A ∂T ∂ V − + = _
dt
∂θ
∂θ
that
∂θ
(d) 0
Upon substituting Eqs. (b) and (c) into Eq. (d), we obtain
m
ð
1
+m
2Þ
€x + m L€θ cos θ − m Lθ_ sin θ + kx = f ðtÞ m L θ€ + m L€x cos θ − m gL sin θ = 0 2
2
2
(e)
2
2
2
2
Static-Equilibrium Positions Setting the accelerations and velocities and the time-dependent forcing Eqs. (e), the equations governing the equilibrium positions
xo
and
θo
f( t)
to zero in
of the system are
kxo = 0
m gL sin θo = 0
(f)
2
From Eqs. (f), the equilibrium positions of the system are obtained as
xo; θo Þ = ð0; 0Þ
ð
and
xo; θoÞ = ð0; π Þ
ð
(g)
368
Multiple Degree-of-Freedom Systems: Governing Equations where the
first
of Eqs. (g) corresponds to the bottom position of the pendulum and the
°
second of Eqs. (g) corresponds to the pendulum being rotated by 180 .
Linearization “small”
Considering
oscillations about the equilibrium position (0,0) and linearizing the
equations of motion given by Eqs. (e) along the lines of what was illustrated with the use of Eqs. (g) of Example 7.3, we obtain
m
ð
+ m €x + m L€θ + kx = f L €θ + m L€x + m gLθ = 2Þ
1
m
2
2
which in matrix form reads as
°m
1
+m
mL
2
2
2
2
m L m L 2
€θ
2
2
+
0
(h)
0
2
!& €x ' ° k
t
ð Þ
0
m gL
!& x ' & f t '
2
θ
=
ð Þ
(i)
0
In Eq. (i), the gravity loading appears explicitly in the equations governing
“small” oscil-
lations about the static-equilibrium position. In this case, the governing equations are coupled due to the non-zero off-diagonal terms in the inertia matrix.
EXAMPLE 7.8
12
Bell and clapper
Consider the bell and clapper shown in Figure 7.10. We shall derive the governing equations of motion for this system by assuming that the bell and clapper do not come into contact. From the
figure, it is seen that the bell rotates about the fixed point O
the clapper rotates about the point inertia of
J
1
O
2
1
and
that moves with the bell. The bell has a rotational
about its centroid, and the clapper has a rotational inertia
J
2
about its
Figure 7.10. Bell and clapper.
d2 O2 a
O1 d1
x b1
m1
b2 m2
i
2 1
j
Bell
y
12
H. M. Hansen and P. F. Chenea,
Mechanics of Vibrations, John Wiley & Sons, New York, 1953, pp. 133–137.
7.2 Governing Equations
b
centroid. The locations of the centroids
r
given by the positions
r r
1
2
=d = −a
sin
1
ð
b
2
of the clapper from
O
are
1
r
and
1
of the bell and
1
369
, respectively, as 2
θ1i + d1
θ1 + d2
sin
θ1 j
cos
sin
θ2Þi + ð − a cos θ1 + d2
sin
θ1 j
cos
(a)
θ2Þ j
The corresponding velocities are
v
1
v
2
= d θ_ ± = −aθ_ 1
cos
1
1
θ1 i − d1θ_ 1
cos
θ1 + d2θ_ 2
cos
² ±
θ2 i + aθ_ 1
θ1 − d2θ_ 2
sin
sin
²
(b)
θ2 j
The kinetic energy of the system is
T=
1
=
1
2
2
m ðv ⋅ v 1
m
+ =
1
1Þ
1
¹±
1
d θ_
m
2
1
1
2
1 2
cos
¹±
− aθ_
2
±m d
1
1
+
2
1
1
m ðv 2
θ1
2
2Þ
² + ±d θ_ 2
1
+J +m a
2
2
1
²θ_
2 1
+
1 2
J θ_
sin
θ1 + d2θ_ 2
cos
1
⋅v +
1
θ1
cos
2 1
2
J θ_ 2
1
2
θ2
2
2
1
² º+
±m d
1
+
2
2 2
J θ_ 1
² + ±aθ_ 2
2 2
+J
2
²θ_
2
1
sin
1
2
+
2
1 2
J θ_ 2
2 2
θ1 −d2 θ_ 2
−m
2
ad θ_ θ_ 2
1
2
sin
1
If
we
assume
that
there
is
1
cos
θ1 − m 2gðd 2
viscous
damping
cos
at
²º 2
θ1 − θ2Þ
cos ð
The potential energy of the system with respect to the datum at point
V = − m gd
θ2
O
1
(c)
is
θ2 − a cos θ1Þ
O
O
and
1
,
2
(d)
then
the
dissipation
function is
D= ct
where
1
ct
and
2
1 2
ct θ_ 1
2 1
+
1 2
If we choose the generalized coordinates and
Q
2
=M
t)
2(
=
2
are the torsion damping coef
respectively.
q
1
±
ct θ_
1
− θ_
ficients
=θ
1
²
2
2
(e)
for the bell and clapper motions,
and
q
2
=θ
, assume that
2
Q
1
=M
1
t
( )
0, then the Lagrange equations become
0 1 d @ ∂T A ∂T ∂D ∂V − + _ + =M dt ∂θ_ ∂θ ∂ θ ∂θ 0 1 d @ ∂T A ∂T ∂D ∂V − + _ + = _ dt ∂θ ∂θ 1
∂θ2
1
2
1
∂θ2
1ð
tÞ
1
0
2
(f)
370
Multiple Degree-of-Freedom Systems: Governing Equations Upon substituting Eqs. (c), (d), and (e) into Eq. (f) and performing the indicated operations, we obtain
±m d 1
±m d 2
+ J +m
2
1
1
² + J €θ
2
2
2
2
²
a € θ 2
2
1
− m ad €θ + ct + ct 2
ð
2
θ1 − θ2 Þ − m2 ad 2θ_ 2 2
cos ð
2
θ_ 1 − ct2θ_ 2 + ðm1 d1 − m 2aÞg
2Þ
1
− m ad €θ θ − θ + m ad θ_ − ct θ_ + ct θ_ + m gd θ = 2
2
2
cosð
1
1
2
2Þ
1
2
2
2
sin
2
2
2
2 1
sinð
θ1 − θ2 Þ
sin ð
sin
θ1 − θ2Þ
θ1 = M1ðtÞ
(g)
0
To obtain the equations governing “small” oscillations about the equilibrium position
θj
=
θj
0, we assume that sin
≈ θj
and cos
θj
≈
1. Then Eqs. (g) lead to the following
coupled linear equations
±m d 1
2
1
+J + m a
2
1
2
²€θ
− m ad €θ + ct + ct θ_ − ct θ_ + m d − m a gθ = M ± ² − m ad €θ + m d + J €θ − ct θ_ + ct θ_ + m gd θ = 1
2
2
2
ð
2Þ
1
1
2
ð
2
1
1
2
2
Þ
2
1
2
2
2
1
2
2
2
2
2
1
2
2
2
tÞ
1ð
(h)
0
13
EXAMPLE 7.9
Three coupled nonlinear oscillators: Lavrov’s device
We shall derive the governing equations of the device shown in Figure 7.11. This device consists of a rigid bar of mass the
xy
m o that is supported on soft springs k
that can translate in
plane and rotate about its center of mass with angular displacement
φ.
The
J. Two pendulums, which are of m , are attached to the bar a distance a on
rotary inertia of the bar about its center of mass is
L
different lengths
1
and
L
and the same mass
2
either side of the bar’s center of mass. The positions of the pendulums are given by
r r
1
2
= x+L = x+L ð
1
sin
ð
2
sin
φ1 − a cos φÞ i + ðy − L1
φ2 + a cos φÞ i + ðy − L2
cos cos
φ1 − a sin φÞj
φ2 + a sin φÞj
(a)
j
y k
a
mo , J
k
a
O L1 1
2
L2
x
i
m
m Figure 7.11.
13
Lavrov’s device.
Regular and Chaotic Oscillations , Springer, Berlin, 2001, pp. 67–71. The case where the mass does N pendulums has been studied by A. Vyas and A. K. Bajaj, “Dynamics of autoparametric vibration absorbers using multiple pendulums”, Journal of Sound and Vibration , 246(1) (2001) 115– 135.
P. S. Landa,
not rotate and there are
7.2 Governing Equations where and
r
2
r
1
371
O to the mass of the pendulum of length L O to the mass of the pendulum of length L .
is the position vector from point
is the position vector from point
1
2
Hence, the velocity of each mass is
v v
1
2
= x_ + L φ_ = x_ + L φ_ ð
1
1
cos
ð
2
2
cos
φ1 + aφ _ sin φÞi + ðy_ + L1φ_ 1
sin
φ2 − aφ _ sin φÞi + ðy_ + L2φ_ 2
sin
φ1 − aφ _ cos φÞj
φ2 + aφ _ cos φÞj
(b)
The system kinetic energy is
T=
1
=
1
=
2
2
m ox_
±
m o x_
+
1
+
1
1 2
2
2
+
2
1 2
2
²
+ y_ +
2
2
h
1 2
m ðx_ +L φ_
1
m ðx_ +L φ_
2
1
h
2
±
m T x_
2
2
1
ð
²
+ y_ +
+ mx_ L φ_ + maφ_ L φ_ ð
+
m oy_
1
1
cos 1
1 2
J φ_
2
+
1 2
mð v ⋅ v
1Þ
1
+
1 2
mðv
2
⋅v
2Þ
J φ_
2
cos
φ1 +aφ_ sin φÞ
+ y_ +L φ_
1
cos
φ2 −aφ_ sin φÞ
+ y_ +L φ_
2
1 2
2
2
JT φ_
2
+
φ1 + L2φ_ 2
1 2
±
m L φ_
cos
ð
1
ð
2
2
1
1
2
+ L φ_ 2
2
2
2
φ − φ1Þ − L2φ_ 1
φ1 − aφ_ cos φÞ
sin
φ2 + aφ_ cos φÞ
2
2
i i
²
φ2 Þ + my_ ðL1φ_ 1
sinð
sin
sin
φ − φ2ÞÞ
φ1 + L2φ_ 2
sin
φ2 Þ
sinð
(c)
where
mT = m o + 2m JT = J + 2ma
2
The system potential energy is
V = − mgL
1
cos
φ1 − mgL1
cos
φ2 +
¶
1 2
k ðy +a sin φÞ
2
+
1 2
k ðy −a sin φÞ
2
+ mT gy
(d)
The Lagrange equations, Eqs. (7.7), for this undamped and unforced system become
d ∂T dt ∂x_
−
∂T ∂x
+
∂V = Qx = 0 ∂x
d ∂T dt ∂ y_
−
∂T ∂y
+
∂V = Qy = 0 ∂y
d ∂T dt ∂ φ_
−
∂T ∂φ
+
∂V = Qφ = 0 ∂φ
¶
¶
¶
d ∂T dt ∂ φ_
−
∂T ∂ φ1
+
∂V = Qφ ∂ φ1
=
0
d ∂T dt ∂ φ_
−
∂T ∂ φ2
+
∂V = Qφ ∂ φ2
=
0
¶ 1
2
1
2
(e)
372
Multiple Degree-of-Freedom Systems: Governing Equations since
=
D
0. Upon substituting Eqs. (c) and (d) into Eqs. (e) and performing the indi-
cated operations, we obtain
±
mT € x + mL φ €
mT
±€ y + mL φ € 1
1
φ1 + φ_ 1
±€ φ € + maL φ
sin
1
1
JT
2
² ±€ φ + mL φ
cos
1
1
φ1 − φ_ 21
cos
2
1
φ1
sin
± − maL φ€ ±€ φ + maL φ 2
€ mL φ 1
+ mL
mL φ €
+ mL
2 2
€ x cos φ
1ð
1
x cos φ €
2ð
2
+ €y
1
+ €y
2
sin
sin
1Þ
φ2 Þ
2
1
± − maL φ€ 2
2
φ2 − φ_ 22
² φ +
cos
sin
2 2
cos
1Þ
sinð
2
sinð
2
2
2
m=
²=
2
Þ
2
2
1Þ
2
sin
cosð
2
cosð
sinð
2
Þ
0
Þ
0
(f)
1Þ
1
2
0
cos
cosð
sin
2
2
When
φ2
+ φ_ ky + mT g = ² φ− φ + ak φ φ ²= φ − φ − φ_ φ −φ ² φ − φ + φ_ φ −φ + mgL φ = ² φ − φ + φ_ φ −φ + mgL φ =
φ2
cosð
2
2
2
sin
2
φ − φ1 Þ − φ _1
sinð
² + mL ±φ€
sin
1
0
Þ
2
0
0, we have the case of bounce and pitch motions discussed in Example 7.3,
except that we have removed the restriction that the center of mass of the bar must only move in the vertical direction.
EXAMPLE 7.10
Governing equations of a rate gyroscope revisited
We revisit the gyro-sensor presented in Example 7.4 and obtain the equations of motion by using Lagrange’s equations. First, we construct the kinetic energy
T,
as follows.
From Eq. (A.16) of Appendix A, we have that
v = ðx_ − ωz yÞe
1
where
e
1
and
e
2
are the unit vectors
fixed
+ y_ + ωz x e ð
Þ
2
in the rotating plane. Therefore, the kinetic
energy is
T=
1
=
1
The potential energy
Due to the rotation
2
2
mðv ⋅ vÞ =
±
m x_
2
L. Meirovitch, 1980,
ωz ,
ibid.
2
mðx_ −ωz yÞ
2
²
+
1 2
mðy_ +ωz xÞ
2
+ y_ + mωz −yx_ + xy_ + 2
ð
Þ
1 2
±
mωz y 2
2
+x
2
²
(a)
V and dissipation function D are, respectively, V=
1
D=
1
2
2
kx x cx x_
2
2
+
1
+
1
2
2
kyy
2
(b)
cy y_
2
the kinetic energy is not in the standard form shown in Eq. (7.8)
and we have an example of a
14
1
nonnatural system.
14
7.2 Governing Equations
373
After using Eqs. (a), Eqs. (b), and Eqs. (7.7), and recognizing that the generalized coordinates are
q
1
=x
and
q
1
=y
and that the associated generalized forces are
Q
1
= fx
and
Q
2
=
0
(c)
we carry out the indicated operations in Eqs. (7.7) to obtain equations that are identical to those given by Eqs. (c) and (d) in Example 7.4.
15
EXAMPLE 7.11
Governing equations of hand-arm vibrations
In Chapter 3, we considered the development of the governing equation of motion of a hand-arm system modeled as a rigid body with one degree of freedom. In this example, a multi-degree-of-freedom system model is used to study hand-arm vibrations. The model shown in Figure 7.12 has are
x ,x ,x ,y 1
2
3,
3
θ3. The angle ~ θ shows the nominal position of the system about which _ the vibrations. It is assumed that ~ θ = 0 throughout this discussion. The
and
we wish to study
x,x
coordinates
1
five degrees of freedom and the associated generalized coordinates
, and
2
x
3
describe the longitudinal motions of the hand, the forearm, and
the elbow, respectively. The coordinate
θ3
coordinate
y
3
describes the vertical motion of the elbow and the
describes the rotation of the elbow joint about the nominal position. In
xo represents the external disturbance acting on the hand. m , the forearm is treated as a mass of mass m , and rotary inertia J g. In and the upper arm is treated as a rigid body of mass m addition, translation springs with stiffness of k through k and a torsion spring of stiffness k t are used to represent system fl exibility. Similarly, translation dampers with
Figure 7.12,
The hand is treated as a mass of mass
1
2
3
1
3
5
3
k5
y j
xo
k1
c1
i
Figure 7.12.
A
k2
m1
x
o
x1
x2
m3, J3g k3
y3
m2 c2
y4
k4 x4
~
3
x3 Lg c3
c5
c4
L
kt3 , c t3
five degree-of-freedom model used to study planar vibrations of the human ~θ: fi
hand-arm system. The nominal position of the upper arm is speci ed by
Source: C. Thomas, S. Rakheja, R. B. Bhat and I. Stiharu, “A study of the model behaviour of the human hand-arm system, ” Journal of Sound and Vibration , 191 (1996) 171– 176. Reprinted by permission of Federation of European Biochemical Societies. Copyright
©
1996, with permission
from Elsevier Science.
15
“A study of the model behaviour of the human hand-arm sys” Journal of Sound and Vibration , 191(1) (1996) 171–176.
C. Thomas, S. Rakheja, R. B. Bhat and I. Stiharu, tem,
374
Multiple Degree-of-Freedom Systems: Governing Equations damping coef
ficients
of
c
c
through
1
5
fi
and a torsion damper with damping coef cient
ct
3
are also used.
first note that the position vector of the center
To construct the different quantities, we
m
of mass of the rigid mass
from the origin
3
±
rg = x
3
+ Lg
cos
o
is given by
±~θ + θ ²²i + ±y 3
3
+ Lg
±~θ + θ ²²j
sin
and, therefore, the velocity of the center of mass is
±
V g = x_
3
− Lgθ_
x
In addition, the displacements
±~θ + θ ²²i + ±y_
sin
3
4
3
and
x
y
4
3
are related to
y
4
3
cos
3
sin
3
cos
x ,y
±~θ + θ ² ±~θ + θ ²
=x + L =y + L
4
+ Lg θ_ 3
(a)
3
±~θ + θ ²²j
(b)
3
, and
3
θ3 as
follows:
3
(c)
3
Then, making use of Eq. (A.29) of Appendix A and Eqs. (a) through (c), the kinetic energy of the system is
T=
1
m x_
+
2
|fflfflffl{zfflfflffl} 2
1
1
1
2
m x_ 1
+
2 1
2
± ² + J θ_ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} m Vg ⋅ V g
1
1 2
±y_ +L θ_ g
m x_ 2
2 2
+
1 2
m
3
2
2
3
2
g
3
Kinetic energy of the upper arm
h± x_ − L θ_ g
3
±~θ + θ ²² i + 2
cos
3
1
3
±~θ + θ ²²
of the forearm
+
3
2
Kinetic energy
of the hand
1
2
|fflfflffl{zfflfflffl} 2
Kinetic energy
=
+
m x_
3
1 2
2
sin
3
J g θ_ 3
3
2
(d)
3
and the potential energy of the system is given by
V=
1
k ðx
− xo
+
2
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} 2
1
1
Þ
1
2
2
± ±~θ +θ ²² k x +L |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ± ±~θ + θ ²² m g y +L |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
1 2
4
associated with spring
1
2
cos
3
3
3
+
2
1Þ
1
3
g
sin
k
4
+
k ðx
−x
3
3
2Þ
2
k
± ±~θ+ θ ²² k y +L |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 5
3
sin
kt ~θ+ θ 3
2
3
Potential energy
associated with spring
2
2
3
associated with spring
kt
3
2
3
Potential energy associated with spring
Potential energy associated with gravity loading
1
Potential energy
k
1
3
± ² |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}
+
2
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} 2
Potential energy
k
Potential energy associated with spring
+
−x
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} 2
Potential energy associated with spring
+
k ðx
k
5
(e)
7.2 Governing Equations
D is given by
The dissipation function
D=
1
c ðx_
− x_ o
+
2
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} 2
1
1
Þ
Dissipation associated with damper
+
c
1
c ð x_
− x_
2
1Þ
with damper
± ±~θ+ θ ²² c x_ −Lθ_ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 3
3
sin
c
c ðx_
− x_
2
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} 2
3
3
2Þ
+
1
Dissipation associated with damper
2
c
2
3
3
cos
2 3
Dissipation
damper
c
ct
3
(f)
3
Dissipation associated with damper
4
3
associated with
3
2
5
ct θ_
|fflffl{zfflffl} 2
± ±~θ +θ ²² + c y_ +Lθ_ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
3
Dissipation associated with damper
c
1
1
2
4
+
2
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}
2
2
Dissipation associated
1
1 2
375
5
From the form of Eqs. (d) through (f), it is clear that we need to make use of Eqs. (7.7) directly to obtain the governing equations of motion. In terms of the chosen coordinates for this
five degree-of-freedom system, we have
¶
d ∂T dt ∂ x_
−
∂T ∂x1
+
∂D ∂ x_ 1
+
∂V = Q1 ∂ x1
d ∂T dt ∂ x_
−
∂T ∂x2
+
∂D ∂ x_ 2
+
∂V = Q2 ∂ x2
d ∂T dt ∂ x_
−
∂T ∂x3
+
∂D ∂ x_ 3
+
∂V = Q3 ∂ x3
d ∂T dt ∂y_
−
+
∂D ∂ y_ 3
+
3
∂T ∂y3
∂V = Q4 ∂y3
∂ θ3
3
1
¶ 2
¶ 3
¶
d dt
(g)
0 1 @ ∂T_ A − ∂∂θT + ∂D_ + ∂∂θV = Q ∂ θ3
5
3
where the generalized forces are given by
Q i = 0 i = 1; 2; …; 5
(h)
since the external disturbance acting on the hand has already been taken into account in determining the kinetic energy, the potential energy, and the dissipation function. In addition, gravity loading has been accounted for in the system potential energy. After substituting Eqs. (d) through (f) and Eq. (h) into Eq. (g) and carrying out the different
376
Multiple Degree-of-Freedom Systems: Governing Equations differentiations in Eqs. (g), we arrive at the following governing equations of motion of the hand-arm system:
+ k + k x − k x + c + c x_ − c x_ = k xo + c x_ o − k x − k x + c + c x_ − c x_ − c x_ = h ± ² ±~θ + θ ²i + k + k x − k x ~ m x€ − Lg € θ θ + θ − Lg θ_ ± ² ±~θ + θ ² = + k L ~θ + θ + c + c x_ − c x_ − c Lθ_ h ±~θ + θ ² − L θ_ ±~θ + θ ²i + k ³y + L ±~θ + θ ²´ m € y + Lg € θ g ³ ±~θ + θ ²´ = − m g + c y_ + Lθ_ ±J + m L ²€θ + m L ³− €x ±~θ + θ ² + €y ±~θ + θ ²´ + k ±~θ + θ ² g g g t ± ² ³ ± ² ´ ± ² ~ ~ − k L θ + θ x + L θ + θ + k L ~θ + θ ³ ± ²´ ± ²³ ±~θ + θ ²´ × y + L ~θ + θ + ct θ_ − c L ~θ + θ x_ − Lθ_ ± ²³ ± ²´ ±~θ + θ ² = + c L ~θ + θ y_ + L ~θ + θ + m gLg m x€ m € x + ðk + k Þx 1
2
3
2
3
2
1
2
3
ð
2
3
1
3
3
ð
4Þ
3
3
3
3
3
3
3
sin
4
sin
3
ð
3
3
cos
3
3
cos
3
2
2
1
3
3
3
2
3
3
cos
3
5
3
3
sin
3
3
sin
4
3
3
cos
3
3
3
3
2
sin
3
3
cos
1
4Þ
4
5
2Þ
2
3
2
3
3
3
3
5
sin
1
3Þ
2
5
2
ð
2
3
sin
3
2
cos
3
3
2Þ
1
2
cos
3
3
2
cos
4
3
3
sin
3
ð
1
0
0
(i)
3
cos
3
3
cos
3
1
3
3
sin
3
1
3
0
Linearization We now linearize the nonlinear system of equations given by Eqs. (i) about the nominal position
~ θ
“small”
to describe
oscillations about this position. To this end, we use
Eqs. (g) of Example 7.3 to determine that
±~θ + θ ² ≈ ±~θ + θ ² ≈
sin
3
cos
sin
~θ + θ
cos
3
~θ − θ
~θ
cos
3
(j)
~θ
sin
3
Making use of Eqs. (j) in Eqs. (i) and retaining only linear terms, we obtain the following system of linear equations:
m € x
þ ð
x m €
þ ð
1
1
k
þ
k Þx 2
1
k Þx m x€ − Lg € θ sin ~ θ 2
³
2
3
k
1
2
þ
3
3
þ
4
g
1
k
3
þ
´
3
3
3
1
3
þ
4
4
²
5
5
m Lg € θ 2
3
3
þ
m Lg 3
c
3
3
2
−c
1
3
2
´ ~θ
sin
cos
~θ þ €y
3
þ
4
3
2
4
3
4
sin
3
cos
2
3
3
sin
3
þ
5
3
sin
±
5
cos
sin
3
2
2
3
þ
5
cos
3
þ
sin 3
cos
Lθ
3
3
´
~θ
þ
¼ 0
~θ
Lθ_
kt θ
3
Lθ_
k
3
3
cos
²
~θ
(k) þ
´ ~θ þ
3
x_
~
5
þ ð
1
sin
3
~θ þ c t θ_
cos
1
2
θ ± c y_
3
3
1
− c x_ − c
cos
cos
− − Lθ θ −k L θ ³ Lθ ~ ~θ y θ kL −kLθ ³ ´ c L ~θ³y_ ~θ − c L ~θ x_ − Lθ_ ~ ~θ − kt ~θ − m gLg θ θ = − m gLg k L sin ~ θ½ x
4
2
k xo þ c x_ o
¼
2
c Þx_
þ
4
3
x_
2
− k x −k ~θ = − k L 3
3
3
2
þ ð
3
³ − €x
c Þ x_
þ
k Þx
³
2
3
3
3
2
þ ð
ð
2
c Þx_ − c x_ − c Lθ_ sin € cos ~θ þ k y þ Lθ þ Lgθ − m g − k L sin ~θ
c ³ m € y
±J =
2
3
þð
3
2
´
3
−k x − c −k x −k x
4
3
3
3
−k
cos
L
5Þ
´
~θ 2
sin
~θ cos ~θ
7.2 Governing Equations
377
Equation (k) is assembled in the following matrix form
2m 66 66 66 4
0
0
0
0
m
0
0
0
0
m
0
1
2
0
0
3
3
1
2
−c
c
2
+c −c
2
0 0
1
3
2
3
8> k x + c x_ >> < ~ = >> −m−gk −Lk L θ >: ^ Q o
−k
o
1
0
cos
4
3
5
0
4
sin
5
9> >> = > ~θ > >;
4
2
g
g
sin
3
cos
3
g
3
2
5
L sin ~θ
3
0
1
0
2
sin
5
3
cos
5
3
55
4
k k L cos
0
0
1
0
3
0
0
0
−k k +k
3
0
5
0
3
0
4
5
4
2
0
0
4
c ~ −c L sin θ c L cos
0
2
0
0
−k k +k −k
2
0
3
0
0
g
3
−c c +c
3
3
2k +k 66 − k 6 + 66 64
3
3
2
1
3
m ~ 0 − m Lg sin θ m Lg cos ~θ c +c −c 0 0
2 66 6 + 66 64
0
3
9> >> = >> >; 38> x_ 9> 77>>< x_ >>= 7 − c L ~θ 77> x_ > ~θ 7 5>>: y_ >>; cL ~θ _ c 38>θ x 9> 77>>< x >>= 7 − k L ~θ 77> x > ~θ 7 5>>: y >>; k L ~θ θ k
38> €x 77>>< €x 7 ~θ 7 €x −m L 7>> ~θ 7 m L 5>: €y €θ J +m L 0
5
3
0
1
0
2
sin
3
cos
3
55
3
(l)
where
c k ^ Q
55
55
5
= ct + c L ± = kt + k L = − m gLg
~θ + c L cos ~θ sin ~ θ − cos ~ θ +k L ~ − k t θ~ + ðk − k ÞL cos θ
2
3
4
2
3
4
3
2
sin
5
2
2
3
²
4
2
5
5
±
cos
2
sin
2
² − m gL
~θ − sin ~θ 2
~θ cos ~θ
3
g
sin
~θ
It is clear from Eq. (l) that the system inertia, damping, and stiffness matrices are symmetric
matrices.
Equation
(l)
equations that is used to study
is
a
system
“small”
of
five
second-order
ordinary
motions about the nominal position
differential
~θ. It is noted
that this position is not an equilibrium position, but rather an operating point where
~θ_ = 0.
It is also noted that depending on the application, linearization of a nonlinear sys-
tem may need to be carried out about a reference position different from the equilibrium position.
378
Multiple Degree-of-Freedom Systems: Governing Equations
7.3
FREE RESPONSE CHARACTERISTICS
In this section, we examine the natural frequencies and mode shapes of undamped and damped systems. As in the case of single degree-of-freedom systems, when the forcing is absent,
the
responses
exhibited
by
a
multi-degree-of-freedom
system
are
called
free
responses. In Sections 7.3.1 and 7.3.2, undamped systems are considered and characteristics such as natural frequencies and mode shapes are determined. Following this, the modes of damped systems are examined in Section 7.3.3, and the notion of proportional damping is introduced. Subsequently, conservation of energy during free oscillations of a multi-degree-of-freedom system is examined. Throughout Section 7.3, characteristics of free responses of multi-degree-of-freedom systems are examined without explicitly determining the solution for the response. In Section 7.4, we study the vibrations of rotating shafts. In Section 7.5, we brie
fly
discuss how the stability of a multi-degree-of-
freedom system is assessed.
7.3.1
Undamped Systems: Natural Frequencies and Mode Shapes For illustration, consider the system of equations given by Eq. (7.3), which govern the motion of a linear multi-degree-of-freedom system. Setting the damping, the circulatory
qi by xi , we obtain
and gyroscopic terms, and the external forces to zero, and replacing
M ±fx€g + ½K ± fxg = f0 g
½ Since
the
system
given
by
equations with constant coef
Eq.
(7.22)
ficients,
is
a
linear
(7.22) system
of
ordinary
differential
the solution for Eq. (7.22) is assumed to be of the
16
form
xðtÞg = fX geλt
(7.23a)
f where
t
λ can be complex X} are given by
is time, the exponent
and the constant vector {
x
f g
8x t >>< x t => >: ⋮ x t 1ð
Þ
2ð
Þ
Nð
Þ
9 >>= >>;
x
valued, and the displacement vector { }
8X >>< X => >: ⋮ X
1
and
Xg
f
2
N
9 >>= >>;
(7.23b)
On substituting Eq. (7.23a) into Eq. (7.22), we obtain
³ K +λ ½
16
±
2
´
M ± fX geλt = f0g
½
For a solution of the form of Eq. (7.23a), we note that the ratio of any two elements time independent. This type of motion is called the same time dependence.
synchronous motion
(7.24)
x j(t)/x k(t) =Xj /Xk
is always
because both generalized coordinates have
7.3 Free Response Characteristics
Eigenvalue Problem
fi
find that
Since Eq. (7.24) should be satis ed for all time, we
³ K +λ ½
2
±
The system of Eq. (7.25) is a system of
fied for X
= X = ⋯ = XN =
1
2
½
´ M X ± f
g
379
=
f0g
(7.25)
N linear algebraic equations. This system is satis-
0, which is a trivial solution. Since we seek nontrivial solu-
…, X N, Eq. (7.25) represents an eigenvalue problem. The unknowns are λ , X , X , …, XN , and since there are only N equations to solve for these (N + 1) unknowns, at best, what one could do is solve for λ , and the ratios X / X , X / X , …, XN /X . The quantity λ is referred to as the eigenvalue and the vector {X X … X N}T is called the eigenvector. The eigenvalues λ are determined by fi nding the roots of the characteristic equation tions for
X
, 1
X
17
, 2
2
1
2
2
2
1
3
1
2
1
1
2
2
³ K +λ
det ½
´=
M±
2
±
½
0
(7.26)
N × N matrices, the expansion of Eq. (7.26) N in λ for an N degree-of-freedom system. Alternatively, one as an N th order polynomial in λ with N roots or eigenvalues
Since the stiffness and mass matrices are is a polynomial of degree 2
2
can view this polynomial
λ1; λ2 ; …; λN . 2
2
2
The associated eigenvectors are solutions of the equations
K ± + λj ½ M ±±fX gj = f0g 2
½½
X }j
where {
(7.27)
is the eigenvector associated with the eigenvalue
λ2j
and we have a total of
N
eigenvectors.
first
To solve for the eigenvectors associated with the
λ 2; 2
value
and so forth, we construct the eigenvectors
8> X >< X = >>: ⋮
11
Xg
f
1
21
XN
1
which can be written as
Xg
f
1
eigenvalue
=X
8> >< X >>:
1
X
21 =
11
⋮
11
XN =X 1
11
9> >= >>;
;
9> >= >>;
;
8> X >< X = >>: ⋮
12
Xg
f
2
22
XN
Xg
f
2
=X
12
2
9> >= >>;
8> >< X >>:
λ21;
8> X >< X = >>: ⋮
N
1
⋯
;
X gN
;
f
N
2
XNN
1
X
22 =
⋮
12
XN =X 2
12
9> >= >>;
;
⋯
9> >= >>;
X gN = X
;f
the second eigen-
1
N
(7.28a)
8> >< X >>:
1
N =X1 N
2
⋮
XNN =X
N
1
9> >= >>;
(7.28b)
17
A]{X} = λ′ [B]{X}, the problem of fi nding the constants λ′ for which the vector {X} is eigenvalue or characteristic value problem. A scalar version of this problem was discussed
In general, for a system [ nontrivial, is called an
for single degree-of-freedom systems in Section 4.3.
380
Multiple Degree-of-Freedom Systems: Governing Equations
X j ≠ 0 for j = 1, 2, …, N. In X } , which is associated with the eigenvalue λ , is called the first eigenvector, first eigenmode, or first mode shape, and {X } , which is associated with the eigenvalue λ , is called the second eigenvector, second eigenmode, or second mode shape, In writing Eqs. (7.28b), it has been assumed that
1
2
Eqs. (7.28), {
1
1
2
2 2
and so on. Due to the nature of the eigenvalue problem, the eigenvectors {
X} j
are arbitrary to a
non-zero scaling constant. A convenient normalization that is often used to remove this arbitrariness is
X
11
=
1;
X
=
1;
…; X
N
=
1
(7.29)
X
=
1;
…; X
N
=
1
(7.30)
12
1
Another choice for normalization is
X
21
=
1;
22
2
and so forth. Physically, the eigenvectors provide information about the relative spatial positions of the different inertial elements in terms of the generalized coordinates. Thus,
λi,
for free oscillation at
each mass
m j moves
a
fixed amount
the resulting modes are called
normal modes. modal matrix [
The mode shapes are placed in a
Φ =» X
½
±
f
g
1
⋯
Xg
f
2
X gN
f
Φ
], which is
2X ¼ = 666 X 64 ⋮
11
X
21
X
XN
where {
mk. The process normalization, and
relative to
of normalizing the mode shapes (eigenvectors) of a system is called
⋯ ⋯ ⋱ ⋯
12
22
⋮
XN
1
2
X
1
X
2
N
N
⋮
XNN
3 77 77 5
(7.31)
X} j is the mode shape associated with the jth eigenvalue λ j . When the normaliza2
tion of the mode shapes is carried out according to Eqs. (7.29), then the modal matrix takes the form
2 66 X Φ = 66 4
½
±
1
X
21 =
⋮
1 11
X N =X 1
11
X
X
22 =
12
⋮
XN =X 2
12
⋯ ⋯ ⋯ ⋯
1
X
N =X 1N
2
⋮
XNN =X
1
3 77 77 5
(7.32)
N
When the normalization is carried out according to Eqs. (7.30), then the modal matrix takes the form
2X 66 Φ = 66 4
½
±
X
11 =
21
X
X
12 =
1
1
⋮
⋮
X N =X 1
21
22
XN =X 2
22
⋯ ⋯ ⋯ ⋯
X
N =X 2N
1
1
⋮ XNN =X
2
N
3 77 77 5
(7.33)
7.3 Free Response Characteristics
381
and so forth. When the eigenvectors are normalized so that their magnitude is one, the corresponding normalization equation is
‖ fX gj ‖ =
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi X
1
j
2
+ X j + ⋯ + XNj = 2
2
2
1
(7.34)
However, regardless of the choice of the normalization, the ratios of the different components in an eigenvector are always preserved.
M]
K ],
For real and symmetric matrices [
and [
the eigenvalues
X} j are also real.
18
and the associated eigenvectors {
λ2
of Eq. (7.25) are real
Hence, it is common to write
λ2 = ð j ωÞ2 = −ω2 where
ω
(7.35)
is a positive quantity. It will be shown later that
quencies of the
N
Eqs. (7.26) and (7.27), we
³ K −ω
det ½
Nth
order polynomial in
the natural frequencies
ωj
ω1, ω2, …, ωN
N
λ2
N natural
fre-
from Eq. (7.35) into
±
ω2,
2
½
´=
M±
0
(7.36)
and that the eigenvectors {
X }j
associated with
are determined from ½½
For a system with
is one of the
find that the natural frequencies are determined by solving the
characteristic equation
which is an
ω
degree-of-freedom system. On substituting for
K ± + ωj ½M ±±fX gj = f0g 2
(7.37)
degrees of freedom, Eq. (7.36) provides the
N natural frequencies X} , {X } , …, {X }N .
and Eq. (7.37) provides the associated eigenvectors {
1
2
The natural frequencies are ordered so that
ω1 ≤ ω2 ≤ ⋯ ≤ ωN Hence, the
first natural frequency is lower than or equal to the second natural frequency,
and so forth. To illustrate how the eigenvalues and eigenvectors are determined for a multi-degreeof-freedom system, we use two degree-of-freedom systems. However, the discussion provided below is valid for any linear multi-degree-of-freedom system.
Free Oscillations of Two Degree-of-Freedom Systems N= fi M K
Setting
2 in Eq. (7.37), and using the de
obtain
°
−ω
2
°m
1
0
0
m
2
! °k +
+k −k
1
2
2
nitions of [
−k k +k
] from Eqs. (7.5b), we
!!& X ' & '
2
2
] and [
3
X
1 2
=
0 0
(7.38a)
which is rewritten as
18
For a comprehensive discussion of eigenvalue problems associated with structural and mechanical systems, see L. Meirovitch, 1980,
ibid.
382
Multiple Degree-of-Freedom Systems: Governing Equations
°k
1
+k −ω m −k
1
2
k
2
!& X ' & '
−k +k −ω m
2
2
X
2
2
3
2
=
1 2
0
(7.38b)
0
In this case, the characteristic equation given by Eq. (7.36) translates to
det
°k
1
+ k −ω −k
2
2
!
m
1
k
2
which, when expanded, takes the form
±k
+k −ω m 2
1
2
1
²±k
−k +k −ω m = 2
2
2
3
²
+k −ω m − k = 2
2
0
(7.39a)
0
(7.39b)
2
3
2
2 2
Equation (7.39b) is rewritten as
m m ω 1
4
2
−
k
½ð
1
+k
m
2Þ
2
+
ð
+k
k
2
m ±ω
3Þ
ω.
which is a fourth-order polynomial in treat it as a quadratic polynomial in
ω1
with the natural frequencies equations:
°k
+ k − ωj m −k
and
2
1
1
2
2
1
ω
2
+
2
1
+k
k
2Þ ð
2
+k −k = 3Þ
2
2
0
(7.39c)
Due to the form of this equation, one can
2
. From Eq. (7.37), the eigenvectors associated
ω2
are determined from the following system of
−k + k − ωj
!& X ' & '
2
k
k
ð
2
3
X
m
2
=
j
1
j
2
0 0
j = 1; 2
(7.40)
Next, we present an example to show the explicit details of determining the natural frequencies and mode shapes of a two degree-of-freedom system before examining a nondimensional form of the system given by Eq. (7.38a). The nondimensional form is better suited for examining the in
fluences
of the different parameters on the system nat-
ural frequencies and mode shapes.
EXAMPLE 7.12
Natural frequencies and mode shapes of a two
degree-of-freedom system
We
shall
illustrate
how
the
algebraic
system
given
by
Eq.
(7.38b)
is
solved
fic
determine the natural frequencies and mode shapes associated with a speci degree-of-freedom system. The modal matrix [
k
We choose the stiffness parameters
m
2
1,
k
2,
Φ
and
to
two
] of the system is also constructed.
k
and the mass parameters
3
m
1
and
so that
k
1
=k =k =k 2
3
and
m
=m =m
1
2
(a)
Thus, making use of Eqs. (a) in Eq. (7.38b), we obtain
°
k −ω m −k
2
2
−k k−ω m
2
2
!& X ' & ' 1
X
2
=
0 0
(b)
7.3 Free Response Characteristics
383
To determine the natural frequencies of the system, we make use of Eq. (7.36) and Eqs. (a) to arrive at
° det
!
k −ω m −k
−k = k−ω m
2
2
2
2
0
(c)
On expanding Eq. (c), the result is the characteristic equation
± k −ω m ²
2
2
2
−k = 2
0
(d)
which yields 2 Thus, the two roots are given by
ω1 =
rffiffiffikffi m
k−ω m=
±k
2
and
ω2 =
(e)
rffiffiffiffiffik 3
m
=
rad s
(f)
which are the two natural frequencies of the system and they have been ordered so that
ω1 < ω2.
To determine the associated mode shapes, we make use of Eqs. (7.40), (b), and (f). Therefore, to determine the mode shape associated with obtain
°
which, upon using the
From the
k−ω m −k
!& X '
−k k −ω m
2
2
X
2
2
°
−k
k −k
k
!& X '
21
=
11
X
21
=
11
first of Eqs. (f), reduces to
ω1,
we set
ω
=ω
1
in Eq. (b) to
& ' 0 0
& ' 0
(g)
0
first of Eq. (g), we find that kX
11
− kX = 21
0
or
X X
21
=
1
(h)
11
The same ratio of modal amplitudes is also obtained from the second equation of the system
(g),
as
expected.
Then,
choosing
the
Eqs. (7.28b) and (h) lead to
Xg
f
Thus, mass
m
1
and mass
m
2
1
=X
11
normalization
given
by
Eq.
(7.29),
& ' 1 1
move in the same direction with the same displacement.
(i)
384
Multiple Degree-of-Freedom Systems: Governing Equations To determine the second mode shape associated with obtain
°
k−ω m −k
−k k−ω
2
2
2
2
2
2
m
ω2,
ω
we set
=ω
2
in Eqs. (b) to
!& X ' & ' X
=
12 22
0 0
which, upon using the second of Eqs. (f), reduces to
° −k −k
From the
!& X ' & '
−k −k
=
12
X
22
0
(j)
0
first of Eq. (j) we find that X X
=−
22
1
(k)
12
Thus,
mass
m
1
and
mass
m
2
move
in
the
opposite
directions
but
with
the
same
magnitude. The ratio of the modal amplitudes shown in Eq. (k) could also have been determined from the second of Eq. (j). Again choosing the normalization given by Eq. (7.29), Eqs. (7.28b) and (k) lead to
=X
Xg
f
& ' 1
−
12
2
(l)
1
For the normalization chosen, the modal matrix is obtained from Eqs. (7.31), (i), and (l) as
Φ=
½
°
±
1
!
1
−
1
(m)
1
In Example 7.12, the natural frequencies and mode shapes were determined for a two degree-of-freedom system with a speci
fic
set of parameters. In order to explore the
natural frequencies and mode shapes associated with arbitrary system parameters, we introduce
several
nondimensional
parameters
and
then
solve
the
system
given
by
Eq. (7.38b) in terms of these nondimensional parameters.
Eigenvalue Problem in Terms of Nondimensional Parameters In order to rewrite Eq. (7.38b) in terms of nondimensional mass, stiffness, and frequency parameters, we introduce the following quantities:
ω2nj =
kj ; mj
ωr =
ωn2 ωn1
j = 1; 2
=
ffiffiffiffiffi m
1 p
r
skffiffiffiffi 2
k
1
mr = ;
m m
and
2
Ω=
;
1
k
32
=
k k
ω ωn 1 (7.41)
3 2
7.3 Free Response Characteristics On substituting the different quantities from Eqs. (7.41) into Eq. (7.38b), we
³
the resulting system is
´
+ ωr mr − Ω X − mr ωr X = ³ ´ − ωr X + ωr + k − Ω X = 2
1
2
2
2
1
°
2
1
ð1
32 Þ
+ ωr mr − Ω − ωr 2
− m r ωr +k −Ω
2
2
ωr 2 ð1
2
0
2
0
X
2
32 Þ
find
that
(7.42a)
!& X ' & '
which, in matrix form, is 1
2
2
385
=
1 2
0 0
(7.42b)
For the eigenvalue formulation given by Eq. (7.42a) or (7.42b), the eigenvalue is and the corresponding eigenvector is {
X}
tion for {
Ω
2
X}. This system of equations has a nontrivial solu-
fi
only when the determinant of the coef cient matrix from Eq. (7.42b) is
°
!
zero. Thus, from Eqs. (7.42b), we arrive at
det
³
1
+ ωr m r − Ω − ωr 2
2
2
ω2r ð1
2
´³ω
− mr ωr = + k −Ω
which gives the characteristic equation 1
+ ωr
2
mr − Ω
2
r
2
ð1
2
32 Þ
´
0
+ k − Ω − mr ωr = 2
32 Þ
4
(7.43)
0
(7.44)
Equation (7.44) is Eq. (7.39c) rewritten in terms of the nondimensional quantities given by Eqs. (7.41). It is important to note that the nondimensionalization introduced in Eqs. (7.41) led to the compact form of the characteristic equation, Eq. (7.44), which enables one to readily identify the parameters on which the natural frequencies depend. Expanding Eq. (7.44) leads to
Ω
4
−a Ω +a = 2
1
2
0
(7.45)
where
a
1
a
2
= + ωr + mr + k ³ ± ² = ωr + k + ωr mr 2
1
2
ð1
1
32
32 Þ
(7.46)
2
1
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ° qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi!
Then the two positive roots of the characteristic equation given by Eq. (7.45) are
Ω
1;2
=
1 2
a
1
∓
a
2 1
−
a
4
(7.47)
2
Ω are ordered such that Ω < Ω . The frequency ω associated with Ω is called the fi rst natural frequency coeffi cient and the frequency ω associated with Ω is called the second natural frequency coeffi cient. We see that when the interconnecting spring k is absent from the system shown in Figure 7.1, we can set k = 0 (ωr = 0) in Eq. (7.39b) and, as expected, the resulting and the frequency ratios
Ω
1
and
19
2
1
1
2
1 2
2
2
2
19
When the eigenvalues are determined numerically by using commercial software, these eigenvalues are not necessarily ordered in terms of magnitude from the lowest to the highest.
386
Multiple Degree-of-Freedom Systems: Governing Equations
1.0 3 0.5 2
Ω1
0.0 2.0
wr
1.5
1 1.0
mr
0.5 0 Figure 7.13.
0.0
Variation of the
first nondimensional natural frequency of the two degree-of-
freedom system shown in Figure 7.1 as a function of
mr and ωr
when
k
3
=
0.
system is uncoupled. The two natural frequencies are, respectively, the natural frequen-
pffiffiffiffiffiffiffiffiffiffiffiffi k m
pffiffiffiffiffiffiffiffiffiffiffiffi k m
cies of two independent single degree-of-freedom systems. One natural frequency is
ωn 1 =
and
k
3
=
1=
1
and the other natural frequency is
ωn2 =
3=
2:
When
k
1
≠
0,
k
≠
2
0,
20
0, appropriate expressions are similarly obtained.
Based on Eqs. (7.45), (7.46), and (7.41), we see that, in general, the natural frequencies of this system are functions of the three nondimensional parameters we assume that
k
32
=
0 (i.e.,
quency ratios as functions of
mr, ωr,
and
k
32.
If
k = 0), then we can graph the first and second natural fremr and ωr. The results are shown in Figures 7.13 and 7.14, 3
and they lead to the following design guideline.
Design Guideline:
For a two degree-of-freedom system with two springs and two
masses, as the mass ratio frequency
Ω
1
decreases
mr
and
=
m /m
the
2
1
increases, the
second
first
nondimensional
nondimensional natural natural
frequency
Ω
2
increases. Thus, increasing the ratio of the two masses tends to drive the two natural frequencies away from each other. For a constant mass ratio, we see that an increase in the stiffness ratio
20
k /k 2
We shall continue to include
k
1
3
fi
increases both
and
c
3
Ω
1
and
Ω
.
2
in determining the necessary equations, but when we numerically evaluate
related expressions, these coef cients are frequently set to zero.
7.3 Free Response Characteristics
387
6 3
4
Ω2
2 2
0 2.0
w
r
1.5
1 1.0 0.5 0 Figure 7.14.
mr
0.0
Variation of the second nondimensional natural frequency of the two degree-of-
freedom system shown in Figure 7.1 as a function of
mr and ωr
when
k
3
=
0.
INTERACTIVE GRAPHIC 7.1: NATURAL FREQUENCIES OF TRANSLATING TWO DEGREE-OF-FREEDOM SYSTEMS The fi rst and second natural frequencies of a translating two degree-of-freedom system are displayed. The variation of these frequencies with respect to ωr and mr can be explored when k32 = 0. In this interactive graphic, the following can be noted. • •
For any value of m r, as ωr increases the value of Ω1 becomes constant at a value of ωr > 1.5 whereas Ω2 increases linearly with increasing ωr . For any value of ωr > 1, the value of Ω1 changes moderately with respect to variation in mr . For values of ωr < 1, the value of Ω1 becomes almost constant with respect to the variation 0 whereas the value of Ω2 approaches 1. in m r and approaches zero as ωr As ωr increases beyond one, the value of Ω1 continually decreases as m r increases whereas the value of Ω2 continually increases.
→
•
388
Multiple Degree-of-Freedom Systems: Governing Equations We return to Eqs. (7.42a) and determine the eigenvectors associated with For
Ω
= Ωj
, we have
³
´
+ ωr mr − Ωj X j − mr ωr X j = h i −ωr X j + ωr + k − Ωj X j = 2
2
1
2
2
1
where
X
j and
1
frequency
X
ωn1Ωj.
2
1
ð1
2
0
2
0
2
32 Þ
Ω
1
and
Ω
.
2
(7.48)
j are the respective displacements of the two masses oscillating at the
2
Since we can solve only for the ratio
X j /X 1
j or
2
X j /X 2
j, from the
1
first of
Eqs. (7.48), we arrive at
X X
1
j
2
j
=
m r ωr 1 + ωr mr − Ωj 2
2
j = 1; 2
2
(7.49)
and from the second of Eqs. (7.48) we arrive at
X X
j
1
j
2
=
ωr 2ð 1 + k32 Þ − Ωj 2
j = 1; 2
ωr 2
(7.50)
Although Eqs. (7.49) and (7.50) have algebraically different forms, they can be shown to be identical by making use of Eqs. (7.47). It is remarked again that the nondimensionalization introduced in Eqs. (7.41) enables us to determine the dependence of the mode shapes on the various system parameters, as seen from the compact forms of Eqs. (7.45) through (7.50).
k
For a special case of interest, we let
32
Figure 7.1. Then, Eqs. (7.46) lead to
a a
1
→ +ω →ω Ω ) 1
→ω →
Ω Ω
→ →ω
1
2
2
r
2
1
1
2 2
r
(7.51)
for
ωr ≤ 1
(7.52)
2
for
ωr > 1
(7.53)
r
Substituting these limiting values into Eq. (7.50), we these two regions are as follows. For
½
and for
ωr
>
ωr
Φ=
≤
°
±
½
1
0
Φ= − ±
1
−
1
°
1
1 for the system shown in
j are determined from Eqs. (7.47) to be
)
and
≪
2
1
2
mr
r
and the associated natural frequency ratios 2
0 and
2
2
Ω Ω
=
1
ω2r
1=
ω2r
0
1
1
that the modal matrices in
! (7.54a)
1
1=
find
! (7.54b)
7.3 Free Response Characteristics
389
The modal matrix plays a key role in determining the response of vibratory systems. In the next section, properties of mode shapes are examined. In the remainder of this section, we illustrate the determination and interpretation of natural frequencies and mode shapes through different examples. In Section 7.3.2, we summarize the results and include additional systems.
EXAMPLE 7.13
Rigid-body mode of a railway car system
A special case of interest is when
k
1
=k = c =c =c = 3
1
2
3
0
(a)
for the system shown in Figure 7.1; that is, we have two masses connected by a spring as shown in Figure 7.15. This system is used to model two interconnected railway cars, a truck towing a car, and other such systems. Through this example, we illustrate what is meant by a
rigid-body mode of oscillation.
°k
−ω m −k 2
2
1
k
2
2
!& X ' & '
In this case, Eq. (7.38b) reduces to
−k −ω m
=
1
2
X
2
2
2
0
(b)
0
fies to
and the characteristic equation given by Eq. (7.39c) simpli
±k
2
2
or
²± −ω m k 1
2
² −ω m −k 2
2
±
ω2 m 1m2 ω2 − k2
²m ð
1
2
2
=
+m = 2Þ
0
(c)
0
The roots of this equation are
ω1 = 0
ω2 = ωn2 where
ωn2
and
mr
are de
fined
pffiffiffiffiffiffiffiffiffiffiffiffi +m 1
(d)
r
in Eqs. (7.41). From Eq. (b), the corresponding mode
shape ratios are
X X X X
11 21
12 22
x1 m1
= =
k
k
1
k
k
−ω m
=
2
2
2
1
1
2
x2 k2
=
2
−ω m 2
2
1
k k − ωn m
2
2
2
2
1 ð1
+ mr
Figure 7.15.
Þ
1
− + mr ð1
=m r
Þ
= − mr
Two railway cars with a spring
interconnection.
m2
=
1
(e)
390
Multiple Degree-of-Freedom Systems: Governing Equations
first
The
mode shape, which corresponds to
ω1
=
0, is a
rigid-body mode,
that is, one in
which there is no relative displacement between the two masses. The second mode shape, which corresponds to out
of
phase;
that
ω2,
is,
indicates that the displacements of the two masses are always
when
m
1
moves
in
one
m
direction,
2
moves
in
the
opposite
direction. In general, the presence of a rigid-body mode is determined by examining the stiffness matrix [
K]. If the size of the square matrix is n and the rank of the matrix is m, n − m) zero eigenvalues and, correspondingly, (n − m ) rigid-body modes.
then
there are (
EXAMPLE 7.14
Natural frequencies and mode shapes of a two-mass-three-spring
system
For the system shown in Figure 7.1, let 20 N/m, and
k
3
=
=
m
1
1.2 kg,
m
2
=
2.7 kg,
k
1
=
10 N/m,
k
2
=
15 N/m. This example is identical in spirit to Example 7.12, except
that we will use the nondimensional quantities in carrying out the computations. We shall
find
the natural frequencies and mode shapes of this system and illustrate how the
mode shapes are graphically illustrated. The modal matrix is also constructed. First, we compute the quantities
ωn 1 =
sffiffiffiffiffiffi 10
1:2
=
2:887 rad=s;
=
mr =
sffiffiffiffiffiffi
ωn2 =
20
2:7
=
2:722 rad=s (a)
ωr =
2:722 2:887
0:943;
From Eqs. (a) and (7.46), we
a
1
a
2
1
ð1
1:2
=
2:25;
and
k
32
=
15 20
=
0:75
find that
= + + + ³ ± = + × + 1
2:7
2:25
0:75
0:75Þ
1
2:25
× ×
2
0:943
2
0:943
²=´ ×
4:556 2
0:943
=
(b) 2:889
Making use of Eqs. (b) and Eqs. (7.47), we obtain the nondimensional natural frequency ratios
Ω
=
Ω
=
1
2
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¹ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiºffi 0:5
×
0:5
×
4:556
−
4:556
+
2
4:556
− × 4
2:889
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¹ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiºffi
Noting that the natural frequencies
ωj
2
4:556
= ωn Ωj 1
− ×
, we
4
2:889
=
0:873 (c)
=
1:948
find from Eqs. (a) and (c) that
ω1 = 2:887 × 0:873 = 2 :519
ω2 = 2:887 × 1:948 = 5 :623 rad=s
(d)
7.3 Free Response Characteristics
m1
m2 1.0
0.893 m1
m2
2.518
First mode at
1
1.0
m1
m2
Figure 7.16.
391
Second mode at
2
Mode shapes for system shown in Figure 7.1 for parameters of Example 7.14.
Next, the mode shape ratios are computed from Eqs. (7.49), (a), and (c) to be
X X
=
X X
=
11 21
12 22
2:25 1
+
2:25
×
+
2:25
×
2
0:943 2
0:943
2:25 1
× ×
−
2
0:873
2
−
0:893 (e)
2
0:943
0:943
=
2
1:948
=−
2:518
After choosing the normalization given by Eqs. (7.30), we set modal matrix as
Φ=
½
m
1
0:893
±
We see that for oscillations in the with
°
−
1
first
2:518
2
=
1 and construct the
! (f)
mode, both masses move in the same direction,
moving an amount that is 0.893 times that of
m
j
2
1
m
mode, the masses move in opposite directions, with direction as
X
1
m
. For oscillation in the second
2
moving 2.518 times as far in one
moves in the opposite direction. It is common, as seen in this example,
that the mode shape ratios are positive in the
X}
we go to the second mode. The modes {
1
first mode, X}
and {
2
and there is a sign change when
are illustrated in Figure 7.16.
INTERACTIVE GRAPHIC 7.2: TWO DEGREE-OF-FREEDOM SYSTEMS WITH TRANSLATION: NATURAL FREQUENCIES AND MODE SHAPES The natural frequency coefficients and animations of the mode shapes associated with the first and second natural frequencies of a translating, two degree-of-freedom system are displayed. The variations of these frequencies and associated mode shapes with respect to ωr and m r can be explored. In this interactive graphic, the following should be noted. •
The motions of the masses are in phase for the mode shape associated with the lowest natural frequency, that is, the ratio X 11/X 21, is positive. On the other hand, the motions of the
392
Multiple Degree-of-Freedom Systems: Governing Equations
• •
masses are out of phase for the mode shape associated with the second natural frequency, that is, the ratio X 12/X 22, is negative. The presence or absence of k32 does not affect these phase relations; however, it does affect their numerical values. The effect of the spring with spring constant k3 is to increase each of the natural frequencies. This spring is absent when k32 = 0. For the system with two masses and one spring, the lowest frequency is zero and the associated mode is a rigid-body mode; in this mode, there is no relative movement between the masses. In the motions associated with the second mode, the masses are out of phase and their ratio is equal to −m r.
EXAMPLE 7.15
Natural frequencies and mode shapes of a pendulum
attached to a translating mass
Consider the system shown in Figure 7.17. We shall derive the governing equations of
θ
motion for small | | and then use these equations to determine the natural frequencies and mode shapes associated with this system. The equations of motion are obtained by using Lagrange’s equations. We shall also illustrate how the expressions for the kinetic energy
and
the
potential
energy
are
appropriately
truncated
to
obtain
the
linear
equations of motion.
System with two degrees of
freedom.
a
j
x
k
z
Figure 7.17.
o
y
i
c.g. L
M
Jo x k
m
Considering the translation of the mass
m
and the rotation of the rigid bar
M,
the
system kinetic energy is
T=
1 2
mx_
2
+
1 2
Joθ_
2
(a)
and the system potential energy is
V=
1 2
k ðx −L sin θÞ
2
+ Mga − ð1
cos
θÞ
(b)
7.3 Free Response Characteristics
393
where the datum for computing the potential energy due to gravity loading has been chosen at the center of mass of the pendulum. For
“small”
oscillations about
θ
=
0,
Taylor-series expansions lead to
sin
θ ≈θ
cos
θ2
θ≈ 1−
(c)
2
21
and the potential energy given by Eq. (b) is approximated as
V = k ðx −LθÞ 1
2
2
+
1
Making use of Eqs. (a) and (c) in Eqs. (7.7) for
Qx =
0 and
Qθ =
Mgaθ
2
q
1
=x
0, we obtain
°m
0
!& x€ ' °
0
Jo
+
€θ
k − kL − kL kL + Mga
2
(d)
and
q
2
=θ
and recognizing that
!& x ' & ' θ
2
=
0
(e)
0
Comparing Eq. (e) with Eq. (7.22), we substitute a solution of the form
& x ' &X' =
θ
into Eq. (e), choose
λ2
= –ω
tion from Eq. (7.38b)
−ω where
ω2
2
°m 0
0
Jo
2
(or
λ
= jω
!& X ' ° +
o
Θo
o
eλt
Θo
(f)
) and arrive at the following eigenvalue formula-
k − kL − kL kL + Mga
!& X ' & ' o
Θo
2
is the eigenvalue and the eigenvector is given by {
=
0
(g)
0
Xo Θo} T.
Introducing the notations
21
ω2n1 =
k ; m
Jr =
mL Jo
ω2n2 = 2
;
Ko ; Jo
Ω=
ω ωn 1
K o = kL
2
;
and
+ Mga
ωr =
ωn 1
For obtaining the linear equations of motion, retaining up to quadratic terms in the functions cient, as illustrated in this example.
(h)
ωn 2
T
and
V
fi
is suf -
394
Multiple Degree-of-Freedom Systems: Governing Equations
°±
Eq. (g) is written as
1
where {
−Ω − Jr
2
²
±ω −− Ω ² 1
r
Xo LΘo} T is the eigenvector and Ω
2
=
o
LΘ o
2
2
°±
!& X ' & ' 0
(i)
0
is the eigenvalue.
²
!
To determine the nontrivial solution of Eq. (i), we obtain a solution to 1
det
±
Thus,
−Ω − Jr
−Ω
1
2
2
±ω −− Ω ² 1
2
²±ω
r
±
2
r
2
²
− Ω − Jr =
²Ω
2
=
0
(j)
0
(k)
Expanding Eq. (k) results in the characteristic equation
Ω
4
− + ωr 1
whose roots are
Ω
=
2;1
ffiffi
1 p
2
2
+ ωr − Jr = 2
0
(l)
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ± ² ffiffi 1
2
+ ωr ± 2
1
−ωr + 2
2
Jr
4
(m)
where the nondimensional natural frequencies have been ordered so that
Ω
1
: k′ >; 2 ϑ+α 6 ≈ 64 − ϑ + α
tors
½
zð
y ðθ Þ
Þ±
cos ð
½
φθ
Þ
sinð
´R
x ðφÞ±
ϑ + αÞ + sinðϑ + αÞ
φ sinðϑ + αÞ − θ cosðϑ + αÞ
ϑ + αÞ + cosðϑ + αÞ
φ cosðϑ + αÞ + θ sinðϑ + αÞ
cosð
−φθ
Þ
8> i 9> < = >: kj >;
sinð
−φ
θ
1
38 9 = 77>< ij > 5>: > ; k (7.114)
where we have assumed
“small”
φ
rotation angles
and
θ
about zero in arriving at the
composite rotation matrix. In addition, if it is assumed that the static unbalance angle
α
=
8< i′ 9= 2 ϑ : k′j′ ; ≈ 4 − θ ϑ
0 and that we can neglect terms with cos
sin
φθ ,
φ sin ϑ − θ cos ϑ φ cos ϑ + θ sin ϑ
ϑ cos ϑ −φ sin
1
The position vector from the disc center of mass ence frame
38< i 9= 5: j ; k
then Eq. (7.114) becomes
P
(7.115)
to the origin of the inertial refer-
O is given by RPO = R CO + R PC = xi + yj + zk + εi ′
After using the
(7.116)
first row of Eq. (7.115) in Eq. (7.116), we obtain
RPO = xi + yj + zk + ε½ cos ϑi + sin ϑj + ð φ sin ϑ − θ cos ϑÞk± = ð x + ε cos ϑÞi + ð y + ε sin ϑÞj + ðz + εφ sin ϑ − εθ cos ϑÞk It is recalled that
ϑðtÞ
is the rotation angle corresponding to the rotation speed of the
shaft. We obtain the velocity of the center of mass
³
´ ³
P
with respect to the
´
time derivative of Eq. (7.117); this leads to
V PO = x_ − εϑ_ sin ϑ i + y_ + εϑ_ cos ϑ j _ − θ_ cos ϑ + ϑθ _ + φ_ + z_ + ε ϑφ The
(7.117)
³
±±
²
±
²
sin
O by
²´
ϑ k
taking the
(7.118a)
z-component of the velocity contains two terms. The first term is the velocity of the C in the axial direction. The second term is the velocity due to the eccentricity and
point
rotations of the cross-section of the shaft. If
³
neglected in comparison to the
ε
is small, then this second term can be
first term. This leads to
´ ³
´
V PO ≈ x_ − εϑ_ sin ϑ i + y_ + εϑ_ cos ϑ j + zk _ Then, the translational kinetic energy is given by
Tt =
1 2
mðV PO ⋅ V POÞ =
1 2
n
m x_
2
+ y_ + z_ + ε 2
2
2
ϑ_
2
+
εϑ_ ½ −x_
2
sin
(7.118b)
o
ϑ + y_ cos ϑ±
(7.119)
7.4 Rotating Shafts on Flexible Supports
427
To determine the rotational kinetic energy, we need to determine the angular speeds about the axis of symmetry
8> ω < >: ωω
Y Z
in the disc. First, noting that
α
C–x′, C–y′, and C–z′ axes as follows
9> 8> φ_ 9> = ³ ´< = + R ϑ =R ϑ R θ >; >: >; 8 9 >< φ_ ϑ θ + θ_ ϑ >= = > − φ_ ϑ θ + θ_ ϑ > : ; − φ_ θ + ϑ_
angular speeds about the
X
fixed
½
zð
Þ±
yð
Þ
0
½
zð
cos
sin
0, we determine the
8> 9> 8> 9> < = < = >: θ_ >; + >: ϑ_ >; 0
0
Þ±
0
cos
=
0
0
sin
cos
(7.120)
cos
sin
where the rotation matrices are given by Eq. (7.113). To determine the angular speeds of the disc about the principal axes of the disc, we need to rotate the
χ
8> ω < >: ωω
the angle
1
2
3
C–y′
8 9> 8 9 = ³ ´>< ω >= ³ ´>< φ_ ϑ >; = R χ >: ωω >; = R χ >: − φ_ ϑ 8> ±φ_ ϑ θ + θ_ ϑ² χ − ±−φ_ φ_ < _ => : ±φ_ ϑ− φ_ θ +ϑ θ_ θϑ+² θ χ +ϑ±φ_
about the
9> = θ + θ_ ϑ >; θ + ϑ_ ² χ9 θ + ϑ_ >= ² χ >; θ + ϑ_
axis. This rotation results in the angular speeds
X
yð
Cx′y′z′ frame through
Þ
cos
yð
Y
Þ
cos
sin
cos
sin
sin
cos
cos
cos
cos
sin
sin
ϑ
cos
sin
cos
sin
sin
cos
Z
cos
θ + θ_
sin
sin
(7.121)
cos
χ is a “small” rotation about zero and that the rotation angles φ “small” rotations about zero as well, then Eq. (7.121) can be simplified to
If it is assumed that and
θ
are
ω1 ≈ φ_ cos ϑ + θ_ sin ϑ − χ ϑ_ ω2 ≈ − φ_ sin ϑ + θ_ cos ϑ ω3 ≈ χ φ_ cos ϑ + θ_ sin ϑ + ϑ_ + φθ _
³
The rotational kinetic energy
Tr = where and
Jt
Jp
1 2
´
Tr is obtained from
f
ω1
ω2
2J 4
t
ω3 g
0
0
0
Jt
0
0
0
Jp
(7.122)
38< ω 9= 5 ω :ω ; 1
(7.123)
2 3
is the principal inertia of the disc about two of the axes in the plane of the disc is
the
principal
inertia
of
the
disc
about
the
third
axis.
Upon
substituting
Eq. (7.122) into Eq. (7.123) and performing the matrix multiplications, we obtain
Tr =
1
≈
1
2
2
+
»J ±ω t
n ¹
2 1
J t φ_
±
2
²
+ ω + Jpω 2 2
+ θ_ + χ 2
χ ϑ_ Jp − Jt
2
2
²±φ_
ϑ_
2
2
¼
3
º
cos
+ Jp
¹_
ϑ + θ_
ϑ
2
sin
+ ϑ
φθ _ ϑ_
2
²¼
º (7.124)
428
Multiple Degree-of-Freedom Systems: Governing Equations where, by assuming “small” rotational speeds
χ, θ, θ_ , and φ_ and _ϑ and ϑ_ 2 are retained.
posed of the products of the terms containing
θ_
and
φ_ ,
the cubic and higher terms com-
their squares have been neglected. However,
Lagrange’s Equations and Equations of Motion Next, we use Lagrange’s equations to determine the equations governing the rigid-body motions of the disc. Since there are no potential energy terms and no damping, the potential energy
V
=
0, and the Rayleigh dissipation function
D
=
0 in Eqs. (7.7). To
F x, F y, and F z are defined along Mφ , Mθ, and Mϑ are de fi ned about the x axis obtained after a φ rotation about the O–x axis, about the intermediate y axis, and C–z axes, respectively. Thus, the forces are Q x = Fx , Qy = Fy, and Q z = F z, and the moments are given by determine the generalized forces, we note that the forces
the
O–x, O–y,
8Q >< >: QQ
φ θ ϑ
and
O–z
directions, respectively, and the moments
8 9 >= ³ ´ >< M >; = R θ >: M M yð
Þ
T
φ θ ϑ
9 2 > = 6 = > ; 4−
cos
θ
0 sin
θ
0
sin
1
0
0
cos
θ θ
38> M 75< M >:
φ θ
Mϑ
9 8 M +M θ 9 >= >= >< M >; ≈ >: M − M θ >; φ
ϑ
θ
ϑ
where we have made use of Eq. (7.113) and also assumed a “small”
θ
(7.125)
φ
rotation about 0.
After using the generalized forces and the generalized moments given by Eq. (7.125) and noting
the
dependence
of
Tt
and
Tr
equations can be written as
¶
d ∂ Tt dt ∂x_
∂T − t ∂x
= Fx
d ∂ Tt dt ∂y_
−
∂ Tt ∂y
= Fy
d ∂ Tt dt ∂ z_
−
∂ Tt ∂z
= Fz
¶ ¶
on
¶
d ∂ Tr dt ∂ φ_
−
the
∂Tr ∂φ
different
r
r
dt
ð
r
θ
tÞ
∂ϑ
∂ϑ
the
Lagrange
= Mφ + Mϑ θ
0 1 d @∂ T A ∂ T − =M dt ∂θ_ ∂θ 0 1 d @∂ T A ∂ T + T − _ r
variables,
(7.126)
= Mϑ − Mφθ
After substituting Eqs. (7.119) and (7.123) into Eq. (7.126) and performing the indicated operations, we obtain the following equations governing the translational motions
¹ ¹ €
m€x = mε €ϑ sin ϑ + ϑ_ m€ y = mε
−ϑ
cos
2
ϑ + ϑ_
m€z = F z
cos
2
º +F º
ϑ
sin
ϑ
x
+ Fy
(7.127a)
(7.127b)
(7.127c)
7.4 Rotating Shafts on Flexible Supports
²¹€ϑ
º
429
For the rotational motions, we obtain the governing equations
±
€ + J pθ_ ϑ_ = χ Jt − Jp Jtφ € + Jpϑθ
±J
t
+ Jt χ = Mϑ − Mφθ 2
p
We
can
2
substitute
Eq. (7.128a),
ϑ − ϑ_
2
ϑ
+ Mφ + Mϑ θ
± ²¹ ϑ − ϑ_ ϑº + M J€ θ − J ϑ_ φ _ = χ J − J ϑ€ ± ²± ² + mε ϑ€ + mε €y ϑ − x€ ϑ + χ J − J φ€ ϑ + €θ
Mϑ
the small angle
p
t
ð
for
Mϑ
cos
sin
from
Eq.
2
sin
p
cos
Þ
p
(7.128c)
into
is multiplied by the small angle
θ,
except for the term
sin
Jp θ€ϑ,
θ.
cos
θ
t
cos
Eq.
(7.128a).
sin
(7.128a)
(7.128b)
² ϑ
We
(7.128c)
note
that
in
When Eq. (7.128c) is multiplied by
all the other terms can be neglected, since
²¹€ϑ
º
they are higher-order terms. Therefore, Eq. (7.128a) becomes
±
Jt φ € + Jpθ_ ϑ_ = χ Jt − Jp
cos
ϑ − ϑ_
2
ϑ
sin
+ Mφ
(7.129)
If the only external forces acting on the rigid body are those due to the reactions of the elastic springs shown in Figure 7.23, then, in general,
kxx x + kxθ θ = Fx kθ xx + k θθ θ = − Mθ
(7.130)
and
kyy y − k yφ φ =
− Fy
−k φy y + kφφ φ = Mφ
(7.131)
k αβ are the spring constants for the coupled motion of the system. The quantities kxx and k yy represent the stiffness due to translations only of the shaft in the x and y directions, respectively, and kφφ and k θθ represent the stiffness of the system due to rotations only about the corresponding axes, respectively. The quantities kx θ, k θx , k yφ, and kφ y represent the stiffness due to the coupling of translation and rotation of the shaft. Also, from Maxwell’s reciprocal theorem, we have that kx θ = kθx and kyφ = kφ y. These where
spring constants are determined subsequently. In view of Eqs. (7.130) and (7.131), Eq. (7.127c) is uncoupled from all the other equations. In addition, we have already used Eq. (7.128c) to obtain Eq. (7.129). Thus, there
remain
four
equations
that
determine
the
motion
of
the
system
shown
in
Figure 7.22. After using Eqs. (7.130) and (7.131) in Eqs. (7.127a), (7.127b), (7.128b), and Eq. (7.129), we arrive at
¹ ¹
m€ x + kxx x + k xθ θ = m ε €ϑ sin ϑ + ϑ_ m€ y + kyy y − kyφ φ = m ε
−ϑ€
Jtφ € + Jp θ_ ϑ_ − kφyy + k φφφ = χ Jt − Jp Jt € θ − Jpϑ_ φ_ + kθ x x + kθθ
t
p
cos
ϑ + ϑ_
± ²¹€ϑ ± ²¹€ϑ θ =χ J − J cos
2
ϑ
º
2
sin
ϑ
º
cos
ϑ − ϑ_
sin
ϑ
sin
ϑ + ϑ_
cos
ϑ
2
2
º º
(7.132)
430
Multiple Degree-of-Freedom Systems: Governing Equations
ωs,
We now assume that the shaft rotates at constant speed and
€ = 0, ϑ
that is,
υ
= ωst
. Then,
ϑ_ = ωs
and Eqs. (7.132) become
m€x + kxx x + kxθ θ = mεωs
ωs t
(7.133a)
m €y + kyy y − k yφ φ = mεωs sin ωs t
(7.133b)
2
cos
2
± ² _ − k y + k φ = − χω J − J Jφ € + J θω ± ² _ + k x + k θ = χω J − J J€ θ − J ϑω t
p
t
s
p
φy
s
2
φφ
θx
t
s
2
θθ
p
t
s
p
sin
cos
ωs t
ωs t
(7.133c)
(7.133d)
The spring constants used in Eqs. (7.133) are determined from a static-force balance and a static-moment balance. By using a procedure similar to that shown in Figure 7.5 for
“small” θ, we find that a force balance along the O–x
direction gives
F x ≈ ðx − aθÞkLx + ðx + bθÞkRx ≈ ðkLx + kRx Þx + ð − akLx + bkRx Þθ
(7.134)
and that a moment balance in the disc plane yields
Mθ ≈ − aðx − aθÞkLx + bðx + bθÞkRx
≈ − akLx + bkRx x + ð
where
a
and
b
are de
Þ
±a k 2
Lx
+b
2
²
kRx θ
(7.135)
fined in Figure 7.23. If we drop the terms with velocities and accel-
erations from Eqs. (7.133a) and (7.133d) and compare the resulting equations with Eqs. (7.134) and (7.135), respectively, we see that
k xx = kLx + kRx kxθ = kθ x = − ak Lx + bk Rx
(7.136)
k θθ = a kLx + b k Rx 2
In a similar manner, we
2
find that kyy = kLy + kRy kyφ = kφy = −ak Ly + bk Ry
(7.137)
kφφ = a kLy + b kRy 2
2
kLx
We shall restrict ourselves to the symmetrical case where
= k Rx = kLy = kRy = k
.
Then, from Eqs. (7.136) and (7.137), we obtain
k k k
1 2
3
where
L
= k xx = kyy = k = k xθ = k θx = k yφ = kφy = kLA = k θθ = kφφ = kL A 2
1
(7.138)
2
2
is the length of the shaft and
A
1
=
−a + b L
and
A
2
=
a
2
+b
2
L
2
(7.139)
7.4 Rotating Shafts on Flexible Supports We note that quantity
k
1
A
1
=
0 when the disc is in the center of the shaft, that is, when
k
represents the stiffness due to translation only,
to rotation only, and
k
2
3
a
431
=b
. The
represents the stiffness due
represents the stiffness due to the coupling of translation and
k
rotation. When the system is restricted to translate only, then,
2
=k = 3
0.
In an attempt to identify key parameters that determine the natural frequencies of the system, we introduce the following quantities
X = x=L; Jmt =
Y = y=L;
Jt mL
;
2
Jmp =
ω1 =
Jp mL
2
skffiffiffiffi 1
m
Ωs =
;
τ = ωs t ;
;
ωs ω1 (7.140)
ε
^ε =
L
After using Eqs. (7.138) and (7.140) in Eqs. (7.133), we obtain the following nondimensional form of Eqs. (7.133):
€ + X + ðA X € + Y − ðA Y Jmt φ € + Jmp Ωs θ_ − ðA
1 =2Þ
Jmt € θ − Jmp Ωs φ_ + ð A
Y + ðA
1 =2Þ
θ = ^ε Ω2s
cos
φ =^ ε Ω2s
sin
1 =2Þ
1 =2Þ
(7.141a)
Ωs τ
± ² φ = −χ Ω J − J ± ² θ = χΩ J −J 2
2 =2Þ
X + ðA
Ωs τ
s
2
2 =2Þ
mt
mt
s
Here, an over dot indicates the derivative with respect to
mp
τ.
sin
cos
mp
(7.141b)
Ωs τ
(7.141c)
Ωs τ
(7.141d)
The right-hand side of the
first two equations contains the force components generated by the unbalance due to the eccentricity between the center of mass of the disc and the center of the disc that lies on the axis of rotation. The right-hand side of the last two equations contains rotational inertia components associated with the angular misalignment of the axis of symmetry of the disc with the axis of rotation of the shaft.
Special Cases Case 1: Translation Only When
the
system
only
translates
in,
say,
the
x
direction,
Eq. (7.141a) reduces to Eq. (5.32a) without damping (i.e., when
then
ζ
=
k
2
=
k
3
=
0
and
0).
Case 2: Translation and Rotation in One Plane When the system can both rotate and translate in only one plane, say the and the shaft is not spinning, then reduce,
after
notational
Eqs. (b) of Example 7.16.
differences
Y
=
are
φ
=
Ωs
=
accounted
CX–CZ
plane
0 and Eqs. (7.141a) and (7.141d) for
and
damping
is
omitted,
to
432
Multiple Degree-of-Freedom Systems: Governing Equations
Case 3: Centrally Located Mass When the disc is located at the center of the shaft
=b
a
and, therefore,
A
1
=
0. In this
case, the Eqs. (7.141a) and (7.141b) uncouple from each other and from Eqs. (7.141c) and (7.141d). Thus, Eqs. (7.141a) and (7.141b) become, respectively,
€ + X = ^ε Ωs X
2
€ + Y = ^ε Ωs Y
2
Each of
Eqs. (7.142) is analogous to
cos
Ωs τ
(7.142a)
sin
Ωs τ
(7.142b)
the equation obtained for a single degree-of-
freedom system subjected to a rotating unbalanced load given by Eq. (5.32a) with
ζ
=
0
in Eq. (5.32a). The solutions to Eqs. (7.142a) and (7.142b) are, respectively,
X=
^ε Ωs cos Ωs τ 1 − Ωs
(7.143a)
Y=
^ε Ωs sin Ωs τ 1 − Ωs
(7.143b)
2
2
and 2
2
The nondimensional displacement of the center of the disc denoted by
r= which, for a given value of
Ωs ,
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p
X
2
+Y = 2
^ε Ωs 1−Ω s
r is given by
2
(7.144)
2
is a constant. Thus, the displaced center of the shaft
O fixed in the inertial reference frame. This type of response is known as synchronous whirl. When Ωs = 1, that is, when ωs = ω , then the shaft is said to be rotating at its critical speed. rotates in a circle of radius
r
about the point
1
General Solution To obtain an analytical solution to Eqs. (7.140), we introduce the following complex variables
33
ξ =X
+ jY η = θ − jφ We multiply Eqs. (7.141b) and (7.141c) by (7.141b) together and subtract the modi
(7.145)
j and then add Eqs. (7.141a) and the modified
fied
Eq. (7.141c) from Eq. (7.141d) and use
Eq. (7.145) to obtain
€ξ + ξ + A η =2 = ^ε Ω ejΩs τ s
± =χ J
2
1
Jmt € η − jJ mp Ωs η_ + A ξ=2 + A η =2 1
33
See Appendix E.
2
mt
− Jmp
²Ω e 2
s
jΩs τ
(7.146)
7.4 Rotating Shafts on Flexible Supports
Free Whirling of an Undamped System The
free
whirling
of
an
undamped
Eq. (7.146) and assuming that
η
and
system
ξ
can
be
obtained
by
setting
^ε = χ = 0
(7.147)
η = ηo ejΩ τ = Θ oejΩτ − jΦoejΩτ
where
=
Ω
ω/ ω1
and
ω
is
the
vibration
³D^ ´ Z c
³D^ ´= "
where
c
1
−Ω
f
frequency.
og
=
#
Eqs.
(7.147)
in
f
&ξ ' o
ηo
2
Ωn
The natural frequencies
Zo g =
and
2
1
using
(7.148)
1
A =2
After
0
A =2 A =2 − Jmt Ω + Jmp Ωs Ω
2
in
are of the form
ξ = ξoejΩτ = Xo ejΩ τ + jY oejΩτ
Eqs. (7.146), we arrive at
433
are those values that satisfy the determinant
∣ D^ c ∣=
0, that
is, when they are the solutions to
Jmt Ω
4
− Jmp Ωs Ω − 3
After solving numerically
34
Jmt + A =2ÞΩ
ð
2
2
+ Jmp Ωs Ω + A = − A = = 2
2
2 1
4
0
for the four roots of Eq. (7.149) as a function of
(7.149)
Ωs,
we
obtain the results shown in Figure 7.24. This diagram is referred to as a Campbell
3 a/L = 0.1 a/L = 0.5 2
Ω
1
0
–1
0.0
0.2
Figure 7.24.
disc with
34
0.4
0.6
Ωs
0.8
1.0
Campbell diagram for a rigid shaft rotating at
Jmt = 3 and Jmp = 5.
Using Matlab,
roots; using Mathematica, NRoots .
1.2
Ωs
1.4
for two different locations of the
434
Multiple Degree-of-Freedom Systems: Governing Equations
Ωs ,
diagram. For each value of
two of these roots are positive and the other two are
negative. The positive roots correspond to whirling in the forward direction, and the negative roots correspond to whirling in the reverse or backward direction.
7.5
STABILITY
Extending the notion of bounded stability presented in Section 4.3, a linear multidegree-of-freedom and subjected to
stable
system
finite
described
A
has a
response {
x}
finite
the
generalized
initial conditions and
if
‖ fxg‖ = where
by
finite
coordinates
x
,
1
x
,
2
…, xN
forcing functions is said to be
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x
2 1
+ x +⋯ + xN ≤ A 2
2
for all time
2
(7.150)
value. This is a boundedness condition, which requires the system
fied,
to be bounded for bounded system inputs. If this condition is not satis
then the system is said to be
EXAMPLE 7.24
unstable.
Stability of an undamped system with gyroscopic forces
We revisit Examples 7.4 and 7.22 and determine under what conditions the gyro-sensor
ωz .
is unstable during free oscillations at angular speeds to determine whether
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x
2
+ y ≤A 2
Based on Eq. (7.150), we need
for all time
(a)
is true. For free oscillations of the gyro-sensor based on the discussion of Section 7.3.3, the motion is described by
& x t ' & X' =
ð Þ
yðtÞ
where
λ
Y
eλt
(b)
are the eigenvalues. If one or more of the eigenvalues have a positive real part,
then Eq. (a) will not be true for all time. We can use this as a basis to determine the angular speeds
ωz
for which the system will be either unstable or stable.
For the special case treated in Example 7.22, the eigenvalues are given by Eqs. (g). From these eigenvalues, we arrive at the following angular speed range for instability
rffiffiffiffi ωz
2
which is not possible.
k k > + ωz m m
2
→
0
>
rffiffiffiffi
k − ωz m
¶
2
(c)
7.5 Stability
EXAMPLE 7.25
435
Wind-induced vibrations of a suspension 35
bridge deck: stability analysis
Consider a section of a deck of a suspension bridge that is subjected to a wind with a speed
V.
Let
us
assume
that
a
section
of
the
deck
fied
Figure 7.25, where the linear springs are speci the deck. The wind produces a lifting force as shown in the
figure.
F
can
be
modeled
as
shown
in
in terms of stiffness per unit length of
in the vertical direction and a moment
M,
The vertical motion of the deck is independent of the torsional
motion. If the deck has a mass per unit length
m
and a mass moment of inertia
J
about
the longitudinal axis of the deck, then the governing equations of motion in the vertical and torsional directions are, respectively,
m€y + Ky = F J θ€ + KL θ = M
(a)
2
where we have used Eq. (k) of Example 7.3 with
k
1
=
k
2
=
K/2
and
L
1
=
L
2
=
L.
Through this example, we shall illustrate how to carry out a stability analysis for a system with two degrees of freedom. Under some simplifying assumptions, the wind-induced lifting force and moment can be expressed as
F = cF ρLV
M = cM ρL V 2
where
cF
and
cM
are constants and
ρ
y_ θ− V
2
"
¶
y_ θ− V
2
¶
m€y + Ky = cF ρLV
y_ θ− V
2
J€ θ + KL θ = cM ρL V 2
y x
K/2 O V Figure 7.25.
35
M. Roseau,
z
K/2
K/2 L L
#
(b)
is the density of air. Upon substituting Eqs. (b)
into Eqs. (a), we obtain the coupled equations
K/2
L − θ_ V
2
" 2
K/2
y M
O
¶
y_ θ− V
¶
L − θ_ V
#
K/2 x
F
Model of a bridge deck subjected to a wind with a speed
V.
Vibrations in Mechanical Systems: Analytical Methods and Applications , –
1983, pp. 236 243.
(c)
Springer-Verlag, Berlin,
436
Multiple Degree-of-Freedom Systems: Governing Equations Next, we introduce the following quantities
w= MF = Ωy = 2
y ; L
cF ρL m ω2y ω2o
V ; L
ωo = 2
K ; m
cM ρL J
MM =
;
KL J
ω2θ = 4
2
τ = ωo t
;
ω2θ
Ωθ = 2
;
ω2y =
ω2o
into Eqs. (c) and obtain
€ θ + MM
w € + MF w_ + Ωyw − MF θ = 0
± θ_ + Ω
²θ + M 2
2
θ
− MM
_ Mw
=
(d)
0
where the over dot now indicates the derivative with respect to
τ.
To investigate the stability of the system, we assume a solution of the form
wðτ Þ = Woeλτ
(e)
θðτÞ = Θ oeλτ
We substitute Eqs. (e) into Eqs. (d) to obtain
"λ
2
+ MF λ + Ωy 2
MM λ
λ2
Setting the determinant of the coef fourth-order polynomial in
− MF + MM λ + Ωθ − MM ficients
¹
=
o
Θo
2
λ
λ4 + ðMF
#& ' & ' W
2
3
(f)
0
in Eq. (f) to zero, we obtain the following
¹ + MM λ + ºΩy + Ωθ±+ MM MF²− MM + Ωy MM + Ωθ MF λ + Ωy Ωθ − MM = Þ
0
2
2
2
2
2
º
λ2
0
(g)
The stability conditions for this system are determined by examining certain combina-
fi
tions of the coef cients of this polynomial using the Routh-Hurwitz stability criterion.
36
36
The polynomial given by Eq. (g) is of the form
λ4 + b1 λ3 + b2 λ 2 + b3 λ 2 + b3 λ + b4 = 0 where
bj
≠
0,
j=
1, 2, 3, 4. For none of the roots of the polynomial to have positive real parts, the following
fi
criteria must be satis ed
b
1
>
0;
½½ ½½ bb
1 3
1
b
2
½½ ½½ >
½½ b ½½ b ½½
1
0;
3
0
1
b b
2 4
0
b b
1 3
½½ ½½ > ½½
½½ b ½½ ½½ b ½½
1
0;
3
0 0
1
b b
2 4
0
0
b b
1 3
0
0 1
b b
2 4
½½ ½½ ½½ > ½½
0
7.5 Stability
EXAMPLE 7.26
A basic model
37
437
Disc brake squeal
that has been used to understand why disc brakes squeal is that shown
in Figure 7.26. The system is subjected to a normal force that preloads the disc and the pad so that they do not separate. Between the brake pad and the brake disc is a spring
k
ficient
. The brake disc is a weightless surface that has a constant coef
3
The elongations of the top springs
Δj
Figure
on
7.27
and
the
forces
acting
of friction
μ.
are determined from the geometry shown in the
mass
are
shown
in
Figure
7.28.
From
Figure 7.27, it is seen that
Δ Δ
1
2
= −x =x
π − α1Þ + y sinð π − α1Þ = x cos α1 + y sin α1
cos ð
cos
α2 + y sin α2
Figure 7.26. A basic model that can be used to understand
y k1
(a)
brake squeal.
k2
α1 α2
m
x
Brake pad
k3 Brake disc FN
FN
m
Figure 7.27.
Δ2
k2
y sinα 2
x cosα 2
Displacement components of one spring.
α2 y
α2 x
37
N. Hoffmann, M. Fischer, R. Allgaier and L. Gaul, coupling
–
197 205.
type
instability
in
friction
induced
“A
minimal model for studying properties of the mode-
” Mechanics Research Communications,
oscillations,
29
(2002)
438
Multiple Degree-of-Freedom Systems: Governing Equations F2 = k2 Δ2
F 1 = k1Δ1
Figure 7.28. Forces on the mass.
mx my m
F N = k3 y
FN =m k3y
x-direction
From Figure 7.28, a summation of forces in the
(horizontal direction)
gives
− m€x + F m €x + k Δ α +k x
cosð
1
m€ x + k ðx cos α 1
1
+y
1
sin
α1Þ cos
2ð
1
π − α1 Þ − F2
cos
1
α1 + k 2Δ2
α2 + μk3 y = 0
cos
α2 − μk3 y = 0
cos
α2 + y sin α2 Þcos α2 − μk3 y = 0
cos
m€x + k x + ðk 11
12
− μk
(b)
y= 0
3Þ
where
k k
11
12
=k =k
A summation of forces in the
1
cos
1
sin
2
α1 + k2
α1
cos
cos
2
α1 + k 2
α2 sin
α2
cos
y-direction (vertical direction) yields
− m€y − F
1
π − α1Þ − F2
sinð
m€y + F sin α + F m€y + k Δ sin α + k Δ 1
m€ y + k ðx cos α 1
1
+y
(c)
α2
1
sin
1
1
1
2
sin
2
sin
2
sin
α2 − k3 y = 0
α2 + k3 y = 0
α2 + k3 y = 0
(d)
α1Þsin α1 + k2 ðx cos α2 + y sin α2Þsin α2 + k3 y = 0
m€y + k x + k y = 0 12
22
where
k
22
=k
1
2
sin
α1 + k2
2
sin
α2 + k 3
(e)
Thus, the system is described by the following coupled equations
m€ x + k x + ðk − μk Þy = 0 m€y + k x + k y = 0 11
12
3
12
22
(f)
It is conjectured that brake squeal will occur when the system goes into oscillatory motion. This transition to instability is determined by assuming a solution to Eqs. (f) of the form
x = X oeλt
y = Y oeλt
(g)
7.5 Stability Then, substituting Eq. (g) into Eq. (f) yields
+k
mλ
2
k
k mλ
11
12
− μk +k
¶& ' X o
3
Yo
2
12
22
=
0
439
(h)
Before proceeding, we introduce the following notation into Eq. (h)
ωn =
rkffiffiffiffi 1
m
K
rad=s;
k k
=
2
2
K
;
3
=
1
+ ~k
2
~k
~k
11
12
^λ
2
12
3
;
^λ = λ
1
and obtain
^λ
k k
¶( )
− μK + ~k
Xo
3
Yo
22
=
ωn
(i)
0
(j)
α2
(k)
where
~ k ~ k ~ k
11
12
22
= = =
cos sin
α1
2
sin
α1 + K 2
2
cos
cos
2
α1 + K 2
α1 + K2
2
sin
α2 sin
α2
cos
α2 + K 3
To reduce the number of variables, we shall assume that the springs are at equal angles with respect to the horizontal, that is,
α1
+α =π 2
. In this case, noting that
α1Þ = sinð π − α2Þ = sinðα2Þ
sinð
α1Þ = cosðπ − α2Þ = − cosð α2Þ
(l)
cos ð
Eqs. (k) become
k~ k~
11
12
k~
22
= = =
cos sin
2
α2ð 1 + K2 Þ
α2
α2ðK 2 − 1Þ
cos
(m)
α2 ð1 + K 2Þ + K 3
2
sin
It is noted from Eqs. (f) and (h) that one of the equations becomes independent of the other when
k~
12
=
sin
α2
cos
α2ðK 2 − 1Þ = 0
(n)
or
μ= The
x and y
k~ K
12
(o)
3
equations completely uncouple from each other when
~ k
12
− μK = 3
sin
α2
cos
α2ðK 2 − 1Þ − μK3 = 0
(p)
440
Multiple Degree-of-Freedom Systems: Governing Equations It is seen from Eq. (n) that one situation in which the system uncouples is when that
is,
the
two
springs
at
the
top
x
equation is uncoupled from the
have
equal
stiffness
equation, while the
x
values.
In
this
K
2
case,
=
1;
y y.
the
equation is dependent on
Thus, one must examine Eqs. (o) and (p) to understand the nature of the coupling.
λ
The values of
are determined by setting the determinant of the coef
ficients
of Eq. (j)
to zero. Then the determinant of Eq. (j) can be written as
^λ
4
+ a ^λ + a = 2
2
0
0
(q)
where
a
2
= +K +K > 1
2
3
0
(r)
and
a
0
=K
α2 Þ + K 3ð1 + K 2Þcos2 α2 + μK3 ðK2 − 1Þsinðα2Þcosðα2Þ
2
2
sin ð 2
The solution to Eq. (q) is
(s)
pffiffiffiffiffiffiffi
^λ = ± b
(t)
1;2
where
b
1;2
=
1
¹
2
When
ds
>
b
1,2
p
ffiffiffiffid < a
−
a
4
0
are complex quantities. The system decays to zero when
s
0,
ds
(u)
2
In general,
pffiffiffiffiº
2
2
ds = a
−a ±
2
;^ λ
Re
³^λ´ ≤
0.
is imaginary and the system goes into oscillatory motion with
constant amplitude. This oscillatory state can be considered as an unstable motion. The value of
μ
for which
ds
=
μ c.
0 is denoted as
μ
the critical value of friction. For
> μc
This value, which is given below, is
, the system remains in an oscillation state. The
goal in the elimination of brake squeal is to have
μc
as large as possible, so that the
system does not go into an oscillatory state. The value of (s), and (u) as
μc = provided that
K
3
≠
+K +K −
ð1
0 and
2
3Þ
2
±K
4
K ðK
4
K
2
≠
1.
3
2
2
2
is obtained from Eqs. (r),
α2Þ + K 3ð1 + K2Þcos2 α2
sin ð2
−
μc
α2 Þcosðα2 Þ
1Þsinð
² (v)
7.6 Summary
441
Table 7.2. Effects of various parameters on the values of μc μc α1 α2
K K K K
The
effects
2 3
2 3
= = = =
of
3
= ° = °
α1 = α2 =
145 35
150 30
0.549
0.385
0.872
0.717
°
°
3 5 8.5
several
different
combinations
of
parameters
on
μc
are
given
in
Table 7.2.
7.6
SUMMARY
In this chapter, the derivation of the governing equations of a multi-degree-of-freedom system was illustrated by making use of force-balance and moment-balance methods and Lagrange’s equations. Linearization of the equations governing a nonlinear system was also addressed. Means to determine the natural frequencies and mode shapes were introduced, and the notions of orthogonality of modes, modal mass, modal stiffness, proportional damping, modal damping factor, node of a mode, and rigid-body mode were introduced. The conditions under which conservation of energy, conservation of linear momentum, and conservation of angular momentum hold during free oscillations of a multiple degree-of-freedom system were determined. Finally, the notion of stability of a linear multi-degree-of-freedom system was introduced. Interactive graphics materials have also been provided to explore the natural frequencies and associated mode shapes of two types of two degree-of-freedom systems.
442
Multiple Degree-of-Freedom Systems: Governing Equations
Exercises
Section 7.2.1 7.1
Consider the in
Figure
“small”
7.9
and
amplitude motions of the pendulum-absorber system shown
derive
the
equations
of
motion
by
using
force-balance
and
moment-balance methods. 7.2
Derive the equations of the hand-arm system treated in Example 7.11 by using force-balance and moment-balance methods for
“large”
“small”
and
oscillations
about the nominal position. 7.3
A container of mass
mc
is suspended by two taut cables of length
as shown in
To. Inside the container, a mass m is elasti-
Figure E7.3. The tension in the cables is cally supported by a spring
L
k.
(a) Determine the equivalent spring constant for the cable-mass system and sketch the equivalent vibratory system. (b) For
the equivalent system determined in
part (a),
determine the
equations
governing the motion of this system.
mc
Figure E7.3.
Cable
Cable
L
7.4
k
m
L
A two degree-of-freedom system with a nonlinear spring element is described by the following system of equations:
αðx2 −x1Þ
= =
Determine the system equilibrium positions. Assume that
m
m € x 1
1
+k
1
x
1
+ αx + k m €x + k 3
2
positive and
α
x
1
2ð
2
2ð
1
x
2
−x +k −x +k
αðx1 −x2Þ
3
2Þ
2
1Þ
2
3
0 0
,
1
m
,
2
k
, and
1
k
2
are
is negative.
Section 7.2.3 7.5
Derive the equations of motion of the systems shown in Figures E7.5a and E7.5b and present the resulting equations in each case in matrix form.
Exercises k2
k1
k4
k3
m1
Figure E7.5.
m3
m2
c1
443
c4
c3
c2
(a) k4 k1 m1
c4
k2
m2
c1
k5 m3
k3
c5
(b)
7.6
Derive the equations of motion for the model of an electronic system tained in a package
m
m
Figure E7.6.
m2
k2
c2
k1
c1
m1
x1 (t)
x3
7.7
Derive the equations of motion of the vehicle model shown in Figure E7.7.
Automobile body Suspension
m1 c1
Axle and wheel mass Tire stiffness and damping
Figure E7.7.
y1 Absorber
k1
m3 k3
k2
y3 c3
m2
y2 c2 ye
con-
, as shown in Figure E7.6, and present them in matrix
1
form.
x2 (t)
2
444
Multiple Degree-of-Freedom Systems: Governing Equations Derive the equations of motion of the pendulum absorber shown in Figure E7.8
7.8
for
large
oscillations
equilibrium
position
and
then
linearize
corresponding
to
these
the
equations
bottom
position
about
the
of
pendulum.
the
static-
Present the linearized equations in matrix form.
Figure E7.8.
M
F o cos t
x
ct j
l
i
l
.
c
k m mg
Derive the equations of motion of the system shown in Figure E7.9 by using
7.9
Lagrange’s equations. Present the equations in matrix form.
m1
x1
Figure E7.9.
x2
k1 k2
k4 k3
m2
c1
7.10
c2
Replace each of the linear springs
k
2
and
k
3
shown in Figure E7.9 by a nonlinear
spring whose force-displacement characteristic is given by
±
F ðxÞ = k x + αx
3
²
and determine the resulting equations of motion. 7.11
Derive
the
equations
of
the
milling
model
shown
in
Figure
7.4b
by
using
Lagrange’s equations. 7.12
Derive the equations of motion of the system shown in Figure E7.12, which is an extended version of a two degree-of-freedom system shown in Figure 7.3. Let
Mo(t)
be the external torque that acts on the disc whose motion is described by
the angular variable
φ1.
Exercises Oil housing
Figure E7.12.
c t2 ct1
c t3
Jo2
Jo1 kt1
Jo3
kt2
kt3
k
1
Mo(t)
2
The
7.13
445
experimental
described
by
the
3
arrangement model
shown
for in
an
airfoil
Figure
mounted
E7.13.
in
Determine
a
wind
the
tunnel
equations
is of
k, the kt, G is the center of the mass of the airfoil located attachment point O′, m is the mass of the airfoil, and JG is
motion governing this system when the stiffness of the translation spring is stiffness of the torsion spring is a distance
l
from the
the mass moment of inertia of the airfoil about the center of mass.
O m, JG
Figure E7.13.
k O'
G
kt
l
A multistory building is described by the model shown in Figure E7.14. Derive
7.14
the equations of motion of this system and present them in matrix form. Are the mass and stiffness matrices symmetric?
x4
m4 k4
k4 m3
k3
x3 k3
m2 k2
x2 k2
m1 j
k1
Figure E7.14.
x1 k1 i
7.15
Obtain the governing equations of motion for large oscillations of the system shown in Figure E7.15 in terms of the generalized coordinates
x
,
1
x
, and
2
θ.
446
Multiple Degree-of-Freedom Systems: Governing Equations
k
The spring
3
is attached at the midpoint of the bar. Linearize the resulting system
of equations for “small” motions about the system equilibrium position and present the resulting equations in matrix form. Point
L1
L2
x1
Figure E7.15.
G
m1 , JG k1
k3
k2 x2
m2
7.16
G is the center of mass of the bar.
A pair of shafts are linked by a set of gears as shown in Figure E7.16a. An equivalent system is determined as shown in Figure E7.16b, where
^ kt
= kt TR 2
2
and the transmission ratio
2
TR
=
ωout/ωin.
J^ i = Ji TR; 2
Determine the governing
equations of motion of the system shown in Figure E7.16b in terms of the generalized coordinates
J3
^ 2; and ϕ ^4 ϕ 1; ϕ
^ i = ϕi =TR ; j = 2; 4: ϕ
J4
kt2
Figure E7.16.
1
out
J1
where
2
4
in
kt1
J2
J1
kt1
kt2
J4
J3 + J2 (a)
7.17
(b)
The mass
m shown
Wrapped
around
unstretched length
in Figure E7.17 slides in a gravity the
L.
rod
is
a
massless
spring
of
field
along a massless rod.
constant
k
that
has
an
One end of the rod is attached to a vertically moving pivot
that oscillates harmonically as
zðtÞ = zo cos ωt The position of
the mass along the
rod is
u(t),
which is measured from the
unstretched position of the spring. Use Lagrange’s equations to obtain the nonlinear equations of motion. Figure E7.17.
g
L
z(t)
y u(t)
k (t) x
m
Exercises 7.18
447
Consider the autoparametric vibration absorber shown in Figure E7.18. The system composed of mass
m
1
and spring
k
is externally excited by a harmonically
1
oscillating force
f ðtÞ = fo cos ωt The oscillations of this primary system are to be attenuated by attaching another
m
system composed of a mass length tance
3
that is attached to a rigid rod of mass
l
. The base of the rod is pivoted on mass 2
ls from the pivot is a spring k
m
m
2
and
. Attached to the rod at a dis1
.
2
(a) Determine expressions for the kinetic energy and the potential energy of the system.
2 ¶ dθ + X −γ 4
3 5=F
(b) Show that the nonlinear equations of motions for this system are
d X dτ 2
2
2
2
−γ
cos
dτ
1
d θ dτ
2
θ+
d X ω sin θ + dτ ω 2
2
2
d θ sin θ dτ 2
2
2
sin
θ
X = x=l
2;
2 2
o
cos
Ωτ
θ =0
cos
1
where
τ = ω1 t ; ω21 = γ1 =
k m +m
1
1
ls
f(t)
m1
l3
l2
x(t)
k1
2 1
ω ω1
+m
;
ω22 =
;
3
+ ml l +m +m 2
3 3= 2
2
Fo =
fo k l
;
1 2
3
Figure E7.18.
m3
m2
2
m 2 m 1
(t)
k2
Ω=
;
γ2 =
k ls
2
2
+m l m + ml l ml +m l
m l
1 2
2 2 2 =3
l
2 2
ð
2
2
2 3 3
2
2 2 =3
3 3= 2 Þ 2 3 3
448
Multiple Degree-of-Freedom Systems: Governing Equations 7.19
For
the
system
shown
in
Figure
E7.19,
Assume that the length of the pendulum is
x1
determine
the
equations
of
motion.
L.
Figure E7.19.
r
m2
g
m1 x2
7.20
Consider the system that is shown in Figure E7.20, where a uniform rod of length
L L
2
m
and mass
is suspended from a massless rod of length
is connected to the rod of length
1
L
2
L
. The rod of length
1
by a frictionless hinge. Determine the
equations of motion.
O
Figure E7.20. i
L1
j
1
L2
g
2
m
7.21
Derive the governing equations of the frictionless system shown in Figure E7.21. Assume motions about the static-equilibrium position.
Figure E7.21.
k2
k1
m, j o
y
O r R k3
Exercises
449
Section 7.3.1 7.22
Determine the natural frequencies and mode shapes associated with the system
m
shown in Figure E7.22 for and
k1
7.23
k
3
=
1
=
m
10 kg,
2
=
20 kg,
k
1
=
100 N/m,
k
2
=
100 N/m,
50 N/m.
k2
m1
k3
m2
Figure E7.22.
Determine the natural frequencies and mode shapes associated with the system
m
shown in Figure E7.23 for
=
1
–3
10
kg,
m
2
=
0.01 kg, and
=k =
k
1
2
2 kN/m.
Include plots of the mode shapes.
Figure E7.23.
y2 (t)
m2
k2
m1
y1(t)
k1
7.24
k
For the system in Exercise 7.10, remove the nonlinear spring
xo,
displacements about
set the damping coef
ficients c
1
=c = 4
, consider small
3
0, and consider the
free oscillations of this system. Choose the values of the parameters as follows:
=
10 kg,
m
=
2
2 kg,
=k =k=
k
1
4
10 N/m, and
linear spring is initially compressed by
xo
=
α
=
–2
2 m
m
1
. Assume that the non-
0.05 m. Determine the natural fre-
quencies and mode shapes associated with free oscillations about the equilibrium position. 7.25
Consider a system with the following inertia and stiffness matrices:
M± =
°
½
2
0
0
6
!
kg;
½
K± =
°
10 000
−k
If the modes of the system are given by
Xg
f
1
=
&
0:5414 1
'
and
Xg
f
2
=
−k
k
&−
!
N=m
22
5:5575 1
'
450
Multiple Degree-of-Freedom Systems: Governing Equations
X}
and the natural frequency associated with {
fi
1
is 19.55 rad/s, then determine
the unknown coef cients in the stiffness matrix and the other natural frequency of the system. Determine whether the system shown in Figure E7.26 has any rigid-body modes.
7.26
x1
x2 k1
m1
x3 k3
m2
Figure E7.26.
m3
Let the system shown in Figure E7.26 represent a system of three railroad cars
7.27
with masses
m
1
=m =m = 2
3
1200 kg and interconnections
k
1
=k =
4800 kN/m.
3
Determine the natural frequencies and mode shapes of this system and plot the corresponding mode shapes. A
7.28
flexible
structural system is represented by the model shown in Figure E7.28.
Determine the governing equations of motion of this system, and from these three equations, determine the eigenvalues and eigenvectors associated with free oscillations of this system. Find the locations of the nodes for the different mode shapes.
x1 j
x2
m
k
i
x3
2m
k
Figure E7.28.
m
Consider the system shown in Figure E7.29 in which the three masses
7.29
m , m , and m 1
2
3
flexural rigidity EI. The inverse of the stiffness matrix for this system ½ K ±, which is called the flexibility matrix, is given by are located on a uniform cantilever beam with
K ±−
½
1
=
L 3EI 3
2 4
27
14
4
14
8
2:5
4
2:5
1
3 5
Determine the following: (a) the stiffness matrix of the system, (b) the governing equations of motion, and (c) when
m
1
=m =m =m 2
3
, determine the natural fre-
quencies and mode shapes of the system. For part (c), let express the natural frequencies in terms of
y1 L
m1
y2 L
m2
y3 L
m3
Figure E7.29.
α.
α
=
EI/mL
3
3
and
Exercises 7.30
The stiffness matrix for the system shown in Figure E7.14 is given by the following matrix
2 EI 6 66 − K = L 4
−
24
½
±
3
12
If all the masses of the
m
3
=m =m
−
0
0
−
12
0
12
24
−
0
12
0
−
12
3 77 75
12
floors of the four-story building are equal, that is, m
1
=m =
4
= EI mL
α
/
α.
3
and express the natural frequencies in
Repeat Example 7.16 for the following values of the nondimensional ratios: 0.5,
L
21
2
, then determine the system eigenvalues and eigenvectors and plot the
eigenvectors of this system. Let terms of
12
24
0
7.31
451
=
2.0, and
ωr
=
1.5.
ωr
=
2.0 and
k
21
=
Jr = 1.0.
7.32
Repeat Example 7.15 when
7.33
An elastically supported machine tool with a total mass of 4000 kg has a resonance frequency of 80 Hz. An 800 kg absorber system with a natural frequency of 80 Hz is attached to the machine tool. Determine the natural frequencies and mode shapes of this system.
7.34
A six-cylinder, four-cycle engine driving a generator is modeled
38
by using an
eight degree-of-freedom system shown in Figure E7.34. Free oscillations of this system are described by the following system
» ¼+ K
€ J± ϕ
½ where
J±
½
1
2
1
3
2
2 66 66 6 = 66 66 66 4 4
3
C. Genta,
ibid.
=
f0g
0
0
0
0
0
0
0
0
21
0
0
0
0
0
0
0
0
21
0
0
0
0
0
0
0
0
21
0
0
0
0
0
0
0
0
21
0
0
0
0
0
0
0
0
21
0
0
0
0
0
0
0
0
98
0
0
0
0
0
0
0
0
49
6
5
7
6 Generator
38
t ± fϕg
21
5
4
½
8
7
3 77 77 77 77 77 75
Figure E7.34.
8
kg
⋅
2
m
452
Multiple Degree-of-Freedom Systems: Governing Equations and
2 66 − 66 66 = 66 66 66 64
−
51 51
Kt ±
0
−
102
−
0
½
51
51
51
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
−
102
−
0
51
51
−
0
0
102
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
−
51
51
102
−
51
−
51
−
117
−
66
66
0
−
81
−
0
15
15
3 77 77 77 77 × 77 77 75
6
10
N
⋅
=
m rad
15
Determine the natural frequencies and mode shapes associated with the system and plot the mode shapes. Does the system have any rigid-body modes? 7.35
Two identical discs of rotary inertia
JO
and radius
r
are mounted on identical
shafts and undergo torsional oscillations. Each shaft has a torsional spring con-
k t.
stant
At a radial distance
a
from their respective centers, the rotors are con-
nected via a translation spring of constant
⋅
2
10 kg m ,
r
=
0.25 m,
kt
=
⋅
600 N m/rad,
k, as shown in k = 1000 N/m,
Figure E7.35. If and
a
=
JO
=
0.15 m, then
determine the natural frequencies of the system.
Jo
Jo Top view
a
r
Figure E7.35.
a
r
k
Side view
7.36
kt
A cable
kt
fixed at one
end and carrying a mass
m at
the other end is stretched over
two pulleys, as shown in Figure E7.36. The pulleys have rotary inertia about their respective rotation centers, and the corresponding radii are
J r
1
1
and and
J r,
2
2
respectively. The stiffness of the various sections of the cables is provided in the
figure.
fi
Assume that there is suf cient friction so that the cable does not slip on
the pulleys. (a) Determine the equations governing this vibratory system and place them in matrix form. (b) If
k
1
m
=
=
12 kg,
k
3
=
25
J
1
×
=
J = 0.3 kg m , r = 120 mm, r and k = 40 × 10 N/m, determine 2
0.2 kg m , 3
10
N/m,
frequencies and mode shapes.
2
2
1
3
2
2
=
160 mm,
the natural
Exercises k2
453
Figure E7.36.
J1
J2
r1
r2
k1
k3 m
An electrical motor and pump system operates at 1800 rpm, and they are elasti-
7.37
cally mounted to a support structure. The mass of the system is
m
1
=
25 kg and
the effective viscous-damping factor of the mount is 0.15. Unfortunately, it is
ωn1
found that 1800 rpm coincides with the natural frequency
of the system and
the horizontal amplitude of the system is excessive. To decrease the magnitude of the horizontal amplitude, it is decided that rather than change the stiffness of the support a second mass
m
2
=
0.25 kg will be added to the system by attaching
m
2
to the end of a cantilever beam, as shown in Figure E7.37. The cantilever beam is a solid circular rod 6 mm in diameter. Use
E
=
1.96
×
11
10
N/m
2
for the Young’s
modulus of elasticity of the beam material. What should be the length of the rod so that the natural frequencies of the modi
ωn1
±
fied
system are not in the range
4%?
x2
m2
Figure E7.37.
kc k1
7.38
x1
m1
Consider the coupled pendulum system shown in Figure E7.38. (a) Use Lagrange’s equations
to
derive the
equations of
effects of gravity. Put the results in matrix form. (b) For and
m L
1
2
system.
= =
10 kg,
m
2
=
15 kg,
m
3
=
5 kg,
k
1
=
k
2
=
motion. Include
100 N/m,
L
1
=
the
0.5 m,
0.4 m, determine the natural frequencies and mode shapes of the
454
Multiple Degree-of-Freedom Systems: Governing Equations Figure E7.38.
m1 m2
L1 k1
7.39
L2
k2
m3
One model that has been used to study the vibratory motion of motor vehicles is shown in Figure E7.39. The body of the vehicle has a mass
JG
m
1
and a rotary inertia
about an axis through the center. The elasticity of the tires is represented by
springs
k
, and the elasticity of the suspension by springs
2
assemblies is
k
. The mass of the tire
1
m
2.
(a) Determine the matrix form for the governing equations of the system. (b) Obtain the natural frequencies and mode shapes for the case where
m
800 kg,
2
⋅
=
25 kg,
k
1
=
60 kN/m,
k
2
=
20 kN/m,
L
=
m JG 1
1.4 m, and
= =
2
180 kg m .
x1 m1 , JG
Figure E7.39.
G
2L k1 x2
k1
m2
x3
m2
k2
7.40
k2
A tractor-trailer is hauling a large cylindrical drum that is elastically supported by a spring
k
2
as shown in Figure E7.40. The drum rolls on the
floor
of the trailer
without slipping. The trailer is attached to the tractor by a system that has an equivalent spring
k
that of the drum is
x1
. The mass of the tractor is
1
m
. 3
x2 m1
k1
Figure E7.40.
k2
r m3
m2
m
, that of the trailer is
1
m
, and
2
Exercises
455
(a) Obtain the equations of motion for this system.
m = m = m, and m = 2m/3, then obtain the cies in terms of k/ m and the corresponding mode shapes.
(b) If
7.41
k
1
=k =k
,
2
1
2
3
natural frequen-
Determine the characteristic equation for the system shown in Figure E7.41 and solve this equation for the special case when
m
3
=m
x1
x3
k1 k2
m1
k
1
=k =k =k 2
3
and
m
1
=m = 2
. Determine whether the system has any rigid-body modes.
k3
m2
Figure E7.41.
m3
x2
7.42
Two discs that roll without slipping on a by a spring with constant
k,
flat
surface are connected to each other
as shown in Figure E7.42. Determine the natural fre-
quencies and mode shapes of the system.
x1 J1 , m1
x2
Figure E7.42.
k r1
r2
J2 , m2
Section 7.3.2 7.43
Using Lagrange’s equations for “small” oscillations about the static-equilibrium position, obtain the (a) Case 7 (b) Case 8 (c) Case 9 (d) Case 10 (e) Case 11 (f) Case 12 (g) Case 13 (h) Case 14 (i) Case 15 (j) Case 16 (k) Case 17 (l) Case 18 (m) Case 19
k ij and the m ij for the following cases appearing in Table 7.1:
456
Multiple Degree-of-Freedom Systems: Governing Equations Section 7.3.3 7.44
Show that the linear independence of eigenvectors given by Eq. (7.72) is true by making use of the orthogonality property of eigenvectors.
7.45
Is
it
possible
for
8< =:
eigenvectors?
Xg
f
a
1
three
1
degree-of-freedom
9= ;
1=2
8< − 9= =: ;
Xg
f
2
Xg
;
1
to
have
the
following
8< 9= =: ; 0
1
;
1
system
f
3
2 1
1=2
Section 7.3.4 7.46
Determine the modal mass, modal stiffness, and modal damping factors associated with the system whose mass matrix, stiffness matrix, and damping matrix
°
!
are given by the following:
½
7.47
M± =
1
0
0
2
;
½
K± =
°
−
1
−
1
1
2
! ;
½
C± =
°
−
3
−
2
2
!
6
Show that the system treated in Example 7.21 is proportionally damped by making use of Eqs. (7.99).
7.48
To describe the vertical motions of an automobile, the two degree-of-freedom sys-
quarter-car model. kg, k = 30 kN/m,
tem shown in Figure E7.48 is used. This model is known as a If the parameters of the system are
k
1
=
300 kN/m, and
c
1
=
m
1
=
80 kg,
m
2
=
1100
2
5000 N/(m/s), determine whether the system is propor-
tionally damped.
Automobile body (sprung mass) Suspension
c2
k2
Axle and wheel mass (unsprung mass) Tire stiffness
7.49
Figure E7.48.
m2
m1 k1
The modal matrix and the damping matrix for a two degree-of-freedom system are, respectively,
½
Φ= ±
°
1
−
1
1
1
! and
Is this system proportionally damped?
½
C± =
°
5
0
0
0
!
Exercises 7.50
457
Mo(t) = 0. Jo = 4 kg⋅ m , Jo = 1 kg⋅m and the torsional stiffness of each shaft be as follows: k t = k t = 10 N⋅ m/rad and k t = 5 N⋅ m/rad. In addition, let the damping coeffi cients be such that ct = 0.5 N⋅m/rad/s, ct = 2 N ⋅m/rad/s, and ct = 0.5 N⋅ m/rad/s. Determine the following. For the system shown in Figure E7.12, assume that the drive torque
Let
Jo
1
=
⋅
2
1 kg m ,
2
2
2
3
1
2
3
1
2
3
(a) Is the system proportionally damped? (b) If the damping associated with the second
⋅
flywheel changes
⋅
from 2 N m/rad/s
to 1.8 N m/rad/s, can the system be approximated as a system with modal damping? 7.51
The eigenvalues associated with a damped three degree-of-freedom system are given by the following:
λd 1
1; 2
λd 2
1; 2
λd 3
1; 2
=− =− =−
∓j ∓j ∓j
0:1
0:995
0:75 0:4
1:299
1:960
Determine the system natural frequencies, modal damping factors, and damped natural frequencies.
Section 7.3.5 7.52
Consider free oscillations of the gyro-sensor treated in Example 7.4 and determine whether energy is conserved in the system.
7.53
Consider the three degree-of-freedom system shown in Figure E7.26 and determine
whether
the
linear
momentum
and
the
total
energy
of
this
system
are
conserved.
Section 7.5 7.54
The eigenvalues determined for two different three degree-of-freedom systems are as follows: (a)
λd1
1;2
λd2
1;2
λd3
1;2
λd1
1;2
λd2
1;2
λd3
1;2
(b)
= −a ∓ jb = −a ∓ jb = −a ∓ jb 1
1
a
2
2
a
3
3
a
= −a ∓ jb = −a ∓ jb = a −b 3;
1
1
2
2
3
1
2
3
> > >
> a > a > a
3
Determine which of these systems is stable.
0;
b
0;
b
0;
b
2
3
> > >
> b > b >
1
0;
2
0;
0;
1
b
3
0 0 0
1
0
2
0
0
Multiple Degree-of-Freedom Systems:
8
General Solution for Response and Forced Oscillations
8.1 Introduction 8.2 Normal-Mode Approach
page
458 460
8.2.1 General Solution
460
8.2.2 Response to Initial Conditions
465
8.3 Response to Arbitrary Forcing and Initial Conditions: Direct Numerical Approach 8.4 Response to Harmonic Forcing and the Frequency-Response Function 8.4.1 Frequency-Response Function 8.5 Vibration Absorbers
8.1
472 475 475 489
8.5.1 Undamped Vibration Absorber
489
8.5.2 Damped Linear Vibration Absorber
492
8.5.3 Centrifugal Pendulum Vibration Absorber
504
8.5.4 Bar Slider System
508
8.5.5 Pendulum Absorber
511
8.5.6 Particle Impact Damper
515
8.5.7 Vibration Absorbers: A Summary
525
8.6 Vibration Isolation: Transmissibility Ratio
525
8.7 Systems with Moving Base
536
8.8 Summary
538
Exercises
539
INTRODUCTION
In the previous chapter, we studied systems with multiple degrees of freedom and discussed the ways in which one can obtain the governing equations of motion. In addition, free-response characteristics of damped and undamped systems were examined. In this chapter, we build on that material and employ different approaches to obtain the solution for the responses of multi-degree-of-freedom systems. We shall use three solution methods to determine the forced and free response. The solution method is selected based on whether the system is linear or nonlinear, the forcing
8.1 Introduction
459
is a transient excitation or harmonic excitation, the nature of damping, and whether the response to initial conditions needs to be determined.
•
Linear systems For
arbitrary
(transient
or
harmonic)
forcing
and/or
fication
speci
of
initial
conditions
Normal-mode approach (restricted to systems with proportional damping) Numerical methods in the time domain (no restriction on the form of damping) Harmonic forcing and zero initial conditions
Direct method,
based on frequency-response functions, system response assumed
to be harmonic
•
Nonlinear systems
Numerical methods
in the time domain irrespective of type of forcing, transient or
harmonic, and with speci
Analytical methods
fied or zero initial conditions
for certain types of nonlinear systems, including weakly non-
linear systems For the normal-mode approach, the form of the damping matrix places a restriction on the domain of applicability of the method. As discussed in Chapter 7 and elaborated further in Section 8.2, the damping matrix needs to have a special structure in order to use the normal-mode approach. However, whenever this approach is applicable, the solution obtained by using it provides insight into the spatial responses of the inertial elements. This type of information is indispensable for system designs where one is minimizing or maximizing the response to external excitations and where one is determining where to place an actuator and/or a sensor. For harmonic forcing of a linear system, one can assume that the system response is also a harmonic motion. This assumption allows us to obtain algebraic expressions from which the damped natural frequencies can be determined, and expressions for the different frequency-response functions can be obtained. For
both
linear and 1
obtained numerically
nonlinear
systems, the
solutions in
the
time
domain
can
be
for transient forcing and/or harmonic forcing and for given initial
conditions; in many cases, it is the most direct path to numerical results when the time domain response is sought. For the nonlinear systems presented in this book, only numerical solutions are pursued, although methods to determine analytical approximations exist for
“weakly”
nonlinear systems.
2
There are other approaches that can be used to obtain the solutions to multi-degreeof-freedom systems. One such method is the Laplace transform, which is applicable to linear systems of
differential equations with
constant
ficients.
coef
The procedure to
determine the solution follows along the lines of what is illustrated in Appendix C for
1 2
These solutions are readily obtained using such programs as Matlab and Mathematica. A. H. Nayfeh and D. T. Mook,
ibid.
460
Multiple Degree-of-Freedom Systems: Forced Oscillations single degree-of-freedom systems, and there is no restriction on the form of damping. This method becomes cumbersome to use as the number of degrees of freedom increases. Another method is the state-space formulation (see Appendix F), which is based on a standard solution procedure from the theory of ordinary differential equations wherein the governing equations are rewritten as a set of
space
first-order
equations called the
state-
form. With this form, analytical solutions can be obtained in linear cases and
numerical solutions can be obtained in both linear and nonlinear cases. This approach is applicable to all forms of damping and forcing. In this chapter, we shall show how to do the following.
•
Use normal modes to determine the responses of multiple degree-of-freedom systems to initial displacements, initial velocities, and external forces.
•
Determine the frequency-response functions for two degree-of-freedom systems.
•
Examine different types of vibration absorbers and identify and choose the appropriate parameters to obtain optimal performance.
•
Isolate
the
force
transmitted
to
the
stationary
boundary
of
two
degree-of-freedom
systems.
•
Analyze the responses of two degree-of-freedom systems with a moving base. In addition, the following interactive graphics have been created to better visualize
natural frequencies, modes shapes, and response of two degree-of-freedom systems. Interactive Graphic 8.1: Two degree-of-freedom system: displacement responses of masses to initial conditions Interactive Graphic 8.2: Two degree-of-freedom system: displacement responses of masses to transient forcing Interactive Graphic 8.3: Two degree-of-freedom system: frequency-response functions Interactive Graphic 8.4: Frequency-response function for an optimal choice of parameters Interactive Graphic 8.5: Particle impact damper Interactive Graphic 8.6: Transmissibility
ratio
(
TR)
variation
with
respect
to
system
parameters
8.2
NORMAL-MODE APPROACH
8.2.1
General Solution In Section 7.3, we discussed the characteristics associated with free oscillations of multiple degree-of-freedom systems such as natural frequencies and mode shapes. In this section, we determine the general solution for the response of a linear multi-degree-offreedom system and present explicit forms for the responses due to both forcing and initial conditions. It will be shown that the normal mode approach provides a way to understand the role that modes play in the response of systems to initial conditions and
8.2 Normal-Mode Approach
461
to externally applied forces and when the modes can be used to describe the subsequent response in the presence of damping. For two degree-of-freedom systems, however, when it comes to obtaining numerical results, it is often more expedient to solve the coupled systems of equations by using a direct numerical approach when arbitrary for-
fi
cing is applied and/or when initial conditions are speci ed. This approach is discussed in Section
8.3.
In
Section
8.2.2,
we
explicitly
illustrate
the
fied.
approach for the case when initial conditions are speci
use
of
the
normal
mode
When the forcing is harmo-
nic, the system of equations become algebraic; this case is examined in Section 8.4. We shall determine the response of the system given by Eq. (7.3), which is repeated below after replacing
qi (t) by xi(t), {Q} by {F }, and dropping the gyroscopic and circula-
tory force terms as
M ±fx€g þ ½ C ±fx_ g þ ½K ±fxg ¼ fF g
½
(8.1)
In general, Eq. (8.1) is a coupled system of equations. One way to solve this system is to uncouple them by using an appropriate coordinate transformation. One candidate trans-
Φ
formation is the modal matrix [
] given by Eq. (7.31), which has the desirable property
M] and [K]. Thus, we assume a solution to Eq. (8.1)
of being orthogonal to the matrices [ of the form
Φ |fflffl{zfflffl} η t |{z}
xðtÞg ¼
f
½
±
f ð Þg
Modal
(8.2)
Modal
matrix amplitudes
ηt
where { ( )} is a column vector of generalized (modal) coordinates that are to be determined. On substituting Eq. (8.2) into Eq. (8.1), we obtain
½
Φ
M ±½
€η
Φ
Pre-multiplying Eq. (8.3) by [
ΦTM Φ
½
±
½
±½
Φ
C ±½
±f g þ ½
T
]
K ±½
Fg
η_
ΦT K Φ
η
η_
K D± fηg ¼ ½
ΦT
η
±f g ¼ f
(8.3)
results in
ΦTC Φ
η €
±f g þ ½
Φ
η_
±f g þ ½
±
½
±½
±f g þ ½
±
½
±½
ΦT
±f g ¼ ½
±
Fg
f
(8.4)
Upon using Eqs. (7.68), we obtain
ΦTC Φ
MD± f€ ηg þ ½
½
±
½
±½
±f g þ ½
MD]−
We now pre-multiply Eq. (8.5) by [
1
η €
MD ±−
1
ΦT C Φ
½
±
MD] −
1
where we have assumed that [
½
±½
η_
±f g þ
(8.5)
MD],
and use Eq. (7.69) to
±ω ² η 2
D
Qg
f g ¼ f
(8.6)
Q } so that
exists and reintroduced the force vector {
Qg ¼ ½ MD ±−
f
Fg
f
, the inverse of [
arrive at
f g þ ½
±
1
ΦT
½
±
Fg
f
(8.7)
462
Multiple Degree-of-Freedom Systems: Forced Oscillations Equations (8.6) and (8.7) are written in expanded form as
2 66 64 ⋮ 1
0
0
1
38> €η t 77>< €η t 7 ⋮ ⋮ 5> >: €η t
… …
0 0
⋮ ⋱
0
…
0
1
2ω 66 64 ⋮
2
Þ
9> >= >>;
… …
0
1
ω2 2
0
ð Þ
Nð
1
2
þ
ð Þ
0
1
þ ½
1
0
2
⋮
… ωN 2
0
MD±−
1
38> η t 77>< η t 75> ⋮ >: η t
0
⋮ ⋱
8> η_ t >< η_ t C Φ >>: ⋮ η_ t 8Q t 9 >>< >> Q t = >>: ⋮ >>; Q t
ΦT
½
ð Þ
ð Þ
Nð
Þ
±
9 >>= >>;
½
±½
2
±
ð Þ
ð Þ
Nð
¼
1ð
Þ
2ð
Þ
Nð
Þ
9> >= >>; (8.8)
Þ
From the form of Eq. (8.8), it is clear that these equations in the transformed coordinates can be uncoupled into
N individual second-order differential equations if the matrix MD ±−
1
½
ΦTC Φ
½
±
½
±½
±
is a diagonal matrix. As discussed in Section 7.3.3, this is possible when the system is proportionally damped. For this case, [
C]
is given by Eq. (7.83) and we obtain from
Eqs. (7.99) and (7.102) that
2 6 Φ C Φ 664 T
C D± ¼ ½
½
±
½
±½
^ 11 ζ1 ω1 M
2
^ 22 ζ 2 ω2 M
0
± ¼
2
⋮
⋮
0 where
^ kk M
2 66 66 64
is the modal mass of the
½
MD ±− ½C D± ¼ 1
^ =M
1
¼
2 66 66 64
0
11
⋮
^ NN ζ N ωN M
…
^ =M
1
22
…
0
…
0
⋮
⋮
⋱
0
0
…
^ 11 ζ1 ω1 M
3 77 75
(8.9)
2
^ NN =M
3 77 77 75
1
…
0
^ 22 ζ2 ω2M
…
0
⋮
⋮
⋱
⋮
0
0
…
0
0 2
ζ1 ω1
2
0
0
⋱
0
⋮ 2
0
kth mode. From Eqs. (7.69b) and (8.9), it follows that
0
2 66 × 66 64
… …
0
0
ζ 2 ω2
2
…
0
…
0
⋮
⋮
⋱
0
0
…
⋮ ζ N ωN
2
^ NN ζ N ωN M
3 77 77 75
2
3 77 77 ± 75 ¼
ζω ÞD
ð2
²
(8.10)
8.2 Normal-Mode Approach
463
Hence, for a system with proportional damping, from Eqs. (8.6) and (8.10), we arrive at the uncoupled set of equations
€η
f g þ
in which the
±
ζω ÞD
ð2
² η_ ±ω ² η 2
f g þ
Qg
f g ¼ f
D
jth equation has the form η j ðtÞ þ 2ζj ωj η_ j ðtÞ þ ωj ηj ðtÞ ¼ Q j ðtÞ j € 2
In Eqs. (8.12),
ζj
¼ 1; 2;
is the damping factor associated with the
frequency associated with the
jth
ηj ,
nates
…; N
(8.12)
jth mode and ωj is the natural
mode. As discussed in Section 7.3.3, since the damping
matrix in Eq. (8.9) is diagonal, the system is said to have
the
(8.11)
modal damping.
The coordi-
in which the governing equations of motion are uncoupled, are referred to as
principal coordinates or modal coordinates.
The equation describing the oscillation in the
j th
mode has the form of the governing
equation of a single degree-of-freedom system, namely, Eq. (4.1). Hence the solution for
j
the response of the th mode is given by Eq. (C.13) of Appendix C if 0
η j ðtÞ
¼
³ ´ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Aj e−ζj ωj t sin ωdj t þ φdj
ð t
þ
1
ωdj
e−ζ j ωj t−ξ ð
Þ
³ω
sin
dj ðt
−ξ
≤ ζj
>< x t >>: ⋮ x t 1ð
Þ
2ð
Þ
Nð
the
9 >>= >>; Φ ¼ ½
Þ
system
response
η t
±f ð Þg ¼
can
be
constructed
ηj (0)
and initial
and initial veloci-
by
we return to
making
2 X X … X 3 8> η t 9> 66 X X … X 77 >< η t >= 64 ⋮ ⋮ ⋱ ⋮ 75 > ⋮ > >: η t >; X X … X |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflffl{zfflfflfflfflfflffl} 12
1
N
1
21
22
2
N
2ð
N1
N2
Nð
Modal matrix
of
ð Þ
11
NN
use
Þ
(8.17)
Þ
Modal amplitudes
where
xi ð tÞ ¼
X N
j¼1
X ij η j ðtÞ
|fflfflfflffl{zfflfflfflffl} X i η ð tÞ
¼
1
X η t |fflfflffl ffl{zfflfflfflffl} i2
þ
1
2
ð Þ
⋯
þ
Component of response
Component of response
in the first mode
in the second mode
X η t |fflfflfflffl ffl{zfflfflfflfflffl} iN N ð
þ
i ¼ 1; 2; … ; N
Þ
Component of response in the
N th
mode
(8.18)
Equation (8.17) is rewritten in the compact form
xðtÞg ¼
f
X N
X η t |fflfflfflfflffl ffl{zfflfflfflfflfflffl}
f
j ¼1
g
j jð
Þ
(8.19)
Oscillation in the
j th
mode
We see from Eq. (8.19) that the displacement of each mass
xi( t)
is composed of response
components in the different modes of the vibratory system. Thus, the response of a multi-degree-of-freedom system is a weighted combination of the individual modes of the system. The modal amplitude provides the weighting for each mode.
Response of a Damped System On substituting Eq. (8.13) into Eq. (8.19), we see that the response of the linear multidegree-of-freedom system given by Eq. (8.1) takes the form
xðtÞg ¼
f
X N
X gj
f
j¼1
2 4A e− j
ζj ωj t
³ω
sin
dj t
þ
φdj
´
ð t
þ
1
ωdj
e−ζj ωj t−ξ ð
Þ
³ω
sin
´ t −ξ Q
dj ð
Þ
jð
3 ξ d ξ5 Þ
0
(8.20)
8.2 Normal-Mode Approach
x(t)} are
where the elements of {
xi ðtÞ ¼
2 X 4 − X Ae N
ij
ζj ωj t
j
³ω
sin
dj t
þ
j¼1
φdj
´
t
þ
ωdj
e−ζj ωj t−ξ ð
Þ
³ω
´ t−ξ Q
dj ð
sin
Þ
3 ξ d ξ5
jð
Þ
0
Response of an Undamped System ζj = ωdj = ωj
When the system is undamped, we set , and obtain
xt
ð
1
465
X N
f ð Þg ¼
X gj
f
j¼1
2 4A ³ω t j
j
sin
þ
0 in Eq. (8.20), note from Eq. (8.14b) that
´ φ j
ð ³ ω t
þ
1
ωj
sin
´ t−ξ Q
dj ð
Þ
3 ξ d ξ5
jð
Þ
(8.21)
0
where from Eqs. (8.14a), we
find that
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! u u η_ tη 2
Aj
¼
2
j
j ð0Þ
ð0Þ þ
ωj
(8.22)
φj
8.2.2
¼ tan
−1 ωj η j ð0Þ η_ j ð 0Þ
Response to Initial Conditions
F } = {0} in Eq. (8.1), then the modal forced Q } = {0} in Eq. (8.7), and the response of the system follows from Eq. (8.20) as
When the forcing is zero, that is, {
vector
{
xðtÞg ¼
f
X N
j ¼1
³ω t φ ´ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} X gj Aj e−ζj ωj t
f
sin
dj
þ
(8.23)
dj
Free oscillation component in the
j th where
Aj
and
φdj
mode
are determined from Eqs. (8.14a) in terms of the initial displacement
and the initial velocity. For 0
≤ ζj
0,
φdj
466
Multiple Degree-of-Freedom Systems: Forced Oscillations The initial conditions and
η_ j ð0Þ
and
η j(0)
can now be determined from Eq. (8.2) in the
following manner. We note from Eq. (8.19) that
X N
x
f ð0Þ g ¼
X N
x_
f ð0Þ g ¼
X gj ηj ð0Þ
f
j ¼1
(8.24)
X gj η_ j ð0Þ
f
j ¼1
X gTi ½M ±; that is,
If we pre-multiply Eqs. (8.24) by f
X gTi ½M ± fxð0Þ g ¼
f
T _ ð0Þ g fX gi ½M ± fx
X N
j ¼1
X N
¼
X gTi ½M ±fX g j ηj ð0Þ
f
j ¼1
(8.25)
T _ j ð0Þ fX gi ½M ±fX g j η
and make use of the orthogonality of the modes given by Eq. (7.62) and the modal mass given by Eqs. (7.67), we obtain
^ jj X gTj ½M ±fxð0Þg=M
ηj ð 0Þ
¼ f
η_ j ð 0Þ
¼ f
(8.26)
^ jj X gTj ½M ±fx_ ð0Þg=M
If the modes are normalized with respect to the mass matrix, then
^ jj M
¼ 1.
Free Oscillations for a Special Case
x_ ð0Þg ¼ nth mode shape that is,
Now we consider the special case where the initial velocity f is provided an initial displacement in its
xð0Þg ¼ ao fX gn
f
From Eqs. (7.62) and (7.67), we
1
we
find
(8.27)
find that ^ jj M
where the
f0g, and the system
X gTj ½M ± fX gn
f
¼
δnj
Kronecker delta function δ nj = 0, n ≠ j and δ nj = 1, n
(8.28)
j
= . Then, from Eqs. (8.26),
that the initial modal displacement and initial modal velocity are given by,
respectively,
aoδ nj
ηj ð 0Þ
¼
η_ j ð 0Þ
¼ 0
(8.29)
8.2 Normal-Mode Approach
467
Therefore, the only mode for which the associated initial condition is not equal to zero is the one corresponding to the has a non-zero response is
ηn( t).
nth
mode shape, and it follows that the only mode that
Then, from Eq. (8.19)
xðtÞg ¼ fX gn η nðtÞ
f
where
ηn
(8.30)
is found from Eq. (8.13) as
η nðtÞ
An e–ζ nωn t sinðωdn t þ φdn Þ
¼
(8.31)
and from Eqs. (8.14a), (8.14b), and (8.29), we obtain
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ! u u ζ ω η η a tη q ffiffiffiffiffiffiffiffiffiffiffi ffi q ffiffiffiffiffiffiffiffiffiffiffi ffi ω 2
An
¼
2
n ð 0Þ
n n ð0Þ
n
þ
n ð0Þ
¼
dn
1
− ζn 2
1
qffiffiffiffiffiffiffiffiffiffiffiffi φdn
− ωdn ηn ð0Þ
−
1
¼ tan
¼ tan
ζ nωnη nð0Þ
1
1
In other words, the system only oscillates in its
o
¼
− ζn 2
(8.32)
− ζn 2
ζn
nth mode shape; none of the other modes
is excited.
EXAMPLE 8.1
Undamped free oscillations of a two degree-of-freedom system
We return to the system shown in Figure 7.1 and consider the case when the forcing and the damping are absent. We shall conditions are as follows:
xð0Þg ¼
f
find
&x ' 10
x
the response of the system when the initial
x_ ð0Þg ¼
and
f
20
& x_ ' 10
x_
(a)
20
The response in this case is determined by making use of the general solution given by Eqs. (8.21) and (8.22) for the undamped system, that is,
X
³ω t
2
xðtÞg ¼
f
j ¼1 where
X gj A j
f
j
sin
þ
φdj
´
(b)
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u η_ tη 2
Aj
¼
2
j ð0Þ
þ
j ð0Þ
ω2j
ωj ηj ð0Þ φdj ¼ tan− 1
η_ j ð0Þ
(c)
468
Multiple Degree-of-Freedom Systems: Forced Oscillations
η j(0)
The initial conditions from the modes
ηj ð0Þ η_ j ð0Þ
Thus,
from
Eq.
(b)
the
1
¼
&x ' 10
x
& x_ '
T fX g ½M ± j
^ jj M
oscillations
weighted sum of the free oscillations of the
EXAMPLE 8.2
are determined from Eqs. (8.25) as
X gTj ½M ±
1
undamped
η_ ð0Þ
f
^ jj M
¼
and
20
(d)
10
x_
of
20
the
system
are
described
as
the
first and second modes of the system.
Damped and undamped free oscillations of a two
degree-of-freedom system
We now continue with Examples 7.14 and 7.19 and assume that damping is present in the form of Eq. (7.104) with
=
ζ
0.05. The damped and undamped free responses of the
system are determined for the initial conditions
&x ' & 10
x
¼
20
'
0:1
−
m
& x_ ' & ' 0
10
and
¼
x_
0:1
m=s
(a)
0
20
The response of the system has the form of Eq. (8.21) for the undamped case and Eq. (8.20) for the damped case. Based on the information in Example 7.12 and from Eqs. (a) and (d) of Example 8.1, we the initial modal displacements
η 1ð 0Þ
^ X gT ½M ±fxð0Þg=M
¼ f
¼
η 2ð 0Þ
ηj (t)
−
1
¼
−
that the initial modal velocities
η_ j ð0Þ
"
)!
are
¼
1 3:658
f0:893
1:2
T 2
22
¼
1 10 :311
0
f
0
#(
1g
"
0:045
^ X g ½M ±fxð0Þg=M
¼ f
11
find
−
2:518
−
2:7
1:2
0
1g 0
0:1
2 :7
¼ 0 and
0:1
#(
0:1
−
)!
(b)
0:1
0:056
where from Eq. (b) of Example 7.14 we have determined that
Φ
½
·
0:893
± ¼ 1
−
2:518
¸
1
Response for the Damped Case For
ζ
=
0.05 and from Eq. (d) of Example 7.14, we obtain
ζω 1
¼ 0:05
ζω 2
¼ 0:05
× ×
2:519 ¼ 0:126 (c) 5:623 ¼ 0:281
8.2 Normal-Mode Approach From Eqs. (8.14), we compute the damped natural frequencies
qffiffiffiffiffiffiffiffiffiffiffiffi −ζ qffiffiffiffiffiffiffiffiffiffiffiffi
ωd 1
¼
ω1
ωd 2
¼
ω2
φd 1
¼ tan
2
1
1
−ζ
2
− ωd η
−
0:05
¼ 5:623
1
ð0Þ
1
−
−
1
− 2
1
2
0:05
2
ð0Þ
−
¼
−
−
1
−
2
0:05
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ! u u ζω η tη 2
A
1
2
¼
1
1
ð0Þ þ
1
ð0Þ
ωd 1
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! u u ζω η tη
¼
2
2
2
¼
2
2
ð0Þ þ
2 ð0Þ
ωd 2
0:05
¼ 5:616
rad=s
ð0Þ
1
1
φdj
as
−ζ
2
ζη1ð0Þ
2
ð0Þ
(d)
rad
pffiffiffiffiffiffiffiffiffiffiffi 1
−ζ
2
ζη2ð0Þ
1:621
rad
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi u u − × t− 2
¼ 0:045
A
=−
0:05
rad=s
1:621
1
Aj as
and the amplitudes
1
− η tan
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
− η
and the phases
¼ 2:516
2
=−
0:05
ζω2 η2 ð0Þ
¼ tan
¼ tan
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
− ωd η tan
¼
1
2
1
ζω1 η1 ð0Þ
¼ tan
φd 2
1
1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi − pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi − pffiffiffiffiffiffiffiffiffiffiffi
¼ 2:519
ωdj
469
2
ð
0:045Þ
0:045
þ
0:126
2 :516
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi u u − × t−
(e)
2
þ
2
ð
0:056Þ
0:056
þ
0:281
5:616
¼ 0:056
Qj
Then, from Eq. (8.13), in the absence of forcing (i.e.,
=
0), we obtain the modal
responses and compute them based on Eqs. (c), (d), and (e), as follows:
η 1ðtÞ η 2ðtÞ
¼ ¼
A e−ζω
1
1
A e−ζω
2
2
t t
ωd 1t þ φd 1Þ
sinð
ωd 2t þ φd 2Þ
sinð
e−
¼ 0:045 ¼
0:126
− 0:056e
0:281
t t
t − 1:621Þ
sinð2:516
(f)
t − 1:621Þ
sinð5:616
The free response of the damped system is determined from Eq. (8.23) and Eqs. (f). Thus,
x ðtÞ ¼ η ð tÞX 1
11
1
¼
1
−
1
11
¼ 0:045
¼
þ
A X e−ζω
×
η2 ðtÞX 12 t
sinð
0:893
0:056
− 0:040e
×
t
2
× e− × e−
2:518
0:126
ωd 1t þ φd1 Þ þ A 2X 12 e−ζω 0:126
t
ωd 2t þ φd 2Þ
sinð
t − 1:621Þ
sinð2:516
0:281
t
t − 1:621Þ
sin ð5:616
t − 1:621Þ − 0:140e−
sinð2:516
t
0:281
t
t − 1:621Þ
sin ð5:616
m
(g)
470
Multiple Degree-of-Freedom Systems: Forced Oscillations and
x ðtÞ ¼ η ðtÞX 2
1
¼ ¼
A X e−ζω t sinðωd t þ φd 1
1
21
1
− 0:045 × 1 × e − 0:045e
0:126
t
0:126
t
0:281
1
t
1Þ
þ
A X e−ζω 2
2
22
t
ωd 2 t þ φd 2 Þ
sinð
t − 1:621Þ
sinð2:516
× × e−
þ 0:056 ¼
η 2ðtÞ X22
þ
21
t − 1:621Þ
sinð5:616
t − 1:621Þ
sinð2:516
− þ 0:056e
t − 1:621Þ
t
0:281
sinð5:616
(h)
m
Response for the Undamped Case The results for the undamped case are computed as follows. From Eqs. (b) and (e), with
ζ
=
0, we have
A A
1
¼ 0:045
2
¼ 0:056
(i)
and from Eqs. (d), we obtain
Then, from Eq. (8.23) with
x ðtÞ ¼ A X 1
1
¼ ¼
− −
11
ζj
ωd 1
¼
ω1
¼ 2:519
ωd 2
¼
ω2
¼ 5:623
φd 1
¼
φd 2
¼
−π −π
(j) =2 =2
= 0 and Eqs. (i) and (j), we obtain
ω1t − π =2Þ þ A2 X12
sinð
0:045
×
ω2t − π=2Þ
sinð
t
0:893 cos ð2:516 Þ þ 0:056
×
t
t
2:518 cos ð5:623 Þ
t
0:040 cosð2:516 Þ þ 0:140 cos ð5:623 Þ
(k)
m
and
x ðtÞ ¼ A X 2
1
¼ ¼
− −
21
sinð
0:045
ω1 t − π =2Þ þ A 2X 22
× × 1
t
cos ð2:516 Þ
t
0:045 cosð2:516 Þ
−
−
ω2t − π=2Þ
sinð
0:056
× × 1
t
t
cosð 5:623 Þ
0:056 cosð5:623 Þ
m
(l)
In the damped case, the response given by Eqs. (g) and (h) consists of exponentially decaying sinusoidal oscillations in each mode. In the undamped case, the response given by Eqs. (k) and (l) consists of harmonic oscillations in each mode. The time histories given by Eqs. (g) and (h) for the damped case are plotted in Figure 8.1, and the time histories given by Eqs. (k) and (l) for the undamped case are plotted in Figure 8.2. Comparing Figures 8.1 and 8.2, we see that in the damped case, the displacements
x (t) 1
and
x (t) 2
approach zero as
t
→∞
; that is, the system eventually
settles down to the equilibrium position. This is not true in the undamped case. See Interactive Graphic 8.1 in Section 8.3.
8.2 Normal-Mode Approach
Figure 8.1. Displacements of a two degree-of-freedom system with
ζ
=
0.05 when both
masses are subjected to equal, but opposite, initial displacements. Solid line line
471
x (t); 1
dashed
x (t). 2
Figure 8.2. Displacements of a two degree-of-freedom system with
subjected to equal, but opposite, initial displacements. Solid line
ζ
=
0 when both masses are
x (t); dashed line x (t). 1
2
472
Multiple Degree-of-Freedom Systems: Forced Oscillations
8.3
RESPONSE TO ARBITRARY FORCING AND INITIAL CONDITIONS: DIRECT NUMERICAL APPROACH
The response of a two degree-of-freedom system to arbitrary forcing can be obtained from Eqs. (8.13) and (8.19) by setting
Aj
=
0; that is, only the response to the forcing is
determined. As can be seen from this equation, the solution requires an integration involving the expression for each forcing function. In this case, it is deemed more expedient to use the direct approach and solve Eq. (8.1), or equivalently Eq. (7.1a), numerically.
fj(t) = Fjgj (t), where F j is the magnitude gj (t) the corresponding shape factor, also that c = k = 0, and we intro-
Before evaluating Eq. (7.1a), we assume that of the force and
3
3
duce the nondimensional quantities given in Eq. (7.41) and the additional quantities
τ δ st
where
¼
ωn1 t;
^ xj
¼
xj
^j F
;
δst
g
is given by Eq. (3.6) and
¼
Fj ; mj g
2
ζj
¼
cj
j
mj ωnj
¼ 1; 2
(8.33)
is the gravity constant. Then Eq. (7.1a) can be writ-
ten as
d x^ dτ 2
1
2
ζ1
þ ð2
ζ2 mr ωr Þ
þ 2
d^x dτ
1
þ
d x^ dτ
³
1 þ
2
2
2
ζ 2 ωr
þ 2
´
mr ωr ^x 2
1
d x^ dτ
2
þ
−
ζ2 mr ωr
2
d^x dτ
2
ω2r x ^2 − 2ζ 2ωr
− mr ωr ^x
2
¼
F^ g ð τÞ
1
¼
F^ g ð τÞ
2
d^x dτ
1
− ωr ^x 2
1
2
1
2
(8.34)
If the initial displacement of
mj
is
Xj
and the initial velocity of
in terms of the nondimensional quantities we
xj ð0Þ ¼ ^
Xj δst
¼
The numerical evaluation
g (τ) 2
ζ2.
=δτ
( ) and
g (τ ) = 1
Xo j ; ;
find that
d x^j ð 0Þ Vj ¼ dτ δst ωn
¼
Vo
;
g (τ) 1
=
is
Vj , j
=
1, 2, then
¼ 1; 2
1
3
of Eq. (8.34) for an impulse force on
m
, that is, when
2
ωr, m r, ζ1, and g2(τ) = u(τ ) and ζ2.
0, is given in Figure 8.3 for several combinations of
The numerical evaluation of Eq. (8.34) for a step force on
and
j
j
mj
m
0, is given in Figure 8.4 for several combinations of
, that is, when 2
ωr , m r, ζ1,
To assist in the exploration of the various combinations of initial conditions and forces that can be applied to the masses, two interactive graphics have been created. Interactive Graphic 8.1 is used for the initial conditions and Interactive Graphic 8.2 is used for the application of either an impulse force or a step force on either of the masses. Their respective descriptions follow.
3
Using Matlab,
ode45; using Mathematica, NDSolveValue.
8.3 Response to Arbitrary Forcing and Initial Conditions: Direct Numerical Approach
2F /) t( 2,1 x
2F /) t( 2,1 x
2F /) t( 2,1 x
2F /)t( 2,1 x
2F /) t( 2,1x
2F /) t( 2,1x
(a)
(b)
m mr = 0.1 and
Figure 8.3. Normalized displacement responses of a two degree-of-freedom system when
is subjected to an impulse force, (b)
473
mr
=
0.5. Solid line
ζ1
=ζ = 2
0.2,
k
x^ ðτÞ; dashed line ^x ðτ Þ: 2
3
=c = 3
0, and
τ
= ωn t 1
: (a)
2
1
INTERACTIVE GRAPHIC 8.1: TWO DEGREE-OF-FREEDOM SYSTEM: DISPLACEMENT RESPONSES OF MASSES TO INITIAL CONDITIONS The time responses of the masses of a translating, two degree-of-freedom system to initial conditions are displayed. The variations in the responses with respect to ωr , m r, ζ1, ζ2, k32, and c32 can be explored through the animations. In this interactive graphic, the following can be noted. •
•
For small values of m r and ωr , the periods of the oscillations of the two masses are different, but as either of these parameter values increases the periods become equal and the extension and compression of the connecting spring is only due to their relative magnitudes. With the damping factor of both systems the same, the values of m r and ωr greatly affect the initial oscillations of the two masses and they can be vastly different.
474
Multiple Degree-of-Freedom Systems: Forced Oscillations
2F /) t( 2,1 x
2F /) t( 2,1 x
2F/) t( 2,1 x
2F/) t( 2,1 x
2F/) t( 2,1 x
2F/) t( 2,1 x
(a) Figure 8.4.
(b)
Solid line
m is mr = 0.1 and (b) mr = 0.5.
Normalized displacement responses of a two degree-of-freedom system when
ζ1 = ζ2 = line ^ x1 ðτ Þ:
subjected to a step force,
x^ ðτÞ; dashed 2
0.2,
k
3
=c = 3
0, and
τ
= ωn t 1
: (a)
2
INTERACTIVE GRAPHIC 8.2 TWO DEGREE-OF-FREEDOM SYSTEM: DISPLACEMENT RESPONSES OF MASSES TO TRANSIENT FORCING The time responses of the masses of a translating, two degree-of-freedom system are displayed when either of the masses is subjected to an impulse force or a step force. The variations in the responses with respect to ωr , m r, ζ1, and ζ 2 can be explored through the animations. In this interactive graphic, the following can be noted. •
When ζ1 and ζ2 have small values and are the same, for an impulse force the displacement responses of the masses are distinctly different depending on which mass the impulse is
8.4 Response to Harmonic Forcing
•
8.4
475
applied when mr and ωr are each less than 0.5. The same is true for the suddenly applied constant force. When ζ1 and ζ2 have small values and are the same, for an impulse force the displacement responses of the masses are very similar irrespective of which mass the impulse is applied when m r and ωr are each greater than 0.9. The same is true for the suddenly applied constant force.
RESPONSE TO HARMONIC FORCING AND THE FREQUENCY-RESPONSE FUNCTION
In this section, we
first determine the response of a general physical system described by
Eqs. (7.11) to harmonic forcing and use a direct approach to obtain the forced response. From this response, we illustrate how the frequency-response function is constructed.
finition
The de
of the frequency-response function remains similar to that for a single
degree-of-freedom system except that due to the presence of more than one inertial element, one deals with more than one frequency-response function. Next we consider the response of undamped systems and introduce the notion of resonance in multiple degree-of-freedom systems. As in the case of vibratory systems described by linear single degree-of-freedom systems, the responses of vibratory systems described by linear multi-degree-of-freedom systems
can
be
determined
if
the
frequency-response
functions
for
the
systems
are
known. For this reason, the material presented here provides a basis for the discussions of vibration absorbers in Section 8.5, transmissibility ratio in Section 8.6, and the system with the moving base model in Section 8.7.
8.4.1
Frequency-Response Function To obtain the frequency-response function, we start with Eqs. (7.11) and (7.12), that is,
µm
11
m
21
m m
12
22
¶& €x ' µ c 1
€ x
2
þ
11
c
21
c c
12
¶& x_ ' µ k 1
x_
22
þ
11
k
21
2
k k
12
22
¶& x ' & f t ' 1
x
2
¼
1ð
Þ
f ðtÞ 2
(8.35) It is assumed that the system is subject to harmonic excitation of the form
f ðtÞ ¼ F e j ωt f ðtÞ ¼ F e j ωt 1
1
2
2
(8.36)
476
Multiple Degree-of-Freedom Systems: Forced Oscillations where
ω
is
the
excitation
frequency.
With
excitation
of
this
form,
we
assume
a
solution as
x ðtÞ ¼ X ð j ωÞe j ωt 1
1
(8.37)
x ðtÞ ¼ X ð j ωÞe j ωt 2
2
Upon substituting Eqs. (8.36) and (8.37) into Eq. (8.35), we arrive at
a a
j ωÞX ð j ωÞ þ a ð j ωÞX ð j ωÞ þ a
j ωÞX ð j ωÞ ¼ F ð j ωÞX ð j ωÞ ¼ F
11 ð
1
12 ð
2
1
N
21
1
22
2
2
N
(8.38)
where
aipð j ωÞ = −mipω Solving Eqs. (8.38) for
X
1
and
X
X ð j ωÞ ¼ 1
X ð j ωÞ ¼ 2
2
þ
jcip ω þ kip
i ; p ¼ 1; 2
N=m
(8.39)
, we obtain
2
1
D o ð j ωÞ 1
D o ð j ωÞ
a
ð
22 ð
a
ð
11 ð
j ωÞ F
1
j ωÞ F
2
−a
12 ð
j ωÞF
2Þ
m (8.40)
−a
21 ð
j ωÞF
1Þ
m
where
D o ð j ωÞ ¼ a
11 ð
fine
We now de
F
2
=
j ωÞa
22 ð
j ωÞ − a
12 ð
j ωÞa
21 ð
j ωÞ
2
2
N =m
the frequency-response functions in the following manner. We
(8.41)
first
set
0 in Eqs. (8.40) and rearrange it to obtain
βG 11 ð jωÞ
¼
β
X ð j ωÞ F 1
¼
1
βG 21 ð jωÞ
¼
X ð j ωÞ β F 2
1
where
βG11
βG21
and
dimensional parameter manner, we set
F
1
=
β
a ð j ωÞ D o ð j ωÞ 22
a ð j ωÞ = −β Do ð j ωÞ
(8.42)
21
are the frequency-response functions and we have introduced a
β
(N/m) so that
βG 11
and
βG 21
have no dimensions. In a similar
0 in Eqs. (8.40) and rearrange it to obtain
βG 12 ð j ωÞ
¼
β
X ð j ωÞ F 1
2
βG 22 ð j ωÞ
¼
X ð j ωÞ β F 2
2
= −β
a ð j ωÞ Do ð j ωÞ 12
a ð j ωÞ ¼ β D o ð j ωÞ 11
(8.43)
8.4 Response to Harmonic Forcing
¹¹
¹¹
477
The magnitudes of the frequency-response functions are given by
HipðωÞ ¼ β G ipð j ωÞ
±G ±G
and the associated phase responses are given by
φip ð j ωÞ
− tan
¼
Im
1
Re
i ; p ¼ 1; 2
² ² jω
ip ð j ωÞ ip ð
i; p
Þ
(8.44)
¼ 1; 2
(8.45)
Application to the System Shown in Figure 7.1 Consider the system shown in Figure 7.1, which is represented by Eq. (7.1b). If we compare Eq. (7.1b) with Eq. (8.35), we
m c k
11
11 11
find that
m ; c þc ; ¼ k þ k ; ¼
m c k
1
¼
1
22
2
1
2
m; c þc ; ¼ k þ k ; ¼
¼
22 22
2
2
3
2
3
m c k
12
12 12
m ¼0 c = −c ¼ k = −k ¼
21
¼
21
(8.46)
2
21
2
and, therefore, it follows from Eq. (8.39)
a a a a
j ωÞ = −m ð j ωÞ = −m ð j ωÞ = −m ð j ωÞ = −m
11 ð
12 21 22
ω2 2 12 ω 2 21 ω 2 22 ω 11
jc þ jc þ jc þ jc þ
ω þ k11 = − m1ω2 þ j ðc1 ω þ k12 = − jc2ω − k 2 12 ω þ k21 = − jc2ω − k 2 21 ω þ k22 = − m2ω2 þ j ðc2 22 11
þ
c Þω þ k 2
1
þ
k
2
(8.47) þ
c Þω þ k 3
2
þ
k
3
At this point, it is convenient to introduce the nondimensional quantities given by Eq. (7.41) and the additional quantities
2
ζj
¼
cj m j ωnj
j
¼ 1; 2
and
c
32
¼
c c
3
(8.48)
2
into Eqs. (8.47) to arrive at
a a a a
j ωÞ ¼ k A ð jΩÞ ð j ωÞ = − k m r Bð jΩÞ ð j ωÞ = − k m r Bð jΩÞ ð j ωÞ ¼ k m r E ð jΩÞ
11 ð
12
21
22
1
1
(8.49)
1
1
where
Að jΩÞ ¼
−Ω
2
ζ1
þ 2ð
þ
ζ 2mr ωr Þ jΩ þ 1 þ mr ω2r
Bð jΩÞ ¼ 2ζ ωr jΩ þ ωr
2
(8.50)
2
Eð jΩÞ ¼
−Ω
2
ζ 2ωr ð1 þ c32 Þ jΩ þ ωr ð1 þ k32 Þ 2
þ 2
Then, Eq. (8.41) becomes
D o ð j ωÞ ¼ a
11 ð
¼
j ωÞa
22 ð
k m r Dð jΩÞ 2
1
j ωÞ − a
12 ð
j ωÞa
21 ð
jωÞ ¼ k mr A ð jΩÞ Eð jΩÞ − k m r B ð jΩÞ 2
2
1
1
2
2
(8.51)
478
Multiple Degree-of-Freedom Systems: Forced Oscillations where
Dð jΩÞ ¼ Ω
−
jΩ ζ ω c m ³ ³ ´ −Ω c ζ m ω ζ ζ ω ³ ³ ζmω´ ζ jΩω c ζ 4
3
2
2
þ
r
rð
2
2
4
32
þ 2
ð
r r
2
32
k m r ωr
4
32
32
2
2
k ωr
2
þ
r
þ 4
þ
2
þ
32
þ
1
r
2
2
r
ζ1 Þ
þ 1Þ þ
r
þ
þ
k ωr þ
2
k
32
mr ωr
³ζ m ω
2
32
þ
r
2
2
2
r
´
´
ζ1 ζ2 ωr þ ω2r
þ 4
ζ1 ωr
þ
ζ 1 ωr
þ
´
þ 1
ω2r
(8.52)
The displacements given by Eqs. (8.40) become 1
X ð jΩÞ ¼
k D ð jΩÞ
1
ð
Eð jΩÞF
1
þ
Bð jΩÞF
2Þ
1
(8.53)
1
X ð jΩÞ ¼
k m r Dð jΩÞ
2
mr Bð jΩÞ F
ð
1
þ
Að jΩÞF
2Þ
1
H
ωÞ
¼
k jG
j ωÞj ¼
H
ωÞ
¼
k jG
j ωÞj ¼
11 ð
21 ð
=k
21 ð
ωÞ
¼
k jG
j ωÞj ¼
H
ωÞ
¼
k jG
j ωÞj ¼
22 ð
β
1
11 ð
H
12 ð
where it is seen that
1
1
1
12 ð
22 ð
find that
¹¹ ¹ ¹¹E jΩ ¹¹¹ ¹D jΩ ¹ ¹¹ ¹ ¹¹B jΩ ¹¹¹ ¹D jΩ ¹ ¹ ¹¹ ¹¹B jΩ ¹¹¹ ¹D jΩ ¹ ¹¹ ¹¹ A jΩ ¹¹ ¹ ¹m D jΩ ¹¹
After using Eqs. (8.49) in Eqs. (8.42) to (8.44), we
ð
Þ
ð
Þ
ð
Þ
ð
Þ
ð
Þ
ð
Þ
ð
Þ
ð
r
(8.54)
Þ
.
1
Special Cases We shall now consider several special cases of the preceding results.
Case 1: c1 = c2
= c3 = 0
For this case, which is the undamped case, we set
c
1
=c =c = 2
3
c
0 (
32
=ζ =ζ = 1
2
0) in
Eqs. (8.50) and (8.52), respectively, and obtain
A ð jΩÞ ¼
−Ω
2
þ 1þ
mr ωr
2
Bð jΩÞ ¼ ωr
2
E ð jΩÞ ¼ and
Dð jΩÞ
→D
ud ð jΩÞ
¼
Ω
4
−Ω
2
³
1 þ
−Ω
(8.55) 2
þ
ωr ð1 þ k 32 Þ 2
´
ω2r ð1 þ m r þ k32 Þ
þ
ω2r
³
1 þ
k
32
³
1 þ
mr ωr
2
´´
(8.56)
It is noted that when Eq. (8.56) is set equal to zero, it is identical to Eq. (7.45), that is, it is the characteristic equation of the system.
8.4 Response to Harmonic Forcing Considering the particular case when for
H
H
and
11
21
¹¹ ¹¹ ¹¹ X ¹¹ ¹F =k ¹ ¹¹ ¹¹ ¹¹ X ¹¹ ¹F =k ¹ 1
2
1
=
0 and
F
≠
2
F
≠
1
0, the frequency-response functions
¹¹ ¹¹ − Ω ω ¹ D jΩ ¹¹ ¹¹ ω ¹¹ ¹ ¹D jΩ ¹¹
¼
2
H
ωÞ
11 ð
¼
¼
2
þ
r
k
ð1 þ
¹¹ ¹¹ ¹
ud ð
32 Þ
Þ
(8.57)
2
H
ωÞ
21 ð
¼
1
r
ud ð
Þ
is given by Eq. (8.56) and we have used Eqs. (8.42). The frequency-
response functions
F
0 and
1
1
Dud( jΩ)
2
are obtained by substituting Eqs. (8.55) into Eq. (8.54), which results in
1
where
=
F
H
and
12
H
22
can be obtained in a similar manner for the case when
0.
From Eqs. (8.57), it is evident that whenever the excitation frequency
Xj
finite;
become in
=
ω
natural frequency of the system, that is, when responses
479
ω1
ω
or
=
ω2,
ω
is equal to a
the displacement
hence, each of these frequency relationships between the
resonance relation and at resonance. In Chapter 5, it was
excitation frequency and a system natural frequency is called a these excitation frequencies the system is said to be in
seen that for a linear single degree-of-freedom system there is one excitation frequency at which we have a resonance. Since a linear two degree-of-freedom system has two natural frequencies, there are two excitation frequencies at which we can have a resonance.
N degrees of freedom has N natural frequencies,
By extension, since a linear system with there are
N
excitation frequencies at which we can have a resonance. It is also noted
from the stability discussion of Section 7.5 that the response of an undamped multiple degree-of-freedom system is unbounded when excited at one of its resonances. From Eqs. (8.57), it is clear that for a given excitation frequency one can choose the ratio
ωr
and
k
32
so that
Ω
2
¼
ω2r ð1 þ k 32 Þ
(8.58)
first inertial element is
For this choice of system parameters, then, the displacement of the
fi
zero at the excitation frequency that satis es Eq. (8.58). This observation is used to design an undamped vibration absorber, which is the subject of Section 8.5.1. In general, the excitation frequencies at which the response amplitude of an inertial element is zero are referred to as the frequencies at which the
zeros of the forced response occur for that inertial element.
The frequency-response functions given by Eqs. (8.57) are plotted as a function of the nondimensional excitation frequency 1, and
k
32
=
1. In this case,
a
1
=
Ω
for the following parameter values:
=
a
4 and
2
¹¹ ¹¹ω k ¹¹ Ω − a Ω ¹¹ ¹¹ ω ¹¹Ω − a Ω
H
11 ð
ΩÞ ¼
r ð1
þ
32 Þ
4
2
1
H
21 ð
ΩÞ ¼
2
r
4
2
1
¹¹ − Ω ¹¹ a ¹¹ ¹¹ ¹¹ a ¹¹ 2
þ
¼
2
¼
þ
=
1,
ωr
=
3 and from Eqs. (7.41), (8.56), and (8.57),
¹¹ ¹¹ ¹¹Ω
the frequency-response functions have the following forms:
2
mr
2
¹¹ ¹¹ ¹¹Ω
4
−Ω
2
4
−
−
Ω
4
1
Ω
4
2
2
2
¹¹ ¹¹ ¹¹
þ 3
¹¹ ¹¹ ¹¹
þ3
(8.59)
480
Multiple Degree-of-Freedom Systems: Forced Oscillations
Frequency-response functions of an undamped two degree-of-freedom system when
Figure 8.5.
m
harmonic forcing is applied to mass
Graphs
of
H
Ω)
(
11
H
and
Ω)
21(
Ω) Ω
(
21
Eqs. (7.47), it is found that
H
1
fi
as given by Eqs. (8.59).
are
=
1
Ω
become in nite valued for
plotted
as
Ω
¼
1 and 1
=
2
1 and
Furthermore, from Eqs. (8.59) it is seen that at
Case 2: c3 = k3 = 0 For this case, we set
c
32
=k = 32
A ð jΩÞ ¼
ffiffi
functions p
ffiffi
Ω
of
in
Figure
3. Thus, it is seen that p
ffiffi
Ω ¼p 3. This is reflected Ω ¼ 2, H (Ω) = 0. 2
8.5.
H
11
0 and Eqs. (8.50) and (8.52), respectively, become
−Ω
2
ζ 1 þ ζ 2m r ωr ÞjΩ þ 1 þ m r ω2r
þ 2ð
2
(8.60)
2
−Ω
2
ζ2 ωr jΩ þ ωr
þ 2
2
and
Dð jΩÞ
→D
3ð
jΩÞ ¼ Ω
4
−
jΩ ðζ ωr ðm r þ 1Þ þ ζ
2
and
in Figure 8.5.
Bð jΩÞ ¼ 2ζ ωr jΩ þ ωr E ð jΩÞ ¼
From
Ω)
( 11
3
jΩωrð ζ
þ 2
1Þ
2
2
þ
−Ω
2
³m ω r
2
r
ζ 1 ζ2 ωr
þ4
þ
ω2r
´
þ1
ζ 1ωr Þ þ ω2r (8.61)
In Figures 8.6 and 8.7, the nondimensional amplitude response functions
k
3
=c = 3
0 are plotted as a function of the excitation frequency ratio
frequency ratio
ωr .
Ω
Hij(Ω)
for
and the system
These plots are graphs of the functions given by Eqs. (8.54) with
A ( jΩ), B( jΩ), E( jΩ),
and
D ( jΩ)
given by Eqs. (8.60) and (8.61). Since the system is
damped, the amplitude responses of the inertia elements
m
1
and
m
2
have
finite
values
8.4 Response to Harmonic Forcing Ω1
0.0
Ω2
wr
3.5
0.5 wr
)Ω( 11H
2.0 1.5
2
1.0 0.5
0 0.0
0.5
0.0 0.0
1.0
1.5
Ω Ω1
0.0 wr
0.5
1.0
1.5
2.0
Ω
2.0
Ω2
wr
3.5
0.5
=1
Ω 1 = 0.854
Ω 2 = 1.17
3.0
1.0
2.5 )Ω(11H
2.0
1.5 3
1.5
2
1.0
1
0.5
0 0.0
0.5
1.0
1.5
Ω
0.0 0.0
wr
3.5
2.0
= 1.4
Ω 1 = 0.922
Ω 2 = 1.52
2.5 )Ω (11H
2.0
1.5 3
1.5
2
1.0
1
0.5 0.5
0.0 0.0
1.0
1.5
Ω
k
3
=c = 3
0,
ζ1
=ζ = 2
0.15,
0.5
1.0
1.5
Ω
2.0
Figure 8.6. Magnitudes of frequency-response function
Figure 7.1 when
1.5
3.0
1.0
0 0.0
1.0
Ω
0.5
wr
0.5
2.0
Ω2
Ω1
0.0
H11(Ω)
Ω 2 = 1.03
2.5
1
H11(Ω)
= 0.6
Ω 1 = 0.584
3.0
1.0
1.5 3 H11(Ω)
481
mr
=
H
11(
0.1, and
Ω) for the two masses shown in
ωr
=
0.6, 1.0, and 1.4.
2.0
482
Multiple Degree-of-Freedom Systems: Forced Oscillations Ω1
0.0
Ω2
wr
4
0.5
= 0.6
Ω1 = 0.536
Ω 2 = 1.12
wr
1.0
3 )Ω( 11H
1.5 H 11(Ω)
2
3 2 1 0
1 0.0
0.5
1.0
Ω
Ω1
0.0
0 0.0 1.5
2.0
Ω2
w
4
1.0
2.0
r=
1
Ω1 = 0.707
Ω 2 = 1.41
) Ω(11H
2
3 2 1 0
1.5
3
1.5 H 11(Ω)
1.0
Ω
0.5
wr
0.5
1 0.0
0.5
1.0
1.5
Ω1
Ω2
Ω
0.0
0 0.0
0.5
1.0
1.5
2.0
Ω
2.0
wr
4
0.5
= 1.4
Ω1 = 0.764
Ω 2 = 1.83
wr
1.0
3 )Ω(11H
1.5 H 11(Ω)
2
3 2 1 0
1 0.0
0.5
1.0
Ω
Figure 8.7.
0 0.0 1.5
k
3
=
c
3
=
0,
ζ 1 = ζ2
2.0
=
0.15,
1.0
1.5
2.0
Ω
H (Ω) for the two masses shown in mr = 0.5, and ωr = 0.6, 1.0, and 1.4.
Magnitudes of frequency-response function
Figure 7.1 when
0.5
11
8.4 Response to Harmonic Forcing for all values of the excitation frequency. For
“small ”
values of
ωr ,
483
it is seen that the
response of one of the inertia elements is pronounced at and close to the lower resonance, while the response of the other inertia element is pronounced at and close to the higher resonance. As the excitation frequency ratio resonance
value,
the
responses
of
both
inertia
Ω
is increased past the higher
elements
are
relatively
uniform,
as
was the case for responses of damped single degree-of-freedom systems subject to harmonic forces.
INTERACTIVE GRAPHIC 8.3: TWO DEGREE-OF-FREEDOM SYSTEM: FREQUENCY-RESPONSE FUNCTIONS The various frequency-response functions of a translating two degree-of-freedom system are displayed. Through Figures 8.6 and 8.7, the variation of H11 with respect to some of the system parameters can be seen. The variations of the functions H11, H12, H21, and H22 with respect to ωr, mr , ζ1, and ζ2 when k32 = c32 = 0 can be explored by using this graphic. In this interactive graphic, the following should be noted. •
•
•
For small and equal values of ζ1 and ζ2, mr and ωr have strong interacting effects on the frequency-response functions regarding both the appearance and “disappearance” of amplitude peaks in the vicinity of the uncoupled natural frequencies and the rapid growth of their values. For a given value of mr , the surfaces indicate that dominant frequency regions can differ markedly with the frequency-response function whereas in some cases they can exhibit great similarity. For a lightly damped system, the examination of H11 can be used to find a set of parameters at which the system can operate as a vibration absorber; for example, ζ1 = ζ2 = 0.06, mr = 0.73, and ωr = 1 or ζ 1 = 0.1, ζ2 = 0.0, m r = 0.1, and ωr = 1.
Case 3: c1 = c2 = c3 = k3 = 0 and ki = ki (1 + jηi) This is the case of a structurally damped system, that is, in the frequency domain,
ki where
ηi
are
constants
¼
k i ð1 þ j η i Þ
representing
i ¼ 1; 2
frequency-independent
(8.62)
damping
that
is
used
to
model the magnitude of the damping. In the current notation, we note that
Ω
→ αΩ
and
1
ωr
→ ω αα r
2
(8.63)
1
where
α2i
¼ 1 þ
j ηi
i ¼ 1; 2
(8.64)
484
Multiple Degree-of-Freedom Systems: Forced Oscillations Then Eqs. (8.50) and (8.52), respectively, become
→α
1
Að jΩÞ Bð jΩÞ
→α
2
1
Bsd ð jΩÞ
2
(8.65)
1
→α
1
Eð jΩÞ
Asd ð jΩÞ
1
2
Esd ð jΩÞ
1
where
−Ω
A sd ð jΩÞ ¼
2
Bsd ð jΩÞ ¼ ωr α
2
Esd ð jΩÞ ¼
2
2
−Ω
þ
α21
þ
ωr α 2
mr ωr α 2
þ
2 2
(8.66)
2 2
2
and
→α
1
Dð jΩÞ where
Dsd ð jΩÞ ¼ Ω
4
−Ω
2
³m ð
r
4
Dsd ð jΩÞ
(8.67)
1
þ 1Þ
ω2r α22
þ
α21
´
þ
ω2r α21α22
(8.68)
The displacements given by Eq. (8.53) become
X
1
¼
1
k D sd ð jΩÞ
Esd ð jΩÞF
ð
þ
1
Bsd ð jΩÞF
2Þ
1
(8.69)
X
2
¼
1
k m r Dsd ð jΩÞ
m rBsd ð jΩÞF
ð
1
þ
Asd ð jΩÞF
2Þ
1
From Eqs. (8.54), the frequency-response functions become
H
ωÞ
¼
H
ωÞ
¼
11 ð
21 ð
H
ωÞ
¼
H
ωÞ
¼
12 ð
22 ð
¹¹ ¹¹ E jΩ ¹¹ ¹ ¹D jΩ ¹¹ ¹¹ ¹ ¹¹B jΩ ¹¹¹ ¹D jΩ ¹ ¹¹ ¹¹ B jΩ ¹¹ ¹ ¹D jΩ ¹¹ ¹¹ ¹ ¹¹ A jΩ ¹¹¹ ¹m D jΩ ¹ sd ð
Þ
sd ð
Þ
sd ð
Þ
sd ð
Þ
sd ð
Þ
sd ð
Þ
sd ð
r
sd ð
Þ
Þ
(8.70)
8.4 Response to Harmonic Forcing
485
Application of Results to Other Systems The
frequency-response
functions
given
by
Eqs.
(8.44)
and
the
associated
phase
responses given by Eqs. (8.45) can be applied directly to many of two degree-of-freedom systems shown in Table 7.1 in the following manner. For those systems in the table that contain linear translational springs, one places, in the most general case, a viscous damper in parallel with each one. Then, if the element as
k
11
=k
k
11
of the stiffness matrix was given
, the corresponding element of the damping matrix 1
fi
c
11
is
c
11
=c
, where
1
c
1
is
the damping coef cient of a viscous damper placed in parallel with the spring of stiffness
k
. For example, Eq. (8.l) could have been obtained directly from Case 1 of Table 7.1 in 1
this manner. The displacements are obtained from Eqs. (8.40).
EXAMPLE 8.3.
Interaction of a rigid structure with its soil foundation
The response of civil structures to earthquakes is an important part of building, bridge, and
highway
design.
It
has
been
found
that
the
interaction
of
the
soil
with
the
foundations of these structures greatly affects the structure’s response. The modeling of the
foundation
typically
involves
describing
wave
propagation
in
the
soil
and
the
modeling of the soil is complicated by the fact that it is usually inhomogeneous, layered, and
fairly
fi
dif cult
to
experimentally
quantify.
In
order
to
obtain
insight
into
the 4
interaction phenomena while avoiding the mathematical complexity, several researchers
have successfully represented the soil-structure foundation interaction by using a two degree-of-freedom
system
model
and
relating
the
results
of
the
more
sophisticated
models to obtain a reasonably accurate correlation of the discrete model with those of the more complex models. The two degree-of-freedom model is shown in Figure 8.8. To simplify matters, we shall restrict our analysis to the case where the rigid foundation is subject to rocking (rotation) only and assume that the foundation is a rigid circular disc that has a mass moment of inertia
Jo
⋅
2
(kg m ). In Figure 8.8,
the soil under the rigid plate and
Mo(t), θ 1 J0
Jθ
Cθ
⋅ ⋅ ⋅
2
(kg m ) is the equivalent rotational inertia of
(N m s/rad) represents the damping in the soil.
Figure 8.8. Spring-dashpot equivalent of a rigid plate on soil.
r0
Cθ
θ2
4
Jθ
J. P. Wolf,
Kθ
“Spring-dashpot-mass models for foundation vibrations,”
Dynamics , 26 (1997) 931– 949. J. W. Meek and J. Geotechnical Engineering, 118(5) (1992) 667–685.
Earthquake Engineering and Structural Journal of
P. Wolf, “Cone models for homogeneous soil. I, ”
486
Multiple Degree-of-Freedom Systems: Forced Oscillations Finally,
Kθ
(N
⋅
m/rad) represents the rotational stiffness of the soil. If
ratio of the soil and it is assumed that
>
ν
ν
is the Poisson’s
1/3, then it has been shown that
K θ ¼ 12ρcs I =z C θ ¼ 2ρcs I Jθ ¼ ρI z 2
0
0
0
0
z
¼
0
π
9
8
(a)
0
ð1
−ν r Þ
m
0
pffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffi
cs ¼ G=ρ (m/s) is the shear wave speed and G is the soil’s shear modulus, cd ¼ Ec =ρ (m/s) is a dilatational wave speed where Ec = 2G(1 − ν)/(1 − 2ν ) is the soil’s constrained modulus, I (m ) is the area moment of inertia of the rigid disc of radius r about an axis perpendicular to the plane of the fi gwhere
ρ
3
(kg/m ) is the density of the soil,
4
0
0
ure, that is,
I
0
π
¼
r
4
(b)
0
4
In addition, for this disc rocking about an axis perpendicular to the plane of the plate and through its center,
J
¼
0
where
m f is the mass of the disc and h
mf 12
³r
2
3
0
þ
h
2
´
(c)
is the height of the disc.
To derive the governing differential equations of the model, we shall use the energy approach and the Lagrange equations. The total kinetic energy energy
T,
the total potential
V, and the total dissipation D , respectively, are T
¼
V
¼
1 2 1 2
Joθ_
2
þ
1
Kθ θ
1 2
Jθ θ_
2 2
2
(d)
1
³
1 D ¼ Cθ θ_ 2
1
−θ_
´
2
2
The applicable Lagrange equations follow from Section 7.2.3 as
0 1 @ ∂T_ A− ∂∂θT ∂θ 0 1 d @ ∂T A ∂ T − _
d dt
dt
1
∂θ2
þ
1
∂ θ2
∂D
∂ θ_
þ
1
∂V ∂θ1
¼
M⋅
∂ω
∂θ_ 1 (e)
þ
∂D
∂ θ_
2
þ
∂V ∂θ2
¼ 0
8.4 Response to Harmonic Forcing
487
where
M
Mo ðtÞk ω ¼ θ_ k ¼
(f)
1
Then, using Eqs. (d) and (f) in Eq. (e), we obtain
Jo
d θ dt 2
1
2
þ
³
C θ θ_
d θ Jθ dt
− θ_
1
´
þ
2
³ − C θ_
2
2
θ
2
Kθ θ
− θ_
1
´
∂ θ_ 1
k ¼ Mo ðtÞ (g)
¼ 0
2
Mo ðtÞ ¼ moej ωt ;
We consider the case where
∂ θ_ 1
Moð tÞk ⋅
¼
1
ω
where
is the frequency of oscillation.
Then we assume a solution of the form
θi ðtÞ
Θi ð j ωÞe j ωt
¼
i ¼ 1; 2
(h)
After substituting Eq. (h) into Eqs. (g), we obtain
³− ω J 2
o
þ
´
j ωCθ þ Kθ Θ ð j ωÞ − j ωCθ Θ ð j ωÞ ¼ mo
−j ωCθΘ
1ð
j ωÞ þ
³ −ω J 1
2
´ j ωC Θ
2
θ þ
Θ
Using the second equation of Eqs. (i) to solve for equation of Eqs. (i), we
find that
Kθ − ω Jo − 2
³
ω2Jθ
1 þ
ω2 Jθ2=C θ2
³
´
2ð
θ
2,
j ωÞ ¼
and using that result in the
j ωCθ ω Jθ =Cθ 1 þ ω J =C θ θ
þ
2
2
2
2
2
(i)
0
first
´!
2
Θ ð j ωÞ ¼ mo 1
(j)
fi
Introducing the following de nition
a
¼
ωJθ
Cθ
¼
ωρI0z0
ωz0
¼
ρcs I0
2
cs
2
¼
ωr0 9π
cs
where we have used Eqs. (a), Eq. ( j) can be written as
Kθ
µ
1
−
a ^ J−
a
2
3
ja
2
3ð1 þ
a
2
Þ
þ
3
3ð1 þ
a
2
16
¶ Þ
ð1
−ν
Þ
Θ ð j ωÞ ¼ mo 1
(k)
(l)
where
^J
¼
and we have used Eqs. (a) to determine that
Cθ Jθ K θ 2
¼
ρ2 c2s I02
4
ρI0 z0
Jo Jθ
µ
(m)
z 12ρc I s 0
2
0
¶ ¼
1 3
(n)
488
Multiple Degree-of-Freedom Systems: Forced Oscillations The dynamic stiffness of the soil-foundation interaction is de
fined as
Sθ ðaÞ ¼ K θ ðk θ ðaÞ þ jacθ ðaÞÞ
(o)
where
kθ ðaÞ ¼ 1 − cθ ðaÞ ¼
a ^ J−
a
2
3
2
3ð1 þ
a
2
Þ (p)
a
2
3ð1 þ
a
2
Þ
We see from Eqs. (l) and (o) that the rotational response of the system is given by
¹¹Θ ¹¹ m
¹¹ ¹¹ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k a a c a
j ωÞ = o Kθ
1ð
1
¼
2
θ
ð Þ þ
2
2
θ
(q)
ð Þ
and its phase response by
φðaÞ
− acθ ðaÞ
¼ tan
1
kθ ðaÞ
Equations (q) and (r) have been used to obtain Figure 8.9.
Figure 8.9.
for
J^
Amplitude and phase responses of a circular rigid foundation resting on soil
¼ 0:4.
(r)
8.5 Vibration Absorbers 8.5
489
VIBRATION ABSORBERS
Vibration absorbers are used in many applications, which include power transmission lines, automobiles, aircraft, optical platforms, and rotating machinery. We start the discussion by
first
introducing the undamped vibration absorber and then broadening the
discussion to include damping and different types of dampers. In Sections 8.5.2 and 8.5.3, we discuss the design of two very different types of vibration absorbers that are based on linear-system principles. In Sections 8.5.4 to 8.5.6, we introduce nonlinear vibration absorbers and give an indication of the types of absorber designs that are pos5
sible.
In Section 8.5.7, we summarize the results and introduce a few additional exam-
ples of vibration absorbers.
8.5.1
Undamped Vibration Absorber Consider the model of a physical system shown in Figure 8.10. This model is an idealization of a primary undamped mechanical system composed of ondary
mechanical
system
mechanical system is called a
of m vibration absorber.
composed
and
2
k
2
is
m
1
and
k
attached.
1
to which a secThe
The generalized coordinates
secondary
x
1
and
x
2
represent the displacements measured from the static-equilibrium position of the system.
Single Degree-of-Freedom System When the secondary system is absent and a harmonic force
f ð tÞ ¼ F 1
is applied to
m
1,
1
sin
ωt
(8.71)
the governing equation of motion is that of a single degree-of-freedom
system given by
m €x 1
Secondary system (absorber)
1
þ
k x 1
1
¼
F
1
sin
Figure 8.10.
m2
x2 (t)
m1
x1 (t)
ωt
(8.72)
Undamped vibration absorber system.
k2
k1
f1(t)
i
j
5
The models presented in this section also have been used to estimate the effects of sloshing dampers (tanks with a
fluid
“Wind-induced vibration of tower and practical applications of ” Journal of Wind Engineering and Industrial Aerodynamics, 33 (1990) 263–272; G. So and note on a natural sloshing absorber for vibration control,” Journal of Sound and Vibration,
in them). See, for example, K. Fujii et al.,
tended sloshing damper,
“A – 1127.
S. E. Semercigil, 269 (2004) 1119
490
Multiple Degree-of-Freedom Systems: Forced Oscillations For the harmonically forced single degree-of-freedom system given by Eq. (8.72), the resulting forced response is given by Eqs. (5.137) and (5.14) with
F k
x ðtÞ ¼
1
1
1
sin 1
ζ
=
0. Thus
ωt
−Ω
(8.73)
2
It is seen from Eq. (8.73) that if the excitation frequency is in the vicinity of the natural frequency
ωn1
≈
Ω
of the primary system, that is, when
1, then the resulting response is
undesirable. However, if we are required to operate the system at
Ω
=
close to it, then a secondary system can be introduced in order to have a
1 (
ω
= ωn
1)
or
finite amplitude
response at the resonance of the single degree-of-freedom system as illustrated below.
Two Degree-of-Freedom System The displacements of the two degree-of-freedom system are obtained from Eqs. (8.57). Thus,
X
F ωr − Ω Do k 2
1
2
1
¼
1
(8.74)
F ωr X ¼ k Do 2
1
2
1
find that
where from Eq. (8.56) we
Do
¼
Ω
−
4
Ω
= ω ωn /
1þ
ωr
It is recalled from Eqs. (7.41) that natural frequencies and
±
²
ω2r ð1 þ m r Þ Ω2
= ωn
/
2
X
1
=
(8.75)
is the ratio of the system’s uncoupled
is the excitation frequency ratio.
1
Since we are forcing the primary system at that
ωn1
ω2r
þ
Ω
=
1, we see from the
first
of Eqs. (8.74)
0 when
ωr
¼
Ω¼1
(8.76)
or, equivalently, from Eq. (7.41) that
ωn1 Evaluating
Do
at
Ω
=
¼
ωn2
ω ¼ ωn1
and
1 from Eq. (8.75) we
find
(8.77)
D o = −m r ωr = −k =k 2
that
2
1:
Then,
Eq. (8.74) leads to
X
1
¼ 0
X
2
¼
F 1 k Do 1
1
¼
−F k ðk =k 1
1
2
1Þ
¼
−F
1
k
(8.78)
2
Although the harmonic excitation of the primary mass is at a frequency equal to the primary system’s natural frequency, the choice of the secondary system’s parameters are such that by having
ωn1
=
ωn2,
the response of the primary system becomes zero.
8.5 Vibration Absorbers
F1 sin
n1 t
X2) sin
k2 (X1
F1 sin
n1 t
Free-body diagram of mass
Figure 8.11.
n1 t
for
ωr
=
491
1 and
Ω = 1.
m
1
m1
k1X1 sin
0
n1t
To understand why the response of the primary mass
m
1
is zero when the excitation fre-
quency is at the natural frequency of the single degree-of-freedom system, we consider the free-body diagram shown in Figure 8.11. In this diagram, the spring forces have been evaluated from Eqs. (8.78). From this
figure,
it is seen that the spring force generated by the
secondary system, the absorber, is equal and opposite to the excitation force at is, the primary mass
m
1
does not experience any effective excitation at
ondary system is designed so that
ωr
= ωn
ωn1
/
2
=
Ω
=
Ω=
1; that
1 when the sec-
1. If one considers the two degree-of-
freedom system from an input energy-output energy perspective, all of the energy input to the system at the excitation frequency
=
Ω
1 goes into the secondary system, the absorber.
In other words, the secondary system absorbs all of the input energy. The natural frequencies of the two degree-of-freedom system are given by the roots of
Do
=
0. We
find from Eq. (8.75) that when ωr = 1, the natural frequencies are solutions of Ω
4
−
ð2 þ
m r ÞΩ
2
þ 1 ¼ 0
(8.79)
Although the presence of the absorber is good for the system in terms of attenuating the response of the primary mass at
ω
= ωn
, there are still two resonances given by the solu-
1
tion to Eq. (8.79). At these two resonances, the system response is unbounded since we have an undamped system. This unbounded response can be eliminated by the inclusion of damping, which is considered in Section 8.5.2.
EXAMPLE 8.4
Absorber for a diesel engine
6
An engine of mass 300 kg is found to experience undesirable vibrations at an operating speed of 6000 rpm. If the magnitude of the excitation force is 240 N, design a vibration absorber for this system so that the maximum amplitude of the absorber mass does not exceed 3 mm. To design the vibration absorber for this system, we make use of the parameters given above
to
determine
the
absorber stiffness
k
2
and
the
absorber mass
m
.
2
From the
information provided, the excitation frequency is given by
ω¼
6
S. S. Rao,
π
ð2
rad=revÞð6000 rev=minÞ 60 s=min
¼ 628:32
rad=s
Mechanical Vibrations , Addison-Wesley, Reading, MA, 1995, Chapter 9.
(a)
492
Multiple Degree-of-Freedom Systems: Forced Oscillations and the excitation magnitude is
F
1
The amplitude
X
2
¼ 240 N
(b)
of the absorber mass has to be such that
∣X ∣ ≤ × 3
2
−
10
3
m
(c)
It is assumed that the system is operating at the natural frequency of the engine, that is,
ω
= ωn
1,
and that the absorber is designed so that
ωn1
= ωn
2
(the absorber’s natural
frequency is the same as the engine’s natural frequency). From Eq. (a) and Eqs. (7.41), we arrive at
ω2n2
¼
k m
2
¼
2
ω2n1
¼
2
ω2
¼ ð628:32Þ
2
2
rad =s
(d)
Equation (d) provides us with one of the equations needed to determine the parameters
k
2
and
m
2
of the absorber. To determine the other equation, we make use of Eq. (b) and
Eq. (8.78) to arrive at
j
X
2j
¼
F k
1
¼
2
From Eqs. (c) and (e), we
240
(e)
k
2
find that the absorber stiffness should be such that j
X
2j
¼
F k
1
¼
≤
240
2
k
2
0:003 m
or
k
2
≥ × 80
3
10
N=m
(f)
Making use of Eq. (d), we determine the absorber mass to be
m
2
¼
k
2
(g)
2
ð628:32Þ
k
Choosing
8.5.2
2
=
100
×
3
10
N/m so that Eq. (f) is satis
fied leads to m
2
=
0.253 kg.
Damped Linear Vibration Absorber As an illustration of a linear vibration absorber, we consider the system shown in Figure 8.12. In this system, it is assumed that the primary system with damping has attached to it a secondary system, the vibration absorber. A harmonic disturbance
f (t ) 1
acts on the primary system and it is assumed that there is no force acting on the absorber mass
m
.
2
The design question that is posed is the following: how can a combination of parameters
k ,c 2
2,
and
m
2
of the secondary system be chosen so that the response amplitude
8.5 Vibration Absorbers x1 , f1(t)
k1
493
x2
k2
m1
m2 c2
c1
Figure 8.12.
System with vibration absorber.
fied
of the primary system is at a minimum (or zero) in the speci
frequency range of the
excitation? To answer this question, we start from the responses given by Eqs. (8.53) with
F
=
2
0 and obtain
X ð jΩÞ Eð jΩÞ ¼ F =k D ð jΩÞ 1
1
where
¼
−Ω
2
þ
1
þ j 2ζ ωr Ω D ð jΩÞ
ω2r
2
(8.80)
D( jΩ) is given by Eq. (8.61) and we have used Eq. (8.60).
In Section 8.5.1, we considered an undamped primary system and an undamped vibration absorber, and it was shown that when this primary system is subjected to a harmonic excitation at the natural frequency of the primary system, the vibration absorber can be designed to provide an
equal and
opposite force to
the excitation force on the
primary system. Therefore, the effective excitation experienced by the primary system is canceled at this excitation frequency. In the present case, the primary and secondary systems
are
damped.
The
question
is
whether
the
same
canceling
effect
can
be
accomplished.
Special Case of Absorber System: ζ2 = 0 E jΩ =
From Eqs. (8.80), we see that when by
X ( jΩ) is also zero. When ζ 1
2
(
)
0 the response of the primary system given
is zero, it is seen from Eq. (8.80) that
X
1
=
0 if
Ω ¼ ωr
(8.81a)
ω ¼ ωn 2
(8.81b)
or equivalently from Eqs. (7.41) that
Equation (8.81a) is the same as that previously determined from Eq. (8.76). Thus, if we choose
k
2
and
m
2
fi
so that Eq. (8.81b) is satis ed, then we can have a zero of the forced
response of the primary system at the chosen excitation frequency. The implication of this observation is as follows. Suppose that a harmonic excitation is imposed on the primary mass
m
1
at a frequency
ω
=
ωn1,
where
ωn1
is the natural frequency of the
undamped primary system, that is, the system of Figure 8.12 without the secondary system. In the absence of the secondary system, since we are exciting the primary system at its undamped natural frequency, we expect the response of this linear system to be
“large.”
With the inclusion of an undamped secondary system, and for the choice of
k
2
494
Multiple Degree-of-Freedom Systems: Forced Oscillations and
ω
m
2
= ωn
find
satisfying Eqs. (8.81), we
. The absorber parameters 1
k
2
that the response of the primary system is zero at
and
ωn2 where
ωn2
m
2
¼
need to be chosen so that
ω ¼ ωn1
(8.82)
is the natural frequency of the undamped, uncoupled secondary system. As
should be expected, another equation or condition apart from Eq. (8.82) will be needed to determine the absorber parameters. In Section 8.4.1, we found that the response of the primary mass is zero despite having an excitation acting directly on it, since the force produced by the absorber on the primary mass is equal and opposite to it. To see if this holds true for this case, we use Eqs. (8.38), (8.49), and (8.60) with
Að
³ jΩ
Þ 1
−Ω
2
ζ2
=
´
j ζ Ω þ mr ωr X ð jΩÞ − m rωr X ð jΩÞ ¼ F =k 2
þ 2
³ m ω
− mr ωr X
1ð
jΩÞ þ
r
´ −Ω X 2
1
1
2
In Eqs. (8.83),
0, to obtain
2
2
r
2
2ð
1
jΩÞ ¼
1
(8.83)
0
X ( jω) and X ( j ω) are the complex amplitudes of the frequency responses m and m , respectively. From the second of Eqs. (8.83), if the 1
of the inertia elements
2
1
2
vibration absorber is chosen to satisfy Eq. (8.81a), then
X ð jΩÞ ¼ 1
and, hence, it follows from the
0
(8.84a)
first of Eqs. (8.83) and Eqs. (7.41) that
X ð jΩÞ = − 2
F ð jΩÞ k mr ωr 1
2
1
=−
F ð jΩÞ k 1
(8.84b)
2
which is identical to Eq. (8.78). Thus, the force generated by the spring mary mass the mass
m
the forcing
m
1
1
k
2
on the pri-
opposes the disturbing force and is equal to it in magnitude. As a result,
does not move and instead all of the energy provided to the system through
f (t) is absorbed by the secondary system. 1
Although we would like to take advantage of the zero of the forced response at a selected excitation frequency, it is not possible to realize this in practice. This can occur because of poor frequency tuning of the absorber after installation or frequency detuning over time caused by changes to the primary system stiffness and inertia characteristics, or by changes to the absorber system stiffness and inertia characteristics. The possible variations in the system parameters pose the question: how can one design an absorber to be effective over a broad frequency range? Before answering this question, we return to the free oscillation problem. With the introduction of the secondary system, the absorber, the two degree-of-freedom system has two natural frequencies. The two natural frequencies of the undamped two degree-of-freedom system are given by the roots of the characteristic equation, which is
Do
¼
Ω
4
−
³
1þ
´
m r ωr þ ωr Ω 2
2
2
þ
ω2r
¼ 0
(8.85)
8.5 Vibration Absorbers
495
2,1Ω Figure 8.13.
Nondimensional natural frequencies of a two degree-of-freedom system with a
vibration absorber as a function of the mass ratio
m r for selected values of ωr.
The solution of Eq. (8.85), which is a quadratic equation in
Ω
1;2
¼
Ω
2
, is
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi · ¸ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ³ ´ ω m ∓ ω m − ω 1 2
1þ
2
r ð1
rÞ
þ
2
1 þ
r ð1
þ
rÞ
The variations of the two nondimensional natural frequencies Figure 8.13 with respect to the mass ratio
mr = m / m 2
1
2
Ω
4
1
2
(8.86)
r
and
Ω
2
are plotted in
and the frequency ratio
ωr .
As the
mass ratio is increased, the separation between the two natural frequencies increases. In the presence of a harmonic disturbance acting on the mass of the primary system, we would like the response of mass
m
1
not to be large when we are operating at frequencies
close to either of the two resonance frequencies of the undamped two degree-of-freedom system. To address this, we return to the forced oscillation problem.
General Case of Absorber System In Figures 8.14 and 8.15, the nondimensional amplitude response
H
ΩÞ ¼ jk G
11 ð
1
11ð
jΩÞj
(8.87)
determined from Eq. (8.80) is plotted as a function of the nondimensional excitation frequency ratio
Ω.
The mass ratio
primary system expressed by is
varied.
The
ratio
ωr
is
ζ1
mr,
the ratio frequency
are held
decreased
Figure 8.14 to Figure 8.15. In both
fixed
from
figures,
ωr ,
and the damping of the
in each case, and the damping factor
1
to
0.97
(i.e.,
by
3%)
in
going
ζ2
from
we have one scenario with an undamped
vibration absorber and two scenarios with two different damped vibration absorbers.
496
Multiple Degree-of-Freedom Systems: Forced Oscillations
)Ω (11 H Figure 8.14.
ωr
=
1,
mr
=
Amplitude response of primary mass 0.1, and
ζ1
=
m
1
0.1 for several values of
with a damped vibration absorber when
ζ2 .
) Ω(11H Figure 8.15.
ωr
=
The
0.97,
Amplitude response of primary mass
mr = 0.1, and ζ
1
responses
seen
freedom systems
in
=
Figures
excited
m
1
with a damped vibration absorber when
0.1 for several values of
by
a
8.14
and
harmonic
8.15
ζ2.
are
excitation,
characteristic where
the
of
two
degree-of-
excitation frequency
range includes the two resonance frequencies. Based on Eq. (8.81a), it is clear that when
ζ2
=
0, the response of the primary system is zero. However, when
if the nondimensional excitation frequency
Ω
≠
ζ2
≠
0 and/or
1, the response of the primary system
8.5 Vibration Absorbers may not have a
“small ”
497
magnitude. Therefore, we would like to choose the absorber
system parameters so that the response of the primary system is as small as possible over a wide frequency range that includes 7
For the general case,
Ω
=
1.
ζ1
when the damping factor
of the primary system is not zero,
the choice of the secondary system parameters cannot be put in explicit form such as Eq. (8.82) and one has to resort to numerical means. However, when 8
obtain
ζ1
=
0, one can
optimal values for the secondary system natural frequency and the secondary
H
system damping ratio to tailor the amplitude response
Ω)
(
11
of the primary system.
The optimal values for the absorber system parameters are obtained when the values of the peaks shown in Figure 8.15 are equal, that is, when
H
ΩA Þ ¼ H
11 ð
11ð
ΩB Þ
(8.88)
Upon carrying out the derivation, the following optimal values are obtained:
ωr;opt The ratio
ωr
¼
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 1 þ
and
mr
ζ 2;opt
mr
3
¼
8ð1 þ
is optimal or there is optimal tuning when the
9
(8.89)
mr Þ
3
first of Eqs. (8.89) is satisfied.
The second of Eqs. (8.89) provides the optimal damping value for the secondary system. It is important to note that the mass ratio is the sole determining factor for the optimal
ζ1
values. For a general case, when
≠
0, although the optimal condition expressed by
Eq. (8.88) is valid, explicit forms such as Eqs. (8.89) cannot be obtained.
7
The response of an absorber system that is limited by an elastic or rigid stop can be found in the work of A. M.
“Non-linear correction of vibration protection system containing tuned dynamic absor” Journal of Sound and Vibration , 239(2) (2001) 335–356. The attachment of a spring-mass-damper system to
Veprik and V. I. Babitsky, ber,
beams, plates, and shells is treated in the following series of articles: E. O. Ayorinde and G. B. Warburton,
8 9
“ Optimum absorber parameters for simple systems,” Earthquake Engineering and Structural Dynamics, 8 (1980) 196 –217; E. O. Ayorinde and G. B. Warburton, “Minimizing structural vibrations with absorbers,” Earthquake Engineering and Structural Dynamics , 8 (1980) 219–236; G. B. Warburton, “Optimum absorber parameters for minimizing vibration response, ” Earthquake Engineering and Structural Dynamics, 9 (1981) 251 –262; and G. B. Warburton, “Optimum absorber parameters for various combinations of response and excitation parameters, ” Earthquake Engineering and Structural Dynamics , 10 (1982) 381–401. J. P. Den Hartog, ibid., pp. 87 –113. Equation (8.89), which can be used to determine the optimum values of the absorber parameters, is based on the displacement response of the mass
m
1.
If one is interested in determining the optimum values of the absorber para-
meters based on the velocity response or acceleration response of the mass optimal values are
ωr;opt
¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m 1 þ 1 þ
r =2
mr
and
ζ 2;opt
See G. B. Warburton, 1982,
ibid.
ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi m
1 ¼ p 1 þ
r
and
r
1 þ
8ð1 þ
and for the acceleration response, the optimal values are
ωr; opt
1
then for the velocity response, the
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ³ m m ´ffi m 3
¼
m,
r
þ5
¼
r =24
m r Þð1 þ mr =2 Þ
2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ζ 2;opt
2
mr mr =2Þ
3
8ð1 þ
498
Multiple Degree-of-Freedom Systems: Forced Oscillations
H11(Ω B)
H11 (ΩA)
)Ω (11H
H11( ΩC)
Figure 8.16.
fication of amplitude response function characteristics to be minimized.
Identi
We shall now determine optimal secondary system parameters for a damped absorber (i.e.,
ζ2 ≠
0) so that there is an operating region including
the amplitude of the primary system
m
1
Ω=
1 wherein the variation of
as a function of frequency can remain relatively
constant. That is, we would like to reduce the absorber’s sensitivity to small variations in frequency. This can be realized through optimization solution techniques.
find
to Figure 8.16, the goal is to
between
Referring
the secondary system parameters for which the peak
B are equal and are as “small” as possible, while the minimum these peaks C is as close to A and B as possible in terms of magnitude. In other
amplitudes
A
10
and
words, we would like to
find
the system parameters that minimize each of the following
three maximum values simultaneously: is given by Eq. (8.54) with
k
32
=c = 32
H
ΩA), H
(
11
(
11
ΩB),
and 1/
H
(
11
ΩC),
where
H
11
Ω)
(
0. This can be stated as follows.
H
ΩA Þg
H
ΩB Þg
min f
ω r; ζ2
min f
ω r; ζ2
11 ð
11 ð
H
min f1=
ω r; ζ2
ΩC Þg
(8.90)
11 ð
subject to:
ωr > 0 ζ2 ≥ 0
For illustration, we assume that available numerical procedures
10
ζ1
=
0.1 and
mr
=
0.1, 0.2, 0.3 and 0.4 and use readily
to determine the optimum values of
ζ2,opt
and
ωr,opt.
E. Pennestri, “An application of Chebyshev’ s min-max criterion to the optimal design of a damped dynamic vibration absorber,”
11
11
Using Matlab,
Journal of Sound and Vibration , 217(4) (1998) 757–765.
fminimax from the Optimization Toolbox.
8.5 Vibration Absorbers
Figure 8.17.
499
Optimum values for the parameters of a vibration absorber and the resulting
amplitude responses of
m
1
for
ζ1
=
0.1. The dashed lines are the responses obtained for the
optimum values given by Eqs. (8.89).
The results are shown in Figure 8.17 for four different values of the mass ratio
mr.
It is
seen that these results satisfy the condition given by Eq. (8.88). Comparing the results shown in the four cases in Figure 8.17, it is clear that the last case is preferable, since the primary system’s response is attenuated to the smallest magnitude over the chosen frequency range. For a given primary system mass, when
mr
=
0.1, the secondary system is
the smallest and the displacement amplitude of the secondary system is likely to be the largest. Since there are restrictions on the amplitude of motion of the secondary system, if the displacement amplitude of the secondary system is required to be less than a certain value, then this requirement can be included as a constraint for Eqs. (8.90). In terms of design, there are still other choices one can come up with if Eq. (8.90) is replaced with another set of objectives. However, here, we have pointed out that for a practical design of an absorber, one will have to resort to numerical means and take advantage of algorithms such as that used in the present context. Furthermore, it is noted that when one carries out design, usually there is more than one design solution to be reckoned with. In addition, the number of degrees of freedom is likely to be higher than two in a practical situation.
500
Multiple Degree-of-Freedom Systems: Forced Oscillations A variation of the attachment of the absorber damper.
Figure 8.18.
c2 x2
m2 k2
f(t) m1
x1 k1
Upon comparing the results obtained from Eqs. (8.89), we see that there are differ-
mr
ences. For example, when
=
0.1, Eqs. (8.89) give that
ωr, opt
=
0.909 and
ζ2, opt
=
0.168, whereas, from Figure 8.17 we see that the numerical optimization procedure gives
ωr ,opt
=
0.714 and tion
ζ2,opt
0.862 and
ζ2,opt
procedure
=
=
0.199. Similarly, for
mr
=
0.4, Eqs. (8.89) give that
ωr ,opt
=
0.234, whereas, from Figure 8.17 we see that the numerical optimiza-
gives
ωr,opt
=
0.655
ζ2,opt
and
=
0.343.
However,
it
is
seen
from
Figure 8.17 that the variations in the amplitude response function around the natural frequencies obtained by using the optimum values from Eqs. (8.89) are consistently greater than those obtained on the basis of Eqs. (8.90).
INTERACTIVE GRAPHIC 8.4: FREQUENCY-RESPONSE FUNCTION FOR AN OPTIMAL CHOICE OF PARAMETERS The optimized frequency-response function H11 of a translating two degree-of-freedom system is displayed along with the associated optimum values of ωr and ζ2 with respect to the choices of m r and ζ1. In this interactive graphic, the following should be noted. •
•
For all combinations of mr and ωr , the optimized value for ζ2 is always greater than that obtained from Eq. (8.89) and the optimized value for ωr is always less than that obtained from Eq. (8.89). The peak values of the frequency-response function obtained using Eq. (8.89) are always greater than those obtained using the current optimization requirements. If the damper
c
2
is not connected to mass
m
1
and instead is connected to a
fixed
end,
as shown in Figure 8.18, then the governing equations take the form
m €x m €x 1
1
2
2
k þ k Þx − k x ¼ f ð tÞ _ þ k ðx − x Þ ¼ 0 þc x
þð
1
2
2
2
1
2
2
2
2
1
(8.91)
1
After using the approach that was employed to obtain Eqs. (8.53), we arrive at the frequency-response function
8.5 Vibration Absorbers
H
11 ð
jΩÞ ¼
¹¹ ¹ ¹¹ X jΩ ¹¹¹ ¹F jΩ k ¹ ¹¹ ¹¹³ ´³ −Ω ω ¹¹ − Ω ω − Ω ´ − Ω ω m 1ð
1ð
Þ
Þ=
1
2
¼
2
1
2
2
r
sffiffiffiffiffiffiffiffiffiffiffiffi
12
¼
1
1
− mr
2
þ
2
It has been found that for this function,
ωr;opt
501
r
2
r
j ζ Ωωr j ζ Ωωr
þ 2
r
2
þ 2
2
³
1
−Ω
2
þ
¹¹ ¹¹ ´ ω m ¹¹ 2
r
(8.92)
r
the optimum values are
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ζ 2;opt
and
¼
mr
3
8ð1
− mr
(8.93)
=2Þ
A comparison of the responses for the two types of attachment of the damper to the absorber mass for
fi
mr = 0.1 is given in Figure 8.19. It is seen that the responses are similar,
but with the con guration of the absorber shown in Figure 8.18, one obtains about a 10% decrease in the maximum magnitude of the displacement for the same mass ratio.
fl
In designing a vibration absorber, there are other aspects such as the static de ection of the two degree-of-freedom system, stresses in the absorber springs, displacement of the absorber mass, etc. that one has to deal with. Many practical hints for designing absorbers are available in the literature.
13
opt opt
)Ω (11 H Optimized amplitude response of m for two types of damper attachments when mr = 0.1. Solid lines are for the case when one end of the damper c is attached to the fixed end as shown in Figure 8.18; dashed lines are for the case when c is attached to m and m .
Figure 8.19.
1
2
2
12
M. Z. Ren,
–
“A
variant design of the dynamic vibration absorber,
762 770. 13
H. Bachmann et al.,
ibid., Appendix D.
1
” Journal of Sound and Vibration,
2
245(4) (2001)
502
Multiple Degree-of-Freedom Systems: Forced Oscillations
Design Guideline:
In many practical situations, a vibration-absorber system should
have a ratio of the absorber mass to the primary system mass of about one-tenth, a damping factor for the absorber system of about 0.2 when the primary system is lightly damped, and an uncoupled natural frequency of the absorber system that is about 15% less than that of the primary system.
EXAMPLE 8.5
Absorber design for a rotating system with mass unbalance
A rotating system with an unbalanced mass is modeled as a single degree-of-freedom system with a mass of 10 kg, a natural frequency of 40 Hz, and a damping factor of 0.2. The system requires an undamped absorber so that the natural frequencies of the resulting two degree-of-freedom system are outside the frequency range of 30 Hz to 50 Hz. In addition, the absorber is to be designed such that for force magnitudes of up to 6000 N at 40 Hz the steady-state amplitude of the absorber’s response will not exceed 20 mm. The parameters provided for the primary system are
m
1
ωn1 ζ1
¼ 10 kg ¼ 40
×
π
2
¼ 251:33 rad=s
(a)
¼ 0:2
For this system, the absorber must be such that the resulting two degree-of-freedom system’s lowest natural frequency
ω1
and the highest natural frequency
ω2
have, respectively,
the following limits
ω1 < 30 × 2π
ω2 > 50 × 2π
¼ 188:50 rad=s (b) ¼ 314:16 rad=s
or
Ω Ω where
Ω
1
and
Ω
2
1
2
¼
¼
ω1 ωn 1 ω2 ωn 1
> @ A k− k k k− B z C B
> k− k k 0 9 z 81 ; z_ : c c − > > A @ z_ C c− c c c− B B
> c− c c 0 9 z_ 81 ; z€ : z€ m > > @ z€ A €z m BB = >
€z m 0 9 z€ 81
g þ θ soc θ_ x_ þ
3 ,2 ,1 =
; €y m − > : > y m− € = < > > 9 €y m − 8
Þ ð
nis
2
gM þ θ soc θ_ Þr − Rðm þ xk þ θ nis €θÞr − Rðm þ x €Þm þ Mð
θ
Lg m −
0 ¼
0 ¼
snoitauqe gninrevoG
)secrof cimsies dna dniw morf sgnidliub llat ni noitom yrotallicso sserppus :esu lacipyT( rebrosba muludnep rednilyc gnilloR
01
9
8
529
0 ¼ 0 ¼
0 ¼
θ
x € − gð þ θ_ z_ 2 þ €θÞz þ Lð
nisÞ
2
0
º
u
k
( ) φ (! gm 2
nis Þ
gm Lgm þ lg om
l+l
=
þ
u €
2
Lm 2
l om þ Lm
2
2
³
m ol om
m
1c
)t (f
u
1k
l
c
1k
M
θ
1k om
k
x
z +L
ϕ
)2 esaC ot secuder siht ,0 = )t (z nehW( 4 rebrosba muludneP
snoitauqe gninrevoG
m 2
´
snoitauqe deziraeniL
) φ€ (! Lm
1
L , k2 = k( snoitauqe gninrevoG
€ l om þ Lm g Lm þ l omð þ € uLm þ φ_ u_ um2 þ φ € um þ φ
gm þ uk þ φ_ um − € um þ φ €Lm
t M
Þ ð
¼
θ_ þ θ nis € θ Þz þ Lðm − θ nis θ_ z_ m 2 − θ soc z€m þ xk þ xc þ x €Þm þ Mð
θ soc gm − θ_ Þz þ Lðm − θ soc € xm þ z 1k þ z_ 1c þ €zm 2
) nis
φ soc ugm þ φ
φ
gm − θ soc
»
¼
0 ¼
t M
Þ ð
)o
muludnep a rof rebrosbA
21
11
).tnoc( .1.8 elbaT
530
8.6 Vibration Isolation: Transmissibility Ratio
Machinery
m2
Floor
Figure 8.32. Representative two degree-of-freedom system for a
x2 , f2 (t)
k2
531
vibration transmission model.
c2 m1
x1
k1
c1
Ground
flooring and m
2
is the mass of the machinery as shown in Figure 8.32. In some instances,
the machinery is connected to a seismic mass, which in turn is connected via springs to the structure, usually the ground. The objective is to determine the various parameters of the system so that the force transmitted to the ground or the structure that is supporting the
flooring
31
is as small as
practical. To do this, we need to examine the ratio of the magnitude of the force transmitted to the ground
fbase(t)
to the magnitude of the force applied to the machinery
f (t ). 2
From Eq. (3.10), the force acting on the base is
F base ðtÞ ¼ k x 1
For harmonic oscillations of the form
Fbase ð tÞ ¼ ðk
1
þ
þ
1
=X
x ( t) 1
1(
dx dt
c
1
1
(8.133)
jω)e jωt, Eq. (8.133) can be written as
j ωc ÞX ð j ωÞ ejωt ¼ F base ð j ωÞe jωt 1
1
where
Fbase ð j ωÞ ¼ ð k
þ
1
F base ð jΩÞ ¼ k
1ð 1
X
1
1
1
find that
In terms of nondimensional quantities, we
Using the expression for
j ωc ÞX ð j ωÞ
j ζ ΩÞX ð jΩÞ
þ 2
1
1
given by Eq. (8.53) with
Fbase ð jΩÞ F
j ζ ΩÞ
¼ ð1 þ 2
1
2
where The
1
=
0, Eq. (8.134) becomes
Bð jΩÞ Dð jΩÞ
B( jΩ)and D( jΩ), respectively, are given by Eqs. (8.60) and (8.61). transmissibility ratio is defined as TR ¼
31
F
(8.134)
J. A. Macinante,
¹¹F ¹¹
¹¹ ¹¹
base ð jΩÞ
F
2
(8.135)
Seismic Mountings for Vibration Isolation, John Wiley & Sons, New York, 1984, Chapter 8.
532
Multiple Degree-of-Freedom Systems: Forced Oscillations Therefore,
TR ¼
¹¹ ¹¹
j ζ ΩÞ
ð1 þ 2
1
Bð jΩÞ D ð jΩÞ
¹¹ ¹¹
(8.136)
The results obtained from Eq. (8.136) are plotted in Figure 8.33 for the case where
ζ 1 = ζ2
=
TR = 0.05 and 0.1. For each combination of these values, Ω for mr = 0.1 and 0.8 is presented. It is seen that linear with respect to Ω irrespective of the values of mr and
0.05 and 0.1 and for
the relationship of
ωr
as a function of
this relationship is virtually that the value of decreases as
mr
mr
affects the slope of this relationship by about 10%; a value that
increases. Comparing these results with the
TR
of-freedom system given in Figure 5.23 we see that to get a values of the damping factor given above, we need to have
Ω
>
value of a single degree-
TR
of 10% and for the
3.3 in a single degree-of-
freedom system. On the other hand, in the case of a two degree-of-freedom system, we
Ω
see from Figure 8.33 that for
>
1.5 and
other words, the natural frequency of support/ obtain a for
ωr
>
flooring
m
2
ωr
irrespective of the value for
TR < 10% with mr =
0.1 and
=
0.5 we will attain the same levels. In
by itself has to be less than half of that for the
m r.
ζ 1 and ζ2
From these curves, it is seen that to
less than 0.1, approximately
Ω>
1
+ ωr
0.5.
2.5 2.0
z1
TR = 0.05 = 0.05 z 2 = 0.05
2.5
m r = 0.1 m r = 0.8
2.0 rw
rw
1.5 1.0 0.5 0.0 1.5
z1
TR = 0.05 = 0.1 z2 = 0.1
mr = 0.1 mr = 0.8
1.5 1.0 0.5
2.0
2.5
3.0
0.0 1.5
3.5
2.0
2.5
2.5 2.0
z1
TR = 0.1 = 0.05 z 2 = 0.05
2.5
mr = 0.1 mr = 0.8
2.0
1.0 0.5
z1
TR = 0.1 = 0.1 z 2 = 0.1
mr = 0.1 mr = 0.8
1.0 0.5
2.0
2.5
3.0
0.0 1.5
3.5
Ω Figure 8.33.
3.5
1.5 rw
rw
1.5
0.0 1.5
3.0
Ω
Ω
Values of
ωr
2.0
2.5
3.0
3.5
Ω that affect the transmissibility ratio for two degree-of-freedom systems
for various combinations of
mr, ζ , ζ 1
2,
and
TR.
8.6 Vibration Isolation: Transmissibility Ratio
533
INTERACTIVE GRAPHIC 8.6: TRANSMISSIBILITY RATIO ( TR ) VARIATION WITH RESPECT TO SYSTEM PARAMETERS The transmissibility ratio (TR) of a two degree-of-freedom system is displayed. The variation of TR with respect to several system parameters can be explored. In this interactive graphic, the following should be noted. As the selected value of TR decreases, so does the range of ωr at which these decreasing values can be attained
Impulse Force to Base We now examine how the system parameters can be selected to attenuate the peak magnitude of the force transmitted to the ground when an impulsive force is applied to the mass
m
2.
This situation arises, for example, in factories where it is necessary to isolate forges and
punch presses. To determine the force transmitted to the ground, we use Eq. (8.34) and set
f
1
=
0 and
d y dτ
= Foδ t
f
( ). Then we use Eq. (8.34),
2
2
1
2
ζ1
þ ð2
ζ 2 m r ωr Þ
þ2
dy dτ
1
d y dτ
þ
2
2
2
³
1 þ
´
mr ωr y 2
1
dy þ 2ζ ωr dτ
−
ζ 2mr ωr
2
dy dτ
2
− mrωr y 2
2
dy þ ωr y − 2ζ ωr dτ
2
2
1
2
2
2
¼ 0 (8.137)
− ωr y 2
1
¼
δ ðτ Þ
where
yi
¼
xi ð ωn F o Þ=ð k m r Þ 1
and we have used the fact that
δ (t)
1
= δ τ ωn = ωn δ τ ( /
1)
1
( ).
The force acting on the base is given by Eq. (8.133), which in the present notation can be written as
F base ðτÞ ¼
ωn1Fo
µ
¶
1
mr
1
1
µ
then, the time-varying transmissibility ratio is
TRð τÞ ¼
dy y þ 2ζ dτ
Fbase ð τÞ 1 ¼ y ωn F o mr
1
1
32
Equation (8.137) is evaluated numerically
dy þ 2ζ dτ
¶
1
and the result is used in Eq. (8.138)
to obtain the representative set of results shown in Figure 8.34 for
32
Using Matlab,
ode45; using Mathematica, NDSolveValue.
(8.138)
1
ζ1
=ζ = 2
0.1,
mr
=
534
Multiple Degree-of-Freedom Systems: Forced Oscillations
) t( RT
) t ( RT
) t( RT
) t( RT
) t( RT
) t ( RT Figure 8.34.
system:
ζ1
Magnitude of an impulse force transmitted to the base of a two degree-of-freedom
=ζ = 2
0.1 and
mr
1.0, and the six values of
=
1.0.
ωr
as indicated in the
figure.
For this value of
that the peak transient transmission reduction is nearly proportional to
m
the introduction of the seismic mass magnitude of the force applied to
1
mr,
ωr .
it is seen
We see that
can provide a larger attenuation of the peak
m
2 , when compared to that of a single degree-of-
freedom system. Recall from Figure 6.3 that for a single degree-of-freedom system, the
pffiffiffiffiffiffiffiffiffiffiffi k =k ;
most attenuation that could be attained was about 18% for with
k
2
mr
≪k
=
. 1
1.0, from Eq. (7.41) it is seen that
ωr
¼
2
1
ζ
=
0.25. For these results
therefore, to keep
ωr
small,
8.6 Vibration Isolation: Transmissibility Ratio
EXAMPLE 8.8
535
Design of machinery mounting to meet a transmissibility ratio
requirement
A 150 kg machine is to be operated at 500 rpm on a platform that is modeled as a single degree-of-freedom system with the following mass, stiffness, and damper values:
m
1
=
k
3000 kg,
1
= × 2
6
10
c
N/m, and
1
=
7500 N/(m/s). The machinery mounting is to
be designed so that the transmissibility ratio does not exceed 0.1. From the given values, we have that
mr ¼
m m
2
150 kg
¼
ωn1
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi v sffiffiffiffiffi u u k t× 1
m
c ζ ¼ 2m ωn 1
ω
N=m
3000 kg
¼ 25:82 rad=s
¼ 1
2
×
7500 N=ðm=sÞ 3000 kg
π
ð500 rpmÞð2
¼
ωn1
10
(a)
1
1
6
2
¼
1
Ω¼
¼ 0:05
3000 kg
1
×
25:82 rad=s
¼ 0:048
rad=revÞ=ð60 s=minÞ
25:82 rad=s
¼ 2:03
We pick a mounting with negligible damping factor, that is, one for which
ζ2 ≈ 0
(b)
We now use Eq. (8.136) to determine the value of ratio that is less than 0.1 at ing factor, we
Ω=
ωr
that provides a transmissibility
2.03. For the choice of mounting with negligible damp-
find from Eqs. (8.60) and (8.61) that Bð jΩÞ ≈ ωr
2
Dð jΩÞ ≈ Ω
4
−j
¹¹ ¹¹ ¹Ω − j
ζ 1 Ω3 −
2
³
1 þ
m r ωr
2
þ
´
ω2r Ω2
2
2
1
On substituting Eqs. (c) into Eq. (8.136), we arrive at
TR ¼
4
ζ 1 Ω3 −
2
³
´
j ζ ΩÞωr 1 þ mr ωr þ ωr Ω ð1 þ 2
1
2
2
2
2
Making use of Eqs. (a) by substituting the values for arrive at
TR ¼
¹¹ ¹¹
0:64
−
ð0:05 þ 0:00982 0:0404 33
Equation (e) is solved numerically this value is
ωr
=
Using Matlab,
2
1
Ω, ζ
, and
1
2
j þ ð− 0:166 þ 0:00982
for the value of
ωr
¹¹ ¹¹ ω ¹
mr
(d)
2
r
into Eq. (d), we
¹¹ ¹ j ω¹ Þ
(e)
2
r
that gives a value of
TR
=
0.1;
0.9741. Thus, since
ωn 2 33
j ζ ωr Ω þ
þ 2
j Þωr
(c)
j ζ ωr Ω þ ωr
þ 2
¼
ωr ωn1
fzero; using Mathematica, FindRoot.
(f)
536
Multiple Degree-of-Freedom Systems: Forced Oscillations the largest value of
k
2
¼
k
is
2
m ωn 2
2
¼ 94:9
8.7
2
×
m ð ωr ωn
¼
2
3
10
2
1Þ
¼ ð150 kgÞ
×
ð 0:9741
×
2
25:82 rad=sÞ
(g)
N=m
SYSTEMS WITH MOVING BASE
Another way to determine the ef
ficacy
of a two degree-of-freedom isolation system is to
compare the magnitude of the peak (maximum) displacement of the peak displacement of the ground.
34
m
2
to the magnitude of
To analyze this type of situation, we consider the
base supporting the two degree-of-freedom system to be a moving base as shown in Figure 8.35. In analyses of such systems, one usually assumes that the masses are initially at rest and that there are no applied forces directly on the inertial elements, and
x (t) is given. Noting the 3
similarity of Figure 8.35 to Figure 3.6 to, we modify Eq. (7.1a)
in the same manner that was done to obtain Eq. (3.27) and arrive at
m
1
d x dt 2
1
2
c
þ ð
1
þ
c
2Þ
dx dt
1
k
þ ð
1
k Þx
þ
2
d x m dt 2
2
−c
1
2
2
dx dt
dx þ c dt
2
2
2
2
2
−k
x
2
2
−c
1
dx dt
3
dx þ k x −c dt
1
2
2
2
−k
1
x
¼ 0
3
(8.139)
−k
2
x
¼ 0
1
After using the nondimensional quantities of Eqs. (7.41) and (8.48), Eqs. (8.139) are rewritten as
d x dτ 2
1
2
ζ1
þ ð2
ζ2 mr ωr Þ
þ 2
dx dτ
1
þ
³
1 þ
´
mr ωr x 2
d x dτ
1
−
ζ2 mr ωr
dx dτ
dx dτ
þ
2
2
2
2
ζ 2 ωr
þ 2
2
2
− m r ωr x − 2
2
dx dτ
ω2r x2 − 2ζ 2ωr
−x
¼ 0
− ωr x
¼ 0
3
ζ1
2
dx dτ
1
3
2
1
(8.140)
x3(t) k1 Moving base
34
k2
m1 c1
Figure 8.35.
x1 (t)
x2(t) m2
c2
Two degree-of-freedom system with moving base.
This analysis also has application in determining the dynamic response of seated humans. See, for example,
“Analysis of Journal of Sound and Vibration, 226(3) (1999) 585–606.
X. Wu, S. Rakheja and P.-E. Boileau,
relationships between biodynamic response functions,
”
8.7 Systems with Moving Base
537
To compare the responses of the single degree-of-freedom system with a moving base to that of a two degree-of-freedom system with a moving base, we assume that the displacement of the base is a half-sine wave, which can be obtained from Case 8 of Table 6.2 as
x ðτÞ ¼ X o sinðΩoτ Þ½ uðτÞ − uðτ − τ oÞ ±
(8.141)
3
(a) Figure 8.36.
(b) Displacement response of
system base for
mr = 0.1, and ζ
1
m
=ζ = 2
2
when a half sine wave displacement is applied to the
0.1: (a)
ωr
=
0.05 and (b)
ωr
=
0.2.
538
Multiple Degree-of-Freedom Systems: Forced Oscillations where
Ωo = ωo/ωn x
1
Assuming that numerically
35
= π τo = /
(0) 3
,
τo = ωn1to,
and
to
= π ωo /
. We shall determine the response of
m. 2
0 for convenience, Eqs. (8.140) with Eq. (8.141) are solved
and the results are shown in Figure 8.36. We see that as the duration of
the half sine wave pulse decreases, the amplitude of
m
2
decreases. This behavior is oppo-
site to what takes place during the base excitation of a single degree-of-freedom system, where
as
the
pulse
duration
decrease the
peak
displacement
(Recall Figure 6.19.) We see, then, that the interposition of per act as a mechanical
filter,
m
1
of
the
mass
increases.
and its spring and dam-
decreasing the amount of relatively high frequency energy
generated by the half sine wave pulse from being transferred to
m
2.
Thus, a two degree-
of-freedom system with appropriately chosen parameters can be an effective isolator of ground motions compared to a single degree-of-freedom system.
8.8
SUMMARY
In this chapter, we illustrated how the responses of linear multiple degree-of-freedom systems could be determined by using the normal-mode approach and by direct numerical evaluation of the governing equations. The responses of multiple degree-of-freedom systems to harmonic, step, and impulse excitations were also examined. The notions of resonance and frequency-response functions introduced in the earlier chapters for single degree-of-freedom systems were revisited and discussed for multiple degree-of-freedom systems. For linear vibratory systems, it was shown how the frequency-response functions can be used to design vibration absorbers and address issues such as base isolation and transmission ratio. A brief introduction to the design of vibration absorbers based on
nonlinear-system
behavior
was
also
presented.
In
addition,
interactive
graphics
materials have been included to explore the responses of different systems in the time and frequency domains. A wide range of systems with vibration absorbers has also been included.
35
Using Matlab,
ode45; using Mathematica, NDSolveValue.
Exercises
539
Exercises
Section 8.2.1 8.1
Consider the two degree-of-freedom system shown in Figure E8.1, where a forcing
f
is imposed on the mass
2
=F
2
ωt
sin
m
. For
2
m
1
=m =
1 kg,
2
k
1
=
2 N/m,
k
2
=
1 N/m, and
f
2
N, use the normal-mode approach to determine the solution of the sys-
tem. Assume that all of the initial conditions are zero.
x1
k1
8.2
x2
k2
m1
Figure E8.1.
f2
m2
k
1 kg,
1
=
2 N/m,
k
2
=
1 N/m,
c
1
=
0.4 N/m/s,
c
2
=
f
0.2 N/m/s, and
=m =f t
m
Consider the two degree-of-freedom system shown in Figure E8.2, where
=
2
1
2
( ). 2
Starting with Eq. (7.1b), uncouple these equations by using the modal matrix and present the uncoupled system of equations.
x1
k1
x2
k2
m1 c1
8.3
Figure E8.2.
f2
m2 c2
Use the normal-mode approach to determine a solution for the response of the system discussed
in Exercise 8.2 when a
imposed on mass
harmonic forcing
m
f
2
=
t
10 sin(20 )
N
is
. Assume that all of the initial conditions are zero and ignore 2
the transient portion of the response. 8.4
Use the normal-mode approach to determine a solution for the response of the three degree-of-freedom system shown in Figure E8.4 when an impulse is imposed on mass
m
1
=m =m =m 2
3
x1 k1
8.5
m1
m
2.
m2
k
1
m
. 2
=Fδt 2
( )
=k =k =k 2
3
.
x3 k3
Repeat Exercise 8.4 with an impulse mass
2
Assume that all of the initial conditions are zero and that
and that
x2 k2
f
Figure E8.4.
m3
f
1
= F δt 1
( ) imposed on mass
m
1
instead of
540
Multiple Degree-of-Freedom Systems: Forced Oscillations Section 8.2.2
Consider the three degree-of-freedom system shown in Figure E8.6. The different
8.6
kg
⋅
kt
2
m ,
1
= kt =
⋅
10 N
2
2c 64
m/rad, and
3 75
=
Jo
system parameter values are as follows:
1
kt
3
=
1 kg
5 N
2J 6 α4
⋅
⋅
where
α
0
0
0
ct
0
0
0
2
ct
O1
¼
3
0
0
JO
0
0
= 0.5. When the external moment
Mo(t)
2
=
4 kg
⋅
2
m ,
Jo
3
=
1
m=rad. Assume that the damping
3 75
matrix is proportional to the inertia matrix; that is,
t1
Jo
2
m ,
0 0
2
JO
3
is absent, determine the response
of this system by using the normal-mode approach for each of the following sets of initial conditions and discuss the participation of different modes in each case. (a) (b)
ϕ1(0) ϕ1(0)
=ϕ =ϕ
2(0)
(0)
3
=ϕ =
=
3(0)
1 rad, and
ϕ2(0)
1 rad,
=–
ϕ_ 1 ð0Þ
0.5 rad,
Oil housing
ϕ_ 2 ð0Þ ¼ ϕ_ 3 ð0Þ ¼ 0 rad=s; _ 1ð0Þ ¼ ϕ_ 2ð0Þ ¼ ϕ_ 3ð0Þ ¼ and ϕ ¼
0 rad=s:
Figure E8.6.
ct3
c t2 ct1 Jo1
Jo2
kt1
Jo3
kt2
Mo(t)
k
kt3
1 2
8.7
3
Consider the damped system of Exercise 8.2 and set the forcing amplitude to zero. Examine the free oscillations of this system by using the normal mode for the following sets of initial conditions and discuss the participation of different modes in each case: (a) (b)
x x
1 ð 0Þ 1 ð 0Þ
¼
x
=−
2 ð0Þ
¼ 0:2 m
0:2 m;
x
2 ð0Þ
x_
and
1 ð0Þ
¼ 0:2 m;
¼
and
x_ ð0Þ ¼ 0 m=s; x_ ð0Þ ¼ x_ ð0Þ ¼ 2
1
2
0 m=s:
Section 8.4.1 8.8
Consider the two degree-of-freedom system shown in Figure E8.2 where
=
1 kg,
k
mine the
1
=
2 N/m,
k
response of
2
=
1 N/m,
c
1
=
⋅
0.4 N s/m, and
this damped system to a
assume the excitation to be of the form
f ðtÞ ¼ F ejωt 2
2
c
2
=
⋅
m
1
=m
2
0.2 N s/m. To deter-
harmonic excitation, one
can
Exercises which is referred to as a
complex-valued excitation
because of the
ej ωt
541
term. To
determine the steady-state solution for the response of this system, assume a solution of the form
&x t ' &X 1ð
Þ
¼
x ðtÞ
'
j ωÞ j ω t e X ðj ωÞ 1ð
2
2
and then solve for the response amplitudes
X (j ω) 1
and
X (j ω) 2
from the governing
equations of motion.
flywheels
Consider the system of
8.9
M (t) o
=M
1
cos
ωt,
shown in Figure E8.9. For an applied moment
determine a solution for the response of the system by using
the direct approach presented in Section 8.4.1.
Flywheel 2
Figure E8.9.
Flywheel 1 Jo1
Jo2
kt1
k
kt2
Mo (t)
1 2
m,m
Consider the system shown in Figure E8.10 in which the three masses
8.10
m
3
1
2,
and
flexural rigidity EI. The inverse is called the flexibility matrix, is
are located on a uniform cantilever beam with
of the stiffness matrix for this system ½ given by
K ±−
½
2 L 4 EI 3
1
¼
3
K ±,
which
27
14
4
14
8
2:5
4
2:5
1
If the masses of the system are all identical, that is,
3 5 m
1
=m =m =m 2
, then deter-
3
mine the response of this system when it is forced sinusoidally at the location of mass
m
2
with a forcing amplitude
y1 L
8.11
m1
y2 L
m2
y3 L
F
2
and an excitation frequency
ω.
Figure E8.10.
m3
Consider the three degree-of-freedom system shown in Figure E8.4. Assume that a harmonic forcing the mass
m
1
f
2
=F
2
cos
ωt
is imposed on the mass
m
2.
Determine whether
can have a zero response at any of the excitation frequencies.
542
Multiple Degree-of-Freedom Systems: Forced Oscillations 8.12
For the system given in Exercise 8.1, carry out the following: (a) determine a solution for the response of this system by using the direct approach discussed in Section
H
8.4.1
Ω)
22(
frequency 8.13
and
(b)
construct
the
frequency-response
functions
H
Ω)
(
12
and
and plot these functions as a function of the nondimensional excitation
Ω.
For the system given in Exercise 8.2, carry out the following: (a) determine a solution for the response of this system by using the direct approach discussed in Section
H
21(
Ω)
8.3
Based
(b)
construct
the
frequency-response
functions
H
11
Ω)
(
and
and plot these functions as a function of the nondimensional excitation
frequency 8.14
and
on
Ω. the
response
amplitudes
determined
in
Exercise
8.8,
construct
the
frequency-response functions
G G Ω
where
12 ð 22 ð
jΩÞ ¼ X ð jΩÞ=F jΩÞ ¼ X ð jΩÞ=F 1
2
2
2
is the nondimensional excitation frequency. Plot graphs of the ampli-
tudes and phases of the each of these functions as a function of
Ω,
compare them
to the plots of frequency-response functions determined in Exercise 8.1, and discuss their differences and similarities.
Section 8.5.1 8.15
An industrial-fan system, a model of which is shown in Figure E8.15, is found to experience undesirable vibrations when operated at 600 rpm. It is assessed that these undesirable vibrations are due to a mass unbalance in the fan, and it is estimated the mass unbalance point
O
=
mo
2 kg is located at a distance
ε
=
0.5 m from the
shown in Figure E8.15. Design a vibration absorber for this industrial-
fan system with the restriction that the mass of the absorber cannot exceed 75 kg.
Figure E8.15.
mo O
t M
k Industrial fan
c
Exercises 8.16
543
For the industrial-fan system of Exercise 8.15, if the original restriction on the absorber mass is removed, design the absorber with the following new restrictions: (a) the natural frequencies of the system with the absorber should lie outside the range of 575 rpm to 625 rpm and (b) the response amplitude of the absorber mass should not exceed 25 mm.
8.17
A machine has a mass of 150 kg and a natural frequency of 150 rad/s. An absorber mass of 30 kg and a spring-damper combination is to be attached to this machine, so that the machine can be operated in as wide a frequency range as possible around the machine’s natural frequency. Determine the optimal parameters for the absorber.
8.18
In order to attenuate oscillations of telecommunication towers, tuned vibration absorbers are typically used. The second natural frequency of a representative telecommunication tower is 0.6 Hz and the third natural frequency of this tower is 1.5 Hz. Design two optimal absorbers, one to be effective in a frequency range that includes 0.6 Hz and the other to be effective in a frequency range that includes 1.5 Hz. The mass ratio for each absorber can be picked to lie in the range 0.02 to 0.05.
8.19
An optical platform is found to be experience undesirable vibrations at a frequency of 100 Hz. The associated magnitude of the disturbance acting on the platform is estimated to be 100 N. Design a spring-mass system as an absorber for this system with the constraint that the absorber response amplitude cannot exceed 5 mm.
Section 8.5.3 8.20
A rotary engine experiences disturbances at the sixth harmonic of the rotating speed of the system. Determine the pendulum length of a centrifugal absorber for this system.
Section 8.5.4 8.21
Determine the free responses of the bar-slider system described by Eqs. (8.115) for the following parameter values:
m
=
0.5,
ζ
=
0.20, and
ωc
2.5. Assume that in each case, the motions are initiated from
=
0.5, 1.5, 2, and
θ(0)
=
0.5 rad with
all of the other initial conditions being zero. Graph the time histories for the radial displacements of the slider and the angular motions of the bar. Discuss the effectiveness
of
the
nonlinear
vibration
absorber
in
suppressing
the
angular
motions of the bar.
Section 8.5.5 8.22
Determine
the
free
responses
of
the
pendulum-absorber
Eqs. (8.124) for the following parameter values: 0.05,
mr
=
0.10 and
fo
=
Ω
=
1,
ωr
system
=
0.5,
described
ζ
=
0.10,
ζt
by
=
0.4. Assume that in each case, the motions are initiated
544
Multiple Degree-of-Freedom Systems: Forced Oscillations from
φ(0)
=
0.2 rad with all of the other initial conditions being zero. Graph the
time histories for the vertical motions of the system and the angular motions of the pendulum. Discuss the effectiveness of the nonlinear vibration absorber in suppressing the vertical translations.
Section 8.5.7 8.23
Use Lagrange’s equations to obtain the equations of motion for the following cases of Table 8.1: (a) Case 3 (b) Case 6 (c) Case 8 (d) Case 9 (e) Case 11 (f) Case 12
Section 8.6 8.24
A 200 kg machine is to be operated at 500 rpm on a
flexible platform that can be
modeled as a single degree-of-freedom system with the following mass, stiffness, and damper values:
m
1
=
3000 kg,
k
1
=
2
×
6
10
N/m, and
c
1
=
7500 N/(m/s).
Determine the properties of the machinery mounting so that the transmissibility ratio does not exceed 0.1.
9
Vibrations of Beams
page
9.1 Introduction 9.2 Governing Equations of Motion
545 547
9.2.1 Preliminaries from Solid Mechanics
548
9.2.2 Potential Energy, Kinetic Energy, and Work
550
9.2.3 Derivation of the Equations of Motion
557
9.2.4 Beam Equations for a General Case
559
9.3 Free Oscillations: Natural Frequencies and Mode Shapes 9.3.1 Introduction
566 566
9.3.2 General Solution for Natural Frequencies and Mode Shapes for Beams with Constant Cross-Section
570
9.3.3 Orthogonality of the Mode Shapes
580
9.3.4 Natural Frequencies and Mode Shapes of Constant Cross-Section Beams Without In-Span Attachments: Effects of Boundary Conditions
583
9.3.5 Effects of Stiffness and Inertial Elements Attached at an Interior Location
599
9.3.6 Effects of an Axial Force and an Elastic Foundation on the Natural Frequency
616
9.3.7 Tapered Beams
9.1
617
9.4 Forced Oscillations
622
9.5 Summary
639
Exercises
640
INTRODUCTION
In Chapters 3 through 8, vibrations of systems with
finite
degrees of freedom were trea-
ted. As mentioned in Section 2.5, elements with distributed inertia and stiffness properties, such as beams, are used to model many physical systems described in Section 2.5. As noted previously,
continuous systems,
distributed-parameter systems, which are also referred to as spatially
fi
have an in nite number of degrees of freedom. Apart from beams,
distributed systems that one could use in vibratory models include strings, cables, bars undergoing axial vibrations, shafts undergoing torsional vibrations, membranes, plates, and shells. Except for the last three systems mentioned, the descriptions of all of the other
546
Vibrations of Beams systems require one spatial coordinate. The equations of motion, which govern vibratory systems with
finite
degrees of freedom, are ordinary differential equations, and these
equations are in the form of an initial-value problem. By contrast, the equations of motion governing distributed-parameter systems are in the form of partial differential equations, with boundary conditions and initial conditions, and the determination of the solution for the vibratory response of a distributed-parameter system requires the use of additional mathematical techniques. However, notions such as natural frequencies, mode shapes, orthogonality of modes, and normal-mode solution procedures used in the context of degree-of-freedom systems apply equally well to in in
finite
finite
degree-of-freedom system has an in
finite
finite
degree-of-freedom systems. An
number of natural frequencies and a
mode shape associated with the free oscillations at each one of these frequencies. In this chapter, the free and forced vibrations of beams are considered at length, and in Appendix G, vibrations of bars, shafts, and strings are considered. As illustrated by the diverse examples of Section 2.5, vibratory models of many physical systems require the use of beam elements. In addition to these examples, other examples where beam elements are used to model physical systems include models of rotating machinery, ship hulls, aircraft wings, and vehicular and railroad bridges. Propeller blades in a turbine and the rotor blades of a helicopter are modeled by using beam elements. Since the vibratory behavior of beams is of practical importance for these different systems, the focus of this chapter will be on beam vibrations. In each of the applications cited above, and in many others, the beams are acted upon by dynamically varying forces. Depending on the frequency content of the forcing, the forcing has the potential to excite the beam at one or more of its natural frequencies. One of the frequent requirements of a design engineer is to create an elastic structure that responds minimally to the imposed dynamic loading, so that large displacement amplitudes, high stresses and structural fatigue are minimized, and wear and radiated noise are decreased. The governing equations of motion for beams are obtained by using the mechanics of elastic
beams
and
Hamilton’s
principle.
Free
oscillations
of
unforced
and
undamped
beams are treated and various factors that in
fluence the natural frequencies and modes are
examined.
treatment
This
examination
includes
the
of
inertial
elements
and
springs
attached at an intermediate location and variation in beam geometry. The limitations of the models used in the previous chapters are also pointed out in the context of systems where a
flexible structure supports systems with one or two degrees of freedom. The use of
the normal-mode approach to determine the forced response of a beam is also presented. In this chapter, we shall show how to do the following.
•
–
Determine the natural frequencies and mode shapes of Bernoulli Euler beams of constant cross-section for a wide range of boundary conditions.
•
Determine the conditions under which the mode shapes are orthogonal for the given mass and stiffness distributions.
9.2 Governing Equations of Motion •
547
–
Determine the natural frequencies and mode shapes of Bernoulli Euler beams with attached local stiffness and inertia elements.
•
Determine
the
natural
frequencies
and
mode
shapes
of
–
Bernoulli Euler
beams
of
variable cross-section.
•
Determine the responses of Bernoulli
–Euler
beams to initial displacements, initial
velocities, and external forcing. In addition, the following interactive graphics have been created to better visualize the natural frequencies, modes shapes, and response of a beam as a function of its boundary conditions, geometry, and in-span and boundary attachments. Interactive Graphic 9.1: Natural frequencies, mode shapes, and node points of beams with various boundary conditions Interactive Graphic 9.2: Natural frequencies and mode shapes of beams with in-span attachments Interactive Graphic 9.3: First mode shape of a double-tapered free-free beam Interactive Graphic 9.4: Response of a cantilever beam to an impulse force
9.2
GOVERNING EQUATIONS OF MOTION
In this section, we obtain the governing equations of an elastic beam undergoing small transverse vibrations for arbitrary loading conditions and boundary conditions. A beam
figuration is shown in Figure 9.1. The x-axis runs along the y-axis and z-axis run along transverse directions to the x-axis. End moments of magnitude M are shown acting along the j direction, and it is assumed that the beam displacement is confined to the x–z plane. The displacement w (x ,t) denotes
element in a deformed con span of the beam, and the
the transverse displacement at a location along the beam.
A
j
i
M
–z
x
so
b
k
A
s
w(x, t)
a
B
Center line (Neutral axis)
R
B
z
Figure 9.1. Deformation of a beam subjected to end moments.
M
548
Vibrations of Beams The
derivation
of
Hamilton principle.
the
governing
equations
To use this principle, one
of
first
motion
is
based
on
the
extended
needs to determine the system poten-
tial energy, the system kinetic energy, and the work done on the system. For determining the system kinetic energy, each element of length
Δx
is treated like a rigid body. For
determining the system potential energy, stress-strain relationships in the beam material are used. To this end, preliminaries from solid mechanics are presented in Section 9.2.1, and then the expressions for the kinetic energy, potential energy, and work are obtained in Section 9.2.2.
9.2.1
Preliminaries from Solid Mechanics In Figure 9.1, it is seen that the face of the beam located toward the center of curvature will be contracted while that on the opposite face will be extended; that is, face be extended, while face
BB
AA
will
will be contracted. The line passing through the centroids of
central line. Here, a fiber along the central line neutral axis. assumed to be described by Bernoulli –Euler beam
the cross-section of the beam is called the
is assumed to experience zero axial strain. Hence, this central line is the The deformation of the beam is theory,
1
which is applicable to thin elastic beams whose length to radius of gyration
ratio is greater than 10. In keeping with this theory, it is assumed that the neutral axis remains unaltered, that the plane sections of the beam normal to the neutral axis remain plane and normal to the deformed central line, and that the transverse normals such as
BA
experience zero strain along the normal direction. For a
fiber located at a distance z
from the neutral axis, as shown in Figure 9.1, the strain experienced along the length of the beam is given by
ε where
R
is the
¼
Δs − Δso Δso
radius of curvature, Δso
is the length of a
fiber
=−
z R
(9.1)
is the length of a
that is located at a distance
z
fiber
along the neutral axis,
Δs
from the neutral axis, and we have
used geometry to write
Δs
¼ ð
R − zÞΔ θ
Δso
and
From Hooke’s law, the corresponding axial stress
σ where
E
Eε = −
RΔθ
acting on the
(9.2)
fiber is
Ez R
(9.3)
is the Young’s modulus of the material. According to the convention shown in
Figure 9.1, a positive displacement
1
¼
σ
¼
E. P. Popov,
w
is in the direction of the unit vector
k.
Therefore,
Engineering Mechanics of Solids, Prentice Hall, Upper Saddle River, NJ, 1990, Chapter 6.
9.2 Governing Equations of Motion
549
fibers above the neutral axis experience a positive σ, which denotes tension, and fibers below the neutral axis experience a negative σ, which denotes compression.
the
At an internal section of the beam, a moment balance about the
ðð y
2
M= −
y
1
where
y
1
and
y
2
b
σ z dz dy ¼
−a
the
y-axis leads to
EI R
(9.4)
are the spatial limits corresponding to integration along the
y-direction,
we have used Eq. (9.3), and
ðð
y
I
2
1
the
y-axis,
I
z dz dy 2
¼
y The quantity
b
−a
represents the area moment of inertia of the beam’s cross-section about
which is through the centroid. In general, the limits of the double integral
in Eq. (9.5) do not have to be constants; that is,
y
2
=
(9.5)
a
=
a(x), b
=
b(x), y
1
=
y ( x), 1
and
y ( x). In this case, the area moment of inertia varies along the length, and therefore, I = I(x). The curvature κ = 1/ R, which is assumed to be positive for concave 2
in general,
curvature downwards, is
κ
¼
1
R
¼
∂2w ∂ x2
" ± ² #− ∂w 2
1þ
If it is assumed that the slope is small, that is, the neutral axis at location
3
=2 (9.6)
∂x
∣∂w=∂x∣≪
1, where
x, then Eq. (9.6) simplifies to κ
¼
1
R
≅
∂2w ∂x2
∂w ∂x /
is the slope of
(9.7)
Upon substituting Eq. (9.7) into Eqs. (9.1) and (9.4), we obtain
ε ¼−z
∂2 w ∂x2
∂2 w M = EI 2 ∂x
(9.8)
Thus, the magnitude of the strain and bending moment are proportional to the second spatial derivative of the beam displacement. The statement that the bending moment is linearly proportional to the second spatial derivative of the beam displacement is the
Bernoulli–Euler law,
thin beams.
which is the underlying basis for the theory of linear elastic
550
Vibrations of Beams f(x, t)
j
V
V
so
x
o
i k
V
M
w M
M
x z Figure 9.2.
Deformation of an element of a beam subjected to a transverse load.
Equation (9.8) was obtained by considering only the effects of moments on the ends of the beam. If, in addition, there is a transverse load
f (x, t),
then there are vertical shear
forces within the beam that resist this force. In Figure 9.2, if the sum of the moments about point
o is taken along the j direction, and if the rotary inertia of the beam element
is neglected, the result is
M þ ðV
þ
ΔV ÞΔx
¼
M
þ
ΔM
which leads to
In the limit
Δx
→
ΔM Δx
¼
V
0, the shear force increment
→
lim
Δx
0
ΔM Δx
þ
ΔV
→
∂M ∂x
¼
¼
which, after making use of Eq. (9.8), results in
V
¼
∂M ∂x
ΔV 0, and we have
V
±
(9.9a)
²
∂ ∂2 w ¼ EI 2 ∂x ∂x
Thus, the shear force is equal to the change of the bending moment along the Consequently, if
9.2.2
(9.9b)
x-axis.
M(x) is constant along x, then V = 0.
Potential Energy, Kinetic Energy, and Work We shall construct the system potential energy, the system kinetic energy, and determine the work done by external forces for the system shown in Figure 9.3. The quantities de
fined in this figure will be used in this and subsequent sections.
Potential Energy The potential energy of a deformed beam has contributions from different sources, including the strain energy. For a beam undergoing axial strains due to bending, if the
9.2 Governing Equations of Motion
551
mo f (x,t)
ki
ML,JL
MR , JR
ko
Mi
k tR
ktL
p(x,t)
p(x,t)
kf
kL
w(x,t)
kR
Ls Lm Lo L
Figure 9.3. Beam on an elastic foundation, under axial and transverse loads, with discrete in-span
elements, and with discrete elements at the boundaries.
strain energy is the only contribution to the system potential energy, the beam’s poten2
tial energy is written as
ðð ð Ly b
U ðtÞ ¼
2 0
ð
y
1
¼
1 2
2
σε z dy dz dx ¼
−a
L
EI
ð ð ð Ez Ly b
2
1
!
∂2 w ∂x2
1 2
0
y
1
−a
R
2
2
dy dz dx
2
dx
(9.10)
0
where we have used Eqs. (9.1), (9.3), and (9.7).
3
Kinetic Energy Assuming that the translation kinetic energy of the beam is the only contribution to the system kinetic energy, one can write it as
ð ð ð ±∂w² Ly
T ðt Þ ¼
1
A
=
A (x)
b
ρ
2 0
where
2
y
1
−a
∂t
ð ±∂w² L
2
dy dz dx ¼
1
ρA
2
∂t
2
dx
(9.11)
ρ(x)
is the mass density
0
is the cross-sectional area of the beam and
ρ
=
of the beam’s material. If the rotary inertia of the beam element is also taken into account, an additional term corresponding to the rotational kinetic energy will have to be included in Eq. (9.11), as described by Eq. (A.30) of Appendix A.
2
See,
for
example,
Chapters 1 3
–3.
Since the symbol
V
I.
S.
Sokolonikoff,
Mathematical Theory of Elasticity,
is used for the shear force, the symbol
the notation used in the earlier chapters.
U
McGraw
Hill,
New
York,
1956,
is used for the potential energy, which differs from
552
Vibrations of Beams
Mi
The kinetic energy of a concentrated mass
TMi
(kg) located at
±
given by
²
Mi ∂ wð Lm ; tÞ ¼ 2 ∂t
x
=
Lm,
0
≤ Lm ≤ L
, is
2
which can be written as
ð ±∂w x t ² L
TMi
¼
1
ð
Mi
2
; Þ
2
∂t
δ ðx − Lm Þdx
(9.12)
0
where
δ( x)
is
the
delta
function
introduced
in
Chapter
6.
Equation
(9.12)
can
be
extended for multiple concentrated or discrete mass elements, as needed.
Work The work done by the applied transverse conservative load per unit length
ð
fc(x,t) is given by
L
Wf ðtÞ ¼ fc ðx; tÞwðx; t Þdx
(9.13a)
0
If gravity is the only distributed conservative load acting on the beam, then
f cð x; tÞ ¼ ρðxÞAðxÞg
(9.13b) 4
If the beam is also under the action of an axial tensile force
p(x,t), as shown in Figure 9.3,
then the length of the central line no longer remains constant, but extends to a new length. If we assume that the deformation is small in magnitude and does not affect the loading
p(x,t), then the change in length of an element of the beam is ( Δs − Δx) where
4 5
flow, and structural columns.
Axial loads are common in rotating blades, pipes with Note that from geometry, we have
Δs
2
¼
which leads to
Δs ¼ Δx Δw/Δx | ≪
For small slope, that is, |
Δw
2
≈
Δx
2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ±Δw ²ffi 2
1þ
1, this leads to
Δs Δx
þ
1þ
1
±Δw ²
2
Δx
or equivalently,
Δs − Δx ≈
Δx
2
1
±Δw ²
2
Δx
2
Δx
5
9.2 Governing Equations of Motion
Δs ≈
" 1 þ
1
±Δw² #
2
553
2
Δx
Δx
(9.14)
6
Therefore, the external work of the axial force is given by
ð
L
Wp = −
1 2
pðx; tÞ
±∂w²
2
∂x
dx
(9.15)
0
where we have used Eq. (9.14). Since the tensile force acts to oppose the beam transverse displacement
w,
the work done has a minus sign. If the axial force is compressive, then
p(x,t) is replaced by −p( x, t).
7
Next, we consider the linear elastic foundation
on which the beam is resting, as
shown in Figure 9.3. The transverse displacement of the beam creates a force in the foundation with the magnitude
ff(x,t)
=
k fw( x, t),
where
kf
is the spring constant per unit
area of the foundation. This spring force opposes the motion of the beam. The external work done by the elastic foundation is
ð
L
Wk ðtÞ = −
1
k f w ðx; tÞdx 2
2
(9.16)
0
where, again, we have introduced a minus sign to account for the fact that the foundation force acting on the beam opposes the beam displacement. Alternatively,
ð
L
1 2
kf w ðx; tÞdx 2
0
can be included in the system potential energy as an additional term in Eq. (9.10). When a discrete elastic translation spring with spring constant the beam at
x = Ls, 0
< Ls < L
ki
(N/m) is attached to
, the energy stored in the spring is
Uki
¼
1 2
ki w ðLs ; tÞ 2
(9.17)
which can be written as
ð
L
Uki
¼
1 2
k i w ðx; t Þδðx − Ls Þdx 2
(9.18)
0
6
To construct this integral, it is noted that the work done on a segment is
Δx 7
→
0;
Δx is replaced by dx.
−p x t (
, )(
Δs
− Δx
), and in the limit as
This type of elastic foundation is frequently used to model structures on soil and it is often referred to as a
Pasternak foundation.
554
Vibrations of Beams
Single Degree-of-Freedom System Attachment Let a single degree-of-freedom system be attached at displacement of the mass be denoted by constant
ko is Uko
¼
1 2
z(t).
x
=
Lo,
0
≤
Lo
≤
L
and let the
Then the energy stored in the spring with
³
k o wð Lo ; tÞ− zðtÞ
´
2
(9.19)
z(t) and the other end w(Lo,t ) at the attachment
since one end of the spring is subjected to the motion of the mass of the spring is subjected to the displacement of the beam point. Equation (9.19) can be written as
ð ³ L
U ko
¼
1
ko wðx ; tÞ −zðtÞ
2
´
2
δðx − LoÞdx
(9.20)
0
The kinetic energy of the mass
m o (kg) of a single degree-of-freedom system is given by Tmo
where
z
± ²
m o ∂z ¼ ∂t 2
2
(9.21)
is the displacement of the mass.
For the single degree-of-freedom system, we use Eqs. (9.19) and (9.21) to obtain the
± ²
difference between the kinetic energy and the potential energy as
LSDOF
¼
Tmo − U ko
¼
m o ∂z 2 ∂t
2
−
1 2
³
System Lagrangian With the objective of constructing the system Lagrangian
G B(x, t, w, w_ , w , w_ , w 0
0
00
´
k o wð Lo; t Þ− zðtÞ
LT,
2
(9.22)
we construct the function
) from the expression
ð ³ L
G B x; t; w; w_ ; w ; w_ ; w 0
0
00
´
dx ¼ T ðtÞ − U ðtÞ þ Wc ðtÞ
(9.23)
0
where we have introduced the compact notation
w_ ¼ In Eqs. (9.24),
∂w ; ∂t
w_ , w , and w 0
00
w
0
¼
∂w ; ∂x
w
00
¼
∂2w ; ∂ x2
w_
0
¼
∂2w ∂ x∂t
(9.24)
represent the beam velocity, beam slope, and beam curvature,
respectively. From Eqs. (9.11) and (9.12), it is found that the kinetic energy is given by
ð
L
T ðtÞ ¼
1 2
ρA ðx Þw _ ðx; tÞ dx þ
ð³ 0
L
¼
1 2
0
ð
L
2
1 2
0
Mi w_ ð x; tÞ δðx − LmÞdx 2
´
ρAð xÞ þ Mi δðx − LmÞ w _ 2ðx; tÞ dx
(9.25)
9.2 Governing Equations of Motion
555
From Eqs. (9.10), (9.18), and (9.20), the potential energy is given by
ðµ
³
L
U ðtÞ ¼
1 2
´
¶
x; tÞ þ ki w ðx; tÞ δð x − Ls Þ þ ko wðx; t Þ− zðtÞ δ ðx − Lo Þ dx
EI ðx Þw
002
2
ð
2
(9.26)
0
From Eqs. (9.13a), (9.15), and (9.16), the external work is given by
ð
WcðtÞ ¼ f ðx; tÞwðx ; tÞdx −
1
Wc(t)
kf
ð
L
w ðx; tÞdx 2
2
0
(9.27)
0
is the work done by conservative forces and it has been constructed
Wf( t)
assuming that the work that the work
pðx; t Þw ðx; tÞ dx − 02
2
0
In Eq. (9.27),
ð L
L
done by the external loading
fc( x, t)
is conservative and
Wp(t ) done by the axial loading p(x, t) is conservative.
8
After collecting the spatial integrals given by Eqs. (9.25) to (9.27), we Eq. (9.23) that
³
G B x; t ; w; w_ ; w ; w_ ; w 0
0
00
´
In Eq. (9.28), there are no
w_
³
2
−
1
−
1
0
´
from
ρA ðx Þ þ Mi δð x − Lm Þ w _ 2ðx; tÞ − EI ðxÞw002ð x; t Þ
1
¼
find
2
2
³
kf
þ
³
1 2
´
ki δð x − Ls Þ w ðx; tÞ þ f ðx; tÞ wð x; tÞ
ko wðx; tÞ −zðtÞ
´
2
2
δðx − LoÞ − pðx; tÞ w02ðx; tÞ 1
(9.28)
2
terms, as we neglected the rotary inertia of the beam cross-
section in the development. On the boundaries of the beam at
x
= 0 and
x
=
L,
one can have discrete external
elements that contribute to the total kinetic energy and the total potential energy of the system. Consider, for example, the beam shown in Figure 9.3. At the left boundary
x
(
kL and a linear torsion spring L), there is a linear translation with stiffness ktR . There is also an
= 0), there is a linear translation spring with stiffness
with stiffness
ktL.
Similarly, at the right boundary (
spring with stiffness
kR
and a linear torsion spring
x
=
ML and rotary inertia JL at the left boundary and an inertia MR and rotary inertia JR at the right boundary. Taking the difference
inertia element with mass element with mass
between the kinetic energy of the inertia element at the left boundary and the potential
8
Conservative forces were
first
mentioned in Section 2.3.1, where it was noted that forces expressed in terms of
a potential function, such as a spring force or a gravity loading, are conservative. On the other hand, dissipative
loads
such
nonconservative.
as
those
due
to
dampers
and
time-dependent
loads
such
as
harmonic
excitations
are
556
Vibrations of Beams energy of the stiffness elements at the left boundary in Figure 9.3, we obtain the discrete Lagrangian function
³
G t; w ; w_ ; w 0
0
0
´
0
0
1
ML w_
2
|fflfflfflffl{zfflfflfflffl}
¼
2
0
1
JL w_
0
−
2
|fflfflffl{zfflfflffl}
þ
2
0
1
kL w
2
|fflffl{zfflffl} 2
0
−
1
ktL w
0
2
|fflfflffl{zfflfflffl} 2
(9.29)
0
Translational
Rotational
Potential
Potential
kinetic energy
kinetic energy
energy of
energy of
of rigid body
of rigid body
translation
torsion spring
spring
x = 0; that w(0,t) indicates the displacement at the boundary x = 0, w ¼ ∂wð0; tÞ=∂x indicates the slope at the boundary x = 0, and so on. Also, the translational velocity of the center of mass of the rigid body is denoted as w _ , the angular rotation of the torsion spring is _ . Similarly, the denoted as w , and the angular velocity of the rigid body is denoted as w discrete Lagrangian function corresponding to the right boundary x = L is given by where the subscript 0 has been used to denote that the quantity is evaluated at
is,
w
0
0
=
0
0
0
0
0
0
³
GL t; wL ; w_ L ; wL ; w_ L 0
0
´
¼
1
MR w_ L 2
|fflfflfflffl{zfflfflfflffl} 2
1
JR w_ L 0
2
|fflfflffl{zfflfflffl}
þ
2
−
1
kR wL 2
|fflfflffl{zfflfflffl} 2
−
1
ktR wL 0
2
|fflfflfflffl{zfflfflfflffl}
2
Translational
Rotational
Potential
Potential
kinetic energy
kinetic energy
energy of
energy of
of rigid body
of rigid body
translation
torsion spring
(9.30)
spring
where the subscript that is,
wL
=
w(L, t)
L
x = L; L, wL ¼ ∂ wð L; tÞ=∂x
has been used to denote that the quantity is evaluated at
indicates the displacement at the boundary
indicates the slope at the boundary
x
=
L,
x
0
=
and so on. Also, the translational velocity of
the center of mass of the rigid body is denoted as
w_ L ,
the angular rotation of the torsion
wL , and the angular velocity of the rigid body is denoted as w_ L . 0
0
spring is denoted as
We now form the Lagrangian of the conservative system as
ðh ³ L
LT
¼
GB x; t; w; w_ ; w ; w_ ; w 0
0
´
00
þ
i
G δ ðxÞ þ G L δðx − LÞ dx 0
(9.31)
0
The
first
term inside the integral on the right-hand side of Eq. (9.31) represents the con-
tributions to the system kinetic energy, the system potential energy, and the work done in the beam interior 0
3.4) and
ηs
Ω /π 1
≈
2.5 when
η
>
2500 (i.e.,
= 0.5. From Case 1 of Table 9.3, we see that a clamped-
clamped beam without attachments has a second natural frequency of with a node point at
Ki
Ω /π 1
= 2.4998
= 0.500. Therefore, by placing a stiff spring at or near the node
point of this second natural frequency, one effectively creates a system whose lowest natural frequency is now equal to that of the second natural frequency of a beam without
9.3 Free Oscillations: Natural Frequencies and Mode Shapes 4
2.0
1.7
605
2.3 2.2
1.9 2.1
3
2.4
1.8 )iK(01gol
2.5 2.0
0.5 0.4
4
0.3 h
s
1.5
2 1.6
Ω1/π 1
3
0.2
2 0.1
1 0.0
log10(Ki)
0 0.0
0
0.1
0.2
(a) Figure 9.10.
0.3 h
0.4
0.5
s
(b)
fi
(a) Variation of the lowest natural frequency coef cient of a clamped-clamped beam
restrained by a spring with nondimensional constant
K i attached at ηs
and (b) contour curves of
fi
constant values of the lowest natural frequency coef cient of the surface in (a). boundary conditions are symmetric, we only need to consider the region 0
Note : Since the
≤ ηs ≤
0.5.
attachments. This is accomplished because the beam is forced to assume the mode shape associated with the second natural frequency of the beam without attachments. From the contour curves shown in Figure 9.10b, we see that, as in the case for a cantilever beam, certain natural frequencies can be obtained from a range of combinations of the values of
Ki
and
ηs .
We also note that as the spring is placed closer to the center of the
beam, the magnitude of
Ki that is required to maintain a constant frequency decreases.
ficients are clearly evident from Figures 9.8 to However, since numerical values are somewhat difficult to obtain from these fi gwe have also presented in Table 9.7 the lowest natural frequency coeffi cients for
The trends of the natural frequency coef 9.10. ures,
many different combinations of boundary conditions and in-span locations.
Beams with In-Span Mass Mi p
We start with a cantilever beam. For this case, we use Case 3 ( Eq. (9.159) and replace
B(Ωn)
with
m i Ωn 4
to obtain the surface shown in Figure 9.11. In
Figure 9.11a, we see that the effect of the mass is to lower the which reaches its minimum value at value of
ηm
= 3) of Table 9.6 in
first
natural frequency,
= 1.0, the free end of the beam, irrespective of the
mi. From the contour curves shown in Figure 9.11b, we see that as with the beam
restrained with an in-span spring, certain natural frequencies can be obtained from a range of combinations of the values of
mi and ηm.
Beam Hinged at Both Ends For this case, we use Case 2 (
p = 2) of Table 9.6 in Eq. (9.159) and replace B(Ωn) with m i Ωn 4
to obtain the surface shown in Figure 9.12. In Figure 9.12a, we again see that the effect of
606
Vibrations of Beams
Table 9.7.
Lowest natural frequency coef
ficient for a beam with an in-span spring
for many different combinations of boundary conditions and in-span locations
Ω
1/
Hinged-hinged
Clamped-clamped
0.4
0.5
η
1
1
1
10
1.0316
1.0432
η
Ki
0
s
= 0.3
π Clamped-free
0.4
0.5
η
1.5056
1.5056
1.5056
1.0477
1.5145
1.5212
s
= 0.3
0.75
1.0
0.5969
0.5969
0.5969
1.5242
0.6442
0.7422
0.84
s
= 0.5
50
1.1296
1.1737
1.1908
1.5473
1.5782
1.5919
0.7515
0.9964
1.0825
100
1.2143
1.2856
1.3151
1.583
1.6393
1.6649
0.8177
1.1552
1.1588
500
1.4645
1.6374
1.7687
1.7519
1.9247
2.0298
0.9423
1.456
1.2315
1000
1.5411
1.744
1.996
1.8412
2.0724
2.2736
0.9691
1.4699
1.2407
2.0
0.25
0.15
1.5 0.35
0.20
1.0
0.6 )i m(01gol
0.4
Ω 1 /π
0.2 0.2
0
0.4 h
s
0.6
1
0.40 0.55
–0.5
0.50
log 10(mi)
0.8 1.0
–1.0 0.0
2
0.6
0.8
1.0
ficient of a cantilever beam with a
mi attached at ηm and (b) contour curves of constant values of the lowest
natural frequency coef
ficient of the surface in (a).
the mass is to lower the
ηm
0.4
(b)
(a) Variation of the lowest natural frequency coef
mass with mass ratio
0.2
hs
(a) Figure 9.11.
0.30
0.0
–1
0.0
0.45 0.5
first
natural frequency, which reaches its minimum value at
= 0.500, the center of the beam for all values of
mi.
From the contour curves shown in
Figure 9.12b, we see that as with the beam restrained with an in-span spring, certain natural frequencies can be obtained from a range of combinations of the values of
m i and ηm.
Beam Clamped at Both Ends For this case, we use Case 1 (
p
= 1) of Table 9.6 in Eq. (9.159) and replace
B(Ωn)
with
to obtain the surface shown in Figure 9.13. The variations in the fi rst natural coefficient for this system are similar to those obtained for the cantilever beam and the beam
m i Ωn 4
hinged at both ends.
9.3 Free Oscillations: Natural Frequencies and Mode Shapes 2.0
607
0.5 0.3 0.9
1.5
0.4 1.0 )im(01 gol
1.0 0.8
Ω1 /π
0.6 0.4 0.0 0
0.2 h
s
0.3
0.8
–0.5
log10(mi)
1
–1.0
0.4 0.5
0.6
0.5
0.0
–1 0.1
0.7
0.0
2
0.1
0.2
(a) Figure 9.12.
0.3 h
0.4
0.5
s
(b)
fi
(a) Variation of the lowest natural frequency coef cient of a hinged-hinged beam
mi attached at ηm and (b) contour curves of constant values of the lowest natural frequency coefficient of the surface in (a). Note: Since the boundary conditions are
with a mass with mass ratio
symmetric, we only need to consider the region 0
≤ ηm ≤
0.5.
2.0
0.6 1.0 0.5
1.5 1.5
0.8
1.0
1.0 0.6
Ω1 /π
0.1
h
s
0.3
1
0.9 1.3 1.2
–0.5
0
0.2
0.5 0.0
0.2 –1
0.0
log10(m i)
1.4
–1.0
0.4
0.0
0.5
2
0.1
0.2
0.3
0.4
0.5
hs
(a) Figure 9.13.
0.7
1.1
)im(01 gol
1.4
(b)
fi
(a) Variation of the lowest natural frequency coef cient of a clamped-clamped beam
mi attached at ηm and (b) contour curves of constant values of the lowest natural frequency coefficient of the surface in (a). Note: Since the boundary conditions are
with a mass with mass ratio
symmetric, we only need to consider the region 0
≤ ηm ≤
0.5.
ficients.
Figures 9.11 to 9.13 clearly show the trends of the natural frequency coef
fi
However, since numerical values are somewhat dif cult to obtain from these
ficients
have also presented in Table 9.8 the lowest natural frequency coef
different combinations of the boundary conditions and in-span locations.
figures
we
for many
608
Vibrations of Beams
Table 9.8.
Lowest natural frequency coef
ficient for a beam with an in-span mass for
many different combinations of boundary conditions and in-span locations
Ω Hinged-hinged
π
Clamped-clamped
0.4
0.5
η
1.0000
1.0000
1.0000
0.1
0.9694
0.9591
0.5
0.8783
1.0
0.8049
5.0
Clamped-free
0.4
0.5
η
1.5056
1.5056
1.5056
0.9553
1.4621
1.4339
0.8496
0.8401
1.3234
0.7697
0.7586
1.208
0.5930
0.5579
0.5474
0.5061
0.4748
0.4656
mi
η
0.0
10
1/
m
= 0.3
0.75
1.0
0.5969
0.5969
0.5969
1.4228
0.5901
0.5735
0.5484
1.2490
1.2246
0.5661
0.5106
0.4520
1.1211
1.0943
0.5412
0.4641
0.3972
0.8806
0.8013
0.7783
0.4364
0.3383
0.2769
0.7497
0.6803
0.6603
0.3804
0.2883
0.2342
m
= 0.3
m
= 0.5
Beams with an In-Span Single Degree-of-Freedom System Mo Ωo = 2. In the lower right-hand cor-
We shall consider the two sets of parameters shown in Figure 9.14. In Figure 9.14a, = 0.5 and
Ωo
ner of each
= 1 and in Figure 9.14b,
figure
Mo
= 4.0 and
is a table in which a number of natural frequency coef
ficients
are dis-
played for the following different scenarios:
Ωn sdof
¼ Beam with single degree-of -freedom system at
Ωn none
¼ Beam with no attachments
;
;
Ωn Ko ;
Ωn Mo ;
Ωn K o ;
þ
Mo
¼ Beam with spring at ¼ Beam with mass at
ηo
ηo
ηo
¼ 1
¼ 1
¼ 1
¼ Beam with mass and spring at
ηo
¼ 1
We note that when a single degree-of-freedom is added to a cantilever beam it causes the combined system to order its natural frequency coef and
Ω
1,
none
→Ω
ficients
so that
Ωo
→Ω
sdof
1,
sdof. This is similar to what happens when two single degree-of-freedom
2,
systems are combined. Recall Figures 7.13 and 7.14. In the case of the beam, however, there is a major difference: the mode shapes of the beam at in that neither one has any node point away from We also note that when
Mo
= 0.5 and
observation also holds for the case when ining the values for
1
1
1
Ω
sdof and
1,
sdof are similar
2,
= 0.
Ωo = 1, Ω sdof ≈ Ω Ko and Ω sdof ≈ Ω Ko . This Mo = 4.0 and Ωs = 2. We explain this by exam2;
1;
3;
2;
Zn/X n. For the former case, we see that associated with the value for Z /X = 1.96 whereas that ratio associated with Ω sdof is the value
Ω sdof is the value Z /X = −0.0426. Hence, 1,
η
Ω
1
2,
relative to the beam displacement at this point, the end of the
spring attached to the mass is virtually stationary and essentially
fixes
the spring at this
end. Consequently, the system acts as if it were a beam that has one end of a spring attached at this point and the other to ground. In the latter case, this reasoning is also
9.3 Free Oscillations: Natural Frequencies and Mode Shapes
609
(a)
(b) Figure 9.14.
ficients and mode shapes for a cantilever beam with a single
Natural frequency coef
degree-of-freedom system at its free end: (a)
M o = 0.5 and Ωo = 1; (b) M o = 4.0 and Ωo = 2.0.
610
Vibrations of Beams correct, except that the ratio is not as small as that for the agreement is slightly poorer than that of the
first case.
first
case and, therefore, the
It is cautioned that the discussion above cannot be applied universally. Other combinations of the parameter values may lead to different explanations.
INTERACTIVE GRAPHIC 9.2: NATURAL FREQUENCIES AND MODE SHAPES OF BEAMS WITH IN-SPAN ATTACHMENTS In this interactive graphic, the natural frequencies and mode shapes of beams with different in-span attachments and with different boundary conditions are displayed. The variations in the natural frequencies and mode shapes with respect to the attachment ’s location and parameters can be explored by using this graphic. In this interactive graphic, the following should be noted. • • •
•
The attachment of a mass always lowers the natural frequencies and its placement and magnitude can greatly influence the mode shapes, especially for a cantilever beam. The attachment of a spring always increases the natural frequencies irrespective of the boundary conditions. Spring stiffness and the spring’s placement are influential parameters in determining mode shapes and the natural frequencies; appropriate location and magnitude can turn the first mode into the second mode of a beam without attachments. A single degree-of-freedom system adds an additional degree of freedom to the coupled system. The mode that it affects depends on the values of M o, Ωo, and the closeness of Ωo to one of the natural frequencies of the beam without any attachment. The natural frequency which is associated with the single degree-of-freedom system is determined by the magnitude of Z n/Yn. From the table in the fi gure, it is apparent that when the ratio |Zn/Yn| is extremely small, the beam acts as if it were restrained by a spring fixed to ground at that point.
Comparison with Two Springs in Series Approximation As discussed in Chapter 2, in many situations where an inertia element is attached to a beam, the beam stiffness is taken into account to establish an equivalent single degree-offreedom system. This situation is revisited in the context of the beam system shown in Figure 9.3 to point out when it is reasonable to neglect the beam inertia and when it is not. First, we consider the determination of the natural frequency of an equivalent single degree-of-freedom system. This is done by using the static stiffness values given for Cases 4, 5, and 6 of Table 2.3 for the cantilever, pinned-pinned and clamped-clamped beams, respectively. In each case, the approximation obtained for the
first natural frequency of the
system shown in Table 2.3 is compared to the natural frequency obtained when the inertia of the beam is taken into account. We note from the discussion on spring combinations in
9.3 Free Oscillations: Natural Frequencies and Mode Shapes
611
series shown in Figure 2.6c and from Eq. (a) of Example 2.3 that the equivalent spring constant
ke for a spring ko attached to a beam of spring constant kbeam ke
±
¼
1
ko
þ
²−
1
1
k beam
k beam α
ko 1 þ k o =kbeam
(9.161)
find that
From Table 2.3, we
where
¼
is
¼
αbc EI
L
(9.162)
3
is a function of the boundary conditions and is given as follows.
Clamped-clamped
3
αbc
¼
η 0ð1 − η0Þ
αbc
¼
η 0ð1 − η0Þ
3
(9.163a)
3
Hinged-hinged 3 2
(9.163b)
2
Clamped-free
αbc In Eqs. (9.163),
η0
¼
3
(9.163c)
η30
is the attachment point of a spring with spring constant
k o.
Then,
from Eq. (9.161), we have
ke
¼
ko 1 þ ko L =α bc EI 3
¼
ko 1 þ Ko =αbc
(9.164)
Thus, the natural frequency of the equivalent single degree-of-freedom system obtained from the equivalent spring constant is given by
ωn To determine the
¼
sffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ± ² k k e
mo
mo
ωe
Ω
1
is the
1þ
K o=αbc
(9.165)
first natural frequency when the beam’s inertia is taken into account, find that the first natural frequency is given by
we use Eq. (9.49) and
where
1
o
¼
first
¼
Ω
2 1
ffi rffiffiffiffiffiffiffiffiffiffi EI mbL
(9.166)
3
nondimensional frequency coef
ficient
obtained by solving the
appropriate characteristic equation for a beam with a single degree-of-freedom system attached at
ηo .
The percentage error
ε
between the natural frequency for a single
612
Vibrations of Beams degree-of-freedom in Figure 9.3 is
ε¼
system
18
100
ωn ωe
0 @ Ω 0 @Ω
¼ 100
¼ 100
When
Ko
is
very
Eq. (9.167) simpli
1
Ω
2 1
2
o 2
and
αbc
natural
frequency
of
the
beam
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 1 0 v u tm k L @ u − − A K =α m EI Ω ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! 1 v u u tK − A M K =α sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 − A % !
1
2
1
1 þ
o
1 þ
1
large,
is,
o
1
bc
bc
(9.167)
1
K o=αbc
that
1þ
system
1
o
1
1
o
1
o
3
b o
1
¼ 100
when
the
fies to ε
Since
first
the
¼ 100
1
Ω
mass
is
ffi ! rffiffiffiffiffiffi α bc
Mo
2 1
−
1
directly
attached
to
%
the
beam,
(9.168)
is a function of the beam boundary conditions and
ηo
is the location of where
the single degree-of-freedom system is attached, the percentage error is also a function of these quantities in addition to
Ko
and
Mo.
We have numerically evaluated Eqs. (9.167) to determine the values of the mass and stiffness
Ko
fi
at speci c values of
less than a speci
fic
ηo
Mo
that are required for the error to be equal to or
value for each of the following boundary conditions: (i) clamped-
clamped, (ii) hinged-hinged, and (iii) clamped-free. For the clamped-clamped and hingedhinged boundary conditions we have selected condition we have selected Figures 9.15 to 9.17 for
ε
ηo
ηo
= 0.5 and for the clamped-free boundary
= 1.0. The results obtained from Eq. (9.167) are shown in
= 2.5% and for
ε=
5%. It is seen from these
figures that the sin-
gle degree-of-freedom system interacts with the beam in a complex way and that in order to approximate the system with two springs in series and stay within these error bounds, one must take into account both the properties of the single degree-of-freedom system on the beam and the beam’s boundary conditions. From these
figures,
it is seen that for a
given set of boundary conditions, the mass of the single degree-of-freedom system must be substantially larger for an error of the magnitude of the stiffness
ε
= 2.5% than for an error of
ε
= 5%, irrespective of
K o.
The results obtained from Eq. (9.168), which is for the case of a mass directly attached to the beam, are shown in Table 9.9 for
ε
= 2.5% and for
ε
= 5%. It is seen that at the
locations given, in order to keep the error below 5% one must use a mass ratio of 2.5 when the mass is located at the free end of a cantilever beam. When the mass is placed 18
For another approach to this topic see: M. Gürgöze, “ On the representation of a cantilever beam carrying a tip mass by an equivalent spring-mass system,”
Journal of Sound and Vibration, 282 (2005) 538–542.
9.3 Free Oscillations: Natural Frequencies and Mode Shapes
Figure 9.15.
system at
ηo
Figure 9.16.
system at
ηo
Values of
Mo
for a beam clamped at each end and carrying a single degree-of-freedom
= 0.5 for which the error is less than 5% and less than 2.5% as a function
Values of
Mo
613
Ko.
for a beam hinged at each end and carrying a single degree-of-freedom
= 0.5 for which the error is less than 5% and less than 2.5% as a function
Ko.
at the midpoint of a clamped-clamped beam, the mass ratio should be greater than 3.5 and for a hinged-hinged beam the mass ratio should be greater than 5. It has been found from additional numerical work that these mass ratio values must be increased as the mass is located away from these respective locations, which are the locations where the beams have their minimum stiffness. Finally, it is noted that when the error is to be less than or equal to 2.5%, the mass ratios are on the order of twice those needed to maintain 5%.
614
Vibrations of Beams
Figure 9.17.
ηo
Values of
Mo
for a cantilever beam carrying a single degree-of-freedom system at
= 1 for which the error is less than 5% and less than 2.5% as a function
Table 9.9.
Values of a concentrated mass Mo located at
η
o
Ko.
for beams with various
end conditions for which the error is less than 2.5% and less than 5% Boundary condition
η
Clamped-clamped
0.5
Hinged-hinged
o
0.5
Clamped-free
1.0
EXAMPLE 9.3
ε
Mo
5 %
3.63
2.5 %
7.35
5 %
4.75
2.5 %
9.60
5 %
2.31
2.5 %
4.66
Determination of the properties of a beam supporting rotating
machinery
An engineer has to mount to the center of a steel beam a piece of rotating machinery that weighs 50 kg and will spin at 1800 rpm. The beam is 3 m long, weighs 100 kg, and is clamped at both ends. Refer to the system shown in Figure 9.18. In order to ensure that the beam is not excited at its fundamental natural frequency, the engineer would like the beam’s
first natural frequency to
be three times that of the excitation frequency.
We shall determine the minimum radius of gyration of the beam’s cross-section so that these requirements are met. The beam’s Young’s modulus is and its density is
ρ
3
pffiffiffiffiffiffiffiffi E ρ
= 7850 kg/m , therefore,
cb
¼
=
¼ 5000
m=s
E
¼ 1:9625
×
11
10
2
N/m ,
9.3 Free Oscillations: Natural Frequencies and Mode Shapes
615
Beam clamped at each boundary and excited by a
Figure 9.18.
rotating machine mounted at its midpoint.
3m The excitation frequency
fe due to the rotating machinery is fe
1800 rev= min
¼
¼ 30 Hz
60 s= min
(a)
Therefore, we need to determine the beam geometry so that the
f
≥
1
fe
3
first natural frequency
¼ 90 Hz
(b)
From Eq. (9.98) and Eq. (b), we have that
f
cb rΩ 2π L
2
¼
1
≥
1
2
fe
3
(c)
which, upon using Eq. (a), leads to
r≥
πL2f e
6
πL2
180
¼
cb Ω
2
cbΩ
1
From the given parameters, we from Table 9.8 that
Ω
1
= 1.2246
find
π.
that
Since
L
γo
1
Mo/mo
=
(d)
2
= 3 m and
= 50/100 = 0.5 and, therefore,
cb
= 5000 m/s, we obtain from
Eq. (d) that
r≥
≥
180 5000
×
6:88
×π×
2
3
π
ð1 :2246 Þ
2
¼ 0:0688
m (e)
cm
If the cross-section of the beam is rectangular, with width
2
¼
and depth
h, then
I bh h ¼ ¼ A 12bh 12 3
r
b
2
(f)
and the depth of the beam is determined from Eqs. (e) and (f) to be
ffiffi
p
h ≥ 2r
3 ¼ 2
×
6:88
×
ffiffi
p
3 ¼ 23:83 cm
Notice that for a beam of rectangular cross-section, the width of the beam
(g)
b
does not
affect the natural frequency. However, it does affect the static displacement of the beam. The equivalent stiffness of a beam clamped at each boundary is provided in Case 6 of Table 2.3. From this expression, it is seen that the displacement is proportional to which depends on the width
b.
I,
616
Vibrations of Beams
9.3.6
Effects of an Axial Force and an Elastic Foundation on the Natural Frequency We shall determine the effects that an axial force have on the natural frequency coef
ficient.
19
p(x,t )
and an elastic foundation
kf
Axial forces arise in beam models of many
vibratory systems including rotating machinery, where the centrifugal forces are the source of the axial forces, and in vertical structures such as water towers. If we assume that
p(x,t)
=
po
is a constant, that the beam has a uniform cross-section and uniform
material along its length, and that
d Y dη 4
ki = mi = ko = f = 0, then Eq. (9.56) becomes
− P^
4
d Y dη 2
2
þ
·K
f
−Ω
4
¸Y
¼ 0
(9.169)
where, from Eq. (9.51), we set
^ ¼ poL P EI
2
(9.170)
Hinged-Hinged Beam
find a general solution to Eq. (9.169), we shall only obtain the solution to a ficient for us to illustrate the effects that po and kf have on the natural frequency coeffi cient. From Eq. (2b) of Table 9.2 and Case 2 of Table 9.3, the spatial function that satis fies the boundary condi-
Rather than
beam that is hinged at each of its ends. This will be suf
tions for a beam hinged at both ends is
Y ðηÞ ¼ sinðnπηÞ n
¼ 1; 2;
…
(9.171)
The substitution of Eq. (9.171) into Eq. (9.169) leads to the characteristic equation
nπ Þ
4
ð which yields
Ωn where
Ωn
is the
nth
þ
^ ðnπÞ P
2
þ
Kf
−Ω
4
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
4
nπ Þ
ð
4
þ
^ ðnπÞ P
natural frequency coef
2
þ
Kf
ficient.
¼ 0
n
Since
(9.172a)
¼ 1; 2;
Kf
>
…
(9.172b)
0, we see that the presence
of the elastic foundation always increases the natural frequencies of the beam, and that a tensile axial force ( axial force (
^ < 0) P
^ > 0) P
always increases the natural frequencies while a compressive
always decreases them. For compressive axial forces, one must make
sure that the buckling limits of the beam are not exceeded. The buckling limits are also a function of the boundary conditions.
19
For a more complete treatment of this topic, see the following: F. J. Shaker,
”
and frequencies of beams,
“Effect of axial load on mode shapes
Lewis Research Center Report NASA TN D-8109, December 1975; G. C. Nihous,
“On the continuity of the boundary value problem for vibrating free-free straight beams under axial load,” Journal of Sound and Vibration, 200(1) (1997) 110–119; and M. A. De Rosa and M. J. Maurizi, “The in fluence of concentrated masses and pasternak soil on the free vibrations of euler beams – exact solution, ” Journal of Sound and Vibration, 212(4) (1998) 573–581. For an example of an axial force in a MEMS application, see A. Singh, R. Mukherjee, K. Turner, and S. Shaw, “MEMS implementation of axial and follower end forces,” Journal of Sound and Vibration, 286 (2005) 637–644.
9.3 Free Oscillations: Natural Frequencies and Mode Shapes 9.3.7
617
Tapered Beams We shall now remove the assumption that the beam has a constant cross-section and 20
consider tapered beams
whose cross-section varies with the position along the length
of the beam, as shown in Figure 9.19. This permits us to model such systems as 21
ing rods,
22
baseball bats, and chimneys.
in Eq. (9.56). Then Eq. (9.56) becomes
d dη
2
±
2
d Y ið ηÞ dη 2
We assume that
²
− Ωt a η Y 4
2
ð Þ
mi
=
ki = p
=
kf
=
¼ 0
fly fish-
f(x,t)
= 0
(9.173)
where we have used the notation of Eqs. (9.49) and introduced the quantity
Ωt
4
¼
ρAo ω2 L4
(9.174)
EI o
and
Io Ao Consider
the
double-tapered
beam
boho 3
¼
¼
12
(9.175)
boho
with
rectangular
cross-section
shown
in
x–z plane and the x–y plane. Let us x–z plane and β = b / bo < 1 in the x–y
Figure 9.19a, that is, a beam that tapers in both the assume that the taper ratios
α
=
h / ho 1
b1
> q_ >> ( _ ) >>>< ⋮ >>>= ±Y_ ² = Y = q_ >> €q >> Y_ >> €q >> >> ⋮ >> >: >; 2
2
In Eq. (F.7), the (2
½
±
½0 ±
M ±−
½
1
1
)
K ± + ½H ± ±
½½
1
N
2
1
(F.8)
2
€qN
and the (2
N
×N
½ ±
−M− ½
±
1
B±
) matrix ½
I
½0 ±
−M−
are given by
1
2
f
(F.7)
±Y_ ²
C ± + ½G ± ±
are, respec-
#
½½
(F.9)
690
Appendix F where
½0 ± is an (
×
N
N)
null matrix; that is, a matrix whose elements are all zero.
Equation (F.7) is referred to as the state-space form
1
of Eq. (F.1).
The initial displacements and the initial velocities of the inertial elements are given by
8> q >< q => >: ⋮ q
1 ð0Þ
Y
f
1 ð0Þg
2 ð0Þ
N ð0Þ
9> >= >> ;
8> q_ >< q_ => >: ⋮ q_ 1
Y
and
f
2 ð0Þg
2
ð 0Þ ð 0Þ
N ð0Þ
9> >= >> ;
(F.10)
The form of Eqs. (F.7) to (F.10) are well suited to numerical evaluation by standard numerical procedures.
2
A±
It is noted that since the matrix ½
is not symmetric, its eigenva-
lues are complex valued.
TRANSFER FUNCTIONS AND FREQUENCY-RESPONSE FUNCTIONS
The transfer function of a single degree-of-freedom system is de
fined in Section 5.4.4. To
determine the transfer function of a two degrees-of-freedom system, we start with the governing equations of motion given by Eq. (7.1b) that is,
m
1
d x dt 2
1
2
d x m dt 2
2
2
2
+
c
ð
1
+c
2Þ
dx dt
1
dx + ðc + c Þ dt
2
2
3
+
ð
k
1
+k
x
2Þ
1
−c
2
dx dt
−k
2
−k
1
3
2
x
2
=f
tÞ
1ð
(F.11)
dx + ð k + k Þx − c dt 2
2
2
2
x
1
=f
tÞ
2ð
Introducing the nondimensional quantities from Eqs. (7.41) and the following additional quantities for the nondimensional time, damping factor, and damping coef
ficient
ratio,
respectively,
τ = ωn1t ;
2
ζj =
cj ; m j ωnj
and
c
32
=
c c
3
(F.12)
2
Eqs. (F.11) are rewritten as
d x dτ 2
2
2
+
ζ 1 + 2ζ 2m r ωr Þ
ð2
d x dτ 2
1
2
dx dτ
1
³
+ + mr ωr
dx + 2ζ ωr ð1 + c Þ dτ
2
2
32
2
1
´x
1
−
ζ 2m r ωr
2
dx dτ
2
− mr ωr x = 2
2
dx + ωr ð1 + k Þ x − 2ζ ωr dτ
1
2
32
2
2
− ωr x = 2
2
f ðτÞ k 1
1
f ðτÞ k mr
(F.13)
2
1
1
It is noted that the state-space form of a system is not unique; one can obtain different state-space forms by considering equations different from Eq. (F.5). 2
For example, using Matlab,
ode45; using Mathematica, NDSolveValue.
State-Space Formulation
691
Carrying out the Laplace transforms of the different terms on each side of Eqs. (F.13) and making use of Laplace transform pair 2 in Table B.1 of Appendix B, we arrive at
A ðsÞX ðsÞ − Bð sÞ X ðsÞ 1
=
2
F ðsÞ k F ðsÞ k mr 1
1
−C
s X ð sÞ + Eð sÞ X ðsÞ
ð Þ
1
=
2
(F.14)
2 1
ficients A(s), B(s), C(s), and E(s) are given by
where the coef
AðsÞ = s
+
2
ζ 1 + ζ2 mr ωr Þs + 1 + m r ω2r
2ð
Bð sÞ = 2ζ mr ωr s + mr ωr
2
2
(F.15)
C ðsÞ = Bð sÞ=mr = 2ζ ωr s + ωr
2
2
Eð sÞ = s
+
2
ζ 2ωr ð1 + c32 Þs + ω2r ð1 + k 32 Þ
2
and we have assumed that the initial conditions are zero. In Eqs. (F.14), are the Laplace transforms of
x (τ) 1
Laplace transforms of the force inputs, Solving for
F (s)
respectively, and
2
1
X (s ) and X (s) F (s) are the 1
and
2
2
f (τ) and f ( τ), respectively. 1
2
X (s) and X ( s) from Eqs. (F.14) yields 1
2
X ðsÞ = 1
1
k D ðsÞ 1
X ðsÞ = 2
4
k D ðsÞ
1
2
2
2
1
2
2
r
2
1
2
(F.16)
F ðsÞC ðs Þ + F ðsÞA ðs Þ=mr ±
½
1
2
2
2
ζ2 ωr ð1 + c32 Þ±s3
2
r
2
32
2
1
2
r
32
r ðζ 1 2
+ζ
2
ωr m r Þ + 2c32 ζ 2ωr
³
1
+ mr ωr
2
2
1
x (τ ) and x (τ ) are determined by X (s ) and X (s) given by Eqs. (F.16), that is, 1
1
µ
2
2
xj ð τÞ = L− Xj ð sÞ 1
´¶s (F.17)
r r
The desired displacement responses inverse Laplace transforms of
¶
ζ1 ζ2 ωr + ω2r k32 + 4ζ 2ωr c32 ðζ 1 + ζ2 ωr m r Þ s2
4
µζω+ ζω + k ω µ ³ ´¶ + ω + k +m ω
+
2
2
+ ζ + ζ ωr m r + µ + + m r ωr + ωr + ½2
1
D (s) is given by
where the denominator
D ðsÞ = s
F ðsÞE ðsÞ + F ðsÞBðsÞ=mr ±
½
2
1 1
2
x (τ ),
and
¶
for
j = 1; 2
executing the
(F.18)
Equations (F.16) will be used to determine four transfer functions, one pair associated with the forcing applied to one inertial element and the other pair associated with the forcing applied to the other inertial element. An impulse force can be used to determine
G ij( s) where the subsubscript j refers to the force
the transfer functions. We shall determine the transfer functions script
i
refers to the response (or output) location and the
692
Appendix F (or
input)
location.
Therefore,
to
impulse forcing is applied to mass
determine
m
the
first
pair
of
transfer
functions,
an
, that is, 1
f ðτ Þ = F oδðτÞ 1
(F.19)
f ðτ Þ = 0 2
Then, we have
3
G
X ðsÞ F ðsÞ
sÞ =
=
1
11 ð
m N
1
G
(F.20)
X ðsÞ F ðsÞ
sÞ =
=
2
21 ð
m N
1
where
F ðsÞ = F o
(F.21)
1
find that
Making use of Eqs. (F.16), (F.20), and (F.21), we
sÞ =
k G
11 ð
1
E ðsÞ D ðsÞ 2
2
sÞ =
k G
21 ð
1
C ðs Þ D ðsÞ
(F.22)
2
Similarly, we determine the other pair of transfer functions by applying an impulse forcing to mass
m
, that is,
2
f ðτ Þ = 0 1
(F.23)
f ðτ Þ = F oδðτÞ 2
Then, we have
G G
sÞ =
X ðsÞ F ðsÞ
sÞ =
X ðsÞ F ðsÞ
12 ð
22 ð
1
=
m N
2
2
(F.24)
=
m N
2
where
F ðsÞ = F o 2
(F.25)
3
It should be clear from the form of Eqs. (F.20) that excitations other than impulse excitations can also be used to determine the transfer functions
Gjk (s).
State-Space Formulation Making use of Eqs. (F.16), (F.24), and (F.25), we
kG 1
kG 1
D (s)
Note that the polynomial
2
find that
sÞ =
Bð sÞ mr D ðsÞ
sÞ =
A ðs Þ mr D ðsÞ
12 ð
22 ð
693
2
(F.26)
2
appears in the denominator of each transfer function
given by Eqs. (F.22) and (F.26). To obtain
s
the frequency-response function, we replace
with
Eqs. (F.13) were written in terms of the nondimensional time
τ
j ω.
However, since
instead of
t
before the
Laplace transforms were executed, the frequency-response functions are given by
ωn1)
or
Gil ( jΩ),
=
Ω
where
ω/ωn1
G il( jω/
is the nondimensional frequency ratio. (See Laplace
transform pair 1 in Table B.1 of Appendix B.) Hence, the frequency-response functions are determined from Eqs. (F.22) and (F.26) to be
k G
11 ð
1
j ΩÞ =
E ð j ΩÞ D ð j ΩÞ 2
2
k G
21 ð
1
j ΩÞ =
C ð j ΩÞ D ð j ΩÞ 2
k G
12 ð
1
j ΩÞ =
B ð jΩ Þ m r D ð j ΩÞ
=k
(F.27) 1
G
jΩÞ
21 ð
2
k G
22 ð
1
j ΩÞ =
Að jΩÞ m r D ð j ΩÞ 2
where the terms in the numerators and the denominators are given by
AðjΩÞ =
−Ω + 2
ζ 1 + ζ 2m r ωr ÞjΩ + 1 + mr ω2r
2ð
BðjΩÞ = 2 ζ mr ωr jΩ + mr ωr
2
2
C ðjΩÞ = 2 ζ ωr jΩ + ωr = Bð jΩÞ =mr 2
2
E ðjΩÞ = 2
D ðjΩÞ 2
− Ω + ζ ωr + c jΩ + ωr + k = Ω − j ζ + ζ ωr m r + ζ ωr + c Ω µ − + mr ωr + ωr + ζ ζ ωr + ωr k + ζ ωr c µ + j ζ ωr + ζ ωr + k ωr ζ + ζ ωr m r + c µ ³ ´¶ + ωr + k + mr ωr 2
4
2
½2
2
1
2
2
1
2
2
2
32
2
32 Þ
2
2
1
2
ð1
2
1
32 Þ
ð1
2
32 Þ±
2
4
2
1
ð1
1
2
32
2
2
32
ð
1
2
(F.28)
3
4
Þ
2
32
¶
ζ 1 + ζ 2ωr mr Þ Ω2
32 ð
2
ζ 2 ωr
³
1
+ mr ωr
2
´¶Ω
2
Equations (F.27) and (F.28) agree with Eqs. (8.50) to (8.54). When the damping is absent,
D ( jΩ) 2
given by the last of Eqs. (F.28) reduces to the left side of the characteris-
tic equation Eq. (7.45) when the spring
k
3
is absent in Eq. (F.28).
694
Appendix F The magnitudes of the frequency-response functions are given by
Hil ðΩÞ = k ∣Gil ðjΩÞ∣ i ; l = 1; 2 1
(F.29)
and the associated phase responses are given by
φil ðΩÞ = tan−1 Notice that
Hil is a nondimensional quantity.
G il ðjΩÞ± Re½G il ðjΩÞ± Im½
(F.30)
APPENDIX G Natural Frequencies and Mode Shapes of Bars, Shafts, and Strings
In this appendix, we shall give the governing equations and boundary conditions for the longitudinal oscillations of uniform bars, the torsional oscillations of uniform circular shafts, and the transverse oscillations of strings under constant tension. For each of these systems, we shall provide the frequency equations from which the natural frequency
fi
coef cients and corresponding mode shapes for several combinations of boundary conditions can be obtained. This material complements the material presented in Chapter 9, wherein natural frequencies and mode shapes of beams have been treated. For the beam system, we have to specify four boundary conditions, whereas for the bars, shafts, and strings we need to specify only two boundary conditions.
GENERAL SOLUTION FOR THE VIBRATIONS OF BARS, SHAFTS, AND STRINGS
The equations governing the longitudinal vibrations of a uniform bar, a uniform circular shaft, and a string under constant tension are given in Table G.1. It is seen from these equations that the equations for all three systems are of the form
a
∂2φ ∂ x2
a shaft is being considered,
φ
(
(
φ(x, t)
(
(
(
(
( , )
)
φ
, ),
, ),
applied force is absent that is, tions of the form
∂2 φ ¼ cðx; tÞ ∂ t2
(G.1)
= u x t c = f x t a = AE φ = θ x t c = τ x t a = JG b = Jρ = yxt c = pxt a = T b = Aρ cxt = ffiffiffiffiffiffi ffi ω = Φ x ejωt j −
where, when a bar is being considered,
is being considered,
−b
, ),
( , ),
, ),
, and
and
, ),
b = A ρ.
When
, and when a string
and
. When the externally
0, and the system is undergoing harmonic oscillap
, where
¼
1 and
is the frequency of oscillation,
Eq. (G.1) becomes
d Φα ðηÞ þ Ωα Φα ðηÞ ¼ 0 dη 2
2
(G.2)
2
where we have introduced the nondimensional quantities given in Table G.1. When we are considering a bar, then ering a shaft, then string,
Φα(η )
Φα( η)
= Φstring η
( )
= =
Φα(η ) = Φbar(η ) = U (η) and Ωα ¼ Ωbar . When we are considΦshaft( η) = Θ (η) and Ωα ¼ Ωshaft. When we are considering a Y( η) and Ωα ¼ Ωstring . In Table G.1, the quantity cbar is the 2
2
2
2
2
2
696
Appendix G
Table G.1.
finitions
Equations of motion for bars, circular shafts, and strings and de
of quantities used in Appendix G Torsional motion of Longitudinal motion of a bar
AE
∂ u ∂x2 2
− Aρ
a circular shaft
∂ u ¼ f ð x; tÞ ∂t2 2
JG
∂ θ ∂ x2 2
− Jρ
Transverse motion of a string
∂ θ ∂ t2 2
¼
τ ðx; tÞ
T
∂2 y ∂x2
− ρA
∂2 y ¼ p ðx ; tÞ ∂t 2
Definitions: dimensional quantities Symbol
Description
Symbol
Description
Symbol
Description
u( x,t)
Axial displacement (m)
θ( x,t)
Rotation (rad)
y (x, t)
Transverse displacement (m)
x t A
x t J
Axial location (m) Time (s) Cross-sectional
Axial location (m) Time (s) 4
Polar inertia (m )
x t A
Location (m) Time (s) Cross-sectional
2
2
area (m )
E
area (m )
G
Young’s modulus 2
L
Shear modulus
T
Tension (N)
L
Length of string (m)
2
(N/m )
(N/m )
L
Length of bar (m)
Length of shaft (m)
U(x) Mo
Θ( x)
Spatial component of
u(x, t) = U (x)ejωt
Spatial component of
Jo
Attached mass (kg)
θ( x,t)
= Θ x ejωt
Attached mass
⋅
Y (x)
( )
Spatial component of
Mo
y (x,t) = Y(x )ejωt
Attached mass (kg)
2
(kg m )
kbar
kshaft
Attached translational spring (N/m)
mbar
∂u/∂x E∂u/ ∂x
= ρAL
kstring
Attached torsion
⋅
spring (N m/rad)
, mass of bar (kg)
Axial strain (m/m) 2
Axial stress (N/m )
J shaft GJ∂θ/ ∂x Gr∂θ /∂x
= ρJL
Attached translational spring (N/m)
⋅
m string
2
(kg m )
⋅
Moment (N m)
∂ y ∂x T∂y ∂x /
2
Shear stress (N/m )
/
= ρAL
, mass of
string (kg) Slope (rad) Transverse tension (N)
ρ ω
Frequency (rad/s)
f(x, t)
Applied force (N/m)
3
Density (kg/m )
ρ ω τ( x,t)
Frequency (rad/s)
ρ ω
Frequency (rad/s)
Applied torque
p (x, t)
Applied force
r
Shaft radius (m)
3
Density (kg/m )
·
(N rad)
3
Density (kg/m )
(N/m)
Definitions: nondimensional quantities Symbol
Description
Symbol
Description
Symbol
Description
η
=xL = Lpcffiffiffiffiffiffiffiffi bar = E ρ = ωtbar = kbarL AE = Mo mbar
η
= xL =p L cffiffiffiffiffiffiffiffi shaft = Gρ = ωtshaft = kshaftL = Jo Jshaft
η
= xL = Lpcffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi string ffi = T ρA = ωt string = kstringL T = Mo mstring
tbar cbar Ω bar Kbar γbar
/
/
=
(
)/(
/
)
tshaft cshaft Ω shaft Kshaft γshaft
/
/
=
(
)/(
/
GJ)
t string cstring Ω string K string γ string
/
/
=ð
(
Þ
)/
/
Natural Frequencies of Bars, Shafts, and Strings speed at which a disturbance travels longitudinally within the bar and
t bar
697
is the time
that it takes for a disturbance to travel the length of the bar. Similar interpretations are valid for the shaft and the string. The solution to Eq. (G.2) is
Φα ðηÞ ¼ A cosðΩα ηÞ þ B sinðΩαη Þ
(G.3)
The various boundary conditions that are appropriate to these systems are summarized in Table G.2.
Table G.2.
Boundary conditions at
η
= 1 for bars, shafts, and strings
Bars Case
Boundary condition
Dimensional form
Nondimensional form for u(η,t) = U(η)e jωt
1
Clamped
u¼
U
2
Free
3
Free with mass
4
Free with spring
5
Free with mass and
∂u ∂x
0
¼ 0
∂u ∂x ∂u EA ∂x
= − Mo
∂u ∂x
= − Mo
EA
EA
∂2u ∂t 2
= − kbaru ∂2u ∂t 2
− kbar u
¼ 0
∂U ∂η
¼ 0
∂U ∂η
¼
∂U ∂η
= − KbarU
∂U ∂η
¼
γ bar Ω2 U
±γ
bar Ω
2
− Kbar
²U
spring
Shafts Case
Boundary condition
Dimensional form
θ(η,t) = Θ(η)e jωt
6
Clamped
θ
Θ¼0
7
Free
8
Free with mass
9
Free with spring
10
Free with mass and spring
¼ 0
∂θ ∂x
∂θ ∂x ∂θ GJ ∂x
GJ
∂Θ ¼ 0 ∂η
¼ 0
GJ
∂θ ∂x
Nondimensional form for
θ ∂t2
∂Θ 2 ¼ γ shaft Ω Θ ∂η
= − kshaft θ
= − Kshaft Θ
= − Jo ∂
= − Jo
2
∂2 θ ∂t2
∂Θ ∂η
− kshaftθ
±
²
∂Θ 2 ¼ γshaft Ω − Kshaft Θ ∂η
698
Appendix G
Table G.2. (cont.) Strings Boundary condition
Case 11
Clamped
12
Free
y¼0
T T
13
∂y ¼ ∂x
T
T
T
T
T
∂y ∂x ∂y T ∂x
T
T
T
Free with mass
14
T
Free with spring
15
Free with mass and
Nondimensional form for y(η,t) = Y(η)e jωt
Dimensional form
∂y ∂x
Y
¼ 0
∂Y ∂η
¼ 0
∂Y ∂η
¼
= − kstring y
∂Y ∂η
= − Kstring Y
= − Mo ∂ y − kstring y
∂Y ∂η
¼
0
= − Mo
∂2 y ∂t2
2
∂t
2
γstring Ω2 Y
±γ
string Ω
2
− Kstring
²Y
spring
CHARACTERISTIC EQUATION FOR BARS
We shall obtain the frequency equation for Cases 5, 10, and 15 of Table G.2 that is, the
η
case where the system is undergoing harmonic oscillations and it is clamped at while at
η
=
find that the boundary condition at η = 0 is Φα ð0Þ ¼ 0
and that the boundary condition at
=
η
∂Φ αð 1Þ ∂η
¼
(G.4a)
1 is
±γ Ω α
2
− Kα
²Φ
α ð1Þ
where, from Tables G.1 and G.2, we see that for a bar
γα
0
1 it is free with an attached mass that is constrained by a spring. Then,
from Table G.2, we
shaft
=
= γshaft
and
γα
(G.4b)
= γbar
Kα = K shaft; and for a string γα = γ string and Kα
and
Kα
= Kstring
= Kbar
; for a
. It should be
realized that Eq. (G.4b) can be reduced to the other four boundary conditions appearing in Table G.2 by considering the limit as zero or in
finity
γα
goes to zero or in
finity
as the case may be. For example, if the end at
Eq. (G.4b) is divided by
Kα and the limit as K α
→∞
η
=
and/or
Kα
goes to
1 is clamped, then
is taken. This results in
Φ α(1) =
0.
Upon substituting Eq. (G.3) into Eq. (G.4), we obtain the characteristic equation or the natural frequency equation
cot
Ωn α − γ α Ωn α þ ;
;
Kα Ωn α ;
¼ 0
(G.5)
Natural Frequencies of Bars, Shafts, and Strings The corresponding mode shape is given by
±
Φn α ðηÞ ¼ sin Ωn α η ;
where
Ωn,α
²
(G.6)
;
fi
is the natural frequency coef cient and
fi
n
699
…
= 1, 2
, indicates the
frequency coef cient. From Table G.1, we see that the natural frequency
nth
fn,α
natural
expressed
in Hz is given by
fn α
Ωn α π tα ;
¼
;
n ¼ 1; 2;
Hz
2
…
(G.7)
In order to obtain the frequency equations for other boundary conditions, we use the limiting procedure introduced above (see also Section 9.2.4). The results of this proce1
dure are summarized
in Table G.3. In addition, we have given in Table G.3 some repre-
Ωn, α
sentative numerical values for
and their corresponding mode shapes. These values
can then be used to obtain the values for the particular system of interest by substituting the appropriate values of
tα, γα, and Kα
fi
as de ned in Table G.1.
COMPARISON TO A SINGLE DEGREE-OF-FREEDOM SYSTEM: BARS
The natural frequency of a single degree-of-freedom system composed of a mass
Mo
sus-
pended from a bar whose spring constant is given by Case 1 in Table 2.3 is
sffiffiffiffiffiffiffiffiffiffi
ωsdof
AE LM o
¼
rad=s
(G.8)
The natural frequency of this same system when it is determined from Eq. (G.5) when
Kbar =
0 is
ω1;bar
Ω bar ¼ tbar 1;
cbar Ω ¼ L
bar
1;
Ω
bar
1;
¼
L
sffiffiffi E ρ
rad=s
(G.9)
fi
where we have used the de nitions appearing in Table G.1. The difference in their numerical values can be represented by the percentage error
εbar
¼ 100
¼ 100
ωsdof ω1;bar
Ω
1
bar
! 1
as
1
¼ 100
− ffiffiffiffiffiffiffi γ
1 p
bar
1;
−
!
εbar
0 sffiffiffiffiffiffiffiffiffiffisffiffiffi 1 AE ρ @Ω L LM − A% E bar
1;
o
%
1
The special cases agree with those that can be found in the literature. See, for example, R. Blevins,
Natural Frequency and Modes Shape, Van Nostrand Reinhold, New York, 1979, pp. 182ff.
(G.10)
Formulas for
enoN
25046.0
978.0
3731.1
379.0 ,784.0
4550.2
889.0 ,656.0 ,923.0
2530.3
enoN
9372.0
719.0
4090.1
679.0 ,884.0
1940.2
989.0 ,956.0 ,033.0
3330.3
7448.0
enoN
2637.1
enoN
675.0
176.2
766.0
947.0 ,473.0
5136.3
8.0 ,4.0
628.0 ,155.0 ,572.0
758.0 ,175.0 ,682.0
n
1
5.0
=
5.1
2 n
1
5.2
3 n
=
5.3
2
enoN
=
5.0
4 n
3
766.0 ,333.0
=
57.0 ,5.0 ,52.0
4
fo snoitanibmoc suoirav rof sepahs edom dna ,stneic
†
stniop edoN
dna
sepahs edoM
γ
)1 = α
K( π/α ,n Ω stniop edoN
5 = α
†
γ π/α ,n Ω
sepahs edoM
stniop edoN
)1 = α (
†
sepahs edoM
K( π/α ,n Ω stniop edoN
)5 = α
†
stniop edoN
π/α ,n Ω
sepahs edoM
†
π/α ,n Ω
sepahs edoM
0 ¼
αK
α ;n Ω
;
;
Ω α γ − α n Ω toc
Ω α γ − α n Ω toc
htiw eerf-depmalC
ssam
htiw eerf-depmalC
ssam dna gnirps
.ylno stniop edon roiretni fo snoitacoL
þ α ;n
0 ¼ α ;n
htiw eerf-depmalC
gnirps
Ω toc þ α ;n
eerf-depmalC
0 ¼ α; n
0 ¼ α; n
αK
Ω soc
depmalc-depmalC
α ;n Ω
Ω nis
snoitidnoc yradnuoB
0 ¼
noitauqe citsiretcarahC
)6.G( .qE morf sepahs
fifeoc ycneuqerf larutan ,snoitauqe citsiretcarahC .3.G elbaT
edom dna )5.G( .qE morf deniatbo era seicneuqerf larutan ;1.G elbaT ni nevig smetsys eerht eht rof snoitidnoc yradnuob
†
700
15
10
5
0
0
5
10
15
20
ga
where the
first
fi
natural frequency coef cient
Eq. (G.5) for each value of
Ω
bar is obtained from the solution of
1,
γbar.
A plot of Eq. (G.10) is given in Figure G.1, where it is seen that for the error to be less than 5%,
γ bar
>
3.3 and for the error to be less than 2%,
γbar
>
8.3.
COMPARISON TO A SINGLE DEGREE-OF-FREEDOM SYSTEM: SHAFTS
We now consider the determination of the natural frequency for the torsional oscillations of a single degree-of-freedom system composed of a mass with polar mass moment of inertia
Jo
that is attached to a shaft whose spring constant is given by Case 3 in
Table 2.3. In this case, we have that
ωsdof
sffiffiffiffiffiffiffiffi ¼
GJ LJ o
rad=s
(G.11)
The natural frequency of this same system when it is determined from Eq. (G.5) when
Kshaft =
0 is
ω1;shaft
Ω shaft ¼ tshaft 1;
cshaftΩ ¼ L
1;
shaft
¼
Ω
shaft
1;
L
sffiffiffiffi G ρ
rad=s
(G.12)
702
Appendix G
finitions
where we have used the de
appearing in Table G.1. The difference in their
numerical values can be represented by the percentage error
εshaft
ωsdof
¼ 100
ω1;shaft
−
Ω
shaft
1
p
1;
as
0 sffiffiffiffiffiffiffi ffisffiffiffiρffi 1 L GJ @Ω − A LJ G
¼ 100
− ffiffiffiffiffiffiffiffiffi γ
1
¼ 100
!
εshaft
!
1
shaft
o
1;
1
(G.13)
shaft
A plot of Eq. (G.13) is the same as that given in Figure G.1, except that replaced by
γ bar
is
γ shaft.
TRANSVERSE VIBRATIONS OF STRINGS WITH IN-SPAN MASS AND SPRING ATTACHMENTS
The equation governing the transverse vibration of a uniform string of length is stretched with a tension
L
1
T
(N) and is carrying a mass
(m) and restrained by a spring
T where
y
=
∂2 y ∂x2
y(x,t)
−
³
ρA þ
Mo
´∂ y 2
1Þ
∂ t2
−
A
(G.14)
p(x,t)
is the externally 2
is the area of the string cross-section (m ) and
density (kg/m ). When the externally applied force is absent, that is, shaft is undergoing harmonic oscillations of the form
ω
(m) is
1
3
and
2
1
k string δ ðx − L Þy ¼ pðx; tÞ L
is the transverse displacement of the string,
applied force per unit length,
y(x,t)
that
(kg) at an interior location
kspring (N/m) at the same location L
Mo δð x − L L
L (m)
p( x, t)
= Y x ejωt (
)
=
ρ
is its
ffiffiffiffiffiffiffi −
0, and the p
, where
j
¼
1
is the frequency of oscillation, Eq. (G.14) becomes
∂ 2Y ð ηÞ 2 þ Ω ½1 þ γ s δð η − η 1 Þ±Y ðηÞ − Ks δ ðη − η 1 Þ Y ðηÞ s ∂ η2
¼ 0
where η = L / L, for convenience we have introduced the compact notation Ks Ωs = Ωspring, and γ s = γ spring, and the other quantities are defined in Table G.1. 1
1
(G.15)
= Kspring
,
To solve Eq. (G.15), we employ the Laplace transform. Upon taking the Laplace transform of Eq. (G.15) and rearranging terms, we obtain
Y ðsÞ ¼
2
E. B. Magrab,
sY ð 0Þ s þ Ωs 2
2
þ
Y s
2
0
ð0Þ
þ
Ωs
2
þ
±K − Ω γ ²
Vibrations of Elastic Structural Members, –
Alphen aan den Rijn, 1979, pp. 66 72.
s
2
s s
e−sη Y ðη s þ Ωs 1
2
2
1
Þ
(G.16)
Sijthoff & Noordhoff International Publishing Co.,
Natural Frequencies of Bars, Shafts, and Strings where
s
is
the
Laplace
=
η
the slope of the string at
Y ðsÞ
Y(0)
is the Laplace transform parameter,
η
is the displacement at
=
0,
703
Y (0)
0, the prime denotes the derivative with respect to
transform
Y(η ).
of
Using
transform
pairs
3,
18,
and
0
η,
19
is
and
from
Table B.1 in Appendix B, the inverse transform of Eq. (G.16) is
Y ðηÞ ¼ Y ð0Þ cos Ωs η þ
Ωs ηÞ Y Ωs
sinð
0
ð0Þ þ
µK
s
Ωs
− Ωs γ s
¶
Y ðη
Ωs ðη − η
1Þ
sinð
1 ÞÞ
uðη − η
1Þ
(G.17)
where
u(η) is the unit step function.
We shall consider only the case where the string is clamped at both ends, that is, when
Y(0) =
Y (1) = 0. Then, Eq. (G.17) simplifies to
0 and
Y ðηÞ ¼ To determine
Ωs η Þ Y Ωs
sinð
Y (0) 0
Eq. (G.18), we
0
ð0Þ þ
µK
s
Ωs
− Ωs γ s
¶
Y ðη
1
Ωs ðη − η
Þ sinð
1
u η−η
ÞÞ ð
Y(1)
in Eq. (G.18), we use the boundary condition
find that
µ
¶
Ωs K Ωs γ s − s Y ðη Y ð0Þ ¼ sin Ωs Ωs 0
1
Þ sinð
Ωs ð 1 − η
1
1
Þ
=
(G.18)
0. Then, from
ÞÞ
(G.19)
Upon substituting Eq. (G.19) into Eq. (G.18) and collecting terms, we obtain
Y ð ηÞ ¼
·
Ωsη Þ sinðΩs ð1 − η sin Ωs
sin ð
1 ÞÞ
−
Ωs ðη − η
sinð
1 ÞÞ
uðη − η
¸µ
1Þ
¶
Ks Ωs γ s − Y ðη Ωs
1Þ
To obtain the characteristic equation, we note that Eq. (G.20) must be valid at Therefore, setting
η
sin
where
Ωs n n ,
=η
η
=η
.
1
in Eq. (G.20) yields
1
Ωs n −
= 1, 2,
(G.20)
µ
;
…
Ks Ωs n γ s − Ωs n ;
¶ ± ² ± Ω η Ω s;n
sin
1
sin
s ;n ð1
;
, are the natural frequency coef
−η
ficients.
²
1
Þ
¼ 0
(G.21)
The corresponding mode
shape is obtained from Eq. (G.20) as
YnðηÞ ¼
± Ω η² ± Ω
sin
sin
s ;n
Ωs n
sin
s ;n ð1
−η
1
² − ±Ω
Þ
sin
s;n ð η
−η
1
²u η − η
Þ
ð
1
Þ
(G.22)
;
where we have arbitrarily set the scale factor
µ
¶
Ks Y ðη Ωs γ s − Ωs
1
Þ ¼ 1
(G.23)
704
Appendix G When
Mo
=
0 and
kstring =
0, the frequency equation given by Eq. (G.21) reduces to
Ωs n
sin
;
¼ 0
(G.24)
which is the characteristic equation for a string clamped at both ends. The mode shape is given by Eq. (G.6). When the spring becomes very stiff so that
Ks
±Ω η ² ±Ω
sin
s;n
1
sin
→∞
s;n ð1
−η
, Eq. (G.21) becomes
²
1Þ
¼ 0
(G.25)
…
which is the characteristic equation for a string rigidly supported at equation are
Ωs n = nπ/(2η ,
) and 1
Ωs n ,
= nπ
/(2[1
−η
]), 1
n
=
1, 3, 5,
.
η1.
The roots of this
APPENDIX H Evaluation of Eq. (9.120)
We shall evaluate the integral appearing on Eq. (9.120), which is
ð" 1
I
4
=
0
d Yk dη
2
2
!
2
ð" 1
+
2
d Yn i ðηÞ dη
d dY k S ðηÞ Yn dη dη
- Yn d dη
2
!
d Yk i ðηÞ dη 2
2
!# dη
2
d dY n - Yk S ð ηÞ dη dη
!# dη
(H.1)
0
First, it is noted that
d dY k S ðηÞYn dη dη d dY n S ðηÞYk dη dη and, therefore,
d dY k S ðηÞ Yn dη dη
!
! !
d dY k S ðηÞ = Yn dη dη d dY S ðηÞ n = Yk dη dη
d dY n S ðηÞ - Yk dη dη
!
=
! !
+S
η
ð Þ
(H.2)
dY dY + S ðηÞ k n dη dη
d dY k S ðηÞY n dη dη
!
-S η
ð Þ
d dY n S ðηÞY k dη dη
=
"
dY k dY n dη dη
d dY k S ðηÞ Y n dη dη
!
dY k dY n dη dη
+S η
ð Þ
dY - Yk n dη
dY k dY n d η dη
#!
(H.3)
706
Appendix H Then,
ð" 1
d dY k S ðηÞ Yn dη dη
!
!#
d dY - Yk S ðηÞ n dη dη
"
ðd 1
dη =
"
dY k S ðηÞ Yn dη dη
dY - Yk n dη
0
0
=
dY k dY - Yk n Yn dη dη
S ðηÞ
#! dη
!#
1
0
(H.4)
fine
Next, we de
gnðηÞ = i ðηÞ snðηÞ = Introducing Eq. (H.5) into the
first
dgn dη
d Yn dη 2
2
d d Yn i ðηÞ dη dη 2
=
!
(H.5)
2
first
part of the
integral on the right-hand side of
Eq. (H.1), we arrive at
ð 1
d Yk dη
2
2
!
ð 1
d Yn ds i ðηÞ d η = Y k n d η = Y k sn dη dη 2
2
0
±± - ðs 1
1
n
0
± ð dY
0
= Yk sn ± -
1
1 0
dY k dη dη
0
k dgn dη d η dη
8< ± = Y s ± - :g
9= ±± ð dY ± ± - g dd ηY d η; dη ± !± ± ð d Y ±± d Y dY ±± i η ±± - i η d η d η ±± + i η dη 0
k n
1 0
d = Yk dη
n
2
k
n
0
2
ð Þ
1
1
k
2
0
1
n
ð Þ
2
1
1
2
n
k
2
0
0
d Yn d Yk dη dη dη 2
ð Þ
2
2
2
0
(H.6) Upon interchanging the subscripts, we Eq. (H.1) is
ð 1
d Yn dη
2 2
find
that the second part of the
!
!±± ±± ± ± ð dY ±± ± + i η dd ηY dη ± 2
2
integral of
1
d Yk d d Yk ið ηÞ d η = Yn ið ηÞ dη dη dη 2
first
2
0
- i ðηÞ d Yk dη 2
2
0
1
1
2
n
ð Þ
0
0
k
2
d Yn dη dη 2
2
(H.7)
Evaluation of Eq. (9.120) Combining Eqs. (H.6) and (H.7), we obtain
ð" 1
d Yk dη
d Yn i ðηÞ dη
2
2
!
d - Yn dη
2
2
0
d d Yn i ðηÞ = Yk dη dη 2
2
=
"
1
2
ð Þ
d Y dY - i ðηÞ n k dη d η 2
1
2
2
"
2
ð Þ
0
k
2
2
1
2
k
ð Þ
0
2
2
n
0
2
2
n
2
dη
0
1
d d Yk i ðηÞ dη dη
- Yn
2
d Y k dY n d Y n dY k dη dη dη dη 2
i ðηÞ
1
1
n
2
0
±± ð ±± + i η d Y d Y d η ± d η dη ± ð dY ±± ± - i η dd ηY dd ηY dη ± !# 1
d Y + i ðηÞ k dη
k
2
2
dη
2
2
2
0
d d Yn iðηÞ dη dη
+
2
!± d Y ±± i η ±± dη !
d - Yn dη Yk
!±± ±± ±
!#
d Yk i ðηÞ dη
2
2
2
!#
0
1
(H.8)
2
0
Then, using Eqs. (H.8) and (H.4), Eq. (H.1) takes the form
I
4
=
"
d d Yn i ðηÞ Yk dη dη
=
"
S ðηÞ Y n
" Yk
+ =
!
2
+ =
2
"
d d Yn iðηÞ dη dη 2
2
iðηÞ
2
2
1 0
1 0
-
!# " 1
+
2
!
!#
i ðηÞ
d Yk dY n d Yn dY k dη dη dη dη 2
2
2
!#
1
2
0
0
1
0
dY - SðηÞ n dη 2
Y kð ηÞΓ nðηÞ±
½
2
d Y k dY n d Yn dY k dη dη dη dη
Y kð ηÞΓ nðηÞ±
½
d d Yk -Y n iðηÞ dη dη
dY k dY - Yk n dη dη
707
!# " 1
-
!#
d d Yk i ðηÞ dη dη 2
Yn
!
2
0
dY - SðηÞ k dη
!#
1
0
1
2
"
+
Y nðηÞΓk ðηÞ±
² + Y
½
0
1 0
# "
#
³ ² η - Y
³ η
dY n gk ðηÞ dη
Y nðηÞΓk ðηÞ±
½
1
0
′ n ðηÞgk ð
1
-
0
Þ
1 0
dY k gnð ηÞ dη ′ k ðηÞgn ð
1
0
Þ
1
where the prime is used to denote the derivative with respect to Eq. (H.5), and
Γn ðηÞ =
´
d d Yn i ðηÞ dη dη 2
2
µ
-S η
ð Þ
dY n dη
(H.9)
0
η, gn(η )
is given by
(H.10)
ANSWERS TO SELECTED EXERCISES
Chapter 2
E2.2
+ Jm + Jm
JO = Jml ðβ Þ + Jms 0
b
e
where 1 Jme = m ee 2; 3 Jml ðβÞ
2 l =m4 + 12
Jms = m s b2;
2
l
β = π − φ − cos−1
E2.6 ke = k E2.10 ke =
±4h ²
2
a
;
±1
h=
! 2
l 2
Jmb =
1 mbb2 3
3 + a − alcos β 5 ! 2
ð Þ
b − acos φ rðφÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L2 − ða=2Þ2
²
1 −1 + k 213 k4
where k213 = k2 + k13 ; k13 =
±1 k1
+
²
1 −1 3E1I1 3E2I2 ; k1 = ; and k 2 = ; k3 L3 L3
E2.12 ke = k123 cos2 θ1 + k 4 cos2ðπ − θ2Þ + k56 cos2ðθ3 − π Þ where k123 =
±
1
k 1 +k 2
+
²
±
²
1 −1 1 1 −1 and k56 = + k3 k5 k 6
Answers to Selected Exercises
E2.17 a) k e = ðk1 + k2Þ + 3ðk1 + k2Þ αx20 b) x0 = 0.01 m and ke = 100060 N/m E2.19 The dual tire relation is Fdual -tire = 79:33 + 966:57 δ + 22:01δ 2 N and that for the wide-base tire is Fwide- base = 136:90 + 787: 04δ + 6: 26δ2 N E2.20 ke = 2ρgAo +
2nAo Po Lo
Chapter 3
E3.1 m€ x + cx_ + kx = 0 E3.2 m€ x + cx_ − 2kx = 0 E3.4 ωn = ωo
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 + ð1 + a=LÞ
Ω2 ω2o
where ωo =
rffiffiffiffi k b
m
and k b =
3EI L3
E3.7 r4o − r4i > 7:272 × 10−7 m4 E3.11 ωn = 15.26 rad/s and ζ = 0.305 E3.16 (a) For tap water ωntw =
E3.23
rffiffiffiffiffiffig ffi
sw h
(b) For salt water ωnsw = 1.095ωntw
ωn = ζ=
skffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi +k 1
23
;
m c1 + c2 + c3 2m ωn
k 23 =
1 1 + k2 k3
!−
1
709
710
Answers to Selected Exercises
³
´
1 E3.29 (a) Jo + mL2 € θ + ca2θ_ + π d 2ρgL2θ = ðm + mb =2ÞgL 4 (b) c = E3.32
Ld a2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ³J + mL ´ ρπ g ffi 2
o
ð
Þ
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ωn
2kcos2γ m
=
rad=s
Chapter 4
E4.3 xð tÞ = 0:002sinð100 :24tÞ m E4.6 θðtÞ = 4: 08te−1:25t rad
E4.7 δ=
πζ p2ffiffiffiffiffiffiffiffiffiffiffi ffi
1 − ζ2
E4.10 Mass m2 separates from mass m1 since g Xoω2n
= 0 981 ≤ 1 :
E4.12 ζ = 0.0367 and ωd = 5.288 rad/s E4.17 For uncracked concrete xmax = 0.5429 m and for cracked concrete xmax = 0.5426 m; therefore, the maximum displacement remains virtually unchanged E4.18 xð tÞ = 1:67e−0 9048tsinð75: 39t + 0:642Þ mm :
Chapter 5
E5.3 Steady-state amplitude is 1.9 mm and the phase is −0.882 rad E5.5
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ωn
=
ω2