Introduction to the Control of Dynamic Systems [1 ed.] 9781600860812, 9781563470837

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Introduction to the Control of Dynamic Systems Frederick O. Smetana North Carolina State University Raleigh, North Carolina

IEDUCATION SERIES J. S. Przemieniecki Series Editor-in-Chief Air Force Institute of Technology Wright-Patterson Air Force Base, Ohio

Published by American Institute of Aeronautics and Astronautics, Inc. 370 L'Enfant Promenade, SW, Washington, DC 20024-2518

American Institute of Aeronautics and Astronautics, Inc., Washington, DC Library of Congress Cataloging-in-Publication Data Smetana, Frederick O., 1928Introduction to the control of dynamic systems / Frederick O. Smetana. p. cm.—(AIAA education series) Includes bibliographical references and index. 1. Automatic control. 2. Dynamics. I. Title. II. Series. TJ213.S486 1994 629.8—dc20 94-8085 ISBN 1-56347-083-7

Copyright © 1994 by the American Institute of Aeronautics and Astronautics, Inc. All rights reserved. Printed in the United States of America. No part of this publication may be reproduced, distributed, or transmitted, in any form or by any means, or stored in a database or retrieval system, without prior written permission of the publisher.

Data and information are for informational purposes only. AIAA is not responsible for any injury or damage resulting from use or reliance, nor does AIAA warrant that use or reliance will be free from privately owned rights.

Texts Published in the AIAA Education Series Re-Entry Vehicle Dynamics Frank J. Regan, 1984 Aerothermodynamics of Gas Turbine and Rocket Propulsion Gordon C. Gates, 1984 Aerothermodynamics of Aircraft Engine Components Gordon C. Gates, Editor, 1985 Fundamentals of Aircraft Combat Survivability Analysis and Design Robert E. Ball, 1985 Intake Aerodynamics J. Seddon and E. L. Goldsmith, 1985 Composite Materials for Aircraft Structures Brian C. Hoskins and Alan A. Baker, Editors, 1986 Gasdynamics: Theory and Applications George Emanuel, 1986 Aircraft Engine Design Jack D. Mattingly, William H. Reiser, and Daniel H. Daley, 1987 An Introduction to the Mathematics and Methods of Astrodynamics Richard H. Battin, 1987 Radar Electronic Warfare August Golden Jr., 1988 Advanced Classical Thermodynamics George Emanuel, 1988 Aerothermodynamics of Gas Turbine and Rocket Propulsion, Revised and Enlarged Gordon C. Gates, 1988 Re-Entry Aerodynamics Wilbur L. Hankey, 1988 Mechanical Reliability: Theory, Models and Applications B. S. Dhillon, 1988 Aircraft Landing Gear Design: Principles and Practices Norman S. Currey, 1988 Gust Loads on Aircraft: Concepts and Applications Frederic M. Hoblit, 1988 Aircraft Design: A Conceptual Approach Daniel P. Raymer, 1989 Boundary Layers A. D. Young, 1989 Aircraft Propulsion Systems Technology and Design Gordon C. Gates, Editor, 1989 Basic Helicopter Aerodynamics J. Seddon, 1990 Introduction to Mathematical Methods in Defense Analyses J. S. Przemieniecki, 1990

Space Vehicle Design Michael D. Griffin and James R. French, 1991 Inlets for Supersonic Missiles John J. Mahoney, 1991 Defense Analyses Software J. S. Przemieniecki, 1991 Critical Technologies for National Defense Air Force Institute of Technology, 1991 Orbital Mechanics Vladimir A. Chobotov, 1991 Nonlinear Analysis of Shell Structures Anthony N. Palazotto and Scott T. Dennis, 1992 Optimization of Observation and Control Processes Veniamin V. Malyshev, Mihkail N. Krasilshikov, and Valeri I. Karlov, 1992 Aircraft Design: A Conceptual Approach Second Edition Daniel P. Raymer, 1992 Rotary Wing Structural Dynamics and Aeroelasticity Richard L. Bielawa, 1992 Spacecraft Mission Design Charles D. Brown, 1992 Introduction to Dynamics and Control of Flexible Structures John L. Junkins and Youdan Kim, 1993 Dynamics of Atmospheric Re-Entry Frank J. Regan and Satya M. Anandakrishnan, 1993 Acquisition of Defense Systems J. S. Przemieniecki, Editor, 1993 Practical Intake Aerodynamic Design E. L. Goldsmith and J. Seddon, Editors, 1993 Hypersonic Airbreathing Propulsion William H. Heiser and David T. Pratt, 1994 Hypersonic Aerothermodynamics John J. Bertin, 1994 Mathematical Methods in Defense Analyses Second Edition J. S. Przemieniecki, 1994 Tailless Aircraft in Theory and Practice Karl Nickel and Michael Wohlfahrt, 1994 Introduction to the Control of Dynamic Systems Frederick O. Smetana, 1994 Published by American Institute of Aeronautics and Astronautics, Inc., Washington, DC

Foreword This latest addition to the AIAA Education Series covers the important topic of the dynamic behavior and control that must be considered in any design of moving mechanical systems whether they be simple cam followers or complex aircraft structures. Introduction to the Control of Dynamic Systems by Frederick O. Smetana is a comprehensive text that develops the basic concepts for the analysis of dynamic systems and for control of such systems to achieve desired

performance. The author describes how such systems can be modeled, and introduces the reader to both analog and digital controls. The most welcome feature of this textbook, however, is the multitude of examples discussed including various machinery components, rotating machinery, engine balancing, shaft whirl, various structural components, and complete aircraft or spacecraft structural systems. The text also describes modeling of other systems such as electrical, thermal, pneumatic, and hydraulic systems. The material is organized into six chapters followed by twelve appendices on special topics and additional information supporting the main chapters. These appendices are particularly useful for students with less than comprehensive background in elementary dynamics. The text can serve as a source of information both for engineering students and practicing engineers to acquire new areas of expertise or to maintain currency. Inaugurated in 1984, the AIAA Education Series embraces a broad spectrum of theory and application of different disciplines in aerospace, including aerospace design practice. The Series has now been expanded to include defense science, engineering, and technology and is intended to serve as both teaching texts for students and reference materials for practicing engineers and scientists.

J. S. Przemieniecki Editor-in-Chief AIAA Education Series

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Preface One of the engineering challenges of the coming decade will be the design of an increasing assortment of programmable manufacturing and assembly machines (robots). Each new generation of machines will be expected to operate faster and to be lighter than the previous generation. Whereas many present-day robots are sufficiently stiff and operate sufficiently slowly that they can be considered to be rigid and their response can be considered to track commands perfectly, future machines will, of necessity, be structurally more flexible and will require feedback control to insure proper performance of their tasks. If the newly graduated engineer is to function effectively in this environment, he must approach the design assignment with the capability for analyzing the response of a flexible system to commands and for rationally determining how the system's performance can be brought to the desired level. Many analysis and/or design assignments in the process, aerospace, and allied industries will require the use of the same skills. This book is intended to help future engineers acquire the beginnings of this capability. To that end it addresses the following questions: 1) How does one represent a physical system with a mathematical model (usually a system of linear differential equations), and how does one interpret the solution of that mathematical model in physical terms? 2) How does one solve the mathematical model he creates in a rigorous yet rapid manner? 3) How may this capability be applied to the analysis of the motion of—or the force created by—a dynamic system? 4) How may one alter the component values in a dynamic system in order to cause it to exhibit the behavior one desires? 5) How does the fact that system performance data are measured as a series of samples rather than continuously alter the techniques one can use to analyze system behavior? Students who take an introductory course in vibrations or one in automatic controls have usually already taken a sophomore-junior course in dynamics. Such courses are generally divided into two parts: 1) kinematics, a study of the geometry of motion, and 2) kinetics, a study of the relations among the forces acting on a body, the mass of the body, and the motion of the body. In most engineering schools in the United States, kinetics is taught from the point of view that given a body's motion, one seeks to discover the forces producing that motion at a particular instant of time. In this text we take an inverse view: given the forces applied to a body over an extended period of time, what is its motion? The former approach requires at most a second derivative of the body's trajectory equation(s) evaluated at the time of interest in order to supply the inertia term(s) needed for the force balance; in this approach a differential equation or system of equations must be solved to discover the trajectory. In the author's experience, students usually fail to see these two views as opposite sides of the same coin.

The author has always believed that most upper division engineering courses are about 5-10% physics and 90-95% how to get practical numerical results from these physical concepts. Prior to Sputnik, these numerical results were often obtained from correlations of empirical data. Then came the great push to make engineering more "science-based." What came to be regarded as fundamentals to be taught in engineering courses were frequently specialized solution techniques for the describing system of differential equations, applicable primarily to the particular conditions being investigated; or, what might be termed solution indicators, techniques that would indicate the nature of the solutions without actually having to obtain them. Although all engineers are exposed to a course on differential equations early in their collegiate careers, few are called upon to solve such equations regularly as part of their baccalaureate training. There seem to be two reasons for this: 1) it is very time-consuming when done manually, and 2) the vast majority of engineering students simply never become sufficiently good at it to warrant the assignment. This avoidance of differential equations seems to carry over into the workplace as well. This situation is quite unfortunate since a graphical view of the time solution(s) of the describing equation(s) is, in the author's view, probably the most powerful tool available for helping the neophyte grasp the significance of changes in parameter values or the addition or deletion of terms in the equation(s). Whereas the experienced engineer will easily make the connection mentally between the results provided by solution indicators and the time history, the classroom neophyte is likely to regard this practice as black magic. The difficulty in solving differential equations and the lack of skill at solving them exhibited by most new engineering graduates may be in part responsible for the unfortunate tendency in some companies to forgo the level of quantitative analysis many engineering problems would seem to call for. With the increasing availability of small computers and suitable software, hopefully this will change. In most companies, engineering is an expensive overhead item. If the engineer is to earn his salary, he must seek to become even more productive, that is, he must always seek to solve problems assigned him more rapidly and with greater precision. Inevitably, this means increasing use of "canned" routines on the computer because he himself will seldom have either the time or the necessary mathematical skills to write the codes. What he must have is a thorough understanding of the physical basis of his "canned" codes and the ability to determine whether the computer results are reasonable—not necessarily correct, but reasonable. Unfortunately, it seems to the author that so much of the usual course time is devoted to techniques (fundamentals to some) that little time can be devoted to insuring that a thorough understanding of the physics exists, and even less time is devoted to discussing the significance, applicability, or limitations of the results. It is usually assumed that if the student becomes proficient in the use of the techniques, the significance, limitations, and so forth of the results will become apparent to him. There is a curious phenomenon associated with the textbooks for such upper division engineering courses. Although textbooks are presumably written to facilitate student learning, they are selected by the course instructors. Savvy textbook authors therefore often seem to pitch their texts at the instructor rather than at the student. As a result, their books, while perhaps treating a wide variety of topics in fairly elegant fashion, are often too brief in the development of the theory and lack discussion of the significance of the results obtained for any but the best students to grasp the technique presented or to understand why they should learn it. Many

instructors, aware of the prodigious amount of material to be learned and the rate at which knowledge is growing, are reluctant to tailor their presentations to the level and rate at which the majority of their students appear to be learning. This is especially true with regard to really new material, particularly when it is of a highly mathematical nature. The consequence of the instructor's failure to present the material at a level and rate that most students can grasp is that students are able to work a few problems for which models have been given in the text or in class, recite the topics the course was supposed to cover, and do little else. They are unable to apply the concepts to the same problems when the problems have been altered in subtle ways or to treat problems that are logical extensions of textbook examples. While it is easy to recognize these deficiencies in textbooks and presentations, it is not so easy to devise alternate approaches that can demonstrate measurably better results. This book is the result of the author's efforts in this direction over a 13-year period. From the beginning he felt that the computer should somehow be an integral part of any plan to maximize student learning in relation to the student's investment of time in the course. To that end he felt that if some of the drudgery could be taken out of solving drill problems, perhaps the time saved could be used to better understand the modeling and solution process and to think about the significance of the solutions. To that end he devised a set of computer programs which, if given the correct model data, could solve any problem likely to be assigned. First implemented on the mainframe in 1981, they were ported to the PC in 1984 and then to departmental workstations beginning in 1988. During the course of time many additional features were added and the programs were consolidated as the capabilities of the platforms on which they were run increased. Today, a single version runs on all major platforms. The programs are described in detail in the accompanying user's manual. Many competing software packages have evolved in the past 13 years. Some have received strong support from major textbook publishers. The author's university [North Carolina State University (NCSU)] has acquired some of these packages and the author has tried them, as have his students. Students unfamiliar with any package seem to adapt to the author's or to commercial packages with about equal ease (or difficulty). Unfortunately, students usually have little appreciation for the precision to which problems can (or should) be solved. Hence, differences in the results given by one package as compared with the results from another are frequently ignored. For some problems it really does not matter; however, the author's background has been the determination of vehicle and particle trajectories where precision matters a great deal. All commercial packages with which the author is familiar either use numerical integration or approximate the forcing function by a succession of steps. The author's package employs evaluations of analytical solutions generated by the program and accepts a variety of analytical forcing functions. The coefficients of the analytical solution are also available on the printout. Time histories are available either as seven digit listings or as graphs made with the printer character set and as PostScript files. The algorithms in the author's package have been set up so that solutions to systems with fewer than 10 eigenvalues are computed using double precision, and solutions to systems with more than nine eigenvalues are computed using multiple precision arithmetic where the numbers are represented as having 240 decimal digits each. Problems with higher-order eigenvalues are handled exactly, and cases where the eigenvalues are separated by less than l.OD-5 rad/s are treated as higher order.

Various other procedures have been incorporated into the package to ensure the attainment of very high accuracy and to reduce typical student errors. Numerical integration has three deficiencies when applied to linear problems: 1) the results are step size dependent; students often are unaware of this and have little idea how to determine a suitable step size; 2) numerical integration cannot give correct infinite time solutions; and 3) since it is a time domain solution, no information on the eigenvalues or the frequency response is obtained. State transition methods built around a step input have the same deficiencies as the first and third described above. The methods coded in commercial packages are intended to yield results rapidly with a minimum of code. Yet for problems with fewer than nine eigenvalues, a single input, and zero initial conditions, there is little actual difference in the execution time compared with the author's package. The author's package can handle multiple forcing functions and initial conditions; it does so by using superposition of the solutions for each of the forcing functions and for each of the initial conditions and not as part of a single solution generation as would be the case for numerical integration. One other difference between the philosophy of commercial packages and the author's may be noted. The commercial packages generally provide a rather limited amount of output. The author's package, designed in the days of batch jobs on a mainframe, seeks to provide a large amount of output for a very limited amount of input, the thought being that the student can study this at his leisure to determine the validity and/or significance of the result. The output files are large and sometimes numerous. If the student wishes to examine the files on line he may quickly delete any he does not wish to print and edit out parts of those he does wish to print. The author has retained this approach through the years because he personally finds it difficult to make necessary decisions regarding options and parameter values interactively. With commercial packages it is usually necessary to run several programs to generate the same material that is provided by SOLV94. There is one additional advantage of the author's package as compared with most commercial packages: the package is distributed as source code and the user is free to modify it as he desires. Unnecessary features are easily deleted. Students can also use it as a tool with which to learn FORTRAN programming since the algorithms coded are simply those which are employed to solve the problems manually. Presumably, students would be encouraged to learn to solve such problems manually before being granted the privilege of solving them with a computer. The book was begun at the suggestion of an editor from McGraw-Hill who was looking for a more modern replacement for the 1967 text by R. H. Cannon. After examining Cannon, the author decided that the first third of that text was already covered adequately by a number of standard dynamics texts. The other twothirds covered material usually taught under the titles of vibrations and automatic controls. The author had already used his software package with the beginning course on automatic controls and felt that with a few enhancements it would also be suitable for use with a first course in vibrations. Many reviewers of the initial outline felt that too few topics were presented and so, to satisfy these criticisms, the book grew. Other modifications and additions occurred as a result of changes in publishers and in curricula at NCSU. Several of the appendices were added, for example, to accommodate students entering from or going to revised tracks. The selection of topics for this book was guided to some extent by a desire to keep the number of software programs the student might be expected to learn

to use to a minimum. The author also recognized that there will likely never be near-universal agreement as to which topics the course should cover. Even at NCSU course content varies noticeably from instructor to instructor. It also varies from department to department. Recent texts by electrical engineers seem to place strong emphasis on selecting computationally efficient algorithms that may be implemented using embedded controllers and on methods for finding the gains in the algorithms which are needed to achieve a specified performance. Mechanical and aerospace engineers, on the other hand, frequently use the tools and vocabulary of the control engineers to describe the behavior of complex dynamic systems. Only after this task is completed is the mechanical or aerospace engineer interested in what kind of controller may be used to modify some aspect of the performance of the system and how this controller may be realized. The chemical engineer can usually represent the systems he deals with by relatively simple mathematics. His interests therefore tend more to issues of precision, reliability, and cost. A single generalized feedback-feedforward algorithm may suffice for his purposes. The level of mathematics used in these courses is a significant factor in the ability of the student to grasp the analysis and design concepts presented. Some book authors seem to be aware that the theory and methods of solution they present are substantially above the level of understanding of the target audience and they therefore may devote as much as a third of the text to a condensed exposition of the relevant mathematics, or they may default to a completely numerical approach. This author is uncomfortable with both of those approaches. In the author's view students unfamiliar with linear algebra are not going to learn it on their own time from such a condensed treatment. Numerical approaches, while quite useful to the practicing engineer, are not the best way for a neophyte to understand the value of having both a frequency and a time domain representation of the system. The author is also painfully aware of the fact that a majority of engineers never have to solve a differential equation during their entire professional careers. It is for these reasons that he has chosen to base the treatment primarily on transform methods and to provide with the text a tool to take the drudgery out of solving systems of linear differential equations. This text, being an entry-level text, restricts itself primarily to linear problems or to linear approximations of nonlinear problems. However, the effect of a nonlinear description on the nature of the problem solution is not overlooked. When several methods of analysis utilize essentially the same information, the text will usually treat one in detail and mention some of the others only briefly. To present them all in the same detail would enlarge the book considerably. While such an encylopedia may be appreciated by practicing professionals, it tends to overwhelm the student and is unkind to his pocketbook. Perhaps the most significant choice made in the selection of the topics to be included involved the location and depth of the treatment of state variables. Some professionals who have grown up with the computer and its ever-increasing ability to manipulate systems of first-order linear equations urged that this treatment be introduced early in the book and that it become the dominant treatment. The author has resisted for a number of reasons. 1) He believes that most subjects are best learned in the order in which they developed chronologically. 2) He believes that most students enter the courses better equipped to perform the simple algebraic operations required of a transform approach than the more numerous and more abstract matrix operations associated with a state variable approach. 3) He

believes that the frequency domain lore that flows naturally from a transform-based approach is extremely valuable in promoting understanding of systems behavior and the role played by controllers. This lore does not naturally present itself in a time domain based state variable approach. 4) While the state variable approach offers both an easier route to a high level of control of complex systems and a simpler, more computationally efficient means for determining the time domain performance of such systems, these advantages evaporate for relatively simple systems that are controlled and analyzed just as rapidly using transform-based methods. 5) Transform-based methods can provide a solid understanding of the physics involved and can serve as a reality check as the student studies the state variable approach. The abstraction involved in the state variable approach does not seem to favor the inverse process. 6) To ignore the transform approach and to begin one's study of the analysis and design of automatic control systems with the state variable approach seems rather like building a house without giving much thought to its foundations. In some cases one may obtain an acceptable structure in less time. The probability for disaster, however, is also heightened. The book is divided into six chapters and 12 appendices. Chapter 1 deals with the representation or modeling of dynamics problems by systems of differential equations. The chapter begins the treatment with Newton's Second Law of Motion as the basis for devising suitable equations. It then considers how one might linearize the resulting equations. As an aside, a simple problem is solved exactly to illustrate the difficulty involved in treating nonlinear equations. It then moves on to consider the formulation of the mathematical model for a number of onedegree-of-freedom physical schematics. The solutions for these models are either stated or developed by classical techniques. The development then considers twodegree-of-freedom systems. The emphasis is to show that writing the equations is not much more difficult than it is for single-degree-of-freedom systems. Computergenerated solutions are presented and the physical significance of these solutions is discussed in detail. The developments move on to consider three-, four-, and five-degree-of-freedom systems pointing out at all times that writing the equations follows a very easily-grasped process, that the use of canned software makes it easy to get solutions for such systems in a short time, and that the real challenge is in interpreting the results in physical terms. Considerable time is devoted to an attempt to aid the student in developing a feel for the interactions of the motions since this is a task that the engineer will have to continue to do manually. Discussion of how one actually solves the differential equations he generates is purposely put off until Chapter 2. Chapter 1 also allocates a section to the creation of mathematical models for certain electrical, pneumatic, hydraulic, and thermal systems. This chapter, if used in its entirety, could well serve as the core for a course on system modeling. Chapter 2 treats the solutions of the math model equations, primarily by the method of Laplace transforms. However, it also discusses numerical integration of the differential equations as an outgrowth of the method of representing functions by Taylor series expansions. Chapter 3 treats the application of lumped parameter models to the analysis of vibrations problems, whereas Chapter 4 examines the tools used to analyze and modify the dynamics of continuous feedback control systems. Chapter 5 has several tasks: 1) discussing the important ways in which sampled data systems differ from continuous data systems; 2) discussing the natural transform-based approach for solving difference equations; and 3) discussing an

alternate transform which in the limit as the sampling rate approaches infinity is indistinguishable from the usual method of analyzing continuous data systems. Chapter 6 also has multiple tasks: 1) discussing the state variable description of dynamic systems and their solution by the state transition method; 2) discussing the idea of simulation as a means for modeling and selecting suitable control parameters; 3) discussing various types of nonlinearities encountered in practice; 4) discussing an alternate representation of system elements, the signal flow graph; 5) discussing some concepts unique to state variable representations such as observability and controllability; 6) discussing some other methods for modeling elements in sampled data systems and for solving difference equations; and 7) describing a method for selecting gains in an automatic control system design, the assignment of system pole locations. The appendices treat a method for determining the stability of systems of linear differential equations that does not require solution of the characteristic equation of the system or generation of the time domain solution; the Nyquist diagram, an alternate way of presenting frequency response information; the representation of time domain functions by Fourier series and some of the conclusions one can draw from a Fourier analysis of a time function; the representation and effect of short time delays in the propagation of signals through a dynamic system; the derivation of the description of the motion of a body in cartesian space as applied to an airplane; the development of a description of the motion of point masses in a central force field with application to simple two-body problems in astronautics; the analysis of constrained motions of a body; the energy description of motion and some applications of the concept; the application of Laplace transforms with fractional powers to a number of engineering problems; some additional broad based design problems; discussions of some hardware considerations in controller design; and methods to verify the solution of a systems of differential equations. The appendices can be used either as additional topics for the vibration or automatic controls course or as suggested reading for the student. The material is arranged so that the book can be used for a vibrations course (Chapters 1, 2, 3, and Appendix C) or an automatic controls course separately (Chapters 1, 2, 4, and Appendices A, B, and D) or as a two-course sequence, in which case parts of Chapters 5 and 6 plus Appendices A-G could be substituted for a repeat of Chapters 1 and 2. It may also serve as the text for an introductory course in digital controls. Given the wide variety of topics covered by such courses in practice, the instructor may wish to supplement the material presented here with topics of his own choosing, for example, the analysis of linkages in the vibrations course or a detailed discussion of various design algorithms in either a continuous or digital controls course. If class time is devoted to a discussion of the end-ofchapter problem solutions and some of those in Appendix L, there is actually sufficient material for three one-semester courses. The end-of-chapter problems (plus those in the appendices) range from simple drill problems to rather challenging problems requiring the integration of knowledge from a variety of sources and/or extensive computational activity. The book is designed so that use of the computer is not required, however. The accompanying software is specifically intended to address the problems in the text. To that end, it uses the same nomenclature and methods featured in the book. The software is distributed without charge in source form to adopting professors and purchasers of the book. Executable versions are available for several platforms and may be had at a nominal charge for the files on diskettes. They can also

be obtained by anonymous ftp at no charge. The author intends to mail, tain the software personally for several years and will make upgrades available at a Lominal charge upon request. When the software use is assigned as part of a class it has been found that students become familiar with it more rapidly if the supervising faculty member also understands its operation himself so that he can answer questions as to what error messages mean, what the results indicate, where certain important numbers appear in the printouts, etc. Students have demonstrated considerable difficulty in reading and interpreting manuals correctly, particularly if the material is very new. They quickly become frustrated and just give up. They feel more comfortable learning to use programs if vexing questions can be answered quickly by the instructor. During this book's 9-year gestation period, the author received help and/or encouragement from a great many people. They are really too numerous to mention at this point. To all those who contributed such help and encouragement: thank you very much. The fault for such mistakes as remain is strictly the author's. He is also personally responsible for about 85% of the drawings and sketches. These were created with individual PostScript programs, most of which were written from scratch. It was a good way to learn the language, but proved to be rather time consuming.

Frederick 0. Smetana August 1994

Table of Contents Chapter 1. Modeling of Dynamic Systems by Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Introduction....................................... 1 1.2 Newton's Second Law of Motion and Its Application to Physical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Examples of Single-Degree of Freedom Systems . . . . . . . . . . . . . 7 1.4 Simplification of the Mathematical Model of a Dynamic System for Convenience in Solving . . . . . . . . . . . . . . . . . . . . . . . . . 11 .5 Free Vibration of a Single-Degree of Freedom System. . . . . . . . . 21 .6 Forced Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 .7 Examples of Single-Degree of Freedom Forced Vibrations . . . . . . 37 Coupled Systems: Examples of Two-Degree of Freedom Forced Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 .9 Examples of Multi-Degree of Freedom Systems . . . . . . . . . . . . . 50 .10 Overview of the Motions of Complicated Systems . . . . . . . . . . . . 80 .11 Modeling Other Physical Systems . . . . . . . . . . . . . . . . . . . . . . . 81 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 P r o b l e m s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Chapter 2. Methods of Solution of Linear Equations of Motion . . . . 121 2.1 I n t r o d u c t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 2.2 Numerical Integration of First-Order Differential Equations . . . . 122 2.3 Solving Ordinary Differential Equations by Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3 1 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 P r o b l e m s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Chapter 3. Applications to the Analysis of Mechanical Vibrations . . . 3.1 Introduction..................................... 3.2 Transmission of Machinery Vibrations . . . . . . . . . . . . . . . . . . . 3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Determination of Natural Frequencies and Damping Ratios of More Complex Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Energy Dissipated in Viscous Damping . . . . . . . . . . . . . . . . . . 3.6 Decay of Vibratory Motion Due to Other Types of Damping . . . . 3.7 Response to Periodic Forces . . . . . . . . . . . . . . . . . . . . . . . . . .

161 161 161 168 173 175 177 193

3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17

Frequency Response and BODE Plots . . . . . . . . . . . . . . . . . . . Arbitrary Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Response Spectrum of Certain Dynamic Systems . . . . . . . . . . . Vibration Absorber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Untuned Viscous (Houdaille) Damper . . . . . . . . . . . . . . . . . . . Centrifugal Pendulum Vibration Absorber . . . . . . . . . . . . . . . . Concepts of Flexibility and Stiffness . . . . . . . . . . . . . . . . . . . . Rotary Balance and Balancing . . . . . . . . . . . . . . . . . . . . . . . . Closure........................................ Advanced Topic: The Vibration of Continuous Media . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems.......................................

202 203 209 213 217 221 224 229 262 263 267 267

Chapter 4. Modifying System Dynamic Behavior to Achieve Desired Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 4.1 Introduction and Historical Perspective . . . . . . . . . . . . . . . . . . 283 4.2 Definitions and Criteria for Acceptable Dynamic Performance . . 290 4.3 When Dynamic Analysis Is Not Necessary . . . . . . . . . . . . . . . . 295 4.4 Transfer Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 4.5 Block Diagrams and Their Algebra . . . . . . . . . . . . . . . . . . . . . 300 4.6 Feedback Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 4.7 Frequency Response and BODE Plots . . . . . . . . . . . . . . . . . . . 314 4.8 Root Loci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 4.9 Steady-State Feedback of Unity Feedback Systems . . . . . . . . . . 343 4.10 Use of Compensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 P r o b l e m s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 Chapter 5. Introduction to Digital Control . . . . . . . . . . . . . . . . . . . 405 5.1 Introductory Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 5.2 Laplace Transform Analysis of Sampled Data Systems . . . . . . . 408 5.3 Examples and Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 5.4 Frequency Response of Systems Excited by Pulses Passed Through Data Holds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423 5.5 Sampled Data and Difference Equations . . . . . . . . . . . . . . . . . 437 5.6 Z Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 5.7 System Time Response by Use of Inverse Z T r a n s f o r m s . . . . . . . 456 5.8 Use of the SAMPLE89 Computer Program . . . . . . . . . . . . . . . 470 5.9 Use of DIGBOD Computer Program . . . . . . . . . . . . . . . . . . . . 471 5.10 Use of REZPON and SOLVZ Computer Codes with Transformed Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474 5.11 The w' Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476 5.12 BODE Plots in the wr Domain . . . . . . . . . . . . . . . . . . . . . . . . 483 5.13 C l o s u r e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 5.14 Table of Transform Equivalents . . . . . . . . . . . . . . . . . . . . . . . 491 5.15 Z Transforms of Number Sequences, €(k) . . . . . . . . . . . . . . . . 493

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498 P r o b l e m s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498 Chapter 6. Introduction to State-Space Analysis of Dynamic Systems............................................. 6.1 Introduction..................................... 6.2 State-Space Representation of Dynamic Systems . . . . . . . . . . . 6.3 Integration of State Equations for Continuous Systems . . . . . . . . 6.4 Signal Flow Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 State-Space Representation of Discrete-Time Systems . . . . . . . . 6.7 Conversion from Continuous-System Equations to Discrete State Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Discrete Signal Implementation of PID Controllers . . . . . . . . . . 6.9 Design by Pole Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 C l o s u r e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems....................................... Appendix A. Routh-Hurwitz Stability Analysis . . . . . . . . . . . . . . . . . A.I Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

507 507 508 551 559 563 563 566 568 571 572 572 573 583 585 586 587 588

Appendix B. Nyquist Diagram: Its Construction and Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 B.I I n t r o d u c t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 B.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 B.3 I n t e r p r e t a t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 B.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 590 B.5 Significance of Frequency Value Where a Nyquist Curve Crosses 1.0 at Angle 180 deg . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 B.6 The NYQUST Computer Program . . . . . . . . . . . . . . . . . . . . . 595

Appendix C. Representation of Periodic Functions by Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C.I Fourier Series Representation of Pulse Trains . . . . . . . . . . . . . . C.2 Note on the Use of Local Linearization to Solve Nonlinear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C.3 The Fast Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

597 602

603 612 613

Appendix D. Effects of Small Time Delays on Continuous System Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615

Appendix E. Modeling the Motion of Bodies in Space . . . . . . . . . . . . . 619 E. 1 Derivation of the Equations of Motion for the Flight of a Rigid Aircraft with Controls Fixed . . . . . . . . . . . . . . . . . . . . . . . 619 E.2 Note on "Static" Longitudinal Stability of Airplanes . . . . . . . . . 639 E.3 Equations of Motion with Reference to a Fixed Position on the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 641 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645 Appendix F. Equations of Motion of a Body in a Central Force Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F.I Example: Geosynchronous Satellite . . . . . . . . . . . . . . . . . . . . F.2 Example: Solar Gravitational Constant . . . . . . . . . . . . . . . . . . Problems.......................................

647 649 650 651

Appendix G. Dynamics of Cam Followers . . . . . . . . . . . . . . . . . . . . . G.I CAM30 Program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G.2 Constant Acceleration Profile . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems.......................................

653 654 657 659 660

Appendix H. Integral Representation of Motion: Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663 H. 1 Example: Mass Falling onto a Spring . . . . . . . . . . . . . . . . . . . 663 H.2 Example: Aircraft Performance Problem . . . . . . . . . . . . . . . . . 664 Appendix I. Verification of Solutions to Differential Equations . . . . . . 1.1 Back Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Comparisons with Simple Physical Models . . . . . . . . . . . . . . . Problems.......................................

667 667 668 669

Appendix J. Problems Involving Laplace Transforms with Fractional Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671 J.I Bessel's Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671 J.2 Heat Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673 J.3 Table of Some Laplace Transforms . . . . . . . . . . . . . . . . . . . . . 674 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675

Appendix K. Some Hardware Considerations . . . . . . . . . . . . . . . . . . 677 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 681 Appendix L. Additional Design Problems . . . . . . . . . . . . . . . . . . . . . L.I Design Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . L.2 Design Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . L.3 Design Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . L.4 L.5

683 683 684 686 Design Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686 Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 687

L.6 L.7

Typical Quiz Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 690 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 697 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706

Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707

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Motto of the Work: Verachtet mir die Meister nicht und ehrt mir ihre Kunst! —R. Wagner Act III Meistersinger von Nurnberg

1

Modeling of Dynamic Systems by Linear Differential Equations

1.1 Introduction For purposes of our present discussion, a system is an assembly of components which performs a specific task. We monitor the behavior of a system in the conduct of its task by observing the values of its states or state variables. If the states (e.g., position, velocity, acceleration, temperature, pressure, heat flux, and power level) can change with time at a sufficiently rapid rate so as to affect system performance, the system is said to be dynamic. It is the task of engineers to solve the problems posed by their employers. Often these problems involve the creation of a machine or process to perform a specific task or the improvement of the performance of an existing machine or process. Cost-effective engineers do not just conceive and build new machines or modify existing machines or processes according to a flash of intuition; rather, they attempt to characterize or model the performance of the system mathematically. Then, by solving the mathematical model, they can predict performance or the results of modifications to the system. The degree to which a math model accurately represents the physical system determines to a large extent the validity of its performance prediction. Unfortunately, the more accurate and complete the math models, the more intractable they usually are. It is therefore the task of engineers to attempt to model accurately only the more significant (in terms of performance under specific circumstances) characteristics—leaving most secondary effects undescribed mathematically—if they are to obtain a reasonably valid prediction of system performance at an acceptable cost in time and resources. The point at which the modeling effort becomes unacceptably costly has changed with time. In the past, technician labor was relatively less expensive, interest rates were low, mathematical prowess in the engineering work force was not abundant, and extensive computations were time-consuming and labor-intensive. Thus, there was considerable recourse to trial-and-error design and development. In recent times, the only one of these factors to change favorably has been the time and cost of extensive calculations. The modern digital computer has made it possible routinely to undertake computations that before 1950 could not have been completed in a lifetime. Computer technology has become so powerful that today's personal computers provide the engineering student with a greater problem-solving capability than was available in most moderate sized corporate engineering departments prior to 1950. It is the student's challenge to learn to use this resource effectively because in the future more and more of the design and development task will shift from experimentation to modeling and simulation. One might characterize the steps the engineer takes in preparing the analysis of a dynamic system as follows:

2

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

(1) Decide which aspect of system performance is to be investigated. (2) Decide on a mathematical representation for this aspect of the system starting from first principles. (3) Decide how much the model can be simplified without losing essential validity. (4) Develop a solution for the simplified math model. (5) Assess the validity and utility of the solution to the math model. (6) Modify the model and resolve as necessary. Step 4 is the only one in the entire process that does not require the application of liberal quantities of what might be called engineering judgment, an attribute that still distinguishes humans from machines. The student will appreciate that, over the years, many specialized techniques have been developed that are easier to apply in specific cases than is solving the model directly, yet these techniques still enable one to discover the essential features of the solution to that particular class of dynamics problem. In the interests of economy, students were asked to learn a generous selection of these methods along with at least the principles of more general methods. The advent of the computer, however, permits one to employ fewer, more general methods economically for a great variety of problems. For this reason, this book devotes most of its attention to a single powerful method for solving linear dynamics problems, the method of Laplace transforms, so that the time normally devoted to learning more specialized techniques can be devoted to those facets of analysis that machines have not yet mastered: formulation of the math model and the interpretation and assessment of the results. In this book, we will illustrate the solution technique in detail with simple examples and exercises and attempt to indicate some of the pitfalls one may encounter in applying it to more complex problems. We will then ask the student to employ the computer to solve problems considerably more complex (and perhaps more realistic) than are normally assigned at this level so that, free of the need to carry out seemingly endless computations manually, the student can devote more time to the real tasks of engineering: formulation of the problem in mathematical terms and assessment of the validity and significance of the solution.

1.2 Newton's Second Law of Motion and Its Application to Physical Systems About 300 years ago, Sir Issac Newton deduced three laws governing the motion of a mass of substance that have been found to be valid for all situations in our common experience. As a result, they form the basis of all our efforts to model the behavior of dynamic systems. The second of these laws states that the rate of change of momentum (product of body mass and velocity) is proportional to the force applied to the body and takes place in the direction in which the force acts. The mathematical expression of this law is

MODELING OF DYNAMIC SYSTEMS

3

We will restrict our consideration to those situations where the body mass does not change significantly during the time the analysis proceeds. As a result, we may write Eq. (1) as

F =m^

(2)

It can be shown that if the magnitude of the momentum remains constant and only its direction changes, the force is normal to the momentum.1 If the direction of

the momentum remains constant and its magnitude changes, force and momentum have the same direction. We see from Eq. (2) that the description of the position of a point mass in space at any time in response to the action of a force is d2* Fx=m--

(3a)

d2y

Fy=™^

(3b)

d2z Fz=m—

(3c)

where Fx, Fy, and Fz are the components of that force along the coordinate axes. Fx, Fy, and Fz are, in general, functions of time, position, and velocity. A body can often be treated as an ensemble of point masses distributed over a finite volume (take, for example, an airplane or a submarine); however, we must then consider the rotation of the other mass points with respect to the point whose translation is described by Eqs. (3). Because of their rotation, the other points in the body will be acted on by forces larger or smaller than those described by Eqs. (3). The time rate of change of the moment of momentum of a mass point with respect to the origin O is equal to the moment of the force with respect to the same point. Denoting moment of momentum by H = r x rav, where r is the position vector of the mass with respect to the origin, we have

— = r x F=M dt

(4)

where M is a moment. You may remember from your study of rigid-body dynamics that the left-hand side of Eq. (4), when applied to a set of Cartesian axes fixed in space and expanded into component form, becomes . . . = PIXX + PIXX - QIxy - QIxy - RIXZ - RIXI dt dHy ~dT dH7 • • - = RIZZ + RIZZ - PIXZ - PIXZ - Qhy - QIxy =

Qlyy

+ QIyy

-

RIyZ

-

RIyZ

-

PIXy

~ P IXy

(5a) (5b)

(5c)

4

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

where P, Q, and R represent the angular velocities about the x, y, and z axes, respectively; Ixz, Iyz, and Ixy are the corresponding products of inertia. If P, Q, R, Ixx, lyy, Izz, Iyz, 7^, and Ixy can &\\ change in time in some fashion not known beforehand, we must consider that these quantities are all dependent variables while time is the independent variable in the equations. We have seen in Eqs. (3), (4), and (5) that the forces and moments that cause a body to move are functions of time, body position and orientation, and the time rate of change of body position and orientation. Newton's equations of motion are therefore differential equations. In the general case, these equations are nonlinear. Now, differential equations are nonlinear if (1) They contain products or powers, including noninteger powers, of the dependent variables. (2) The dependent variables appear in transcendental functions (trigonometric, exponential, hyperbolic, etc.). For example, the general second-order differential equation d2x dx A——2 + B— + Cx=Q dt dt is nonlinear if A, B, and/or C is a function of x or if A, B, and/or C is a function like sin*, ex, and sinhjc. A, B, and/or C may be functions of /, the independent variable, however, and the equation remains linear. The solutions of nonlinear differential equations usually cannot be obtained analytically. They must be obtained numerically, a process which is very time-consuming if done manually, and their character is dependent on the magnitude of the forcing function. This makes it difficult to draw conclusions regarding the effect of a change in one of the parameter values in the equation(s) since a change in the behavior of the solution(s) may also result from a change in the magnitude of the forcing function. For these reasons, we will restrict ourselves principally to the study of linear differential equations. In Sec. 1.4.1, we will consider means by which we can linearize an equation or system of equations and the restrictions that this imposes. We will begin our consideration with situations in which the equations are linear. If a mass is axisymmetric and rotates about that axis, its moment of inertia is constant with time. It possesses only the angular velocity component along the axis which, however, may change as time changes. In such situations, the right-hand side of Eqs. (5) is linear. In Eq. (5a), for example, Q, R, and ixx are zero and Ixx is constant. Similarly, if a machine or member of a machine can be represented by a point mass and all its motion is substantially along one line, then its acceleration is just the second derivative of position. No products of velocities remain, which is a requisite for a linear differential equation. Whether such a model of a physical system is justified is usually determined after the fact: Does its solution match the essential behavior of the physical system with acceptable accuracy? Even if the match is not satisfactory quantitatively but does illustrate system behavior qualitatively, the analysis may be worth performing for the insight it provides. It is our purpose here to aid the student in acquiring some skill at formulating the mathematical problem that models a simplified representation of a physical system. From Eqs. (3) and (4), we see that, in order to produce a motion, the application of some unbalanced force (or torque) is necessary. As long as these forces are

MODELING OF DYNAMIC SYSTEMS

5

directly proportional to the position or the velocity of the mass, terms representing them can be included in the differential equation and it will remain linear. (Recall that the position of the mass is the dependent variable and its time derivative is the velocity of the mass.) We may cite as examples of devices that produce such

forces: (1) The linear spring, where the ratio of the force required to compress or extend the spring to the distance the spring compresses or extends is constant, that is, F = kx

AF with k = —— = const AJC

(2) The linear damper, which produces a resistive force directly proportional to the velocity at which the piston is moving, that is, F = Ci, with C = AF/Ai =const (see Fig. 1-1). Although these are examples of what we would probably call mechanical devices, in nature examples abound of other types of systems that do not appear to have any relation or similarity to mechanical devices but can be represented by the same mathematical expressions. Consider an airplane in forward flight rotating as well about an axis stretching from wing tip to wing tip. It will encounter a resistance to its angular motion usually taken to be proportional to the magnitude of the pitching velocity and opposite to it in direction. The resistive force generally does not act through the center of motion and therefore creates a damping moment about the center of motion. An additional aerodynamic moment is present that is proportional to the angular displacement of the aircraft from its equilibrium flight position. The value of the proportionality of the "spring constant," however, depends on the local dynamic pressure (product of the atmospheric density and flight velocity squared). For this reason, among others, maneuvers that extend over substantial changes in flight velocity cannot be analyzed accurately by linear equations. For small changes in forward velocity, however, the moments are similar to those produced by a linear damper and a linear spring acting as shown in Fig. 1-lb. An electrical circuit consisting of discrete inductors, capacitors, and resistors is another example. The inductance is analogous to the mass or moment of inertia and the capacitance to the spring constant. The student may recall from physics

spring

mass

H/WH damper

(a) linear

(b) angular

Fig. 1-1 Two arrangements of mass-spring-damper systems.

6

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

class that, for a series L, R, and C circuit, the time rate of change of electrical charge is given by

In fluid systems, the flow rate is analogous to electrical current flow, the pressure to voltage. Since fluids have small but finite masses, they have inertia; fluid interaction with walls or protrusions results in frictional losses. Fluid reservoirs can be considered analogous to electrical capacitors. And so it goes in example after example. We discuss these analogies in more detail in Sec. 1.11. Two arrangements of physical components that we will often employ to model more complex systems are shown in Fig. 1-1. For Fig. 1-1 a, assume motion along one line, with the positive direction to the right. Then, the mass times acceleration term is mx. This, according to Nev/ton's second law, is equal to the unbalanced forces present in the problem. No\v, as the spring is compressed, it exerts a force to the left equal to kx, where k is the spring constant. The damper also exerts a force to the left equal to ic, where c is the damping constant and x represents the position of the mass at any time. In addition to these forces, there is an external force /(r) applied to the mass to disturb it from its equilibrium position. The forcing function f ( t ) i s known a priori and has some specified variation with time, for example, impulsive, sinusoidal, or constant. When the relationship among these forces is written in equation form, we have or

mx = -kx - ex + f ( t )

(6a)

mx + cx+kx = f ( t )

(6b)

Similarly, for components shown in Fig. 1-lb, the product of the polar moment of inertia J and the angular acceleration 6 must be equal to the net applied torque. Note that, as shown, both the spring and the damper oppose the angular motion of the mass. The angular displacement of the mass 0 becomes a linear displacement rO at the points at which the spring and damper are attached. The spring force is therefore krO and the damper force is crO. If we set k = kr and c = cr, then the equation of motion for the motion of the mass in the second figure is

J0 + c0+k0 = T(t)

(7)

/, c, and k are constant and T(t) is the forcing function expressed in the form of a

torque or moment, which can vary in time in some prescribed fashion. Except for

the numerical values and units of the constants, Eqs. (6) and (7) are the same. Our study then will focus on ways to solve this equation and the equations resulting from more complex arrangements of the same components in order to learn the effect of varying parameter values and component arrangements on the position, velocity, or acceleration of the mass or masses as functions of time or of the forces transmitted by such motions to the supporting structure. Later, we will be interested in devising means to modify the system so as to cause the motions to follow desired paths.

MODELING OF DYNAMIC SYSTEMS

7

The examples that follow illustrate the modeling of a variety of physical systems by idealized components of three types (mass, spring, and damper or, alternately, inductor, capacitor, and resistor). What is perhaps not so apparent to the neophyte is that the modeling of most systems can be achieved to the degree of realism desired by the judicious choice of the number and arrangement of such components. Unfortunately, we cannot always tell beforehand how elaborate a model is needed to achieve a specified degree of accuracy. This is where physical insight and experience frequently help reduce the number of attempts. Were it possible to codify this insight and experience, we would now do so. Since it is not, we must hope that by studying these relatively simple examples in sufficient variety, number, and detail, we can begin to gain an understanding of the physical behavior of dynamic systems. Then, when confronted with a real-world situation, we might say to ourselves: This looks a lot like a problem I had in school. I'll start with the model I used then and see how close it comes to describing the real thing. If my description doesn't work, maybe the way my model fails, or the degree to which it fails, will suggest ways to change it so that it comes closer to the real system.

1.3

Examples of Single-Degree of Freedom Systems

1.3.1 Example: Simply Supported Beam A simply supported beam with a concentrated load acting on the midspan is shown in Fig. 1-2. If the mass of the beam is negligible compared to the load, write the equation of motion.

Solution. From the theory of strength of materials, the deflection at the midspan of a simply supported beam due to a concentrated load P at its center is 8 = PL3/(48El). Here, 8 is the deflection, L the length between supports, / the moment of inertia, and E the modulus of elasticity. For small deflections, the beams acts like a linear spring, which means that k = P/8 = 48£7/L3. The required equation of motion is, therefore, 48£7 mx + —^-x = f ( t )

This has the same form as Eq. (6), with the damping coefficient c = 0.

Fig. 1-2

•- —•»

Simply supported prismatic beam loaded at its center.

(8)

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

cp

m 1

I

(

2x

iT

w W

Fig. 1-3 U-tube manometer.

1 .3.2 Example: Manometer A manometer has a uniform bore of cross-sectional area A. If the column of liquid of length L and density p is set in motion as shown in Fig. 1-3, write the equation of motion. Solution. The mass of the fluid is simply L Ap. The difference in the height of the two columns is 2x. The pressure associated with this height difference is pg(2x), so that the force resulting from the height difference is 2Axpg. The force acts to return the column to equilibrium so that the equation of motion is pLAx = —2Axpg

(9)

Dividing Eq. (9) by pAL yields (10)

as the required equation. 1.3.3 Example: Simple Pendulum A simple pendulum is shown in Fig. 1-4. Determine the equation of motion (a) if the rod mass is negligible compared to the bob mass and (b) if the rod mass is not negligible compared to the bob mass. Solution. We find that when the bob moves through an angle 0, it rises by an amount give by L(l — cos#). See Fig. 1-4. To hold the bob in this position requires that a force F be applied as shown in Fig. 1-5. The weight of the bob is mg. If the tension in the rod is T, the equations for static equilibrium are F sin 9 -f T cos 0 = mg

TsmO - FcosO = 0

Fig. 1-4 Simple pendulum.

(Ha) (lib)

MODELING OF DYNAMIC SYSTEMS

Fig. 1-5 Forces on a displaced pendulum.

Multiply Eq. (1 la) by sin 9 and Eq. (1 Ib) by cos 0 and subtract: F = mgsmO

(12)

If 0 is small, sinO = 0 and F — mg9. F can then be assumed to be horizontal. If F is removed, the tension applied to the body by the rod exerts an unbalanced force equal to F horizontally to the left, that is, opposite to the direction of motion. The equation of motion is, therefore, Q

(13a)

Since x = L@, this equation can be written as mLO + mgO = 0,

0 + -0 = 0

(13b)

or jc + -x = 0

(13c)

For the second part of the problem, the mass of the rod, mi, can be assumed to lie at a point halfway between the ends. The restoring force associated with this mass is m2gO Because the forces do not act at the same point, we sum moments about the rod support and set this equal to For the bob, / = ml2. For the rod, / = w 2 L 2 /3. The sum is

The torque due to the bob is mgLO, whereas that due to the rod is rri2gOL/2. Thus, the equation of motion for the combination becomes

L2(m + ro2/3)0 + (mgL + m2gL/2)0 = 0 or m + w 2 /3 J L

This, of course, reduces to Eq. (13b) if w 2 /w n - 1.

MODELING OF DYNAMIC SYSTEMS

29

is 14141 —

Thus, the decaying exponential has squeezed the sine wave effectively to zero in less than a quarter-cycle. 1 .5.2 Example: Natural Frequency of a Cantilever Beam Loaded at Its Free End Consider the cantilever beam loaded at its free end as shown in Fig. 1-10. The deflection of the free end of this beam is, according to the theory of strength of materials,

3EI

(79)

Following more or less conventional notation practice, y is the deflection, E the modulus of elasticity, / the moment of inertia, and mg the weight of the block. If k is to represent the proportionality between load mg and deflection y, we have 3EI

(80)

Since the natural frequency is defined by yfc/m, its value for the present case is con =

I3EI mL 3

(8 la)

or, in hertz,

- JL I3EI3

^"^V^Z

(81b)

1.5.3 Example: Period of a Cylinder Oscillating on a Cylindrical Surface A cylinder of weight w and radius r rolls without slipping on a cylindrical

surface of radius R as shown in Fig. 1-11. Find its period of oscillation.

m L —*

Fig. 1-10

Cantilever beam loaded at its free end.

30

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Fig. 1-11 Cylinder on cylindrical surface.

Comment: This problem is not the same as that for the simple pendulum because of the angular motion of the cylinder. Some of the restoring force provided by gravity which normally would be used to translate the cylinder, goes instead to changing its angular velocity. As a result, the frequency of translational oscillation is lower than that of the pendulum, Solution. For this problem we will multiply Newton's second law by djc/df: duck dx m — — = F— dt dt dt

(82)

Setting dx/dt = v on the left side and multiplying by dt yield mvdv = Fdx

(83)

Integrating both sides then gives

mv2

= I Fdx

(84)

a relationship of perfectly general validity. The left-hand side is the familiar expression for the kinetic energy of the cylinder. We have chosen this form of Newton's law because the cylinder has both translational and rotational motion and it is often easier to sum these effects in the form of energy that is a scalar quantity. Just as for the simple pendulum, the restoring force is — mg

while the displacement of the cylinder centroid is (R - r)dO The velocity of the cylinder center is (R - r)9

so that its kinetic energy of translation is ~[(R-r)9}2

MODELING OF DYNAMIC SYSTEMS

31

The rotational velocity of the cylinder about its own axis we designate as 0. In a fixed frame of reference, the rotational velocity is 0 — 0. But since r0 = R0, D

0 = -0

(85)

r

The moment of inertia of the cylinder about its own axis is mr2

so that the kinetic energy of rotation is

\ i2

^2 r / j? Summing the two energy components yields

|02 \(R - r)2 + (fl - r) 2 =

(R - r)2e2m

(86)

The right-hand side of Eq. (84) may now be written as

- / mgs'mO(R - r)d0 Taking the derivative of both sides with respect to time results in the expression 3 -m(R-ry2ee = -mgsinO(R - r)0

(87)

Dividing by m(R — r)0 gives 3 -(R — r)0 + g sinO = 0 2

or -r)

sin oo, it becomes a relatively simple matter to complete the plot of the steady-state portion of xk/FQ as a function of a)/a)n with f as a parameter. Such a plot is shown in Fig. 1-12. 1.7 Examples of Single-Degree of Freedom Forced Vibrations In this section, we will be concerned with developing the differential equations for simplified models of physical systems. We will not be particularly concerned at this point with calculating solutions of these equations, although the techniques

38

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

discussed earlier will serve us adequately. In the next chapter, however, we will discuss in considerable detail general methods for solving such equations and describe the use of computer programs that utilize these methods to solve linear differential equations. 1.7.1 Example: Simplified Vehicle Suspension Here, we will consider only the vertical motion of a single wheel, or that of a solid axle with a wheel at each end, supporting a mass, as depicted in Fig. 1-13. We will consider the wheel mass to be negligible when compared to that of the mass being supported. The wheel (or axle) is connected to the mass by a spring and a linear damper (shock absorber). We will assume that the road consists of sinusoidal undulations having an amplitude yo. The equation for the road is, therefore, y = yo sin( V/L)t, where V is the velocity of the vehicle and L is the distance from the beginning of one period of the sine wave to the beginning of the next. Write the differential equation describing the vertical motion of the vehicle mass.

Solution. The vertical position of the mass in a fixed frame of reference is x. Hence, the product of mass and acceleration is mx. The spring force, however, depends on the extension of the spring, which is x — y, and the damping force depends on the rate of change of the extension, x - y. But Vt

y = y0 sin —

and

Vt The required equation of motion is then

Vt cV Vt mx = —kx — ex + y§k sin — + yo — cos — Li

or

LJ

Li

Vt cV Vt mx + ex + kx = yok sin — + yo — cos — Li

L

L

Fig. 1-13 Simplified vehicle suspension, stage 1.

MODELING OF DYNAMIC SYSTEMS

39

1 . 7.2 Example: Simple Pendulum Restrained by Spring and Damper Suppose we have a system such as the one represented in Fig. 1-14. Assume that the rod holding the pendulum is infinitely stiff but massless. Write the differential equation of motion for the pendulum bob with the forcing function represented by

Solution. A product of the moment of inertia of the mass about its pivot point and its angular acceleration is equal to the unbalanced torque. The moment of inertia of the bob about the pivot is mL2 if we ignore the moment of inertia of the bob about its own centroid. Two forces are applied to the pendulum rod: one due to the spring, which is L\0k, and one due to the damper, which is L20c. In addition, there is a restoring torque due to the change in height of the bob which, for small 0, is given by mgOL. Newton's second law is now written as

mL20 = -L26k - L2cO - mgOL + f ( t ) or

(1 14a)

Ll2cO + L\kO + mgOL = f ( t )

which is the required result. As a matter of interest, we can compare the oscillation frequency of the bob with that found for a simple pendulum, Eq. (50). We find the damped natural frequency of the oscillation by setting the right-hand side to zero and using the quadratic formula to find the roots of the characteristic equation*: -Ljc 2m L2

L\k+mgL~\ 2

2mL

or

Fig. 1-14 Simple pendulum restrained by spring and damper. *We will examine the reason this can be done, as well as what is meant by the characteristic equation, in the next chapter.

40

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

from which

(

}

the damped natural frequency. This arrangement differs from the simple pendulum in that L i and c are nonzero. With these set to zero, we see that

the value for a simple pendulum. This exercise illustrates a practice that should be an integral part of each engineer's problem-solving methodology: Check for agreement with the known results of limiting cases. If the results cannot be reduced to the expected expression when the parameters that cause the configuration to differ from the simple case are set to zero, there is obviously a mistake in the development. On the other hand, correct reduction to the expected results does not necessarily guarantee that the expression for the more general case is correct. 1.7.3 Example: Dropped Package A package having a mass m is dropped from a height h. The package has some resiliency, which will be modeled by a linear spring with constant equal to k. There is also some internal damping, which we will represent by a linear damper. Find the equation describing the motion of the package starting with the moment of contact with the ground. Use Fig. 1-15 as the basis for your analysis.

Solution. The force of impact is transmitted to the mass through the spring and the damper. Following the procedure developed previously, we write our differential equation as

mx + cx + kx = 0

(116)

In this case, the initial conditions are not zero. There is an initial velocity, which we can find, assuming no air resistance, by observing that h=l-gt2

y

Fig. 1-15

(Ilia)

Falling package.

MODELING OF DYNAMIC SYSTEMS

and

41

(nib)

i>o = gt

From Eq. (117b), t = VQ/g. Thus,

or

In the solution to Eq. (116) represented by Eq. (106), we must set FQ = 0 and X(Q) = 0. We obtain immediately

(119) Since

1 m

and

then (120)

where

The solution, Eq. (119) or (120), describes the package upon hitting the ground as a damped oscillating system. This is a realistic description as long as the internal forces that keep it together as a single system are not exceeded by the force kx. Note also that we have neglected the deformation of the package due to its own weight.

42

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

1.8 Coupled Systems: Examples of Two-Degree of Freedom Forced Vibrations Often, the model that seems more appropriate for representing some physical system requires that a mass move in two different directions at once or that two or more masses move in the same direction in such a manner that the motion of one mass influences the motion of the other. We can generalize the model shown in Fig. 1-13 to represent a vehicle suspension to include both front and rear axles, as shown in Fig. 1-16. Such a model, we might imagine, gives a better representation of physical reality than the simpler one shown in Fig. 113. But now the mass rotates through an angle as well as translates. For this reason, we will need two equations to describe the two degrees of freedom of the mass. We will write these in terms of the variables x and 9. These equations will, in general, be coupled; that is, the vertical motion will affect the angular motion, and vice versa. The angular motion can be thought of as being about the center of gravity, and so we will use the e.g. as the point for which we write our equations. The body has a mass m and a vertical displacement x. The sum of the unbalanced forces applied to the body is equal to mx. By unbalanced forces we mean those that exist after the weight has been supported by the springs. As in Fig. 1-13, the forces exerted by the springs and dampers depend on their extensions, which will not, in general, be the same. We can write an equation expressing this equality as

mx = -ki(x - L{0 - vi) - ci(x - L\0 - y\) -k2(x + L20 - v2) - c2(x + L20 - y2)

(121)

We can also take moments about the e.g. and equate them to the product of moment of inertia about the e.g. and the angular acceleration. This gives JO =k}(x-

- k2(x + L20 - y2)L2

(122)

Fig. 1-16 Simplified vehicle suspension, stage II.

MODELING OF DYNAMIC SYSTEMS

43

As a system of two differential equations in two variables, x and 0, we have

mx + (ci + c2)x + (ki + &2)* + (L2c2 - Lic00 + (L2£2 - £i*i)0

(123a)

- y\k\L\

(123b)

for the differential equations describing the motion. Note that, in general, there can be two forcing functions for each equation. Recall that forcing functions are generated by the motion of each wheel with respect to the inertial reference. Each wheel can generate both a rocking motion and a vertical motion, all four of which can be applied to the body at the same time. It is obvious that problem complexity increases considerably when the vehicle is represented as having two degrees of freedom. Since an actual vehicle, such as an automobile, is many orders of magnitude more difficult to describe mathematically, it is important that engineers always look for the simplest model that will exhibit the behavior they seek to study. Systems of equations like (123) are considerably more difficult to solve than are single equations. We will study two techniques in detail in the next chapter. In both cases, it is usually simpler to carry out the operations if numerical values are substituted for the symbols in the equations before the solution is undertaken. Typical numbers for a vehicle like the one shown in Fig. 1-16 might be: k\ k2 LI L2

= 250.0 Ib/in. J = 30186.335 lb-s2 - in. = 270.0 Ib/in. m = 10.35 Ib-s 2 /m. weight = 4000 Ib = 48.0 in. c\ = 10.0 Ib-s/in. = 72.0 in. c2 = 12.0 Ib-s/in.

where the units are those often employed by automotive engineers. With these numbers inserted in Eq. (123), we obtain

x + 2.1256* + 50.24* + 3840 + 718.846* = (lOjh + 12J2 + 250y{ + 270j2)/10.35

(124a)

for the first equation and

0 + 2.824(9 + 65.45(9 + 0.13169* + 0.24647* = (864j2 - 480ji + 19440j2 - 12000^0/30186.335

(124b)

for the second equation. Most of the problems we will undertake to solve in this book may be termed single-input, single- or multiple-output problems. Another way of saying this is that, for each system we consider, we will permit only one excitation, for which we will then determine the time histories of all the dependent variables. For example, an aircraft in lateral-directional flight has three degrees of freedom, represented

44

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

by the lateral velocity, the bank angle, and the yaw angle. We can calculate the way each will change with time if either the rudder or the ailerons are moved.

To apply to the present problem the solution methods that we will develop in the next chapter, we can take advantage of the principle of superposition, first to excite y\ alone and solve for x\(t) and Q\(t)\ then, to excite y2 alone and solve for ;c2(0 and 02(0; and, finally, to add the two solutions to obtain the result of the dual excitation. Notice that we have not yet specified the functional form of the excitation of either "wheel." It could be harmonic, as in the Example, Sec. 1.7.1, with continuous undulations in the "road" so that the phase relations between the excitations are dependent on the "wheelbase," or it could be of an entirely different nature, such as a "curb" encountered first by one wheel and then by the other. We will consider the mathematical representation of other types of excitation in the next chapter. We will now move on to consider some simpler examples of two-degree of freedom problems. 1.8. 1 Example: Double Pendulum For the arrangement shown in Fig. 1-17, develop the homogeneous equations of motion. Solution. We begin by assuming that both 0\ and 92 are small enough that sin 0i = 0i sin 02 = 02 cos 0i = 1 cos 02 = 1

We seek to write Newton's second law for each mass in the form JO = torque. For mass 1 , the effective restoring torque created when both masses are elevated as a result of rotation is (mi The moment of inertia of mass 1 about its hinge point is simply

m\L\

Fig. 1-17 Double pendulum.

MODELING OF DYNAMIC SYSTEMS

45

Since ra2 effectively adds mass to mi insofar as it determines the torque necessary to move the combination through 9\ , we write (mi

as the net value of J .

We now ask the question: If 9\ is zero, what is the force on mi due to motion by mi that can cause mi to rotate? A force balance of the rod connecting mi and m 2 shows that if there is a horizontal force equal to L 2 m 2 0 2 causing m 2 to move, there must be an equal and opposite force applied to m\. Thus, we have, for the equation of motion of mass 1,

or

(mi + m2)L2lOi + ro 2 L 2 Li0 2 + (mi + m^gliOi = 0

(125a)

(mi + /n 2 )Li0i + m 2 L 2 0 2 + (mi + m2)g6>! = 0

(125b)

For mass 2, the restoring force due to elevation of the mass is m2g i. We could also have noted that the absolute velocity of m 2 is LI#I + L2#2, so that the acceleration in a fixed reference frame is L\0\+ L2#2. The equation of motion for mass 2 is, therefore,

(126) 1.8.2 Example: Horizontal Mass-Spring System with Suspended Simple Pendulum Suppose we have the system depicted in Fig. 1-18. Write the equations of motion. We might use such a model to describe the motion of a partially full railway tank car connected by the spring to other cars in the train. The fluid in the car has no viscosity.

—> x

AAAAA/M

M

Fig. 1-18 Sliding mass-spring with suspended pendulum.

46

is

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Solution. For the pendulum, the absolute velocity in the horizontal direction LO

so that the acceleration is x + LO. For small 0, the restoring force acting on m is mgO. The equation of motion of the pendulum bob is, therefore, or

m(x + LO) = -mgO

(127a)

x + LO + gO = 0

(127b)

The effective mass being acted by the spring is m -f M. The action of the pendulum introduces an additional force on m equal in magnitude tomLO so that the equation of motion for M is (M + m)x + raL# + kx = 0

(128)

1.8.3 Example: In-Line Mass-Spring-Damper System For the system depicted in Fig. 1-19, write the equations of motion. The mass is excited sinusoidally. Solution. We write an equation of motion for the plate connecting k\ to k2 even though we will later assume that mass to be zero. We find it convenient to do this because we need the position of the plate to determine the forces due to k\, k2, and c. The forces acting on the fictitious mass are —k\x\, —cx\, and k2(x2 ~~ *i)Thus, the equation of motion is or

m\x\ = —k\x\ — cx\ + k2(x2 — x\)

(129a)

m\x

(129b)

cx\ -

= 0

For the mass, we have only the force —k2(x2 — x\), so that its equation of motion is

x2 + k2(x2 — x\) = FQ s'mcot

Fig. 1-19 In-line mass-spring-damper.

(130)

MODELING OF DYNAMIC SYSTEMS

47

1.8.4 Example: Harmonically Excited Mass-Spring with Suspended Mass-Spring, a System of Two Coupled Equations A system is represented by Fig. 1-20. Write the equations of motion.

Solution. For M, we have Mx\ = — k\x\ + k2(x2 — x\)

(131a)

Afjci + (k\ +k2)x\ -k2x2 = Fosinoot

(13Ib)

mx2 = —k2(x2 — x\)

(132a)

mx2 4- k2x2 - k\x\ = 0

(132b)

or

and, form, we have or

Discussion. Because the steady-state solutions of these equations illustrate a rather interesting application of engineering analysis and are easily obtained by classical techniques, we will depart in this section from the practice of developing only the descriptive equations and proceed to obtain their solutions. Take as trial solutions jci = A sin cot and x2 = B sin cot. Then,

M(—Aco2 sin cot) + (k\ + k2) A sin cot — k2B sin cot = FQ sin cot (133a) m(—Bco2 sincot) + k2B sincot — k2A sin cot = 0 (133b) Dividing by sin cot yields

-MAoo2 + (Jti + k2)A - Bk2 = F0 9

-mBoo2 + k2B -

F sinwt

Fig. 1-20 Two coupled mass springs.

(134a)

48

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

or

A[-Mco2 +k{+ k2] - Bk2 = FQ 9

B[-ma)2 + k2] - k2A = 0

(134b)

Multiplying the first equation by k2 - mco2 and the second by k2 gives A[ki + k2- Mco2][k2 - ma)2] - k2(k2 - ma)2)B = FQ(k2 - ma)2) (135a) Bk2(k2 - ma)2) -k^A = 0 (135b) We can then add in order to solve for A:

A = ———————2-—————2-——2[ki +k2- Ma) ][k2 - ma) ] -k

(136a)

from which B is easily found: B=

———————— 2 ———— 2——— 2 [fci + k2 - Mco ][k2 - m(D } - k

(136b)

Equation (136a) shows that the vibrations of a mass-spring system can be effectively damped if a second mass m and spring k2 are hung below the original mass and ^/(k2/m) is the frequency at which the system is being forced. This is the idea behind one type of vibration absorber. We will have more to say regarding this concept in Chapter 3.

1.8.5 Example: Two Pendulums on the Same Axis Connected by a Torsional Spring In Fig. 1-21 below are two pendulums pivoted about a common axis. They are each connected to one end of a linear torsional spring. Initially 0\ = —02. Both are released simultaneously. Write the differential equations for the resulting motion.

Solution. The equations are similar to those we derived earlier, that is,

except that there is now a term — (01-02) added to the first equation and a term — 02-01) m2

MODELING OF DYNAMIC SYSTEMS

49

Torsional Spring With Spring Constant k

Fig. 1-21 Two pendulums on the same axis connected by a torsional spring.

added to the second. Thus,

g LI

k mi

(137)

§2 + --02+—— (02-0^=0

L2

m2

The nonzero initial conditions, 0\ and #2 at time = 0, are employed to evaluate the constants in the solution and are not part of the formulation of the equations. 1.8.6 Closure Except in one case where the solutions were easy to obtain and we wished to point out a significant conclusion that could be drawn from them, we did not attempt to solve the systems of differential equations we generated in the foregoing example problems. There are several reasons for not doing this at this time: (1) Complete solutions, following only those methods already discussed, are obtained rather laboriously. (2) We have some more universal techniques available that form the basis of computer programs. (3) We will use these techniques to solve the problems both manually and using the computer programs in the next chapter. (4) We will discuss the significance of the results at the time we obtain them. Before moving on to consider these solution techniques, however, we will attempt to improve our skills at problem formulation by treating systems that

50

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

require three, four, or even five equations in order to describe them properly. Students should appreciate that, with the availability of computer programs, problems that cannot be attacked manually because of their computational complexity can now be solved rather routinely. However, there is an old adage in the computer business: If one puts an incorrect formulation or incorrect numbers into the program, the results will be meaningless ("Garbage in equals garbage out"). Thus, because of the increased potential for undertaking more complex problems, the use of computers makes it even more important that students grasp firmly the principles involved in formulating the differential equations to be solved and that they be able to devise highly-idealized limiting cases, the solution of which they already know or can find easily by hand, with which to check the reasonableness of the computer solution.

1.9 Examples of Multi-Degree of Freedom Systems The following ten examples show how certain physical systems can be modeled by sets of three, four, or five simultaneous linear differential equations. The arguments necessary to apply Newton's second law of motion to the individual cases are given or at least indicated. The student should observe that certain types of problems can generally be approached in similar fashion, regardless of the number of degrees of freedom. Although this chapter deals with only the formulation of the mathematical description of engineering problems, discussions of the physical significance of the computer results have also been provided. Inputs to the computer program (SOLVE) needed to obtain these results are also given. Students should not concern themselves with the mechanics of the solution technique at this time; this will be covered in detail in succeeding chapters. Rather, solutions are provided at this point to help students acquire an understanding of the physical behavior of the systems described by these mathematical models. Later, students should employ this understanding to determine the extent to which a relatively simple linear model will describe the system they wish to analyze. Some of the exercises in the use of the computer program will refer to these examples; for these students will be able to check their results easily.

1.9.1 Example: A Three-Degree of Freedom System Determine the position of the three masses in Fig. 1-22 as functions of time if Wi = l , m 2 = l , / W 3 = l, and k = I and all three masses are excited impulsively with an amplitude = 10. The units for m are kilograms in the Systeme International (SI) and slugs in U.S. Customary Units (USCU); for k, the units are newtons per meter in SI and pounds per foot in USCU; for the impulse amplitude, the units are meter-seconds in SI and feet-seconds in USCU. The corresponding displacements are then in meters and feet. Note that the computer program does not check units. It assumes that all the input data for the same problem use consistent units. Solution. As usual, we can write the equation of motion for mass 1 as

-kxi + k(x2 - jci) + k(x3 - *i) + 105(0

MODELING OF DYNAMIC SYSTEMS

51

Fig. 1-22 Three-mass, three-spring system.

or

m\x\

The equations of motion for masses 2 and 3 are obtained in similar fashion: X2 — — k(x2 — x\) + k(x^ — ^2) — kx% + 106(0 — JCi) — k(x?> — X2) -

Rearranging, we obtain, for the system,

m\x\ + 3kx\ — kx2 — kx^ = 106(0 -kx3 -kxi = 106 (t) 1712X2 -kxi -kx2 = 106(0

(138a) (138b) (138c)

Discussion. With the appropriate numerical values inserted, the coefficient matrix of Eqs. (138), when transformed into the Laplace domain,* is -1 (139) *The student should accept this statement without proof at this time. Transforming differential equations into algebraic equations by the method of Laplace transforms as the first step in their solution will be treated in detail in Chapter 2.

52

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

SOLVE then yields

a>\ = 1 .0 rad/s (02 - 2.0 rad/s coi = 2.0 rad/s

Notice that two roots are the same. When two or more roots are the same, the system is said to be degenerate. In this example, all the masses are excited impulsively, and the system thereafter vibrates freely, that is, without being forced. As a result of the degeneracy, all the masses oscillate at the same frequency, 1 .0 rad/s, in phase.

7.9.2 Example: A Three-Degree of Freedom System Given the system shown in Fig. 1-23, determine the displacement as a function of time of the three masses, assuming that mass 1 is excited by a force FQ sin cot in the ^-direction. The car wheels are frictionless on the contact surface. Assume that k\ = k2 — £3 — £4 = 1, c\ — c2 = c^ = €4 = 1, m\ = m2 — m^ = m$ = I in some compatible system of units. FQ = 1.0.

Typically, k is in newtons/meter, c is in newton- seconds/meter, m is in kilograms, and FO is in newtons. The displacement jc will then have units of meters. In both SI and USCU, Jk/m is in radians/second.

Solution. Mass 1 has applied to it two forces on the left, —k\x\ and —c\x\, and two forces on the right, k2(x2 - x\) and c2(x2 - *i)- The equation of motion for mi is, therefore, m\x\ = —k\x\ — c\x\ + c2(x2 — x\) + k2(x2 — x\) -f /(/) or

+ (fci -f k2)xi - k2x2 - c2x2 = f ( t )

(140a) (140b)

Fig. 1-23 Horizontal three-degree of freedom mass-spring-damper system.

MODELING OF DYNAMIC SYSTEMS

53

For mass 2, we have two forces on the right, £3 (.£3 — x2) and A^fe — *2), and two forces on the left, — k2(x2 — x\) and — c2(x2 — x\). Thus,

m2x2 = —k2(x2 — Xi) — c2(x2 — xi) 4~ k^(x^ — x2) -f or tf*2*2 + (C2 H~ C-$)X2 + (k2 + ^3)^2 — k2Xi — &3*3 — C2Xi — €3X3 = 0

For mass 3, all the forces can be assumed to act to the left. We have then

— x2) — cifa — i2) — £4*3 — C4^3

(142a)

or i -f (^3 + ^4)^3 — A:3jc2 — C3jc2 = 0

(142b)

Discussion. In the Laplace domain, the coefficient matrix is ^2) c2S - k2 0

-c2S-k2 m2S2 + (c2 4- cs +(£2+^3) -c3S-ki

0

(143)

With the numbers substituted, this becomes

-S-l 0

0 S-l

(144)

-S-l

and the right-hand side terms are £(1, 1) = 1.0, 5(2, 1) = 0.0, and B (3, 1) = 0.0 With these data, SOLVE returns:

= 0.7653668647 rad/s, = 1.4142135624 rad/s, = 1.847759065 rad/s,

= 0.3826834324 = 0.7071067812 = 0.9238795325

The output is characterized by a well-damped transient response that is almost indiscernible on top of the steady-state response. Since all the transient modes decay, the steady-state response is characterized by a single frequency, 1.0 rad/s. The amplitude of mass 2 is somewhat less than that of mass 1, and its motion lags that of mass 1 by about 0.5 s. Similarly, mass 3 has a smaller amplitude than mass 2, and its motion lags that of mass 2 by about 0.4 s.

1.9.3 Example: A Three-Degree of Freedom System Suppose one has the three pendulums shown in Fig. 1-24. If k = 1 Ib/ft and m = 1 slug, determine Oi(t), 02(t), and #3(0 when the leftmost mass is excited impulsively. Take a = 10.0 ft and g = 32.2. ft/s.

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

54 T

a

-i a

VwwVwvA Fig. 1-24

Interconnected pendulums.

Solution. We will restrict our consideration to small departures from the equilibrium vertical position so that sin 9 = 9. With this assumption, the horizontal displacement of mass 1 can be expressed as 2a9\ (See Fig. 1-25.) The inertial force associated with the motion of the mass is then given by m[d2(2aOi )/df 2 ] = 2am9\. Hence, the moment for this force about the pivot point is 4a2mO\. The horizontal component of the weight of the pendulum mass is mg sin 9\* so that the moment about the pivot point associated with this force is 2mga9\. The compression of the spring is, of course, a(92 - 0\), so that the moment of this force about the pivot point is ka1(92 — 9\). This last moment is counterclockwise, whereas the other two are clockwise when the mass motion is to the right. The sum of these torques must be zero at all times t. Thus, the equation of motion of mass 1 is (2mga

- ka202 = f ( t )

(145)

The other two equations are obtained in a similar fashion and read

4a2m92 + (2mga + 2ka2)02 - ka29i - ka293 = 0 (2mga + ka2)93 - ka292 = 0

(146) (147)

Discussion. With the numerical values substituted, the coefficient matrix in the Laplace domain becomes

Fig. 1-25 Free-body diagram of pendulum. *A free-body diagram of the pendulum mass shows that to hold the mass in the deflected position, a force F must be applied. Summing forces gives V : Pcos6> + Fsin2 or 0*3 because of the overwhelming influence of o)\. In free vibration, for example, the oscillation at o)\ is responsible for over 90% of the total motion. This illustrates the limitation of a lumped parameter model such as this for representing the vibrations of continuous fixed-fixed beams. Only the fundamental mode of the beam is obtained with reasonable accuracy using a model with this number of masses. Other modes can be expected to be defined as the number of masses in the model is increased.

1.0 r—r—r

-1.0

i—i 1.0

-1.0

10

TIME, seconds Fig. 1-29 Three-mass string system, excited sinusoidally. W = 7.65367 rad/s. Solid = mass 1; dots = mass 2; dashes = mass 3. Displacement of masses 1 and 3 are identical.

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

58

1.9.5 Example: A Three-Degree of Freedom System Suppose we have a triple pendulum like the one in Fig. 1-30. Find the position of the three masses as functions of time if mass 3 is excited impulsively with a magnitude equal to 100.0 Ib/s. Let mi = mi = m 3 — 1 slug and LI = L2 = L 3 = 10.0 ft; g = 32.2 ft/s2. Solution. For this problem, there are two forces that act on a mass in the horizontal direction: the gravitational or restoring force, which attempts to move the mass back to its equilibrium position, and the inertia, which is the force necessary to place an accelerating body in static equilibrium. The magnitude of the restoring force that acts on each of the masses may be deduced from Figs. 1-311-33. F acts in the direction opposite to the restoring force. The force necessary to move the masses to the right in Fig. 1-31 is L\(m2 + m2 -f- w 3 )$i. If mass 2 is moving relative to mass 1, there is an additional force L2(m2 + m3)6*2. Similarly, if mass 3 is also moving relative to mass 2, there is a third force, L 3 m 3 6> 3 . Hence, the equation of motion for mass 1 becomes LI (mi

L 2 (m 2 + w3)02 + L3m36>3 + (m\

=0 (154)

The force necessary to move mass 2 to the right, as we have seen, is L 2 (m 2 + mi)02. If mass 3 also moves relative to mass 2, we have an additional force L 3 m 3 0 3 . For mass 2 to move, however, mass 1 much also move. The force necessary to move mass 1 is an additional force applied to mass 2 and is given by LI (mi -f m2}9\ . The equation of motion for mass 2 is then

L\(m\ + ro2)0'i -f L 2 (m 2 + m3)3 + (m2 + m3)g02 = 0

(155)

Following similar arguments, we may obtain the equation of motion for mass 3: -f- L2m26>2 + Lim^ + m3g03 = f ( t )

Fig. 1-30 Triple pendulum.

(156)

MODELING OF DYNAMIC SYSTEMS

Tsine,= F T 0056,= (m1 + m 2+ m 3)9

g(m 1+ m2 + m3) F1 = g(m^ + m 2+ m3 ) 6.,

Fig. 1-31

Forces on mass 1 triple pendulum.

Tsine 2 = F

T cosE^ = (m2+ m ^ g

Fig. 1-32

Forces on mass 2 triple pendulum.

F3 = g m 3 e 3

Fig. 1-33

Forces on mass 3 triple pendulum.

59

60

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Discussion. The coefficient matrix in the Laplace domain is then

S2L\(m\+m2) 2

(157) 2

S Llml

S L2m2

With the numerical values substituted, we have

30S2 + 96.6 20S2 10S2 20S2 20S2 + 64.4 WS2 2 2 10S 10S 10S2 + 32.2

(158)

The B vector in the Laplace domain matrix formulation is 5(1,1) =0.0 5(2,1) = 0.0 5(3,1)- 1.0

(159)

With this input, SOLVE yields

coi = 1.157062692 rad/s o)2 = 2.718010809 rad/s co3 = 4.500402556 rad/s As one might imagine with three natural frequencies fairly close together, the angular positions of the pendulum bobs as functions of time are not well represented by pure sine waves. Masses 1 and 3 respond primarily at 2.718 rad/s, while the response for mass 2 shows equal contributions from the 1.157 rad/s mode and the 2.718 rad/s mode, plus a contribution from the 4.5 rad/s mode that is twice as large as that of the other two. According to Figs. 1-31-1-33, positive displacement is to the right, the direction in which mass 3 is impulsively excited to begin the oscillations. It moves to a peak of 6.586 at 0.525 s. Mass 2 has receded from its first negative peak and has a displacement of -1.68 at this time. Mass 1 is moving toward its initial negative peak and has a displacement of -1.0176. It appears that masses 2 and 1 lag behind mass 3 and then begin to catch up. The complete behavior is shown in Fig. 1-34. The reader should not be too concerned with the magnitude of the numbers shown in Fig. 1-34. A large impulse was used to excite the system so that a greater number of digits would appear in the fixed-format printout. Actually, a response of 6.586 rad would indicate that the bob had executed more than a complete circle, a situation far outside the range for which a linear model is valid. However, the relative magnitudes of the mass positions are valid if the excursion of mass 3 is kept to about 0.1 rad. The linear model will yield the same relative magnitudes regardless of the size of the forcing function.

61

MODELING OF DYNAMIC SYSTEMS

2

3

4

5

TIME, seconds Fig. 1-34 Triple-pendulum system, mass 3 excited impulsively. Solid = mass 1; dots = mass 2; dashes = mass 3.

1.9.6 Example: A Four-Degree of Freedom System In Fig. 1-35, assume that all contacting surfaces are smooth and frictionless. Now, determine the position of all the masses as functions of time when mass 3 is excited sinusoidally by a force in the x direction given by f ( t ) = 1.0 sin lOf. The parameter values are m\ w2 ra3 ra4

= 6.5 = 3.0 = 1.0 = 1.0

ki = 1.0 k2 = 2.0 &3 = 6.0 k4 = 5.0

ci = 10.0 c2 = 5.0

in consistent units.

Solution. We begin by writing the equations of motion of the individual masses. For mass 1, there is a force due to the spring k\ acting to the left (negative direction) of magnitude k\x\. The force due to spring ki acts to the right and has a

I—>X1

I —>X 2

I —>

HM/WH M,

K

/// X

4

Fig. 1-35 Four-degree of freedom horizontal mass-spring-damper system.

62

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

magnitude given by k2(x2 - x\). The same is true with respect to the force due to spring £4. When m\ moves in the positive x direction the damper c\ acts to resist that motion with a force to the left of magnitude c\x\ . The equation of motion for mass 1 is, therefore,

m\x\ — -k\x\ + k2(x2 -

- *i) - c\x\

(160)

For mass 2, there is a force to the left whose magnitude is k2(x2 — x\) and a force to the right whose magnitude is ^3(^3 — ^2). The equation of motion is, therefore,

- x2)

m2x2 = -k2(x2 -

(161)

For mass 3, there is a force to the left whose magnitude is ^3(^3 — x2). Hence, the homogeneous equation of motion for mass 3 is

(162)

-x2)

Finally, for mass 4, there is a force to the left whose magnitude is &4(jc4 - x\) and a second force to the left whose magnitude is c2x4. As a result, the equation motion is

(163)

- x\) - c2x4 Grouping Eqs. (160-163) yields - k2x2 - £4*4 = 0

m\x\ -\- c\x\ H- (^i + k2 m2x2 + (k2 -f

(164a)

- k^ = 0

(164b)

—0

(164c)

-f

(164d)

= 0

Discussion. The coefficient matrix in the Laplace domain is easily obtained:

-*2

m 2S2 -f (k2

0

(165)

0

0

0

0

With the numerical values substituted, this becomes 6.5S2 + lOS + 8

-2 0

- 5

-2

0

-5

2

-6

0

_6

2

0

3S + 8

S +6

0

0

(166)

MODELING OF DYNAMIC SYSTEMS

63

When the forcing function is applied, the coefficients on the right-hand side become 5(1,1) = 0.0

5(2, 1) = 0.0

(167)

5(3, 1) = 1.0 5(4, 1) = 0.0 Although the frequency equation can be obtained manually for this system with

some difficulty, we propose to follow through to a complete solution using the

SOLVE program. With these numerical values, SOLVE yields the following natural frequencies: coi =0.70343146 rad/s o)2 = 1.710818 rad/s

o)3 = 2.8606497 rad/s Two additional roots are real (rather than complex), one at —0.0686275 and the

other at —3.7830143. This indicates that the damping provided has been sufficient

to suppress one of the natural oscillations entirely. However, even a system with all real roots will still follow an oscillatory forcing function. It is in the phase relationship between input and output that the effect of damping will usually be seen. Since the forcing frequency is much higher than any of the system's natural frequencies, we would expect the amplitudes of the resulting oscillatory motions to be relatively small. This is indeed the case.

4

6

10

TIME, seconds Fig. 1-36 Four-mass system, mass 3 excited at 10 rad/s. Solid = mass 1; dots = mass 2; long dashes = mass 3; dashes = mass 4.

64

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

The results are shown in Fig. 1-36.

1.9.7 Example: A Four-Degree of Freedom System Suppose we have the system depicted in Fig. 1-37. Let m = 1 and k = 1 in units such that k/m is expressed in rad/s. Assume that the left mass is excited impulsively in the y direction and that the right mass is excited impulsively in the x direction. What are the positions of the masses as functions of time? Solution. Each mass has two degrees of freedom and thus requires two equations of motion. If we assume that the displacements in each direction are small, we can decouple the motion in the y direction from those in the x direction so that they are independent. For the left mass, mass 1, the ;c equation of motion contains a force term 4kx, acting to the left and a second force term k(x2 — x\) acting to the right. Thus, we have

mx\ = —4kx\

— x\)

(168a)

or Jci + (5k)x\ - 2kx2 = 0

(168b)

The x equation for mass 2 is similar:

5kx2 — 2kx\ = 0

(169)

The y equation for mass 1 shows two forces, both acting down, applied to the mass:

my\ + 4ky\ = 0

(170)

The y equation for mass 2 is similar: yi + 6ky2 — 0

Fig. 1-37 Four-degree of freedom mass-spring system.

(171)

MODELING OF DYNAMIC SYSTEMS

65

Discussion. The coefficient matrix in the Laplace domain is then -2k 0 0

-2k S2 + 0 0

0 0 mS2 + 0

0 0 0

(172)

With numerical values substituted, we obtain

S2 + 5 -2 0 -2 S2 + 5 0 0 0 S2 + 0 0 0

0 0 0

(173)

The forcing functions for this system are then /3 (0 = 1.05(0 /2(0 = 1.05(0 /i (0 = 0.0 /4(0 = 0.0

(174a) (174b) (174c) (174d)

With these numerical values, SOLVE yields

coi = 1.73025 rad/s o)2 = 2.000 rad/s &>3 = 2.64575 rad/s 0. It does not refer to conditions that have existed for some

time prior to the initiation of our consideration, that is, for t < 0 because since the Laplace transform is not defined for such times or for the conditions existing at t = 0, that is, initial conditions. Application of the initial value theorem to a step function shows that the change in height occurs at t = 0. (The step function has an infinite slope at t = 0+ and a zero slope everywhere else.) 2.3.2 Short Table of Laplace Transform Pairs In the preceding section, we indicated the derivation of some simple Laplace transform pairs (the function in the time domain and its corresponding expression in the Laplace domain). Table 2-3 below lists these, as well as some others for which a derivation will not be given. A sketch of the function in the time domain is included for some of the pairs in the table.

2.3.3 Inverse Laplace Transforms When an expression in the Laplace domain (say, a differential equation that has been transformed from the time domain) is solved for the transform of the dependent variable, we may recognize that the inverse transform of this rearranged expression is the time solution of the differential equation. We illustrate this concept with the following example. Suppose that one has the differential equation c mx + ex + kx = —y -f ky (29a) with m = 100, c = 500, k = 1000, and y = 0.1 sin lOf. Assume that the initial conditions may be taken to be zero as well. Then, the Laplace transform of Eq. (29a) is

100S2* (S) + 500Sx (S) + lOOOx (S) = »3J^fnn i " A\J\J

(500)

U + (1000) c°2 o ~T~ !m A\j\)

METHODS OF SOLUTION OF LINEAR EQUATIONS OF MOTION

Table 2-3

Laplace transform pairs

F(S)

/(O

Unit double impulse [8 (t)]2

139

1

i(

o

^ t

////

Unit impulse 5(0

f

i

0

Unit step w(r)

i

1

1 Delayed unit step u(t — T)

-\

1

c

t

f!l s

X

Decaying exponential e

T Ramp t

(Table 2-3 continued on next page.)

140

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Table 2-3

/(O

te~

(continued) Laplace transform pairs

F(S)

(5

Continuous sine wave

Continuous cosine wave cos&tf

Decaying sine wave e~at sin CD/

Decaying cosine wave e~ai cos CD/

(5 + a)2 +

+ a)2 + w2

(Table 2-3 continued on next page.)

METHODS OF SOLUTION OF LINEAR EQUATIONS OF MOTION

Table 2-3

141

(continued) Laplace transform pairs

F(S)

/(O

t"e~

(S + a)n

sin a)t — cot cos cot

(S2 + co2)2

sin at ct(p2 - a2) sin fit

where x(S) is the Laplace transform of x ( t ) , and so forth. Rearranging and simplifying, we. obtain Q.5S+10 S2 + 100

(30)

or x(S) =

0.5S+10 (S2 + 100) (S2 + 5S + 10)

(31)

It is this form that we must transform back into the time domain in order to obtain x(t). The easiest way to do this is to search our transform pair table for a form that matches x(S). This would immediately indicate the correct form for x ( t ) . Failing that, we can split the right side of Eq. (31) into a sum of terms such that each term in the sum appears in the transform pair table. One such expansion for Eq. (31) is x (o) _ = ~

424

424

3.8 424

_45_ 424 15

(32)

Each of the four terms in Eq. (32) appears in Table 2-3. Inversion now yields* The reader may wonder why the two arbitrary constants that appear in the complementary solution when the equation is solved in the more classical manner are absent here. It is not because the initial

142

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

3.8

2.6

3.8 _5, 15 35.5 /T _5, —-e 2 ' c o s A / — r + —— A / —:e '' 424 V 4 424 V 15

(33)

If we substitute Eq. (33) into Eqs. (29), we obtain

3.8\/25\ /15~ 424JUJ C ° S VT'

^J 380 260 100 —— cos 10* + —— sin 10* + e 424 424

15 _ 5 , / 3 . 8 \ cos /T5r l(K/l5/3.8\ _55 ,sin /TJ' A/V ~r A/V —4 * VTTT 424/ 4 + ——— 4 V^rr 424K /

~ ~r^ 4

.

+ 500 -——

38

t

-e~2'cos J— f + J—

26

1

—• sin IQt - —- cos IQt 424 424 J

424

= 50.0cos I0t+ lOOsinlOr which shows that Eq. (33) is indeed a solution of Eqs. (29). The reader is encouraged to verify this result as an exercise. How did we find the numerator terms on the right side of Eq. (32)? The method is called a partial fraction expansion. We set up a sum of the denominator terms we wish to use with unknown numerators: 0.55+10

AS + £

CS + D

(34)

conditions are zero but rather because the particular solution is already included in Eq. (33). If the initial conditions were not zero, there would be additional terms in the numerator of Eq. (32) that would cause the coefficients of Eq. (33) to change.

METHODS OF SOLUTION OF LINEAR EQUATIONS OF MOTION

1 43

Then, we combine the right-hand side of Eq. (34) to obtain

(AS + B)(S2 +5S+ 10) + (CS + D)(S2 + 100) AS3 + 5 AS2 + WAS + BS2 + 5BS + 10 B CS3 + 1QOCS -f PS2 + 100D

The numerator of Eq. (35) must be equal to the numerator of the left side of Eq. (34). We can therefore write four equations to find the four unknown coefficients of the numerator of Eq. (35) by equating coefficients of like powers:

S3 : A + C = 0 S2 : 5A + B + D = 0 S : 10A + 5£ + 100C = 0.5 : IOB + 100D = 10

(36)

To solve,* first, divide the fourth equation by 10. Then, substitute the value for C from the first equation into the third equation. We have, then,

5A + B + D = 0 -90A+55=0.5 B + 10D = 1

(37)

If we now subtract the third equation from the first and subtract five times the third equation from the second, we get 5 A - 9 D = -1 - 90A - 50D = -4.5

(38)

then, we multiply the first equation by 18 and add the two equations. We obtain

-162D-50D = -22.5 or

D-^-45 ° ~ 212 ~ 424

(39) (39)

By back substitution of Eq. (39) into the third equation of Eqs. (37), we have

450 424

26 424

*For this purpose, we may also employ Cramer's rule, which is discussed in Sec. 2.3.4.

(40) V ;

1 44

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Then, we can find A from the first of Eqs. (37): l 26 45 A _ _1_9_ / 5_-_ _3_ '8_ j D] _\_ _ _ _ _ _1 A _ _l [r _BB _m

C is simply -A. Another method for finding the numerator coefficients in an expansion of

Eq. (31), called the method of residues, is better suited to computer programming. In this method, Eq. (3 1) is expanded in terms of its first-order poles, that is, factors of its denominator, in the complex S plane:

0.55* +10 100)(52 + 55+10)

KI 5-10./

K2 S+lQj

To evaluate the residues Kj at the zeros of the denominator, we multiply the left side of Eq. (42) by S + Pj and evaluate the resulting expression when S — — Pj•. Applying this procedure to Eq. (42), we have K\ =

0.5(10;) + 10 20 j (- 100 + 50; + 10) 5 + 10 200(-5 - 9j)

5; + 10 20; (507 - 90)

-200(5 + 9;) (5 - 9j)

50 + 45 - 657 _ 95 - 657 -200(25 + 81) ~ -200(106) 95 - 657 _ -95 + 657 21200 ~ 21200 0.5(-107) + 10 -207 (-100 - 507 + 10) 10-57 200(-5 + 97)

(43)

-57 + 10 -207 (-90 - 507)

(10-57X-5-97) 200(-5 + 97)(-5 - 97)

-50 - 45 - 657 _ -95 - 65 j 200(25 + 81) ~~ 21200

(44)

METHODS OF SOLUTION OF LINEAR EQUATIONS OF MOTION

0.5

-2-5 +

7 + 10

-

"2-5

"

145

2 7

/? '

!50 150

328.125 + 384.375 + 36.30921887.; - 3473.5819397

712.5 + 3437.2727; 159000

(45)

(46)

x(t) is obtained by recognizing that the inverse Laplace transform of a simple pole, Kj/(S + P/)> is Kje~pjt and that the right-hand side of Eq. (42) is simply a sum of four such poles. Each may be inverted separately and the result summed to yield =

-95 + 65; 10), -95-65; oionn oionn 21200 21200 712.5-3437.27277 _ 25 ,

10,,

————————————e e 159000 712.5 + 3437.2727J _ 2 5,

159000

e

6

(47)

Now,

fl5 15 15 15 \5 = cos,/—r + / s i n , / — r + cos,/ —f — / sin,/ —r = 2cos,/ —r V4 V4 V4 V4 V4

1 46

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

and e

( ) _ e- (

= 2j

sin

r

(48)

Through the use of Eq. (48), Eq. (47) becomes

-95(2)

712.5(2)

25 ,

15

, - 2 x • in 3437.2727(2)(y2) _ 25 , . ——21200 ° ) S m l 0 t ———— or

1425


(5) = 69.%8e(S) (0.00135 + 0.0235)^(5) + (S2 + 1.925)0(5) = 26.\8e(S)

(56)

For simplicity, first divide all three equations by 8e(S). Then, to solve for u(S)/8e(S), w(S)/8e(S)9 and 9(S)/8e(S), we apply Cramer's rule of linear algebra for the solution of three linear equations with three unknowns. We recall that the procedure consists of writing the unknowns as a ratio of two determinants. The denominator determinant is the coefficients of the three dependent variables. The numerator determinant is the same except that the coefficients of the forcing function terms have been substituted for the column containing the coefficients of u(S)/8e(S) when we want to solve for u(S)/8e(S) and so on for w(S)/8e(S) and 0(S)/8e(S).

u(S)

w(S)

-0.0016 32.2 0 69.8 (5+1.43) -6605 26.1 (0 0135 + 0.0235) (S 2 + 1.925) 32.2 (5 + 0.0097) -0.0016 -6605 (5+1.43) 0.0955 (0.0135 + 0.0235) (52 + 1.925) 0

(57)

(5 + 0.0097) 0 32 0.0955 69.8 -6605 0 26.1 (52 + 1.925) (58) (S + 0.0097) -0.0016 32.2 0.0955 (5+1.43) -6605 0 (0.00135 + 0.0235) (52 + 1.925)

METHODS OF SOLUTION OF LINEAR EQUATIONS OF MOTION

149

+ 0.0097) 0.0955

0(5) = _______0 8e (S) ~ "(5 + 0.0097) 0.0955 0

-0.0016 0 (5+1.43) 69.8 (0.00135 + 0.35) 26.1) -0.0016 32.2 (5+1.43) -6605 (0.00135 + 0.0235) (52 + 19.25)

2.3.5 Factoring the Denominator To prepare Eqs. (57-59) for inversion, we must first write them in the manner of Eqs. (50). The reader will be instantly aware that a lot of algebra is required to expand these determinants. We will simply state the results on the assumption that algebraic multiplication, addition, and subtraction are widely known, if distasteful skills. w(5) 8e (5)

0.111752-8105- 1149 5 + 4.21775 + 18.2965752 + 0.181375 + 0.07226 4

3

(60)

w(S) _ 69.853 + 1736Q.691452 + 168.3921515 + 80.2601013 8e(S) ~ 544.217753 + 18.2965752 + 0.181375 + 0.07226

(61)

(9(5) _ 26.QQ9246852 + 35.93498235 + 0.3501102 = 4 8e (5) 5 + 4.217753 + 18.2965752 + 0.181375 + 0.07226

(62)

A condition that a solution of Eqs. (55) exists is that the denominator determinants of Eqs. (57-59) be equal to zero. The expansion of this determinant set equal to zero is called the characteristic equation of the system. The roots, or zeros, of this characteristic equation describe the unforced motion of the system. It is the numerator of these ratios of polynomials that provides the additional information needed to distinguish the behavior of one dependent variable from that of another and that permits the effects of the forcing function to be included in the resulting motion. To find the roots of a characteristic equation of fourth-order manually is a tedious job. An appropriate technique, Lin's method, is described in Ref. 1. In this book, we will not ask the student to extract the roots of a fourth- or higher-order polynomial. The LOCUS portion of the SOLVE computer program will find the roots of a polynomial of order up to 99 very rapidly. There are exact methods for finding the roots of cubic polynomials. Again, however, these are quite tedious. Newton's method for finding the real roots of polynomials is relatively easy to use and converges rapidly. Its use with cubic polynomials is particularly appropriate because such a polynomial must have at least one real root. To use this method, we write

P7+1 j+l = PjJ - -4-4

(63)

where Pj represents the current guess at the root value, F(Pj) is the characteristic

equation evaluated at Pj, and F'(Pj) is the derivative of the characteristic equation

1 50

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

with respect to S evaluated at P/. The following example illustrates this process. Let the characteristic equation read 10 = 0

(64)

Its derivative with respect to S is 352 + 8S + 9 = 0

(65)

Suppose that we choose as our first guess for the value of a real root

S = -3

(66)

Then, the Newton formula would read

The next iteration then yields -12.7037037 + 21.7777777 -21 + 10 16.33333333 - 18.6666666 + 9 -1.92592593 6.6666666 = -2.044444 = —2.33 —

The third iteration gives

_ -8.545272971 + 16.71901234 - 18.4 + 10 Sj+l - -2.044444 12.53925925 - 16.3555555 -0.226260631 = -2.044444 - — Sj+l = -2.044444 + 0.43648387 = -2.000796056 By repeating the process, we can approach the current value of —2.0 with as much precision as desired. For Eqs. (60-62), the roots or poles of the denominator can be found to be 5! 52 53 54

= -0.0045114 + 0.0627563./ = -0.0045114 - 0.0627563; = -2.1043377 + 3.7183857; = -2.1043377 - 3.7183857;

(

}

Equations (60-62) also contain the term 8e(S) on the left side. Before the equations can be inverted, this must be given an explicit value and moved to the right side. If, for example, we wish 8e(t) to be a step function, then 8e(S) — A/S. A is the height of the step. If we take A = 0.01, then we multiply Eqs. (60-62) by 0.01/5 to get w(S), tu(S), and 6(S). This gives us^ve roots on the right-hand side

METHODS OF SOLUTION OF LINEAR EQUATIONS OF MOTION

151

that we must find residues for before we invert. The expansion for Eq. (60) then appears as follows: 39.774979- 1.66679107 S + 0.0045114 - 0.0627563;

39.774979 + 1.6667910; S + 0.0045114 + 0.0627563;

0.02585855 + 0.0054929145; 0.02585855 - 0.0054929145; S + 2.1093377 - 3.7183857; ~ S + 2.1093377 + 3.7183857; 79.498337 + 0.000024898;

(68)

The imaginary part of the last residue should be zero. That it is not indicates that the residue computations need to be performed with greater precision. We could also write Eq. (68) symbolically as u( } =

K\

+

K2

p2)

+

KT,

+

K^

+

KS

(S + P3) (S + P4) (S + PS)

}

The inversion to the time domain is then readily written symbolically as je-p>'

(70)

The relations for w(t) and 0(t) are similar with only the K being different for the different dependent variables. The K will also differ if the forcing function is different. 2.3.6 Discussion It should be evident that the process of solving a system of linear, inhomogeneous differential equations by the method of Laplace transforms is straightforward, with no intuition required as to the form of the solution. Unfortunately, as the reader has just witnessed, it is also very laborious and time-consuming. For this reason, engineers sought ways to deduce the essential features of the solution without actually having to solve the equation(s). And, although we shall study some of these techniques in later chapters for the insight they can provide into the effect of changing system parameters, we will also make use of computer programs that can perform the entire process of applying Cramer's rule to a system of up to five equations in the Laplace domain, finding the roots of the characteristic equation, finding the residues of these roots for each equation, evaluating its time solution at up to 2600 points and, finally, plotting these results via the printer or viewing them on the screen, all in the space of less than two minutes per equation on today's personal computer. We will discuss the use of these programs in a separate user's manual. There is an alternate method for solving a system of differential equations which, as will be shown, may take better advantage of the strengths of digital computers than does the method of Laplace transforms. This is the state transition

152

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

matrix method. We will illustrate this method by the following example. Assume that the system to be solved consists of two state differential equations:

x\ =* 2 *2 = -2*i -3*2 + r(t) In matrix form these equations can be written as

.

r o

11

» -* -3

The matrix immediately to the right of the equals sign we designate as A. The third matrix we designate as B. We then form the quantity

S -1 2 5+ 3 The inverse of this matrix is (SI -A)" 1 -

1 S2 + 3S 4- 2

rs +23 s11

\

L-

We define the inverse Laplace transform of the previous form as 1e~* — e~2t

e~* — e~2t

-2e-'+2e-»

-c- + 2*-*

The solution of the system of equations is then given by x(0 = >(t - r)Bdr

The first term represents the contribution from the initial conditions while the second term represents the contribution due to the forcing function. The integrals represented by the second term can be evaluated as follows.

and

- I'e-^ =-e-^ Jo 2

The net result is [0.5 + 0.5e~2t [ e-. As a result, the machine submits the supporting structure to a vibratory force in the direction shown, which

Fig. 3-1 Vibrating machine transmitting force to its environment.

161

1 62

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

is additional to the machine's weight. Let us represent the force generated by the machine in this direction as

F = F0sincot

(1)

What force is experienced by the rigid supporting structure? To determine this, we begin by writing the equation of motion of the machinery mass along the vertical axis x as

m x + ex + kx = FQ sin cot

(2)

As we have seen, the solution to this equation has a transient part and a steadystate part. The transient part decays at a rate expressed by the factor

We will assume that, when t > 6(2ra/c), this factor can be taken to be 0.0, which is to say that the transient solution has then decayed to zero (actually is less than 0.248% of its initial value). For t > 6(2m/c), we assume that x ( t ) contains only the steady-state solution*: F0(k-mco2)

. FQCCO sin cot — ————— -———— cos cot (k - mco ) + (ca)) (k - mco2) + (coo)2 2

2

(3)

t > I2m/c or

x = —-=====^====== sin(cot — 0) J (k — mco2) + (c&>)

(4)

where 1 ' / ik —^ mco2) J

^

The force applied to the machine by the springs is kx and that applied by the damper is ex. According to Newton's third law, these same forces are also applied in the opposite direction to the supporting structure. By the use of Eq. (4), it will be seen that the two forces are 90 deg out of phase; the magnitude of the force, FT, transmitted to the supporting structure is, therefore, FT =

m)

+

*The reader will note that Eq. (3) is the last two or the steady-state terms of Eq. (106) in Chapter 1 (which provides the complete solution for this problem).

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

163

or

where — mo)2) + (ca))2

is the amplitude of the machine's displacement as given by Eq. (4). The amplitude of its velocity, i m , is, therefore, _ Xm —

With these expressions, we may write FT as F^k

FT =

This can be rearranged to show the magnitude of the force transmitted to the supporting structure in terms of the magnitude of the force generated by the machine:

If we define ^k/m as the natural frequency con and c/2vkm as the damping ratio f , Eq. (6) can be written

(

a>\

\" 2*-^) ^"'/i / A

FT

r ^i

\

2

(7)

rd L

«J

A graph of Eq. (7) for various values of £ is shown in Fig. 3-2. In the figure caption, input refers to the force generated by the machine and output to the force felt by the support. The accompanying output minus input phase relationships can be developed from Eqs. (3) and can be shown to be the same as that given by Eq. (2). These relationships are depicted in Fig. 3-3. There are, of course, many situations in which we want to reduce transmission to the surroundings of the vibrations generated within a machine. We cite three

164

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

LLJ Q

FREQUENCY RATIO

Fig. 3-2 Input/output magnitude relationships, Eq. (7). Damping ratio 0.05 for greatest magnitude; other values: 0.1,0.15, 0.2,0.25,0.3,0.45,0.55.

FREQUENCY RATIO Fig. 3-3 Input/output phase relationships. Plot of Eq. (20). Damping ratio increasing downward.

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

1 65

examples: (1) Automobiles: By supporting the engine on specially designed compliant mounts, designers have isolated the engine from the vehicle frame, enabling passengers to enjoy a smoother ride; additionally, vehicle parts are not vibrated as often or as intensely, which extends their useful lives. (2) Submarines: Isolating the propulsion unit from the hull frames reduces the sound radiated into the ocean, making the submarine more difficult to detect. (3) Factories and offices: Workers become less productive and more prone to injury in environments where sound or vibration levels are high. Even when these high-level vibratory frequencies are subaudible (less than 63 rad/s), they can cause internal organs in the human body to vibrate in place, resulting in malaise, nausea, and other signs of distress. Thus, preventing vibrations generated by otherwise beneficial machines from intruding on our physical and mental health has become a pervasive engineering concern. Figures 3-2 and 3-3 indicate that, to reduce the transmission of vibrating forces from the machine to the environment, it is necessary to mount the machine on springs sufficiently soft that the square root of the ratio of their effective spring constant to the machine's mass is much less than 0.7 times the frequency at which the machine is vibrating, that is,

Jkjm « Q.la)

(8)

Another means of reducing the transmission of vibratory forces to the environment is to add mass to the machine. Notice, in Fig. 3-2, that, if co/a)n > \/2> an undamped spring is superior to a damped spring for reducing the transmission of force. Nevertheless, some damping is desirable to help limit the force excursion encountered when co passes through resonance (that is, when CD = a)n) after the machine is started. Stops are sometimes used to limit the displacement excursions that occur under these conditions. At resonance, the maximum displacement of an unrestrained system is found from Eq. (4) to be

Notice that increasing the machine's mass or reducing the constant of the supporting springs increases the amplitude of the machine's vibratory motion at resonance and lowers its natural frequency. Also, notice that the vibratory amplitude at resonance is inversely proportional to the damping. A system with no damping is said to be marginally stable because, once excited, it continues to oscillate with no means of absorbing the energy. At resonance, a system with no damping will continue to grow in amplitude once the excitation energy is added in phase to that already present in the system. One is often tempted, when mounting machines that generate considerable vibration, to fasten them securely to the floor of the building. The result is that frequently the entire building shakes perceptibly because the floor is not really as massive as one might imagine and the steel supporting the floor may be quite elastic. If some "tuned" machine support is not used, the usual practice is to mount the machine on a massive piece of reinforced concrete that is supported directly by the earth and isolated from the rest of the building structure.

166

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

The foregoing results in graphic form permit the relative transmission to be read directly. For that purpose, we first define

A=^ R

(Ida)

= l ~ Y0

(10b)

/ =£

(100

A is called the statical deflection since it indicates the compression of the supporting springs due only to the weight of the machine when it is not operating. R is a dimensionless quantity that might be termed the relative reduction; when positive and multiplied by 100, it is the percent reduction in vibratory amplitude of the machine. A negative value for the relative reduction indicates that the vibratory amplitude has been magnified by the choice of supports. Such a situation occurs when 0 < &>/&>„ < \/2. In Eq. (lOc), / is the vibratory frequency in hertz. We have shown in conjunction with Fig. 3-2 that the maximum reduction in vibratory amplitude occurs when f = 0 and when co/con > \/2. For this case, the amplitude of the sine wave in Eq. (4) can be written

x = (co2Fo/k m/k)

- 1

But, from Eqs. (10), xk/FQ = 1 - /?, co2 = (2;r/)2, and m/k = A/g. Using these definitions, the expression for the vibratory amplitude can be rewritten as

(2*/)2A or

(2;r/)2A Then, o\

\\

/ ? - ff\

Hz

(11)

Equation (11) plots as a straight line on log-log paper, with / as the ordinate and A as the abscissa for any constant value of the relative reduction R. Thus, if we know the frequency in hertz at which the machine has been vibrating, we can readily determine how much the vibratory amplitude will be reduced by the use of a specific supporting spring by measuring its statical deflection A with the machine in place and then consulting the graph.

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

167

The same graph can also be used to plot the relation between the vibrating frequency /, the statical deflection A, and the relative magnification — R of the vibratory amplitude. Such magnification occurs, it will be recalled, when c = 0 and a)/ct)n < \/2. By restricting co/con to values of 1.0 or less, however, we can avoid the ambiguity of having the same relative magnification at two frequencies. When a}/con < 1.0, the vibratory amplitude can be written as 1 - (a)2m/k)

Through the use of Eq. (10), we obtain

\-R =

1 (2;r/)2A

or -1 +

l-R

Then,

and

Note that, when R = 0 (that is, when the vibratory amplitude at the supports is the same as at the machine), the vibratory frequency is zero. When R = —1.0 (that is, when the vibratory amplitude at the supports is twice that at the machine),

Similarly, when the vibratory amplitude at the supports is 50% greater than at the machine (R = —0.5),

Thus, it will be seen that, if

then,

R = -oo

168

3.3

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Examples

The concepts of vibratory amplitude, the force transmitted by a machine to its supports, and the support motion felt by the machine are illustrated by the following examples. 3.3.7 Example: Vibratory Amplitude Suppose we have a machine that has a mass of 100 kg. It is supported by four springs, each of which has a stiffness of 175 kN/m. The machine generates an oscillatory force that can be described by the expression F = 350 sin(2 * 500 N. Determine the amplitude of the machine's vibratory displacement or motion and the effect on this displacement of installing a damper. The damper constant is 3347.2 N - s/m.

Solution. Springs in parallel, as in the present case, generate the same force as a single spring with a constant n times as great, where n is the number of springs in parallel. Thus, the total stiffness is 700 kN/m. The force applied by a 100 kg mass to a spring is (100)(9.81) = 981 N. The statical deflection of the springs is, therefore, 981/700,000 — 1.4014 mm, while the natural frequency of the spring-mass system, o)n, is (700, 000/100)5 = 83.666 rad/s = 13.32 Hz. From the expression given for the oscillatory force generated by the machine, we see that the system is excited at 50 Hz. When these data are substituted in Eq. (4), we find that the machine vibrates at 50 Hz with an amplitude of 0.038195 mm without the damper installed and 0.0379 mm with the damper installed. The effect of the vibration is to vary the support spring deflection from 1.3632 to 1.4396 mm with respect to their original or unloaded condition at a rate of 50 times/s. One can use such results, for example, to determine how much clearance must be provided in order to permit the internal parts of the machine to vibrate without their striking the machine's case.

3.3.2 Example: Vibratory Forces Transmitted Determine the force transmitted to the support by the machine in the preceding example. Solution. From Eq. (7), we obtain FT

— —

0

/1 + (2*0.2*50/13.32) 2

niQ 10.0 *&> V m or, for this example, when k > 100.0 * m * [2n * 50] = 9859.604 kN/m.

3.3.3 Example: The Inverse Problem—Support Motion Felt by the Machine Now, suppose the support is shaking and the machine is not operating. What force is felt by the machine? Solution. If the position of the support is y and the position of the machine remains ;c, the equation of motion becomes

mx = c(y — x) + k(y — jc)

(13a)

y = YQsina)t

(13b)

with

If we make the transformation

z =x - y

(14)

then Eqs. (13) become = ma)2YQsina)t

tnz + cz -f kz = — my

(15)

Equation (15) now has the same form as Eq. (2); we may therefore write the solution as 7 — —-

mco YQ

— sin(r/)f — 0) - ma>2)2 + (cco)2 |

K — war

,

(16a)

(16b)

If we let z = Zej(a)t-^ and y = YGeja)t and substitute them into Eq. (15), we obtain mZe-J+ejaa(ja>)2 + cZe~j) + kZe~j(l)ejOJt = eja)

or ^ [—ma)2 + jca) -f k\ = mco2YQ

170

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

from which

kt — mo/ + jcco

(17)

Since x = z + y = (Ze-'* + Yo) eju" =

I" ma? \_k — mco2 + jcco

+

k - mo? + jcail k — mco2 + jcco] (18)

. — mcoz + jcco

the steady-state amplitude is

X=

,

Y

(k - mco2)2 + (co))2

:Y0

(19)

while the phase angle is ~

i F

mcco3

-7———— [k(k - mco2^) ———— + (cco)2T]

The ratio X/¥Q is seen to be the same as FT/FQ in Eq. (7), indicating that the problem of isolating a machine from the motion of its support is identical to that of isolating the support from the vibratory forces generated in the machine. The difference is that when the vibration originates in the machine, the support "feels" a force but is assumed not to move. When the support moves as a result of an externally applied force, the machine responds with a motion whose amplitude is related to that of the support by Eq. (19). Since \co2X/co2YQ\ is equivalent to a ratio of forces, we see that the force transmitted to the machine relative to the force exciting the massless support is also given by Eq. (19). Although the analysis given in this section has been limited to consideration of a single mass moving in one direction, the principles can be applied to more general systems. 3.3.4 Rapid Determination of Natural Frequency and Damping Ratio If the graphic record of an oscillation is assumed to have been generated by a simple second-order system, that record can be fit mathematically by the expression

x = Xe-^»r sin V 1 - $2a)nt + 0

(21)

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

5

10

171

15

TIME, seconds Fig. 3-4 Decaying sine wave.

where X is the amplitude of the oscillatory envelope at t = 0 . Such a record is shown in Fig. 3-4. If we let x\ be the value of the first peak and *2 the value of the second peak, or the values of any two successive peaks, we may define a quantity 8 as (22a)

8 = L

The period of the oscillation r, as shown in Fig. 3-4, is related to a)n by T —

2n

(22b)

Through the use of Eqs. (22b) and (21), we may write Eq. (22a) as 8 =

Erv

X exp [ - £confa + r)] sin (V1 - f 2con fa + r] + 0

Since the angle n frl + T) + i

is 2n rad larger than the angle

(23)

172

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

the sine functions for the two angles are the same. Thus, =

exp(-fft> n f)

^+

=

+

=

from which 27T

(24)

If f < 0.4, -/I ~~ f 2 > 0.9165. Thus, if the record is lightly damped, a good approximation for the damping ratio is (25)

With r known and f as just determined, 27T

(26)

While these simple expressions are not valid for higher-order systems, they may often be used with little error to estimate the natural frequency of a mode that dominates the system response. Theoretically, any two points on the record separated by a full period of the oscillation may be used to determine 8. However, locating off-peak values of x\ and x^ that are separated in time by exactly r s is more difficult than determining peak values; further, the distance on the record from the equilibrium value of jc, XQ, to x\ and X2 can be measured more accurately (as a percentage of the measured value) as (x\ - XQ) and (x2 — XQ) increase. The concept just developed may be carried further to show the damping ratio required to cause the response amplitude to decay by one-half in n cycles. Then, x\/X2 is 2.0 and Eq. (23) becomes

0.69314 =

2nt;n

(27)

For the case in which f < 0.4,

n=

0.1103

(28)

or 0.1103

to within ±8.35%.

(29)

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

173

3.4 Determination of Natural Frequencies and Damping Ratios of More Complex Systems In Sec. 3.2, we saw that the damping ratio and natural frequency of simple second-order systems were evident by inspection of the differential equation since it had the form = f(t)

(30)

When the system is more complex, its mathematical representation may require a higher-order equation, such as

v J M , ( Y _ R+, r L

, ^ • (KLJ + D2m)0 = KpDm(/)p

Td/+U

or several simultaneous equations of various orders. In such circumstances, we cannot identify the damping ratio and natural frequency of the various contributions to the motion simply by inspection of the coefficients of the equation(s). Once the equations are transformed into the Laplace domain, however, the motions that contribute to the solution of a particular problem can be identified through the following process. (This process has been coded as the CRAM and LOCUS computer routines, which accompany this book.) In the Laplace domain, the equations are algebraic although the coefficients of each transformed variable may be polynomials in the Laplace operator S. With Cramer's rule, as discussed in Chapter 2, such an equation or system of equations can be solved for each dependent variable in the problem in terms of a ratio of polynomials in S. For example, if we have the system of differential equations 0 f(t)

(31)

their transforms read

(32) Then,

o

-i 2

1 5 + l -F(S) S +2 -1 2

(33a)

174

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

S2 + 2 0 2

-1

; +2 -1

1

-i

F(S)

(33b)

S2 +

Notice that, for a given problem, the denominator polynomial in each ratio will be the same. One way of saying that the dependent variables in the problem are linearly independent of one another is that the time history of one variable will not simply be the same as the time history of another variable multiplied by a scaling factor. If such dependence were to exist, one row of the denominator determinant in Eqs. (33) would be the same as another row multiplied by a constant. If such dependence does occur, we have too many equations for the number of functionally distinct variables, and a unique solution to the problem is usually not possible without reformulation. If each term in the denominator determinant does not vanish identically, a necessary and sufficient condition that the system of algebraic equations has a solution is that the expansion of the denominator determinant be equal to zero. Thus, for our example, we have

f 1) - 1 = 0

(34a)

or

S4 + 3S2 + 1 = 0

(34b)

The roots of this characteristic equation, as it is called, describe the nature of the individual contributions to the overall system behavior. The roots are, in general, complex, with the real parts describing exponential decays or expansions and the imaginary parts, if present, describing the frequencies of oscillations. For our example, we may rewrite the characteristic equation as

(S + 1.618y)(5 - 1.6187)(5 + 0.618y)(S - 0.6187) = 0

(35)

It will be seen that the four roots have no real parts and, thus, the responses will experience no decays or expansions. The roots occur as two conjugate pairs, meaning that when the system is excited, its free vibration consists of the sum of two oscillatory motions, one at 1.618 rad/s and the other at 0.618 rad/s. In general, the impulse responses have the form

x i ( t ) = KI sin(1.618r) + AT 2 sin(0.618f) x2(t) = K3 sin(1.6180 + #4 sin (0.6180

(36)

where K\ / K3 and K2 •£ K^. As the system complexity increases, finding the characteristic equation and its roots becomes more difficult. In these circumstances, use of an applicable

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

175

computer program is recommended in order to reduce computational time and minimize oppurtunities for errors.

3.5 Energy Dissipated in Viscous Damping In this section, we will make use of the law of energy conservation to aid us in computing the decay in the oscillatory amplitude of mass-spring systems that results from the introduction of viscous damping. Briefly, the law states that the energy in a system, including that added through the action of an external agent, either remains in the system or leaves in some form. A conservative system is one in which there is no dissipation and no gain or loss in energy. In such a system, the sum of the potential energy and the kinetic energy is always constant. The systems we wish to examine are nonconservative in that energy is lost to viscous damping and energy may be added through the action of a forcing function. The kinetic or potential energy converted to heat as a result of damping is the product of the damping force and the distance over which it acts. For one complete cycle, the work expended, W^, is

Wd = I Fddx

(37)

For viscous damping, the force is ex. Since we are considering only oscillatory systems at this point, the displacement-time relation is of the form x = X sin(a)t - 0)

(38)

from which we have

x = Xcocos(cot — 0)

(39a)

djc = X cos(cot - )d(a)t)

(39b)

and

Then the work lost to viscous damping over one cycle of the oscillation at any co is /•27T

Wd = coX2c I

JQ

[

cos2(a)t - 0)d(otf)

1 1 ~\2n -cot + - sin 2cot = nccoX2 2 4 J0

(40)

We have frequently noted that frictional dissipation has its greatest effect at resonance, where the ratio of the response amplitude to the forcing function amplitude is inversely proportional to the damping ratio. To determine the work done at resonance, we set

176

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

and c =:

Then, from Eq. (40), Wd = n2^Vkm^/kJmX2 = 2^nkX2

(41)

There is an interesting graphical representation of the fractional work done during one cycle of oscillation. We begin by writing the velocity in the form

= ±a)XJ 1 - — = ±WX 2 - jc2

(42)

The positive sign should be used when x is increasing; the negative sign should be used when jc is decreasing. The force is therefore given by (43)

from which we may obtain (44a) or

(44b) This is the equation of a plane ellipse. It is readily graphed by plotting Fj against x as shown in Fig. 3-5. The values of the variables c, CD, and X are given a priori. The semimajor axis is X and the semiminor axis ccoX. Since the area of an ellipse is n times the product of the semimajor and semiminor axes, or nca)X2 [which is the same result as that obtained in Eq. (40)], we see that the area of

Fig. 3-5 Graph of Eq. (44).

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

177

the ellipse in Fig. 3-5 represents the frictional work done during one cycle of the oscillation. Note that the direction of the frictional force also depends on whether x is increasing or decreasing. As a result, the graph of Fj vs x is a loop, called

a hysterisis loop, rather than a straight line. The upper branch is for increasing x and the lower branch for decreasing x. Other types of damping will produce other loop shapes. If, for example, we take c(x}2 as the damping term with c a constant, the equation of motion is nonlinear, and the form of the solution depends on the magnitude of the forcing function. This type of damping is called aerodynamic damping because its best-known manifestation is the frictional force, that is, drag, experienced by objects moving through the air. The drag experienced by a body is given by CD = \PAV2

where

CD = drag coefficient p = atmosphereic density V — velocity A = some reference area CD depends directly on body shape and orientation. When these and the atmospheric density are held constant, CD decreases approximately as V~* up to the speed at which shock waves first form on the body. Since this is a large velocity (> 600 ft/s), CD is usually taken to be constant for a given shape and orientation to the airflow. The results of numerical integration of the equation of motion when a c(x)2 term replaces the linear, viscous damping term are shown in Figs. 3-6 and 3-7. The forcing function remains a sine wave. Its amplitude is 0.1 for the results in Fig. 3-6 and 10.0 for the results in Fig. 3-7. Since these are forced oscillations, they decay only to steady-state values. Figures 3-6b and 3-7b are of particular interest. First, they show that the expected ellipselike loop in the force-displacement graph is not achieved until sometime after the forcing function is applied. In fact, the larger the forcing function, the longer this takes. Second, the figure is not an ellipse but rather like the trace of a distorted lemon. Compared with a true ellipse, the force in this figure approaches its maximum more rapidly (while x is still negative) and also diminishes more rapidly as x approaches its maximum positive displacement. Notice, too, that the figure begins in the center of the graph, where both Fd and x are zero at t = 0. We examine the vibratory response of systems with other than linear viscous damping in more detail in the next section.

3.6 Decay of Vibratory Motion Due to Other Types of Damping In all of the analyses to this point, we have considered the force generated by a damper to be linearly proportional to the velocity of the mass, that is, of the term ex

178

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS 0

O4

_'

r- .

i

0.02

O l cos2 cot

jc — Xcosa>ntd(cont) With these expressions, the energy loss becomes

-2 /

Jn/2

cX2(ji>2n cos2 o)nt (X cos contd(a)nt)) /»37T/2

= -2cX3co2l I

Jji/2

[

cos3 contd(cont)

1 I 3jr/2 8 - sinco n t (cos2 cont + 2) =3

\7i/2

3

(50)

188

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

We note that, for linear viscous damping, the relation at resonance between the response amplitude and the excitation force amplitude was given by Eq. (4) as ccon

The energy loss per cycle for viscous damping is n2n

I Jo

cxdx = xX2a)nn

Thus, for equal energy loss, the equivalent viscous loss coefficient is

(51) Substituting this value into Eq. (4) yields a relationship between response amplitude and the excitation force amplitude when the damping is of the aerodynamic drag type: (52a) or

Examination of Eq. (52b) shows that the amplitude of the system's response at resonance can be expected to grow only as the square root of the magnitude of the exciting force. Qualitatively, this behavior is readily demonstrated by integrating Eq. (49) using the technique described in Sec. 3.6.1 for two different values of F0. The results are shown in Figs. 3-11 and 3-12. Close examination of the numerical solutions shows that what appears to be the steady-state response at the higher force is greater than that predicted by Eq. (52) by about 4.5%. This is a consequence of the fact that the steady-state solutions for these two cases do not travel the relative paths that were assumed in the foregoing analysis. In nonlinear systems, the path traveled by the mass in a given time from the onset of excitation is not (/ r 02 // r oi) times the path traveled when the amplitude of the excitation is FO, . Nor is it

times the path length of the first response. In the example shown, F02 is 25.0 and FO, is 1.0. Table 3-1 shows the distance traveled by each mass during a specific time interval and the ratios of these distances. Although the two forcing functions are in phase, the smaller response first crosses the line X = 0 on the way down 2.62 s after the onset of excitation while the larger response does not cross this line until 2.90 s have elapsed. The first response lags the forcing function by a phase of 80 deg and the second by

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS 0.75

—i

i i

i

|

i

•

•

•

|

-

i

• i

|

i

i

i

189

•

O.5O —

O.25 —

CD

-0.25 —

-0.50 —

-0.75 O

2

4

6

8

1O

TIME, seconds

Fig. 3-11 System with aerodynamic damping. AMP = 1.0, C = 1.0, k = 3.0, W = 1.732.

107.76 deg. Later, both responses lag the forcing function by 90 deg, as expected. During the "steady" oscillation, the wave shapes are also slightly different from one another and, significantly, slightly different from a sine wave. For example, at 7T/4, a sine wave has a value 0.707 times its peak amplitude. Here the smaller wave reaches only 0.69 times its peak amplitude and the larger wave 0.65 times its peak amplitude at 1/8 wavelength. Obviously, the waves are not true sine waves. The larger the wave, the more it departs from the true sine shape, and this accounts for the difference in the computed and predicted wave amplitudes. Another aspect of this nonlinear behavior is shown in Fig. 3-13. Here, the system is permitted to respond freely to an initial displacement. Notice that the motion is characteristic of an overdamped system (f > 1.0) until the displacements and velocities become small, at which point the motion becomes oscillatory. This is what one would expect since, at large velocities, the damping force can exceed the spring force while, at low velocities, the spring force will be larger than the damping force.

3.6.3 Internal Damping The force associated with the internal damping* of vibrating materials is such

that the energy dissipated has been found to be independent of the frequency of oscillation but to increase as the square of the amplitude of the forcing function.

*The reference here is to structural materials like steel and aluminum which, when vibrated in a vacuum, exhibit the type of internal energy dissipation mechanism described in the text.

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

190

4 —

O ] [a2 + a>2]

(73)

and (l/m)e1°*da>

27T J.

(74)

*Strictly speaking, the Laplace transform is not defined for negative values of the forcing function.

200

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Now, define

*

m

\2m/

n

With this definition, Eq. (74) can be written x(t) =

a2 2nm -oo [co + ja][co- ja][a) + a)f + j — ^ [co-cof + j — ^

The evaluation of the integral proceeds according to the following rules2 derived from complex function theory: (1) The integral is equal to -2nj ]T (residues of the integrand at all poles on or below the real axis); t > 0. (2) The integral is equal to 2nj ^ (residues at all poles above the real axis); t y- + (fco n — a)2

(76b)

Residue of pole at o>/ - j£con: ———-JeWe-w'eJ*

4nma)f

^

1

a + (cof -\- j£a)0)n)

Residue of pole at -co/ - j^con: n20-SVntpjtt>ft

47tma)f

a2 -f- (cof -f j^coa)n)

(76d)

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

201

Thus, for t < 0,

aeat 2m[^ + (^+0) 2 ]

(77)

and, for t > 0, sin

x(t) =

2m

(78a) (78b)

0 = sin~

Equation (77) indicates that the displacement of the mass increases exponentially during the time that t < 0. At t = 0, there is a discontinuity in the rate of change of the force. This discontinuity generates a transient, damped vibration that is superimposed on the exponential decaying displacement. The latter results from the forcing function [right-hand side of Eq. (72)]. If the constant a in the forcing function is permitted to approach oo,

for t < 0, and

(79)

x(t) =

for t > 0. These results show that, while the displacement of the mass is zero for t < 0, x is not zero at t = 0; thus, the mass has an initial velocity equal to 1/m. This initial velocity is a consequence of a force applied to the mass that is nonzero only at t = 0 (an impulse). We can determine that the impulse in this case is a unit impulse since impulse as a concept is defined as the time integral of a force. Rearrangement of Newton's second law yields

//«*-/

mdv

(80)

In the present case, Av = 1/m at t = 0. Thus, f ( t ) Ar = 1. The example illustrates that, for cases in which there are no initial conditions [and f ( t ) = 0 for t < 0], the Fourier transform can be obtained from the Laplace transform by the substitution of jco for 5.

202

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

3.7.1 Applications of Fourier Transforms to Experimental Data By comparing the Fourier transform of a system's response to the Fourier transform of its forcing function at specific frequencies, it is possible to determine a system's transfer function experimentally. If the system is not too complex, a mathematical model of this transfer function is readily constructed. This technique has been used since 1950 to determine, for example, the values of parameters called an aircraft's stability derivatives from flight-test data. The utility that Fourier transforms provide in diagnosing system behavior has led to the development of algorithms called fast Fourier transforms and to electronic devices that employ them to analyze experimental data, virtually in real time. These algorithms approximate the time function being analyzed by a sequence of adjacent rectangular pulses and replace the integral

f

f(t)e~jcotdt

J• / — —coo

by a large sum of the type

(U*~M'"Af

(81)

in order to evaluate F(jco) at various co. A more accurate but slower method is that used in Appendix C to evaluate the coefficients of a Fourier series. This method consists of three steps: (1) Represent f ( t ) in an interval by a cubic spline, (2) integrate

(Ajt3 + Bjt2 + Cjt + Dj) e~ja}tdt

(82)

analytically for each interval, and (3) sum over all intervals. The resulting expressions can then be evaluated at any desired value of co smaller than some very high limit.

3.8 Frequency Response and BODE Plots In the previous section, we saw that we could express the Fourier transform of the system response as a product of the Fourier transform of the input or forcing function and the system transfer function with jco substituted for S. A corollary of this result is that we can find G(jco), termed the system frequency response, by taking the ratio of the Fourier transforms of the system response and the system forcing function and plotting the amplitude ratio and phase angle differences vs frequency. This technique can be used to develop the system transfer function from experimental data. By comparing transfer functions derived from experiment with those used to model the system, it is often possible to determine: (1) The adequacy of the mathematical model. (2) Coefficient values of some of the physical components in the system, such as the spring constant and damping constant. By comparing experimental transfer functions generated for different amplitudes and/or durations of the forcing function, it is also possible to assess the extent to

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

203

which the system is linear or the extent to which certain coupling phenomena are present. In aircraft analysis, for example, most mathematical models assume that movement of the elevator (the aerodynamic surface that initiates a pitching motion) does not excite motion in the lateral-directional plane. For some aircraft, however, it has been noted that pitching oscillations resulting from elevator movements also lead to yawing and rolling oscillations at the same frequency. This is termed gyroscopic coupling between the longitudinal and lateral-directional modes of the aircraft. It is the result of precession of the gyroscopic mass represented by a jet engine rotating at high speed along the aircraft's longitudinal axis. Aerodynamic asymmetries can also cause motions in one plane to initiate motion in an orthogonal plane. The Fourier transform of a sine wave with frequency co\ that begins at t = 0 and ends at t = T is

/' Jo

o-J"' Qin™, frit = ~J

e

(&M———i~

,

JW^WI~,

^3)

co

In developing this result, we have assumed that the time function sinco\t is zero for / > T. It can be shown that the magnitude of the Fourier transform exhibits a sharp peak when plotted against co. The peak, which is much larger than the magnitude of Eq. (83) at other frequencies, occurs when co = co\. The peak magnitude is T/2 if T = (2n/co\)n, where n is the number of cycles of the time function. It follows, therefore, that when T is very large, that is, when there are many cycles of the time function, the activity is essentially concentrated at one frequency. A stable system (or marginally stable system not excited at resonance) will ultimately respond to excitation by a continuous sine wave with a sine wave at the same frequency. The amplitude and phase relationship of the responses are functions of the excitation frequency. As we have mentioned, the graphs of these amplitude ratio and phase angle differences vs frequency are called frequency responses. These graphs can use linear scales, as we have done earlier in this chapter, or log-log and semilog scales, which are employed in Chapter 4. That segment of the computer program that accompanies this book and that evaluates the analytical frequency expression at 100-200 frequencies and plots these data is called BODE. Typical output is shown in Fig. 3-20. Additional details on the interpretation of the plots may be found in Chapter 4. 3.9 Arbitrary Excitation In our discussion of periodic excitation in Sec. 3.7, we indicated that, by the use of the Fourier transform, we could, in fact, let the period of the excitation become infinite. Then it really does not matter what form the excitation has so long as it is continuous. There is yet another scheme by which we could determine the response to an arbitrary forcing function, which we shall now discuss. We begin by considering how a system responds to an impulse. In Sec. 3.7, we saw that the effect of such excitation on a system was identical to that produced by a velocity initial condition (and no forcing function) when the magnitude of the impulse is equal to the product of the initial velocity and the system mass. We can also demonstrate this for the system shown in Fig. 3-21.

204

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

180. A R

-70. 0. P H A S E A N G L E

d

e

9

r e e s

-270.

.001

0.01

0.1 1.0 FREQUENCY, radians/sec

10.0

Fig. 3-20a BODE plot for transfer function. (lOOOS* + 90005 + 20,000)/($5 + 300S* + 20,00053)

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

205

-360. u.

p

H A S E A N G L E

d

e

g

r e e s

\

•270. .()01

s s.

-!!^

** •• • ••• IIM^ 1.0 0.1 0.01 FREQlJENCY, radians/sec X 100,000

10.0

Fig. 3-20b BODE plot for transfer function. (1000S2 + 90005 + 20,000)/(S5 + 300S4 + 20,00053)

206

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Fig. 3-21 Impulsively excited massspring-damper system.

The differential equation of motion is

mx + ex + kx = F/€ = F, = 0,

where

0 < t €

=r Jo

(84)

Fdt

and 0

in the limit. F has the units of force * time and F/€ has the units of force. From Table 2-3, the Laplace transform of the force term in the differential equation is F. Hence, the Laplace transform of the system response is mS2 + cS + k

F/m S2 + (c/m)S + k/m

(85)

and the time solution is given by x(t) = ——{exp[— (ct/2m)}} sincot mo}

(86)

where

The equivalent differential equation for the system with no forcing function reads

mx + ex -f kx = 0 i;(0) = F/m

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

207

The Laplace transform for this is m[S2x(S)-Sx(Q)-x

c [Sx(S) - jc(0)] + kx(S) = 0

or F/m

x(S) =

If the application of the impulse is delayed by a time interval r, then, it is readily seen that the time response is x(t -r) = —— mco

-[c(f - r)/2m]} sin[o>(f - r)]

(87)

Now, suppose the excitation consists of a number of impulses. Because the system is linear, the overall response can be calculated by computing the response to a single, unit impulse and then performing the following tasks: (1) Multiply this response by the strength of the first impulse. (2) Multiply this response by the strength of the second impulse. (3) Add the results obtained in step (2), shifted by r seconds, to the results obtained in step (1). (4) Continue in this fashion until all the impulses have been accounted for. Let us suppose also that the strengths of these successive impulses are related to the values /(r) of some arbitrary time function that occurs every 6 s, as indicated in Fig. 3-22. Then, the strength of the yth impulse Fj is /(r/)£. If we designate the response of the specific system to which a unit impulse has been applied as h(t), the response of this system to an impulse whose strength is Fj is

Fjh(t -

(88)

The response of this system to the arbitrary time function is, thus, the sum of all such impulse responses which, in the limit as € -> 0, becomes

= /

Fig. 3-22

f ( r ) h ( t - r)dr

Impulse sample of arbitrary forcing function.

(89)

208

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

The right-hand side of Eq. (89) is called the convolution integral. In fact, any integral of the form

f

J -c

X(x)Y(y-x)dx

is, in the mathematical literature, called a convolution of x with y or a convolution

integral.

In developing Eq. (89), we have shown that the response of a system to an arbitrary excitation can be constructed by representing the excitation as the superposition of an infinite number of impulses whose strengths vary in the manner of the excitation. Implementation of the process represented by the convolution integral on a computer requires replacement of the integration by a summation. We might

proceed as follows: Suppose that the array H(J) contains the unit impulse response of a particular system evaluated every DT s for TMAX s. DT corresponds to € in Fig. 3-22. r represents a number of intervals DT in width. H is dimensioned as N — DINT(TMAXTDT). Also, suppose that the forcing function evaluated at every DT s is contained in the array F(J). HH will contain the overall result. Then the necessary code to generate the HH array is DO 1 J = 1, N

1

HH(J) = O.ODO DO 2 I = 1, N DO 2 J = 1, N IF(J.LT.I) GO TO 2 HH(J)

2

= HH(J)

+ F(I)

* H(J - I + 1) * DT

CONTINUE

The execution time and memory requirements for such a program depend heavily on DT and TMAX. TMAX should be sufficiently large that the transient response resulting from the impulse sample in Fig. 3-22 will have had time to decay. DT is generally made as small as memory space and execution time permit so that N ranges between 1000 and 10,000. As an example of the use of this computational technique in particular and the convolution process in general, we will apply it to a situation whose result we can determine in another way: a mass-spring-damper, excited by a step change in force applied to the mass. The differential equation for this situation is

mx -j- ex + kx = FQ = 0

for for

t >0 t 0. Using the Laplace transform, we can show that the solution of the differential equation is L/l - f 2cDnt - tan'1 (—L

(91)

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

209

or

x ( t ) =0.4 1 -

0.9798

cos (4.899f - tan'1 (0.204124))

(92)

Alternately, we can apply the series approximation of the convolution integral as generated by the preceding code segment. Table 3-2 shows x computed in this way for three values of DT, along with an evaluation of Eq. (92) for the same times. It will be seen that, for DT < 0.01, the error is roughly proportional to DT. However, for the same TMAX, the computation time increases as (1/DT)2. A decrease in DT by a factor of 10 or 100 thus represents a practical limit for microcomputers running at 1 million instructions/s. We should note also that a step is the most difficult function for a series representation to fit because of the discontinuity at t = 0. Significantly better results will be obtained when the initial slope of the function is finite. Better results will also be obtained if the H array is loaded with the response to a unit pulse of width DT rather than with the impulse response. A computer program based on the use of pulses rather than impulses, SAMPLE84, is discussed in Chapter 5.

3.10 Response Spectrum of Certain Dynamic Systems A response spectrum is a plot of the peak response of a single-degree of freedom (mass-spring) oscillator as a function of its natural frequency for a specific excitation. Design engineers considering a system able to withstand dynamic as well as static loads find this concept particularly useful. As an example, consider a mass supported by a spring. A force or load, whose time history is shown in Fig. 3-23, is applied to the mass. What then is the maximum excursion of the mass? The constant force in Fig. 3-23 after t = t\ leads to a constant deflection x = Fo/k. The actual deflection as a function of time is obtained by solving the Table 3-2

Evaluation of convolution integral

Time

;c,DT = 0.1

;c,DT = 0.01

*,DT = 0.001

jc, Analytical

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1

0.0 0.08690758 0.22568355 0.37613024 0.50274673 0.58175546 0.60425357 0.57549217 0.51114543 0.43194295 0.35815331 0.30516985

0.0 0.05019381 0.16880542 0.31734561 0.45720954 0.55900668 0.60704774 0.60041633 1.55049446 0.47620743 0.39845680 0.33512393

0.0 0.04630677 0.16270682 0.31066425 0.45162571 0.55557883 0.60616404 0.60183001 0.55349367 0.47985753 0.40184812 0.33756912

0.0 0.04587247 0.16201351 0.30991291 0.45099378 0.55518508 0.60605285 0.60197502 0.55381645 0.48025441 0.40221778 0.33783465

210

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS FA

Fo

t

Fig. 3-23 Time history of excitation load.

differential equation. With the ramp-step forcing function (shown in Fig. 3-23), we need to invert

S2 (S2 + k/m) Fo/(tik) S2

Fo/(fim) S2 + k/m

(93)

in order to determine the deflection of the mass from t = 0 to t = t\. Thus, during this period,

F0 FQ fm . k x(t) = —t-— TsmJ-t t\k t\k\/ k Vm

(94)

The forcing function shown in Fig. 3-23 can be thought of as a ramp, (Fo/t\)t, for 0 < t < 1 1 and as the sum of a positive ramp and a delayed negative ramp for t >ti:

Thus, the deflection of the mass for t > t\ can be obtained from Eq. (94) by subtracting from it the deflection [i.e., Eq. (94)] delayed by t\ s: m . FQ , ,, . x(t) = —t - —(t - ti) - — -smJ-t + — - sin J-[t t\ky k m t\k V k t\k t\k

1-

sin l-(t-ti) V m_

(95)

r

m

The extremum points of x(t) occur when dx/dt = 0 or when Ik sin J — (t - t\) V m

jk ' sinJ— t \ m =0 m

(96)

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

211

Hence, k !~k\ Hk k . k . k • cos,/ — t = J-\ cos J —t - cosJ —t\ + sin J — t - sin-,/ —t\ m Vm V m I Vm m Vm Vm

or

k

• J/*—1\ - tan J — t 1 = c o s . //* — f i + sin V m Vm Vm

(97)

from which we may obtain the time at which maximal deflection occurs as

m

tan

1 -c

= Vi "

/*, m k

(98)

smJ—f V m

The earliest the first peak can occur is when t = n^/(m/k) since this is the time the first peak occurs when the excitation is a pure step, that is, when t\ = 0. When t\ > 0, Eq. (98) shows the first peak occurs later. When t\ is large enough that the angle ^/(k/m)t\ lies in the third quadrant, both sine and cosine are negative and Eq. (98) fails. For this case, we can employ the triangle shown in Fig. 3-24 to find the sine and cosine values. In Fig. 3-24, we define H2 = Sin2

k_t

—ti + cos 2 ./—1\ = 2 - 2 c o s J —

V m

in order to find

/* t cos,/— V m

-sin,/-* (99a)

2 1-

- sin^k/m t 1

Fig. 3-24 Relation among displacement components.

212

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

and

,— Ik m

_

(99b)

*max ~~

With these values inserted in the displacement equation, we have

1+

1 — cosj—f Vm

1 — cos J — f \l m

t

— 1\ m

sin",/—1\ m —

L \

1+

ml

1-COS,—

1 — cos J—1\ m

(100)

By expanding the cosine function in a series, we can show that when ^/(k/m)t\ = 0, the dynamic peak is twice the static deflection F$/k. The factor in brackets in

Eq. (100) evaluated for other values of *J(k/m)t\ is shown in Fig. 3-25, the normalized response spectrum for this particular excitation. Other excitations will, of course, yield different spectra. One way these can be generated is to solve the differential equation for a range of Jk/m values and of the forcing function parameter t\ using the computer. Then, for the peak response of each solution, we can plot (xk/FQ)max against the value o f t \ / r . Each spectrum need be generated only once since both axes are nondimensionalized. On the ordinate, x is normalized and nondimensionalized by dividing x by F^/k and, on the abcissa, t\ is normalized and nondimensionalized by dividing t\ by •c = 2n^/k/m. In virtually all cases, the maximum dynamic response will be less than or equal to twice the static deflection. There are also cases, for example, in the analysis of the effects of earthquakes on structures, in which the peak velocity rather than the peak displacement is the quantity of interest. The velocity spectrum can be obtained in a manner similar to that used to obtain the deflection response spectrum if the differential equation(s) are written so as to yield velocity as the solution or one of the solutions. In these

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

213

Normalized Time, t/T Fig. 3-25 Normalized response spectrum to ramp-step excitation; t indicates the time at which excitation becomes constant, r - 2irl+Jk/m.

cases, the friction between components can materially affect the peak velocities achieved. It is, therefore, usual practice to consider that the oscillator includes a damper. 3.11 Vibration Absorber Suppose that we have a suspended spring-mass system excited by a harmonic force FQ sin cot. Connected to the underside of the mass is a second spring-mass system. The ratio of the spring constant to the mass for the lower system is to2. The arrangement of elements is shown in Fig. 3-26. Can we, in fact, reduce the motion of mi to zero by the proper selection of k2 and m2l We begin our investigation by writing the differential equations for the motion of the two masses:

m\x\ = Fosincot — k\x + k2(x2 — x\)

(lOla)

m2x2 = k2(xi -x2)

(lOlb)

In the Laplace domain, these equations become

[m2S2 + k2] X2(S) -

= 0

(102)

214

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Fig. 3-26

Vibration damper.

from which

-k2 0

-fa

m2S2+k2 + fa

(103a)

+ [mifa + ^2 (^i + fa)] S2 + fafa

(103b)

The roots of the fourth-order polynomial can be found by letting S2 = y and solving for y in the resulting quadratic and then finding the square root of the two y: = 0

(104a)

[mifa + (k\ +fa)m 2 ] , v[mifa-f (/ci+fa)m2] 2 2m\m2 2m\m2 (104b)

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

215

To continue, it is easier to use numerical values. For convenience, let k\ = 5, k2 = I , mi =5, and m2 = 1. With these numbers,

y = -1.5582576,

-0.6417424

(105)

and S = ±7(1.248301874)

S = ±j (0.801088278)

It is also possible to rearrange the quartic in the denominator of the transfer function as follows: First, divide the equation by m\m2, and set mi

The result is

Next, add and subtract the term (k2/k\)a)la)l2. The results can be written ' 4- co11 + coU2 — I (S2 k\)

/vi

= 0

(107)

Consequently, the transfer function for x \ ( S ) / F ( S ) becomes 1

xi(S) F(S)

,_,

.

O X

m

- (k2/ k^c

(108)

Since we are considering a steady-state situation, we let S = ja>, and we divide numerator and denominator by co^co2^ This enables us to obtain the expression

^1^1 ^o

(109) 1-

From this expression, we see that we have indeed satisfied the design requirement that x\ be zero when u> — u>22. However, the system now exhibits two natural

216

INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

frequencies [approximately 1.25 and 0.8 rad/s with the numbers used to obtain Eq. (105)], one on either side of the original natural frequency at which the response, in the absence of damping, becomes infinite. The second transfer function, X2(S)/F(S), indicates that to provide the force needed to balance the force on mass 1, the amplitude of ra2 must be (—FQ/^), when *i is motionless. Thus, the size of the absorber is determined by its allowable amplitude in order to generate the required balancing force. The reader should be aware that when Eq. (101) is submitted to the computer routines supplied with this book, the solutions do not appear to be related to the result given by Eqs. (109). The result for x\, shown in Fig. 3-27a, indicates a response with a significant oscillatory amplitude characterized by the beating between two closely spaced frequencies. When normalized, the peak amplitude is around 2.0 and not the 0.0 expected from the analysis leading to Eq. (109). Is there a "bug" in the computer program or an error in the analysis? Neither. Careful examination of the computer program printout shows that, for the BODE plot, the amplitude ratio is 10~74 db (the smallest value the computer can generate using logarithms) when CD = 1.0 rad/s. In the RESPON portion of the printout, the residues of the poles at ±y is 0.0. However, there are significant imaginary residues for the poles at ±0.8010887' and ±1.2487. Thus, the oscillations seen in the plot are due to these two modes. They are excited by the onset of the sinusoidal forcing function which generates significant amplitudes at other frequencies as well (see Sec. 3.8). Once excited, these frequencies cannot decay because no damping term has been provided in the equations.

0.4

-0.4

i

Damper Attached Fig. 3-27a Response of mi to resonant excitation.

TMAX= 59.7sec

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

217

TMAX= 59.7sec

Fig. 3-27b Damper (mi) response.

This dichotomy illustrates the problem we can sometimes encounter in constructing simplified models of physical reality. It has long been known that a vibration absorber built as depicted in Fig. 3-26 does, in fact, work as indicated. However, real springs all have small amounts of internal damping, which normally does not need to be included in the math model. In this case, the damping will eventually cause the oscillation at 0.801088 rad/s and 1.248 rad/s to decay, leaving x\ at zero and m^ oscillating only at a>. Since the internal damping has little effect on any portion of the response except the ultimate decay of transient oscillations, it is easy to see why early analysts chose to model the arrangement as having no damping and then forced the result to agree with experimental observations by considering excitation at only a single frequency.

3.12 Untuned Viscous (Houdaille) Damper The tuned vibration absorber of Sec. 3.11 is effective only when the system represented by the upper mass-spring in Fig. 3-26 operates continuously at a single speed. Many machines, such as automobile engines, must frequently change speeds. Therefore a damper that can smooth crankshaft torsional vibrations is often needed. Such a device is shown in Fig. 3-28. Analysis of the characteristics of this damper proceeds as follows. Two masses are involved, and so there will be two equations of motion. Assume the left end of the crankshaft to be fixed. The shaft has a torsional stiffness K, which generates a torque linearly proportional to the angular displacement 9. The end of the shaft and the pulley structure have a polar moment of inertia J. A damping moment is created if the mass inside the pulley does not move in unison with the pulley. The

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INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

Mass inside pulley structure surrounded by a viscous fluid. The cavity is sealed. End of crankshaft

Pulley structure (often) used to drive fan belt.

Fig. 3-28 Untuned viscous damper.

damping moment is -c(0 — ^), where ty is the angular velocity of the damping mass. The damping constant c is determined by (1) the surface of the damping mass, (2) the gap between the damping mass and the pulley structure, and (3) the viscosity of the damping fluid. The differential equation for the shaft end and pulley is

JO = -KO - c(0 - VO + M0e~-ja}t

(HOa)

JO = -KO + c(Vr - 0) + MQe~ja)t

(HOb)

or

The differential equation for the damping mass is then + c(Vr - 0) = 0

(HI)

We can approach the solution of this system of differential equations in two ways: (1) We can write the forcing function as MO sin&tf and solve the equations with a computer program for various values of co, K/ J, Jj/J, and c/(2\/KJ). (2) We can look for an analytical amplitude expression that is a function of the same parameters. Proceeding with the second approach, we assume solutions for the two equations as

= 6oejat

(H2a)

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

219

and substitute them into Eqs. (110) and (111) with the following result:

K

K \ co)~\ co) MO __^ + 7 _j, 0 + 7 _ V f o = _ 9

-*+'* 1

CO)

.CCO

• = J-r6o

The second equation can be solved for V'o to yield .CO)

fa = ~,——Jd

.

(113)

Jd /

V

This result can then be substituted into the first equation to obtain

or IVIQ

L w

• 1 , — = [\lv ( K - J ai >2\) +, jco)\ + / . 2 . •—r 0o (—Jd + Jca)) 2 2 _ -Jdco (K - Jco ) - jJdco)3 + jco) (K - Jo)2} - c2o)2 + c2 =

(-W + 7

Hence, MO ~ /jo;2 (AT - Jo)2) + yc^ [/a;2 - K + W]

(115)

If we now let con = *JK/J, t; = c/2^KJ, and /x = /£///, we have

Mo

l2

(116) as the normalized amplitude ratio. Examination of this expression shows that \KOQ/MQ\ is a function of three parameters, a>/o)n, IJL, and £. Thus, in order to graph it easily, it is necessary to fix two parameters while varying the third. For

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INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

example, we could set ^ and then graph \KOQ/MQ\ vs co/con for specified values off. Two values of £, zero and infinity, are of special interest. When £ = 0,

1

K0Q

(117)

When £ = oo,

(118) When £ = 0, the equation has the same form as that for a mass on a shaft restrained by a torsional spring. The damper mass is then uncoupled from its housing. When £ = oo, the coupling is solid and both masses oscillate together at a natural frequency given by JK/(J + J 5

Fig. 3-35 Frame with rigid corners.

3.14.4 Frame with Rigid Corners The principle of superposition can be applied to such structures when moments are applied at the beam-column joints in addition to transverse loads. For example, a structure such as that shown in Fig. 3-35 has beam-column joints that are assumed to remain square (i.e., 90 deg). The problem has three flexible members and three loads. We may, therefore write the relations between loads and deflection as F Mi M2

k\\

(150) 33

where 8 represents the linear deflection of the two columns at their tops. 0\ represents the slope at the left side of the beam with respect to the vertical, and 02 represents the slope at the right side of the beam with respect to the vertical. As a result of the application of the force alone, we generate k\\, £21, and £31. With the application of the moment M\ alone, & 3 1,^32, and &23 are generated. Similarly, with the application of M^, alone, £31, £32, and £33 can be determined. The stiffness matrix representing the situation in which the three loads are applied simultaneously is obtained by adding the three matrices produced by the application of the individual loads.

3.15 Rotary Balance and Balancing Many vibrations problems that engineers are asked to deal with arise in rotating or reciprocating machinery. In this section, we will consider ways to treat a number of such problems: (1) Harmonic excitation generated in a rotating machine. (2) The meaning of static balance. (3) Bearing reactions in a dynamically unbalanced rotating structure. (4) Dynamically balancing a rotating machine. (5) Forces and moments generated by a slider-crank mechanism. (6) Dynamic balancing of an ensemble of slider-crank mechanisms through an appropriate choice of crank-throw angle, bore spacing, and bank angle. (7) Conditions under which a rotating shaft will whirl, that is, rotate in a bowed shape about an axis through the bearing centers.

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INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

3. 15. 1 Rotating Imbalance Suppose we have a structure that we wish to excite harmonically. Suppose, also, that we wish to use only rotary motion for this purpose rather than reciprocating motion, rotary motion being less expensive and simpler mechanically. How might we do it? One way is to mount a circular portion of the system mass on bearings and either add or remove mass at one point just inside the circular boundary. A typical arrangement is shown in Fig. 3-36. The mass that is added or removed is located at a distance e from the bearing centers. If the circular portion of the machine is now rotated at a constant angular velocity CD, the effect is to rotate the center of the mass of the system at the same rate. The vertical location of the center of the mass will, therefore, vary in a sinusoidal manner. Designate the vertical location of the bearing centers as x. When the machine is not running, x = 0. Designate the added or deleted mass as m. Its vertical position in a fixed coordinate system is, therefore, x + e sin cot. If M is the original mass in the system plus the added mass, M — m is the original mass. If, on the other hand, an eccentric mass distribution is to be obtained by deleting m, we simply change the sign on m in the following development. The equation of motion of the system is d2 (M — m)x + m—r(x + esincot) = -ex - kx or

MX -f ex 4- kx = meco2 sin cot

(151b)

We note that if mco2e is set equal to FQ, the equation is identical to Eq. (4) of this chapter, developed earlier to describe the motion of a mass excited by a sinusoidal force. We may, therefore, use the steady-state solution obtained in that instance to write

meco

- Mco1}2 + (co))2

\

Fig. 3-36

(152a)

added or removed mass circular portion of system mass bearing

Single-degree of freedom system excited by rotation of an eccentric mass.

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

231

and 0 = tan~

(152b)

The amplitude can be normalized by multiplying it by M/me to obtain XM em

(!}„

\

1-

(153a)

——

= tan"

(153b) 1

This result is graphed in Fig. 3-37. Another way to obtain the harmonic excitation is to use two counterrotating weights (or holes) as shown in Fig. 3-38. In this mode, the vertical components of the harmonic forces simply add if the weights have the same angular velocity and begin their rotation from a horizontal line through the bearing centers. The horizontal force components will then cancel. For this reason, such an arrangement is the preferred implementation when we wish to construct a "shaker" or add vertical forces to a machine.

0

0.5

1.0

1.5

2.0

2.5

3.0

Frequency Ratio. Damping Ratio .05, .15, .25, .35, .45, .55, .65, 1.0 Fig. 3-37a

Normalized force transmitted to machine as a result of rotary imbalance.

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INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

0

0.5

1.0

1.5

2.0

2.5

3.0

Frequency Ratio. Damping Ratio 0.05, 0.15,... 0.65 Fig. 3-37b Normalized force transmitted to machine by rotary imbalance.

Fig. 3-38

Counterrotating weights used to generate vertical harmonic force.

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

233

3.15.2 Static Imbalance of Rotors In this and the following two sections, we consider an ensemble of masses distributed about an axle supported at each end by a bearing. The masses are said to be in static balance if, for all elements in the system, the products of each mass and its distance from the axis of rotation sum vectorially to zero. For example, in the system shown in Fig. 3-39, assume that m\—mi and the magnitude of e\ equals the magnitude of £2. Then, the vector sum of the mass-distance product is

m\(ei) + m 2 (£ 2 cos30deg j - £ 2 sin30deg/) = em (I — sin 30 deg )/ + em cos 30 degj

(154)

which is obviously not zero. However, if a third mass, m 3 , is attached to the axle in the same plane as mi and m 2 and at an angle of 120 deg measured counterclockwise from mi andifm 3 £ 3 =m^i,then, me(l - 1/2 - 1/2) +

= 0

(155)

and the condition for static balance is satisfied. Note that the location and magnitude of the required mass-distance product can be determined empirically as follows: Allow the system to come to equilibrium. The two weights will stop moving when they are below the axle and each is 60 deg from a vertical line. Then, add weight along the vertical line but above the axle. Disturb the system rotationally. The system will be statically balanced when sufficient weight has been added that, after a disturbance, it comes to rest in no preferred orientation. Note also that it is not possible to determine m 3 and £3 independently. Only their product can be determined from the condition of balance. Without disturbing the static balance of the system, m 3 can be moved along a line parallel to the axle. However, if we then rotate the system about its axle, unbalanced centrifugal forces will generate a rocking moment that is felt at the bearings. This dynamic imbalance will disappear if m 3 is returned to a plane normal to the axle that also contains mi and m 2 .

Fig. 3-39 Example of system in static imbalance.

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INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

As a result of these investigations, we conclude that, when a system contains unbalanced masses that all lie in a single plane, the imbalance is static since it can be identified and corrected without spinning the system. When the imbalance appears in two or more planes along the axle, the system is dynamically unbalanced and, as we shall see in Sec. 3.15.4, requires the placement of a mass-distance product at a specific angular location in each of two planes to achieve balance.

3.75.3 Bearing Reactions in Dynamically Unbalanced Systems In this section, we consider the bearing reactions generated by the rotation of a dynamically unbalanced system of masses. We will take as an example of such a system a round bar that contains a number of 90-deg bends, as shown in Fig. 3-40. We will assume that the torque applied to the bar can be expressed by

T(t) = lQ + 5t

(156)

N-m. We pose the problem of determining the ;c and y bearing reactions* at t = 3.0s.

Solution. In addition to Newton's law expressing the balance between the forces and the inertias in the two planes normal to the z axis,1" namely, j(a)x

(157a)

>(5)>

(157b)

the system must satisfy the general angular momentum equations for plane motion at bearing A and at bearing B. For bearing A, these are: (MA)X = -Ixz(b + Iyza)2 (MA)y = -Iyzu - Ixza)2

(158)

T = L7co

We begin our analysis by fixing an inertial reference to the system at bearing A with the z axis along a line between bearing centers, the x axis vertical, and the y axis to the right when we are looking from A to B. We then compute the moments *The word reactions here refers to the forces that the bearings exert on the bar. By Newton's third law of motion, these are the same forces magnitudewise but opposite in direction that the bar applies to the bearings. tThe equations are to be read as: The sum of the x (or y) components in the problem (bearing reactions and rod weight) is equal to the sum of the jc (or y) components of the j(ma) products. For purposes of analysis, the acceleration components of each of the nine segments of the bar are computed separately, multiplied by the appropriate mass, and then summed.

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

235

APPLIED TORQUE IS CLOCKWISE LOOKING INTO THE PAGE

Fig. 3-40 Round bar with eight 90-deg bends. All dimensions are in meters; the bar is 0.02-m diameter; the weight is 70 N/m.

and products of inertia of the nine segments with respect to this axis system. Using the dimensions indicated in Fig. 3-40, we find:

Mass

z-

y-

Distance

Distance Segment 3

Segment 2

[70(0.4)1

L 9.81 J

(0.7)

(0.2) +

0= 1.142 kg-m2

(159a)

Segment 4

Z ~

Mass

Distance Segment 6

[70(0.3)1

X-

Distance Segment 7

0 = 1.220kg-m2

981 Segment 8

(159b)

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INTRODUCTION TO THE CONTROL OF DYNAMIC SYSTEMS

+ Segment 1

Segments 2 and 4

Segment 3

Segments 6 and 8

= 1.083kg-m

Segment 5

Segment 7

Segment 9

2

(159c)

When t = 3 s, the angular acceleration is T

10 4- 15

a) = — = —-— = 23.084 rad/s2 Izz 1.083

(160)

The angular velocity is found by integrating Eq. (156) to obtain

0

l

zz

At t\ = 3 s, a) = 48.48 rad/s. On insertion of these results in Eq. (158), we have (MA)X = 1.22(23.084) + 1.142(48.48)2 = 2655.89 N - m

(MA)y = 1.142(23.084) - 1.22(48.48)2 = -2893.74 N • m

(161)

These are the total moment components acting on bearing A; they are equal to the sum of the moments due to the bearing reaction at B and the moments due to segment weights. In the x — z plane, we have (Af A ) = -2893.74 = 1.40B, - (0.3)(70)(0.3/2) - (0.4)(70)(0.3) -(0.4)(70)(0.5) - (0.4)(70)(0.7) - (0.1)(70)(0.75) - (0.3)(70)(0.8) -(0.3)(70)(0.95) - (0.3)(70)(1.1) - (0.3)(70)(1.25)

(162)

Therefore, Bx = -1969.46 N

(163)

(MA)X = 2655.89 N - m = lABy

(164)

In the y-z, plane we have

APPLICATION TO THE ANALYSIS OF MECHANICAL VIBRATIONS

237

from which

By = 1897.06 N

(165)

We now solve for the bearing reaction at A by using Eqs. (157): -(2.8)(70) + A, + Bx

,03X48.48,' A;c = -906N

z r =