Mechanical Vibrations: An Introduction [1st ed.] 9783030450731, 9783030450748

Table of contents :
Front Matter ....Pages i-xiv
Introduction (György Szeidl, László Péter Kiss)....Pages 1-29
Impact (György Szeidl, László Péter Kiss)....Pages 31-54
Some Vibration Problems (György Szeidl, László Péter Kiss)....Pages 55-97
Introduction to Multidegree of Freedom Systems (György Szeidl, László Péter Kiss)....Pages 99-139
Some Problems of Multidegree of Freedom Systems (György Szeidl, László Péter Kiss)....Pages 141-181
Some Special Problems of Rotational Motion (György Szeidl, László Péter Kiss)....Pages 183-199
Systems with Infinite Degrees of Freedom (György Szeidl, László Péter Kiss)....Pages 201-241
Eigenvalue Problems of Ordinary Differential Equations (György Szeidl, László Péter Kiss)....Pages 243-304
Eigenvalue Problems of Ordinary Differential Equation Systems (György Szeidl, László Péter Kiss)....Pages 305-345
Eigenvalue Problems Described by Degenerated Systems of Ordinary Differential Equations (György Szeidl, László Péter Kiss)....Pages 347-378
Back Matter ....Pages 379-448

Citation preview

Foundations of Engineering Mechanics

György Szeidl László Péter Kiss

Mechanical Vibrations An Introduction

Foundations of Engineering Mechanics Series Editors Vladimir I. Babitsky, School of Mechanical, Electrical and Manufacturing Engineering, Loughborough University, Loughborough, Leicestershire, UK Jens Wittenburg, Karlsruhe, Germany

More information about this series at http://www.springer.com/series/3582

György Szeidl László Péter Kiss •

Mechanical Vibrations An Introduction

123

György Szeidl Institute of Applied Mechanics University of Miskolc Miskolc-Egyetemváros, Hungary

László Péter Kiss Institute of Applied Mechanics University of Miskolc Miskolc-Egyetemváros, Hungary

ISSN 1612-1384 ISSN 1860-6237 (electronic) Foundations of Engineering Mechanics ISBN 978-3-030-45073-1 ISBN 978-3-030-45074-8 (eBook) https://doi.org/10.1007/978-3-030-45074-8 © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

The present book is based on the lectures delivered for those Hungarian and foreign M.Sc. students who have to take the course Mechanical Vibrations. When we began to write it we faced with the following problem: what level of preliminary knowledge could be expected. Hungarian students in mechanical engineering with a B.Sc. degree have some familiarity with tensor algebra and analysis. However, this is, in general, not the case for foreign students because of the differences in curricula. We had to compromise. Mainly in Chap. 1 which is a summary of the fundamental results of dynamics. In order to avoid the difficulties caused by the differences in the preliminary knowledge of the students we preferred matrix notations to tensorial ones. In spite of that the parts of the text which are bounded with two thin horizontal lines contain some materials which are presented in vectorial and tensorial notations. For the first reading, these parts of the text can be left out of consideration. As we have mentioned above, Chap. 1 is a sort of Introduction which is devoted to the fundamentals of dynamics. The students are, in principle, familiar with the content of Chap. 1. As regards the content of the chapter Introduction, the necessary kinematic relations with an emphasis on rigid body motions are all covered. Then the equivalence of the effective forces to the external forces, the principle of impulse and momentum as well as the principle of work and energy are presented in the text. Proofs are, sometimes, omitted or are left for problems to be solved. Chapter 2 deals with impact. First, we consider central impact. Since the graphical solutions are very illustrative the use of Maxwell’s diagrams is also shown. As regards the issue of eccentric impact, we clarify the way how to reduce the equations providing the solution into such a form which coincides with the equations valid for central impact. The chapter contains various exercises (for which the solutions are also included in the text) and is closed by the problems that are left for the independent work of the students. Some simple but important vibration issues are presented in Chap. 3 which is, among others, a short introduction to the vibratory problems of single degree of freedom systems. Undamped and damped free vibrations and forced vibrations are considered. As regards the applications, some simple machine foundation problems v

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are modeled and solved. It is worth emphasizing that the machine foundation problems are not simple at all and the one degree of freedom model we have presented is only aimed at drawing the attention of the readers to these important problems. Chapter 4 is an Introduction to multidegree of freedom systems. We establish Lagrange’s equation of the second kind for a system of particles first. They can, however, be applied to establishing the equations of motion for such systems which involve rigid bodies as well. Special emphasis is laid on spring–mass systems with two degrees of freedom. Solutions are presented for various free and forced vibration problems. It is also shown how to tune a system to avoid resonance. The general theory of multidegree of freedom system is considered in Chapter 5. The eigenvalue problem that provides the eigenfrequencies for a vibrating finitedegree of freedom system is also presented. It is shown what properties these eigenvalue problems have including the fundamental characteristics of the eigenvalues and eigenvectors. Some simple solution procedures are suggested. The concept of the Rayleigh quotient is introduced. The case of forced vibrations is investigated at the end of the chapter. Chapter 6 covers some phenomena of the rotating motion. We discuss the most important properties of a flywheel which can be used to store kinetic energy and to make the rotational motion smoother by reducing the speed of fluctuations. Stability problems caused by a change in the load torque is also investigated. If the shaft is not rigid further problems occur. We present Laval’s theorem and the gyroscopic effect of the rotational motion. Chapter 7 is devoted to the vibration problems of systems with infinite degrees of freedom. First, we consider the longitudinal vibrations of rods. Then solutions are presented for the transverse vibration of a string and for the torsional vibrations of rods with circular cross section. This is followed by the analysis of transverse vibrations of beams. Chapter 8 is concerned with the eigenvalue problem of ordinary differential equations. We present the definition of the Green functions and reduce some eigenvalue problems to homogeneous Fredholm integral equation with the Green function as kernel. A solution algorithm is suggested by the use of which numerical solutions are given for some vibration problems of circular plates subjected to constant radial in plane load and for the vibratory behavior of beams loaded by an axial force. Eigenvalue problems of ordinary differential equation systems are discussed in Chap. 9. The concept of the Green function matrix is introduced. By utilizing the Green function matrices the eigenvalue problems described by ordinary differential equation systems can be reduced to eigenvalue problems governed by homogeneous Fredholm integral equation systems. The solution algorithm presented in Chap. 8 is generalized for such eigenvalue problems. The applications are related to the vibration problems of Timoshenko beams. Vibration problems of curved beams with a centerline of constant radius are governed by degenerated differential equation systems. Chapter 10 provides a definition for the Green function matrices concerning the degenerated differential

Preface

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equation systems. These matrices are determined for pinned-pinned, fixed-fixed, and pinned-fixed heterogeneous curved beams. By utilizing the Green function matrices the eigenvalue problems that describe the vibratory behavior of these beams are reduced to Fredholm integral equation systems. Numerical solutions are also presented in graphical format. Appendix A is a very short introduction to tensor algebra. It presents those tools which are needed to understand some proofs in Chap. 1 and Appendix C which is a collection of solutions to some selected problems. Appendix B gives the definition of the Dirac function. Miskolc-Egyetemváros, Hungary

György Szeidl László Péter Kiss

Acknowledgments In preparing the manuscript, serious assistance was provided by Dr. Gábor Csernák who read the first version of the manuscript and suggested various changes in order to make it better. We gratefully acknowledge his assistance. In structuring and correcting the manuscript we got valuable assistance from Mrs. Sudhany Karthick at Springer. Her help is highly appreciated. Since this book is a textbook it is important to acknowledge our debts to our teachers. Imre Kozák introduced us to the modern literature on applied mechanics. We could always turn to him for advice when difficulties arose in our research work. We are also grateful to Miss Lisa Lui (Zhejiang Guanbao Industrial Co.), Mr. Tony Griffiths of lathes.co.uk and Mr. Brian Le Barron (Vibratech TVD 180 Zoar Valley Rd. Springville, New York, USA) for allowing us to include some photos in the book—see Figs. 3.22, 4.11 and 4.24. László Péter Kiss is especially grateful to his parents and sister, simply for everything they mean to him. There is further dedication to his little nephew, Balázs. Gyorgy Szeidl owes his wife Babi a debt of gratitude for her support and encouragement through her persistent asking the question “Did you deal with your book today?”. This always turned his attention to writing instead of dealing with less important things. Special thanks are due to his daughter Agnes and son Adam for their continuous support. There is also an opportunity here to make some remarks for editing the text of the book: (a) Vectors and tensors are typeset in boldface. Matrices are also typeset in boldface but the letters are underlined. (b) Equations are numbered sequentially in each chapter. (c) The formulae that we regard important are, in general, framed. (d) Bibliographies are presented at the end of the chapters. The main text is followed by Appendices and Index.

Contents

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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Model Creation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Kinematic Relations—A Summary . . . . . . . . . . . . . . 1.2.1 Motion of a Material Point (Particle) . . . . . . 1.2.2 Motion of a Rigid Body . . . . . . . . . . . . . . . 1.3 Fundamental Relations of Kinetics—A Summary . . . 1.3.1 The First Moment of a Mass Distribution . . 1.3.2 Momentum Distribution—Vector System of Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Moment of Momentum for a Rigid Body . . 1.4 Parallel Axis Theorem . . . . . . . . . . . . . . . . . . . . . . . 1.5 Fundamental Theorems of Kinetics . . . . . . . . . . . . . 1.5.1 Introductory Remarks . . . . . . . . . . . . . . . . . 1.5.2 Force Systems . . . . . . . . . . . . . . . . . . . . . . 1.5.3 Systems of Effective Forces . . . . . . . . . . . . 1.5.4 The Fundamental Theorem of Dynamics . . . 1.5.5 Kinetic Energy of a Rigid Body. Power . . . . 1.5.6 Principle of Work and Energy . . . . . . . . . . . 1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Impact . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 What Is Meant by Impact . . . . . . . . . 2.2 Central Impact . . . . . . . . . . . . . . . . . 2.3 Eccentric Impact . . . . . . . . . . . . . . . . 2.3.1 Impact of Two Bodies Which Motion . . . . . . . . . . . . . . . .

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Impact of Two Bodies if One or About a Fixed Point . . . . . . . . . 2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

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Two Bodies Rotate ................. ................. .................

Some Vibration Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Single Degree of Freedom Systems . . . . . . . . . . . . . . . . . . . 3.1.1 Introductory Remarks . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Machine Foundations and Vibration Isolation . . . . . . . . . . . . 3.2.1 Introductory Remarks . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Minimal Model with One Degree of Freedom . . . . . 3.2.3 Machine Foundation Under the Action of a Rotating Unbalance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4 Displacement Excitation on a Machine Foundation . . 3.3 Forced Vibrations of Undamped Systems Under Special Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 First We Shall Assume That xn ¼ xf . . . . . . . . . . . 3.3.2 Second We Shall Assume That jxf  xn j ¼ e  1 . 3.3.3 Third We Shall Assume That the System Is Subjected to a Constant Force . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Introduction to Multidegree of Freedom Systems . . . . . . . . . . . . 4.1 Lagrange’s Equations of Motion of the Second Kind . . . . . . 4.1.1 Generalized Coordinates . . . . . . . . . . . . . . . . . . . . . 4.1.2 Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . 4.1.3 Derivation of Lagrange’s Equation of Motion of the Second Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Applications of Lagrange’s Equations . . . . . . . . . . . . . . . . . 4.2.1 Calculations of Generalized Forces . . . . . . . . . . . . . 4.3 Multidegree of Freedom Systems with Solutions . . . . . . . . . . 4.3.1 Examples for Multidegree of Freedom Systems . . . . 4.3.2 Two Degree of Freedom Systems . . . . . . . . . . . . . . 4.3.3 Free Vibrations of Two Degree of Freedom Spring–Mass Systems . . . . . . . . . . . . . . . . . . . . . . . 4.3.4 Forced Vibrations of Two Degree of Freedom Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.5 Vibration Absorbers . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Some Problems of Multidegree of Freedom Systems . . . . . 5.1 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Spring–Mass Systems . . . . . . . . . . . . . . . . . . 5.1.2 The General Form of the Equation of Motion 5.2 Eigenvalue Problem of Symmetric Matrices . . . . . . . . 5.2.1 Eigenvalues and Eigenvectors . . . . . . . . . . . . 5.2.2 Calculation of the Eigenvectors . . . . . . . . . . . 5.2.3 Orthogonality of the Eigenvectors . . . . . . . . . 5.2.4 Repeated Roots . . . . . . . . . . . . . . . . . . . . . . 5.2.5 The Natural Frequencies Are Real Numbers . 5.2.6 Uncoupled Equations of Motion . . . . . . . . . . 5.3 Rayleigh Quotient . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Definition and Properties . . . . . . . . . . . . . . . 5.3.2 Matrix Iteration . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Convergence of the Iteration Algorithm . . . . . 5.3.4 Properties of the Iteration Algorithm . . . . . . . 5.3.5 Calculation of the Largest Eigenvalue . . . . . . 5.4 Forced Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Forced Harmonic Vibrations . . . . . . . . . . . . . 5.4.2 Nonharmonic Inhomogeneity . . . . . . . . . . . . 5.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Some Special Problems of Rotational Motion . . . . . . . 6.1 Flywheels—Rotational Speed Fluctuation . . . . . . . 6.2 The Effect of a Change in the Load Torque on the Speed of Shafts . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Rotating Shafts—Balancing . . . . . . . . . . . . . . . . . 6.3.1 Determination of the Reactions . . . . . . . . 6.3.2 Balancing in Two Planes . . . . . . . . . . . . 6.3.3 Elastic Shafts, Stability of Rotation. Laval’s Theorem . . . . . . . . . . . . . . . . . . 6.3.4 Stability of Rotation. Gyroscopic Effects . 6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Systems with Infinite Degrees of Freedom . . . . . . 7.1 Equilibrium Equations for Spatial Beams . . . 7.2 Longitudinal Vibrations of Rods . . . . . . . . . 7.2.1 Equations of Motion . . . . . . . . . . . . 7.2.2 Solution by Separation of Variables . 7.3 Transverse Vibration of a String . . . . . . . . . 7.3.1 Equations of Motion . . . . . . . . . . . . 7.4 Torsional Vibrations . . . . . . . . . . . . . . . . . .

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7.5

Flexural Vibrations of Beams . . . . . . . . . . . . . . . . . . . . . 7.5.1 Equilibrium Equations . . . . . . . . . . . . . . . . . . . . 7.5.2 Equation of Motion and Solutions for the Axially Unloaded Beam . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.3 Orthogonality of the Eigenfunctions . . . . . . . . . . 7.5.4 Forced Vibration of Beams . . . . . . . . . . . . . . . . . 7.5.5 Vibration of Axially Loaded Beams . . . . . . . . . . 7.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Eigenvalue Problems of Ordinary Differential Equations . . . . 8.1 Differential Equations and Boundary Conditions . . . . . . . . 8.2 Self-adjointness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Orthogonality of the Eigenfunctions in General Sense . . . . 8.4 On the Reality of the Eigenvalues . . . . . . . . . . . . . . . . . . 8.5 Boundary Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Sign of Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Determination of Eigenvalues . . . . . . . . . . . . . . . . . . . . . 8.8 The Green Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9 Calculation of the Green Function . . . . . . . . . . . . . . . . . . 8.10 Symmetry of the Green Function . . . . . . . . . . . . . . . . . . . 8.11 Calculation of the Green Functions for Some Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.11.1 Simple Beam Problems . . . . . . . . . . . . . . . . . . . . 8.11.2 Preloaded Beams . . . . . . . . . . . . . . . . . . . . . . . . 8.11.3 Preloaded Circular Plates . . . . . . . . . . . . . . . . . . 8.12 Eigenvalues with Multiplicity . . . . . . . . . . . . . . . . . . . . . 8.13 Properties of the Rayleigh Quotient . . . . . . . . . . . . . . . . . 8.14 Some Concepts of Integral Equations . . . . . . . . . . . . . . . . 8.15 Transformation to an Algebraic Eigenvalue Problem . . . . . 8.15.1 Classical Solution . . . . . . . . . . . . . . . . . . . . . . . . 8.15.2 Solution Using the Boundary Element Technique . 8.16 Vibration of the Preloaded Beams . . . . . . . . . . . . . . . . . . 8.16.1 Formulation of the Governing Equations . . . . . . . 8.17 Computational Results for Fixed-Pinned and Fixed-Fixed Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.17.1 Data for the Computations . . . . . . . . . . . . . . . . . 8.17.2 Computational Results . . . . . . . . . . . . . . . . . . . . 8.18 Vibration of Preloaded Circular Plates . . . . . . . . . . . . . . . 8.18.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . 8.18.2 Integral Equation of the Eigenvalue Problem . . . . 8.19 Computational Results . . . . . . . . . . . . . . . . . . . . . . . . . . 8.19.1 Data for the Computations . . . . . . . . . . . . . . . . . 8.19.2 Clamped Plate . . . . . . . . . . . . . . . . . . . . . . . . . .

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8.19.3 Simply Supported Plate 8.19.4 Spring Supported Plate . 8.20 Problems . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . 9

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Eigenvalue Problems of Ordinary Differential Equation Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Vibrations Described by Differential Equation Systems . . . . 9.1.1 Engesser–Timoshenko Beam . . . . . . . . . . . . . . . . . 9.1.2 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . 9.2 Eigenvalue Problem for a Class of Differential Equations . . 9.2.1 The Differential Equation System . . . . . . . . . . . . . 9.2.2 Orthogonality of the Eigenfunction Vectors . . . . . . 9.2.3 On the Reality of the Eigenvalues for Eigenvalue Problem (9.45) . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.4 Rayleigh Quotient . . . . . . . . . . . . . . . . . . . . . . . . 9.2.5 Determination of Eigenvalues . . . . . . . . . . . . . . . . 9.2.6 The Green Function Matrix . . . . . . . . . . . . . . . . . . 9.2.7 Calculation of the Green Function Matrix . . . . . . . 9.2.8 Symmetry of the Green Function Matrix . . . . . . . . 9.2.9 Green Function Matrices for Some Beam Problems 9.2.10 Solution of Static Boundary Value Problems . . . . . 9.2.11 Multiple Eigenvalues of ODES . . . . . . . . . . . . . . . 9.2.12 Properties of the Rayleigh Quotient of ODES . . . . 9.2.13 Eigenvalue Problems Governed by a System of Fredholm Integral Equations . . . . . . . . . . . . . . . . . 9.2.14 Calculation of Eigenvalues . . . . . . . . . . . . . . . . . . 9.3 Free Vibrations of Timoshenko Beams . . . . . . . . . . . . . . . . 9.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

300 301 303 304

. . . . . . .

. . . . . . .

. . . . . . .

305 305 305 311 314 314 317

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

318 318 319 320 321 324 325 333 334 335

. . . . .

. . . . .

. . . . .

336 338 340 344 345

. . . . .

. . . . .

347 347 347 348 355

10 Eigenvalue Problems Described by Degenerated Systems of Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Vibration of Heterogeneous Curved Beams . . . . . . . . . . . . . 10.1.1 Introductory Remarks . . . . . . . . . . . . . . . . . . . . . . . 10.1.2 Equilibrium Equations . . . . . . . . . . . . . . . . . . . . . . 10.1.3 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Green Function Matrix for Degenerated Differential Equation Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Existence and Calculation of the Green Matrix Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 Green Function Matrices for Some Curved Beam Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 357 . . 357 . . 360 . . 364

xiv

Contents

10.2.4 Free Vibrations of Curved Beams . . . . . . . . . . . . . . . . 370 10.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 Appendix A: A Short Introduction to Tensor Algebra . . . . . . . . . . . . . . . 379 Appendix B: Some Useful Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . 393 Appendix C: Solutions to Selected Problems. . . . . . . . . . . . . . . . . . . . . . . 399 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445

Chapter 1

Introduction

1.1 Model Creation When one studies a dynamical phenomenon there arises the question what model should be applied. It is a fundamental expectation that the model established should reflect those features of the phenomenon in question which are the most important for the examination to be carried out. When making a model the following facts should, therefore, be taken into account: (a) what are the main objectives of the examinations (calculations) since the model should be based on them; (b) what are the limits of the applied simplifications when the model is established. For instance: – A particle is such a body the motion of which can be described by the motion of one point of the body considered. The earth is a particle when its motion is considered around the sun. – The rigid body is such a body for which the distance between any two but different points within the body remains constant during the motion. This means that a rigid body does not deform under the action of the applied forces. – A solid body is capable of deformation. – A mass distribution can be (i) discrete (then the considered dynamical system consists of particles and has a finite degree of freedom—the degree of freedom is the minimum number of coordinates required to describe completely the motion of a particle or body) or (ii) continuous (then the dynamical system has infinite degrees of freedom). – As regards the material behavior that can be either (i) linear (the springs in the system follow Hooke’s law) or (ii) non-linear.

© Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_1

1

2

1 Introduction

– The geometrical (kinematic) relations between the displacements and deformations can also be (i) linear or (ii) non-linear. – The constraints in the dynamical system can be either integrable (or holonomic) or non-integrable (nonholonomic)—these concepts are to be discussed later.

1.2 Kinematic Relations—A Summary 1.2.1 Motion of a Material Point (Particle) Figure 1.1 shows the Cartesian coordinate system (x yz)—the unit vectors of the coordinate axes are denoted by i ( = x, y, z)—the particle with mass m and its path on which s is the arc coordinate. The position vector of the particle is r(t) = r[s(t)] where t is time. The function r(t) is called motion law of the particle. The unit tangent and normal of the path are denoted by t and n, respectively. Let v be the velocity of the particle. It is obvious that v=

dr ds dr = = v(s)t(s), dt ds dt

(1.1a)

where v(s) = ds/dt

(1.1b)

is the speed of the particle. Given the velocity of the particle a=

dv dt ds dv = t+v dt dt ds dt

is the acceleration. Since

(1.2a)

dt n = ds ρ

Fig. 1.1 Resolution of the acceleration for particle m

(1.2b)

path s =0 S (t)

z

m r(t)= r[s(t )]

t

n a

iy

y

ix

x

a an

iz

O

at

v

C

1.2 Kinematic Relations—A Summary

3

in which ρ is the radius of curvature, Eq. (1.2a) can be rewritten in the following form: (1.2c) a = at t + an n, where at =

dv v2 and an = dt ρ

(1.2d)

are the acceleration components tangential to the path and perpendicular to it. If the motion of the particle is not constrained it has three degrees of freedom.  Typical units: [m/s] for the velocity and m/s2 for the acceleration. Fig. 1.2 The i-th particle in a system of particles

mi

z

vi

ri (t)

ai

O x

y

Considering a system of particles with masses m i (i = 1, . . . , N )—see Fig. 1.2 for details—the velocity and acceleration for the i-th particle can be given by vi = r˙ i =

dri dt

(1.3a)

ai = v˙ i =

dvi , dt

(1.3b)

and

where ri (t) is the motion law of the i-th particle.

1.2.2 Motion of a Rigid Body Figure 1.3 shows, among others, the rigid body B and the points A = B = G of the body with position vectors r A , r B , rG . The position vector of the point B with respect to the point A (the vector from A to B) is r AB . Velocities and accelerations at the points A and B are denoted by v A , a A and v B , a B . Further let ω [1/s] and α [1/s2 ] be the angular velocity and angular acceleration of the body. Let t be a given (fixed) point of time. Then v = v(r) ,

r∈B

4

1 Introduction

Fig. 1.3 Characteristic values for the states of velocity and acceleration in a rigid body

V G z

rG B

rB O

y

rAB

rA

x

vB

aB A

vA

aA

is the velocity distribution within body B—referred to as the velocity state of the body at time t. As is well known—see [1–3], for instance, v B = v A + ω × r AB = v A + r B A × ω , r AB = r B − r A = −r B A .

(1.4)

Hence, v A and ω uniquely determine the velocity state of the body B. For a given point of time t, the function a = a(r) , r ∈ B is the acceleration distribution within body B—referred to as the acceleration state of the body at time t. Since ˙ =α r˙ AB = r˙ B − r˙ A = v B − v A = ω × r AB and ω

(1.5)

a B = v˙ B = a A + α × r AB + ω × (ω × r AB ) .

(1.6a)

it follows that

Consequently, a A , ω, and α uniquely determine the acceleration state of the body B. Assume that the body is in a plane motion for which ω = ωz (t)iz = ω(t)iz , α = αz (t)iz = α(t)iz and r AB ⊥ iz —r AB is in the plane of motion. Making use of (A.1.10) we can write ω × (ω × r AB ) = (ω · r AB ) ω − (ω · ω) r AB = −ωz2 r AB = −ω 2 r AB    =0

thus a B = a A + αiz × r AB − ω 2 r AB .

(1.6b)

1.3 Fundamental Relations of Kinetics—A Summary

5

1.3 Fundamental Relations of Kinetics—A Summary 1.3.1 The First Moment of a Mass Distribution Figure 1.4 shows (a) a system of particles (for which the mass distribution is not continuous) and (b) a rigid body denoted by B which occupies volume V (for body B the mass distribution is, in general, a continuous—or a piecewise continuous— function of the position vector, ρ [kg/m3 ] stands for the density, ρ is the position vector of dm with respect to the material point B). V

mi ri(t ) z

z

r

y

A

x

y

A

x

rBA

d m = dV

rBA

B

B

Fig. 1.4 A system of particles and a rigid body

The first moment of a mass distribution with respect to the point A (about the point A) is denoted by Q A and is defined by the following relations: QA =

N 

m i ri ,

QA =

r ρ(r) dV .

(1.7a)

V

i=1

The total mass is given by the relations m=

N 

mi ,

m=

ρ(r) dV .

(1.7b)

V

i=1

  The unit of Q A is kg m . If ρ = ρ(r) = constant then the body is inhomogeneous. If ρ = ρ(r) = constant, i.e., ρ is independent of r then the body is homogeneous. It is obvious that



QB =

ρ ρdV = V



r ρdV +r B A

(r+r B A ) ρdV = V

V

ρdV = Q A +m r B A V

(1.8) is the first moment of the body with respect to the point B.

6

1 Introduction

The mass center (or the center of gravity) is the point (denoted by G) for which QG = 0 .

(1.9a)

For G = B it follows from Eq. (1.8) that r AG = −rG A =

QA . m

(1.9b)

1.3.2 Momentum Distribution—Vector System of Momenta A vector is called (a) free vector if its point of application can freely be selected, i.e., it is freely movable in space, (b) fixed vector if it has a unique point of application, and (c) sliding vector if it has a unique line of action on which its point of application can freely be selected. A system of vectors with fixed points of application is referred to as vector system. However, it may happen that some effects of the vectors that constitute the vector system mentioned are independent of their points of application1 and in this respect they behave as if they were sliding vectors. mi ri(t )

V

pi= m i vi

z

z

r dp= v dm = v dV

G x

y

A

A

vG

x

y

Fig. 1.5 Momentum vectors

The momentum vector is denoted by p and, for a single particle, is defined as p = mv [kgm/s]—m and v are the mass and the velocity of the particle. Resultants of [the momentum vectors pi ] and {the infinitesimal momentum vectors dp = vρdV } are given by p=

N  i=1

pi =

N 

m i vi

p=

i=1

v ρdV = m vG . V

Equation (1.10)2 follows from the manipulation 1 Forces

acting on (a rigid body)[deformable] body are (sliding)[fixed] vectors.

(1.10)

1.3 Fundamental Relations of Kinetics—A Summary



7



p=

v ρdV = ↑ = V

(1.4)

(vG + ω × ρ)ρdV = ↑ = (1.9a) ρ dV vG + ω × ρρ dV . =  V    V  

V

QG =0

m

We remark that some textbooks denote the momentum vector by L—see, for instance, book [1].

1.3.3 Moment of Momentum for a Rigid Body Figure 1.6 shows body B in the absolute (not moving) coordinate system (x yz). The points A and G are those of the body, the point G is the mass center. The coordinate system (ξηζ) with origin at G and the coordinate system (ξ  η  ζ  ) with origin at A move together with the body in such a way that coordinate axes (a) x, ξ, ξ  ; (b) y, η, η  ; and (c) z, ζ, ζ  remain parallel to each other, i.e., the Greek coordinate systems moving with the body do not change their orientation. In Fig. 1.6, the position vector of the infinitesimal mass element dm with respect to the mass center is denoted by ρ. The moment of momentum with respect to the point G is defined as Fig. 1.6 Various coordinate systems (fixed and moving with the body) V

dp= v dm = v dV m

z

rG

rAG O

n

y

rA A x

vA

HG =

ρ × vρ dV . V

It follows from Eq. (1.4) that v = vG + ω × ρ .

m

G

n

8

1 Introduction

Consequently, HG =

ρ × (vG + ω × ρ) ρ dV = ρ ρ dV × vG + ρ × (ω × ρ) ρ dV . = V  V   V

QG =0



Hence, HG =

ρ × (ω × ρ) ρ dV .

(1.11)

V

Remark 1.1 For our later consideration we shall introduce matrix notations. Let a and b be two vectors. Their matrices are written as ⎡ ⎤ ⎡ ⎤ ax bx a = a = ⎣ ay ⎦ and b = b = ⎣ by ⎦ . (1.12) (3×1) (3×1) az bz Further let

⎡

⎤ 0 −az a y a× = a× = ⎣ az 0 −ax ⎦ (3×3) −a y ax 0

(1.13)

be a skew matrix which is given in terms of ax , a y , and az . It is obvious that there belongs a similar skew matrix to any other vector. Consider now the cross product c = a × b.

(1.14a)

It is not too difficult to check that the column matrix c can be calculated as ⎡

⎤⎡ ⎤ 0 −az a y bx c = c = a× b = ⎣ az 0 −ax ⎦ ⎣ b y ⎦ . (3×1) (3×3) (3×1) −a y ax 0 bz

(1.14b)

Making use of the calculation rule for triple cross products from (A.1.10) we can transform Eq. (1.11) into the following form:

 2  ρ ω − ρ (ρ · ω) ρdV .

HG =

(1.15)

V

The matrices of ρ, ω, and HG in the coordinate system (ξ, η, ζ) are given by ⎡ ⎤ ξ ρ = ⎣η⎦ , (3×1) ζ

⎡

⎡ ⎤ ⎤ ωξ HGξ ω = ⎣ ωη ⎦ and HG = ⎣ HGη ⎦ . (3×1) (3×1) ωζ HGζ

(1.16)

1.3 Fundamental Relations of Kinetics—A Summary

9

A comparison of Eqs. (1.15) and (1.16) yields ⎫ ⎧ ⎡ ⎤ ⎡ ⎤ ⎨ 100 ξ  ⎬    2 HG = ρdV ω , ξ + η2 + ζ 2 ⎣ 0 1 0 ⎦ − ⎣ η ⎦ ξ η ζ ⎭ V ⎩ 001 ζ that is,

⎡

HG =

or

V

⎤ η 2 + ζ 2 −ξη −ξζ ⎣ −ηξ ξ 2 + ζ 2 −ηζ ⎦ ρdV ω −ζξ −ζη ξ 2 + η 2

⎡

⎤ ⎡ ⎤⎡ ⎤ HGξ Jξ −Jξη −Jξζ ωx ⎣ HGη ⎦ = ⎣ −Jηξ Jη −Jηζ ⎦⎣ ω y ⎦ , HGζ −Jζξ −Jζη Jζ ωz         HG (3×1)

(1.17a)

ω

J

G (3×3)

(3×1)

where JG is the matrix of inertia in which the diagonal elements

 2  η + ζ 2 ρdV , Jη =

Jξ = V





ξ +ζ 2

2





 2  ξ + η 2 ρdV

ρdV , Jζ =

V

V

(1.17b) are the moments of inertia with respect to the axes ξ, η, and ζ while the off-diagonal elements



Jξη = Jηξ =



ξη ρdV , Jηζ = Jζη = V

ηζ ρdV , Jζξ = Jξζ = V

ζξ ρdV V

(1.17c) are referred to as products of inertia. Let n and m be unit vectors at G—see Fig. 1.6. Further let n and m be perpendicular to each other. Then Jn = n T JG

n

(1×3) (3×3) (3×1)

and Jnm = Jmn = − m T JG

n = − n T JG

(1×3) (3×3) (3×1)

m

(1.17d)

(1×3) (3×3) (3×1)

are the moment of inertia with respect to the axis n, and product of inertia with respect to the axes n and m. Since the matrix of inertia JG is a symmetric matrix it holds that JG = J GT . (3×3)

(3×3)

The eigenvalue problem JG

n =J n ,

(3×3) (3×1)

(3×1)

nT n = 1,

(1×3) (3×1)

(1.18)

10

1 Introduction

in which J and n are the unknowns, has at least three real solutions for the vector n. If the number of solutions for n is equal to three the solution vectors ni (i = 1, 2, 3) are mutually perpendicular to each other:  ni · n j =

1 if i = j 0 if i = j

(i, j = 1, 2, 3) .

(1.19)

Fig. 1.7 Principal directions

n3 n3

n1

G

n1 n2

n2

If the number of solutions for n is more than three then the number of solutions is infinite: one can, however, always select such three solutions which are mutually perpendicular to each other. The direction the solution for ni determines is called principal direction. The eigenvalue (called principal moment of inertia) that belongs to the principal direction determined by ni is denoted by JGi . It is well known that ni JGi



 is referred to as principal

axis of inertia. moment of inertia.

The matrix of inertia is a diagonal one in the coordinate system (n 1 n 2 n 3 ) constituted by the principal axes: ⎤ ⎡ JG1 0 0 JG = ⎣ 0 JG2 0 ⎦ , (1.20) (n 1 n 2 n 3 ) 0 0 JG3 where we have assumed that JG1 ≥ JG2 ≥ JG3 ,

(JGi > 0 – JG is positive definite!) .

(1.21)

(3×3)

As regards the eigenvalue problem of symmetric tensors the reader is referred to Sect. A.2.6 in the Appendix (Fig. 1.7).

1.3 Fundamental Relations of Kinetics—A Summary

11

By using tensorial notations (1.15) can be written as     ρ2 ω − ρ (ρ · ω) ρdV = ρ2 1 − ρ ◦ ρ ρdV · ω, HG = V

V

where 1 is the unit tensor and ◦ stands for the operation sign of the dyadic product. Here   ρ2 1 − ρ ◦ ρ ρdV JG = (1.22) V

is the tensor of inertia at G. Using the tensor of inertia we can give HG as a product: HG = J G · ω .

(1.23)

The matrix JG = JG of the tensor of inertia J G is given by Eqs. (1.17). (3×3)

Let A be a point of the rigid body (A = G). The moment of momentum about the point A—see Fig. 1.6 for details—is defined by the equation HA =

ρˆ × vρ dV .

(1.24)

V

It can be proved that the matrix of H A can be calculated as HA (3×1)

= m r AG× v A + J A (3×3) (3×1)

ω = Q A× v A + J A

(3×3) (3×1)

ω ,

(1.25a)

⎤ v Aξ  v A = ⎣ v Aη ⎦ (3×1) v Aζ  ⎤

(1.25b)

(3×3) (3×1)

(3×3) (3×1)

where ⎡

0

 mr AG× = Q A× = m ⎣ ζ AG (3×3) (3×3) −η AG ⎡

⎤  −ζ AG η AG 0 −ξ AG ⎦ ,  ξ AG 0

Jξ  −Jξ  η −Jξ  ζ  J A = ⎣ −Jη ξ  Jη −Jη ζ  ⎦ (3×3) −Jζ  ξ  −Jζ  η Jζ 

⎡

(1.25c)

in which  • ζ AG , η AG , ξ AG are the coordinates of the mass center in the primed coordinate system; • v Aξ  , v Aη , v Aζ  are the velocity components in the same coordinate system; • Jξ  , Jη , Jζ  are the moments of inertia with respect to the axis ξ  , η  , ζ  ; and • Jη ξ  , Jξ  ζ  , Jη ζ  are the corresponding products of inertia.

12

1 Introduction

If the body rotates about the point A, which is regarded as a fixed point, then v A = 0, consequently Eq. (1.25a) simplifies to HA (3×1)

= JA

ω .

(1.26)

(3×3) (3×1)

It also holds that H A = HG + r AG × p .

(1.27)

1.4 Parallel Axis Theorem It can be proved that ⎡

⎤ ηG2 A + ζG2 A −ξG A ηG A −ξG A ζG A J A = J G + m ⎣ −ηG A ξG A ξG2 A + ζG2 A −ηG A ζG A ⎦ (3×3) (3×3) −ζG A ξG A −ζG A ηG A ξG2 A + ηG2 A   

(1.28a)

J

GA (3×3)

or J

A (3×3)

= J G + mJ G A , (3×3)

(1.28b)

(3×3)

where m is the mass of the body B, while ξG A , ηG A , and ζG A are the coordinates of the point A with respect to the point G in the coordinate system (ξηζ). Equations (1.28) constitute the well-known parallel axis theorem.

1.5 Fundamental Theorems of Kinetics 1.5.1 Introductory Remarks As is well-known kinetics is a study of motion and its causes. In the present section, we shall give a short summary of the most important concepts and theorems of kinetics. After successfully completing a course in dynamics, the students are expected to have a familiarity with the fundamental concepts and theorems of dynamics. On the other hand, a repetition is the mother of knowledge and refreshes the things you learnt earlier.

1.5 Fundamental Theorems of Kinetics

13

1.5.2 Force Systems Figure 1.8a shows a system of N particles. The j-th particle exerts a force Fi j (i, j = 1, . . . , N ) on the i-th particle: Fi j = −F ji (law of action and reaction), Fii = 0 (a particle does not exert a force on itself). The forces Fi j are internal forces. The forces exerted on a particle by the bodies (and/or the particles) that do not belong to the system are referred to as external forces. The resultant of the external forces acting on particle m i is denoted by Fi .

a ri

z

V

b

mi

P

Fi z

Fij

M r AG

y x

A

G

Fji rj

mj

F

Fj

x

y

A

Fig. 1.8 Force systems acting on a system of particles and a rigid body

Figure 1.8b shows a rigid body B. The body is loaded at the points P ( = 1, . . . , K ): an external force F and an external couple M are acting on the body at the point P —the couples can have values of zero. The force couple system equivalent to the external loads [at A—system of particles] (at G—rigid body) is given by the equations R=F=

N 

Fi ,

R=F=

MA =

i=1

ri × Fi ,

F ,

=1

i=1 N 

K 

MG =

K  =1

ρ × F +

K 

(1.29) M .

=1

It is customary to denote the equivalent force couple system by putting the resultant and the moment about the point considered (called simply resultant moment) into a pair of square brackets in which they are separated from each other by a comma: [R, M A ] or [R, MG ]. For the rigid body B, the moment of the external forces about the point A can be calculated as (1.30) M A = MG + r AG × R,

14

1 Introduction

where the right side is the moment of the external forces about the point G plus the moment of the resultant R—which is attached mentally to the point G—about the point A. Remark 1.2 It is assumed here and later in Sect. 1.5.4 that the external force system is a special one since it includes the forces F and couples M only. This assumption does not violate the generality: the fundamental theorem of dynamics presented in the subsection cited above remains valid for any loading type: for example, in the case of distributed loads as well.

1.5.3 Systems of Effective Forces The effective forces for [the system of particles] (the rigid body B) shown in Figs. 1.8 and 1.9 are defined by the relations [Ki = m i ai ] (dK = adm = a ρdV ). Note that dK is the intensity of the effective forces distributed on the volume V of the body.

a ri

z

V

b

mi i =m i a i

d = a dm = a dV z

r AG

y

A

x

G x

y

A

Fig. 1.9 Effective forces

The resultant of the effective forces for the system of particles is K=

N 

Ki =

i=1

N 

m i ai .

(1.31a)

i=1

The resultant of the effective forces acting on the rigid body B can be calculated as



K=

a ρdV = ↑ = (1.6a)

V

  ρdV aG + α × ρ ρdV + ω × ω × ρ ρdV = maG  V    V    V  

=

(aG + α × ρ + ω × (ω × ρ)) ρdV = V

m

QG =0

QG =0

1.5 Fundamental Theorems of Kinetics

15

or simply K = m aG .

(1.31b)

 Remark 1.3 Observe that we have taken into account that (a) V ρdV = m and have utilized the fact that the first moment of a rigid body with respect to the mass center vanishes QG = 0. A comparison of (1.10) and (1.31) yields K=

dp = p˙ . dt

(1.32)

The moment of the effective forces is, in general, denoted by D. For the system of particles, the moment of the effective forces about the fixed point A is DA =

N 

ri × Ki =

i=1

N 

ri × m i ai =

i=1

N 

ri ×

i=1

=

dpi = ↑ = dt vi ×pi =0

N d  d ri × pi = H A . dt i=1 dt

(1.33)

Remark 1.4 It is worth emphasizing that the sum N 

ri × pi

i=1

in the above equation is the moment of momentum with respect to the point A concerning the system of particles considered. For the rigid body B shown in Fig. 1.9b, the moment of the effective forces about the mass center G is given by



DG =

ρ × dK =

ρ × a ρdV = V

V

= ↑ =

ρ × (aG + α × ρ + ω × (ω × ρ)) ρdV = ρ ρdV × aG + ρ × (α × ρ) ρdV + ρ × [ω × (ω × ρ)] ρdV . = (1.6a)

V

V

V

V

(1.34) Before examining the three integrals on the right side of the above equation we shall introduce the following matrix notations for the moment of the effective forces D G (taken about G), the angular acceleration α, and the angular velocity ω:

16

1 Introduction

⎤ DGξ D G = D G = ⎣ DGη ⎦ , (3×1) DGζ ⎡ 0 ω × = ω × = ⎣ ωζ (3×3) −ωη ⎡

⎤ αξ α = α = ⎣ αη ⎦ , (3×1) αζ ⎤ −ωζ ωη 0 −ωξ ⎦ . ωξ 0 ⎡

(1.35)

• For the first integral on the right side of Eq. (1.34), it holds that 

V

ρ ρdV × aG = QG × aG = ↑ = 0 . (1.9a)  

(1.36a)

QG

• Recalling Eqs. (1.11) and (1.17a), we can come to the conclusion that the final form of the second integral ρ × (α × ρ) ρdV V

on the right side of Eq. (1.34) can be obtained from (1.17a) if we write α for ω. Consequently, equation ⎡

⎤⎡ ⎤ Jξ −Jξη −Jξζ αx ⎣ −Jηξ Jη −Jηζ ⎦⎣ α y ⎦ = J α G −Jζξ −Jζη Jζ αz     

(1.36b)

α

J

G (3×3)

(3×1)

is the matrix form of the second integral. • As regards the third integral on the right side of Eq. (1.34) take into account that ρ × [ω × (ω × ρ)] = ω × [ρ × (ω × ρ)] . (The validity of this equation can be checked with ease if we make use of Eq. (A.1.10) for calculating the value of the triple cross products and take into account that ω × ρ, which we regard as a single factor when applying the expansion of the cross product, is perpendicular both to ω and to ρ.) Upon substitution of the above equation into the third integral, we get

V

ρ × [ω × (ω × ρ)] ρdV =  ω× ω×



V

ρ × (ω × ρ) ρdV .   JG ω

Observe that the above equation shows the result in matrix notation too.

(1.36c)

1.5 Fundamental Theorems of Kinetics

17

Making use of Eqs. (1.36) for the moment of the effective forces about the mass center, we get DG (3×1)

= JG

α + ω× J G

(3×3) (3×1)

ω = JG

(3×3) (3×3) (3×1)







α + ω× H G .

(3×3) (3×1)

(1.37)

(3×3) (3×1)

HG

(3×1)

For completeness, here, we give the above equation in tensorial notation as well: D G = J G · α + ω × J G · ω = J G · α + ω × HG .

(1.38)

It can be proved that DG =

d HG . dt

(1.39)

Let A be a point of the rigid body B (A = G). The moment of the effective forces about the point A—see Fig. 1.10 for details—is defined by the equation DA =

ρˆ × aρ dV .

(1.40)

V

It can be proved by using the definition given above that DA (3×1)

= Q A× a A + J A (3×3) (3×1)

α + ω× J A

(3×3) (3×1)

ω ,

(1.41a)

(3×3) (3×3) (3×1)

Fig. 1.10 Effective forces on a rigid body depicted for the calculation of D A V z

d = a dm = a dV rG G rAG

O

y

rA A x

vA

aA

18

1 Introduction

where ⎡

0

 Q A× = mr AG× = m ⎣ ζ AG (3×3) −η AG

⎡ ⎤ ⎤  −ζ AG a Aξ  η AG 0 −ξ AG ⎦ , a A = ⎣ a Aη ⎦ . ξ AG 0 a Aζ 

(1.41b)

Here we give, again for completeness, Eq. (1.41a) in tensorial notation as well: D A = mr AG × a A + J A · α + ω × J A · ω = Q A × a A + J A · α + ω × J A · ω .

  

(1.42)

QA

If the body rotates about the point A, which is now regarded as a fixed point, then v A = a A = 0. Hence, Eq. (1.41a) simplifies to DA (3×1)

= JA

α + ω× J A

(3×3) (3×1)

ω .

(1.43)

(3×3) (3×3) (3×1)

Assume that the body B is in plane motion. Let the coordinate plane x y be the plane of motion. If the body is in plane motion the velocities and accelerations are all parallel to the plane of motion. Hence, the moment of momentum as well as the moment of the effective forces should be perpendicular to that plane. If there are no constraints to prevent the motion of the body B in the direction z (or which is the same perpendicularly to the plane of motion) the only way this condition can be satisfied is that the products of inertia Jζξ = Jζη = Jζ  ξ  = Jζ  η all vanish. Since for the plane motion considered ω = ωz iz = ωiz and α = αz iz = αiz it follows that ⎡

⎤ 0 J G α = ⎣ 0 ⎦ and (3×3) (3×1) Jζ α

⎡

⎤ 0 JG ω = ⎣ 0 ⎦ . (3×3) (3×1) Jζ ω

(1.44)

Consequently, ω× J G ω (3×3) (3×3) (3×1)

= 0,

(1.45)

and Eq. (1.37) simplifies to D G = Jζ α iz .

(1.46)

If the body in plane motion rotates about the fixed point A we get in the same way that (1.47) D A = Jζ  α iz .

1.5 Fundamental Theorems of Kinetics

19

1.5.4 The Fundamental Theorem of Dynamics Assume that the rigid body B is loaded at the point P ( = 1, . . . , K ) by an external force F and an external couple M —see Fig. 1.11. They together constitute the system of external forces (in a general sense since couples are also included in this system) acting on the body. Figure 1.11 shows the system of effective forces as well. The fundamental theorem of dynamics states that the system of effective forces is statically equivalent to the system of external forces. Consequently, it holds that m

[aρ dV ] = [F1 , . . . , F , . . . , F K ; M1 , . . . , M , . . . , M K ] , where the letter m over the equality sign expresses the fact that the two force systems are statically equivalent (they have the same moment space).

V

V

a dV m

=

r

z

P z

M r AG

r AG

G x

F

G y

A

y

A

x

Fig. 1.11 Equivalence of the external and effective forces

We have already seen—recall Eq. (1.29)—that the equivalent force couple system [R, MG ] for the external loads at the point G is given by R=

K 

F ,

MG =

=1

K 

ρ × F +

=1

K 

M .

(1.48a)

=1

As regards the effective forces, the equivalent force couple system at G is of the form [K, D G ]. It follows from Eqs. (1.31b) and (1.37) that K and D G can be calculated as K = m aG ,

(3×1)

(3×1)

DG (3×1)

= JG

α + ω× H G .

(3×3) (3×1)

(1.48b)

(3×3) (3×1)

In tensorial notation, we can write K = m aG ,

D G = JG · α + ω × HG .

(1.48c)

20

1 Introduction

For two force systems to be statically equivalent, it is necessary and sufficient that the resultants and moment resultants are equal: K = m aG = R ,

D G = MG .

(1.49a)

These equations constitute the equations of motion for a rigid body. If the body rotates, say, about the fixed point A then the second equation takes the form D A = MA .

(1.49b)

˙ G (see Eqs. (1.39)) Eqs. (1.49a) can Since K = p˙ (see Eq. (1.32)) and D G = H also be written in the form p˙ = R ,

˙ G = MG . H

(1.49c)

Let t1 and t2 be two different points of time (t1 < t2 ). Integrate equations (1.49c) with respect to time. We obtain p(t2 ) − p(t1 ) = p2 − p1 =

t2

R dt, t1



t2

HG (t2 ) − HG (t1 ) = HG2 − HG1 =

MG dt t1

or



t2

p2 = p1 +

R dt ,

HG2 = HG1 +

t1

t2

MG dt .

(1.49d)

t1

Equations (1.49d) are known as the principles of impulse and momentum for a rigid body [1].

1.5.5 Kinetic Energy of a Rigid Body. Power The kinetic energy is denoted by E. It is obvious from Fig. 1.12 that dE = Thus E=

1 2

v2 ρdV = ↑ = V

(1.4)

1 1 2 v dm = v2 ρdV . 2 2 1 2



  v · v A + ω × ρˆ ρdV = ↑ ↑ = (1.10) (1.24) V 1 1 v ρdV · v A + ω · ρˆ × v ρdV = 2 V 2      V  p

HA

1.5 Fundamental Theorems of Kinetics

21

Fig. 1.12 Representation of the velocities for calculating the kinetic energy

or shortly E=

1 (v A · p + ω · H A ) . 2

(1.50)

Special cases: • If A = G then

2 = m v TG v G . vG · p = ↑ = m vG (1×3) (3×1)

(1.10)

Let us give ω in the form ω = ω nω , |nω | = 1. Then we can write ω · HG = ↑

↑ = ω 2 nTω J G n ω ,

(1.17a)(1.17d)1

(1×3) (3×3) (3×1)







JGω

where JGω is the moment of inertia with respect to an axis parallel to ω and passing through the mass center G. Consequently,    1 2 1 T 2 T m vG vG + ω nω JG nω = m vG + JGω ω 2 . E= 2 2 (1×3) (3×3) (3×1) (1×3)(3×1) (1.51a) If ω = 0 the above equation simplifies to E=

1 2 m vG . 2

(1.51b)

22

1 Introduction

• If v A = 0 then a similar line of thought yields E=

1 2 T 1 ω n ω J A n ω = J Aω ω 2 , 2 2 (1×3) (3×3) (3×1)   

(1.52)

J Aω

where J Aω is the moment of inertia with respect to an axis parallel to ω and passing through the point A.

Fig. 1.13 Forces and velocities on a rigid body

Assume that the rigid body B is loaded at the points P ( = 1, . . . , K ) by an external force F and an external couple M —see Fig. 1.13 for details. The equivalent force couple system at the point A is given by R=F=

K  =1

F ,

MA =

K  =1

M +

K 

r A P × F .

(1.53)

=1

As is well known PF = F · v and PM = M · ω

(1.54)

are the powers of the force F and couple M , where v is the velocity at P and ω is the angular velocity of the body. Hence, the power of this force system is the sum

1.5 Fundamental Theorems of Kinetics

P=

K 

(PF + PM ) =

=1

= vA ·

K 

F +

=1

 K  =1

K 

23

K    F · v A + ω × r A P + M · ω =

=1

M +

K 

=1



r A P × F · ω = R · v A + M A · ω .

(1.55a)

=1

If A = G Eq. (1.55a) is of the form P = R · vG + MG · ω ,

(1.55b)

in which MG is the moment of the external forces about the mass center G.

1.5.6 Principle of Work and Energy It can be shown that d E =P. dt

(1.56)

This equation is the differential form of the principle of work and energy. For A = G, Eq. (1.50) assumes the form E=

1 1 1 (vG · p + ω · HG ) = vG · p + ω · HG 2 2 2       Ev

Eω

As regards the time derivative of Ev , we get E˙v =

1 1 1 1 ↑ = vG · p˙ = R · vG . aG · p + vG · p˙ = vG · maG + vG · p˙ = ↑  2  2  2 2 (1.48c)(1.49a)1 p˙

mvG

R

For the time derivative of Eω , a similar line of thought yields E˙ω =

=

1 1 1 1 ˙G = ↑ ↑ = α · JG · ω + ω · D G = ↑ α · HG + ω · H 2 2 2 2 (1.17a) (1.39) (1.49a)2

1 ω· 2

JG · α   

=D G −ω×HG

+

↑ =

(1.48c)2

1 1 1 1 MG · ω = ω · D G − ω · (ω × HG ) + MG · ω = MG · ω . 2 2 2 2    =0

Consequently, d E = E˙ = R · vG + MG · ω = P , dt which is a proof of Eq. (1.56).

(1.57)

24

1 Introduction

Let t1 and t2 be two different points of time (t1 < t2 ). Integrate equation (1.56) with respect to time and take into account that the time integral of the power P is the work, which is denoted, in general, by W , done by the external forces in the time interval [t1 , t2 ]. We get the integral form of the principle of work and energy: E2 − E1 =

t2

Pdt = W12 .

(1.58a)

t1

Remark 1.5 If the external forces have a potential U then W12 = U1 − U2 . Consequently E2 + U2 = E1 + U1 . (1.58b) This is the principle of the energy conservation. Exercise 1.1 A cylinder with weight m cyl g (g is the gravitational acceleration) is lifted up by the crane shown in Fig. 1.14. Let Je be the moment of inertia of the engine with respect to its axis. Further let m e g be the weight of the engine. The simply supported beam AB is a rigid one and the rope is inextensible. Using the notations of Fig. 1.14—M is the lifting moment, ϕ is the angle of rotation about the axis of the engine (ϕ = 0 if z = h o )—determine the motion equation of the cylinder. Fig. 1.14 A crane is lifting up a weight

engine

R C

M B

A

z ho

m cyl

It is clear from Fig. 1.14 that z = h o − Rϕ from where z˙ = −R ϕ. ˙ The total kinetic energy is the sum of the kinetic energies the engine and the cylinder have. Using (1.51a), we get

1.5 Fundamental Theorems of Kinetics

E=

25

 1 1 1 1 m cyl z˙ 2 + Je ϕ˙ 2 = m cyl R 2 + Je ϕ˙ 2 = m gen ϕ˙ 2 2 2 2 2  

(1.59)

m gen

in which m gen is referred to as generalized mass. The power of the external forces is due to the lifting moment M and the weight of the cylinder:   ˙ P = M ϕ˙ + m cyl g z˙ = M − m cyl g R ϕ.

(1.60)

Upon substitution of Eqs. (1.59) and (1.60) into the principle of work and energy (1.56), we obtain   m gen ϕ˙ ϕ¨ = M − m cyl g R ϕ˙ or m gen ϕ¨ − M = −m cyl g R .

(1.61)

If the lifting moment M is constant we have a simple linear differential equation (DE for short) with constant coefficients. It is worth mentioning that M, in general, depends on the angular velocity: M = M(ϕ). ˙ Then, Eq. (1.61) is, in most cases, a non-linear differential equation. Exercise 1.2 Assume that the beam and the rope in Exercise 1.1 are elastic. Further assume that their masses are negligible when compared to those of the engine and the cylinder. Determine the kinetic energy of the system (Fig. 1.15). Fig. 1.15 A crane on an elastic beam is lifting up a weight

engine

R C

M

A

B y z

m cyl

Let y be the vertical displacement at the middle of the beam. The elastic elongation of the rope is denoted by η. The total kinetic energy of the system is a sum: the kinetic energy of the engine plus that of the cylinder. Making use of (1.51a) and (1.52) we can write E=

1 1 1 m cyl (˙z + η) ˙ 2 + Je ϕ˙ 2 + m e y˙ 2 = 2 2 2 1 1 1 = m cyl (−R ϕ˙ + η) ˙ 2 + Je ϕ˙ 2 + m e y˙ 2 = 2 2 2  2 2  1 1 1 = m cyl R ϕ˙ − 2R ϕ˙ η˙ + η˙ 2 + Je ϕ˙ 2 + m e y˙ 2 2 2 2

26

1 Introduction

⎡ ⎤⎡ ⎤ 2 ϕ˙  m cyl R + Je 0 −m cyl R 1 ⎦ ⎣ y˙ ⎦ ϕ˙ y˙ η˙ ⎣ 0 0 me E= 2 η˙ 0 m cyl −m cyl R   

from where

M (3×3)

in which M = MT is the mass matrix. Exercise 1.3 The mass m, the mass center G, i.e., r G , and the geometric data are all known for the crankrod shown in Fig. 1.16. Perform measurements to determine the moment of inertia Ja of the crankrod with respect to the axis about which it rotates. A

i

y

i rG G

Solution to the problem raised is based on the equation of motion that describes the rotation of the crankrod about A. We shall assume small rotations. Since 1 E = Ja ϕ˙ 2 2

i

is the kinetic energy of the crankrod and x

iR

P = W · vG = mg ix · r G ϕ˙ iϕ =   = mg ix · r G ϕ˙ − sin ϕ ix + cos ϕ i y =   

Fig. 1.16 A crankrod

iϕ

= −mg r G ϕ˙ sin ϕ is the power of the weight W = mg ix of the crankrod it follows from the principle of work and energy (1.56) that Ja ϕ˙ ϕ¨ = −mg r G ϕ˙ sin ϕ or Ja ϕ¨ + mg r G sin ϕ = 0 . Small rotation has been assumed. Then |ϕ| 1 and sin ϕ = ϕ − · · · ≈ ϕ. Consequently ϕ¨ +

mg r G ϕ = ϕ¨ + ωn2 ϕ = 0 Ja    ωn2

ϕ3 3!

+

ϕ5 5!

+

(1.62)

1.5 Fundamental Theorems of Kinetics

27

is the linearized equation of motion. Its general solution is of the form ϕ = C1 sin ωn t + C2 cos ωn t = = Ao cos φ sin ωn t + Ao sin φ cos ωn t = Ao sin (ωn t + φ)       C1

C2

in which C1 and C2 are integration constants (they depend on the initial conditions), Ao is the amplitude, and φ is the so-called phase angle. The solution obtained shows that the crankrod performs harmonic vibration with an angular velocity ωn , consequently the period of vibrations τn is  Ja 2π τn = = 2π . ωn mg r G

(1.63)

After we have measured the period of vibrations τn Eq. (1.63) can be used to calculate the moment of inertia sought: Ja =

1 mg r G τn2 . 4π 2

(1.64)

Exercise 1.4 Generalize the line of thought presented in the previous exercise for the case of an arbitrary rigid body.

a

b

c

Fig. 1.17 A rigid circular tray

Figure 1.17a shows a rigid circular tray and the way it is hung up. Figure 1.17b shows the body B on the tray. Its mass center G is located on the vertical center line n (|n| = 1) of the tray. The tray and the body can perform small torsional vibrations. The equation of motion established for the torsional vibrations makes it possible to determine the moment of inertia Jn = Jξ of body B.

28

1 Introduction

It is not too difficult to check that the following approximative kinematic relations hold: r ϑ = r ϕ −→ ϑ= ϕ,   r  x =  (1 − cos ϑ) =  1 − cos ϕ 

and

from where we obtain x˙ = r ϕ˙ sin

r2 r ϕ= ↑ ∼ = ϕ˙ ϕ . r    ϕ 1

Let m tray be the mass of the tray. Let us denote the moment of inertia of the tray with respect to the axis n by Jtray . Then E=

  1 1 Jn + Jtray ϕ˙ 2 + m B + m tray x˙ 2 2 2

is the kinetic energy of the system. In practice Jn = Jξ Jtray and the part of the kinetic energy due to the vertical motion can be neglected. Consequently, we may assume that ∼ 1 Jn ϕ˙ 2 . E= 2 The power of the external forces is given by   r2 P = − m B + m tray g x˙ = −mg x˙ = −mg ϕ˙ ϕ .     m

With the knowledge of E and P we get the equation of motion from the principle of work and energy (1.56): r2 Jn ϕ¨ + mg ϕ = 0  or which is the same ϕ¨ +

mg r 2 ϕ = 0. J  n 

(1.65)

ωn2

Since the torsional vibrations are harmonic it follows that τn =

2π , ωn

i.e.,

Jn = Jξ =

mg r 2 2 τ . 4π 2  n

(1.66)

1.5

Fundamental Theorems of Kinetics

29

Assume now that the coordinate system (ξηζ) is rigidly attached to the mass center G of the body. Then, as we have just seen, Jn = Jξ . Put the body B onto the tray in such a way that (a) (b)

n = iξ , iη , iζ  1 n = iξ + iη 2  1 n = iη + iζ 2  1 n = iζ + iξ 2

then we get

Jn = Jξ , Jη , Jζ ;

1 2 1 Jn = nT J G n = 2 (1×3)(3×3)(3×1) 1 Jn = nT J G n = 2 (1×3)(3×3)(3×1) Jn = nT J G

n =

(1×3)(3×3)(3×1)

then we get



 Jξ + Jη − 2Jξη ,



 Jη + Jζ − 2Jηζ ,



 Jζ + Jξ − 2Jζξ .

By solving the last three equations for Jξη , Jηζ , and Jζξ , we obtain the missing three elements of the matrix JG .

1.6 Problems Problems 1.1 Prove that Jn and Jmn are the moment of inertia and product of inertia. Problems 1.2 Prove that the moment of momentum about the point A is given by Eq. (1.25a). Problems 1.3 Prove the parallel axis theorem (1.28). Problems 1.4 Show that Eq. (1.39) is true. (The proof is based on the Coriolis theorem. In this respect, the reader is referred to Appendix C.)

References 1. F.P. Beer, Jr. E.R. Johnston, D.F. Mazurek, P.J. Cornwell, E.R. Eisenberg, Vector Mechanics for Engineers Statics and Dynamics (McGraw-Hill, New York, 2010) 2. I. Sályi. Engineering Mechanics I. Elements of Kinematics (in Hungarian) (Tankönyvkiadó, Budapest, 1960) 3. G. Csernák. Dynamics (in Hungarian). Akadémiai Kiadó (Publisher of the Hungarian Academy of Sciences, Budapest, 2018)

Chapter 2

Impact

2.1 What Is Meant by Impact The collision between two bodies, which occurs in a very short time period and during which the two bodies exert relatively large impulsive forces on each other (if there are no constraints these forces are much greater than the forces exerted on the two bodies by other bodies—then the effects of the former forces can be neglected) is called impact. A typical example of impact is a bat striking a ball. Figure 2.1a, b shows bodies B1 and B2 just before collision and at the beginning of impact. The outward unit normal vectors of bodies B1 and B2 are denoted by n1 and n2 . The common normal to the surfaces in contact is the line of impact, see Fig. 2.1b, c. If the mass centers G 1 and G 2 are located on this line then we speak about central impact, otherwise the impact is eccentric—the impact shown in Fig. 2.1 is central, the issue of eccentric impact will be considered later. The impact is called direct if v1 ||v2 ||n1 = −n2 —see Fig. 2.1b—and is oblique if the velocity of one or both of the bodies is not parallel to the line of impact—see Fig. 2.1c. When clarifying how to solve impact problems we shall apply the following assumptions: (a) Bodies B1 and B2 are rigid except a small neighborhood of the point A1 = A2 = A, i.e., the point at which they collide— Ai = A is on body Bi (i = 1, 2). (b) The normals n1 = −n2 do not change during impact. (c) If there are no constraints the contact forces F12 = −F21 —here (F12 )[F21 ] is the force exerted by body (B2 )[B1 ] on body (B1 )[B2 ]—are much greater than the other forces acting on bodies B1 and B2 . Then the effects of the other forces can, therefore, be neglected. (d) Bodies B1 and B2 are smooth, i.e., the lines of actions of the contact forces coincide with the common normal (with the line of impact).

© Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_2

31

32

2 Impact

a

b n1

A1

n2

A2

v1

v2

G2

A 1=A 2=A

v2

G2

B2

v1

B2

G1

G1

B1

B1

c

A 1=A 2=A G1

G2

v2

B2

v1

B1 Fig. 2.1 Two bodies just before collision and at the beginning of the impact

Remark 2.1 If the impact is central the contact forces F12 = −F21 have no moment about the mass centers G 1 and G 2 . Consequently, the angular velocities ω 1 and ω2 the bodies have when impact begins will remain unchanged during impact. For central impact the velocities at the mass centers before and after impact are denoted by v1 , v2 and v1 , v2 .

2.2 Central Impact Figure 2.2 shows the two bodies when the impact begins. If v1 · n1 > v2 · n1 then body B1 and B2 will deform. Let us denote the velocities during impact by v1c and v2c —c is the first letter in the word collision. Deformation ends when v1c · n1 = v2c · n1 ,

(2.1)

which means that the velocity components on the line of impact are equal. This quantity is denoted by cn . Hereupon a period of restitution takes place. By the end of the period of restitution (depending on the magnitude of the contact forces and the material properties of the two bodies) both bodies will have regained their original shape and form (the impact is then perfectly elastic) or will stay permanently deformed (the impact is then plastic). If there is no restitution at all the impact is called perfectly plastic. We shall assume that t = 0 when the impact begins and t = τ when it ends. During impact time is denoted by tc ∈ [0, τ ].

2.2 Central Impact

33

Fig. 2.2 Two bodies when impact begins

t G1

v1 n1

A n2

G2 v2

B1 v1

n

B2 v1 n 1 v2 n 1

For a system of particles, the position vector of the mass center is given by Eq. (1.9b). If the system consists of two bodies only then r OG = rG = (m 1 r1 + m 2 r2 )/(m 1 + m 2 ), from where m 1 r1 + m 2 r2 = (m 1 + m 2 )rG , m 1 v1 + m 2 v2 = (m 1 + m 2 )vG , m 1 a1 + m 2 a2 = (m 1 + m 2 )aG = K = R = F12 + F21 = 0 .

(2.2)

Equation (2.2)3 is a consequence of Eq. (1.49a)2 which is a necessary condition for the equivalence of the external forces and the effective forces. Equation (2.2)2 yields m 1 v1 +m 2 v2 = (m 1 +m 2 )vG = p1 +p2 = p = m 1 v1c +m 2 v2c = constant.

(2.3)

This equation reflects that the momentum of the system is preserved during impact. For our later considerations, we rewrite the above equation into two different forms:

vG =

m1 m2 m1 + m2 m2 v1 + v2 = v1 + (v2 − v1 ) = m1 + m2 m1 + m2 m1 + m2 m1 + m2 m2 m1 = v1 + (2.4a) (v2 − v1 ) = v2 + (v1 − v2 ) m + m2 m + m2      1  1

or vG =

m1 m2 m1 + m2 m2 v1c + v2c = v1c + (v2c − v1c ) = m1 + m2 m1 + m2 m1 + m2 m1 + m2 m2 m1 = v1c + (2.4b) (v2c − v1c ) = v2c + (v1c − v2c ) . m1 + m2 m1 + m2      

34

2 Impact

G1c

vy = v t

G1 m2 m1 m2

v2

v1

v1c v1

G

vG

v1 vx = vn

v2c

v2

v2

G2c

G2 m1 m1 m2

v1

v2

Fig. 2.3 Velocity diagram

Figure 2.3 represents the velocities v1 , v2 , vG , and v1c , v2c as well as the points G1 , G2 , G, and G1c , G2c in the coordinate system vx = vn , v y = vt . It follows from Eqs. (2.4)—observe what directions the terms that are underbraced have—that the points G1 , G2 , G constitute a straight line. The points G1c , G2c , and G are also on a straight line. The two straight lines intersect each other at the point G. It also holds that GG1 GG1c m2 = = . (2.5) GG2 GG2c m1 It is, however, an open issue what directions the line segments G1 G1c and G2 G2c have.

t

a G1

v1

b

t

t F12 n

n1 = n A

G2

F21

n

A n 1 = n v2

B1

B2

Fig. 2.4 The plane coordinate system (n; t)

Consider now the motion of body B1 —see Fig. 2.4a. It is clear that m 1 a1 = m 1 v˙ 1 = F12 = F12 n1 ,

F12 < 0 .

(2.6a)

2.2 Central Impact

35

Assume that t = 0 when the impact begins and let tc be the time during impact. Integrating Eq. (2.6a) from t = 0 to tc , we get  m 1 (v1c − v1 ) = n1   

tc

F12 dt .

(2.6b)

t=0

p1

Hence, v1 (tc ) = v1c − v1 || n1 .

(2.6c)

For body B2 a similar line of thought yields m 2 a2 = m 2 v˙ 2 = F21 = F21 n1 ,  m (v − v2 ) = n1    2 2c p2

tc

F21 > 0 ,

(2.7a)

F21 dt ,

(2.7b)

t=0

v2 (tc ) = v2c − v2 || n1 .

(2.7c)

For our later considerations, it is worth introducing the plane coordinate system (n, t); n1 = n, |t| = 1—see Fig. 2.4a. As is mentioned in Remark 2.1 velocities at the end of restitution are denoted by primed letters: v1 , v2 . On the basis of all that has been said so far Fig. 2.5—referred to as Maxwell’s diagram—shows how the velocities change during impact [1]. Fig. 2.5 Velocity diagram with velocity changes

t

v1n cn

v1n’ v1’

v1(t c )

G1’ G1c

v1

G1

v1t

G n

v2t

v2’

v2 v2n

G2

v2(t c) v2n’

G2c

G2’

36

2 Impact

Note that the velocity changes in Fig. 2.5 are, in accordance with Eqs. (2.6c) and (2.7c), parallel to n. However, we do not know the velocity changes, that is, the locations of the points G1 and G2 in Fig. 2.5. In the sequel, we shall use the coordinate system (n, t) introduced above. In this coordinate system (2.8) v = vn n + vt t is, in general, the form of a velocity vector. (a) It follows from Eqs. (2.6c) and (2.7c) that  = v1t , v1t

 v2t = v2t ,

(2.9a)

which means that the velocity components perpendicular to the line of impact remain unchanged. (b) The straight line perpendicular to the line of impact (to n) through the point G— Fig. 2.5—determines cn , i.e., the end of deformation since v1cn = v2cn = cn . Recalling Eq. (2.3) we can write

and

(m 1 + m 2 ) cn = m 1 v1n + m 2 v2n

(2.9b)

  + m 2 v2n = m 1 v1n + m 2 v2n . m 1 v1n

(2.9c)

Equations (2.9a) (two equations) (2.9b) (one equation), (2.9c) (one equation) involve five unknowns:     v1n , v1t , v2n , v2t , cn .

Since we have four equations for five unknowns a further equation is needed. It is an experimental experience that  = e (v1n − cn ) , cn − v1n  cn − v2n = e (v2n − cn ) ,

e ∈ [0, 1],

(2.9d)

where e is the coefficient of restitution. We remark that Eqs. (2.9d) uniquely determine the locations of points G1 and G2 in Fig. 2.5. It is worth now drawing a figure which may serve as a basis for the graphical solution of the impact problem. It is clear from Fig. 2.6, which is a summation of our results—recall Figs. 2.3, 2.5 and Eqs. (2.9d)—that the solution to the impact problem can be obtained by performing the following steps: (a) Construct the straight line G1 G2 and then locate the point G on that line.

2.2 Central Impact

37

t

v1n cn e (v1n cn)

v’1n

v1n cn

G1c

G1’

G1

v1

v1’

m2 m1 m2

v1t

G

vG

vG

1 m1 m2

v 2n

v 1n

m 1 v1

m 2 v2

v 1n

v 2n

n

v2t

v2’

v2

G2 v2n

G2c

cn v2n

m1 m1 m2

G 2’ e(cn v2n)

v2’n Fig. 2.6 Velocity diagram with the graphical solution

(b) Draw a straight line perpendicularly to n through the point G. This makes possible to find the points G1c and G2c which lie on the horizontal lines that pass through the points G1 and G2 . (c) Mark off the distance (|e (v1n − cn ) |) [|e (v2n − cn ) |] starting at (and to the left of the point G 1c if v1n > v2n or to the right if v1n < v2n ) [and to the right of the point G2c if v1n > v2n or to the left if v1n < v2n ]. Now you have the points G1 and G2 which determine the unknown velocities v1 and v2 . If e = 0 this step is unnecessary: the unknown velocities are determined by the points G1c and G2c . By subtracting Eq. (2.9d)2 from Eq. (2.9d)1 , we obtain   − v1n = e (v1n − v2n ) . v2n

(2.9e)

 Equations (2.9a), (2.9c) and (2.9e)—four equations—involve four unknowns (v1n ,   v2n , and v2t ): we have as many equations as there are unknowns. If e = 1 the impact is perfectly elastic—the two bodies regain their original shape. If e = 0 the impact is perfectly plastic—there is no restitution at all. If e ∈ (0, 1) the impact is partially elastic. With m 1 v1n + m 2 v2n (2.10) cn = m1 + m2

 , v1t

equation system (2.9c) and (2.9d) can be rewritten into the following form:

38

2 Impact   m 1 v1n + m 2 v2n = (m 1 + m 2 ) cn ,   v2n − v1n = e (v1n − v2n ) .

(2.11)

It is not too difficult to check that the solutions to this equation system are m2 m1  = cn + e (v1n − v2n ) , v2n (v1n − v2n ) . m1 + m2 m1 + m2 (2.12) We remark that these solutions can be read off Fig. 2.6 as well. If e = 1 there is an energy loss which is given by the following equation:  v1n = cn − e

Eloss =

 m1m2 1 1 − e2 (v1n − v2n )2 . 2 m1 + m2

(2.13)

The proof of this equation is left for Problem 2.1. Exercise 2.1 Measure the coefficient of restitution e.

B1

m1

a

Position 1

B1

m1

Position 3

h b

H

v1n’ = e v1n

Position 2

t m2=

v1n

B2

t m2=

n

v1n’

B2

O =G v1n

n

Fig. 2.7 A body bouncing off the ground

Let us drop body B1 (Particle 1) with mass m 1 from a height h onto body B2 . It is assumed that the material properties are the same for the two bodies which collide— see Fig. 2.7a. Since body B2 is, in fact, a half-space it follows that m 2 = ∞. Then  v2n = v2n = 0 and Eq. (2.9d) yields  = ev1n . − v1n

(2.14)

2.2 Central Impact

39

As regards the velocity v1n the principle of work and energy (1.57) says E2 − E1 = W12 , where E1 = 0 ,

E2 =

1 2 m 1 v1n , 2

W12 = m 1 gh .

Consequently, 1 2 m 1 v1n = m 1 gh , ⇒ 2

|v1n | =



2gh.

(2.15a)

 till it reaches the After impact body B1 bounces back with an initial velocity v1n height H (H ≤ h). It is obvious that  | = 2g H . (2.15b) |v1n

A comparison of Eqs. (2.14) and (2.15) gives the following result:

 | H |v1n = . e= |v1n | h

(2.16)

This equation says that the measurement of e requires the measurement of distances. Since m 2 = ∞ it follows from Eq. (2.4a) that vG = 0 , ⇒

G = O.

Under these conditions Maxwell’s diagram simplifies to the diagram shown in Fig. 2.7b. Exercise 2.2 A ball strikes a rigid wall. After impact the ball has a velocity v1 = −3ix − 4i y [m/s]. The coefficient of restitution e = 0.5. Determine the velocity the ball had before impact.

y =t m2= A x =n

n=i x v1’

B2

t O =G

4 m/s

G1

v 1t = v ’1t =

B1

v1’

v1

4 3

v1’n= ev1n= 3 m/s

n

6 v 1n= v1’n /( e) = 6 m/s

Fig. 2.8 Maxwell’s diagram for the ball that strikes a rigid wall

The solution can be read off Fig. 2.8: v1 = 6ix − 4i y [m/s].

40

2 Impact

Exercise 2.3 A sphere rebounds horizontally as shown in Fig. 2.9 after striking an inclined plane with a vertical velocity v1 of magnitude 0.5 m/s. Determine the velocity after impact and the coefficient of restitution if tan α = 0.75. Fig. 2.9 Maxwell’s diagram for a sphere rebounding horizontally

z

n

B1 G1

v1’

O =G

m2= A

B2

v1t = v’1t

n v1x ’

v’1n

v1

x

v1z

t

t v1n

It follows from Maxwell’s diagram that  |v1x | = tan α , |v1z |

  |v1x | = v1x = |v1z | tan α = 0.5 · 0.75 = 0.375 m/s .

On the other hand, 1 = 0.5 × 0.8 = 0.4 m/s , |v1n | = |v1z | cos α = |v1z | √ 1 + tan2 α tan α    |v1n | = |v1x | sin α = |v1x |√ = 0.375 × 0.6 = 0.225 m/s . 1 + tan2 α Consequently, see Eq. (2.16) e=

 0.225 | |v1n = = 0.5625. |v1n | 0.4

Exercise 2.4 A ball is thrown with a velocity vo = 4ix + 4i y [m/s]. What velocity will have had the ball after rebounding from the plane if the air resistance can be neglected and the coefficient of restitution e = 0.75? If there is no air resistance v1x = v1t = vox = 4 m/s ,

−v1y = voy = v1n = 4 m/s .

It follows from Maxwell’s diagram—see Fig. 2.10 for details—that  = −ev1n = −0.75 × 4 = −3 m/s ; v1n

v1 = 4ix + 3i y [m/s] .

2.2 Central Impact

41

y

y

v1t =vox

v1’

vo

m1 m 2=

v1n’ = e v1n

O =G

t =x

t =x

v1 v1n= voy

n n Fig. 2.10 Maxwell’s diagram for a thrown ball

Exercise 2.5 Two balls with masses m 1 = 2 kg and m 2 = 6 kg collide—see Fig. 2.11 for details. The velocities v1 and v2 are known. Determine the velocities v1 and v2 if (a) e = 0 (perfectly plastic impact), (b) e = 1 (perfectly elastic impact), (c) e = 0.6 (partially elastic impact). v1 = 4ix + 4i y [m/s] ,

v2 = −4i y [m/s] .

t =y v1

A

G1

B1 t

G2 v2

n =x

B2 t

t 1

2

3

v1’

3

v1’

v1

4

1

n

O

0.8

3

v1

v1’

v1

n

O

1

n

O

2

v2 v2’

G

4

G v2’

v2

a

1

1

Fig. 2.11 Impact of two balls with graphical solution

v2

b

G v2’ 1 0.6

c

42

2 Impact

The velocity vG is obtained from Eq. (2.4): vG =

 1 1 8ix + 8i y − 24i y = ix − 2i y [m/s] . (m 1 v1 + m 2 v2 ) = m1 + m2 8

After having determined vG we can construct three Maxwell’s diagrams: The solutions in [m/s] can be read off from Fig. 2.11a–c. (a) (b) (c) v1 = ix + 4i y , v1 = −2ix + 4i y , v1 = −0.8ix + 4i y , v2 = ix − 4i y , v2 = 2ix − 4i y , v2 = 1.6ix − 4i y . Exercise 2.6 Two balls with masses m 1 = 2 kg and m 3 = 6 kg collide—see Fig. 2.12 for details. The velocities v1 and v2 are known, v1 = 6ix [m/s], v2 = 2ix [m/s]. Determine the velocities v1 and v2 if (a) e = 0 (perfectly plastic impact), (b) e = 1 (perfectly elastic impact).

t =y v1 A

n =x

G2

G1

B1

v2

t

B2

t

v1 4 4

v1 4

6

5

4

’v1

O

4

v2

v2’

2

a

8

v1’

G v G

6

5

G n

v G

O

n

4

v2

v2’

2

b

Fig. 2.12 Direct impact of two balls the graphical solution

Since the impact is direct—the two velocities are parallel to each other—the construction of Maxwell’s diagram can be made clearer if we add fictitious velocity components perpendicular to the line of impact to the velocities v1 and v2 . Let these

2.2 Central Impact

43

velocity components be 4i y and −4i y . Then v1 = 6ix + 4i y [m/s] ,

v2 = 2ix − 4i y [m/s] .

Therefore, vG =

 1 1 18ix + 12i y + 2ix − 4i y = (m 1 v1 + m 2 v2 ) = m1 + m2 4 = 5ix + 2i y [m/s] .

We can now construct two Maxwell’s diagrams. On the basis of the diagram the real solutions are as follows: (a) (b) v1 = 5ix , v1 = 4ix v2 = 5ix , v2 = 8ix

[m/s] , [m/s] .

It is worth mentioning that the fictitious velocity components can be selected arbitrarily.

2.3 Eccentric Impact 2.3.1 Impact of Two Bodies Which Perform Free Plane Motion Figure 2.13 shows the two bodies when the impact begins. Let vi and ω i be the velocity at the mass center G i and the angular velocity for body Bi (i = 1, 2) when impact begins. We remark that primed letters denote these quantities right after the impact. Fig. 2.13 Eccentric impact of two bodies in plane motion

t

n = t =1

t

B2

v1

G1

1

A

v2

n1 = n 2

B1

G2

n

44

2 Impact

The assumptions we make are the same as those on Sect. 2.2. The state of velocity before and after impact is uniquely determined by (a) v1 , ω 1 , v1 , ω 1 (body B1 ) and (b) v2 , ω 2 , v2 , ω 2 (body B2 ). It is worth, however, emphasizing that the angular velocities will also change since the contact forces have, now, moment about the mass centers. Since the velocity states of the bodies before impact are known, the primed quantities are the unknowns: v1 (two unknowns), ω 1 (one unknown—the angular velocity is perpendicular to the plane of motion), v2 (two unknowns), ω 2 (one unknown). It is our main objective to formulate the problem in such a way that the equations that provide the solutions be of the same structure as Eqs. (2.9a), (2.9c), (2.9d) valid for central impact. Then the problem can be solved by using Maxwell’s diagram. To achieve this goal some manipulations are to be carried out. Fig. 2.14 Motion characteristics of body B1

K1 rG K

1 1

1

t

P

rG P v 1 1

= n =1 n =0 rG K = 1 1

1

rG A = r1 +d 1 n 1

G1 T1

r1

1

rG A 1

d1

F12= F n

A

n

n

B1 First, we shall examine the motion of body B1 shown in Fig. 2.14. It follows from Eqs. (1.32), (1.31b) and (1.39), (1.46), (1.49a)2 that p˙ 1 = m 1 a1 = m 1 v˙ 1 = F12 ˙ G 1 = JG 1 α1 = rG 1 A × F12 , H

(2.17a) (2.17b)

where vG 1 is the velocity at G 1 and α1 is the angular acceleration if tc ∈ [0, τ ]. Integrating these equations with respect to time, remember that t = 0 when the impact begins and t = τ when it ends, we have   m 1 v1 − v1 = −n     JG 1 ω 1 − ω 1 = rG 1 A × −n



τ

F dt = p1 ,

(2.18a)

0



τ

F dt

= rG 1 A × p1 .

(2.18b)

0

Let P be an arbitrary point of B1 . For our later considerations, we shall determine the velocity at P before and after impact. Making use of Eq. (1.4) we can write

2.3 Eccentric Impact

45

v P = v1 + ω 1 × rG 1 P ,

vP = v1 + ω 1 × rG 1 P .

(2.19)

Subtract now the first equation from the second one. The result is   vP − v P = v1 − v1 + ω 1 − ω 1 × rG 1 P .

(2.20)

If we multiply throughout by m 1 JG 1 and take into account (2.18b), we obtain       m 1 JG 1 vP − v P = JG 1 m 1 v1 − v1 + m 1 JG 1 ω 1 − ω 1 × rG 1 P       p1

rG 1 A ×p1

in which  

    m 1 rG 1 A × p1 × rG 1 P = m 1 rG 1 A · rG 1 P p1 − rG 1 A rG 1 P · p1 , hence       m 1 JG 1 vP − v P = JG 1 + m 1 rG 1 A · rG 1 P p1 − m 1 rG 1 A rG 1 P · p1 . (2.21) This equation expresses that the velocity difference at any point within the body is directly proportional to p1 . (a) If P is on the line  then rG 1 P · p1 = 0 (the two vectors are perpendicular to each other) and Eq. (2.21) simplifies to     m 1 JG 1 vP − v P = JG 1 + m 1 rG 1 A · rG 1 P p1 .

(2.22)

 (b) It  v1 − v1 is perpendicular to the line . In addition the cross product  is clear that ω 1 − ω 1 × rG 1 P is also perpendicular to the line  if P is on the line . Consequently, there is a point on the line —this point is denoted by K 1 —for which the velocity difference vK 1 − v K 1 vanishes

  vK 1 − v K 1 = v1 − v1 + ω 1 − ω 1 × rG 1 K 1 = 0 . Then Eq. (2.22) yields

  JG 1 + m 1 rG 1 A · rG 1 K 1 p1 = 0,  =0

from where—see Fig. 2.14—it follows that JG 1 + m 1 (r1  + d1 n) · (−ρ1 ) = JG 1 − m 1r1 ρ1 = 0, which means that ρ1 =

JG 1 m 1 r1

(2.23)

46

2 Impact

is the vertical distance between the points G 1 and K 1 . Observe that here we have used the notations of Fig. 2.14. Point K 1 is the point of B1 with no velocity change. For this reason, it is referred to as the swaying center. (c) If P = T1 Eq. (2.22) yields     m 1 JG 1 vT1 − vT1 = JG 1 + m 1 rG 1 A · rG 1 T1 p1 =   

= JG 1 + m 1 (r1  + d1 n) · r1  p1 = JG 1 + m 1r12 p1 = JT1 p1    JT1

or

 JG  m 1 1 vT1 − vT1 = p1 , JT   1

m˜ 1 = m 1

JG 1 . JT1

(2.24)

m˜ 1

This means that the velocity difference at the point T1 is parallel to p1 . The physical content of this equation is the same as that of Eq. (2.6b) valid for central impact. m˜ 1 is the mass reduced to point T1 (Fig. 2.15). Fig. 2.15 Motion characteristics of body B2

B2

t F21= F n

A

v2

n

T2 2

r2

G2 2

K2

As regards body B2 a similar line of thought results in ρ2 = and

 JG  m 2 2 vT2 − vT2 = p2 = −p1 , JT2    m˜ 2

where

JG 2 m 2 r2

(2.25)

2.3 Eccentric Impact

47

m˜ 2 = m 2

JG 2 . JT2

(2.26)

After adding Eqs. (2.24)1 –(2.26)1 , we have     m˜ 1 vT1 − vT1 + m˜ 2 vT2 − vT2 = 0 or

m˜ 1 vT1 + m˜ 2 vT2 = m˜ 1 vT1 + m˜ 2 vT2 = (m˜ 1 + m˜ 2 ) vG ,

(2.27)

which expresses that the momenta that belong to the reduced masses are conserved. It also holds that      (2.28) vT2 − vT1 · n = e vT1 − vT2 · n . On the basis of Eqs. (2.27) and (2.28), one can construct Maxwell’s diagram or can calculate the solution by using the following scalar equations: vT 1 t = vT1 t ,

vT 2 t = vT2 t ;

(2.29a)

m˜ 1 vT1 n + m˜ 2 vT2 n = m˜ 1 vT 1 n + m˜ 2 vT 2 n ;

(2.29b)

  vT 2 n − vT 1 n = e vT1 n − vT2 n .

(2.29c)

After having determined the solution for vT1 we can obtain the velocity v1 at the mass center G. Comparison of Eqs. (2.26) and (2.18a) results in the equation     m˜ 1 vT1 − vT1 = p1 = m 1 v1 − v1 from where v1 = v1 +

 m˜ 1   vT1 − vT1 , m1

JG m˜ 1 = 1. m1 JT1

(2.30)

As regards body B2 we should write 2 for 1 in Eq. (2.30). The last issue is how to determine ω 1 . Substituting p1 from Eq. (2.26) into Eq. (2.18a), we get     JG 1 ω 1 − ω 1 = rG 1 A × p1 = m˜ 1 rG 1 A × vT1 − vT1 , which can be solved for the unknown angular velocity: ω 1 = ω 1 +

  m˜ 1 rG 1 A × vT1 − vT1 . JG 1

(2.31)

48

2 Impact

Change 1 to 2 in Eq. (2.30) for body B2 . Exercise 2.7 The slender rod B2 with mass m 2 = 3.5 kg and length 2 = 1.5 m is at rest when it is hit by the ball with mass m 1 = 3 kg. The velocity of the ball B1 is v1 = 12 ix [m/s]. Determine the state of velocity of the rod right after impact if e = 0.8 and r2 = 2 /3 = 0.5 m. Fig. 2.16 Impact of slender rod and ball

t =y

B2

2

K2

2 2

B1

G2 r2 v1

G1

2

2

n =x

T2

A

It follows from Eq. (2.25) that ρ2 =

1 m 2 2 JG 2 2 = 0.375 m . = 12 2 2 = m 2 r2 4 m2 3

Using the parallel axis theorem, we obtain JT2 = JG 2 + m 2 r22 =

7 2 1 m 2 22 + m 2 2 = m 2 22 . 12 9 36

With this value Eq. (2.26) yields m˜ 2 = m 2

JG 2 1 36 3 m 2 = m 2 = 1.5 kg . = JT2 12 7 7

It is obvious now that m˜ 1 = m 1 = 3 kg, m˜ 2 = 1.5 kg, vT1 t = v1t = 0, vT1 n = v1n = 12 m/s, vT2 t = 0, vT2 n = 0 and e = 0.8. With the knowledge of these values it is  = 0, vT 2 t = 0. Upon substitution of the known clear from Eqs. (2.29a) that vT 1 t = v1t values into Eqs. (2.29b) and (2.29c), we have 3 × 12 + 0 = 3vT 1 n + 1.5vT 2 n , vT 2 n − vT 1 n = 0.8 × (12 − 0) = 9.6 from where

vT 2 n = 14.4 m/s and vT 1 n = 4.8 m/s .

2.3 Eccentric Impact

49

Since the point K 2 remains at rest during impact it is the instantaneous center of rotation when the impact ends. Consequently, ω2 =

vT 2 n r2 + ρ2

=

1 14.4 = 15.457 . 0.5 + 0.375 s

2.3.2 Impact of Two Bodies if One or Two Bodies Rotate About a Fixed Point There are some changes if a body, say body B1 , rotates about a fixed axis. Figure 2.17 shows such a body, i.e., body B1 under the assumption that it rotates about the hinge at O. It is obvious that Fig. 2.17 Body rotating about the hinge at O

t O 1

rOT

1

rOA 1

T1

G1

vT1

vA A

F12= F n

B1

v A = ω1 × rO A , from where

vT1 = ω 1 × r O T1

and

  vT1 − v A = ω 1 × r O T1 − r O A   

| · n,

||n

which means that

  vT1 − v A · n = 0 .

It is also obvious JO α1 = r O A × F12 from where by performing time integration, we get      JO ω 1 − ω 1 = r O A × −n

0



τ

F dt

= r O A × (−p2 ) .

n

50

2 Impact

If we now cross multiply the above equation from right by r O T1 , we obtain   JO ω 1 × r O T1 − ω 1 × r O T1 = − (r O A × p2 ) × r O T1 =     = − r O A · r O T1 p2 + r O T1 · p2 r O A       =0

ρ21

or − p2 =

   JO   v − vT1 = m˜ 1 vT1 − vT1 , ρ21 T1

m˜ 1 =

JO . ρ21

(2.32)

After adding this equation to Eq. (2.26)2 we arrive at (2.27). Equation (2.28) is still valid. Consequently, solutions of impact problems with bodies rotating about a fixed point can be calculated in the same way as shown in Sect. 2.3.1. Exercise 2.8 A bullet with mass m 2 is fired with a horizontal velocity v2 = −v2 n into the side of a panel suspended from a hinge at B. Knowing that the panel is initially at rest determine the angular velocity of the panel immediately after the bullet is embedded and the impulsive reaction R B at the hinge B assuming that the bullet is embedded in τ sec. (τ 1). Fig. 2.18 A panel hit by a bullet

z ’ 1

B

t

1 1

B2

G1 T1

v2 n = x

n m2

B1

m1

We shall assume that the necessary data of the panel—m 1 , J B , 1 and ρ1 —are all known. According to Eq. (2.32) m˜ 1 = J B /ρ21 . Since the bullet is embedded it follows that  = vT 1 n ; v2 = v2n

e=0.

2.3 Eccentric Impact

51

 Equation (2.29b) can now be utilized to find the unknowns v2 = v2n = vT 1 n . We can write m˜ 1 vT n + m 2 vn = m˜ 1 vT 1 n + m 2 v2 ,   1  =0

or

m 2 vn = (m˜ 1 + m 2 ) vT 1 n

from where

v2 = vT 1 n =

m2 v2 . m˜ 1 + m 2

It can be checked with ease that ω1 =

vT 1 n ρ1

=

v2 . ρ1

We remark that ω1 is shown in Fig. 2.18. Let p and p be the momenta of the whole system before and after impact. Using the principle of impulse and momentum (1.49d)1 , we can write p = p +

 0

where

τ

R B dt ∼ t , = (R Bx ix + R Bz iz ) 

(2.33)

τ

1 p ∼ = −m 1 v1 ix = −m 1 ω1 1 ix = −m 1 v2 ix ρ1

(m 1  m 2 !)

and p = −m 2 v2 ix . Hence, R Bz = 0 and R Bx =

  1 m1 1 1 1 = m 2 v2 m 2 v2 − m 1 v2 1− τ ρ1 τ m˜ 1 + m 2 ρ1

are the components of the impulsive reaction at B. Exercise 2.9 The panel in the previous exercise will sway to the left till it reaches an extreme position. Determine what angle ϕ belongs to this position. Using the principle of work an energy (1.58a), we can write E2 − E1 = W12 = U12 = U1 − U2 ,

52

2 Impact

Fig. 2.19 The swaying panel

z

Position 2

B G1

1

G1 T1 Position 1

x

B1

where E1 =

  2 v 1 1   2 1 1 (m˜ 1 + m 2 ) m 2 m2 m 2 v2 + J B 22 = m 2 v2 m 2 v22 = 2 2 2 (m˜ 1 + m 2 )2 m˜ 1 + m 2 2  2 ρ1 Eo

and E2 = 0. On the other hand, U1 = 0 and U2 = mgρ1 (1 − cos ϕ) , where m = m1 + m2  m1 , m1  m2 . Thus,

1 m2 m 2 v22 = m 1 gρ1 (1 − cos ϕ)    m˜ 1 + m 2 2 2 sin2



from where v2 = 2

ϕ 2

m 1 gρ1 (m˜ 1 + m 2 ) ϕ sin 2 2 m1

is the equation for ϕ. Conversely, if we measure ϕ we can determine the velocity v2 of the bullet.

2.4 Problems

53

2.4 Problems Problem 2.1 Prove formula (2.13) which provides the energy loss during impact. Problem 2.2 Solve Exercise 2.5 by using Eqs. (2.9a), (2.9c), and (2.9e). Problem 2.3 Two balls with masses m 1 = 2 kg and m 2 = 6 kg collide—see Fig. 2.20 for details. The velocities v1 and v2 the balls have after impact are known. Determine the velocities v1 and v2 before impact if e = 1 (perfectly elastic impact). v1 = −2ix + 4i y [m/s] , v2 = 2ix − 4i y [m/s] .

Fig. 2.20 Impact of two balls

v1’

t =y G2

B1

G1

A

v2

n =x

B2 v2’

v1 t

30

o

n

Problem 2.4 Two identical hockey pucks are moving on a hockey rink at the same speed of 4 m/s in parallel but opposite directions till they strike each other. Determine the velocity of each puck after impact if the coefficient of restitution e = 1.

v2 = v1 Fig. 2.21 Impact of two hockey pucks

Problem 2.5 Figure 2.22 shows two pendulums. Each pendulum has the same mass and length. Pendulum 1 is released from a horizontal position with no initial velocity. It swings from the horizontal position till it hits pendulum 2 which is at rest in a vertical position. Find the velocities after impact if e = 1. What happens to the system after impact?

54 Fig. 2.22 Impact of two pendulums

2 Impact

m Pendulum 1

b

a

m

m

m

v1 Pendulum 2

Problem 2.6 Assume that the pendulums in the previous problem have different masses, i.e., m 1 = m 2 —the lengths of the pendulums are the same. Assume further that e = 0.9. Determine the velocity of Pendulum 2 after impact. What equations can be used to determine the angle ϑ Pendulum 2 swings through after impact? Problem 2.7 A slender rod of length  and mass m is dropped onto the rigid supports B and D. Since the support B is slightly higher than support D the rod strikes support B first. Then the velocity of the rod is v0 = −vo i y , vo > 0. Assume that the impact is perfectly elastic both at B and D. Determine the angular velocity of the rod and the velocity at its mass center right after the rod (i) strikes support B, (ii) strikes support D, and (iii) strikes support B again.

Fig. 2.23 Slender rod dropped onto rigid supports

y D

B G vo

voiy

Reference 1. I. Sályi, Kinetics (in Hungarian) (Tankönyvkiadó, Budapest, 1961)

x

Chapter 3

Some Vibration Problems

3.1 Single Degree of Freedom Systems 3.1.1 Introductory Remarks Let us denote the coordinate that describes the motion by q(t). Vibration is a mechanical phenomenon whereby oscillations occur about an equilibrium point. More precisely the motion q(t) is said to be vibration if • q(t) is limited and describes backward and forward movement hence • the derivative dq(t)/dt constantly changes its sign.

q (t ) n

n

t

Fig. 3.1 Periodic function

The vibration is periodic if q(t) = q(t + τn ) , in which τn is the period (Fig. 3.1), and random if dq(t)/dt changes its sign a lot of times (e.g., movement of a tire on a gravel road). For periodic vibrations, the number of cycles per unit time defines the frequency and the maximum of |q(t)| is called amplitude of the vibration. © Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_3

55

56

3 Some Vibration Problems

Mechanical vibration takes place when a system is displaced from a position of stable equilibrium. Then the system tends to the equilibrium position under the action of restoring forces (spring, gravitational pendulum). When reaching this position the system has a velocity which carries it beyond that position. The process can be repeated infinitely, i.e., the system keeps moving back and forth across the equilibrium position. When the motion is maintained by the restoring force only the vibration is said to be free vibration; if a periodic external force is applied to the system then the resulting motion is forced vibration. When the effect of friction can be neglected the vibrations are said to be undamped, otherwise we speak about damped vibrations. (Each real system is actually damped to some degree.)

3.1.2 Classification 3.1.2.1

Undamped Free Vibrations

Figure 3.2a shows a spring–mass system which, in accordance with its name, consists of a spring and a mass (a particle). The horizontal displacement of the mass center G is denoted by x. If x = 0 the system is at rest, i.e., there is no force in the spring. It is obvious that (x > 0)[x < 0] is the (elongation) [shortening] of the spring. Figure 3.2b represents graphically how the spring force Fk (the force exerted on the spring) may depend on x. If the function Fk (x) is linear it can be given in the form Fk = k x =

1 x f

(3.1)

in which k [N/m] is the spring constant (or spring stiffness) and f is referred to as spring flexibility. The spring characteristic Fk (x) is (degressive)[progressive] if the derivative dFk /dx is (decreasing)[increasing]. We shall, in general, assume that the spring characteristic is linear. c

b

a m

k

progressive spring

Fk

G

F

degressive spring

x Fk

linear spring

x

F G

Fig. 3.2 Spring mass system with spring characteristic

x

3.1 Single Degree of Freedom Systems

57

If we displace the particle from its equilibrium position—see Fig. 3.2a—then it moves under the action of the restoring force F = −Fk . Our aim is to find the motion law (the function) x(t). Using Newton’s second law, we get the equation of motion ma = m x¨ = F = −kx . This equation is a homogeneous ordinary differential equation of order two: x¨ + ωn2 x = 0 ,

ωn2 =

k . m

(3.2)

Here ωn [1/s] is the natural circular (or angular) frequency (or eigenfrequency for short). The general solution of the differential equation (3.2) takes the form x = A cos ωn t + B sin ωn t,

(3.3)

where A and B are undetermined integration constants. Let xo and vo be the initial displacement and velocity of the particle m. The undetermined constants of integration A and B can be determined from the initial conditions: t = to = 0 t = to = 0

x(to ) = x(0) = xo = (A cos ωn t + B sin ωn t)|t=0 = A , x(t ˙ o ) = v(0) = vo = ωn (−A sin ωn t + B cos ωn t)|t=0 = ωn B . (3.4) With the integration constants A = xo and B = vo /ωn vo x = xo cos ωn t + sin ωn t = D sin (ωn t + φ)  ωn  D sin φ

(3.5a)

D cos φ

is the solution in which  D=

 xo2 +

vo ωn

2 ,

φ = arctan

ωn xo vo

(3.5b)

are the amplitude and the phase angle. Equation τn = τ = 2π/ωn

(3.6a)

gives the period of the spring–mass system for which the unit is, in general, 1/s. This definition is the same as that given by equation (1.63). Its reciprocal is denoted by f n and is called natural frequency: f n = 1/τn = ωn /2π .

(3.6b)

The unit of the natural frequency is 1/s. It is called hertz and is abbreviated as Hz. Figure 3.3 shows the displacement time curve.

58

3 Some Vibration Problems

x (t ) y

t =0 n

t

A =D sin

t x

B =D cos D D n

Fig. 3.3 Displacement time curve

Remark 3.1 The kinetic energy of the particle and the potential energy stored in the spring are given by the following relations: E=

1 2 m x˙ , 2

U=

1 2 kx , 2

(3.7)

where x = D sin (ωn t + φ) , It is obvious that Emax =

x˙ = Dωn cos (ωn t + φ) .

1 2 2 ω D m, 2 n

Umax =

(3.8)

1 2 kD . 2

(3.9)

are maximum of E and U . Since E = 0 if U = Umax and U = 0 if E = Emax it follows from the principle of energy conservation (1.58b) that Emax = Umax .

(3.10)

Hence ωn2 =

Umax k = . m Emax

(3.11)

Exercise 3.1 Liquid vibrates in the U-pipe shown in Fig. 3.4. Find the eigenfrequency of the oscillating liquid column. Fig. 3.4 Vibrating liquid in a U pipe

x x

a

b

3.1 Single Degree of Freedom Systems

59

Figure 3.4a, b shows (the equilibrium position of the liquid) [the liquid in motion]. Let ρ and A be the density of the liquid and the cross-sectional area of the pipe. Further let x be the displacement of the liquid column in motion. It can be seen that the change in potential energy consists of two parts: (a) the potential energy of the raised liquid column plus (b) the potential energy of the depressed liquid column. Hence x x U = ρ Agx + ρ Agx = ρ Agx 2 . 2 2 The kinetic energy of the moving liquid is given by the following equation: E=

1 ρ A x˙ 2 . 2

Let us assume that the motion of the liquid is harmonic. Then x = D cos ωn t and U = ρ Ag D 2 cos2 ωn t ,   

E=

Umax

1 ρ A ωn2 D 2 sin2 ωn t . 2    Emax

With Umax and Emax Eq. (3.11) yields the sought eigenfrequency: ωn2 =

Umax k 2g = . = m Emax 

Remark 3.2 For our later considerations we shall introduce some notational conventions. The beam shown in Fig. 3.5 is subjected to vertical unit forces at the points Pi and P j . The vertical displacement at P j due to the unit force at Pi is denoted by f ji —here the first subscript identifies the point where we consider the displacement, whereas the second subscript identifies the point where the unit force is acting. In accordance with this rule f ii is the displacements at Pi due to the unit force exerted on the beam at the same point and f i j is the displacement at Pi under the action of the unit force at P j , etc. The displacements f ji , f i j f ii , and f j j are called flexibility influence coefficients (or simply flexibilities).

w

f jj

a

Fig. 3.5 Cantilever beam

f ji

1 Pj

f ii

b

f ij

1 Pi

x

c

60

3 Some Vibration Problems

Table 3.1 Eigenfrequencies of beams with one concentrated mass m

x

ωn2 =

b

a

m

ωn2 =

x

ωn2 =

3I E3 ma 3 b3

ωn2 =

3I E mb2 

ωn2 =

3I E mb2 (1 − a/4)

m

b

a

m

m

a

x

b

a

12I E3 + b)

x b

a

3I E ma 2 b2

x

b

ma 3 b2 (3

Exercise 3.2 The cantilever beam shown in Fig. 3.6 carries a mass attached to its right end. We shall assume that m is much greater than the mass of the beam. We shall denote the flexibility influence coefficient at the right end of the beam by f 11 . It is known that f 11 =

3 , 3I E

(3.12)

where  is the length of the beam, I is the moment of inertia, and E is Young’s modulus. Find the eigenfrequency of the oscillating beam. Fig. 3.6 Cantilever beam with a mass at its right end

w

wB B m

x

3.1 Single Degree of Freedom Systems

61

If the beam oscillates the displacement w B at the right end of the beam is due to the force of inertia −m w¨ B . We can, therefore, write w B = − f 11 m w¨ B .

(3.13)

After rearranging this equation, we get m w¨ B +

1 w B = m w¨ B + k11 w B = 0 , f 11

k11 =

1 3I E = 3 . f 11 

(3.14)

This is the equation of motion of an undamped free system with one degree of freedom in which k11 = 1/ f 11 is the spring constant (or spring stiffness). Thus, it follows from Eq. (3.2)2 that 3I E k11 = ωn2 = . (3.15) m m3 Based on the solution presented in Exercise 3.2, Table 3.1 contains the square of the eigenfrequencies for some beam supports.

3.1.2.2

Damped Free Vibrations

We have mentioned that the vibrations are always damped to some degree. By damping we mean an influence within or upon an oscillatory system that has the effect of reducing, restricting, or preventing its oscillations. Damping is caused, in general, by various friction forces such as dry friction or fluid friction—if a rigid body moves in a fluid then the forces exerted on it slacken the speed of motion. It is worth mentioning that damping is always associated with energy dissipation. We shall assume viscous damping which is caused by fluid friction. If the velocity of the particle is under a certain and not too high limit the damping force exerted on the body is directly proportional and opposite in direction to the velocity. Then the damping force Fc can be given in the form ˙ (3.16) Fc = −c x, where c [Ns/m] is called damping coefficient. Fig. 3.7 Viscous damper

dashpot

fluid

plunger

Figure 3.7 shows the sketch of a vibration damper and its main parts.

62

3 Some Vibration Problems

Fig. 3.8 Damped free spring-mass system

m

k

c

G x F

Fc G

Figure 3.8 is that of a damped free system with one degree of freedom if x˙ > 0 which means that Fc < 0. The equation of motion takes the form m x¨ + c x˙ + k x = 0,

(3.17a)

which can be rewritten as x¨ + 2β x˙ + ωn2 x = 0 ,

2β =

c k , ωn2 = . m m

(3.17b)

Let us assume that x = eλt is the solution to the differential equation (3.17b). Here λ is the characteristic value. Upon substitution of the solution into Eq. (3.17b) we get the characteristic equation λ2 + 2βλ + ωn2 = 0

(3.18a)

from where λ1,2 = −β ±



β 2 − ωn2 = −β ± μ ,

μ=

β 2 − ωn2 .

(3.18b)

Assume that the two roots are different. Then x = Aeλ1 t + Beλ2 t = Ae−(β−μ) t + Be−(β+μ)t

(3.19)

is the general solution. The behavior of the system depends on the amount of damping. The undetermined constants of integration A and B can be determined by utilizing the initial conditions. We speak about critical damping if we have double roots, that is, μ = 0. Then the damping coefficient, which is now called critical damping coefficient, cannot be arbitrary. Let us denote it by cn . It can be calculated if we utilize the condition μ = 0 and Eq. (3.17b)2,3 :

c 2 k n β 2 = ωn2 → = 2m m from where

3.1 Single Degree of Freedom Systems

63

cn = 2m

k = 2mωn . m

(3.20)

Depending on what sign the expression β 2 − ωn2 under the square root has, we distinguish three cases: (a) If β > ωn (c > cn ) then μ is real. Since β > μ > 0 the roots λ1 = −β + μ and λ2 = −β − μ are negative real numbers. Consequently, the solution (3.19) is non-vibratory and tends to zero as t tends to infinity. In accordance with (3.4) let xo and vo be the displacement and velocity at t = 0. It can be checked with paper and pencil calculations that

 1 − (vo − λ2 xo ) eλ1 t + (vo − λ1 xo ) eλ2 t = λ2 − λ1  1 = (vo + (β + μ)xo ) e−(β−μ)t − (vo + (β − μ)xo ) e−(β+μ)t 2μ x=

(3.21)

is the solution in terms of the initial values. Figure 3.9 shows the quotient x/xo for a given system in the time interval t ∈ [0, 2] if (i) vo > 0, (ii) vo = 0, and finally (iii) if vo < 0. It should be mentioned that there exist such negative initial velocities for which the solution does not change sign. Since solution (3.21) is not oscillatory the system is overdamped. In other words, we are faced with the case of heavy damping.

x xo vo

0

1.0

vo = 0

0.5

vo

0

t

1.0

2.0

Fig. 3.9 Dependence of x/x0 on time −v0 is the parameter

(b) If β = ωn (c = cn ) we have double roots and the solution assumes the form x = (A + Bt) e−β t = [xo + (βxo + vo ) t] e−β t .

(3.22)

64

3 Some Vibration Problems

This solution is also non-vibratory. Its graphical representation is more or less the same as that shown in Fig. 3.9. The damping is called critical since the condition c = cn separates the vibratory and non-vibratory solutions. (c) If β < ωn (c < cn ) we speak about light damping: the system is underdamped. The solutions for λ are complex and conjugate: λ1,2 = −β ± i ωn2 − β 2 = −β ± iωd

(3.23a)

in which ωd2 = ωn2 − β 2 =



c 2 k k 1− = − m 2m m

c k 4m 2 m   



  c2 = ωn2 1 − 2 cn

cn2

hence

 ωd = ωn 1 −

 c2 = ↑ = ωn 1 − ζ 2 . 2 cn ζ = ccn

(3.23b)

Here ωd is the natural circular frequency (eigenfrequency) of the damped vibrations and c (3.23c) ζ = cn is the damping factor. The real solution for x is of the form   vo + βxo sin ωd t x = e−βt (A cos ωd t + B sin ωd t) = e−βt xo cos ωd t + ωd (3.24a) or x = e−βt

 xo cos ωd t + 

D sin φ

vo + βxo sin ωd t ω  d 



= e−βt D sin (ωd t + φ) ,

D cos φ

(3.24b) where

 D = xo 1 + 



 vo + βxo 2 ωd xo and φ = arctan . ωd xo vo + βxo  

(3.25)

Dˆ

Solution (3.24) describes a vibratory motion with diminishing amplitude and the time interval

3.1 Single Degree of Freedom Systems

τd =

65

1 1 2π 2π  = = τn  > τn , ωd ωn 1 − ζ 2 1 − ζ2 

(3.26)

τn

which separates two successive points where the solution touches one of the limiting curves—see Fig. 3.10 for details—is called period of the damped vibration.

x xo

d

d

1.0

De t

t

-1.0 6.0

4.0

2.0 Fig. 3.10 Dependence of x/x0 on time

The logarithmic decrement is defined by the equation δ = ln = ln

e−βt D sin (ωd t + φ) x (t) = ln −β(t+τ ) = ↑ = d D sin [ω (t + τ ) + φ] x (t + τd ) e ωd τd =2π d d

e−βt D sin (ωd t + φ) −β(t+τ d ) D sin (ω t + φ + e d

= ln e−βt eβ(t+τd ) = ln e 2m τd = c

2π )

c τd . 2m (3.27)

With δ, τd (these two quantities are measurable), and m, we can determine the damping coefficient.

3.1.2.3

Undamped Forced Vibrations

Excitations play an important role in various engineering applications since they are the sources of the forced vibrations of a system. These vibrations are often caused by a periodic force (by a force of excitation) exerted on the system. It may also occur that the system considered is connected elastically to such machine part which performs an alternating motion. Then we speak about displacement excitation which results in, again, forced vibrations. We shall assume that the force of excitation

66

3 Some Vibration Problems

Fig. 3.11 Excitation on a spring mass system

Ff =Ffo cos k

ft

m G x

Fk

F

Ff G

is a harmonic one. Consider now the spring–mass system shown in Fig. 3.11. The system is subjected to a harmonic external force F f = F f o cos ω f t where F f o is the amplitude of the excitation force while ω f is referred to as forced circular frequency. Remark 3.3 Here and in the sequel a harmonic vectorial quantity (F f in the present case) is depicted as is shown in Fig. 3.11 since the two extreme values of the vector differ from each other in sign. Using Newton’s second law, we get the equation of motion in the form m x¨ = −F + F f = −kx + F f o cos ω f t

(3.28a)

or after dividing throughout by m it is x¨ + ωn2 x = where

Ff o cos ω f t , m

ωn2 =

k , m

(3.28b)

Ff o Ff o k = = ωn2 δ f . m k m 

(3.28c)

δf

Here δ f is the elongation due to the amplitude F f o of the excitation force. Using this notation, we can rewrite Eq. (3.28b) in the following form: x¨ + ωn2 x = ωn2 δ f cos ω f t ,

ωn2 =

k , m

δf =

Ff o . k

(3.29)

The above equation is an inhomogeneous differential equation with constant coefficients. As is well known its general solution can be expressed in the form x = xhom (t) + xpart (t),

(3.30a)

xhom (t) = A cos ωn t + B sin ωn t

(3.30b)

where

is the solution of the homogeneous differential equation

3.1 Single Degree of Freedom Systems

67

x¨ + ωn2 x = 0 and

xpart = xm cos ω f t

(3.30c)

is the particular solution that satisfies the inhomogeneous equation—it never satisfies the homogeneous equation. The unknown amplitude xm can be obtained from the condition that the particular solution should satisfy the inhomogeneous equation. Upon substitution of the particular solution (3.30c) into (3.29)1 , we have −ω2f xm cos ω f t + ωn2 xm cos ω f t = ωn2 δ f cos ω f t from where the unknown amplitude is xm =

ω2f ωn2 1 1 2 δ = δ = δ , η = . f f f 2 ω 1 − η2 ωn2 ωn2 − ω2f 1 − ω2f

(3.31)

n

With xm the general solution takes the form xm

   δf x(t) = A cos ωn t + B sin ωn t + cos ω f t .    1 − η2    transient function

(3.32)

steady-state vibration

The transient vibration is a free vibration of the system which soon disappears due to the damping inherent in the system. The second term represents the steady-state vibrations caused and maintained by the force of excitation. Its frequency coincides with that of the excitation force. The amplitude xm depends on the frequency ratio η. The magnitude of the quotient xm /δ f is called magnification factor:    xm  1 m f =   = . δf |1 − η2 |

(3.33)

The quotient xm /δ f has the following properties: (a) If η = ω f /ωn ∈ [0, 1) then xm /δ f is a positive quantity and so is the amplitude xm of the steady-state vibrations. Then we say that the vibration is in phase with the excitation. (b) If η = ω f /ωn → 1 then for (η < 1) [η > 1] the quotient xm /δ f tends to (+∞) [−∞] and so does the amplitude xm of the steady-state vibrations. This phenomenon is known as resonance which is to be avoided since it may result in harmful vibrations. (c) If η = ω f /ωn > 1 then the quotient xm /δ f is a negative quantity and so is the amplitude xm of the steady-state vibrations. Then we say that the vibration is 180o out of phase with respect to the excitation.

68

3 Some Vibration Problems

xm

a

mf

f

=

b

xm f

1.0 =

f n

1.0

-1.0

=

1.0 1.0

2.0

2.0

f n

3.0

3.0

Fig. 3.12 Magnification factor (Resonance curve)

Figure 3.12 represents the magnification factor and its magnitude against the dimensionless parameter η.

3.1.2.4

Forced Vibrations with Damping

If the damped single degree of freedom system shown in Fig. 3.8 is subjected to a harmonic excitation force F f = F f o cos ω f t we speak about forced-damped vibrations. Fig. 3.13 Excitation on a damped spring-mass system

Ff =Ffo cos

ft

c

k m G x F

Fc

Ff

G The equation of motion for forced vibrations with damping can easily be established if we make use of the notations from Fig. 3.13 which depicts a harmonically excited and damped single degree of freedom system: m x¨ + c x˙ + kx = F f o cos ω f t.

(3.34a)

After dividing throughout by m and utilizing then the notations (3.17b)1,2 and Eq. (3.28c)2 , we get

3.1 Single Degree of Freedom Systems

69

x¨ + 2β x˙ + ωn2 x = ωn2 δ f cos ω f t.

(3.34b)

Assume that the system is underdamped. The general solution is again the sum of the general solution to the homogeneous equation, which is given by Eq. (3.24b), and a harmonic particular solution:   x(t) = e−βt (A cos ωd t + B sin ωd t) + xm cos ω f t + φ .       transient function

steady-state vibration

Since the transient function is usually negligible it is our aim to determine both the amplitude xm of the steady-state vibrations and the phase angle φ. Substituting the particular solution   (3.35) xpart = xm cos ω f t + φ into the equation of motion (3.34b), we obtain       −ω2f xm cos ω f t + φ − 2βω f xm sin ω f t + φ + ωn2 xm cos ω f t + φ = = ωn2 δ f cos ω f t. Making ω f t + φ equal to 0 (ω f t = −φ) and π/2 (ω f t = π/2 − φ), we can write   2 ωn − ω2f xm = ωn2 δ f cos φ , −2βω f xm =

ωn2 δ f cos (π/2

− φ) =

ωn2 δ f

(3.36a) sin φ .

(3.36b)

Squaring equations (3.36) and adding them, we have xm2 from where

  2 ωn2 − ω2f + 4β 2 ω2f = ωn4 δ 2f

   xm  1 m f =   =    δf   2 2  2β ω f 2  1 − ωf + ωn2 ωn ωn

(3.37)

is the magnification factor in which ωf = ↑ = η, ωn (3.31)2

2β c 1 1 2c = ↑ = = 2c = ↑ = = ↑ = 2ζ . ωn (3.17b)2 m ωn 2mωn (3.20) cn (3.23c) (3.38)

Thus 1 m f =  . 2 1 − η2 + (2ζ η)2

(3.39)

70

3 Some Vibration Problems = 0.0

mf = xm f

= 0.125 = 0.185

3.0

2.0 = 0.25

1.0

=

0.5 = 1.0 =

2.0

1.0

0.5

=

f n

Fig. 3.14 Magnification factor for a damped system

Figure 3.14 shows the magnification factor m f = |xm /δ f | for different values of the damping factor ζ . It is clear from Fig. 3.14 that the amplitude xm of the forced vibrations has a local maximum if η ≈ 1 which is the place of resonance for the undamped system. The greater the difference η − 1.0 for η > 1 the smaller the amplitude xm . If we increase the damping factor ζ we can also reduce the amplitude xm . To get the phase angle we have to divide the left and right sides of Eq. (3.36b) by the left and right sides of Eq. (3.36a). We obtain tan φ = −

2βω f 2βω f = = ↑ =− 2 ωn2 − ω2f η= ω f ωn 1 − η2 ωn

=

↑

c (3.20)→2β= mc = cn /2ω n

from where φ = arctan

2ζ η . η2 − 1

=2ζ ωn

=−

2ζ η 1 − η2

(3.40a)

(3.40b)

Exercise 3.3 A rigid block with mass m = 100 kg moves horizontally as shown in Fig. 3.15. The block is moved xo = 30 mm to the right from its equilibrium position and then released. Knowing that k1 = 5 kN/m, k2 = 3.1 kN/m determine the period of the vibration, the maximum velocity, and acceleration of the block for each spring arrangement.

3.1 Single Degree of Freedom Systems

k1

71

m F1= k1x m =

a k2

m x..

F2= k2 x

x xo m

k1

k2

b

1

2

x xo x 1 +x 2 = x

1+x 1

2 +x 2

Fig. 3.15 Springs attached in parallel and series

The two springs can be replaced by a single one for each spring arrangement. (a) The two springs are attached in parallel. The force exerted by the two springs on the block should be the same as the force exerted on the block by the single spring that replaces the original two. Let us denote the spring constant of the equivalent spring by k. It is obvious that the following equation should be satisfied for any x during the motion: k1 x + k2 x = (k1 + k2 ) x . This means that k = k1 + k2 ,

1 1 1 = + , f f1 f2

(3.41)

where we have utilized the definition (3.1) of the spring flexibility. In other words, if two springs are attached in parallel the stiffness of the equivalent spring is the sum of the spring constants the two springs have. Since the initial velocity vo is zero it follows from (3.5a), (3.2), and (3.6a) that  1 k k1 +k2 8.1 · 103 kg m/m s2 = = =9 , x = xo cos ωn t, ωn = m m 100 kg s 2π 2π τn = s = 0.698 s . = ωn 9 Making use of the solution, we get

72

3 Some Vibration Problems

v = x˙ = −xo ωn sin ωn t and a = v˙ = −xo ωn2 cos ωn t hence

and

vmax = xo ωn = 30 · 9

m mm = 0.27 s s

amax = xo ωn2 = 30 · 81

mm m = 2.43 2 s2 s

are the maximum velocity and acceleration. (b) The two springs are attached in series. The equivalent spring stiffness can easily be obtained from the fact that the total elongation x is the sum of the elongations the two springs have under the action of the spring force F. Making use of (3.1), we can write   1 F F 1 1 =F . x1 + x2 = + =F + k1 k2 k1 k2 k Consequently 1 1 1 = + , k k1 k2

f = f1 + f2 .

(3.42)

In other words, the reciprocal of the equivalent spring stiffness is the sum of the reciprocals of the two original spring constants. Utilizing the data given it is not too difficult to check that k= ωn =

 k = m

1 1 k1

+

1 k2

=

kN k1 k2 5 · 3.1 = 1. 914 , = k1 + k2 5 + 3.1 m

1 2π 2π 1.914 · 103 = 4.374 , τn = = 1.436 s , = 100 s ωn 4.374

vmax = xo ωn = 30 · 4.374

and amax = xo ωn2 = 30 · (4.374)2

mm m = 0.131 s s m mm = 0.573 2 . s s

Exercise 3.4 Figure 3.16 shows a uniform rod of length  and mass m. The rod is supported by a pin at A and a spring of constant k at D and is connected at B to a dashpot of damping coefficient c. Knowing m, , k, and c determine, for small oscillations, (a) the equation of motion regarding the polar angle ϕ as an unknown, (b) the circular frequency ωn for the undamped system, and (c) the critical damping coefficient cn . The moment of inertia of the rod with respect to the axes passing through the point A perpendicularly to the plane of motion, the forces Fk and Fc exerted on the rod by the spring are all given by the following equations: JA =

1 2 m , 3

Fk = k

ϕ , 2

Fc = cϕ˙ .

3.1 Single Degree of Freedom Systems

/2

A

73

m

/2

D

B

k

c

y RAx x Fk

R Ay

Fc

Fig. 3.16 Vibrating uniform rod

Since the moment of the effective forces about the point A should be the same as that of the external forces, we can write utilizing Eqs. (1.47) and (1.49b) that J A ϕ¨ = −Fc  − Fk

 2

from where by dividing throughout by J A it follows that k2 1 2 m ϕ¨ +  c2 ϕ˙ + ϕ=0 4 3    mg

cg

or

ϕ¨ +

3k 3c ϕ˙ + ϕ = 0, m 4m

kg

which is the equation of motion in two formally different (but equivalent) forms. Recalling (3.2)2 and (3.20) we obtain the circular frequency of the undamped system and the critical damping coefficient:  kg 3k 3k 1 2 2 km ωn = = , cgn = 2ωn m g = m =  , mg 4m m 3 3 cgn km cn = 2 = .  3

3.2 Machine Foundations and Vibration Isolation 3.2.1 Introductory Remarks Machines (machine tools) are put, in general, onto a foundation in order (a) to reduce the harmful effects of those vibrations which are transmitted from the machine (the source of vibration) to the environment or/and

74

3 Some Vibration Problems

(b) to reduce the harmful vibrations that are transmitted from the environment (neighboring machines—a forging hammer, for instance) to the machine considered. As regards their effects, vibrations can be either harmful (e.g., pneumatic hammer, chain saw, ventilator, forging machine) or useful (e.g., riddle machine, bolting machine, compression machine, etc.). Vibration reduction can be achieved in various ways. Some of the most important techniques are listed below: • Mass balancing by changing the mass distribution. • Active vibration reduction by changing the parameters of the vibrating system appropriately (e.g., the spring constants, the damping coefficients, etc.). This can be achieved by measuring a time signal of the vibration which is then processed and used to control an actuator, i.e., the device that changes the system parameters. • Passive vibration reduction by means of preventing vibration transmission from the neighborhood by applying (a) some vibration isolation devices and/or (b) damping elements which are capable of absorbing vibration energy, etc. As regards the issue of machine foundations it is worth emphasizing that there are a number of books devoted to this important engineering problem: we cite here only two [1, 2].

3.2.2 Minimal Model with One Degree of Freedom The machine foundation can be modeled, in many cases, as a rigid body which has six degrees of freedom. However, a model with one degree of freedom (block foundation) is often sufficient for clarifying some basic questions of vibration isolation as regards the effect of periodic excitation. Figure 3.17 depicts a simple one degree of freedom model for machine foundation. The block foundation is a rigid body which can move

x sd

a

b

Ff = Ffocos

ft

=0 x=0

x

m

block foundation

k

m k

x, Fig. 3.17 Single degree of freedom model for machine foundation

x,

3.2 Machine Foundations and Vibration Isolation

75

only vertically. Let xsd be the static deflection of the foundation caused by the weight of the machine that is put onto the foundation (Fig. 3.17 does not show the machine) and the weight of the foundation itself. Let m be the total mass of the foundation and the machine. If the foundation and the machine vibrates together vertically, the displacement that belongs to this motion is denoted by x. We shall assume that the ground is an elastic body: k stands for the ground rigidity. The total vertical displacement is given by the following equation: ξ = xsd + x =

mg +x. k

(3.43)

(a) If there is no excitation—see Fig. 3.17a—the equation of motion is as follows: m ξ¨ + kξ = mg,

(3.44a)

where ξ¨ = x¨sd + x, ¨ x¨sd = 0 and kξ = kxsd + kx = mg + kx. Hence x¨ + ωn2 x = 0 ,

ωn2 =

k m

(3.44b)

is the final form of the equation of motion. Observe that this equation coincides with Eq. (3.2). With xsd —this quantity can be measured easily—the ground rigidity is given by mg . (3.45) k= xsd Hence

ωn =

k = m



g . xsd

(3.46a)

This means that the static deflection determines the natural circular frequency of the free vibrations for the system constituted by the foundation and the machine. As regards the natural frequency of the free vibrations, we get √ √ g 1 981 1 1 ωn 5 = fn = = = √ √ √ τn 2π 2π xsd 2π xsd xsd

(3.46b)

provided that xsd is given in cm. (b) If there is an excitation—see Fig. 3.17b—then the equation of motion is the same as Eq. (3.29) which, for the sake of completeness, is repeated and supplemented here x¨ + ωn2 x = ωn2 δ f cos ω f t , ωn2 =

Ff o F f o xsd k g , δf = = = . m xsd k mg (3.47)

76

3 Some Vibration Problems 2.0 f

1.5

1.0

0.5

0.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.4

2.2

2.0

2.6

n

overtuned system

3.0 f

2 f

2.8

f

n

n

undertuned system

Fig. 3.18 Resonance curve

It follows from the resonance curve—see Fig. 3.18—that the amplitude√xm of the steady-state vibrations satisfies the inequality |xm /δ f | < 1 if ω f > 2 ωn √ (η > 2). √ √ For given m and ω f , the condition ω f > 2 ωn = 2k/m can only be satisfied by changing (if necessary) the ground rigidity k. The system is [overtuned]{undertuned} if [ω f < ωn ]{ω f > ωn }. The force exerted on the ground is given by Fgr = kx = kxm cos ω f t = Fgr o cos ω f t

(3.48a)

from where 1 1 Fgr o = kxm = k δ f   2  = F f o |1 − η2 |   ωf  F f o 1 −   ωn 

(3.48b)

√ is the amplitude of the force Fgr which is less than F f o if ω f > 2ωn . (c) Figure 3.19 is a sketch of the machine foundation if damping is taken into account. The equation of motion is the same as Eq. (3.34b) which is supplemented here by two formulae for ωn and δ f each taken form (3.47). Thus, we have x¨ + 2β x˙ + ωn2 x = ωn2 δ f cos ω f t ,    xm (3.49) F f o xsd g ωn2 = . , δf = xsd mg

3.2 Machine Foundations and Vibration Isolation

77

Ff = F fo cos

x sd

ft

=0 x=0

x m block foundation

c

k

x, Fig. 3.19 Model for the machine foundation with damping

The steady-state solution—see Eq. (3.35)—is given by   xpart = xm cos ω f t + φ

(3.50)

in which—it is worth reminding the reader of Eqs. (3.38), (3.39), and (3.40) here— ωf 1 c xm = δ f  , η= , ζ = 2 ωn cn 1 − η2 + (2ζ η)2 and tan φ = −

2ζ η . 1 − η2

(3.51a)

(3-42b)

The force exerted on the ground reflects the effects of the spring and the damping:     Fgr = c x˙ + kx = kxm cos ω f t + φ − cω f xm sin ω f t + φ =    c   = xm k 1 cos ω f t + φ − ω f sin ω f t + φ . (3.52) k Here it will be assumed that c ω f = Agr sin γ k

1 = Agr cos γ , from where 1+ in which

c k

ωf

2

= A2gr

(3.53)

(3.54a)

78

3 Some Vibration Problems

c c cn ωf = ωf = ↑ = k cn k (3.23c), (3.20) ωf 2mωn ω f = ↑ = 2ζ = 2ζ η. =ζ k ωn (3.2)2 Hence

Agr =

1+

c k

and tan γ =

ωf

2

=



1 + 4ζ 2 η2

c ω f = 2ζ η . k

(3.54b)

(3.55a)

(3.55b)

A comparison of Eqs. (3.52), (3.53), and (3.55a) yields   Fgr = xm k Agr cos ω f t + φ + γ =    Fgr o



  1 + 4ζ 2 η2 = δ f k  cos ω f t + φ + γ .  2 (3.51a),(3.55a) 1 − η2 + (2ζ η)2

=

↑

(3.56)

Let us introduce the following notations: 1 , V1 (η, ζ ) =  2 1 − η2 + (2ζ η)2  1 + 4ζ 2 η2 V2 (η, ζ ) =  . 2 1 − η2 + (2ζ η)2

(3.57)

Making use of the notations introduced we get, in simple forms, the amplitudes xm and Fgr o of the vertical displacement x and the force Fgr exerted on the ground by the machine and its foundation: xm = δ f V1 (η, ζ ) ,

Fgr o = δ f k V2 (η, ζ ) . 

(3.58)

Ff o

Figures 3.20 and 3.21 depict the functions V1 (η, ζ ) and V2 (η, ζ ). Observe that Fig. 3.20 is actually the same as Fig. 3.14. Remark 3.4 If the system is overtuned, i.e., ωn > ω f (η < 1) then the amplitude of vibrations xm is greater than the deflection δ f caused by the amplitude F f o of the excitation force. Remark 3.5 If the system is undertuned, i.e., ωn < ω f (η > 1) then the amplitude xm of the vibrations can be made smaller than the deflection δ f caused by the amplitude F f o of the excitation force.

3.2 Machine Foundations and Vibration Isolation

79

= 0.0

V1 ( , )

= 0.075

= 0.1

5.0 = 0.15

= 0.25 = 0.5

1.0 1.0 f

2.0

n

f

n

=

f n

=

f n

undertuned system

overtuned system

Fig. 3.20 V1 (η; ζ ) against η (ζ is a parameter)

V2 ( , ) = 0.075 = 0.1

5.0 = 0.15 = 0.25

= 0.5

1.0 1.0

2

2.0

Fig. 3.21 V2 (η; ζ ) against η (ζ is a parameter)

Exercise 3.5 A sawing machine1 (see Fig. 3.22) is excited (due to the motion of the sawing tool, i.e., the saw and its frame): F f = F f o cos ω f t, F f o = 16 · 104 N, f f = 600/min (ω f = 2π f f = 62.83 1/s). Determine the mass of the machine and foundation if (a) the force exerted on the ground cannot be greater than 0.05F f o and if (b) the inequality xm < 1 mm should also be satisfied. 1 Photo

courtesy of Zhejiang Guanbao Industrial Co., China.

80

3 Some Vibration Problems

Fig. 3.22 Sawing machine

Under the assumption of an undamped system (ζ = 0) we can use the model shown in Fig. 3.17. If we neglect the transient function the solution for x follows from (3.32), (3.58)1 , (3.57)1 , and (3.47)3 :   x = xm cos ω f t + φ ,

1 , xm = δ f V1 (η, 0) = δ f  1 − η2 

δf =

Ff o . k

The force exerted on the ground is obtained from (3.56)2 , (3.58)2 , and (3.57)2 :   Fgr = Fgr o cos ω f t + φ + γ , where

1 . Fgr o = δ f k V2 (η, 0) = F f o V1 (η, 0) = F f o  1 − η2   Ff o

Since Fgr o  0.05F f o we have to choose an undertuned system: ω f > ωn , η = ω f /ωn > 1. Then 1 Fgr o = 0.05F f o = F f o 2 η −1 from where

and

η2 = 1 +

1 = 21 , 0.05

ωn =

η=

√

21 = 4.583

ωf 62.83 1 = = 13.71 . η 4.583 s

Since xm = 1 we can write xm = 1 = δ f

Ff o 1 1 = η2 − 1 k η2 − 1

and it follows that k=

Ff o 16.0 · 104 N kg = = 0.8 · 104 = ↑ = 0.8 · 107 2 . η2 − 1 20.0 mm 1N=1000 kg mm s s2

3.2 Machine Foundations and Vibration Isolation

81

Using now (3.28b)2 we find the total mass Ff o 0.8 · 107 ∼ k = = = 42558 kg . ωn2 η2 − 1 13.712

m=

Equation (3.49)2 yields the static deflection xsd =

g 9.81 · 103 = ≈ 52 mm . ωn2 13.712

If the fulfillment of the more rigorous inequality Fgr ≤ 0.005F f o is the requirement in the manner of the previous solution, we obtain η=

√ ωf ωf 62.83 1 = = 1.405 , = 2001 = 44.73 , ωn = ωn η 44.73 s 7 k 0.8 · 10 ∼ m= 2 = = 4.055112 × 106 kg . ωn 1.4052

This mass is very large, and therefore it is not allowable.

3.2.3 Machine Foundation Under the Action of a Rotating Unbalance Figure 3.23 shows a block foundation of mass M. The foundation carries a rotating machine—its total mass is denoted by m—which is unbalanced since the mass center does not coincide with the center of rotation. The effect of the ground on the foundation is modeled by a spring of stiffness k and a viscous vibration damper with damping coefficient c. It will be assumed that the unbalance can be represented by a mass m u which rotates on a circle of radius e (sometimes called the eccentricity) with the angular velocity ω f of the rotating machine. We assume further that the rotating Fig. 3.23 Rotating machine on block foundation

mu

e

x M

k 2

c

k 2

82

3 Some Vibration Problems

machine and its foundation is constrained to move vertically. We have, therefore, a single degree of freedom system. Let x be the vertical displacement. When the system is at rest x = 0, it is clear that ϕ = ωf t ,

vux = x˙ − eω f sin ω f t

are the polar angle and the vertical velocity component of the unbalanced mass m u . The equation of motion is as follows: (M + m − m u ) x¨ + m u

d vux = −c x˙ − kx dt

or (M + m) x¨ + c x˙ + kx = m u eω2f cos ω f t   

(3.59a)

Ff o

in which F f o is the amplitude of the excitation force. If the rotating machine is balanced e = 0, then there is no excitation. Otherwise, the amplitude of the excitation force F f o depends on m u and its distance e from the axis of rotation as well as on the angular velocity ω f . For small e, however, the effect of the excitation force can be neglected. If we divide throughout by M = M + m we can manipulate Eq. (3.59a) into a form similar to that of Eq. (3.34b) x¨ +

2 k k m u eω f c x˙ + x= cos ω f t M M M   k     ωn2

2β

ωn2

(3.59b)

δf

or x¨ + 2β x˙ + ωn2 x = ωn2 δ f cos ω f t in which 2β =

c , M

ωn2 =

k , M

δf =

(3.59c)

m u eω2f Ff o = . k k

(3.60)

Recalling (3.50) and (3.51), we have the steady-state motion in the form   xpart = xm cos ω f t + φ , where xm = δ f 

1 − η2

1 2

+ (2ζ η)

2

= ↑ = δ f V1 (η, ζ ) , η = (3.58)

ωf c , ζ = ωn cn

is the amplitude of the steady-state motion. By utilizing the transformation δf =

m u eω2f k

=

2 mu 1 2 mu e ω f mu e 2 η = ωf = M k M ωn2 M M

(3.61)

3.2 Machine Foundations and Vibration Isolation

we can rewrite the amplitude xm into the form mu e 2 mu e η V1 (η, ζ ) = V3 (η, ζ ) , M M η2 . V3 (η, ζ ) = η2 V1 (η, ζ ) =  2 2 2 1 − η + (2ζ η) xm =

Fig. 3.24 Dimensionless amplitude of the rotating machine

Fig. 3.25 Dimensionless force exerted on the ground by the rotating machine

83

84

3 Some Vibration Problems

Figure 3.24 is obtained by plotting the dimensionless quotient xm M = V3 (η, ζ ) e mu against η. For small η (for small ω f ), the amplitude xm of the motion is nearly 0: then the system is overtuned. For relatively large η, the amplitude xm becomes a constant of value e m u /M independently of the amount of damping and the angular velocity ωf. Condition dV3 (η, ζ ) =0 dη

−→

  1 − η2 1 − 2ζ 2 = 0

√ yields that the amplitude xm has a finite maximum if 0 < ζ < 1/ 2 = 0.707 and η2 =

1 . 1 − 2ζ 2

√ The greater ζ ∈ (0, 1/ 2) is the smaller the maximum amplitude is. Remark 3.6 If we want to reduce xm it is worthwhile to overtune the system. For internal combustion engines (e.g., four-stroke engines), the frequency spectrum contains subharmonic frequencies. Overtuning is, however, possible for one (say a subharmonic) frequency only. If the wheels of your car are unbalanced you have to speed up to increase ω f when your car begins to shake. This is the way to avoid shaking of your car. Remark 3.7 The rotating machine can be balanced if we place a balancing mass (denoted by m b ) onto the machine diametrically opposite to m u . Let h be the distance of the balancing mass m b from the axis of rotation. The rotating machine is balanced if h m b = e m u since then the mass center of the two masses is located on the axis of rotation. The amplitude of the force exerted on the ground by the foundation is obtained from (3.58)2 and (3.57)2 : Fgr o

 1 + 4ζ 2 η2 = δ f k V2 (η, ζ ) = δ f k  , 2 1 − η2 + (2ζ η)2

where utilizing (3.61), we get δfk = Hence

m u ek 2 η . M

(3.62)

3.2 Machine Foundations and Vibration Isolation

Fgr o

85

 η2 1 + 4ζ 2 η2 m u ek 2 m u ek m u ek  η V2 (η, ζ ) = V4 (η, ζ ). = =  2 M M M 1 − η2 + (2ζ η)2    V4 (η,ζ )

(3.63) Figure 3.25 depicts the dimensionless quotient Fgr o M = V4 (η, ζ ) e mu k against η. Remark 3.8 Since lim η→∞ V4 (η, ζ ) = ∞ it follows that the force exerted by the foundation on the ground can be reduced effectively only if we design an overtuned system.

3.2.4 Displacement Excitation on a Machine Foundation Excitation can be caused not only by a force but by giving a prescribed motion to an elastic element of the system. If we apply again a single degree of freedom model for describing the behavior of a block foundation the displacement excitation means that the motion of the ground on which the block lies is prescribed due to an external effect. An analysis which clarifies how such an excitation influences the Fig. 3.26 Displacement excitation

m block foundation

G x c

k

sd = s do cos

ft

behavior of the foundation could help us to reduce the harmful vibrations that are transmitted from the environment. Figure 3.26 shows the model we apply to examine this phenomenon. It is not too difficult to check that the change in length of the spring is x − sd . Consequently, the velocity difference is given by x˙ − s˙d . In terms of these quantities, the equation of motion takes the form m x¨ + c (x˙ − s˙d ) + k (x − sd ) = 0

86

3 Some Vibration Problems

or m x¨ + c x˙ + kx = c˙sd + ksd . If we divide throughout by m and apply the notations (3.17b)2,3 , we get   x¨ + 2β x˙ + ωn2 x = sdo −2βω f sin ω f t + ωn2 cos ω f t .

(3.64)

Let us manipulate the right side of this equation into a cos function. The transformation is based on the equation   (3.65) − 2βω f sdo sin ω f t + ωn2 sdo cos ω f t = ωn2 δ f cos ω f t + γ       A cos γ

A sin γ

in which A and γ are the unknowns. If we square both sides of the formulas A sin γ = 2βω f sdo ,

A cos γ = ωn2 sdo

and then add them to each other we get A = sdo



ωn2

2



+ 2βω f

2

 =

sdo ωn2

 1+

2β ω f ωn ωn

2

in which according to (3.38) ωf 2β = η and = 2ζ . ωn ωn

(3.66)

With this result we get on the base of (3.65) that A = sdo ωn2



from where δ f = sdo

1 + (2ηζ )2 = ωn2 δ f

(3.67a)

1 + (2ηζ )2 .

(3.67b)

It is obvious that tan γ = 2βω f /ωn2 = 2ηζ .

(3.68)

Upon substitution of (3.65) and (3.67a), the equation of motion (3.64) can be rewritten in the form     2 2 2 x¨ + 2β x˙ + ωn x = ωn δ f cos ω f t + γ = ωn sdo 1 + (2ηζ )2 cos ω f t + γ . (3.69) The amplitude of the steady-state solution   xpart = xm cos ω f t + γ + φ

(3.70)

3.2 Machine Foundations and Vibration Isolation

87

    (see Eqs. (3.35) and (3.34b)— ω f t + γ + φ corresponds to ω f t + φ in (3.35)) follows from (3.39) by utilizing (3.67b) and (3.57)  δf sdo 1 + (2ηζ )2 x m =  = ↑ =  = ↑ = sdo V2 (η, ζ ) . 2 2 1 − η2 +(2ζ η)2 (3.67b) 1 − η2 + (2ζ η)2 (3.57) (3.71) Remark 3.9 For passive vibration reduction, it is worth designing an undertuned system for which ωn is much smaller than ω f , i.e., the mass of foundation is relatively large and the spring stiffness is small (Fig. 3.27). Exercise 3.6 The switch box of a machine tool is to be protected against the vibrations that are transmitted from the environment. The machine and the switch box are on the second floor. The vibrations of the floor have an amplitude of sdo = 20 µm = 20 × 10−6 m. The natural frequency is f f = 960 rpm = 16/s. The amplitude of the switch box vibrations cannot be more than xm = 2 µm = 2 · 10−6 m. The mass of the switch box is 300 kg. It is put onto four identical springs. Determine the spring constants if the damping is negligible. switch box

x

sd

second floor model

sd Fig. 3.27 Switch box of a machine tool

We have a displacement excitation of the form sd = sdo cos ω f t. If there is damping the solution is given by (3.70). Since now we have no damping ζ = 0 and it follows from (3.71) that xm = sdo V2 (η, ζ ) |ζ =0 = sdo

1 1 = ↑ = sdo 2 |1 − η2 | xm cn which means that we have heavy damping. Prove that the body never passes (again) through its position of equilibrium (a) if it is released with vo = 0 initial velocity from an arbitrary position xo or (b) if it is started from its equilibrium position (for which naturally xo = 0) with an arbitrary initial velocity vo . (Hint: the proofs should be based on Eq. (3.21).) Problem 3.3 Show that the logarithmic decrement δ—see Eq. (3.27)—can be given in the form 2π c/cn . δ= 1 − (c/cn )2

96

3 Some Vibration Problems

Problem 3.4 Show that the logarithmic decrement can also be given as δ=

x(t) 1 ln , k x(t + kτd )

where k is a natural number and k ≥ 2. For k = 1, we get back definition (3.27). Problem 3.5 Figure 3.34 shows two support arrangements for a slender uniform rod of length  and mass m. The rod is subjected to a harmonic force of excitation at its middle point. Examine which support arrangement results in a steady-state response function with a smaller amplitude. /2

/4

/4

m

A

Ff = Ffocos

c

k

ft

1 /2

/4

/4

m

A

Ff = Ffocos

k

c

ft

2 Fig. 3.34 Models of vibrating slender rods

Problem 3.6 A block with mass m 1 = 20 kg is dropped from a height of 2.5 m onto the m 2 = 10 kg pan of a spring scale. Determine the maximum deflection of the pan if the impact is perfectly plastic. The spring constant is k = 20 kN/m.

m1 h m2

k

n Fig. 3.35 Block m 1 dropped onto block m 2

t

References

97

References 1. F.E. Richart, J.R. Hall, R.D. Woods, Vibrations of Soils and Foundations (Prentice Hall, Inc., Upper Saddle River, 1970) 2. H. Dreisig, F. Holzweißig, Dynamics of Machinery (Springer, Berlin, 2010)

Chapter 4

Introduction to Multidegree of Freedom Systems

4.1 Lagrange’s Equations of Motion of the Second Kind 4.1.1 Generalized Coordinates If a single particle moves freely in space it has three degrees of freedom which means that three Cartesian coordinates are needed to give its position in space. If a rigid body moves freely in the three-dimensional space we need six independent quantities to give the position of the body uniquely. Consider this issue in a bit more detail. The rigid body B shown in Fig. 4.1 moves freely in space. The Cartesian coordinate system (x yz) is the absolute coordinate system. The Greek Cartesian coordinate system (ξηζ) is attached to the mass center G and moves together with the body in such a way that the orientation of the coordinate axes remains unchanged, i.e., ξ||x, Fig. 4.1 Coordinate systems for a body moving freely in space

z

r AG G

A

z AG yAG

y xAG

x

© Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_4

99

100

4 Introduction to Multidegree of Freedom Systems

η||y, and ζ||z during the motion. The third coordinate system (ξ  η  ζ  ) is also attached to the body at the mass center G but, in contrast to the coordinate system (ξηζ), it rotates together with the body. When the motion begins it will be assumed that the two Greek coordinate systems coincide. During the motion the angles formed by the axes ξ and ξ  , η and η  , ζ and ζ  will be denoted by ϕξ  ξ , ϕη η , ϕζ  ζ , respectively. It is obvious that the six time functions x AG (t), y AG (t), z AG (t), ϕξ  ξ (t), ϕη η (t), ϕζ  ζ (t) uniquely determine the position of the rigid body in space: which shows that the degree of freedom is really six for the free motion of a rigid body. It follows from the mentioned two examples that the degree of freedom is, in fact, the number of independent coordinates we need to describe the motion. Observe that the quantities ϕξ  ξ (t), ϕη η (t), ϕζ  ζ (t) are not Cartesian coordinates; they are, in fact, angle coordinates. For the sake of generality, the quantities needed to describe the motion uniquely will also be referred to as generalized coordinates which are, in general, denoted by the letter q. We can now say that we need the six generalized coordinates q1 = x AG , q2 = y AG , q3 = z AG and q4 = ϕξ  ξ , q5 = ϕη η , q6 = ϕζ  ζ

(4.1)

to describe the free motion of the rigid body considered. Motions of particles and bodies are not always free motions since in many cases there are various restrictions, called constrains, which force the bodies to move in a predetermined manner. If there are constraints the motion is a constrained motion. Exercise 4.1 Examine what coordinates could be used to describe the motion of the spherical pendulum (a mass m on an inextensible cord of length r ) shown in Fig. 4.2. Fig. 4.2 Spherical pendulum

z

A

y

r cos

x

y

r cos

sin

x

r cos

cos

r z

r sin

m Observe that the set of coordinates we can use to describe the motion is not unique. The position of the mass m at any point of time may be given in Cartesian coordinates x, y, and z—they are, however, not independent on each other—or in terms of the spherical coordinates r = r , ψ, and φ: x = r cos φ cos ψ ,

y = r cos φ sin ψ , z = −r sin φ .

(4.2)

4.1 Lagrange’s Equations of Motion of the Second Kind

101

If we use spherical coordinates only two, i.e., φ and ψ are independent since the length r is constant in time. If we use Cartesian coordinates, we can choose, for instance, x and y as independent coordinates while z is the solution of the constraint equation  (4.3) f c (x, y, z, t) = f c (x, y, z) = r − x 2 + y 2 + z 2 = 0, which expresses that r is constant. For this choice z is a superfluous coordinate which can be eliminated by utilizing the constraint equation (4.3). Since we have two independent coordinates q1 = φ and q2 = ψ (or q1 = x and q2 = y) the spherical pendulum represents a system of two degrees of freedom. Equation (4.3) is a geometric constraint equation. The constraint equation of the form f c (x, y, z, t) = 0

(4.4)

is called holonomic. Exercise 4.2 Particles m 1 and m 2 are connected with a weightless rod of constant length . Assume that these two mass systems move in such a manner in the coordinate plane x y that the velocity at the midpoint C is parallel to the rod. What are the constraint equations? Fig. 4.3 Weightless rod with two masses at the endpoints

y

m2

C

y2 y1

m1 x 2 x1

A

x

Since the system moves in the coordinate plane x y and the length of the rod is constant the following equations should be satisfied: f c1 (x1 , y1 , z 1 , x2 , y2 , z 2 , t) = z 1 = 0, f c2 (x1 , y1 , z 1 , x2 , y2, z 2 , t) = z 2 = 0, f c3 (x1 , y1 , z 1 , x2 , y2 , z 2 , t) =  − (x2 − x1 )2 + (y2 − y1 )2 = 0 .

(4.5a)

Constraints (4.5a) are all holonomic. It is obvious on the base of Eq. (1.4) that the velocities linearly depend on the distances. Consequently, vC x =

1 1 (x˙1 + x˙2 ) , vC y = ( y˙1 + y˙2 ) 2 2

102

4 Introduction to Multidegree of Freedom Systems

are the two velocity components at the midpoint C. If vC is parallel to the rod then vC y y2 − y1 = vC x x2 − x1 from where it follows that f c4 (x1 , y1 , z 1 , x2 , y2 , z 2 , x˙1 , y˙1 , z˙ 1 , x˙2 , y˙2 , z˙ 2 , t) =

y˙1 + y˙2 x˙1 + x˙2 − = 0 . (4.5b) y2 − y1 x2 −x1

The constraint equations that involve velocities are called nonholonomic constraint equations provided that the velocity components, which are various time derivatives, cannot be removed form the constraint equation by integration. Equation (4.5b) is a nonholonomic constraint equation. It can also be called kinematic constraint equation since it expresses that the velocity at C is parallel to the rod.

4.1.2 Principle of Virtual Work A virtual displacement (or displacement variation) is an infinitesimal change in the corresponding coordinate which may be chosen arbitrarily irrespective of the time provided that it is consistent with the constraints, i.e., it does not violate the constraint equations of the system. Exercise 4.3 Determine the virtual displacements for the spherical pendulum shown in Fig. 4.2. Since ψ and φ may change arbitrarily (independently of the constraint equation) it follows from Eq. (4.2) that ∂x ∂x δφ+ δψ = −r (sin φ cos ψ δφ + cos φ sin ψ δψ) , ∂φ ∂ψ ∂y ∂y δφ+ δψ = r (− sin φ sin ψδφ + cos φ cos ψ δψ) , δy = ∂φ ∂ψ ∂z ∂z δφ + δψ = −r cos φ δφ δz = ∂φ ∂ψ

δx =

(4.6)

are the three virtual displacements. We remark that the constraint condition r −

 (x + δx)2 + (y + δ y)2 + (z + δz)2 = 0

should also be satisfied by the virtual displacements. Consider a system of N (N ≥ 2) particles in motion—see Fig. 4.4. We assume that the system has n degrees of freedom 2 ≤ n < 3N . Here and in the sequel we

4.1 Lagrange’s Equations of Motion of the Second Kind Fig. 4.4 A system of particles in motion

103

y

m R r

mj rj

A

Rj

y

x

will also assume that the constraint equations are all holonomic, i.e., they can be given in the form: f ci (x1 , y1 , z 1 , x2 , y2 , z 2 , . . . , x N , y N , z N , t) = 0, i = 1, 2, . . . , 3N − n . (4.7) Let R j be the resultant of the forces acting on particle m j . It can be resolved into two parts (4.8) R j = F j + f jc , where f jc is the constraint force exerted on the jth particle. It is a further assumption that the constraint forces do no work at all (for example, if the constraints are smooth contacting surfaces then the constraint forces (contact forces) are perpendicular to the velocity, and therefore they have no power and do no work). The constraints are called ideal ones if the constraint forces do no work. According to Newton’s second law, the motion of the jth particle is described by the dynamic equilibrium equation F j + f jc − m j r¨ j = 0 ,

j = 1, . . . , N .

(4.9)

Let δr j be the virtual displacement of the jth particle. If we dot multiply Eq. (4.8) by δr j and add the dot products obtained to each other, we get N    F j + f jc − m j r¨ j · δr j = 0

(4.10a)

j=1

in which f jc · δr j = 0—the constraint forces do no work—hence N    F j − m j r¨ j · δr j = 0,

(4.10b)

j=1

where δW =

N  j=1

F j · δr j and δWinertia = −

N  j=1

m j r¨ j · δr j

(4.11)

104

4 Introduction to Multidegree of Freedom Systems

are the virtual works done by the forces F j and the forces of inertia m j r¨ j throughout the virtual displacements δr j . Equation (4.10b) is known as the principle of virtual work (or the weak form of the dynamic equilibrium equations) which says that the sum of the two virtual works is equal to zero (4.12) δW + δWinertia = 0 .

4.1.3 Derivation of Lagrange’s Equation of Motion of the Second Kind It is worth mentioning that these equations were derived by Joseph Louis Lagrange [1, 2]. Since the system has n degrees of freedom the position vector of the jth particle can be given in terms of the generalized coordinates q1 , q2 , . . . , qn in the form (4.13) r j = r j (q1 , q2 , . . . , qn ; t) . The virtual displacements of particle j can now be calculated in the same manner as in Exercise 4.3:  ∂r j ∂r j ∂r j ∂r j δq1 + δq2 + · · · + δqn = δq . ∂q1 ∂q2 ∂qn ∂q =1 n

δr j =

(4.14)

Making use of the previous equation one can rewrite the virtual work δW in such a manner which makes possible to introduce the concept of generalized forces: δW =

N  j=1

F j · δr j =

N  j=1

Fj ·

 n  ∂r j =1

∂q

 δq

⎛ ⎞ n N   ∂r j⎠ ⎝ Fj · δq . (4.15) = ∂q  j=1 =1

  Q

Here Q =

N 

Fj ·

j=1

∂r j ∂q

(4.16)

is the generalized force which belongs to the generalized coordinate q . (We have as many generalized forces as there are generalized coordinates.) The work δW can now be given in terms of the generalized forces: δW =

n  =1

Q  δq .

(4.17)

4.1 Lagrange’s Equations of Motion of the Second Kind

105

Consider now, for the sake of our further consideration, the velocity of particle j. Recalling (4.13), we can write  ∂r j dr j ∂r j ∂r j ∂r j ∂r j ∂r j q˙1 + q˙2 + · · · + q˙n + q˙ + = = . (4.18) dt ∂q1 ∂q2 ∂qn ∂t ∂q ∂t =1 n

vj = Hence

∂ r˙ j ∂r j ∂v j = = , ∂ q˙ ∂ q˙ ∂q

which shows that the derivative

(4.19)

∂v j ∂ q˙

is the velocity that belongs to a unit generalized coordinate velocity. Let us now substitute (4.19) into (4.16). We get

Q =

N 

∂v j Fj · . ∂ q˙ j=1  

(4.20)

P j

In other words, the generalized force is the sum of the powers that are calculated as the product of a force and a velocity which belongs to a unit coordinate velocity. In what follows, our aim is to manipulate δWinertia in Eq. (4.11) into a more suitable form. To this end, substituting (4.14) into the expression m j r¨ j · δr j , we have m j r¨ j · δr j =

n  =1

m j r¨ j ·

∂r j δq , ∂q

(4.21)

where ∂r j d = ∂q dt  d m j r˙ j = dt

m j r¨ j ·

  ∂r j d ∂r j − m j r˙ j m j r˙ j · = ↑ = ∂q dt ∂q ∂r j = ∂r˙ j ∂q ∂ q˙    ∂ r˙ j ∂ r˙ j 1 d ∂ ∂ − m j r˙ j m j r˙ j · r˙ j = · = − ∂ q˙ ∂q dt ∂ q˙ ∂q 2    =

Ej

d ∂ ∂ − dt ∂ q˙ ∂q

 Ej

in which E j is the kinetic energy of particle j. If we insert this result into (4.21), we get

106

4 Introduction to Multidegree of Freedom Systems

 n   d ∂ ∂ E j δq . m j r¨ j · δr j = − dt ∂ q˙ ∂q =1 Let us now sum over the N particles. We obtain δWinertia

⎞ ⎛  N n   d ∂ ∂ ⎝ ⎠ =− m j r¨ j · δr j = − − E j δq dt ∂ q˙ ∂q j=1 j=1 =1   N 

(4.22)

E

in which E is the total kinetic energy of the system. To complete the derivation, we shall add Eqs. (4.17) and (4.22). If we take (4.12) into account, we get δW + δWinertia =

n  

Q −

=1

d ∂E ∂E + dt ∂ q˙ ∂q

 δq = 0

from where with regard to the arbitrariness of δq it follows d ∂E ∂E − = Q , dt ∂ q˙ ∂q

 = 1, 2, . . . , n .

(4.23)

This equation is known as Lagrange’s equation of motion of the second kind (or Lagrange’s equations of the second kind). Generalizations: 1. Assume that the system with n degrees of freedom consists of N rigid bodies. It is worthy of mentioning that Lagrange’s equations of motion of the second kind are valid for such systems as well. 2. Assume that the external forces exerted on the jth rigid body are replaced by the equivalent force couple system [F j , M j ] at the point P j of the body. Let v j be the velocity of the point P j . Further let ω j be the angular velocity of the jth body. With the knowledge of these quantities we can calculate the generalized forces Q  by using the following equation: Q =

N  j=1

∂v j  ∂ω j + Mj · , ∂ q˙ ∂ q˙ j=1 N

Fj ·

 = 1, 2, . . . , n .

(4.24)

Assume that the system considered is a conservative one. Then there exists a potential function U (q1 , q2 , . . . , qn ) such that the work done by the forces acing on the system is equal to the negative of the potential function: W (q1 , q2 , . . . , qn ) = −U (q1 , q2 , . . . , qn ),

(4.25)

4.1 Lagrange’s Equations of Motion of the Second Kind

107

while the virtual work is given by δW (q1 , q2 , . . . , qn ) = −

n  ∂U (q1 , q2 , . . . , qn )

∂q

=1

δq .

(4.26)

Comparison of Eqs. (4.17) and (4.25) yields formula Q = −

∂U (q1 , q2 , . . . , qn ) ∂q

(4.27)

for the generalized forces. Thus, we can write −∂U/∂q for Q  in Eq. (4.23) obtaining in this manner a further form of Lagrange’s equations: d ∂E ∂E ∂U − + = 0, dt ∂ q˙ ∂q ∂q

 = 1, 2, . . . , n .

(4.28)

The Lagrangian can then be defined by the relation L = E −U .

(4.29)

The dimension of the Lagrangian is, of course, energy. Since U is independent of the generalized coordinate velocities q˙ Lagrange’s equations (4.28) can be rewritten into the form d ∂L ∂L − = 0,  = 1, 2, . . . , n . (4.30) dt ∂ q˙ ∂q

4.2 Applications of Lagrange’s Equations 4.2.1 Calculations of Generalized Forces Spring forces. Figure 4.5a shows the rod AB which is supported by a spring at the right end. Let q be the displacement at the point B in the direction of the spring. It is obvious that the strain energy stored in the spring and the force exerted by the spring on the rod are given by U=

1 2 kq , 2

Fk = −kqn .

(4.31)

Utilizing the power P = Fk · v B of the spring force we obtain that the work done by the force in the time interval [0, t] is

108

4 Introduction to Multidegree of Freedom Systems

Fig. 4.5 Rods supported by springs

y

n

a

yB

n =1

q x

A

B

k

y b

kv

q x B

A 

t

W =



t

P dt =

0

 Fk · v B dt =

0

↑

vB =

∂r B ∂q

= q˙

0

t

∂r B Fk · q˙ dt = ∂q    dq



q

Q k dq 0

Qk

(4.32a) or 

t



t



q

1 kq dq = − kq 2 = −U . 2 n·v B =q˙ 0 0 0 (4.32b) Comparison of Eqs. (4.32a) and (4.32b) yields the generalized spring force: W =

Fk · v B dt =

−kqn · v B dt = ↑ = −

Qk = −

∂U . ∂q

(4.33)

Note that the above line of thought is, in fact, a proof of equation (4.27) for this simple case. Rod AB in Fig. 4.5b is supported by a volute spring of spring constant kv at the left end. The angle of rotation is denoted by q. If we rotate the rod about the axis at A 1 M A = −kv qiz (4.34) U = kv q 2 , 2 are the strain energy stored in the volute spring and the restoring moment exerted by the volute spring on the rod. The angular velocity of the rod is ω = qi ˙ z . Since the power of the restoring moment is given by the relation P = M A · ω = −kv q q˙ it follows that the work done by the restoring force in the time interval [0, t] can be calculated as

4.2 Applications of Lagrange’s Equations

 W =

t



t

M A · ω dt =

0

0

109

 −kv q qdt ˙ =− 

q 0

1 kv q dq = − kv q 2 = −U . (4.35) 2

dq

On the other hand,

 W =

q

Q k dq .

(4.36)

0

Comparison of Eqs. (4.35) and (4.36) results in relation (4.33) again.

y

n

q

vB

n =1

vn = ( vB n) n

x

B

A c

Fig. 4.6 Rod supported by a vibration damper

Damping forces. Figure 4.6 shows the rod AB which is now supported by a vibration damper at the right end—viscous damping is assumed. The unit vector n lies on the axis of the vibration damper. On the base of Eq. (4.20) we can write for the generalized damping force that Q c = Fc ·

∂v B , ∂ q˙

(4.37)

where Fc is the damping force, v B is the velocity of the point B, and q is the generalized coordinate (the displacement component at B parallel to n). The damping force is given by the equation (4.38) Fc = −cvn in which c is the damping factor and vn is that velocity component of v B which is parallel to n. It is not too difficult to check if we utilize Eq. (4.19) that  vn = (v B · n) n =

   ∂v B ∂r B q˙ · n n . q˙ · n n = ↑ = ∂q ∂ q˙ ∂r B ∂v B  ∂q = ∂ q˙ vB

(4.39)

110

4 Introduction to Multidegree of Freedom Systems

Substituting first (4.38) and then (4.39) into (4.37), we get Q c = −c vn ·

   2 ∂v B ∂v B ∂v B ∂v B ˙ = −c q˙ · n · n = −c · n q˙ = −cg q, ∂ q˙ ∂ q˙ ∂ q˙ ∂ q˙

  cg

(4.40) where



∂v B cg = c ·n ∂ q˙

2 (4.41)

is the generalized damping coefficient. Generalized excitation forces. For a single degree of freedom system subjected to only one excitation force ∂v (4.42a) Q f = Ff · ∂ q˙ is the formula for the generalized excitation force in which F f is the excitation force and v is the velocity of its point of application. For a multidegree of freedom system Q f =

N  j=1

Ffj ·

∂v j , ∂ q˙

 = 1, . . . , n

(4.42b)

is the th generalized excitation force where F f j is the force of excitation exerted on the jth particle (or body), v j is the velocity of the point where the force F f j is applied. Remark 4.1 For a single degree of freedom system ∂E d ∂E − = Qk + Qc + Q f , dt ∂ q˙ ∂q

Qk = −

∂U ∂q

(4.43)

is the equation of motion where E is the total kinetic energy, Q k , Q c , and Q f are the generalized spring force, damping force, and excitation force. Since the spring force has, in general, a potential U it can be given as the negative of the derivative ∂U/∂q. Exercise 4.4 Figure 4.7 shows a single degree of freedom system. The solid disk of mass m and radius R is hinged at A. The vibration damper with a damping coefficient c is attached to the mass center. The motion of the right end of the spring (the spring constant is k), which is attached to the top of the disk, is prescribed. Derive the equation of motion by choosing ϑ for the generalized coordinate q and determine the circular frequency of the undamped free vibrations. Assume that ϑ is small.

4.2 Applications of Lagrange’s Equations

111

y k

o cos

ft

m

c

G

R

x

A Fig. 4.7 Displacement excitation on a spring mass system

The kinetic energy of the disk is given by E=

1 ˙2 JAϑ , 2

JA =

3 m R2 . 2

(4.44)

Using the length change ηo cos ω f t − 2Rϑ of the spring we get the potential energy in the form 2 1  (4.45) U = k ηo cos ω f t − 2Rϑ . 2 The corresponding generalized force is −

  ∂U 2 = −k ηo cos ω f t − 2Rϑ (−2R) = 2Rkηo cos ω f t − 4R

 k ϑ =   ∂ϑ kg

Q fo

=Q f o cos ω f t + kg ϑ ,

   Qf

(4.46)

Qk

where Q f o and kg are the amplitudes of the generalized excitation force and the generalized spring constant while Q f and Q k are the generalized excitation and spring forces. The generalized damping force is obtained from (4.40) by taking into ˙ x , n = ix . We get account that vG = ϑRi 

∂vG Q c = −c · ix ∂ ϑ˙

2

ϑ˙ = −c R 2 ϑ˙ = −cg q˙ 

(4.47)

cg

in which cg is the generalized damping coefficient. Upon substitution of (4.44), (4.46), and (4.47) into (4.43) the equation of motion is obtained

112

4 Introduction to Multidegree of Freedom Systems 2 J A ϑ¨ − 0 = 2Rkηo cos ω f t − 4R c R 2 ϑ˙  k ϑ −     mg

or

kg

Q fo

(4.48a)

cg

m g ϑ¨ + cg ϑ˙ + kg ϑ = Q f o cos ω f t,

(4.48b)

where m g is the generalized mass. For the circular frequency of the undamped free vibrations, Eqs. (3.2)2 and (4.48b) yield  ωn =

kg = mg



4R 2 k = 3 m R2 2



8k . 3m

(4.49)

Exercise 4.5 Figure 4.8 depicts a simple model of a gear box (a geared torsional system). When investigating the torsional vibrations of the shafts we apply the following assumptions: (a) shafts 1 and 2 are elastic. They behave if they were linear volute springs for which k1v =

I p1 G 1 , 1

k2v =

I p2 G 2 2

(4.50)

are the spring constants where I pi , G i , and i (i = 1, 2) are the polar moments of inertia, the shear moduli, and the lengths of the shafts. It will be assumed the masses of the shafts are much less than those of the gears. Therefore, their effect on the kinetic energy will be neglected. (b) Gears 1, 2  , 2  , and 3 are rigid, their moments of inertia with respect their axes of symmetry are denoted by J1 , J2 , J2 and J3 . Let ϑ1 , ϑ1 , ϑ1 , and ϑ3 be the angles of rotation of the gears. Find the equations of motion provided that a harmonic torsional moment M f = M f o cos ω f t is acting on gear 3.

R1 J2

k1 J1

1

2

R3

R2 R2

k2

Mf

M fo cos

2

1

J2 Fig. 4.8 A simple model for a gear box

J3 2

3

f

t

4.2 Applications of Lagrange’s Equations

113

The generalized coordinates q1 , q2 , and q3 are defined by the following relations: q1 = R2 ϑ1 ,

q2 = R2 ϑ2 = R2 ϑ2 ,

q3 = R2 ϑ3 .

(4.51)

Making use of the angular velocities q˙1 ϑ˙ 1 =  , R2

q˙2 ϑ˙ 2 =  , R2

q˙2 ϑ˙ 2 =  , R2

q˙3 ϑ˙ 3 =  R2

(4.52)

it is easy to calculate the total kinetic energy of the system: E=

1  ˙ 2 1   ˙  2 1   ˙  2 1  ˙ 2 J1 ϑ1 + J2 ϑ2 + J2 ϑ2 + J3 ϑ3 = 2 2  2 2     J J1 J J3 1 2 2 2 2 2 =   q˙ +   2 +   2 q˙2 +   2 q˙3 2 R 2 1 R R R2 2 2  2      m1

m2

(4.53a)

m3

in which m1 = 

J1 2 , R2

m2 = 

J2 J2 +   2 , 2 R2 R2

m3 = 

J3 R2

2

(4.53b)

are the generalized masses. Hence, E=

 1 m 1 q˙12 + m 2 q˙22 + m 3 q˙32 2

(4.53c)

is the total kinetic energy. The potential energy (strain energy) stored in the shafts is given by U=

 2 1  2 1 k1v ϑ2 − ϑ1 + k2v ϑ3 − ϑ2 = 2 2 1 k2v 1 k1v =  2 (q2 − q1 )2 +  2 (q3 − q2 )2 ,  2 R 2 R  2 2     k1

where k1 = 

(4.54a)

k2

k1v k2v  and k2 =   2  2 R2 R2

(4.54b)

are generalized spring constants. Consequently, U=

1 1 k1 (q2 − q1 )2 + k2 (q3 − q2 )2 . 2 2

(4.54c)

114

4 Introduction to Multidegree of Freedom Systems

If we utilize Eq. (4.24) we get the generalized excitation force Q f3=Q f =Mf

M f ∂ q˙3 ∂ω3 ∂ ϑ˙ 3 =Mf =  = ∂ q˙3 ∂ q˙3 R2 ∂ q˙3 Mfo = cos ω f t = Q f o cos ω f t . R  2 

(4.55)

Q fo

Substituting now Eqs. (4.53c), (4.54c), and (4.55) into (4.23) by taking relation (4.27) into account, we get d ∂E ∂E ∂U − =− → m 1 q¨1 = k1 (q2 −q1 ) , dt ∂ q˙1 ∂q1 ∂q1 d ∂E ∂E ∂U − =− → m 2 q¨2 = −k1 (q2 −q1 )+k2 (q3 −q2 ) , dt ∂ q˙2 ∂q2 ∂q2 d ∂E ∂E ∂U − =− + Q f → m 3 q¨3 = −k2 (q3 −q2 )+ Q f o cos ω f t dt ∂ q˙3 ∂q3 ∂q3

(4.56a)

or m 1 q¨1 + k1 (q1 − q2 ) = 0 , m 2 q¨2 − k1 (q1 − q2 ) + k2 (q2 − q3 ) = 0 ,

(4.56b)

m 3 q¨3 − k2 (q2 − q3 ) = Q f o cos ω f t . Equations (4.56b) can be rewritten in matrix form ⎡

m1 ⎣ 0 0

⎤⎡ ⎤ ⎡ ⎤⎡ ⎤ k1 −k1 q¨1 q1 0 0 0 m 2 0 ⎦ ⎣ q¨2 ⎦ + ⎣ −k1 k1 + k2 −k2 ⎦ ⎣ q2 ⎦ = q¨3 0 −k2 k2 q3 0 m3

   

    M (3×3)

q¨

(3×1)

K

q

(3×3)

(3×1)

⎡

⎤ 0 ⎦ 0 =⎣ Q f o cos ω f t

 

(4.56c)

Q (3×1)

in which M,

(3×3)

K,

(3×3)

q (3×1)

and

Q (3×1)

are the mass and stiffness matrices, the generalized displacement matrix, and the loading matrix.

4.2 Applications of Lagrange’s Equations

m1

k1

115

m2

k2

m3

G2

G1 q1

G3 q2

q3

Fig. 4.9 Three degree of freedom spring-mass system

It is not too difficult to check that the equations of motion (4.56) are those of a spring–mass system shown in Fig. 4.9.

4.3 Multidegree of Freedom Systems with Solutions 4.3.1 Examples for Multidegree of Freedom Systems Figure 4.10 shows a few multidegree of freedom systems: (a) a coupled pendulum, (b) a branched geared system, (c) a belt drive (pulley A drives pulley B via a belt), and (d) an unbranched geared system (in this respect we remind the reader of Exercise 4.5). The examples shown in Fig 4.10 represent that the dynamical behavior of some machine parts can be understood only if the model established for the investigation has more than one degree of freedom: it is not too difficult to check that (a) the coupled pendulum has two, (b) the branched geared system has four, (c) the belt drive has again two, and finally the unbranched geared system has three degrees of freedom. We shall consider systems with two degrees of freedom first. a

b

m1

m2 Branched geared system

Coupled pendulum c

d B

A

Driving Fig. 4.10 Multidegree of freedom systems

Unbranched geared system

116

4 Introduction to Multidegree of Freedom Systems

4.3.2 Two Degree of Freedom Systems Coordinate coupling. A vibrating system for which two coordinates (or two generalized coordinates) are needed to describe the motion is referred to as a two degree of freedom system. The equations of motion for two degree of freedom system are, in general, coupled. This means that both generalized coordinates appear in each equation. In the most general case, the equations of motion for an undamped free system assume the following form:        q¨1 q1 m 11 m 12 k k 0 + 11 12 = , m 21 m 22 q¨2 k21 k22 q2 0        



M (2×2)

q¨

(2×1)

K

q

(2×2)

(2×1)

(4.57)

where M is the mass matrix and K is the stiffness matrix. These matrices are symmetric, i.e., M = MT and K = K T . We speak about (a) mass or dynamical coupling if the mass matrix is nondiagonal and (b) stiffness or static coupling if the stiffness matrix is nondiagonal. For undamped systems, it is always possible to find such generalized coordinates for which the mass and stiffness matrices are both diagonal. Then the equations of motion are decoupled and can be solved independently of each other. The coordinates in this coordinate system are called principal or normal coordinates. Exercise 4.6 Figure 4.11 shows a lathe machine1 on its foundation which is modeled as a rigid bar with a mass center not coinciding with the geometrical center of the bar. Investigate how to choose the coordinates so that the equations of motion are statically coupled, dynamically coupled, and both statically and dynamically coupled.

G

=

=

k2

k1 =

G

k2

k1 1

Fig. 4.11 Lathe machine and its model 1 Photo

courtesy of Tony Griffiths of lathes.co.uk

2

4.3 Multidegree of Freedom Systems with Solutions Fig. 4.12 Data for static coupling

117

x x

0

x

1

xG

x

2

G 1

2

Static coupling. Let x and ϑ be the two generalized coordinates. Making use of Eq. (1.51a) and the data of Fig. 4.12 we get the kinetic energy of the bar and the strain energy stored in the two springs: E=

1 1 m vG2 + JG ϑ˙ 2 , 2  2

U=

1 1 k1 (x − 1 ϑ)2 + k2 (x + 2 ϑ)2 , 2 2

x˙ G =x˙

where m and JG are the mass of the bar and the moment of inertia with respect to the mass center. Substituting these quantities into Lagrange’s equation (4.28), we obtain d dt d dt

∂E ∂E ∂U − + = m x¨ + k1 (x − 1 ϑ) + k2 (x + 2 ϑ) = 0 , ∂ x˙ ∂x ∂x ∂E ∂U ∂E + = JG ϑ¨ − 1 k1 (x − 1 ϑ) + 2 k2 (x + 2 ϑ) = 0 . − ∂ϑ ∂ϑ ∂ ϑ˙

These equations can be rewritten into a matrix form 

m 0 0 JG

       x¨ k1 + k2 2 k2 − 1 k1 x 0 + = 2 k2 − 1 k1 21 k1 + 22 k2 ϑ 0 ϑ¨

from where it is clear that the mass matrix is a diagonal matrix. The system is, therefore, dynamically uncoupled but statically coupled.

Fig. 4.13 Data for dynamic coupling

3

4

x xC

0

xC

3

C

y xC

4

e G 1

2

x Dynamic coupling. There exists such a point C on the geometric centerline of the bar (Fig. 4.13) where a vertical force exerted on the bar yields pure vertical translation,

118

4 Introduction to Multidegree of Freedom Systems

the bar does not rotate. Then the two spring forces acting on the bar have no moment about the point C. This condition results in the following equation: k1 xC 3 = k2 xC 4 .   spring force

spring force

Consequently, 3 k1 = 4 k2 . As regards the kinetic energy of the bar it is worth recalling Eq. (1.51a) again in which vG = x˙C + eϑ˙ hence 2 1 1 1  1 E m vG2 + JG ϑ˙ 2 = m x˙C + eϑ˙ + JG ϑ˙ 2 = 2 2 2 2   1 1 2 = m x˙C + 2me x˙C ϑ˙ + JG + me2 ϑ˙ 2 = 2 2

  JC

 1 1 2 m x˙C + 2me x˙C ϑ˙ + JC ϑ˙ 2 . = 2 2 The strain energy stored in the springs is U=

1 1 k1 (xC − 3 ϑ)2 + k2 (xC + 4 ϑ)2 . 2 2

With E and U , Lagrange’s equations (4.28) yield ∂E ∂U d ∂E − + = dt ∂ x˙C ∂xC ∂xC = m x¨C + meϑ¨ + k1 xC − 3 k1 ϑ + k2 xC + 4 k2 ϑ = 0 , ∂U d ∂E ∂E + = − ˙ dt ∂ ϑ ∂ϑ ∂ϑ = JC ϑ¨ + me x¨C − 3 k1 xC + 23 k1 ϑ + 4 k2 xC + 22 k2 ϑ = 0 . By taking into account that the terms underlined cancel each other, we get the matrix equation 

m me me JC



x¨C ϑ¨





0 k + k2 + 1 0 23 k1 + 24 k2



xC ϑ



  0 = , 0

which shows that the system is now dynamically coupled but statically uncoupled.

4.3 Multidegree of Freedom Systems with Solutions Fig. 4.14 Data for static and dynamic coupling

119

0

x xC

xC

C

xC

1

1

2

G 1

2

Static and dynamic coupling. Let now the point C be located at the left end of the bar. It is not too difficult to check using the data of Fig. 4.14 that E=

2 1 1 1  1 m vG2 + JG ϑ˙ 2 = m x˙C + 1 ϑ˙ + JG ϑ˙ 2 = 2 2 2 2   1 1 2 = m x˙C +2m1 x˙C ϑ˙ + JG +m21 ϑ˙ 2 = 2 2   JC

 1 1 2 m x˙C +2m1 x˙C ϑ˙ + JC ϑ˙ 2 = 2 2 and U=

1 1 k1 x 2 + k2 (xC + ϑ)2 , 2 C 2

 = 1 + 2

by the use of which, we get d ∂E ∂E ∂U − + = dt ∂ x˙C ∂xC ∂xC = m x¨C + m1 ϑ¨ + k1 (xC + 1 ϑ) + k2 (xC + ϑ) = 0 , d ∂E ∂E ∂U + = − dt ∂ ϑ˙ ∂ϑ ∂ϑ = JC ϑ¨ + m1 x¨C +1 k1 (xC + 1 ϑ)+k2 (xC +ϑ) = 0 or in matrix form         x¨C m m1 k1 + k2 1 k1 + k2 xC 0 + = . m1 JC 1 k1 + k2 21 k1 + 2 k2 ϑ 0 ϑ¨ This system is a dynamically and statically coupled one.

120

4 Introduction to Multidegree of Freedom Systems

4.3.3 Free Vibrations of Two Degree of Freedom Spring–Mass Systems 4.3.3.1

Equations of Motion

Figure 4.15 shows the typical spring–mass systems with two degrees of freedom. System A is free at its ends, System B is fixed at the left end while System C is fixed at both ends. One can check with ease by using Lagrange’s equations that the equations of motion are A

B m1

m2

k1

G1

m1

ko

G2 x1

k1

G1 x2

m2 G2

x1

x2

C m1

ko

k1

m2

k2

G2

G1

x1

x2

Fig. 4.15 Spring-mass systems of two degrees of freedom

A.



B

C.



m1 0 0 m2 m1 0 0 m2





      x¨1 k1 −k1 x1 0 + = , x¨2 −k1 k1 x2 0

(4.58a)

      x¨1 k + k1 −k1 x1 0 + o = , x¨2 −k1 k1 x2 0

(4.58b)

       x¨1 x1 m1 0 k + k1 −k1 0 + o = , x¨2 −k1 k1 + k2 x2 0 m2 0    

     

M (2×2)

q¨

(2×1)

K

q

0

(2×2)

(2×1)

(2×1)

(4.58c)

which can be given in terms of the mass matrix M, stiffness matrix K, and the matrix q of generalized displacements:

4.3 Multidegree of Freedom Systems with Solutions

q¨ + K

M

(2×2) (2×1)

4.3.3.2

121

q = 0 .

(2×2) (2×1)

(4.59)

(2×1)

General Solution

Solution of differential equation (4.59) is sought in the form q = A cos ωn t, (2×1)

(4.60)

(2×1)

where the matrix AT = [A1 |A2 ] contains the amplitudes for harmonic motion of the two masses. This motion is called normal mode oscillation. After substituting solution (4.60) into the differential equation (4.59) and taking into account that q¨ = −ωn2 A cos ωn t = −ωn2 q ,

q˙ = −ωn A sin ωn t ,

(4.61a)

we get     A = 0 , K − ωn2 M K − ωn2 M A cos ωn t = 0 or (2×2)

(2×2) (2×1)

(4.61b)

(2×1)

which is a homogeneous linear equation system with the unknowns A1 and A2 . Solution for A which is different from the trivial one exists if and only if the determinant of the coefficient matrix vanishes   det K − ωn2 M = 0 .

(4.62)

Introduce the notation ωn2 = λ and then expand the determinant. In this manner, three characteristic equations are obtained:    k1 − m 1 λ −k1     −k1 k1 − m 2 λ  = λ [m 1 m 2 λ − k1 (m 1 + m 2 )] = 0

A.

or m1

1 m 2 λ − (m 1 + m 2 ) = 0 , k1

(4.63a)

(4.63b)

B.    ko + k1 − m 1 λ −k1   =  −k1 k1 − m 2 λ  = m 1 m 2 λ2 − [k1 (m 1 + m 2 ) + ko m 2 ] + ko k1 = 0 or

1 1 m 1 m 2 λ2 − [k1 (m 1 + m 2 ) + ko m 2 ] λ + 1 = 0 , ko k1

(4.64a)

(4.64b)

122

4 Introduction to Multidegree of Freedom Systems

C.     ko + k1 − m 1 λ −k1 =   −k1 k1 + k2 − m 2 λ  = m 1 m 2 λ2 − [(ko + k1 ) m 2 + (k1 + k2 ) m 1 ] + (ko + k1 ) (k1 + k2 ) − k12 = 0 . (4.65)

4.3.3.3

Solutions for System A

Characteristic equation (4.63a) has the following solutions for λ: λo = 0 ,

λ1 =

m1 + m2 . m 1 k11 m 2

(4.66a)

If λ = λo = 0 then ωn = 0. Consequently, q¨ = −ωn2 q = 0

(4.66b)

q = C1 + C2 t

(4.66c)

from where we have (2×1)

(2×1)

(2×1)

in which C1 and C2 are undetermined integration constant matrices. The motion described by Eq. (4.66c) is the well-known rigid body motion. Since neither the left end nor the right end of system A is fixed it can really perform a translational motion (a non-vibratory motion). If λ = λ1 equation system (4.61b) takes the form 

k 1 − λ1 m 1 −k1 −k1 k 1 − λ1 m 2



A11 A21



  0 = 0

(4.67a)

or   m1 + m2 k1 − k1 A11 − k1 A21 = 0, m2   m1 + m2 A21 = 0, −k1 A11 + k1 − k1 m1

(4.67b)

where the second subscript, which is here 1, shows that A = A1 belongs to λ1 . If we multiply Eq. (4.67b)1 by m 2 and Eq. (4.67b)2 by m 1 we arrive at the same result: −k1 m 1 A11 − k1 m 2 A21 = 0 from where it follows that

4.3 Multidegree of Freedom Systems with Solutions

A21 m1 =− A11 m2

or

123

A21 m 2 + A11 m 1 = 0 .

(4.68)

Consequently, Eq. (4.67b) is not independent. Comparison of (4.60) and (4.68) yields the solution      A11 m2 cos ωn1 t = A cos ωn1 t , ωn1 = λ1 q= (4.69) A21 −m 1 in which A is an arbitrary constant.

f1

m1

1/k 1

m2

G1

G2

A

A11

m2 nodal point

f A21

G1 m1

m1

G2

f1

1/k 1

f1

1/k 1

m2

Fig. 4.16 Single degree of freedom systems equivalent to a two degree of freedom system

Assume that the distance between the two mass centers G 1 and G 2 is the flexibility f 1 = 1/k1 —see Fig. 4.16 for details. Since the masses m 1 and m 2 move in opposite directions during motion there exists such a point—called nodal point—between the two masses which remains at rest. Figure 4.16 shows how the amplitude of the motion (as a function of the flexibility f ) changes between the two masses. The nodal point divides the two degree of freedom system into two single degree of freedom systems which have the same natural circular frequency. The bottom part of Fig. 4.16 shows the two single degree of freedom systems. Recalling that ωn2 = k/m for a single degree of freedom system we may write 2 = ωn1

1 k1 m1 + m2 k1 1 = = = = and f 1 + f 1 = f 1 (4.70)  1 m 1 f1 m1 m2 m 2 f 1 m 1 k1 m 2

124

4 Introduction to Multidegree of Freedom Systems

from where it follows that f 1 =

1 1 m2 ,  = k1 m 1 + m 2 k1

f 1 =

1 1 m1 .  = k1 m 1 + m 2 k1

(4.71)

Exercise 4.7 Determine the natural frequencies and the solution q for system B if m1 = m2 = m ,

and

ko = k1 = k.

(4.72)

Under these conditions Eq. (4.64a) yields    2k − λm −k   = m 2 λ2 − 3mkλ + k 2 = 0   −k k − λm 

(4.73)

or

k2 k λ+ 2 = 0. m m √ Introduce a new variable defined by the relation ωˆ = k/m. Using this quantity we can rewrite the above equation into the form λ2 − 3

λ2 − 3 ωˆ 2 λ + ωˆ 4 = 0 , hence λ1,2 = ωˆ

23∓

⎧ √ k ⎨  3 − 5 /2 = 0.3820 9−4 = . √ 2 m ⎩ 3 + 5 /2 = 2.618 0

√

(4.74)

Since determinant (4.61b) is that of the coefficient matrix in Eq. (4.61b) the amplitudes A ji (i, j = 1, 2) should satisfy the following linear equation system: 

2k − λi m −k −k k − λi m



A1i A2i

 =

  0 . 0

(4.75)

If i = 1 we have to solve the equations (2k − λ1 m) A11 − k A21 = 0 , −k A11 + (k − λ1 m) A21 = 0, which are, however, not independent equations. Therefore, it is sufficient to consider the first equation. If we divide throughout by m, we get 

√  k k k 3− 5 2 − m A11 − A21 = 0 m m 2 m

4.3 Multidegree of Freedom Systems with Solutions

fo

Aj1

1/k

m

125

f1

1/k

m

A21

A11

1.618

virtual nodal point

1.0

f

real nodal point

A12

Aj2

1.0

f real nodal point

real nodal point

0.618

A22

Fig. 4.17 Amplitude functions

√ A21 1+ 5 = = 1.618 0 . A11 2

from where

If i = 2 the first equation of (4.75) is of the form (2k − λ2 m) A12 − k A22 = 0 . Dividing throughout again by m, we obtain the equation 

which yields

√  k k k 3+ 5 2 − A12 − A22 = 0, m m 2 m √ A22 1− 5 = −0.6180 . = A12 2

Assume that A11 = A12 = 1. Under this condition Fig. 4.17 shows the amplitudes as functions of f , i.e., the two nodal modes. When the system vibrates in its first mode there exists only one real nodal point. The straight line obtained by joining the points A11 = 1.0 and A12 = 1.618 intersects the axis f outside the interval [0, 2 f o ]. This nodal point is referred to as virtual nodal point. When the system vibrates in its second mode there are two real nodal points. One of them is located in the interval [ f o , 2 f o ]. The system can, therefore, be replaced by two single degree of freedom systems as shown in Fig. 4.18.

126

4 Introduction to Multidegree of Freedom Systems

a

fo

b

f1

m

f1

f1

f1

m

f1

Fig. 4.18 Equivalent single degree of freedom systems

Systems (a) and (b) vibrate with the same natural frequency  ωn2 =

√ 3+ 5 ωˆ , 2

ωˆ =



 k/m =

1 . fo m

As regards system (a) we can write using Eq. (3.2), in which k + k1 should be substituted for k on the base of Eq. (3.41) that 2 ωn2 =

√   1 1 3+ 5 2 k + k1 1 ωˆ = = +  . 2 m m fo f1

Hence, √ k 3+ 5 1 1 = + m 2 m fo m f 1

−→

√ 1 3+ 5 1  = k = . 1 f 1 2 fo

We obtain in the same manner for system (b) that 2 ωn2

√ 3+ 5 1 1 = = 2 m f1 m f 1

−→

√ 1 3+ 5 1  = k1 = . f 1 2 f1

It can easily be checked that f 1 + f 1 = f 1 = f o . As regards system C in Fig. 4.15 the methods that should be applied to finding solutions are basically the same as those we have utilized in connection with systems A and B.

4.3 Multidegree of Freedom Systems with Solutions

127

4.3.4 Forced Vibrations of Two Degree of Freedom Systems It will be assumed that the spring–mass system considered is subjected to a harmonic excitation force   F1 cos ω f t = F f o cos ω f t . (4.76) Ff = 0 (2×1) (2×1) It is also assumed that the motion can be described by the matrix equation 

m1 0 0 m2



      x¨1 k11 k12 x1 F1 cos ω f t . + = x¨2 k21 k22 x2 0

(4.77)

Depending on how the elements of the symmetric stiffness matrix K are selected, this equation is the equation of motion for one of the systems shown in Fig. 4.15. The excitation force is acting on mass m 1 . We seek the steady-state solution in the form     x1 Af1 (4.78) = cos ω f t = A f cos ω f t . x2 Af2 (2×1) Substituting it into Eq. (4.77), we obtain 

k12 k11 − m 1 ω 2f k21 k22 − m 2 ω 2f

or



K − ω 2f M





Af (2×2) (2×1)

(2×2)

Af1 Af2



 =

F1 0



= Ffo .

(4.79a)

(4.79b)

(2×1)

We should notice that       K − ω 2 M  = k11 − ω 2 m 1 k22 − ω 2 m 2 − k 2 = f f f 12 (2×2)

(2×2)

  2  2 2 ω f − ωn2 , = m 1 m 2 ω 2f − ωn1

(4.80)

2 2 where ωn1 = λ1 and ωn2 = λ2 (λ1 ≤ λ2 ) are the roots of the characteristic equation

   K −λ M  = 0, (2×2)

(2×2)

i.e., they are the squares of normal mode frequencies (or the natural frequencies) of the free system. Consequently,      k22 − m 2 ω 2f −k12 F1 2   adj K − ω f M 2 −k k − m ω 0 21 11 1 f (2×2) (2×2)  Ffo =   Af =  2  K −ω M  2 2 f ω 2f − ωn2 m 1 m 2 ω 2f − ωn1 (2×1) (2×1) (2×2)

(2×2)

(4.81)

128

4 Introduction to Multidegree of Freedom Systems

or Af1 =

k22 − m 2 ω 2f   F1 , 2 2 ω 2f − ωn2 m 1 m 2 ω 2f − ωn1

(4.82a)

Af2 =

−k21   F1 . 2 2 2 ω 2f − ωn2 m 1 m 2 ω f − ωn1

(4.82b)

Exercise 4.8 Examine the steady-state vibrations of the system shown in Fig. 4.19. Plot its frequency response curve.

F1 cos ko

ft

k

k1 m1

m

G1

k2

k m2

m

k

G2

x1

x2

Fig. 4.19 Excitation on a two degree of freedom spring-mass system

Comparison of Eqs. (4.58c) and (4.77) yields the equations of motion in the form 

m 0 0 m



      2k −k F1 x¨1 x1 + = cos ω f t , x¨2 x2 0 −k 2k

(4.83)

where it is taken into account that k11 = k22 = 2k ,

k12 = k21 = −k .

Consequently,   2     −k 2 4 2 2   K − ω 2 M  =  2k − m ω f 2  = m ω f − 4kω f + 3k = f  −k 2k − m ω (2×2) (2×2) f    k 3k ω 2f − =0 (4.84a) = m 2 ω 2f − m m is the characteristic equation and its product form since 2 ωn1 =

k m

and

2 ωn2 =

3k m

(4.84b)

are the roots. Equation (4.82a), therefore, becomes Af1 =

2k − m ω 2f C1 C2   + 2 , F1 = 2 2 2 2 ωn1 − ω f ωn2 − ω 2f m 2 ωn1 − ω 2f ωn2 − ω 2f

(4.85)

4.3 Multidegree of Freedom Systems with Solutions

where2 C1 =

129

  2 F1 2k − m ωn1 F1 2k − k  2  = F1 = 2 m (3k − k) 2m m 2 ωn2 − ωn1   2 F1 2k − m ωn2 F1  = , C2 = 2  2 2 2m m ωn1 − ωn2

and

hence, Af1

F1 = 2m

!

1 1 + 2 2 2 ωn1 − ω f ωn2 − ω 2f

"

F1 = 2k

!

1

"

1

2 + 2 .   1 − ω f /ωn1 3 − ω f /ωn1 (4.86a)

Treating A2 in the same manner, we have Af2

F1 = 2k

!

1

1

"

2 − 2   1 − ω f /ωn1 3 − ω f /ωn1

.

(4.86b)

Figure 4.20 shows the frequency response curve (the dimensionless quotient A f i /(F√ 1 /2k) (i = 1, 2) against ω f /ωn1 ). Resonance occurs if ω f = ωn1 or ω f = ωn2 = 3 ωn1 . 2Afi k F1

2Af 2 k F1

2Af 1 k F1

4.0 3.0 2.0 1.0

2Af 1 k F1 2Af 2 k F1

f

3 1.0

1.0 2.0

n1

2.0 2Af 2 k F1

3.0 2Af 1 k F1

3.0 4.0 Fig. 4.20 Frequency response curve

2 To

2 − ω 2 ] {ω 2 − ω 2 } and set ω to [ω ]{ω }. solve for [C1 ] {C2 } multiply (4.85) by [ωn1 f n1 n2 n2 f f

130

4 Introduction to Multidegree of Freedom Systems

4.3.5 Vibration Absorbers 4.3.5.1

The Two Degree of Freedom Tuned Mass Damper

Consider a single degree of freedom system with mass m 1 and spring constant ko (Fig. 4.21). We shall assume that the system is subjected to a harmonic force of excitation F f = F f o cos ω f t. It is our aim to find a way for reducing the amplitude of the steady-state vibrations. To this end, a second spring–mass system is attached to the main mass m 1 for which we prescribe that its natural circular frequency coincides with the circular frequency of the excitation force, i.e., it holds that main mass

ko

F1 cos

ft

k1

m1

m2

G1 the first spring mass system

G2

x1

the second spring mass system

x2

Fig. 4.21 Two degree of freedom tuned mass damper

2 ω12 =

k1 = ω 2f . m2

(4.87)

The equation of motion 

       m1 0 x¨1 x1 k + k1 −k1 F1 + o = cos ω f t x¨2 −k1 k1 x2 0 0 m2    

      K

(2×2)

R q

(2×1)

K

(2×2)

q

(2×1)

(4.88)

Ffo (2×1)

can be obtained from Eq. (4.77) if in the former we set k11 = ko + k1 , k12 = k21 = −k1 , k22 = k1 .

(4.89)

Consequently, we can use Eq. (4.82) to determine amplitude A1 . After substituting relations (4.89) into Eq. (4.82), we get    k1 − m 2 ω 2f k1 2   adj K − ω f M k1 ko + k1 − m 1 ω 2f (2×2) (2×2)  =  A =   K − ω2 M   ko + k1 − m 1 ω 2f (2×1) −k1 f  (2×2) (2×2)  −k1 k1 − m 2 ω 2 f

from where

     

F1 0



4.3 Multidegree of Freedom Systems with Solutions

131

k1 − m 2 ω 2f  A1 =  F1 = ↑ ↑ = ko k1 2 2 ko + k1 − m 1 ω 2f k1 − m 2 ω 2f − k12 m 1 =ω11 m 2 =ω12  ω2 k1 1 − ω2f = # 12 $ F1 ω 2f ω2 k1 1 − ω2f − kko1 k o k 1 1 + ko − ω 2 11

or A1 ko = F1 1+

1− k1 ko

−

ω 2f 2 ω11

12

ω 2f 2 ω12



1−

ω 2f 2 ω12

−

(4.90)

k1 ko

in which ω11 is the natural circular frequency of the original spring–mass system, and ω12 is that of the second spring–mass system—see Eq. (4.87). The natural circular frequencies of the two degree of freedom system obtained by adding a single degree of freedom system to the original single degree of freedom system are denoted in the usual way by ωn1 and ωn2 . A1k o F1

5.0

1.0 f n1 12

=0.5

1.0

n2 12

=2.0

5.0

12

Fig. 4.22 Dimensionless amplitude as a function of ω f

Figure 4.22 shows the quotient |A1 ko /F1 | against the quotient ω f /ω12 for the case ω11 = ω12 , k1 /ko = 0.5. It follows form Eq. (4.90) and it can also be seen in Fig. 4.22 that the amplitude of the steady-state vibrations A1 is zero when ω12 = ω f . It is an important question how to select mass m 2 . Let A2 be the amplitude of mass m 2 . Further let Fk1 be the amplitude of the force exerted by spring k1 on mass m 2 .

132

4 Introduction to Multidegree of Freedom Systems

k1

Fig. 4.23 Inner forces

m2 G2

k1

Fk1

Fk1

Fk1 m 2 G2

Figure 4.23 shows the forces acting on the second spring and the absorber mass m 2 . Since the left end of the spring k1 is at rest the equation of motion for the absorber mass is of the form m 2 x¨2 + k1 x2 = 0

2 (x¨2 = −ω12 x2 = −ω 2f x2 ) .

This equation is always valid. If x2 = A2 we can write m 2 x¨2 + k1 x2 = − ω 2f A2 m 2 + Fk1 = −k1 A2 + Fk1 = 0    

−ω 2f

A2 m 2

k1 A2 =Fk1

(4.91)

k1 m2

from where A2 =

Fk1 . k1

These results show that the spring constant k1 and the mass m 2 depend on the allowable values of A2 and Fk1 .

4.3.5.2

Untuned Viscous Damper

An untuned viscous damper consists of a disk (pulley) which can rotate freely because it is either mounted on bearings inside the housing of the damper or for the simple reason that the clearance between the housing and the disk (pulley) is filled with silicone fluid. The latter is preferred to oil for two reasons: (a) because of its high viscosity index and (b) the fact that its viscosity is not sensitive to the temperature changes. These dampers are used to limit the amplitudes of torsional vibrations in crankshafts (or shafts) for which the excitation frequency is, in general, not constant. They have the advantage in contrast to the tuned two degree of freedom mass damper that they reduce the amplitude of steady-state vibrations in a wide interval of the

4.3 Multidegree of Freedom Systems with Solutions

133

Damper housing Pulley=Inertita ring

M fo e i

ks

ft

Silicone fluid

Fig. 4.24 Untuned viscous damper

excitation frequency ω f . Figure 4.24 shows both the mechanical model of the damper and shaft and the picture of a real damper.3 In the later case, the shaft is not shown. To derive the equations of motion we shall introduce the following notations: let J p be the moment of inertia of the disk (pulley) about the axis of rotation. Further let Js be the moment of inertia of the shaft about the same axis. The angle of rotation of the disk (pulley) and the housing are denoted by φ and ϕ, respectively. The torsional stiffness of the shaft is k and the harmonic excitation is the torque M f eiω f t where i is the imaginary unit. The damper torque is proportional to the relative angular velocity ϕ˙ − φ˙ of the housing with respect to the disk and it is given by   Mc = c ϕ˙ − φ˙ ,

(4.92)

where c is the damping coefficient. Making use of the notations introduced it is not too difficult to check that the equations of motion for the housing and the disk are as follows:   Js ϕ¨ + c ϕ˙ − φ˙ + kϕ =M f o eiω f t ,   J p φ¨ − c ϕ˙ − φ˙ = 0 .

3 Photo

courtesy of Vibratech TVD 180 Zoar Valley Rd. Springville, New York USA.

(4.93a) (4.93b)

134

4 Introduction to Multidegree of Freedom Systems

These equations can be rewritten in matrix form: 

Js 0 0 Jp

          ϕ¨ ϕ˙ c −c k0 ϕ M f o eiω f t . + + = 0 −c c 00 φ φ¨ φ˙

(4.94)

If there is no damping we get the equation of motion of the free vibrations: 

Hence,

Js 0 0 Jp

       ϕ¨ k0 ϕ 0 + = . 00 φ 0 φ¨

   k − ω 2 Js   0  n  = −ωn2 J p k − ωn2 Js = 0  0 −ωn2 J p 

is the characteristic equation and ωn1 = 0

and

ωn2 = ωn =

 k/Js = 0

(4.95)

are the two natural frequencies. Let us assume that     ϕ ϕo = eiω f t φo φ

(4.96)

is the steady-state solution in which ϕo and φo are the two unknown amplitudes. Upon substitution of the steady-state solution into the equations of motion (4.94) we arrive at the following system of linear equations for the unknown amplitudes ϕo and φo :      icω f − ω 2f Js + k −icω f ϕo Mfo = . (4.97) −icω f icω f − ω 2f J p φo 0 The solution is  −icω f icω f − ω 2f Js + k    adj  −icω f icω f − ω 2f J p Mfo ϕo   = =  icω f − ω 2f Js + k  φo 0 −icω f    −icω f icω f − ω 2f J p    icω f − ω 2f J p icω f   icω f icω f − ω 2f Js + k Mfo  =  ,  icω f − ω 2f Js + k  0 −icω f    −icω f icω f − ω 2f J p  

where the numerator is the adjugate of the coefficient matrix in (4.97) and

(4.98)

4.3 Multidegree of Freedom Systems with Solutions

135

    icω f − ω 2f Js + k −icω f = d =  2 −icω f icω f − ω f J p       = −ω 2f J p k − ω 2f Js − icω f ω 2f J p − k − ω 2f Js

(4.99)

is the denominator. Recalling (3.20)—Js corresponds to m—we get the critical damping coefficient: cn = 2Js ωn . Hence c=

c cn = ζcn = 2ζ Js ωn cn

(4.100)

is the damping coefficient in terms of the damping factor ζ. Let us introduce the dimensionless quantities μ=

Jp Js

and

η=

ωf ωn

by the use of which we can manipulate the denominator d into a more suitable form:      d = − ω 2f J p k − ω 2f Js − icω ω 2f J p − k − ω 2f Js =      ω 2f ω 2f 2 Jp 2 Jp = −Js kω f 1 − k − i2ζ Js kωn ω f ω f k − 1 − k = Js Js Js Js Js %    &  = −Js kωn ω f ημ 1 − η 2 + i2ζ μη 2 − 1 − η 2 . (4.101a) For the first element of the adjugate in (4.98), we can write in a similar manner that  icω f −

ω 2f

J p = −Js ωn ω f

ω f Jp − i2ζ ωn Js

 .

(4.101b)

μη − i2ζ kϕo  % &   = 2 Mfo ημ 1 − η + i2ζ μη 2 − 1 − η 2

(4.102a)

Making use of Eq. (4.101) we obtain from solution (4.98) that

or

    kϕo  (μη)2 + 4ζ 2       2 . M  = 2 fo (ημ)2 1 − η 2 + 4ζ 2 μη 2 − 1 − η 2

(4.102b)

  Figure 4.25 shows the graph kϕo /M f o  for μ = 1 and different values of ζ— keep in mind that the relation ζ ∈ [0, 1] should be satisfied. It is a further issue how to reduce the peak values of the graph. To find an optimum we have to clarify the effect of the optimum conditions

136

4 Introduction to Multidegree of Freedom Systems

k o M fo =1.0

6 =1.00

=0.10

=0.75

4

=0.50

2

0

=0.25

0.0

0.4

0.2

0.6

1.2

1.0

0.8

1.4

1.6

1.8

2.0

Fig. 4.25 The amplitude |kϕo /M f o | against η

η 2 + 4 μζ 2 d       =0 d ζ 2 η 2 1 − η 2 2 + 4 ζ 22 1 − (1 + μ) η 2 2 2

(4.103a)

μ

and

η 2 + 4 μζ 2 d       =0 d η 2 η 2 1 − η 2 2 + 4 ζ 22 1 − (1 + μ) η 2 2 2

(4.103b)

μ

on the system parameters. Let us introduce the following notations for the numerator and denominator in (4.103): u = η2 + 4

ζ2 μ2

and

 2 2 ζ2  v = η 2 1 − η 2 + 4 2 1 − (1 + μ) η 2 . μ

(4.104)

Using these notations, the optimum conditions (4.103) can be rewritten in the following forms: du dv v  2 = u  2 d ζ d ζ

and

du dv v  2 = u  2 . d η d η

Consider now optimum condition (4.105)1 in which du 4  = 2 , μ d ζ2

2 dv 4    = 2 μη 2 − 1 + η 2 . 2 μ d ζ

(4.105)

4.3 Multidegree of Freedom Systems with Solutions

137

Hence,        2 ζ2  ζ2  2 2 2 2 2 2 2 η 1 − η + 4 2 1 − (1 + μ) η − η + 4 2 μη − 1 + η 2 = 0 μ μ which yields after performing some manipulations that   −η 4 μ μη 2 − 2 + 2η 2 = 0 is the first optimum condition. Consequently,  ηopt =

2 2+μ

(4.106)

is the optimal value of η. Consider now optimum condition (4.105)2 by taking the relations   ζ2 du d 2   =   η +4 2 =1 μ d η2 d η2 and    2 2 dv ζ2  d   =   η 2 1 − η 2 + 4 2 1 − (1 + μ) η 2 = μ d η2 d η2 2     ζ2  = 1 − η 2 − 2 1 − η 2 η 2 − 8 2 1 − (1 + μ) η 2 (1 + μ) μ into account. We get in this manner optimum condition (4.105)2 in the form  2 2 ζ2  η 2 1 − η 2 + 4 2 1 − (1 + μ) η 2 − μ   2   ζ 2  2 η +4 2 1 − η2 − 2 1 − η2 η2 − μ    ζ2  −8 2 1 − (1 + μ) η 2 1 + η 2 = 0 . μ

(4.107)

If we substitute here ηopt from Eq. (4.105) we find by shaking the second optimum condition down that    2 μ + 4ζ 2 + 2ζ 2 μ −μ2 + 4ζ 2 + 6ζ 2 μ + 2ζ 2 μ2 = 0

138

4 Introduction to Multidegree of Freedom Systems

from where we get ζ=√

μ 2 (1 + μ) (2 + μ)

(4.108)

since ζ 2 cannot be negative.

4.4 Problems Problem 4.1 Prove Eq. (4.24). Problem 4.2 A cart of mass m 1 and a rod of mass m 2 and length 2L are connected to each other by a pin joint. The cart is subjected to a force of excitation F f = F1 cos ω f t. Find the equations of motion in terms of q and ϑ by using Lagrange’s equations. Linearize the equations of motion and give them in matrix form.

Fig. 4.26 Vibrating system constituted by a cart and rod

F1 cos

y

k

ft

q m1

G1 c

x 2L

G2

m2

Problem 4.3 Determine the natural frequencies and the mode shapes for system C if m 1 = m 2 = m and ko = k1 = k2 = k—see Fig. 4.15. (Hint: Make use of Eq. (4.65).) Problem 4.4 Solve Problem 4.3 if ko = k2 = k and k1 = μ k, where μ is a positive integer.

References

139

References 1. J.L. Lagrange, Mécanique Analytique, 2nd edn. (Cambridge University Press, Cambridge, 2009). The first edition was published in Paris (1788) 2. J.L. Lagrange, Analytical Mechanics. Translated from the Mécanique analytique, novelle edition of 1811 and 1815, eds. by R.S. Cohen. Translated by A. Boissonnade, V.N. Vagliente, vol. 191. Boston Study of Philosophy and Science (Springer Netherlands, Dordrecht, 1997). https://doi. org/10.1007/978-94-015-8903-1

Chapter 5

Some Problems of Multidegree of Freedom Systems

5.1 Equations of Motion 5.1.1 Spring–Mass Systems Figure 5.1 depicts (a) a free spring–mass system with five degrees of freedom, (b) a four degree of freedom spring–mass system fixed at the left end and free at the right end, and finally (c) a three degree of freedom system fixed at both ends. These mo

ko

m1

k1

G1

Go xo

m2

k2

m3 G3

G2 x2

x1

m4

k3

G4 x3

x4

(a) Free system mo

ko

m1

k1

G1

m2

k2

m3 G3

G2 x2

x1

m4

k3

G4 x3

x4

(b) System fixed at the left end mo

ko

m1

k1

G1

m2

k2

k3

m4

G3

G2 x1

m3

x2

x3

(c) System fixed at both ends Fig. 5.1 Spring-mass systems © Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_5

141

142

5 Some Problems of Multidegree of Freedom Systems

systems represent the three possible categories for the spring–mass systems. The equation of motion for the free spring–mass system is given by the equation ⎡

mo ⎢0 ⎢ ⎢0 ⎢ ⎣0 0

0 m1 0 0 0 ⎡

0 0 m2 0 0

0 0 0 m3 0

⎤ 0 0⎥ ⎥ 0⎥ ⎥ 0⎦ m4

⎡ ⎢ ⎢ ⎢ ⎢ ⎣

x¨o x¨1 x¨2 x¨3 x¨4

⎤ ⎥ ⎥ ⎥+ ⎥ ⎦

⎤ ko −ko 0 0 0 ⎢−ko ko +k1 −k1 0 0 ⎥ ⎢ ⎥ ⎥ k +k −k 0 0 −k +⎢ 1 1 2 2 ⎢ ⎥ ⎣ 0 0 −k2 k2 +k3 −k3 ⎦ 0 0 0 −k3 k3

⎡ ⎢ ⎢ ⎢ ⎢ ⎣

xo x1 x2 x3 x4

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

⎡ ⎤ 0 ⎢0⎥ ⎢ ⎥ ⎥ =⎢ ⎢0⎥ ⎣0⎦ 0

(5.1)

which can be rewritten in matrix form as M

q¨ + K

(5×5) (5×1)

q = 0 ,

(5×5) (5×1)

(5.2)

(5×1)

where M is the mass matrix, K is the stiffness matrix, and q is the matrix of the generalized displacements. Comparison of Figs. 4.15, and 5.1 shows that the equations of the three spring–mass systems in Fig. 5.1 have the same structure as those in Fig. 4.15. Remark 5.1 For the free system, the sum of the elements in a row or column in the stiffness matrix is equal to zero. Remark 5.2 If m o tends to infinity the left end of spring ko becomes fixed and system (a) changes into system (b). The equations of motion are obtained from the equations of motion (5.1) by canceling the first row in each matrix and the first column in the mass and stiffness matrices. The result is presented in frames in Eq. (5.1). For completeness, it is repeated here: ⎡ m1 ⎢0 ⎢ ⎣0 0

0 m2 0 0

0 0 m3 0

⎤⎡ ⎤ ⎡ 0 0 ko +k1 −k1 x¨1 ⎢x¨2 ⎥ ⎢ −k1 k1 +k2 −k2 0⎥ ⎥⎢ ⎥ + ⎢ 0 ⎦ ⎣x¨3 ⎦ ⎣ 0 −k2 k2 +k3 x¨4 0 0 −k3 m4

⎤⎡ ⎤ ⎡ ⎤ 0 0 x1 ⎢x2 ⎥ ⎢0⎥ 0 ⎥ ⎥ ⎢ ⎥=⎢ ⎥ . −k3 ⎦ ⎣x3 ⎦ ⎣0⎦ k3 x4 0

(5.3)

Remark 5.3 If m 4 tends to infinity the right end of spring k4 becomes fixed and system (b) changes into system (c) which is fixed at both ends. The equations of motion are obtained from the equations of motion (5.1) by canceling the first and last row in each matrix as well as the first and last column in the mass and stiffness matrices. The result is presented again in frames drawn by dashed lines in Eq. (5.1). For completeness, we give it here:

5.1 Equations of Motion

143

⎡

⎤⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ m1 0 0 x¨1 ko +k1 −k1 x1 0 0 ⎣ 0 m 2 0 ⎦ ⎣x¨2 ⎦ + ⎣ −k1 k1 +k2 −k2 ⎦ ⎣x2 ⎦ = ⎣0⎦ . 0 x¨3 0 −k2 k2 +k3 x3 0 0 m3

(5.4)

5.1.2 The General Form of the Equation of Motion For a free n degree of freedom system (n ≥ 2) with no damping the equation of motion has the following general form: ⎡

m 11 m 12 . . . ⎢m 21 m 22 . . . ⎢ ⎢ .. .. . . ⎣ . . . m n1 m n2 . . . 

M (n×n)

⎤⎡ ⎤ ⎡ q¨1 m 1n k11 ⎢q¨2 ⎥ ⎢k21 m 2n ⎥ ⎥⎢ ⎥ ⎢ .. ⎥ ⎢ .. ⎥ + ⎢ .. . ⎦⎣.⎦ ⎣ . m nn

q¨n    q¨



k12 k22 .. .

... ... .. .

⎤⎡ ⎤ ⎡ ⎤ q1 k1n 0 ⎢q2 ⎥ ⎢0⎥ k2n ⎥ ⎥⎢ ⎥ ⎢ ⎥ .. ⎥ ⎢ .. ⎥ = ⎢ .. ⎥ . . ⎦ ⎣ . ⎦ ⎣.⎦

kn1 kn2 . . . knn qn

  

(n×1)

(5.5)

0  

K

q

0

(n×n)

(n×1)

(n×1)

Remark 5.4 In contrast to the equations that describe the motions of the considered spring–mass systems—see, for instance, (5.1)–(5.4)—the elements of the matrix q are now denoted by q ( = 1, . . . , n) since there is no guarantee that the general displacements q are all horizontal vector components, i.e., parallel to each other. Let v, (|v| = 0) be an (n × 1) matrix. Further let A be an (n × n) symmetric matrix.

⎫ ⎧ ⎫ ⎧ T ⎪v Av > 0⎪ ⎪ ⎪ ⎪ ⎪ positive definite ⎪ ⎨ T ⎬ ⎨ ⎬ ⎪ positive semidefinite v Av ≥ 0 for any v. if A is said to be vT A v ≤ 0 ⎪ ⎪ ⎪ ⎪ ⎪ negative semidefinite ⎪ ⎪ ⎪ ⎩ T ⎭ ⎩ ⎭ negative definite v Av < 0 If none of the above relations is satisfied the matrix A is indefinite. Remark 5.5 The mass matrix is a symmetric and positive definite matrix: M = MT ,

E=

1 T q˙ M q˙ > 0 . 2

(5.6)

Equation (5.6)2 shows that the positive definiteness is a consequence of the fact that the kinetic energy E is a positive quantity.

144

5 Some Problems of Multidegree of Freedom Systems

Remark 5.6 Let us assume that the kinetic energy can be given in the form (5.6)2 but the mass matrix is not symmetric. We define the following two matrices: Ms =

 1 M + MT , 2

Ma =

 1 M − MT 2

which obviously satisfy the relations M = Ms + Ma ,

M s = MTs ,

M a = −MTa .

and

Because of these properties M s and M a are called the symmetric and antisymmetric (or skew) parts of M. As is well known it holds for any v that vT M a v = 0 . Consequently, E=

1 1 1 1 T q˙ M q˙ = q˙ T M s q˙ + q˙ T M a q˙ = q˙ T M s q˙ 2 2 2  2 =0

which shows that the antisymmetric matrix M a has no effect on the kinetic energy. On the basis of this fact, we shall assume that the mass matrix is always symmetric. Remark 5.7 The stiffness matrix is also symmetric but positive semidefinite matrix: K = KT ,

U=

1 T q Kq ≥ 0. 2

(5.7)

Equation (5.7)2 shows that the positive semidefiniteness is a consequence of the fact that the strain energy U stored in the elastic elements of the system (in the springs) is a positive quantity except the case when the body (the elastic elements) has a rigid body motion.

AQ1

Exercise 5.1 The cantilever beam of length  = 3a is subjected to the vertical forces Q 1 , Q 2 , and Q 3 . Find the vertical displacements q1 , q2 , and q3 at the points P1 , P2 , and P3 , i.e., at the points where the forces are applied. Use the results obtained to prove that the stiffness matrix is symmetric (Fig.5.2).

Fig. 5.2 Cantilever beam with the forces acting on it

q

a

Q3

P1

P2

P3

a

Q2

a Q1

x

5.1 Equations of Motion

145

In accordance with Remark 3.2 the vertical displacement at the point Pi due to a vertical unit force applied at the point P j (i, j = 1, 2, 3) is denoted by f i j and is called flexibility. Since the bending problem of an elastic beam is governed by linear equations we can apply the principle of superposition to determine the displacements in question: q1 = f 11 Q 1 + f 12 Q 2 + f 13 Q 3 , q2 = f 21 Q 1 + f 22 Q 2 + f 23 Q 3 , q3 = f 31 Q 1 + f 32 Q 2 + f 33 Q 3 .

(5.8a)

These equations can be rewritten in matrix form ⎡

⎤ ⎡ q1 f 11 ⎣ q2 ⎦ = ⎣ f 21 q3 f 31   

⎤⎡ ⎤ Q1 f 12 f 13 f 22 f 23 ⎦ ⎣ Q 2 ⎦, f 32 f 33 Q3

  

q

f

Q

(3×1)

(3×3)

(3×1)

(5.8b)

or concisely as q = f

Q ,

(5.8c)

(3×3) (3×1)

(3×1)

where f is the flexibility matrix. Multiply throughout by the inverse of the flexibility matrix. The result is −1

f (3×3)

q = K (3×1)

q = Q

(3×3) (3×1)

(3×1)

f (3×3)

−1

= K

−1

(5.9a)

(3×3)

⎡

⎤ ⎡ ⎤⎡ ⎤ Q1 k11 k12 k13 q1 ⎣ Q 2 ⎦ = ⎣ k21 k22 k23 ⎦ ⎣ q2 ⎦ . Q3 k31 k32 k33 q3

or

(5.9b)

It follows from the structure of Eqs. (5.9) that K is the stiffness matrix. Maxwell’s reciprocity theorem states that the flexibility matrix is symmetric f = fT

−→

f i j = f ji (i, j = 1, 2, 3) .

(5.10)

For the proof of this theorem we shall assume that the cantilever beam is subjected to two forces Q I and Q J only where the capital letters mean that the subscripts I, J ∈ (1, 2, 3) are fixed. Maxwell’s theorem is based on the fact that the work done by the two forces is independent of the loading order. Let W I I and W J J be the works done by the forces when we apply them one by one independently of each other. First let I , J be the loading order. Then W I J denotes the work done by the force Q I throughout the displacements caused by the force Q J . The total work for this loading order is given by

146

5 Some Problems of Multidegree of Freedom Systems

W = WI I + WI J + W J J =

1 1 f I I Q 2I + Q I f I J Q J + f J J Q 2J . 2 2

(5.11a)

For the reverse loading order W J I is the work done by the force Q J throughout the displacements caused by the force Q I . Consequently, W = W J J + W J I + WI I =

1 1 f I I Q 2I + Q I f J I Q J + f J J Q 2J 2 2

(5.11b)

is the total work for the second loading order. Comparison of Eqs. (5.11) yields the symmetry condition to be proved: fI J = fJ I .

(5.12)

Conclusion: If the flexibility matrix is symmetric then so is the stiffness matrix.

5.2 Eigenvalue Problem of Symmetric Matrices 5.2.1 Eigenvalues and Eigenvectors Let us introduce the notations   qT M q = q, q

M

,

  qT K q = q, q

K

(5.13)

which reflect that the products are performed on the matrices M and K. Since the mass matrix is positive definite and the stiffness matrix is positive semidefinite it holds that     >0, q, q ≥ 0 . (5.14) q, q M

K

It is well known from the theory of differential equations that the equations of motion M

q¨ + K

(n×n) (n×1)

q = 0

(n×n) (n×1)

(5.15)

(n×1)

have one and only one solution (uniqueness theorem) which satisfies the initial conditions     = qo , q˙ (t) = vo . (5.16) q (t)   to =0 to =0   initial velocities

initial displacements

Assume that q = A exp (iωt) , (n×1)

(n×1)

(5.17a)

5.2 Eigenvalue Problem of Symmetric Matrices

147

where i is the imaginary unit, ω is an unknown parameter (the natural circular frequencies sought), and A is the unknown amplitude vector. It is obvious that q¨ = −ω 2 A exp (iωt) = −ω 2 q .

q˙ = iωA exp (iωt) = iω q ,

(5.17b)

Upon substitution of q and q¨ into the equations of motion (5.15), we obtain 

K −λ M

(n×n)



A = 0 ,

(n×n) (n×1)

(n×1)

λ = ω2

(5.18)

which is a generalized algebraic eigenvalue problem with λ = ω 2 as an eigenvalue and A as an eigenvector. Solution for A (different from the trivial one) exists if and only if   (5.19) pn (λ) = K − λ M = 0 , where pn (λ) is a polynomial of degree n. Let λ be (λ1 ≤ λ2 ≤ λ3 ≤ · · · ≤ λn ) the th root (the th eigenvalue) and A be the corresponding solution (the th eigenvector or eigenmatrix). Remark 5.8 According to a fundamental algebraic theorem there exist n (not necessarily different) roots of the equation pn (λ) = 0. Let ν be an arbitrary scalar (ν = 0). It is clear that 

K − λ M

     νA = K − ω2 M νA = 0

(5.20)

which shows √ that νA is also a solution of Eq. (5.18). If we have already determined ω = ± λ and A then q = A exp (iω t) = A (cos ω t + i sin ω t)

(5.21a)

q = A exp (−iω t) = A (cos ω t − i sin ω t)

(5.21b)

or

are solutions to the equations of motion (5.15). Since both the real part and the imaginary parts of the solutions (5.21) satisfy Eq. (5.15) it follows that the general and real solution of this equation (under the condition λ = λk ;  = k, , k = 1, 2, 3, . . . , n) assumes the following form: q=

n    a A  cos ω t + b A  sin ω t ,

(5.22)

=1

where the undetermined constants of integration a and b can be determined from the initial conditions (5.16).

148

5 Some Problems of Multidegree of Freedom Systems

Since the mass matrix M is positive definite it has an inverse M−1 . Multiplying Eq. (5.15) from the left by M−1 , we obtain M−1 M q¨ + M−1 K q = I q¨ + D q = 0,     I

(5.23)

D

where I is the unit matrix and D is referred to as dynamic matrix. (n×n)

(n×n)

After substituting the solution (5.17a) for q and its second time derivative (5.17b)2 into Eq. (5.23) we arrive at the result that A satisfies the equation 

D −λ I

(n×n)



A = 0 ,

(n×n) (n×1)

(n×1)

λ = ω2

(5.24)

which is nothing but the well-known algebraic eigenvalue problem.

5.2.2 Calculation of the Eigenvectors It is possible to find the eigenvectors by utilizing the adjugate matrix of the equation system (5.24). For brevity we shall introduce the following notation for the coefficient matrix: (5.25) B = D − λI. The inverse of B in terms of the adjugate matrix is given by 1 B−1 =   adj B , B

  det B = B = B .

(5.26)

Multiplying now throughout from the left side by B B, we have B B B−1 = B I = B adj B

(5.27)

which can be rewritten in terms of D and λ I       I D − λ I  = D − λ I adj D − λ I .  

(5.28)

B

Since the determinant B is zero if λ = λ it is worth substituting λ for λ in the above equation. We obtain     D − λ I adj D − λ I = 0 ,  = 1, 2, . . . , n. (n×n)

(n×n)

(n×n)

(5.29a)

5.2 Eigenvalue Problem of Symmetric Matrices

149

Recalling that the th eigenvector (the th mode shape) is a solution of the equation   D − λ I A  = 0 (n×n)

(n×1)

(5.29b)

(n×1)

  we should recognize each column in the adjugate matrix adj D − λ I is directly proportional to the eigenvector A  .

5.2.3 Orthogonality of the Eigenvectors Now we shall show that two eigenvectors corresponding to two different eigenvalues are mutually orthogonal with respect to the mass and stiffness matrices. Let λ and λk be two different eigenvalues. The corresponding eigenvectors A  and A k satisfy the equations   (5.30a) K − λ M A  = 0 and

  K − λk M A k = 0 .

(5.30b)

Left multiplying Eq. [(5.30a)] ((5.30b)) by [ATk ] (AT ), we obtain

and

    A k , A  K = λ A k , A  M

(5.31a)

    A k , A  K = λk A k , A  M ,

(5.31b)

where we have taken into account that     A k , A  K = A , A k K ,

    A k , A  M = A , A k M

(5.32)

which is a consequence of the fact that the matrices M and K are symmetric. Thus, subtracting Eq. (5.31b) from Eq. (5.31a) the left sides cancel out and we get   0 = (λ − λk ) A k , A  M .  

(5.33)

=0

Since λ = λk the above product vanishes if and only if the general orthogonality condition   (5.34a) Ak, A M = 0 is fulfilled.

AQ2

150

5 Some Problems of Multidegree of Freedom Systems

It follows from Eqs. (5.32) with regard to the orthogonality condition (5.34) that the other orthogonality condition   Ak, A K = 0

(5.34b)

also holds. Remark 5.9 If the product 

Ak, A

 I

= ATk A  = 0

(5.35)

taken on the unit matrix I is zero then we say that the column matrices A k and A  are orthogonal in the classical sense. Equations (5.34a) and (5.34b) are the general orthogonality conditions we wanted to prove. For k = , Eqs. (5.31) result in the following formula:     A  , A  K = λ A  , A  M .

(5.36)

The length (or norm) of A  with respect to the mass matrix M is defined as     A  = A , A  M . 

(5.37)

It is often useful to normalize the eigenvectors. This can be done by dividing the eigenvector A  with its length: A A ψ  =    =   . A   A , A  M

(5.38)

For the normalized eigenvectors, it holds that   ψ k, ψ  

where δk =

1 if k =  0 if k = 

M

= δk ,

k,  = 1, 2, 3, . . . , n

(5.39a)

(5.39b)

is the Kronecker delta. Equation (5.39a) is a consequence of the orthogonality condition (5.34a). Let us now divide Eq. (5.31a) by the norm (5.38). If we take the relations (5.39a) and (5.39b) into account we get   = λ δk . ψ k, ψ  K

(5.39c)

5.2 Eigenvalue Problem of Symmetric Matrices

151

For the sake of our later considerations, we shall introduce the following notations:   A  , A  M = m  ,   A  , A  K = k .

(5.40)

Exercise 5.2 Consider the two degree of freedom system shown in Fig. 5.3. Determine the corresponding eigenvalues and eigenvectors. ko

k

k1

k2

k

G2

G1 m1

k

m

m2

2m

q1

q2

Fig. 5.3 Spring-mass system with two degrees of freedom

It follows from (4.58c) that the equations of motion in matrix notation are of the form         m 0 q¨1 q1 2k −k 0 + = . (5.41) 0 2m q¨2 q2 −k 2k 0         q¨

M (2×2)

(2×1)

K

q

(2×2)

(2×1)

After substituting the solution and its second derivative from (4.58c) and left multiplying then by   1/m 0 M−1 = , (5.42) 0 1/2m we obtain      k   2 m − λ − mk A1 0 = , D − λI A = k k A 0 − 2m − λ 2 m

λ = ω2 .

(5.43)

The characteristic equation is the determinant of the coefficient matrix   D − λ I  = λ 2 − 3 k λ + 3 m 2



k m

2



k = λ− m

√

3+3 2



k λ+ m

√ 3−3 2

=0

from where k λ1 = − m

√ k 3−3 = 0.633 97 , 2 m

k λ2 = m

√

k 3+3 = 2. 366 03 2 m

are the roots. There are two possibilities for calculating the eigenvectors. (a) We can find them from Eq. (5.43) if we back-substitute λ1 and λ2 (the solution steps are

AQ3

152

5 Some Problems of Multidegree of Freedom Systems

basically the same as those in Exercise 4.7). (b) We can also use the adjugate matrix   adj D − λ I for their calculation. Here we will select the second possibility so that we can show how to apply the adjugate matrix: 

  adj D−λ I = adj

 k  k 2 mk −λ − mk −λ m m = ,  = 1, 2 . k k k − 2m −λ 2 mk −λ m 2m

If  = 1 we get   k adj D−λ I = m where #$

√

1−(3− 3)/2

%2

!

+0.52

√

1− 3−2 0.5

"

3

1

√ 2− 3−2 3

= 0.619 66 and

=

k m



 0.366 03 1.000 00 , 0.500 00 1.366 02

#$

%2 √ 2−(3− 3)/2 +1 = 1.692 93

are the norms of the two columns. Now we can normalize each column to unity. We obtain ⎡ ⎤ 1     k 0.590 69 0.590 69 A1 ⎦ k 0.590 69 ⎣ which means that A 1 = 1 = . m 0.806 89 0.806 89 m 0.806 89 A 2

For  = 2, we find in the same manner that ⎡ A2 = ⎣

2

⎤

A1 ⎦ 2

A1

=

k m



−0.939 071 0.343 724

 .

With the two eigenvectors 2

   & ' m 0 −0.939 07 0.590 69 0.806 89 = 0 2m 0.343 72   ' −0.939 07 k2 & k2 0.590 69 0.806 89 = = (−0.590 69 × 0.939 07 + 0.687 46 m m + 0.806 89 × 0.687 46) = 0   A 1, A 2 M =



k m

is their product on M. This result shows that the orthogonality condition (5.34a) is also fulfilled.

5.2 Eigenvalue Problem of Symmetric Matrices

153

5.2.4 Repeated Roots Assume that repeated roots are found when we solve the characteristic equation which can be given in terms of the distinct roots as the product   pn (λ) = K − λ M =

 d = a (λ−λ1 )d1 (λ−λ2 )d2 (λ−λ3 )d3 . . . λ−λμ μ = 0, a = constant,

(5.44)

where the number dk (k = 1, 2, . . . , μ), μ ≤ n is the algebraic multiplicity of the root λk . If there are repeated roots at least one multiplicity is greater than one. It is also well known that the multiplicities satisfy the condition d1 + d2 + d3 + · · · + dμ = n. We assume that the number of linearly independent eigenvectors which belong to a repeated root is the same as the algebraic multiplicity of the root itself.1 For a repeated root, we shall show that it is always possible to find such eigenvectors which are mutually orthogonal to each other on M. Let λ L be a repeated root of (5.44) with a multiplicity r (2 ≤ r < n). The corresponding normalized eigenvectors are denoted by ψ 1L , ψ 2L , . . . , ψ r L where the second subscript L shows that these eigenvectors belong to the repeated eigenvalue λ L . Since the eigenvectors ψ 1L , ψ 2L , . . . , ψ r L are normalized it holds that   ψ k L , ψ L

M

= δk .

(5.45)

The linear combinations ˆ = ψ 1L

r  k=1

ˆ c1k ψ k L , ψ = 2L

r 

ˆ = c2k ψ k L , . . . , ψ rL

k=1

r 

cr k ψ k L ,

(5.46)

k=1

in which the constants c1k , c2k , etc. are, in general, non-zero constants are clearly eigenvectors. The normalized eigenvectors ψ k L (x) k ∈ 1, 2, . . . , r are not orthogonal to each other. In the sequel, we shall construct an orthogonal set by selecting the coefficients c1k , c2k , etc. appropriately. The procedure we detail below is called Gram–Schmidt orthogonalization [2]. Let ˆ = ψ 1L (5.47a) ψ 1L ˆ ∗ in the following manner: by the use of which we define ψ 2L ˆ∗ =ψ −ψ ˆ (ψ , ψ ˆ )M . ψ 2L 2L 1L 2L 1L

1 If

(5.47b)

the matrix in question is symmetric then it is a normal matrix [1] for which the assumption is fulfilled.

154

5 Some Problems of Multidegree of Freedom Systems

ˆ (ψ , ψ ˆ ) M is the projection of ψ on ψ ˆ . Since In this equation, the term ψ 1L 2L 1L 2L 1L ˆ ∗ ,ψ ˆ ) M = (ψ , ψ ˆ ) M − (ψ , ψ ˆ ) M (ψ , ψ ˆ )M = 0 (ψ 2L 1L 2L 1L 1L  1L 1L  2L =1

ˆ ∗ and it follows that ψ 2L $ % ˆ ,ψ ˆ )M = 1 (ψ 2L 2L

ˆ ∗ /(ψ ˆ ∗ ,ψ ˆ ∗ )M , ˆ =ψ ψ 2L 2L 2L 2L

(5.47c)

ˆ . are both orthogonal to ψ 1L

ˆ are the two sought eigenfunctions. ˆ and ψ If the multiplicity r = 2 then ψ 1L 2L If the multiplicity r is greater than two we proceed in a manner similar to the ˆ . Let definition of ψ 2L     ˆ ˆ ˆ ˆ ˆ∗ =ψ −ψ −ψ ψ 3L , ψ ψ 3L , ψ ψ 3L 3L 1L 1L M 2L 2L M

(5.48a)

    ˆ ˆ ˆ ˆ ψ 3L , ψ ψ 3L , ψ and ψ are the in which the column matrices ψ 1L 1L M 2L 2L M ˆ ˆ projections of ψ on ψ and ψ . Since 3L



ˆ ∗ ,ψ ˆ ψ 3L 1L



1L



M

ˆ = ψ 3L , ψ 1L

2L

 M

    ˆ ,ψ ˆ ˆ ψ 3L , ψ − ψ − 1L 1L M 1L M 

 =1



   ˆ ,ψ ˆ ˆ − ψ ψ 3L , ψ = 0, 1L 2L M 2L M 

 =0



ˆ ˆ ∗ ,ψ ψ 3L 2L

 M

      ˆ ˆ ,ψ ˆ ˆ ψ 3L , ψ = ψ 3L , ψ − ψ − 2L M 1L 2L M 1L M 

 =0

    ˆ ,ψ ˆ ˆ − ψ ψ 3L , ψ =0 2L 2L M 2L M 

 =1

ˆ ∗ and we can come to the conclusion that ψ 3L  ∗  ˆ∗ / ψ ˆ ,ψ ˆ∗ ˆ =ψ ψ 3L 3L 3L M , 3L ˆ . ˆ and ψ are orthogonal to ψ 1L 2L

$

ˆ ,ψ ˆ ψ 3L 3L

 M

% =1

(5.48b)

5.2 Eigenvalue Problem of Symmetric Matrices

155

The steps leading to (5.47a), (5.47c), and (5.48b) can easily be generalized: ˆ∗ ψ i+1,L = ψ i+1,L −

i 

  ˆ ˆ ψ , ψ , ψ kL i+1,L kL M

k=1

ˆ ψ i+1,L

  ∗ ˆ∗ ˆ ˆ∗ =ψ i+1,L / ψ i+1,L , ψ i+1,L M ,

$

ˆ ˆ ψ ,ψ i+1,L i+1,L

 M

% =1 .

(5.49)

We have already mentioned that the above procedure is attributed to Gram [3–5] and Schmidt [6, 7]. It is also worth citing the review paper by Björck et al. [8] here.

5.2.5 The Natural Frequencies Are Real Numbers We prove that the natural frequencies are real numbers. Suppose first that the eigenvalue λ is not real, i.e., it is a complex number and we will show that this assumption leads to a contradiction. If λ is complex then so is the corresponding eigenvector since the matrix K is a real matrix in equation K A  = λ M A 

(5.50)

to be satisfied by A —see (5.18) for a comparison. Let λ be of the form λ = a + ib,

(5.51a)

where a and b are non-zero real numbers. It is obvious that the complex conjugate of λ , i.e., the number (5.51b) λ¯  = a − ib is also an eigenvalue and the complex conjugate of A is an eigenvector. Consequently, it holds that ¯  = λ¯  M A ¯ . (5.52) KA ¯  and AT and recalling the notational conLeft multiplying (5.50) and (5.52) by A vention (5.13), we can write that T

and

    ¯ , A  ¯  , A  = λ A A K M

(5.53a)

    ¯  = λ¯  A  , A ¯ , A , A K M

(5.53b)

156

5 Some Problems of Multidegree of Freedom Systems

where for symmetry reasons it holds that 

¯ , A  A

 K

  ¯ = A , A K

  ¯ , A  A

and

M

  ¯ = A , A . M

(5.54)

Subtract (5.53b) from (5.53a). If we take (5.51) and (5.54) into account, we get      ¯  = 2bi A  , A ¯ =0 λ − λ¯  A  , A M M   ¯  = 0 since the mass matrix is positive definite. Hence, in which A  , A M b=0

(5.55)

and that was to be proved.

5.2.6 Uncoupled Equations of Motion We have already seen in Exercise 4.6 that the equations of motion, depending on how we select the coordinate system, can be statically or dynamically coupled. We shall now investigate the issue how to transform the equations of motion to an uncoupled form. It is obvious on the base of (5.21a) and (5.39) that the matrix of the generalized displacements can be given in the form q = x1 (t)A 1 + x2 (t)A 2 + · · · + xn (t)A n =

n 

x (t)A 

(5.56)

=1

which is a linear combination of the eigenvectors. By introducing matrix notations: xT = [ x1 | x2 | x3 | · · · | xn ] ,

(1×n)

!

"

⎡

⎢ A1 A2 · · · An     = ⎢ A =   ⎢ ⎣ (n×1) (n×1) (n×1) (n×n)

A11 A21 .. .

A12 A22 .. .

A13 A23 .. .

(5.57) ··· ··· .. .

⎤

A1n A2n ⎥ ⎥ .. ⎥ , . ⎦

(5.58)

An1 An2 An3 · · · Ann

where Ak is the kth element of the eigenvector A  and we can rewrite Eq. (5.56) as a matrix equation (5.59) q = A x . (n×1)

(n×n) (n×1)

5.2 Eigenvalue Problem of Symmetric Matrices

157

If we substitute this solution into the differential equation (5.5), we get M

A

x¨ + K

(n×n) (n×n) (n×1)

A

x = 0 .

(n×n) (n×n) (n×1)

(5.60a)

(n×1)

Left multiplying the above equation by A T , we get (n×n)

AT M

x¨ + A T

A

(n×n) (n×n) (n×n) (n×1)

K

A

x = 0

(n×n) (n×n) (n×n) (n×1)

(5.60b)

(n×1)

in which ⎡

AT

(n×n)

A 1T ⎢ A 2T ⎢ M A =⎢ . ⎣ .. (n×n) (n×n)

⎤ ⎥ & ' ⎥ ⎥ M A1 A2 · · · An = ↑ = ⎦ (n×n) 

 (5.13) (n×n)

AT  n  (n×n)

   ⎡ A A , A , A 1 1 1 2 M    M · · · ⎢ A 2, A 1 A 2, A 2 M · · · M ⎢ =⎢ .. .. . ⎣ .   .  ..  A n, A 1 M A n, A 2 M · · ·  ⎡ A 1, A 1 M  0  ··· ⎢ 0 A , 2 A2 M · · · ⎢ =⎢ .. .. .. ⎣ . . . 0

0

 ⎤  A , A 1 n M  A 2, A n M ⎥ ⎥ ⎥= ↑ = .. ⎦ (5.39a) .   A n, A n M ⎤ 0 ⎥ 0 ⎥ ⎥= ↑ = .. .  ⎦ (5.40)1  · · · A n, A n M ⎡ ⎤ m 11 0 · · · 0 ⎢ 0 m 22 · · · 0 ⎥ ⎢ ⎥ =⎢ . .. . . .. ⎥ ⎣ .. . . ⎦ . 0 0 · · · m nn

(5.61a)

is the first term on the left side. After performing similar steps we obtain for the second term on the left side that

158

5 Some Problems of Multidegree of Freedom Systems

⎡

AT

(n×n)

A 1T ⎢ A 2T ⎢ K A =⎢ . ⎣ .. (n×n) (n×n)

⎤ ⎥ & ' ⎥ ⎥ M A1 A2 · · · An = ↑ = ⎦ (n×n) 

 (5.13) (n×n)

AT  n 

⎡ ⎢ ⎢ =⎢ ⎣

A 1, A 1 0 .. . 0

(n×n)

 K

⎤ 0  ··· 0  ⎥ A 2, A 2 K · · · 0 ⎥ ⎥= ↑ = .. . .. .. ⎦ (5.40)2 . .   0 · · · A n, A n K ⎤ ⎡ k11 0 · · · 0 ⎢ 0 k22 · · · 0 ⎥ ⎥ ⎢ =⎢ . . . .. ⎥ . . . . ⎣ . . . . ⎦

(5.61b)

0 · · · knn

0

This means that the equation of motion (5.60b) is an uncoupled differential equation system: ⎡

0 ··· m 22 · · · .. . . . . 0 0 ···

m 11 ⎢ 0 ⎢ ⎢ .. ⎣ .

0 0 .. .

⎤⎡

⎤ x¨1 ⎥ ⎢ x¨2 ⎥ ⎥⎢ ⎥ ⎥ ⎢ .. ⎥ + ⎦⎣ . ⎦ x¨n

m nn ⎡

k11 ⎢ 0 ⎢ +⎢ . ⎣ ..

0 k22 .. .

0

··· ··· .. .

0 0 .. .

0 · · · knn

⎤⎡

⎤ ⎡ ⎤ x1 0 ⎥ ⎢ x2 ⎥ ⎢ 0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎥ ⎢ .. ⎥ = ⎢ .. ⎥ ⎦⎣ . ⎦ ⎣ . ⎦ xn

(5.62)

0

constituted by the independent differential equations of order two m  x¨ + k x = 0 , where

  m  = A  , A  M ,

 = 1, 2, . . . , n,

(5.63)

  k = A  , A  K .

(5.64)

If we recall Eq. (5.36) which relates the products of eigenvectors on the mass and stiffness matrices to each other we obtain     A  , A  K = λ A  , A  M     k

m 

(5.65)

5.2 Eigenvalue Problem of Symmetric Matrices

from where 2 λ = ωn =

159

k . m 

(5.66)

Consequently, (5.63) can also be given in the form 2 x = 0 , x¨ + ωn

 = 1, 2, . . . , n

(5.67)

which is the equation of motion of an undamped single degree of freedom system.

5.3 Rayleigh Quotient 5.3.1 Definition and Properties Let M and K be  the  mass and stiffness matrices of an n degree of freedom system. Further let A (A = 0) be an arbitrary column matrix of size n × 1. We assume that A is the amplitude of the generalized displacements. The Rayleigh quotient2 is defined by the equation [11, 12]   A, A K  . R=  A, A M

(5.68a)

Remark 5.10 If the motion is harmonic with the natural circular frequency ω—this is, the case, in general, for undamped free vibrations—then the maximum kinetic and potential energy of the system are given by Emax =

 1 2 ω A, A M , 2

Umax =

 1 A, A K . 2

Since Emax = Umax we can write   A, A K  ω =R=  A, A M 2

(5.68b)

which is nothing but the Rayleigh quotient. This explains the background of definition (5.68a).

2 Also

known as the Rayleigh–Ritz ratio, named after Lord Rayleigh (John William Strutt (1842– 1919)) and Walther Ritz (1878–1909) [9, 10].

160

5 Some Problems of Multidegree of Freedom Systems

The Rayleigh quotient has the following properties:     1. Because A, A K ≥ 0 and A, A M > 0 the Rayleigh quotient is a real nonnegative quantity.   2. The Rayleigh quotient is independent of A (the length of A). Really if we give A in the form A = νe, we get

    e = e, e = 1, I

    A, A I , ν = A =

      A, A K e, e K ν 2 e, e  =   K . R=  = A, A M e, e M ν 2 e, e M

(5.69)

(Consider the hypersphere of unit radius and centered at the origin of the ndimensional space. Then ω 2 = λ = R is a function of e, i.e., it varies on the surface of the hypersphere.) 3. The Rayleigh quotient is bounded—it has both a lower limit and an upper one.   – Since M is positive definite and its elements are all finite it follows that e, e M is bounded. – Since   K is positive semidefinite and its elements are all finite it follows that e, e K is also bounded. – Consequently,   e, e K 2   ω =λ=R= e, e M is bounded and has lower and upper limits. 4. The [lower] {upper} limit of the Rayleigh quotient is the [smallest] {greatest} eigenvalue [λ1 ]{λn }. The system of eigenvectors ψ  ( = 1, 2, . . . , n) is complete (constitutes a basis in the n-dimensional space). Hence, any A can be given as a linear combination of the eigenvectors ψ  : A = c1 ψ 1 + c2 ψ 2 + c3 ψ 3 + · · · + cn ψ n ,

(5.70)

where the weight c is the component of A in the direction ψ  . Recalling that   ψ , ψ 

M

= 1,

  ψ k, ψ 

for the Rayleigh quotient, we get

M

= 0 (k = ) and



ψ , ψ 

 K

= λ

5.3 Rayleigh Quotient

161



 A, A K  = ω =λ=R=  A, A M 2

  c1 ψ 1 + c2 ψ 2 + · · · + cn ψ n , c1 ψ 1 + c2 ψ 2 + · · · + cn ψ n K = = c1 ψ 1 + c2 ψ 2 + · · · + cn ψ n , c1 ψ 1 + c2 ψ 2 + · · · + cn ψ n M       c12 ψ 1 , ψ 1 + c22 ψ 2 , ψ 2 + · · · + cn2 ψ n , ψ n K  K  K = =  2 2 2 c1 ψ 1 , ψ 1 + c2 ψ 2 , ψ 2 + · · · + cn ψ n , ψ n M

M

M

λ1 c12 + λ2 c22 + · · · + λn cn2 = c12 + c22 + · · · + cn2 from where it follows that ω 2 = λ = R = λ1 or ω 2 = λ = R = λn

λ2 2 c λ1 2 2 c1 + c22

c12 +

λ1 2 c λn 1 c12

+ +

+ ··· + + ··· +

λn 2 c λ1 n cn2

λ2 2 c + · · · + cn2 λn 2 c22 + · · · + cn2

(5.71a)

.

(5.71b)

Equation (5.71a) proves that 2 ωmin = λmin = Rmin = λ1 .

(5.72a)

We get in the same manner from Eq. (5.71b) that 2 = λmax = Rmax = λn . ωmax

(5.72b)

5. The Rayleigh quotient has a further and very useful property. Let ψ  ( = 1, 2, . . . , n) be an eigenvector. Further let A = ψ + εx

(5.73)

    be an approximation of ψ  where (a) |ε| 1 and (b) ψ  , x = ψ  , x = K M 0—this assumption does not violate generality. Then       λ A = R A = λ  + O ε2 which means the error we have made is of order two. Really we can write

(5.74)

162

5 Some Problems of Multidegree of Freedom Systems

  ψ  + ε x, ψ  + ε x     K = λ A =R A =  ψ  + ε x, ψ  + ε x M

λ

=0

 

      ψ  , ψ  + 2ε x, ψ  + ε2 x, x K K K  =   = ψ , ψ  + 2ε x, ψ  + ε2 x, x M  M  M =1 =0   λ + ε2 x, x K   = = ↑ ≈ 1 1 + ε2 x, x M ≈1−ε2 (x, x) M 1+ε2 (x, x) M        ≈ λ + ε2 x, x K 1 − ε2 x, x M =   &        ' = λ + ε2 x, x K − λ x, x M − ε2 x, x K x, x M = λ + O ε2 .

5.3.2 Matrix Iteration Our aim is to find the first eigenvalue and eigenvector by the use of an iteration procedure. Later on we shall generalize the procedure for the higher eigenvalues and eigenvectors. The iteration procedure starts by assuming an initial value (a first approximation) for the displacement A. Initialization Setting the initial value A 1 = input (0)

Normalization #  Aˇ 1 = A 1, A 1 M (0)

(0)

(0)

ψ1 (0)

= A 1 / Aˇ 1

ψ1 (1)

= A 1 / Aˇ 1

(0)

(0)

Estimation   λ 1 = ω 21 = ψ 1 , ψ 1 K (0)

(0)

(0)

(0)

Iteration Steps Step 1. Solution for A 1 (1)

K A1 = M ψ 1 (1)

(0)

Normalization #  Aˇ 1 = A 1, A 1 M (1)

(1)

(1)

Estimation   λ 1 = ω 21 = ψ 1 , ψ 1 K (1)

(1)

(1)

(1)

(1)

(1)

5.3 Rayleigh Quotient

163

Step 2. Solution for A 1

Normalization #  A 1, A 1 M Aˇ 1 =

(2)

K A1 = Mψ 1 (2)

(2)

(1)

(2)

ψ1 (2)

(2)

= A 1 / Aˇ 1 (2)

(2)

Estimation   λ 1 = ω 21 = ψ 1 , ψ 1 K (2)

Step  Solution for A 1

(2)

(2)

Normalization #  Aˇ 1 = A 1, A 1 M

()

K A1 = M ψ ()

(2)

1 (−1)

()

()

ψ 1 = A 1 / Aˇ 1

()

()

()

()

Estimation   λ 1 = ω 21 = ψ 1 , ψ 1 K ()

()

()

()

Convergence is deemed to have been reached if the inequalities      ˇ 1    A λ 1    1 − (−1)  < ελ and 1 − (−1)  < ε A    λ 1  Aˇ 1    () ()  are satisfied where ελ and ε A are the prescribed relative errors. We shall now proceed and generalize the previous algorithm with the aim of determining the second eigenvalue and eigenvector. To this end, we introduce new mass and stiffness matrices in such a way that they will be orthogonal to the first eigenvector of the system. First we normalize the eigenvector A 1 to unity—after we have performed the previous iteration A 1 and λ1 are known with a high accuracy: A1 ˆ 1 =  A1 A   = A  . 1 A 1, A 1 I

(5.75)

ˆ 1 we define the new mass and stiffness matrices by the following Making use of A relations: ˆ1 A ˆ 1T M2 = M1 − M1 A

(n×n)

(n×n)

(n×n) (n×1)(1×n)

 

M1 (n×n)

= M ;

(5.76a)

K1 (n×n)

= K .

(5.76b)

(n×n)

(n×n)

ˆ1 A ˆ 1T K2 = K1 − K1 A

(n×n)

(n×n)

(n×n) (n×1)(1×n)

  (n×n)

(n×n)

164

5 Some Problems of Multidegree of Freedom Systems

Since the set of eigenvectors is complete any A can be given in the form   ˆ 1 + γ A 2 + c3 A 3 + c4 A 4 + · · · , A = βA where β, γ, c3 , c4 , etc. are scalars and it holds that    ˆ 1 , γ A 2 + c3 A 3 + c4 A 4 + · · · = βA M    ˆ 1 , γ A2 + c3 A3 + c4 A3 + · · · = βA = 0. K

˜ It is easy to check that Let us denote the sum A 2 +c3 A 3 +c4 A 4 +· · · by A.   ˆ 1 + γA ˜ = M 2 A = M 2 βA   ˆ 1 − M 1A ˆ 1A ˆ 1T A ˆ 1 + γM 2 A ˜ = γM 2 A ˜. = β M 1A   =1 

 ˆ 1 =0 βM 2 A

ˆ 1 (or ψ ). Hence, the mass matrix M 2 is orthogonal to A 1 We remark that the same statement is valid for the stiffness matrix K 2 , i.e., K 2 is also ˆ 1 and ψ . In the light of this result, it is obvious that the procedure orthogonal to A 1 we have just established for the first eigenvalue and eigenvector is also applicable to determining the second eigenvalue and eigenvector. Here we repeat the th step only: Step  Solution for A 2 ()

K 2 A2 = M 2 ψ 1 ()

(−1)

Normalization #  A 2, A 2 M Aˇ 2 = ()

()

()

ψ2 ()

= A 2 / Aˇ 2 ()

()

Estimation   λ 2 = ω 22 = ψ 2 , ψ 2 K ()

()

()

()

5.3.3 Convergence of the Iteration Algorithm The convergence properties of the matrix iteration are discussed below. It follows from Eq. (5.30a) that the normalized eigenvector ψ  satisfies the equation K ψ  = λ M ψ 

λ = ω2 .

(5.77)

5.3 Rayleigh Quotient

165

If we multiply this equation by K−1 we get a relationship we shall utilize in the sequel: ψ = K−1 M ψ  . (5.78) λ It is well known that the set of eigenvectors is complete, i.e., any vector in the ndimensional space can be given as a linear combination of the eigenvectors ψ 1 , ψ 2 , . . . , ψ n . Hence, for the initial value of the first eigenvector, we can write Aˇ 1 =

A 1 = c10 ψ 1 + c20 ψ 2 + · · · + c2n ψ n ,

(0)

ψ1 = (0)

#

(0)

A 1, A 1

(0)

(0)

 M

 1 1  c10 ψ 1 + c20 ψ 2 + · · · + c2n ψ n , A1 = Aˇ 1 (0) Aˇ 1

(0)

(5.79)

(0)

where on the base of (5.39b)   c0 = A 1 , ψ  K

(5.80)

(0)

is the th constant in the linear combination giving A 1 . (0)

For the first approximation, the following equation system should be solved for A 1:

(1)

K A 1 = M ψ 1. (1)

(0)

The solution is of the form A 1 = K−1 M ψ 1 = K−1 M

(1)

(0)

 1  c10 ψ 1 + c20 ψ 2 + · · · + cn0 ψ n = Aˇ 1

(0)

  1 c10 K−1 M ψ 1 + c20 K−1 M ψ 2 + · · · + cn0 K−1 M ψ n . =     Aˇ 1   (0)

ψ1 λ1

ψ2 λ2

With c1 =

(5.81)

ψn λn

1 c0 Aˇ 1

(0)

for the final form of the first approximation we can write

(5.82)

166

5 Some Problems of Multidegree of Freedom Systems

#  c11 c21 cn1 A 1, A 1 M , ψ1 + ψ2 + ··· + ψn , Aˇ 1 = (1) λ1 λ2 λn (1) (1)   1 1 c11 c21 cn1 ψ1 = A1 = ψ1 + ψ2 + ··· + ψn . λ2 λn Aˇ 1 (1) Aˇ 1 λ1 (1)

A1 =

(1)

(1)

(5.83)

(1)

As regards the second approximation we have to solve the equation system K A1 = M ψ 1 (2)

(1)

for A 2 . The solution is of the form (1)

A 1 = K−1 M ψ 1 = K−1 M

(2)



=

(1)

1 Aˇ 1

(1)



c11 c21 c11 ψ + ψ + ··· + ψ λ1 1 λ2 2 λn n

 =

 1 c11 −1 c21 −1 cn1 −1 K M ψ1 + K M ψ2 + ··· + K M ψn = λn   Aˇ 1 λ1   λ2  

(1)

ψ1 λ1

ψ2 λ2

=

ψn λn

c12 c22 cn2 ψ1 + 2 ψ2 + ··· + 2 ψn , 2 λn λ1 λ2

where c2 =

1 c1 . Aˇ 1

(5.84)

(5.85)

(1)

It is now obvious that the i-t approximation assumes the following form: A1 =

(i)

c1i c2i cni ψ + ψ + ··· + i ψn , λn λi1 1 λi2 2

ci =

1 c,i−1 . Aˇ 1

(5.86)

(i)

Here the constants ci are limited and λ1 < λ2 < · · · < λn−1 < λn . Hence, for a sufficiently large i it holds that c2i c3i cni c1i i i ··· i λn λi1 λ2 λ3

(5.87)

which means that the very first term is the significant one in the right side of (5.86)1 , and consequently A1 (i)

tends to be the first eigenvector.

5.3 Rayleigh Quotient

167

If A1 tends to be the first eigenvector the Rayleigh quotient tends to be the first (i)

eigenvalue. If we want to determine the second eigenvector and eigenvalue we have to solve the eigenvalue problem K 2 ψ  = λ M 2 ψ 

λ = ω2

(5.88)

for which λ2 is the smallest eigenvalue and ψ 2 is the corresponding eigenvector. It is obvious the previous proof remains valid for this problem word by word, i.e., the matrix iteration yields the second eigenvector and the second eigenvalue. By now there is no need for a further explanation to see that the previous results remain valid for eigenvalues and eigenvectors of order higher than two.

5.3.4 Properties of the Iteration Algorithm The iteration algorithm we have presented is called matrix iteration (or inverse matrix iteration). Its main advantage is that it converges to an eigenvalue quickly, since both the eigenvalue and the eigenvector that belongs to it are improved in each iteration step. It is worth mentioning here that the convergence rate for the eigenvalue is cubic which means that the number of correct digits in the eigenvalues triples in each iteration step. • The algorithm we have presented in Sect. 5.3.2 assumes that the eigenvalues (natural frequencies) are distinct, that is, there are no repeated roots of the characteristic Eq. (5.44). • The algorithm requires less iteration steps if the eigenvalues λ1 < λ2 < λ3 < · · · < λn are well separated from each other. • The number of iteration steps to be performed for finding λ1 depends on how the trial eigenvector (the initial value of the first eigenvector) resembles the first eigenvector. It is worth, however, emphasizing again that the convergence for the eigenvalue is very fast. Solution to Problem 5.5 in Sect. C.5 clearly shows that. • In case the trial vector (the initial value) is parallel to an eigenvector the solution algorithm does not converge since in each iteration step the approximation of the eigenvector remains parallel to the initial value.

5.3.5 Calculation of the Largest Eigenvalue The matrix iteration in Sect. 5.3.2 can be applied in a slightly modified form to determine the largest eigenvalue and the corresponding eigenvector. Consider the eigenvalue problem M A  = ν K A  ,

 = 1, 2, . . . , n

(5.89)

168

5 Some Problems of Multidegree of Freedom Systems

for which ν is the eigenvalue, A is the eigenvector, and ϕ is the normalized eigenvector: A (5.90) ϕ =    . A , A  K Comparison of the eigenvalue Problem (5.89) to the original eigenvalue Problem (5.50) shows that they are equivalent and A 1 = A n , A 2 = A n−1 , . . . , A n−1 = A 2 , A n = A 1 ; 1 1 1 1 ν1 = , ν2 = , . . . , νn−1 = , νn = . λ1 λn−1 λ2 λ1

(5.91)

This means that the first eigenvector A 1 is equal to the last one for the eigenvalue problem (5.50)—this is A n and the reciprocal of the first eigenvalue ν1 is the eigenvalue λn we want to determine. We can now rewrite the matrix iteration into the following form: Initialization Setting the initial value A 1 = input (0)

Normalization #  ˇ1= A A 1, A 1 K (0)

(0)

(0)

ϕ1 (0)

ˇ1 = A1/ A

ϕ1 (1)

ˇ1 = A1/ A

ϕ1 ()

ˇ1 = A1 / A

(0)

(0)

Estimation   1 1 ν1= = 2 = ϕ 1, ϕ 1 M (0) λn ωn (0) (0) (0)

(0)

Iteration Steps Step 1. Solution for A 1 (1)

M A1 = K ϕ 1 (1)

(0)

Normalization #  ˇ1= A 1, A 1 K A (1)

(1)

(1)

(1)

(1)

Estimation   1 1 ν1 = = 2 = ϕ 1, ϕ 1 M (1) λn ωn (1) (1) (1)

Step  Solution for A 1 ()

M A1 = K ϕ 1 ()

(−1)

(1)

Normalization #  ˇ1= A 1, A 1 K A ()

()

()

Estimation   1 1 ν1 = = 2 = ϕ 1, ϕ 1 M () λn ωn () () ()

()

()

()

5.3 Rayleigh Quotient

169

m

k

k

m

q1

q2

Fig. 5.4 Spring-mass system fixed at the left end

Exercise 5.3 Consider the system shown in Fig. 5.4. Estimate the first natural frequency and the first eigenvector. This problem is the same as that we solved in Exercise 4.7. The mass matrix, the stiffness matrix, the first eigenvalue, and eigenvector are as follows:     10 2 −1 M=m , K=k , 01 −1 1 √     1√ k k 3− 5 1.000 00 . = 0.382 01, A1 = 1+ 5 = λ1 = 1.618 03 m 2 m 2 Following the iteration algorithm given in Sect. 5.3.2 step by step yields: Initialization Setting the initial value: A1 =

(0)

  1 . 1

Normalization: Aˇ 1 =

#

(0)

A 1, A 1

(0)



(0)

M

=

√

2m ,

1 ϕ 1 = A 1 / Aˇ 1 = √ (0) 2m (0) (0)

  1 . 1

Estimation: λ1 (0)

=

ω 21 (0)



= ϕ 1, ϕ 1 (0)

(0)

 K

   ' 2 −1 k k & k 1 11 = 0.5. = = −1 1 1 m 2m m

Iteration Steps Step 1. Solution for A 1 : (1)

⎤ ⎡ #     A m 10 1 2 −1 ⎣ (1)11 ⎦ , K A 1 = M ϕ 1 −→ k = A 0 1 1 −1 1 2 21 (1) (0) (1) √ √ m m 2 3 , A 21 = √ . A 11 = √ (1) (1) k k 2 2 

(5.92)

170

5 Some Problems of Multidegree of Freedom Systems

Normalization: Aˇ 1 =

#

(1)

A 1, A 1

(1)

 M

(1)

m = k

#

4+9 , 2

  1 1 2 ˇ ϕ 1 = A 1/ A 1 = √ √ . (1) m 4+9 3 (1) (1) Estimation: &   k λ 1 = ω 21 = ϕ 1 , ϕ 1 K = (1) (1) m (1) (1)

23

'



2 −1 −1 1 4+9

  2 3

= 0.384 615

k , m

√   m 1.0 2 A1 = √ . 1.5 2 k (1) Step 2. Solution for A 1 : (2)

⎤ ⎡ √     A 11 m 10 2 2 −1 ⎣ (2) ⎦ K A 1 = M ϕ 1 −→ k , =√ A 0 1 3 −1 1 21 4 + 9 (2) (1) (2) √ √ m m 5 8 , A 21 = √ . A 11 = √ (2) (2) 4+9 k 4+9 k 

Normalization: Aˇ 1 =

(2)

#

A 1, A 1

(2)

(2)

 M

=

m k

#

25 + 64 , 4+9

  1 1 5 ϕ 1 = A 1 / Aˇ 1 = √ √ . 8 (2) m 25 + 64 (2) (2) Estimation: λ1 (2)

  = ω 21 = ϕ 1 , ϕ 1 K = (2)

=

(2)

(2)

   & ' 2 −1 1 k k 5 × 58 = 0.382 02 , −1 1 8 m 25 + 64 m √

m A1 = √ 4+9 k (2) 5



1.0 1.6

(5.93a)

 .

(5.93b)

It is worth comparing the results typeset in boldface letters in Eqs. (5.92) and (5.93).

5.3 Rayleigh Quotient

171

Exercise 5.4 Estimate the second (the largest) eigenvector and eigenvalue for the system shown in Fig. 5.4. Since the spring–mass system is the same as that of Exercise 4.7 it holds that      k 2 −1 2 −1 2 −1 , , K=k , M−1 K = −1 1 −1 1 m −1 1

 M=m

(5.94)

√     1√ k k 3+ 5 1.000 000 = 2.618 034, A1 = A2 = 1− 5 = λ2 = (5.95) −0.618 034 m 2 m 2 ν1 =

2 m 1 m = √ = 0.381 966. λ2 k 3+ 5 k

(5.96)

For finding a good estimation, we shall apply the iteration algorithm given in Sect. 5.3.5: Initialization Setting the initial value:   1 A1 = . 1 (0) Normalization: ˇ1= A

#

(0)

A 1, A 1 (0)

(0)

(

 K

=

& ' k 11



2 −1 −1 1 A1 (0)

  √ 1 = k, 1

1 ϕ1 = =√ ˇ1 k A (0)

  1 . 1

(0)

Estimation: ν1 =

(0)

   ' 10   m 1 1 m& 2m 1 11 =2 . = 2 = ϕ 1, ϕ 1 M = = 01 1 λ2 ω2 k k k (0) (0)

(0)

(0)

Iteration Steps Step 1. Solution for A 1 : (1)

M A 1 = K ϕ 1 → A 1 = M−1 K ϕ 1 = (1)

(0)

(1)

ˇ 11 = A

(1)

√

(0)

k , m

√    √   k k 1 2 −1 1 , = 1 m −1 1 m 0 ˇ 21 = 0 . A

(1)

172

5 Some Problems of Multidegree of Freedom Systems

Normalization: ˇ 1= A

#

(1)

A 1, A 1 (1)



(1)

( k = K m

&

10

'



2 −1 −1 1

  k√ 1 2, = 0 m

A1 (1)

1 ϕ1 = =√ ˇ 2k A1 (1)

  1 . 0

(1)

Estimation:      1 1 m& ' 10 1 m m 10 ν 1= = 2 = ϕ 1, ϕ 1 M = = = 0.5 . 0 1 0 (1) λ2 ω2 k 2k k (1) (1) (1)

(1)

Step 2. Solution for A 1 : (2)

M A 1 = K ϕ 1 → A 1 = M−1 K ϕ 1 = (2)

(2)

(1)

(1)

√  √     k k 1 1 2 −1 2 1 =√ , =√ 0 2 m −1 1 2 m −1

ˇ

A 11 (2)

√ √ k , = 2 m

ˇ

A 21 (2)

√ k 1 . = −√ 2 m

Normalization: ˇ1= A

(2)

#

A 1, A 1 (2)

(2)



) " ! *$ %  2 −1 √2 √ k* k√ 1 + 2 − √2 = 6.5 , = 1 K √ −1 1 − 2 m m

ϕ1 = (2)

A1 (2) ˇ

A1 (2)

1

=√ 13k



 2 . −1

Estimation: ν1 (2)

=

  1 1 = 2 = ϕ 1, ϕ 1 M = λ2 ω2 (2) (2) (2) (2)    ' 10 m 1 & m 2 2 −1 = = 0.384 615 38 . 01 −1 k 13 k

5.3 Rayleigh Quotient

173

Step 3. Solution for A 1 : (3)

M A 1 = K ϕ 1 → A 1 = M−1 K ϕ 1 = (3)

(3)

(2)

(2)

√  √     k k 5 1 1 2 −1 2 =√ , =√ −1 13 m −1 1 13 m −3

√ k ˇ 11 = √5 , A (3) 13 m

√ k ˇ 21 = − √3 A . (3) 13 m

Normalization: ˇ 1= A

#

(3)

A 1, A 1 (3)



(3)

( 1 k =√ K 13 m

& ' 5 −3



2 −1 −1 1

ϕ1 = (2)

A1 (2) ˇ

A1 (2)



√  k 89 5 = √ , −3 m 13

=√

1



89k

 5 . −3

Estimation: ν1 =

(3)

  1 1 = 2 = ϕ 1, ϕ 1 M = λ2 ω2 (2) (2)

(2)

(2)

   ' 10 m m 1 & 5 5 −3 = 0.382 022 . = 0 1 −3 k 89 k

For the sake of a comparison, we present the third estimation for the eigenvalue and the eigenvector (for the latter the first element is normalized to unity) in the following table. Table 5.1 Estimated and exact values

Eigenvalue ν1 = 1/λ2 =

Eigenvector A1T = A2T

1/ω22

Estimation

0.382 022 mk

Exact value

0.381 966 mk

& &

1.0 −0.6000 1.0 −0.6180

' '

Though the initial value for the eigenvector did not resemble the solution the results obtained in the third step are quite good.

174

5 Some Problems of Multidegree of Freedom Systems

5.4 Forced Vibrations 5.4.1 Forced Harmonic Vibrations Assume that the vibrating system with n degrees of freedom is subjected to a harmonic force (5.97) F f = F f o cos ω f t . (n×1)

(n×1)

If the vibrations are undamped M

q¨ + K

(n×n) (n×1)

q = F f o cos ω f t

(n×n) (n×1)

(5.98)

(n×1)

is the equation of motion. The steady-state solution is sought in the form q = A f cos ω f t, (n×1)

(5.99)

(n×1)

where A f is the amplitude of the forced vibrations. Upon substitution of this solution into the equation of motion (5.97) we arrive at an inhomogeneous linear equation system for A f :   (5.100) K − ω 2f M A f = F f o . (n×n)

(n×n) (n×1)

(n×1)

Its solution is of the form ' & adj K − ω 2f M (n×n) (n×n)  Ffo = Af =   K − ω2 M  f (n×1) (n×1) (n×n)

(n×n)

& ' adj K − ω 2f M (n×n) (n×n)  Ffo , = ,  n 2 an =1 ωn − ω 2f (n×1)

(5.101)

where          K −ω 2 M  = an ω 2 n +an−1 ω 2 n−1 +an−2 ω 2 n−2 + · · · +a1 ω 2 +a0 = f f f f f (n×n)

(n×n)

n  2  = an ωn − ω 2f =1

(5.102) 2 in which ωn is the th root of the polynomial

5.4 Forced Vibrations

175

 n  n−1 an ω 2f + an−1 ω 2f + · · · + a1 ω 2f + a0 = 0 , i.e., ωn is the th natural circular frequency of the free vibrations. Resonance occurs if ω f tends to ωn ( = 1, 2, . . . , n).

5.4.2 Nonharmonic Inhomogeneity Assume now that there are nonharmonic external and time-dependent forces in the system. If damping is neglected the equation of motion takes the form M

q¨ + K

(n×n) (n×1)

q = F f , F Tf =

(n×n) (n×1)

&

' f 1 (t) f 2 (t) · · · f n (t) .

(5.103)

(n×1)

Recalling that the system of the mass normalized eigenfunctions is complete in the n-dimensional space we can give the generalized displacement vector in terms of the eigenfunctions as q = q1 ψ 1 + q2 ψ 2 + q3 ψ 3 + · · · + qn ψ n =

(n×1)

n 

q (t)ψ

=1



(5.104)

or in a more concise form  q = φ (n×1)

q =

(n×n) (n×1)

ψ1 ψ2 ··· ψn

(n×1) (n×1)

⎡

ψ11 ⎢ ψ21 ⎢ =⎢ . ⎣ ..



(n×1)

ψ12 ψ22 .. .

ψn1 ψn2

ψ13 · · · ψ23 · · · .. . . . . ψn3 · · ·

q = (n×1)

⎤⎡ ⎤ q1 ψ1n ⎢ q2 ⎥ ψ2n ⎥ ⎥⎢ ⎥ .. ⎥ ⎢ .. ⎥ , . ⎦⎣ . ⎦

ψnn

(5.105)

qn

where ψk is the kth element of the eigenvector ψ  . Equation (5.104) in terms of the new independent variable q assumes the form M

φ

q¨ + K

(n×n) (n×n) (n×1)

φ

q = Ff .

(n×n) (n×n) (n×1)

(5.106a)

(n×1)

If we left multiply the above equation by φ T we get (n×n)

φT M

φ

q¨ + φ T K

(n×n) (n×n) (n×n) (n×1)

in which

φ

q = φT F f

(n×n) (n×n) (n×n) (n×1)

(n×n) (n×1)

(5.106b)

176

5 Some Problems of Multidegree of Freedom Systems

⎡

φ

T

M

ψT ⎢ ψ 1T ⎢ 2 =⎢ ⎢ .. ⎣ . ψ nT 

φ

(n×n) (n×n) (n×n)

⎥ ⎥ & ' ⎥ M ψ ψ ··· ψ = ↑ = ⎥ 1 2 n

 notation (5.13) ⎦ (n×n)  (n×n)

(n×n)

 ⎡ ψ 1, ψ 1 M ⎢ ⎢ ⎢ ψ 2, ψ 1 M =⎢ ⎢ .. ⎢ . ⎣  ψ n, ψ 1

⎤



  ψ 1, ψ 2  M ψ 2, ψ 2 M .. .   ψ n, ψ 2

M

M

 ⎤  · · · ψ 1, ψ n M ⎥  ⎥ · · · ψ 2, ψ n ⎥ M⎥= ↑ = ⎥ (5.39a) . .. .. ⎥ .  ⎦  · · · ψ n, ψ n ⎡

1 ⎢0 ⎢ =⎢. ⎣ .. 0 

M

0 1 .. .

··· ··· .. .

⎤

0 0⎥ ⎥ .. ⎥ = I . ⎦ (n×n)

(5.107a)

0 ··· 1

 I

is the unit matrix of size (n × n). As regards the product φ T K

φ , we get in

(n×n) (n×n) (n×n)

a similar manner that

φT K

φ

(n×n) (n×n) (n×n)

⎡ ⎢ ⎢ ⎢ =⎢ ⎢ ⎢ ⎣

ψ 1, ψ 1 0 .. . 0

⎡ T⎤ ψ ⎢ψ 1T ⎥ ⎢ 2⎥ & ' ⎥ =⎢ ↑ = ⎢ .. ⎥ K ψ 1 ψ 2 · · · ψ n =notation (5.13) ⎣ . ⎦ (n×n) (n×n) ψT  n  (n×n)

 K



0



···

0

⎤

⎥ ⎥ 0 ⎥ K ⎥= ↑ = ⎥ (5.39c) .. ⎥ .  ⎦  0 · · · ψ n , ψn ⎡ ⎤K λ1 0 · · · 0 ⎢ 0 λ2 · · · 0 ⎥ ⎢ ⎥ =⎢ . . . . ⎥= K . ⎣ .. .. . . .. ⎦ (n×n)

ψ 2, ψ 2 .. .

··· .. .

0 0 · · · λn

(5.107b)

5.4 Forced Vibrations

177

After back-substituting relations (5.107) in the equation of motion (5.106b), we arrive at an uncoupled system of differential equations I

q¨ + K

(n×n) (n×1)

q = φT F f .

(n×n) (n×1)

(5.108)

(n×n) (n×1)

Consequently, q¨  + λ q = Q  = ψ T F f ,

 = 1, 2, . . . , n ,

(5.109)

(1×n) (n×1)

where Q  is defined by the matrix product on the right side. Since Q  is a function of time there arises the question how to find a solution for the inhomogeneous equation of motion (5.108). This issue will be considered in Remarks 5.11 and 5.12. Remark 5.11 Let ψ(t, τ ) be a function of two variables t and τ . We shall assume that ψ(t, τ ) is differentiable continuously as many times as required. Further let τ1 (t) and τ2 (t) be two continuous time functions. Consider now the integral . Ψ (t) =

τ2 (t)

τ1 (t)

ψ(t, τ ) dτ .

(5.110)

It is known that dΨ (t) dτ2 dτ1 = ψ(t, τ2 ) − ψ(t, τ1 ) + dt dt dt

.

τ2 (t) τ1 (t)

∂ψ(t, τ ) dτ . ∂t

(5.111)

Remark 5.12 Assume that the particular solution of Eq. (5.108) is of the form . q part =

t τ =0

Q  (τ )g(t − τ ) dτ ,

(5.112)

where g(t) is an unknown function. Substituting q part and its second time derivative  / . t d dg(t − τ ) Q  (τ )g(t −τ )|τ =t + dτ = Q  (τ ) dt dt (5.111) τ1 ≡0 τ =0   . t ' d & dg(t −τ )  d2 g(t −τ ) = Q  (τ )g(t −τ )|τ =t + Q  (τ ) + Q (τ ) dτ =   dt dt dt 2 τ =0 τ =t   . t dQ  (t) d2 g(t −τ ) dg(t −τ )  = g(t −τ )|τ =t Q (t)+ Q (τ ) dτ +    dt dt dt 2 q¨  part = ↑

↑ =

τ =t

into Eq. (5.109) we get

τ =0

178

5 Some Problems of Multidegree of Freedom Systems

 d2 g(t − τ ) Q  (τ ) + λ g(t − τ ) dτ + dt 2 τ =0   dg(t − τ )  dQ  (t) + + g(t − τ )|τ =t  Q  (t) = Q  (t) . dt dt τ =t

.

t



This equation is satisfied if  dg  = 1, dt t=0

d2 g + λ g = 0 and g(0) = 0 , dt 2 i.e., if

0 1 g = √ sin λ t . λ

Consequently,

. q part =

(5.113)

0 1 Q  (τ ) √ sin λ (t − τ ) dτ λ τ =0 t

(5.114a)

is the particular solution and . t 0 0 0 1 Q  (τ ) √ sin λ (t − τ ) dτ q = a cos λ t + b sin λ t +

  λ

  τ =0 transient function particular solution

(5.114b) is the total solution in which a and b are undetermined integration constants. These can be calculated by using the initial conditions ˙ = 0) = q(0) ˙ q(t = vo .

q(t = 0) = q(0) = qo ,

(5.115)

Recalling (5.105) we can write q = φ (n×1)

q

(n×n) (n×1)

from where left multiplying by φ T M and taking the relation (n×n) (n×n)

φT M

φ

(n×n) (n×n) (n×n)

= I

(n×n)

into account, we obtain φT M

q = φT M

(n×n) (n×n) (n×1)

φ

q = q

(n×n) (n×n) (n×n) (n×1)

(n×1)

(5.116)

5.4 Forced Vibrations

179

Hence, ˙ = 0) = q˙ o = φ T M v o . q(t = 0) = q o = φ T M q o and q(t

(5.117)

If t = 0 then q part = q˙  part = 0. Consequently, q (t = 0) = a = ψ T M

q

o (1×n) (n×n) (n×1)

q˙  (t = 0) =

√ λ b = ψ T M

, (5.118)

vo .

(1×n) (n×n) (n×1)

In many cases, it is sufficient to determine and then to utilize the first k eigenvectors and natural frequencies.

5.5 Problems Problem 5.1 Given the flexibility matrix of the simply supported uniform beam shown in Fig. 5.5: ⎡

⎤ ⎡ ⎤ f 11 f 12 f 13 9 13 7 3 a ⎣ 13 16 11 ⎦ . f = ⎣ f 21 f 22 f 23 ⎦ = (3×3) 12I E f 31 f 32 f 33 7 11 9 Here I is the moment of inertia and E is the modulus of elasticity. Determine the stiffness matrix and then the equations of motion under the following conditions: (a) the mass of the beam is much less than the masses m 1 , m 2 , and m 3 ; (b) there is no excitation and system is undamped.

q

q1

m1

q2

m2

q3

m3

x a

/4

P1

a

/4

P2

a

/4

P3

a

/4

Fig. 5.5 Simply supported beam with three concentrated masses

Problem 5.2 Determine the eigenvalues (natural frequencies) of the system shown in Fig. 5.5. Assume that m 1 = m 2 = m 3 = m. (Hint: The equation of motion is given in a matrix form by Eq. (C.5.13).)

180

5 Some Problems of Multidegree of Freedom Systems

Problem 5.3 Given the flexibility matrix of the cantilever beam shown in Fig. 5.6: ⎡ ⎤ 2 5 8 a3 ⎣ 5 16 28 ⎦ . f = (3×3) 6I E 8 28 54 Here I is the moment of inertia and E is the modulus of elasticity. Determine the stiffness matrix and then the equations of motion under the following conditions: (a) the mass of the beam is much less than the masses m 1 , m 2 , and m 3 ; (b) there is no excitation and the system is undamped.

Fig. 5.6 Cantilever beam with three concentrated masses

q

q1

m1

q2

m2

q3

m3

x a

/3

P1

a

/3

P2

a

/3

P3

Problem 5.4 Determine the eigenvalues (natural frequencies) of the cantilever beam shown in Fig. 5.5. Assume that m 1 = m 2 = m 3 = m. (Hint: The equation of motion is given in matrix form by Eq. (C.5.15).) Problem 5.5 Find the equations of motion for the system shown in Fig. 5.7. Use the matrix iteration detailed in Sect. 5.3.2 for determining the three eigenvalues (natural frequencies) and eigenvectors.

2k

m

k

G1

m

k

G3

G2 x1

m

x2

x3

Fig. 5.7 Spring-mass system with three degrees of freedom

Problem 5.6 Determine the highest eigenvalue and eigenvector for Problem 5.5 (for the previous problem) by using the matrix iteration detailed in Sect. 5.3.5.

References

181

References 1. R.A. Horn, C.R. Johnson, Matrix Analysis (Cambridge University Press, Cambridge, 1985) 2. G.B. Arfken, H.J. Weber, Mathematical Methods for Physicists, 7th edn. (Elsevier Academic Press, Cambridge, 2005) 3. J.P. Gram, Om Rakkeudviklinger, bestemte ved mindste Kvadraters Methode (On series expansions, determined by the method of least squares) (Host, Copenhagen, 1879) 4. J.P. Gram, Om Beregning af en Bevoxnings ved Hjælp af Provetrær. Tiddskrift Skorburg 6, 137–198 (1883) 5. J.P. Gram, Über die Entwickelung reeller Functionen in Reihen mittelst der Methode der kleinsten Quadrate. J. Reine Angew. Math. 94, 41–73 (1883) 6. E. Schmidt, Zur Theorie der linearen und nichtlinearen Integralgleichungen I, Entwicklung willkülicher Funktionen nach Systemen vorgeschriebener. Math. Ann. 63, 433–476 (1907) 7. E. Schmidt, Zur Theorie der linearen und nichtlinearen Integralgleichungen II, Aufliisung der allgemeinen linearen Integralgleichung. Math. Ann. 64, 161–174 (1907) 8. S.J. Leon, A. Björck, W. Gander, Gram–Schmidt orthogonalization: 100 years and more. Numer. Linear Algebra Appl. 1, 1–40 (2010) 9. J.W. Strutt, L. Rayleigh, On the calculation of Chladni’s figures for a square plate. Philos. Mag. Ser. 6 22, 225–229 (1911) 10. R. Walther, Über eine neue Methode zur Lösung gewisser Variationsprobleme der mathematischen Physik (On a new method for the solution of certain variational problems of mathematical physics). J. Reine Angew. Math. 135, 1–61 (1909) 11. J.W. Strutt, L. Rayleigh, The Theory of Sound, vol. 1 (Macmillen and Co., New York, 1877) 12. J.W. Strutt, L. Rayleigh, The Theory of Sound, vol. 2 (Macmillen and Co., New York, 1878)

Chapter 6

Some Special Problems of Rotational Motion

6.1 Flywheels—Rotational Speed Fluctuation A flywheel is the wheel on the end of a crankshaft or a rotating shaft—see Fig. 6.1 which shows a typical flywheel. A flywheel has, in general, three functions: 1. Storing and providing kinetic energy, for instance, in reciprocating engines since the energy source is intermittent in them. 2. Moderating speed fluctuations of a shaft (or shafts) in an engine through its inertia. Any sudden increase due to a change in the loading of the system can be evened out if we apply a flywheel.

Fig. 6.1 Flywheel © Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_6

183

184

6 Some Special Problems of Rotational Motion

3. Controlling the orientation of a mechanical system. In such applications, the angular momentum of a flywheel is purposely transferred as a torque to the attaching mechanical system when energy is transferred to or from the flywheel, thereby causing the attaching system to rotate into a desired position.1 In the present book, we shall deal with the first two functions of the flywheel only. Fig. 6.2 Rotating shaft with a flywheel

Mt

Fig. 6.3 Torque against the angle of rotation

Mt

A

2

A t

to

t

0

t1

t

t2

T

Figure 6.2 shows a simplified model of a rotating shaft with the flywheel attached to the right end of the shaft. Assume that the torque exerted on the flywheel is a periodic function of the angle of rotation, that is, it satisfies the relation Mt (ϕ) = Mt (ϕ + 2π) .

(6.1)

Further we shall also assume that the total area A+ + A− = 0—see Fig. 6.3. In other words,  2π

Mt (ϕ)dϕ = 0 .

(6.2)

0

Making use of the principle of work and energy (1.58a) we can write  1 1  E1 − Eo = Jw ω12 − ωo2 = (ω1 + ωo ) (ω1 − ωo ) Jw = 2 2

1 Item

3 and Fig. 6.1 are taken from Wikipedia.



ϕ(t1 ) 0

Mt (ϕ) dϕ = A+ , (6.3a)

6.1 Flywheels—Rotational Speed Fluctuation

185

 1 1  E2 − E1 = Jw ω22 − ω12 = (ω2 + ω1 ) (ω2 − ω1 ) Jw = 2 2



ϕ(t2 ) ϕ(t1 )

Mt (ϕ) dϕ = A− ,

(6.3b) where Jw is the moment of inertia of the flywheel and shaft with respect to the axis of rotation, ω is the angular velocity of the flywheel, ωo = ω(t = to = 0), ω1 = ω(t = t1 ), ω2 = ω(t = t2 ). If we add (6.3a) to (6.3b), we get that  1  2 Jw ω2 − ω02 = 0 2 thus

ω2 = ω0 .

(6.4)

Since A+ > 0 and A− < 0 it follows that ω1 = ωmax

and

ω2 = ωmin

(6.5)

are the maximum and minimum of the angular velocity. Let ωmean be the mean value of ω: 1 (6.6) ωmean = (ωmax + ωmin ) . 2 The coefficient of fluctuation of rotational speed δ is defined by the relation δ=

ωmax − ωmin . ωmean

(6.7)

By utilizing (6.5), (6.6), and (6.7) we obtain from (6.3a) that ωmean ωmean δ Jw = A+ or δ=

A+ 2 ωmean Jw

.

(6.8)

In other words, the greater the moment of inertia Jw for a given ωmean and A+ the smaller the coefficient of fluctuation of rotational speed. Table 6.1 Values for the fluctuation of rotational speed Device name Spinning machinery Pump (rotary or motor) Machine tools Car engine Direct current motor Direct drive electric machines

Fluctuation of rotational speed 0.02–0.10 0.03–0.05 0.025–0.02 0.0003–0.005 0.0005–0.001 0.0001–0.0002

Table 6.1 shows the possible values for the fluctuation of rotational speed.

186

6 Some Special Problems of Rotational Motion

6.2 The Effect of a Change in the Load Torque on the Rotational Speed of Shafts The rotational motion is said to be stable if a small change in the load torque results in only a small change in the angular velocity of the steady-state rotational motion. Figure 6.4 shows a model of the rotating shaft, where Md is the driving moment and M is the load torque. The angular velocity and acceleration are denoted by ω and α. It is not difficult to check that the equation of motion is of the form

Md

M

x

Fig. 6.4 Rotating shaft model

a

M

Md

Driving moment

b Load torque

Md

M

M Md

Driving moment

Stable operation point

M

Md Unstable operation point

o

Md

Load torque

o

M

Md

M

Fig. 6.5 M and Md against the angular velocity

Js α = Js ω˙ = Md − M

(6.9)

where Js is the moment of inertia of the shaft and flywheel with respect to the axis x. For steady-state motion Md = M . Hence, α=0

and

ω = ωo .

6.2 The Effect of a Change in the Load Torque on the Rotational Speed of Shafts

187

Here ωo is the angular velocity of the steady-state rotation. Figures 6.5.a and 6.5.b represent the functions Md (ω) and M (ω). The intersection point of the two curves is the operating point (or working point) since the angular acceleration is zero while the angular velocity is constant at this point: ω = ωo = constant. Assume that the angular velocity changes for some reasons in the neighborhood of the operating point. Then, it holds that Js α = Md − M = M .

(6.10)

If ω suddenly changes in the case (a) then  Js α =

M < 0 α < 0 [if ω is increased] M > 0 α > 0 (if ω is decreased)

consequently, ω returns to the steady-state value: the operating point is then said to be stable. If ω suddenly changes in the case (b) then  Js α =

M > 0 α > 0 [if ω is increased] M < 0 α < 0 (if ω is decreased)

consequently ω does not return to the steady-state value: the operating point is then said to be unstable.

6.3 Rotating Shafts—Balancing 6.3.1 Determination of the Reactions Figure 6.6 depicts a rotating shaft of mass m. We shall assume that (a) The shaft is a rigid body. (b) The coordinate axis x coincides with the centerline of the shaft, the axis y is vertical. (c) The Greek coordinate system is attached rigidly to the shaft (it rotates together with the shaft). Its origin is the mass center G and the coordinate axis ξ is parallel to the axis x. (d) The mass center G is not located on the axis x. (e) The position vector ρG lies in the coordinate plane (ηζ). (f) The two bearings are smooth, their middle planes intersect the axis of rotation x at the points A and B; the unknown reaction forces F A and F B pass through the points A and B. (g) The angular velocity ω = ωξ = ωx is constant.

188

6 Some Special Problems of Rotational Motion

y G

A

FA

z

x

B

G

1

FB

2

Fig. 6.6 Rotating shaft model for calculating reaction forces

Since the effective forces are equivalent to the external forces it follows that the resultants should be equal: K = F A + FB + W ,

W = −mg i y ,

(6.11a)

where m is the mass of the shaft. The moments of the two force systems about the points A and B should also be equal: D A = MA , In (6.11b)

D B = MB .

(6.11b)

D A = D G + K × rG A

(6.12)

is the moment of the effective forces about the point A. It also holds: (i) K = maG = −mω 2 ρG ,

(6.13a)

rG A = −ρG − L1 iξ ,

(6.13b)

(ii)

(iii) and on the basis of Eq. (1.37) DG (3×1)

= JG

α + ω× J G

(3×3) (3×1)





=0



⎡

ω = ω× H G =

(3×3) (3×3) (3×1)



⎤⎡



(3×3) (3×1)



HG (3×1)

⎤⎡ ⎤ 0 0 0 Jξ −Jξη −Jξζ ωξ = ⎣ 0 0 −ωξ ⎦ ⎣ −Jηξ Jη −Jηζ ⎦ ⎣ 0 ⎦ = 0 0 ωξ 0 −Jζξ −Jζη Jζ       ω× (3×3)

ω

J

G (3×3)

(3×1)

⎡

⎤ ⎤⎡ 0 0 0 Jξ ωξ = ⎣ 0 0 −ωξ ⎦ ⎣ −Jηξ ωξ ⎦ 0 ωξ 0 −Jζξ ωξ     ω× (3×3)

HG (3×1)

6.3 Rotating Shafts—Balancing

189

or in vectorial notation   D G = ω × HG = ωξ2 iξ × Jξ iξ − Jηξ iη − Jζξ iζ =   = ωξ2 iξ × −Jηξ iη − Jζξ iζ , (6.13c) (iv) and finally   K × rG A = −mωξ2 ρG × −ρG − L1 iξ = mωξ2 L1 ρG × iξ .

(6.13d)

Making use of relations (6.13) Eq. (6.12) yields   D A = ωξ2 iξ × −mL1 ρG − Jηξ iη − Jζξ iζ .

(6.14)

We get in a similar manner that D B = D G + K × rG B ,

(6.15)

where   K × rG B = −mωξ2 ρG × −ρG + L2 iξ = −mω 2 L1 ρG × iξ .

(6.16)

Comparison of Eqs. (6.15), (6.13c) and (6.16) leads to the following result:   D B = ωξ2 iξ × mL2 ρG − Jηξ iη − Jζξ iζ .

(6.17)

On the other hand, M A = r AB × F B +r AG × W =

    = (L1 + L2 ) iξ × F B + L1 iξ + ρG × −mgi y

(6.18)

and M B = r B A ×F A + r BG × W =

    = − (L1 + L2 ) iξ × F A + −L2 iξ + ρG × −mgi y .

(6.19)

Upon substitution of (6.14) and (6.18) into equation D A = M A , we get   ωξ2 iξ × −mL1 ρG − Jηξ iη − Jζξ iζ =

    = (L1 + L2 )iξ × F B + L1 iξ + ρG × −mgi y .

Cross-multiplying from the right by iξ and expanding then the triple cross products by taking the relations 

 −mL1 ρG − Jηξ iη − Jζξ iζ · iξ = 0

and

F B · iξ = 0

190

6 Some Special Problems of Rotational Motion

into account, we find     ωξ2 −mL1 ρG − Jηξ iη − Jζξ iζ = (L1 + L2 ) F B + L1 −mgi y from where FB =

ωξ2   L1 −mL1 ρG − Jηξ iη − Jζξ iζ mgi y + L +L L + L2  1 2 1  independent ofωξ

(6.20)

rotates with the angular velocityωξ

is the reaction force at B. Let us proceed with equation D B = M B . Substituting (6.14) and (6.19), we obtain   ωξ2 iξ × mL2 ρG − Jηξ iη − Jζξ iζ =

    = − (L1 + L2 ) iξ × F A + −L2 iξ + ρG × −mgi y .

Cross-multiplying from the right by iξ we shall find an equation for F A if we utilize that now   and F A · iξ = 0 . mL2 ρG − Jηξ iη − Jζξ iζ · iξ = 0 This way we have     ωξ2 mL2 ρG − Jηξ iη − Jζξ iζ = − (L1 + L2 ) F A + L2 mgi y from where FA =

ωξ2   L2 mL2 ρG − Jηξ iη − Jζξ iζ mgi y − L1 + L2 L1 + L2     independent ofωξ

(6.21)

rotates with the angular velocityωξ

is the reaction at A.  force  Let e = ρG  be the length of the position vector ρG . If e = 0 then aG = 0. Hence, it follows from Eq. (6.11a)—maG = 0—that F A + F B − mg i y = 0 .

(6.22)

This means that the resultant of the external forces acting on the shaft vanishes: we say that the shaft is statically balanced. Remark 6.1 If the mass center of a body rotating about a fixed axis is on the axis of rotation then the body is said to be statically balanced: in this case, the resultant of the external forces acting on the body vanishes.

6.3 Rotating Shafts—Balancing

191

For statically balanced shafts e = 0. Hence FB =

ωξ2   L1 Jηξ iη + Jζξ iζ mgi y − L1 + L2 L1 + L2

(6.23)

FA =

ωξ2   L2 mgi y + Jηξ iη + Jζξ iζ L1 + L2 L1 + L2

(6.24)

and

are the two reactions. Remark 6.2 If e = 0 and Jηξ = Jζξ = 0 the axis of rotation is a principal axis of inertia. Then the shaft is said to be dynamically balanced. For dynamically balanced shafts, the reaction forces FB =

L1 mgi y L1 + L2

FA =

and

L2 mgi y L1 + L2

(6.25)

are independent of time (they do not rotate together with the shaft).

6.3.2 Balancing in Two Planes By [adding](removing) masses [to the shaft](from the shaft) in planes S1 and S2 —see Fig. 6.7—we can eliminate dynamic unbalance: we can set Jηξ and Jζξ to zero. Now it is our aim to clarify how to balance the shaft dynamically, i.e., to establish the equations that show the effects of the various parameters on the solution. We shall

y S1 m1 1

S2

m1 G

G

G

A 2

m2

m2

s1 1

Fig. 6.7 Rotating shaft—balancing in two planes

B

s2 2

x

192

6 Some Special Problems of Rotational Motion

use the notations of Fig. 6.7, which are basically the same as those in Fig. 6.6. The new notations are as follows: m 1 and m 2 are the masses (to be added to the shaft or to be removed from the shaft) in the planes S1 and S2 which are perpendicular to the axis of rotation. The horizontal distances of the two planes from the coordinate plane (ηζ) in the absolute, i.e., the Latin and the Greek (the rotating), coordinate systems are naturally the same and are denoted by s1 and s2 . The position vectors ρ1 and ρ2 are those of the masses m 1 and m 2 . The products m 1 ρ1 and m 2 ρ2 are called unbalances. We shall derive equations for the unbalances from the conditions the dynamically balanced shaft should satisfy. The line of thought is based on the equations of dynamic equilibrium. The system of effective forces can be replaced at the mass center G by the resultant ˆ = −mωξ2 ρG + −m 1 ωξ2 ρ1 + −m 2 ωξ2 ρ2 = K + K1 + K2 K       K – see (6.13a)

K1

(6.26)

K2

and the moment resultant ˆG = D

D G + r1 × K1 + r2 × K2 ,  see (6.13c)

where r1 = −ρG − s1 iξ + ρ1

and

r2 = −ρG + s2 iξ + ρ2 .

Consequently,       ˆ G = ωξ2 iξ × −Jηξ iη − Jζξ iζ + −ρG − s1 iξ + ρ1 × −m 1 ωξ2 ρ1 + D     (6.27) + −ρG + s2 iξ + ρ2 × −m 2 ωξ2 ρ2 . ˆ = 0, i.e., If the shaft is dynamically balanced then K m 1 ρ1 + m 2 ρ2 + mρG = 0

(6.28a)

ˆ G = 0. Thus and D   iξ × −Jηξ iη − Jζξ iζ + m 1 ρG × ρ1 + m 1 s1 iξ × ρ1 + m 2 ρG × ρ2 − m 2 s2 iξ × ρ2 = 0 .

Cross-multiplying from the right by iξ and after expanding the triple cross products taking into account that iη , iζ , ρ1 , ρ2 , and ρG are perpendicular to iξ , we obtain − Jηξ iη − Jζξ iζ + m 1 s1 ρ1 − m 2 s2 ρ2 = 0 .

(6.28b)

Equations (6.28a) and (6.28b) can be solved for the unknown unbalances.

6.3 Rotating Shafts—Balancing

193

After multiplying Eq. (6.28a) by s2 and then adding it to Eq. (6.28b), we get ms2 ρG + m 1 ρ1 (s1 + s2 ) − Jηξ iη − Jζξ iζ = 0 from where m 1 ρ1 =

 1  Jηξ iη + Jζξ iζ − ms2 ρG . s1 + s2

(6.29a)

If we multiply Eq. (6.28a) by s1 and subtract Eq. (6.28b) from the result we arrive at the relation ms1 ρG + m 2 ρ2 (s1 + s2 ) + Jηξ iη + Jζξ iζ = 0 . Hence, m 2 ρ2 = −

 1  Jηξ iη + Jζξ iζ + ms1 ρG . s1 + s2

(6.29b)

Remark 6.3 In principle, we know m, ρG , Jηξ , and Jζξ . With the knowledge of these quantities we can select the locations of the planes S1 and S2 freely. As soon as we have selected s1 and s2 the unbalances m 1 ρ1 and m 2 ρ2 are uniquely determined. Remark 6.4 It may occur that m 1 (or m 2 ) is negative (mass should be removed). To avoid a negative mass the locations of the planes S1 and S2 should be selected carefully when we design the rotating shaft.

6.3.3 Elastic Shafts, Stability of Rotation. Laval’s Theorem Since a rotating shaft is, in general, unbalanced the forces of inertia (the centrifugal force) will cause it to bend out. When the shaft rotates at an angular velocity equal to the natural circular frequency of transverse oscillations this vibration becomes large and shows up as a whirling of the shaft. Our aim is to examine this phenomenon. First, we shall consider a simple (one degree of freedom) model. Assumptions and notations: (i) The mass of the shaft can be neglected. (ii) a = b—the midplane of the disk (or the gear) is a plane of symmetry of the structure. (iii) The mass and the moment of inertia of the disk (with respect to the vertical axis of symmetry) are m and Jx . (iv) The mass center G does not coincide with the geometric center C. The radius from C to the mass center G is denoted by e. This distance is, however, much smaller than the radius of the disk.

194

6 Some Special Problems of Rotational Motion

x

x

a G

C

C G

b =a y

y z

z

e C

C

y

G

C r

e

y G

Fig. 6.8 Laval’s model for investigating stability of rotation

(v) The Latin coordinate system (x yz) is the absolute coordinate system. The Greek coordinate system (ξηζ) is rigidly attached to the rotating shaft—it rotates together with the shaft. The coordinate axis ξ is perpendicular to the middle plane of the disk—for symmetry reasons, it is now parallel to the coordinate axis x. (vi) The angular velocity ω = ωx = ωξ is constant. (vii) The radial displacement of the geometric center C in the Greek coordinate system is η C . (viii) The shaft is a simply supported beam with a circular cross section of diameter d, and E is the Young modulus. For a simply supported shaft of circular cross section, the radial displacement of the geometrical center C due to a radial force F of the same direction is ηC =

F = Ff , k

f =

1 L3 = , k 48I E

I =

d 4π . 64

(6.30)

6.3 Rotating Shafts—Balancing

195

In the present case, F is the centrifugal force: F = (η C + e) mω 2 . Consequently, ηC =

F (η C + e) mω 2 = k k



or

(6.31)

ηC 1 −

ω2

 =e

k m

mω 2 k

from where ω2

ω2 2 χ2 ωcr  e = |η C | =   e = |η C | =  e   |1 − χ2 | ω 2  ω 2    1 − k  1 − ω 2    cr m k m



in which

k 2 = ωcr m

χ=

and

ω2 2 ωcr

(6.32a)

(6.32b)

are the critical angular velocity and a dimensionless parameter. It follows from (6.32a) that r 1 |η C + e| = = , (6.33) e e |1 − χ2 | where r is the total radial displacement of the mass center G—see Fig. 6.8. The quotient r/e is a function of the dimensionless parameter χ, which is shown in Fig. observed that the rotational motion is smoother if χ > √ 6.9. Laval (1845–1913) √ 2, i.e., if ω > 2 ωcr . r2.0 e ro e

1.5

1.0

r e

1

0.5

0.0

0.0 0.2 0.4 0.6 0.8 1.0 1.2

2

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

To be avoided

Fig. 6.9 Dimensionless radial displacement against χ

196

6 Some Special Problems of Rotational Motion

Accordingly, Laval’s theorem: (a) the critical angular velocity is given by  ωcr =

k ; m

(6.34)

(b) the total radial displacement r is less than e if ω > If r = ro is a prescribed value and

√ 2ωcr .

(i) ω < ωcr then χ < 1 and e ω2 =1− 2 , ro ωcr

 e ω = ωcr 1 − ; ro

(6.35a)

(ii) ω > ωcr then χ > 1 and ω2 e = 2 − 1, ro ωcr

 ω = ωcr 1 +

e . ro

(6.35b)

For more details concerning the behavior of rotating shafts, the reader is referred to the following works: [1, 2].

6.3.4 Stability of Rotation. Gyroscopic Effects If the arrangement is not symmetric, i.e., a = b the middle plane of the disk will not be perpendicular to the axis ξ. In order to take this fact into account, a more accurate model is needed. Assume that the shaft is simply supported, its cross section is circular with a diameter d, I = d 4 π/64 and E is Young’s modulus of the shaft. Let f 11 and ϕ11 be the deflection and the angle of rotation at G due to a unit force and a unit couple both acting at G—see Fig. 6.10. The deflection f 11 m and angle of rotation ϕ11 m at G caused by a unit couple and a unit force at G satisfy the symmetry condition f 11 m = ϕ11 m . It can be shown that f 11 =

a 2 b2 , 3I EL

f 11 m = ϕ11 m =

ab (b − a) a 2 − ab − b2 , ϕ11 m = . 3I EL 3I EL

(6.36a)

For a = b, it follows that f 11 =

L3 , 48I E

f 11 m = ϕ11 m = 0 ,

ϕ11 m = −

L . 12I E

(6.36b)

If the beam is subjected to the force F and couple M then yG = f 11 F + f 11 m M

and

are the deflection and angle of rotation at G.

ϕG = ϕ11 F + ϕ11 m M

(6.37)

6.3 Rotating Shafts—Balancing

197

y 11m

11

f 11m

f 11

G

F

a

x

M

b

L 2

y

i2

G

i1

1 G

x

G

Fig. 6.10 Deformed rotating shaft

Figure 6.10 shows the deformed rotating shaft assuming that (a) e = 0, (b) the angular velocity ω = ωx ix = ω ix is constant. The coordinate system (x yz) is the absolute coordinate system. The Greek coordinate system (ξηζ) with origin at G is attached to the disk (gear) in such a manner that the coordinate axis ξ remains parallel to the coordinate axis ξ when the shaft rotates. The principal axes (123) of the disk (gear) at G are given by the unit vectors i1 , i2 and i3 = i1 × i2 —the latter is not seen in Fig. 6.10 since that shows the shaft at such a point of time when i3 is perpendicular to the plane of Fig. 6.10. The moments of inertia with respect to these axes are denoted in the usual way by J1 > J2 = J3 . It is clear that this coordinate system rotates also with the shaft. The following transformations are performed in the coordinate system of the principal axes. Since ω = ω ix = ω (cos ϕG i1 − sin ϕG i2 )

(6.38)

it follows that ⎤⎡ ⎤ ⎡ ⎤ cos ϕG J1 0 0 J1 cos ϕG H G = J G ω = ω ⎣ 0 J2 0 ⎦ ⎣− sin ϕG ⎦ = ω ⎣−J2 sin ϕG ⎦ . (3×1) (3×3) (3×1) 0 0 0 0 J3 ⎡

(6.39)

198

6 Some Special Problems of Rotational Motion

If we take into account that α = ω ˙ = 0 and |ϕ|  1 we obtain from (1.37) (or (1.38)) that D G = ω × H G =ω 2 (cos ϕG i1 − sin ϕG i2 ) × (J1 cos ϕG i1 − J2 sin ϕG i2 ) = = ω 2 (−J2 sin ϕG cos ϕG + J1 sin ϕG cos ϕG ) i3 = = ω 2 (J1 − J2 ) sin ϕG cos ϕG i3 = ω 2 Jd ϕG i3 .       Jd

≈ϕG

(6.40)

≈1

The shaft is elastic. The system of effective forces at G is equivalent to the following force couple system: F = −F iη = −mηG ω 2 iη ,   F

MG = ω 2 Jd ϕG i3 ,  

(6.41)

−M

where F is the resultant of the forces of inertia and M is the moment due to the forces of inertia. Upon substitution of F and M into (6.37), we obtain ηG = f 11 F + f 11 m M = f 11 mω 2 ηG − f 11 m ω 2 Jd ϕG , ϕG = ϕ11 F + ϕ11 m M = ϕ11 mω 2 ηG − ϕ11 m ω 2 Jd ϕG 

 1 − f 11 mω 2 ηG + f 11 m ω 2 Jd ϕG =0 ,   −ϕ11 mω 2 ηG + 1 + ϕ11 m ω 2 Jd ϕG =0.

or

(6.42)

2 2 2 2 Let us introduce the following notations: f 11 m = 1/ωcr o , ω /ωcr o = χ (if a = b 2 2 we get k = 1/ f 11 = 1/ f and ωcr o = ωcr see (6.32b)). Making use of these notations, we obtain the following equation system from (6.42):

f 11 m Jd 2 χ ϕG =0 , (1 − χ2 )ηG + f 11 m   ϕ11 2 ϕ11 m Jd 2 χ ϕG =0 . − χ ηG + 1 + f 11 f 11 m

(6.43)

Solutions different from the trivial ones for ηG and ϕG exist if and only if  f 11 m Jd 2   χ   1 − χ2   f 11 m d= ϕ =  − 11 χ2 1 + ϕ11 m Jd χ2    f 11 f 11 m     ϕ11 ϕ11 m Jd 4 ϕ11 m Jd 2 = 1 − χ2 1 + χ + χ =0 f 11 m f 11 f 11 m from where we get

(6.44a)

6.4 Problems

199

  2 2 m χ2 + f 11 m = 0. (ϕ11 ϕ11 m − f 11 ϕ11 m ) Jd χ4 + f 11 ϕ11 m Jd − f 11

(6.44b)

The above equation can easily be solved for the unknown χ.

6.4 Problems Problem 6.1 Find the critical angular velocity of the rotating shaft shown in Fig. 6.11. Assume that (a) the eccentricity of the mass center of the rotating disk e = 0, (b) the angular velocity ω = ωx ix = ω ix is constant.

y

G

x

L

Fig. 6.11 Disk at the left end of the rotaing shaft

Problem 6.2 Find the critical angular velocity of the rotating shaft shown in Fig. 6.10 if b = 4a.

References 1. A. Föppl, Das Problem der Lavalschen Turbinenwelle (Der Civilingenieur, 1895), pp. 333–342 2. G.L. Robert, J.P. Vincent, Dynamics of Rotating Shafts. The Shock and Vibration Monograph Series (1969)

Chapter 7

Systems with Infinite Degrees of Freedom

7.1 Equilibrium Equations for Spatial Beams First, we shall consider such problems for which the physical quantities depend only on one spatial coordinate and time. These investigations are based on the equilibrium conditions a spatial beam should satisfy. Figure 7.1 shows the centerline AB of a beam portion before deformation and the beam portion together with its centerline after deformation. The arc coordinate (ξ)[s] is measured on the centerline of the beam (before)[after] deformation. Let us denote the arc coordinate on the left cross section (on cross section A) by so , on the right cross section (on cross section B) by s. We remark that the arc coordinate s will be regarded as if it were a parameter for the further investigations. The unit tangent to the deformed centerline is denoted by t(s). We shall assume that the loads exerted on the beam are equivalent to a distributed Fig. 7.1 A beam portion with the external and internal forces acting on it

f(s ) B

FC(s ) t(s )

(s )

d

-MC(s o)

MC(s ) B

u( )

z

A

r

so -FC(s o ) A

y O x

© Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_7

201

202

7 Systems with Infinite Degrees of Freedom

force system f(s) and a distributed moment load µ(s) both acting on the deformed centerline. The resultant and moment resultant of the inner forces (stresses) at the center of cross sections A and B are denoted by −FC (so ), −MC (so ) and FC (s), MC (so ). Since the beam is in equilibrium, the total system of forces acting on the beam should have a zero resultant and a zero moment resultant (say) about the origin O. The resultant is zero if  s f(ξ)dξ + FC (s) = 0 . (7.1a) − FC (so ) + so

Let u(s) be the displacement on the centerline. A vanishing moment resultant (taken about the origin) follows from the following equation:  − (r(so ) + u(so )) × FC (so )− MC (so ) +

s

[(r(ξ) + u(ξ)) × f(ξ) + µ(ξ)] dξ+

so

+ (r(s) + u(s)) × F S (s) + M S (s) = 0 .

(7.1b)

Let us derive the above equations with respect to s. Since the quantities regarded at so are constants (so is fixed), we get by taking into account that the derivative of an integral with respect to the upper limit is the integrand taken at the upper limit that dFC (s) + f(s) = 0 ds

(7.2a)

and   dMC (s) d (r(s)+u(s)) dFC (s) + × FC (s)+µ(s)+(r(s)+u(s)) × + f(s) = 0, ds ds ds    =0

or

dMC (s) d (r(s) + u(s)) + × FC (s) + µ(s) = 0 . ds ds

(7.2b)

Equations (7.2a) and (7.2b) are the equilibrium conditions sought. It is obvious that they are valid for the deformed beam. For small displacements and deformations, however, we do not make a difference between the initial and deformed states of the beam. Then d (r(s) + u(s)) dr(s) ≈ = t(s) (7.3) ds ds and Eq. (7.2b) simplifies to dMC (s) + t(s) × FC (s) + µ(s) = 0 . ds

(7.4)

7.2 Longitudinal Vibrations of Rods

203

7.2 Longitudinal Vibrations of Rods 7.2.1 Equations of Motion Figure 7.2 shows the portion of a uniform rod with length x. The rod (or bar) is made of homogeneous material. If the vibrations are longitudinal, the resultant of the inner forces is an axial force 1

z

y B

A O

x

F

x Fig. 7.2 Uniform rod and its element

FC (s, t) = FC (x, t) = N (x, t)ix

(7.5)

and MC (s, t) = MC (x, t) = µ(s, t) = µ(x, t) = 0 , t = ix , t(s) × FC (s, t) = ix × N (x, t)ix = 0 .

(7.6)

Thus Eq. (7.4) is an identity. As regards Eq. (7.2a), the distributed load for a unit length f(s) = f(x) is the force of inertia. For longitudinal vibrations, each cross section moves horizontally as if it were a rigid body. Therefore u = u(x)ix is the displacement field in the rod and f(s) = − ρA  dm

∂ 2 u(x, t) ∂t 2

(7.7)

is the force of inertia for a unit length where ρ is the density and A is the area of the cross section. Upon substitution of Eqs. (7.5) and (7.7) into Eq. (7.2a), we get ∂2u ∂N − ρA 2 = 0 . ∂x ∂t

(7.8)

204

7 Systems with Infinite Degrees of Freedom

It is known from mechanics of materials that the axial force N can be given in terms of the normal stress σx x : (7.9) N = Aσx x , where according to Hooke’s law σx x = Eεx x = E

∂u . ∂x

(7.10)

Here E is Young’s modulus and εx x = ∂u/∂x is the axial strain. Consequently N = AE

∂u . ∂x

(7.11)

Upon substitution of N into (7.8), we obtain the equation of motion for the axial displacement u:

∂u ∂2u ∂ AE = ρA 2 . (7.12) ∂x ∂x ∂t This equation is a partial differential equation of order two. Boundary conditions should be imposed on u or N . In addition, the solution for u should also satisfy the initial conditions prescribed on u and its time derivative u. ˙

7.2.2 Solution by Separation of Variables After rearranging Eq. (7.12) we have c2

∂2u ∂2u = , ∂x 2 ∂t 2

c2 =

E > 0. ρ

(7.13)

We seek the solution in the form (separation of variables) u(x, t) = U (x)γ(t) .

(7.14)

Substituting this solution into Eq. (7.13) yields 1 d2 γ(t) c2 d2 U (x) = = −ω 2 . U (x) dx 2 γ(t) dt 2

(7.15)

Since the left side of this equation is independent of t, whereas the right side is independent of x we can come to the conclusion that each side must be the same

7.2 Longitudinal Vibrations of Rods

205

constant. Letting this constant be −ω 2 (ω is an unknown parameter here), we arrive at two ordinary differential equations: one equation for U (x), the other equation for γ(t): −

 ω 2 ρω 2 d2 U (x) ; = λU (x), λ = = dx 2 c E

d2 γ(t) + ω 2 γ(t) = 0 . dt 2

(7.16)

γ(t) = C sin ωt + D cos ωt ,

(7.17)

The general solutions are given by the equations U (x) = A sin

√ √ λ x + B cos λ x ,

where the unknown integration constants (A, B) and [C, D] depend on the (boundary) and [initial] conditions. It follows from (7.14) and (7.11) that U (x) and AE U (1) (x) are, in fact, the amplitudes of the longitudinal motion and the axial force along the rod. The boundary conditions that can be imposed are categorized as essential and natural boundary conditions. As we have already mentioned (i) either the amplitude of the motion, i.e., U (x) can be prescribed at an end of the rod (this type of boundary conditions is the essential boundary condition) or (ii) the amplitude of the axial force, i.e., N (x) = AE U (1) (x) can be prescribed at the same end of the rod (this type of boundary conditions is called natural boundary condition). Exercise 7.1 Consider a uniform rod fixed at its left end and free at its right end—see Fig. 7.3. Determine the solution if u(x, 0) = u(x) and u(x, ˙ 0) = g(x) are the initial conditions.

z

y x

O

x =b =

x =a =0 Fig. 7.3 Uniform rod fixed at its left end

It is evident that u(0, t) = U (0)γ(t) = 0 and N (, t) = AE U (1) ()γ(t) = 0. Consequently, differential equation (7.16)1 is associated with the boundary conditions  √ √

U (0) = A sin λ x + B cos λ x =B=0, x=0  (7.18) √ √ √ √ √

= λ A cos λ  = 0 . U (1) () = λ A cos λ x − B sin λ x x=

206

7 Systems with Infinite Degrees of Freedom

Since A and λ cannot be zero for a nontrivial solution concerning U (x), we have cos

√

λ = 0.

(7.19)

This equation is called frequency equation. It is satisfied if 

λk  = ±(2k − 1)

π , 2

k = 1, 2, 3, . . .

(7.20a)

are the roots from where π2 λk = 2 (2k − 1)2 , 4



ωk = ↑ = c λ k = (7.16)2



 E λk = ρ

E π (2k − 1) . ρ 2 (7.20b)

Observe that the number of roots λk is infinite. We remark that A can be arbitrary since the differential equation (7.16)1 , is homogeneous. In what follows we shall set it to 1. The roots λk are called eigenvalues, and the values of ω are the natural frequencies or eigenfrequencies. The reason for calling ωk the kth natural frequency is that solution (7.17)3 describes a harmonic motion with this natural frequency. The function  (2k − 1)π x (7.21) Uk (x) = sin λk x = sin 2 is the kth eigenfunction, or the kth normal mode. For k = 1 we speak about fundamental frequency and fundamental mode. Fig. 7.4 Three eigenfunctions of the longitudinally vibrating rod

U 2(x )

U 1(x )

1.0 0.5 x

0.0 -0.5 -1.0 U3(x )

Figure 7.4 shows the first three eigenfunctions. The eigenfunctions Uk (x) have the following properties:

7.2 Longitudinal Vibrations of Rods







207



Uk (x) Un (x)dx =

sin λk x sin λn x dx =   (2n−1)π  1 if k = n, (2k −1)π x sin x dx = sin = k, n = 1, 2, 3 . . . 2 2 2 0 if k = n. 0 (7.22) 0

0



If the functions Uk (x) and Un (x) satisfy this equation—known as orthogonality condition—then we say that they are orthogonal to each other. This result means that the eigenfunctions with different subscripts are orthogonal functions. The functions  (2k − 1)π 2 sin x (7.23) ψk (x) =  2 are called normalized eigenfunctions. Recalling (7.14), (7.17)2 , and (7.21), we can give the solution that belongs to ωk in the form u k (x, t) = Uk (x)γk (t) = sin λk x (Ck sin ωk t + Dk cos ωk t).       Uk (x)

(7.24)

γk (t)

The function u k (x, t) is the kth mode of vibration. The points at which Uk (x) (or u k (x, t)) is equal to zero are the nodal points or simply nodes. It is obvious that γ˙ k (t) = ωk (Ck cos ωk t − Dk sin ωk t) .

(7.25)

The total solution can be obtained by the use of the principle of superposition: u(x, t) =

∞ 

∞ 

u k (x, t) =

k=1

k=1 ∞ 

=

Uk (x) γk (t) = (7.26) sin λk x (Ck sin ωk t +Dk cos ωk t) .

k=1

Its time derivative is given by u(x, ˙ t) =

∞  k=1

u˙ k (x, t) =

∞ 

Uk (x) γ˙ k (t) =

k=1

=

∞  k=1

ωk sin λk x (Ck cos ωk t − Dk sin ωk t) .

(7.27)

208

7 Systems with Infinite Degrees of Freedom

The unknown integration constants can be determined by using the initial conditions:

u(x) = u(x, 0) = g(x) = u(x, ˙ 0) =

∞  k=1 ∞ 

Uk (x) γk (0) =

∞ 

Dk Uk (x) , k=1 ∞ 

Uk (x) γ˙ k (0) = −

k=1

ωk Ck Uk (x) .

(7.28a)

(7.28b)

k=1

Multiply throughout by Un (x) (n = 1, 2, 3, . . .) and integrate the result obtained with respect to x from zero to . We have 



u(x)Un (x) dx =

0





∞ 

 Dk

0

Uk (x) Un (x)dx = ↑ = (7.22)

0

k=1

g(x)Un (x) dx = −



∞ 





ωk Ck

Uk (x) Un (x)dx = ↑ = (7.22)

0

k=1

 Dk 1 if k = n , 2 0 if k = n

=−

 ωk Ck 1 ifk = n . 0 ifk = n 2

Hence Dn =

2 





u(x)Un (x) dx ,

0

Cn =

2 ωn





g(x)Un (x) dx .

(7.29)

0

Exercise 7.2 Assume that the rod shown in Fig. 7.3 is subjected to an axial force No at its free end (at the right end). Determine the solution of the equation of motion if the force is suddenly removed. Due to the force No the axial strain is constant εx x =

σx x No du(x) = = dx E AE

from where u(x) = u(x, 0) =

No x AE

(7.30)

is the initial displacement field in the rod. When the force is suddenly removed, the rod is at rest; therefore, the initial velocity g(x) is zero. We can now utilize Eq. (7.29) to determine the integration constants Dn and Cn . We get

7.2 Longitudinal Vibrations of Rods

2 Dn = 



 0

209

2No u(x)Un (x) dx = ↑ = AE (7.30), (7.21)

and Cn =

2 ωn









(2n − 1)π x dx = 2 0 8No  (−1)n (7.31a) = AE π 2 (2n − 1)2 x sin

g(x)Un (x) dx = 0 .

(7.31b)

0

With the integration constants equation (7.26) yields ∞ 8No   (−1)n u(x, t) = ↑ = cos AEπ 2 n=1 (2n − 1)2 (7.20b)2



E π (2n − 1)t. ρ 2

(7.32)

7.3 Transverse Vibration of a String 7.3.1 Equations of Motion Figure 7.5 shows a string (or cable). We shall assume that (i) the string is uniform (the cross section is constant), (ii) it is made of homogeneous and linearly elastic material for which E is Young’s modulus, and ρ is the density, (iii) the string is flexible, i.e., it has no resistance to bending (MC = 0 in Eq. (7.2b)), (iv) its statical load is vertical (f(s) = f (x)iz —for free vibrations f (x) = 0), (v) each cross section moves vertically (u = w(x, t)iz is the displacement field), (vi) the displacements and deformations are small, (vii) the string is uniformly stretched by a constant axial force (tensile force) N , and (viii) its weight can be neglected.

Fig. 7.5 String fixed at both ends

z

s = =x x

x = a =0

x =b =

210

7 Systems with Infinite Degrees of Freedom

Under these conditions r(s) + u(s) = xix + w(x, t)iz , µ = 0, f(x, t) = −ρA

∂ 2 w(x, t) iz ∂t 2

(7.33)

are the position vector of the deformed string, the distributed moment load, and the forces of inertia per unit length. Hence, Eq. (7.2b) yields dMC (s) d (r(s) + u(s)) ∂ (xix + w(x, t)iz ) + × FC (s) + µ(s) = ×FC (x, t) = ds ds

∂x ∂w(x, t) = ix + (7.34a) iz × FC (x, t) = 0 ∂x in which the inner force FC (x, t) is resolved into a horizontal (the constant tensile force) and a vertical component: FC (x, t) = N ix + FC z (x, t)iz .

(7.34b)

Let us derive Eq. (7.34a) with respect to x. We get

∂FC (x, t) ∂w(x, t) ∂ 2 w(x, t) i = 0. (7.35) i × i + F (x, t)i + i (N )+ z x Cz z x z × ∂x 2 ∂x ∂x If we recall Eq. (7.2a) and take into account that for free vibrations f(s) = f(x, t) is given by (7.33)3 , we get ∂ 2 w(x, t) ∂FC (x, t) = ρA iz . ∂x ∂t 2

(7.36)

Upon substitution of this relation into Eq. (7.35) N

∂ 2w ∂ 2 w(x, t) i − ρA iy y ∂x 2 ∂t 2

is the result. Hence, c2

∂ 2w ∂ 2w = , ∂x 2 ∂t 2

c2 =

N >0 ρA

(7.37)

is the equation of motion. We remark that boundary conditions should be imposed on w. In addition, the solution for w should also satisfy the initial conditions prescribed on w and its time derivative w. ˙ We seek the solution again in the form of a product (separation of variables) W (x, t) = W (x)γ(t) . Substituting this product into (7.37), we obtain

(7.38)

7.3 Transverse Vibration of a String

211

c2 d2 W (x) 1 d2 γ(t) = = −ω 2 . W (x) dx 2 γ(t) dt 2

(7.39)

Here, the left side is independent of t, whereas the right side is independent of x; hence, each side must be the same constant. Denoting this constant by −ω 2 (ω is an unknown parameter here), we arrive at two ordinary differential equations: −

 ω 2 ρAω 2 d2 W (x) ; = λW (x), λ = = 2 dx c N

d2 γ(t) + ω 2 γ(t) = 0 . dt 2

(7.40)

G(t) = C sin ωt + D cos ωt .

(7.41)

The general solutions are given by the equations W (x) = A sin

√ √ λ x + B cos λ x ,

The unknown integration constants (A, B) and [C, D] can be determined by using the boundary and initial conditions. Assume that the string is fixed at both ends. Then, differential equation (7.37) is associated with the following boundary conditions: w(0, t) = W (0) = 0,

w(, t) = W () = 0.

(7.42)

Let w(x) and g(x) x ∈ [0, ] be given functions. The initial conditions w(x, 0) = w(x),

w(x, ˙ 0) = g(x)

(7.43)

should also be satisfied by the solution w(x, t). Remark 7.1 Compare now the equations that describe the longitudinal vibrations of a rod and the transverse vibrations of a string. As regards the equations of motion, we can get (7.37) from (7.12) if we write w for u and c2 = N /ρE for c2 = E/ρ. The differential equation for the amplitude function (7.40)1 and the solution (7.41)1 can be obtained from (7.16)1 and (7.17)1 if we write W for U . The differential equations (7.40)2 , (7.16)2 for the time function γ(t) and the solutions (7.41)2 , (7.17)2 coincide formally with each other. Note that we have denoted the unknown integration constants in the same way for both problems, i.e., by A, B, C, and D. Exercise 7.3 Assume that the string is fixed at its ends. Determine the solution if the initial conditions are given by Eq. (7.41). After substituting solution (7.41) into boundary conditions (7.42), we get  √

√ =B=0, W (0) = A sin λ x + B cos λ x x=0  √ √ √ √ √

W () = λ A sin λ x + B cos λ x = λ A sin λ  = 0 . x=

Since A = 0 in (7.44)2

(7.44)

212

7 Systems with Infinite Degrees of Freedom

sin

√ λ = 0

(7.45)

is the frequency equation. Hence  λk  = ±kπ ,

k = 1, 2, 3, . . .

(7.46a)

are the roots from where k 2 π2 λk = 2 , 





kcπ kπ ωk = ↑ = c λ k = =   (7.40)2

N . ρA

(7.46b)

The number of roots λk is naturally infinite. The function Wk (x) = sin

 kπ x λk x = sin 

(7.47)

is the kth eigenfunction (or the kth normal mode). Fig. 7.6 Three eigenfunctions of the vibrating string

W2(x )

W1(x)

1.0 0.5 0.0 -0.5 -1.0 W3(x )

Figure 7.6 shows the first three eigenfunctions. It can be checked with ease that the eigenfunctions satisfy the following orthogonality condition: 







Wk (x) Wn (x) dx = sin λk x sin λn x dx = 0 0    nπ kπ /2 if k = n, x sin x dx = sin = k, n = 1, 2, 3 . . . 0 if k = n.   0

(7.48)

7.3 Transverse Vibration of a String

213

The normalized eigenfunctions are given by the equation  ψk (x) =

kπ 2 sin x.  

(7.49)

Since B = 0, it follows from (7.38) and (7.41) that the solution corresponding to ωk is of the form wk (x, t) = Wk (x)γk (t) = sin λk x (Ck sin ωk t + Dk cos ωk t) .       Wk (x)

(7.50)

γk (t)

The function wk (x, t) is the kth mode of vibration, and the points at which Wk (x) = wk (x, t) = 0 are the nodal points. By applying the principle of superposition, we get the total solution w(x, t) =

∞ 

∞ 

wk (x, t) =

k=1

sin λk x (Ck sin ωk t + Dk cos ωk t)

(7.51)

ωk sin λk x (Ck cos ωk t − Dk cos ωk t) .

(7.52)

k=1

and its time derivative as well: w(x, ˙ t) =

∞ 

∞ 

w˙ k (x, t) =

k=1

k=1

The initial conditions w(x) = w(x, 0) = g(x) = w(x, ˙ 0) =

∞  k=1 ∞ 

Wk (x) γk (0) =

∞ 

Dk Wk (x) , k=1 ∞ 

Wk (x) γ˙ k (0) = −

k=1

(7.53a)

ωk Ck Wk (x)

(7.53b)

k=1

make it possible to determine the unknown integration constants if we take the orthogonality conditions (7.48) into account in the following integrals: 



w(x)Wn (x) dx =

0

∞ 

 Dk

k=1



 0

g(x)Wn (x) dx = −



Wk (x) Wn (x)dx = ↑ = (7.48)

0

∞  k=1





ωk Ck 0

Dk 2



1 if k = n , 0 if k = n

Wk (x) Wn(x)dx = ↑ = (7.48)

ωk Ck =− 2



1 if k = n . 0 if k = n

214

7 Systems with Infinite Degrees of Freedom

Hence Dn =

2 





w(x)Wn (x) dx ,

Cn =

0

2 ωn





g(x)Wn (x) dx .

(7.54)

0

7.4 Torsional Vibrations Figure 7.7 represents a shaft. We shall assume that (i) the shaft is uniform (the cross section is constant), (ii) it is made of homogeneous and linearly elastic material for which G is the shear modulus of elasticity and ρ is the density, (iii) its statical load is a distributed moment load µ(s) = μx (x)ix acting on its longitudinal axis (for free vibrations μx (x) = 0), (iv) each cross section rotates in its own plane as if it were a rigid body—the rotation vector is denoted by ϕ(x)ix , and (v) the rotations and deformations are small. 1 2

r

(x,t)

t2

dA

dA

r

s = =x z A

d

2 x

=r 2

y

B

(x,t) dA t2

M x(x,t) = J pG

(x,t)

x x

O

x

MC(s,t )=M x(x,t) i x

Fig. 7.7 Shaft of uniform cross section and its element

Under these conditions, the intensity of the distributed load f(s) = f(x) and the resultant of the inner forces FC (s) = FC (x) are equal to zero. Consequently, equilibrium equation (7.2a) is identically satisfied, whereas the second equilibrium equation, i.e., Eq. (7.2b) simplifies to the following form: dMC (x) + µ(x) = 0 . dx

(7.55)

For the vibration problems considered, MC = Mx (x, t)ix is the torsional moment which is related to the angle of rotation ϕ(x) via the equation

7.4 Torsional Vibrations

215

Mx (x, t) = G J p

dϕ(x, t) , dx

(7.56)

where J p is the areal polar moment of inertia. It is clear from Fig. 7.7 that the moment of the effective forces for a unit length of the axis of rotation can be given as 



Dx =

Dx = A

∂2ϕ r 2ρ d A 2 , ∂t  A  

(7.57)

Ip

where I p is the mass polar moment of inertia. If the diameter of the shaft is constant I p = ρJ p . The distributed moment load is due to the effective forces and in accordance with the d’Alembert principle can be given in the following form: μx (x, t) = −I p

∂2ϕ ∂2ϕ = −J p ρ 2 . 2 ∂t ∂t

(7.58)

Since both MC and µ are parallel to the axis x from Eq. (7.55), we get dMx (x, t) + μx (x, t) = 0 . dx

(7.59)

Substitution of (7.56) and (7.58) into this equation results in the equation of motion: ∂ ∂x

G Jp

∂ϕ ∂x

= Ip

∂ 2ϕ . ∂t 2

(7.60)

If the diameter of the shaft is constant this equations becomes a bit simpler c2

∂ 2ϕ ∂ 2ϕ = , ∂x 2 ∂t 2

c2 =

G > 0. ρ

(7.61)

It should be mentioned that boundary conditions have to be imposed on ϕ or torsional moment Mx . In addition, the solution for ϕ should satisfy the initial conditions prescribed on ϕ and its time derivative ϕ. ˙ Assume that ϕ(x, t) = Φ(x)γ(t) . (7.62) Then we get in the same manner as earlier—see, for instance, the steps leading to Eq. (7.40)—that −

 ω 2 ρω 2 d2 Φ(x) = λΦ(x) , λ = = ; dx 2 c G

d2 γ(t) + ω 2 γ(t) = 0 . (7.63) dt 2

216

7 Systems with Infinite Degrees of Freedom

The general solutions are given by the equations Φ(x) = A sin

√

λ x + B cos

√ λx ,

γ(t) = C sin ωt + D cos ωt .

(7.64)

Remark 7.2 Let us introduce the notations y(x) = U (x) = W (x) = Φ(x) and operators K (y) = −y (2) (x) and M(y) = y(x). Further let x = a and x = b be those x coordinates that belong to the left and right extremities of the rod (or shaft): b > a and b − a =  is the length of the rod (or shaft). Making use of the notations introduced differential equations (7.16)1 , (7.40)1 and (7.63)1 can be rewritten in a common form: K [y(x)] = λM[y(x)] . (7.65) This differential equation is, in general, associated with the following boundary conditions: ⎫ y(a) = 0 , y(b) = 0 (the ends of the rod or shaft are fixed); or ⎬ y(a) = 0 , y (1) (b) = 0 (the left end is fixed, the right end is free); or ⎭ y (1) (a) = 0 , y (1) (b) = 0 (the ends of the rod or shaft are free).

(7.66)

Differential equation (7.65) and boundary conditions (7.66)1 (or (7.66)2 , or (7.66)3 ) constitute a boundary value problem for which both the differential equation and the boundary conditions are homogeneous, and λ is an unknown parameter. This problem is also referred to as an eigenvalue problem with lambda as the eigenvalue. Table 7.1 shows some typical values for the three eigenvalue problems. The second and third rows in the table are based on the solution of Exercises 7.1 and 7.3. Data in the fourth row are taken from the solution of Problem 7.2. Table 7.1 Characteristic values for the eigenvalue problems considered Boundary conditions

y(0) = 0 y() = 0

Frequency equation

sin

√

λ = 0

y(0) = 0 cos √λ = 0 y (1) () = 0 y (1) (0) = 0 sin √λ = 0 y (1) () = 0

Normalized eigenfunctions (mode shapes)

ψk (x) = ψk (x) =



ψk (x) =



2 

2 

2 

Natural frequencies ωk

kπ x 

k 2 π2 2

ckπ 

(2k−1)π x 2

π 2 (2k−1)2 42

cπ(2k−1) 2

kπ x 

k 2 π2 2

ckπ 

sin

sin



Eigenvalues λk

sin

7.5 Flexural Vibrations of Beams

217

7.5 Flexural Vibrations of Beams 7.5.1 Equilibrium Equations 7.5.1.1

Assumptions

Figure 7.8 depicts a beam. We shall assume that the followings are true: (i) the beam is uniform (the cross section is constant), (ii) the coordinate plane x z is a plane of symmetry, (iii) the beam is made of homogeneous and linearly elastic material for which E is Young’s modulus, ν is the Poisson number, and G is the shear modulus of elasticity, (iv) the displacements and deformations are small, (v) for the arc coordinates it holds that s = ξ = x, ds = dξ = dx, (vi) the vertical static load f(s) = f z (x)iz is exerted on the centerline x (under this load the deformed centerline remains in the coordinate plane x z)—for free vibrations f z (x) = 0, (vii) the distributed moment load µ = μ y (x)i y is zero, (viii) the beam is preloaded by a constant axial force N which is (positive) [negative] if it is a (tensile) [compressive] force, and (ix) it will be assumed that the displacement field of the beam is of the form u = (u(x) + ϕ y (x)z)ix + w(x)iz ,

ϕ y (x) = −dw/dx,

(7.67)

MC(s )=M y(x ) i y N =constant

P

FC(s )=N i x –V z(x ) i z

z s = =x

u

y

V z(x ) w

y

z dA

z

r

y

x Po

x

u

Fig. 7.8 Beam and its element with the inner forces

218

7 Systems with Infinite Degrees of Freedom

which means that the cross section of the beam rotates as if it were a rigid body and remains perpendicular to the deformed centerline1 —ϕ y is the rotation of the centerline about the direction y, (x) since the deformations are small the derivative du/dx satisfies the inequality |du/dx|  1 which means that it can be neglected when it is compared to 1.

7.5.1.2

Equilibrium Conditions

Under these conditions FC = N ix − Vz (x)iz ,

MC = M y (x)i y

(7.68)

are the resultant and moment resultant of the inner forces acting on a cross section. The bending moment M y is positive if its vector points in the positive y direction, the shear force Vz is positive if it points in the negative z direction—see Fig. 7.8 for details. With FC Eq. (7.2a) yields −

dVz + fz = 0 . dx

(7.69)

As regards Eq. (7.2a), we may write dMC (x) d (r(x) + u(x)) + × FC (x) + µ(x) = dx  dx d (x + u) ix + wiz dM y iy + × (N ix − Vz (x)iz ) + µ(x) = 0 = dx dx

(7.70)

where d du  1, (x + u) = 1 + dx dx

N = constant ,

and µ = 0 .

(7.71)

Calculate now the cross products in Eq. (7.70) and dot multiply the result by i y . We have dM y dw +N + Vz = 0 . (7.72) dx dx After deriving this equation with respect to x we can substitute (7.69) for dVz /dx. The result is 1 This

assumption is known as Euler–Bernoulli beam hypothesis.

7.5 Flexural Vibrations of Beams

219

d2 M y d + dx 2 dx



dw N dx

+ fz = 0 .

(7.73)

This equation is the equilibrium equation the moment resultant of the internal forces (the bending moment M y ) should satisfy if the beam is preloaded by an axial force N [1, 2]. We shall denote the aerial moment of inertia of the cross section A with respect to the centroidal axis y by I . It is obvious on the basis of Fig. 7.8 that  z2 d A .

I = A

In accordance with assumption (ix), the positive ϕ y rotates the axis z (the centerline) in clockwise sense as shown in Fig. 7.8. The bending moment M y and the shear force Vz —see Eq. (7.72)—can all be given in terms of the vertical displacement w [3]. The corresponding equations are as follows: My = I E

dϕ y d2 w d3 w dw = −I E 2 , Vz = I E 3 − N . dx dx dx dx

(7.74)

Upon substitution of (7.74)2 for M y in Eq. (7.73), we have d2 dx 2



d2 w IE 2 dx



d − dx



dw N dx

= fz .

(7.75a)

If the beam is a uniform one and the axial force is independent of x equilibrium equation (7.75a) is simplified: IE

d4 w d2 w − N 2 = fz . 4 dx dx

(7.75b)

Equations (7.75) make it possible to determine the deflection w(x) if the beam is preloaded by the axial force N .

7.5.1.3

Boundary Conditions

It is obvious that Eqs. (7.75) should be associated with appropriate boundary conditions which depend on the supports at the endpoints of the beam. If an endpoint of the beam (a) cannot move vertically then the deflection w is zero. (b) cannot rotate then the rotation ϕ y is zero. (c) can freely rotate then the bending moment M y is zero.

220

7 Systems with Infinite Degrees of Freedom

(d) is free then the shear force Vz is zero. (e) is (vertically) [rotationally] restrained by a (spring) [torsional spring] the sum of the (forces) [moments] shown in Fig. 7.9 should be zero—the spring constants are denoted by k z , k zr and kγ , kγr , respectively.

kz w

Vz

Vz

k zr w

k

y

My My

k

r

y

Fig. 7.9 Spring supported beam ends with inner forces

In accordance with these rules, Table 7.2 shows the boundary conditions for the beams that are (i) simply supported, (ii) fixed at the left end and hinged at the right end, (iii) fixed at both ends, and (iv) fixed at the left end and free at the right end. The axial force N is rigid which means that it is always parallel to the axis x. If the left or right endpoint of the beam is restrained (vertically) (rotationally) by a (spring) [torsional spring], then the boundary conditions are of the form Vz + k z w = 0 , Vz − k zr w = 0 ;

M y − kγ ϕ y = 0 , M y + kγr ϕ y = 0 . (7.76)

Table 7.2 Boundary conditions for some beams Support arrangements

Boundary conditions w(0) = 0 M y (0) = 0

w() = 0 M y () = 0

w(0) = 0 ϕ y (0) = 0

w() = 0 M y () = 0

w(0) = 0 ϕ y (0) = 0

w() = 0 ϕ y () = 0

w(0) = 0 ϕ y (0) = 0

M y () = 0 Vz () = 0

7.5 Flexural Vibrations of Beams

221

7.5.2 Equation of Motion and Solutions for the Axially Unloaded Beam 7.5.2.1

Equation of Motion

If the beam is unloaded, the axial force N is zero and the distributed load f z is the force of inertia: ∂2w (7.77) f z = −ρA 2 . ∂t Upon substitution of Eq. (7.77) into equilibrium equation (7.75a) we get the equation of motion

∂2w ∂2w ∂2 I E + ρA = 0, (7.78) ∂x 2 ∂x 2 ∂t 2 which can be rewritten in the following form: c2

∂4w ∂2w + 2 = 0, ∂x 4 ∂t

c2 =

IE >0 ρA

(7.79)

if the beam is uniform. 7.5.2.2

Solutions

By applying the method of separation of variables, we may write

Thus

w(x, t) = W (x)γ(t) .

(7.80)

1 d2 γ(t) c2 d4 W (x) = − = ω2 W (x) dx 4 γ(t) dt 2

(7.81)

in which ω 2 is a constant (It will turn out later that ω is the circular frequency of the vibrations). Equation (7.81) can be rewritten in the form  ω 2 ρAω 2 d4 W (x) = β4; = λW (x), λ = = dx 4 c IE

d2 γ(t) + ω 2 γ(t) = 0. (7.82) dt 2

The general solution of Eq. (7.82)3 is given by γ(t) = C sin ωt + D cos ωt .

(7.83)

As regards Eq. (7.82)1 , we shall assume that W (x) = A eχx

(7.84)

222

7 Systems with Infinite Degrees of Freedom

is a particular solution where A and χ are constants. If we substitute it into (7.82)1 , we get the characteristic equation χ4 − λ = χ4 − β 4 = 0.

(7.85)

Hence √ 4 χ1 = β, χ2 = −β, χ3 = iβ, χ4 = −iβ; β = λ =



ω = c



ρAω 2 IE

(7.86a)

= A1 eβx +A2 e−βx +A3 (cos βx +i sin βx)+A4 (cos βx −i sin βx)

(7.86b)

4

in terms of which W (x) = A1 eβx +A2 e−βx +A3 eiβx +A4 e−iβx =

is the complex general solution to the differential equation (7.82)1 in which A1 , A2 , A3 , and A4 are undetermined integration constants. Since the real and imaginary parts in Eq. (7.86b) are real and linearly independent particular solutions of the differential equation (7.82)1 it follows that the real general solution can be given as a linear combination of these particular solutions: W (x) = A1 cos βx + A2 sin βx + A3 cosh βx + A4 sinh βx ,

(7.87)

or 1 1 (cosh βx + cos βx) + A2 (sinh βx + sin βx) + 2 2 1 1 +A3 (cosh βx − cos βx) + A4 (sinh βx − sin βx) = 2 2 = A1 S(βx) + A2 T (βx) + A3 U (βx) + A4 V (βx)

W (x) = A1

where the functions

1 (cosh βx + cos βx) , 2 1 T (βx) = (sinh βx + sin βx) , 2 1 U (βx) = (cosh βx − cos βx) , 2 1 V (βx) = (sinh βx − sin βx) 2

(7.88a)

S(βx) =

(7.88b)

are called Krylov–Duncan functions [4–6]. We remark that A1 , A2 , A3 , and A4 are, in each case, different constants. They can be determined by utilizing the boundary conditions.

7.5 Flexural Vibrations of Beams

7.5.2.3

223

Properties of the Krylov–Duncan Functions

It can be checked by making paper-and-pencil calculations that the Krylov–Duncan functions and their derivatives at x = 0 form a unit matrix shown in Table 7.3: Table 7.3 The unit matrix that belongs to the Krylov–Duncan functions Krylov–Duncan functions and their derivatives at x = 0



d d2 d3

S(0) = 1 =0 S(βx) = 0 S(βx) =0 S(βx)

2 dx 2 3 dx 3 βdx β β x=0 x=0 x=0 T (0) = 0

d =1 T (βx)

βdx x=0

d2

T (βx) =0

β 2 dx 2 x=0

d3

T (βx) =0

β 3 dx 3 x=0

U (0) = 0

d =0 U (βx)

βdx x=0

d2

U (βx) =1

β 2 dx 2 x=0

d3

U (βx) =0

β 3 dx 3 x=0

V (0) = 0

d =0 V (βx)

βdx x=0

d2

V (βx) =0

β 2 dx 2 x=0

d3

V (βx) =1

β 3 dx 3 x=0

For the sake of completeness, the first four derivatives of the Krylov–Duncan functions are presented in Table 7.4. Table 7.4 Derivative rules Derivatives of the Krylov–Duncan functions S(βx)

S (1) (βx) = βV (βx)

T (βx)

S (2) (βx) = β 2 U (βx)

S (3) (βx) = β 3 T (βx)

S (4) (βx) = β 4 S(βx)

T (1) (βx) = β S(βx)

T (2) (βx) = β 2 V (βx)

T (3) (βx) = β 3 U (βx)

T (4) (βx) = β 4 T (βx)

U (βx) U (1) (βx) = βT (βx)

U (2) (βx) = β 2 S(βx)

U (3) (βx) = β 3 V (βx)

U (4) (βx) = β 4 U (βx)

V (βx) V (1) (βx) = βU (βx)

V (2) (βx) = β 2 T (βx)

V (3) (βx) = β 3 S(βx)

V (4) (βx) = β 4 V (βx)

Making use of the formulae detailed in Table 7.3, the integration constants A1 , A2 , A3 , and A4 in (7.88a) can all be given in terms of the initial parameters: A1 = W (βx)|x=0 ,

1 d2 W (βx)

A3 = 2 , β dx 2 x=0

1 dW (βx)

, β dx x=0

1 d3 W (βx)

A4 = 3 . β dx 3

A2 =

(7.89)

x=0

On the basis of Table 7.4, we can give the solution and its derivatives in an ordered form:

224

7 Systems with Infinite Degrees of Freedom

W (x) = A1 S(βx) + A2 T (βx) + A3 U (βx) + A4 V (βx) , dW (x) = β (A1 V (βx) + A2 S(βx) + A3 T (βx) + A4 U (βx)) , dx (7.90) d2 W (x) = β 2 (A1 U (βx) + A2 V (βx) + A3 S(βx) + A4 T (βx)) , 2 dx d3 W (x) = β 3 (A1 T (βx) + A2 U (βx) + A3 V (βx) + A4 S(βx)) . dx 3 For the sake of our later considerations, we give some useful relations here: T 2 (βx) − V 2 (βx) = sinh βx sin βx , 1 U (βx)T (βx) − V (βx)S(βx) = (cosh βx sin βx − cos βx sinh βx) , 2 1 U 2 (βx) − V (βx)T (βx) = (1 − cosh βx cos βx) , 2 1 S 2 (βx) − T (βx)V (βx) = (cosh βx cos βx + 1) . 2 7.5.2.4

(7.91)

Boundary Conditions

By utilizing the data in Table 7.2 and Eqs. (7.74)1,2 , we can give the common boundary conditions in terms of the deflection W : 1. For a fixed (clamped) end, the deflection and the rotation are equal to zero: W = 0,

ϕy = −

dW = 0. dx

(7.92a)

2. For a pinned endpoint, the deflection and the bending moment are equal to zero: W = 0,

M y = −I E

d2 W = 0. dx 2

(7.92b)

3. For a free end, the bending moment and the shear force are equal to zero: M y = −I E

d2 W = 0, dx 2

Vz = I E

d3 W = 0. dx 3

(7.92c)

We shall assume in accordance with Table 7.2 that x = 0 at the left end of the beam with length . Comparison of the boundary conditions considered at the left end of the beam, the first column in Table 7.3, and Eqs. (7.90) shows that two from the integration constants A1 , A2 , A3 , and A4 are always zero.

7.5 Flexural Vibrations of Beams

7.5.2.5

225

Eigenfrequencies of Simply Supported (Pinned-Pinned) Beams

By utilizing the boundary conditions (7.92a)1 , (7.92b)2 taken at the left end of the beam, Eqs. (7.90)1,3 , and the first column in Table 7.3, we get W (βx)|x=0 = A1 = 0 ,

d2 W (βx)

= β 2 A3 = 0, dx 2 x=0

(7.93)

which shows that the integration constants A1 and A3 are equal to zero. Hence W (βx) = A2 T (βx) + A4 V (βx)

(7.94)

is the solution. Boundary conditions (7.92a)1 , (7.92b)2 taken now at the left end of the beam yield two homogeneous linear equations for the integration constants A2 and A4 : W (βx)|x= = A2 T (β) + A4 V (β) = 0 ,

d W (βx)

= β 2 (A2 V (β) + A4 T (β)) = 0 dx 2 x= 2

or

A2 T (β) + A4 V (β) = 0 ,

(7.95)

A2 V (β) + A4 T (β) = 0 .

Solutions for A2 and A4 different from the trivial ones exist if and only if the determinant of this equation system vanishes:

T (β) V (β) 2 2

↑ = sinh β sin β = 0 .

V (β) T (β) = T (β) − V (β) =(7.91) 1

(7.96)

This equation is the frequency equation (or characteristic determinant). Its roots are given by βk  = kπ , k = 1, 2, 3, . . . (7.97) With βn Eq. (7.82)2 results in two formulae, one for the eigenvalue λ the other for the eigenfrequency ω: λk = (βk )4 =

kπ 



4 ,

ωk = (βk )2

IE = ρA4



kπ 

2 

IE . ρA

(7.98)

From now on the subscript k shows that the quantity considered belongs to the root βk .

226

7 Systems with Infinite Degrees of Freedom

Since determinant (7.96) is zero, Eqs. (7.95) are not independent. Consequently, the integration constants are also not independent. From Eq. (7.95)1 , we get A4k = −A2k

T (βk ) sinh βk x + sin βk x = −A2k = −A2k . V (βk ) sinh βk x − sin βk x

Hence W (βk x) = A2k (Tk (βk x) − Vk (βk x)) =

1 1 = A2k (sinh βk x + sin βk x) − (sinh βk x − sin βk x) = 2 2 kπ x = A2k Wk (x), = A2k sin βk x = A2k sin  where the function Wk (x) = sin

kπ x 

(7.99)

(7.100)

is the kth eigenfunction (or the kth normal mode) that belongs to the eigenvalue λk = (βk )4 . Remark 7.3 Note that eigenfunction (7.100) coincides with eigenfunction (7.47) which describes the kth normal mode of a vibrating string. Let us assume that the initial conditions are given by the following equations: w(x, 0) = w(x) ,

w(x, ˙ 0) = g(x),

(7.101)

where w(x) and g(x) are given functions. By repeating the line of thought presented in Exercise 7.3, it is easy to prove that w(x, t) =

∞ 

sin λk x (Ck sin ωk t + Dk cos ωk t)

(7.102)

k=1

is the total solution where 2 Dk = 

 0



2 w(x)Wk (x) dx and Ck = ωk





g(x)Wk (x) dx .

(7.103)

0

Table 7.5 contains the frequency equations, the eigenfunctions, and the values (formulae) of the product βk  for six support arrangements. Proofs have been given only for simply supported and fixed-pinned beams in Sect. 7.5.2.5 and Exercise 7.4. Proofs for the other four cases are left to Problem 7.4.

7.5 Flexural Vibrations of Beams

227

Table 7.5 Frequency equations, eigenfunctions and the values of βk  Beam with Frequency Eigenfunction its supports equation Wk (x)

Value of βk  k = 1, 2, . . .

Simply supported

sin βk  = 0

sin βk x

βk  = kπ

Fixed-pinned

tan βk  = tanh βk 

cosh βk x − cos βk x− −bk (sinh βk x − sin βk x) cosh βk  − cos βk  bk = sinh βk  − sin βk 

β1  = 3.92660 β2  = 7.06858 βk  ≈ π4 + kπ

Fixed-fixed

cosh βk  cos βk  = 1

cosh βk x − cos βk x− −bk (sinh βk x − sin βk x) cosh βk  − cos βk  bk = sinh βk  − sin βk 

β1  = 4.73004 β2  = 7.85321 βk  ≈ π2 +kπ

Fixed-free

cosh βk  cos βk  = −1 cosh βk x − cos βk x− −bk (sinh βk x − sin βk x) cosh βk  + cos βk  bk = sinh βk  + sin βk 

β1  = 1.87510 β2  = 4.69409 βk  ≈ (k −0.5)π

Pinned-free

tan βk  = tanh βk 

β1  = 3.92660 β2  = 7.06858 βk  ≈ π4 +kπ

sinh βk x + bk sin βk x sinh βk  bk = sin βk 

0 for rigid body rotation

Free-free

cosh βk  cos βk  = 1

cosh βk x + cos βk x− −bk (sinh βk x + sin βk x) cosh βk  − cos βk  bk = sinh βk  − sin βk 

β1  = 4.73014 β2  = 7.85321 βk  ≈ π2 +kπ 0 for rigid body motion

It is also worth to mention that the relative error of the approximate relations presented in the last column of Table 7.5 reaches its maximum for the fixed-free beam if k = 3:   (β3 )exact × 100 = δmax = 1 − (β3 )approximate



7.854557 7.854557 × 100 = 1 − × 100 = 0.0098% . (7.104) = 1− 2.5 × π 7.853982

228

7 Systems with Infinite Degrees of Freedom

Exercise 7.4 Find the eigenfrequencies of a beam with length  if the beam is fixed at the left end (x = 0) and pinned at the right end (x = )—see Table 7.2 which shows, among others, this beam as well. The solution is based on Sect. 7.5.2.5. Making use of the boundary conditions W (βx)|x=0 = A1 = 0 ,

dW (βx)

= βA2 = 0 , dx x=0

W (βx)|x= = A3 U (β) + A4 V (β) = 0 ,

d W (βx)

= β 2 (A3 S(β) + A4 T (β)) = 0, dx 2

(7.105a) (7.105b)

2

(7.105c)

x=

we obtain the following equation system for the non-zero integration constants A3 and A4 : A3 U (β) + A4 V (β) = 0 , (7.106) A3 S(β) + A4 T (β) = 0 . We can get non-zero solutions for the integration constants A3 and A4 if the frequency equation (characteristic determinant) is zero:

U (β) V (β)

↑ =

S(β) T (β) = U (β)T (β) − V (β)S(β) =(7.91) 2 =

1 (cosh β sin β − cos β sinh β) = 0 . 2

(7.107)

It follows from here that tan β = tanh β .

(7.108)

The roots of this equation are given by βk  ≈

π + kπ . 4

(7.109a)

For completeness, we list the first four roots here: β1  = 3.926602 , β2  = 7.068583 , β3  = 10.210176 , β4  = 13.351768 . (7.109b) According to Eq. (7.98), the kth eigenfrequency of the beam is  ωk = (βk )

2

IE . ρA

(7.110)

7.5 Flexural Vibrations of Beams

229

The integration constants A3 and A4 are not independent of each other. Equation (7.106)1 yields A4k = −A3k

U (βk ) cosh βk  − cos βk  = −A3k . V (βk ) sinh βk  − sin βk 

(7.111)

Consequently,

U (βk ) V (βk x) = W (βn x) = A3k U (βk x) − V (βk ) 1 cosh βk  − cos βk  cosh βk x − cos βk x − = A3k (sinh βk x − sin βk x)) = 2 sinh βk  − sin βk  = A3k Wk (x), (7.112) where U (βk ) V (βk x) = V (βk ) cosh βk  − cos βk  = cosh βk x − cos βk x − (sinh βk x − sin βk x) sinh βk  − sin βk 

Wk (x) = U (βk x) −

(7.113)

is the kth eigenfunction.

7.5.3 Orthogonality of the Eigenfunctions Let us introduce the notations y(x) = W (x), yk (x) = Wk (x) and the operators K (y) = y (4) (x) and M(y) = y. Further, let x = a = 0 and x = b =  be those x coordinates that belong to the left and right extremities of the beam. Making use of the notations introduced differential equation (7.82) can be rewritten in the following form: K [y(x)] = λM[y(x)] . (7.114) This differential equation is, in general, associated with the following conditions:  y(a) = 0 , y (2) (a) = 0 (simply supported beam), y(b) = 0 , y (2) (b) = 0  y(a) = 0 , y (1) (a) = 0 (fixed-pinned beam), y(b) = 0 , y (2) (b) = 0  y(a) = 0 , y (1) (a) = 0 (fixed-fixed beam), y(b) = 0 , y (1) (b) = 0

boundary

(7.115a) (7.115b) (7.115c)

230

7 Systems with Infinite Degrees of Freedom

y(a) = 0 , y (1) (a) = 0 (2) y (b) = 0 , y (3) (b) = 0 y(a) = 0 , y (2) (a) = 0 (2) y (b) = 0 , y (3) (b) = 0 y(a)(2) = 0 , y (3) (a) = 0 y (2) (b) = 0 , y (3) (b) = 0

 (fixed-free beam),

(7.115d)

(pinned-free beam),

(7.115e)

(free-free beam).

(7.115f)

 

Differential equation (7.114) and boundary conditions (7.115) constitute six boundary value problems for which both the differential equation and the boundary conditions are homogeneous, and λ is an unknown parameter. These problems are referred to, in accordance with all that has been said so far, as eigenvalue problems with lambda as the eigenvalue. The integrals  

b

yn (x)K [yk (x)] dx = (yn , yk ) K

a b

(7.116a) 

yn (x)M[yk (x)] dx = (yn , yk ) M

(7.116b)

a

define the products of the eigenfunctions yn (x) and yk (x) taken on the operator (K (y)) [M(y)]. As regards the product (yn , yk ) K , we may write

b  b dyn (x) d3 yk (x) d4 yk (x) d3 yk (x)

yn (x) dx = y (x) − dx = n dx 4 dx 3 a dx dx 3 a a

b

b  b 2 d yn (x) d2 yk (x) d3 yk (x)

dyn (x) d2 yk (x)

− + dx = = yn (x)

3 2 dx dx dx dx 2 dx 2 a a a      

 (yn , yk ) K =

b

=0

 = a

=0

b

d yn (x) d2 yk (x) dx , dx 2 dx 2 2

k, n = 1, 2, 3, . . . (7.117)

in which due to boundary conditions (7.115) the underbraced terms are equal to zero. Consequently, it holds that 

b

d2 yn (x) d2 yk (x) dx = dx 2 dx 2



d2 yk (x) d2 yn (x) dx = (yk , yn ) K . dx 2 dx 2 a a (7.118) Hence, the product (yn , yk ) K is a commutative operation: (yn , yk ) K =

b

(yn , yk ) K = (yk , yn ) K .

7.5 Flexural Vibrations of Beams

231

According to its definition  (yn , yk ) M =

b

yn (x)yk (x)dx = (yk , yn ) M ,

(7.119)

a

which shows that the product (yn , yk ) M is also a commutative operation: (yn , yk ) M = (yk , yn ) M . Since the eigenfunctions yk (x) and yn (x) satisfy differential equation (7.114), we may write d4 yn (x) d4 yk (x) = λ y (x) and = λn yn (x) . (7.120) k k dx 4 dx 4 Multiply the first equation by yn (x), the second by yk (x) then integrate the result over the interval [a, b]. With regard to definitions (7.116), we have (yn , yk ) K = λk (yn , yk ) M , (yk , yn ) K = λn (yk , yn ) M .

(7.121a) (7.121b)

Subtract now Eq. (7.121b) from Eq. (7.121a). In a view of the commutativity of the products (yn , yk ) K and (yn , yk ) M , we obtain  (λk − λn ) (yk , yn ) M = (λk − λn )

b

yn (x)yk (x)dx = 0,

(7.122)

a

where we have assumed that λk − λn = 0 if k = n. This means that 

b

yn (x)yk (x)dx = 0 ,

k = n .

(7.123)

a

If this equation is satisfied then we say—in accordance with what we said concerning Eq. (7.22)—that the functions yn (x) and yk (x) are orthogonal to each other. We define the norm of the eigenfunction yk (x) by the following equation: 



|yk (x)| =

b

yk (x)yk (x) dx .

(7.124)

a

Comparison of Eqs. (7.122) and (7.124) yields 

b

 yn (x)yk (x)dx =

a

|yk (x)|2 if k = n, 0 if k = n.

Equation ψk (x) =

yk (x) |yk (x)|

k, n = 1, 2, 3 . . .

(7.125)

(7.126)

232

7 Systems with Infinite Degrees of Freedom

defines the normalized eigenfunction ψk (x). Since ψk (x) is normalized, it holds that 

b

ψk (x)ψk (x) dx = 1 .

(7.127)

a

Hence 

b

 ψn (x)ψk (x)dx =

a

1 if k = n, 0 if k = n.

k, n = 1, 2, 3 . . .

(7.128)

It follows from Eqs. (7.117) and (7.119) that (yn , yn ) K > 0 ,

(yn , yn ) M > 0 .

(7.129)

If we make k and n equal in (7.121), we get (yn , yn ) K = λn (yn , yn ) M . Consequently λn =

(yn , yn ) K > 0, (yn , yn ) M

(7.130)

(7.131)

which shows that the non-zero eigenvalues are all positive quantities.

7.5.4 Forced Vibration of Beams If the beam is subjected to a time-dependent transverse load and vibrates—we assume that the axial force in the beam is zero—then the total load consists of two parts, the first is the force of inertia the second is the time-dependent transverse load: f y = −ρA

∂2w + f y (x, t) . ∂t 2

(7.132)

If we substitute f y into equilibrium equation (7.75a), we get the equation of motion for the case of forced vibrations:

∂2 ∂2w ∂2w I E + ρA = f y (x, t) . (7.133a) ∂x 2 ∂x 2 ∂t 2 For a uniform beam, this equation is a bit simpler: IE

∂2w ∂4w + ρA = f y (x, t) . ∂x 4 ∂t 2

(7.133b)

7.5 Flexural Vibrations of Beams

233

When seeking solution to this problem, we shall utilize some results presented in Sect. 5.4.2 which is devoted to a similar problem, i.e., to the forced vibrations of systems with finite degrees of freedom. Since the set of normalized eigenfunctions ψk (x) is complete in the Infinitedimensional function space, we shall assume that the solution is of the form w(x, t) = q1 (t)ψ1 (x) + q2 (t)ψ2 (x) + q3 (t)ψ3 (x) + · · · = =

∞ 

qk (t)ψk (x),

(7.134)

k=1

where the functions qk (t) are the unknowns. Substituting solution (7.134) into Eq. (7.133b), we get IE

∞  d4 ψk (x)

dx 4

k=1

qk (t) + ρA

∞ 

ψk (x)

k=1

d2 qk (t) = f y (x, t) . dt 2

(7.135)

Since the eigenfunctions ψk (x) satisfy Eq. (7.82), it also holds that d4 ψk (x) − λk ψk (x) = 0 , dx 4

λk =

ρAωk2 . IE

(7.136)

We can rewrite Eq. (7.135) if we take these relations into account: ∞ 

ωk2 ψk (x)qk (t) +

k=1

∞ 

ψk (x)

k=1

f y (x, t) d2 qk (t) . = 2 dt ρA

(7.137)

Multiply this equation by ψn (x) and integrate the result over the interval [a = 0, b = ]. We get ∞  k=1

 ωk2

b

ψn (x)ψk (x)dx qk (t)+

a

∞   k=1



b

a b

= a

ψn (x)ψk (x)dx

d2 qk (t) = dt 2

f y (x, t) ψn (x) dx . ρA

(7.138)

f y (x, t) ψn (x) dx ρA

(7.139)

Let us introduce the notations ωn2

= λˆ n ,

 Q n (t) = a

b

and take relations (7.128) into account. Then, we obtain a differential equation for the function qn (t): (7.140) q¨ n (t) + λˆ n qn (t) = Q n (t),

234

7 Systems with Infinite Degrees of Freedom

where Q n (t) is the generalized force that belongs to the function qn . This equation coincides formally with Eq. (5.109) we derived for vibrating systems with finite degrees of freedom. Hence, its solution is given by Eq. (5.114b) which can be rewritten in the following form: 

qn =an cos λˆ n t + bn sin   transient function

   t 1 ˆλn t + Q n (τ )  sin λˆ n (t − τ ) dτ =  τ =0 λˆ n   

= an cos ωn t + bn sin ωn t +

1 ωn



particular solution t

τ =0

Q n (τ ) sin ωn (t − τ ) dτ .

(7.141)

Here, an and bn are unknown integration constants. They can be determined by using the initial conditions. Exercise 7.5 The beam shown in Fig. 7.10 is subjected to a harmonic excitation force (7.142) f z (x, t) = F1 δ(x − 1 ) sin ω f t, where δ(1 − x) is the Dirac function. Find the particular solution, i.e., the steadystate vibrations of the beam. Fig. 7.10 Beam subjected to harmonic excitation

f z x, t

z

F1 x

1

cos

f

t x

1

In view of (7.100), it is not too difficult to check that the normalized eigenfunctions for a simply supported beam are given by the following equation:  ψn (x) =

nπ 2 sin x.  

(7.143)

Making use of Eq. (7.139), we can determine the generalized force Q n (t) for the function qn :  Q n (t) =

b=

a=0

f y (x, t) F1 ψn (x) dx = ρA ρA

It follows from (7.141) that

   2 nπ x, dx = δ(x −1 ) cos ω f t sin  0   F1 2 nπ1 = (7.144) sin cos ω f t . ρA  

7.5 Flexural Vibrations of Beams

qn part =

where

1 ωn

1 ωn





t

τ =0

t

235

Q n (τ ) sin ωn (t − τ ) dτ =   t nπ1 1 F1 2 sin = cos ω f τ sin ωn (t − τ ) dτ , ρA   ωn 0

cos ω f τ sin ωn (t − τ ) dτ =

0

cos ω f t cos ωn t ω f + 2 . 2 2 ωn − ω f ωn − ω 2f ωn

Note that the second term can be dropped since that is a solution to the homogeneous part of Eq. (7.140). Hence qn part

F1 = ρA



nπ1 cos ω f t 2 sin .   ωn2 − ω 2f

It is worth mentioning that the same result can be obtained for qn part if we consider an undamped single degree of freedom system subjected to the harmonic excitation (7.144). With qn part , we may give the total solution: ∞

2F1  1 nπ1 nπx w(x, t) = sin sin cos ω f t . ρA n=1 ωn2 − ω 2f  

7.5.5 Vibration of Axially Loaded Beams Assume that the axial force N is constant (independent of x). Then, with regard to (7.77), the equation of motion follows from Eq. (7.75b): IE

∂2w ∂2w ∂4w ∓ N 2 + ρA 2 = 0, 4 ∂x ∂x ∂t

(7.145)

where N is now the magnitude of the axial force. Hence, the sign of N in the above equation is (negative) [positive] if the axial force is (tensile) [compressive] force. It is obvious that the solution has the same mathematical form as that of Eq. (7.78) which is given by Eq. (7.80). Thus c2 W (x)



d4 W (x) N d2 W (x) ∓ 4 dx I E dx 2

from where we have

=−

1 d2 γ(t) IE , (7.146) = ω˜ 2 , c2 = γ(t) dt 2 ρA

236

7 Systems with Infinite Degrees of Freedom

2 d4 W (x) N d2 W (x) ω˜ ∓ = W (x) = λW (x) , 4 2 dx I E dx c d2 γ(t) ρAω˜ 2 + ω˜ 2 γ(t) = 0 . λ= = β4 ; IE dt 2

(7.147)

We seek a particular solution to Eq. (7.147)1 in the form W (x) = A eχx .

(7.148)

After substituting it into Eq. (7.147)1 , we get χ4 ∓

N 2 ρAω˜ 2 χ − = 0. IE IE

(7.149)

If the axial force is compressive, the roots for χ are

χ1 = β1c

  

 N N 2 ρAω˜ 2  + , = − + 2I E 2I E IE

χ2 = −β1c

(7.150a)

χ3 = −iβ2c .

(7.150b)

and

χ3 = iβ2c

  

 N N 2 ρAω˜ 2  + , =i + 2I E 2I E IE

If the axial force is tensile, the roots for χ are different:

χ1 = β1t = β2c

  

 N N 2 ρAω˜ 2  + , = + 2I E 2I E IE

χ2 = −β2c

(7.151a)

and    



2

ρAω˜ 2 , χ4 = −iβ2t = −iβ1c . IE (7.151b) We can now express the solution for W both for the case of χ3 = iβ2t = iβ1c = i

N + − 2I E

N 2I E

+

(a) compressive axial force: W (x) = A1 cosh β1c x + A2 sinh β1c x + A3 cos β2c x + A4 sin β2c x , (7.152a)

7.5 Flexural Vibrations of Beams

237

(a) tensile axial force: W (x) =A1 cosh β1t x + A2 sinh β1t x + A3 cos β2t x + A4 sin β2t x = =A1 cosh β2c x + A2 sinh β2c x + A3 cos β1c x + A4 sin β1c x . (7.152b) The simply supported uniform beam shown in Fig. 7.11 is subjected to an axial force N . In the sequel, it is our aim to find a relation between the eigenfrequencies ω˜ k and the axial force N [7, 8]. The boundary conditions the solution for W should satisfy are given in Table 7.2 where relation (7.74) between M y and W should also be taken into account: W N

x

Fig. 7.11 Simply supported beam and the axial force acting on it

d2 W

= 0, dx 2 x=0

d2 W

M y () = −I E = 0. dx 2

W (0) = 0 ,

M y (0) = −I E

W () = 0 ,

(7.153a) (7.153b)

x=

First, we shall consider the case of the compressive axial force. After substituting solution (7.152a) into boundary conditions (7.153a), we have W (0) = A1 + A3 = 0 ,

d W

2 2 = β1c A1 − β2c A4 = 0 . dx 2 2

x=0

Since the determinant of this homogeneous linear equation system is not zero, it follows that A1 = A3 = 0 . (7.154) Boundary conditions (7.153b) result in the following equations: W () = A2 sinh β1c  + A4 sin β2c  = 0 ,

d W

2 2 = β1c A2 sinh β1c  − β2c A3 sin β2c  = 0 . dx 2 2

x=

Solutions for A2 and A4 different from the trivial ones can be obtained from this equation system if (7.155a) sinh β1c  sin β2c  = 0 .

238

7 Systems with Infinite Degrees of Freedom

If the axial force is a tensile one, a similar line of thought leads to the following characteristic determinant: sinh β2c  sin β1c  = 0 .

(7.155b)

Since (sinh β2c ) [sinh β1c ] is not zero if (β2c  = 0) [β1c = 0] it follows from Eqs. (7.155) that sinh β2c  = 0 . (7.156) sin β1c  = 0 , Hence, β1c  = kπ ,

β2c  = kπ

k = 1, 2, 3, . . .

(7.157)

After substituting β1c and β2c and squaring the result, we get  N + − 2I E or



N 2I E

2

2 ρAω˜ 2k kπ = + , IE 



N 2I E

2 +

 N + 2I E

ρAω˜ 2k = IE



kπ 



N 2I E

2 ±

2

2 ρAω˜ 2k kπ = + , IE 

N , 2I E

where the last sign on the right side is (positive) [negative] if N is (tensile) [compressive]. Square this equation and divide it by (kπ/)4 . We obtain ω˜ 2k N 4 = 1 ± 2 . I E kπ kπ EI ρA  

(7.158)

It is worth rewriting this equation by taking the following facts into account: (a) According to Eq. (7.98), the square of the kth eigenfrequency of the unloaded and simply supported beam is ωk2 =

IE ρA



kπ 

4 .

(7.159a)

(b) As regards the stability problem of the simply supported beam, the kth critical compressive force (buckling load) is given by the following equation: Nk crit = E I

kπ 

2 .

(7.159b)

(c) The strains due to the axial force and its critical values on the longitudinal axis of the beam can be calculated as

7.5 Flexural Vibrations of Beams

239

εx =

N ; AE

εx k crit =

Nk crit . AE

(7.159c)

Making use of Eqs. (7.159), we may rewrite Eq. (7.158) in the following form: ω˜ 2k N εx =1± =1± . Nk crit εx k crit ωk2

(7.160)

Remark 7.4 If I E tends to zero and the last sign is positive in Eq. (7.158)—then the axial force is tensile—we have  kπ N . (7.161) ω˜ k =  ρA This result coincides with the eigenfrequencies of a taut string—compare Eqs. (7.161) and (7.46b)2 . Remark 7.5 Equation (7.160) shows that the square of the eigenfrequencies of the simply supported beam subjected to an axial force is a linear function of the force. Figure 7.12 depicts the quotient ω˜ 21 /ω12 against the quotient N /N1 crit . It is clear from the figure that ω˜ 21 = ω12 if N = 0. If the force is compressive, then ω˜ 21 = 0 for N = N1 crit . 2 1 2 1

N is tensile 2.0

1.5

1.0

N is compressive 0.5

0.0

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

N N 1 crit

Fig. 7.12 The effect of the axial force on the first circular frequencies

It is also an important question what happens to the simply supported beam if the compressive force is greater than N1 crit . In this respect, we refer the reader to book [1].

240

7 Systems with Infinite Degrees of Freedom

7.6 Problems Problem 7.1 The uniform rod shown in Fig. 7.13 is fixed at its ends. The rod vibrates longitudinally. Assume that u(x, 0) = u(x) and u(x, ˙ 0) = g(x) are the initial conditions. Determine the eigenfunctions, the natural frequencies, and the function that describes the motion. z x

x = a =0

x=b =

Fig. 7.13 Uniform rod fixed at its ends

Problem 7.2 The uniform rod shown in Fig. 7.14 is free at its ends. We assume that the rod vibrates longitudinally. Determine the eigenfunctions and the natural frequencies. What is the solution if the rod is subjected to the same tensile forces No at its ends (the rod is in equilibrium) and the forces are suddenly removed. z x

x = a =0

x=b =

Fig. 7.14 Uniform rod free at its ends

Problem 7.3 Prove that ω 2 (or λ) is a positive quantity in Eq. (7.82)2 . Problem 7.4 Derive the frequency equation, the eigenfunctions, and the eigenvalues (or the products βk ) for fixed-fixed, fixed-free, pinned-free, and free-free beams. Problem 7.5 Given the flexibility matrix of the simply supported uniform beam shown in Fig. 7.15, ⎤ ⎡ ⎤ f 11 f 12 f 13 12.5 19.5 8.5 3  ⎣ 19.5 40.5 19.5 ⎦ . f = ⎣ f 21 f 22 f 23 ⎦ = (3×3) 1944I E f 31 f 32 f 33 8.5 19.5 12.5 ⎡

Let , I , E, A, and ρ be the length of the beam, the area moment of inertia, the modulus of elasticity, the cross-sectional area, and the density. Determine the stiffness matrix

7.6 Problems

241

and then the natural frequencies provided that m 1 = m 2 = m 3 = m/3 = ρA/3 where m is the total mass of the beam and compare the results obtained with those valid for a continuous beam. q

q1

m1

q2

m2

q3

m3

x /6

P1

/6

P2

/6

/6

/6

P3

/6

Fig. 7.15 Beam with three concentrated messes

References 1. L.N. Virgin, Vibration of Axially Loaded Structures (Cambridge University Press, Cambridge, 2007) 2. A. Bokaian, Natural frequencies of beams under compressive axial loads. J. Sound Vib. 126, 49–65 (1988). https://doi.org/10.1016/0022-460X(88)90397-5 3. S.P. Timoshenko, D.H. Young, Elements of Strength of Materials (Van Nostrand Reinhold, New York, 1977) 4. A.N. Krylov, Vibration of Ships (in Russian) (GI Red Sudistroit Lit., Moscow, 1936) 5. W.J. Duncan, Free and forced oscillations of continuous beams treatment by the admittance method. Lond. Edinb. Dublin Philos. Mag. J. Sci. Ser. 7 34(228), 49–63 (1943). https://doi.org/ 10.1080/14786444308521329 6. W.J. Duncan, Mechanical Admittances and their Applications to Oscillation Problems (H.M.S.O, London, 1947) 7. A.E. Galef, Bending frequencies of compressed beams. J. Accustical Soc. Am. 44, 643 (1968). https://doi.org/10.1121/1.1911144.339 8. V. Birman, I. Elishakoff, J. Singer, On the effect of axial compression on the bounds of simple harmonic motion. Isr. J. Technol. 20, 254–258 (1982)

Chapter 8

Eigenvalue Problems of Ordinary Differential Equations

8.1 Differential Equations and Boundary Conditions Consider the ordinary differential equation1 K [y] = λ M [y] ,

(8.1a)

where y(x) is the unknown function and λ is a parameter (the eigenvalue sought). The differential operators K [y] and M [y] are defined by the following relationships: K [y] =

κ   (n) (−1)n f n (x)y (n) (x) , n=0

μ   (n) M [y] = (−1)n gn (x)y (n) (x) ,

dn (. . .) = (. . .)(n) ; dx n (8.1b) κ>μ≥1

n=0

in which the real function ( f ν (x)) [gν (x)] is differentiable continuously (κ) [μ] times and gμ (x) = 0 if x ∈ [a, b] . (8.1c) f κ (x) = 0 , Note that 2κ, which is the order of the differential operator on the left side of (8.1a), is greater than 2μ, which is the order of the differential operator on the right side. Let x ∈ [a, b], a > b, b − a =  be the interval in which we seek the solution of differential equation (8.1). We shall assume that differential equation (8.1) is associated with 2κ homogeneous boundary conditions of the form

1 The

line of thought in section Eigenvalue problems of ordinary differential equations is based mainly on the book [1] by Lothar Collatz (1910–1990). © Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_8

243

244

8 Eigenvalue Problems of Ordinary Differential Equations

Ur [y] =

2κ    αnr y n−1 (a) + βnr y n−1 (b) = 0 ,

r = 1, 2, . . . , 2κ,

(8.2)

n=1

where αnr and βnr are real constants. Remark 8.1 Differential equation (8.1) and boundary conditions (8.2) constitute an eigenvalue problem in which λ is the eigenvalue to be determined and the solution that belongs to λ is called eigenfunction. If μ = 0 the right side of differential equation (8.1) is (8.3) M [y] = λg0 (x)y(x) and then the eigenvalue problem is called simple. The eigenvalue problems that provide the eigenfrequencies for the longitudinal and torsional vibrations of rods as well as for the transverse vibrations of strings and beams—we remind the reader of Remark 7.2 and Sect. 7.5.3 here—are all simple ones. If μ > 0 we speak about generalized eigenvalue problem. Remark 8.2 The boundary conditions should be linearly independent of each other. It is obvious that a linear combination of the boundary conditions is also a boundary condition. By selecting suitable linear combinations, derivatives with an order higher than κ − 1 can be removed from some boundary conditions. If they are removed in every possible way we may have, altogether, say, e boundary conditions which do not involve derivatives higher than κ − 1. These boundary conditions are called essential boundary conditions. The further 2k − e boundary conditions are the natural boundary conditions [2]. The functions u(x) and v(x) (u(x), v(x) are not identically equal to zero if x ∈ [a, b]) are called (a) admissible functions if they satisfy the essential boundary conditions, (b) comparison functions if they satisfy both the essential and the natural boundary conditions, and (c) eigenfunctions if they satisfy differential equation (8.1) and boundary conditions (8.2).

8.2 Self-adjointness The integrals  (u, v) K = a

b

 u(x) K [v(x)] dx ,

(u, v) M =

b

u(x) M[v(x)] dx

(8.4)

a

taken on the set of the comparison functions u(x), v(x) are products defined on the operators K and M. Remark 8.3 For the sake of comparison, it is worth recalling a similar definition set up for vibrating systems with finite degrees of freedom by Eq. (5.13). It will turn out

8.2 Self-adjointness

245

that products (8.4) play such a role in the sequel which resembles that of the products (5.13) in the algebraic eigenvalue problems of oscillating systems with finite degree of freedom. Let us now detail the product (u, v) K . Making use of (8.1b), we may write  (u, v) K =

b

u(x) a

κ   (n) (−1)n f n (x)v (n) (x) dx

(8.5)

n=0

in which the integral  Iuv = (−1)n

b

 (n) u(x) f n (x)v (n) (x) dx

a

can be manipulated into a more suitable form if we perform partial integrations. We get  (n−1) b  Iuv = (−1)n u(x) f n (x) v (n) (x) +  + (−1)n−1 (1)

− u(x)



b



(n−1)

u(x) f n (x)v (n) (x)

a

(n−2)

f n (x) v (n) (x)

a

  (n−1) dx = (−1)n u(x) f n (x) v (n) (x) −

b  (n−3) + u(x)(2) f n (x) v (n) (x) − ··· + a

 +

b

n−1   (n−1−r ) b  (n) (n) (n+r ) (r ) (n) f n (x) v (x) u (x) f n (x)v (x) dx = (−1) u(x) +

a

r =0

a



b

+

u

(n)

(x) f n (x)v

(n)

(x) dx.

(8.6)

a

Hence

κ n−1 b   (n−1−r ) (n+r ) (r ) (n) f n (x) v (x) (−1) u(x) + (u, v) K = n=0 r =0

+

κ  

a b

u (n) (x) f n (x)v (n) (x) dx .

(8.7a)

a

n=0

It follows from Eq. (8.7a) that (u, v) M

μ n−1 b   (n−1−r ) (n+r ) (r ) (n) = (−1) u(x) gn (x) v (x) + n=0 r =0

+

μ   n=0

a

a b

u (n) (x)gn (x)v (n) (x) dx .

(8.7b)

246

8 Eigenvalue Problems of Ordinary Differential Equations

These results are naturally valid for the products (v, u) K and (v, u) M if we write u for v and conversely v for u. Eigenvalue problem (8.1), (8.2) is said to be self-adjoint if the products (8.4) are commutative, i.e., it holds that (u, v) K = (v, u) K

and

(u, v) M = (v, u) M .

(8.8)

Conditions (8.8) are called conditions of self-adjointness. It follows from Eqs. (8.7a) and (8.7b) that

(u, v) K − (v, u) K =

κ  n−1   (n−1−r ) (−1)(n+r ) u(x)(r ) f n (x) v (n) (x) − n=0 r =0

 (n−1−r ) b −v(x)(r ) f n (x) u (n) (x) =0 a

(8.9a)

and (u, v) M − (v, u) M

μ n−1   (n−1−r ) = (−1)(n+r ) u(x)(r ) gn (x) v (n) (x) − n=0 r =0

 (n−1−r ) b −v(x)(r ) gn (x) u (n) (x) =0 a

(8.9b)

if the eigenvalue problem (8.1), (8.2) is self-adjoint. Remark 8.4 According to Eqs. (8.9), self-adjointness of eigenvalue problem (8.1), (8.2) depends on what value the solutions have on the boundary of the interval [a, b], i.e., on the boundary conditions. Remark 8.5 The three eigenvalue problems defined by differential equation (7.65) and boundary conditions (7.66) are all self-adjoint. The same is valid for the six eigenvalue problems defined by differential equation (7.114) and boundary conditions (7.115).

8.3 Orthogonality of the Eigenfunctions in General Sense Let us assume that the eigenvalue problem (8.1), (8.2) is self-adjoint. Further, let yk and y be two different solutions to the eigenvalue problem (two linearly independent eigenfunctions) which, therefore, satisfy the differential equations K [yk ] = λk M[yk ] , and boundary conditions

K [y ] = λ M [y ]

(8.10)

8.3 Orthogonality of the Eigenfunctions in General Sense

Ur [yk ] =

247

2κ    αnr ykn−1 (a) + βnr ykn−1 (b) = 0 , n=1

2κ    αnr yn−1 (a) + βnr yn−1 (b) = 0, Ur [y ] =

r = 1, 2, . . . , 2κ

(8.11)

n=1

where λk and λ denote the eigenvalues that belong to the eigenfunctions yk and y . We shall assume that λk = λ . Multiply Eq. (8.10)1 by y (x), Eq. (8.10)b by yk (x) then integrate the result over the interval [a, b]. With regard to definitions (8.4), we get (y , yk ) K = λk (y , yk ) M ,

(yk , y ) K = λ (yk , y ) M .

(8.12)

Consider now the difference (8.12)1 –(8.12)2 . Since we have assumed that the eigenvalue problem (8.1), (8.2) is self-adjoint the products (y , yk ) K and (y , yk ) M are commutative. Hence, the difference is (λk − λ ) (yk , y ) M = 0,

(8.13)

where λk − λ = 0. This means that (yk , y ) M = 0

k =  .

(8.14a)

In view of this equation, it follows from (8.12) that (yk , y ) K = 0

k =  .

(8.14b)

If Eqs. (8.14) are satisfied, then the functions yk (x) and y (x) are orthogonal to each other in general sense. For simple eigenvalue problems M[y(x)] = g0 (x) y(x) .

(8.15)

Consequently, the orthogonality condition assumes the form  (yk , y ) M =

b

yk (x) g0 (x) y (x) dx = 0 .

(8.16)

a

√ √ For g0 (x) > 0, x ∈ [a, b], the functions g0 (x)yk (x) and g0 (x)y (x) are orthogonal in the traditional sense. In view of Eq. (8.12), we may write on the basis of (8.14) that (yk , y ) K =

λn (yk , y ) M if k = , 0 if k = .

k,  = 1, 2, 3, . . .

(8.17)

248

8 Eigenvalue Problems of Ordinary Differential Equations

Remark 8.6 Note that the line of thought in this subsection is basically the same as in Sect. 7.5.3.

8.4 On the Reality of the Eigenvalues We shall prove that the eigenvalues of the self-adjoint eigenvalue problems are all real numbers. We assume that the eigenvalue λ is complex, i.e., it can be given in the form λ = a + ib—a and b are real numbers—and we shall show that this assumption leads to a contradiction. Let y (x) be the eigenfunction that belongs to the eigenvalue λ . It is clear that λ and y (x) satisfy the differential equation K [y (x)] = λ M[y (x)] .

(8.18)

If y (x) is a real function, the left side of this equation is also real, whereas the right side is complex since λ is complex. This means that the eigenfunction which belongs to a complex eigenvalue should also be complex. Hence, y (x) can be given in the form (8.19) y (x) = u  (x) + iv (x), where u  (x), v (x) are real functions and i is the imaginary unit. Let us multiply Eq. (8.18) by the complex conjugate of the eigenfunction y (x); then, integrate the result in the interval [a, b]. We get ( y¯k , y ) K = λ ( y¯k , y ) M .

(8.20)

If we express the left side of this equation in terms of u  (x) and v (x), we obtain ( y¯k , y ) K = (u  (x) − iv (x), u  (x) + iv (x)) K =   (u  (x), u  (x)) K + (u  (x), v (x)) K − (v (x), u  (x)) K i + (v (x), v (x)) K = = (u  (x), u  (x)) K + (v (x), v (x)) K , (8.21) where, because of the commutativity of the product (u  (x), v (x)) K , the coefficient of the imaginary unit is zero. This result shows that the left side of Eq. (8.20) is not complex but a real number. As regards the product ( y¯k , y ) M , we may write by substituting M for K in the above equation that ( y¯k , y ) M = (u  (x), u  (x)) M + (v (x), v (x)) M ,

(8.22)

which is also a real number. Consequently, we can rewrite (8.20) in the following form:

8.4 On the Reality of the Eigenvalues

249

  (u  (x), u  (x)) K + (v (x), v (x)) K = a (u  (x), u  (x)) M + (v (x), v (x)) M +   + ib (u  (x), u  (x)) M + (v (x), v (x)) M , (8.23) where the left side is real, whereas the right side is complex which is, therefore, a contradiction. For   (u  (x), u  (x)) M + (v (x), v (x)) M = 0, the equality in (8.23) is possible only if b = 0, i.e., if the eigenvalue λ is a real number. We can prove the reality of the eigenvalues more formally if we take into account that y¯ and λ¯  satisfies Eq. (8.18): K [ y¯ (x)] = λ¯  M[ y¯ (x)] .

(8.24)

If we multiply (8.24) by y (x), integrate the result in the interval [a, b] and follow the steps leading to (8.23), we get   (u  (x), u  (x)) K + (v (x), v (x)) K = a (u  (x), u  (x)) M + (v (x), v (x)) M −   − ib (u  (x), u  (x)) M + (v (x), v (x)) M . (8.25) Let us now consider the difference of Eqs. (8.23) and (8.25). We have   2ib (u  (x), u  (x)) M + (v (x), v (x)) M = 0

(8.26)

  from where it follows that b = 0 if (u  (x), u  (x)) M + (v (x), v (x)) M = 0.

8.5 Boundary Expressions For the comparison function u(x) = v(x), we may rewrite (8.7a) and (8.7b) in the following forms: (u, u) K =

κ  

(u, u) M =

μ   n=0

where

u (n) (x) f n (x)u (n) (x) dx + K 0 [u(x)]

(8.27a)

u (n) (x)gn (x)u (n) (x) dx + M0 [u(x)],

(8.27b)

a

n=0

and

b

a

b

250

8 Eigenvalue Problems of Ordinary Differential Equations

κ n−1 b   (n−1−r ) (n+r ) (r ) (n) f n (x) u (x) K 0 [u(x)] = (−1) u(x) n=0 r =0

(8.28a)

a

and

μ n−1 b  (n−1−r )  (n+r ) (r ) (n) (−1) u(x) gn (x) u (x) . M0 [u(x)] = n=0 r =0

(8.28b)

a

Here K 0 [u(x)] and M0 [u(x)] are the Dirichlet boundary expressions [3].

8.6 Sign of Eigenvalues The eigenvalue problem (8.1), (8.2) is said to be positive definite if the eigenvalues are positive, positive semidefinite if one eigenvalue is zero and the other eigenvalues are all positive, negative semidefinite if one eigenvalue is zero and the other eigenvalues are all negative, and finally negative definite if the eigenvalues are negative. If we equalize  and k in (8.12), we get (y , y ) K = λk (y , y ) M . Hence

 λk =

b

(y , y ) K = ab (y , y ) M

y (x) K [y (x)] dx

(8.29)

.

(8.30)

y (x) M[y (x)] dx

a

This equation shows that the sign of λ is a function of the products (y , y ) K and (y , y ) M . Assume that (8.31) (u, u) K > 0 , and (u, u) M > 0 for any comparison function u(x). Then the eigenvalue problem (8.1), (8.2) is positive definite (or full definite). The Rayleigh quotient is defined by the equation 

b

(u, u) K = ab R[u(x)] = (u, u) M a

u(x) K [u(x)] dx (8.32) u(x) M[u(x)] dx

8.6 Sign of Eigenvalues

251

in which u(x) is a comparison function. Upon substitution of (8.27a) and (8.27b) for the numerator and denominator in (8.30), we get κ  

R[u(x)] =

(u, u) K n=0 = μ   (u, u) M n=0

b

u (n) (x) f n (x)u (n) (x) dx + K 0 [u(x)]

a

.

b

u

(n)

(x)gn (x)u

(n)

(8.33)

(x) dx + M0 [u(x)]

a

If the eigenvalue problem (8.1), (8.2) is self-adjoint, then K 0 [u(x)] = M0 [u(x)] = 0 .

(8.34)

Assume that the eigenvalue problem (8.1), (8.2) is self-adjoint and M[y] = (−1)μ [gμ (x)y (μ) ](μ) .

(8.35)

Then the eigenvalue problem belongs to the single term class of eigenvalue problems [3]. If μ = 0 in (8.35), the considered self-adjoint eigenvalue problem is simple.

8.7 Determination of Eigenvalues Let us denote the linearly independent particular solutions of the differential equation K [y] = λ M [y] by z  (x, λ) ( = 1, 2, . . . , 2κ). With z  (x, λ), the general solution is of the form 2κ  y(x) = A z  (x, λ), (8.36) =1

where the undetermined integration constants A can be obtained from the boundary conditions: 2κ  Ur [z  (x, λ)]A = 0 , r = 1, 2, . . . , 2κ . (8.37) =1

Since these equations constitute a homogeneous linear equation system for the unknowns A solutions different from the trivial one (nontrivial solutions) exist if and only if the determinant of the system is zero:   (λ) = det Ur [z  (x, λ)] = 0 .

(8.38)

This is the equation which should be solved for finding the eigenvalues λ. We remark that (λ) is often referred to as characteristic determinant.

252

8 Eigenvalue Problems of Ordinary Differential Equations

In the sequel, we shortly outline, without entering deeply into details, some properties the eigenvalues have. Assume that the coefficients of the differential equation considered and the initial conditions—if there are initial conditions at all—are continuous and differentiable functions of the parameter λ. Then the solution is an analytical function of λ—see Sect. 5.7 in [1] or [4] for further details. Differential equation (8.10) obviously meets the above conditions: (a) it is a linear function of λ and (b) there are no initial conditions. It then follows that z  (x, λ) is an analytical and single-valued (regular) function of λ, i.e., an entire function of λ for all fixed x ∈ [a, b]. Hence, it holds that: If (λ) is identically equal to zero then each λ is an eigenvalue. Otherwise, function (λ) has an infinite sequence of isolated zero points which can be ordered according to their magnitudes: 0 ≤ |λ1 | ≤ |λ2 | ≤ |λ3 | ≤ · · ·

(8.39)

Exercise 8.1 Consider the eigenvalue problem − y (2) (x) = λy(x) , y(0) − y(1) = 0 ,

y (1) (0) + y (1) (1) = 0 .

Find the eigenvalues λ [1]. Making use of the notation k 2 = λ, we can give the general solution of this differential equation in the form y(x) = A1 cos kx + A2 sin kx in which A1 and A2 are undetermined integration constants. Consequently

1 − cos k   − sin k

= k 1 − sin2 k + cos2 k = 0 (λ) =

−k sin k k (1 + cos k) is the characteristic determinant. This equation shows that any λ = k 2 is an eigenvalue. Exercise 8.2 The transverse vibrations of a free-free beam are described, according to (7.82), by the differential equation ρAω 2 d4 W (x) = β4, = λW (x) , λ = dx 4 IE

(8.40)

which is associated with the following boundary conditions:

W (2) (x) x=0 = 0 ,

W (2) (x) x= = 0 ,

W (3) (x) x=0 = 0 ,

W (3) (x) x= = 0 .

(8.41)

8.7 Determination of Eigenvalues

253

Find the eigenvalues λ. Since W (x) = constant is such a solution for which (λ) is identically equal to zero, it follows that λ = 0 is an eigenvalue. The non-constant solutions are given by (7.87), and it is shown in Sect. C.7—see the solution to Problem 7.4—that   k = 1, 2, 3, . . . (λ) = cosh βk  cos βk  − 1 = cosh 4 λk  cos 4 λk  − 1 = 0 , is the characteristic determinant from where we get  4

λ1  = 4.730141,

 4

λ2  = 7.853205,

 4

π + kπ, k = 3, 4, 5, . . . 2

λk  ≈

8.8 The Green Functions Consider the inhomogeneous ordinary differential equation L[y(x)] = r (x),

(8.42a)

where the differential operator of order κ is defined by the following equation: L[y(x)] =

κ 

pn (x) y (n) (x) .

(8.42b)

n=0

Here, κ ≥ 1 is a natural number, and the functions pn (x) and r (x) are continuous if x ∈ [a, b] (b > a, b − a = ) and pκ (x) = 0. We shall assume that the inhomogeneous differential equation (8.42) is associated with the homogeneous boundary conditions Ur [y] =

κ    αnr y n−1 (a) + βnr y n−1 (b) = 0 ,

r = 1, 2, 3, . . . , κ

(8.43)

n=1

which are formally the same as boundary conditions (8.2). Solution of the boundary value problem (8.42), (8.43) is sought in the form  y(x) =

b

G(x, ξ)r (ξ) dξ,

(8.44)

a

where G(x, ξ) is the Green2 function [5–7] defined by the following four properties: 1. The Green function is a continuous function of x and ξ in each of the triangles a ≤ x ≤ ξ ≤ b and a ≤ ξ ≤ x ≤ b. In addition, it is κ times differentiable with 2 George

Green (1793–1841).

254

8 Eigenvalue Problems of Ordinary Differential Equations

respect to x and the derivatives ∂ n G(x, ξ) = G (n) (x, ξ) , ∂x n

(n = 1, 2, . . . , κ)

are also continuous functions of x and ξ in the triangles a ≤ x ≤ ξ ≤ b and a ≤ ξ ≤ x ≤ b. 2. Let ξ be fixed in [a, b]. The Green function and its derivatives G (n) (x, ξ) =

∂ n G(x, ξ) , ∂x n

(n = 1, 2, . . . , κ − 2)

(8.45)

should be continuous for x = ξ:   lim G (n) (ξ + ε, ξ) − G (n) (ξ − ε, ξ) = ε→0   = G (n) (ξ + 0, ξ) − G (n) (ξ − 0, ξ) = 0 (n = 0, 1, 2, . . . , κ − 2) . (8.46a) The derivative G (κ−1) (x, ξ) should, however, have a jump   lim G (κ−1) (ξ + ε, ξ) − G (κ−1) (ξ − ε, ξ) =

ε→0

  = G (κ−1) (ξ + 0, ξ) − G (κ−1) (ξ − 0, ξ) =

1 pκ (ξ)

(8.46b)

if x = ξ. 3. Let α be an arbitrary but finite non-zero constant. For a fixed ξ ∈ [a, b], the product G(x, ξ)α as a function of x (x = ξ) should satisfy the homogeneous differential equation L [G(x, ξ)α] = 0 . 4. The product G(x, ξ)α as a function of x should satisfy the boundary conditions Ur [G(x, ξ)α] = 0 ,

r = 1, 2, 3, . . . , κ .

Remark 8.7 The above definitions are basically the same as that of Bocher [5, 6] who generalized an earlier result concerning ordinary differential equations of order two. If there exists the Green function defined above for the boundary value problem (8.42), (8.43) then the function y(x) given by (8.44) satisfies differential equation (8.42) and boundary conditions (8.43). Using (8.44), we can determine the first κ − 1 derivatives of y(x): (1)



y (x) = a

b

(1)

(2)



G (x, ξ)r (ξ) dξ, y (x) = a

b

G (2) (x, ξ)r (ξ) dξ, . . .

(8.47a)

8.8 The Green Functions

...,

y

(κ−1)

255



b

(x) =  =

G (κ−1) (x, ξ)r (ξ) dξ =

a x

G

(κ−1)



b

(x, ξ)r (ξ) dξ +

a

G (κ−1) (x, ξ)r (ξ) dξ .

(8.47b)

x

When calculating the κ-th derivative, we have to take, however, into account the additive resolution of the (κ − 1)-th derivative as well as discontinuity (8.46b). We have  b y (κ) (x) = G (κ) (x, ξ)r (ξ) dξ+ a

+ G (κ−1) (x, ξ)r (ξ) ξ=x−0 − G (κ−1) (x, ξ)r (ξ) ξ=x+0 =  b   G (κ) (x, ξ)r (ξ) dξ + G (κ−1) (x, x − 0) − G (κ−1) (x, x + 0) r (x), = a

where x > x − 0 and x < x + 0. Hence y (κ) (x) =



b

G (κ) (x, ξ)r (ξ) dξ +

a

r (x) . pκ (x)

(8.48)

If we substitute solution (8.44) and derivatives (8.47), (8.48) back into the differential equation (8.42), we get L[y(x)] =

κ  n=0

pn (x) y

(n)

 (x) = a

b



κ 

 pn (x) G

(κ)

(x, ξ) r (ξ) dξ+

n=0

+

r (x) pκ (x) = r (x) . pκ (x)

(8.49)

Here, the expression within the braces is zero due to Property 3 of the definition— Properties 1 and 2 were taken into account when we determined the derivatives of the Green function. Equation (8.49) shows that integral (8.44) is really a solution of differential equation (8.42). According to Property 4, the Green function satisfies the boundary conditions. Hence, integral (8.44) is really the solution of the boundary value problem (8.42), (8.43) as well.

8.9 Calculation of the Green Function Let us denote the linearly independent particular solutions of the homogeneous ordinary differential equation L[y(x)] = 0 (8.50)

256

8 Eigenvalue Problems of Ordinary Differential Equations

by z 1 (x) , z 2 (x) , z 3 (x) , . . . , z κ (x) .

(8.51)

Since the general solution is a linear combination of the particular solutions, it can be given in the following form (Fig. 8.1): y(x) =

κ 

A z  (x),

(8.52)

=1

where the coefficients A are arbitrary integration constants. Since the Green function satisfies the homogeneous differential equation (8.50) in each of the triangular domains a ≤ x ≤ ξ ≤ b and a ≤ ξ ≤ x ≤ b, it follows that it can be given as a Fig. 8.1 Triangular domains

b x x y

a a

b

x

linear combination of the particular solutions z  (x), i.e., by Eq. (8.51). The integration constants A should, however, be different in the two triangular domains. For this reason, we shall assume that G(x, ξ) = G(x, ξ) =

κ  =1 κ 

(a (ξ) + b (ξ)) z  (x) ,

x ≤ ξ; (8.53)

(a (ξ) − b (ξ)) z  (x) ,

x ≥ξ

=1

where the coefficients a (ξ) and b (ξ) are unknown functions. Continuity conditions (8.46a) yield the following equations: κ 

b (ξ) z (n) (ξ) = 0 ,

n = 0, 1, 2, . . . , κ − 2

=1

As regards discontinuity condition (8.46b), we get

(8.54a)

8.9 Calculation of the Green Function

257

κ    1 = G (κ−1) (ξ + 0, ξ) − G (κ−1) (ξ − 0, ξ) = −2 b (ξ) z (κ−1) (ξ) . pκ (ξ) =1

Hence

κ 

b (ξ) z (κ−1) (ξ) = −

=1

1 . 2 pκ (ξ)

(8.54b)

Continuity and discontinuity conditions (8.46) have resulted in the inhomogeneous linear system of Eqs. (8.54) for the unknowns b (ξ) ( = 1, 2, . . . , κ). Its determinant assumes the following form: z 1 (ξ) z 1(1) (ξ)

z 2 (ξ) . . . z κ (ξ) z 2(1) (ξ) . . . z κ(1) (ξ) . .................. z 1(κ−1) (ξ) z 2(κ−1) (ξ) . . . z κ(κ−1) (ξ)

(8.55)

This determinant is the Wronskian3 [8] of differential equation (8.42) which is not zero since the particular solutions z  (x) are linearly independent. This means that we can always find a unique solution for the coefficients b (ξ) in representation (8.53) of the Green function. Making use of Property 4 of the Green function, we can find equations for the coefficients a (ξ): Ur [G] = Ur

κ 

 (a (ξ) ± b (ξ)) z  (x) = 0 ,

r = 1, 2, 3, . . . , κ

(8.56)

=1

where the sign is (positive) [negative] if the Green function in the boundary condition Ur is taken at (x = a) [x = b]. Since the boundary conditions are linear functions of the solution y(x)—in the present case of the Green function—it follows that κ 

a Ur [z  ] = ∓

=1

κ 

b Ur [z  ] ,

r = 1, 2, 3, . . . , κ .

(8.57)

=1

This inhomogeneous linear equation system has a unique solution for the coefficients a (ξ) if and only if its determinant is not zero: U1 [z 1 ] U1 [z 2 ] . . . U1 [z κ ] U2 [z 1 ] U2 [z 2 ] . . . U2 [z κ ] = 0 . |Ur [z  ]| = .............. Uκ [z 1 ] Uκ [z 2 ] . . . Uκ [z κ ]

3 Józef

Höené-Wro´nski (1776–1853).

(8.58)

258

8 Eigenvalue Problems of Ordinary Differential Equations

8.10 Symmetry of the Green Function Consider the following two inhomogeneous boundary value problems: L[u(x)] = r (x) ,

Ur [u(x)] = 0 ;

(8.59a)

L[v(x)] = s(x) ,

Ur [v(x)] = 0,

(8.59b)

where the differential operator L is given by Eq. (8.42b), u(x) and v(x) are the unknown functions, while r (x) and s(x) are continuous inhomogeneities in the interval x ∈ [a, b]. We shall assume that boundary value problems (8.59a) and (8.59b) are self-adjoint. Then (u, v) L − (v, u) L = 0, in which according to Eq. (8.44)  u(x) =

b

 G(x, ξ)r (ξ) dξ

and

v(x) =

a

b

G(x, ξ)s(ξ) dξ .

a

Thus  (u, v) L − (v, u) L =  = a

b

b

(u L[v] − v L[u]) dx =

a b

s(x) [G(x, ξ) − G(ξ, x)] r (ξ) dξdx = 0 .

(8.60)

a

Since both r (x) and s(x) are arbitrary continuous and non-zero functions in the interval [a, b], it follows that the last integral in Eq. (8.60) can be zero if and only if G(x, ξ) = G(ξ, x) .

(8.61)

In words: the Green function of self-adjoint boundary value problems is a symmetric function.

8.11 Calculation of the Green Functions for Some Practical Problems 8.11.1 Simple Beam Problems Consider a uniform beam and assume that N = 0, which means that there is no axial force in the beam. Then according to Eq. (7.75b) the equilibrium problems of uniform beams are governed by the differential equation:

8.11 Calculation of the Green Functions for Some Practical Problems

L(w) =

d 4w f z (x) = f˜z (x), = dx4 IE

259

(8.62)

where f z (x) is the intensity of the distributed load, I is the areal moment of inertia, and E is the Young’s modulus. For a simply supported beam of length  (a = 0, b =), this equation is associated with the following boundary conditions: U1 = w(0) = 0 , U2 = w (2) (0) = 0 , U3 = w() = 0 , U4 = w (2) () = 0. (8.63) Boundary value problem (8.62), (8.63) is self-adjoint. Our aim is to determine the Green function. The linearly independent particular solutions of equation L(w) = 0 are given by w1 = 1, w2 = x, w3 = x 2 , w4 = x 3 .

(8.64)

We seek the Green function in the form given by Eq. (8.53) where the functions b1 (ξ), . . . , b4 (ξ) are solutions of the linear equation system (8.54): ⎡

1 ⎢0 ⎢ ⎣0 0 from where we get

ξ 1 0 0

ξ2 2ξ 2 0

⎤ ⎤⎡ ⎤ ⎡ 0 ξ3 b1 ⎥ ⎢ ⎥ ⎢ 3ξ 2 ⎥ ⎥ ⎢ b2 ⎥ = ⎢ 0 ⎥ 6ξ ⎦ ⎣ b3 ⎦ ⎣ 0 ⎦ b4 6 − 21

⎡

⎤ ⎡ ⎤ b1 ξ3 2⎥ ⎢ b2 ⎥ ⎢ ⎢ ⎥ = 1 ⎢ −3ξ ⎥ . ⎣ b3 ⎦ 12 ⎣ 3ξ ⎦ b4 −1

(8.65)

Boundary conditions (8.63) result in the following equation system: a1 w1 (0) + a2 w2 (0) + a3 w3 (0) + a4 w4 (0) = = −b1 w1 (0) − b2 w2 (0) − b3 w3 (0) − b4 w4 (0) , a1 w1(2) (0) + a2 w2(2) (0) + a3 w3(2) (0) + a4 w4(2) (0) = = −b1 w1(2) (0) − b2 w2(2) (0) − b3 w3(2) (0) − b4 w4(2) (0) , a1 w1 () + a2 w2 () + a3 w3 () + a4 w4 () = = b1 w1 () + b2 w2 () + b3 w3 () + b4 w4 () , a1 w1(2) () + a2 w2(2) () + a3 w3(2) () + a4 w4(2) () = = b1 w1(2) () + b2 w2(2) () + b3 w3(2) () + b4 w4(2) ()

260

8 Eigenvalue Problems of Ordinary Differential Equations

from where after substituting the particular solutions (8.64) and the results for the coefficients b1 (ξ), . . . , b4 (ξ) we obtain ⎡

1 ⎢0 ⎢ ⎣1 0 Thus

0 0  0

0 2 2 2

⎡ ⎤⎡ ⎤ ⎤ −ξ 3 a1 0 ⎢ ⎢ ⎥ ⎥ −6ξ 0⎥ ⎥ ⎢ a2 ⎥ = 1 ⎢ 3 ⎥ 3 ⎦⎣ 2 2 3⎦ . ⎦ ⎣  a3 12 ξ − 3ξ  + 3ξ −  6 a4 6ξ − 6 ⎤ ⎡ ⎤ 3 a1  2 −ξ   2 ⎥ ⎢ ⎢ a2 ⎥ ⎢ ⎥ = 1 ⎢ ξ 2ξ − 3ξ + 4 ⎥ ⎣ a3 ⎦ 12 ⎣ ⎦ −3ξ a4 2ξ −  ⎡

and G(x, ξ) = (a1 + b1 ) + (a2 + b2 ) x + (a3 + b3 ) x 2 + (a4 + b4 ) x 3 =    x x  3 ξ − 3ξ 2  + 2ξ2 + x 2 ξ − x 2  = = ( − ξ) 2ξ − ξ 2 − x 2 6 6

x ≤ξ

Since boundary value problem (8.62), (8.63) is self-adjoint, the Green function is symmetric. Consequently, we can use symmetry Eq. (8.61) to give the Green function for x ≥ ξ:   ξ G(x, ξ) = x ≥ ξ. ( − x) 2x − ξ 2 − x 2 6 Remark 8.8 Assume that f z = δ (x − ξ) in boundary value problem (8.62), (8.63) for which we have determined the Green functions above. This means that the simply supported beam shown in Fig. 8.2 is subjected to a unit force at the point with coordinates ξ. It is also obvious that the right side of Eq. (8.62) is then Fig. 8.2 Simply supported beam loaded by a vertical unit force

z

fz x

x

x,

x x

δ (x − ξ) . f˜z (x) = IE Making use of Eq. (8.44), we can now give the solution for the deflection w(x) in the following form:

8.11 Calculation of the Green Functions for Some Practical Problems

w(x) =

  0

G(x, η) f˜z (η) dη =

261

  1 1 G(x, η)δ(η − ξ) dη = G(x, ξ) = G(x, ξ), IE 0 IE

=0

1 2 x (3ξ − x) 6I E =

w (3) () w (2) ()

w() =

w(0) = w (1) (0) = 0 Fixed-free

=0

x2 ( − ξ)2 (3ξ − x − 2ξx) 6I E3 w (1) ()

w(0) = w (1) (0) = 0 Fixed-fixed

=0

  x2 (−ξ) 6ξ2 −22 x −3ξ 2 −2ξx +xξ 2 12I E3 w (2) () w() =

w() =

w(0) = w (1) (0) = 0 Fixed-pinned

=0

  x ( − ξ) 2ξ − ξ 2 − x 2 6I E w(0) = w (2) (0) = 0 Simply supported

w (2) ()

Boundary conditions Beam with its supports

Table 8.1 Green’s functions for some beams

Green’s functions for DE I Ew (4) = f z (G (x, ξ) ∀x ≤ ξ)

(8.66) where we have taken integral (B.1.2) into account. It follows from this equation that (a) the Green function for the differential equation I Ew(4) (x) = f z (x), which is associated with boundary conditions (8.63) (simply supported beam), is G(x, ξ) = G(x, ξ)/I E; (b) the Green function G(x, ξ) is the deflection G(x, ξ) = G(x, ξ)/I E of the beam at the point x due to a vertical unit force exerted on the beam at the point ξ. Thus, we can also calculate the Green function for beam problems by determining the deflection at the point x caused by the unit force exerted on the beam at the point ξ.

262

8 Eigenvalue Problems of Ordinary Differential Equations

Table 8.1 contains the Green function G(x, ξ) for four typical support arrangements including the case of the simply supported beam. Calculation of the Green functions for fixed-pinned, fixed-fixed, and fixed-free beams is left for Problem 8.1, and the details are presented in Sect. C.8. Given the Green functions G(x, ξ) for simply supported (pinned-pinned), fixedpinned, fixed-fixed, and fixed-free beams—see Table 8.1 for details—solution for the deflection w(x) due to the load f z is given by the relation 



w(x) =

G(x, η) f z (η) dη .

(8.67)

0

8.11.2 Preloaded Beams Equilibrium problems of beams preloaded by an axial force are governed by the differential equation L(w) =

N d2 w 1 d4 w ∓ = fz , 4 dx I E dx 2 IE

(8.68)

which is associated with boundary conditions w(0) = w (2) (0) = w() = w (2) () = 0

(8.69)

for the simply supported beam shown in Fig. 8.3. It can easily be seen that boundary value problem (8.68), (8.69) is self-adjoint. Fig. 8.3 Simply supported beam loaded by a vertical unit force—the preload is an axial force

z

fz x

x

x,

N

x

x

If the force is a compressive one, the sign of N is positive in (8.68). Then, the four linearly independent particular solutions of the homogeneous differential equation L(w) = 0 are w1 = 1, w2 = x, w3 = cos px, w4 = sin px,  p = N /I E , N > 0 .

(8.70) (8.71)

8.11 Calculation of the Green Functions for Some Practical Problems

263

Our aim is to find the Green function both for compressive and tensile axial forces. First, we shall assume that the axial force is a compressive one. It follows on the basis of Remark 8.8 that the Green function is the deflection at the point x ∈ [0, ] due to a unit force exerted on the beam at the point ξ ∈ [0, ]. Since the point ξ divides the interval [0, ] into two parts, we shall assume that G(x, ξ) = A1 (ξ) + A2 (ξ) x + A3 (ξ) cos px + A4 (ξ) sin px,

x ≤ ξ; (8.72a)

G(x, ξ) = B1 (ξ) + B2 (ξ) x + B3 (ξ) cos px + B4 (ξ) sin px,

x ≥ ξ . (8.72b)

The Green function should satisfy (a) the boundary conditions at the left end of the beam: G(0, ξ) = 0 (2)

G (0, ξ) = 0

(the deflection is zero),

(8.73a)

(the bending moment is zero);

(8.73b)

(b) the continuity and discontinuity conditions if ξ = x: G(ξ − 0, ξ) = G(ξ + 0, ξ) (1)

(1)

G (ξ − 0, ξ) = G (ξ + 0, ξ)

(continuity of the deflection),

(8.74a)

(continuity of the rotation—see (7.74)1 ), (8.74b)

G (2) (ξ − 0, ξ) = G (2) (ξ + 0, ξ) (continuity of the bending moment—see (7.74)1 ), (8.74c) G (3) (ξ − 0, ξ) − G (3) (ξ + 0, ξ) = −1/I E (8.74d) (discontinuity condition by taking into account the fact that the right side of equation is δ(x − ξ)/I E for the Green function);

(c) and the boundary conditions at the right end of the beam: G(, ξ) = 0 (2)

G (, ξ) = 0

(the deflection is zero),

(8.75a)

(the bending moment is zero).

(8.75b)

Substitution of (8.72) into Eqs. (8.73)–(8.75) results in a system of linear equations for the unknown functions A1 (ξ), . . . , A4 (ξ) and B1 (ξ), . . . , B4 (ξ). The coefficient matrix is of the form

264

8 Eigenvalue Problems of Ordinary Differential Equations

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ A=⎢ ⎢ ⎢ ⎢ ⎢ ⎣

1 0 1 0 0 0 0 0

0 0 ξ 1 0 0 0 0

1 − p2 cos pξ − p sin pξ − p 2 cos pξ p 3 sin pξ 0 0

0 0 sin pξ p cos pξ − p 2 sin pξ − p 3 cos pξ 0 0

0 0 −1 0 0 0 1 0

0 0 −ξ −1 0 0  0

⎤ 0 0 ⎥ 0 0 ⎥ − cos pξ − sin pξ ⎥ ⎥ p sin pξ − p cos pξ ⎥ ⎥ p 2 cos pξ p 2 sin pξ ⎥ ⎥ − p 3 sin pξ p 3 cos pξ ⎥ ⎥ ⎦ cos p sin p 2 2 − p cos p − p sin p

Utilizing the notations XT =



 A1 A2 A3 A4 B1 B2 B3 B4 ,

and FT =



 0 0 0 0 0 −1/I E 0 0 ,

the equation system to be solved can be given in the following form: A X = F.

(8.76)

Making use of the solutions A1 = 0 ,

A2 =

ξ− , p 2 I E

A3 = 0 ,

A4 = −

sin [ p(ξ − )] p 3 I E sin p

(8.77)

we can determine the sought Green functions from Eq. (8.72a): G(x, ξ) =

1 p 3 I E sin

p

[x p (ξ − ) sin p −  sin [ p(ξ − )] sin px] ,

x ≤ξ

(8.78) The Green function for the case of x ≥ ξ can be calculated either by substituting the solutions ξ

ξ

sin pξ 1 , B4 = − 3 sin px cos p 3 p IE p I E sin p (8.79) into (8.72b), or by writing ξ for x and x for ξ in (8.78). Hence B1 = −

p2 I E

, B2 =

p 2 I E

, B3 =

G(x, ξ) = Dc [x p (ξ − ) sin p −  sin [ p(ξ − )] sin px] , G(x, ξ) = Dc [ξ p (x − ) sin p −  sin [ p(ξ − )] sin pξ] ,

x ≤ ξ; (8.80a) x ≥ ξ (8.80b)

where Dc = 1/ p 3 I E sin p .

(8.80c)

8.11 Calculation of the Green Functions for Some Practical Problems

265

If the force is a tensile one w1 = 1, w2 = x, w3 = cosh px, w4 = sinh px,  p = N /I E , N > 0

(8.81) (8.82)

are the four linearly independent particular solutions of the differential equation L(w) = 0. Then the Green function assumes the form G(x, ξ) = A1 (ξ) + A2 (ξ) x + A3 (ξ) cosh px + A4 (ξ) sinh px,

x ≤ ξ; (8.83a)

G(x, ξ) = B1 (ξ) + B2 (ξ) x + B3 (ξ) cosh px + B4 (ξ) sinh px,

x ≥ ξ. (8.83b)

Continuity and discontinuity conditions (8.73)–(8.75) yield again a system of linear equations for the unknowns A1 (ξ), . . . , A4 (ξ) and B1 (ξ), . . . , B4 (ξ). Making use of the solutions A1 = 0 , A2 = −

ξ− sinh [ p(ξ − )] , A3 = 0 , A4 = 3 , p 2 I E p I E sinh p

ξ ξ , B2 = − 2 , p2 I E p I E sinh pξ 1 B3 = − 3 , B4 = 3 sinh pξ cosh p p IE p I E sinh p

(8.84a)

B1 =

(8.84b)

from (8.83), one obtains the Green function: G(x, ξ) = Dt [x p ( − ξ) sinh p −  sinh [ p(ξ − )] sinh px] , x ≤ ξ . (8.85a) G(x, ξ) = Dt [ξ p ( − x) sinh p −  sinh [ p(ξ − )] sinh pξ] , x ≥ ξ, (8.85b) where Dt = 1/ p 3 I E sinh p.

(8.85c)

It can be checked with ease that Green functions (8.78) and (8.83) satisfy symmetry condition (8.61). Tables 8.2 and 8.3 contain the Green functions of pinned-pinned, fixed-pinned, and fixed-fixed beams both for compressive and for tensile preloads. Given the Green functions G(x, ξ) for the axially preloaded simply supported (pinned-pinned), fixed-pinned, and fixed-fixed beams in Tables 8.2 and 8.3 solution for the deflection w(x) due to the transverse load f z is given by Eq. (8.67).

8 Eigenvalue Problems of Ordinary Differential Equations 266

w(0) = w (1) (0) = 0

w() = w (2) () = 0

w(0) = w (1) (0) = 0

w() = w (2) () = 0

w(0) = w (2) (0) = 0

Boundary conditions

Table 8.2 Green’s functions for compressive preload

Beams preloaded by a compressive force

Simply supported

Fixed-pinned

Fixed-fixed

w() = w (1) () = 0

Green’s functions

for DE w 4 + pw (2) = f z /I E (G (x, ξ) ∀x ≤ ξ)

Dc [ξ p (x − ) sin p −  sin [ p(ξ − )] sin pξ]

Dc = 1/ p 3 I E sin p

   Dc p (ξ −) sin p− sin [ p(ξ −)] (1 + cos px)−    − p (ξ −) cos p−sin [ p(ξ −)] (sin px − px)

Dc = 1/ p 3 I E ( p cos p − sin p)   D −ξ p+sin [ p(ξ −)]+p cos [ p(ξ −)]−sin pξ+ c  +sin p+(ξ −) p cos p (1−cos px) −  − 1−cos [ p(ξ −)]+(ξ −) p sin p−   − cos p+cos pξ (sin px − px)

Dc = 1/ p 3 I E (2 cos p + p sin p − 2)

Fixed-fixed

Fixed-pinned

Simply supported

Beams preloaded by a tensile force

w() = w (1) () = 0

w(0) = w (1) (0) = 0

w() = w (2) () = 0

w(0) = w (1) (0) = 0

w() = w (2) () = 0

w(0) = w (2) (0) = 0

Boundary conditions

Table 8.3 Green’s functions for tensile preload





Dt = 1/ p 3 I E (2 − 2 cosh p + p sinh p)

Dt = 1/ p 3 I E ( p cosh p − sinh p)  Dt −ξ p+sinh [ p(ξ −)]+p cosh [ p(ξ −)]−sinh pξ+  +sinh p+(ξ − ) p cosh p (1−cosh px) −  − 1−cosh [ p(ξ −)]−(ξ −) p sinh p−   −cosh p+cosh pξ (sinh px − px)

   Dt p (ξ −) sinh p− sinh [ p(ξ −)] (1−cosh px)−    − sinh [ p(ξ −)]− p (ξ −) cosh p (sinh px − px)

Dt = 1/ p 3 I E sinh p

Dt x p ( − ξ) sinh p −  sinh [ p(ξ −)] sinh pξ

for DE w 4 − pw (2) = f z /I E (G (x, ξ) ∀x ≤ ξ)

Green’s functions

8.11 Calculation of the Green Functions for Some Practical Problems 267

268

8 Eigenvalue Problems of Ordinary Differential Equations

8.11.3 Preloaded Circular Plates 8.11.3.1

The Governing Equations of the Problem

Figure 8.4 shows (1) a clamped circular plate, (2) a simply supported circular plate, and (3) a simply supported circular plate which is elastically restrained on its boundary by a linear torsional spring with a spring constant kγ . The thickness and the outer radius of the plates are denoted by 2b and R◦ 2b. The material of the plates is homogeneous and isotropic with Young’s modulus E and Poisson’s number ν. We shall assume that the plates are preloaded by a constant radial load f R in their middle plane. The preload can be either compressive or tensile—Figure 8.4 depicts the case of compressive radial load. Moreover, the plates are subjected to a vertical distributed load pz . It is a further assumption that the vertical load pz , the deformations, and the stress state are all axisymmetric, i.e., they are functions of the radial coordinate R only. Fig. 8.4 Circular plates preloaded by a compressive radial load

1. 2. 3.

fR

z z

fR

R

fR

R

fR

R

Ro

Ro fR

fR

z

For this reason, we shall use the cylindrical coordinate system (R, θ, z) when calculating the Green functions that belong to the boundary value problems of the three support arrangements shown in Fig. 8.4. For our later considerations, we shall introduce some notations: I1 = 8b3 /12 is the moment of inertia, E 1 = E/(1 − ν 2 ) is the modified Young’s modulus, F = Ro2 f R /I1 E 1 and Fν = F/(1 − ν) are the dimensionless in-plane load and the modified dimensionless in-plane load, K = Ro kγ /I1 E 1 and Kν = 1 − K/(1 − ν) are the dimensionless spring constant and the modified dimensionless spring constant, and r = R/Ro and ξ are dimensionless independent variables. The bending moment and shear force per unit length are denoted by M R and Q R . The deflection w due to the load pz (r ) acting perpendicularly to the middle plane of the plate should meet the differential equation [9, 10] ˜ w ˜ ± F w ˜ = 

Ro4 pz , I1 E 1

˜ = 

d2 1 d + dr 2 r dr

(8.86)

8.11 Calculation of the Green Functions for Some Practical Problems

269

where the sign preceding F is positive for compression and negative for tension. It is known [11] that 1 dw ϕθ = − , Ro dr and QR

I1 E 1 MR = − 2 Ro

Ro3 d = I1 E 1 dr





d2 w ν dw + dr 2 r dr

 (8.87)

d2 1 d + ± F w(r ) dr 2 r dr

(8.88)

are the rotation about the axis θ, the bending moment, and the shear force in terms of the deflection. Depending on what the supports are, Eq. (8.86) should be associated with appropriate boundary conditions. As regards the outer boundary, it is clear from Fig. 8.4 that for a clamped plate w|r =1 = 0 ,

dw

=0 dr r =1

(8.89)

are the boundary conditions. If the plate is simply supported then the boundary conditions are of the form   2 d w ν dw

+ =0. (8.90) w|r =1 = 0 , dr 2 r dr r =1 k

Fig. 8.5 Circular plate element

R

MR

dw Ro dr

Ro z

For the plate supported by a torsional spring, it follows from Figs. 8.4 and 8.5 that

M R |r =1 + kγ ϕθ r =1 =



w|r =1 = 0 ,



dw

d w ν dw Ro + + kγ =0 dr 2 r dr r =1 I1 E 1 dr r =1 2

(8.91)

are the boundary conditions. It is also clear that the deflection at the center of the plate should meet the conditions w|r =0 = finite ,

dw

= 0. dr r =0

(8.92)

270

8.11.3.2

8 Eigenvalue Problems of Ordinary Differential Equations

Definition of the Green Function

Assume that the plate is subjected to a uniform load distributed on the circle with radius Ro ξ (ξ is also a dimensionless coordinate)—see Fig. 8.4 . The resultant of the total load is assumed to be 1. The deflection due to the load at r is denoted by G(r, ξ) and is referred to as the Green function. It is obvious that the Green function should satisfy the homogeneous equation in (8.86) if 0 ≤ r < ξ and ξ < r ≤ 1.

8.11.3.3

Green Functions for Compressive f R

As is well known, the general solution of the homogeneous equation in (8.86) assumes the form √ √ (8.93) w(r ) = c1 + c2 ln r + c3 Jo ( Fr ) + c4 Yo ( Fr ), √ where c√ 1 , c2 , c3 , and c4 are undetermined constants of integration, while Jo ( Fr ) and Yo ( Fr ) are Bessel functions. Since the Green function should meet conditions (8.92), it can be given as √ G(r, ξ) = A1 + A3 J0 ( Fr ) ,

√

0≤r ≤ξ,

√

G(r, ξ) = B1 + B2 ln r + B3 J0 ( Fr ) + B4 Y0 ( Fr ) ,

(8.94a) ξ ≤ r ≤ 1,

(8.94b)

where the constants of integration A1 , A3 and B1 , B2 , B3 , B4 are to be determined from the continuity and discontinuity conditions prescribed at r = ξ and from the boundary conditions which are imposed on the boundary r = 1. Note that the continuity conditions G(ξ − 0, ξ) = G(ξ + 0, ξ) , (1)

(1)

(2)

(2)

G (ξ − 0, ξ) = G (ξ + 0, ξ) , G (ξ − 0, ξ) = G (ξ + 0, ξ)

(8.95a) (8.95b) (8.95c)

and the discontinuity condition 2π Ro ξ [Q R (ξ + 0) − Q R (ξ − 0)] = 2π Ro ξ Q R (ξ + 0) = 1 ,

(8.96)

in which Q R (ξ − 0) = 0 from the vertical equilibrium are all independent of the supports. Here (a) the derivatives with respect to r are denoted in the same manner as the derivatives with respect to x in (8.45) and (b) according to (8.88) it holds that QR

Ro3 d = I1 E 1 dr



d2 1 d + F G(ξ, r ) . + dr 2 r dr

(8.97)

8.11 Calculation of the Green Functions for Some Practical Problems

271

In the sequel, we shall determine the Green function for the elastically restrained and simply supported plate. When calculating the derivatives of we have to utilize the √ the Green functions, √ derivatives of the Bessel functions Jo ( Fr ) and Yo ( Fr ). The latter derivatives are given by Eqs. (B.2.7). Thus √ √ 0≤r ≤ξ, G (1) (r, ξ) = −A3 F J1 ( Fr ) √ √ √ √ 1 G (1) (r, ξ) = B2 − B3 F J1 ( Fr ) − B4 F Y1 ( Fr ) r ⎡ G (2) (r, ξ) = A3 F ⎣

J1

√

Fr

√ Fr

⎡

⎤ √ − J0 Fr ⎦ ,

(8.98a) ξ ≤ r ≤ 1 , (8.98b)

0≤r ≤ξ,

(8.99a)

√

⎤ √ Fr 1 G (2) (r, ξ) = −B2 2 + B3 F ⎣ √ − J0 Fr ⎦ + r Fr ⎛ √ ⎞ √ Y1 Fr + B4 F ⎝ √ − Y0 Fr ⎠ , ξ ≤ r ≤ 1 , (8.99b) Fr J1

 √ F 3/2  √ J3 ( Fr ) − 3J1 ( Fr ) = 4  √ √ 1 F 3/2 = A3 J2 ( Fr ) − J1 Fr , √ 4 Fr

G (3) (r, ξ) = −A3

0≤r ≤ξ

(8.100a)

and G (3) (r, ξ) = =

  √ √ 2 F 3/2  √ F 3/2  √ J Y B − B ( ( Fr ) −B ( ( Fr )−3J Fr )−3Y Fr ) = 2 3 3 1 4 3 1 r3 4 4  √ √ 1 F 3/2 2 J2 ( Fr )− J1 Fr + = 3 B2 + B3 √ r 4 Fr  √ √ 1 F 3/2 Fr , ξ≤r ≤1 (8.100b) + B4 √ Y2 ( Fr )−Y1 4 Fr

are the derivatives of the Green function.

272

8 Eigenvalue Problems of Ordinary Differential Equations

After substituting the Green function and its derivatives into the continuity condition (8.95a) and combining then the continuity conditions (8.95b) and (8.95c), we have √ √ √ A1 + A3 J0 ( Fξ) = B1 + B2 ln ξ + B3 J0 ( Fξ) + B4 Y0 ( Fξ) , √ √ √ √ √ √ 1 −A3 F J1 Fξ = B2 − B3 F J1 Fξ − B4 F Y1 Fξ ξ √ √ √ A3 J0 ( Fξ) = B3 J0 ( Fξ) + B4 Y0 ( Fξ) .

(8.101a) , (8.101b) (8.101c)

It follows from (8.97) that %

1 R 3K d 1 − 2 + 2 + F ln r

+ Q R (ξ + 0) = B2 I1 E 1 dr r r r =ξ   

√

  √ √ √ F d J1 ( Fr )

J2 ( Fr )− J0 ( Fr ) −F √ + B3 +F J0 ( Fr )

dr 2 Fr r =ξ   

√

  √ √ √ d Y1 ( Fr ) F

Y2 ( Fr )−Y0 ( Fr ) −F √ +F Y0 ( Fr ) = + B4

dr 2 Fr r =ξ %

1 d 1 − 2 + 2 + F ln r

+ = B2 dr r r r =ξ



 √

 √ F √ d J1 ( Fr )

J2 ( Fr ) + J0 ( Fr ) − F √ + B3

+ dr 2 Fr

'( )

& =0 r =ξ





√  √ F √ d Y1 ( Fr )

1 + B4 Y2 ( Fr ) + Y0 ( Fr ) − F √ = B2 F .

dr 2 ξ Fr

'( )

& =0 r =ξ

Consequently, discontinuity condition (8.96) leads to the equation B2 =

Ro2 1 1 1 1 Ro3 ξ= ξ= . 2 R f R o I1 E 1 2πξ Ro F 2π f R I1 E 1 I1 E1 2πξ

(8.101d)

The last two equations for the integration constants are obtained from the boundary conditions (8.91): √ √ B1 + B3 J0 ( F) + B4 Y0 ( F) = 0 ,

 2  d G(r, ξ) ν dG(r, ξ)

dG(r, ξ)

+ = − K ,

dr 2 r dr dr r =1 r =1

(8.102a) (8.102b)

8.11 Calculation of the Green Functions for Some Practical Problems

where K=

Ro k γ I1 E 1

273

(8.103)

is the earlier mentioned dimensionless spring constant. The left side of boundary condition (8.102b)2 can be manipulated into a more suitable form if we take into account that % 1 1 dG(r ) 1 1 d2 G(r ) 1 dG(r ) − (1 − ν) = B2 − 2 + 2 − (1 − ν) 2 + + dr 2 r dr r dr r r r √ √  %  √ √ F J1 ( Fr ) J1 ( Fr ) + B3 J2 ( Fr ) − J0 ( Fr ) − F √ + (1 − ν)F √ + 2 Fr Fr & '( ) √ −F J0 ( Fr )

√ √ %  √ F √ Y1 ( Fr ) Y1 ( Fr ) Y2 ( Fr ) − Y0 ( Fr ) − F √ = + (1 − ν)F √ 2 Fr Fr & '( )

+ B4

√ −F Y0 ( Fr )



√ √ 1 J1 ( Fr ) = −B2 (1 − ν) 2 + B3 (1 − ν)F √ − F J0 ( Fr ) + r Fr 

√ √ Y1 ( Fr ) − F Y0 ( Fr ) + B4 (1 − ν)F √ Fr from where for r = 1 we get

1 dG(r, ξ)

d 2 G(r, ξ) +ν = −B2 (1 − ν)+ dr 2 r dr r =1  



√ √ √ √ J1 ( F) Y1 ( F) + B3 (1−ν)F √ −F J0 ( F) +B4 (1 − ν)F √ −F Y0 ( F) . F F (8.104) As regards the right side of (8.102b)2 , we have

√ √ √ √

K  d B2 − B3 F J1 ( F) − B4 FY1 ( F) . =− − K G(r, ξ)

dr 1−ν r =1 (8.105) If we equalize (8.104) and (8.105), we obtain  − B2 1 −

    √ √ √ K F K + B3 J1 ( F) F 1 − − J0 ( F) + 1−ν 1−ν 1−ν    √ √ √ F K − + B4 Y1 ( F) F 1 − Y0 ( F) = 0 . (8.106) 1−ν 1−ν

274

8 Eigenvalue Problems of Ordinary Differential Equations

√ √ √ Introducing the notations √ √ Jnr = Jn ( Fr √ ), Jnξ = Jn ( Fξ), Jn1 = Jn ( F), Ynr = Yn ( Fr ), Ynξ = Yn ( Fξ), Yn1 = Yn ( F) (n = 0, 1) and ⎡

1 √J0ξ ⎢ 0 − F J1ξ ⎢ ⎢0 J0ξ A=⎢ ⎢0 0 ⎢ ⎣0 0 0 0

⎤ −1 − ln ξ −J0ξ −Y0ξ √ √ ⎥ 0 − 1ξ F J1ξ FY1ξ ⎥ ⎥ 0 0 −J0ξ −Y0ξ ⎥, ⎥ 0 1 0 0 ⎥ ⎦ 1 0 √ J01 √ Y01 0 −Kν J11 FKν − Fν J01 Y11 FKν − Fν Y01

  X T = A1 A2 B1 B2 B3 B4 ,   F T = 0 0 0 1/2π f R 0 0 , F K Fν = , Kν = 1 − 1−ν 1−ν

(8.107)

we can rewrite Eqs. (8.101), (8.102a), and (8.106) in the following form: A X = F.

(8.108)

The solutions are given by the following relations: Dc A1 = ξ ln ξ

√

F Dc A2 = J0ξ

√

    FKν J11 J1ξ Y0ξ − Y1ξ J0ξ − Fν J01 J1ξ Y0ξ − Y1ξ J0ξ −   −Kν J01 ξ J1ξ Y0ξ − Y1ξ J0ξ + Kν J0ξ (J11 Y01 − J01 Y11 ) , (8.109a)

√ √   FKν Y11 − Fν Y01 + Kν Fξ J1ξ Y0ξ − J0ξ Y1ξ +  √ +Y0ξ Fν J01 − FKν J11 , (8.109b)

   Dc B1 = Kν J0ξ (J11 Y01 − J01 Y11 ) + J01 ξ J0ξ Y1ξ − J1ξ Y0ξ , B2 =

1 , 2π f R

(8.109c) (8.109d)

√ √ √   F Dc B3 = −J0ξ FKν Y11 − Fν Y01 + Kν Fξ J1ξ Y0ξ − J0ξ Y1ξ , (8.109e)  √ √ (8.109f) F Dc B4 = −J0ξ −Fν J01 + FKν J11 , where  √   Dc = 2 −J01 Fν + FKν J11 J1ξ Y0ξ − J0ξ Y1ξ ξπ f R .

(8.110)

8.11 Calculation of the Green Functions for Some Practical Problems

275

Let us substitute solutions (8.109), (8.110) into (8.94) and take the equations J1ξ Y0ξ − J0ξ Y1ξ =

2 √ πξ F

J11 Y01 − J01 Y11 =

2 √ , π F

which follow from (B.2.7e) into account. After some further manipulations, we get the sought Green function:  √   Kν J01 − J0ξ − J0r + π2 J0r J0ξ Fν Y01 − FKν Y11 G(r, ξ) = + √ 2π f R (Fν J01 − FKν J11 ) ln ξ − π2 J0r Y0ξ + , 0 ≤ r ≤ ξ, 2π f R  √   Kν J01 − J0r − J0ξ + π2 J0ξ J0r Fν Y01 − FKν Y11 + G(r, ξ) = √ 2π f R (Fν J01 − FKν J11 ) ln r − π2 J0ξ Y0r + , ξ ≤ r ≤ 1. 2π f R

(8.111)

If K = Ro kγ /I1 E 1 → 0 in (8.111) we get the Green function of the simply supported plate: √ J01 − J0ξ − J0r + π2 J0ξ J0r (Fν Y01 − F Y11 ) G(r, ξ) = + √ 2π f R (Fν J01 − F J11 ) ln ξ − π2 J0r Y0ξ + , 0≤r ≤ξ, 2π f R  √ J01 − J0r − J0ξ + π2 J0r J0ξ Fν Y01 − F Y11 + G(r, ξ) = √ 2π f R (Fν J01 − F J11 ) ln r − π2 J0ξ Y0r + , ξ≤r ≤1. 2π f

(8.112)

If K = Ro kγ /I1 E 1 tends to infinity in (8.111) we get the Green function of the clamped plate:

G(r, ξ) =

G(r, ξ) =

    J11 ln ξ + √1F J0r + J0ξ − J01 + π2 J0r J0ξ Y11 −Y0ξ J11 2π f R J11 J11 ln r + √1F





,

0≤r ≤ξ,

J0ξ + J0r − J01 + π2 J0ξ

(J0r Y11 −Y0r J11 )

2π f R J11 ξ≤r ≤1.

(8.113) ,

276

8 Eigenvalue Problems of Ordinary Differential Equations

8.11.3.4

Green Functions for Tensile f R

If the in-plane load is a tensile one w(r ) = c1 + c2 ln r + c3 I0

√ √ Fr + c4 K 0 Fr

(8.114)

is the general solution we need when determining the Green function—c √ 1 , c2 , c3 , Fr and and c4 which are again undetermined constants of integration while I0 √ K0 Fr are Bessel functions. With the knowledge of the general solution, we shall assume that the Green function, which satisfies conditions (8.90), can be given in the following form: √ G(r, ξ) = A1 + A3 I0 ( Fr ) ,

r ξ

(8.115b)

in which the integration constants are denoted in the same as for the case of compressive f —however, this may not cause any misunderstanding. If we substitute the Green function (8.115) into continuity conditions (8.95), discontinuity condition (8.96), and boundary conditions (8.91), we obtain a system of linear equations for A1 , A3 , B1 , . . . , B4 . Here, we present the result, i.e., the equation system only, the corresponding √ calculations√are left for Problem √ 8.5. By introducing √ = I ( Fr ), I = I ( Fξ), I = I ( F), K nr = K n ( Fr ), the notations I nr n nξ n n1 n √ √ K nξ = K n ( Fξ), K n1 = K n ( F) (n = 0, 1), ⎡

1 √ I0ξ ⎢ 0 F I1ξ ⎢ ⎢ 0 I0ξ A=⎢ ⎢0 0 ⎢ ⎣0 0 0 0

⎤ −1 − ln ξ −I0ξ √ √−K 0ξ 1 ⎥ 0 −ξ − F I1ξ F K 1ξ ⎥ ⎥ 0 0 −I0ξ −K 0ξ ⎥, ⎥ 0 1 0 0 ⎥ ⎦ K√ 1 0 I√ 01 01 0 −Kν Fν I01 − F I11 Kν Fν K 01 + F K 11 Kν

and utilizing the notations given by Eqs. (8.107) we can give the equation system in the form equation system (8.108) has. The solutions are given by the following equations: √

    FKν I11 I0ξ K 1ξ + I1ξ K 0ξ − Fν I01 I0ξ K 1ξ + I1ξ K 0ξ +   + Kν I01 ξ I0ξ K 1ξ + I1ξ K 0ξ − Kν I0ξ (I01 K 11 + K 01 I11 ) , (8.116a)

A1 Dt = ξ ln ξ

√

FDt A2 = I0ξ

√

√   FKν K 11 + Fν K 01 − Kν Fξ I0ξ K 1ξ + I1ξ K 0ξ −  √ − K 0ξ Fν I01 − FKν I11 , (8.116b)

8.11 Calculation of the Green Functions for Some Practical Problems

277

   B1 Dt = Kν −I0ξ (I11 K 01 + I01 K 11 ) + I01 ξ I0ξ K 1ξ + I1ξ K 0ξ , B2 =

1 , 2π f R

(8.116c) (8.116d)

√ √ √   FDt B3 = I0ξ FKν K 11 + Fν K 01 − Kν Fξ I0ξ K 1ξ + I1ξ K 0ξ , (8.116e)  √ √ FDt B3 = I0ξ −Fν I01 + FKν I11 , where Dt = 2

√

FKν I11 − Fν I01



(8.116f)

√  I0ξ K 1ξ + I1ξ K 0ξ ξπ f F .

(8.117)

Let us substitute solutions (8.116), (8.117) into the definition of the Green function (8.115). If we take the relations 1 I0ξ K 1ξ + I1ξ K 0ξ = √ ξ F

1 I11 K 01 + I01 K 11 = √ F

into account after some manipulations, we get  √   Kν I0ξ + I0r − I01 − I0r I0ξ Fν K 01 + FKν K 11 + G(r, ξ) = √ 2π f R (Fν I01 − FKν I11 ) ln ξ + I0r K 0ξ + , 0 0. k − 1 + χ ˆ + 2π k + 2π k χr ˆ y y 44 r y4

(9.44)

Hence, it is a positive quantity. Since the first term on the right side of (9.43) is always greater than the second one, it follows that λ > 0 if χˆ is positive. The smaller eigenvalue belongs to the bending vibration mode of the beam, while the greater one corresponds to the vibration mode characterized by shear [6].

9.2 Eigenvalue Problem for a Class of Differential Equations 9.2.1 The Differential Equation System Determination of the eigenfrequencies of a vibrating Timoshenko beam has led to an eigenvalue problem governed by the homogeneous differential equation system (9.34a) and the homogeneous boundary conditions presented in Row 1 of Table 9.1. Consider now the class of those eigenvalue problems which is governed by the homogeneous ordinary differential equation system (referred to as eigenvalue problem of ODES) (9.45a) K[ y ] = λ M[ y ] and the homogeneous boundary conditions

9.2 Eigenvalue Problem for a Class of Differential Equations

Ur [ y ] = 0 ,

315

r = 1, 2, . . . 2κ

(9.45b)

where yT (x) = [y1 (x)|y2 (x)| . . . |yn (x)]; n ≥ 2 is the unknown function vector, 2κ and 2μ (κ > μ) are the order of the differential operators K and M—see Eq. (9.46) below—while λ is a parameter (the eigenvalue sought). Let K ν (x), (ν = 0, 1, ..., 2κ)

and

(n×n)

M ν (x), (ν = 0, 1, ..., 2μ, μ < κ) (n×n)

be continuous otherwise arbitrary quadratic matrices in the interval x ∈ [a, b], a < b. It is assumed that the matrices K 2κ (x) and M 2μ (x) have inverse for all x ∈ [a, b]. The differential operators K[ y ] and M[ y ] are defined by the following relationships: 2κ  dν (. . .) K[ y ] = K ν (x)y(ν) (x) , = (. . .)(ν) ; ν dx ν=0 (9.46) 2μ  M ν (x)y(ν) (x) . M[ y ] = ν=0

Let A νr (n×n)

and

B νr (n×n)

ν, r = 1, 2, . . . , 2κ

be constant quadratic matrices. The homogeneous boundary conditions prescribed by (9.45b) can be given in the following form: Ur [ y ] =

2κ  

 A νr y(ν−1) (a)+ B νr y(ν−1) (b) = 0 , r = 1, 2, . . . 2κ

(9.47)

ν=1

where the rows of the matrix U r should be linearly independent of each other. Eigenvalue problem (9.45) is said to be simple if M[ y ] = M 0 (x)y(x) .

(9.48)

The function vectors (or column matrices) u(x) and v(x) (|u(x)| and |v(x)| are not identically equal to zero if x ∈ [a, b]) are called (a) comparison function vectors if they satisfy boundary conditions (9.45b) and (b) eigenfunction vectors (or simply eigenfunctions) if they satisfy differential equation (9.45a) and boundary conditions (9.45b). The integrals  u, v K =



b

 u (x) K [ v(x)] dx , u, v M =



T

a

a

b

uT (x) M [ v(x)] dx

(9.49)

316

9 Eigenvalue Problems of Ordinary Differential Equation Systems

taken on the set of the comparison function vectors u(x), v(x) are products defined on the operators K and M. Eigenvalue problem of ordinary differential equations (9.45) is said to be selfadjoint if the products (9.49) are commutative, i.e., if it holds that   u, v K = v, u K

and

  u, v M = v, u M .

(9.50)

Exercise 9.3 Show that the eigenvalue problem defined by Eq. (9.34a) and the boundary conditions in Row 1 of Table 9.1 is self-adjoint. It is obvious that this eigenvalue problem is governed by the differential equation K[ y ] =

K ν (x) y(ν) (x) =

ν=0

  (1)      (2)  y1 y1 y1 0 χˆ 0 0 χˆ 0 = + + y2 y2 y2 −χˆ 0 0 −χˆ 01  

 

 

 =

2 

K2

K1

= λM[ y ] = λ

K0

  −1 0 y1 0 −r y2 y2  



(9.51a)

M0

associated with the boundary conditions y1 (0) = y1 () = 0

and

y2 (0) = y2 () = 0 .

(9.51b)

Hence, 

    uT K 2 v(2) dx + uT K 1 v(1) dx + uT K 0 v dx = 0 0 0       T (1) T (1)  (1) u K 2 v dx + uT K 1 v(1) dx+ = u K2 v 0 − 0 0   

 (1) (1)  T u K 0 vdx = χv ˆ 1 u 1 + v2 u 2  − χu ˆ 1 v2 0 − + 0   0

 u, v K =









 (1) (1) u 1 v2 + v1 u 2 dx −

− χˆ 0    − χv ˆ 2 u 2 dx = v, u K

0

=0



 (1) (1) χv ˆ 1(1) u (1) dx− 1 + v2 u 2 (9.52a)

0

and  u, v M =

 0



 uT M 0 v dx = −

 0

  v1 u 1 + r y2 v2 u 2 dx = v, u M ,

(9.52b)

9.2 Eigenvalue Problem for a Class of Differential Equations

317

which means that the boundary value problem considered is really self-adjoint. It can be proved that the eigenvalue problems defined by differential equation (9.34a) and the boundary conditions in Row 2, 3 and 4 of Table 9.1 are all selfadjoint. The proof is left for Problem 9.1.

9.2.2 Orthogonality of the Eigenfunction Vectors The line of thought in the present section is basically the same as that of Sect. 8.3. It is assumed that the eigenvalue problem (9.45) is self-adjoint. Let yk and y be two different eigenvector functions (k,  = 1, 2, 3, . . .), which therefore satisfy the differential equation systems K[ yk ] = λk M[ yk ] ,

K[ y ] = λ M[ y ]

(9.53)

where λk and λ are the eigenvalues that belong to the eigenfunction vectors yk and y —λk = λ . Multiply Eq. (9.53)1 by y (x), and Eq. (9.53)2 by yk (x). Integrating the result over the interval [a, b] yields



 y , yk = λk y  , y k K

M





yk , y = λ y k , y 

,

K

M

.

(9.54)

 

Since the products y , yk and y , yk are commutative, the difference (9.54)1 K M (9.54)2 is of the form 

= 0, (9.55) (λk − λ ) yk , y M

where λk − λ = 0. Hence 

yk , y

M

=0

k =  .

(9.56a)

In view of this equation, it follows from (8.12) that

 yk , y =0 K

k =  .

(9.56b)

Fulfillment of Eq. (9.56) proves that the eigenvector functions yk (x) and y (x) are orthogonal to each other in general sense. For M[ y ] = 1 y where 1 is the n × n unit matrix, the eigenvector functions are orthogonal in traditional sense. With regard to (9.54), it follows on the basis of (9.56) that

 yk , y = K



 λ y k , y  0

M

if k = , if k = .

k,  = 1, 2, 3 . . .

(9.57)

318

9 Eigenvalue Problems of Ordinary Differential Equation Systems

9.2.3 On the Reality of the Eigenvalues for Eigenvalue Problem (9.45) We shall prove that the eigenvalues are real numbers if the eigenvalue problem of ODES (9.45) is self-adjoint. Assume first that the eigenvalue is not real but complex. If the eigenvalue λ is complex, then so is the corresponding eigenvector function which can, therefore, be given in the following way: y (x) = u  (x) + iv  (x) ,

λ = a + ib,

where u  (x) and v  (x) are real vector functions, while a and b are real constants. It is also obvious that the complex conjugates y¯  and λ¯ satisfy differential equation (9.45a). On the basis of (9.54) we can write

y¯  , y

 K

 = λ y¯  , y

M

y , y¯ 

 K



= λ¯ y , y¯ 

M

.

(9.58)



Subtract (9.58)2 from (9.58)1 by taking into account that the products y¯  , y K 

and y¯  , y are commutative. We have M

 2ib y , y¯ 

M

  = 2ib u  , u  M + v  , v  M = 0 .

  This means that b = 0, i.e., the eigenvalue λ is real if u  , u  M + v  , v  M = 0.

9.2.4 Rayleigh Quotient For  = k Eq. (9.54) yields

from where it follows that

y , y

 K

 = λ y  , y 

y , y

M

(9.59)



K . λ = y , y

(9.60)

M

On the basis of Eq. (9.60), the Rayleigh quotient is defined on the set of comparative function vectors as  b  uT (x) K [u(x)] dx u, u K =  ab . (9.61) R[u(x)] =  u, u M uT (x) M [u(x)] dx a

9.2 Eigenvalue Problem for a Class of Differential Equations

319

The eigenvalue problem (9.45) is positive definite if the eigenvalues are positive, positive semidefinite if one eigenvalue is zero and the other eigenvalues are all positive, negative semidefinite if one eigenvalue is zero and the other eigenvalues are all negative, and finally negative definite if the eigenvalues are negative. Assume that   (9.62) u, u K > 0 , and u, u M > 0 for any comparison function vector u(x). Then the eigenvalue problem (9.45) is positive definite (or full definite). Exercise 9.4 Find the Rayleigh quotient for the eigenvalue problem defined by Eq. (9.34a) and the boundary conditions in Row 1 of Table 9.1. Comparison of (9.61) and (9.52) yields  u, u K = R[u(x)] =  u, u M



 0

    (1) 2 (1) 2 2 χˆ (u (1) + (u dx ) + (u ) ) + 2 χu ˆ u 2 2 1 2 1 .     2 2 2 (u 1 ) + r y (u 2 ) dx 0

(9.63)

9.2.5 Determination of Eigenvalues The general solution of the differential equation K[ y ] = λ M[ y ] can be given in the following form: 2κ×n  y(x) = z  (x, λ)A , (9.64) =1

where z  (x, λ) stand for the linearly independent particular solutions while the undetermined integration constants are denoted by A . The latter constants can be obtained from the boundary conditions: 2κ×n 

Ur [z (x, λ)] A = 0 ,

r = 1, 2, . . . 2κ .

(9.65)

=1

Since these equations constitute a homogeneous linear equation system for the unknown A solutions different from the trivial one (nontrivial solutions) exist if and only if the determinant of the system is zero:   (λ) = det Ur [z (x, λ)] = 0 .

(9.66)

This is the equation which should be solved to find the eigenvalues λ. We remark that (λ) is often referred to as characteristic determinant.

320

9 Eigenvalue Problems of Ordinary Differential Equation Systems

9.2.6 The Green Function Matrix Consider the inhomogeneous ordinary differential equation system L [ y(x)] = r(x),

(9.67a)

where the differential operator of order κ is given by the equation L[ y ] =

κ 

P ν (x)y(ν) (x),

(9.67b)

ν=0

in which yT (x) = [y1 (x)|y2 (x)| . . . |yn (x)]; n ≥ 2 is the unknown function vector, κ is the order of the differential operator, r(x) = [r1 (x)|r2 (x)| . . . |rn (x)] is a known inhomogeneity and the coefficients P ν (x) , (n×n)

ν = 0, 1, ..., κ , x ∈ [a, b], a < b

are continuous otherwise arbitrary quadratic matrices. It is assumed that P κ (x) has an inverse for all x ∈ [a, b]. If there exist no inverse, the differential equation system is called degenerated [7]. Let A νr (n×n)

and

B νr (n×n)

ν, r = 1, 2, . . . , κ

be constant quadratic matrices. The differential equation system (9.67) is associated with the following homogeneous boundary conditions: Ur [ y ] =

κ    A νr y(ν−1) (a) + B νr y(ν−1) (b) = 0 , r = 1, 2, . . . κ

(9.68)

ν=1

which are basically the same as the boundary conditions given by Eq. (9.47). Solution of the boundary value problem (9.67), (9.68) can be given in the form  y(x) =

b

G(x, ξ)r(ξ) dξ,

(9.69)

a

where G(x, ξ) is the Green function matrix [8] defined by the following four properties: 1. The Green function matrix is a continuous function of x and ξ in each of the triangles a ≤ x ≤ ξ ≤ b and a ≤ ξ ≤ x ≤ b. In addition, it is κ times differentiable with respect to x and the derivatives

9.2 Eigenvalue Problem for a Class of Differential Equations

∂ n G(x, ξ) = G(n) (x, ξ) , ∂x n

321

(n = 1, 2, . . . , κ)

are also continuous functions of x and ξ in the triangles a ≤ x ≤ ξ ≤ b and a ≤ ξ ≤ x ≤ b. 2. Let ξ be fixed in [a, b]. The Green function matrix and its derivatives G(n) (x, ξ) =

∂ n G(x, ξ) , ∂x n

(n = 1, 2, . . . , κ − 2)

(9.70)

should be continuous for x = ξ:   lim G(n) (ξ + ε, ξ) − G(n) (ξ − ε, ξ) =   = G (ξ + 0, ξ) − G(n) (ξ − 0, ξ) = 0 , (n = 0, 1, 2, . . . κ − 2). (9.71a) ε→0 (n)

The derivative G(κ−1) (x, ξ) should, however, have a jump   lim G(κ−1) (ξ + ε, ξ) − G(κ−1) (ξ − ε, ξ) = ε→0   = G(κ−1) (ξ + 0, ξ) − G(κ−1) (ξ − 0, ξ) = P−1 κ (ξ)

(9.71b)

if x = ξ. 3. Let αT = [α1 |α2 | . . . |αn ] , αν = 0 (ν = 1, 2, . . . , n) be an arbitrary constant matrix with finite elements. For a fixed ξ ∈ [a, b], the product G(x, ξ)α as a function of x (x = ξ) should satisfy the homogeneous differential equation   L G(x, ξ)α = 0 . 4. The product G(x, ξ)α as a function of x should satisfy the boundary conditions   Ur G(x, ξ)α = 0 ,

(r = 1, 2, 3, . . . , κ).

(9.72)

9.2.7 Calculation of the Green Function Matrix The definition of the Green function matrix given in the previous subsection is a constructive one which means that it provides the means that are needed to calculate the Green function matrix. The general solution of the homogeneous differential equation system L [ y(x)] = r(x) can be given in the form

(9.73)

322

9 Eigenvalue Problems of Ordinary Differential Equation Systems

y=

 κ 



Y  (x) C  (n×n) =1 (n×n)

e ,

(9.74)

(n×1)

where each column of the matrices Y  satisfies the homogeneous differential equation system (9.73), C is a constant quadratic matrix, while e is a constant column matrix (a constant vector). Recalling the third property of the definition and the structure of the general solution (9.73) the Green function matrix can be expressed in the following manner [8]: κ    Y  (x) A  (ξ) ± B  (ξ) , (9.75) G(x, ξ) = (n×n)

=1

where the sign is [positive](negative) if [x ≤ ξ](ξ ≤ x). The above choice automatically assures the fulfillment of the first property of the definition. On the basis of the second property, the following equations can be set up for calculating the elements of the matrices B  (ξ): κ 

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬

Y  (ξ) B  (ξ) = 0 ,

=1 κ 

Y(1)  (ξ) B  (ξ) = 0 ,

=1

···

⎪ ⎪ ⎪ ⎪ ⎪ (κ−2) ⎪ Y  (ξ) B  (ξ) = 0 , ⎪ ⎪ ⎪ ⎪ =1 ⎪ ⎪ ⎪ ⎪ κ ⎪  (κ−1) 1 −1 ⎪ ⎪ Y  (ξ) B  (ξ) = − P κ (ξ) ,⎪ ⎭ 2

(9.76)

κ 

=1

where 0 is the n × n zero matrix. Let us denote the ν-th column (ν = 1, 2, . . . , n) of the matrix B  by B ν : (n×1)

B (n×n)

= [ B 1 | B 2 | · · · | B n ]. (n×1) (n×1)

(9.77)

(n×1)

The matrix B Tν

(1×(κ×n))

  = BT1ν | . . . | BTiν | . . . |BTκν , (1×n)

(1×n)

(1×n)

i = 2, 3, . . . , κ − 1

is that of the unknowns for a fixed ν. The i-th column (i = 1, 2, . . . , n) of the matrix Y  is denoted by η i :

(9.78)

9.2 Eigenvalue Problem for a Class of Differential Equations

Y (n×n)

323

= [ η 1 | η 2 | · · · | η n ]. (n×1) (n×1)

(9.79)

(n×1)

The matrix B ν is the solution of a linear equation system of the form W B ν = P ν,

(9.80a)

where ⎡

η 11 · · · η 1n · · · η 1 · · · η n · · · η κ1 · · · η κn

⎤

⎥ ⎢ ⎢ η (1) · · · η (1) · · · η (1) · · · η (1) · · · η (1) · · · η (1) ⎥ ⎢ 11 1n 1 n κ1 κn ⎥ W=⎢ ⎥ ⎥ ⎢ .................................. ⎦ ⎣ (κ−1) (κ−1) (κ−1) (κ−1) (κ−1) · · · η · · · η · · · η · · · η · · · η η (κ−1) 11 1n 1 n κ1 κn

(9.80b)

and P ν is the transpose of the ν-th row in the matrix: 

 . 0 0 · · · 0 − 21 P−1 κ 1 2 κ−1 κ

(9.80c)

Note that the coefficient matrix W is the same for each B ν . It is also worth mentioning that the first three properties of the definition have all been satisfied. After having determined the matrices B  , we can proceed with the calculation of the matrices A  by using the fourth property of the definition. Let α be the ν-th unit vector in the n × n space:   αT = 0 0 · · · 1 · · · 0 . 1 2 ν n

(9.81)

With the above αT , the fourth property leads to the following equations: Ur

 n 

 Y  (x)A  (ξ)α = ∓Ur

=1



n 

 Y  (x)B  (ξ)α

r = 1, 2, . . . , κ . (9.82)

=1

The matrix  ATν (1×(κ×n))

=

AT1ν | . . . | ATiν | . . . |ATκν (1×n) (1×n) (1×n)

 ,

i = 2, 3, . . . , κ − 1

(9.83)

is that of the unknowns for a fixed ν in α. Making use of (9.78) and (9.82), the equation system to be solved is of the form     U r Y 1 , Y 2 , . . . , Y n A ν = ∓U r Y 1 , Y 2 , . . . , Y n B ν r = 1, 2, 3, . . . , κ (9.84)

324

9 Eigenvalue Problems of Ordinary Differential Equation Systems

  where U r Y 1 , Y 2 , . . . , Y n is a matrix with size n × (κ × n). If the boundary conditions are linearly independent, the determinant of equation system (9.84) is different from zero. Then equation system (9.84) is solvable, i.e., there is a unique solution for Aν (for the matrices A ). This means that there exists the Green function matrix of the boundary value problem (9.67), (9.68).

9.2.8 Symmetry of the Green Function Matrix The present section is a formal generalization of Sect. 8.10. Consider the following two inhomogeneous boundary value problems: L [u(x)] = r(x) , L [v(x)] = s(x) ,

U r [u(x)] = 0 ; U r [v(x)] = 0,

(9.85a) (9.85b)

where L is defined by Eq. (9.67b), u(x) and v(x) stand for the unknown functions, while r(x) and s(x) are continuous inhomogeneities in the interval x ∈ [a, b]. It is assumed that the boundary value problems (9.85a) and (9.85b) are self-adjoint. Hence, (u, v) L − (v, u) L = 0, in which on the basis of Eq. (9.69) it holds that 

b

u(x) =

 G(x, ξ)r(ξ) dξ

a

and

v(x) =

b

G(x, ξ)s(ξ) dξ .

a

Consequently,  (u, v) L −(v, u) L =  = a

b



b

 T u L[v] − v T L[u] dx =

a b

  sT (ξ) GT (x, ξ) − G(ξ, x) r(ξ) dξdx = 0,

(9.86)

a

where both r(x) and s(x) are arbitrary continuous and non-zero vector functions in the interval [a, b]. Thus GT (x, ξ) = G(ξ, x) . (9.87) If this equation is satisfied, the Green function matrix is called cross-symmetric.

9.2 Eigenvalue Problem for a Class of Differential Equations

325

9.2.9 Green Function Matrices for Some Beam Problems 9.2.9.1

Pinned-Pinned Timoshenko Beam

Find the Green function matrix for the boundary value problem governed by the differential equation system  AGκ y

d2 w dψ y + d xˆ 2 d xˆ

 + fz = 0 ,

d2 ψ y IE − AGκ y d xˆ 2



 dw + ψy + μy = 0 d xˆ (9.88a)

and the boundary conditions w(0) = ψ (1) y (0) = 0,

w() = ψ (1) y () = 0,

(9.88b)

where, for the sake of our later considerations, the earlier independent variable x is denoted by x. ˆ According to (9.15) and Table 9.1, these equations describe the mechanical behavior of a pinned-pinned Timoshenko beam if the axial force N is zero. Introducing the dimensionless variables xˆ = x,

y1 =

w , 

y2 = ψ y , χ =

AGκ y 2 = χ ˆ 2, IE

(9.89)

where x is a new dimensionless independent variable differential equations (9.88a) and boundary conditions (9.88b) can be rewritten in the following form: 

d2 y1 dy2 χ + dx 2 dx and



3 =− f z = r1 , , IE

y1 (0) = y2(1) (0) = 0,

  2 d2 y2 dy1 = − + y μ y = r2 − χ 2 dx 2 dx IE (9.90a) y1 (1) = y2(1) (1) = 0 .

(9.90b)

By using matrix notation, differential equations (9.90a) can be rewritten in the following form: 

  (2)    (1)      χ0 y1 y1 y1 r 0 χ 0 0 = 1 . + + y2 y2 y2 r2 01 −χ 0 0 −χ  

 

 

 

K2

K1

K0

(9.91)

r

On the basis of Eq. (9.25), the general solution of the homogeneous equation assumes the following form: 

y1 y2





     1 2 − 2 x − 13 x 3 1 x C1 C3 = + . x χ2 + x 2 C2 C4 0 −1  

 

Y 1 (x)

Y 2 (x)

(9.92)

326

9 Eigenvalue Problems of Ordinary Differential Equation Systems

The Green function matrix of differential equation (9.91) can be expressed as G(x, ξ) = (2×2)

2 

  Y  (x) A  (ξ) ± B  (ξ) ,

(9.93)

=1

where the sign is [positive](negative) if [x ≤ ξ](ξ ≤ x). For the sake of the further calculations, it is worth partitioning the matrices Y  , B  , and A  : ⎤   Y Y ⎣ 11 12 ⎦ ⎡ Y =





Y 21 Y 22

  1 { Y 1 = , 1 { Y 2  

(9.94a)

(2×2)

⎡

⎤     A A 11 12 A A ⎣ ⎦ 1 2 , A =   = (2×1) (2×1) A21 A22

(9.94b)

⎡

⎤     B B 11 12 ⎦ B B ⎣ 1 2 . B =   = (2×1) (2×1) B 21 B 22

(9.94c)

It follows from Eq. (9.76) that the matrices B  should satisfy the following equations: 2  =1

2  =1





⎡

1

1

⎤

1 ξ ⎣ B 11 B 12 ⎦ + 1 1 0 −1 B 21 B 22 ⎤ ⎡  2 2    1 2 − 2 ξ − 13 ξ 3 ⎣ B 11 B 12 ⎦ 00 = , + 2 2 ξ χ2 + ξ 2 00 B 21 B 22

Y  (ξ) B  (ξ) =





⎡

1

1

⎤

0 1 ⎣ B 11 B 12 ⎦ + 1 1 00 B 21 B 22 ⎤ ⎡   2 2   2 1 χ1 0 −ξ −ξ ⎣ B 11 B 12 ⎦ = − . + 2 2 1 2ξ 2 01 B 21 B 22

Y(1)  (ξ) B  (ξ) =

(9.95a)

(9.95b)

By introducing the new variables 1

1

2

2

a = B 1i , b = B 2i c = B 1i d = B 2i for i = 1 we get the equation system

(9.96)

9.2 Eigenvalue Problem for a Class of Differential Equations

⎡

1 ⎢0 ⎢ ⎣0 0

327

⎤⎡ ⎤ ⎡ ⎤ 0 ξ − 21 ξ 2 − 13 ξ 3 a ⎢ ⎥ ⎢ ⎥ −1 ξ χ2 + ξ 2 ⎥ ⎥⎢b⎥ ⎢ 0 ⎥ 2 ⎦⎣ c ⎦ = ⎣− 1 ⎦ 1 −ξ −ξ 2χ d 0 0 1 2ξ

from where 1

1 1 1 1 1 ξ − ξ 3 , b = B 21 = ξ 2 − , 2χ 12 4 2χ 2 1 1 d = B 21 = − . = ξ, 2 4

a = B 11 = 2

c = B 11

(9.97a)

If i = 2 we have in a similar way: ⎡

1 ⎢0 ⎢ ⎣0 0

⎤⎡ ⎤ ⎡ ⎤ ξ − 21 ξ 2 − 13 ξ 3 0 a 2 ⎢ ⎥ ⎢ ⎥ −1 ξ χ + ξ 2 ⎥ ⎥⎢b⎥ ⎢ 0 ⎥ 2 ⎦⎣c ⎦ = ⎣ 0 ⎦ 1 −ξ −ξ d − 21 0 1 2ξ

from where

1

1 1 2 1 ξ , b = B 22 = − ξ, 4 2 2 1 = − , d = B 22 = 0 . 2

a = B 12 = 2

c = B 12

(9.97b)

Note that the matrices B  are independent of the boundary conditions. According to Eq. (9.72) the product G(x, ξ)α should satisfy boundary conditions (9.90b). If αT = [1|0], the boundary conditions yield the following equation system: 2 

2    Y 1 (x)x=0 A 1 (ξ) = − Y 1 (x)x=0 B 1 (ξ),

=1

=1

2 

 Y 1 (x)x=1 A 1 (ξ) = +

2 

  Y(1) 2 (x)

2 

=1 2 

=1

=1 2  =1

 Y 1 (x)x=1 B 1 (ξ),

  Y(1) (x)  2

x=0

A 1 (ξ) = −

x=1

A 1 (ξ) = +

x=0

B 1 (ξ),

  Y(1) (x)  2

x=1

B 1 (ξ).

=1 2  =1

(9.98)



 Y(1) 2 (x)

If αT = [0|1], the boundary conditions result in the equation system

328

9 Eigenvalue Problems of Ordinary Differential Equation Systems 2 

2    Y 1 (x)x=0 A 2 (ξ) = − Y 1 (x)x=0 B 2 (ξ),

=1 2 

=1 2    Y 1 (x)x=1 A 2 (ξ) = + Y 1 (x)x=1 B 2 (ξ),

=1 2 

=1

  Y(1) (x)  2

=1 2 

  Y(1) (x)  2

=1

x=0

A 2 (ξ) = −

x=1

A 2 (ξ) = +

2 

x=0

B 2 (ξ),

  Y(1) (x)  2

x=1

B 2 (ξ).

=1 2 

(9.99)



 Y(1) 2 (x)

=1

Since the coefficient matrices on the left sides are the same in (9.98) and (9.99), the two equations can be united: ⎡1 ⎤ ⎡ ⎡ ⎤ A1i ⎤ ⎥ −a 10 0 0 ⎢ 1 ⎥ ⎢ ⎢ 1 1 − 1 − 1 ⎥ ⎢ A2i ⎥ ⎢ a + b − 1 c − 1 d ⎥ ⎢ ⎥ ⎢ 2 3 ⎥⎢ 2 3 ⎥ i = 1, 2 . ⎣0 0 1 0 ⎦⎢ 2 ⎥ = ⎣ ⎦ −c ⎢ Ai1 ⎥ 00 1 2 ⎣2 ⎦ c + 2d A2i The solutions are as follows: 1 2 A2i = 2a + b − c , 3

1

A1i = −a , 2

2

A1i = −c ,

A2i = c + d

from where by substituting (9.97a) and (9.97b) we obtain 1

1 1 ξ + ξ3 , 2χ 12 1 = − ξ, 2

1

ξ ξ 1 1 1 + − + ξ2 − ξ3 2χ χ 3 4 6 1 1 = ξ− 2 4

A11 = −

A21 = −

2

2

A11 and

A21

1 1 A12 = − ξ 2 , 4 2 1 A11 = , 2

1

1 1 2 1 ξ − ξ+ , 2 2 3 1 =− . 2

(9.100a)

A22 = 2

A22

Hence,     G(x, ξ) = Y 1 (x) A 1 (ξ) ± B 1 (ξ) + Y 2 (x) A 2 (ξ) ± B 2 (ξ) =     1 1 3 ξ + 12 ξ − 41 ξ 2 − 2χ 1 x = ± 1 0 −1 − 2χ + χξ − 3ξ + 41 ξ 2 − 16 ξ 3 21 ξ 2 − 21 ξ + 13

(9.100b)

9.2 Eigenvalue Problem for a Class of Differential Equations

 ±  +

− 21 x 2 − 13 x 3 x

2 χ

1 1 3 1 2 ξ − 12 ξ 4ξ 2χ 1 2 1 ξ − 2χ − 21 ξ 4

 

− 21 ξ

+ x2

1 ξ 2

−

1 2 1 4

− 21

329

* +



 ±

1 ξ 2 − 41

− 21

* (9.101)

0

is the Green function matrix. Since the boundary value problem defined by differential equation (9.91) and boundary conditions (9.90b) is self-adjoint —we remind the reader of Exercise 9.3—it follows that the above Green function matrix is crosssymmetric. Exercise 9.5 A concentrated mass is attached to the simply supported beam shown in Fig. 9.2. It is assumed that the mass m is much greater than that of the beam. The effect of the beam mass can, therefore, be neglected. Find the circular frequency of the mass if it vibrates vertically given that the effect of the rotation can be left out of consideration. Fig. 9.2 Simply supported beam with a concentrated mass

z

m x

a

P1

b

If we use the Euler–Bernoulli beam theory, the flexibility at the point x = a can be determined using the Green function given for simply supported beams in Table 8.1: 



f 11 =





G (a, ξ) δ (a − ξ) dξ =

0

0

Hence, k11 =

 a (−ξ) 2ξ −ξ 2 −a 2 δ (a −ξ) dξ = 6I E a 2 b2 ab  2a − 2a 2 = . = 6I E 3I E 1 3I E = 2 2 f 11 a b

is the corresponding rigidity and ω2 =

3I E k11 = 2 2 m a b m

(9.102)

is the square of the circular frequency sought. The solution steps are the same within the framework of the Timoshenko beam theory. The flexibility can be calculated utilizing the Green function matrix (9.101) and taking into account the fact what form the right side of the differential equations

330

9 Eigenvalue Problems of Ordinary Differential Equation Systems

(9.90a) have. Recalling that ξ and x are now dimensionless coordinates and using the relations xa = a/, r1 (ξ) = −δ (xa − ξ) 3 /I E, r2 (ξ) = 0, ξˆ = ξ we have 

   1  r1 (ξ) G 11 (xa , ξ) G 12 (xa , ξ) y1 (xa ) = dξ = G 21 (xa , ξ) G 22 (xa , ξ) y2 (xa ) r2 (ξ) 0 ⎡



⎤    G ξˆ ξˆ ξˆ 11 x a ,  G 12 x a ,  3 dξˆ δ x − a  ⎣



⎦ =− ˆξ ˆξ IE 0  0 G 21 xa ,  G 22 xa , 

from where    1 3 ξˆ  ξˆ dξˆ w (xa ) = f 11 = − = G 11 xa , δ xa − y1 (xa ) =   IE 0       *    − χ1 xa + 16 xa3  1 2 1 3  −xa 1 3  1 xa =− + = x − x − xa 1 2 a 3 a x − x3a − 16 xa3  IE 2 a χ =

1 1 ab   1 3 xa χab + 32 = (1 − xa ) −χxa2 + χxa + 3 =  3I E χ  3I E χ   2 1 ab ab + 3 . =  3I E χ

Hence, f 11

  2 ab ab + 3 = 3I E χ

k11 =

and

1 3I E .

= 2 f 11 ab ab + 3 χ

Consequently, ω2 =

3I E k11 =

= 2 m mab ab + 3  χ

9.2.9.2

χ=

↑

AGκ y 2 IE

↑

↑=

E=2G(1+ν) r y2= I A

3I E .

6r 2 (1+ν) mab ab + y κ y (9.103)

Fixed-Pinned Timoshenko Beam

According to Table 9.1 differential equation (9.91) is associated with the following boundary conditions: y1 (0) = y2 (0) = 0,

y1 (1) = y2(1) (1) = 0 .

(9.104)

When calculating the Green function matrix, it is sufficient to determine the matrices A1 and A2 since the matrices B1 and B2 are independent of the boundary conditions. Without entering into details, we have

9.2 Eigenvalue Problem for a Class of Differential Equations

331

      1 x G(x, ξ) = Y 1 (x) A 1 (ξ) ± B 1 (ξ) + Y 2 (x) A 2 (ξ) ± B 2 (ξ) = × 0 −1   1 2 1 1 3 ξ + 12 ξ ξ − 2χ 4  3 × ± ξ 1 2ξ χ + ξ 2 χ2 − 3ξ 2 χ − 12ξ + 2χ + 6 21 χ+3 − 4χ(χ+3) (3ξ + χ − 3) *    1 2 1 1 3 ξ − ξ ξ − 21 x 2 − 13 x 3 2χ 12 4 × + ± 1 2 2 1 x + x2 ξ − 2χ − 21 ξ χ 4    +  1 −2 3ξ − ξχ + 3ξ 2 χ − ξ 3 χ −2 3χξ 2 − 6χξ + χ + 3 × ± 3χξ (ξ − 2) 4 (χ + 3) −χξ 3 + 3χξ 2 + 6ξ − χ − 3 ,  1 ξ −1 . (9.105) ± 21 2 −4 0 9.2.9.3

Fixed-Fixed Timoshenko Beam

The Green function matrix is as follows:     G(x, ξ) = Y 1 (x) A 1 (ξ) ± B 1 (ξ) + Y 2 (x) A 2 (ξ) ± B 2 (ξ) =     1 1 3 ξ + 12 ξ − 14 ξ 2 − 2χ 1 x  = ± ξ 1 0 −1 8ξ 3 χ+ξ 2 χ2 −48ξ +2χ+24 2(χ+12) − 4χ(χ+12) (12ξ +χ) *    1 1 3 1 2 ξ − 12 ξ 4ξ − 21 x 2 − 13 x 3 2χ × + ± 1 2 1 x χ2 + x 2 ξ − 2χ − 21 ξ 4   +  1 1 ξχ−12ξ 2 −4ξ 2 χ+2ξ 3 χ − 21 χ−24ξ −8ξχ+6ξ 2 χ+12 2 ± × 3ξχ (ξ − 1) χ+12 − 41 4χξ 3 −6χξ 2 −24ξ +χ+12  1 , ξ −1 ± 21 2 . (9.106) −4 0 The paper-and-pencil calculations that lead to the above result are again omitted. 9.2.9.4

Fixed-Free Timoshenko Beam

For completeness, the Green function matrix for the fixed-free Timoshenko beam is also presented here:     G(x, ξ) = Y 1 (x) A 1 (ξ) ± B 1 (ξ) + Y 2 (x) A 2 (ξ) ± B 2 (ξ) =   *    1 2 1 1 3 1 1 3 ξ + 12 ξ − 41 ξ 2 ξ − ξ ξ − 2χ 1 x 4  2 ± 2χ1 2 121 + = 1 1 0 −1 − 4χ ξ ξ − 2χ − 21 ξ χξ + 2 2 4  + 1  1 2   1 , −2x − 13 x 3 ξ − 21 ξ − 21 2 2 + ± . (9.107) 2 x + x2 − 41 0 − 14 0 χ

332

9 Eigenvalue Problems of Ordinary Differential Equation Systems

9.2.9.5

Axially Loaded and Pinned-Pinned Timoshenko Beam

Find the Green function matrix if the pinned-pinned Timoshenko beam is axially loaded. It is obvious on the basis of Eq. (9.15) and Table 9.1 that the Green function matrix in question belongs to the boundary value problem governed by the differential equations 

 d2 w dψ y + fz = 0 , + AGκ y dx 2 dx   d2 ψ y dw dw − AGκ y + ψ y + μ y = 0, ±N IE dx 2 dx dx

(9.108a)

which are associated with the boundary conditions w(0) = ψ (1) y (0) = 0,

w() = ψ (1) y () = 0.

(9.108b)

Note that Eq. (9.108a) differs from Eq. (9.88a) in one term only while boundary conditions (9.88b) and (9.108b) coincide with each other. Recalling definition (9.90a) of r1 , r2 and introducing dimensionless variables by the use of Eq. (9.89) differential equations (9.108a) can be rewritten in the following form: 

  (2)    (1)      χ0 y1 y1 y1 r 0 χ 0 0 = 1 , + + y2 y2 y2 r2 01 −χ ± N 0 0 −χ 

 

  

K2

K1

(9.109)

K0

where N = n2 =

N 2 . IE

(9.110)

(a) Green function matrix if the axial force is a compressive one. If the axial force is compressive, the sign of N is negative in differential equation (9.109). Recalling (9.20) and (9.21), it is obvious that the general solution of the homogeneous equation is given by the following equation: 

y1 y2



   1   1 x C1 C3 cos nx − n1 sin nx n = + . 0 − χ1 (χ + N ) C2 C4 sin nx cos nx 

  



(9.111)

Y 2 (x)

Y 1 (x)

The relatively long calculation of the Green function matrix is left for Problem 9.3. Here the final result is presented only:     G(x, ξ) = Y 1 (x) A 1 (ξ) ± B 1 (ξ) + Y 2 (x) A 2 (ξ) ± B 2 (ξ) =   *    ξ ξ 1 − 2n1 2 − χ(κ−1) 1 x 2χ(κ−1) n2 = ± + 1 1 0 −κ 0 − 2ξ−1 2 2χ(κ−1)

n

χ(κ−1)

9.2 Eigenvalue Problem for a Class of Differential Equations

1

cos nx − n1 sin nx n sin nx cos nx

1 + 2

 

κ sin nξ − 2χ(κ−1)

333



cos nξ 2n ± κ(2 sin nξ cos n+cos nξ sin n) 2 cos n cos nξ+sin n sin nξ 2χ(κ−1) sin n 2n sin n  κ sin nξ * − cos2nnξ 2χ(κ−1) ± κ cos nξ , (9.112a) sin nξ 2χ(κ−1) 2n

where κ=

1 (χ + N ) . χ

(9.112b)

(b) Green function matrix if the axial force is a tensile one. Then the sign of N is positive in differential equation (9.109). On the basis of Eqs. (9.23) and (9.24), it follows that the general solution of the homogeneous equation assumes the following form: 

   1     1 x C1 y1 − n cosh nx − n1 sinh nx C3 = + . 0 − χ1 (χ−N ) C2 y2 C4 sinh nx cosh nx 

  

(9.113)

Y 2 (x)

Y 1 (x)

Calculation of the Green function matrix is detailed in the solution of Problem 9.3. The result is as follows:     G(x, ξ) = Y 1 (x) A 1 (ξ) ± B 1 (ξ) + Y 2 (x) A 2 (ξ) ± B 2 (ξ) = *      ξ ξ 1 − n12 − 2χ(κ−1) 1 x 2χ(κ−1) 2n 2 + = ± 2ξ−1 1 0 −κ 0 − 2χ(κ−1) − n12 2χ(κ−1)  cosh nx sinh nx  − n − n + × sinh nx cosh nx   κ sinh nξ cosh nξ ×

2χ(κ−1) 2n ± κ(cosh nξ sinh n−2 sinh nξ cosh n) sinh nξ sinh n−2 cosh nξ cosh n 2χ(κ−1) sinh n n sinh n  κ sinh nξ cosh nξ * − 2χ(κ−1) − 2n ± κ cosh nξ sinh nξ , 2χ(κ−1) 2n

where κ=

1 (χ − N ) . χ

(9.114)

(9.115)

9.2.10 Solution of Static Boundary Value Problems Let us assume that the Green function matrices are known for the boundary value problems related to pinned-pinned, fixed-pinned, fixed-fixed, and fixed-free Timo-

334

9 Eigenvalue Problems of Ordinary Differential Equation Systems

shenko beams provided that there are no axial force acting on the beams. Assume further that the Green function matrix of the axially loaded and pinned-pinned Timoshenko beam is also known. If the dimensionless distributed load i.e., r(x), which is the right side of Eq. (9.91) (or Eq. (9.109)), is known then, on the basis of Eq. (9.69), the solution for y can be given in a closed form:  y(x) =

1

G(x, ξ)r(ξ) dξ .

(9.116)

0

9.2.11 Multiple Eigenvalues of ODES Let the eigenvalue problem (9.45) of ODES be self-adjoint and positive definite. Then the eigenfunction vectors yi (x) (i = 1, 2, 3, . . .) fulfill the condition

 yi , yi > 0. M

It is clear that the normalized eigenfunction vectors -  ψ i = yi / yi , yi

(9.117a)

M



ψi , ψi

satisfy the equation

M

= 1.

(9.117b)

Comparison of Eqs. (9.54)1 , (9.56), and (9.117) shows that the normalized eigenfunction vectors ψ k (x) also satisfy the equation

ψi , ψn

+

 K

=

λn if i = n, 0 if i = n.

i, n = 1, 2, 3, . . .

(9.118)

The eigenvalue λ L (L ∈ 1, 2, 3, . . .) of the eigenvalue problem (9.45) is a multiple one with multiplicity r > 1 if the number of the linearly independent eigenfunction vectors that belong to λ L is r . The normalized eigenfunction vectors with multiplicity r are denoted by ψ k L (x) (k ∈ 1, 2, . . . , r ) where the second subscript is that of the eigenvalue λ L . It is obvious that K[ ψ k L ] = λ L M[ ψ k L ] , Ur [ ψ k L ] = 0 ,

(k = 1, 2, . . . , r ).

(9.119)

Let c1k , c2k (k = 1, 2, . . . , r ), etc. be given weights or coefficients. The linear combinations ˆ 1L = ψ

r  k=1

ˆ 2L = c1k ψ k L , ψ

r  k=1

ˆ rL = c2k ψ k L , . . . ψ

r  k=1

cr k ψ k L

(9.120)

9.2 Eigenvalue Problem for a Class of Differential Equations

335

are also eigenfunction vectors of the eigenvalue problem. The normalized eigenfunctions ψ k L (x) are, however, not orthogonal to each other. Recalling the Gram–Schmidt orthogonalization presented in Sect. 8.12, it can be seen that the procedure ˆ 1L = ψ 1L , ψ  ˆ kL , ˆ k L ψ i+1,L , ψ ˆ∗ i = 1, 2, . . . , r − 1 ψ ψ i+1,L = ψ i+1,L − M k=1    ∗  ˆ∗ ˆ ˆ∗ ˆ i+1,L ˆ i+1,L , ψ ˆ i+1,L = ψ / ψ ,ψ , ψ =1 ψ i 

i+1,L

i+1,L

i+1,L M

M

(9.121) yields orthogonal eigenfunction vectors.

9.2.12 Properties of the Rayleigh Quotient of ODES We shall assume that the eigenvalue problem of ODES (9.45) is positive definite and simple. Then it holds that   (9.122) M y = M0 (x)y(x),   where the products u, u K , u, u M regarded on the set of the comparison vectors u are positive. It can be proved—we remind the reader of the proofs given in Sect. 8.13 and in the book [9]—that the Rayleigh quotient  u, u K R[u(x)] =  u, u M

(9.123)

has the following properties: 1. The Rayleigh quotient R[u(x)] is a positive quantity. 2. Let .  b |u| = uT u dx .

(9.124)

a

The Rayleigh quotient is independent of |u|. 3. The Rayleigh quotient has a positive lower limit. 4. The lower limit of the Rayleigh quotient is the smallest eigenvalue λ1 . Exercise 9.6 Estimate the first eigenvalue for the vibrating pinned-pinned Timoshenko beam. As is well known, the Rayleigh quotient provides an upper limit for λ1 . Hence, (9.63) gives the estimation

336

9 Eigenvalue Problems of Ordinary Differential Equation Systems

 R[u(x)] =

 0

    2 2 2 χˆ (u (1) + (u (1) ˆ (1) 1 ) + (u 2 ) 2 ) + 2χu 1 u 2 dx > λ1 ,     2 2 2 (u 1 ) + r y (u 2 ) dx

(9.125)

0

in which u 1 and u 2 should be comparative functions. It can be checked with ease that the functions πx πx π , u 2 = − cos u 1 = sin    are comparative ones. Note that u 1 is the first eigenfunction of the vibrating pinned-pinned Euler– Bernoulli beam, while u 2 is the corresponding rotation. Substitution of u 1 and u 2 into (9.125) yields     

π 4   π4 (1) 2 (1) 2 2 2 πx dx = χˆ (u (1) +(u dx = ) +(u ) ) + 2 χu ˆ u sin 2 2 1 2 1   23 0 0 for the numerator and    

π 2     1  2 πx πx dx +r y2 dx =  +π 2 r y2 (u 1 )2+r y2 (u 2 )2 dx = sin2 cos2    2 0 0 0 for the denominator of the Rayleigh quotient. Thus R[u(x)] =

π4  > λ1 2 2 + π 2 r y2

(9.126)

is the estimation for the first eigenvalue. It is worth mentioning that estimation (9.126) (a) is independent of κ y and (b) coincides with the first eigenvalue of the vibrating pinned-pinned Euler–Bernoulli beam if r y2 is set to zero.

9.2.13 Eigenvalue Problems Governed by a System of Fredholm Integral Equations Equation

 y(x) = λ a

(2×1)

b

K(x, ξ) y(ξ) dξ (2×2)

(9.127)

(2×1)

is a Fredholm integral equation system in which x, ξ ∈ [a, b], b > a, y(x) is the unknown function vector (unknown function matrix), K(x, ξ) is the given kernel function matrix (or simply kernel):  K=

 K11 (x, ξ) K12 (x, ξ) , K21 (x, ξ) K22 (x, ξ)

(9.128)

9.2 Eigenvalue Problem for a Class of Differential Equations

337

while λ is a parameter (the eigenvalue sought). It will be assumed that (a) the kernel is differentiable with respect to x and ξ, (b) the kernel satisfies the inequality K(x, ξ) =

m 

a(x) b(ξ),

=1 (2×1) (1×2)

m≥1

(9.129)

in which a(x) and b(ξ) are differentiable, and (c) the kernel is cross-symmetric: K(x, ξ) = KT (ξ, x) .

(9.130)

Under these conditions, (a) the number of different eigenvalues is infinite, (b) the eigenvalues are real numbers, and (c) the number of solutions for y(x)—they are called eigenfunction vectors (or simply eigenfunctions)—that belong to an eigenvalue is finite (this number is the multiplicity of the eigenvalue and is denoted by r ). The kernel is positive definite if for any continuous and non-zero v(x) it holds that  b b v T (x) K(x, ξ) v(ξ) dξ dx > 0 . (9.131) a

a

(1×2)

(2×1)

(2×1)

If the kernel is positive definite, the eigenvalues are positive numbers: 0 < λ1 < λ2 < λ3 < · · · The eigenfunction vector that belongs to λ L (L = 1, 2, 3, . . .) is denoted by y L if the multiplicity of λ L is one, by yk L if the multiplicity of λ L is r (k = 1, 2, . . . , r ). The eigenfunction vectors can be normalized: ψ i (x) =

yi (x)

-  , |yi (x)| = yi , yi ,

I |yi (x)|  

b yi , yi = yiT (x) yi (x) dx, i = 1, 2, 3, . . . I

(9.132)

a

where ψ i is the normalized eigenfunction vector. Let λi and λ be two not necessarily different eigenvalues each with only one eigenfunction vector. It can be proved that the orthogonality condition  a

b

ψ iT (x) ψ  (x) dx =



ψi , ψ = I



1 if i =  , 0 if i =  .

i,  = 1, 2, 3, . . .

(9.133)

holds. If r > 1 the number of linearly independent eigenfunction vectors which belong to the considered eigenvalue is r . These eigenfunctions are, in general, not orthogonal to each other. They can, however, be made orthogonal by applying the Gram–Schmidt orthogonalization procedure presented in Sect. 9.2.11.

338

9 Eigenvalue Problems of Ordinary Differential Equation Systems

9.2.14 Calculation of Eigenvalues The boundary element technique can also be applied to reducing the eigenvalue problem governed by the Fredholm integral equation system  y(x) = (2×1)

b

K(x, ξ) y(ξ) dξ ,

a

(2×1)

λ = 1 ,

(9.134)

(2×2)

to an algebraic eigenvalue problem. The procedure detailed below is based on Sect. 8.15.2. The interval [a, b] is the same as the one shown in Fig. 8.6. The number of elements in this interval is n e = 5. The elements and their lengths are also denoted in the same way as earlier, i.e., by L1 , . . . , Ln e . The nodal points where the unknown function vector y(x) is considered are taken at the extremes (or ends) and the midpoint of an element. The nodal point numbering is shown in Fig. 8.6—the numbering scheme has remained unchanged. Let xe and yek be the matrices of the nodal coordinates and the unknown function vectors at the nodes of the element Le :

  e T  e e e   e e  yk2 , k = 1, 2, 3 x = x1 x2 x3 , y ek T = yk1

(9.135)

It is assumed that the approximation over the elements is quadratic. If this is the case, the shape functions Nk (η) are given by Eq. (8.179). Making use of the notations ⎡

⎤ y e1 ye = ⎣ y e2 ⎦ (6×1) y e3

 N k (η) =

and

(2×2)

 Nk (η) 0 , 0 Nk (η)

(9.136)

the unknown function matrix over the element Le can be approximated by the equation ⎡ e⎤ y1   e   ⎣ y e2 ⎦ = N ye , y ξ (η) = N 1 (η) N 2 (η) N 3 (η) (9.137a) (2×6) (6×1) (2×1) y e3 where ⎡

⎤ x1e 1  ξ e (η) = N1 (η) N2 (η) N3 (η) ⎣ x2e ⎦ = η x3e − x1e + x2e . 2 x3e 



(9.137b)

Note that the derivation of the latter equation is omitted since it should obviously coincide with Eq. (8.181). According to relations (8.182), it also holds that dξ e = J dη ,

J =

Le 1 e x3 − x1e = . 2 2

(9.138)

9.2 Eigenvalue Problem for a Class of Differential Equations

339

Substituting (9.137a) and (9.138) into (9.134) and integrating, then element by element yields

y(x) =

ne   e=1

⎡

⎤ y e1 K [x, ξ (η)] N1 (η) N2 (η) N3 (η) J (η)dη ⎣ y e2 ⎦ = Le y e3 n be  η=1  K [x, ξ (η)] N (η)J (η)dη ye . (9.139) = 



e=1

η=−1 (2×2)

(2×6)

(6×1)

If the node coordinates are denoted by xi (i = 1, 2, . . . , 2n e + 1) and the previous equation is taken at these points we obtain

y(xi ) =

e=1

⎡

⎤ y e1 K [xi , ξ(η)] N1 (η) N2 (η) N3 (η) J (η)dη ⎣ y e2 ⎦ , η=−1 y e3

ne  

η=1





(i = 1, 2, . . . , 2n e + 1) .

(9.140)

For the sake of simplicity it is worth introducing further notations:  krie =

(2×2)

η=1

K [xi , ξ(η)] Nr (η)J (η)dη ,

η=−1 (2×2)

(2×2)

y (xi ) = y i , (2×1)

(2×1)

(i = 1, 2, 3, . . . , 2n e + 1; r = 1, 2, 3),

(9.141)

where i identifies the global number of the node, e is the number of the element over which the integral is taken, and r is the number of the interpolation function matrix. With notations (9.141), Eq. (9.140) assumes the following form: 

 i,n e −1 in e in e i1 i1 i2 i2 i2 i3 i3 e × + kin ki1 1 k2 k3 + k1 k2 k3 + k1 k1 . . . k3 1 k2 k3 ⎡ ⎤ y1 ⎢ y2 ⎥   ⎢ ⎥ 0 ⎢ ⎥ × ⎢ · · · ⎥ − yi = , (i = 1, 2, 3, . . . , 2n e + 1), 0 ⎣ y2n e ⎦ y2n e +1

(9.142)

in which the column matrices yi are the unknowns. Note that the number of Eq. (9.142) coincides with the number of unknowns. By introducing the notations i1 i1 i2 i2 i2 i3 ki1 = ki1 1 , ki2 = k2 , ki3 = k3 +k1 , ki4 = k2 , ki5 = k3 +k1 , . . . , in e e ki,2n e = kin 2 , ki,2n e+1 = k3

(9.143a)

340

9 Eigenvalue Problems of Ordinary Differential Equation Systems



and 1 =

(2×2)

 10 , 01

 0 =

(2×2)

 00 , 00

  0 0ˆ = , 0 (2×1)

(9.143b)

the structure of equation system (9.142) becomes easy to survey: ⎧⎡ ⎤ ⎡ k11 k12 · · · k1,2n e+1 1 ⎪ ⎪ ⎨⎢ ⎥ ⎢ k k · · · k 0 2,2n e+1 ⎥ ⎢ 21 22 ⎢ ⎣ ⎦ − ⎣ ............. ⎪ ⎪ ⎩ 0 kn e 1 kn e 2 · · · k2n e+1,2n e+1

⎤⎫ ⎡ ⎤ y1 0 ⎪ ⎪ ⎬⎢ ⎥ 0⎥ ⎥ ⎢ y2 ⎥ = ⎦⎪ ⎣ · · · ⎦ ⎪ ⎭ 1 y2n e+1 ⎤ 0ˆ ⎢ 0ˆ ⎥ ⎥ =⎢ ⎣· · · ⎦ . 0ˆ

0 ··· 1 ··· ..... 0 ··· ⎡

(9.144)

Equation (9.144) is again an algebraic eigenvalue problem—similar to Eq. (8.188)— with  = 1/λ as eigenvalue. After solving it numerically, we have both the eigenvalues r = 1/λr and the corresponding eigenvectors yrT (1×(4n e +2))

  T T T T y2r y3r · · · y2n , (r = 1, 2, 3, . . . , 2n e + 1). = y1r e+1,r

As regards the issue of numerical integration, it is worthy introducing the notation φ(η) = K [xi , ξ(η)] Nr (η)J (η) .

(9.145)

With (9.145) one can rewrite integral (9.141) in the following form:  krie

=

η=1 η=−1

 K [xi , ξ(η)] Nr (η)J (η) dη =

η=1

η=−1

φ(η) dη

(9.146)

for which it is easy to apply the Gaussian quadrature rules presented in Sect. B.3.

9.3 Free Vibrations of Timoshenko Beams By the use of dimensionless variables on the base of Eq. (9.89), the motion equations (9.29) can be rewritten in the following form:   AGκ y 2 d2 y1 ρAω 2 4 dy2 = −  y1 , + IE dx 2 dx IE   AGκ y 2 dy1 d2 y2 ρAω 2 4 I N 2 dy1 = − − + y  ± y2 2 dx 2 I E dx IE dx IE A2

(9.147)

9.3 Free Vibrations of Timoshenko Beams

341

 dy2 d2 y1 = −λy1 , + dx 2 dx   d2 y2 dy1 dy1 2 − χ + y ±N 2 = −λr y2 , dx 2 dx dx 

or

χ

(9.148a)

where y1 and y2 are dimensionless amplitude functions, x is a dimensionless coordinate and χ=

AGκ y 2 N 2 ρAω 2 4 I , N = , λ=  , r2 = . IE IE IE A2

(9.148b)

If there is no axial force, the amplitude function should satisfy the equations 

d2 y1 dy2 χ + 2 dx dx

 = −λy1

  d2 y2 dy1 + y2 = −λr2 y2 , −χ dx 2 dx

(9.149)

which can be rewritten in matrix form as well: 

  (2)       (1)    χ0 y1 y1 y1 −1 0 y1 0 χ 0 0 =λ . + + y2 y2 y2 y2 01 0 −r2 −χ 0 0 −χ  

 

 

 

K2

K1

K0

M0

(9.150) It is clear that the left side of the above equation coincides with the left side of Eq. (9.91). The expression    −1 0 y1 λ 0 −r2 y2 on the right side corresponds to r in the solution (9.116). Hence, 

     1 −1 0 G 11 (x, ξ) G 12 (x, ξ) y1 y1 (x) =λ dξ , 2 y ξ) G ξ) 0 −r y2 (x) G (x, (x, 21 22 2 0

where for pinned-pinned, fixed-pinned, fixed-fixed, and fixed-free beams the Green function matrices are given by Eqs. (9.101), (9.105), (9.106), and (9.107), respectively. Introduction of the new variables Y and K as shown by the equation 

10 0r

   1  1 0 −G 11 (x, ξ) −G 12 (x, ξ) 1 y1 (x) =λ y2 (x) 0 r −G 21 (x, ξ) −G 22 (x, ξ) 0 0  

 



Y(x)

K(x,ξ)

 0 1 r 0



  0 y1 dξ y2 r 

Y(ξ)

results in an eigenvalue problem governed by the Fredholm integral equation 

1

Y(x) = λ 0

K(x, ξ) Y(ξ) dξ .

(9.151)

342

9 Eigenvalue Problems of Ordinary Differential Equation Systems

The eigenvalue problem governed by the homogeneous integral equation (9.151) can be solved numerically by applying the solution procedure presented in Sect. 9.2.14. A FORTRAN 90 program was written by applying the solution procedure mentioned. Some results are presented graphically in Figs. 9.3, 9.4, 9.5, and 9.6. Let ωˇ k (k = 1, 2) be the first two natural circular frequencies of the Timoshenko beam considered. The first natural frequency of the same beam within the framework of the Euler–Bernoulli beam theory is denoted by ω1 . These figures show the quotient ωˆ k /ωk against r = r y /, where r y is the radius of gyration and  is the length of the beam. A further parameter for the graphs shown in Figs. 9.3, 9.4, 9.5, and 9.6 is the quotient E = E/Gκ y . It follows from Eq. (9.148b) which define the dimensionless parameters χ and r2 that χ = 1/Er2 . If r = 0, the solution for ωˆ coincides with that of the Euler beam theory. It follows from this fact by taking the data of Table 7.5 and Eq. (7.86a)5 into account that the solution curves in Figs. 9.3, 9.4, 9.5, and 9.6 start from the ordinates (points) [0, 1], [0, 4] (Pinned-Pinned Beam); [0, 1], [0, (7.0685/3.9266)2 = 3.2406] (Fixed-Pinned Beam); [0, 1], [0, (7.8532/4.7300)2 = 2.7565] (Fixed-Fixed Beam); [0, 1], [0, (4.6941/1.8751)2 = 6.2668] (Fixed-Free Beam).

Fig. 9.3 Quotient ωˆ k /ωk against r = r y / for simply supported Timoshenko beam

9.3 Free Vibrations of Timoshenko Beams

Fig. 9.4 Quotient ωˆ k /ωk against r = r y / for fixed-pinned Timoshenko beam

Fig. 9.5 Quotient ωˆ k /ωk against r = r y / for fixed-fixed Timoshenko beam

343

344

9 Eigenvalue Problems of Ordinary Differential Equation Systems

Fig. 9.6 Quotient ωˆ k /ωk against r = r y / for fixed-free Timoshenko beam

The graphs shown are drown by joining the points that represent the numerical solutions by short straight lines. There is a closed-form solution for the eigenvalue λ for pinned-pinned beams—see Problem 9.4 and Eq. (C.9.48) in its solution. The points depicted by diamonds in Fig. 9.3 represent the analytical solution. They fit onto the numerical solution with a very good accuracy.

9.4 Problems Problem 9.1 Prove that the eigenvalue problems defined by differential equation (9.34a) and the boundary conditions shown in Rows 2, 3, and 4 of Table 9.1 are all self-adjoint. Problem 9.2 A concentrated mass is attached to the right end of the beam shown in Fig. 9.7. It is assumed that the mass m is much greater than that of the beam. Assume

Fig. 9.7 Fixed beam with a concentrated mass at the right end of the beam

z

m x P1

9.4 Problems

345

that the mass vibrates vertically. Prove that the square of its circular frequency is given by 3I E k11 = 3 (Euler–Bernoulli beam theory) ω2 = m  m or by ω2 =

3

3I E

 2r 2 1 + 2y (1+ν) κy

(Timoshenko beam theory)

(9.152)

Problem 9.3 Detail the calculation of the Green function matrix for axially loaded and pinned-pinned Timoshenko beams in order to prove the validity of the solutions (9.112) and (9.114). Problem 9.4 Find the frequency equations for the free vibrations of pinned-pinned Timoshenko beams.

References 1. S.P. Timoshenko, On the correction for shear of the differential equation for transverse vibrations of prismatic bars. Lond. Edinb. Dublin Philos. Mag. J. Sci. 41(245), 744–746 (1921). https:// doi.org/10.1080/14786442108636264 2. F. Engesser, Die knickfestigkeit grader stëabe. Central Bauvervaltung 11, 483–486 (1981) 3. G.R. Cowper, The shear coefficient in Timoshenko’s beam theory. J. Appl. Mech. 33(2), 335– 340. (1966). https://doi.org/10.1115/1.3625046 4. G.R. Cowper, On the accuracy of Timoshenko’s beam theory. ASCE, J. Eng. Mech. Div. 94(6), 1447–1453 (1968) 5. A. Marinetti, G. Oliveto, On the Evaluation of the Shear Correction Factors: a Boundary Element Approach (2009) 6. M. Levinson, D.W. Cooke, On the frequency spectra of Timohsenko beams. J. Sound Vib. 84(3), 319–324 (1982). https://doi.org/10.1016/0022-460X(82)90480-1 7. G. Szeidl, Effect of Change in Length on the Natural Frequencies and Stability of Circular Beams. in Hungarian. Ph.D. thesis. Department of Mechanics, University of Miskolc, Hungary (1975), 158 pp 8. J.G. Obádovics, On the Boundary and Initial Value Problems of Differential Equation Systems. Ph.D. thesis. Hungarian Academy of Sciences (1967), (in Hungarian) 9. L. Collatz, Eigenwertaufgaben mit Technischen Anwendungen. Russian Edition in 1968 (Akademische Verlagsgesellschaft Geest & Portig K.G., 1963)

Chapter 10

Eigenvalue Problems Described by Degenerated Systems of Ordinary Differential Equations

10.1 Vibration of Heterogeneous Curved Beams 10.1.1 Introductory Remarks As regards the vibration problem of curved beams, it is worthwhile to mention that Den Hartog [1, 2] is known to be the first to deal with the free vibrations of circular beams. Other early but relevant results, considering the inextensibility of the centerline, were achieved in [3, 4]. A more recent research by Qatu and Elsharkawy [5] presents an exact model and numerical solutions for the free vibrations of laminated arches. With the differential quadrature method, Kang et al. [6] determine the eigenfrequencies for the in- and out-of-plane vibrations of circular Timoshenko arches with rotatory inertia and shear deformations included. Tüfekçi and Arpaci [7] present exact solutions for the differential equations which describe the in-plane free harmonic vibrations of extensible curved beams. Krishnan and Suresh [8] tackle the very same issue with a shear-deformable finite element (FE) model. Paper [9] by Ecsedi and Dluhi analyzes some dynamic features of non-homogeneous curved beams and closed rings assuming cross-sectional heterogeneity. Elastic foundation is taken into account in [10]. Survey paper [11] by Hajianmaleki and Qatu collects a bunch of references up until the early 2010s in the topic investigated. Kovács [12] considers layered arches with both perfect and even imperfect bonding between any two adjacent layers. The present chapter is aimed at clarifying the vibratory properties of circular beams made of heterogeneous material. The differential equation system that describes these vibrations is a degenerated one. The numerical solution will be based on the use of the Green function matrices which should be redefined since the differential equation system is, as has just been mentioned, a degenerated one.

© Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8_10

347

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10.1.2 Equilibrium Equations 10.1.2.1

Fundamental Assumptions

The following assumptions will be applied when the equations that govern the vibration problem of heterogeneous curved beams are derived: (i) the beam is uniform, i.e., the cross section is constant; (ii) the radius of the centerline is constant and the cross section is symmetric with respect to the plane of the centerline; (iii) the material of the beam is heterogeneous which means that the modulus of elasticity depends on the cross-sectional coordinates only; thus, it is independent of the coordinate measured on the centerline of the beam (cross-sectional heterogeneity)—since the beam is heterogeneous, a precise definition for the centerline will be given later; (iv) the displacements and deformations are small; (v) the centerline of the beam keeps its own plane during the deformations (vibrations); (vi) the normal stress parallel to the centerline satisfies the relation σξ  ση , σζ — see Fig. 10.1 for the coordinate directions; (vi) the work done by the shear stresses τ can be neglected when it is compared to that of the normal stress σξ ; (vii) the beam is subjected to a distributed load lying in the plane of the centerline and/or to concentrated forces. Fig. 10.1 Curved beam element and the selected coordinate systems

The coordinate system selected is the polar one shown in Fig. 10.1. The unit vector eξ is tangent to the centerline, eη is perpendicular to the plane of the centerline while the unit vectors eξ , eη , and eζ form a right-hand triad, hence eζ = eξ × eη . Consequently, eζ is perpendicular to the centerline in its plane. It is clear from Fig. 10.1 that R is the radius of the centerline, s is the arc coordinate on the centerline, and ϕ is the polar angle. It also holds that deξ 1 = − eζ ; ds R

deζ 1 = eξ . ds R

(10.1)

10.1 Vibration of Heterogeneous Curved Beams

349

In accordance with assumption (iii), the modulus of elasticity should satisfy the relation E(η, ζ) = E(−η, ζ) which means that E is an even function of the η-coordinate. The E-weighted first moment of the cross section with respect to the axis η is defined by the equation  Q eη =

E(η, ζ)ζd A.

(10.2)

A

The location of the centerline—that can also be called E-weighted centerline— is determined uniquely by the requirement Q eη = 0. The E-weighted centerline intersects the cross section at the point Ce called E-weighted center. For our later considerations, it is worth introducing the following three quantities [13, 14]:  Ae R = A

R E(η, ζ)d A = R+ζ

   ζ ζ2 1 − + 2 − ... E(η, ζ)d A ∼ = R R A  ∼ E(η, ζ)d A  Ae , (10.3a) = A

 QeR = A

   ζ ζ2 1 − + 2 − ... ζ E(η, ζ)d A ∼ = R R A   1 ∼ ζ E(η, ζ)d A − ζ 2 E(η, ζ)d A (10.3b) = R A A      

R E(η, ζ)ζd A = R+ζ

Q eη =0

Ieη

and  Ie R = A

R E(η, ζ)ζ 2 d A = R+ζ

  A

 ζ ζ2 1 − + 2 − ... ζ 2 E(η, ζ)d A ∼ = R R  ∼ ζ 2 E(η, ζ)d A = Ieη , (10.3c) = A

where Ae and Ieη are the E-weighted cross-sectional area and the E-weighted moment of inertia. In the sequel, it will be assumed that Ae R = Ae , Q e R = −Ieη /R and Ie R = Ieη . Note that these quantities characterize the geometry of the cross section and the material distribution over the cross section. The nabla operator in the considered polar coordinate system has the following form: 1 ∂ ∂ ∂ eη + eζ . eξ + (10.4) ∇= ∂s 1 + Rζ ∂η ∂ζ

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10.1.2.2

Strain State

As regards the deformations, it is assumed that the cross section rotates as if it were a rigid body—Euler–Bernoulli hypothesis. Then u = [u o (s) + ψoη (s)ζ]eξ + wo (s)eζ

(10.5)

is the displacement field where u o and wo are the tangential and normal displacement components on the centerline, while ψoη is the rotation of the cross section about the η-axis. As is well known 1 ψ = − (u × ∇) 2 is the rotation field in the beam. Hence ψ|ζ=0

1 = − (u × ∇) . 2 ζ=0

(10.6)

Substituting (10.4), (10.1) and taking the relation eζ × eξ = eη into account yields 1 uo 1 dwo 1 eη − eη + ψoη eη 2 R 2 ds 2 for the right side of (10.6). As regards the left side, we have ψ|ζ=0 = ψoη eη . Thus ψoη =

dwo uo − R ds

(10.7)

is the rotation on the centerline. The strain tensor is given by the equation ε=

1 (u ◦ ∇ + ∇ ◦ u) 2

from where εξ = eξ ·

1 1 ∂u 1 1 1 ∂u · eξ = + (u ◦ ∇ + ∇ ◦ u) · eξ = eξ · 2 2 ∂s 1 + Rζ 2 1 + Rζ ∂s   du o wo dψoη 1 ∂u R · eξ = + + ζ = R + ζ ds R ds 1 + Rζ ∂s

10.1 Vibration of Heterogeneous Curved Beams

351

is the axial strain in the coordinate direction ξ. Since ζ = 0 on the η-axis εoξ =

wo du o + ds R

(10.8)

is the axial strain on the centerline. The curvature κo of the centerline is defined by the equation   d dwo uo dψoη =− − . (10.9) κo = ds ds ds R Consequently, εξ =



1 1+

10.1.2.3

ζ R

du o wo d + −ζ ds R ds



dwo uo − ds R

 =

1 1+

ζ R



εoξ + ζκo . (10.10)

Inner Forces

In accordance with Fig. 10.2, the positive inner forces, i.e., the axial force N , the shear force V , and the bending moment M acting on the cross section, are shown in Fig. 10.2. Since σξ  ση , σζ , the simple Hooke’s law is applicable, i.e., σξ = E(η, ζ)εξ . With (10.10) 



N=

σξ d A = εoξ A

A

R E(η, ζ)d A + κo R+ζ

 A

R E(η, ζ)ζd A R+ζ

(10.11)

is the axial force. It is clear that ds = Rdϕ. With regard to this relation, it holds dk 1 dk 1 (. . .) = (. . .) = k (. . .)(k) k k k ds R dϕ R

k = 1, 2, 3, . . .

(10.12)

By substituting (10.3) for the integrals and changing over to derivatives taken with respect to the polar angle Eq. (10.11) for the axial force can be manipulated into a more suitable form. After performing some rearrangement, we have

Fig. 10.2 Inner forces acting on the cross section

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10 Eigenvalue Problems Described by Degenerated Systems…

Ieη N= 3 R



 

(1) Ae R 2 (2) − 1 u o + wo + wo + wo = Ieη

 (1)

Ieη (2) = 3 m u o + wo + wo + wo , R

where m=

Ae R 2 − 1. Ieη

(10.13)

(10.14)

As regards the bending moment, the following equation can be deduced: 



ζσξ d A = εoξ

R E(η, ζ)ζd A + κo R+ζ



R E(η, ζ)ζ 2 d A . A A A R+ζ (10.15) Making use of (10.3) and then (10.8), (10.9) formula (10.15) for the bending moment can be rewritten in the following form: M=

M =−

10.1.2.4

Ieη (2) wo + wo . 2 R

(10.16)

Equilibrium Conditions

It follows from equilibrium equations (7.2a) and (7.2b) that the resultant of the inner forces FC = N eξ − V eζ and the moment resultant of the inner forces M = Meη should satisfy the following equations: 

   dN dV deξ deζ + f ξ eξ + N − − f ζ eζ − V = 0, ds ds ds ds

 dM d eη + V eη + (u o eξ + wo eζ ) × (N eξ − V eζ ) + μeη = 0, ds ds

(10.17a)

(10.17b)

where the intensity of the distributed load acting on the centerline is denoted by f ξ and f ζ , while μ is the intensity of the moment load distribution. Substitution of the derivatives given by Eqs. (10.1) into (10.17a) results in two scalar equations: N dV + − fζ = 0 , ds R

V dN − + fξ = 0 . ds R

(10.18)

Using again (10.1) for the derivatives in (10.17b) and performing then the cross product yields a scalar equation if the result is dot multiplied by eη :

10.1 Vibration of Heterogeneous Curved Beams

353

    du o wo uo dM dwo +V +V + − +N +μ = 0. ds ds R ds R       εoξ

(10.19)

−ψoη

Within the framework of the linear theory, the quadratic terms can be dropped in (10.19). In the sequel, it will be assumed that the intensity of the moment load distribution μ is zero. Under these conditions, Eq. (10.19) becomes much simpler: dM + V = 0. ds

(10.20)

This equation makes possible the elimination of the shear force V in Eq. (10.18). It can be checked with ease that eliminating V in (10.18) results in the following two equations: 1 dM d2 M N dN + + fξ = 0 , + fζ = 0 . − (10.21) 2 ds R ds ds R Equations for the dimensionless displacement components Uo = u o /R and Wo = wo /R can be obtained by substituting (10.13) for N and (10.16) for M in (10.21). After changing the derivatives with respect to s to those with respect to ϕ and performing some rearrangement, the following differential equation system (presented here in matrix form) is obtained:

00 01



Uo Wo

(4)

(2)

(1) 



−m 0 0 −m Uo Uo + + + Wo Wo 0 2 m 0



(0)

 R3 fξ 0 0 Uo + . = Wo 0 m+1 Ieη f ζ

(10.22)

Equation (10.22) should be supplemented with boundary conditions. If the support is pinned-pinned, the displacement and the bending moment are zero. If the supports are fixed, the displacement and the rotation are zero. Table 10.1 shows the three support arrangements for which the determination of the first four natural frequencies is one of our main objectives in the present chapter. Table 10.1 contains the boundary conditions too; they are presented in terms of the displacement components here by using Eqs. (10.7) and (10.16). Note that the central angle of the beam is ϑ¯ = 2ϑ.

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Table 10.1 Boundary conditions for curved beams Support arrangements

Boundary conditions

0

1.

Uo (−ϑ) = 0 Wo (−ϑ) = 0 (2) Wo (−ϑ) = 0

Uo (ϑ) = 0 Wo (ϑ) = 0 (2) Wo (ϑ) = 0

2.

0

Uo (−ϑ) = 0 Wo (−ϑ) = 0 Wo(1) (−ϑ) = 0

Uo (ϑ) = 0 Wo (ϑ) = 0 Wo(1) (ϑ) = 0

3.

0

Uo (−ϑ) = 0 Wo (−ϑ) = 0 (2) Wo (−ϑ) = 0

Uo (ϑ) = 0 Wo (ϑ) = 0 (1) Wo (ϑ) = 0

Remark 10.1 Differential equation system (10.22) can be rewritten in the following matrix form: (10.23) K [ y(x)] = r(x), where K[ y ] =

4 

K ν (x)y(ν) (x) ,

K 3 (x) = 0 .

(10.24)

ν=0

On the basis of (10.22) and (10.24), differential equation system (10.23) can be detailed as

 (4)

 (2)

 (1) 00 y1 y1 y1 −m 0 0 −m + + + y2 y2 y2 01 0 2 m 0          K4

K2

K1

 (0)  y1 0 0 r + = 1 y2 r2 0 m+1   

(10.25)

K0

in which x = ϕ and

y1 y2



=

Uo Wo



,

r1 r2

 =

R3 Ieη



fξ fζ

 .

(10.26)

It is worth mentioning that r is a dimensionless distributed load. Since the matrix K4 has no inverse differential equation system (10.25) is degenerated.

10.1 Vibration of Heterogeneous Curved Beams

355

10.1.3 Equations of Motion For the problem of free vibrations the distributed loads are forces of inertia. Thus f ξ = −ρa A

∂2uo ∂ 2 Uo = −Rρ A a ∂t 2 ∂t 2

f ζ = −ρa A

∂ 2 wo ∂ 2 Wo = −Rρ A , a ∂t 2 ∂t 2 (10.27)



where ρa =

ρ(η, ζ)d A ,

ρ(η, ζ) = ρ(−η, ζ)

(10.28)

A

is the average of the density ρ over the cross section of area A and t denotes time. Note that now the displacement components are functions of ϕ and t. With (10.27) equation (10.22) assumes the following form:

00 01

(4)

(2)



−m 0 0 Uo Uo + + Wo Wo 0 2 m



(0) ρa A 0 0 Uo + = −R 4 Wo 0 m+1 Ieη



(1) Uo + Wo

 ∂ 2 Uo . ∂t 2 Wo −m 0



(10.29)

For harmonic vibrations, it is assumed that the solutions are Uo (ϕ, t) = U (ϕ) sin ωt ,

Wo (ϕ, t) = W (ϕ) sin ωt,

(10.30)

where U (ϕ) and W (ϕ) are the amplitude functions and ω is the circular frequency of the vibrations. Substituting solution (10.30) into (10.29) yields the following differential equation system for U (ϕ) and W (ϕ):

00 01



U W

 (2)

 (1) −m 0 0 −m U U + + + 0 2 m 0 W W

 (0)

  0 0 U 10 U + =λ , 0 m+1 W 01 W

(4)



λ = R4

ρa Aω 2 , Ieη

(10.31a)

(10.31b)

where λ is the dimensionless eigenvalue. It is clear from Table 10.1 that the amplitude functions should also satisfy the following boundary conditions: Pinned-pinned beam: U (−ϑ) = W (−ϑ) = W (2) (−ϑ) = 0,

U (ϑ) = W (ϑ) = W (2) (ϑ) = 0. (10.32)

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10 Eigenvalue Problems Described by Degenerated Systems…

Fixed-fixed beam: U (−ϑ) = W (−ϑ) = W (1) (−ϑ) = 0,

U (ϑ) = W (ϑ) = W (1) (ϑ) = 0. (10.33)

Pinned-fixed beam: U (−ϑ) = W (−ϑ) = W (2) (−ϑ) = 0,

U (ϑ) = W (ϑ) = W (1) (ϑ) = 0. (10.34)

The homogeneous differential equation (10.31) and the also homogeneous boundary conditions (10.32), (10.33), and (10.34) determine three eigenvalue problems with λ as the eigenvalue. All that has been said about eigenvalue problems governed by homogeneous differential equation systems and boundary conditions in Sects. 9.2.1, 9.2.2, 9.2.3, and 9.2.4 remain valid for the three eigenvalue problems determined by equations (10.31) and (10.32), (10.33), and (10.34). There is, however, a difference since the governing differential equation (10.32) is degenerated—the coefficient matrix P 4 in (10.25) has no inverse in contrast to P κ in (9.67b) for which the existence of the inverse is a requirement. For this reason, the definition of the Green function matrix should be rethought. Remark 10.2 Using the notations introduced by Eqs. (10.23), (10.24), and (10.25), the eigenvalue problems defined by differential equation (10.32) and boundary conditions (10.32), (10.33), and (10.34) can be rewritten as follows: Differential equation: K [ y(x)] =

4 

K ν (x)y(ν) (x) = λM 0 (x)y(x) ,

ν=0

M 0 (x) =





  U 10 y , , y= 1 = y2 W 01

(10.35)

where x = ϕ is the dimensionless independent variable. Differential equation (10.35) is associated with the following boundary conditions: Pinned-pinned beam: y1 (x)|x=±ϑ = y2 (x)|x=±ϑ = y2(2) (x)

x=±ϑ

= 0.

(10.36)

= 0.

(10.37)

Fixed-fixed beam: y1 (x)|x=±ϑ = y2 (x)|x=±ϑ = y2(1) (x)

x=±ϑ

Pinned-fixed beam: y1 (x)|x=±ϑ = y2 (x)|x=±ϑ = y2(2) (x)

x=−ϑ

= y2(1) (x)

x=ϑ

= 0.

(10.38)

10.2 Green Function Matrix for Degenerated Differential Equation Systems

357

10.2 Green Function Matrix for Degenerated Differential Equation Systems 10.2.1 Definition Consider the degenerated system of ordinary differential equations   (κ) κ  ν 0 0 y1 (ν) κ P(x)y (x) = + ···+ L[y(x)] = y 0 P 2 22 ν=0   (k+1) ⎡ k k ⎤ (k) 0 0 y y1 P P + + ⎣ 11 k 12 ⎦ 1 + ···+ k+1 y y 0 P 22 2 2 0 P 22 ⎡ ⎤ s s

(s)

 y1 P P r 11 12 ⎣ ⎦ + s + ··· = 1 , s y r 2 2 P P 21

(10.39)

22

where κ > k > s > 0 and n is the number of unknown functions (or the size of y), ν

j is the size of y2 , and the matrices P and r are continuous for x ∈ [a, b]; a < b. It will be assumed that κ

k

• P22 and P11 are invertible if x ∈ [a, b] • the system of differential equations (10.39) is associated with linear homogeneous boundary conditions of the form U μ [y] =

κ    A νμ y(ν−1) (a) + B νμ y(ν−1) (b) = ν=1

⎧⎡ ⎫ ⎡11 12 ⎤ ⎤ 12



(ν−1) (ν−1) ⎬ κ ⎨ 11  y y (a) (b) A A B B ⎣ νμ νμ ⎦ 1 + ⎣21 νμ 22 νμ ⎦ 1 = = y2 (a) y2 (b) ⎩ 21 22 ⎭ ν=1 A νμ A νμ B νμ B νμ ⎤ ⎡ 0 ((n− j)×1) ⎦, (10.40a) =⎣ 0 ( j×1)

where μ = 1, . . . , κ and for ν > k the constant matrices A νμ and B νμ fulfill the conditions 11

21

11

21

A νμ = A νμ = B νμ = B νμ = 0.

(10.40b)

• l¯ = κn − [(n − j)k + κ j] rows are identically zero in the hypermatrix ⎡

⎤ A 11 . . . A κ1 B 11 . . . B κ1 Pf = ⎣······ ······ ······ ······ ······ ······⎦ . A 1κ . . . A κκ B 1κ . . . B κκ

(10.40c)

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10 Eigenvalue Problems Described by Degenerated Systems…

By introducing appropriate new variables, differential equation system (10.39) can be replaced by (n − j)k + κ j differential equations of order one for which it is possible to construct the corresponding Green function. In the present case, however, there is no need for this transformation since the definition presented here is based on the original equation [15]. Condition (10.40b) expresses that y(ν) 1 cannot appear in the boundary conditions if ν ≥ k. Condition (10.40c) is a restriction on the number of boundary conditions. Solution to the degenerated boundary value problems (10.39), (10.40a) is sought in the same way as for the non-degenerated boundary value problem defined by Eqs. (9.67), (9.68), i.e., in the form of an integral:  y(x) =

b

G(x, ξ)r(ξ) dξ = ⎡ ⎤⎡ ⎤ r1 (ξ)  b G 11 (x, ξ) G 12 (x, ξ) ⎢ ((l− j)×(l− j)) ((l− j)× j) ⎥ ⎢ ((l− j)×1) ⎥ = ⎣ ⎦⎣ ⎦ dξ, G 21 (x, ξ) G 22 (x, ξ) r2 (ξ) a a

( j×(l− j))

( j× j)

(10.41)

( j×1)

where G(x, ξ) is again a Green function matrix defined by the following four properties: 1. The Green matrix function is a continuous function of x and ξ in each of the triangles a ≤ x ≤ ξ ≤ b and a ≤ ξ ≤ x ≤ b. The functions

 G 11 (x, ξ), G 12 (x, ξ) G 21 (x, ξ), G 22 (x, ξ) are (k times) [κ times] differentiable with respect to x and the derivatives ∂ ν G(x, ξ) = G(ν) (x, ξ), ∂x ν ∂ ν G2i (x, ξ) = G(ν) 2i (x, ξ), ∂x ν

(ν = 1, 2, . . . , k), (ν = 1, 2, . . . , κ; i = 1, 2)

are continuous functions of x and ξ. 2. Let ξ be fixed in [a, b]. Though the derivatives G(ν) 11 (x, ξ) (ν = 1, 2, . . . , k − 2),

G(ν) 12 (x, ξ) (ν = 1, 2, . . . , k − 1),

G(ν) 21 (x, ξ) (ν = 1, 2, . . . , κ − 1),

G(ν) 22 (x, ξ) (ν = 1, 2, . . . , κ − 2)

are continuous for x = ξ, the higher derivatives, i.e., (κ−1) (x, ξ) G(k−1) 11 (x, ξ) and G 22

have a jump on the diagonal given by the following two equations:

10.2 Green Function Matrix for Degenerated Differential Equation Systems

359

  k (k−1) −1 lim G(k−1) 11 (ξ + ε, ξ) − G 11 (ξ − ε, ξ) = P 11 (ξ), ε→0   κ (ξ + ε, ξ) − G(κ−1) (ξ − ε, ξ) = P−1 lim G(κ−1) 22 22 22 (ξ).

ε→0

3. Let α be an arbitrary otherwise constant vector. For a fixed ξ ∈ [a, b] the vector G(x, ξ)α as a function of x (x = ξ) should satisfy the homogeneous differential equation  L G(x, ξ)α = 0 . 4. The vector G(x, ξ)α as a function of x should satisfy the boundary conditions  Uμ G(x, ξ)α = 0, (μ = 1, . . . , n) . Remark 10.3 It is worth comparing the definition presented in Sect. 9.2.6 to the above definition. The difference appears in Properties 1 and 2; Properties 3 and 4 are basically the same as earlier. Remark 10.4 If there exists the Green matrix function defined above for the boundary value problem (10.39), (10.40a) then the vector (10.41) satisfies differential equation (10.39) and the boundary conditions (10.40a). The part of the statement concerning the boundary conditions follows immediately from the comparison of the fourth property of the definition. As regards the second part of our statement, substitute the representation (10.41) into (10.39) by utilizing and G(κ−1) are discontinuous if x = ξ. In this way, we have that the matrices G(k−1) 11 22 L(y) =

κ  ν P(x)y(ν) (x) = ν=0

 0







 r1 (ξ) dξ+ r2 (ξ) 0 P 22 (x) a  

 0 0 r1 (x) ! " κ + + r2 (x) 0 P 22 (x) G(κ−1) (x, x − 0) − G(κ−1) (x, x + 0) 22 22  



 b 0 0 0 0 r1 (ξ) + dξ+ k+1 (k+1) (k+1) 0 P 22 (x) a G 21 (x, ξ) G 22 (x, ξ) r2 (ξ) ⎡ ⎤ k k 

  b (k) P (x) P (x) (x, ξ) r1 (ξ) G 11 (x, ξ) G(k) 11 12 12 ⎣ ⎦ dξ+ + k (k) (k) k r2 (ξ) P 22 (x) P 22 (x) a G 21 (x, ξ) G 22 (x, ξ) n ! " 

 (k−1) P 11 (x) G(k−1) (x, x − 0) − G (x, x + 0) 0 r1 (x) 11 11 + + 0 0 r2 (x) =

+ ...

κ

0

b

0

G(κ) 21 (x, ξ)

0

G(κ) 22 (x, ξ)

(10.42)

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10 Eigenvalue Problems Described by Degenerated Systems…

Taking into account Property 1 of the definition and the fact that due to Property 2 the sum of the integrals on the right side of (10.42) vanishes, substitute the value of the jump. This transformation results in the right side of (10.39), i.e., the representation (10.41) of the solution really satisfies the differential equation (10.39). Let r(ξ) = eδ(ξ − ξ f ) where e is a constant vector attached mentally to the fixed point ξ f ∈ [a, b] and δ(ξ − ξ f ) is the Dirac delta function. Assume that each element of e is equal to 1. It follows from (10.41) that 

b

y(x) =

G(x, ξ)δ(ξ − ξ f ) e dξ = G(x, ξ f ) e .

(10.43)

a

This means that the i-th column in the matrix G(x, ξ f ) is the solution that belongs to δ(ξ − ξ f )ei , where ei is the column matrix in which the i-th element is 1 while the other elements are all equal to zero.

10.2.2 Existence and Calculation of the Green Matrix Function In accordance with (9.74), the general solution of differential equation L(y) = 0 can be given again in the form  y=

κ 

Y  (x) C  (n×n) =1 (n×n)

 e ,

(10.44)

(n×1)

where Ci is a constant non-singular matrix and e is a constant vector. For the sake of distinguishing the columns in the matrices Y i , they are denoted by η iν (ν = 1, . . . , l). In contrast to (9.74) l¯ = κn − [(n − j)k + κ j] columns are, however, identically equal to zero in the matrices Yi . Consider now the hypermatrix ⎡

⎤ U1 [η 11 ] . . . U1 [η 1n ] . . . U1 [η iν ] . . . U1 [η κn ] D = ⎣ ······ ······ ······ ······ ······ ······ ······ ⎦ . Uκ [η 11 ] . . . Uκ [η 1n ] . . . Uκ [η iν ] . . . Uκ [η κn ]

(10.45)

Taking into account what has been said about the boundary conditions—here we think of the structure of the matrix P f –and recalling that l¯ columns are identically zero—one can come to the conclusion that l¯ rows and columns are identically zero ˇ be the matrix obtained by removing the zero rows and in the hypermatrix D. Let D columns. Then we can establish the following statement: ˇ = 0 then there exists a uniquely determined Green matrix function If det(D) which meets Properties 1. to 4. of the definition. In addition, the solution given by (10.41) is the only solution of the boundary value problem (10.39), (10.40a) for an arbitrary right side r. The proof of our statement is similar to that given in [16].

10.2 Green Function Matrix for Degenerated Differential Equation Systems

361

To prove the statement, we have to detail the calculation of the Green matrix function. The line of thought presented in the sequel is basically the same as the one presented in Sect. 9.2.7. The notations we are going to use are more or less also the same as those in the subsection mentioned. With regard to the third property of the definition, the Green function matrix G(x, ξ) is sought in the following form: G(x, ξ) =

n 

 Y  (x) A  (ξ) ± B  (ξ) ,

(10.46)

=1

where A  (ξ) and B  (ξ) are n × n matrices and the sign is [positive](negative) if [x ≤ ξ](ξ ≤ x). As regards the structure of (10.46), it coincides formally with Eq. (9.75). After partitioning the matrices Y  and B 

 n − j { Y 1 j { Y 2   

j

n− j

     B B ,  1 2  n×n

n×n

the following equation system can be obtained from Property 2 of the definition:

# 

# # Y 1 B 1 # Y 1 B 2 = 0 0  Y 2 B 1  Y 2 B 2

# (1) 

# (1) Y 1 B 1  Y 1 B 2 0 # (1) # (1) = 0  Y 2 B 1  Y 2 B 2

0 0 0 0

 ,

(10.47a)

,

(10.47b)



··· = ··· 

# (k−1)   # (k−1) k 1 −1 Y 1 B 1  Y 1 B 2  − P 0 # (k−1) # (k−1) = , 2 11 0 0  Y 2 B 1  Y 2 B 2  # (k)  # (k) 00 ,  Y 2 B 1  Y 2 B 2 = # 

Y(κ−1) B 1 2

··· = ···   # (κ−1) n 1 −1 , = Y B 0 − 2 P22 2  2

(10.47c) (10.47d)

(10.47e)

where ( = 1, 2, . . . , κ). Note that the sizes of the zero sub-matrices on the right sides of the above equations are the same as those of the corresponding sub-matrices on the left sides. Let the ν-th column vector of B  be denoted by B ν . For a given ν, the matrix  B Tν (1×(κ×n))

=

 BT1ν | . . . | BTiν | . . . |BTκν (1×n) (1×n) (1×n)

, i = 2, 3, . . . , κ − 1

(10.48)

362

10 Eigenvalue Problems Described by Degenerated Systems…

is that of the unknowns. The system of equations (10.47) has the same coefficient matrix for each unknown vector B ν : W B ν = P ν, (10.49) where ⎡

η 11 1 . . .

η 1n 1 . . .

η iν 1 . . .

...

η 1n 2 . . .

η iν 2 . . .

η (κ−1) ... 11 2

η (κ−1) ... 1n 2

(κ−1) η iν ... 2

⎢ ⎢ η 11 2 ⎢ ⎢ ⎢ ⎢ (k−1) ⎢η 11 1 W=⎢ ⎢ η (k−1) ⎢ 11 2 ⎢ (k) ⎢η ⎢ 11 1 ⎢ ⎣

η κ1 1 . . .

η κn 1

η (κ−1) ... κ1 2

η (κ−1) κn 2

⎤

⎥ η κn 2 ⎥ ⎥ ······ ······ ······ ······ ······ ······ ······ ······⎥ ⎥ ⎥ (k−1) (k−1) (k−1) (k−1) ⎥ ... η 1n 1 . . . η iν 1 . . . η κ1 1 . . . η κn 1 ⎥ ⎥ (k−1) ... η (k−1) ... η iν ... η (k−1) ... η (k−1) ⎥ 1l 2 2 κ1 2 κn 2 ⎥ (k) (k) (k) (k) ... η 1n 1 . . . η iν 1 . . . η κ1 1 . . . η κn 1 ⎥ ⎥ ⎥ ······ ······ ······ ······ ······ ······ ······ ······⎦

in which



η iν 1 = η iν 2

η iν



η κ1 2 . . .

}n− j } j

and P ν is the ν-th column in the transpose of the matrix formulated from the right sides of Eqs. (10.47): columns in the blocks ⎡

n

n

j

j

j

0

0 ...

0

k

⎣0 ...

0

0 ...

0

1

n− j

k−1

T − 21 (P−1 11 )

0

0 k

⎤ 0 n

0

T − 21 (P−1 22 )

κ−1

κ

0 ... k+1

j

⎦

n-j j

block number Note the size of the coefficient matrix W is ((kn + (κ − k) j) × κn). It also holds that l¯ = κn − [(n − j)k + κ j] columns in the matrix W are identically equal to zero. If we remove the zero columns, the column number will be equal ˇ to the row number, i.e., a square matrix is obtained which will be denoted by W. Consequently, the elements in the unknown matrices B ν with the same index are also set to zero. The matrix obtained is denoted by Bˇ ν , and it is the solution of the equation system

10.2 Green Function Matrix for Degenerated Differential Equation Systems

ˇ Bˇ ν = P ν . W

363

(10.50)

Introducing new variables, the system of ordinary differential equation L[y] can be replaced by a system of ordinary differential equations of order one. The new ˇ coincides with the Wronsky variables can always be chosen in such a way that det(W) determinant of the system of differential equations of order one. Since the Wronsky determinant differs from zero and the vector P ν has at least one non-zero element, the linear equation system (10.50) is soluble, i.e., there exists a solution different from the trivial one. If the matrices B  have been determined, the functions A  can be calculated from Property 4 of the definition. The calculation steps are similar to those presented at the end of Sect. 9.2.7. Let α be the ν-th unit vector in the space of size n × n. Then, it follows from the fourth property that Uμ

 n  =1

 Y  (x)A  (ξ)α = ∓Uμ



n 

 Y  (x)B  (ξ)α , μ = 1, 2, . . . , κ .

=1

(10.51) Let Aiν be the ν-th column vector in Ai . The matrix defined by the equation  T T T | . . . |Aiν | . . . |Anν ATν = A1ν is, in fact, that of the unknowns which belong to the index ν. Due to the choice of α, Eq. (10.51) can be rewritten in the following form:   U μ Y 1 (x)| . . . |Y n (x) A ν = ∓U μ Y 1 (x)| . . . |Y n (x) B ν μ = 1, . . . , κ. (10.52) Taking into account the linearity of Uμ and the structure of the column vectors of Y i , it can readily be seen that the coefficient matrix on the right side of the equation system (10.52) coincides with the matrix D given by Eq. (10.45). For the same reasons as before (for obtaining a solvable linear equation system) the elements of A ν with the same index as the elements of B ν set to zero have, are also set to zero. The result is denoted by Aˇ ν . Taking into account the structure of D, we can remove the identically zero columns ˇ multiplied by Aˇ ν . and rows in the left side of equation—this results in the matrix D As regards the right side, a similar reasoning shows that the right side, which is a ˇ Bˇ ν . This means that we have known quantity, has a similar form with the left side: D to solve the equation system ˇ Bˇ ν ˇ Aˇ ν = ∓D (10.53) D ˇ is not equal to zero. If this for Aˇ ν . The equation system (10.53) is solvable if det D condition is satisfied, then there exists the Green function matrix.

364

10 Eigenvalue Problems Described by Degenerated Systems…

10.2.3 Green Function Matrices for Some Curved Beam Problems 10.2.3.1

Pinned-Pinned Curved Beam

In this subsection, it is our aim to determine the Green function matrix for pinnedpinned curved beams. Differential equation (10.25) and boundary conditions (10.36) constitute our point of departure. This means that we want determine the Green function matrix for the boundary value problem governed by the differential equation  (4)

 (2)

 (1) y1 y1 y1 00 −m 0 0 −m + + + y2 y2 y2 01 0 2 m 0         

4

2

1

P=K 4

P=K 2

P=K 1

 (0)  y1 0 0 r + = 1 y2 r2 0 m+1   

(10.54)

0

P=K 0

associated with the boundary conditions y1 (x)|x=±ϑ = y2 (x)|x=±ϑ = y2(2) (x)

x=±ϑ

= 0.

(10.55)

The general solution to the homogeneous part of Eq. (10.54) is given by the equation y=

 4 



Yi Ci i=1 (2×2) (2×2)

e ,

(10.56)

(2×1)

where Ci is a constant non-singular matrix, e is a constant column matrix and

 cos x 0 Y1 = , sin x 0



 − sin x 0 Y2 = , cos x 0



  − sin x + x cos x (m +1)x − cos x − x sin x 1 , Y4 = . Y3 = x sin x −m x cos x 0

(10.57a) (10.57b)

It follows from Eq. (10.46) that the Green function matrix can be given in the following form:

10.2 Green Function Matrix for Degenerated Differential Equation Systems

G(x, ξ) =    (2×2)

4 

365

 Y  (x) A  (ξ) ± B  (ξ) =

=1

 $ 1 1   1 1 % cos x 0 A11 A12 ± B 11 B 12 = + sin x 0 0 0 0 0

 $ 2 2   2 2 % − sin x 0 A11 A12 ± B 11 B 12 + + cos x 0 0 0 0 0 ⎧⎡ ⎤ ⎡ ⎤⎫ 3 3

⎨ 3 3 ⎬ − sin x + x cos x (m + 1)x ⎣ A11 A12 ⎦ ⎣ B 11 B 12 ⎦ ± + + 3 3 x sin x −m ⎭ ⎩ 2 2 A21 A22 B 21 B 22 ⎧⎡ ⎤ ⎡ ⎤⎫ 3 3

⎨ 3 3 ⎬ A A B B − cos x − x sin x 1 ⎣ 11 12 ⎦ ± ⎣ 11 12 ⎦ . (10.58) + 3 3 x cos x 0 ⎩ 3 3 ⎭ A21 A22 B 21 B 22

The sign is [positive](negative) if [x ≤ ξ] (ξ ≤ x). In accordance with (10.47), Property 2 of the definition leads to the following equation system for the non-zero elements of the matrices B  : ⎡ ⎡

cos ξ ⎢ sin ξ ⎢ ⎢ − sin ξ ⎢ ⎢ cos ξ ⎢ ⎣ − sin ξ − cos ξ

− sin ξ − sin ξ + ξ cos ξ (1 + m)ξ cos ξ ξ sin ξ −m − cos ξ −ξ sin ξ 1+m − sin ξ ξ cos ξ + sin ξ 0 − cos ξ −ξ sin ξ + 2 cos ξ 0 sin ξ −ξ cos ξ − 3 sin ξ 0

− cos ξ − ξ sin ξ ξ cos ξ −ξ cos ξ −ξ sin ξ + cos ξ −ξ cos ξ − 2 sin ξ ξ sin ξ − 3 cos ξ

1

B ⎤ ⎢ 2 11 1 ⎢ ⎢ B 11 ⎢ 0⎥ ⎥⎢ 3 ⎥ 0⎥⎢ ⎢ B 11 ⎢3 0⎥ ⎥ ⎢ B 21 ⎦ 0 ⎢ ⎢4 0 ⎢ ⎣ B 11 4

⎡

⎤ 1 B 12 ⎥ 2 ⎥ B 12 ⎥ ⎥ 3 ⎥ B 12 ⎥ ⎥= ⎥ 3 B 22 ⎥ ⎥ ⎥ 4 B 12 ⎥ ⎦ 4

B 21 B 22 ⎤

0 0 ⎢ 0 0 ⎥ ⎥ ⎢ ⎥ ⎢ 1 ⎥ ⎢ 0 ⎥ ⎢ 2m =⎢ ⎥ . (10.59) ⎢ 0 0 ⎥ ⎥ ⎢ ⎢ 0 0 ⎥ ⎣ 1⎦ 0 − 2

ˇ the first column in the matrix of Here, the coefficient matrix corresponds to W, unknowns to Bˇ 1 and the second to Bˇ 2 in (10.50). As regards the right side, the first column corresponds to P 1 and the second to P 2 in (10.50). 2

4

Now the matrices P11 and P22 are scalars, and their values are −m and 1. The solutions are given by the following equations:

366

10 Eigenvalue Problems Described by Degenerated Systems… 1

1 1 sin ξ − ξ cos ξ , 2 4 1 1 = ξ sin ξ + cos ξ , 4 2 1 = cos ξ , 4 1 = , 2m 1 = − sin ξ , 4 1 ξ = − (m + 1) , 2 m

B 11 = 2

B 11 3

B 11 3

B 21 4

B 11 4

B 21

1 1 1 B 12 = − cos ξ − ξ sin ξ , 4 4 2 1 1 B 12 = sin ξ − ξ cos ξ , 4 4 3 1 B 12 = sin ξ , 4 3

(10.60)

B 22 = 0 , 4 1 B 12 = cos ξ , 4 4 1 B 21 = . 2

Note that the solutions obtained for the non-zero elements of the matrices B are independent of the boundary conditions. For the sake of a simplification in writing equations, it is worth introducing the following notations: 1

2

3

3

4

4

a = B 1i ; b = B 1i ; c = B 1i ; d = B 2i ; e = B 1i ; f = B 2i , (i = 1, 2). (10.61) By substituting the Green function matrix (10.58) into boundary conditions (10.55) and taking the above notation convention also into account, the following equation system is obtained for calculating the non-zero elements of the matrices A  : ⎡

cos ϑ ⎢ cos ϑ ⎢ ⎢− sin ϑ ⎢ ⎢ sin ϑ ⎢ ⎣ sin ϑ − sin ϑ ⎡

sin ϑ − sin ϑ cos ϑ cos ϑ − cos ϑ − cos ϑ

sin ϑ − ϑ cos ϑ − sin ϑ + ϑ cos ϑ ϑ sin ϑ ϑ sin ϑ 2 cos ϑ − ϑ sin ϑ 2 cos ϑ − ϑ sin ϑ

⎡ ⎤ 1 A1i ⎥ ⎤⎢ 2 ⎥ −(m + 1)ϑ − cos ϑ−ϑ sin ϑ 1 ⎢ ⎢A ⎥ ⎢ 1i ⎥ ⎥ (m + 1)ϑ − cos ϑ−ϑ sin ϑ 1⎥⎢3 ⎥ ⎢ ⎥ −m −ϑ cos ϑ 0⎥ A ⎥ ⎥⎢ ⎢3 1i ⎥= −m ϑ cos ϑ 0⎥ ⎢ ⎥⎢A ⎥ ⎥ 0 2 sin ϑ+ϑ cos ϑ 0⎦⎢ 2i ⎥ ⎢4 ⎥ 0 −2 sin ϑ−ϑ cos ϑ 0 ⎢A1i ⎥ ⎣ ⎦ 4 A2i

−a cos ϑ − b sin ϑ − c (sin ϑ − ϑ cos ϑ) + d(m + 1)ϑ + e (cos ϑ + ϑ sin ϑ) − ⎢a cos ϑ − b sin ϑ + c (− sin ϑ + ϑ cos ϑ) + d(m + 1)ϑ − e (cos ϑ + ϑ sin ϑ) + ⎢ ⎢ a sin ϑ − b cos ϑ − cϑ sin ϑ + dm + eϑ cos ϑ =⎢ ⎢ a sin ϑ + b cos ϑ + cϑ sin ϑ − dm + eϑ cos ϑ ⎢ ⎣ −a sin ϑ + b cos ϑ − c (2 cos ϑ − ϑ sin ϑ) − e (2 sin ϑ + ϑ cos ϑ) −a sin ϑ − b cos ϑ + c (2 cos ϑ − ϑ sin ϑ) − e (2 sin ϑ + ϑ cos ϑ)

⎤ f f⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

(10.62) Here, the first two equations are obtained from the boundary conditions y1 (x)|x=±ϑ = 0, the second two equations from the boundary conditions y2 (x)|x=±ϑ = 0, while the last two equations from the boundary conditions y2(2) (x)|x=±ϑ = 0. ˇ the matrix of Note that the coefficient matrix in (10.62) corresponds to D, ˇ ˇ ˇ unknowns to A ν , while the right side to the product D B ν in (10.53).

10.2 Green Function Matrix for Degenerated Differential Equation Systems

367

With D1 and D2 defined by the equation D1 = sin2 ϑ,

D2 = mϑ + 2 (m + 1) ϑ cos2 ϑ − 3m sin ϑ cos ϑ,

(10.63)

the solutions for the non-zero elements of the matrices A are 

2 3 3 1 2 B 1i sin ϑ cos ϑ + 2 B 1i ϑ − B 2i m (2 sin ϑ + ϑ cos ϑ) , A1i = 2D1 1



1 1 A1i = B 1i 2(1 + m)ϑ sin ϑ cos ϑ − m sin2 ϑ + 2m cos2 ϑ + D2  4

4 2 2 + B 1i 3mϑ + 2ϑ − 2m − B 2i m (ϑ sin ϑ − 2 cos ϑ) ,

(10.64a)

2

1 A1i = D2 3



(10.64b)

1 4

B 1i m − B 1i m cos2 ϑ − 2m sin2 ϑ + 2(1 + m)ϑ sin ϑ cos ϑ +  4 + B 2i m cos ϑ ,

(10.64c) 3

A2i =

 4 4 2 1 B 1i cos ϑ + B 1i (ϑ sin ϑ − cos ϑ) + B 2i cos2 ϑ , D2 4

A1i =

1 2D1

  3 3 −2 B 1i sin ϑ cos ϑ + B 2i m sin ϑ ,

2 3 1 −2 B 1i sin ϑ − 2 B 1i (sin ϑ + ϑ cos ϑ) + A2i = 2D1  3

2 2 + B 2i mϑ cos ϑ + 3m sin ϑ (cos ϑ + ϑ sin ϑ) + 2ϑ sin ϑ .

(10.64d)

(10.64e)

4

(10.64f)

Here, the solutions are presented in terms of the non-zero elements of the matrices B  by using definition (10.61) of the quantities a, . . . , f . With regard to the fact that the solutions (10.60) and (10.64) obtained for the non-zero elements of the matrices B  and A  are relatively long formulas, we do not present the whole Green function matrix here.

10.2.3.2

Fixed-Fixed Curved Beam

The solution steps are the same as those applied to calculating the elements of the Green function matrix for pinned-pinned beams. Since the matrices B  are boundary

368

10 Eigenvalue Problems Described by Degenerated Systems…

condition-independent quantities, the solutions given by Eq. (10.60) are valid for fixed-fixed beams too. It is therefore sufficient to deal with the calculation of the non-zero elements of the matrices A  . For fixed-fixed beam, the first four boundary conditions are the same as those for the pinned-pinned beam. The last two boundary conditions are, however, different: y2(1) (x)|x=±ϑ = 0. Hence, the first four equations in (10.62) will not change. After utilizing the last to boundary conditions, we arrive at the following equation system: ⎡

cos ϑ ⎢ cos ϑ ⎢ ⎢− sin ϑ ⎢ ⎢ sin ϑ ⎣ cos ϑ cos ϑ

sin ϑ − sin ϑ cos ϑ cos ϑ sin ϑ − sin ϑ

sin ϑ − ϑ cos ϑ −(m + 1)ϑ −cos ϑ−ϑ sin ϑ − sin ϑ + ϑ cos ϑ (m + 1)ϑ −cos ϑ−ϑ sin ϑ ϑ sin ϑ −m −ϑ cos ϑ ϑ sin ϑ −m ϑ cos ϑ − sin ϑ − ϑ cos ϑ 0 cos ϑ − ϑ sin ϑ sin ϑ + ϑ cos ϑ 0 cos ϑ − ϑ sin ϑ

⎡1 ⎤ A1i ⎤ ⎢2 ⎥ ⎥ 1 ⎢ ⎢A1i ⎥ ⎥ 1⎥ ⎢ ⎥ ⎢3 ⎥ ⎥ A 0⎥ ⎢ 1i ⎥ ⎢3 ⎥ ⎥= 0⎥ ⎢ ⎢A2i ⎥ ⎥ 0⎦ ⎢ ⎢4 ⎥ ⎥ 0 ⎢ A ⎣ 1i ⎦ 4

A2i

⎡

−a cos ϑ − b sin ϑ − c (sin ϑ − ϑ cos ϑ) + d(m + 1)ϑ + e (cos ϑ + ϑ sin ϑ) − ⎢a cos ϑ − b sin ϑ + c (− sin ϑ + ϑ cos ϑ) + d(m + 1)ϑ − e (cos ϑ + ϑ sin ϑ) + ⎢ a sin ϑ − b cos ϑ − cϑ sin ϑ + dm + eϑ cos ϑ ⎢ =⎢ a sin ϑ + b cos ϑ + cϑ sin ϑ − dm + eϑ cos ϑ ⎢ ⎣ −a cos ϑ − b sin ϑ + c (sin ϑ + ϑ cos ϑ) − e (cos ϑ − ϑ sin ϑ) a cos ϑ − b sin ϑ + c (sin ϑ + ϑ cos ϑ) + e (cos ϑ − ϑ sin ϑ)

⎤ f f⎥ ⎥ ⎥ ⎥. ⎥ ⎦

(10.65) By using the notations D1 = ϑ − sin ϑ cos ϑ ,

(10.66a)

D2 = m sin ϑ (ϑ cos ϑ − 2 sin ϑ) + (m + 1) ϑ + ϑ cos ϑ sin ϑ, 2

(10.66b)

the solutions can be given in the following forms: 1

A1i =

2

A1i =

3

A1i =



2 3 3 1 − B 1i cos2 ϑ + B 1i ϑ2 + B 2i m (cos ϑ − ϑ sin ϑ) , D1

1 4 1 1 B 1i (m + 1) ϑ sin2 ϑ + 2 B 1i m sin ϑ cos ϑ − 2 B 1i mϑ+ D2  4 4 + B 1i (m + 1) ϑ3 + B 2i m (ϑ cos ϑ + sin ϑ) ,

(10.67a)

(10.67b)

4 1 1 B 1i (m + 1) ϑ + B 1i (m + 1) ϑ cos2 ϑ− D2 4

4

−2 B 1i m sin ϑ cos ϑ + B 2i m sin ϑ ,

(10.67c)

10.2 Green Function Matrix for Degenerated Differential Equation Systems 3

A2i =

369



1 4 4 1 2 B 1i sin ϑ − 2 B 1i ϑ cos ϑ + B 2i (ϑ + sin ϑ cos ϑ) , D2 1 A1i = D1 4



2

3



3

B 1i − B 1i sin ϑ − B 2i m cos ϑ , 2

(10.67d)

(10.67e)

2 3 3 1 2 B 1i cos ϑ − 2 B 1i ϑ sin ϑ + B 2i (m + 1) ϑ2 − A2i = D1 4

3

3



− B 2i (m + 1) ϑ sin ϑ cos ϑ − 2 B 2i m cos ϑ . 2

(10.67f)

Note that the solutions are given again in terms of the non-zero elements of the matrices B  by utilizing relations (10.61). With (10.60) and (10.67), the Green function matrix for fixed-fixed beams can be given in a closed form. 10.2.3.3

Pinned-Fixed Curved Beam

It is obvious that the non-zero elements of the matrices B can be obtained from Eq. (10.60). As regards the non-zero elements of the matrices A , the reader is referred to Problem 10.3. Its solution in Sect. C.10 contains the formulas for the non-zero elements of the matrices A —see Eq. (C.10.56) for details. With (10.60) and (C.10.56), the Green function matrix can also be given in a closed form for pinned-fixed curved beams. 10.2.3.4

Closing Remarks Concerning the Calculation of the Green Function Matrices

It is proven in the solution of Problem 10.1 that the differential equation (10.54), which coincides with left side of the differential equation (10.35), is self-adjoint. Hence, the Green function matrices we have determined for the pinned-pinned, fixedfixed, and pinned-fixed curved beams should be cross-symmetric, i.e., they should satisfy the requirement GT (x, ξ) = G(ξ, x) . Fulfillment of this equation is checked numerically—a Fortran 90 program is written for the computation of the three Green function matrices and the computational results prove that the Green function matrices mentioned are really cross-symmetric. Let us assume that the dimensionless load is either a tangential or a normal unit load applied to the curved beam at the point ξ, i.e.,

r=

δ(ξˇ − ξ) 0



 or

r=

 0 , δ(ξˇ − ξ)

(10.68)

370

10 Eigenvalue Problems Described by Degenerated Systems…

where δ is the Dirac function, ξˇ is a parameter (ξˇ ∈ [−ϑ, ϑ]), and ξ is fixed. Then, it follows from (10.41) that 

b



a

or

 a

b



ˇ G 12 (x, ξ) ˇ G 11 (x, ξ) ˇ ˇ G 21 (x, ξ) G 22 (x, ξ)

ˇ G 12 (x, ξ) ˇ G 11 (x, ξ) ˇ G 22 (x, ξ) ˇ G 21 (x, ξ)





 

G 11 (x, ξ) δ(ξˇ − ξ) dξˇ = G 21 (x, ξ) 0

(10.69a)

 

0 ˇ = G 12 (x, ξ) . d ξ G 22 (x, ξ) δ(ξˇ − ξ)

(10.69b)

In words: (a) G 11 (x, ξ) and G 21 (x, ξ) are dimensionless tangential and normal displacements at the point x due to a tangential dimensionless unit load applied at the point ξ; (b) G 12 (x, ξ) and G 22 (x, ξ) are dimensionless tangential and normal displacements at the point x due to a normal dimensionless unit load applied at the point ξ. This is the geometrical meaning of the elements in the Green function matrix. For a pinned-fixed curved beam, Fig. 10.3 depicts graphically the four dimensionless displacement components and the dimensionless unit load if x < 0 and ξ > 0. Fig. 10.3 Displacement components due to unit forces

10.2.4 Free Vibrations of Curved Beams The eigenvalue   problems determined by differential equation system (10.35)—the left side K y of this equation is given by Eq. (10.24)—and boundary conditions (10.36), (10.37), and (10.38) describe the free vibrations of pinned-pinned, fixedfixed, and pinned-fixed curved beams. This means that the eigenvalues—if they are known—make it possible to calculate the circular frequencies by the use of Eq. (10.31b), while the corresponding eigenfunctions coincide with the mode shapes of the free vibrations at the considered circular frequency. According to Eq. (10.41), the solutions of the boundary value problems determined by the differential equation system K[y] = r

10.2 Green Function Matrix for Degenerated Differential Equation Systems

371

repeated here on the basis of (10.24) for completeness, and boundary conditions (10.36), (10.37), and (10.38) can be given in closed forms provided that we know the Green function matrices for each of the mentioned boundary conditions:  y(x) =

ϑ −ϑ

G(x, ξ)r(ξ) dξ .

(10.70)

Here r is the dimensionless load. It follows from (10.35) that r(ξ) = λM0 (x) y(x) = λy(x) is the dimensionless load for the vibration problems of curved beams. By substituting λy(x) for r in (10.70), a homogeneous Fredholm integral equation system is obtained

 

  ϑ

y1 (ξ) y1 (x) G 11 (x, ξ) G 12 (x, ξ) =λ dξ, y2 (x) y2 (ξ) −ϑ G 21 (x, ξ) G 22 (x, ξ)

(10.71)

which can be solved numerically by applying the solution algorithm presented in Sect. 9.2.13—now G(x, ξ) is the kernel function. Remark 10.5 Consider a straight beam with the same length  as that of the curved beam. It is also assumed that material distribution over the cross sections of the straight beam is the same as that of the curved beam which means that the straight beam is also heterogeneous. The free vibrations of such beams are described by the differential equation ρa Aω 2 d4 W = λW , λ = , (10.72) dx 4 Iey where W is the amplitude of the deflection, ρa is the average density on the cross section, and  z 2 E(y, z)d A , E(y, z) = E(−y, z) . (10.73) Iey = A

We remark that the derivation of these equations is left for Problem 10.5. Note that differential equation (10.72) which describes the free vibration of heterogeneous straight beams coincides with differential equation (7.82) which describes the free vibrations of homogeneous straight beams. In addition, the boundary conditions for pinned-pinned, fixed-fixed, and pinned-fixed beams are also the same independent of the fact whether the straight beam is homogeneous or not. Hence, the results presented for the eigenvalues in Sect. 7.5.2 remain valid for heterogeneous beams too except one thing: the product I E should be replaced by Iey = Ieη . The latter equality reflects that the cross section and the material distribution over the cross section are the same for the straight and curved beams.

372

10 Eigenvalue Problems Described by Degenerated Systems…

Let λˆ i and ωˆi be the i-th eigenvalue and eigenfrequency (i = 1, . . . , 4) of a straight beam (pinned-pinned, fixed-fixed, or pinned-fixed). The first eigenfrequency of the pinned-pinned beam is, however, denoted by ωˆ 1 pp . For our later considerations, we determine the quotient Ci char = ωˆ i /ωˆ 1 pp . The calculations are based on equation (7.109b) and the last column of Table 7.5: Ci char

ωˆ i = = ωˆ 1 pp



βi β1 pp

2

 =

βi  β1 pp 

2 =

βi2 . π2

(10.74)

The results are gathered in Table 10.2: Table 10.2 Typical values of Ci,char

Beams i =1 i =2 i =3 i =4 Pinned-pinned beams 1 4 9 16 Fixed-fixed beams 2.260 6.265 12.25 20.25 Pinned-fixed beams 1.562 5.062 10.56 18.06 Since ωˆ 1 pp =

2

π2 &

it also holds that ωˆ i = Ci char ωˆ 1 pp =

ρa A Ieη

,

Ci char π 2 & . 2 ρIaeηA

(10.75)

(10.76)

The eigenvalue problem governed by the Fredholm integral equation (10.71) is solved numerically. The computations are based on the algorithm presented in Sect. 9.2.14) and coded in Fortran 90. Recalling that (10.31b) gives ωi in terms of λi and taking into account that  = 2ϑ = ϑ¯ Figs. 10.4, 10.5, and 10.6 show the quotient ωi Ci,char = ωˆ i

&

√

λi ρa A 2 Ieη R

2 & π

ρa A 2 Ieη 

√ ϑ¯ 2 λi = π2

(10.77)

against ϑ¯ = 2ϑ and m is a parameter. The eigenfrequencies of the curved beams we have considered are therefore compared to the first eigenfrequency of straight beams with the same length and same material composition. It is worth emphasizing that the material composition is incorporated into the model via the parameter m—the figures clearly represent its effect— Ieη and ρa . For the quotients (10.77) depicted in Figs. 10.4, 10.5, and 10.6, the dimensionless quantity m serves as a parameter. The selected values of m are 750, 1 000, 1 300,, 1 750, 2 400, 3 400, 5 000, 7 500, 12 000, 20 000, 35 000, 60 000, 100 000, and 200 000. It should also be mentioned that Figs. 10.4, 10.5, and 10.6 are taken

10.2 Green Function Matrix for Degenerated Differential Equation Systems

373

25

20

16 15

10 9

5 4

1 0

1

2

3

4

5

6

Fig. 10.4 The first four natural circular frequencies against the central angle for pinned-pinned curved beam

from paper [14] with a permission from the Miskolc University Press. The results obtained for pinned-pinned and fixed-fixed beams are identical to those published in [15] for homogeneous beams. In this work, the eigenfrequencies are, however, computed by utilizing the frequency determinant. For ϑ¯ = 2π, the eigenfrequency ω2 of the pinned-pinned beam is zero. This result reflects the fact that the pinned-pinned beam can rotate freely about the two supports which, in this case, coincide.

374

10 Eigenvalue Problems Described by Degenerated Systems…

35

30

25

20.25

15 12.25 10

6.265

2.260 0

1

2

3

4

5

6

Fig. 10.5 The first four natural circular frequencies against the central angle for fixed-fixed curved beam

It is also an interesting phenomenon that the ratios for the even eigenfrequencies do not depend on m. Another important property is that a frequency shift can be observed: in terms of magnitude, the first/third frequency becomes the second/fourth one if the central angle is sufficiently great.

10.2 Green Function Matrix for Degenerated Differential Equation Systems

375

Fig. 10.6 The first four natural circular frequencies against the central angle for pinned-fixed curved beam

Note that the results obtained for pinned-fixed beams are larger than those for pinned-pinned beams and smaller than the results for fixed-fixed beams. This difference follows from the fact that the rigidity of the pinned-fixed beams is larger than that of the pinned-pinned beams and is smaller than the rigidity of fixed-fixed beams.

376

10 Eigenvalue Problems Described by Degenerated Systems…

Table 10.3 Data comparison for pinned-pinned beam

2ϑ π/2 π/2 π/2 π/2 π π π π

ω1 ω2 ω3 ω4 ω1 ω2 ω3 ω4

Current model Ref. [7] 38.41 38.28 89.77 89.08 172.18 169.75 245.82 243.05 6.33 6.32 19.32 19.28 39.05 38.87 63.79 63.29

Table 10.4 Data comparison for fixed-fixed beam

2ϑ π/2 π/2 π/2 π/2 π π π π

ω1 ω2 ω3 ω4 ω1 ω2 ω3 ω4

Current model Ref. [7] 63.1 62.62 117.5 115.85 218.2 213.28 249.8 247.96 12.24 12.21 26.92 26.80 50.07 49.70 76.85 75.95

The numerical results obtained and presented graphically in Figs. 10.4, 10.5, and 10.6 are compared with the findings of [7]. This article incorporates axial extension, rotatory inertia, and transverse shear effects into the model as well. Homogeneous, rectangular cross section is assumed with the following data: ρa = 7 800 kg/m3 , A = 0.01 m2 , Ieη = 1.66 · 106 Nm2 . The natural circular frequencies are compared in Table 10.3 for pinned-pinned and in Table 10.4 for fixed-fixed supports as long as m = 10 000. The correlation happens to be really good. If there is a vertical load applied to the curved beam at the crown point, the eigenfrequencies of the vibrations will change. The effect of this load is clarified partly in thesis [13] and in papers [17–19]. Measurement data for the free vibrations of fixed-fixed beams are also published in [17]. They coincide with the computed values with a good accuracy.

10.3 Problems Problem 10.1 Prove that the eigenvalue problems defined by differential equation (10.35) and boundary conditions (10.36), (10.37), and (10.38) are all self-adjoint. Problem 10.2 Find the Rayleigh quotient for the eigenvalue problems considered in the previous problem.

10.3 Problems

377

Problem 10.3 Consider the pinned-fixed curved beams and find the equation system for the non-zero elements of the matrices A which are included in the Green function matrix. Determine the solutions too. Problem 10.4 Find the Rayleigh quotient by utilizing the solution of Problem 10.2 if 2x π 2ϑ π u1 = − sin x, u 2 = cos x. (10.78) π π 2ϑ 2ϑ These functions are comparative for pinned-pinned curved beams. Problem 10.5 Show that the free vibrations of straight beams with cross-sectional heterogeneity are described by differential equation (10.72).

References 1. J.P. Den Hartog, Mechanical Vibrations. Civil, Mechanical and Other Engineering Series, 1st edn.: 1934 (Dover Publications, Mineola, 1985) 2. J.P. Den Hartog, Vibration of frames of electrical machines. J. Appl. Mech. 50, 1–6 (1828) 3. E. Volterra, J.D. Morrel, On the fundamental frequencies of curved beams. Bull. Polytech. Inst. Jassy 7(11), 1–2 (1961) 4. K. Federhofer, Dynamik des Bogenträgers und Kreisringes [Dynamics of Arches and Circular Rings] (Springer, Wien, 1950) 5. M.S. Qatu, A.A. Elsharkawy, Vibration of laminated composite arches with deep curvature and arbitrary boundaries. Comput. Struct. 47(2), 305–311 (1993). https://doi.org/10.1016/00457949(93)90381-M 6. K. Kang, C.W. Bert, A.G. Striz, Vibration analysis of shear deformable circular arches by the differential quadrature method. J. Sound Vib. 181(2), 353–360 (1995). https://doi.org/10.1006/ jsvi.1995.0258 7. E. Tüfekçi, A. Arpaci, Exact solution of in-plane vibrations of circular arches with account taken of axial extension, transverse shear and rotatory inertia affects. J. Sound Vib. 209(5), 845–856 (1997). https://doi.org/10.1006/jsvi.1997.1290 8. A. Krishnan, Y.J. Suresh, A simple cubic linear element for static and free vibration analyses of curved beams. Comput. Struct. 68(5), 473–489 (1998). https://doi.org/10.1016/S00457949(98)00091-1 9. I. Ecsedi, K. Dluhi, A linear model for the static and dynamic analysis of non-homogeneous curved beams. Appl. Math. Model. 29(12), 1211–1231 (2006). https://doi.org/10.1016/j.apm. 2005.03.006 10. F.F. Çalim, Forced vibration of curved beams on two-parameter elastic foundation. Appl. Math. Model. 36(3), 964–973 (2012). https://doi.org/10.1016/j.apm.2011.07.066 11. M. Hajianmaleki, M.S. Qatu, Vibrations of straight and curved composite beams: a review. Compos. Struct. 100, 218–232 (2013). https://doi.org/10.1016/j.compstruct.2013.01.001.340 12. B. Kovács, Vibration analysis of layered curved arch. J. Sound Vib. 332(18), 4223–4240 (2013). https://doi.org/10.1016/j.jsv.2013.03.011 13. L.P. Kiss, Vibrations and stability of heterogeneous curved beams. Ph.D. thesis. Institute of Applied Mechanics, University of Miskolc, Hungary (2015), 142 pp 14. L.P. Kiss, Green’s functions for nonhomogeneous curved beams with applications to vibration problems. J. Comput. Appl. Mech. 12(1), 19–41 (2017). https://doi.org/10.32973/jcam.2017. 002

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15. G. Szeidl, Effect of change in length on the natural frequencies and stability of circular beams in Hungarian. Ph.D. thesis. Department of Mechanics, University of Miskolc, Hungary (1975), 158 pp 16. J.G. Obádovics, On the boundary and initial value problems of differential equation systems. Ph.D. thesis. Hungarian Academy of Sciences (1967) (in Hungarian) 17. L. Kiss et al., Vibrations of fixed-fixed heterogeneous curved beams loaded by a central force at the crown point. Int. J. Eng. Model. 27(3–4), 85–100 (2014) 18. L. Kiss, G. Szeidl, Vibrations of pinned-pinned heterogeneous circular beams subjected to a radial force at the crown point. Mech. Based Des. Struct. Mach.: Int. J. 43(4), 424–449 (2015). https://doi.org/10.1080/15397734.2015.1022659 19. L.P. Kiss, G. Szeidl, Vibrations of pinned-fixed heterogeneous circular beams pre-loaded by a vertical force at the crown point. J. Sound Vib. 393, 92–113 (2017). https://doi.org/10.1016/j. jsv.2016.12.032

Appendix A

A Short Introduction to Tensor Algebra

A.1 A.1.1

Notational Conventions Some Vector Operations

Let a and b be two non-zero vectors. The scalar product s of the two vectors is denoted in the usual way: s = a · b. (A.1.1) The operation sign is the reason for calling it dot product. The scalar product is a commutative vector operation: a · b = b · a. If a · b = 0, then a and b are perpendicular to each other. This condition is known as condition of perpendicularity. The zero vector is perpendicular to any other vector. The cross product c of the vectors a and b is also denoted in the usual way: c = a × b.

(A.1.2)

The cross product is not commutative: a × b = −b × a. If c = a × b = 0, then a and b are parallel to each other. This condition is that of the parallelism for a and b. The zero vector is parallel to any other vector. The dyadic product of two vectors is denoted by W = a ◦ b.

(A.1.3)

The left side of this equation is the name we have given to the product. We remark that any name can be given to a dyadic product. These names are, in general, typeset by boldface and italic letters.

© Springer Nature Switzerland AG 2020 G. Szeidl and L. P. Kiss, Mechanical Vibrations, Foundations of Engineering Mechanics, https://doi.org/10.1007/978-3-030-45074-8

379

380

Appendix A: A Short Introduction to Tensor Algebra

Properties: 1. If we dot multiply it from left or right by c, we get W · c = (a ◦ b) · c = a (b · c) , c · W = c · (a ◦ b) = (c · a) b .

(A.1.4a) (A.1.4b)

Let d be a non-zero vector. With regard to (A.1.4a) and (A.1.4b), it is clear that the product d · W · c = d · (a ◦ b) · c = (d · a) (b · c) (A.1.4c) is a scalar. 2. Equations (A.1.4) show that the dyadic product is not a commutative operation. 3. The dot product of the two dyadic products a ◦ b and c ◦ d is also a dyadic product defined by the relation (a ◦ b) · (c ◦ d) = (a ◦ d) b · c .

(A.1.5)

4. The inner product a ◦ b and c ◦ d is a scalar: (a ◦ b) · · (c ◦ d) = (a · c) (b · d) .

(A.1.6)

5. If we cross multiply W from left or right by c, we get two dyadic products: W × c = (a ◦ b) × c = a ◦ (b × c) , c × W = c × (a ◦ b) = (c × a) ◦ b .

(A.1.7a) (A.1.7b)

The box product of the vectors a, b, and c is defined by the following equation: [abc] = (a × b) · c.

(A.1.8)

The box product has the following properties: 1. The placement of the operation signs is interchangeable: (a × b) · c = a · (b × c) .

(A.1.9a)

2. Cyclic interchangeability: [abc] = [bca] = [cab] = − [bac] = − [cba] = − [acb] .

(A.1.9b)

If the three vectors are measured from the same point, the box product results in the volume of the parallelepiped determined by the three vectors. The box product is zero if the three vectors lie in the same plane. The triple cross products of the vectors a, b, and c can be expended as a × (b × c) = (a · c) b − (a · b) c ,

(a × b) × c = (a · c) b − (b · c) a . (A.1.10)

Appendix A: A Short Introduction to Tensor Algebra

A.1.2

381

Base Vectors

Let g1 , g2 , and g3 be three non-zero vectors. It will be assumed that   g1 g2 g3 = γ◦ = 0,

γ◦∗ = 1/γ◦ .

(A.1.11)

Then, the three vectors (if measured from the same point) have no common plane. It is obvious that any vector v can be given in the form v = v 1 g1 + v 2 g2 + v 3 g3 ,

(A.1.12)

where v  ( = 1, 2, 3) is the component of v in the direction g . This statement follows form the fact that the above equation system, which is detailed here ⎡

⎤ ⎡ ⎤⎡ 1⎤ vx gx1 gx2 gx3 v ⎣ v y ⎦ = ⎣ g y1 g y2 g y3 ⎦ ⎣ v 2 ⎦ , vz gz1 gz2 gz3 v3

(A.1.13)

always has a unique solution for v  . The triplet g is called basis in the 3D space since any vector v can be given in terms of g . The dual basis (The dual base vectors) is (are) defined by the following equations: g1∗ =

g2 × g3 g3 × g1 g1 × g2 , g2∗ = , g3∗ = . γ◦ γ◦ γ◦

(A.1.14a)

It can be shown that g1 =

g2∗ × g3∗ g3∗ × g1∗ g1∗ × g2∗ , g = , g = . 2 3 γ◦∗ γ◦∗ γ◦∗

(A.1.14b)

The base vectors gk and g∗ have the following properties: 1 if k =  , 0 if k =  .

=

g1∗

  g1 g2 g3 g2 × g3 γ◦ = g1 · = = = 1, γ◦ γ◦ γ◦

gk · For example g1 ·



g∗

k,  = 1, 2, 3

(A.1.15)

or g2 ·

g3∗

  g1 g2 a2 g1 × g2 g1 × g2 g1 · (g2 × g2 ) = g2 · = · g2 = = = 0. γ◦ γ◦ γ◦ γ◦

382

Appendix A: A Short Introduction to Tensor Algebra

Since ix × i y = iz , i y × iz = ix , iz × ix = i y , and it follows that

  ix i y iz = 1,

ix = i∗x , i y = i∗y , iz = i∗z .

(A.1.16a)

(A.1.16b)

This means that the basis and the dual basis is the same in the Cartesian coordinate system. If we utilize relation (A.1.15), we get from Eq. (A.1.12) that v = v · g∗ .

(A.1.17)

In the Cartesian coordinate system (x yz), it also holds that vm = v · im ,

A.2 A.2.1

m = x, y, z .

(A.1.18)

Tensors Classification

A scalar field, which describes a physical state of a body, is said to be a tensor field of order zero. (For example, the temperature field in a solid body.) A vector field, which describes a physical state of a body, is said to be a tensor field of order one. (For example, the displacement field of a solid body.) A common property of tensors of order zero and one is that they are independent of the coordinate system, i.e., they do not change if we translate and rotate the coordinate system (if the coordinate system performs a rigid body motion). In the following subsection, we shall deal with tensors of order two.

A.2.2

Homogeneous Linear Vector-Vector Functions: Tensors of Order Two

Consider the vector-vector function w = f(v),

(A.2.19)

in which w is the dependent variable and f denotes the prescription made on the independent variable v. Function (A.2.19) is a mapping of the vectors v (measured from the origin Ov ) onto the vectors w (measured from the origin Ow —see Fig. A.1):

Appendix A: A Short Introduction to Tensor Algebra

383

z

z

v

object vector

y

Ov

image vector

w y

Ow

x

x

Fig. A.1 Mapping of v onto w

The vector-vector function w = f(v) is a homogeneous linear function if the following functional equation is satisfied: w = f(v) = f(vx ix + v y i y + v y iz ) = vx f(ix ) + v y f(i y ) + vz f(iz ) =    wx

wy

wz

= w x v x + w y v y + wz vz , where

(A.2.20)

wx = wx x ix + w yx i y + wzx iz , w y = wx y ix + w yy i y + wzy iz , wz = wx z ix + w yz i y + wzz iz

(A.2.21)

are the images of the base vectors ix , i y , and iz . Mapping (A.2.20) is uniquely determined if we know the nine scalars wx x , w yx , . . . , wzz . Mapping (A.2.20) is degenerated if the 3D space (the vectors v) is (are) mapped onto a plane, or a straight line, or the origin Ow . Since vx wx = wx (ix · v) = (wx ◦ ix ) · v ,  vx

v y w y = w y (i y · v) = (w y ◦ i y ) · v ,  vy

(A.2.22)

vx wz = wz (iz · v) = (wz ◦ iz ) · v,  vz

we can rewrite Eq. (A.2.20) in the following form:

 w = wx ◦ ix + w y ◦ i y + wz ◦ iz · v,

(A.2.23)

where the sum W = w x ◦ i x + w y ◦ i y + wz ◦ iz

(A.2.24)

is called tensor of order two (or simply tensor). With the knowledge of W mapping (A.2.20) takes the form w = W ·v. (A.2.25)

384

Appendix A: A Short Introduction to Tensor Algebra

Because of its definition mapping (A.2.20) is independent of the coordinate system. Hence, so is the tensor W . The sum (A.2.24) is, however, the presentation of the tensor W in the Cartesian coordinate system (x yz). It is obvious that the image vectors wx , w y , and wz can be given in terms of W as wx = W · ix ,

wy = W · iy ,

wz = W · iz .

(A.2.26)

If we dot multiply these equations by im and take relations (A.1.4c) and (A.2.21) into account, we get m, n = x, y, z . (A.2.27) wmn = im · W · in , We remark that the simple dyad (A.1.3) is also a tensor: 

A = a ◦ b = a ◦ bx ix +b y i y +bz iz = abx ◦ ix + ab y ◦ i y + abz ◦ iz .    image of ix

image of i y

image of iz

(A.2.28) Observe that tensor a ◦ b is degenerated since the image vectors A · c = (a ◦ b) · c = a (b · c) are all parallel to a independently of the value c has.

A.2.3

Special Tensors

The zero tensor 0, for which the images of the unit vectors ix , i y , and iz are all zero vectors, maps any vector onto a zero vector. The unit tensor 1 maps each vector onto itself. It follows from the transformation

 v = vx ix + v y i y + vz iz = ix (ix · v) +i y i y · v + iz (iz · v) =

 = i x ◦ i x + i y ◦ i y + iz ◦ iz · v that 1 = i x ◦ i x + i y ◦ i y + iz ◦ iz . Consider the vector triplet n ,

|n | = 1 ,

nk · n =



1 0

if k =  , if k = 

(A.2.29)

k,  = 1, 2, 3 .

(A.2.30)

Since the unit vectors n1 , n2 , and n3 are mutually perpendicular to each other, they are said to [constitute] (determine) [an orthonormal basis] (a Cartesian coordinate system) in the 3D space. It is obvious that the unit tensor assumes the form 1 = n1 ◦ n1 + n2 ◦ n2 + n3 ◦ n3 in this basis.

(A.2.31)

Appendix A: A Short Introduction to Tensor Algebra

385

The inverse of the tensor   wo = w x w y w z  = 0

W = w x ◦ i x + w y ◦ i y + wz ◦ iz ,

is denoted by W −1 and is defined by the following equation: W · W −1 = W −1 · W = 1 .

(A.2.32)

  Since wo = wx w y wz = 0, there is a dual base for the image vectors wx , w y , and wz : w y × wz wx × w y wz × w x w∗x = , w∗y = , wz∗ = . (A.2.33) wo wo wo The inverse W −1 can be given in terms of the dual base: W −1 = ix ◦ w∗x + i y ◦ w∗y + iz ◦ wz∗ .

(A.2.34)

Proof  

W −1 ·W = ix ◦ w∗x + i y ◦ w∗y + iz ◦ wz∗ · wx ◦ ix + w y ◦ i y + wz ◦ iz =





 = ix ◦ ix w∗x · wx + ix ◦ i y w∗x · w y + ix ◦ iz w∗x · wz +    =1 =0 =0

∗ 

∗ 

 + i y ◦ ix w y · wx + i y ◦ i y w y · w y + i y ◦ iz w∗y · wz +    =0 =1 =0





 + iz ◦ ix wz∗ · wx + iz ◦ i y wz∗ · w y + iz ◦ iz wz∗ · wz =    =0

=0

=1

= i x ◦ i x + i y ◦ i y + iz ◦ iz = 1 . The transpose of the tensor W is defined by the following equation: W T = i x ◦ w x + i y ◦ w y + iz ◦ wz .

(A.2.35)

Note that

  W · v = wx (ix · v) + w y i y · v + w y i y · v =

 = (v · ix ) wx + v · i y w y + (v · iz ) wz = v · W T . If we dot multiply the above equation by the vector u, we get the following relation: u · W · v = v · W T · u, which should hold for any v and u.

(A.2.36)

386

Appendix A: A Short Introduction to Tensor Algebra

Let p and q be two non-zero vectors. Further, let S = u ◦ v and T = p ◦ q be two tensors. Some properties of the transpose are detailed below:

WT

T

=W,

(A.2.37a)

(S · T )T = T T · ST .

(A.2.37b)

Property (A.2.37a) is obvious. As regards property (A.2.37a), compare the following two equations:  T  T (S · T )T = (u ◦ v) · (p ◦ q) = (v · p) u ◦ q = (v · p) q ◦ u , T T · ST = (q ◦ p) · (v ◦ u) = (v · p) q ◦ u . The tensor W is symmetric if W = WT .

(A.2.38)

For a symmetric W , Eq. (A.2.36) yields v · W ·u = u· W ·v.

(A.2.39)

If we write im for v, in for u (m, n = x, y, z) in the above equation and take (A.2.27) into account, we get for the symmetric tensor W that wmn = im · W · in = in · W · im = wnm .

(A.2.40)

The tensor W is skew if W = −W T .

(A.2.41)

For the skew tensor W , it follows from Eq. (A.2.36) that v · W · u = −u · W · v .

(A.2.42)

If we now write im for v, in for u (m, n = x, y, z) in (A.2.42), and take (A.2.27) into account, we obtain for the skew tensor W that wmn = im · W · in = −in · W · im = −wnm hence wmm = 0 . The tensor W is said to be ⎫ positive definite ⎪ ⎪ ⎬ positive semidefinite if, for any u, negative semidefinite ⎪ ⎪ ⎭ negative definite

⎧ u·W ⎪ ⎪ ⎨ u·W ⎪u · W ⎪ ⎩ u·W

and it is indefinite if none of the above relations is satisfied.

·u>0 ·u≥0 . ·u≤0 ·u