Vibration of Discrete and Continuous Systems [3rd ed. 2019] 978-3-030-04347-6, 978-3-030-04348-3

Table of contents :
Front Matter ....Pages i-xiii
Introduction (Ahmed Shabana)....Pages 1-53
Lagrangian Dynamics (Ahmed Shabana)....Pages 55-105
Multi-degree of Freedom Systems (Ahmed Shabana)....Pages 107-199
Vibration of Continuous Systems (Ahmed Shabana)....Pages 201-278
The Finite Element Method (Ahmed Shabana)....Pages 279-342
Methods for the Eigenvalue Analysis (Ahmed Shabana)....Pages 343-378
Back Matter ....Pages 379-409

Citation preview

Mechanical Engineering Series

Ahmed Shabana

Vibration of Discrete and Continuous Systems Third Edition

Mechanical Engineering Series Series Editor Francis A. Kulacki Department of Mechanical Engineering University of Minnesota Minneapolis, MN, USA

The Mechanical Engineering Series presents advanced level treatment of topics on the cutting edge of mechanical engineering. Designed for use by students, researchers and practicing engineers, the series presents modern developments in mechanical engineering and its innovative applications in applied mechanics, bioengineering, dynamic systems and control, energy, energy conversion and energy systems, fluid mechanics and fluid machinery, heat and mass transfer, manufacturing science and technology, mechanical design, mechanics of materials, micro- and nano-science technology, thermal physics, tribology, and vibration and acoustics. The series features graduate-level texts, professional books, and research monographs in key engineering science concentrations.

More information about this series at http://www.springer.com/series/1161

Ahmed Shabana

Vibration of Discrete and Continuous Systems Third Edition

Ahmed Shabana Richard and Loan Hill Professor of Engineering Department of Mechanical & Industrial Engineering University of Illinois at Chicago Chicago, IL, USA

Originally published with the title: Theory of Vibration Volume 2 ISSN 0941-5122 ISSN 2192-063X (electronic) Mechanical Engineering Series ISBN 978-3-030-04347-6 ISBN 978-3-030-04348-3 (eBook) https://doi.org/10.1007/978-3-030-04348-3 Library of Congress Control Number: 2018962488 © Springer Nature Switzerland AG 1997, 2019 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, express 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

Dedicated to the memory of Professor M.M. Nigm.

Preface

The theory of vibration of single and two degree of freedom systems is covered in detail in the book Theory of Vibration: An Introduction, which can serve as a textbook for a first one-semester undergraduate course focused on mechanical vibrations. This book, Vibration of Discrete and Continuous Systems, contains material for a one-semester graduate course that covers the theory of multi-degree of freedom and continuous systems. An introduction to the finite element method is also presented in this book. In the two volumes, Theory of Vibration: An Introduction and Vibration of Discrete and Continuous Systems, the author attempts to cover only the basic elements of the theory of vibration that students should learn before taking more advanced courses on this subject. Each volume, however, represents a separate entity and can be used without reference to the other. This gives the instructor the flexibility of using one of these volumes with other books in a sequence of two courses on the theory of vibration. For this volume to serve as an independent text, several sections from the first volume are used in Chaps. 1 and 4 of this book.

Contents of the Book The book contains six chapters and an appendix. In the appendix, some of the basic operations in vector and matrix algebra, which are repeatedly used in this book, are reviewed. The contents of the chapters can be summarized as follows: Chapter 1 covers some of the basic concepts and definitions used in dynamics, in general, and in the analysis of single degree of freedom systems, in particular. These concepts and definitions are also fundamental in the vibration analysis of multidegree of freedom and continuous systems. Chapter 1 is of an introductory nature and can serve to review the materials covered in an elementary vibration course. The chapter also discusses the importance of the linear theory of vibration covered in this book in the analysis of highly nonlinear systems. vii

viii

Preface

In Chap. 2, a brief introduction to Lagrangian dynamics is presented. The concepts of generalized coordinates, virtual work, and generalized forces are first introduced. Using these concepts, Lagrange’s equation of motion is then derived for multi-degree of freedom systems in terms of scalar energy and work quantities. The kinetic and strain energy expressions for vibratory systems are also presented in a matrix form. Hamilton’s principle is discussed in Sect. 2.7 of this chapter, while general energy conservation theorems are presented in Sect. 2.8. Section 2.9 provides a discussion on the use of the principle of virtual work in dynamics and D’Alembert’s principle, while Sect. 2.10 explains the use of the basic methods described in this chapter in developing the vibration equations of complex physics and engineering systems. Matrix methods for the vibration analysis of multi-degree of freedom systems are presented in Chap. 3. The use of both Newton’s second law, Lagrange’s equation of motion, and virtual work for deriving the equations of motion of multi-degree of freedom systems is demonstrated. Applications related to angular and torsional oscillations are provided. The case of the undamped free vibration is first presented, and the orthogonality of the mode shapes is discussed. Forced vibration of the undamped multi-degree of freedom systems is discussed in Sect. 3.6. The vibration of viscously damped multi-degree of freedom systems using proportional damping is examined in Sect. 3.7, and the case of general viscous damping is presented in Sect. 3.8. Coordinate reduction methods using the modal transformation are discussed in Sect. 3.9. Numerical methods for determining the mode shapes and natural frequencies are discussed in Sects. 3.10 and 3.11. Section 3.12 discusses the use of the modal decomposition technique in the analysis of nonlinear systems. Chapter 4 deals with the vibration of continuous systems. Free and forced vibrations of continuous systems that have infinite number of degrees of freedom are discussed. The partial differential equations of longitudinal, torsional, and transverse vibrations of continuous systems are derived. The orthogonality relationships of the mode shapes are developed and are used to define the modal mass and stiffness coefficients. The use of both elementary dynamic equilibrium conditions and Lagrange’s equations in deriving the equations of motion of continuous systems is demonstrated. This chapter also explains using approximation methods to reduce the number of coordinates of continuous systems to a finite set. In Chap. 5, an introduction to the finite element (FE) method is presented. The assumed element displacement field, connectivity between elements, and formulation of the element mass and stiffness matrices using the FE method are discussed. The procedure for assembling the element matrices in order to obtain the structure equations of motion is outlined. The convergence of the FE solution is analyzed, and the use of higher-order and spatial elements in the vibration analysis of structural systems is demonstrated. In Sect. 5.10 of this chapter, a brief introduction to the absolute nodal coordinate formulation (ANCF) is provided. ANCF finite elements can be used for solving large rotation and deformation problems as well as liquid sloshing problems as discussed in Sect. 5.10. In Sect. 5.11, sub-structuring techniques used to obtain reduced order models are discussed.

Preface

ix

Chapter 6 is devoted to the eigenvalue analysis and to a more detailed discussion on the similarity transformation. The results presented in this chapter can be used to determine whether or not a matrix has a complete set of independent eigenvectors associated with repeated eigenvalues. The definition of Jordan matrices and the concept of the generalized eigenvectors are introduced, and several computer methods for solving the eigenvalue problem are presented.

Acknowledgment I would like to thank many of teachers, colleagues, and students who contributed, directly or indirectly, to this book. In particular, I would like to thank my students and visiting scholars Drs. D.C. Chen, W.H. Gau, Emanuele Grossi, Hao Ling, Carmine Pappalardo, Mohil Patel, Brian Tinsley, and Lingmin Xu, who have made major contributions to the development and proofreading of this book. Special thanks to my Ph.D. student Brynne Nicolsen for her help in preparing the figures of the book. The editorial and production staffs of Springer-Verlag deserve also special thanks for their cooperation and their thorough professional work. Finally, I thank my family for the patience and encouragement during the time of preparation of this book. Ahmed A. Shabana Chicago, IL, USA

Contents

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Dynamic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Single Degree of Freedom Systems . . . . . . . . . . . . . . . . . . . . . . 1.4 Oscillatory and Nonoscillatory Motion . . . . . . . . . . . . . . . . . . . . 1.5 Other Types of Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Forced Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Impulse Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Response to an Arbitrary Forcing Function . . . . . . . . . . . . . . . . . 1.9 Linear Theory of Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 6 11 17 24 28 36 39 42

2

Lagrangian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Generalized Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Virtual Work and Generalized Forces . . . . . . . . . . . . . . . . . . . . 2.3 Lagrange’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Generalized Inertia Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Strain Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Conservation Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Virtual Work and D’Alembert’s Principle . . . . . . . . . . . . . . . . . 2.10 Vibration of Complex Nonlinear Systems . . . . . . . . . . . . . . . . .

. . . . . . . . . . .

55 55 59 64 71 74 81 84 88 93 95

3

Multi-degree of Freedom Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Undamped Free Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Orthogonality of the Mode Shapes . . . . . . . . . . . . . . . . . . . . . . 3.4 Rigid-Body Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Forced Vibration of the Undamped Systems . . . . . . . . . . . . . . .

. . . . . . .

107 108 118 123 133 140 143

xi

xii

Contents

3.7 3.8 3.9 3.10 3.11 3.12

Viscously Damped Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . General Viscous Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . Approximation and Numerical Methods . . . . . . . . . . . . . . . . . . Matrix-Iteration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Method of Transfer Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . Nonlinear Systems and Modal Decomposition . . . . . . . . . . . . .

. . . . . .

145 151 160 165 179 189

4

Vibration of Continuous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Free Longitudinal Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Free Torsional Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Free Transverse Vibrations of Beams . . . . . . . . . . . . . . . . . . . . 4.4 Orthogonality of the Eigenfunctions . . . . . . . . . . . . . . . . . . . . . 4.5 Forced Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Inhomogeneous Boundary Conditions . . . . . . . . . . . . . . . . . . . 4.7 Viscoelastic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Approximation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.10 Galerkin’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.11 Assumed-Modes Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . .

201 202 214 221 232 239 247 250 252 255 265 269

5

The Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Assumed Displacement Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Comments on the Element Shape Functions . . . . . . . . . . . . . . . . 5.3 Connectivity Between Elements . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Formulation of the Mass Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Formulation of the Stiffness Matrix . . . . . . . . . . . . . . . . . . . . . . 5.6 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Convergence of the FE Solution . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Higher-Order Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Spatial Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.10 Large Rotations and Deformations . . . . . . . . . . . . . . . . . . . . . . . 5.11 Sub-structuring Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . .

279 281 287 290 296 300 309 313 317 321 329 333

6

Methods for the Eigenvalue Analysis . . . . . . . . . . . . . . . . . . . . . . . 6.1 Similarity Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Polynomial Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Equivalence of the Characteristic Matrices . . . . . . . . . . . . . . . . 6.4 Jordan Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Elementary Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Generalized Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Jacobi Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Householder Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 QR Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

343 344 346 350 354 359 361 365 367 372

. . . . . . . . . .

Contents

Appendix A: Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.1 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.2 Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.3 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.4 Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xiii

379 379 381 387 395

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405

1

Introduction

The theory of vibration of single degree of freedom systems serves as one of the fundamental building blocks in the theory of vibration of discrete and continuous systems. As will be shown in later chapters, the concepts introduced and the techniques developed for the analysis of single degree of freedom systems can be generalized to study discrete systems with multi-degree of freedom as well as continuous systems. For this book to serve as an independent text, several of the important concepts and techniques used in the analysis of single degree of freedom systems are briefly discussed in this chapter. First, the methods of formulating the kinematic and dynamic equations are reviewed in the first three sections. It is also shown in these sections that the dynamic equation of a single degree of freedom system can be obtained as a special case of the equations of the multi-degree of freedom systems. The free vibrations of the single degree of freedom systems are reviewed in Sections 1.4 and 1.5, and the significant effect of viscous, structural, and Coulomb damping is discussed and demonstrated. Section 1.6 is devoted to the analysis of the forced vibrations of single degree of freedom systems subject to harmonic excitations, while the impulse response and the response to an arbitrary forcing function are discussed, respectively, in Sections 1.7 and 1.8. Section 1.9 discusses the importance of the linear theory of vibration covered in this book in the analysis of nonlinear systems.

1.1

Kinematics of Rigid Bodies

The dynamic equations of motion of multi-degree of freedom mechanical and structural systems can be formulated using the Newtonian or the Lagrangian method. Using either method, the formulation of the kinematic position and velocity equations is a necessary step which is required in developing the inertia, elastic, and applied forces. When the approach of the Newtonian mechanics is used, one must also formulate the kinematic acceleration equations in addition to the position # Springer Nature Switzerland AG 2019 A. Shabana, Vibration of Discrete and Continuous Systems, Mechanical Engineering Series, https://doi.org/10.1007/978-3-030-04348-3_1

1

2

1 Introduction

Fig. 1.1 Coordinates of the rigid body

and velocity equations. In the Lagrangian method described in the following chapter, the equations of motion are formulated using the scalar energy quantities and, as such, the formulation of the acceleration equations is not necessary. In the case of a general rigid body displacement, the kinematic position, velocity, and acceleration equations can be expressed in terms of six independent parameters: three parameters describe the absolute translation of a reference point on the rigid body, and three rotation coordinates define the body orientation in a selected global frame of reference. Using these six independent parameters, the position, velocity, and acceleration vectors of an arbitrary point on the rigid body can be systematically developed. Figure 1.1 depicts a rigid body that has a body coordinate system denoted as XbYbZb. The global position vector of an arbitrary point on this rigid body can be described using the translation of the reference point Ob as r ¼ R þ Aub

ð1:1Þ

where R is the global position vector of the reference point as shown in the figure, A is the transformation matrix that defines the orientation of the body in the global coordinate system, and ub is a constant vector that defines the location of the arbitrary point with respect to the reference point. The vector ub is defined in terms of its constant components as ub ¼ ½x

y

zT

ð1:2Þ

Planar Kinematics In the case of planar motion, the rotation of the rigid body can be described by one parameter only, while the translation can be described by two parameters. In this case, the vector R has two time-varying components, while the transformation matrix is a function of one angle that defines the rotation of the rigid body about the axis of rotation. Without any loss of generality, we select the axis of rotation to be the Z axis. If the angle of rotation is denoted as θ, the transformation matrix A can be written in this special case as

1.1 Kinematics of Rigid Bodies

3

2

cosθ 6 A ¼4 sinθ 0

3 0 7 05

sinθ cosθ 0

ð1:3Þ

1

while the vector R is defined as  R ¼ Rx

Ry

0

T

ð1:4Þ

Spatial Kinematics In the case of a general three-dimensional displacement, the vector R is a function of three time-dependent coordinates, while the transformation matrix is a function of three independent parameters that define the rotations about the three axes of the body coordinate system. A simple rotation ϕ about the Xb-axis is defined by the matrix 2 3 1 0 0 6 7 Ax ¼ 4 0 cosϕ sinϕ 5 ð1:5Þ 0 sinϕ cosϕ A simple rotation θ about the body Yb-axis can be described using the rotation matrix 2 3 cosθ 0 sinθ 6 7 Ay ¼4 0 1 0 5 ð1:6Þ sinθ

0

cosθ

Similarly, a simple rotation ψ about the body Zb-axis is described by the rotation matrix 2 3 cosψ sinψ 0 6 7 Az ¼ 4 sinψ cosψ 0 5 ð1:7Þ 0 0 1 The three independent rotations ϕ, θ, and ψ, shown in Fig. 1.2, can be used to define an arbitrary orientation of a rigid body in space. Since these angles are defined about the moving-body axes, the sequence of the three transformations previously defined

Fig. 1.2 Three-dimensional rotations

4

1 Introduction

can be used to define the transformation matrix that defines the final orientation of the body as A ¼ Ax Ay Az 2

cosθsinψ

cosθcosψ

6 ¼ 4 sinϕsinθcosψ þ cosϕsinψ

sinϕsinθsinψ þ cosϕcosψ

cosϕsinθcosψ þ sinϕsinψ

3

sinθ

cosϕsinθsinψ þ sinϕcosψ

7 ð1:8Þ sinϕcosθ 5 cosϕcosθ

The columns of this transformation matrix define three unit vectors along the axes of the body coordinate system. It can be demonstrated that this matrix satisfies the following orthogonality condition: AT A ¼ AAT ¼ I

ð1:9Þ

where I is the 3  3 identity matrix. The orthogonality of A implies that the inverse of A is equal to its transpose. Furthermore, since the transformation matrix in planar kinematics is a special case of the spatial transformation matrix, the planar transformation matrix also must satisfy the preceding orthogonality condition. In the spatial kinematics, different sequences and angles of rotations can be used to define the body orientation. Therefore, different transformation matrices that depend on the selection of the sequence and angles of rotations can be obtained, as demonstrated by the following example. Example 1.1

In this section, the rotations ϕ, θ, and ψ are defined about the axes of the body coordinate system. If these rotations are performed about the axes of the global coordinate system, a different sequence of multiplication must be used to define the final transformation matrix. In this case, the final orientation of the body is determined by using a sequence of three rotations about the fixed axes. The transformation matrix A is defined in this case as follows: A ¼ Az Ay Ax with which, upon using the previously obtained expressions for the simple rotation matrices Ax, Ay, and Az, one obtains 2 3 cosθcosψ cosϕsinψ þ sinϕsinθcosψ sinϕsinψ þ cosϕsinθcosψ 6 7 A ¼ 4 cosθsinψ cosϕcosψ þ sinϕsinθsinψ sinϕcosψ þ cosϕsinθsinψ 5 sinθ

sinϕcosθ

cosϕcosθ ð1:10Þ

It can be shown that this transformation matrix is also orthogonal.

1.1 Kinematics of Rigid Bodies

5

Velocity Vector The velocity vector can be obtained by differentiating the position vector with respect to time. This leads to _ ub _ þA v ¼ r_ ¼ R

ð1:11Þ

The velocity vector also can be written in the familiar form _ þωu v¼R

ð1:12Þ

where ω ¼ [ωx ωy ωz]T is the angular velocity vector and u ¼ Aub. One can ~ u, where prove the cross-product identity ω  u ¼ ω 2 3 0 ωz ωy ~ ¼ 4 ωz 0 ωx 5 ω ð1:13Þ ωy ωx 0 The velocity vector can then be written as _ þω ~ Aub v¼R

ð1:14Þ

_ or ω _ T . Using this identity ~A¼ A ~ ¼ AA It follows from Eqs. 1.11 and 1.14 that ω and the transformation matrix of Eq. 1.8, it can be shown that the angular velocity vector ω can be expressed in terms of the angles ϕ, θ, and ψ and their time derivatives as (Goldstein 1950; Greenwood 1988; Roberson and Schwertassek 1988; Shabana 2010A, 2013) ω ¼ Gβ_

ð1:15Þ

where 2

1 6 G ¼4 0 0

3 0 sinθ 7 cosϕ sinϕcosθ 5, sinϕ

β ¼ ½ϕ

θ

ψ T

ð1:16Þ

cosϕcosθ

Note that the components of the angular velocity vector are not the time derivatives of the orientation coordinates. The angular velocity vector can also be defined in the body coordinate system as ωb ¼ ATω, where ωb is the angular velocity vector defined in the body coordinate system. This vector can be written in terms of the orientation coordinates and their time derivatives as ωb ¼ Gb β_

ð1:17Þ

Using the angles that define the transformation matrix of Eq. 1.8, one can show that 2 3 cosθcosψ sinψ 0 6 7 Gb ¼ 4 cosθsinψ cosψ 0 5 ð1:18Þ sinθ

0

1

6

1 Introduction

Example 1.2

Using the different definitions of the sequence of rotations and the resulting transformation matrix defined in Example 1.1 and using the identity _ AT , the angular velocity vector can be defined in terms of the orienta~ ¼A ω _ where tion coordinates and their derivatives as ω ¼ Gβ, 2 3 cosθcosψ sinψ 0 6 7 G ¼ 4 cosθsinψ cosψ 0 5, β ¼ ½ ϕ θ ψ T sinθ 0 1 It also can be shown, in this case, that the angular velocity vector defined in the _ where body coordinate system is given by ωb ¼ Gb β, 2 3 1 0 sinθ Gb ¼ 4 0 cosϕ sinϕcosθ 5 0 sinϕ cosϕcosθ Angular Acceleration Vector The angular acceleration vector α is defined as the _ which can time derivative of the absolute angular velocity vector ω, that is, α ¼ ω, also be written in terms of the derivatives of the orientation coordinates as _ β_ α ¼ Gβ€ þ G

ð1:19Þ

The angular acceleration vector can also be defined in the body coordinate system as αb ¼ ATα. This equation can be written as   _ ωb þ Aω _b αb ¼ AT ω_ ¼ AT A ð1:20Þ The following identities can be verified (Roberson and Schwertassek 1988; Shabana 2010A): _ ¼ω ~ b ωb ¼ ωb  ωb ¼ 0 ~ b, ω AT A ð1:21Þ It follows, upon the use of these identities and the orthogonality property of the _ b. transformation matrix, that αb ¼ ω

1.2

Dynamic Equations

The three-dimensional motion of a rigid body can be described using six equations which can be written in terms of the velocities and accelerations. In the Newton– Euler formulation, a centroidal body coordinate system that has an origin rigidly attached to the body center of mass is used. The Newton–Euler equations can be

1.2 Dynamic Equations

7

written as (Goldstein 1950; Greenwood 1988; Roberson and Schwertassek 1988; Shabana 2010A)        € mI 0 FR 0 R ð1:22Þ ¼ þ 0 Ib αb Fα ωb  ðIb ωb Þ where m is the total mass of the rigid body, Ib is the inertia tensor, and FR and Fα are the vectors of forces and moments acting at the center of mass of the rigid body. The inertia tensor is defined as 2 3 I xx I xy I xz 6 7 Ib ¼4 I xy I yy I yz 5 ð1:23Þ I xz I yz I zz where ð I xx ¼

ð ð ρðy2 þ z2 Þ dV, I yy ¼ ρðx2 þ z2 Þ dV, I zz ¼ ρðx2 þ y2 Þ dV, V V V ð ð ð I xy ¼  ρxydV, I xz ¼  ρxz dV, I yz ¼  ρyz dV V

V

9 > > = > > ;

V

ð1:24Þ in which V and ρ are, respectively, the volume and mass density of the rigid body; Ixx, Iyy, and Izz are the mass moments of inertia; and Ixy, Ixz, and Iyz are the products of inertia. In the Newton–Euler equations, the vector R is the global position vector of the center of mass of the body. As a result of using the center of mass as the reference point, there is no inertia or dynamic coupling between the translation and the rotation of the body in the Newton–Euler formulation. Constrained Motion A mechanical system that consists of nb rigid bodies can have at most 6 nb independent coordinates. The number of independent coordinates or the degrees of freedom of the system is nd ¼ 6nb  nc, where nd is the number of the system degrees of freedom and nc is the number of constraint functions that describe the relationships between the 6nb coordinates. The dynamics of a mechanical system that has nd degrees of freedom can be described using nd independent differential equations only. In order to demonstrate this fact, we consider the case in which the rigid body rotates about a fixed axis. Without any loss of generality, we assume that the axis of rotation is the Zb-axis of the body coordinate system. In this special case, the angular velocity and acceleration vectors are ωb ¼ ½ 0

0

ψ_ T ,

αb ¼ ½ 0

0 ψ€ T

ð1:25Þ

In this case, the vector of gyroscopic moment can be written as ωb  ðIb ωb Þ ¼ ψ_ 2 ½ I yz

I xz

0 T

and the Newton–Euler equations of the rigid body can be written as

ð1:26Þ

8

1 Introduction

2

m

60 6 6 60 6 6 60 6 6 40 0

0

0

0

0

m

0

0

0

0

m

0

0

0

0

I xx

I xy

0

0

I xy

I yy

3 2 3 3 2 €x F Rx 0 R 7 7 6€ 7 6 6 0 7 76 R 7 6 F Ry 7 6 0 7 76 y 7 6 7 7 6 6 € 7 6 F Rz 7 6 0 7 0 7 z 7 6 76 R 7 7 26 76 7¼6 7 ψ_ 6 7 6 αbx 7 6 F αx 7 6 I yz 7 I xz 7 76 7 6 7 7 6 76 7 6 7 7 6 I yz 54 αby 5 4 F αy 5 4 I xz 5

0

0

I xz

I yz

I zz

0

32

αbz

F αz

ð1:27Þ

0

Since the rotation is about a fixed axis, there is only one independent orientation coordinate, and the rigid body has four independent coordinates. The accelerations of the rigid body can be expressed in terms of the independent accelerations as 3 3 2 2 €x 1 0 0 0 R 72 € 3 6€ 7 6 6 Ry 7 6 0 1 0 0 7 R 7 x7 7 6 6 7 6 0 0 1 0 76 6R € €y 7 R 76 6 z7 6 7 ð1:28Þ 76 7¼6 6 6 €z 7 6 αbx 7 6 0 0 0 0 74 R 5 7 7 6 6 7 7 6 6 4 αby 5 4 0 0 0 0 5 ψ€ αbz

0

0

0

1

This matrix can be written compactly as € ¼ Br q €i q

ð1:29Þ

where 2

€x R

3

7 6 7 6R 6 €y 7 7 6 6€ 7 6 Rz 7 7, € ¼6 q 7 6 6 αbx 7 7 6 7 6 6 αby 7 5 4 αbz

2 2€ 3 Rx 6 7 6R €y 7 7 €i ¼6 q 6 7, 7 6R € 4 z5 ψ€

1

6 60 6 6 60 6 Br ¼6 6 60 6 6 60 4 0

0 1 0 0 0 0

0 0

3

7 0 07 7 7 1 07 7 7 7 0 07 7 7 0 07 5 0 1

ð1:30Þ

Substituting Eq. 1.29 into the Newton–Euler equations and premultiplying by the transpose of the matrix Br, one obtains 2 32 € 3 2 3 F Rx m 0 0 0 Rx 6 0 m 0 0 76 7 €y 7 7 6 R 6 76 6 F Ry 7 6 ¼6 ð1:31Þ 6 76 7 7 7 € z 5 4 F Rz 5 4 0 0 m 0 54 R 0

0

0

I zz

ψ€

F αz

1.2 Dynamic Equations

9

If we further assume that the body does not translate along the Z-axis, the number of degrees of freedom is reduced by one, and we obtain the Newton–Euler equations which govern the planar motion of the rigid body as 2 32 3 2 3 €x m 0 0 F Rx R 6 76 € 7 6 7 ð1:32Þ 4 0 m 0 54 R y 5 ¼4 F Ry 5 0

0

I zz

ψ€

F αz

In the planar Newton–Euler equations, only one mass moment of inertia, Izz, is required, while in the spatial Newton–Euler equations, six inertia coefficients are required in order to form the inertia tensor of Eq. 1.23. Example 1.3

We consider the case of the spherical pendulum shown in Fig. 1.3. The rod is assumed to be uniform with mass m and length l. The centroidal coordinate system of the rod is assumed to have principal axes such that the products of inertia of the rod are all equal to zero. In this case, the inertia tensor of the rod is defined as 2 3 0 I xx 0 6 7 Ib ¼4 0 I yy 0 5 0 0 I zz Since point O is assumed to be a fixed point, the absolute acceleration of this point is equal to zero, and therefore, € þ Aðαb  ubO Þ þ Aðωb  ðωb  ubO ÞÞ ¼ 0 aO ¼ R

Fig. 1.3 Spherical pendulum

(continued)

10

1 Introduction

where  ubO ¼ 0

0

l 2

T

is the position of point O with respect to the center of mass of the rod. It follows that € ¼ A~ R u bO αb þ γ where γ ¼ Aðωb  ðωb  ubO ÞÞ ¼ A~ ω 2b ubO Using the acceleration constraint equations, the accelerations of the rod can be expressed in terms of the angular accelerations as " # " # " # € γ A~ u bO R ¼ αb þ ¼ Br αb þ γt I 0 αb where " Br ¼

A~ u bO I

# ,

γt ¼

" # γ 0

Substituting into Eq. 1.22 and premultiplying by the transpose of the matrix Br, one obtains  T  T T ~ bO ~ bO þ Ib αb þ m~ u bO AT γ ¼ u AT FR þ Fα  ωb  ðIb ωb Þ ð1:33Þ m~ u bO u It can be shown that 3

2

ml2 0 6 I xx þ 4  T  6 6 ~ bO þ Ib ¼ 6 m~ u bO u ml2 6 0 I þ yy 4 4 0 0 2 3 ωby ωbz 2 ml 6 7 T m~ u bO AT γ ¼ 4ωbx ωbz 5 4 0

0 7 7 7 7 0 7 5 I zz

(continued)

1.3 Single Degree of Freedom Systems

11

It can also be shown that the vector of gyroscopic forces is given in terms of the components of the angular velocity vector as  3 2 ωby ωbz I yy  I zz 6 7 7 ωb  ðIb ωb Þ ¼6 4 ωbx ωbz ðI zz  I xx Þ 5   ωbx ωby I xx  I yy If the gravity is the only force acting on the pendulum, one has FR ¼ ½ 0

0

mg T ,

Fα ¼ ½ 0 0

0 T

It follows that l T ~ bO AT FR ¼ mg ½ a32 a31 u 2

0 T

where aij are the elements of the transformation matrix A. Therefore, the equations of motion of the spherical pendulum, defined by Eq. 1.33, can be written explicitly in a matrix form as 3 2 ml2 2 3 0 0 7 αbx 6 I xx þ 4 76 6 7 76 6 2 74 αby 7 6 ml 5 7 6 0 5 0 I yy þ 4 4 αbz 0 0 I zz 2 3 ωby ωbz 26 7 ml ¼ 6 ωbx ωbz 7 4 5 4 0  3 2 3 2 a32 ωby ωbz I yy  I zz 7 6 7 l6 mg 6 þ6 ωbx ωbz ðI zz  I xx Þ 7 a31 7 4 5 4 5 2   ωbx ωby I xx  I yy 0

1.3

Single Degree of Freedom Systems

Single degree of freedom systems are special cases of multi-degree of freedom systems or continuous systems. The dynamics of these systems is governed by one differential equation which can be obtained by imposing more motion constraints. Clearly, a more detailed and accurate model requires, in many applications, the use

12

1 Introduction

of more degrees of freedom. Nonetheless, the theory of vibration of single degree of freedom systems remains one of the fundamental building blocks in the analysis of multi-degree of freedom and continuous systems. In order to demonstrate how the mathematical model of a single degree of freedom system can be obtained as a special case of a multi-degree of freedom model, we use Eq. 1.32, which defines Newton–Euler equations for the planar motion of rigid bodies. If we assume that one point on the rigid body is fixed such that its velocity remains equal to zero, as in the case of a simple pendulum, one has _ ub ¼ 0 _ þA v¼R

ð1:34Þ

9  T R ¼ Rx Ry , ub ¼ ½x yT > > = " # sinψ cosψ > _ ¼ ψ_ > A ; cosψ sinψ

ð1:35Þ

where in this special case

Note that R is the global position vector of the center of mass of the body, and ub is the position vector of the fixed point with respect to the center of mass of the body. The two velocity constraints of the fixed point can be differentiated with respect to time. The result of the differentiation can be used to write the acceleration vector of the center of mass in terms of the angular acceleration as " # " # " # €x xcosψ  ysinψ xsinψ þ ycosψ R €¼ R ð1:36Þ ¼ ψ€ þ ψ_ 2 €y xsinψ þ ycosψ xcosψ þ ysinψ R These constraint equations can be used to write the accelerations of the body in terms of the independent angular acceleration and velocity as 2 3 2 3 2 3 €x xcosψ  ysinψ xsinψ þ ycosψ R 6 7 6 7 €y 7 € ¼6 ð1:37Þ q 4R 5 ¼ ψ€4 xcosψ þ ysinψ 5þ ψ_ 2 4 xsinψ þ ycosψ 5 ψ€

1

0

The coefficient vectors of the angular velocity and acceleration on the right-hand side of this equation are orthogonal as the result of the orthogonality of the tangential and normal components of the acceleration of the center of mass. Substituting the preceding equation into Eq. 1.32, which defines the Newton–Euler matrix equation in the case of planar motion, and premultiplying by the transpose of the coefficient vector of the angular acceleration defined in the preceding equation, one obtains the independent differential equation of motion of the single degree of freedom system as   I zz þ ml2 ψ€ ¼ F αz þ F Rx ðxsinψ þ ycosψ Þ þ F Ry ðxcosψ þ ysinψ Þ ð1:38Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi where l ¼ x2 þ y2 .

1.3 Single Degree of Freedom Systems

13

Example 1.4

It is shown in Example 1.3 that the equations of motion of the spherical pendulum are 3 2 ml2 0 0 72 αbx 3 6 I xx þ 4 7 6 76 6 7 2 74 αby 5 6 ml 7 6 0 0 I þ yy 5 α 4 4 bz 0 0 I zz 2 3 ωby ωbz 7 ml2 6 ¼ 6 ωbx ωbz 7 4 5 4 0  3 ωby ωbz I yy  I zz l6 7 6 7 mg 4 a31 5þ4 ωbx ωbz ðI zz  I xx Þ 5 2   ωbx ωby I xx  I yy 0 2

a32

3 2

The case of a simple pendulum can be obtained as a special case of the spherical pendulum by using the following assumptions: αby ¼ αbz ¼ 0,

ωby ¼ ωbz ¼ 0

Using these assumptions, the matrix equation of motion of the spherical pendulum reduces to

ml2 l I xx þ αbx ¼ mga32 2 4 Since the rotation is assumed to be about the X-axis only, one has a32 ¼ sinϕ,

αbx ¼ ϕ€

The equation of free vibration of the single degree of freedom simple pendulum reduces to l Io ϕ€ þ mg sinϕ ¼ 0 2 where Io is the mass moment of inertia about point O and is defined as Io ¼ I xx þ

ml2 4

14

1 Introduction

Fig. 1.4 Multi-body system

Linear and Nonlinear Oscillations The results of the preceding example demonstrate that the dynamics of simple single degree of freedom systems can be governed by second-order nonlinear differential equations. Nonlinearities can be due to nonlinear restoring, damping, or inertia forces. Because of the kinematic constraints, the inertia forces associated with the system degrees of freedom can take a complex nonlinear form. In order to demonstrate this, we consider the multibody system shown in Fig. 1.4. The system consists of two homogeneous circular cylinders, each of mass m and centroidal mass moment of inertia I, and a connecting rod AB of mass mr and length l. The cylinders, which have radius r, are assumed to roll without slipping. Because of the rolling conditions, the velocities of the centers of mass of the cylinders are equal and given by  T vO ¼ vC ¼ r θ_ 0 ð1:39Þ The absolute velocities of points A and B are equal since the angular velocity of the connecting rod is equal to zero. These velocities, which are also equal to the absolute velocity of the center of mass of the rod, are defined as " # ðracosθÞθ_ vB ¼ vA ¼ vO þ vAO ¼ vO þ ω  rAO ¼ ð1:40Þ _ aθsinθ where vAO is the velocity of point A with respect to point O, ω is the angular velocity vector of the cylinders, and rAO is the position vector of point A with respect to point O and is defined as rAO ¼ ½ asinθ

acosθ T

ð1:41Þ

The absolute acceleration of the center of mass of the cylinders and the acceleration of the center of mass of the rod ar, which is equal to the accelerations of points A and B, are

1.3 Single Degree of Freedom Systems

15

Fig. 1.5 Dynamics of the subsystems

9  T aO ¼ aC ¼ r θ€ 0 > > " # " #= 2 arx ðra cosθÞθ€  aθ_ sinθ > > ar ¼ ¼ aA ¼ aB ¼ ; ary aθ€ sinθ þ aθ_ 2 cosθ

ð1:42Þ

The multi-body system shown in Fig. 1.4 has one degree of freedom, and therefore, one must be able to obtain one differential equation that does not include the constraint forces. This equation can be obtained by using D’Alembert’s principle (Greenwood 1988; Roberson and Schwertassek 1988; Shabana 2010A). To this end, we consider the free body diagram of the right cylinder shown in Fig. 1.5a, where Rix and Riy denote the components of the reaction forces acting on cylinder i as the result of its connection with the rod. By taking the moments of the inertia forces of the right cylinder about point D2 and equating the results with the moments of the applied and joint forces, one obtains I θ€ þ mr 2 θ€ ¼ R1x ðr  acosθÞ þ R1y asinθ

ð1:43Þ

16

1 Introduction

One can also consider the forces acting on the connecting rod AB. Using the free body diagram shown in Fig. 1.5b and taking the moments of the inertia and applied forces about point A, one obtains l l mr ary ¼ R1y l  mr g 2 2

ð1:44Þ

Next, we consider the equilibrium of the subsystem shown in Fig. 1.5c. We apply again D’Alembert’s principle by taking the moments of the inertia and applied forces about point D1. The result is

l þ asinθ I θ€ þ mr2 θ€  mr arx ðr  acosθÞ þ mr ary 2

ð1:45Þ l ¼ R1x ðr  acosθÞ  R1y ðl þ asinθÞ  mr g þ asinθ 2 Adding the first and third of the last preceding three equations, subtracting the second equation from the result, and using the expression for the acceleration of the center of mass of the connecting rod, one can eliminate the joint reaction forces and obtain the equation of motion of the multi-body system shown in Fig. 1.4 as   1  2 I þ mr 2 þ mr r 2 þ a2  2racosθ θ€ þ mr raθ_ 2sinθ þ mr gasin θ ¼ 0 ð1:46Þ 2 Since for a cylinder I ¼ mr2/2, the preceding equation reduces to   3 2 1  2 2 2 mr þ mr r þ a  2racosθ θ€ þ mr raθ_ 2 sinθ þ mr gasinθ ¼ 0 2 2

ð1:47Þ

This is a nonlinear differential equation that governs the dynamics of the single degree of freedom multi-body system shown in Fig. 1.4. If the assumption of small oscillations is used, the preceding equation becomes a linear second-order ordinary differential equation given by  3 1 2 mr2 þ mr ðraÞ2 θ€ þ mr gaθ ¼ 0 ð1:48Þ 2 2 In this book, the focus will be on the analysis of linear systems. The vibration analysis of complex nonlinear multi-body systems is discussed in the multi-body literature (Roberson and Schwertassek 1988; Shabana 2013). In the remainder of this chapter, we briefly review the theory of vibration of single degree of freedom systems, which is important in the vibration analysis of multi-degree of freedom systems as well as continuous systems. The cases of free and forced vibrations as well as the effect of the damping on the dynamics of single degree of freedom systems will be examined.

1.4 Oscillatory and Nonoscillatory Motion

1.4

17

Oscillatory and Nonoscillatory Motion

In this section, we study the effect of viscous damping on the free vibration of single degree of freedom systems. The differential equation of such systems will be developed, solved, and examined, and it will be seen from the theoretical development and the examples presented in this section that the damping force has a pronounced effect on the stability of the systems. Figure 1.6a depicts a single degree of freedom system consisting of mass m supported by a spring and a damper. The stiffness coefficient of the spring is k, and the viscous damping coefficient of the damper is c. If the system is set in motion because of an initial displacement and/or an initial velocity, the mass will vibrate freely. At an arbitrary position x of the mass from the equilibrium position, the restoring spring force is equal to kx and the viscous damping force is proportional to the velocity and is equal to cx_ , where the displacement x is taken as positive downward from the equilibrium position. Using the free body diagram shown in Fig. 1.6b, the differential equation of motion can be written as m€x ¼ mg  cx_  kðx þ ΔÞ

ð1:49Þ

where Δ is the static deflection at the equilibrium position. Since the damper does not exert force at the static equilibrium position, the condition for the static equilibrium can be written as mg ¼ kΔ. Substituting this equation into Eq. 1.49 yields m€x ¼ cx_  kx, or m€x þ cx_ þ kx ¼ 0

ð1:50Þ

This is the standard form of the second-order differential equation of motion that governs the linear free vibration of damped single degree of freedom systems. A solution of this equation is in the form x ¼ Ae pt. Substituting this solution into the differential equation yields (mp2 + cp + k)Ae pt ¼ 0, from which the characteristic equation is defined as

Fig. 1.6 Damped single degree of freedom system

18

1 Introduction

mp2 þ cp þ k ¼ 0

ð1:51Þ

The roots of this equation are given by p1 ¼ 

c 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ c2  4mk , 2m 2m

c 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 ¼   c2  4mk 2m 2m

ð1:52Þ

Define the following dimensionless quantity: ξ¼

c Cc

ð1:53Þ

where ξ is called the damping factor, and Cc is called the critical damping coefficient defined as pffiffiffiffiffiffi Cc ¼ 2mω ¼ 2 km ð1:54Þ where ω is the system circular or natural frequency defined as pffiffiffiffiffiffiffiffiffi ω ¼ k∕m

ð1:55Þ

The roots p1 and p2 of the characteristic equation can be expressed in terms of the damping factor ξ as qffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffi 2 p1 ¼ ξω þ ω ξ  1, p2 ¼ ξω  ω ξ21 ð1:56Þ If ξ is greater than one, the roots p1 and p2 are real and distinct. If ξ is equal to one, the root p1 is equal to p2 and both roots are real. If ξ is less than one, the roots p1 and p2 are complex conjugates. The damping factor ξ is greater than one if the damping coefficient c is greater than the critical damping coefficient Cc. This is the case of an overdamped system. The damping factor ξ is equal to one when the damping coefficient c is equal to the critical damping coefficient Cc. In this case, the system is said to be critically damped. The damping factor ξ is less than one if the damping coefficient c is less than the critical damping coefficient Cc, and in this case, the system is said to be underdamped. In the following, the three cases of overdamped, critically damped, and underdamped systems are discussed in more detail. Overdamped System In the overdamped case, the roots p1 and p2 of Eq. 1.56 are real. The response of the single degree of freedom system can be written as xðt Þ ¼ A1 ep1 t þ A2 ep2 t

ð1:57Þ

where A1 and A2 are arbitrary constants. Thus the solution, in this case, is the sum of two exponential functions, and the motion of the system is nonoscillatory. The velocity can be obtained by differentiating Eq. 1.57 with respect to time as x_ ðt Þ ¼ p1 A1 ep1 t þ p2 A2 ep2 t

ð1:58Þ

1.4 Oscillatory and Nonoscillatory Motion

19

The constants A1 and A2 can be determined from the initial conditions. For instance, if x0 and x_0 are, respectively, the initial displacement and velocity, one has from Eqs. 1.57 and 1.58 x0 ¼ A1 + A2 and x_0 ¼ p1 A1 þ p2 A2 , from which A1 and A2 are A1 ¼

x0 p2  x_0 , p2  p1

A2 ¼

x_0  p1 x0 p2  p1

ð1:59Þ

provided that ( p1p2) is not equal to zero. The displacement x(t) can then be written in terms of the initial conditions as xðt Þ ¼

    1  x0 p2  x_0 ep1 t þ x_0  p1 x0 ep2 t p2  p1

ð1:60Þ

Example 1.5

The damped mass–spring system shown in Fig. 1.6 has mass m ¼ 10 kg, stiffness coefficient k ¼ 1000 N/m, and damping coefficient c ¼ 300 Ns/m. Determine the displacement of the mass as a function of time. Solution. The natural frequency ω of the system is rffiffiffiffi rffiffiffiffiffiffiffiffiffiffi k 1000 ω¼ ¼ ¼ 10 rad∕s m 10 The critical damping coefficient Cc is C c ¼ 2mω ¼ 2ð10Þð10Þ ¼ 200 Ns∕m The damping factor ξ is given by ξ¼

c 300 ¼ 1:5 ¼ C c 200

Since ξ > 1, the system is overdamped and the solution is given by xðt Þ ¼ A1 ep1 t þ A2 ep2 t where p1 and p2 can be determined using Eq. 1.56 as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi p1 ¼ ξω þ ω ξ2  1 ¼ ð1:5Þð10Þ þ ð10Þ ð1:5Þ2  1 ¼ 3:8197 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi p2 ¼ ξω  ω ξ2  1 ¼ ð1:5Þð10Þ  ð10Þ ð1:5Þ2  1 ¼ 26:1803 The solution x(t) is then given by xðt Þ ¼ A1 ep1 t þ A2 ep2 t ¼ A1 e3:8197t þ A2 e26:1803t The constants A1 and A2 can be determined from the initial conditions.

20

1 Introduction

Critically Damped Systems For critically damped systems, the damping coefficient c is equal to the critical damping coefficient Cc. In this case, the damping factor ξ is equal to 1, and the roots p1 and p2 of the characteristic equation are equal and are given by p1 ¼ p2 ¼ p ¼ ω. The solution, in this case, is given by xðt Þ ¼ ðA1 þ A2 t Þeωt

ð1:61Þ

where A1 and A2 are arbitrary constants. It is clear from the above equation that the solution x(t) is nonoscillatory, and it is the product of a linear function of time and an exponential decay. The form of the solution depends on the constants A1 and A2 or, equivalently, on the initial conditions. The velocity x_ can be obtained by differentiating Eq. 1.61 with respect to time as x_ ðt Þ ¼ ½A2  ωðA1 þ A2 t Þeωt

ð1:62Þ

The constants A1 and A2 can be determined from the initial conditions. For instance, given the initial displacement x0 and the initial velocity x_0 , Eqs. 1.61 and 1.62 yield, respectively, x0 ¼ A1 and x_0 ¼ A2  ωA1 , from which A1 ¼ x0 and A2 ¼ x_0 þ ωx0 . The displacement can then be written in terms of the initial conditions as    xðt Þ ¼ x0 þ x_0 þ ωx0 t eωt ð1:63Þ Example 1.6

The damped mass–spring system shown in Fig. 1.6 has mass m ¼ 10 kg, stiffness coefficient k ¼ 1000 N/m, and damping coefficient c ¼ 200 Ns/m. Determine the displacement of the mass as a function of time. Solution. The natural frequency ω of the system is rffiffiffiffi rffiffiffiffiffiffiffiffiffiffi k 1000 ω¼ ¼ ¼ 10 rad∕s m 10 The critical damping coefficient Cc is given by C c ¼ 2mω ¼ 2ð10Þð10Þ ¼ 200 Ns∕m The damping factor ξ is given by ξ¼

c 200 ¼1 ¼ C c 200

Since ξ ¼ 1, the system is critically damped and the solution is given by Eq. 1.61 as xðt Þ ¼ ðA1 þ A2 t Þe10t where the constants A1 and A2 can be determined from the initial conditions.

1.4 Oscillatory and Nonoscillatory Motion

21

Underdamped Systems In the case of underdamped systems, the damping coefficient c is less than the critical damping coefficient Cc. In this case, the damping factor ξ is less than one, and the roots of the characteristic equation p1 and p2, defined by Eq. 1.56, are complex conjugates. Let us define the damped natural frequency ωd as qffiffiffiffiffiffiffiffiffiffiffiffi ωd ¼ ω 1  ξ2 ð1:64Þ Using this equation, the roots p1 and p2 of the characteristic equation given by Eq. 1.56 are defined as p1 ¼ ξω þ iωd ,

p2 ¼ ξω  iωd

ð1:65Þ

In this case, one can show that the solution x(t) of the underdamped system can be written as xðt Þ ¼ Xeξωt sinðωd t þ ϕÞ

ð1:66Þ

where the amplitude X and the phase angle ϕ are constant and can be determined from the initial conditions. The solution x(t) is the product of an exponential decay and a harmonic function. Unlike the preceding two cases of overdamped and critically damped systems, the motion of the underdamped system is oscillatory. The velocity x_ ðt Þ is obtained by differentiating Eq. 1.66 with respect to time. This leads to x_ ðt Þ ¼ Xeξωt ½ξωsinðωd t þ ϕÞ þ ωd cosðωd t þ ϕÞ

ð1:67Þ

The peaks of the displacement curve can be obtained by setting x_ ðt Þ equal to zero, that is, Xeξωti ½ξωsinðωd t i þ ϕÞ þ ωd cosðωd t i þ ϕÞ ¼ 0

ð1:68Þ

where ti is the time at which peak i occurs. The above equation yields pffiffiffiffiffiffiffiffiffiffiffiffiffi 1  ξ2 ωd ¼ tanðωd t i þ ϕÞ ¼ ð1:69Þ ξω ξ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Using the trigonometric identity sinθ ¼ tanθ∕ 1 þ tan 2 θ, Eq. 1.69 yields qffiffiffiffiffiffiffiffiffiffiffiffiffi sinðωd t i þ ϕÞ ¼ 1  ξ2 ð1:70Þ Equations 1.66 and 1.70 can be used to define the displacement of the peak i as qffiffiffiffiffiffiffiffiffiffiffiffiffi xi ¼ 1  ξ2 Xeξωti ð1:71Þ Undamped Vibration In the case of undamped systems, c ¼ 0, and Eqs. 1.66 and 1.67 reduce in this special case to

22

1 Introduction

xðt Þ ¼ Xsinðωt þ ϕÞ,

x_ðt Þ ¼ ωXcosðωt þ ϕÞ

ð1:72Þ

Example 1.7

The damped mass–spring system shown in Fig. 1.6 has mass m ¼ 10 kg, stiffness coefficient k ¼ 1000 N/m, and damping coefficient c ¼ 10 Ns/m. Determine the displacement of the mass as a function of time. Solution. The circular frequency ω of the system is rffiffiffiffi rffiffiffiffiffiffiffiffiffiffi k 1000 ω¼ ¼ ¼ 10 rad∕s m 10 and the critical damping factor ξ is given by Cc ¼ 2mω ¼ 2ð10Þð10Þ ¼ 200 Ns∕m Therefore, the damping factor ξ is given by ξ¼

c 10 ¼ 0:05 ¼ C c 200

The damped natural frequency ωd is given by qffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ωd ¼ ω 1  ξ2 ¼ 10 1  ð0:05Þ2 ¼ 9:9875 rad∕s Substituting ω, ξ, and ωd into Eq. 1.66, the solution for the undamped single degree of freedom system can be expressed as x ¼ Xe0:5t sin ð9:9875t þ ϕÞ where X and ϕ are constants which can be determined from the initial conditions.

Equivalent Coefficients The linear differential equation of free vibration of the damped single degree of freedom system can, in general, be written in the following form: me€x þ ce x_ þ k e x ¼ 0

ð1:73Þ

where me, ce, and ke are equivalent inertia, damping, and stiffness coefficients and the dependent variable x can be a linear or angular displacement. In this general case, me, ce, and ke must have consistent units. The natural frequency ω, the critical damping coefficient Cc, and the damping factor ξ are defined, in this case, as

1.4 Oscillatory and Nonoscillatory Motion

rffiffiffiffiffiffi ke ω¼ , me

C c ¼ 2me ω ¼ 2

23

pffiffiffiffiffiffiffiffiffiffi me ke ,

ξ¼

ce Cc

ð1:74Þ

These definitions reduce to the definitions of Eqs. 1.55, 1.54, and 1.53 in the simple case of damped mass–spring systems. The use of these general definitions is demonstrated by the following example.

Example 1.8

Assuming small oscillations, obtain the differential equation of the free vibration of the pendulum shown in Fig. 1.7. Determine the circular frequency, the critical damping coefficient, and the damping factor of this system, assuming that the rod is massless.

Fig. 1.7 Angular oscillations

Solution. As shown in the figure, let Rx and Ry be the components of the reaction force at the pin joint. The moments of the externally applied forces about O are   _ M a ¼ ðklsinθÞlcosθ clθcosθ l cosθ  mglsinθ For small oscillations, sin θ  θ and cos θ  1. In this case, Ma reduces to _ M a ¼ kl2 θ  cl2 θmglθ One can show that the moment of the inertia (effective) forces about O is given by M eff ¼ ml2 θ€ (continued)

24

1 Introduction

Therefore, the second-order differential equation of motion of the free vibration is given by kl2 θ  cl2 θ_  mglθ ¼ ml2 θ€ or ml2 θ€ þ cl2 θ_ þ ðkl þ mgÞlθ ¼ 0

ð1:75Þ

which can be written in the general form of Eq. 1.73 as me θ€ þ ce θ_ þ ke θ ¼ 0 where me ¼ ml2 ,

ce ¼ cl2 ,

k e ¼ ðkl þ mgÞl

where the units of me are kgm2 or, equivalently, Nms2, the units of the equivalent damping coefficient ce are Nms, and the units of the equivalent stiffness coefficient ke are Nm. The natural frequency ω is rffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ke ðkl þ mgÞl kl þ mg ω¼ ¼ ¼ rad∕s ml me ml2 The critical damping coefficient Cc is rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kl þ mg 2 C c ¼ 2me ω ¼ 2ml ¼ 2 ml3 ðkl þ mgÞ ml The damping factor ξ of this system is ξ¼

1.5

ce cl2 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C c 2 ml3 ðkl þ mgÞ

Other Types of Damping

Thus far, we have considered the case of a viscous damping force which is proportional to the velocity. In many cases, such simple expressions for the damping forces are not directly available. It is, however, possible to obtain an equivalent viscous damping coefficient by equating expressions that represent the dissipated energy during the motion. In this section, we consider the case of structural damping which is sometimes referred to as hysteretic damping and the case of Coulomb or dry friction damping.

1.5 Other Types of Damping

25

Structural Damping The influence of the structural damping can be seen in the vibration of solid materials, which, in general, are not perfectly elastic. When solids vibrate, there is an energy dissipation due to internal friction, which results from the relative motion between the particles during deformation. It was observed that there is a phase lag between the applied force F and the displacement x, as shown by the hysteresis loop curve in Fig. 1.8, which clearly demonstrates that the effect of the force does not suddenly disappear when the force is removed. The energy loss during one cycle can be obtained as the enclosed area in the hysteresis loop and can be expressed mathematically using the following integral: ð ΔD ¼ F dx ð1:76Þ It was also observed experimentally that the energy loss during one cycle is proportional to the stiffness of the material k and the square of the amplitude of the displacement X and can be expressed in the following simple form: ΔD ¼ πcs kX 2

ð1:77Þ

where cs is a dimensionless structural damping coefficient and the factor π is included for convenience. Equation 1.77 can be used to obtain an equivalent viscous damping coefficient, if simple harmonic oscillations in the form x ¼ X sin (ωt + ϕ) is assumed. The force exerted by a viscous damper can then be written as Fd ¼ ce x_ ¼ ce Xωcos ðωt þ ϕÞ and the energy loss per cycle can be written as ð ð ΔD ¼ F d dx ¼ ce x_ dx

ð1:78Þ

ð1:79Þ

Since x_ ¼ dx∕dt, we have dx ¼ x_ dt. Substituting this equation and Eq. 1.78 into Eq. 1.79 yields

Fig. 1.8 Hysteresis loop

26

1 Introduction

ΔD ¼

ðτ

ce x_ dt ¼ 2

0

ðτ

ce ω2 X 2 cos 2 ðωt þ ϕÞdt

0

ð1:80Þ

¼ πce ωX 2 where τ is the periodic time defined as τ ¼ 2π/ω. Equating Eqs. 1.77 and 1.80 yields the equivalent viscous damping coefficient as ce ¼

kcs ω

ð1:81Þ

Coulomb Damping Consider the mass–spring system shown in Fig. 1.9. In the case of Coulomb damping, the friction force always acts in a direction opposite to the direction of the motion of the mass, and this friction force can be written as Ff ¼ μN

ð1:82Þ

where μ is the coefficient of sliding friction and N is the normal reaction force. If the motion of the mass is to the right, x_ > 0, and the friction force Ff is negative, as shown in Fig. 1.9b. If the motion of the mass is to the left, x_ < 0, and the friction force Ff is positive, as shown in Fig. 1.9c. Therefore, the vibration of the system is governed by two differential equations which depend on the direction of motion. From the free body diagram shown in Fig. 1.9b, it is clear that if the mass moves to the right, the differential equation of motion is m€x ¼ kx  Ff ,

x_ > 0

ð1:83Þ

Similarly, the free body diagram of Fig. 1.9c shows that the differential equation of motion, when the mass moves to the left, is m€x ¼ kx þ Ff ,

x_ < 0

ð1:84Þ

Equations 1.83 and 1.84 can be combined in one equation as m€x þ kx ¼ ∓Ff

Fig. 1.9 Coulomb damping

ð1:85Þ

1.5 Other Types of Damping

27

where the negative sign is used when the mass moves to the right and the positive sign is used when the mass moves to the left. Equation 1.85 is a nonhomogeneous differential equation, and its solution consists of two parts: the homogeneous solution or the complementary function and the forced or the particular solution. Since the force Ff is constant, the particular solution xp is assumed as xp ¼ C, where C is a constant. Substituting this solution into Eq. 1.85 yields xp ¼ Ff ∕k. Therefore, the solution of Eq. 1.85 can be written as 9 Ff xðt Þ ¼ A1sinωt þ A2cosωt  , x_  0 > = k ð1:86Þ > Ff ; xðt Þ ¼ B1sinωt þ B2 cosωt þ , x_ < 0 k pffiffiffiffiffiffiffiffiffi where ω is the natural frequency defined as ω ¼ k∕m, A1 and A2 are constants that depend on the initial conditions of motion to the right, and B1 and B2 are constants that depend on the initial conditions of motion to the left. Consider the case in which the mass was given an initial displacement x0 to the right and zero initial velocity. The second equation  in Eq.  1.86 (x_ < 0) can then be used to yield the algebraic equations x0 ¼ B2 þ Ff ∕k and ωB1 ¼ 0, which imply   that B1 ¼ 0 and B2 ¼ x0  Ff ∕k , that is,

Ff Ff ð1:87Þ xð t Þ ¼ x0  cosωt þ k k and



Ff x_ ðt Þ ¼ ω x0  sinωt k

ð1:88Þ

The direction of motion will change when x_ ¼ 0. The above equation then yields the time t1 at which the velocity starts to be positive. The time t1 can be obtained from this equation as ω(x0  (Ff∕k))sinωt1 ¼ 0, or t 1 ¼ π∕ω. At this time, the displacement is determined from Eq. 1.87 as xðt 1 Þ ¼ x

π  ω

¼ x0 þ

2Ff k

ð1:89Þ

which shows that the amplitude in the first half-cycle is reduced by the amount 2Ff ∕k, as the result of dry friction. In the second half-cycle, the mass moves to the right and the motion is governed by the first equation in Eq. 1.86 ( x_  0) with the initial conditions x(π∕ω) ¼ x0 + (2Ff∕k) and x_ ðπ∕ωÞ ¼ 0. Substituting these initial conditions into the first equation in Eq. 1.86 yields A1 ¼ 0 and A2 ¼ x0  3(Ff∕k). The displacement x(t) in the second half-cycle can then be written as

Ff Ff ð1:90Þ xðt Þ ¼ x0  3 cosωt  k k

28

1 Introduction

Fig. 1.10 Effect of the Coulomb damping

and the velocity is

Ff x_ ðt Þ ¼  x0  3 ωsinωt k

ð1:91Þ

Observe that the velocity is zero at time t2 when t2 ¼ 2π/ω ¼ τ, where τ is the periodic time of the natural oscillations. At time t2, the end of the first cycle, the displacement is



2π 4Ff xð t 2 Þ ¼ x ¼ x0  ð1:92Þ ω k which shows that the amplitude decreases in the second half-cycle by the amount 2Ff∕k, as shown in Fig. 1.10. By continuing in this manner, one can verify that there is a constant decrease in the amplitude of 2Ff∕k every half-cycle of motion. Furthermore, unlike the case of viscous damping, the frequency of oscillation is not affected by the Coulomb damping. It is also important to point out, in the case of Coulomb friction, that it is not necessary that the system comes to rest at the undeformed spring position. The final position will be at an amplitude Xf, at which the spring force F ¼ kXf is less than or equal to the friction force.

1.6

Forced Vibration

Figure 1.11 depicts a viscously damped single degree of freedom mass–spring system subjected to a forcing function F(t). By applying Newton’s second law, the differential equation of motion can be written as

1.6 Forced Vibration

29

Fig. 1.11 Forced vibration of single degree of freedom systems

m€x þ cx_ þ kx ¼ F ðt Þ

ð1:93Þ

where m is the mass, c is the damping coefficient, k is the stiffness coefficient, and x is the displacement of the mass. Equation 1.93, which is a nonhomogeneous second-order ordinary differential equation with constant coefficients, has solution that consists of two parts: the complementary function xh and the particular solution xp, that is, x ¼ xh + xp, where the complementary function xh is the solution of the homogeneous equation m€xh þ cx_h þ kxh ¼ 0

ð1:94Þ

The complementary function xh is sometimes called the transient solution, since in the presence of damping, this solution dies out. Methods for obtaining the transient response were discussed in the preceding sections. The particular solution xp represents the response of the system to the forcing function and is sometimes called the steady-state solution because it exists long after the transient vibration disappears. The transient solution contains two arbitrary constants, while the steady-state solution does not contain any arbitrary constants. Therefore, the complete solution x ¼ xh + xp contains two arbitrary constants, which can be determined by using the initial conditions. In the analysis presented in this section, the case of harmonic excitation in which the forcing function F(t) can be expressed in the form F(t) ¼ F0 sin ωft is considered. Substituting this force function into Eq. 1.93 yields m€x þ cx_ þ kx ¼ F 0 sinωf t

ð1:95Þ

The steady-state solution xp can be assumed in the form xp ¼ A1sinωf t þ A2 cosωf t

ð1:96Þ

which yields the following expressions for the velocity and acceleration: x_ p ¼ ωf A1cosωf t  ωf A2 sinωf t and €xp ¼ ω2f xp . Substituting xp , x_ p , and €xp into Eq. 1.95 and rearranging terms yields

30

1 Introduction



      k  ω2f m A1  cωf A2 sinωf t þ cωf A1 þ kω2f m A2 cosωf t ¼ F 0 sinωf t

This equation yields the following two algebraic equations in A1 and A2:   ) k  ω2f m A1  cωf A2 ¼ F 0   cωf A1 þ k  ω2f m A2 ¼ 0 Dividing these two equations by the stiffness coefficient k yields ) ð1  r 2 ÞA1  2r ξA2 ¼ X 0 2rξA1 þ ð1  r 2 ÞA2 ¼ 0

ð1:97Þ

ð1:98Þ

ð1:99Þ

where r¼

ωf , ω

ξ¼

c c , ¼ C c 2mω

X0 ¼

F0 k

ð1:100Þ

in which Cc ¼ 2mω is the critical damping coefficient. The two algebraic equations of Eq. 1.99 can be solved to obtain the constants A1 and A2 as A1 ¼

ð1  r 2 ÞX 0 ð1 

r 2 Þ2

þ ð2rξÞ

A2 ¼

,

2

ð2rξÞX 0 ð1  r 2 Þ2 þ ð2rξÞ2

ð1:101Þ

The steady-state solution xp of Eq. 1.96 can then be written as xp ¼



X0 ð1 

r 2 Þ2

2

þ ð2rξÞ

  1  r 2 sinωf t  ð2rξÞcosωf t

ð1:102Þ

which can be written as X0 xp ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sinðωf t  ψ Þ ð1  r 2 Þ2 þ ð2rξÞ2

ð1:103Þ

where ψ is the phase angle defined by 1

ψ ¼ tan



2rξ 1  r2

ð1:104Þ

Equation 1.103 can be written in a more compact form as xp ¼ X 0 βsin ðωf t  ψ Þ

ð1:105Þ

where β is the magnification factor defined in the case of damped systems as 1 β ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1  r 2 Þ2 þ ð2rξÞ2

ð1:106Þ

1.6 Forced Vibration

31

Fig. 1.12 Magnification factor

If the damping factor ξ is equal to zero, the magnification factor β reduces to β¼

1 1  r2

ð1:107Þ

When r ¼ 1, that is, ωf ¼ ω, the magnification factor in the case of undamped systems approaches infinity. This case is known as resonance. The magnification factor β and the phase angle ψ are shown, respectively, in Figs. 1.12 and 1.13 as functions of the frequency ratio r for different damping factors ξ. It is clear from these figures that for damped systems, the system does not attain infinite displacement at resonance since for ωf ¼ ω, which corresponds to the case in which the frequency ratio r ¼ 1, the magnification factor reduces to β¼

1 2ξ

ð1:108Þ

Furthermore, at resonance, the magnification factor β does not have the maximum value, which can be obtained by differentiating β of Eq. 1.106 with respect to r and setting the result equal to zero. This leads to an algebraic equation which can be solved for the frequency ratio r at which the magnification factor β is maximum. By so doing, one can show that the magnification factor β is maximum when pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ¼ 1  2ξ2 . At this value of the frequency ratio, the maximum magnification factor is given by

32

1 Introduction

Fig. 1.13 Phase angle

Fig. 1.14 Transmitted force

βmax ¼

1 pffiffiffiffiffiffiffiffiffiffiffiffiffi 2ξ 1  ξ2

ð1:109Þ

Force Transmission From Eq. 1.103 and Fig. 1.12, it is clear that by increasing the spring stiffness k and the damping coefficient c, the amplitude of vibration decreases. The increase in the stiffness and damping coefficients, however, may have an adverse effect on the force transmitted to the support. In order to reduce this force, the stiffness and damping coefficients must be properly selected. Figure 1.14 shows a free body diagram for the mass and the support system. The force transmitted to the support in the steady state can be written as Ft ¼ kxp þ cx_p . From Eq. 1.105, x_p is x_p ¼ ωf X 0 β cos ðωf t  ψ Þ. Therefore, the force transmitted Ft can be written as

1.6 Forced Vibration

33

Ft ¼ kX 0 β sinðωf t  ψ Þ þ cωf X 0 β cosðωf t  ψ Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   ¼ X 0 β k 2 þ ðcωf Þ2 sin ωf t  ψ where ψ ¼ ψ  ψt and ψt is a phase angle defined as

cω  f ψt ¼ tan1 ¼ tan 1 ð2rξÞ k Equation 1.110 can also be written as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   Ft ¼ X 0 kβ 1 þ ð2rξÞ2 sin ωf t  ψ Since X0 ¼ F0∕k, the above equation can be written as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   Ft ¼ F 0 β 1 þ ð2rξÞ2 sin ωf t  ψ   ¼ F 0 βt sin ωf t  ψ where qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ð2rξÞ2 2 βt ¼ β 1 þ ð2rξÞ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1  r 2 Þ2 þ ð2rξÞ2

ð1:110Þ

ð1:111Þ

ð1:112Þ

ð1:113Þ

ð1:114Þ

Note that βt represents the ratio between the amplitude of the transmitted force and the amplitude of the applied force. βt is called the transmissibility and is plotted in Fig. 1.15 versus the frequency ratio r forp different values of the damping factor ξ. It is ffiffiffi clear from Fig. 1.15 that βt > 1 for r < 2, that is, in this region, the amplitude of the transmitted force is greater than the amplitude of the applied force. Furthermore, pffiffiffi for r < 2, the transmitted force pffiffiffi to the support can be reduced by increasing the damping factor ξ. For r > 2, βt < 1, and in this region, the amplitude of the transmitted force is less than the amplitude of the applied force. In this region, the amplitude of the transmitted force increases by increasing the damping factor ξ. Work Per Cycle Equation 1.105, which defines the steady-state response to a harmonic excitation in the presence of damping, implies that, for a given frequency ratio r and a given damping factor ξ, the amplitude of vibration remains constant. This can be achieved only if the energy input to the system, as the result of the work done by the external harmonic force, is equal to the energy dissipated as the result of the presence of damping. In order to see this, we first evaluate the work of the harmonic force as dWe ¼ F ðt Þdx ¼ F ðt Þx_ dt, where We is the work done by the external force per cycle and is given by ð 2π∕ωf ð 2π∕ωf F ðt Þx_ dt ¼ F 0 sinωf t X 0 βωf cosðωf t  ψ Þdt We ¼ 0 0 ð1:115Þ ð 2π ¼ F0X0 β

sinωf tcosðωf t  ψ Þdðωf t Þ ¼ πF 0 X 0 βsinψ

0

34

1 Introduction

Fig. 1.15 Transmissibility

Similarly, one can evaluate the energy dissipated per cycle, as the result of the damping force, as Wd ¼

ð 2π∕ωt

cx_ x_ dt

0

¼ cX 20 β2 ωf

ð 2π

ð1:116Þ cos ðωf t  ψ Þdðωf t Þ ¼ 2

0

πcX 20 β2 ωf

Note that the input energy to the system is a linear function of the amplitude of the steady-state vibration X0β, while the energy dissipated as the result of the damping force is a quadratic function of the amplitude. Since at the steady state they must be equal, one has We ¼ Wd, or πF 0 X 0 βsinψ ¼ πcX 20 β2 ωf , which defines the magnification factor β as β¼

F 0 ∕X 0 sinψ cωf

ð1:117Þ

Using the definition of X0 and the phase angle ψ given, respectively, by Eqs. 1.100 and 1.104, the magnification factor β can be written as β¼

F 0 ∕X 0 k 2rξ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sinψ ¼ cωf cωf ð1  r 2 Þ2 þ ð2rξÞ2

and since cωf∕k ¼ 2rξ, the above equation reduces to

ð1:118Þ

1.6 Forced Vibration

35

1 β ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1  r 2 Þ2 þ ð2rξÞ2

ð1:119Þ

which is the same definition of the magnification factor obtained by solving the differential equation. It is obtained here from equating the input energy, resulting from the work done by the harmonic force, to the energy dissipated as the result of the damping force. In fact, this must be the case since the change in the strain energy in a complete cycle must be equal to zero owing to the fact that the spring takes the same elongation after a complete cycle. This can also be demonstrated mathematically by using the definition of the work done by the spring force as Ws ¼

ð 2π∕ωf

kxx_ dt

0

¼ kX 20 β2

ð 2π

ð1:120Þ sinðωf t  ψ Þcosðωf t  ψ Þdðωf t Þ ¼ 0

0

Example 1.9

A damped single degree of freedom mass–spring system has mass m ¼ 10 kg, spring coefficient k ¼ 4000 N∕m, and damping coefficient c ¼ 40 Ns∕m. The amplitude of the forcing function F0 ¼ 60 N, and the frequency ωf ¼ 40 rad∕s. Determine the displacement of the mass as a function of time. Determine also the transmissibility and the amplitude of the force transmitted to the support. Solution. The circular frequency of the system is rffiffiffiffi rffiffiffiffiffiffiffiffiffiffi k 4000 ω¼ ¼ ¼ 20 rad∕s m 10 The frequency ratio r is given by r¼

ωf 40 ¼2 ¼ ω 20

The critical damping coefficient Cc is defined as C c ¼ 2mω ¼ 2ð10Þð20Þ ¼ 400 Ns∕m The damping factor ξ is then given by ξ¼

c 40 ¼ 0:1 ¼ C c 400

which is the case of an underdamped system. One can then write the complete solution in the following form: (continued)

36

1 Introduction

xð t Þ ¼ xh þ xp ¼ Xeξωt sin ðωd t þ ϕÞ þ X 0 β sin ðωf t  ψ Þ where ωd is the damped circular frequency qffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ωd ¼ ω 1  ξ ¼ 20 1  ð0:1Þ2 ¼ 19:8997 rad∕s The constants X0, β, and ψ are X0 ¼

F0 60 ¼ 0:015 m ¼ 4000 k

1 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi β ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ r ¼ 0:3304 2 2 2 ð1  r 2 Þ þ ð2rζ Þ 1  ð2Þ2 þ ð2  2  0:1Þ2 ψ ¼ tan1

!

2rξ 2 ð 2 Þ ð 0:1 Þ ¼ tan 1 ¼ tan 1 ð0:13333Þ ¼ 0:13255 rad 1r 2 1  ð2Þ2

The displacement can then be written as a function of time as xðt Þ ¼ Xe2t sinð19:8997t þ ϕÞ þ 0:004956 sinð40t  0:13255Þ The constants X and ϕ can be determined using the initial conditions. The transmissibility βt is defined as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ð2rξÞ2 1 þ ð2  2  0:1Þ2 βt ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:35585 h i2 2 2 ð1  r 2 Þ2 þ ð2rξÞ2 1  ð2Þ þ ð2  2  0:1Þ The amplitude of the force transmitted is given by jFt j ¼ F 0 βt ¼ ð60Þð0:35585Þ ¼ 21:351 N

1.7

Impulse Response

An impulsive force is defined as a force which has a large magnitude and acts during a very short time duration such that the time integral of this force is finite. If the impulsive force F(t) shown in Fig. 1.16 acts on the single degree of freedom system shown in Fig. 1.17, the differential equation of motion of this system can be written as m€x þ cx_ þ kx ¼ F ðt Þ. Integrating this equation over the very short interval (t1, t2), one obtains

1.7 Impulse Response

37

Fig. 1.16 Impulsive force

Fig. 1.17 Single degree of freedom system under the effect of impulsive force F(t)

ð t2 t1

m€x dt þ

ð t2 t1

cx_ dt þ

ð t2 t1

kx dt ¼

ð t2

F ðt Þ dt

ð1:121Þ

t1

Since the time interval (t1, t2) is assumed to be very small, we assume that x does not change appreciably, and weÐ also assume that the change Ð t in the velocity x_ is finite. t One, therefore, has lim t12 cx_ dt ¼ 0 and lim t12 kx dt ¼ 0. Therefore, if t1 t 1 !t 2 Ðt Ðtt1 !t2 approaches t2, Eq. 1.121 yields t12 m€x dt ¼ t12 F ðt Þdt. Using this equation and €x ¼ d x∕ _ dt, one can write ð x_2 ð t2 m dx_ ¼ F ðt Þdt ð1:122Þ x_1

t1

where x_1 and   x_Ð2t are, respectively, the velocities at t1 and t2. Equation 1.122 yields m x_2  x_1 ¼ t12 F ðt Þdt, which defines ð 1 t2 Δx_ ¼ x_ 2  x_1 ¼ F ðt Þdt ð1:123Þ m t1 where Δx_ is the jump discontinuity in the velocity of the mass due to the impulsive force. The time integral in Eq. 1.123 is called the linear impulse I and is defined by

38

1 Introduction

I¼

ð t2

F ðt Þdt

ð1:124Þ

t1

In the particular case in which the linear impulse is equal to one, I is called the unit impulse. Equation 1.123 can be written as Δx_ ¼ x_2  x_1 ¼

I m

ð1:125Þ

This result indicates that the effect of the impulsive force, which acts over a very short time duration on a system which is initially at rest, can be accounted for by considering the motion of the system with initial velocity I/m and zero initial displacement. That is, in the case of impulsive motion, we consider the system vibrating freely as the result of the initial velocity defined by Eq. 1.125. For example, the free vibration of the underdamped single degree of freedom system shown in Fig. 1.17 is governed by the equations ) xðt Þ ¼ Xeξωt sin ðωd t þ ϕÞ ð1:126Þ x_ ðt Þ ¼ ξωXeξωt sin ðωd t þ ϕÞ þ ωd Xeξωt cos ðωd t þ ϕÞ where X and ϕ are constants to be determined from the initial conditions, ω is the natural frequency, pffiffiffiffiffiffiffiffiffiffiffiffiffi ξ is the damping factor, and ωd is the damped natural frequency ωd ¼ ω 1  ξ2 . As the result of applying an impulsive force with a linear impulse I at t ¼ 0, the initial conditions are x(t ¼ 0) ¼ 0 and x_ ðt ¼ 0Þ ¼ I∕m. Since the initial displacement is zero, x(t) of Eq. 1.126 yields ϕ ¼ 0. Using the definition of x_ ðt Þ in Eq. 1.126 and the initial velocity, one can show that xð t Þ ¼

I ξωt e sin ωd t mωd

ð1:127Þ

which can be written as xðt Þ ¼ IH ðt Þ

ð1:128Þ

where H(t) is called the impulse response function and is defined as H ðt Þ ¼

1 ξωt e sin ωd t mωd

ð1:129Þ

Example 1.10

Find the response of the single degree of freedom system shown in Fig. 1.17 to the rectangular impulsive force shown in Fig. 1.16, where m ¼ 10 kg, k ¼ 9000 N/m, c ¼ 18 Ns/m, and F0 ¼ 10,000 N. The force is assumed to act at time t ¼ 0, and the impact interval is assumed to be 0.005 s. (continued)

1.8 Response to an Arbitrary Forcing Function

39

Solution. The linear impulse I is given by I¼

ð t2

F ðt Þdt ¼

t1

ð 0:005

10,000 dt ¼ 10,000ð0:005Þ ¼ 50 Ns

0

The natural frequency of the system ω is given by rffiffiffiffi rffiffiffiffiffiffiffiffiffiffi k 9000 ω¼ ¼ ¼ 30 rad∕s m 10 The critical damping coefficient Cc is Cc ¼ 2mω ¼ 2ð10Þð30Þ ¼ 600 The damping factor ξ is ξ¼

c 18 ¼ 0:03 ¼ C c 600

The damped natural frequency ωd is qffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ωd ¼ ω 1  ξ2 ¼ 30 1  ð0:03Þ2 ¼ 29:986 rad∕s The system response to the impulsive force is then given by I ξωt e sin ωd t mωd 50 eð0:03Þð30Þt sin 29:986t ¼ ð10Þð29:986Þ

xð t Þ ¼

¼ 0:1667e0:9t sin 29:986t

1.8

Response to an Arbitrary Forcing Function

In this section, we consider the response of the single degree of freedom system to an arbitrary forcing function F(t), shown in Fig. 1.18. The procedure described in the preceding section for obtaining the impulse response can be used as a basis for developing a general expression for the response of the system to an arbitrary forcing function. The arbitrary forcing function F(t) can be regarded as a series of impulsive forces F(τ) acting over a very short-lived interval dτ. The force F(τ) then produces the short duration impulse F(τ) dτ, and the response of the system to this impulse for all t > τ is given by dx ¼ F ðτÞdτ H ðt  τÞ

ð1:130Þ

where H(t) is the impulse response function defined by Eq. 1.129. It follows that

40

1 Introduction

Fig. 1.18 Arbitrary forcing function F(t)

dx ¼ F ðτÞdτ

1 ξωðtτÞ e sin ωd ðt  τÞ mωd

ð1:131Þ

In this equation, dx represents the incremental response of the damped single degree of freedom system to the incremental impulse F(τ) dτ for t > τ. In order to determine the total response, we integrate Eq. 1.131 over the entire interval ðt xðt Þ ¼ F ðτÞ H ðt  τÞdτ ð1:132Þ 0

or 1 xð t Þ ¼ mωd

ðt

F ðτÞeξωðtτÞ sin ωd ðt  τÞdτ

ð1:133Þ

0

Equation 1.132 or Eq. 1.133 is called the Duhamel integral or the convolution integral. It is important to emphasize, however, that in obtaining the convolution integral, we made use of the principle of superposition which is valid only for linear systems. Furthermore, in deriving the convolution integral, no mention was given to the initial conditions and, accordingly, the integral of Eq. 1.132, or Eq. 1.133, provides only the forced response. If the initial conditions are not equal to zero, that is, x0 ¼ x(t ¼ 0) 6¼ 0 and/or x_0 ¼ x_ ðt ¼ 0Þ6¼ 0, then Eq. 1.133 must be modified to include the effect of the initial conditions. To this end, we first define the homogeneous solution and determine the arbitrary constants in the case of free vibration as the result of these initial conditions and then use the principle of superposition to add the homogeneous function to the forced response. Special case A special case of the preceding development is the case of an undamped single degree of freedom system. In this case, ωd ¼ ω and ξ ¼ 0, and the impulse response function of Eq. 1.129 reduces to H ðt Þ ¼

1 sin ωt mω

ð1:134Þ

1.8 Response to an Arbitrary Forcing Function

41

The forced response, in this special case, is given by ð 1 t xð t Þ ¼ F ðτÞ sinωðt  τÞdτ mω 0

ð1:135Þ

If the effect of the initial conditions is considered, the general solution is given by ð 1 t x_0 xðt Þ ¼ sinωt þ x0 cosωt þ F ðτÞ sinωðt  τÞdτ ð1:136Þ mω 0 ω Example 1.11

Find the forced response of the damped single degree of freedom system to the step function shown in Fig. 1.19.

Fig. 1.19 Step function

Solution. The forced response of the damped single degree of freedom system to an arbitrary forcing function is ðt 1 xð t Þ ¼ F ðτÞeξωðtτÞ sinωd ðt  τÞdτ mωd 0 In the case of a step function, the forcing function F(t) is defined as F ðt Þ ¼ F0 ;

t>0

that is, xð t Þ ¼ ¼

1 mωd F0 mωd

ðt

F0 eξωðtτÞ sinωd ðt  τÞdτ

0

ðt

eξωðtτÞ sinωd ðt  τÞdτ

0

" # F0 eξωt ¼ 1  pffiffiffiffiffiffiffiffiffiffiffiffiffi cosðωd t  ψ Þ k 1  ξ2 (continued)

42

1 Introduction

where the angle ψ is defined as ψ ¼ tan

1

ξ pffiffiffiffiffiffiffiffiffiffiffiffiffi 1  ξ2

!

The response of this system is shown in Fig. 1.20.

Fig. 1.20 Response of damped single degree of freedom system to a step forcing function

1.9

Linear Theory of Vibration

This book is concerned, for the most part, with the linear theory of vibration of multidegree of freedom and continuous systems. The modal decomposition approach developed in this book allows for representing the multi-degree of freedom and continuous systems using decoupled equations, each of which is in a form similar to the mathematical model used for the single degree of freedom system. In the modal decomposition approach discussed in this book, it is assumed that the system dynamics is governed by linear ordinary or partial differential equations. Using the linearity assumption, an eigenvalue problem is formulated and solved for the system natural frequencies and mode shapes. The number of the system natural frequencies is equal to the number of degrees of freedom. Accurate prediction of the natural frequencies is necessary in the design of a large number of civil, mechanical, and aerospace systems. It is important, however, to point out that the dynamics of many mechanical and aerospace systems is governed by highly nonlinear equations of motion because of geometric nonlinearities. These geometric nonlinearities, which are attributed to the large rotations of the system components, lead to a highly nonlinear inertia matrix that includes nonlinear dynamic coupling between different displacement modes. The linear vibration theory covered in this book has been fundamental and successfully used in the analysis of these highly nonlinear systems. For such systems, the

1.9 Linear Theory of Vibration

43

strength and stress calculations are often performed using the floating frame of reference (FFR) formulation. In the FFR formulation, a body coordinate system is introduced for each component in the system. This coordinate system, which shares the large displacement of the component, is used to define a local linear vibration problem. While the motion of the component is governed by highly nonlinear equations, the vibration of the components is defined with respect to the component coordinate system and is governed by a linear system of equations. The local linear vibration problem is used in the nonlinear FFR formulation to formulate an eigenvalue problem similar to the eigenvalue problem developed in this book. The linear FFR eigenvalue problem is solved for the mode shapes that are used to define the component deformation basis vectors. It is important, however, to point out that for the highly nonlinear systems, the solution of the local linear eigenvalue problem does not define the system natural frequencies as it is the case with the systems whose dynamics is governed by linear equations. The frequencies of oscillations of the nonlinear systems can only be determined by performing a time domain analysis of the solution of the nonlinear problem. Therefore, one, in general, should not linearize the equations of motion to define the natural frequencies of the nonlinear systems. This is clear from the wellknown rotating beam example in which the linearized equations predict instability and false resonance frequency equal to the frequency associated with the first bending mode of vibration of the beam (Schilhans 1958; Johnson 1980; Kane et al. 1987). This instability is not experimentally observed and a nonlinear model that captures the geometric stiffening resulting from the coupling between the axial and bending deformations does not predict this false instability. Therefore, the eigenvalues of the linearized equations of many geometrically nonlinear mechanical and aerospace systems should not be interpreted as the natural frequencies of the nonlinear system. Furthermore, one needs to distinguish between the oscillation frequencies associated with certain coordinates and the system natural frequencies (Shabana 2010B). Another example that was used in the literature to shed light on the interpretation of the frequencies predicted using the linearized equations is the wheelset of a railroad vehicle. The motion of such a wheelset is governed by highly nonlinear equations of motion that include the gyroscopic moments (Shabana et al. 2011). These equations can be formulated using Euler angles as orientation parameters, as discussed in this chapter, or using any other set of orientation parameters. The equations of motion in this case, which include the nonlinear gyroscopic moments, can be linearized at different time points that correspond to different configurations, and at these points, the eigenvalue problems can be solved to determine the frequencies and mode shapes. Figure 1.21 shows the mode shapes and frequencies predicted using the linearized equation when Euler angles are used. The results presented in Fig. 1.21 are obtained at one time instant, and these results change as the wheelset configuration changes. Another alternative to formulating the equations of motion of the wheelset is to use another set of orientation parameters such as the four Euler parameters (Roberson and Schwertassek 1988; Shabana 2013). The four Euler parameters are

44

1 Introduction

Fig. 1.21 Wheelset mode shapes predicted using linearized Euler angles equations (Shabana et al. 2011). (a) First mode (0 Hz); (b) second mode (0 Hz); (c) third mode (0 Hz); (d) fourth mode (0 Hz); (e) fifth mode (0.31 Hz); (f) sixth mode (0.62 Hz); (g) seventh mode (209.7 Hz); (h) eighth mode 8 (219.0 Hz)

related by a kinematic constraint equation. Figure 1.22 shows the mode shapes and frequencies predicted using the linearized equations of the wheelset when Euler parameters are used (Shabana et al. 2011). While linearization techniques may not work well for highly nonlinear systems, the linear vibration theory discussed in this book remains an integral part of the formulation of the nonlinear equations of motion of complex mechanical and aerospace systems. As previously mentioned in this section, the mode decomposition concept is fundamental in the FFR formulation widely used in the analysis of nonlinear physics and engineering systems. Therefore, good understanding of the linear vibration theory is necessary for the analysis of the nonlinear oscillations.

Problems

45

Fig. 1.22 Wheelset mode shapes predicted using linearized Euler parameters equations (Shabana et al. 2011). (a) First mode (0 Hz); (b) second mode (0 Hz); (c) third mode (0 Hz); (d) fourth mode (0 Hz); (e) fifth mode (0.41 Hz); (f) sixth mode (3.1 Hz); (g) seventh mode (209.7 Hz); (h) eighth mode (214.4 Hz)

Problems 1.1. A single degree of freedom mass–spring system consists of a 10 kg mass suspended by a linear spring which has a stiffness coefficient of 6  103 N∕m. The mass is given an initial displacement of 0.04 m, and it is released from rest. Determine the differential equation of motion, and the natural frequency of the system. Determine also the maximum velocity. 1.2. The oscillatory motion of an undamped single degree of freedom system is such that the mass has a maximum acceleration of 50 m∕s2 and has natural frequency of 30 Hz. Determine the amplitude of vibration and the maximum velocity.

46

1 Introduction

1.3. A single degree of freedom undamped mass–spring system is subjected to an impact loading which results in an initial velocity of 5 m∕s. If the mass is equal to 10 kg and the spring stiffness is equal to 6  103 N∕m, determine the system response as a function of time. 1.4. The undamped single degree of freedom system of Problem 1.1 is subjected to the initial conditions x0 ¼ 0.02 m and x_0 ¼ 3 m∕s. Determine the system response as a function of time. Also, determine the maximum velocity and the total energy of the system. 1.5. A single degree of freedom system consists of a mass m which is suspended by a linear spring of stiffness k. The static equilibrium deflection of the spring was found to be 0.02 m. Determine the system natural frequency and the response of the system as a function of time if the initial displacement is 0.03 m and the initial velocity is zero. What is the total energy of the system? 1.6. The system shown in Fig. P1.1 consists of a mass m and a massless rod of length l. The system is supported by two springs which have stiffness coefficients k1 and k2, as shown in the figure. Derive the system differential equation of motion assuming small oscillations. Determine the natural frequency of the system.

Fig. P1.1

1.7. If the two springs k1 and k2 in Problem 1.6 are to be replaced by an equivalent spring which is connected at the middle of the rod, determine the stiffness coefficient ke of the new spring. 1.8. If the system in Problem 1.6 is given an initial angular displacement θ0 counterclockwise, determine the system response as a function of time assuming that the initial angular velocity is zero. Determine also the maximum angular velocity. 1.9. In the system shown in Fig. P1.2, m ¼ 5 kg, k1 ¼ k5 ¼ k6 ¼ 1000 N∕m, k3 ¼ k4 ¼ 1500 N∕m, and k2 ¼ 3000 N∕m. The motion of the mass is assumed to be in the vertical direction. If the mass is subjected to an impact such that the motion starts with an initial upward velocity of 5 m∕s, determine the displacement, velocity, and acceleration of the mass after 2 s.

Problems

47

Fig. P1.2

1.10. The system shown in Fig. P1.3 consists of a mass m and a massless rod of length l. The system is supported by a spring which has a stiffness coefficient k. Obtain the differential equation of motion and discuss the stability of the system. Determine the stiffness coefficient k at which the system becomes unstable.

Fig. P1.3

48

1 Introduction

1.11. The system shown in Fig. P1.4 consists of a uniform rod which has length l, mass m, and mass moment of inertia about its mass center I. The rod is supported by two springs which have stiffness coefficients k, as shown in the figure. Determine the system differential equation of motion for small oscillations. Determine also the system natural frequency.

Fig. P1.4

1.12. A damped single degree of freedom mass–spring system has m ¼ 4 kg, k ¼ 3000 N∕m, and c ¼ 300 Ns∕m. Determine the equation of motion of the system. 1.13. A damped single degree of freedom mass–spring system has m ¼ 0.5 kg, k ¼ 1000 N∕m, and c ¼ 10 Ns∕m. The mass is set in motion with initial conditions x0 ¼ 0.05 m and x_ ¼ 0:5 m∕s. Determine the displacement, velocity, and acceleration of the mass after 0.3 s. 1.14. A viscously damped single degree of freedom mass–spring system has a mass m of 2 kg, a spring coefficient k of 2000 N∕m, and a damping constant c of 5 Ns∕m. Determine (a) the damping factor ξ, (b) the natural frequency ω, (c) the damped natural frequency ωd, and (d) the spring coefficient needed to obtain a critically damped system. 1.15. An overdamped single degree of freedom mass–spring system has a damping factor ξ ¼ 1.5 and a natural frequency ω ¼ 20 rad∕s. Determine the equation of motion and plot the displacement and velocity versus time for the following initial conditions: (a) x0 ¼ 0, x_0 ¼ 1 m∕s; (b) x0 ¼ 0:05 m, x_0 ¼ 0; and (c) x0 ¼ 0:05 m, x_0 ¼ 1m∕s. 1.16. Repeat Problem 1.15 if the system is critically damped.

Problems

49

1.17. A mass equal to 0.5 kg attached to a linear spring elongates it 0.008 m. Determine the system natural frequency. 1.18. For the system shown in Fig. P1.5, let m ¼ 0.5 kg, a ¼ 0.2 m, l ¼ 0.4 m, k ¼ 1000 N∕m. Determine the damping coefficient c if the system is to be critically damped. If the system has an initial angular velocity of 5 rad∕s counterclockwise, determine the angular displacement and angular velocity after 0.3 s. Assume small oscillations.

Fig. P1.5

1.19. For the system shown in Fig. P1.6, let m ¼ 0.5 kg, l ¼ 0.5 m, a ¼ 0.2 m, and k ¼ 3000 N∕m. Determine the damping coefficient c if: (a) the system is underdamped with ξ ¼ 0.09, (b) the system is critically damped; and (c) the system is overdamped with ξ ¼ 1.2.

Fig. P1.6

50

1 Introduction

1.20. The system shown in Fig. P1.7 consists of a uniform bar of length l, mass m, and mass moment of inertia I. The bar is supported by a spring and damper which have stiffness and damping coefficients k and c, respectively. Derive the differential equation of motion and determine the system natural frequency and the critical damping coefficient.

Fig. P1.7

1.21. A single degree of freedom mass–spring system has the following parameters: m ¼ 0.5 kg, k ¼ 2 103 N∕m, coefficient of dry friction μ ¼ 0.15, initial displacement x0 ¼ 0.1 m, and initial velocity x_0 ¼ 0. Determine: 1. The decrease in amplitude per cycle 2. The number of half-cycles completed before the mass comes to rest 3. The displacement of the mass at time t ¼ 0.1 s 4. The location of the mass when the oscillation stops 1.22. A spring–mass system is subjected to a harmonic force which has an amplitude 30 N and frequency 20 rad∕s. The system has mass m ¼ 5 kg and stiffness coefficient k ¼ 2 103 N. The initial conditions are such that x0 ¼ 0, x_0 ¼ 2 m∕s. Determine the displacement, velocity, and acceleration of the mass after 0.5, 1, and 1.5 s. 1.23. A single degree of freedom mass–spring system has a mass m ¼ 3 kg, a spring stiffness k ¼ 2700 N∕m, and a damping coefficient c ¼ 18 Ns∕m. The mass is subjected to a harmonic force which has an amplitude F0 ¼ 20 N and frequency ωf ¼ 15 rad∕s. The initial conditions are x0 ¼ 1 cm and x_0 ¼ 0. Determine the displacement, velocity, and acceleration of the mass after time t ¼ 0.5 s. 1.24. For the system shown in Fig. P1.8, let m ¼ 3 kg, k1 ¼ k2 ¼ 1350 N∕m, c ¼ 40 Ns∕m, and y ¼ 0.04 sin15t. The initial conditions are such that x0 ¼ 5 mm and x_0 ¼ 0. Determine the displacement, velocity, and acceleration of the mass after time t ¼ 1 s.

Problems

51

Fig. P1.8

1.25. For the system shown in Fig. P1.9, let m ¼ 3 kg, k1 ¼ k2 ¼ 1350 N/m, c ¼ 40 Ns/m, and y ¼ 0.02sin15t. The initial conditions are x0 ¼ 5 mm and x_0 ¼ 0. Determine the displacement, velocity, and acceleration of the mass after time t ¼ 1 s.

Fig. P1.9

1.26. Derive the differential equation of motion of the system shown in Fig. P1.10, assuming small angular oscillations. Determine also the steady-state response of this system.

Fig. P1.10

52

1 Introduction

1.27. In Problem 1.26, let the rod be uniform and slender with mass m ¼ 0.5 kg and l ¼ 0.5 m. Let k ¼ 2000 N/m, c ¼ 20 Ns/m, and F ¼ 10sin10t N. The initial conditions are θ0 ¼ 0 and θ_0 ¼ 3 rad∕s. Determine the displacement equation of the beam as a function of time. 1.28. Determine the forced response of the undamped single degree of freedom mass–spring system to the forcing function shown in Fig. P1.11.

Fig. P1.11

1.29. Determine the forced response of the damped single degree of freedom mass– spring system to the forcing function shown in Fig. P1.11. 1.30. Determine the response of the damped single degree of freedom mass–spring system to the forcing function shown in Fig. P1.12.

Fig. P1.12

Problems

53

1.31. Obtain the response of the damped mass–spring system to the forcing function shown in Fig. P1.13.

Fig. P1.13

2

Lagrangian Dynamics

The differential equations of motion of single and multi-degree of freedom systems can be developed using the vector approach of Newtonian mechanics. Another alternative for developing the system differential equations of motion from scalar quantities is the Lagrangian approach where scalars such as the kinetic energy, strain energy, and virtual work are used. The use of the Lagrangian approach is convenient in developing the dynamic relationships of multi-degree of freedom systems. Important concepts and definitions, however, have to be first introduced. In the first section of this chapter, we introduce the concept of the system generalized coordinates, and in Sect. 2.2, the virtual work is used to develop the generalized forces associated with the system generalized coordinates. The concepts and definitions presented in the first two sections are then used in Sects. 2.3, 2.4, 2.5, and 2.6 to develop Lagrange’s equation of motion for multi-degree of freedom systems in terms of scalar quantities such as the kinetic energy, strain energy, and virtual work. An alternate approach for deriving the dynamic equations of motion using scalar quantities is Hamilton’s principle which is discussed in Sect. 2.7. Hamilton’s principle can be used to derive Lagrange’s equation and, consequently, both techniques lead to the same results when the same set of coordinates is used. The use of conservation of energy to obtain the differential equations of motion of conservative systems is discussed in Sect. 2.8, where general conservation theorems are developed and their use is demonstrated using simple examples.

2.1

Generalized Coordinates

In the Lagrangian dynamics, the configuration of a mechanical system is identified by a set of variables called coordinates or generalized coordinates that completely define the location and orientation of each component in the system. The configuration of a particle in space is defined using three coordinates that describe the translation of this particle with respect to the three perpendicular # Springer Nature Switzerland AG 2019 A. Shabana, Vibration of Discrete and Continuous Systems, Mechanical Engineering Series, https://doi.org/10.1007/978-3-030-04348-3_2

55

56

2

Lagrangian Dynamics

axes of the inertial frame. No rotational coordinates are needed to describe the motion of the particle and, therefore, the three translational coordinates completely define the particle position. This simplified description of the particle kinematics is the result of the assumption that the particle has such small dimensions that a point in the three-dimensional space is sufficient to describe its position. This assumption is not valid, however, when rigid bodies are considered. The configuration of an unconstrained rigid body, as demonstrated in the preceding chapter, can be completely described using six independent coordinates, three coordinates describe the location of the origin of the body axes and three rotational coordinates describe the orientation of the body with respect to the fixed frame. Once this set of coordinates is identified, the global position of an arbitrary point on the rigid body can be defined. In this chapter and subsequent chapters, the vector q will be used to denote the system generalized coordinates. If the system has n generalized coordinates, the vector q is given by q ¼ ½q1 q2    qn T

ð2:1Þ

In this book, it is assumed that the generalized coordinates q1, q2, . . ., qn are independent, and as such, the number of system generalized coordinates is equal to the number of system degrees of freedom. Figure 2.1 shows some examples of vibratory systems, which have different types or numbers of generalized coordinates. The system shown in Fig. 2.1a is a single degree of freedom system, and accordingly, the vector of generalized coordinates q reduces to a scalar since q ¼ q1 ¼ x. The configuration of this system can be completely defined in terms of the generalized coordinate q1. The configuration of the two degree of freedom system shown in Fig. 2.1b, on the other hand, is completely defined by the coordinates q1 ¼ x1 and q2 ¼ x2, that is, q ¼ ½ q1 q2 T ¼ ½ x1 x2 T . Similarly, the vector of the generalized coordinates of the system shown in Fig. 2.1c is q ¼ ½ q1 q2 T ¼ ½ θ1 θ2 T . The position coordinates of an arbitrary point p in this system can be written in terms of these generalized coordinates as xp ¼ l1 sin θ1 + d sin θ2 and yp ¼  l1 cos θ1  d cos θ2, where d is the position of point p with respect to point O2 measured along the rod axis. The system shown in Fig. 2.1d has more than two degrees of freedom. One can verify that the vector of generalized coordinates of this system is given by q ¼ ½ q1 q2 q3 T ¼ ½ θ1 θ2 θ3 T . It is important to emphasize that the set of generalized coordinates q1, q2, . . ., qn is not unique. For instance, the configuration of the system shown in Fig. 2.1c can also be described by the generalized coordinates q1 ¼ x1 and q2 ¼ x2 as shown in the figure. In many cases, simple relationships can be found between different sets of generalized coordinates of the system. For example, in Fig. 2.1c, one can verify that x1 ¼ l1 sin θ1 and x2 ¼ l1 sin θ1 + l2 sin θ2. The inverse relationship can also be obtained as sin θ1 ¼ x1/l1 and sin θ2 ¼ (x2  x1)/l2 or θ1 ¼ sin1(x1/l1) and θ2 ¼ sin1((x2  x1)/l2). The coordinates θ1 and θ2 or x1 and x2 in the example of

2.1 Generalized Coordinates

57

Fig. 2.1 Generalized coordinates

Fig. 2.1c are independent since they can be arbitrarily changed. Any coordinate which can be expressed in terms of the other coordinates is not an independent one and cannot be used as a degree of freedom. Similarly, a coordinate which is specified as a function of time is not a degree of freedom since it has a specified value and cannot be changed arbitrarily. For instance, consider the system shown in Fig. 2.1e, which has the two degrees of freedom q1 ¼ x and q2 ¼ θ. If q1 is specified as a function of time, say q1 ¼ x ¼ X1sinωft where X1 and ωf are, respectively, given amplitude and frequency, then the system has only one degree of freedom which is q2 ¼ θ. The relationship q1 ¼ x ¼ X1sinωft can be considered as a kinematic constraint imposed on the motion of the system, and every such kinematic constraint relationship, as demonstrated in the preceding chapter, eliminates one of the system degrees of freedom. In other words, the number of the system degrees of freedom is equal to the number of coordinates minus the number of kinematic constraint equations provided that these kinematic constraint equations are linearly independent.

58

2

Lagrangian Dynamics

Example 2.1

Discuss different alternatives for selecting the generalized coordinates of the two degree of freedom system shown in Fig. 2.2.

Fig. 2.2 Two degree of freedom system

Solution. Since the system has only two degrees of freedom, mainly the vertical displacement and the rotation, only two generalized coordinates are required to identify the configuration of the system. As shown in Fig. 2.2, one set of coordinates is given by q ¼ ½ q1

q2  T ¼ ½ y c

θ T

where yc is the vertical displacement of the center of mass of the beam and θ is the angular oscillation of the beam. The coordinates yc and θ can be used as the system degrees of freedom since they are independent and are sufficient to define the configuration of the system. For instance, given an arbitrary point p at a distance lp from the center of mass of the beam, the vertical displacement of the arbitrary point p can be expressed in terms of the coordinates yc and θ as yp ¼ yc þ lp sin θ As a second choice for the generalized coordinates, one may select the following set: q ¼ ½ q1

q2  T ¼ ½ y a

θ T

where ya is the vertical displacement of the end point a. The coordinates ya and θ are also independent and can be used to define the displacement of an arbitrary point on the beam. The vertical displacement of the point p can be expressed in terms of the generalized coordinates ya and θ as (continued)

2.2 Virtual Work and Generalized Forces

59


yp ¼ ya þ

 l þ lp sin θ 2

where l is the length of the beam. Observe that the vertical displacement of the center of mass yc which was previously used as a generalized coordinate can be expressed in terms of the new set of generalized coordinates ya and θ as yc ¼ ya þ ðl=2Þsin θ. An alternate set of generalized coordinates is also q ¼ ½ q1 q2 T ¼ ½ ya yb T , where yb is the vertical displacement of the end point b. In terms of these coordinates, the vertical displacement of the center of mass and the arbitrary point p are given, respectively, by yc ¼

y b þ ya , 2

yp ¼ ya þ ðyb  ya Þ

ðl=2Þ þ lp l

Also, the angular orientation θ can be expressed in terms of the new set ya and yb as sinθ ¼ (yb  ya)/l, or y  y  a θ ¼ sin 1 b l Clearly, for a simple system such as the one used in this example, there are an infinite number of different sets of generalized coordinates. In many applications, proper selection of the generalized coordinates can significantly simplify the dynamic formulation.

2.2

Virtual Work and Generalized Forces

In this section, the important concept of the virtual work is introduced and used to provide definitions for the generalized forces associated with the system generalized coordinates. As will be seen in the following section, the definition of the generalized forces is an important step in the Lagrangian formulation, wherein the dynamic equations of motion are formulated in terms of the generalized coordinates. Since the generalized coordinates are sufficient to completely identify the configuration of a system of particles or rigid bodies, the position vector of an arbitrary point in the system can be expressed in terms of the generalized coordinates as r ¼ rðq1 ; q2 ; . . . ; qn Þ ¼ rðqÞ

ð2:2Þ

60

2

Lagrangian Dynamics

where r is the position vector of an arbitrary point in the system, and q ¼ [q1 q2 . . . qn]T is the vector of generalized coordinates which are assumed to be independent. The virtual displacements refers to a change in the configuration of the system as the result of an arbitrary infinitesimal change in the vector of system generalized coordinates q, consistent with the forces and constraints imposed on the system at the given instant of time t. The displacement is called virtual to distinguish it from an actual displacement of the system occurring in a time interval Δt, during which the time and constraints may be changing. For a virtual change in the system coordinates, Eq. 2.2 yields δr ¼

n X ∂r ∂r ∂r ∂r δq1 þ δq2 þ    þ δqn ¼ δqj ∂q1 ∂q2 ∂qn ∂q j j¼1

ð2:3Þ

where δr is the virtual change in the position vector of Eq. 2.2 and δqi is the virtual change in the generalized coordinate qi. Let F be the vector of forces that act at the point whose position vector is defined by the vector r of Eq. 2.2; then the virtual work due to this force vector is defined by the dot product δW ¼ Fδr ¼ FT δr

ð2:4Þ

which upon using Eq. 2.3 yields ∂r ∂r ∂r δq þ FT δq þ    þ FT δq ∂q1 1 ∂q2 2 ∂qn n n X ∂r ¼ FT δq ∂qj j j¼1

δW ¼ FT

ð2:5Þ

The generalized force Qj associated with the jth system generalized coordinate qj is defined as Qj ¼ FT

∂r , j ¼ 1,2,. . . ,n ∂qj

ð2:6Þ

Using this definition, the virtual work of Eq. 2.5 can be expressed in terms of the generalized forces Qj as δW ¼ Q1 δq1 þ Q2 δq2 þ    þ Qn δqn n X ¼ Qj δqj ¼ QT δq

ð2:7Þ

j¼1

where Q is the vector of generalized forces defined as Q ¼ ½ Q1

Q2

   Q n T .

2.2 Virtual Work and Generalized Forces

61

Example 2.2

Derive the generalized forces due to the spring, damper, and the external force F(t) of the damped single degree of freedom system shown in Fig. 2.3.

Fig. 2.3 Single degree of freedom system

Solution. For this single degree of freedom system, the vector of generalized coordinates q reduces to a scalar given by q1 ¼ x. It follows that δx ¼ δq1. The forces that act on this system are shown in the figure and given by f ¼ F ðt Þ  kx  cx_ The virtual work is then defined as δW ¼ f δx ¼ ðF ðt Þ  kx  cx_Þδx Since in this example x ¼ q1 and x_ ¼ q_1 , in terms of the generalized coordinate q1, the virtual work δW can be written as δW ¼ ðF ðt Þ  kq1  cq_1Þδq1 ¼ Q1 δq1 where Q1 is the generalized force associated with the generalized coordinate q1 and is defined as Q1 ¼ F ðt Þ  kq1  cq_1

Example 2.3

For the two degree of freedom system shown in Fig. 2.4, obtain the generalized forces associated with the generalized coordinates q1 ¼ x1 and q2 ¼ x 2. (continued)

62

2

Lagrangian Dynamics

Fig. 2.4 Two degree of freedom system

Solution. As shown in the figure, the following force   f1 ¼ k1 x1  c1 x_1 þ k 2 ðx2  x1Þ þ c2 x_2  x_1 þ F1 ðt Þ acts at a point whose displacement is x1 ¼ q1. The force f2 given by   f 2 ¼ k2 ðx2  x1Þ  c2 x_ 2  x_1 þ F2 ðt Þ acts at a point whose displacement is x2 ¼ q2. Therefore, the virtual work δW is given by δW ¼ f 1 δx1 þ f 2 δx2 Using the fact that x1 ¼ q1, x_1 ¼ q_1 , x2 ¼ q2, and x_2 ¼ q_2 , the virtual work δW can be expressed in terms of the generalized coordinates as    δW ¼ k1 q1  c1 q_1 þ k2 ðq2  q1Þ þ c2 q_ 2  q_1 þ F1 ðt Þ δq1    þ k2 ðq2  q1Þ  c2 q_ 2  q_1 þ F 2 ðt Þ δq2 or δW ¼ fF1 ðt Þ  ðk1 þ k2 Þq1  ðc1 þ c2 Þq_ 1 þ k2 q2 þ c2 q_ 2 gδq1 þfF 2 ðt Þ  k2 q2  c2 q_ 2 þ k 2 q1 þ c2 q_ 1 gδq2

2.2 Virtual Work and Generalized Forces

63

Example 2.4

For the three degree of freedom pendulum system shown in Fig. 2.5, obtain the generalized forces associated with the generalized coordinates q1 ¼ θ1, q2 ¼ θ2, and q3 ¼ θ3, where F and M are defined, respectively, as F ¼ ½ F1 F 2 T and M ¼ ½ M 1 M 2 M 3 T .

Fig. 2.5 Three degree of freedom system

Solution. The position vector of the point of application of the force F is given by the vector



 r1 l1 sin θ1 þ l2 sin θ2 þ l3 sin θ3 r¼ ¼ r2 ðl1 cos θ1 þ l2 cos θ2 þ l3 cos θ3 Þ where l1, l2, and l3 are the lengths of the rods. The virtual change in the position vector r can be written as

 l1 cos θ1 δθ1 þ l2 cos θ2 δθ2 þ l3 cos θ3 δθ3 δr ¼ l1 sin θ1 δθ1 þ l2 sin θ2 δθ2 þ l3 sin θ3 δθ3 which can be written as

δr ¼

δr 1 δr 2



¼

l1 cos θ1

l2 cos θ2

l1 sin θ1

l2 sin θ2

2 3  δθ1 l3 cos θ3 6 7 4 δθ2 5 l3 sin θ3 δθ3 (continued)

64

2

Lagrangian Dynamics

The virtual work of the component Mi of the moment is Miδθi. Therefore, the virtual work δW of the force vector F and the moment M is given by δW ¼ FT δr þ MT δθ where δθ ¼ ½ δθ1 explicit form by

δθ2 "

δW ¼ ½ F1

F2 

þ½ M 1

δθ3 T . The virtual work δW is then given in a more

l1 cos θ1

l2 cos θ2

l1 sin θ1 l2 sin θ2 2 3 δθ1 6 7 7 M 2 M 3 6 4 δθ2 5 δθ3

2 3 # δθ1 l3 cos θ3 6 7 4 δθ2 5 l3 sin θ3 δθ3

¼ M 1 þ l1 ðF1 cos θ1 þ F 2 sin θ1 Þ M 2 þ l2 ðF1 cos θ2 þ F 2 sin θ2 Þ 2 3 δθ1 7

6 7 M 3 þ l3 ðF1 cos θ3 þ F 2 sin θ3 Þ 6 4 δθ2 5 δθ3 which can also be written as 2

δW ¼ ½ Q1

Q2

3 δθ1 6 7 Q3 4 δθ2 5 ¼ QT δθ δθ3

where the generalized forces Q1, Q2, and Q3 associated, respectively, with the generalized coordinates θ1, θ2, and θ3 are defined as Q1 ¼ M 1 þ l1 ðF1 cos θ1 þ F 2 sin θ1 Þ Q2 ¼ M 2 þ l2 ðF1 cos θ2 þ F 2 sin θ2 Þ Q3 ¼ M 3 þ l3 ðF1 cos θ3 þ F 2 sin θ3 Þ

2.3

Lagrange’s Equation

In this section, we utilize the concepts introduced in the preceding section to develop Lagrange’s equation of motion for a system of particles. The obtained Lagrange’s equation, however, can be applied to the dynamics of rigid bodies as well, since a rigid body can be considered as a collection of a large number of particles. In the

2.3 Lagrange’s Equation

65

following discussion, we assume that the system consists of np particles. The displacement ri of the ith particle is assumed to depend on a set of generalized coordinates qj, where j ¼ 1,2,. . .,n. Hence r i ¼ r i ð q1 ; q2 ; . . . ; qn ; t Þ

ð2:8Þ

Differentiating Eq. 2.8 with respect to time using the chain rule of differentiation yields r_ i ¼

n ∂ri ∂ri ∂ri ∂ri X ∂ri ∂ri ¼ q_ 1 þ q_ 2 þ    þ q_ n þ q_j þ ∂q1 ∂q2 ∂qn ∂t ∂qj ∂t j¼1

ð2:9Þ

The virtual displacement δri can be expressed in terms of the virtual change of the generalized coordinates as δri ¼

n X ∂ri δq ∂qj j j¼1

ð2:10Þ

The dynamic equilibrium of the particle can be defined using Newton’s second law as p_ i ¼ Fi , where pi is the vector of linear momentum, and Fi is the vector of the total forces acting on the particle i. The linear momentum vector pi is defined as pi ¼ mi r_ i , where mi is the mass of the particle i which is assumed to be constant. Consequently, p_ i ¼ mi€ri . It follows that mi€ri  Fi ¼ 0, which, when multiplied by the virtual displacement δri, yields 

T mi€ri  Fi δri ¼ 0,

i ¼ 1,2,. . . ,np

ð2:11Þ

By summing up these expressions for the particles of δri Pand using the definition   Pnp  i i T n i given by Eq. 2.10, one obtains i¼1 m €r  Fi j¼1 ∂r =∂qj δqj ¼ 0, which can also be written as ! np n X i X  i i  i T ∂r δq ¼ 0 m €r  F ∂qj j j¼1 i¼1

ð2:12Þ

Define the generalized force Qj associated with the generalized coordinate qj as Qj ¼

np X i¼1

FiT

∂ri ∂qj

ð2:13Þ

One can also show that np X i¼1

∂ri m €r ∂qj i iT

!

" ! !# np i X d ∂ri i iT ∂r i iT d m r_ ¼  m r_ dt dt ∂qj ∂qj i¼1

It is, however, clear from Eq. 2.9 that

ð2:14Þ

66

2

Lagrangian Dynamics

∂_ri ∂ri ¼ ∂q_j ∂qj

ð2:15Þ

Furthermore, d ∂ri dt ∂qj

! ¼

2 2 n X ∂ ri ∂ ri ∂_ri ¼ q_k þ ∂qj ∂qk ∂qj ∂t ∂qj k¼1

ð2:16Þ

Substituting the results of Eqs. 2.15 and 2.16 into Eq. 2.14 yields #
 np np "
i X X d ∂ 1 i iT i ∂ 1 i iT i i i T ∂r m r_ r_ m r_ r_ m €r ¼  ∂qj 2 ∂qj i¼1 dt ∂q_j 2 i¼1 ! np X d ∂T i ∂T i ¼  dt ∂q_j ∂qj i¼1

ð2:17Þ

where T i is the kinetic energy of the particle i defined as T i ¼ 12 mi r_ i T r_ i . The kinetic Pn p i T . Therefore, Eq. 2.17 can energy of the system of particles is given by T ¼ i¼1 be written in terms of the kinetic energy of the system of particles as ! np i X d ∂T ∂T i i T ∂r m €r ¼ ð2:18Þ  _ dt ∂ q ∂q ∂q j j j i¼1 Substituting Eqs. 2.13 and 2.18 into Eq. 2.12 yields " ! # n X d ∂T ∂T  Qj δqj ¼ 0  dt ∂q_j ∂qj j¼1

ð2:19Þ

Since the generalized coordinates q1, q2, . . ., qn are assumed to be  linearly  indepen  dent, Eq. 2.19 yields a set of n equations defined as d ∂T=∂q_j =dt  ∂T=∂qj  Qj ¼ 0, j ¼ 1,2,. . . ,n, or ! d ∂T ∂T ¼ Qj , j ¼ 1,2,. . . ,n ð2:20Þ  dt ∂q_j ∂qj This equation is called Lagrange’s equation of motion. Clearly, there are as many equations as the number of generalized coordinates. These equations can be derived using scalar quantities, mainly the kinetic energy of the system and virtual work of the applied and elastic forces. The use of Lagrange’s equation for developing the differential equations of motion for single and two degree of freedom systems is demonstrated by the following examples.

2.3 Lagrange’s Equation

67

Example 2.5

Using Lagrange’s equation, develop the differential equation of motion of the single degree of freedom system given in Example 2.2 and shown in Fig. 2.3. Solution. The system of Example 2.2 has one degree of freedom and only one differential equation results from the application of Lagrange’s equation which can be stated in this case as
 d ∂T ∂T ¼Q  dt ∂x_ ∂x It was shown in Example 2.2 that the generalized force Q is given by _ The kinetic energy of this system is given by Q ¼ F ðt Þ  kx  cx. T ¼ 12 mx_ 2 , which yields
 ∂T d ∂T ∂T ¼ mx_ , ¼0 ¼ m€x, ∂x_ dt ∂x_ ∂x Using Lagrange’s equation, one obtains m€x ¼ F ðt Þ  kx  cx_ , or m€x þ cx_ þ kx ¼ F ðt Þ which is the same differential equation obtained for this system by applying Newton’s second law.

Example 2.6

Using Lagrange’s equation, derive the differential equations of motion of the two degree of freedom system given in Example 2.3 and shown in Fig. 2.4. Solution. Since the system has the two degrees of freedom q1 ¼ x1 and q2 ¼ x2 as generalized coordinates, the application of Lagrange’s equation yields the two equations
 d ∂T ∂T ¼ Q1 ,  dt ∂q_1 ∂q1


 d ∂T ∂T ¼ Q2  dt ∂q_ 2 ∂q2

where T is the total kinetic energy of the system and Q1 and Q2 are the generalized forces associated, respectively, with the system generalized coordinates. The kinetic energy T is defined as (continued)

68

2

1 1 1 T ¼ m1 q_ 21 þ m2 q_ 22 ¼ ½ q_ 1 2 2 2

q_ 2 

m1

0

0

m2



Lagrangian Dynamics

q_ 1 q_ 2



It was shown in Example 2.3 that the generalized forces Q1 and Q2 are given by Q1 ¼ F1 ðt Þ  ðk1 þ k 2 Þq1  ðc1 þ c2 Þq_ 1 þ k2 q2 þ c2 q_ 2 Q2 ¼ F 2 ðt Þ  k2 q2  c2 q_ 2 þ k2 q1 þ c2 q_ 1 By using the definition of the kinetic energy, one obtains
 ∂T d ∂T ∂T ¼ m1 q_ 1 , ¼0 ¼ m1 €q1 , ∂q_ 1 dt ∂q_ 1 ∂q1
 ∂T d ∂T ∂T ¼ m2 q_ 2 , ¼0 ¼ m2 €q2 , ∂q_ 2 dt ∂q_ 2 ∂q2 Substituting into Lagrange’s equation for q1 and q2 yields, respectively, the following two differential equations: m1 € q1 ¼ F1 ðt Þ  ðk1 þ k 2 Þq1  ðc1 þ c2 Þq_ 1 þ k2 q2 þ c2 q_ 2 m2 € q2 ¼ F 2 ðt Þ  k2 q2  c2 q_ 2 þ k 2 q1 þ c2 q_ 1 or m1 € q1 þ ðc1 þ c2 Þq_ 1  c2 q_ 2 þ ðk 1 þ k2 Þq1  k2 q2 ¼ F1 ðt Þ m2 €q2 þ c2 q_ 2  c2 q_ 1 þ k2 q2  k2 q1 ¼ F 2 ðt Þ which can be written in a matrix form as "

m1 0

#"

0 m2

€ q1 € q2

#

" þ

c1 þ c2 c2 c2

c2

#"

q_1 q_2

#

" þ

k 1 þ k2 k2 k2

k2

#"

q1 q2

#

" ¼

F1 ðt Þ

#

F 2 ðt Þ

which is the same matrix equation that can be obtained for this two degree of freedom system by applying Newton’s second law.

Example 2.7

Using Lagrange’s equation, derive the differential equation of motion of the two degree of freedom system shown in Fig. 2.6. Use the assumption of small oscillations. (continued)

2.3 Lagrange’s Equation

69

Fig. 2.6 Pendulum with moving base

Solution. We select x and θ as the system generalized coordinates, that is, q ¼ ½ x θ T . The kinetic energy of the system is 1 1 1 T ¼ mx_2 þ mr x_ 2c þ I θ_2 2 2 2 where mr is the total mass of the rod, x_ c is the absolute velocity of the center of mass of the rod, and I is the mass moment of inertia of the rod about its mass center, that is, I ¼ mrl2/12, where l is the length of the rod. The absolute velocity of the center of mass of the rod can be expressed in terms of the generalized velocities as l x_ c  x_ þ θ_ 2 where the assumption of small oscillation is utilized. The kinetic energy T can then be written in terms of the generalized velocities as
 1 2 1 l 2 1 _2 _ T ¼ mx_ þ mr x_ þ θ þ Iθ 2 2 2 2
 1 l 1 l2 ¼ ðm þ mr Þx_2 þ mr x_ θ_ þ I þ mr θ_ 2 2 2 2 4 1 l 1 ¼ ðm þ mr Þx_2 þ mr x_ θ_ þ I O θ_ 2 2 2 2 " #" #

m þ mr mr ðl=2Þ x_ 1 ¼ x_ θ_ 2 mr ðl=2Þ IO θ_ (continued)

70

2

Lagrangian Dynamics

where IO is the mass moment of inertia of the rod about point O and is given by IO ¼ I + mr(l2/4). The virtual work of all the forces acting on the system is given by δW ¼ ½F ðt Þ  kx  cx_ δx  mr gδyc where g is the gravitational constant and δyc is the virtual displacement of the center of mass in the vertical direction which is given by δyc ¼ (l/2)sinθδθ. Therefore, l δW ¼ ½F ðt Þ  kx  cx_ δx  mr g sinθδθ 2 ¼ Qx δx þ Qθ δθ where Qx and Qθ are the generalized forces associated, respectively, with the generalized coordinates x and θ and defined as Qx ¼ F ðt Þ  kx  cx_ l l Qθ ¼ mr g sinθ  mr g θ 2 2 The application of Lagrange’s equation yields the following two equations:
 d ∂T ∂T ¼ Qx ,  dt ∂x_ ∂x


 d ∂T ∂T ¼ Qθ  _ dt ∂θ ∂θ

Differentiation of the kinetic energy with respect to time and with respect to the generalized coordinates yields ∂T l ¼ ðm þ mr Þx_ þ mr θ_ ∂x_ 2
 d ∂T l € ∂T ¼0 ¼ ðm þ mr Þ€x þ mr θ, dt ∂x_ 2 ∂x
 ∂T l d ∂T l _ € ¼ mr x_ þ I O θ , ¼ mr €x þ I O θ, _ _ 2 dt ∂θ 2 ∂θ

∂T ¼0 ∂θ

Substituting these equations in Lagrange’s equation and using the expressions for the generalized forces yields l ðm þ mr Þ€x þ mr θ€ ¼ F ðt Þ  kx  cx_ 2 l l mr €x þ I O θ€ ¼ mr g θ 2 2 or (continued)

2.4 Generalized Inertia Forces

71

l ðm þ mr Þ€x þ mr θ€ þ cx_ þ kx ¼ F ðt Þ 2 l l mr €x þ I O θ€ þ mr g θ ¼ 0 2 2 which can be written in a matrix form as "

2.4

m þ mr

mr ðl=2Þ

mr ðl=2Þ

IO

# 

€x c þ θ€ 0

0 0

 

  

x F ðt Þ x_ k 0 ¼ þ 0 θ_ 0 mr gðl=2Þ θ

Generalized Inertia Forces

Lagrange’s equation as defined by Eq. 2.20 states that the generalized external force Qj associated with the jth generalized coordinate is equal to the generalized inertia force associated with the same coordinate. This generalized inertia force is defined in terms of the system kinetic energy as d ∂T ðQi Þj ¼ dt ∂q_j

! 

∂T , j ¼ 1,2,. . . ,n ∂qj

ð2:21Þ

Therefore, Lagrange’s equation as defined by Eq. 2.20 can be written as ðQi Þj ¼ Qj ,

j ¼ 1,2,. . . ,n

ð2:22Þ

Note that the generalized inertia force (Qi)j was originally defined by Eq. 2.18 as ðQ i Þj ¼

np X i¼1

p ∂ri X ∂_ri ¼ mi€ri T ∂qj ∂q_j i¼1

n

mi€ri T

ð2:23Þ

Therefore, in the Lagrangian formulation, one can use Eq. 2.21 or Eq. 2.23 to obtain the generalized inertia forces associated with the system generalized coordinates. These equations can be applied to systems of particles as well as rigid body systems. In rigid body dynamics, there are inertia forces as well as inertia moments. Therefore, in the case of a system that consists of particles and rigid bodies, Eqs. 2.21 and 2.23 must be modified to ðQi Þj ¼

np X i¼1

and

" # nb i i i X ∂_ r ∂r ∂θ mi€riT þ mi€ricT c þ I i θ€i ∂q_j i¼1 ∂qj ∂qj

ð2:24Þ

72

2

ðQi Þj ¼

np X i¼1

Lagrangian Dynamics

" # nb i _i ∂_ri X i i T ∂_rc i €i ∂θ m €r þ m €rc þI θ ∂q_j i¼1 ∂q_j ∂q_j i iT

ð2:25Þ

where rci is the vector of coordinates of the center of mass of the rigid body i, θi is its angular orientation, mi and Ii are, respectively, its mass and mass moment of inertia about the center of mass, and nb is the total number of rigid bodies in the system. In order to demonstrate the use of the equations presented in this section in the case of rigid body systems, we consider the system shown in Fig. 2.6. It was shown in the preceding example that the use of Eq. 2.21 leads to the following generalized inertia forces associated with the coordinates x and θ, respectively 9 l €> ðQi Þx ¼ ðm þ mr Þ€x þ mr θ > = 2 > l > ; ðQi Þθ ¼ mr €x þ I O θ€ 2

ð2:26Þ

where m is the mass of the block, mr and l are, respectively, the mass and length of the rod, and IO is the mass moment of inertia of the rod about point O. Without linearizing the kinematic relationship, the coordinates of the center of mass of the rod can be written as 2 3 l " # 6 x þ 2 sin θ 7 xc 7 ð2:27Þ rc ¼ ¼6 4 5 l yc  cos θ 2 Differentiating these kinematic equations once and twice with respect to time leads to the velocity 2 3 " # _ l cos θ _ x þ θ 6 7 x_c 2 7, r_ c ¼ ð2:28Þ ¼6 4 5 l y_c _ θ sin θ 2 and the acceleration " €rc ¼

€xc €yc

#

2

3 l 2 l _ € € 6 x þ θ 2 cos θ  θ 2 sin θ 7 7 ¼6 4 5 l 2 l _ € cos θ θ sin θ þ θ 2 2

ð2:29Þ

Note that ∂rc ∂_rc ∂€rc ¼ ¼ ¼ ∂x ∂x_ ∂€x

" # 1 0

2

,

3 l cos θ 7 ∂rc ∂_rc ∂€rc 6 2 7 ¼ ¼ ¼6 4 5 l ∂θ ∂θ_ ∂θ€ sin θ 2

ð2:30Þ

2.4 Generalized Inertia Forces

73

Therefore, the use of Eq. 2.24 leads to ðQi Þx ¼ m€x

∂x ∂rc ∂θ þ mr €rcT þ I θ€ ∂x ∂x ∂x

ð2:31Þ

Since θ is an independent coordinate, ∂θ/∂x ¼ 0 and the preceding equation lead to ðQi Þx ¼ m€x þ mr €rcT 2

∂rc ∂x

3T l l €x þ θ€ cos θ  θ_ 2 sin θ " # 6 7 1 2 2 7 ¼ m€x þ mr 6 4 5 0 l l θ€ sin θ þ θ_ 2 cos θ 2 2 l l ¼ ðm þ mr Þ€x þ mr θ€ cos θ  mr θ_ 2 sin θ 2 2

ð2:32Þ

Assuming small oscillations, the generalized inertia force associated with the coordinate x becomes l ðQi Þx ¼ ðm þ mr Þ€x þ mr θ€ 2

ð2:33Þ

which is the same generalized inertia force previously obtained. Similarly, the generalized inertia force (Qi)θ associated with the coordinate θ can be obtained using Eq. 2.24 as ðQi Þθ ¼ m€x

∂x ∂rc ∂θ ∂rc þ mr €rcT ¼ mr €rcT þ I θ€ þ I θ€ ∂θ ∂θ ∂θ ∂θ

ð2:34Þ

which upon using the definition of €rc and ∂rc/∂θ yields 2

3T 2 3 l € l cos θ  θ_ 2 l sin θ € cos θ x þ θ 6 7 62 7 2 2 7 6 7 þ I θ€ ðQi Þθ ¼ mr 6 4 5 4 5 l l 2 l _ € cos θ sin θ θ sin θ þ θ 2 2 2 2 l l l ¼ mr €x cos θ þ mr θ€ þ I θ€ ¼ mr €x cos θ þ I O θ€ 2 2 4

ð2:35Þ

where IO ¼ I + mr(l2/4). As in the case of the preceding example, we assume small oscillations. In this case € which is the cos θ  1 and consequently (Qi)θ reduces to ðQi Þθ ¼ mr ðl=2Þ€x þ I θ, same as the generalized inertia force obtained by using the kinetic energy. Note also as theP result of Eq. 2.15,  the generalized external force Qj can also be written as b Qj ¼ ni¼1 FiT ∂_r i =∂q_j .

74

2.5

2

Lagrangian Dynamics

Kinetic Energy

As was shown in the preceding section, one alternative for formulating the dynamic equations is to obtain an expression for the kinetic energy of the system and use it with Lagrange’s equation, as defined by Eq. 2.20. In this section, the formulation of the kinetic energy using vector and matrix notation is discussed. System of Particles In many applications, such as the examples presented in the preceding sections, the position vector of the particle i in the system can be expressed in terms of the generalized coordinates as ri ¼ ri(q1, q2, . . ., qn) and the velocity vector as r_ i ¼

n X ∂ri ∂ri ∂ri ∂ri q_ 1 þ q_ 2 þ    þ q_n ¼ q_ ∂q1 ∂q2 ∂qn ∂qj j j¼1

ð2:36Þ

which can be written using matrix notation as 2

r_ i ¼

∂ri ∂q1

∂ri ∂q2



3 q_ 1  7 ∂ri 6 6 q_ 2 7 6 7 ∂qn 4 ⋮ 5

ð2:37Þ

q_n This equation can be written in a compact form as r_ i ¼ Bi q_

ð2:38Þ

where

Bi ¼

∂ri ∂q1

∂ri ∂q2



 ∂ri , ∂qn

q ¼ ½ q1

q2



qn  T

ð2:39Þ

The kinetic energy of a particle i can also be written as T i ¼ ð1=2Þmi r_ iT r_ i , where mi is the mass of the particle i. Substituting Eq. 2.38 into the kinetic energy expression yields T  i  1  T i ¼ mi Bi q_ B q_ 2  1 1  ¼ mi q_ T BiT Bi q_ ¼ q_ T mi BiT Bi q_ 2 2

ð2:40Þ

The kinetic energy of the system of particles can also be obtained as T¼

np X i¼1

Ti ¼

"n # p X     1 q_ T mi BiT Bi q_ ¼ q_ T mi BiT Bi q_ 2 2 i¼1

np X 1 i¼1

which can be written as

ð2:41Þ

2.5 Kinetic Energy

75

1 T ¼ q_ T Mq_ 2

ð2:42Þ

where M is the mass matrix of the system of particles, defined as M¼

np X 

mi BiT Bi



ð2:43Þ

i¼1

The kinetic energy of Eq. 2.42 is a quadratic function in the generalized velocity _ and the mass matrix M is symmetric and may be constant or may depend on vector q, the system generalized coordinates. Rigid Bodies The kinetic energy of rigid bodies can also be expressed in a form similar to Eq. 2.42. For instance, for a rigid body i in the system, let xci and yci be the x-and y-coordinates of the center of mass of the body and let θi be the angular orientation of the body. The kinetic energy of the body can be written as  2 2 2 T i ¼ mi x_ ic þ mi y_ ic þ I i θ_i =2, where mi and Ii are, respectively, the body mass and mass moment of inertia defined with respect to the center of mass of the body. The kinetic energy T i can also be written as 2 i 32 i 3 m 0 0 x_ c

1 i 6 76 7 1 i i _r Ti ¼ ð2:44Þ _x c y_ci θ_i 4 0 mi 0 54 y_ ci 5 ¼ q_ iT r Mr q 2 2 i i 0 0 I θ_ where qir ¼ xci

yci

θi

T

, and 2

mi 6 Mri ¼ 4 0 0

0 mi 0

3 0 7 05 I

ð2:45Þ

i

The absolute velocities x_ci , y_ci , and θ_ i can be expressed in terms of the generalized velocities as x_ ci ¼

n X ∂xci q_ , ∂qj j j¼1

y_ ci ¼

n X ∂yci q_ , ∂qj j j¼1

θ_ i ¼

n X ∂θi q_ ∂qj j j¼1

ð2:46Þ

That is, q_ ri ¼ Bi q_ where the matrix Bi is a 3  n matrix defined as

ð2:47Þ

76

2

2

∂xci 6 ∂q 6 1 6 i 6 ∂yc Bi ¼ 6 6 ∂q 6 1 6 i 4 ∂θ ∂q1

∂xci ∂q2 ∂yci ∂q2 ∂θi ∂q2

Lagrangian Dynamics

3 ∂xci  ∂qn 7 7 7 ∂yci 7 7  ∂qn 7 7 i 7 ∂θ 5  ∂qn

ð2:48Þ

Substituting Eq. 2.47 into Eq. 2.44 leads to an expression for the kinetic energy in terms of the generalized velocities as T i ¼ ð1=2Þq_ T BiT Mri Bi q_ , which can also be written as T i ¼ ð1=2Þq_ T Mi q_ , where Mi is the mass matrix of the rigid body defined as Mi ¼ BiT Mri Bi . Pb i The kinetic energy of a system of rigid bodies is T ¼ ni¼1 T ¼ Pn b T i ð1=2Þ i¼1 q_ M q_ , where nb is the total number of bodies in the system. The Pnb

i _ ¼ system kinetic energy can also be written as T ¼ ð1=2Þq_ T i¼1 M q Pnb i T ð1=2Þq_ Mq_ , where M is the system mass matrix defined as M ¼ i¼1 M . Illustrative Example Consider the system shown in Fig. 2.7a, in which AB is a uniform rod that has mass m, mass moment of inertia I, and length l. The kinetic energy of the rod can be defined using Eq. 2.44. However, the rod has only one degree of freedom since point A slides in the vertical direction only, while point B slides in the horizontal direction only. Because of these two restrictions on the motion of the rod, one has the following kinematic relationships: xA ¼ xc 

l cos θ ¼ 0, 2

yB ¼ yc 

l sin θ ¼ 0 2

ð2:49Þ

where xA and yB are the horizontal and vertical coordinates of points A and B, respectively. The preceding two equations, which show that the center of mass of the rod follows a circular trajectory which has a radius l/2, lead to

Fig. 2.7 Generalized mass

2.5 Kinetic Energy

77



  _

x_ c θ l  sin θ ¼ 2 cos θ y_c

ð2:50Þ

The absolute velocities of the rod can then be expressed in terms of the independent angular velocity as 2 3 l 2 3 x_ c 6  2 sin θ 7 7 6 7 _6 _ 7 ð2:51Þ 4 y_c 5 ¼ θ 6 6 l cos θ 7 ¼ Bθ 4 2 5 _ θ 1 where B is a 3  1 matrix which is recognized as B ¼ ½ ðl=2Þ sin θ ðl=2Þ cos θ 1T . The kinetic energy of the rod can be expressed in terms of the independent angular velocity as
 1 T ml2 _2 l _ 2 2 _ T ¼ B Mr Bθ ¼ I þ ð2:52Þ θ ¼ I oθ 2 2 4 where Io is the mass moment of inertia about point o shown in Fig. 2.7b. In this figure, point o is the point of intersection of the lines of action of the reaction forces FA and FB. The preceding equation shows that the generalized mass matrix associated with the independent coordinate of the rod reduces to a scalar defined by the mass moment of inertia about that point. Note that by using simple trigonometric identities, one can show that the length oc is l/2. Definitions In both cases of systems of particles and rigid bodies, the kinetic energy is a quadratic function in the system velocities, or more precisely, a homogeneous, second-degree quadratic form in the velocities q_ 1 , q_ 2 , . . . , q_ n . The mass matrix M is said to be the matrix of the quadratic form. A quadratic form is said to be positive (negative) if it is equal or greater (less) than zero for all the real values of its variables. A positive (negative) quadratic form is said to be positive (negative) definite if it is equal to zero only when all its variables are equal to zero. A positive (negative) quadratic form which is equal to zero for some nonzero values of its variables is said to be positive (negative) semidefinite. A quadratic form which can be positive or negative for the real values of its variables is said to be indefinite. The matrix of quadratic form is said to be positive (negative) definite, positive (negative) semidefinite, or indefinite according to the nature of the quadratic form. Since there can be no motion of particles and rigid bodies while the kinetic energy remains equal to zero, the kinetic energy is always a positive definite quadratic form. Consequently, the mass matrix is always positive definite. Therefore, the mass matrix is nonsingular, its determinant has always a nonzero value, and its inverse exists.

78

2

Lagrangian Dynamics

Vector Form of Lagrange’s Equation Using the quadratic form of Eq. 2.42, Lagrange’s equation can be expressed in a vector form. First, we introduce the notations ∂T=∂q_ ¼ T q_ ¼ q_ T M, and ∂T/∂q ¼ Tq, where vector subscript implies differentiation with respect to this vector. Using this notation, one can then write Lagrange’s equation in the following vector form: d  T  T T q_  T q ¼ Q dt

ð2:53Þ

  T d Mq_  T q ¼ Q dt

ð2:54Þ

or

where Q is the vector of generalized forces associated with the generalized coordinates and defined as Q ¼ ½ Q1 Q2    Qn T . Observe that Lagrange’s equation can also be written compactly as Qi ¼ Q, where Qi is the vector of the generalized inertia forces expressed in terms of the kinetic energy as Qi ¼

  T d Mq_  T q dt

ð2:55Þ

Example 2.8

Using Lagrange’s equation, derive the differential equation of motion of the system shown in Fig. 2.8.

Fig. 2.8 Double pendulum

(continued)

2.5 Kinetic Energy

79

Solution. The position vectors of the two masses are given by " r ¼ 1

" r ¼ 2

x1 y1 x2

#

" ¼

#

" ¼

y2

#

l1 sin θ1 l1 cos θ1

#

l1 sin θ1 þ l2 sin θ2 l1 cos θ1  l2 cos θ2

The velocity vectors are given by " r_ ¼ 1

" r_ 2 ¼

x_1 y_1 x_2 y_2

#

" ¼

#

" ¼

θ_1 l1 cos θ1 θ_1 l1 sin θ1

#

" ¼

l1 cos θ1

0

#"

θ_1 θ_2

# ¼ B1 q_

l1 sin θ1 0 # " l1 cos θ1 θ_1 l1 cos θ1 þ θ_2 l2 cos θ2 ¼ l1 sin θ1 θ_1 l1 sin θ1 þ θ_2 l2 sin θ2

l2 cos θ2 l2 sin θ2

#"

θ_1

#

θ_2

¼ B2 q_ where " q¼

q1

#

" ¼

q2

θ1

#

θ2

and the coefficient matrices B1 and B2 are given by " B ¼ 1

l1 cos θ1

0

l1 sin θ1

0

"

# ,

B ¼ 2

l1 cos θ1

l2 cos θ2

l1 sin θ1

l2 sin θ2

#

The kinetic energy of the system is T¼

2 2 X 1X 1 1 Ti ¼ mi r_ iT r_ i ¼ m1 r_ 1T r_ 1 þ m2 r_ 2T r_ 2 2 i¼1 2 2 i¼1

1 1 ¼ q_ T m1 B1T B1 q_ þ q_ T m2 B2T B2 q_ 2 2

1 T 1 ¼ q_ m1 B1T B1 þ m2 B2T B2 q_ ¼ q_ T Mq_ 2 2 where the mass matrix M is defined as (continued)

80

2

Lagrangian Dynamics

M ¼ m1 B1T B1 þ m2 B2T B2 " #" # l1 cos θ1 l1 sin θ1 l1 cos θ1 0 ¼ m1 l1 sin θ1 0 0 0 " #" # l1 cos θ1 l1 sin θ1 l1 cos θ1 l2 cos θ2 þm2 l2 cos θ2 l2 sin θ2 l1 sin θ1 l2 sin θ2 " # " # m2 l21 m2 l1 l2 cos ðθ1  θ2 Þ m1 l21 0 ¼ þ m2 l1 l2 cos ðθ1  θ2 Þ m2 l22 0 0 " # ðm1 þ m2 Þl21 m2 l1 l2 cos ðθ1  θ2 Þ ¼ m2 l1 l2 cos ðθ1  θ2 Þ m2 l22 The virtual work of the forces that act on the system is given by δW ¼ m1 gδy1  m2 gδy2 þ M 1 δθ1 þ M 2 δθ2 where the virtual displacements δy1 and δy2 are δy1 ¼ l1 sin θ1 δθ1 δy2 ¼ l1 sin θ1 δθ1 þ l2 sin θ2 δθ2 That is, δW ¼ m1 gl1 sin θ1 δθ1  m2 gðl1 sin θ1 δθ1 þ l2 sin θ2 δθ2 Þ þM 1 δθ1 þ M 2 δθ2 ¼ ðM 1  m1 gl1 sin θ1  m2 gl1 sin θ1 Þδθ1 þ ðM 2  m2 gl2 sin θ2 Þδθ2 which can be written as δW ¼ Qθ1 δθ1 þ Qθ2 δθ2 ¼ QT δq where Qθ1 and Qθ2 are, respectively, the generalized forces associated with the coordinates θ1 and θ2 and are given by " Q¼

Qθ1 Qθ2

#

" ¼

M 1  m1 gl1sin θ1  m2 gl1sin θ1

#

M 2  m2 gl2 sin θ2

Having determined the kinetic energy and the generalized forces, one can use Lagrange’s equation. To this end, we evaluate the following: (continued)

2.6 Strain Energy

81


 d ∂T d  d T  _ €T M þ q_ T M T q_ ¼ q_ M ¼ q ¼ dt ∂q_ dt dt That is, d  T _ q_ T q_ ¼ M€ qþM dt _ is the matrix where M " _ ¼ M

0

  m2 l1 l2 θ_1  θ_2 sinðθ1  θ2 Þ

#   m2 l1 l2 θ_1  θ_2 sinðθ1  θ2 Þ 0

Therefore, #" #
T " ðm1 þ m2 Þl21 m2 l1 l2 cosðθ1  θ2 Þ θ€1 d ∂T ¼ dt ∂q_ θ€2 m2 l1 l2 cosðθ1  θ2 Þ m2 l22 2 3   m2 l1 l2 θ_2 θ_1  θ_2 sinðθ1  θ2 Þ 5 4   _ _ _ m2 l1 l2 θ1 θ1  θ2 sinðθ1  θ2 Þ One can also show that " #
T  T m2 l1 l2 θ_1 θ_2sinðθ1  θ2 Þ ∂T ¼ Tq ¼ ∂q m2 l1 l2 θ_1 θ_2sinðθ1  θ2 Þ Substituting into Lagrange’s equation of motion given by Eq. 2.53 or 2.54, one obtains the differential equations of motion of this system.

2.6

Strain Energy

The generalized elastic forces of the springs, which were obtained in the preceding sections using the virtual work, can also be obtained using the strain energy expression. For example, in the single degree of freedom mass–spring system shown in Fig. 2.3, the strain energy due to the deformation of the spring is given by U ¼ kx2/2. By using this expression for the strain energy, the generalized force of the spring can simply be defined as Qs ¼ ∂U/∂x ¼ kx. For this system, one can write Lagrange’s equation of motion in the following form:

82

2

Lagrangian Dynamics



 d ∂T ∂T  þ Qs  ¼Q dt ∂x_ ∂x

ð2:56Þ

 is the vector of all generalized forces excluding the spring force. Since where Q Qs ¼  ∂U/∂x, the preceding equation can be written as


 d ∂T ∂T ∂U   þ ¼Q dt ∂x_ ∂x ∂x

ð2:57Þ

In general, one may write Lagrange’s equation in the case of n generalized coordinates in the following alternate form d ∂T dt ∂q_j

! 

∂T ∂U j , þ ¼Q ∂qj ∂qj

j ¼ 1,2,. . . ,n

ð2:58Þ

where U ¼ U(q1, q2, . . ., qn) is the strain energy function that depends on the system generalized coordinates. Keeping in mind that the differentiation of the scalar strain energy function with respect to the generalized coordinate vector q ¼ h i T ∂U ∂U ∂U [q1 q2 q3    qn] leads to a row vector, that is, ∂U=∂q ¼ U q ¼ ∂q ∂q    ∂q , 1

2

n

Eq. 2.58 can be written using vector notation as
T
T
T d ∂T ∂T ∂U   þ ¼Q ð2:59Þ dt ∂q_ ∂q ∂q

¼ Q 1 Q 2    Q n T is the vector of all generalized forces excluding where Q the elastic forces which are accounted for using the strain energy function U. We have previously shown that the kinetic energy function T can be expressed as a quadratic form in the vector of system generalized velocities. One can also show that the strain energy function U of linear systems can be expressed as a quadratic form in the vector of generalized coordinates as 1 U ¼ qT Kq 2

ð2:60Þ

where K is the symmetric stiffness matrix of the system given by 2

k 11 6 k 21 K¼6 4⋮ k n1

3

symmetric k 22 ⋮ k n2

⋱ 

7 7 5

ð2:61Þ

k nn

Consequently, the term ∂U/∂q in Lagrange’s equation of Eq. 2.59 is the row vector ∂U/∂q ¼ qTK. Since the stiffness matrix is symmetric, that is, K ¼ KT,

2.6 Strain Energy

83

one has (∂U/∂q)T ¼ Kq. While the kinetic energy is always a positive-definite quadratic form in the velocities, the strain energy can be a positive-semidefinite quadratic form in the coordinates. This situation occurs when the system has rigid body degrees of freedom. The motion of such a system can occur without any change in the potential energy. The stiffness matrix of such a system is always singular, and the system is said to be semidefinite. The dynamics of the semidefinite systems will be discussed in more detail in the following chapter. Example 2.9

Using the strain energy function, determine the stiffness matrix of the two degree of freedom mass–spring system shown in Fig. 2.4. Solution. The strain energy of the spring k1 is 1 1 U 1 ¼ k 1 x21 ¼ k1 q21 2 2 The strain energy of the spring k2 is 1 1 U 2 ¼ k 2 ð x 2  x 1 Þ 2 ¼ k 2 ð q2  q1 Þ 2 2 2 The total strain energy of the system is 1 1 U ¼ U 1 þ U 2 ¼ k1 q21 þ k2 ðq2  q1 Þ2 2 2 1 1 ¼ ðk 1 þ k2 Þq21  k2 q1 q2 þ k2 q22 2 2 which is a quadratic form in the generalized coordinates q1 and q2 and can be written as " 1 U ¼ ½ q1 2

q2 

ðk 1 þ k 2 Þ

k2

k 2

k2

#"

q1 q2

# 1 ¼ qT Kq 2

where K is the stiffness matrix defined as

K¼

k1 þ k2

k2

k2

k2



which is the same stiffness matrix obtained in Example 2.6 using the virtual work.

84

2.7

2

Lagrangian Dynamics

Hamilton’s Principle

Another alternate approach for deriving the differential equations of motion from scalar energy quantities is Hamilton’s principle which states that δ

ð t2

ðT  V Þdt þ

ð t2

t1

δWnc dt ¼ 0

ð2:62Þ

t1

where T is the system kinetic energy, V is the system potential energy, and δWnc is the virtual work of the nonconservative forces. Hamilton’s principle is sometimes written in the following form: δ

ð t2 t1

L dt þ

ð t2

δWnc dt ¼ 0

ð2:63Þ

t1

where L is called the Lagrangian and defined as L¼T V

ð2:64Þ

Hamilton’s principle of Eq. 2.62 or its alternate form of Eq. 2.63 states that the variation of the kinetic and potential energy plus the line integral of the virtual work done by the nonconservative forces during any time interval between t1 and t2 must be equal to zero. The potential energy V is defined as the strain energy minus the work done by the conservative forces, that is, V ¼ U  Wc

ð2:65Þ

where Wc is the work done by the conservative forces. Excluding the forces that contribute to the strain energy U, one may define the virtual work of all other forces, conservative and nonconservative, as δW ¼ δWc þ δWnc

ð2:66Þ

Substituting Eqs. 2.65 and 2.66 into Eq. 2.62 leads to the following alternate form of Hamilton’s principle: δ

ð t2 t1

ðT  U Þdt þ

ð t2

δW dt ¼ 0

ð2:67Þ

t1

When applying Hamilton’s principle, it is assumed that the system coordinates are specified at the two end points t1 and t2, that is, δqðt 1 Þ ¼ δqðt 2 Þ ¼ 0

ð2:68Þ

Relationship Between Hamilton’s Principle and Lagrange’s Equation The use of Hamilton’s principle is equivalent to the application of Lagrange’s equation. In fact,

2.7 Hamilton’s Principle

85

Hamilton’s principle can be used to derive Lagrange’s equation of motion. To this end, we write δ

ð t2

T dt ¼

t1

ð t2 t1

 ð t2
∂T ∂T _ δq þ δq dt δT dt ¼ ∂q ∂q_ t1

ð2:69Þ

Using integration by parts, the second term in the preceding equation yields ð t2
t1


 t ð t2
 2 ∂T ∂T d ∂T δq_ dt ¼ δq  δq dt _ ∂q_ ∂q_ t1 t 1 dt ∂q

ð2:70Þ

Using the assumption of Eq. 2.68, the preceding equation reduces to ð t2


 ð t2
 ∂T d ∂T δq_ dt ¼  δq dt _ ∂q_ t 1 dt ∂q

t1

ð2:71Þ

Substituting Eq. 2.71 into Eq. 2.69 yields δ

ð t2 t1


 ð t2

∂T d ∂T  T dt ¼ δq dt dt ∂q_ t 1 ∂q

ð2:72Þ

Similarly, δ

ð t2 t1

U dt ¼

ð t2
 ∂U δq dt, ∂q t1

ð t2

δW dt ¼

t1

ð t2

 T δq dt Q

ð2:73Þ

t1

 is the vector of generalized forces. where Q Substituting Eqs. 2.72 and 2.73 into Hamilton’s principle of Eq. 2.67 yields
  ð t2

∂T d ∂T ∂U  T  þ Q δq dt ¼ 0  dt ∂q_ ∂q t 1 ∂q

ð2:74Þ

If the coordinates q1, q2, . . ., qn are independent, the integrand in Eq. 2.74 must be identically zero. This leads, after rearranging terms, to Lagrange’s equation defined by Eq. 2.59. Example 2.10

Use Hamilton’s principle to derive the differential equation of motion of the single degree of freedom system of Example 2.2. Solution. The system kinetic energy is T ¼ mx_ 2 =2, and the system strain energy is U ¼ kx2/2. The virtual work of all forces, excluding the spring force, that act on this system is δW ¼ F ðt Þδx  cx_ δx (continued)

86

2

Lagrangian Dynamics

Using Hamilton’s principle of Eq. 2.67, one has δ

ð t2

ðT  U Þdt þ

t1

ð t2

δW dt ¼ 0

t1

where δ

ð t2 t1

ð t2


 1 2 1 2 mx_  kx dt ðT  U Þdt ¼ δ 2 2 ð t2 t1   mx_ δx_  kxδx dt ¼ t1

which, by integrating by parts the first term, leads to δ

ð t2 t1

t ð t2 2    m€xδx þ kxδx dt ðT  U Þdt ¼ mx_ δx  t1

t1

Substituting this expression and the expression of the virtual work into Hamilton’s principle and using Eq. 2.68 lead to ð t2

ðm€x þ kx  F ðt Þ þ cx_Þδx dt ¼ 0

t1

Since δx is an independent coordinate, the above equation leads to m€x þ cx_ þ kx ¼ F ðt Þ which is the same equation of motion obtained in Example 2.5.

Example 2.11

Use Hamilton’s principle to derive the differential equations of motion of the system shown in Fig. 2.9.

Fig. 2.9 Two degree of freedom system

(continued)

2.7 Hamilton’s Principle

87

Solution. The kinetic and strain energies of the system are given by 1 1 T ¼ m1 x_ 21 þ m2 x_ 22 , 2 2

1 1 U ¼ k 1 x21 þ k2 ðx2  x1 Þ2 2 2

The virtual work of the damping and external forces is given by   δW ¼ c1 x_1 δx1  c2 x_2  x_1 δðx2  x1 Þ þ F1 δx1 þ F 2 δx2 The variation of the time integral of (T  U ) can be obtained as  ð t2

1 1 1 2 1 2 2 2 m1 x_ 1 þ m2 x_ 2  k1 x1  k2 ðx2  x1 Þ dt δ ðT  U Þdt ¼ δ 2 2 2 t1 t1 2 ð t2 ¼ ½m1x_1δx_1 þ m2 x_2 δx_2  k 1 x1 δx1  k2 ðx2  x1 Þδðx2  x1 Þdt ð t2

t1

By integrating by parts and using the assumption of Hamilton’s principle that the displacements are specified at the end points t1 and t2, one obtains δ

ð t2

ðT  U Þdt ¼

t1

ð t2

½m1€x1 δx1  m2€x2 δx2  k 1 x1 δx1  k 2 ðx2  x1 Þδðx2  x1 Þdt

t1

By rearranging the terms, one obtains δ

ð t2 t1

ð t2

ðT  U Þdt ¼  f½m1€x1 þ ðk1 þ k2 Þx1  k 2 x2 δx1 t 1

þ m2€x2 þ k2 x2  k2 x1 δx2 dt

Using this equation and the expression of the virtual work δW, Hamilton’s principle yields ð t2 ð t2 δ ðT  U Þdt þ δW dt ¼ 0 t1

t1

That is, 

ð t2 t1

þ



½m1€x1 þ ðk1 þ k2 Þx1  k2 x2 δx1 þ m2€x2 þ k2 x2  k2 x1 δx2 dt

ð t2



 

c1 x_1 δx_1  c2 x_2  x_1 δðx2  x1 Þ þ F1 δx1 þ F 2 δx2 dt ¼ 0

t1

which yields, after rearranging terms, (continued)

88

2

ð t2

Lagrangian Dynamics

½m1€x1 þ ðc1 þ c2 Þx_1  c2 x_2 þ ðk 1 þ k2 Þx1  k2 x2  F1 δx1 dt

t1

þ

ð t2



m2€x2 þ c2 x_2  c2 x_1 þ k2 x2  k2 x1  F 2 δx2 dt ¼ 0

t1

Since x1 and x2 are assumed to be independent coordinates, the preceding equation leads to the following two differential equations: m1€x1 þ ðc1 þ c2 Þx_1  c2 x_2 þ ðk1 þ k2 Þx1  k2 x2 ¼ F1 m2€x2 þ c2 x_2  c2 x_1 þ k 2 x2  k2 x1 ¼ F 2

2.8

Conservation Theorems

In this section, a special case of the preceding development is considered. This is the case of conservative systems in which all the forces acting on the system can be derived from the potential function V. In this case, δWnc ¼ð 0. Substituting this equation into Hamilton’s principle of Eq. 2.62, one obtains δ

t2

ðT  V Þdt ¼ 0, or

t1

equivalently δ

ð t2

L dt ¼ 0

ð2:75Þ

t1

where L ¼ T  V is the Lagrangian of the system. Clearly, for a conservative system, Lagrange’s equation reduces to ! d ∂L ∂L ¼ 0, j ¼ 1,2,. . . ,n  dt ∂q_j ∂qj

ð2:76Þ

or d ∂L dt ∂q_j

! ¼

∂L , ∂qj

j ¼ 1,2,. . . ,n

ð2:77Þ

Observe that the total time derivative of the Lagrangian L is given by n dL X ∂L ∂L €qj þ ¼ q_ _ dt ∂qj ∂qj j j¼1

Substituting Eq. 2.77 into Eq. 2.78, one obtains

! ð2:78Þ

2.8 Conservation Theorems

89

" ! # ! n n X dL X ∂L d ∂L d ∂L ¼ q€ þ q_ q_ ¼ dt ∂q_j j dt ∂q_j j dt ∂q_j j j¼1 j¼1

ð2:79Þ

This equation can be rewritten as n dL d X ∂L  q_ ¼ 0 dt dt j¼1 ∂q_j j

ð2:80Þ

or equivalently n X d ∂L L q_ dt ∂ q_j j j¼1

! ¼0

ð2:81Þ

Since the potential energy V is independent of the velocity, one has ∂L ∂T ¼ ∂q_j ∂q_j

ð2:82Þ

Substituting this equation into Eq. 2.81, one gets n X d ∂T L q_ dt ∂ q_j j j¼1

! ¼0

ð2:83Þ

That is, L

n X ∂T q_ ¼ H ∂ q_j j j¼1

ð2:84Þ

where H is a constant called the Hamiltonian. Equation 2.84 is valid for any conservative single or multi-degree of freedom system. The Hamiltonian H can be written in a more convenient form using the following identity: n X ∂T q_ ¼ 2T ∂q_j j j¼1

ð2:85Þ

Substituting this equation into Eq. 2.84 and using the fact that L ¼ T  V, one obtains H ¼ ðT  V  2T Þ ¼ T þ V

ð2:86Þ

which implies that the Hamiltonian is equal to the total energy of the system. Since, for conservative systems, the Hamiltonian H is constant, one has

90

2

Lagrangian Dynamics

dH d ¼ ðT þ V Þ ¼ 0 dt dt

ð2:87Þ

This equation can be used to develop the dynamic equations of motion of conservative single and multi-degree of freedom systems, as demonstrated by the following examples. Example 2.12

Use the method of conservation of energy to derive the equation of free vibration of the single degree of freedom system of Example 2.2 assuming that the damping coefficient is equal to zero. Solution. The Hamiltonian of the system is given by 1 1 H ¼ T þ V ¼ mx_2 þ kx2 2 2 Applying Eq. 2.87, one obtains dH _ x þ kxx_ ¼ 0 ¼ mx€ dt This equation can be rewritten as   m€x þ kx x_ ¼ 0 Since x_ is not equal to zero for all values of t, one must have m€x þ kx ¼ 0 which is the equation of undamped free vibration of the single degree of freedom system.

Example 2.13

Use the method of conservation of energy to derive the differential equations of the free undamped vibration of the system in Example 2.9. Solution. The kinetic energy of this system is 1 1 T ¼ m1 x_ 21 þ m2 x_ 22 2 2 (continued)

2.8 Conservation Theorems

91

and the strain energy is given by 1 1 U ¼ V ¼ k1 x21 þ k2 ðx2  x1 Þ2 2 2 The Hamiltonian of the system is 1 1 1 1 H ¼ T þ V ¼ m1 x_ 21 þ m2 x_ 22 þ k1 x21 þ k2 ðx2  x1 Þ2 2 2 2 2 The application of Eq. 2.87 leads to   dH ¼ m1 x_1€x1 þ m2 x_2€x2 þ k1 x1 x_1 þ k2 ðx2  x1 Þ x_2  x_1 ¼ 0 dt This equation can be rewritten as ½m1€x1 þ k1 x1  k2 ðx2  x1 Þx_1 þ ½m2€x2 þ k2 ðx2  x1 Þx_2 ¼ 0 Since x_1 and x_2 are independent velocities, their coefficients in the above equation must be equal to zero. This leads to the following two differential equations of motion for the undamped two degree of freedom system: m1€x1 þ ðk1 þ k 2 Þx1  k2 x2 ¼ 0 m2€x2 þ k2 x2  k2 x1 ¼ 0

Nonlinear Systems Conservation theorems can also be applied to nonlinear constrained multi-body systems. In order to demonstrate that, we consider the multi-body system shown in Fig. 2.10, which was considered in the preceding chapter. The system consists of two homogeneous circular cylinders, each of mass m and centroidal mass moment of inertia I, and a connecting rod AB of mass mr and length l. The cylinders, which have radius r, are assumed to roll without slipping.

Fig. 2.10 Multi-body system

92

2

Lagrangian Dynamics

Because of the rolling conditions, the velocities of the centers of mass of the cylinders are equal and given by vO ¼ vC ¼ r θ_

0

T

ð2:88Þ

The velocity of the center of mass of the connecting rod is equal to the absolute velocities of points A and B since the angular velocity of this rod is equal to zero. These velocities are defined as " vr ¼ vB ¼ vA ¼ vO þ vAO ¼ vO þ ω  rAO ¼

ðr  a cosθÞθ_ aθ_ sinθ

# ð2:89Þ

where vAO is the velocity of point A with respect to point O, vr is the absolute velocity of the center of mass of the connecting rod, ω is the angular velocity vector of the cylinders, and rAO is the position vector of point A with respect to point O defined as rAO ¼ ½ a sinθ

a cosθ T

ð2:90Þ

The kinetic and potential energies of the system are
 9 1 T 1 _2 1 T ¼ 2 mvO vO þ I θ þ mr vrT vr = 2 2 2 ; V ¼ mr gað1  cosθÞ

ð2:91Þ

Using the velocity expressions, the Hamiltonian can be written as H ¼T þV    2 1  2 2 2 ¼ I þ mr þ mr r þ a  2 arcos θ θ_ þ mr gað1  cosθÞ 2

ð2:92Þ

It follows that    dH 1  ¼ 2 I þ mr 2 þ mr r 2 þ a2  2racosθ θ€θ_ dt 2 3 _ þmr raθ sinθ þ mr gaθ_ sinθ ¼ 0

ð2:93Þ

Since for a cylinder I ¼ mr2/2, the preceding equation leads to 

  3 2 1  2 2 2 mr þ mr r þ a  2racosθ θ€ þ mr raθ_2sinθ þ mr gasinθ ¼ 0 2 2

ð2:94Þ

which is the same nonlinear differential equation obtained in the preceding chapter by applying D’Alembert’s principle.

2.9 Virtual Work and D’Alembert’s Principle

2.9

93

Virtual Work and D’Alembert’s Principle

In this chapter, a brief introduction to the subject of Lagrangian dynamics is presented, and the important concepts of the generalized coordinates, virtual work, and the generalized forces are introduced. These concepts are then used to derive Lagrange’s equation from Newton’s second law. The final form of Lagrange’s equation is presented in terms of scalar energy quantities such as the strain and kinetic energies. The quadratic forms of the kinetic and strain energies are also examined and used to define the system mass and stiffness matrices. Hamilton’s principle which represents an alternate approach for deriving the dynamic equations of motion using scalar quantities was also discussed. It was also shown that Lagrange’s equation or Hamilton’s principle can be used to develop conservation theorems that can be used to derive the differential equations of motion of conservative systems. The development of Lagrange’s equation using the vector equation of Newton’s second law consists of several steps. At the end of each intermediate step, the resulting equations can be used to derive the dynamic equations of the mechanical system. Equation 2.11, for example, summation over the number of particles leads to the wellknown principle of virtual work in dynamics which can be written as δWi ¼ δWe

ð2:95Þ

where δWi is the virtual work of the inertia forces and δWe is the virtual work of the applied forces. The virtual work of the workless constraint forces is equal to zero and, consequently, such forces do not appear in Eq. 2.95. Equation 2.95, even though it is an intermediate step in deriving Lagrange’s equation, represents a powerful tool which can be used to develop the equations of motion of mechanical systems. The use of Eq. 2.95 can be demonstrated by considering the simple pendulum shown in Fig. 2.11a. The free-body diagram and the inertia-force diagram

Fig. 2.11 Application of the principle of virtual work in dynamics

94

2

Lagrangian Dynamics

are shown in Fig. 2.11b. The position of the center of mass of the rod can be expressed in terms of the degree of freedom of the system θ as xc ¼

l sin θ, 2

l yc ¼  cos θ 2

ð2:96Þ

It follows that 9 l l > cos θδθ, δyc ¼ sin θδθ > = 2 2 > lθ_ lθ_ > ; x_c ¼ cos θ, y_c ¼ sin θ 2 2

δxc ¼

ð2:97Þ

The components of the acceleration of the center of mass are 9 lθ€ l > cos θ  θ_ 2 sin θ > = 2 2 > lθ€ l > €yc ¼ sin θ þ θ_ 2 cos θ ; 2 2 €xc ¼

ð2:98Þ

The virtual work of the inertia forces and moments can then be written as € δW i ¼ m€xc δxc þ m€yc δyc þ I θδθ 

l€ l _2 l cos θδθ ¼ m θ cos θ  θ sin θ 2 2 2 

l€ l _2 l € þm θ sin θ þ θ cos θ sin θδθ þ I θδθ 2 2 2

ð2:99Þ

This equation can be simplified and rewritten as
2 l € € δW i ¼ m θð cos2 θ þ sin2 θÞδθ þ I θδθ 2 "
 # l 2 € ¼ m þ I θδθ 2

ð2:100Þ

The virtual work of the external forces is given by l δW e ¼ mgδyc ¼ mg sin θδθ 2

ð2:101Þ

Note that the virtual work of the reactions Rx and Ry is equal to zero since the virtual change in the position vector of point O is equal to zero. Therefore, the application of Eq. 2.95 leads to "
 # l 2 € ¼ mg l sin θδθ ð2:102Þ m þ I θδθ 2 2

2.10

Vibration of Complex Nonlinear Systems

95

which leads to l I O θ€ þ mg sin θ ¼ 0 2

ð2:103Þ

where IO ¼ I + m(l/2)2. Equation 2.103 can also be obtained by the direct application of Newton’s second law or Lagrange’s equation. It is derived in this section using the principle of virtual work in dynamics. Equation 2.103 can also be obtained using D’Alembert’s principle by equating the moments of the applied and inertia forces about point O. The identity IO ¼ I + m(l/2)2 is the parallel axis theorem which states that the moment of inertia about an arbitrary axis is equal to the moment of inertia about a parallel axis passing through the center of mass plus the total mass of the body multiplied by the square of the distance between the two axes. This identity can also be derived using the kinetic energy of the system. In the example shown in Fig. 2.11, the kinetic energy is 1 1 1 T ¼ mx_2c þ my_2c þ I θ_2 2 2 2

ð2:104Þ

If x_c and y_c are expressed in terms of the generalized velocities, Eq. 2.104 yields
2
2 1 lθ_ 1 lθ_ 1 cos θ þ m sin θ þ I θ_ 2 T¼ m 2 2 2 2 2
2

1 1 l ¼ mθ_2 cos2 θ þ sin2 θ þ I θ_ 2 2 2 2

ð2:105Þ

Since cos2θ + sin2θ ¼ 1, the preceding equation leads to "
 # 1 l 2 1 m þ I θ_ 2 ¼ I O θ_ 2 T¼ 2 2 2

ð2:106Þ

where IO is previously defined.

2.10

Vibration of Complex Nonlinear Systems

In this chapter, the basic approaches used for formulating the vibration equations are discussed, and the relationship between them is explained. Any of these approaches can be used to formulate the vibration equations of multi-degree of freedom mechanical systems. The methods discussed in this chapter also represent the foundation for the new computational multibody system (MBS) methods that can be used to automatically construct and numerically solve the vibration equations of complex and highly nonlinear systems that consist of a large number of interconnected bodies (Roberson and Schwertassek 1988; Shabana 2013). The equations of motion of such systems can always be written as product of a

96

2

Lagrangian Dynamics

mass matrix and an acceleration vector equated to the applied forces. Algebraic equations that impose constraints on the motion of the system can be written in a vector form as C(q, t) ¼ 0. These algebraic equation, as in the case of the pin joint constraints of the pendulum discussed in the preceding section, can be used to eliminate dependent variables and obtain a number of equations equal to the number of the system degrees of freedom. This procedure can be used to develop a general computer algorithm that allows for automatically constructing and numerically solving the nonlinear vibration equations of complex systems. An example of a complex system which can be modeled using the principles covered in this chapter and MBS techniques is the tanker truck shown in Fig. 2.12. The analysis of the motion of tanker trucks is important for several reasons. First, highway vehicles transport more materials than all other modes of transportation, including rail vehicles and shipping vessels. The material transported can be hazardous such as crude oils, and therefore, truck accidents can be deadly and environmentally damaging. Furthermore, tanker trucks are more likely to roll over compared to passenger vehicles because of their high center of gravity. Consequently, it is crucial to thoroughly understand the dynamics of these vehicles for the safety of the drivers and the surrounding environment and infrastructure. This can be accomplished by developing and testing virtual models of tanker trucks, such as the model in Fig. 2.13 (Nicolsen et al. 2017). By using virtual prototyping, the dynamics of the vehicle can be studied and better understood, without the cost and risk of testing physical vehicles. One of the challenges in developing such virtual models is capturing the fluid sloshing effect using a continuum-based model. During curve negotiation, sudden braking, or rapid lane changes, liquid sloshing can have a significant effect on the vehicle vibration and dynamics. Developing a continuum-based approach for modeling liquid sloshing requires, as discussed in Chap. 5, using a large number of degrees of freedom.

Fig. 2.12 Tanker truck

2.10

Vibration of Complex Nonlinear Systems

97

Fig. 2.13 Virtual tanker truck model

The vehicle model shown in Fig. 2.13 consists of 20 bodies that include a cab, a tank, a body representing the fluid inside the tank, a frame, a four-bar steer axle, two rear axles, and 10 tires (Nicolsen et al. 2017). The fluid motion can be modeled using finite element (FE) formulations, which will be discussed in Chap. 5. The number of degrees of freedom of the model can be very large and depends on the FE mesh used to model the fluid sloshing. The chassis and axles are connected by a suspension system composed of linear spring-damper elements. The equations of motion of this system can be written as M€ q ¼ Qe , where M is

the system mass matrix, q ¼ q1T q2T    qnb T , qi is the vector of coordinates € ¼ ½ €q1 €q2 . . . €qn T , of body i, nb is the total number of bodies in the model, q and Qe is the vector that includes the system applied forces as well as the Coriolis and centrifugal inertia forces. The applied forces include gravity forces, the stiffness and damping forces exerted by the suspension system on the axles and frame, and the contact forces exerted by the ground on the tires. The equations of motion and the algebraic constraint equations C(q, t) ¼ 0 that represent the vehicle mechanical joints are solved for the system accelerations at every time step of the simulation. Using the algebraic equations C(q, t) ¼ 0, the independent accelerations can be identified and integrated in time using numerical integration techniques to determine the system independent coordinates and velocities. The dependent coordinates and velocities are determined using the algebraic constraint equations at the position and velocity levels, respectively. Using this procedure, the motion of the vehicle can be varied in order to study its stability and performance under different motion scenarios such as lane changes, traction, braking, and curve negotiations. Figure 2.14 shows the lateral friction force acting on one of the outer tires of the vehicle as a function of time. The results presented in Fig. 2.14, which demonstrate the effect of the oscillatory liquid sloshing during curve negotiation, are obtained using the constrained dynamics procedure briefly discussed in this section (Nicolsen et al. 2017). This procedure can also be used to test different component designs, such as testing the use of leaf springs for the suspension, or testing different types of tires.

98

2

Fig. 2.14 Lateral friction force on an outer tire during curve negotiation ( ANCF model)

Lagrangian Dynamics

Rigid model,

Problems 2.1. The single degree of freedom system shown in Fig. P2.1 consists of a mass m, a damper c, and two springs k1 and k2. Develop expressions for the kinetic and strain energies and the virtual work of the damping force. Use these energy and work expressions to develop the equations of free vibration of the system.

Fig. P2.1

2.2. In the case of free vibration of the system shown in Fig. P2.1, show that the time rate of change of the sum of the kinetic and strain energies is equal to the negative of the damping coefficient multiplied by the square of the velocity, that is,

Problems

99

d ðT þ U Þ ¼ cx_2 dt where x is the displacement of the mass. 2.3. Derive the differential equation of motion of the system shown in Fig. P2.1 by using Newton’s second law and also by using Lagrange’s equation. 2.4. Show that the kinetic energy of a system of n particles can be written as n X 1 1 i i2 T ¼ mvc2 þ mv 2 2 c i¼1

where m is the total mass of the system of particles, vc is the velocity of the center of mass, mi is the mass of the particle i, and vci is the velocity of the particle i with respect to the center of mass. 2.5. In the case of a rigid body, derive the principle of work and energy which states that the change in the kinetic energy of a rigid body is equal to the work of the forces acting on the body. 2.6. Use the principle of work and energy obtained in Problem 2.5 to derive the equation of motion of the system shown in Fig. P2.1. Compare the use of this principle with the result presented in Problem 2.2. 2.7. Obtain expressions for the kinetic and potential energies and the virtual work for the system shown in Fig. P2.2 in the following two cases: (a) small angular rotations; and (b) finite angular rotations. Derive the equations of motion in the two cases using the principle of virtual work in dynamics.

Fig. P2.2

100

2

Lagrangian Dynamics

2.8. Obtain the differential equation of free vibration of the single degree of freedom system shown in Fig. P2.3. Assume small oscillations.

Fig. P2.3

2.9. In Problem 2.8, determine the time rate of change of the sum of the kinetic and strain energies. 2.10. The system shown in Fig. P2.4 consists of a rigid massless bar of length l1 + l2 and two masses m1 and m2 which are rigidly attached to the bar. Obtain the differential equation of free vibration of this system by using Lagrange’s equation. Use the assumption of small oscillations.

Fig. P2.4

Problems

101

2.11. The virtual work of the inertia forces of a rigid body can be written as ð δWi ¼ ρ€rT δr dV V

where V is the volume of the body, ρ is the mass density, and r is the global position vector of an arbitrary point on the rigid body. Using the preceding equation, determine the virtual work of the inertia forces of the rigid body in terms of the acceleration of the center of mass, and the angular acceleration of the body. Assume the case of planar motion. 2.12. Repeat Problem 2.11 in the case of three-dimensional motion. 2.13. For the system shown in Fig. P2.5, determine the mass and stiffness matrices using, respectively, the kinetic and strain energies. Use the energy expressions and Lagrange’s equation to develop the matrix equation of motion of this system.

Fig. P2.5

2.14. Using the assumption of small oscillations, obtain the differential equations of motion of the double pendulum shown in Fig. P2.6 using Lagrange’s equation.

102

2

Lagrangian Dynamics

Fig. P2.6

2.15. Derive the differential equations of motion of the two degree of freedom system shown in Fig. P2.7 using Lagrange’s equation. Assume small oscillations.

Fig. P2.7

2.16. In the system shown in Fig. P2.8, AB is a massless rod which pivots freely about a pin connection at A. If the generalized coordinates are selected to be the angular orientation of the rod θ and the displacement x2 of the mass m2, obtain the differential equations of this system using Lagrange’s equation. Use the assumption of small oscillations.

Problems

103

Fig. P2.8

2.17. By using Lagrange’s equation, derive the equations of motion of the two degree of freedom system shown in Fig. P2.9.

Fig. P2.9

2.18. In the system shown in Fig. P2.10, AB is a rigid bar which pivots freely about a pin connection at A. At equilibrium, the bar AB is in a horizontal position. Derive the differential equation of motion of this system using Lagrange’s equation.

104

2

Lagrangian Dynamics

Fig. P2.10

2.19. Using Hamilton’s principle, obtain the differential equations of motion of the system shown in Fig. P2.9. 2.20. Using Hamilton’s principle, obtain the differential equations of motion of the system shown in Fig. P2.11.

Fig. P2.11

2.21. Obtain the differential equations of motion of the oscillatory system shown in Fig. P2.12 using Lagrange’s equation.

Fig. P2.12

Problems

105

2.22. Use Hamilton’s principle to derive the equations of motion of the system shown in Fig. P2.12. 2.23. Use the principle of virtual work in dynamics to derive the equations of motion of the system shown in Fig. P2.9. 2.24. Use the principle of virtual work in dynamics to derive the nonlinear equation of motion of the system shown in Fig. P2.13. The system consists of two cylinders, each of which has mass m, and a connecting rod that has mass mr and length l. The cylinders are assumed to roll without slipping.

Fig. P2.13

3

Multi-degree of Freedom Systems

The methods of vibration analysis of single degree of freedom systems can be generalized and extended to study systems with an arbitrary finite number of degrees of freedom. Mechanical systems in general consist of structural elements which have distributed mass and elasticity. In many cases, these systems can be represented by equivalent systems which consist of some elements which are bulky solids and can be treated as rigid elements with specified inertia properties, while the other elements are elastic elements which have negligible inertia effects. In fact, the single degree of freedom systems discussed in the preceding chapters are examples of these equivalent models which are called lumped mass systems. We have shown in the preceding chapters that a single degree of freedom system exhibits motion governed by one second-order ordinary differential equation, while a two degree of freedom system exhibits motion governed by two second-order ordinary differential equations. It is expected, therefore, that a system having n degrees of freedom exhibits motion which is governed by a set of n simultaneous second-order differential equations. An example of these systems is shown in Fig. 3.1. In Sect. 3.1 of this chapter, the general form of the second-order ordinary differential equations of motion that govern the vibration of multi-degree of freedom systems is presented, and the use of these equations is demonstrated by several applications. In Sect. 3.2, the undamped free vibration of multi-degree of freedom systems is discussed, and it is shown that a system with n degrees of freedom has n natural frequencies. Methods for determining the mode shapes of the undamped systems are presented, and the orthogonality of these mode shapes is discussed in Sect. 3.3. In this section, we also discuss the use of the modal transformation to obtain n uncoupled second-order ordinary differential equations of motion in terms of the modal coordinates. Sections 3.4 and 3.5 are devoted, respectively, to the analysis of semidefinite systems and the conservation of energy in the case of undamped free vibration. In Sect. 3.6, the forced vibration of the undamped multi-degree of freedom systems is discussed. The solution of the equations of motion of viscously damped multi-degree of freedom systems is obtained in Sect. 3.7 using proportional damping, # Springer Nature Switzerland AG 2019 A. Shabana, Vibration of Discrete and Continuous Systems, Mechanical Engineering Series, https://doi.org/10.1007/978-3-030-04348-3_3

107

108

3

Multi-degree of Freedom Systems

Fig. 3.1 Example of multi-degree of freedom systems

while the case of general viscous damping is covered in Sect. 3.8. In Sect. 3.9, the modal truncation technique frequently used in order to reduce the number of degrees of freedom is discussed, and Sects. 3.10 and 3.11 are devoted to the classical computer and numerical methods used in the vibration analysis of multi-degree of freedom systems. Section 3.12 explains the use of the concept of the modal decomposition technique in the analysis of complex and highly nonlinear practical applications.

3.1

Equations of Motion

Using matrix notation, the general form of the matrix equation of motion of the multi-degree of freedom system is given by M€ q þ Cq_ þ Kq ¼ F

ð3:1Þ

where M, C, and K are, respectively, the mass, damping, and stiffness matrices, q is the vector of coordinates, and F is the vector of forces that act on the multi-degree of freedom system. If the system has n degrees of freedom, the vectors q and F are given, respectively, by q ¼ [q1 q2    qn]T, and F ¼ ½ F 1 F 2    F n T , and the mass, damping, and stiffness matrices, are given in a more explicit form as 2

m11

6 6 m21 6 M¼6 6⋮ 4 2

mn1 k11

6 6 k21 6 K¼6 6⋮ 4 kn1

m12

   m1n

3

⋮

7    m2n 7 7 7, ⋱ ⋮ 7 5    mnn 3    k 1n 7    k 2n 7 7 7 ⋱ ⋮ 7 5

kn2

   k nn

m22 ⋮ mn2 k12 k22

2

c11

6 6 c21 6 C¼6 6⋮ 4 cn1

c12 c22 ⋮ cn2

   c1n

3

7    c2n 7 7 7, ⋱ ⋮ 7 5    cnn

ð3:2Þ

3.1 Equations of Motion

109

where mij, cij, and kij, i, j ¼ 1,2,. . .,n, are, respectively, the mass, damping, and stiffness coefficients. Special Cases Equation 3.1 is the general form of the matrix equation that governs the forced damped vibration of the multi-degree of freedom system. The equations of motion of the free damped vibration can be obtained from Eq. 3.1 by letting F ¼ 0, that is, M€ q þ Cq_ þ Kq ¼ 0

ð3:3Þ

which can be written in a more explicit form as 2

m11

6 6 m21 6 6⋮ 4 mn1

m12

   m1n

32

€q1

3

2

c11

c12



c1n

32

q_ 1

3

76 7 6 76 7    m2n 76 €q2 7 6 c21 c22    c2n 76 q_ 2 7 76 7 þ 6 76 7 76 7 6 7 6 ⋮ ⋱ ⋮ 7 54 ⋮ 5 4 ⋮ ⋮ ⋱ ⋮ 54 ⋮ 5 €qn cn1 cn2    cnn mn2    mnn q_ n 2 32 3 2 3 k 11 k 12    k1n q1 0 6 76 7 6 7 6 k 21 k 22    k2n 76 q2 7 6 0 7 76 7 6 7 þ6 6 ⋮ ⋮ ⋱ ⋮ 76 ⋮ 7 ¼ 6 ⋮ 7 4 54 5 4 5 0 k n1 k n2    knn qn m22

ð3:4Þ

Furthermore, if we assume the case of free undamped vibration, Eq. 3.3 reduces to M€ q þ Kq ¼ 0. Similarly, the case of undamped forced vibration can be described by the following matrix differential equation: M€ q þ Kq ¼ F

ð3:5Þ

The use of Newton’s second law and Lagrange’s equation to derive the differential equations of motion of several multi-degree of freedom systems is demonstrated by the following examples. Rectilinear Motion Figure 3.2a shows a multi-degree of freedom system which consists of n masses connected by springs and dampers. The system has n degrees of freedom denoted as x1, x2,. . .,xn, which can be written in a vector form as x ¼ ½ x1 x2    xn T . Let F be the vector of forces that act on the masses, that is, F ¼ ½ F 1 F 2    F n T . The free-body diagrams of the masses are shown in Fig. 3.2b. By direct application of Newton’s second law, it can be verified that the equations of motion of the masses are given by   m1€x1 ¼ k1 x1 þ k2 ðx2  x1 Þ  c1 x_1 þ c2 x_2  x_1 þ F 1     m2€x2 ¼ k2 ðx2  x1 Þ þ k3 ðx3  x2 Þ  c2 x_2  x_1 þ c3 x_3  x_2 þ F 2 ⋮

110

3

Multi-degree of Freedom Systems

Fig. 3.2 Multi-degree of freedom rectilinear system

    mi €xi ¼ ki ðxi  xi1 Þ þ k iþ1 ðxiþ1  xi Þ  ci x_i  x_i1 þ ciþ1 x_iþ1  x_i þ Fi ⋮  mn€xn ¼ kn ðxn  xn1 Þ  k nþ1 xn  cn x_n  x_n1  cnþ1 x_n þ Fn which can be rearranged and written as 9 m1€x1 þ ðc1 þ c2 Þx_ 1  c2 x_ 2 þ ðk 1 þ k2 Þx1  k 2 x2 ¼ F1 > > > > > m2€x2 þ ðc2 þ c3 Þx_ 2  c2 x_1  c3 x_ 3 þ ðk 2 þ k3 Þx2  k 2 x1  k 3 x3 ¼ F2 > > > > = ⋮ mi €xi þ ðci þ ciþ1 Þx_i  ci x_ i1  ciþ1 x_ iþ1 þ ðki þ kiþ1 Þxi  ki xi1  kiþ1 xiþ1 ¼ Fi > > > > > > ⋮ > > > ; mn€xn þ ðcn þ cnþ1 Þx_ n  cn x_ n1 þ ðkn þ knþ1 Þxn  k n xn1 ¼ Fn ð3:6Þ These equations can be written in a matrix form as M€x þ C_x þ Kx ¼ F, where 2

m11

6 6 m21 M ¼6 6⋮ 4 mn1

m12 m22 ⋮ mn2

   m1n

3

2

m1

7 6    m2n 7 6 0 7¼6 6 ⋱ ⋮ 7 5 4⋮ 0    mnn

0

0



m2

0



⋮

⋮

⋱

0

0



2

3 2 k11 k12 k 13   k1n k1 þ k2 k2 0  6 k k k   k 7 6 k k 2 þ k3 k3  2n 7 2 6 21 22 23 6 K ¼6 7¼6 4⋮ ⋮ ⋮ ⋱ ⋮ 5 4 ⋮ ⋮ ⋮ ⋱ kn1 kn2 k n3   knn

0

0

0

0

3

7 0 7 7 ⋮ 7 5 mn 0 0 ⋮

 k n þ knþ1

3 7 7 7 5

3.1 Equations of Motion

2

111

c11 6c 6 21 C ¼6 4⋮

c12 c22

c13 c23

⋮

⋮

3 2 c1 þ c2    c1n 7 6    c2n 7 6 c2 7¼6 ⋱ ⋮ 5 4 ⋮

cn1

cn2

cn3

   cnn

0

3

c2 c 2 þ c3

0 c3

 

0 0

⋮

⋮

⋱

⋮

0

0

   cn þ cnþ1

7 7 7 5

The differential equations of motion of Eq. 3.6 obtained from the application of Newton’s second law can also be obtained using Lagrange’s equation. To this end, the system kinetic energy, strain energy, and the virtual work of nonconservative forces are defined as 9 1 1 1 1 > T ¼ m1 x_21 þ m2 x_ 22 þ    þ mi x_ 2i þ    þ mn x_ 2n > > > 2 2 2 2 > > > > 1 2 1 1 1 2 2 2> > U ¼ k1 x1 þ k2 ðx2  x1 Þ þ    þ ki ðxi  xi1 Þ þ    þ knþ1 xn > = 2 2 2 2 > δW ¼ F1 δx1 þ F2 δx2 þ    þ Fi δxi þ    þ Fn δxn  c1 x_1 δx1 > > > >     > > >  c2 x_2  x_1 δðx2  x1 Þ      ci x_i  x_ i1 δðxi  xi1 Þ > > > ;      cnþ1 x_ n δx_ n

ð3:7Þ

For this multi-degree of freedom system, Lagrange’s equation can be stated as   d ∂T ∂T ∂U þ ¼ Qi ,  dt ∂x_i ∂xi ∂xi

i ¼ 1,2, . . . , n

ð3:8Þ

where Qi is the generalized force associated with the coordinate xi. In order to determine Qi, the virtual work δW can be written as δW ¼ ½F1  ðc1 þ c2 Þx_1 þ c2 x_ 2 δx1 þ ½F2  ðc2 þ c3 Þx_ 2 þ c2 x_1 þ c3 x_ 3 δx2 þ    þ½Fi  ðci þ ciþ1 Þx_ i þ ci x_ i1 þ ciþ1 x_ iþ1 δxi þ    þ½Fn  ðcn þ cnþ1 Þx_ n þ cn x_ n1 δxn ð3:9Þ from which, one can define the following generalized forces: Q1 ¼ F1  ðc1 þ c2 Þx_1 þ c2 x_ 2 Q2 ¼ F2  ðc2 þ c3 Þx_2 þ c2 x_1 þ c3 x_ 3

9 > > > > > > > > =

⋮ Qi ¼ Fi  ðci þ ciþ1 Þx_i þ ci x_i1 þ ciþ1 x_iþ1 > > > > > > ⋮ > > ; Qn ¼ Fn  ðcn þ cnþ1 Þx_n þ cn x_ n1

ð3:10Þ

112

3

Multi-degree of Freedom Systems

One can also verify that   d ∂T ∂T ¼ mi €xi , i ¼ 1,2,. . . ,n  dt ∂x_i ∂xi

ð3:11Þ

and ∂U ¼ ðk1 þ k2 Þx1  k 2 x2 ∂x1

9 > > > > > > > > > > > > > > > > > =

∂U ¼ ðk2 þ k3 Þx2  k 2 x1  k 3 x3 ∂x2 ⋮ > ∂U > ¼ ðki þ k iþ1 Þxi  k i xi1  kiþ1 xiþ1 > > > > ∂xi > > > > ⋮ > > > > > ∂U > > ¼ ðkn þ knþ1 Þxn  kn xn1 ; ∂xn

ð3:12Þ

Substituting Eqs. 3.10, 3.11, and 3.12 into Eq. 3.8 leads to the same set of n secondorder differential equations of Eq. 3.6. Angular Oscillations Figure 3.3a shows a set of n masses which are rigidly connected to massless rods which have length l. The rods are pinned at their ends as shown in the figure. Let T ¼ [T1 T2    Tn]T be the vector of external torques that act on the rods. By using the free-body diagram shown in Fig. 3.3b and applying Newton’s second law or D’Alembert’s principle with the assumption of small angular oscillations, one can verify that the differential equations of motion of the masses are given by   m1 l2 θ€1 ¼ m1 glθ1  k 1 l2 θ1 þ k 2 l2 ðθ2  θ1 Þ  c1 l2 θ_1 þ c2 l2 θ_2  θ_1 þ T 1     m2 l2 θ€2 ¼ m2 glθ2  k 2 l2 ðθ2  θ1 Þ þ k 3 l2 ðθ3  θ2 Þ  c2 l2 θ_2  θ_1 þ c3 l2 θ_3  θ_2 þ T 2

9 > > > > > > > > > =

⋮     mi l2 θ€i ¼ mi glθi  k i l2 ðθi  θi1 Þ þ k iþ1 l2 ðθiþ1  θi Þ  ci l2 θ_i  θ_i1 þ ciþ1 l2 θ_iþ1  θ_i þ T i > > > ⋮

  mn l2 θ€n ¼ mn glθn  k n l2 ðθn  θn1 Þ  k nþ1 l2 θn  cn l2 θ_n  θ_n1  cnþ1 l2 θ_n þ T n

> > > > > > ;

ð3:13Þ These equations can be rearranged and written in a matrix form as Mθ€ þ C θ_ þ Kθ ¼ T, where θ is the vector of angular oscillations given by θ ¼ ½ θ1 θ2    θn T , and the mass, damping, and stiffness matrices, M, C, and K are given by

3.1 Equations of Motion

113

Fig. 3.3 Angular oscillations

2

m1 l2

6 6 0 6 M¼6 6 0 6 4 ⋮ 0 2

 

0

m2 l2 0   0 m3 l2   ⋮ ⋮ ⋱

0 0 ⋮

0

0

0

0

6 6 K¼6 6 4

7 7 7 7 7 7 5

  mn l2

ðc1 þ c2 Þl2 c2 l2 0 6 2 ðc2 þ c3 Þl2 c3 l2 6 c2 l 6 C¼6 ðc3 þ c4 Þl2 0 c3 l2 6 6 4 ⋮ ⋮ ⋮ 0 0 0 2

3

     

0 0 0

⋱ ⋮   ðcn þ cnþ1 Þl2

 ðk 1 þ k2 Þl2 þ m1 gl k 2 l2 0   2 ðk2 þ k 3 Þl2 þ m2 gl k3 l2 k2 l   ðk3 þ k4 Þl2 þ m3 gl 0 k 3 l2 ⋮ ⋮ ⋮ 0 0 0

3 7 7 7 7 7 7 5

3  0 7  0 7 7  0 7 5 ⋱  ⋮     ðkn þ knþ1 Þl2 þ mn gl

In the case of angular oscillations, appropriate units must be used for the inertia, damping, and stiffness coefficients. For example, the units for the inertia coefficients

114

3

Multi-degree of Freedom Systems

must be kilogramsquare meters (kgm2), for the damping coefficients newton metersseconds (Nms), and for the stiffness coefficients newtonmeters (Nm). The differential equations of motion of the system shown in Fig. 3.3a can also be obtained by applying Lagrange’s equation. The system kinetic and strain energies and the virtual work in the case of small oscillations are defined as n 1 1 1 1 1X T ¼ m1 l2 θ_12 þ m2 l2 θ_22 þ    þ mi l2 θ_i2 þ    þ mn l2 θ_n2 ¼ m j l2 θ_j2 2 2 2 2 2 j¼1

9 > > > > > > > > > > > > > > > > > > > =

1 1 1 U ¼ k 1 l2 θ21 þ k 2 l2 ðθ2  θ1 Þ2 þ    þ ki l2 ðθi  θi1 Þ2 þ    2 2 2 1 2 1 þ k n l ðθn  θn1 Þ2 þ k nþ1 l2 θ2n 2 2 > > > > > δW ¼ T1 δθ1 þ T2 δθ2 þ    þ Ti δθi þ    þ Tn δθn  c1 l2 θ_1 δθ1 > >     > 2 _ 2 >  c2 l θ 2  θ_1 δðθ2  θ1 Þ      ci l θ_i  θ_i1 δðθi  θi1 Þ     > > > > > n > X     > > 2 _ 2 >  cn l θn  θ_n1 δ θ_n  θ_n1  cnþ1 l θ_n δθ_n  mi glθi δθi > ; i¼1

ð3:14Þ It is left to the reader to show that the use of the kinetic energy, strain energy, and virtual work given above in Lagrange’s equation leads to the same differential equations of motion obtained by applying Newton’s second law. Torsional Oscillations Figure 3.4a shows a set of n disks connected by massless shafts. The disk i in the system has mass moment of inertia Ii and is subjected to an

Fig. 3.4 Torsional oscillations

3.1 Equations of Motion

115

external torque Ti. Figure 3.4b shows the free-body diagrams of the disks. The application of Euler equations leads to the following n differential equations of motion: 9 I 1 θ€1 ¼ k 1 θ1 þ k 2 ðθ2  θ1 Þ þ T1 > > > > € I 2 θ2 ¼ k 2 ðθ2  θ1 Þ þ k3 ðθ3  θ2 Þ þ T2 > > > > > = ⋮ ð3:15Þ I i θ€i ¼ ki ðθi  θi1 Þ þ k iþ1 ðθiþ1  θi Þ þ Ti > > > > > > > ⋮ > > ; € I n θn ¼ k n ðθn  θn1 Þ  knþ1 θn þ Tn where θi is the torsional oscillation of the disk i and ki is the torsional stiffness of the massless shaft i defined as ki ¼ GiJi/li, in which Gi,Ji, and li are, respectively, the modulus of rigidity, the second moment of area, and the length of the shaft i. The preceding n differential equations can be written in the following matrix form: Mθ€ þ Kθ ¼ T, where θ ¼ ½ θ 1 θ 2    θ n T , T T ¼ [T1 T2    Tn] , 2 3 I1 0 0    0 6 7 6 0 I2 0    0 7 6 7 M¼6 7 4⋮ ⋮ ⋮ ⋱ ⋮5 0 0 0    In and 2

k1 þ k2

6 k 2 6 6 6 K¼6 0 6 4 ⋮ 0

k 2

0



0

k2 þ k3

k3



0

k 3

k3 þ k4



0

⋮

⋮

⋱

⋮

0

0



kn þ k nþ1

3 7 7 7 7 7 7 5

The differential equations of motion for the torsional system shown in Fig. 3.4a can also be obtained by using Lagrange’s equations. For this system, the kinetic energy, strain energy, and virtual work are defined as 1 1 1 1 T ¼ I1 θ_21 þ I 2 θ_22 þ    þ Ii θ_2i þ    þ I n θ_2n 2 2 2 2

9 > > > > > =

1 1 1 1 U ¼ k1 θ21 þ k2 ðθ2  θ1 Þ2 þ    þ k i ðθiþ1  θi Þ2 þ    þ knþ1 θ2n > > > 2 2 2 2 > > ; δW ¼ T1 δθ1 þ T2 δθ2 þ    þ Ti δθi þ    þ Tn δθn

ð3:16Þ

116

3

Multi-degree of Freedom Systems

These scalar energy and work expressions can be used with Lagrange’s equation in order to obtain the n second-order differential equations of motion of the n-degree of freedom torsional system shown in Fig. 3.4a. Example 3.1

The torsional system shown in Fig. 3.5 consists of three disks which have mass moment of inertia, I1 ¼ 2.0  103 kgm2, I2 ¼ 3.0  103 kgm2, and I3 ¼ 4.0  103 kgm2. The stiffness coefficients of the shafts connecting these disks are k1 ¼ 12  105 Nm, k2 ¼ 24  105 Nm, and k3 ¼ 36  105 Nm. The matrix equation of motion of the free vibration of this system is given by Mθ€ þ Kθ ¼ 0 where, in this example, M,K, and θ are given by 2

0

I1

6 M¼ 40 2

0

2

3

1:0

0

I2

6 7 0 5 ¼ 2  103 4 0

1:5

0

I3

0

0

k1

k1

6 K ¼ 4 k1 0 θ ¼ ½ θ1

0

θ2

0

k1 þ k2

k2

k2

k2 þ k3

3

0

3

7 0 5kgm2 2 2

1

6 7 5 ¼ 12  105 4 1 0

1 3

0

3

7 2 5Nm

2

5

T

θ3 

in which θ1, θ2, and θ3 are the torsional oscillations of the disks.

Fig. 3.5 Three-degree of freedom torsional system

Formulation of the Stiffness and Flexibility Matrices In this section, the stiffness matrix was obtained as the result of application of Newton’s second law or Lagrange’s equation. There are other techniques which can also be used to formulate the stiffness matrix. One of these methods utilizes the definition of the stiffness coefficients. In order to demonstrate the use of this method, let us consider the case

3.1 Equations of Motion

117

of static analysis, in which the equations derived in this chapter reduce to Kq ¼ Q, where K is the stiffness matrix, q is the vector of generalized coordinates, and Q P nis the vector of generalized forces. This equation can also be written as j¼1 k ij q j ¼ Qi , which can be written explicitly as ki1q1 + ki2q2 +    + kiiqi +    + kinqn ¼ Qi, where kij, i, j ¼ 1,2,  ,n are the elements of the stiffness matrix. Note that the stiffness coefficient kii is the force resulting from a unit displacement of the coordinate i while holding all other coordinates equal to zero. The stiffness coefficient kij(i 6¼ j), on the other hand, is the force associated with the coordinate i as the result of a unit displacement of the coordinate j while holding all other coordinates equal to zero. Since the stiffness matrix is symmetric, it is also clear that kij is the force associated with the coordinate j as the result of a unit displacement of the coordinate i. It is left to the reader to try to use this approach to develop the stiffness matrices of the systems presented in this section. The flexibility matrix is defined to be the inverse of the stiffness matrix. The 1 equation for the static equilibrium can be written as q ¼ KP Q, where K1 is n 1 fij Q j ¼ fi1 Q1 þ the flexibility matrix. The equation q ¼ K Q yields qi ¼ j¼1 fi2 Q2 þ    þ fii Qi þ    þ fin Qn , where fij are the flexibility coefficients. Note that the flexibility coefficient fii is equal to the displacement of the coordinate i as the result of the application of a unit load Qi while Qj( j 6¼ i) are all zeros. The flexibility coefficient fij is the displacement of the coordinate j as the result of the application of a unit load Qi ¼ 1 associated with the coordinate i while Qj( j 6¼ i) are all equal to zero. The flexibility matrix is symmetric, and consequently, fij is also equal to the displacement of the coordinate i as the result of the application of a unit load Qj, while all other forces are equal to zero. The flexibility matrix can be found using the unit load approach or by the direct inversion of the stiffness matrix. Use of the methods discussed in this section for formulating the stiffness and the flexibility matrices can be demonstrated using the system shown in Fig. 3.6. If the mass m1 is given a unit displacement while holding the mass m2 fixed, the force acting on the mass m1 is given by k11 ¼ k1 + k2, while the force acting on the mass m2 as the result of a unit displacement of the mass m1 is k12 ¼ k2. Similarly, if the mass m2 is given a unit displacement while holding the mass m1 fixed, the force acting on the mass m1 as the result of this displacement is k21 ¼ k2, and the force acting on m2 is k22 ¼ k2. Therefore, the stiffness matrix K is given by

Fig. 3.6 Evaluation of the stiffness and flexibility coefficients

118

3



k þ k2 K¼ 1 k2

k2 k2

Multi-degree of Freedom Systems



In order to determine the elements of the flexibility matrix, a unit load Q1 ¼ 1 is first applied statically at the mass m1, while the external force acting on the mass m2 is assumed to be equal to zero. In this case, the displacements of the two masses are equal and are given by f11 ¼ f21 ¼ 1/k1. In order to determine f12 and f22, a unit load Q2 ¼ 1 is applied statically to the mass m2 and Q1 is assumed to be zero. Using the concept of equivalent springs (Hartog 1968; Thomson 1988; Timoshenko et al. 1974; Shabana 2018B), it is clear that f22 ¼ (k1þk2)/k1k2, and f12 ¼ 1/k1. Therefore, the flexibility matrix K1 is given by " K

1

¼

1=k1

1=k 1

1=k1

ðk1 þ k2 Þ=k 1 k2

#

Note that KK1 ¼ K1K ¼ I, where I is the 2  2 identity matrix.

3.2

Undamped Free Vibration

The undamped free vibration of multi-degree of freedom systems is governed by the equation M€ q þ Kq ¼ 0. In a manner like that of the case of single degree of freedom systems, we assume a solution in the form q ¼ A sin ðωt þ ϕÞ

ð3:17Þ

where A is the vector of amplitudes, ω is the frequency, and ϕ is the phase angle. Differentiation of Eq. 3.17 twice with respect to time leads to € ¼ ω2 A sin ðωt þ ϕÞ q

ð3:18Þ

Substituting Eqs. 3.17 and 3.18 into the free-vibration equation M€ q þ Kq ¼ 0, one obtains ω2MAsin (ωt þ ϕ)þKAsin (ωt þ ϕ) ¼ 0, which leads to KA  ω2 MA ¼ 0

ð3:19Þ

Natural Frequencies Equation 3.19, which is an eigenvalue problem, can be considered as a system of homogeneous equations in the vector of unknown amplitudes A. This equation can be written in the following form: 

 K  ω2 M A ¼ 0

ð3:20Þ

This equation has a nontrivial solution if and only if the coefficient matrix is singular, that is,

3.2 Undamped Free Vibration

119



K  ω2 M ¼ 0

ð3:21Þ

This equation is called the characteristic equation and can be written in a more explicit form as

k11  ω2 m11

k  ω2 m

21 21

k31  ω2 m31



⋮



kn1  ω2 mn1

k12  ω2 m12

k13  ω2 m13

k22  ω2 m22

k23  ω2 m23

k32  ω2 m32

k33  ω2 m33

⋮

⋮

kn2  ω mn2

kn3  ω2 mn3

2

   k 1n  ω2 m1n

   k 2n  ω2 m2n



   k 3n  ω2 m3n

¼ 0



⋱ ⋮



   k nn  ω2 mnn

ð3:22Þ

The above equation leads to a polynomial of order n in ω2, and as such, the term of the highest order in this polynomial is (ω2)n. The roots of this polynomial denoted as ω21 , ω22 , . . . , ω2n are called the characteristic values or the eigenvalues. If the mass matrix M is positive definite and the stiffness matrix K is either positive definite or positive semidefinite, the characteristic values ω21 , ω22 , . . . , ω2n are real nonnegative numbers. The square roots of these numbers, ω1,ω2. . .,ωn, are called the natural frequencies of the undamped multi-degree of freedom system. Thus, a system with n degrees of freedom has n natural frequencies. Mode Shapes Associated with each characteristic value ωi, there is an ndimensional vector called the characteristic vector or the eigenvector Ai which can be obtained by using Eq. 3.20 as follows: 

 K  ω2i M Ai ¼ 0

ð3:23Þ

This is a system of homogeneous algebraic equations with a singular coefficient matrix since ω2i is one of the roots of the polynomial resulting from Eq. 3.22. Therefore, Eq. 3.23 has a nontrivial solution which defines the eigenvector Ai to within an arbitrary constant. The eigenvector (amplitude) Ai is sometimes referred to as the ith mode shape, normal mode, or principal mode of vibration. As a generalization to the procedure used for the case of the two degree of freedom systems (Thomson 1988; Shabana 2018B), the general solution in the case of undamped free vibration of the multi-degree of freedom system can be written as q ¼ α1 A1 sin ðω1 t þ ϕ1 Þ þ α2 A2 sin ðω2 t þ ϕ2 Þ þ    þ αn An sin ðωn t þ ϕn Þ n X ð3:24Þ ¼ αi Ai sin ðωi t þ ϕi Þ i¼1

where αi, and ϕi, i ¼ 1,2,  , n, are 2n arbitrary constants which can be determined from the initial conditions.

120

3

Multi-degree of Freedom Systems

Initial Conditions By differentiating Eq. 3.24 with respect to time, one obtains q_ ¼

n X

αi ωi Ai cos ðωi t þ ϕi Þ

ð3:25Þ

i¼1

Let q0 and q_ 0 be, respectively, the vectors of initial displacements and velocities. Substituting these initial conditions into Eqs. 3.24 and 3.25, one obtains q0 ¼

n X

αi Ai sin ϕi ,

q_ 0 ¼

i¼1

n X

αi ωi Ai cos ϕi

ð3:26Þ

i ¼ 1,2,. . . , n

ð3:27Þ

i¼1

One can define the following two constants: ai ¼ αi sin ϕi ,

bi ¼ αi cos ϕi ,

In terms of these constants, q0 and q_ 0 of Eq. 3.26 can be written as q0 ¼

n X

ai Ai ,

q_ 0 ¼

n X

i¼1

bi ωi Ai

ð3:28Þ

i¼1

which can be written in a matrix form as Φωb ¼ q_ 0

Φa ¼ q0 , where a ¼ ½ a1 matrices

   an  T , b ¼ ½ b1

a2 2

ω1

0

0



6 60 6 6 ω ¼6 0 6 6⋮ 4

ω2

0



0

ω3



⋮

⋮

⋱

0

0

0



Φ ¼ ½ A1

A2

0

b2

   bn T , and ω and Φ are the 9 > > > > > > > > > > > > > > > > > > > =

3

7 0 7 7 7 0 7 7 ⋮ 7 5 ωn 2

ð3:29Þ

A11 6 6 A12    An  ¼ 6 6⋮ 4

A21

A1n

A2n

A22 ⋮

> 3> >    An1 > > > > 7> >    An2 7 > > 7> > > > ⋱ ⋮ 7 5> > > > ;    Ann

ð3:30Þ

The matrix Φ, whose columns are the eigenvectors, is called the modal matrix. The two equations in Eq. 3.29 represent a system of 2n linear algebraic equations that can be solved for the vectors of unknowns a and b. If the number of these equations is large, numerical techniques can be used on digital computers to solve this system of equations. Once the vectors of the 2n unknowns a ¼ ½ a1 a2    anT

3.2 Undamped Free Vibration

121

and b ¼ ½ b1 b2    bn T are determined, the constants αi and ϕi of Eq. 3.27 can be calculated from the following relationships: αi ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2i þ b2i ,

ai ϕi ¼ tan1 , bi

i ¼ 1,2, . . . , n

ð3:31Þ

These constants can be substituted into Eqs. 3.24 and 3.25 in order to determine the displacements and velocities as functions of time. Example 3.2

In the torsional system of Example 3.1, the symmetric mass and stiffness matrices were given by 2

I1

3

2

0

0

6 M ¼4 0

I2

7 6 0 5 ¼ 2  103 4 0

0

0

I3

2

k1

k1

6 K ¼ 4 k1

0

k1 þ k2 k2

0

3

1

0

0 1:5 0

0

3

7 0 5kgm2 2

2

1

1

0

0

2

5

7 6 k2 5 ¼ 12  105 4 1 k2 þ k3

3

7 3 2 5Nm

and the governing equation of motion of the free vibration is Mθ€ þ Kθ ¼ 0. A solution in the form θ ¼ Asin (ωt+ϕ) is assumed. By substituting this assumed solution into the differential equation, one obtains 

 K  ω2 M A ¼ 0

Substituting for the mass and stiffness matrices, one gets 2

2

1 46  105 4 1 0

3 2 1 0 1 0 3 2 5  ω2  103 4 0 1:5 2 5 0 0

33 0 0 55A ¼ 0 2

which can be rewritten as 22

1 44 1 0

1 3 2

3 2 0 1 2 5  β4 0 5 0

33 0 0 1:5 0 55A ¼ 0 0 2

This system has a nontrivial solution if the determinant of the coefficient matrix is equal to zero. This leads to the following characteristic equation: (continued)

122

3

Multi-degree of Freedom Systems

β3  5:5β2 þ 7:5β  2 ¼ 0 which has the following roots: β1 ¼ 0:3516,

β2 ¼ 1:606,

β3 ¼ 3:542

Since ω2 ¼ 600β, the natural frequencies associated with the roots β1, β2, and β3 are given, respectively, by ω1 ¼ 14:52 rad=s, ω2 ¼ 31:05 rad=s,

ω3 ¼ 46:1 rad=s

For a given root βi, i ¼ 1,2,3, the mode shapes can be determined using the equation 2

1  βi 6 4 1 0

32

3 2 3 Ai1 0 76 7 6 7 54 Ai2 5 ¼ 4 0 5

1 3  1:5βi

0 2

2

5  2βi

0

Ai3

which, by partitioning the coefficient matrix, leads to 

 1 ð3  1:5βi Þ Ai1 þ 0 2

2 5  2βi



Ai2 Ai3



 0 ¼ 0

or 

Ai2



Ai3

 5  2βi 1 ¼ 2 Ai1 2 3βi  13:5βi þ 11

Using the values obtained previously for βi, i ¼ 1,2,3, one has " " "

A12 A13 A22 A23 A32 A33

#

" ¼

#

" ¼

#

" ¼

0:649

#

0:302

A11

0:607 0:679 2:54 2:438

for

ω1 ¼ 14:52 rad=s

# A21

for

ω2 ¼ 31:05 rad=s

A31

for

ω3 ¼ 46:1 rad=s

#

Since the mode shapes are determined to within an arbitrary constant, we may assume Ai1 ¼ 1, for i ¼ 1,2,3. This leads to the following mode shapes: (continued)

3.3 Orthogonality of the Mode Shapes

123

3 3 3 3 2 2 3 2 2 3 2 1 1 1 A11 A21 A31 A1 ¼ 4 A12 5 ¼ 4 0:649 5, A2 ¼ 4 A22 5 ¼ 4 0:607 5, A3 ¼ 4 A32 5 ¼ 4 2:54 5 0:302 0:679 2:438 A13 A23 A33 2

The modal matrix Φ is then defined as 2

1 1 Φ ¼ 4 0:649 0:607 0:302 0:679

3 1 2:54 5 2:438

Figure 3.7 shows the modes of vibration of the system.

Fig. 3.7 Mode shapes

3.3

Orthogonality of the Mode Shapes

An important property of the mode shapes is the orthogonality. This property guarantees the existence of a unique solution to the system of algebraic equations of Eq. 3.29. In this section, we discuss this important property which can be used to obtain a set of n decoupled differential equations of motion for the multi-degree of freedom systems. These decoupled equations are expressed in terms of a new set of coordinates called modal coordinates defined using the modal matrix which is sometimes referred to as the modal transformation.

124

3

Multi-degree of Freedom Systems

For the ith and jth natural frequencies ωi and ωj and the ith and jth mode shapes Ai and Aj, Eq. 3.23 can be written as KAi ¼ ω2i MAi ,

KA j ¼ ω2j MA j

ð3:32Þ

Premultiplying the first equation by the transpose of the vector Aj and taking the transpose of the second equation and postmultiplying the resulting equation by the vector Ai, one obtains, respectively, ATj KAi ¼ ω2i ATj MAi ,

ATj KT Ai ¼ ω2j ATj MT Ai

ð3:33Þ

Since the mass and stiffness matrices are symmetric, the second equation leads to ATj KAi ¼ ω2j A Tj MAi . Subtracting this equation from the first equation of Eq. 3.33 yields

ω2i  ω2j ATj MAi ¼ 0 ð3:34Þ If ω2i and ω2j are distinct eigenvalues, that is, ω2i 6¼ ω2j , and M is a positive definite matrix, Eq. 3.34 implies that ATj MAi ¼ 0 ¼ mi 6¼ 0

for

i 6¼ j

for

i¼j

ð3:35Þ

where mi is a real positive scalar. That is, the eigenvectors associated with distinct eigenvalues are orthogonal with respect to the mass matrix. The positive definiteness of the mass matrix guarantees that ATi MAi is not equal to zero for the nonzero vector Ai. The mode shapes are also orthogonal with respect to the stiffness matrix. This can be proved by writing Eq. 3.32 in the following alternate form: MAi ¼

1 KAi , ω2i

MA j ¼

1 KA j ω2j

ð3:36Þ

Following a procedure similar to the one used to prove the orthogonality of the mode shapes with respect to the mass matrix, one can verify that ATj KAi ¼ 0

for

i 6¼ j

¼ ki

for

i¼j

ð3:37Þ

where ki is a nonnegative scalar. Linear Independence of the Mode Shapes The orthogonality with respect to the mass or stiffness matrix can be used to show that the mode shapes of a multi-degree of freedom system are linearly independent, and as a consequence, any of these mode shapes cannot be written as a linear combination of the others. The linear independence is an important property which guarantees that the modal matrix of

3.3 Orthogonality of the Mode Shapes

125

Eq. 3.30 has a full rank and accordingly a solution of the algebraic equations of Eq. 3.29 does exist. The set of vectors A1, A2,   , An are said to be linearly independent, if the following relationship, β1 A1 þ β2 A2 þ    þ βi Ai þ    þ βn An ¼ 0

ð3:38Þ

where βi, i ¼ 1,2,. . ., n, are scalars holds only when the scalars β1, β2,    βn are all identically equal to zero. In order to prove that this is indeed the case, we premultiply Eq. 3.38 by ATi M and use the orthogonality conditions of Eq. 3.35 to obtain βi AiT MAi ¼ βi mi ¼ 0. Since mi is not equal to zero, we conclude that βi ¼ 0 for any i ¼ 1,2,. . ., n. That is, the mode shapes of the multi-degree of freedom system are indeed linearly independent. Accordingly, the modal matrix Φ of Eq. 3.30 has a full rank and is thus nonsingular. In proving the orthogonality conditions, it was assumed that the eigenvalues ω2i and ω2j are distinct. In some engineering applications, however, the case of repeated roots, in which two or more eigenvalues are equal, may be encountered. For a general eigenvalue problem, the eigenvectors associated with repeated roots may or may not be linearly independent. Let ω2r be an eigenvalue with multiplicity s, thatis,  ω2r , ω2rþ1 , . . . , ω2rþs1 are equal eigenvalues. If the rank of the matrix K  ω2r M is equal to n  s where n is the  number of the system coordinates, it can be proved that linearly the system of equations K  ω2r M Ar ¼ 0 has s nontrivial   independent eigenvectors Ar, Ar+1,. . .,Ar+s1. If the rank of the matrix K  ω2r M is greater than n  s, the vectors Ar, Ar+1,. . .,Ar+s1 associated with the repeated eigenvalues are not totally independent, and in this case, the number of linearly independent eigenvectors is less than the number of the system coordinates. However, if the mass matrix M and the stiffness matrix K are real symmetric, it can be shown that the eigenvectors associated with repeated eigenvalues are linearly independent (Wylie and Barrett 1982). Modal Transformation The orthogonality of the mode shapes can be used to obtain a set of n uncoupled second-order differential equations in terms of a new set of coordinates called modal coordinates. Each of the resulting equations is similar to the equation of the single degree of freedom system. To demonstrate this fact, consider the matrix equation of motion of the undamped free vibration of the multi-degree of freedom system given by M€ q þ Kq ¼ 0. One can write the following coordinate transformation: 2

ϕ11 6ϕ 6 21 q ¼ ΦP ¼ 6 4⋮

ϕ12 ϕ22

ϕ13 ϕ23

32 3 P1    ϕ1n 7 6    ϕ2n 76 P2 7 7 76 7 ⋮ 54 ⋮ 5

ϕn1

ϕn2

ϕn3

   ϕnn

Pn

ð3:39Þ

126

3

Multi-degree of Freedom Systems

Pn or equivalently, qj ¼ i¼1 ϕ ji Pi , where Φ is the modal transformation defined by Eq. 3.30, and P is the vector of modal coordinates. Differentiating Eq. 3.39 twice € þ KΦP ¼ 0. with respect to time and substituting into M€ q þ Kq ¼ 0 yields MΦP T Premultiplying this equation by Φ , one obtains € þ ΦT KΦP ¼ 0 ΦT MΦP

ð3:40Þ

Using the orthogonality of the mode shapes with respect to the mass and stiffness matrices (Eqs. 3.35 and 3.37), the above equation yields € þ Kp P ¼ 0 Mp P

ð3:41Þ

where 2 6 6 6 6 Mp ¼ ΦT MΦ ¼ 6 6 6 4

0

0



0

0

AT2 MA2

0



0

0

0

AT3 MA3



0

⋮

⋮

⋮

⋱

⋮

0



ATn MAn

0

2

m1 60 6 ¼6 60 4⋮ 0

0 m2 0 ⋮ 0

3

AT1 MA1

0 0 m3 ⋮ 0

   ⋱ 

03

7 7 7 7 7 7 7 5

ð3:42Þ

0 0 7 7 0 7 7 ⋮ 5 mn

and 2

AT1 KA1

6 60 6 T Kp ¼ Φ KΦ ¼ 6 60 6 4⋮ 2

k1 6 60 6 ¼6 60 6 4⋮ 0

0 k2

0 0 0

 

0 ⋮

k3 ⋮

 ⋱

0

0



0

0

 0

AT2 KA2 0

0 AT3 KA3

 0  0

⋮

⋮

⋱

0 3 0 7 0 7 7 0 7 7 7 ⋮5 kn

0

   ATn KAn

⋮

3 7 7 7 7 7 7 5 ð3:43Þ

Note that Mp and Kp are diagonal matrices. The matrix Mp is called the modal mass matrix, and the matrix Kp is called the modal stiffness matrix. The scalars mi and ki

3.3 Orthogonality of the Mode Shapes

127

defined by Eqs. 3.35 and 3.37 are called, respectively, the modal mass and stiffness coefficients. Owing to the fact that the modal mass and stiffness matrices Mp and Kp are diagonal, the modal coordinates in Eq. 3.41 are not coupled. That is, Eq. 3.41 contains n uncoupled second-order ordinary differential equations which can be written as € i þ ki Pi ¼ 0, i ¼ 1,2,. . . ,n mi P

ð3:44Þ

This equation is similar to the equation of motion of the undamped single degree of freedom system. Furthermore, it is clear from Eq. 3.32 that the natural frequency ωi can be obtained using the equation ω2i ¼

ATi KAi ki ¼ T Ai MAi mi

ð3:45Þ

That is, the square of the natural frequency of a mode is equal to the modal stiffness divided by the modal mass—a relationship which is similar to the definition used in the case of single degree of freedom systems. The solutions of the differential equations of Eq. 3.44 take the form Pi ¼ Ci sin ðωi t þ ψ i Þ,

i ¼ 1,2,. . . ,n

ð3:46Þ

in which Ci is the amplitude and ψ i is the phase angle. The constants Ci and ψ i can be determined from the initial conditions on the modal coordinates. These initial conditions can be obtained using the transformation of Eq. 3.39 as P0 ¼ Φ1q0, where P0 is the vector of the initial values of the modal coordinates, q0 is the vector of the given initial displacements, and Φ1 is the inverse of the modal matrix. Similarly, P_ 0 ¼ Φ1 q_ 0 . Thus, the initial modal coordinates and velocities can be determined if the initial displacements and velocities are given. The constants Ci and ψ i can be expressed in terms of the initial modal coordinates and velocities as sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 P_i0 þ ðPi0 Þ2 , Ci ¼ ωi

ψ i ¼ tan 1

ωi Pi0 P_ i0

ð3:47Þ

The jth physical coordinate can then be determined using Eqs. 3.39 and 3.46 as qj ¼

n X i¼1

ϕ ji Pi ¼

n X ϕ ji Ci sin ðωi t þ ψ i Þ

ð3:48Þ

i¼1

which is the same solution obtained in the preceding section and defined by Eq. 3.24, in which the coefficients of the mode shapes are the modal coordinates defined by Eq. 3.46.

128

3

Multi-degree of Freedom Systems

Inverse of the Modal Matrix In this section, the initial modal coordinates and velocities are expressed in terms of the inverse of the modal matrix. The orthogonality of the modal matrix with respect to the mass or stiffness matrix can be used to define the inverse of the modal matrix. From Eq. 3.42, one has ΦTMΦ ¼ Mp, where Mp is a diagonal matrix whose inverse can be easily obtained. Premultiplying T this equation by the inverse of the matrix Mp, one obtains M1 p Φ MΦ ¼ I, where I is the n  n identity matrix. By postmultiplying by Φ1, one obtains T 1 M1 ¼ Φ1 , or p Φ MΦΦ T Φ1 ¼ M1 p Φ M

ð3:49Þ

This equation is useful in defining the initial modal coordinates and velocities. Normalized Mode Shapes In many situations, it is desirable to normalize the mode shapes with respect to the mass matrix or with respect to the stiffness matrix. In order to normalize the mode shapes with respect to the mass matrix, each mode is divided by the square root of the corresponding modal mass coefficient, that is, Ai Ai Aim ¼ pffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , mi AiT MAi

i ¼ 1,2,. . . ,n

ð3:50Þ

where Aim is the mode shape i normalized with respect to the mass matrix. It is clear that ATim MAim ¼ ATi MAi =mi ¼ 1, or ΦmT MΦm ¼ I

ð3:51Þ

where I is an n  n identity matrix, and Φm is the modal matrix whose columns are the mode shapes normalized with respect to the mass matrix. Similarly, one can write ATim KAim ¼ ATi KAi =mi ¼ k i =mi ¼ ω2i . It follows that ΦmT KΦm ¼ ωm

ð3:52Þ

where ωm is a diagonal matrix whose diagonal elements are the square of the system natural frequencies. This matrix is defined as 2

ω21

60 6 6 ωm ¼ 6 60 6 4⋮ 0

3

0

0



ω22

0



0

ω23



⋮

⋮

⋱

0 7 7 7 0 7 7 7 ⋮ 5

0

0



ω2n

0

ð3:53Þ

3.3 Orthogonality of the Mode Shapes

129

The mode shapes can also be normalized with respect to the stiffness matrix. To this end, each mode shape is divided by the square root of the corresponding modal stiffness coefficient, that is, Ai Ai Ais ¼ pffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , i ¼ 1,2,. . . ,n ki AiT KAi

ð3:54Þ

where Ais is the ith mode shape normalized with respect to the stiffness matrix. Clearly, in this case, one has ATis KAis ¼ ATi KAi =k i ¼ 1, and ATis MAis ¼ ATi MAi =ki ¼ mi =ki ¼ 1=ω2i , and as a consequence, ΦsT KΦs ¼ I,

ΦsT MΦs ¼ ωs

ð3:55Þ

where Φs is the modal matrix whose columns are the mode shapes normalized with respect to the stiffness matrix, and ωs is the matrix 2

0

0



0

1=ω22 0

0 1=ω23

 

0 0

⋮ 0

⋮ 0

1=ω21

6 6 0 6 ωs ¼ 6 6 0 6 4 ⋮ 0

3

7 7 7 7 7 7 ⋱ ⋮ 5    1=ω2n

ð3:56Þ

The matrix Φm is said to be orthonormal with respect to the mass matrix, while the matrix Φs is said to be orthonormal with respect to the stiffness matrix. Example 3.3

In Example 3.2, the mass and stiffness matrices were given by 2

1 M ¼ 2  103 4 0 0

3 0 0 1:5 0 5kgm2 , 0 2

2

1 K ¼ 12  105 4 1 0

1 3 2

3 0 2 5Nm 5

The natural frequencies were found to be ω1 ¼ 14.52 rad/s, ω2 ¼ 31.05 rad/s, and ω3 ¼ 46.1 rad/s and the modal matrix 2

1:000

1:000

6 Φ ¼ 4 0:649 0:607 0:302 0:679

1:000

3

7 2:541 5 2:438

This modal matrix is orthogonal with respect to the mass and stiffness matrices. One can show the following: (continued)

130

3

2 6 Mp ¼ ΦT MΦ ¼ 2  103 4 2 6 Kp ¼ ΦT KΦ ¼ 2  105 4 2 6 ¼4

Multi-degree of Freedom Systems

1:814

0

0

0

2:475

0

0 3:828

0 0

0

23:86

0

0

ω21 m1

0

0

0

ω22 m2

0

0

0

ω23 m3

3

2

m1

0

7 6 5¼4 0

0

3

7 0 5

m2

22:573 0 0 m3 3 2 3 k1 0 0 0 7 6 7 0 5 ¼ 4 0 k2 0 5 3

497:7

0

0

k3

7 5

where mi and ki are, respectively, the modal mass and stiffness coefficients. The modal matrix can be normalized with respect to the mass matrix M or with respect to the stiffness matrix K. For example, in order to make the modal matrix orthonormal with respect to the stiffness matrix, for each mode shape Ai we use the following equation: Ai Ais ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi AiT KAi It follows that 2

1:000

3

2

0:1143

3

A1 1 6 7 6 7 A1s ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:649 5 ¼ 102 4 0:0742 5 24 8:7497  10 T A1 KA1 0:302 0:0345 2 3 2 3 1:000 0:0458 A2 1 6 7 6 7 A2s ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:607 5 ¼ 102 4 0:0278 5 24 21:845  10 T A2 KA2 0:679 0:0311 2 3 2 3 1:000 0:0102 A3 1 6 7 6 7 A3s ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2:541 5 ¼ 102 4 0:0259 5 24 97:949  10 T A3 KA3 2:438 0:0249 The new modal matrix which is orthonormal with respect to the stiffness matrix is then given by 2

0:1143 Φs ¼ 102 4 0:0742 0:0345

0:0458 0:0278 0:0311

3 0:0102 0:0259 5 0:0249 (continued)

3.3 Orthogonality of the Mode Shapes

131

One can show that ΦsT KΦs ¼ I where I is the 3  3 identity matrix, that is, the modal stiffness coefficients are all equal to 1. One can also show that 2 6 ΦsT MΦs ¼ 103 4

4:743

0

0 0

1:0372 0

2

3

0

1=ω21

7 6 0 5¼4 0 0:4705 0

0 1=ω22 0

0

3

7 0 5 1=ω23

Similarly, if the modal matrix is made orthonormal with respect to the mass matrix, one has ΦmT MΦm ¼ I and 2 6 ΦmT KΦm ¼ 103 4

0:2108

0

0

0:964

0

0

2

3

0

ω21

7 6 5¼4 0 2:1254 0 0

0 ω22 0

0

3

7 0 5 ω23

That is, the modal mass coefficient mi ¼ 1 for i ¼ 1,2,3 and the modal stiffness coefficient ki ¼ ω2i , i ¼ 1,2,3.

Example 3.4

In this example, we consider the free vibration of the system in the preceding example as the result of the initial conditions 2

0:1

3

6 7 θ0 ¼ 4 0:05 5rad, 0:01

2

10

3

6 7 θ_0 ¼ 4 15 5rad=s 20

The equations of motion of this system in terms of the modal coordinates are given by Eq. 3.44. The solution of these equations is given by Eq. 3.46, and the arbitrary constants are defined by Eq. 3.47. The initial modal coordinates and velocities can be obtained using the inverse of the modal matrix T Φ1 ¼ M1 p Φ M. By using this equation and the results of the preceding example, the inverse of the modal matrix Φ1 can be written as (continued)

132

3

2 6 Φ1 ¼ 4

0:2755

0

0

0

0:202

0

32

Multi-degree of Freedom Systems

1:000 0:649

0:302

32

0 0:551

0:044 0:169

0

3

76 76 7 54 1:000 0:607 0:679 54 0 3:0 0 5

0 0:02215 1:000 2:541 2:438 3 0:537 0:333 6 7 ¼ 4 0:404 0:368 0:549 5 2

2:0 0 0

0 4:0

0:216

The initial modal coordinates are given by 2

0:551

6 P0 ¼ Φ1 θ0 ¼ 4 0:404 0:044

0:537 0:368 0:169

0:333

32

0:1

3

2

0:085

3

76 7 6 7 0:549 54 0:05 5 ¼ 4 0:017 5rad 0:216 0:01 0:002

and the initial modal velocities are 2

0:551

0:537

6 P_ 0 ¼ Φ1 θ_0 ¼ 4 0:404

0:368

0:044

0:169

0:333

32

10

3

2

20:225

3

76 7 6 7 0:549 54 15 5 ¼ 4 12:460 5rad=s 0:216

20

2:228

Using the initial modal coordinates and velocities, the modal coordinates can be defined by using Eq. 3.46, or its equivalent form given by Pi ¼ Pi0 cos ωi t þ

P_i0 sin ωi t, i ¼ 1,2,3 ωi

which yields 2

P1

3

2

0:085cos14:52t þ 1:393sin14:52t

3

7 6 7 6 7 P ¼ 4 P2 5 ¼ 6 4 0:017cos31:05t  0:401sin31:05t 5 P3 0:002cos46:1t þ 0:048sin46:1t The physical coordinates θ can then be obtained using the relationship θ ¼ ΦP.

Special Case As was pointed out earlier, in order to determine the initial conditions of the modal coordinates P0 and P_ 0 , it is not necessary to obtain the inverse of the modal matrix. If the number of degrees of freedom is large, an LU factorization of the modal matrix can be used to solve for the initial modal coordinates and velocities. The orthogonality of the mode shapes with respect to the mass or stiffness matrices can also be used as an alternative to find the inverse of the modal matrix. In this later

3.4 Rigid-Body Modes

133

case, as shown previously, the inverse of the modal matrix can be expressed in terms T of the inverse of the diagonal modal mass matrix as Φ1 ¼ M1 p Φ M, where Mp is the diagonal matrix defined in Eq. 3.42. Therefore, the initial conditions associated with the modal coordinates can be obtained using the equations MpP0 ¼ ΦTMq0 and Mp P_ 0 ¼ ΦT Mq_ 0 . Using these equations, one can show that if the vectors of initial coordinate q0 and initial velocity q_ 0 are proportional to a given mode shape, then the multi-degree of freedom system will oscillate in this mode of vibration, a case which is equivalent to the vibration of a single degree of freedom system. In order to demonstrate this, let us consider the special case in which q0 and q_ 0 are proportional to the mode shape i. One can then write q0 and q_ 0 , respectively, as q0 ¼ d1Ai and q_ 0 ¼ d2 Ai , where d1 and d2 are proportionality constants. Therefore, the initial conditions associated with the modal coordinates can be expressed as MpP0 ¼ d1ΦTMAi and Mp P_ 0 ¼ d 2 ΦT MAi . Because of the orthogonality of the mode shapes with respect to the mass matrix, one can verify that the preceding equations yield  m j Pj0 ¼

d1 mj , 0,

j¼i , j 6¼ i

m j P_ j0 ¼

j¼i , j 6¼ i

P_ j0 ¼



d2 mj , 0,

j¼i j 6¼ i

ð3:57Þ

which yields  Pj0 ¼

d1 , 0,



d2 , 0,

j¼i j 6¼ i

ð3:58Þ

€j þ k j Pj ¼ 0, j ¼ 1,2,. . . , n, only one This implies that among the equations mj P equation which is associated with mode i has a nontrivial solution, and the vibration of the multi-degree of freedom system is indeed equivalent to the vibration of a single degree of freedom system. Therefore, every mode shape can be excited independently. Similarly, one can show that if the vector of initial coordinates or velocities is proportional to a linear combination of m mode shapes where m < n, only m mode shapes are excited and only m equations associated with m modal coordinates have nontrivial solutions. In this special case, the motion of the n-degree of freedom system is equivalent to the motion of an m-degree of freedom system.

3.4

Rigid-Body Modes

There are situations in which one or more of the eigenvalues may be equal to zero. This is the case of a semidefinite system, in which the eigenvector associated with the zero eigenvalue corresponds to a rigid-body mode of vibration. In a rigid-body mode, the system can move as a rigid body without deformation in the elastic elements. For a semidefinite system, the stiffness matrix is positive semidefinite and, consequently, its determinant is equal to zero. By using Eq. 3.45, one may € k þ ω2k Pk ¼ 0. If ωk ¼ 0, which is the case of a rewrite Eq. 3.44 for mode k as P

134

3

Multi-degree of Freedom Systems

€ k ¼ 0, which can be integrated to rigid-body mode, the preceding equation leads to P _ obtain the modal coordinate as Pk ¼ Pk0 t þ Pk0 , where Pk0 and P_ k0 are, respectively, the initial modal coordinates and velocities associated with the rigid-body mode. Observe that for the rigid-body mode k, Eq. 3.23 reduces to KAk ¼ 0. Since K is a singular matrix, the preceding system of equations has a nontrivial solution, and there exists a nonzero vector Ak such that AkT KAk ¼ 0. As a consequence, there is no change in the potential energy of the system as the result of the rigid-body motion. Observe also that the rank of the stiffness matrix must be equal to the number of degrees of freedom minus the number of the rigid-body modes of the system. Consequently, the equation KAk ¼ 0 has as many independent nontrivial solutions as the number of the rigid-body modes of the system. In general, the degree of singularity or the rank deficiency of the matrix K defines the number of dependent equations in the system KAk ¼ 0. The number of the remaining independent equations is less than the number of the elements of the vector Ak, and as such, these independent equations can be solved for only a number of unknown elements equal to the rank of the matrix. The other elements can be treated as independent variables that can be varied arbitrarily in order to define a number of independent solutions equal to (n  r), where n and r are, respectively, the dimension and rank of the stiffness matrix. It is important to emphasize at this point that while ATk KAk ¼ 0 for the rigid-body mode k, ATk MAk is not equal to zero. However, the equation ω2k ATk MAk ¼ ATk KAk

ð3:59Þ

is satisfied since ωk ¼ 0. Nonrigid-body modes are sometimes referred to as deformation modes. The complete solution of the equations of free vibration can thus be expressed as a linear combination of the rigid-body and deformation modes. For the purpose of demonstration, consider the four degree of freedom system shown in Fig. 3.8. The equations of motion of this system can be developed using the

Fig. 3.8 Semidefinite system

3.4 Rigid-Body Modes

135

principle of virtual work in dynamics discussed in the preceding chapter. For this system, the virtual work of the inertia and applied forces can be written in the case of small oscillations as )

δWi ¼ m1€x1 δx1 þ m2€x2 δx2 þ m3€x3 δx3 þ m4€x4 δx4

ð3:60Þ

δWe ¼ k1 x1 δx1  k2 ðx2  x1 Þδðx2  x1 Þ  m3 g δy3  m4 g δy4 where x3, x4, y3, and y4 in the case of small oscillations are given by 9 > =

x3 ¼ x1 þ l sin θ1  x1 þ lθ1 x4 ¼ x2 þ l sin θ2  x2 þ lθ2 y3 ¼ l cos θ1 ,

y4 ¼ l cos θ2

ð3:61Þ

> ;

The accelerations €x3 and €x4 in the case of small angular oscillations are €x3 ¼ €x1 þ lθ€1 ,

€x4 ¼ €x2 þ lθ€2

ð3:62Þ

Therefore, the virtual work of the inertia forces is given by   δWi ¼ m1€x1 δx1 þ m2€x2 δx2 þ m3 €x1 þ lθ€1 δðx1 þ lθ1 Þ   þ m4 €x2 þ lθ€2 δðx2 þ lθ2 Þ     ¼ ðm1 þ m3 Þ€x1 þ m3 lθ€1 δx1 þ ðm2 þ m4 Þ€x2 þ m4 lθ€2 δx2     þ m3 l €x1 þ lθ€1 δθ1 þ m4 l €x2 þ lθ€2 δθ2

ð3:63Þ

which can be written in a matrix form as 2 δWi ¼ ½ δx1

δx2

δθ1

6 6 δθ2 6 4

ðm1 þ m3 Þ

0

m3 l

0 m3 l

ðm2 þ m4 Þ 0

0 m 3 l2

0

m4 l

0

¼ δqT Qi

0

32

€x1

3

76 €x 7 76 2 7 76 € 7 54 θ 1 5 2 m4 l θ€2 m4 l 0

ð3:64Þ where q ¼ [x1 x2 θ1 θ2]T and Qi is the vector of generalized inertia forces given by 2 6 6 Qi ¼ 6 4

ðm1 þ m3 Þ

0

m3 l

0

ðm2 þ m4 Þ

0

m3 l

0

m3 l2

0

m4 l

0

The virtual work of the generalized applied forces is

0

32

€x1

3

6 7 m4 l 7 76 €x2 7 76 € 7 0 54 θ 1 5 m4 l2 θ€2

ð3:65Þ

136

3

Multi-degree of Freedom Systems

δWe ¼ k 1 x1 δx1  k2 ðx2  x1 Þδðx2  x1 Þ  m3 glθ1 δθ1  m4 glθ2 δθ2 ¼ ½ðk1 þ k2 Þx1  k 2 x2 δx1  ðk 2 x2  k 2 x1 Þδx2

ð3:66Þ

m3 glθ1 δθ1  m4 glθ2 δθ2 which can be written in a matrix form as 2

δWe ¼ ½ δx1

δx2

δθ1

ðk 1 þ k 2 Þ 6 k2 δθ2 6 4 0 0

k2 k2 0 0

0 0 m3 gl 0

¼ δqT Qe

32 3 0 x1 6 x2 7 0 7 76 7 0 54 θ1 5 ð3:67Þ m4 gl θ2

where Qe is the vector of generalized applied forces defined as 2

ðk 1 þ k 2 Þ 6 k2 Qe ¼ 6 4 0 0

k 2 k2 0 0

32 3 0 0 x1 6 x2 7 0 0 7 76 7 m3 gl 0 54 θ 1 5 0 m4 gl θ2

ð3:68Þ

The principle of virtual work in dynamics states that δWi ¼ δWe. Since the coordinates x1, x2, θ1, and θ2 are independent, one has as the result of the application of the principle of virtual work in dynamics Qi ¼ Qe, which leads to the differential equation M€ q þ Kq ¼ 0, where the mass matrix M and the stiffness matrix K are 2

3 2 3 ðm1 þ m3 Þ 0 m3 l 0 ðk1 þ k2 Þ k2 0 0 6 6 0 ðm2 þ m4 Þ 0 m4 l 7 k2 0 0 7 7, K ¼ 6 k 2 7 M ¼6 2 4 m3 l 5 4 0 0 m3 gl 0 5 0 m3 l 0 0 m4 l 0 0 0 m4 gl 0 m 4 l2 ð3:69Þ In this system, there are inertia coupling terms between the translational and rotational coordinates, while there are no elastic coupling terms. Note that the stiffness matrix is positive definite, and it is nonsingular. Therefore, the system shown in Fig. 3.8 has no rigid-body modes. Any possible motion leads to a change in the system potential energy which in this case is a positive definite quadratic from. Let us consider now the case in which the effect of gravity is not considered. In this case, the stiffness matrix becomes 2

ðk 1 þ k 2 Þ 6 k2 K¼6 4 0 0

k2 k2 0 0

0 0 0 0

3 0 07 7 05 0

ð3:70Þ

The rank of this stiffness matrix is two since there are only two linearly independent rows. Therefore, the stiffness matrix is singular, its determinant is equal to zero, and

3.4 Rigid-Body Modes

137

the potential energy becomes a positive semidefinite quadratic form. That is, there exist possible configurations of the system, different from the zero configuration, such that the system potential energy is equal to zero. In fact, for this system, as the result of neglecting the effect of gravity, there are two rigid-body modes. The natural frequency associated with each of these modes can be determined using the characteristic equation and is precisely equal  to zero.The associated mode shape can be determined using the equation K  ω2i M Ai ¼ 0. Since ω1 ¼ ω2 ¼ 0, for i ¼ 1, 2, one has KAi ¼ 0. Recall that the rank of the 4  4 stiffness matrix is two. One is then guaranteed that the preceding matrix equation has two linearly independent solutions. These two linearly independent solutions can be written as 2 3 0 607 7 A1 ¼ α1 6 4 1 5, 0

2 3 0 607 7 A2 ¼ α2 6 405 1

ð3:71Þ

where α1 and α2 are arbitrary nonzero constants. The physical interpretation of these two solutions is that when the gravity effect is neglected, the rods connected to the masses m1 and m2 can have independent rigid body rotations without any change in the system potential energy. That is, ATi KAi ¼ 0, i ¼ 1, 2. Note also that a linear combination of the vectors A1 and A2 can still be selected as an eigenvector, for example, one may choose A1 and A2 as 2 3 0 607 7 A1 ¼ α1 6 4 1 5, 0

2

3 0 607 7 A2 ¼ 6 4 α2 5 α3

ð3:72Þ

with α1, α2, and α3 as constants. These two vectors can also be used as the rigid-body modes since they are linearly independent and satisfy the equation KAi ¼ 0, i ¼ 1,2. Observe that the condition ATi KAi ¼ 0 remains valid, while ATi MAi 6¼ 0 ði ¼ 1; 2Þ since the kinetic energy is always positive definite. In fact, for the two rigid-body modes A1 and A2, one can show that 2 X

ATi MAi ¼ α21 m3 l2 þ α22 m4 l2

ð3:73Þ

i¼1

which is proportional to the rotational kinetic energy of the two rods, and α1 and α2 are constants. Observe also that the orthogonality conditions ATi MA j ¼ 0, ATi KA j ¼ 0 are still in effect.

) ,

i 6¼ j,

i, j ¼ 1,2,3,4

ð3:74Þ

138

3

Multi-degree of Freedom Systems

Example 3.5

Consider the system of Example 3.1, when k3 ¼ 0. This is the case in which the shaft is free to rotate about its own axis. In this case, the stiffness matrix is given by 2

3 2 0 1 k2 5 ¼ 12  105 4 1 0 k2

k1 k1 þ k2 k2

k1 K ¼ 4 k 1 0

1 3 2

3 0 2 5N  m 2

which is singular since the sum of the second and third rows is the negative of the first row. Consequently, the rows of the stiffness matrix are not linearly independent. The rank of this matrix is equal to two and, therefore, the system has one rigid-body mode. The mass matrix, however, remains the same and is given by 2

I1 M¼40 0

0 I2 0

2 3 0 1 3 0 5 ¼ 2  10 4 0 0 I3

3 0 0 1:5 0 5kg  m2 0 2

The matrix equation of the free vibration of this system is given by Mθ€ þ Kθ ¼ 0. By assuming a solution in the form of Eq. 3.17, one obtains the eigenvalue problem [K  ω2M]A ¼ 0, in which the determinant of the coefficient matrix can be written as

1  β

1

0

1 3  1:5β 2

0

2

¼ 0 2  2β

where β ¼ ω2/600. Therefore, the characteristic equation can be written in terms of β as β(1  β)(β  3) ¼ 0, which has the following roots β1 ¼ 0, β2 ¼ 1, and β3 ¼ 3. The corresponding natural frequencies are ω1 ¼ 0,

ω2 ¼ 24:495 rad=s,

ω3 ¼ 42:426 rad=s

The first natural frequency ω1 corresponds to a rigid-body mode. In order to obtain the mode shapes, we use the following equation: 2

1  βi 4 1 0

1 3  1:5βi 2

3 2 3 32 0 0 Ai1 2 54 Ai2 5 ¼ 4 0 5, i ¼ 1,2,3 0 2ð1  βi Þ Ai3

which, by partitioning of the coefficient matrix, leads to the following equation: (continued)

3.4 Rigid-Body Modes



139

 3  1:5βi 1 Ai1 þ 0 2

2 2ð 1  β i Þ



Ai2 Ai3

¼

 0 0

or 

2 2ð 1  β i Þ

3  1:5βi 2



Ai2 Ai3



 1 ¼ A 0 i1

That is, 

Ai2 Ai3

¼

  1 2ð 1  β i Þ 2 1 A 2 ð3  1:5βi Þ 0 i1 2ð1  βi Þð3  1:5βi Þ  4

For the rigid-body mode, β1 ¼ 0 and, accordingly, 

A12 A13



   1 1 2 2 1 A11 A11 ¼ ¼ 2 2 3 0 1

In this mode of vibration, all the disks have the same amplitude and the system moves as one rigid body. For the second and third natural frequencies, one has the following deformation modes " 1 0 ¼ 4 2 A23 " " # A32 1 4 ¼ 2 2 A33 "

A22

#

" # #" # 0 1 A21 A21 ¼ 0:5 1:5 0 " # #" # 2:0 2 1 A31 A31 ¼ 1:0 1:5 0 2

One may assume that Ai1 ¼ 1 for i ¼ 1,2,3. The resulting modal matrix is given by 2

1:0 1:0 Φ ¼ 4 1:0 0:0 1:0 0:5

3 1:0 2:0 5 1:0

Figure 3.9 shows the modes of vibration of the system examined in this example. Observe that 2

AT1 KA1 ¼ 12  105 ½ 1:0

1:0

1 1:0 4 1 0

1 3 2

32 3 0 1:0 2 54 1:0 5 ¼ 0 2 1:0

while (continued)

140

3

Multi-degree of Freedom Systems

Fig. 3.9 Modes of vibration of the semidefinite system

2

A1T MA1 ¼ 2  103 ½ 1:0

1:0

1 0 1:0 4 0 1:5 0 0

32 3 0 1:0 0 54 1:0 5 ¼ 9  103 2 1:0

that is, the modal stiffness coefficient associated with the rigid-body mode is equal to zero, while the modal mass coefficient is not equal to zero. Therefore, if the system vibrates freely in its rigid-body mode, the strain energy is equal to zero, while the kinetic energy is not equal to zero. Observe also that A1T MAk ¼ 0, and A1T KAk ¼ 0, for k ¼ 2, 3.

3.5

Conservation of Energy

Using the modal expansion, it can be shown that in the case of the undamped free vibration of the multi-degree of freedom systems, the sum of the kinetic and potential energies is a constant of motion. To this end, the kinetic and potential energies of the _ multi-degree of freedom system are written, respectively, as T ¼ q_ T Mq=2 and U ¼ qTKq∕2. By using the modal transformation, the kinetic and potential energies can be expressed in terms of the modal variables as 1 1 T ¼ P_ T ΦT MΦP_ ¼ P_ T Mp P_ 2 2 1 T T 1 T U ¼ P Φ KΦP ¼ P Kp P 2 2

9 > > = > > ;

ð3:75Þ

where Mp and Kp are, respectively, the diagonal modal mass and stiffness matrices, and P is the vector of modal coordinates. In terms of the modal mass and stiffness coefficients mi and ki, the kinetic and potential energies can be written as

3.5 Conservation of Energy

T¼

141 n 1X mi P_ 2i , 2 i¼1

U¼

n 1X k i P2i 2 i¼1

ð3:76Þ

In the case of free vibration, the modal coordinates and velocities are harmonic functions given, respectively, by Pi ¼ Ci sin (ωit+ψ i) and P_ i ¼ Ci ωi cos ðωi t þ ψ i Þ. Substituting these two equations into the expressions of the kinetic and potential energies, one obtains T¼

9 n 1X > mi C 2i ω2i cos 2 ðωi t þ ψ i Þ > > > = 2 i¼1

n 1X k i C 2i sin 2 ðωi t þ ψ i Þ U¼ 2 i¼1

> > > > ;

ð3:77Þ

Adding these two equations and keeping in mind that ki ¼ mi ω2i , one obtains T þU ¼

n   1X ki C 2i cos 2 ðωi t þ ψ i Þ þ sin 2 ðωi t þ ψ i Þ 2 i¼1

ð3:78Þ

which can simply be written as T þU ¼

n n 1X 1X ki C2i ¼ mi ω2i C2i 2 i¼1 2 i¼1

ð3:79Þ

That is, the sum of the kinetic and potential energies of the conservative multi-degree of freedom system is constant. Rayleigh Quotient In the preceding sections, it was shown that the vibration of the multi-degree of freedom system can be represented as a linear combination of its mode shapes. However, a multi-degree of freedom system can vibrate at one of its mode shapes only if this mode is the only mode that is excited. Let us assume that the system is made to vibrate freely at its ith mode shape. In this case, the sum of the kinetic and potential energies of the system is given by Eq. 3.79, which implies that ω2i ¼

ki AT KAi ¼ Ti mi Ai MAi

ð3:80Þ

That is, the square of the natural frequency associated with the mode i can be written in a form of quotient where the numerator is proportional to the potential energy, while the denominator is proportional to the kinetic energy. This quotient is called the Rayleigh quotient. If the vibration of the multi-degree of freedom system is described by an arbitrary vector q, the Rayleigh quotient can be written as

142

3

RðqÞ ¼

Multi-degree of Freedom Systems

qT Kq qT Mq

ð3:81Þ

If q is parallel to any of the eigenvectors, the Rayleigh quotient is precisely the eigenvalue associated with this eigenvector. In the general case, however, q can be expressed P n as a linear combination of the linearly independent mode shapes as q ¼ i¼1 αi Ai . If one assumes that the mode shapes are made orthonormal with respect to the mass matrix, that is, ATi MAi ¼ 1, the Rayleigh quotient can be written as n P T

q Kq i¼1 ¼ n RðqÞ ¼ T q Mq P i¼1

n P

α2i ATi KAi α2i ATi MAi

¼

α2i ω2i

i¼1 n P

i¼1

α2i

ð3:82Þ

Since ω1  ω2  ω3    ωn, one has m P

α2i ω2i

i¼1 m P

i¼1

α2i

  α21 ω21 þ α22 ω22 þ    þ α2m ω2m ω21 α21 þ α22 þ    þ α2m ¼  ¼ ω21 ð3:83Þ α21 þ α22 þ    þ α2m α21 þ α22 þ    þ α2m

for any integer m. It follows that the minimum value of the Rayleigh quotient is the square of the natural frequency associated with the fundamental mode of vibration. If, however, the shape of the ith mode can be assumed, the Rayleigh quotient can be used to provide an estimate of the square of the natural frequency corresponding to this mode. In practical applications, this is seldom done for high-frequency modes, since the shape of these modes cannot be assumed with good accuracy. Therefore, the Rayleigh quotient is often used to obtain an estimate for the fundamental frequency of the system, since in many applications the shape of the fundamental mode can be assumed with sufficient accuracy. Illustrative Example In order to demonstrate the use of the Rayleigh quotient to predict the fundamental frequency, the system of Example 3.1 is considered. The mass and stiffness matrices of this three degree of freedom system were found to be 2

1 M ¼ 2  103 4 0 0

3 0 0 1:5 0 5kgm2 , 0 2

2

1 K ¼ 12  105 4 1 0

3 1 0 3 2 5Nm 2 5

Let us assume that the deformation of the system in its first mode can be described by the harmonic function sin(πx/2l). Using this function and assuming that the distances between the disks are equal, one has an estimate for the elements of the first mode shape as A11 ¼ sin(πl/2l) ¼ 1.00, A12 ¼ sin(π(2l/3)/2l) ¼ 0.8660, and A13 ¼ sin(π(l/3)/2l) ¼ 0.5. Therefore, a rough estimate of the first mode shape is

3.6 Forced Vibration of the Undamped Systems

2

143

1:00

3

6 7 A1 ¼ 4 0:8660 5 0:50 In this case, one has 2

AT1 KA1 ¼ 12  105 ½ 1

0:8660

1 0:5 4 1 0

1 3 2

32 3 0 1:00 2 54 0:8660 5 5 0:5

¼ 12:43  105 2

AT1 MA1 ¼ 2  103 ½ 1

0:8660

1 0 0:5 4 0 1:5 0 0

32 3 0 1:00 0 54 0:8660 5 ¼ 5:2499  103 2 0:5

Therefore, an estimate for the fundamental frequency using the Rayleigh quotient can be obtained as ω21 

AT1 KA1 12:43  105 ¼ ¼ 2:3677  102 AT1 MA1 5:2499  103

which yields ω1 ¼ 15.387 rad/s. The exact fundamental frequency evaluated in Example 3.2 is 14.52 rad/s. That is, with a rough estimate for the first mode shape, the error in the obtained fundamental frequency using the Rayleigh quotient is less than 6%.

3.6

Forced Vibration of the Undamped Systems

The general matrix equation which governs the forced vibration of the undamped multi-degree of freedom systems is given by M€ q þ Kq ¼ F, where M and K are the symmetric mass and stiffness matrices of the system, q is the vector of the displacements, and F is the vector of the forcing functions. This equation is a set of n coupled differential equations which must be solved simultaneously. When the number of degrees of freedom is very large, determining the solution of these equations becomes a difficult task. An elegant approach to overcome this problem is to use the modal superposition technique discussed in the preceding sections. In this technique, the vector of displacements is expressed in terms of the modal coordinates using the modal transformation matrix. The displacement vector, in this case, is written as a linear combination of the mode shapes. By utilizing the orthogonality of the mode shapes with respect to the mass and stiffness matrices, n uncoupled differential equations of motion can be obtained. These equations can be solved using techniques that are similar to the ones used in the solution of the equations of motion of single degree of freedom systems.

144

3

Multi-degree of Freedom Systems

By using the mode superposition technique, the displacement and acceleration € are expressed, respectively, in terms of the modal variables as vectors q and q € where P is the vector of modal coordinates and Φ is the € ¼ ΦP, q ¼ ΦP and q modal transformation matrix whose columns are the mode shapes of the multidegree of freedom systems. Substituting these kinematic displacement and acceleration equations into the equations of motion M€ q þ Kq ¼ F and premultiplying €þ by the transpose of the modal transformation matrix, one obtains ΦT MΦP T T Φ KΦP ¼ Φ F, which can be rewritten as € þ Kp P ¼ Q Mp P

ð3:84Þ

where Mp, Kp, and Q are, respectively, the modal mass and stiffness matrices, and the vector of modal forcing functions defined as Mp ¼ ΦT MΦ,

Kp ¼ ΦT KΦ,

Q ¼ ΦT F

ð3:85Þ

By using the fact that Mp and Kp are diagonal matrices and by using the definition of the elements of Mp and Kp given by Eqs. 3.42 and 3.43, Eq. 3.84 can be written in the following alternate form: € i þ ki Pi ¼ Qi , i ¼ 1,2,. . . ,n mi P

ð3:86Þ

where Qi is the ith element in the vector Q. The differential equations given by Eq. 3.86 have the same form as the differential equations that govern the vibration of the undamped single degree of freedom systems. As discussed in Chap. 1, the solution of the differential equations given by Eq. 3.86 can be expressed as P_i0 Pi ðt Þ ¼ Pi0 cos ωi t þ sin ωi t ð t ωi 1 þ Q ðτÞ sin ωi ðt  τÞdτ, mi ωi 0 i

ð3:87Þ

i ¼ 1,2,. . . ,n

where Pi0 and P_ i0 are the initial conditions associated with the ith modal coordinate. Having determined the modal coordinates from Eq. 3.87, the vector of displacements q can be obtained using the modal transformation q ¼ ΦP. Example 3.6

Find the response of the system of Example 3.1 to the step load given by 2 3 1 Fðt Þ ¼ 104 4 2 5Nm 4 Assume that the system is to start from rest and neglect the damping effect. (continued)

3.7 Viscously Damped Systems

145

Solution. In order to be able to use Eq. 3.84, we first evaluate the vector of modal forces Q defined by Eq. 3.85 as Q ¼ ΦTF. Using the modal matrix Φ obtained in Example 3.2, the vector Q is given by 2

1:0 Q ¼ ΦT F ¼ 4 1:0 1:0

2 3 32 3 3:506 0:302 1 0:679 54 2 5  104 ¼ 104 4 2:930 5 5:670 2:438 4

0:649  0:607 2:54

The equations of motion can then be written in terms of the modal coordinates as € i þ k i Pi ¼ Qi , mi P

i ¼ 1,2,3

where mi and ki, i ¼ 1,2,3, are, respectively, the modal mass and stiffness coefficients defined in Example 3.3. Since Qi is constant, one can verify that the solutions of the above equations, in the case of zero initial conditions, are given by Pi ¼

Qi ð1  cos ωi t Þ ki

Using the results of Examples 3.2 and 3.3, we have 2

P1

3

2

0:0463ð1cos14:52t Þ

3

6 7 6 7 P ¼ 4 P2 5 ¼ 4 0:0062ð1cos31:05t Þ 5 0:0006ð1cos46:1t Þ P3 The vector of torsional oscillations θ is given by 2

1:0

6 θ ¼ ΦP ¼ 4 0:649 2

0:302

1:0

1:0

32

P1

3

76 7 0:607 2:54 54 P2 5 0:679 2:438 P1 þ P2 þ P3

P3 3

6 7 ¼ 4 0:649P1  0:607P2  2:54P3 5 0:302P1  0:679P2 þ 2:432P3

3.7

Viscously Damped Systems

We considered thus far the free and forced vibrations of undamped multi-degree of freedom systems, and it was demonstrated that, for the undamped system, the eigenvalues and eigenvectors are real. This is not, however, the case when damping

146

3

Multi-degree of Freedom Systems

is included in the mathematical model, since for a viscously damped multi-degree of freedom system, the eigenvalues and eigenvectors can be complex numbers. Even though in many applications damping is small, its effect on the system stability and system response in the resonance region may be significant. The general matrix equation that governs the vibration of the viscously damped multi-degree of freedom system is given by M€ q þ Cq_ þ Kq ¼ F (Eq. 3.1) in which C is the damping matrix defined in Eq. 3.2. In general, ΦTCΦ is not a diagonal matrix, and as a consequence, the use of the modal coordinates does not lead to a system of uncoupled independent differential equations. In the remainder of this section, we will consider an important special case in which the damping matrix is proportional to the mass and stiffness matrices. In this special case, a system of uncoupled differential equations expressed in terms of the modal coordinates can be obtained. The case of a general damping matrix is discussed in the following section. Proportional Damping If the damping matrix can be written as a linear combination of the mass and stiffness matrices, the matrix C takes the form C ¼ αM þ βK

ð3:88Þ

where α and β are constants. Substituting Eq. 3.88 into Eq. 3.1 yields M€ q þ ðαM þ βKÞq_ þ Kq ¼ F

ð3:89Þ

The vector of displacement q can be expressed in terms of the modal coordinates by using the mode shapes of the undamped system as q ¼ ΦP, where Φ is the modal transformation matrix and P is the vector of modal coordinates. Using the modal transformation in Eq. 3.89 and premultiplying by ΦT, one obtains € þ ΦT ðαM þ βKÞΦP_ þ ΦT KΦP ¼ ΦT F ΦT MΦP

ð3:90Þ

Using the orthogonality of the mode shapes with respect to the mass and stiffness matrices yields € þ Cp P_ þ Kp P ¼ Q Mp P

ð3:91Þ

in which Mp ¼ ΦT MΦ,

Kp ¼ ΦT KΦ,

Q ¼ ΦT F,

Cp ¼ ΦT ðαM þ βKÞΦ ¼ αMp þ βKp

) ð3:92Þ

Because the matrix C is assumed to be a linear combination of the mass matrix M and the stiffness matrix K, the modal damping matrix Cp is diagonal. The elements on the diagonal of this matrix are denoted as ci, i ¼ 1,2,. . .,n. The modal

3.7 Viscously Damped Systems

147

damping coefficient ci can be expressed in terms of the modal mass and stiffness coefficients mi and ki, respectively, by using Eq. 3.92 as ci ¼ αmi þ βki

ð3:93Þ

Equation 3.91 can be then written as n uncoupled differential equations as € i þ ci P_ i þ k i Pi ¼ Qi , mi P

i ¼ 1,2,. . . ,n

ð3:94Þ

These equations, which are in the same form as the differential equation of motion which governs the forced vibration of damped single degree of freedom systems, can be also written in the following form: € i þ 2ξi ωi P_ i þ ω2i Pi ¼ Qi P mi

ð3:95Þ

where the modal damping factor ξi is defined as ξi ¼

ci 2mi ωi

ð3:96Þ

As discussed in the case of the single degree of freedom systems, the solution of Eq. 3.95 will depend on the modal damping factor ξi. In the case of an underdamped system in which ξi < 1, the solution of Eq. 3.95 is given by  Pi ðt Þ ¼ e

ξi ωi t

þ

ðt

P_i0 þ ξi ωi Pi0 Pi0 cos ωdi t þ sin ωdi t ωdi

ð3:97Þ

Qi ðτÞhi ðt  τÞdτ, i ¼ 1,2,. . . ,n

0

where Pi0 and P_ i0 are, respectively, the initial modal coordinate and velocity associated with mode i and ωdi ¼ ωi hi ð t Þ ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffi 1  ξ2i

9 > > =

> eξi ωi t ; sin ωdi t > mi ωdi

ð3:98Þ

in which ωdi is the damped natural frequency of mode i, and hi is the unit impulse response associated with the ith mode of vibration. Experimental Modal Analysis The concept of proportional damping discussed in this section can also be used in cases in which the damping is small. Despite the fact that in these cases the resulting modal damping matrix is not, in general, diagonal, a reasonable approximation can be made by neglecting the off-diagonal

148

3

Multi-degree of Freedom Systems

Fig. 3.10 Resonance curve

terms. The matrix Cp can then be considered as a diagonal matrix and a set of independent uncoupled differential equations expressed in terms of the modal coordinates can be obtained. This approximation is acceptable in many engineering applications, since structural damping is usually small. Furthermore, there are several experimental techniques which can be used to determine the natural frequencies, mode shapes, and modal damping coefficients. Figure 3.10 shows a resonance curve that can be obtained experimentally for a multi-degree of freedom system by exciting the system using a vibration exciter and measuring the coordinate qi at a selected point. The resonance curve of a multi-degree of freedom system has a number of resonance regions equal to the number of the system degrees of freedom, which is equal to the number of the system natural frequencies. Each resonance region can be considered as a resonance curve of a single degree of freedom system, and, therefore, for a lightly damped system, the frequencies at which the peaks occur define approximate values for the system natural frequencies. The techniques described in the first volume of this book (Shabana 2018B) for the experimental determination of the damping factors of single degree of freedom systems then can be used to determine the damping factor ξi associated with the ith mode shape. Experimental modal analysis techniques also can be used to determine the mode shapes of the multi-degree of freedom systems. In order to demonstrate the € þ Cp p_ þ Kp P ¼ ΦT F. This use of this approach, Eq. 3.91 is rewritten as Mp P equation can also be written in the case of a harmonic excitation as € i þ ci P_ i þ k i Pi ¼ ATi F0 sinωf t, mi P

i ¼ 1,2, . . . ,n

ð3:99Þ

3.7 Viscously Damped Systems

149

where F0 is the vector of force amplitudes. The preceding equation defines the steady-state solution for the modal coordinate i as ATi F0 =ki Pi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin ðωf t  ψ i Þ 2 ð1  r 2i Þ þ ð2r i ξi Þ2

ð3:100Þ

where ri ¼

ωf , ωi

ψ i ¼ tan 1



2r i ξi 1  r 2i

 ð3:101Þ

The vector of physical displacements can be obtained using the modal transformation as q ¼ ΦP ¼

n X

A1 P1 þ A2 P2 þ    þ An Pn

i¼1

¼

n X i¼1

Ai ATi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F0 sinðωf t  ψ i Þ 2 ki ð1  r 2i Þ þ ð2r i ξi Þ2

ð3:102Þ

In the resonance region of a mode i, the solution is dominated by this mode, and as a consequence, one has q¼

Ai ATi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi F0 sinðωf t  ψ i Þ  2 2 ωi  ω2f þ ð2ξi ωi ωf Þ2 mi

ð3:103Þ

In the resonance region, Ai ATi Ai ATi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   2 2 2ξi ω2i ωi  ω2f þ ð2ξi ωi ωf Þ2

ð3:104Þ

It follows that q¼

Ai ATi 1 F0 sinðωf t  ψ i Þ ¼ Bi F0 sinðωf t  ψ i Þ 2ξi ki 2ξi ki

ð3:105Þ

where Bi ¼ Ai ATi . The symmetric matrix Bi has n2 elements, but only n of these elements are unique. There are two methods to determine the n unique elements of the matrix Bi. In the first method, the excitation is made at n nodes, one at a time, and the measurement is taken at one node. The obtained n measurements define, to within an arbitrary constant, the row of the matrix Bi that corresponds to the measured coordinate. In the second method, the excitation is made at only one

150

3

Multi-degree of Freedom Systems

node, and all of the elements of the vector q are measured. In this case, the n measurements can be used to define, to within an arbitrary constant, the column of the matrix Bi that corresponds to the point of excitation. If one row or one column of the matrix Bi is determined, one can use this row or column to determine the entire mode shape vector Ai. For example, if the vector Ai ¼ [a1 a2. . .an]T, the matrix Bi takes the following form: 2

a21

6a a 6 2 1 Bi ¼ 6 4 ⋮ an a1

a1 a2

a1 a3



a22 ⋮

a2 a3 ⋮

 ⋱

an a2

an a3



a1 an

3

a2 an 7 7 7 ⋮ 5

ð3:106Þ

a2n

If, for instance, the excitation is made at the kth nodal point, the measurements of the n elements of the nodal displacement vector q can be used to define the kth column  T of the matrix Bi, given by a1 ak a2 ak    a2k    an ak . Knowing the numerical value (ak)2, one can determine ak, and use other elements of the preceding vector to evaluate all the elements of the mode shape Ai. We also note that, because of the phase angle ψ i, the phase between the force and the displacement must be determined. At resonance, ψ i ¼ π/2 or π/2. This phase must be determined in order to correctly determine the sign of the elements of the matrix Bi. As previously pointed out, the matrix Bi can be determined using Eq. 3.105 to within any arbitrary constant. Once the mode shapes are determined, the modal mass coefficients can be calculated. For example, if the elements of the vector q are measured, all of the other variables in Eq. 3.105 are known except the modal stiffness coefficient ki, which can be calculated using any of the scalar equations of Eq. 3.105. Using this stiffness coefficient and the natural frequency of the mode i, the modal mass coefficient can also be determined. As we previously demonstrated, the mode shapes can be scaled such that either the modal mass or the modal stiffness coefficient is equal to 1. The procedure for the experimental determination of the mode shapes and the modal parameters is outlined in this section in order to demonstrate the basic concepts underlying an important practical tool widely used in industry for the vibration analysis of many mechanical systems. The actual implementation of this procedure requires a more detailed presentation (Brown et al. 1979; Ewins 1984; Shabana 1986) that includes elaborate statistical and error analysis. Figure 3.11 shows an example that demonstrates the wide use of experimental modal identification techniques in construction machine industry. Figure 3.12 shows some the mode shapes and frequencies that were determined experimentally for the hydraulic excavator shown in Fig. 3.11.

3.8 General Viscous Damping

151

Fig. 3.11 Hydraulic excavator (Courtesy of Komatsu Ltd)

3.8

General Viscous Damping

The assumption made in the preceding section that the damping matrix is proportional to the mass and stiffness matrices enables us to use the modal transformation matrix to decouple the equations of motion. The procedure outlined in the preceding section led to n uncoupled differential equations expressed in terms of the modal coordinates. Each of these equations is similar to the differential equation of motion of the viscously damped single degree of freedom system. In this section, we discuss the case of a general viscous damping matrix which cannot be expressed as a linear combination of the mass and stiffness matrices. Before we start our discussion on the case of general damping matrix, we present some matrix results that are used frequently in this chapter. If B is a nonsingular matrix that has λi as an eigenvalue, then 1/λi, is an eigenvalue of the matrix B1 since BAi ¼ λiAi implies that Ai ¼ λiB1Ai, or 1 Ai ¼ B1 Ai λi

ð3:107Þ

This equation shows that 1/λi is an eigenvalue of B1 and Ai is the corresponding eigenvector. Furthermore, it can be shown that Ai is also an eigenvector of the matrix B2. This can be demonstrated by premultiplying the eigenvalue problem by B to yield B2 Ai ¼ λi BAi ¼ λ2i Ai , which implies that λ2i is an eigenvalue of B2, and Ai is an eigenvector for both B and B2. Using a similar procedure, it can be shown that if λi is an eigenvalue of the matrix B, then λik is an eigenvalue of Bk and the matrices B and Bk have the same eigenvector.

152

3

Multi-degree of Freedom Systems

Fig. 3.12 Experimentally identified mode shapes of the chassis

If B is a symmetric matrix, the eigenvectors of B are orthogonal. If Ai is normalized such that ATi Ai ¼ 1, then one has ATi BAi ¼ λi ,

ATi Bk Ai ¼ λik

ð3:108Þ

3.8 General Viscous Damping

153

It follows that if Φ is the matrix of the eigenvectors of the symmetric matrix B and the matrix Bk, then ΦT BΦ ¼ Λ,

ΦT Bk Φ ¼ Λk

ð3:109Þ

where Λ is a diagonal matrix whose diagonal elements are the eigenvalues. Exponential Matrix Forms If a is a scalar, the use of Taylor series shows that ea ¼ 1+a+a2/2! +a3/3! +  . If B is a matrix, one also has eBt ¼ I þ Bt þ

ðBt Þ2 ðBt Þ3 þ þ  2! 3!

ð3:110Þ

This is a convergent series that has the properties eBr eBt ¼ eBðrþtÞ ,

eBt eBt ¼ I,

d  Bt  e ¼ BeBt dt

ð3:111Þ

It is then clear that if B is skew symmetric, then eBt is an orthogonal matrix. It was previously demonstrated in this section that if λi is an eigenvalue and Ai is the corresponding normalized eigenvector of the matrix B, then λik and Ai are the eigenvalue and eigenvector of the matrix Bk. Therefore, if Φ is the matrix of the eigenvectors, one has Bk ¼ ΦΛk Φ1

ð3:112Þ

where Λ is a diagonal matrix whose diagonal elements are the eigenvalues of the matrix B. Using the preceding identity, the exponential form of the matrix Bt can be written as ΦΛ2 Φ1 t 2 ΦΛ3 Φ1 t 3 þ þ  2! 3! ! ðΛt Þ2 ðΛt Þ3 þ þ    Φ1 ¼ ΦeΛt Φ1 ¼ Φ I þ Λt þ 2! 3!

eBt ¼ I þ ΦΛΦ1 t þ

ð3:113Þ

Note that the matrix eBt is never singular since its inverse is defined by the matrix eBt. This is also clear from the determinant

Bt

e ¼ eλ1 t eλ2 t   eλn t ¼ etrðBtÞ This determinant cannot be equal to zero.

ð3:114Þ

154

3

Multi-degree of Freedom Systems

Illustrative Example As an example, we consider the matrix 2

3 2 05 0

λ2 ¼ 5,

λ3 ¼ 1

4 1 B ¼ 41 0 2 0 which has the eigenvalues λ1 ¼ 0,

The corresponding orthonormal eigenvectors are 2

3 0 1 A1 ¼ pffiffiffi 4 2 5, 5 1

2 3 5 1 A2 ¼ pffiffiffiffiffi 4 1 5, 30 2

2

3 1 1 A3 ¼ pffiffiffi 4 1 5 6 2

The matrix Φ whose columns are the orthonormal eigenvectors is 3 5 1 pffiffiffiffiffi pffiffiffi 30 67 7 1 1 7 pffiffiffiffiffi pffiffiffi 7 7 30 67 7 2 2 5 pffiffiffiffiffi pffiffiffi 30 6

2 0

6 6 6 2 6 Φ ¼ 6 pffiffiffi 6 5 6 4 1 pffiffiffi 5 The matrix eΛt is 2

eΛt

1 ¼ 40 0

3 0 0 5 et

0 e5t 0

It follows that 2

eBt ¼ ΦeΛt Φ1

5e5t þ et 6 6 5t 6 t 6 e e ¼6 6 4 5t 6 t e e 3

e5t  et 6 e5t þ 5et þ 24 30 e5t þ 5et  6 15

3 e5t  et 7 3 7 e5t þ 5et  6 7 7 7 15 5 5t t 2e þ 10e þ 3 15

State Space Representation In the remainder of this section, we will present a technique for solving Eq. 3.1 in the case of general viscous damping matrix. To this end, we define the state vector y as  y¼

q q_

ð3:115Þ

3.8 General Viscous Damping

155

Clearly, the state vector has dimension 2n. Differentiation of this vector with respect to time leads to  y_ ¼

q_ € q

ð3:116Þ

Premultiplying Eq. 3.1 which governs the vibration of a viscously damped multi€ þ M1 Cq_ þ degree of freedom system by the inverse of the mass matrix leads to q 1 1 1 1 € ¼ M Cq_  M Kqþ M1 F. M Kq ¼ M F, which can be written as q This equation, with the identity q_ ¼ Iq_ where I is an n  n identity matrix, leads to 

q_ y_ ¼ € q





0 I ¼ M1 K M1 C



 q 0 þ q_ M1 F

ð3:117Þ

which can be written compactly as y_ ¼ By þ Rðt Þ

ð3:118Þ

in which the state vector y is defined by Eq. 3.115 and the 2n  2n matrix B and the 2n-dimensional vector R(t) are given, respectively, by  B¼

0 M1 K

I , M1 C

 R ðt Þ ¼

0

M1 F

ð3:119Þ

Free Vibration If the viscously damped multi-degree of freedom system is vibrating freely, the vector of forcing functions F is identically zero, and accordingly, R(t) ¼ 0. Equation 3.118 then reduces to y_ ¼ By

ð3:120Þ

To this equation, a solution in the form y ¼ Aeμt is assumed, where in this case, the vector A has dimension 2n and μ is a constant. It follows that y_ ¼ μAeμt . Substituting into Eq. 3.120 yields μAeμt ¼ BAeμt, which leads to ½B  μIA ¼ 0

ð3:121Þ

where I in this equation is a 2n  2n identity matrix. Equation 3.121 is an eigenvalue problem in which the coefficient matrix H ¼ B  μI is called the characteristic matrix. Equation 3.121 has a nontrivial solution if and only if the determinant of the coefficient matrix is equal to zero, that is, jHj ¼ jB  μIj ¼ 0

ð3:122Þ

156

3

Multi-degree of Freedom Systems

This equation defines a polynomial of order 2n in μ. The roots of this polynomial, which may be complex numbers, define the eigenvalues μ1, μ2, . . ., μ2n. Associated with each eigenvalue μi, there exists an eigenvector Ai which can be determined to within an arbitrary constant using Eq. 3.121 as BAi ¼ μi Ai

ð3:123Þ

In this case, the modal matrix is defined as Φ ¼ ½ A1

   A2n 

A2

ð3:124Þ

It can be verified that Φ1 BΦ ¼ μ

ð3:125Þ

where Φ is the 2n  2n modal matrix whose columns consist of the 2n eigenvectors Ai, i ¼ 1,2,. . .,2n, and μ is the diagonal matrix 2

μ1 60 6 μ¼6 60 4⋮ 0

0 μ2 0 ⋮ 0

0 0 μ3 ⋮ 0

3 ... 0 ... 0 7 7 ... 0 7 7 ⋱ ⋮ 5 . . . μ2n

ð3:126Þ

Recall that, for any integer r, the following identity holds: Φ1 Br Φ ¼ μr

ð3:127Þ

The modal transformation of Eq. 3.124 can be used to obtain 2n uncoupled firstorder differential equations. To this end, we write y ¼ ΦP,

y_ ¼ ΦP_

ð3:128Þ

where P is a 2n-dimensional vector. Substituting Eq. 3.128 into Eq. 3.120 yields ΦP_ ¼ BΦP or P_ ¼ Φ1 BΦP, which upon using Eq. 3.125 can be written as P_ ¼ μP

ð3:129Þ

This is a system of 2n uncoupled first-order differential equations which can be written as P_ i ¼ μi Pi ,

i ¼ 1,2,. . . ,2n

ð3:130Þ

The solution of these homogeneous first-order differential equations is simple and can be obtained as follows. Equation 3.130 can be written as dPi/dt ¼ μiPi or dPi/Pi ¼ μi dt, which upon integration yields

3.8 General Viscous Damping

157

ln Pi ¼ μi t þ C 0

ð3:131Þ

where C0 is the constant of integration. An alternate form of Eq. 3.131 is Pi ¼ eμi t eC0 . Using the initial conditions, one can show that eC0 ¼ Pi0 and, consequently, Pi ¼ Pi0 eμi t

ð3:132Þ

where Pi0 is the initial condition associated with the coordinate Pi. These initial conditions can be obtained by using Eq. 3.128 as P0 ¼ Φ1y0, where y0 is the state vector which contains the initial displacements and velocities. Having determined the 2n coordinates of Eq. 3.132, the displacements and velocities of the multi-degree of freedom system can be obtained using the transformation of Eq. 3.128. Forced Vibration Equation 3.118 describes the forced vibration of viscously damped multi-degree of freedom systems. This equation with Eq. 3.128 can be used to obtain 2n uncoupled differential equations. In order to demonstrate this, Eq. 3.128 is substituted into Eq. 3.118. This yields ΦP_ ¼ BΦP þ Rðt Þ, or P_ ¼ Φ1 BΦP þ Φ1 Rðt Þ, which upon using Eq. 3.125 can be written as P_ ¼ μP þ Q

ð3:133Þ

where the matrix μ is defined by Eq. 3.126 and Q ¼ ½ Q1 vector

Q2

   Q2n T is the

Q ¼ Φ1 R

ð3:134Þ

Equation 3.133 is a set of 2n uncoupled first-order differential equations which can be written as P_ i ¼ μi Pi þ Qi , i ¼ 1,2, . . . , 2n

ð3:135Þ

The solution of these equations is given by μi t

Pi ¼ e Pi0 þ

ðt

eμi ðtτÞ Qi ðτÞ dτ,

i ¼ 1,2,3, . . . ,2n

ð3:136Þ

0

Define the diagonal matrix T1 as 2

eμ 1 t

6 0 6 6 T1 ðt Þ ¼ 6 6 0 6 4 ⋮ 0 and

0

0

...

e μ2 t

0

...

0

e μ3 t

...

⋮

⋮

⋱

0

0

...

0

3

0 7 7 7 μt 0 7 7¼e 7 ⋮ 5 eμ2n t

ð3:137Þ

158

3

Multi-degree of Freedom Systems

T1 ðt; τÞ ¼ eμðtτÞ

ð3:138Þ

T1 ðt 1 ; t 2 Þ ¼ eμðt1 t2 Þ

ð3:139Þ

or more generally

The matrix T1(t) can also be expressed in a series form as T1 ðt Þ ¼ eμt ¼ I þ tμ þ

t2 2 t3 3 μ þ μ þ  2! 3!

ð3:140Þ

In terms of the matrix T1, Eq. 3.136 can be written as P ¼ T1 ðt ÞP0 þ

ðt

T1 ðt  τÞ QðτÞ dτ

ð3:141Þ

0

This equation is expressed in terms of the vector of modal coordinates P. In order to obtain the space coordinates and velocities, we use Eqs. 3.128 and 3.134 and multiply Eq. 3.141 by the modal matrix Φ to obtain y ¼ ΦP ¼ ΦT1 Φ1 y0 þ

ðt

ΦT1 ðt  τÞΦ1 RðτÞ dτ

ð3:142Þ

0

which can be written compactly as y ¼ Tðt Þy0 þ

ðt

Tðt  τÞ RðτÞ dτ

ð3:143Þ

0

where T is called the state transition matrix and is defined as T ¼ ΦT1 Φ1

ð3:144Þ

The state transition matrix which plays a fundamental role in theory of linear dynamic systems satisfies the following identities: Tðt 1 ; t 1 Þ ¼ I ¼ Tð0Þ, Tðt 1 ; t 2 Þ Tðt 2 ; t 1 Þ ¼ I,

9 Tðt 3 ; t 1 Þ ¼ Tðt 3 ; t 2 Þ Tðt 2 ; t 1 Þ > = ½Tðt 1 ; t 2 Þ1 ¼ Tðt 2 ; t 1 Þ,

Tðt 1 ÞTðt 2 Þ ¼ Tðt 1 þ t 2 Þ

> ;

ð3:145Þ

where I is the identity matrix. Illustrative Example In order to demonstrate the use of the state transition matrix in the vibration analysis of mechanical systems, a simple viscously damped single degree of freedom system is considered. The equation of motion of such a system is given by m€q þ cq_ þ kq ¼ F ðt Þ

3.8 General Viscous Damping

159

where m, c, and k are, respectively, the mass, damping, and stiffness coefficients, F(t) is the forcing function, and q is the degree of freedom of the system. The differential equation of the system can also be written in the following form k c F ðt Þ €q ¼  q  q_ þ m m m F ðt Þ 2 ¼ ω q  2ξωq_ þ m where ω is the natural frequency of the system and ξ is the damping factor, that is rffiffiffiffi k ω¼ , m

c 2mω

ξ¼

One can, therefore, define the state equations as " y_ ¼

"

#

q_ €q

¼

#"

0

1

ω2

2ξω

q

#

q_

" þ

#

0 F ðt Þ=m

From which the matrix B and the vector R(t) of Eq. 3.118 can be defined as 

0 B¼ ω2

1 , 2ξω



0 Rðt Þ ¼ F ðt Þ=m



The characteristic matrix H can then be defined for this system as  H ¼ B  μI ¼

μ ω2

1 2ξω  μ



which defines the characteristic equation as

μ jHj ¼

ω2



¼ μð2ξω þ μÞ þ ω2 ¼ 0 2ξω  μ

1

This is a quadratic equation in μ which can simply be written as μ2 þ 2ξωμ þ ω2 ¼ 0 The roots of this equation are pffiffiffiffiffiffiffiffiffiffiffiffiffi ξ2  1 pffiffiffiffiffiffiffiffiffiffiffiffiffi μ2 ¼ ξω  ω ξ2  1 μ1 ¼ ξω þ ω

The eigenvector Ai associated with the eigenvalue μi can be obtained by using Eq. 3.121 or Eq. 3.123. In this case, one has

160

3

" ½B  μi IAi ¼

Multi-degree of Freedom Systems

#"

μi

1

ω2

2ξω  μi

Ai1

#

Ai2

¼0

which yields Ai2 ¼ μiAi1. The eigenvector Ai can then be written as 

1 Ai ¼ , i ¼ 1,2 μi and the modal matrix Φ of Eq. 3.124 is  Φ ¼ ½ A1

A2  ¼

1 μ1

1 μ2



It follows that Φ1 is the matrix Φ

1

 1 μ2 ¼ μ2  μ1 μ1

1 1



which must satisfy the identity of Eq. 3.125. The state transition matrix T is given in this example by T ¼ ΦT1 Φ1 # " #" μ t #" e1 μ2 1 1 1 0 1 ¼ μ2  μ1 μ1 μ2 0 e μ2 t μ1 1 " # μ1 t μ2 t μ2 t μ1 t μ2 e  μ1 e e e 1 ¼ μ2  μ1 μ1 μ2 ðeμ1 t  eμ2 t Þ μ2 eμ2 t  μ1 eμ1 t It is left to the reader as an exercise to show that the use of Eqs. 3.132 and 3.143 leads, respectively, to the free and forced vibration solution obtained in Chap. 1 in the case of the single degree of freedom systems.

3.9

Approximation and Numerical Methods

It was shown in the preceding sections that the solution of the equations of motion of the multi-degree of freedom system can be expressed in terms of the modal coordinates. It was also shown that the generalized forces associated with the modal coordinates can be obtained by premultiplying the vector of externally applied forces by the transpose of the modal matrix, and as a result, the frequency contents in a generalized force associated with a certain mode depend on the frequency contents

3.9 Approximation and Numerical Methods

161

in the externally applied forces. Clearly, if the modal force has a frequency close to the natural frequency of the corresponding mode, this mode of vibration will be excited and this mode will have a significant effect on the dynamic response of the system. Quite often, the frequency content in a given forcing function has an upper limit, and accordingly, mode shapes which have frequencies higher than this upper limit are not excited, and their effects on the dynamic response of the system can be ignored. In this case, the multi-degree of freedom system can be represented by an equivalent system which has a smaller number of degrees of freedom determined by the number of the significant low-frequency modes of vibration. If the multi-degree of freedom system has only one significant mode of vibration, the system can be replaced by an equivalent system which has only one degree of freedom. Similarly, the multi-degree of freedom system can be represented by a two degree of freedom system if only two modes are significant. The procedure of eliminating insignificant modes is called modal truncation. Consider a system which has n degrees of freedom and assume that only m modes of vibration are significant, where m < n. It was previously shown that the vibration of the n degree of freedom system is governed by n second-order ordinary differential equations which can be written in the form M€ q þ Cq_ þ Kq ¼ Fðt Þ (Eq. 3.1), where M, C, and K are, respectively, the n  n mass, damping, and stiffness matrices, and q and F are, respectively, the n-dimensional vectors of displacements and externally applied forces. Let Φt be the modal matrix which contains only the m significant modes of vibration, that is, Φt has n rows and m columns, and accordingly, it is not a square matrix. The elimination of insignificant mode shapes can be expressed mathematically as q ¼ Φt P

ð3:146Þ

where the vector of modal coordinates P consists in this case of m elements. Substituting Eq. 3.146 into Eq. 3.1 and premultiplying by the transpose of the modal matrix Φt, one obtains € þ Φ T CΦt P_ þ Φ T KΦt P ¼ Φ T Fðt Þ ΦtT MΦt P t t t

ð3:147Þ

Assuming the case of proportional damping and utilizing the orthogonality of the mode shapes with respect to the mass and stiffness matrices, Eq. 3.147 can be rewritten as € þ Cp P_ þ Kp P ¼ Q Mp P

ð3:148Þ

where Mp,Cp, and Kp are square diagonal matrices with dimension m  m and P is the m-dimensional vector of the modal coordinates. Equation 3.148 has m second-order uncoupled differential equations which can be solved for the m modal coordinates as discussed in the preceding sections. The n-dimensional vector of actual displacements can be recovered by using the transformation of Eq. 3.146.

162

3

Multi-degree of Freedom Systems

Example 3.7

For the three degree of freedom torsional system discussed in Example 3.2, it was shown that the mass and stiffness matrices are given by 2

2 M ¼ 103 4 0 0

3 0 0 3:0 0 5 kgm2 , 0 4

2

1 K ¼ 12  105 4 1 0

1 3 2

3 0 2 5 Nm 5

The modal matrix Φ for this system is given by 2

1:000 1:000 6 Φ ¼ 4 0:649 0:607

3 1:000 7 2:541 5

0:302 0:679

2:438

The equation of motion of the undamped system is Mθ€ þ Kθ ¼ F, where θ is the vector of the torsional oscillations of the system. It was also shown that the natural frequencies of this system are ω1 ¼ 14.52 rad/s, ω2 ¼ 31.05 rad/s, and ω3 ¼ 46.1 rad/s. If we assume that the frequency content in the forcing function F is not in the resonance region of the third mode, we may ignore this mode by deleting the third column of the modal transformation Φ. In this case, the reduced number of degrees of freedom m is equal to two and the new modal transformation Φt is given by 2

1:000

6 Φt ¼ 4 0:649 0:302

1:000

3

7 0:607 5 0:679

By using Eq. 3.146, the system equations of motion can be expressed in terms of the modal coordinates as € þ Kp P ¼ Q, Mp P where Mp and Kp are 2  2 diagonal matrices defined as  Mp ¼

ΦtT MΦt

¼ 10

3

 Kp ¼ ΦtT KΦt ¼ 105

3:628 0 7:656 0

0





m1

0



¼ 4:95 0 m2  0 k1 0 ¼ 47:72 0 k2

where mi and ki, i ¼ 1,2 are the modal mass and stiffness coefficients. The generalized modal force vector Q is a two-dimensional vector defined by Q ¼ ΦtT F. (continued)

3.9 Approximation and Numerical Methods

163

Similarly, if the upper limit on the frequency content in the forcing function is much below the second natural frequency of the system, one may replace the three degree of freedom system by an equivalent single degree of freedom system. This is the case in which the effect of the second and third modes of vibrations is ignored. In this case, the modal transformation is reduced to a column vector defined as 2

1:000

3

6 7 7 Φt ¼ 6 4 0:649 5 0:302 where, in this case, m ¼ 1. Using this modal transformation, the equation of motion of the equivalent single degree of freedom system is given by € 1 þ k1 P € 1 ¼ Q1 m1 P where m1 ¼ 3.628  103, k1 ¼ 7.656  105, and Q1 is the scalar defined by Q1 ¼ ΦtT F.

Direct Numerical Integration The method of mode superposition discussed in this chapter is widely used in the vibration analysis of multi-degree of freedom systems. In addition to the physical insight gained from its use, the mode-superposition technique leads to a set of uncoupled differential equations which can be solved in a closed form in a manner similar to the equations that describe the vibration of single degree of freedom systems. Furthermore, if high-frequency modes of vibration do not significantly contribute to the solution of the vibration equations, modal truncation can be used to reduce the number of degrees of freedom as described in this section. In this case, one has to determine only the low-frequency modes of vibration. In some other applications, however, such as in the case of impact loading, the contribution of the high-frequency modes of vibration is significant. If the number of degrees of freedom is very large, the use of the mode-superposition technique is not recommended since solving the eigenvalue problem of a large system is challenging. Another situation where the use of mode superposition is not recommended is in the cases where the vibration equations are nonlinear. As a simple example, consider the multi-degree of freedom system shown in Fig. 3.3 and discussed in Sect. 3.1. In deriving the equations of motion of this system, the assumption of small oscillations was used. This leads to a set of linear differential equations which can be solved using the techniques discussed in the preceding sections. If the displacements of the masses are large such that the assumption of small oscillations is no longer valid, the resulting differential equations of motion are no longer linear, and the mode shapes and natural frequencies of the

164

3

Multi-degree of Freedom Systems

system are no longer constant. Consequently, the use of the mode-superposition technique can be expensive computationally. An alternate approach for the use of the modal-expansion method is the direct numerical integration of the system differential equations. This subject was discussed in Chap. 5 of the first volume of this book (Shabana 2018b), and the method presented there for the analysis of single degree of freedom systems can be generalized to the case of multi-degree of freedom systems. To this end, the equations of motion of the linear or the nonlinear system can be written in the following general form:   _ t M€ q ¼ f 1 q, q,

ð3:149Þ

where f1 is a vector function which can be nonlinear in the coordinates, velocities, and time. In the case of linear systems, the vector function f1 is given by   _ t ¼ Cq_  Kq þ Fðt Þ f 1 q, q,

ð3:150Þ

Assuming that the mass matrix is nonsingular, the preceding two equations can be used to define the accelerations as follows:   _ t € ¼ M1 f 1 q, q, q

ð3:151Þ

y2 T as

One may define the vectors y ¼ ½ y1  y¼

y1



 ¼

y2

q q_

ð3:152Þ

Using the preceding two equations, y_ 2 can be defined as   _ t ¼ M1 f 1 ðy1 ; y2 ; t Þ ¼ M1 f 1 ðy; t Þ € ¼ M1 f 1 q, q, y_ 2 ¼ q

ð3:153Þ

which can be written compactly as y_ 2 ¼ Gðy; t Þ

ð3:154Þ

where G(y, t) ¼ M1f1(y, t). Equation 3.154 can be combined with the equation y_ 1 ¼ y2 to yield the following system of equations: 

y_ 1 y_ ¼ y_ 2



 ¼

y2 Gðy; t Þ

ð3:155Þ

For a system of n degrees of freedom, the preceding equation has 2n first-order differential equations which can be integrated numerically using direct numericalintegration methods such as the Runge–Kutta method (Atkinson 1978; Camhan et al. 1969; Shabana 2018B).

3.10

3.10

Matrix-Iteration Methods

165

Matrix-Iteration Methods

The methods presented thus far for the vibration analysis of multi-degree of freedom systems require the evaluation of the mode shapes and natural frequencies which define the solution of the vibration equations. If the number of degrees of freedom is very large, the solution of the characteristic equation to determine the natural frequencies becomes more difficult, and in these cases, the use of the numerical methods is recommended. It was shown in the preceding sections that the equations of the free undamped vibration of the multi-degree of freedom system can be written as M€ q þ Kq ¼ 0. A solution for this equation can be assumed in the form q ¼ A sin (ωt þ ϕ), leading to ω2 MA ¼ KA

ð3:156Þ

If the mass matrix is assumed to be nonsingular, Eq. 3.156 can be written as ω2 A ¼ M1 KA

ð3:157Þ

λA ¼ Hs A

ð3:158Þ

which can be written as

in which λ ¼ ω2 and Hs ¼ M1K. Equation 3.158 is referred to as the stiffness formulation of the eigenvalue problem of the multi-degree of freedom system. Note that in this formulation, the lowest eigenvalue corresponds to the lowest natural frequency in the system. Alternatively, if the stiffness matrix of the system is nonsingular, Eq. 3.156 leads to ω2 K1 MA ¼ A

ð3:159Þ

μA ¼ Hf A

ð3:160Þ

which can be written as

where μ ¼ 1/ω2 and Hf ¼ K1M. Equation 3.160 is referred to as the flexibility formulation of the eigenvalue problem of the multi-degree of freedom system. In this formulation, the lowest eigenvalue corresponds to the highest natural frequency of the system. In the following, we discuss some numerical methods which can be used to determine the eigenvalues and eigenvectors of the system using either the stiffness or the flexibility formulation. Matrix-Iteration Method This method, which is an iterative procedure referred to as the Stodla method, can be used with either the stiffness or the flexibility formulation. When it is used with the stiffness formulation, the solution

166

3

Multi-degree of Freedom Systems

converges to the mode shape which corresponds to the highest eigenvalue, or equivalently the highest natural frequency. On the other hand, if the flexibility formulation is used, the solution converges to the mode shape which corresponds to the highest eigenvalue which corresponds in this case to the lowest natural frequency. In the remainder of this section, the matrix-iteration method is discussed using the flexibility formulation of Eq. 3.160. In this method, a trial shape associated with the lowest natural frequency is assumed. We denote this trial shape which should represent the best possible estimate of the first mode as A10, where the first subscript refers to the mode number and the second subscript refers to the iteration number. Since the mode shape can be determined to within an arbitrary constant, without any loss of generality, the first element in A10 is assumed equal to 1, that is, 2

1

3

6 ðA Þ 7 10 2 7 ^ 10 ¼ 6 A 6 7 4 ðA10 Þ3 5

ð3:161Þ

⋮ ^ 10 is substituted into the right-hand side In the matrix-iteration method, the vector A of Eq. 3.160. This yields ^ 10 ¼ A11 ¼ μ11 A ^ 11 Hf A

ð3:162Þ

^ 11 such that its first Since the eigenvalue μ is not known at this stage, we select A ^ 11 is the same as element is equal to 1. This in turn will automatically define μ11. If A ^ ^ 11 in the ^ A 10 , then A 10 is the true mode shape. If this is not the case, we substitute A right-hand side of Eq. 3.160. This leads to ^ 11 ¼ A12 ¼ μ12 A ^ 12 ¼ Hf A

1 2^ H A 10 μ11 f

ð3:163Þ

^ 12 is the same or close enough to A11, convergence is achieved, otherwise A ^ 12 is If A substituted into the right-hand side of Eq. 3.160 to determine A13. This process ^ 1j is close enough to A ^ 1ðj1Þ for some integer j. The vector A ^ 1j then continues until A

defines the first mode shape and the scalar μ1j defines the first (highest) eigenvalue. This eigenvalue corresponds to q theffiffiffiffiffiffiffiffiffiffi lowest natural frequency which can be deterffi mined from the equation ω1 ¼ 1=μ1j . Clearly, the stiffness formulation can be used if the interest is in determining the higher mode of vibration.

Convergence It can be proved that the iterative procedure described in this section converges to the first mode shape of vibration of the system. Let A1, A2, . . ., An be the linearly independent true mode shapes of the multi-degree of freedom system. Since these mode shapes are linearly independent, any other n-dimensional vector

3.10

Matrix-Iteration Methods

167

can be expressed as a linear combination of these mode shapes. Therefore, the initial guess A10 can be written in terms of these mode shapes as A10 ¼ α1 A1 þ α2 A2 þ    þ αn An ¼

n X

αk Ak

ð3:164Þ

k¼1

where α1, α2, . . ., αn are scalar coefficients. Since the mode shapes can be determined to within an arbitrary constant, the iteration procedure described in this section leads to the following relationships:   A1 j ¼ Hf A1ðj1Þ ¼ Hf Hf A1ðj2Þ ¼ Hfj A10

ð3:165Þ

Substituting Eq. 3.164 into Eq. 3.165, one obtains A1 j ¼ Hfj

n X

αk Ak ¼

n X

k¼1

αk Hfj Ak

ð3:166Þ

k¼1

It is clear from the eigenvalue problem of Eq. 3.160 that HfAk ¼ μkAk. If both sides of this equation are multiplied by Hf, one obtains H2f Ak ¼ μk Hf Ak ¼ μ2k Ak . If both sides of this equation are multiplied again by Hf, one gets H3f Ak ¼ μ2k Hf Ak ¼ μ3k Ak . One, therefore, has the following general relationship: Hfj Ak ¼ μkj Ak

ð3:167Þ

Substituting this equation into Eq. 3.166 yields A1 j ¼

n X

αk Hfj Ak ¼

k¼1

n X

αk μkj Ak

ð3:168Þ

k¼1

in which μ1 > μ2 > μ3 >    > μn. As the number of iterations j increases, the first term in the summation of Eq. 3.168 becomes larger in comparison with other terms. Consequently, A1j will resemble the first true mode A1, that is, A1 j ¼

n X

αk μkj Ak  α1 μ1j A1

ð3:169Þ

k¼1

as j increases. Example 3.8

For the torsional system of Example 3.1, the symmetric mass and stiffness matrices are given by (continued)

168

3

2

2 M ¼ 103 4 0 0

3 0 0 3 0 5, 0 4

Multi-degree of Freedom Systems

2

1 K ¼ 12  105 4 1 0

1 3 2

3 0 2 5 5

In order to use the flexibility formulation, one has to find the inverse of the stiffness matrix as K1

2 11 1 4 5 ¼ 72  105 2

5 5 2

3 2 25 2

The matrix Hf of Eq. 3.160 is given by 2

11 1 4 5 Hf ¼ K1 M ¼ 3600 2

7:5 7:5 3

3 4 45 4

An initial guess of the trial shape of the first mode is assumed as 2

^ 10 A

3 1 ¼ 4 0:8 5 0:4

Substituting this trial shape into the right-hand side of Eq. 3.160, one obtains ^ 10 Hf A

2 11 7:5 1 4 5 7:5 ¼ 3600 2 3

32 3 2 3 4 1 18:6 1 4 12:6 5 ¼ A11 4 54 0:8 5 ¼ 3600 4 0:4 6

^ 11 , μ11 , and ω11 as which can be used to define A 2

^ 11 A

3 1:00 6 7 ¼ 4 0:6774 5,

μ11 ¼

0:3226

18:6 ¼ 5:1667  103 , 3600

ω11 ¼ 13:912 rad=s

^ 11 into the right-hand side of Eq. 3.160 Substituting the vector A ^ 11 Hf A

2 11 7:5 1 4 5 7:5 ¼ 3600 2 3

32 3 2 3 4 1:00 17:3709 1 4 11:3709 5 ¼ A12 4 54 0:6774 5 ¼ 3600 4 0:3226 5:3226

which yields (continued)

3.10

Matrix-Iteration Methods

2

1:000

169

3

7 ^ 12 ¼ 6 A 4 0:6546 5, 0:3064

μ12 ¼

17:3709 ¼ 4:8253  103 , 3600

ω12 ¼ 14:3959 rad=s

^ 12 as the new trial vector, one obtains Using A ^ 12 Hf A

2 11 7:5 1 4 5 7:5 ¼ 3600 2 3

32 3 2 3 4 1:000 17:1351 1 4 11:1351 5 ¼ A13 4 54 0:6546 5 ¼ 3600 4 0:3064 5:1894

^ 13 , μ13 , and ω13 as which defines A 2

1:000

3

17:1351 7 ^ 13 ¼ 6 ¼ 4:7598  103 , ω13 ¼ 14:4947 rad=s A 4 0:6498 5, μ13 ¼ 3600 0:3028 Observe that the matrix-iteration method is converging to the solution obtained in Example 3.2. If another iteration is made, one has ^ 13 Hf A

2 11 7:5 1 4 5 7:5 ¼ 3600 5 3

32 3 2 3 4 1:000 17:0847 1 4 11:0847 5 ¼ A14 4 54 0:6498 5 ¼ 3600 4 0:3028 5:1606

^ 14 , μ14 , and ω14 as which defines A 2

1:000

3

7 ^ 14 ¼ 6 A 4 0:649 5, 0:302

μ14 ¼

17:0847 ¼ 4:7458  103 , 3600

ω14 ¼ 14:516 rad=s

Analysis of Higher Modes In order to prove the convergence of the matrix-iteration method to the mode shape associated with the lowest natural frequency, the fact that the initial guess for the trial shape can be expressed as a linear combination of the linearly independent true mode shapes was utilized. Clearly, the convergence will be faster if the trial shape is close to the shape of the first mode. Furthermore, the matrix-iteration method is more efficient if the natural frequencies are widely separated, since in the analysis of convergence, the argument that the jth power of the first eigenvalue is much higher than the jth power of the other eigenvalues was made. In order to determine the second mode of vibration, one can simply eliminate the dependence of the trial shape on the first mode of vibration. By using a similar argument to the one previously made, one can prove that this trial shape converges to

170

3

Multi-degree of Freedom Systems

the second mode of vibration, since the jth power of μ2, for large j, is assumed to be much higher than the jth power of μ3, μ4, . . ., μn. Let us assume that the trial shape for the second mode A20 can be expressed as a linear combination of the true mode shapes, that is, A20 ¼

n X

αk Ak

ð3:170Þ

k¼1

where α1, α2, . . ., αn are constant scalars. In order to eliminate the dependence of the trial shape on the first mode of vibration, the orthogonality condition is used. Since the mode shapes are orthogonal with respect to the mass matrix, one has AT1 MA20

¼ AT1 M

" n X

# αk Ak ¼ α1 AT1 MA1 þ α2 AT1 MA2 þ    þ αn AT1 MAn

k¼1

ð3:171Þ

¼ α1 AT1 MA1

If the trial shape is assumed to be independent of the first mode of vibration, the preceding equation reduces to AT1 MA20 ¼ 0

ð3:172Þ

This is a scalar equation, which can be written as b1 ðA20 Þ1 þ b2 ðA20 Þ2 þ    þ bn ðA20 Þn ¼ 0

ð3:173Þ

where b1, b2, . . ., bn are constant scalars, and (A20)k is the kth element of the trial shape A20. The constant scalars b1, b2, . . ., bn are assumed to be known at this stage, since the first mode is assumed to be known. These constants are simply given by bT ¼ ½ b1 b2    bn  ¼ AT1 M. Dividing Eq. 3.173 by b1, one obtains ðA20 Þ1 ¼ s12 ðA20 Þ2 þ s13 ðA20 Þ3 þ    þ s1n ðA20 Þn

ð3:174Þ

in which the first element in the trial mode shape is expressed as a linear combination of the other elements, and s1k ¼ 

bk , b1

k ¼ 2,3,. . . ,n

In addition to Eq. 3.174, one has the following simple equations:

ð3:175Þ

3.10

Matrix-Iteration Methods

171

ðA20 Þ2 ¼ ðA20 Þ2 ðA20 Þ3 ¼

9 > > > =

ðA20 Þ3

⋮

⋱

ðA20 Þn ¼

ðA20 Þn

> > > ;

ð3:176Þ

Combining Eqs. 3.174 and 3.176, one obtains A20 ¼ S1 A20

ð3:177Þ

where S1 is called the sweeping matrix of the first mode and is given by 2

0 6 60 6 S1 ¼ 6 60 6 4⋮ 0

s12 1

s13 0

0

1

⋮ 0

⋮ 0

3    s1n 7  0 7 7  0 7 7 7 ⋱ ⋮5  1

ð3:178Þ

In Eq. 3.177, the dependence of the trial shape on the first mode is eliminated. When Eq. 3.177 is substituted in the right-hand side of Eq. 3.160, one obtains ^ 20 . This leads to the following iterative procedure: A21 ¼ Hf S1 A ^ 2ðj1Þ A2 j ¼ Hf S1 A

ð3:179Þ

This iterative procedure will converge to the second mode of vibration, thus defining μ2 and A2. The scalar μ2 can then be used to determine the second natural frequency. In order to determine the third mode of vibration using the matrix-iteration method, the dependence of the trial shape on the first and second modes must be eliminated. To this end, we use the following two orthogonality conditions: AT1 MA30 ¼ 0,

AT2 MA30 ¼ 0

ð3:180Þ

Using these two conditions, the first two elements in the trial-shape vector A30 can be expressed in terms of the other elements in a manner similar to Eq. 3.174. This leads to the sweeping matrix S2 which has the first two columns equal to zero. That is, the form of this sweeping matrix is as follows: 2

0

6 60 6 S2 ¼ 6 60 6 4⋮ 0

0 0 0

s13 s23

s14 s24

1

0

⋮

⋮

⋮

0

0

0

3    s1n 7    s2n 7 7  0 7 7 7 ⋱ ⋮5  1

ð3:181Þ

172

3

Multi-degree of Freedom Systems

Observe that the sweeping matrix S1 of Eq. 3.178 has a column rank n  1, and the sweeping matrix S2 of Eq. 3.181 has a column rank n  2. Using the sweeping matrix S2, one can define the trial shape vector which is independent of the first and second modes as A30 ¼ S2 A30

ð3:182Þ

^ 30 . This leads By substituting this equation into Eq. 3.160, one obtains A31 ¼ Hf S2 A to the iterative procedure, which can be expressed mathematically as ^ 3ðj1Þ A3 j ¼ Hf S2 A

ð3:183Þ

In order to determine an arbitrary mode i using the matrix-iteration method, one has to determine the sweeping matrix Si1 using the following orthogonality relationships: AT1 MAi0 ¼ 0,

AT2 MAi0 ¼ 0,

...,

ATi1 MAi0 ¼ 0

ð3:184Þ

By using the sweeping matrix Si1, Eq. 3.183 can be generalized for the ith mode as ^ iðj1Þ Aij ¼ Hf Si1 A

ð3:185Þ

This iterative procedure converges to the ith mode since the dependence of the trial shape on mode 1,2,. . ., i  1 has been eliminated using the sweeping matrix Si1. Example 3.9

In Example 3.8, it was shown that the first mode of vibration of the torsional system of Example 3.1 is given by 2

1:000

3

6 7 A1 ¼ 4 0:649 5 0:302 In order to determine the second mode, the orthogonality condition A1T MA20 ¼ 0 of Eq. 3.172 is used. The vector b is given by 2

2 bT ¼ A1T M ¼ ½ 1:000 0:649 0:302 4 0 0 ¼ 103 ½ 2:000 1:947 1:208 

0 3 0

3 0 0 5  103 4

(continued)

3.10

Matrix-Iteration Methods

173

Therefore, s1k, k ¼ 2, 3 of Eq. 3.175 is defined as ½ s12

s13  ¼ ½ 0:9735 0:604 

The sweeping matrix S1 of Eq. 3.178 is given by 2

0 S1 ¼ 4 0 0

s12 1 0

3 2 s13 0 0 5 ¼ 40 1 0

3 0:9735 0:604 1 0 5 0 1

Therefore, 2

11

7:5

4

32

0

0:9735

1 6 76 1 4 5 7:5 4 54 0 3600 2 3 4 0 0 2 3 0 3:2085 2:644 1 6 7 ¼ 2:6325 0:980 5 40 3600 0 1:053 2:792

0:604

Hf S1 ¼

0

3 7 5

1

One may assume the following trial shape for the second mode: 2

^ 20 A

3 1:0 6 7 ¼ 4 1:0 5 1:0

which leads to 2

A21

0 1 6 ^ ¼ Hf S1 A 20 ¼ 40 3600 0 2

3:2085 2:6325

1:053 3 5:8525 1 6 7 ¼ 4 3:6125 5 3600 3:845

32 3 2:644 1:0 76 7 0:980 54 1:0 5 2:792

1:0

^ 21 , μ21 , and ω21 as This vector can be used to define A 2

1:000

3

5:8525 7 ^ 21 ¼ 6 ¼ 1:6257  103 , ω21 ¼ 24:8016 rad=s A 4 0:61726 5, μ21 ¼ 3600 0:65698 (continued)

174

3

Multi-degree of Freedom Systems

^ 21 into the right-hand side of For the second iteration, we substitute A Eq. 3.179. This yields 2 ^ 21 ¼ A22 ¼ Hf S1 A

0

1 6 40 3600 0 2

3:2085 2:6325

1:053 3 3:7175 1 6 7 ¼ 4 2:2688 5 3600 2:4843

2:644

32

3

1:00

76 7 0:980 54 0:61726 5 0:65698

2:792

^ 22 , μ22 , and ω22 as which defines A 2

1:000

3

7 ^ 22 ¼ 6 A 4 0:6103 5,

μ22 ¼

0:6683

3:7175 ¼ 1:0326  103 , 3600

ω22 ¼ 31:119 rad=s

For the third iteration, we have 2 ^ 22 ¼ A23 ¼ Hf S1 A

0

1 6 40 3600 0 2

3:2085 2:6325

1:053 3 3:725 1 6 7 ¼ 4 2:2615 5 3600 2:509

2:644

32

1:000

3

76 7 0:980 54 0:6103 5 0:6683

2:792

^ 23 , μ23 , and ω23 can be defined as. Using this vector, A 2

^ 23 A

3 1:000 ¼ 4 0:607 5, 0:673

μ23 ¼

3:725 ¼ 1:034  103 , 3600

ω23 ¼ 31:087 rad=s

Another iteration shows that 2

^ 24 A

3 1:000 ¼ 4 0:606 5, 0:678

ω24 ¼ 31:05 rad=s

Comparing these results with the result of Example 3.2 demonstrates the convergence of this method to the true second mode of vibration. (continued)

3.10

Matrix-Iteration Methods

175

In order to use the matrix-iteration method to determine the third mode, we use the two orthogonality conditions of Eq. 3.180. This leads to the following sweeping matrix: 2

0 S2 ¼ 4 0 0

0 0 0

3 0:4096 1:0412 5 1

Hence, 2

11

7:5 4

32

0

0

1 6 76 4 5 7:5 4 54 0 0 3600 2 3 4 0 0 2 3 0 0 0:69105 1 6 7 ¼ 4 0 0 1:76475 5 3600 0 0 1:6941

Hf S2 ¼

0:4093

3

7 1:0415 5 1

One may assume a trial shape for the third mode as 2

^ 30 A

3 1:00 ¼ 4 1:00 5 1:00

The first iteration then yields 2 A31 ¼ Hf S2 A30 ¼

0

1 6 40 3600 0 2

0 0

0:69105

32

1:00

3

76 7 1:76475 54 1:00 5

0

1:6941 3 0:69105 1 6 7 ¼ 4 1:76475 5 3600 1:6941

1:00

^ 31 , μ31 , and ω31 as This vector can be used to define A 2

1:000

3

6 7 ^ 31 ¼ 6 2:5537 7, μ31 ¼ 0:69105 ¼ 1:9196  103 , ω31 ¼ 72:1766 rad=s A 4 5 3600 2:4515 (continued)

176

3

Multi-degree of Freedom Systems

The second iteration yields 2

^ 32 A

3 1:000 ¼ 4 2:5537 5, 2:4515

μ32 ¼ 4:7059  104 ,

ω32 ¼ 46:0978 rad=s

Rigid-Body Modes In a semidefinite system, the stiffness matrix is singular. In this case, the inverse of the stiffness matrix does not exist because of the rigid-body modes. The matrix equation of free undamped vibration of such systems can be written as M€ q þ Kq ¼ 0. Since the natural frequencies associated with the rigidbody modes are equal to zeros, these modes can be calculated or identified by inspection. These rigid-body modes will be denoted as Ar1 , Ar2 , . . . , Arnr where nr is the number of the rigid-body modes. The solution vector can be expressed using the mode superposition as q¼

nr X

αl Arl þ

l¼1

nd X

βk Ak

ð3:186Þ

k¼1

where αl and βk are constants, and nd is the number of deformation modes. In order to use the flexibility formulation to determine the deformation modes, the rigid-body modes must be eliminated by writing the solution vector q in terms of the deformation modes only. Therefore, Eq. 3.186 reduces to q¼

nd X

βk Ak

ð3:187Þ

k¼1

Using the orthogonality of the mode shapes and Eq. 3.187, one has the following relationships: ATr1 Mq ¼ 0,

ATr2 Mq ¼ 0,

...,

ATrnr Mq ¼ 0

ð3:188Þ

The scalar equations in Eq. 3.188 can be combined in one matrix equation, which can be written as Bq ¼ 0

ð3:189Þ

where B is the matrix 2

3 ATr1 6 T 7 6 Ar2 7 7 B¼6 6 ⋮ 7M 4 5 ATrnr

ð3:190Þ

3.10

Matrix-Iteration Methods

177

If the relationships of Eq. 3.188 are assumed to be linearly independent, the matrix B has a full row rank. Observe that the dimension of the matrix B is nr  n where n ¼ nr + nd is the total number of generalized coordinates. Equation 3.189 can then be used to write nr coordinates in terms of the other nd coordinates. In this case, Eq. 3.189 can be written in the following partitioned form:  ½ Bd

Bi 

qd qi

¼0

ð3:191Þ

or Bd qd þ Bi qi ¼ 0

ð3:192Þ

where Bd is an nr  nr matrix, and Bi is an nr  nd matrix. The partitioning in Eq. 3.191 is made in such a manner that the matrix Bd is a nonsingular matrix. Such a matrix exists since the matrix B is assumed to have a full row rank. The vector qd of Eq. 3.192 can then be expressed in terms of the vector qi as qd ¼ B1 d Bi qi

ð3:193Þ

Using Eq. 3.193, the vector q of the generalized coordinates can be written as  q¼

qi qd



 ¼



qi B1 d Bi qi

 ¼

I B1 d Bi

qi

ð3:194Þ

which can be written as q ¼ Br qi

ð3:195Þ

where Br is the matrix defined as  Br ¼

I



B1 d Bi

ð3:196Þ

€i . € ¼ Br q Differentiating Eq. 3.195 twice with respect to time, one obtains q Substituting into M€ q þ Kq ¼ 0 and premultiplying by the transpose of the matrix €i þ BrT KBr qi ¼ 0, which can be written as Br one obtains BrT MBr q €i þ Ki qi ¼ 0 Mi q

ð3:197Þ

where Mi ¼ BrT MBr and Ki ¼ BrT KBr . The stiffness matrix Ki is a nonsingular matrix since the rigid-body modes are eliminated. Therefore, the flexibility formulation as described earlier in this section can be used with the matrix-iteration method to determine the mode shapes numerically.

178

3

Multi-degree of Freedom Systems

Example 3.10

In this example, the matrix-iteration method is used to determine the vibration modes of the torsional system of Example 3.5. The mass and stiffness matrices of this system are given by 2

2 0 M ¼ 103 4 0 3 0 0

3 0 0 5, 4

2

1 K ¼ 12  105 4 1 0

1 3 2

3 0 2 5 2

This system has one rigid-body mode which occurs when all the disks have the same displacement. Therefore, nr ¼ 1 and 2

3 1:00 Ar ¼ 4 1:00 5 1:00 The matrix B of Eq. 3.190 can then be determined as 2 B ¼ ArT M ¼ ½ 1:000

2

6 1:000 1:000 4 0 0

¼ 103 ½ 2:000 3:000

0 0

3

7 3 0 5  103 0 4

4:000 

Substituting this matrix into Eq. 3.189 with the vector q ¼ ½ θ1 one obtains

θ2

θ 3 T ,

2θ1 þ 3θ2 þ 4θ3 ¼ 0 or 2

½2 3

3 θ1 4 4 θ 2 5 ¼ 0 θ3

From which θ1 can be expressed in terms of θ2 and θ3 as θ1 ¼ 1:5θ2  2θ3 Therefore, the vector q can be written in terms of θ2 and θ3 as 2

3 2 θ1 1:5 q ¼ 4 θ2 5 ¼ 4 1 0 θ3

3 2:00  θ 0 5 2 θ3 1

This defines the matrix Br of Eq. 3.195 as (continued)

3.11

Method of Transfer Matrices

179

2

1:5 Br ¼ 4 1 0

3 2:00 0 5 1

The matrices Mi and Ki of Eq. 3.197 are given by 2 32 3 2 0 0 1:5 2:00 1:5 1 0 4 0 3 0 54 1 0 5 Mi ¼ BrT MBr ¼ 103 2:00 0 1 0 0 4 0 1  7:5 6:0 ¼ 103 6:0 12 2 32 3  1 1 0 1:5 2:00 1:5 1 0 4 Ki ¼ BrT KBr ¼ 12  105 1 3 2 54 1:0 0:0 5 2:00 0 1 0 2 2 0 1:0  8:25 3:0 ¼ 12  105 3:0 6:0 

Using Mi and Ki, the reduced system of equations can be written as €i þ Ki qi ¼ 0 Mi q or  103

7:5 6:0 6:0 12



 θ€2 5 8:25 þ 12  10 3:0 θ€3

3:0 6:0



θ2 θ3

¼

 0 0

This system has no rigid-body modes, and, therefore, other vibration modes can be determined using the analytical methods or the numerical procedure described in this section. Having determined the modes of θ2 and θ3, Eq. 3.195 can be used to determine the modes of θ1, θ2, and θ3.

3.11

Method of Transfer Matrices

In the preceding section, a numerical procedure for determining the mode shapes and natural frequencies of the multi-degree of freedom systems was discussed. The convergence of this numerical procedure was proved, and numerical examples that demonstrate its use were presented. The basic idea used in the matrix-iteration method or Stodla method, as it is sometimes referred to, is that an assumed mode shape is successively adjusted until the true mode shape is obtained. Once convergence is achieved, the natural frequency associated with this mode can be evaluated. In this section, an alternative method which proceeds in the reverse sequence is

180

3

Multi-degree of Freedom Systems

Fig. 3.13 Holzer’s method

discussed. That is, an initial assumption for the frequency is first made. By successive adjustments, the true frequency is obtained and the mode shape associated with this frequency is simultaneously evaluated. The procedure discussed in this section is suitable for the analysis of mechanical and structural chain systems whose elements are arranged along a basic axis. While such a procedure is not used in practical applications, the concept used to develop such a procedure is important and is worthy of the presentation. In order to demonstrate the fundamental concept of the procedure, we first discuss Holzer’s method. We then generalize this procedure by using a rectilinear mass– spring multi-degree of freedom system. The application of the method to other systems such as torsional systems are also discussed in this section. Holzer’s Method This method is basically a trial-and-error scheme for determining the natural frequencies of multi-degree of freedom systems. While the method has had its widest application in the analysis of multi-degree of freedom torsional systems, the same concept can be applied to the analysis of other systems as demonstrated later in this section. Figure 3.13 shows a torsional system which has n degrees of freedom. Recall that if this system vibrates at one of its natural frequencies, the shape of deformation will be the mode shape associated with this frequency. We have previously shown that in the case of free vibration, the equation of motion of the multi-degree of freedom system is given by M€ q þ Kq ¼ 0, where M and K are, respectively, the symmetric mass and stiffness matrices, and q is the vector of generalized coordinates. A solution of the equation of motion is assumed in this case in the form q ¼ Asin (ωt þ ψ), where A ¼ ½ A1 A2    An T is the vector of amplitudes of the disks. Substituting this solution into the equation of motion yields the following relationships between the amplitudes of the disks:   K  ω2 M A ¼ 0

ð3:198Þ

where 2

and

3 0 0 7 7 7 ⋱ ⋮ 5

I1 0 0 60 I 0 2 6 M¼6 4⋮ ⋮ ⋮

 

0



0

0

In

ð3:199Þ

3.11

Method of Transfer Matrices

2

k11

k12

6k 6 21 K¼6 4⋮

k22 ⋮

kn1

kn2

   k 1n

181

3

2

k1 7 6    k 2n 7 6 k 1 7¼6 ⋱ ⋮ 5 4⋮ 0    k nn

k1 k1 þ k2

0 k2

 

0 0

⋮

⋮

⋱

⋮

0

0

   k ðn1Þ

3 7 7 7 5

ð3:200Þ

Therefore, Eq. 3.198 yields the following scalar equations: 9 I 1 ω2 A1 þ k 12 ðA1  A2 Þ ¼ 0 > > > > I 2 ω2 A2 þ k12 ðA2  A1 Þ þ k 23 ðA2  A3 Þ ¼ 0 > > = I 3 ω2 A3 þ k23 ðA3  A2 Þ þ k 34 ðA3  A4 Þ ¼ 0 ⋮ I n ω An þ kðn1Þn ðAn  An1 Þ ¼ 0 2

> > > > > > ;

ð3:201Þ

where Ij denotes the mass moment of inertia of the disk j and kij is the ijth element of the stiffness matrix K. Since there is no external force or couple acting on this system, adding the preceding equations yields n X

I j ω2 Aj ¼ 0

ð3:202Þ

j¼1

Since the mode shape can be determined to within an arbitrary constant, without any loss of generality, we assume that the amplitude of the first disk is equal to one. Assuming different values of ω, Eq. 3.198 or 3.201 can be solved for the amplitudes A2, A3, . . ., An. For example, from the first equation in Eq. 3.201, the amplitude A2 can be written in terms of the amplitude A1 as A2 ¼ A1  (I1ω2A1/k12). Once A2 is determined A3 can be calculated by using the second equation in Eq. 3.201 as A3 ¼ A2  ((k12(A2  A1)  I2ω2A2)/k23). Continuing in this manner, all the amplitudes corresponding to the assumed value of ω can be determined. The assumed value of ω and the calculated amplitudes can then be substituted into Eq. 3.202. If this equation is satisfied, then ω is a root of this equation, and accordingly, the assumed ω is a natural frequency of the system, and the corresponding amplitudes A1, A2, . . ., An represent the mode shape associated with this frequency. One, therefore, may write Eq. 3.202 as n X

  I j ω2 Aj ¼ f ω2

ð3:203Þ

j¼1

One can then plot f(ω2) as a function of ω2, and the roots of this function which define the natural frequency of the system can be found. The method of transfer matrices can be considered as a generalization of the Holzer’s method. The application of the transfer-matrix method to the vibration analysis of rectilinear mass–spring systems as well as torsional systems is also discussed in this section.

182

3

Multi-degree of Freedom Systems

Fig. 3.14 Transfer-matrix method

Rectilinear Mass–Spring System Figure 3.14 shows a multi-degree of freedom mass–spring system. Let €xi denote the acceleration of the mass i, F ir is the spring force acting on the mass from the right, and F il is the spring force acting on the mass i from the left. The dynamic equilibrium condition for the mass i is given by mi €xi ¼ F ir  F il

ð3:204Þ

If one assumes harmonic oscillation, the displacement of the mass i can be written as xi ¼ Xi sin (ωt þ ϕ), where Xi is the amplitude, ω is the frequency, and ϕ is the phase angle. It follows that €xi ¼ ω2 xi . Substituting this equation into Eq. 3.204 and rearranging the terms in this equation, one obtains F ir ¼ ω2 mi xi þ F il

ð3:205Þ

Furthermore, for the point mass mi, one has xir ¼ xil ¼ xi

ð3:206Þ

Equations 3.205 and 3.206 can be combined into one matrix equation. This yields 

xir F ir





1 ¼ ω2 mi

0 1



xil F il

ð3:207Þ

which can be written compactly as sir ¼ Pi sil

ð3:208Þ

where 

sir

xir ¼ , F ir



sil

xil ¼ , F il



1 Pi ¼ ω2 mi

0 1

ð3:209Þ

The vectors sir and sil are called the state vectors, and Pi is called the point matrix.

3.11

Method of Transfer Matrices

183

If the equilibrium of the massless spring ki is considered, the following relationships can be verified: r xil ¼ xi1 þ

1 r F , ki i1

r F il ¼ F i1

ð3:210Þ

These two equations can be combined in one matrix equation. This leads to "

xil

#

 ¼

F il

1 0

1=ki 1



r xi1

ð3:211Þ

r F i1

which can be written as r sil ¼ Hi si1

ð3:212Þ

where Hi is called the field matrix and is defined as  Hi ¼

1 0

1=k i 1

ð3:213Þ

Substituting the vector sil as defined by Eq. 3.212 into Eq. 3.208, one obtains r sir ¼ Pi Hi si1

ð3:214Þ

r sir ¼ Ti si1

ð3:215Þ

which can also be written as

where Ti is called the transfer matrix defined as  T i ¼ Pi Hi ¼

1

ω2 mi 1 ¼ ω2 mi 

0



1

1

1=ki



0 1 1=ki ð1  ω2 mi =k i Þ

ð3:216Þ

r can be expressed in terms of the state Using a similar procedure, the state vector si1 r vector si2 as r r ¼ Ti1 si2 si1

ð3:217Þ

r which, upon substituting into Eq. 3.215, leads to sir ¼ Ti Ti1 si2 . Continuing in this manner, one obtains

sir ¼ Ti Ti1 Ti2   T1 s0r

ð3:218Þ

where s0r is the state vector at the initial station. One can then write Eq. 3.218 in the simple form

184

3

Multi-degree of Freedom Systems

sir ¼ Tit s0r

ð3:219Þ

Tit ¼ Ti Ti1 Ti2   T1

ð3:220Þ

where the matrix Tit is given by

Equation 3.219, in which the state vector at station i is expressed in terms of the state vector of the initial station, represents two algebraic equations. Usually, a force boundary condition or a displacement boundary condition would be known at each end of the system. When these boundary conditions are substituted into this equation, one obtains the characteristic equation which can be solved for the natural frequencies, which can then be used to determine the mode shapes of the system. This procedure represents one alternative for using Eq. 3.219. Another alternative which is more suited for a system with a large number of degrees of freedom is to use a numerical procedure based on Holzer’s method. The natural frequency of the system and the unknown boundary condition at station 0 are assumed. The unknown values of the displacement and force at station 1 can then be determined by using the transfer matrix T1. This process can be continued until the other end of the system is reached, thus defining the boundary condition at this end. Since either the displacement or the force must be known at this end, the results obtained using the procedure of the transfer matrix can be checked and the error can be determined. If the error is not equal to zero or not relatively small, another value of the natural frequency is assumed and the entire procedure is repeated. This process is continued until the error becomes zero or relatively small. Example 3.11

The use of the transfer-matrix method is demonstrated in this example by using the simple two degree of freedom system shown in Fig. 3.15. We assume for simplicity that m1 ¼ m2 ¼ m and k1 ¼ k2 ¼ k. The boundary conditions at station 0 are x ¼ 0 and F ¼ F0. The boundary conditions at mass 2 are x ¼ x2 and F ¼ 0. By using Eq. 3.215, one has s1r ¼ T1 s0r . That is, 

x1r F 1r



 ¼

1

1=k

ω2 m

ð1  ω2 m=kÞ



x0r F 0r



 ¼

1

1=k

ω2 m

ð1  ω2 m=kÞ



0



F0

Fig. 3.15 A two degree of freedom system

(continued)

3.11

Method of Transfer Matrices

185

Similarly, one has s2r ¼ T2 s1r , which can be written in a more explicit form as 

x2r F 2r



¼

x2 0



 ¼

1 ω2 m

1=k ð1  ω2 m=kÞ



x1r F 1r



which, upon substitution for x1r and F 1r , yields 

x2 0



 ¼ 2 6 ¼4



1

1=k

ω2 m

ð1  ω2 m=kÞ



1

1=k

ω2 m

ð1  ω2 m=kÞ

3

0



F0

 7 0 5 2 F0 ω2 m=k þ ð1  ω2 m=kÞ

ð1  ω m=k Þ   ω2 m 2 ω m 2  k

ð1=kÞð2  ω m=kÞ

2

2

which yields the following two algebraic equations:   F0 ω2 m 2 x2 ¼ k "k # 2 ω2 m ω2 m 0¼ 1  F0 k k The second algebraic equation yields  2  2 2 ω2 m ω2 m ωm ω2 m ¼ þ1¼0 1  3 k k k k or  2 k ω  3ω þ ¼0 m m 2k

4

Observe that this is the characteristic equation of the system, which has the following roots: ω2 ¼

k h pffiffiffii 3∓ 5 2m

That is, ω21

k ¼ 0:38197 , ω1 ¼ 0:618 m

rffiffiffiffi k , m

ω22

k ¼ 2:61803 , m

rffiffiffiffi k ω2 ¼ 1:618 m (continued)

186

3

Multi-degree of Freedom Systems

If we substitute these values for ω1 and ω2 into the first algebraic equation obtained from the application of the transfer-matrix procedure, we obtain for the first mode x2 ¼

  F0 ω2 m F0 F0 2 1 ¼ ð2  0:38197Þ ¼ 1:61803 k k k k

and for the second mode x2 ¼

  F0 ω2 m F0 F0 2 2 ¼ ð2  2:61803Þ ¼ 0:61803 k k k k

Since x1 ¼ F0/k, the first and second mode shapes are given by 

1 , A1 ¼ 1:61803



1 A2 ¼ 0:61803



Torsional System The transfer-matrix method can also be used in the vibration analysis of torsional systems. The point, field, and transfer matrices can be obtained in a manner similar to the case of rectilinear mass–spring systems. Figure 3.16 shows a disk i, the left surface of which is connected to a shaft which has stiffness ki, while the right surface is connected to a shaft which has stiffness ki+1. Assuming harmonic oscillations, the equation of motion of the disk is given by M ir ¼ ω2 I i θi þ M il

ð3:221Þ

where Mi is the moment, Ii is the mass moment of inertia of the disk, and θi is the angular oscillation which can be used to define the equation θi ¼ θir ¼ θil Equations 3.221 and 3.222 can be combined in order to obtain  l  r  θi 1 0 θi ¼ M ir ω2 I i 1 M il

ð3:222Þ

ð3:223Þ

which is in the same form as Eq. 3.207 developed for the rectilinear system. It can, therefore, be written in the form of Eq. 3.208 as

Fig. 3.16 Torsional system

3.11

Method of Transfer Matrices

187

sir ¼ Pi sil

ð3:224Þ

where the state vectors sir and sil and the point matrix Pi are defined by  sir ¼

θir , M ir

 sil ¼



θil , M il

Pi ¼

1 ω2 I i

0 1

ð3:225Þ

If one considers the equilibrium of the shaft ki which is assumed to be massless, we have r θil ¼ θi1 þ

l r M , k i i1

r M il ¼ M i1

ð3:226Þ

which can be combined in one matrix equation as 

θil M il





1 ¼ 0

1=ki 1



r θi1 r M i1

ð3:227Þ

where the field matrix Hi can be identified as  Hi ¼

1 0

1=k i 1

ð3:228Þ

Equation 3.227 can be substituted into Eq. 3.223. This yields r sir ¼ Ti si1

ð3:229Þ

where Ti is the transfer matrix defined in this case as 

1 Ti ¼ Pi Hi ¼ ω2 I i

0 1



1 0

1=ki 1





1 ¼ ω2 I i

1=ki ð1  ω2 I i =ki Þ

ð3:230Þ

If Eq. 3.229 is applied to successive points of the torsional system, one obtains equations similar to Eqs. 3.219 and 3.220 developed for the rectilinear mass–spring system. Example 3.12

In order to demonstrate the use of the transfer-matrix method in the vibration analysis of torsional systems, we consider the simple two degree of freedom system shown in Fig. 3.17 which consists of two disks which have moments of inertia I1 and I2 and two shafts which have torsional stiffness coefficients k1 and k2. For simplicity, we assume that I1 ¼ I2 ¼ I and k1 ¼ k2 ¼ k. The boundary conditions at station 0 is θ ¼ 0 and M ¼ M0, while the boundary conditions at station 2 are θ ¼ θ2 and M ¼ 0. We, therefore, have (continued)

188

3



θ1r M 1r



¼

1 ω2 I

1=k ð1  ω2 I=kÞ



θ0r M 0r



¼

Multi-degree of Freedom Systems

1 ω2 I

1=k ð1  ω2 I=k Þ



0 M0



and 

θ2r M 2r





θ ¼ 2 0





1 ¼ ω2 I

1=k ð1  ω2 I=k Þ



θ1r M 1r



These two matrix equations yield "

θ2 0

#

" ¼

1

1=k

ω2 I

ð1  ω2 I=kÞ

#"

1

1=k

#"

0

#

ω2 I ð1  ω2 I=kÞ M 0 2 3" #  1 2  ω2 I=k ð1  ω2 I=kÞ 0 5 k ¼4 M0 2 2 2 2 2 ω I ð2  ω I=kÞ ω I=k þ ð1  ω I=kÞ

Fig. 3.17 Two degree of freedom system

which yields the following two scalar equations:   M0 ω2 I 2 θ2 ¼ k k ω2 I ω2 I 0¼ 1  M0 k k The second algebraic equation yields the characteristic equation of the system  2 k k ω4  3ω2 þ ¼0 I I which defines the natural frequencies of the system ω1 and ω2 as (continued)

3.12

Nonlinear Systems and Modal Decomposition

rffiffiffi k ω1 ¼ 0:618 , I

189

rffiffiffi k ω2 ¼ 1:618 I

By using a similar procedure to the one described in the preceding example, one can show that the mode shapes of this system are given by 

1 A1 ¼ , 1:61803

3.12



1 A2 ¼ 0:61803



Nonlinear Systems and Modal Decomposition

The concept of the modal decomposition discussed in this chapter is fundamental in many engineering applications, some of which are governed by highly nonlinear equations of motion. Recall that the mode shapes and natural frequencies are determined using constant mass and stiffness matrices of linear system of vibration equations. When the displacements are large, the assumption of linearity cannot be made. Nonlinearities are classified as geometric or material nonlinearities. In some engineering applications, the mass matrix can be highly nonlinear because of the geometric nonlinearities resulting from the finite rotation of the system components. In some other applications, the stiffness matrix can be nonlinear because of the geometric nonlinearities resulting from the large deformations. Material nonlinearities are the result of using nonlinear constitutive models as in the case of using stiffness coefficients that are functions of the coordinates or using materials that do not obey the linear Hooke’s law. When the mass and/or stiffness matrices are not constant, the mode shapes and natural frequencies become configuration dependent. In many important automotive, aerospace, and machine applications, however, the concept of the linear modes can still be used for highly nonlinear systems by creating a local linear problem. Using the local linear problem, an eigenvalue problem can be solved only once to determine the shape of deformation of the flexible components. This approach is widely used in the automotive, aerospace, and machine industries in the durability investigations of multibody systems (MBS) which are highly nonlinear. By introducing a local body coordinate system, a local linear vibration problem can be created while retaining all the geometric nonlinearities resulting from the finite rotations of the flexible bodies. The large displacements of the body are described using the local body coordinate system, while the body deformation is defined with respect to this body coordinate system. This approach is referred to in the literature as the floating frame of reference (FFR) formulation (Shabana 2013). The FFR formulation, widely used for MBS applications, allows using the mode shapes determined by solving the local linear problem only once in the vibration analysis of complex and highly nonlinear mechanical systems. An example of these MBS applications is an off-road vehicle called the High Mobility Multipurpose Wheeled Vehicle (HMMWV). The HMMWV model, shown in Fig. 3.18, is used to demonstrate the physical meaning of the eigenvectors in the

190

3

Multi-degree of Freedom Systems

Fig. 3.18 HMMWV vehicle model

Fig. 3.19 Chassis FE mesh with reference conditions

case of complex and highly nonlinear mechanical systems (Patel et al. 2018). The HMMWV, which is a four-wheel drive vehicle that is used for both military and civilian applications, is capable of operating both on-road and off-road. The vehicle has a 190-horsepower engine, a double “A” arm suspension with coil springs and double-acting shock absorbers, and a recirculating ball-, worm-, and nut-based powerassisted steering. The gross operating mass of the vehicle can vary; however, a good estimate of the vehicle curb mass is approximately 2500 kg. The maximum vehicle on-road speed is 113 km/h. In the MBS mathematical formulation of the HMMWV model, the suspension system can be modeled using rigid bodies, while the chassis and the tires can be considered as flexible bodies. In particular, the chassis can be modeled using the approach of small deformation analysis, and, therefore, its dynamic response can be represented as a superposition of normal modes. In order to obtain a flexible MBS model for the HMMWV chassis, the finite element (FE) method, discussed in Chap. 5 for obtaining the structure mass and stiffness matrices, can be effectively used (Patel et al. 2018). The FE mesh of the HMMWV chassis, shown in Fig. 3.19, is constructed using 435 three-dimensional nonisoparametric beam elements that employ displacements and rotations as nodal coordinates. The beam elements are assumed to have a rectangular cross-section of width W ¼ 0.05 m and height H ¼ 0.1 m. The material properties considered in this illustrative example are modulus of elasticity E ¼ 2.1  1011 Pa, Poisson ratio ν ¼ 0.33,

3.12

Nonlinear Systems and Modal Decomposition

191

Fig. 3.20 First five chassis mode shapes

and mass density ρ ¼ 7200 kg/m3. In the development of the FE mesh for the HMMWV chassis, displacement reference conditions denoted as nodal displacement reference conditions (NDRC) are applied to eight material points shown in Fig. 3.19 that correspond to the geometric locations where the vehicle suspension system and chassis are connected. The application of these conditions ensures a positive definite stiffness matrix for the chassis model, and therefore, there are no rigid-body modes of the chassis with respect to its coordinate system. The first five normal modes of the chassis and the corresponding natural frequencies obtained from the solution of the generalized eigenvalue problem are shown in Fig. 3.20. As will be discussed in Chap. 4, continuous systems such as the HMMWV chassis has an infinite number of degrees of freedom. The FE method, discussed in Chap. 5, allows for obtaining a finite dimensional model. In the FFR formulation, the local linear problem defines the vibration of the chassis with respect to its coordinate system which can have arbitrarily large rigid body displacements. The resulting local linear problem defines vibration equations with constant mass and stiffness matrices. The eigenvalue problem associated with this local linear vibration problem defines a modal transformation matrix, as described in this chapter. Modal truncation methods, as the one previously discussed in this chapter, can be used to significantly reduce the number of coordinates by eliminating high-frequency insignificant modes of vibration. The resulting reduced order linear vibration model can be used systematically to develop the highly nonlinear equations of the chassis which can experience arbitrarily large displacements. The use of the FFR formulation and the linear modes in the vibration analysis of highly nonlinear complex systems is covered in more detail in the MBS literature.

192

3

Multi-degree of Freedom Systems

Problems 3.1. By using Newton’s second law, obtain the differential equations of motion of the multi-degree of freedom system shown in Fig. P3.1.

Fig. P3.1

3.2. By using Lagrange’s equation, obtain the differential equations of motion of the system shown in Fig. P3.1. 3.3. Determine the natural frequencies and the mode shapes for Problem 3.1, assuming k1 ¼ k2 ¼ k3 ¼ k4 ¼ k ¼ 1  103 N/m, and m1 ¼ m2 ¼ m3 ¼ m ¼ 2 kg. 3.4. Repeat Problem 3.3, for the case in which k1 ¼ k2 ¼ k3 ¼ k4 ¼ k and m1 ¼ m, m2 ¼ 2m and m3 ¼ m. 3.5. Obtain the differential equations of free vibration of the three degree of freedom system shown in Fig. P3.2 by using Newton’s second law. Assume small oscillations.

Problems

193

Fig. P3.2

3.6. Use Lagrange’s equation to derive the differential equations of free vibration of the system of Problem 3.5. Obtain also the natural frequencies and the mode shapes of vibration, assuming that m1 ¼ m2 ¼ m3 ¼ 1 kg, l ¼ 0.5 m, and k1 ¼ 0, k2 ¼ k3 ¼ 2  103 N/m. 3.7. Derive the differential equations of motion of the system shown in Fig. P3.3. Determine the natural frequencies and the mode shapes of vibration. Assume that m1 ¼ m2 ¼ m3 ¼ 1 kg, and l ¼ 0.5 m. Assume small oscillations.

Fig. P3.3

194

3

Multi-degree of Freedom Systems

3.8. Assuming small oscillations, determine the natural frequencies and mode shapes of vibration of the system shown in Fig. P3.4. Assume that m1 ¼ 2 kg, m2 ¼ m3 ¼ 0.5 kg, l1 ¼ l2 ¼ 0.5 m, and k1 ¼ 2  103 N/m.

Fig. P3.4

3.9. Figure P3.5 depicts a shaft which supports three disks which have mass moments of inertia I1, I2, and I3. The shaft, which is assumed to have negligible mass, has a circular cross section of diameter d and modulus of rigidity G. Obtain the differential equations that govern the torsional vibration of this system.

Fig. P3.5

Problems

195

3.10. Obtain the natural frequencies and mode shapes of the torsional system of Problem 3.9 assuming that the shafts are made of steel and have equal diameters of 0.01 m, I1 ¼ I2 ¼ I3 ¼ 1  103 kgm2 and l1 ¼ l2 ¼ l3 ¼ 0.5 m. 3.11. Assuming small oscillations, derive the differential equations of motion of the system shown in Fig. P3.6 by using Newton’s second law. Obtain also the differential equations using Lagrange’s equation of motion.

Fig. P3.6

3.12. In Problem 3.11, if m1 ¼ m2 ¼ m3 ¼ m ¼ 1 kg, and k1 ¼ k2 ¼ k ¼ 2  103 N/m, obtain the natural frequencies and mode shapes, assuming that the length of the rod is 0.5 m. 3.13. By using Newton’s second law or Lagrange’s equation, derive the differential equations of motion of the system shown in Fig. P3.7. Assume small oscillations.

Fig. P3.7

196

3

Multi-degree of Freedom Systems

3.14. In Problem 3.3, if m ¼ 3 kg, k ¼ 1000 N/m, and the initial displacements of the masses m1, m2, and m3 are, respectively, 0.01, 0, and 0 m, determine the solution of the free vibration assuming that the masses have zero initial velocities. 3.15. For the system shown in Fig. P3.4, let m1 ¼ m2 ¼ m3 ¼ 2 kg, l1 ¼ l2 ¼ 0.5 m, and k1 ¼ 1000 N/m. The masses m1, m2, and m3 are assumed to have zero initial displacements. Obtain the solution for the free vibration assuming that the rod l2 has an initial angular velocity of 3 rad/s. 3.16. Find the solution of the free vibration of the system of Problem 3.8 in terms of the initial displacements and velocities. 3.17. Obtain the solution for the free vibration of the system in Problem 3.12 in terms of the initial conditions. 3.18. Determine the modal mass and stiffness coefficients of the system of Problem 3.6. 3.19. Determine the modal mass and stiffness coefficients of the system of Problem 3.8. 3.20. Determine the modal mass and stiffness coefficients of the system of Problem 3.10. 3.21. Determine the modal mass and stiffness coefficients of the system of Problem 3.12. 3.22. In Problem 3.12, if the spring k1 is removed, does the new system have a rigidbody mode? If the answer is yes, identify this rigid-body mode. 3.23. Show that, for an n-degree of freedom system, if the vectors of initial coordinates and velocities are proportional to a linear combination of m mode shapes, where m < n, then the motion of the n-degree of freedom system is equivalent to the motion of an m-degree of freedom system. 3.24. In Problem 3.12 determine the natural frequencies and mode shapes if the effect of gravity is neglected. 3.25. By inspection, identify the rigid-body modes of the system shown in Fig. P3.7 if the effect of gravity is neglected. 3.26. Using Lagrange’s equation or Newton’s second law, obtain the differential equations of motion of the forced vibration for the system shown in Fig. P3.8. The moment M is given by M ¼ M0 sin ωft, where M0 ¼ 2 Nm, and ωf ¼ 5 rad/s. Assume small oscillations, m1 ¼ m2 ¼ m3 ¼ 1 kg, k1 ¼ k2 ¼ 10 N/m, c1 ¼ c2 ¼ 10 Ns/m, and l ¼ 0.5 m.

Problems

197

Fig. P3.8

3.27. In Problem 3.26, if c1 ¼ c2 ¼ 0, obtain the steady-state solution. 3.28. The system shown in Fig. P3.9 consists of masses which are connected by elastic elements and a damper. The system is subjected to support excitation as shown in the figure. Derive the differential equations of motion of this system.

Fig. P3.9

3.29. In Problem 3.28, obtain the steady-state solution for the undamped system. Assume m1 ¼ m2 ¼ m3 ¼ 1 kg, k1 ¼ k2 ¼ k3 ¼ 1  103 N/m, Y0 ¼ 0.01 m, and ωf ¼ 4 rad/s 3.30. Assuming small oscillations, derive the differential equations of motion of the system shown in Fig. P3.10, where a1 þ a2 þ a3 þ a4 ¼ l.

Fig. P3.10

198

3

Multi-degree of Freedom Systems

Fig. P3.11

3.31. Obtain the differential equations of motion of the system shown in Fig. P3.11. Assume small oscillations. 3.32. Obtain the differential equations of motion of the system shown in Fig. P3.12. Assume small oscillations.

Fig. P3.12

3.33. Use the state transition matrix to determine the solution of the free vibration equation for the underdamped and overdamped single degree of freedom systems.

Problems

199

3.34. Use the state transition matrix to determine the solution of the equation of the forced vibration of the underdamped single degree of freedom systems. 3.35. Obtain the differential equations of motion of the vehicle system model shown in Fig. P3.13. Assume small angular oscillations.

Fig. P3.13

3.36. If the damping coefficient of the system in Fig. P3.9 is equal to zero and if m1 ¼ 1 kg, m2 ¼ 1.5 kg, m3 ¼ 2 kg, k1 ¼ 6  105 N/m, k2 ¼ 8  105 N/m, and k3 ¼ 10  105 N/m, determine the natural frequencies and mode shapes using the matrix-iteration method. 3.37. Show that the reduced system equations of motion obtained in Example 3.10 lead to the same natural frequencies and mode shapes determined in Example 3.5. 3.38. Use the matrix-iteration method to determine the natural frequencies and mode shapes of the system of Problem 3.3, assuming that m ¼ 2.5 kg. 3.39. Use the matrix-iteration method to determine the natural frequencies and mode shapes of the system of Problem 3.8 assuming that k1 ¼ 1.5  103 N/m. 3.40. Determine the natural frequencies and mode shapes of the system of Problem 3.27 using the matrix-iteration method. 3.41. Determine the mode shapes and natural frequencies of the system of Problem 3.36 using the transfer-matrix method.

4

Vibration of Continuous Systems

Mechanical systems in general consist of structural components which have distributed mass and elasticity. Examples of these structural components are rods, beams, plates, and shells. Our study of vibration thus far has been limited to discrete systems which have a finite number of degrees of freedom. As has been shown in the preceding chapters, the vibration of mechanical systems with lumped masses and discrete elastic elements is governed by a set of second-order ordinary differential equations. Rods, shafts, beams, and other structural components on the other hand are considered as continuous systems which have an infinite number of degrees of freedom. The vibration of such systems is governed by partial differential equations which involve variables that depend on time as well as the spatial coordinates. In this chapter, an introduction to the theory of vibration of continuous systems is presented. It is shown in the first two sections that the longitudinal and torsional vibration of rods can be described by second-order partial differential equations. Exact solutions for these equations are obtained using the method of separation of variables. In Sect. 4.3, the transverse vibrations of beams are examined and the fourth-order partial differential equation of motion that governs the transverse vibration is developed using the assumptions of the elementary beam theory. Solutions of the vibration equations are obtained for different boundary conditions. In Sect. 4.4, the orthogonality of the eigenfunctions (mode shapes or principal modes) is discussed for the cases of longitudinal, torsional, and transverse vibrations, and modal parameters such as mass and stiffness coefficients are introduced. The material covered in this section is used to study the forced vibration of continuous systems in Sect. 4.5, where the solution of the vibration equations is expressed in terms of the principal modes of vibration. Section 4.6 is devoted to the solution of the vibration equations subject to inhomogeneous boundary conditions, while in Sect. 4.7 the effect of damping on the vibration of viscoelastic materials is examined. In Sect. 4.8, the use of Lagrange’s equation and the scalar energy quantities in deriving the differential equations of motion is demonstrated. The last three sections are devoted to the use of approximation methods in the analysis of continuous systems. # Springer Nature Switzerland AG 2019 A. Shabana, Vibration of Discrete and Continuous Systems, Mechanical Engineering Series, https://doi.org/10.1007/978-3-030-04348-3_4

201

202

4.1

4

Vibration of Continuous Systems

Free Longitudinal Vibrations

In this section, we study the longitudinal vibration of prismatic rods such as the one shown in Fig. 4.1. The rod has length l and cross-sectional area A. The rod is assumed to be made of material which has a modulus of elasticity E and mass density ρ and is subjected to a distributed external force F(x, t) per unit length. Figure 4.1 also shows the forces that act on an infinitesimal volume of length δx. The geometric center of this infinitesimal volume is located at a distance x + δx/2 from the end of the rod. Let P be the axial force that results from the vibration of the rod. The application of Newton’s second law leads to the following condition for the dynamic equilibrium of the infinitesimal volume: 2

ρA

∂ u ∂P δx  P þ F ðx; t Þδx δx ¼ P þ 2 ∂t ∂x

ð4:1Þ

which can be simplified to yield ρA(∂2u/∂t2)δx ¼ (∂P/∂x)δx + F(x, t)δx. Dividing this equation by δx leads to 2

ρA

∂ u ∂P þ F ðx; t Þ ¼ ∂t 2 ∂x

ð4:2Þ

The force P can be expressed in terms of the axial stress σ as P ¼ Aσ. The stress σ can be written in terms of the axial strain ε using Hooke’s law as σ ¼ Eε, while the strain displacement relationship is ε ¼ ∂u/∂x. Therefore, the axial force can be expressed in terms of the longitudinal displacement as P ¼ EA

∂u ∂x

ð4:3Þ

Substituting Eq. 4.3 into Eq. 4.2 leads to
 2 ∂ u ∂ ∂u EA ρA 2 ¼ þ F ðx; t Þ ∂t ∂x ∂x

ð4:4Þ

This is the partial differential equation that governs the forced longitudinal vibration of the rod.

Fig. 4.1 Longitudinal vibration of rods

4.1 Free Longitudinal Vibrations

203

Free Vibration The equation of free vibration can be obtained from Eq. 4.4 by letting F(x, t) ¼ 0, that is,
 2 ∂ u ∂ ∂u EA ρA 2 ¼ ∂t ∂x ∂x

ð4:5Þ

If the modulus of elasticity E and the cross-sectional area A are assumed to be constant, the partial differential equation of the longitudinal free vibration of the rod can be written as 2

ρA

2

∂ u ∂ u ¼ EA 2 2 ∂t ∂x

ð4:6Þ

or 2

2

∂ u ∂ u ¼ c2 2 2 ∂t ∂x

ð4:7Þ

sffiffiffiffi E c¼ ρ

ð4:8Þ

where c is a constant defined by

Separation of Variables The general solution of Eq. 4.7 can be obtained using the method of the separation of variables. In this case, one assumes the solution in the form uðx; t Þ ¼ ϕðxÞqðt Þ

ð4:9Þ

where ϕ is a space-dependent function and q is a time-dependent function. The partial differentiation of Eq. 4.9 with respect to time and with respect to the spatial coordinate leads to 2

∂ u ¼ ϕðxÞ€qðt Þ, ∂t 2

2

∂ u ¼ ϕ00 ðxÞqðt Þ ∂x2

ð4:10Þ

where () denotes differentiation with respect to time and (0 ) denotes differentiation with respect to the spatial coordinate x, that is, €qðt Þ ¼ d 2 q=dt 2 and ϕ00 ðxÞ ¼ d2 ϕ=dx2 . Substituting Eq. 4.10 into Eq. 4.7 leads to ϕ€q ¼ c2 ϕ00 q, or c2

ϕ00 €q ¼ q ϕ

ð4:11Þ

Since the left-hand side of this equation depends only on the spatial coordinate x and the right-hand side depends only on time, one concludes that Eq. 4.11 is satisfied only if both sides are equal to a constant, that is,

204

4

c2

Vibration of Continuous Systems

ϕ00 €q ¼ ¼ ω2 q ϕ

ð4:12Þ

where ω is a constant. A negative constant ω2 was selected, since this choice leads to oscillatory motion. The choice of zero or a positive constant does not lead to vibratory motion, and therefore, it must be excluded. For example, one can show that if the constant is selected to be zero, the solution increases linearly with time, while if a positive constant is selected, the solution contains two terms: first is an exponentially increasing function, and the second is an exponentially decreasing function. This leads to an unstable solution which does not represent an oscillatory motion. Equation 4.12 leads to the following two equations: ϕ00 þ ðω=cÞ2 ϕ ¼ 0 q€ þ ω2 q ¼ 0

) ð4:13Þ

The solution of these two equations is given, respectively, by ω ω ) x þ A2 cos x c c qðt Þ ¼ B1 sin ωt þ B2 cos ωt

ϕðxÞ ¼ A1 sin

ð4:14Þ

By using Eq. 4.9, the longitudinal displacement u(x, t) can then be written as uðx; t Þ ¼ ϕðxÞqðt Þ  ω ω  ¼ A1 sin x þ A2 cos x ðB1 sin ωt þ B2 cos ωt Þ c c

ð4:15Þ

where A1, A2, B1, B2, and ω are arbitrary constants to be determined by using the boundary and initial conditions. Elastic Waves Equation 4.15 can also be written in the following alternate form: uðx; t Þ ¼ A sin

ω c

 x þ ϕ1 sin ðωt þ ϕ2 Þ

ð4:16Þ

where A, ϕ1, and ϕ2 are constants that can be expressed in terms of the constants A1, A2, B1, and B2. By using the trigonometric identity 2sinαsinβ ¼ cos (α  β)  cos (α + β), the preceding equation for the displacement u can be written as uðx; t Þ ¼

ω  A ω  A cos x  ωt þ ϕ1  ϕ2  cos x þ ωt þ ϕ1 þ ϕ2 2 c 2 c

which can be written as

ð4:17Þ

4.1 Free Longitudinal Vibrations

205

  uðx; t Þ ¼ f 1 ðkx  ωt þ ϕm Þ þ f 2 kx þ ωt þ ϕp

ð4:18Þ

where the constants ϕm, ϕp, and k are, respectively, ϕm ¼ ϕ1  ϕ2, ϕp ¼ ϕ1 + ϕ2, k ¼ ω/c, and the functions f1(kx  ωt + ϕm) ¼ (A/2) cos (kx  ωt + ϕm) and f2(kx + ωt + ϕp) ¼ (A/2) cos (kx + ωt + ϕp). The function f1(kx  ωt + ϕm) represents a harmonic wave traveling in the positive x direction with a wave velocity c ¼ ω/k, while the function f2(kx + ωt + ϕp) represents another elastic harmonic wave traveling in the opposite direction with the same wave velocity c, where c is defined by Eq. 4.8. Therefore, Eq. 4.7 is often called the wave equation of the longitudinal vibration of the rod. Note that the wave velocity c is constant and depends on the material properties of the rod. The constant k which relates the wave velocity c to the frequency of the harmonic ω is called the wave number. Boundary Conditions and the Orthogonality of the Eigenfunctions In order to demonstrate the procedure for determining the constants in Eq. 4.15, the example shown in Fig. 4.2, where the rod is fixed at one end and is free at the other end, is considered. At the free end, the stress must be equal to zero, that is, σ(l, t) ¼ Eε(l, t) ¼ E(∂u(l, t)/∂x) ¼ 0. Therefore, the boundary conditions at the fixed and free ends are given, respectively, by uð0; t Þ ¼ 0,

u0 ðl; t Þ ¼ 0

ð4:19Þ

The first boundary condition of Eq. 4.19 describes the state of displacement at the fixed end. This type of boundary condition is called geometric boundary condition. On the other hand, the second boundary condition of Eq. 4.19 describes the state of force or stress at the free end of the rod. This type of condition is often referred to as natural boundary condition. That is, the geometric boundary conditions describe the specified displacements and slopes, while the natural boundary conditions describe the specified forces and moments. Substituting Eq. 4.19 into Eq. 4.15 results in 9 uð0; t Þ ¼ ϕð0Þqðt Þ ¼ A2 qðt Þ ¼ 0 =   ω ω ω 0 A1 cos l  A2 sin l qðt Þ ¼ 0 ; u0 ðl; t Þ ¼ ϕ ðlÞqðt Þ ¼ c c c

Fig. 4.2 Fixed-end conditions

ð4:20Þ

206

4

Vibration of Continuous Systems

which lead to the two conditions A2 ¼ 0 and A1 cos(ωl/c) ¼ 0. For a nontrivial solution, these two conditions lead, respectively, to ϕðxÞ ¼ A1 sin

ω x, c

cos

ωl ¼0 c

ð4:21Þ

The second equation in Eq. 4.21, cos(ωl/c) ¼ 0, is called the frequency or the characteristic equation. The roots of this equation are given by ωl π 3π 5π ð2n  1Þπ ¼ , , , ..., , ... c 2 2 2 2

ð4:22Þ

This equation defines the natural frequencies of the rod as ωj ¼

ð2j  1Þπc , 2l

j ¼ 1,2,3,. . .

ð4:23Þ

Using the definition of the wave velocity c given by Eq. 4.8, the jth natural frequency ωj can be defined as ð2j  1Þπ ωj ¼ 2l

sffiffiffiffi E , ρ

j ¼ 1,2,3,. . .

ð4:24Þ

Thus, the continuous rod has an infinite number of natural frequencies. Corresponding to each of these natural frequencies, there is a mode shape or an eigenfunction ϕj defined by the first equation of Eq. 4.21, ϕðxÞ ¼ A1 sin ðω=cÞx, as ϕj ¼ A1 j sin

ωj x, c

j ¼ 1,2,3,. . .

ð4:25Þ

where A1j, j ¼ 1,2,3, . . ., are arbitrary constants. The eigenfunctions satisfy the following orthogonality condition: ðl

ϕi ϕj dx ¼

0

0 hi

if i 6¼ j if i ¼ j

ð4:26Þ

where hi is a constant. The longitudinal displacement u(x, t) of the rod can then be expressed as uðx; t Þ ¼ ¼

1 X

ϕj ðxÞq j ðt Þ

j¼1 1  X j¼1



ωj C j sin ω j t þ D j cos ω j t sin x c

ð4:27Þ

where Cj and Dj are constants to be determined by using the initial conditions.

4.1 Free Longitudinal Vibrations

207

Initial Conditions Let us assume that the rod is subjected to the following initial conditions: u_ ðx; 0Þ ¼ gðxÞ

uðx; 0Þ ¼ f ðxÞ,

ð4:28Þ

Substituting these initial conditions into Eq. 4.27 leads to uðx; 0Þ ¼ f ðxÞ ¼

1 X

D j sin

j¼1

ωj x c

ð4:29Þ

and u_ ðx; 0Þ ¼ gðxÞ ¼

1 X

ω j C j sin

j¼1

ωj x c

ð4:30Þ

In order to determine the constants Dj in Eq. 4.29, this equation is multiplied by sin(ωj/c)x and integrated over the length of the rod. By using the orthogonality condition of Eq. 4.26, one obtains 2 Dj ¼ l

ðl

f ðxÞ sin

0

ωj x dx, c

j ¼ 1,2,3,. . .

ð4:31Þ

Similarly, in order to determine the constants Cj, Eq. 4.30 is multiplied by sin(ωjx/c) and integrated over the length of the rod. Use of the orthogonality condition of Eq. 4.26 leads to 2 Cj ¼ lω j

ðl 0

gðxÞ sin

ωj x dx, c

j ¼ 1,2,3,. . .

ð4:32Þ

It is clear that the solution of the equation of the longitudinal vibration of the rod given by Eq. 4.29 is expressed as the sum of the modes of vibrations which in this case happen to be simple harmonic functions. The contribution of each of these modes to the solution will depend on the degree to which this particular mode is excited. As the result of a sudden application of a force, as in the case of impact loading, many of these modes will be excited, and accordingly their contribution to the solution of the vibration equation is significant. It is important, however, to point out that it is possible for the rod to vibrate in only one of these modes of vibration. For instance, if the initial elastic deformation of the rod exactly coincides with one of these modes and the initial velocity is assumed to be zero, vibration will occur only in this mode of vibration, and the system behaves as a single degree of freedom system. In order to demonstrate this, let us consider the special case in which the initial conditions are given by u(x, 0) ¼ sin (ωkx/c) and u_ ðx; 0Þ ¼ 0. Clearly, the initial displacement takes the shape of the kth mode. Substituting these initial conditions into Eqs. 4.31 and 4.32 and using the orthogonality of the mode shapes, one obtains Dj ¼ 0 if j 6¼ k, Dj ¼ 1 if j ¼ k, and Cj ¼ 0, j ¼ 1,2,. . . . That is, the

208

4

Vibration of Continuous Systems

solution of the equation of longitudinal vibration of Eq. 4.27 reduces in this special case to uðx; t Þ ¼ qk ðt Þ sin ðωk x=cÞ ¼ Dk cos ωk t sin ðωk x=cÞ, which demonstrates that the rod indeed vibrates in its kth mode due to the fact that the initial conditions in this special case cause only the kth mode of vibration to be excited. The continuous system in this special case behaves as a single degree of freedom system. This can be demonstrated by differentiating the displacement twice with respect to 2 x and t, yielding, respectively, ∂2u/∂x2 ¼ (ωk/c)2qk(t) sin (ωkx/c) and ∂ u=∂t 2 ¼ €qk ðt Þ sin ðωk x=cÞ. Substituting these two equations into the partial differential equation of Eq. 4.7, one obtains €qk sin ðωk x=cÞ ¼ ω2k qk sin ðωk x=cÞ, which leads to the differential equation €qk þ ω2k qk ¼ 0. This equation is in the same form as the equation which governs the free vibration of a single degree of freedom system. The frequency of oscillation in this case is ωk. If the initial conditions take a different form, the results will be different, and other modes may be excited as well. For example, one can show that if the initial displacement of the system is a linear combination of several modes, only these modes will be excited. The solution can then be represented by a truncated series which has a finite number of terms instead of the infinite series of Eq. 4.27. In this case, the continuous system can be analyzed as a multi-degree of freedom system with the number of degrees of freedom being equal to the number of modes which are excited. In many applications in practice, this approach is being used in the study of vibration of continuous systems. Quite often, a finite number of modes are excited and the techniques used in the analysis of multi-degree of freedom systems are often used in the analysis of continuous systems as well. It is important, however, to point out that the accuracy of the reduced finite-dimensional multi-degree of freedom model in representing the actual infinite-dimensional continuous system will greatly depend on the judgment of which modes are significant. The analysis of the frequency content of the initial conditions and the forcing functions may help in deciding which modes can be truncated. Other Boundary Conditions From the analysis presented in this section, it is clear that the mode shapes and the natural frequencies of the rod depend on the boundary conditions. Thus far we have considered only the case in which one end of the rod is fixed while the other end is free. Following a similar procedure to the one described in this section, the natural frequencies and mode shapes can be determined for other boundary conditions. If the rod is assumed to be free at both ends, the boundary conditions are given by ∂uð0; t Þ ¼ 0, ∂x

∂uðl; t Þ ¼0 ∂x

ð4:33Þ

Using these boundary conditions and the separation of variables technique, one can show that the frequency equation is given by sin

ωl ¼0 c

ð4:34Þ

4.1 Free Longitudinal Vibrations

209

which yields the natural frequencies of the longitudinal vibration of the rod with free ends as jπc jπ ¼ ωj ¼ l l

sffiffiffiffi E , ρ

j ¼ 1,2,3, . . .

ð4:35Þ

The associated eigenfunctions or mode shapes are given by ϕj ¼ A2 j cos

ω jx c

ð4:36Þ

Another example is a rod with both ends fixed. The boundary conditions in this case are given by uð0; t Þ ¼ 0,

uðl; t Þ ¼ 0

ð4:37Þ

Using these boundary conditions and the separation of variables method, one can show that the frequency equation is given by sin

ωl ¼0 c

ð4:38Þ

which yields the natural frequencies jπc jπ ¼ ωj ¼ l l

sffiffiffiffi E , ρ

j ¼ 1,2,3, . . .

ð4:39Þ

The mode shapes associated with these frequencies are ϕj ¼ A1 j sin

jπx , j ¼ 1,2,3, . . . l

ð4:40Þ

Example 4.1

The system shown in Fig. 4.3 consists of a rigid mass m attached to a rod which has mass density ρ, length l, modulus of elasticity E, and cross-sectional area A. Determine the frequency equation and the mode shapes of the longitudinal vibration. Solution. The end condition at the fixed end of the rod is given by u(0, t) ¼ 0. The other end of the rod is attached to the mass m which exerts a force on the rod because of the inertia effect. From the free-body diagram shown in the figure, the second boundary condition is given by (continued)

210

4

Vibration of Continuous Systems

Fig. 4.3 Longitudinal vibration of a rod with a mass attached to its end

2

m

∂ uðl; t Þ ∂uðl; t Þ ¼ Pðl; t Þ ¼ σ ðl; t ÞA ¼ EA ∂t 2 ∂x

or 2

m

∂ uðl; t Þ ∂uðl; t Þ ¼ EA 2 ∂t ∂x

The longitudinal vibration of the rod is governed by the equation uðx; t Þ ¼ ϕ  ðxÞqðt Þω ω  ¼ A1 sin x þ A2 cos x ðB1 sin ωt þ B2 cos ωt Þ c c Substituting the boundary condition u(0, t) ¼ 0 into this equation yields A2 ¼ 0. Using this condition, one gets uðx; t Þ ¼ ϕðxÞqðt Þ ¼ A1 sin

ωx ðB1 sin ωt þ B2 cos ωt Þ c

which yields 2

∂ u ωx ¼ ω2 A1 sin ðB1 sin ωt þ B2 cos ωt Þ ∂t 2 c ∂u ω ωx ¼ A1 cos ðB1 sin ωt þ B2 cos ωt Þ ∂x c c Therefore, the second boundary condition at the rod end attached to the mass yields mω2 A1 sin

ωl ω ωl ðB1 sin ωt þ B2 cos ωt Þ ¼ EA A1 cos ðB1 sin ωt þ B2 cos ωt Þ c c c

(continued)

4.1 Free Longitudinal Vibrations

211

This equation implies (mωc/EA) sin (ωl/c) ¼ cos (ωl/c), or tan

ωl EA ¼ c mωc

This is the frequency equation which, upon multiplying both of its sides by ωl/c, yields ωl ωl ωlEA lAρ M tan ¼ ¼ ¼ c c mωc2 m m where M is the mass of the rod. The preceding equation can be written as γ tan γ ¼ μ

ð4:41Þ

where γ ¼ ωl/c and μ ¼ M/m. Note that the frequency equation is a transcendental equation which has an infinite number of roots, and therefore, its solution defines an infinite number of natural frequencies. This equation can be expressed for each root as γj tan γj ¼ μ,

j ¼ 1,2,3,. . .

ð4:42Þ

where γ j ¼ ωjl/c, and the eigenfunction associated with the natural frequency ωj is ϕj ¼ A1 j sin

ω jx c

ð4:43Þ

Table 4.1 shows the first 20 roots of Eq. 4.42 for different values of the mass ratio μ. Table 4.1 Roots of Eq. 4.42 Mode number 1 2 3 4 5 6 7 8 9 10

1/μ ¼ 0 1.5708 4.7124 7.8540 10.9956 14.1372 17.2788 20.4204 23.5619 26.7035 29.8451

1/μ ¼ 0.01 1.5552 4.6658 7.7764 10.8871 13.9981 17.1093 20.2208 23.3327 26.4451 29.5577

γ 1/μ ¼ 0.1 1.4289 4.3058 7.2281 10.2003 13.2142 16.2594 19.3270 22.4108 25.5064 28.6106

1/μ ¼ 1 0.8603 3.4256 6.4372 9.5293 12.6453 15.7713 18.9024 22.0365 25.1725 28.3096

1/μ ¼ 10 0.3111 3.1731 6.2991 9.4354 12.5743 15.7143 18.8549 21.9957 25.1367 28.2779

(continued)

212

4

Vibration of Continuous Systems

Table 4.1 (continued) Mode number 11 12 13 14 15 16 17 18 19 20

1/μ ¼ 0 32.9867 36.1283 39.2699 42.4115 45.5531 48.6947 51.8363 54.9779 58.1195 61.2611

1/μ ¼ 0.01 32.6710 35.7847 38.8989 42.0138 45.1292 48.2452 51.3618 54.4791 57.5969 60.7155

γ 1/μ ¼ 0.1 31.7213 34.8371 37.9567 31.0795 44.2048 47.3321 50.4611 53.5916 56.7232 56.7232

1/μ ¼ 1 31.4477 34.5864 37.7256 40.8652 44.0050 47.1451 50.2854 53.4258 56.5663 59.7070

1/μ ¼ 10 31.4191 34.5604 37.7018 40.4832 43.9846 47.1260 50.2675 53.4089 56.5504 59.6919

Example 4.2

The system shown in Fig. 4.4 consists of a prismatic rod which has one end fixed and the other end attached to a spring with stiffness k as shown in the figure. The rod has length l, cross-sectional area A, mass density ρ, and modulus of elasticity E. Obtain the frequency equation and the eigenfunctions of this system.

Fig. 4.4 Longitudinal vibration of a rod with one end attached to a spring

Solution. As in the preceding example, the boundary condition at the fixed end is given by u(0, t) ¼ 0. At the other end, the axial force of the rod is equal in magnitude and opposite in direction to the spring force, that is, P(l, t) ¼ ku(l, t). Since P ¼ EAu0, the preceding boundary condition is given by EAu0 ðl; t Þ ¼ kuðl; t Þ Using the technique of the separation of variables, the solution of the partial differential equation of the rod can be expressed as (continued)

4.1 Free Longitudinal Vibrations

213

uðx; lÞ ¼ ϕðxÞqðt Þ  ω ω  ¼ A1 sin x þ A2 cos x ðB1 sin ωt þ B2 cos ωt Þ c c As in the preceding example, the boundary condition at the fixed end leads to A2 ¼ 0. Therefore, the expression for the longitudinal displacement u may be simplified and written as u ¼ ϕðxÞqðt Þ ¼ A1 sin

ω xðB1 sin ωt þ B2 cos ωt Þ c

and u0 ¼

ω ω A1 cos xðB1 sin ωt þ B2 cos ωt Þ c c

By using the second boundary condition, the following equation is obtained: ω ω ω EA cos l ¼ k sin l c c c or tan

ω EAω l¼ c kc

Multiplying both sides of this equation by ωl/c and using the definition of c ¼ pffiffiffiffiffiffiffiffi E=ρ, one obtains ωl ωl EAω2 l EAω2 lρ ω2 M tan ¼  ¼ ¼ 2 c c kE k kc where M is the mass of the rod. The above equation is the frequency equation which can be written as γ tan γ ¼ 

ω2 M k

where γ ¼ ωl/c. The roots of the frequency equation can be determined numerically and used to define the natural frequencies, ωj, j ¼ 1,2,3,. . . . It is clear in this case that the eigenfunction associated with the jth natural frequency is ϕj ¼ A1 j sin

ω jx c

The frequency equations and mode shapes of the longitudinal vibrations obtained using different sets of boundary conditions are presented in Table 4.2.

214

4

Vibration of Continuous Systems 

Table 4.2 Frequency equations and mode shapes of the longitudinal vibrations

c¼

pffiffiffiffiffiffiffiffi E=ρ ;

γ ¼ ðωl=cÞ; A j ¼ constantÞ Boundary conditions

Frequency equation cos

Fixed-free

ωj ¼

ω jx c

ϕ j ¼ A j cos

ω jx c

ϕ j ¼ A j sin

ω jx c

M m

ϕ j ¼ A j sin

ω jx c

ω2 M k

ϕ j ¼ A j sin

ω jx c

M m

ϕ j ¼ A j cos

ω jx c

ω2 M k

ϕ j ¼ A j cos

ω jx c

ωl ¼0 c

jπc ωj ¼ l sin

Fixed-fixed

ϕ j ¼ A j sin

ð2j  1Þπc 2l

sin

Free-free

ωl ¼0 c

Mode shapes

ωl ¼0 c

jπc ωj ¼ l γ tan γ ¼ μ where μ ¼

Fixed-mass

γ tan γ ¼  Fixed-spring

γ cot γ ¼ μ where μ ¼ Free-mass

γ cot γ ¼ Free-spring

4.2

Free Torsional Vibrations

In this section, we consider the torsional vibration of straight circular shafts as the one shown in Fig. 4.5a. The shaft is assumed to have a length l and a cross section which has a polar moment of inertia Ip. The shaft is assumed to be made

4.2 Free Torsional Vibrations

215

Fig. 4.5 Torsional vibration

of material which has modulus of rigidity G and mass density ρ and is subjected to a distributed external torque defined per unit length by the function Te(x, t). In the analysis presented in this section, the cross sections of the shaft are assumed to remain in their planes and rotate as rigid surfaces about their centers. The equation of dynamic equilibrium of an infinitesimal volume of the shaft in torsional vibration is given by 2

ρI p

∂ θ ∂T δx  T þ T e δx δx ¼ T þ 2 ∂t ∂x

ð4:44Þ

where θ ¼ θ(x,t) is the angle of torsional oscillation of the infinitesimal volume about the axis of the shaft, T is the internal torque acting on the cross section at a distance x from the end of the shaft, and δx is the length of the infinitesimal volume. Equation 4.44 can be simplified and written as 2

ρI p

∂ θ ∂T þ Te ¼ ∂t 2 ∂x

ð4:45Þ

From elementary strength of material theory, the internal torque T is proportional to the spatial derivative of the torsional oscillation θ. The relationship between the internal torque and the torsional oscillation is given in terms of the modulus of rigidity G as T ¼ GI p

∂θ ∂x

ð4:46Þ

Substituting Eq. 4.46 into Eq. 4.45, one obtains
 2 ∂ θ ∂ ∂θ ρI p 2 ¼ GI p þ Te ∂t ∂x ∂x

ð4:47Þ

If the shaft is assumed to have a constant cross-sectional area and a constant modulus of rigidity, Eq. 4.47 can be expressed as

216

4 2

ρI p

Vibration of Continuous Systems

2

∂ θ ∂ θ ¼ GI p 2 þ T e ∂t 2 ∂x

ð4:48Þ

Free Vibration The partial differential equation that governs the free torsional vibration of the shaft can be obtained from Eq. 4.48 if we let Te ¼ 0, that is, 2

ρI p

2

∂ θ ∂ θ ¼ GI p 2 ∂t 2 ∂x

ð4:49Þ

which can be rewritten as 2

2

∂ θ ∂ θ ¼ c2 2 2 ∂t ∂x

ð4:50Þ

where c is a constant that depends on the inertia properties and the material of the shaft and is given by sffiffiffiffi G c¼ ρ

ð4:51Þ

Equation 4.50 is in the same form as Eq. 4.7 that governs the longitudinal vibration of prismatic rods. Therefore, the same solution procedure described in the preceding section can be used to solve Eq. 4.50. Using the separation of variables technique, the torsional oscillation θ can be expressed as θ ¼ ϕðxÞqðt Þ

ð4:52Þ

where ϕ(x) is a space-dependent function and q(t) depends on time. Following the procedure described in the preceding section, one can show that 9 ωx ωx = þ A2 cos c c ; qðt Þ ¼ B1 sin ωt þ B2 cos ωt

ϕðxÞ ¼ A1 sin

ð4:53Þ

where A1, A2, B1, B2, and ω are arbitrary constants to be determined using the boundary and initial conditions. Substituting Eq. 4.53 into Eq. 4.52 yields  ωx ωx þ A2 cos ðB1 sin ωt þ B2 cos ωt Þ θðx; t Þ ¼ A1 sin c c

ð4:54Þ

Boundary Conditions Equation 4.54 is in the same form as Eq. 4.15 which describes the longitudinal vibration of prismatic rods. Therefore, one expects that the natural frequencies and mode shapes will also be in the same form. For example,

4.2 Free Torsional Vibrations

217

in the case of a shaft with one end fixed and the other end free, the natural frequencies and eigenfunctions of torsional vibration are given, respectively, by sffiffiffiffi 9 ð2j  1Þπc ð2j  1Þπ G > > = ¼ ωj ¼ 2l 2l ρ > ω jx > , j ¼ 1,2,3, . . . ; ϕj ¼ A1 j sin c

ð4:55Þ

and the solution for the free torsional vibration is θ¼

1 X

sin

j¼1

 ω jx  C j sin ω j t þ D j cos ω j t c

ð4:56Þ

where the constants Cj and Dj, j ¼ 1,2,3, . . ., can be determined by using the initial conditions as described in the preceding section. In the case of a shaft with free ends, the natural frequencies and eigenfunctions are given, respectively, by jπc jπ ¼ ωj ¼ l l

sffiffiffiffi G , ρ

ϕj ¼ A2 j cos

ω jx , c

j ¼ 1,2,3,. . .

ð4:57Þ

and the solution for the free torsional vibration of the shaft with free ends has the form θ¼

1 X

cos

j¼1

 ω jx  C j sin ω j t þ D j cos ω j t c

ð4:58Þ

where Cj and Dj, j ¼ 1,2,3,. . ., are constants to be determined from the initial conditions. Similarly one can show that the natural frequencies and mode shapes of the torsional vibration in the case of a shaft with both ends fixed are jπc jπ ¼ ωj ¼ l l

sffiffiffiffi G , ρ

ϕj ¼ A1 j sin

ω jx , c

j ¼ 1,2,3,. . .

ð4:59Þ

and the solution for the free torsional vibration is given by θ¼

1 X j¼1

sin

 ω jx  C j sin ω j t þ D j cos ω j t c

ð4:60Þ

As in the case of longitudinal vibrations discussed in the preceding section, the orthogonality of the mode shapes can be utilized in determining the arbitrary constants Cj and Dj using the initial conditions.

218

4

Vibration of Continuous Systems

Example 4.3

The system shown in Fig. 4.6 consists of a disk with a mass moment of inertia Id attached to a circular shaft which has mass density ρ, length l, crosssectional polar moment of inertia Ip, and modulus of rigidity G. Obtain the frequency equation and the mode shapes of the torsional vibration.

Fig. 4.6 Torsional vibration of a shaft with one end attached to a disk

Solution. The two boundary conditions in this case are given by θð0; t Þ ¼ 0,

T ðl; t Þ ¼ GI p θ0 ðl; t Þ ¼ I d θ€ðl; t Þ

The solution for the free torsional vibration can be assumed in the form θðx; t Þ ¼ ϕðxÞqðt Þ  ωx ωx þ A2 cos ¼ A1 sin ðB1 sin ωt þ B2 cos ωt Þ c c Substituting the two boundary conditions into this solution and following the procedure described in Example 4.1, it can be shown that the frequency equation is given by γ tan γ ¼ μ where ωl γ¼ , c

ρI p l μ¼ , Id

sffiffiffiffiffiffi G c¼ ρ

This equation has an infinite number of roots γ j, j ¼ 1,2,3,. . ., which can be used to define the natural frequencies ωj as ωj ¼

γ jc , l

j ¼ 1,2,3,. . .

4.2 Free Torsional Vibrations

219

Example 4.4

The system shown in Fig. 4.7 consists of a straight cylindrical shaft which has one end fixed and the other end attached to a torsional spring with stiffness kt as shown in the figure. The shaft has length l, mass density ρ, cross-sectional polar moment of inertia Ip, and modulus of rigidity G. Obtain the frequency equation and the eigenfunctions of this system.

Fig. 4.7 Torsional vibration of a shaft with one end attached to a torsional spring

Solution. The boundary conditions in this case are given by θð0; t Þ ¼ 0,

T ðl; t Þ ¼ GI p θ0 ðl; t Þ ¼ kt θðl; t Þ

The solution for the free torsional vibration is given by θðx; t Þ ¼ ϕðxÞqðt Þ  ωx ωx þ A2 cos ¼ A1 sin ðB1 sin ωt þ B2 cos ωt Þ c c Substituting the boundary conditions into this solution and following the procedure described in Example 4.2, it can be shown that the frequency equation is given by γ tan γ ¼ μ where ω2 ρlI p μ¼ , kt

ωl γ¼ , c

sffiffiffiffiffiffi G c¼ ρ

The roots γ j, j ¼ 1,2,3,. . ., of the frequency equation can be obtained numerically. These roots define the natural frequencies ωj, j ¼ 1,2,3,. . ., as ωj ¼

γ jc , j ¼ 1,2,3,. . . l

The associated mode shapes are ϕj ¼ A1 j sin

ω jx c

220

4

Vibration of Continuous Systems

Table 4.3 shows the frequency equations and mode shapes of the torsional vibration obtained using different sets of boundary conditions.  Table 4.3 Frequency equations and mode shapes of the torsional vibrations  γ ¼ ωl=c, A j ¼ constant Boundary Conditions

Frequency equation cos

Fixed-free

ωl ¼0 c

ð2j  1Þπc ωj ¼ 2l

sin

ωl ¼0 c

jπc ωj ¼ l

c¼

pffiffiffiffiffiffiffiffiffi G=ρ ,

Mode shapes

ϕ j ¼ A j sin

ω jx c

ϕ j ¼ A j cos

ω jx c

ϕ j ¼ A j sin

ω jx c

ϕ j ¼ A j sin

ω jx c

ϕ j ¼ A j sin

ω jx c

ϕ j ¼ A j cos

ω jx c

ϕ j ¼ A j cos

ω jx c

Free-free sin

Fixed-fixed

ωl ¼0 c

jπc ωj ¼ l γ tan γ ¼ μ where μ ¼

ρI p l Id

Fixed-inertial mass γ tan γ ¼ μ where μ ¼

ω2 ρlI p kt

Fixed-spring γ cot γ ¼ μ where μ ¼

ρI p l Id

Free-inertial mass γ cot γ ¼ μ ω2 ρlI p where μ ¼ kt Fixed-spring

4.3 Free Transverse Vibrations of Beams

4.3

221

Free Transverse Vibrations of Beams

It was shown in the preceding two sections that the differential equations that govern the longitudinal and torsional vibration of rods have the same form. The theory of transverse vibration of beams is more difficult than that of the two types of vibration already considered in the preceding sections. In this section, the equations that govern the transverse vibration of beams are developed and methods for obtaining their solutions are discussed. There are several important applications of the theory of transverse vibration of beams; among these applications are bending vibrations of beam structures, vibrations of rotating shafts and rotors, and the transverse vibration of suspended cables. Elementary Beam Theory In the elementary beam theory, all stresses are assumed to be equal to zero except the normal stress σ which is assumed to vary linearly over the cross section with the y-coordinate of the beam as shown in Fig. 4.8. The normal stress σ can be written as σ ¼ ky

ð4:61Þ

where k is constant, and y ¼ 0 contains the neutral surface along which the normal stress σ is equal to zero. The assumption that all other stresses are equal to zero requires that the resultant of the internal forces be zero and that the moments of the internal forces about the neutral axis equal the bending moment M. Consequently, ð

ð σ dA ¼ 0,

A

yσ dA ¼ M

ð4:62Þ

A

where AÐis the cross-sectional Ð area of the beam. Substituting Eq. 4.61 into Eq. 4.62 yields k Ay dA ¼ 0, and k Ay2dA ¼ M. Since k is a nonzero constant, the first equation implies that the neutral and centroidal axes of the cross section coincide. The second equation can be used to define k as k¼

M Iz

Fig. 4.8 Elementary beam theory

ð4:63Þ

222

4

Vibration of Continuous Systems

Table 4.4 Moments of inertia of the cross sections Shape

Area

A¼

Moment of inertia

πD2 4

Iy ¼ Iz ¼

A ¼ bh

A¼

Iy ¼

bh3 hb3 , Iz ¼ 12 12

bh , 2 Iy ¼

h z ¼ 3

A ¼ πab

bh3 36

Iy ¼

πab3 πba3 , Iz ¼ 4 4

Iy ¼


 r4 1 γ  sin 2γ 2 4

A ¼ γr 2 , 2r sin γ y ¼ 3 γ

πD4 64

Ð where Iz ¼ Ay2dA is the moment of inertia of the cross section about the z-axis of the beam cross section. Table 4.4 shows the area properties of selected shapes of the cross section. Substituting Eq. 4.63 into Eq. 4.61 yields σ¼

My Iz

ð4:64Þ

Using Hooke’s law, the strain ε is given by ε¼

σ My ¼ E EI z

ð4:65Þ

4.3 Free Transverse Vibrations of Beams

223

It is known from simple geometry that for a curve segment defined by the angle dθ, rdθ ¼ ds, where s is the arc length and r is the radius of curvature. It follows that the strain ε in Eq. 4.65 at an arbitrary coordinate y can be written in terms of the radius of curvature and the angle θ as ε ¼ ((r + y)dθ  rdθ)/rdθ, which simplifies to ε ¼ ydθ/rdθ ¼ y/r. The negative sign used in this equation accounts for the definition of the positive rotation about the z-axis. Let v denotes the transverse displacement of the beam. For small deformations (dv/dx  1), it can be shown that 1 d2 v ε M  ¼ ¼ r dx2 y EI z

ð4:66Þ

M ¼ EI z v00

ð4:67Þ

This equation implies that

This equation is known as the Euler–Bernoulli law of the elementary beam theory. Partial Differential Equation In order to determine the differential equation for the transverse vibration of beams, we consider an infinitesimal volume at a distance x from the end of the beam as shown in Fig. 4.9. The length of this infinitesimal volume is assumed to be δx. Let V and M be, respectively, the shear force and bending moment, and let F(x, t) be the loading per unit length of the beam. Neglecting the rotary inertia, the sum of the moments about the left end of the section yields 2

Mþ

∂M ∂V ðδxÞ2 ∂ v ðδxÞ2 δx  M  Vδx  ðδxÞ2  F ðx; t Þ  ρA 2 ¼0 ∂x ∂x ∂t 2 2

ð4:68Þ

Taking the limit as δx approaches zero, the preceding equation leads to V¼

∂M ∂x

Fig. 4.9 Moments and shear forces

ð4:69Þ

224

4

Vibration of Continuous Systems

The dynamic equilibrium condition for the transverse vibration of the beam is obtained by applying Newton’s second law as 2

ρAδx

∂ v ∂V δx  V þ F ðx; t Þδx ¼Vþ 2 ∂t ∂x

ð4:70Þ

where ρ is the mass density and A is the cross-sectional area. Equation 4.70 can be rewritten after simplification as ρA(∂2v/∂t2) ¼ (∂V/∂x) + F(x, t), which upon using Eq. 4.69 yields 2

ρA

2

∂ v ∂ M ¼ þ F ðx; t Þ ∂t 2 ∂x2

ð4:71Þ

The moment M can be eliminated from this equation by using the moment displacement relationship. To this end, Eq. 4.67 is substituted into Eq. 4.71. This leads to 2

ρA

2

∂ v ∂ ¼  2 ðEI z v00 Þ þ F ðx; t Þ 2 ∂t ∂x

ð4:72Þ

If E and Iz are assumed to be constant, Eq. 4.72 becomes 2

ρA

4

∂ v ∂ v ¼ EI z 4 þ F ðx; t Þ ∂t 2 ∂x

ð4:73Þ

In the case of free vibration, F(x, t) ¼ 0, and Eq. 4.73 reduces to 2

4

∂ v ∂ v ¼ c2 4 ∂t 2 ∂x

ð4:74Þ

sffiffiffiffiffiffiffi EI z c¼ ρA

ð4:75Þ

where c is a constant defined as

Separation of Variables Equation 4.74 is a fourth-order partial differential equation that governs the free transverse vibration of the beam. The solution of this equation can be obtained by using the technique of the separation of variables. In this case, we assume a solution in the form v ¼ ϕðxÞqðt Þ

ð4:76Þ

where ϕ(x) is a space-dependent function, and q(t) is a function that depends only on time. Equation 4.93 leads to

4.3 Free Transverse Vibrations of Beams 2

∂ v d 2 qð t Þ ¼ ϕð xÞ ¼ ϕðxÞ€qðt Þ 2 ∂t dt 2

225

9 > > > =

> > ∂ v d 4 ϕð xÞ > iv ; ¼ q ð t Þ ¼ ϕ ð x Þq ð t Þ 4 ∂x4 dx 4

ð4:77Þ

Substituting these equations into Eq. 4.74, one obtains ϕðxÞ€qðt Þ ¼ c2 ϕiv ðxÞqðt Þ, which implies that €qðt Þ ϕiv ðxÞ ¼ ω2 ¼ c2 ϕðxÞ qð t Þ

ð4:78Þ

where ω is a constant to be determined. Equation 4.96 leads to the following two equations:  ω 2 €q þ ω2 q ¼ 0, ϕiv  ϕ¼0 ð4:79Þ c The solution of the first of these two equations can be assumed as q ¼ B1 sin ωt + B2 cos ωt, while for the second equation as ϕ ¼ Aeλx. Substituting ϕ ¼ Aeλx into the second equation of Eq. 4.79 yields [λ4  (ω/c)2]Aeλx ¼ 0, or λ4  (ω/c)2 ¼ 0, which can be written as λ4  η 4 ¼ 0

ð4:80Þ pffiffiffiffiffiffiffiffi where η ¼ ω=c. The of Eq. 4.80 are λ1 ¼ η, λ2 ¼ η, λ3 ¼ iη, and pffiffiffiffiffiffiroots ffi λ4 ¼ iη, where i ¼ 1. Therefore, the general solution of the second equation of Eq. 4.79 can be written as ϕðxÞ ¼ A1 eηx þ A2 eηx þ A3 eiηx þ A4 eiηx

ð4:81Þ

which can be rewritten as eηx  eηx eηx þ eηx þ A6 2 2 eiηx  eiηx eiηx þ eiηx þ A8 þA7 ðiÞ 2 2

ϕð xÞ ¼ A 5

ð4:82Þ

where A1 ¼ (A5 + A6)/2, A2 ¼ (A6  A5)/2, A3 ¼ (A8  iA7)/2, and A4 ¼ (A8 + iA7)/2. Equation 4.82 can then be rewritten, using Euler’s formula of the complex variables, as ϕðxÞ ¼ A5 sinh ηx þ A6 cosh ηx þ A7 sin ηx þ A8 cos ηx

ð4:83Þ

Using this equation and the assumed solution for q ¼ B1sinωt + B2cosωt in Eq. 4.76 yields

226

4

Vibration of Continuous Systems

vðx; t Þ ¼ ðA5 sinh ηx þ A6 cosh ηx þ A7 sin ηx þ A8 cos ηxÞ ðB1sinωt þ B2cosωt Þ

ð4:84Þ

where ω is defined as ω ¼ cη2. Boundary Conditions The natural frequencies of the beam, as well as the constants that appear in Eq. 4.84, depend on the boundary and initial conditions. For example, if the beam is simply supported at both ends, the boundary conditions are v(0, t) ¼ 0, v00 (0, t) ¼ 0, v(l, t) ¼ 0, and v00 (l, t) ¼ 0, which imply that ϕ00 ð0Þ ¼ 0 ϕ00 ðlÞ ¼ 0

ϕð0Þ ¼ 0, ϕðlÞ ¼ 0,

ð4:85Þ

It is clear that in this case, there are two geometric boundary conditions that specify the displacements, and there are two natural boundary conditions that specify the moments at the ends of the beam. Substituting these conditions into Eq. 4.83 yields 9 A6 þ A8 ¼ 0 > > > A  A ¼ 0= 6

8

A5 sinh ηl þ A6 cosh ηl þ A7 sin ηl þ A8 cos ηl ¼ 0 > > > ; A5 sinh ηl þ A6 cosh ηl  A7 sin ηl  A8 cos ηl ¼ 0

ð4:86Þ

These equations are satisfied if A5 ¼ A6 ¼ A8 ¼ 0 and A7 sin ηl ¼ 0

ð4:87Þ

The roots of this equation are ηl ¼ jπ,

j ¼ 1,2,3,. . .

ð4:88Þ

Therefore, the natural frequencies are given by j2 π 2 j2 π 2 ωj ¼ 2 c ¼ 2 l l

sffiffiffiffiffiffiffi EI z , ρA

j ¼ 1,2,3,. . .

ð4:89Þ

and the corresponding modes of vibration are ϕj ¼ A7j sin ηj x,

j ¼ 1,2,3,. . .

ð4:90Þ

The solution for the free vibration of the simply supported beam can then be written as vðx; t Þ ¼

1 X j¼1

ϕj q j ¼

1  X  C j sin ω j t þ D j cos ω j t sin ηj x j¼1

ð4:91Þ

4.3 Free Transverse Vibrations of Beams

227

The arbitrary constants Cj and Dj, j ¼ 1,2,3, . . ., can be determined using the initial conditions by the method described in Sect. 4.1 of this chapter. In the case of a cantilever beam, the geometric boundary conditions at the fixed end are v(0, t) ¼ 0 and v0(0, t) ¼ 0, and the natural boundary conditions at the free end 00 000 are v (l, t) ¼ 0 and v (l, t) ¼ 0. These conditions imply that ϕð0Þ ¼ 0, ϕ00 ðlÞ ¼ 0,

ϕ 0 ð 0Þ ¼ 0 000 ϕ ðlÞ ¼ 0

ð4:92Þ

Substituting these conditions into Eq. 4.83 yields the frequency equation cos ηl cosh ηl ¼ 1

ð4:93Þ

The roots of this frequency equation can be determined numerically. The first six roots are (Timoshenko et al. 1974) η1l ¼ 1.875, η2l ¼ 4.694, η3l ¼ 7.855, η4l ¼ 10.996, η5l ¼ 14.137, and η6l ¼ 17.279.Approximate values of these roots  can be calculated using the equation ηj l  j  12 π. The fundamental natural pffiffiffiffiffiffiffiffi frequencies of the system can be obtained using Eq. 4.75 and η ¼ ω=c as pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ω1 ¼ ω j ¼ η2j c ¼ η2j EI z =ρA. The first six natural frequencies are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3:51563 EI z =ml3 , ω2 ¼ 22:03364 EI z =ml3 , ω3 ¼ 61:7010 EI z =ml3 , ω4 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 120:9120 EI z =ml3 , ω5 ¼ 199:8548 EI z =ml3 , and ω6 ¼ 298:5638 EI z =ml3 . In this case it can be verified that the mode shapes are    j cos ηj x  cosh ηj x , ϕj ðxÞ ¼ A6 j sin ηj x  sinh ηj x þ D

j ¼ 1,2,3, . . . ð4:94Þ

where A6j is an arbitrary constant and j ¼ D

cos ηj l þ cosh ηj l sin ηj l  sinh ηj l

ð4:95Þ

Example 4.5

Neglecting the effect of rotary inertia, find the first four natural frequencies of the transverse vibration of a beam with free ends. The beam has length l, crosssectional area A, moment of inertia Iz, mass density ρ, and modulus of elasticity E. Solution. At the free end of a beam, the bending moment and shear force are equal to zero. Therefore, the boundary conditions in this example are given by ϕ00 ð0Þ ¼ 0, ϕ000 ð0Þ ¼ 0, ϕ00 ðlÞ ¼ 0, ϕ000 ðlÞ ¼ 0 Substituting the first two boundary conditions into Eq. 4.83 yields (continued)

228

4

Vibration of Continuous Systems

ϕ00 ð0Þ ¼ η2 A6  η2 A8 ¼ 0 ϕ000 ð0Þ ¼ η3 A5  η3 A7 ¼ 0 which imply A6 ¼ A8 and A5 ¼ A7. Hence, ϕðxÞ ¼ A5 ðsinh ηx þ sin ηxÞ þ A6 ðcosh ηx þ cos ηxÞ Substituting the third and fourth boundary conditions into this equation, one obtains A5 ðsinh ηl  sin ηlÞ þ A6 ðcosh ηl  cos ηlÞ ¼ 0 A5 ðcosh ηl  cos ηlÞ þ A6 ðsinh ηl þ sin ηlÞ ¼ 0 That is, "

sinh ηl  sin ηl

cosh ηl  cos ηl

cosh ηl  cos ηl sinh ηl þ sin ηl

#"

A5 A6

# ¼

" # 0 0

This system of homogeneous algebraic equations in the unknowns A5 and A6 has a nontrivial solution if and only if the determinant of the coefficient matrix is equal to zero. This leads to the following frequency equation: 

 sinh2 ηl  sin2 ηl  ðcosh ηl  cos ηlÞ2 ¼ 0

Since cosh2 ηl  sinh2 ηl ¼ 1 and cos 2 ηl þ sin 2 ηl ¼ 1, the frequency equation can be simplified to cos ηl cosh ηl ¼ 1 This equation is satisfied if ηl ¼ 0. Therefore, the first natural frequency is ω1 ¼ 0. This frequency is associated with a rigid-body motion of the beam. Thus the associated mode shape is given by ϕ1 ðxÞ ¼ A51 þ A61 x. One can show that the second, third, and fourth natural frequencies are defined by the equations c c ω2 ¼ η2 c ¼ η2 l2 2 ¼ ð4:730Þ2 2 l l c 2c ω3 ¼ ð7:853Þ 2 , ω4 ¼ ð10:996Þ2 2 l l (continued)

4.3 Free Transverse Vibrations of Beams

229

The function ϕ(x) obtained in this example is expressed in terms of the constants A5 and A6. The relationship between these two constants is A6 sinh ηl  sin ηl cosh ηl  cos ηl  ¼ ¼D ¼ cosh ηl  cos ηl sinh ηl þ sin ηl A5 Therefore, the space-dependent function ϕ can be expressed in terms of one constant as    j cosh ηj x þ cos ηj x , ϕj ðxÞ ¼ A5 j sinh ηj x þ sin ηj x þ D

j ¼ 1,2,3,. . .

The first three deformation mode shapes of the beam are shown in Fig. 4.10.

Fig. 4.10 First three deformation mode shapes of a beam with free ends

Example 4.6

Neglecting the effect of the rotary inertia, find the first four natural frequencies of the transverse vibration of the beam in the preceding example if both ends of the beam are fixed. Solution. For a beam with fixed ends, the boundary conditions are ϕð0Þ ¼ 0,

ϕ0 ð0Þ ¼ 0,

ϕðlÞ ¼ 0,

ϕ0 ðlÞ ¼ 0

Substituting these boundary conditions into Eq. 4.83, and following the procedure described in the preceding example, one can show that the frequency equation, in this case, is given by cos ηl cosh ηl ¼ 1 The first four roots of this equation are (continued)

230

4

η1 l ¼ 1:875,

η2 l ¼ 4:694,

Vibration of Continuous Systems

η3 l ¼ 7:855,

η4 l ¼ 10:996

The space-dependent function ϕj(x) is given in this case by    j cos ηj x  cosh ηj x , ϕj ðxÞ ¼ A6 j sin ηj x  sinh ηj x þ D

j ¼ 1,2,3,. . .

j is a constant defined as where D j ¼  D

sin ηj l  sinh ηj l cosh ηj l  cos ηj l ¼ cos ηj l  cosh ηj l sin ηj l þ sinh ηj l

The first three mode shapes of vibration, in the case in which both ends are fixed, are shown in Fig. 4.11.

Fig. 4.11 First three mode shapes of a beam with fixed ends

Table 4.5 shows the frequency equations and eigenfunctions of the transverse vibrations of beams with different end conditions. Effect of the Rotary Inertia In the analysis presented in this section, it was assumed that the cross sections of the beam during deformation remain perpendicular to the neutral axis, and accordingly the rotation of the cross section of the beam is not taken into account. This assumption leads to the simple relationship between the shear force and the moment given by Eq. 4.69. This simple beam theory in which the rotation of the cross section is assumed to be insignificant as compared to the translational displacement is valid only when the height of the beam is small compared to its length. If the height is denoted as h and the length as l, the accuracy of the simple beam theory is acceptable when h/l  1/10. Lord Rayleigh (1894) in his study of wave propagation obtained more satisfactory results when he took into account the effect of the rotary inertia of the cross section. If the assumption of small deformation is made, the rotation of the cross

4.3 Free Transverse Vibrations of Beams

231

Table 4.5 Frequency equations and mode   pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi η ¼ ω=c , c ¼ EI z =ρA , A j ¼ constant Boundary conditions

shapes

of

the

transverse

Frequency equation

vibrations

Mode shapes

sin ηl ¼ 0 ωj ¼ Simply supported

j2 π 2 c l2

cos ηlcosh ηl ¼ 1
 1 η jl  j  π 2 Cantilever beam

ω j ¼ η2j c

Free-free

cos ηlcosh ηl ¼ 1

ϕj ¼ Aj sin ηjx  ϕ j ðxÞ ¼ A j sin η j x  sinh η j x   þD j cos η j x  cosh η j x cos η j l þ cosh η j l sin η j l  sinh η j l  ϕ j ðxÞ ¼ A j sinh η j x þ sin η j x   þD j cosh η j x þ cos η j x Dj ¼

cosh η j l  cos η j l sinh η j l þ sin η j l  ϕ j ðxÞ ¼ A j sin η j x  sinh η j x   þD j cos η j x  cosh η j x Dj ¼ 

cos ηlcosh ηl ¼ 1

cosh η j l  cos η j l sin η j l þ sinh η j l  ϕ j ðxÞ ¼ A j sinh η j x  sin η j x   þD j cosh η j x  cos η j x Dj ¼ 

Clamped-clamped

tanh ηlcot ηl ¼ 1

sinh η j l  sin η j l cosh η j l  cos η j l  ϕ j ðxÞ ¼ A j sinh η j x  sin η j x   þD j cosh η j x  cos η j x Dj ¼ 

Fixed-simply supported ηl

Fixed-mass

cosh ηl sin ηl  sinh ηl cos ηl cosh ηl cos ηl þ 1 M ¼ m

Dj ¼ 

sinh η j l þ sin η j l cosh η j l þ cos η j l

  ϕ j ðxÞ ¼ A j sin η j x þ D j sinh η j x tanh ηl cot ηl ¼ 1 Pin-free

Dj ¼

sin η j l sinh η j l

section about the z-axis can be defined as α ¼ ∂v/∂x. In this case, the sum of the moments given by Eq. 4.68 must be equal to the rotary inertia. This leads to
 2 3 ∂M ∂ α ∂ v  V ¼ ρI z 2 ¼ ρI z ð4:96Þ  ∂x ∂t ∂x∂t 2 where Iz is the second moment of area of the cross section. This equation, upon differentiation with respect to x, leads to

232

4 2

Vibration of Continuous Systems

4

∂V ∂ M ∂ v ¼ þ ρI z 2 2 ∂x ∂x2 ∂x ∂t

ð4:97Þ

Substituting this equation into Eq. 4.71, one obtains 2

ρA

2

4

∂ v ∂ M ∂ v ¼ þ ρI z 2 2 þ F ðx; t Þ ∂t 2 ∂x2 ∂x ∂t

ð4:98Þ

which upon the use of Eq. 4.67 leads to 2

2

2

∂ v ∂ ∂ v ρA 2 ¼  2 EI z 2 ∂t ∂x ∂x

!

4

þ ρI z

∂ v þ F ðx; t Þ ∂x2 ∂t 2

ð4:99Þ

The second term on the right-hand side of this partial differential equation represents the effect of the rotary inertia. Observe that if the effect of rotary inertia is neglected, this equation reduces to Eq. 4.89 which was obtained earlier in this section. It is clear from the results presented in the preceding example that the first mode shape has no nodes. That is, zero displacements occur only at the fixed ends. On the other hand, the second mode has one node, and the third mode has two nodes. The analysis of higher modes shows that as the mode number increases, the number of nodes increases. Consequently, the change in the slope of the curve that describes the shape of a high-frequency mode is much faster than the change in the slope of the curve of the low-frequency modes. Therefore, for high-frequency modes, the rotation of the cross sections can be significant and the use of the simple beam theory that neglects the effect of the rotary inertia can lead to a significant error in the analysis of these modes in particular and in the analysis of the wave motion in general.

4.4

Orthogonality of the Eigenfunctions

In this section, we study in more detail the important property of the orthogonality of the eigenfunctions of the continuous systems. This property can be used to obtain an infinite number of decoupled second-order ordinary differential equations whose solution can be presented in a simple closed form. This development can be used to justify the use of approximate techniques in later sections to obtain a finitedimensional model that represents, to a certain degree of accuracy, the vibration of the continuous systems. Furthermore, the use of the orthogonality of the eigenfunctions leads to the important definitions of the modal mass, modal stiffness, and modal force coefficients for the continuous systems. As will be seen in this section, there are an infinite number of such coefficients since a continuous system has an infinite number of degrees of freedom. Longitudinal and Torsional Vibration of Rods The partial differential equations that govern the longitudinal and torsional vibration of rods have the same form, and consequently, the resulting eigenfunctions are the same for similar end conditions.

4.4 Orthogonality of the Eigenfunctions

233

Therefore, in the following we discuss only the orthogonality of the eigenfunctions of the longitudinal vibration of rods. It was shown, in Sect. 4.1, that the partial differential equation for the longitudinal vibration of rods can be written as (Eq. 4.5)
 2 ∂ u ∂ ∂u EA ρA 2 ¼ ∂t ∂x ∂x

ð4:100Þ

where u ¼ u(x, t) is the longitudinal displacement, ρ and A are, respectively, the mass density and cross-sectional area, and E is the modulus of elasticity. The solution of Eq. 4.100, which was obtained by using the separation of variables technique, can be expressed as u(x, t) ¼ ϕ(x)q(t), where ϕ(x) is a space-dependent function, and q(t) depends only on time and can be expressed as q(t) ¼ B1 sin ωt + B2 cos ωt. By using these equations, the acceleration can be written as ∂2u/∂t2 ¼ ω2ϕ(x)q(t). Hence, Eq. 4.100 can be written as ρAω2ϕ(x)q(t) ¼ (EAϕ0 (x))0 q(t), which leads to ρAω2 ϕðxÞ ¼ ðEAϕ0 ðxÞÞ0

ð4:101Þ

For the jth eigenfunction ϕj, j ¼ 1,2,3, . . ., Eq. 4.101 yields 

 EAϕ0 j ðxÞ 0 ¼ ρAω2j ϕj ðxÞ

ð4:102Þ

Multiplying both sides of this equation by ϕk(x) and integrating over the length, one obtains ðl 0



 EAϕ0 j ðxÞ 0 ϕk ðxÞdx ¼ ω2j

ðl

ρAϕj ðxÞϕk ðxÞdx

ð4:103Þ

0

Integrating the integral on the left-hand side of this equation by parts, one obtains

l ð l ðl

EAϕ0j ðxÞϕk ðxÞ

 EAϕ0 j ðxÞϕ0 k ðxÞdx ¼ ω2j ρAϕj ðxÞϕk ðxÞdx 0

0

ð4:104Þ

0

where l is the length of the rod.   For simple end conditions, such as free ends ϕ0 j ðlÞ ¼ 0 or fixed ends (ϕj(l ) ¼ 0), the first term in Eq. 4.104 is identically zero, and this equation reduces to ðl

0

0

EAϕ j ϕ k dx ¼

0

ω2j

ðl

ρAϕj ϕk dx

ð4:105Þ

ρAϕk ϕj dx

ð4:106Þ

0

Similarly, for the kth eigenfunction, we have ðl 0

0

0

EAϕ k ϕ j dx ¼

ω2k

ðl 0

Subtracting Eq. 4.106 from Eq. 4.105, one obtains

234

4

Vibration of Continuous Systems

 ðl ω2j  ω2k ρAϕj ϕk dx ¼ 0

ð4:107Þ

0

This equation yields ðl

ρAϕj ϕk dx ¼ 0,

0

ðl

0

ðl

EAϕ0j ϕ0k dx

¼0

0

ρAϕ2j dx

¼ m j,

ðl

0

2 EAϕ0j dx

¼ kj

9 > > for j 6¼ k > = > > for j ¼ k > ;

ð4:108Þ

The coefficients mj and kj, j ¼ 1,2,3,. . ., are called, respectively, the modal mass and modal stiffness coefficients. It is also clear from Eq. 4.105 that k j ¼ ω2j m j , j ¼ 1,2,3,. . .. That is, the jth natural frequency ωj is defined by ðl ω2j

kj ¼ ¼ ð0l mj 0

EAϕ0j dx 2

,

j ¼ 1,2,3,. . .

ð4:109Þ

2 ρAϕ0j dx

For torsional systems one can follow a similar procedure to show that the jth natural frequency of the torsional oscillations is given by ðl

GI p ϕ0j dx kj 2 0 ωj ¼ , ¼ ðl mj ρIp ϕj2 dx 2

j ¼ 1,2,3,. . .

ð4:110Þ

0

where G is the modulus of rigidity and Ip is the polar moment of inertia. If the end conditions are not simple, the definitions of the modal mass and stiffness coefficients given by Eq. 4.108 must be modified. To this end, the general relationship of Eq. 4.104 must be used to define the modal coefficients as demonstrated by the following example. Example 4.7

Find the orthogonality relationships of the mode shapes of the longitudinal vibration of the rod shown in Fig. 4.12.

Fig. 4.12 Longitudinal vibration of bars

(continued)

4.4 Orthogonality of the Eigenfunctions

235

Solution. The boundary conditions for this system are 2

uð0; t Þ ¼ 0,

m

∂ uðl; t Þ ∂uðl; t Þ ¼ kuðl; t Þ  EA ∂t 2 ∂x

These conditions yield ϕ(0) ¼ 0 and (k  ω2m)ϕ(l) ¼ EAϕ0(l). The general orthogonality relationship of Eq. 4.104 can be written as EAϕ0j ðlÞϕk ðlÞ  EAϕ0j ð0Þϕk ð0Þ  ðl ¼ ω2j ρAϕj ðxÞϕk ðxÞdx

ðl

EAϕ0j ðxÞϕ0 k ðxÞdx

0

0

By using the boundary conditions of this example, this orthogonality relationship can be written as ðl ðl   2 0 0 2  k  ω j m ϕj ðlÞϕk ðlÞ  EAϕ j ðxÞϕk ðxÞdx ¼ ω j ρAϕj ðxÞϕk ðxÞdx 0

0

or kϕj ðlÞϕk ðlÞ þ

ðl 0

EAϕ0j ðxÞϕ0k ðxÞdx

 ¼

ω2j

mϕj ðlÞϕk ðlÞ þ

ðl

 ρAϕj ðxÞϕk ðxÞdx

0

Similar relationship can be obtained for mode k as kϕj ðlÞϕk ðlÞ þ

ðl 0

  ðl EAϕ0j ðxÞϕ0k ðxÞdx ¼ ω2k mϕj ðlÞϕk ðlÞ þ ρAϕj ðxÞϕk ðxÞdx 0

Subtracting this equation from the one associated with mode j, one obtains the following orthogonality relationships for j 6¼ k: mϕj ðlÞϕk ðlÞ þ kϕj ðlÞϕk ðlÞ þ

ðl 0 ðl 0

ρAϕj ðxÞϕk ðxÞdx ¼ 0 EAϕ0j ðxÞϕ0k ðxÞdx ¼ 0

and for j ¼ k, one has mϕ2j ðlÞ þ kϕ2j ðlÞ þ

ðl 0

ðl

0

ρAϕ2j ðxÞdx ¼ m j

EAϕ0j ðxÞdx ¼ k j 2

(continued)

236

4

Vibration of Continuous Systems

where mj and kj are, respectively, the modal mass and stiffness coefficients. The jth natural frequency of the system can be defined as

ω2j

kj ¼ ¼ mj

kϕ2j ðlÞ þ

ðl

mϕ2j ðlÞ þ

0

EAϕ0 j ðxÞdx

ðl 0

2

ρAϕ2j ðxÞdx

If m and k approach zero, the natural frequency ωj approaches the value obtained by Eq. 4.109 for the simple end conditions.

Transverse Vibration It was shown in Sect. 4.3 that the partial differential equation that governs the free transverse vibration of beams is given by ! 2 2 2 ∂ v ∂ ∂ v ρA 2 ¼  2 EI z 2 ð4:111Þ ∂t ∂x ∂x where v ¼ v(x, t) is the transverse displacement, ρ is the mass density, A is the crosssectional area, E is the modulus of elasticity, and Iz is the moment of inertia of the cross section about the z-axis. The solution of this equation can be written using the separation of variables technique as v ¼ ϕ(x)q(t), where ϕ(x) and q(t) are, respectively, space- and time-dependent functions. The function q(t) is defined as q(t) ¼ B1sinωt + B2cosωt. Substituting v ¼ ϕ(x)q(t) into Eq. 4.111 yields ω2ρAϕ(x)q 00 00 (t) ¼ (EIzϕ (x))00q(t) or ω2ρAϕ(x) ¼ (EIzϕ (x))00. Therefore, for the jth natural frequency ωj, one has  00 EI z ϕ00j ¼ ω2j ρAϕj ð4:112Þ Multiplying this equation by ϕk and integrating over the length, we have ðl  ðl 00 00 2 EI z ϕ j ϕk dx ¼ ω j ρAϕj ϕk dx 0

j ¼ 1,2,3,. . .

ð4:113Þ

0

The integral on the left-hand side of this equation can be integrated by parts to yield ðl  0

EI z ϕ00j

00

l ð l  0

l

00 00 0

ϕk dx ¼ EI z ϕ j ϕk  EI z ϕ j ϕk þ EI z ϕ00j ϕ00k dx 0

0

ð4:114Þ

0

Substituting into Eq. 4.113, one obtains

l ð l ðl  0

l

00 00 0

EI z ϕ j ϕk  EI z ϕ j ϕk þ EI z ϕ00j ϕ00k dx ¼ ω2j ρAϕj ϕk dx, j ¼ 1,2,3,. . . ð4:115Þ 0

0

0

0

This is the general expression for the orthogonality condition of the eigenfunctions of the transverse vibration of beams.

4.4 Orthogonality of the Eigenfunctions

237

One can show that if the beam has simple end conditions such as fixed ends, free ends, or simply supported ends, the orthogonality condition of Eq. 4.115 reduces to ðl

EI z ϕ00j ϕ00k dx ¼ ω2j

ðl

ρAϕj ϕk dx

ð4:116Þ

Similarly, for the kth natural frequency ωk, we have ðl ðl EI z ϕ00j ϕ00k dx ¼ ω2k ρAϕj ϕk dx

ð4:117Þ

0

0

0

0

Subtracting Eq. 4.117 from Eq. 4.116, one obtains the following relationships for the simple end conditions: 9 ðl ðl > 00 00 > ρAϕj ϕk dx ¼ 0, EI z ϕj ϕk dx ¼ 0, for j 6¼ k > = 0 0 ð4:118Þ ðl ðl > > 2 00 2 > ρAϕ dx ¼ m j , EI z ϕ dx ¼ k j , for j ¼ k ; 0

j

0

j

where mj and kj are, respectively, the modal mass and modal stiffness coefficients which from Eq. 4.116 are related by k j ¼ ω2j m j , or ðl

ω2j

EI z ϕ00j dx kj 0 ¼ ¼ ðl mj ρAϕ2j dx 2

ð4:119Þ

0

If the end conditions are not simple, Eq. 4.115 can still be used to define the modal mass and stiffness coefficients as demonstrated by the following example. Example 4.8

Obtain the orthogonality relationships of the transverse vibration of the beam shown in Fig. 4.13.

Fig. 4.13 Transverse vibration

(continued)

238

4

Vibration of Continuous Systems

Solution. The boundary conditions for this system are ∂vð0; t Þ ¼0 ∂x

vð0; t Þ ¼ 0, 2

m

∂ vðl; t Þ ¼ ðEI z v00 ðl; t ÞÞ0  kv, ∂t 2

EI z v00 ðl; t Þ þ kt v0 ðl; t Þ ¼ 0

By substituting these boundary conditions in the general orthogonality relationship given by Eq. 4.115, one can verify that the orthogonality relationships for this example are given for j 6¼ k by ðl ðl 0

ρAϕj ϕk dx þ mϕj ðlÞϕk ðlÞ ¼ 0

0

EI z ϕj00 ϕ00k dx þ kϕj ðlÞϕk ðlÞ þ kt ϕj0 ðlÞϕ0k ðlÞ ¼ 0

and for j ¼ k ðl ðl 0

0

ρAϕj2 dx þ mϕj2 ðlÞ ¼ mj

EI z ϕj00 dx þ kϕj2 ðlÞ þ k t ϕ0j ðlÞ ¼ k j 2

2

where mj and kj, j ¼ 1,2,3,. . ., are, respectively, the modal mass and stiffness coefficients. Therefore, the jth natural frequency ωj is defined as ðl kj ω2j ¼ ¼ mj

EI z ϕj00 dx þ kϕj2 ðlÞ þ k t ϕj0 ðlÞ 0 ðl ρAϕj2 dx þ mϕj2 ðlÞ 2

2

0

If m, k, and kt are equal to zero, the natural frequency ωj becomes the same as the natural frequency defined by Eq. 4.119 for the cases of simple boundary conditions.

Rigid-Body Modes Unrestrained continuous systems have rigid-body modes. An example is the beam with free ends discussed in Example 4.5. It was shown that the first natural frequency of the system is equal to zero and corresponds to the rigidbody mode ϕ1 ð xÞ ¼ c1 þ c2 x

ð4:120Þ

where c1 and c2 are constants. The expression of the rigid-body mode ϕ1 is the sum of translational and rotational motions. The rigid-body translation is represented by

4.5 Forced Vibrations

239

the constant c1, while the contribution of the rigid-body rotation is represented by c2 ¼ ϕ01 ðxÞ. In the case of pure translation c2 ¼ 0, while in the case of pure rotation c1 ¼ 0. The general displacement of the beam, however, is the combination of the rigid-body and deformation modes. A rigid-body mode does not contribute to the change in the system potential energy. In order to demonstrate this fact, we use the orthogonality condition of Eq. 4.118 which is applicable to the case of simple end conditions. The modal stiffness coefficient associated with the rigid-body mode of a beam with free ends is defined as k1 ¼

ðl 0

EI z ϕ100 dx ¼ 2

ðl

2

EI z 0

∂ ðc1 þ c2 xÞdx ∂x2

ð4:121Þ

where E, Iz, and l are, respectively, the modulus of elasticity, the second moment of area of the cross section, and the length of the beam. Clearly, the modal stiffness coefficient associated with the rigid-body mode of the beam is equal to zero, that is, if the beam moves as a rigid body, there is no change in the strain energy. One, however, can show using the orthogonality conditions, that the modal mass coefficient associated with a rigid-body mode is not equal to zero, since there can be no rigid-body motion of the beam with zero kinetic energy. Since the square of a natural frequency is defined to be the modal stiffness coefficient divided by the modal mass coefficient, the natural frequency associated with a rigid-body mode is identically equal to zero. In analogy with the definitions used in the analysis of multi-degree of freedom systems, a continuous system which has a rigid-body mode is said to be positive semidefinite, while a continuous system which does not have a rigid-body mode is said to be positive definite. It is clear from the orthogonality conditions presented in this section for the cases of longitudinal, torsional, and transverse vibrations that the modal mass and stiffness coefficients must be nonnegative numbers. In the case of a positivedefinite system, both the kinetic and strain energies must be positive for nonzero displacements and equal to zero only when the displacements are equal to zero. In the case of a positive-semidefinite system, the kinetic energy of the system must be greater than zero for nonzero velocities and equal to zero only when the velocities are zero. The strain energy of a positive-semidefinite system on the other hand is positive, but it can be equal to zero for a nonzero rigid-body displacement.

4.5

Forced Vibrations

In this section, we use an analytical approach for developing the differential equations of the forced vibrations of continuous systems. As shown in the preceding sections, the vibrations of continuous systems are governed by partial differential equations expressed in terms of variables that are space and time dependent. In this section, the orthogonality of the eigenfunctions (mode shapes) are used to convert

240

4

Vibration of Continuous Systems

the partial differential equation to an infinite number of uncoupled second-order ordinary differential equations expressed in terms of the modal coordinates. These equations are similar to the equations that govern the vibration of single degree of freedom systems. Longitudinal Vibration It was shown in Sect. 4.1 that the partial differential equation that governs the longitudinal forced vibration of rods is given by Eq. 4.4. Using the technique of the separation of variables, the displacement u can be written as uðx; t Þ ¼

1 X

ϕj ðxÞqj ðt Þ

ð4:122Þ

j¼1

where ϕj is the jth space-dependent eigenfunction (mode shape) and qj is the timedependent P modal coordinate. A virtual change in the longitudinal displacement u is δu ¼ 1 j¼1 ϕj δq j . Multiplying Eq. 4.4 by δu and integrating over the length of the rod leads to ðl

2

ρA

0

∂ u δu dx ¼ ∂t 2


 ðl ∂ ∂u EA δu dx þ F ðx; t Þδu dx ∂x 0 ∂x 0

ðl

ð4:123Þ

which upon using the expression for δu yields 1 X 1 ðl h i X  0 ρAϕk ðxÞϕj ðxÞ€qk  EAϕk0 ðxÞ ϕj ðxÞqk  Qj dxδqj ¼ 0 j¼1 k¼1

ð4:124Þ

0

where Qj ¼ F(x, t)ϕj(x). By using the integration by parts, one has ðl 0

l ð l

ðEAϕk0 ðxÞÞ0 ϕj ðxÞdx ¼ EAϕ0k ðxÞϕj ðxÞ

 EAϕ0k ðxÞϕ0j ðxÞdx 0

ð4:125Þ

0

Substituting this equation into Eq. 4.124 yields " 1 X 1 ðlh X j¼1

k¼1 0

l #

ρAϕk ðxÞϕj ðxÞ€qk þ EAϕ0k ðxÞϕj0 ðxÞqk  Q j dx  EAϕ0k ðxÞϕj ðxÞ

qk δq j ¼ 0 i

0

ð4:126Þ Using the boundary conditions and the orthogonality relationship of the eigenfunctions, one can show in general that "ð

l # 1  l X

0 0 0 ρAϕk ðxÞϕj ðxÞ€qk þ EAϕk ðxÞϕj ðxÞqk dx  EAϕk ðxÞ ϕj ðxÞ

qk k¼1

0

¼ mj € qj þ k j q j

0

ð4:127Þ

4.5 Forced Vibrations

241

where mj and kj are, respectively, the modal mass and stiffness coefficients that depend on the boundary conditions and can be defined using the orthogonality relationships of the eigenfunctions. Substituting Eq. 4.127 into Eq. 4.126 yields 1  X mj €qj þ k j qj  Q j δqj ¼ 0

ð4:128Þ

j¼1

Since the virtual changes δqj are linearly independent, Eq. 4.128 yields mj €qj þ k j qj ¼ Q j ,

j ¼ 1,2,3,. . .

ð4:129Þ

These are uncoupled second-order ordinary differential equations which are in the same form as the vibration equations of the single degree of freedom systems. Therefore, their solution can be obtained using Duhamel’s integral as q_ j0 1 qj ¼ qj0 cos ωj t þ sin ωj t þ mj ωj ωj

ðt

Q j ðτÞ sin ωj ðt  τÞdτ,

j ¼ 1,2,3,. . .

0

ð4:130Þ where qj0 and q_j0 are the initial modal displacements and velocities, and ωj is the jth pffiffiffiffiffiffiffiffiffiffiffi natural frequency defined as ω j ¼ k j =mj , j ¼ 1,2,3,. . .. Having determined qj using Eq. 4.130, the longitudinal displacement u(x, t) can be determined by using Eq. 4.122. In the case of free vibration, the modal force Qj is equal to zero and Eq. 4.129 reduces to mj € qj þ k j qj ¼ 0, j ¼ 1,2,3,. . .. The solution of these equations is   qj ¼ qj0 cos ωj t þ q_ j0 =ωj sin ωj t, j ¼ 1,2,3,. . .. Example 4.9

If the rod in Example 4.7 is subjected to a distributed axial force of the form F (x, t), determine the equations of forced longitudinal vibration of this system. Solution. The boundary conditions in this example are 2

uð0; t Þ ¼ 0,

m

∂ u ∂uðl; t Þ ¼ kuðl; t Þ  EA 2 ∂t ∂x

which yield ϕð0Þ ¼ 0,

EAϕ0 ðlÞqðt Þ ¼ kϕðlÞqðt Þ  mϕðlÞ€qðt Þ

That is, ϕk ð0Þϕj ð0Þ ¼ 0 EAϕk ðlÞϕj ðlÞqk ðt Þ ¼

0 kϕk ðlÞϕj ðlÞqk ðt Þ  mϕk ðlÞϕj ðlÞ€qk ðt Þ (continued)

242

4

Vibration of Continuous Systems

The last term on the left-hand side of Eq. 4.127 can be written as

l

EAϕk0 ðxÞϕj ðxÞ

  qk ¼ EAϕk0 ðlÞϕj ðlÞ  EAϕk0 ð0Þϕj ð0Þ qk

0

which upon using the boundary conditions yields

l

0 EAϕk ðxÞϕj ðxÞ

qk 0

¼ kϕk ðlÞϕj ðlÞqk  mϕk ðlÞϕj ðlÞ€qk

Therefore, Eq. 4.127 can be written as 1 ðl X k¼1 0

¼

 ρAϕk ðxÞϕj ðxÞ€qk þ EAϕk0 ðxÞϕj0 ðxÞqk dx þ kϕk ðlÞϕj ðlÞqk þ mϕk ðlÞϕj ðlÞ€ qk

1  X

mϕk ðlÞϕj ðlÞ þ

þ

 ρAϕk ðxÞϕj ðxÞdx €qk

0

k¼1

1  X

ðl

kϕk ðlÞϕj ðlÞ þ

ðl

0

k¼1



EAϕk0 ðxÞϕj0 ðxÞdx

qk

Comparing this equation with the orthogonality relationships obtained in Example 4.7, it is clear that 1  X

 ðl mϕk ðlÞϕj ðlÞ þ ρAϕk ðxÞϕj ðxÞdx € qk 0 k¼1  ðl 1  X þ kϕk ðlÞϕj ðlÞ þ EAϕk0 ðxÞϕj0 ðxÞdx qk ¼ mj €qj þ k j qj 0

k¼1

where mj and kj are the modal mass and stiffness coefficients defined by mj ¼

mϕ2j ðlÞ

þ

k j ¼ kϕ2j ðlÞ þ

ðl

ðl 0

0

ρAϕ2j ðxÞdx

EAϕ0j ðxÞdx, 2

j ¼ 1,2,3,. . .

Therefore, the equations of motion of the forced longitudinal vibration of the rod, expressed in the modal coordinates, are mj €qj þ k j qj ¼ Q j , j ¼ 1,2,3,. . ..

Concentrated Loads If the force F is a concentrated load that acts at a point p on the beam, that is, F ¼ F(t), there is no need to carry out integration in order to obtain the modal forces. In this case, the virtual work of this force is given by

4.5 Forced Vibrations

243

    P δW ¼ F ðt Þδu xp ; t ¼ F ðt Þ 1 which can be written as j¼1 ϕj xp δq j ðt Þ, P Q δq ð t Þ, where Q is the modal force associated with the jth modal δW ¼ 1 j j j j¼1 coordinate and defined as Qj ¼ F(t)ϕj(xp), in which ϕj(xp) is the jth mode shape evaluated at point p on the beam. Initial Conditions As in the case of the multi-degree of freedom systems, the orthogonality conditions can be utilized in determining the initial conditions qj0 and q_ j0 , j ¼ 1,2,3,. . .,. Let f(x, 0) and g(x, 0) be, respectively, the initial displacements and velocities of the rod, that is, f ðx; 0Þ ¼ uðx; 0Þ ¼ gðx; 0Þ ¼ u_ ðx; 0Þ ¼

1 X

9 > > ϕj ðxÞqj0 > > =

j¼1 1 X j¼1

ð4:131Þ

> > ϕj ðxÞq_j0 > > ;

Multiplying these equations by ρAϕk and integrating over the length yields  9 > > ρAϕk ðxÞϕj ðxÞdx qj0 > > = 0 0 j¼1   ðl ð 1 l X > > ρAϕk ðxÞgðx; 0Þdx ¼ ρAϕk ðxÞϕj ðxÞdx q_j0 > > ; ðl

ρAϕk ðxÞ f ðx; 0Þdx ¼

0

1 ð l X

j¼1

ð4:132Þ

0

These two equations can be used to determine the initial modal displacements and velocities. To this end, the orthogonality conditions may be utilized. For example, in the case of simple boundary conditions, one has ðl

ρAϕk ðxÞϕj ðxÞdx ¼

0

if j 6¼ k if j ¼ k

0 mj

ð4:133Þ

Substituting this orthogonality relationship into Eq. 4.132, one obtains qj0 ¼

1 mj

1 q_j0 ¼ mj

ðl

ρAϕj ðxÞf ðx; 0Þdx

0

ðl

ð4:134Þ ρAϕj ðxÞgðx; 0Þdx

j ¼ 1,2,3,. . .

0

If the end conditions are not simple, Eq. 4.132 can still be used to solve for the initial modal displacements and velocities. For instance, if the problem of Example 4.7 is considered, one of the orthogonality conditions was given by mϕj ðlÞϕk ðlÞ þ

ðl 0

ρAϕj ðxÞϕk ðxÞdx ¼

0 mj

if j 6¼ k if j ¼ k

ð4:135Þ

244

4

Vibration of Continuous Systems

where m is the concentrated mass attached to the end of the rod. The preceding equation may be written as 1  X

mϕj ðlÞϕk ðlÞqj þ


ð l

  ρAϕj ðxÞϕk ðxÞdx qj ¼ mk qk

ð4:136Þ

0

j¼1

which can be rewritten as mϕk ðlÞuðl; t Þ þ

1
ð l X j¼1

 ρAϕj ðxÞϕk ðxÞdx qj ¼ mk qk

ð4:137Þ

0

which at t ¼ 0 leads to 9 > > ρAϕj ðxÞϕk ðxÞdx qj0 ¼ mk qk0  mϕk ðlÞf ðl; 0Þ > > = 0 j¼1   ð 1 l X > > ρAϕj ðxÞϕk ðxÞdx q_j0 ¼ mk q_k0  mϕk ðlÞgðl; 0Þ > > ; 1 ð l X

j¼1



ð4:138Þ

0

Substituting these two equations into Eqs. 4.150 and 4.151, one obtains qk0

1 ¼ mk

1 q_k0 ¼ mk

ð l 0

ð l

 ρAϕk ðxÞ f ðx; 0Þdx þ mϕk ðlÞf ðl; 0Þ  ρAϕk ðxÞgðx; 0Þdx þ mϕk ðlÞgðl; 0Þ ,

9 > > > = > > k ¼ 1,2,3,. . . > ;

ð4:139Þ

0

Torsional Vibration The equation for the forced torsional vibration takes a similar form to the equation of forced longitudinal vibration. This equation was given in Sect. 4.2 by Eq. 4.47. The solutionP of Eq. 4.47 can be expressed using the separation of variables technique as θðx; t Þ ¼ 1 j¼1 ϕj ðxÞq j ðt Þ. Following the same procedure as in the case of longitudinal vibration, one can show that the equations of the torsional vibration of the shaft in terms of the modal coordinates are given by mj € qj þ k j qj ¼ Q j , j ¼ 1,2,3,. . ., where mj and kj are, respectively, the modal mass Ðl and stiffness coefficients and Q j ¼ 0 T e ðx; t Þϕj ðxÞdx. Therefore, the mathematical treatment of the linear torsional oscillations of shafts is the same as the one used for the longitudinal vibration of rods. Transverse Vibration The partial differential equation of the forced transverse vibration of the beams was given by Eq. 4.72. By using the technique P of separation of variables, we may write the transverse displacement v as v ¼ 1 j¼1 ϕj ðxÞq j ðt Þ. Multiplying both sides of Eq. 4.72 by the virtual displacement δv and integrating over the length of the beam, we obtain

4.5 Forced Vibrations

ðl " 0

245

! # ðl 2 2 2 ∂ v ∂ ∂ v ρA 2 δv þ 2 EI z 2 δv dx ¼ F ðx; t Þδv dx ∂t ∂x ∂x 0

ð4:140Þ

which upon using the expression for v yields 1 X 1 ðl h i X  00 ρAϕk ðxÞϕj ðxÞ€qk þ EI z ϕ00k ðxÞ ϕj ðxÞqk δqj dx j¼1 k¼1

¼

0

1 ðl X j¼1

ð4:141Þ F ðx; t Þϕj ðxÞδqj dx

0

Integration by parts yields ðl 0

l ð l

  l  00 00 00 00 0

EI z ϕk ðxÞ ϕj ðxÞdx ¼ EI z ϕk 0ϕj  EI z ϕk ϕj þ EI z ϕ00k ϕ00j dx 0

0

ð4:142Þ

0

Substituting this equation into Eq. 4.141 and using the boundary conditions and relationships of the eigenfunctions, one gets P1 the orthogonality qj þ k j qj  Q j δqj ¼ 0, where mj and kj are the modal mass and modal j¼1 m j € stiffness coefficients that depend on the boundary conditions and can be defined using the orthogonality of the mode shapes, and Qj is the modal forcing function Ðl defined as Q j ¼ 0 F ðx; t Þϕj ðxÞdx. Since the modal coordinates qj, j ¼ 1,2,3,. . ., are independent, one obtains the uncoupled second-order ordinary differential equations mj € qj þ k j qj ¼ Q j , j ¼ 1,2,3,. . .. These equations are in a form similar to the one obtained in the preceding section for the two cases of longitudinal and torsional vibrations. Example 4.10

For the system given in Example 4.8 obtain the uncoupled second-order differential equations of transverse vibration. Solution. The boundary conditions for this system are vð0; t Þ ¼ 0,

∂vð0; t Þ ¼0 ∂x

2

∂ vðl; t Þ 0 ¼ ðEI z v00 ðl; t ÞÞ  kvðl; t Þ ∂t 2 EI z v00 ðl; t Þ þ kt v0 ðl; t Þ ¼ 0 m

which yield (continued)

246

4

Vibration of Continuous Systems

ϕ0 ð0Þ ¼ 0,

ϕð0Þ ¼ 0,

mϕðlÞ€qðt Þ ¼ ðEI z ϕ00 ðlÞÞ0 qðt Þ  kϕðlÞqðt Þ, EI z ϕ00 ðlÞqðt Þ þ kt ϕ0 ðlÞqðt Þ ¼ 0 Substituting the boundary conditions at the fixed end into Eq. 4.142 yields ðl 0



l ð l

 0 l 00 EI z ϕ00k ðxÞ ϕj ðxÞdx ¼ EI z ϕ00k ϕj

 EI z ϕ00k ϕj0

þ EI z ϕ00k ϕ00j dx 0

0

0

 0  0 ¼ EI z ϕ00k ðlÞ ϕj ðlÞ  EI z ϕ00k ð0Þ ϕj ð0Þ ðl EI z ϕ00k ðlÞϕj0 ðlÞ þ EI z ϕ00k ð0Þϕj0 ð0Þ þ EI z ϕ00k ϕ00j dx 0



0

¼ EI z ϕ00k ðlÞ ϕj ðlÞ  EI z ϕ00k ðlÞϕj0 ðlÞ þ

ðl

0

EI z ϕ00k ϕ00j dx

Multiplying this equation by qk(t) and using the remaining boundary conditions, we obtain ðl 0



00   EI z ϕ00k ðxÞ ϕj ðxÞdxqk ¼ EI z ϕ00k ðlÞ 0 ϕj ðlÞqk

 EI z ϕ00k ðlÞϕj0 ðlÞqk þ

ðl 0

EI z ϕ00k ϕ00j dxqk ¼ mϕk ðlÞϕj ðlÞ€qk ðt Þ

þ kϕk ðlÞϕj ðlÞqk ðt Þ þ kt ϕk0 ðlÞϕj0 ðlÞqk ðt Þ þ

ðl 0

EI z ϕ00k ϕ00j dxqk ðt Þ

Substituting this equation into Eq. 4.141 yields ( 1 1 ðl h i X X ρAϕk ðxÞϕj ðxÞ€qk ðt Þ þ EI z ϕ00k ϕ00j qk ðt Þ dx þ mϕk ðlÞϕj ðlÞ€qk ðt Þ j¼1

k¼1

0

þkϕk ðlÞϕj ðlÞqk ðt Þ þ

kt ϕk0 ðlÞϕj0 ðlÞqk ðt Þ

1 ðl o X δqj ¼ F ðx; t Þϕj ðxÞdx δqj j¼1

0

By using the orthogonality relationships obtained in Example 4.8, the preceding equation can be written as 1  X mj €qj þ k j qj  Q j δqj ¼ 0 j¼1

or mj €qj þ k j qj ¼ Q j ,

j ¼ 1,2,3,. . . (continued)

4.6 Inhomogeneous Boundary Conditions

247

where mj ¼

ðl 0

kj ¼

ðl

0

Qj ¼

ðl

ρAϕ2j ðxÞdx þ mϕ2j ðlÞ EI z ϕ00j ðxÞdx þ kϕ2j ðlÞ þ kt ϕ0j ðlÞ 2

2

F ðx; t Þϕj ðxÞdx

0

4.6

Inhomogeneous Boundary Conditions

The solution of the vibration equations of continuous systems subject to homogeneous boundary conditions, which are not time dependent, was obtained and examined in the preceding sections. In this section, we discuss the case of inhomogeneous, time-dependent boundary conditions. To this end, we consider the partial differential equation that governs the longitudinal vibration of bars and is given by (∂2u/∂t2)  c2(∂2u/∂x2) ¼ F(x, t) subject to the initial conditions u(x, 0) ¼ f(x) and u_ ðx; t Þ ¼ gðxÞ and the inhomogeneous boundary conditions a1 uð0; t Þ  b1 u0 ð0; t Þ ¼ h1 ðt Þ

)

a2 uðl; t Þ þ b2 u0 ðl; t Þ ¼ h2 ðt Þ

ð4:143Þ

where c, a1, a2, b1, and b2 are constants and h1 and h2 are given functions. In order to solve the partial differential equation subject to the inhomogeneous boundary conditions of Eq. 4.143, the problem may be reduced to one with homogeneous boundary conditions. This can be achieved by introducing an auxiliary assumed function W(x, t) which is twice differentiable and satisfies the boundary conditions of Eq. 4.143. It is not necessary, however, that the function W satisfies the partial differential equation. In terms of the new function W, the function u(x, t) can be written as uðx; t Þ ¼ W ðx; t Þ þ Z ðx; t Þ

ð4:144Þ

where Z(x, t) is an unknown function which will be determined. Substituting Eq. 4.144 into the partial differential equation, the initial conditions, and the boundary conditions, leads to € þ c2 W 00 Z€  c2 Z 00 ¼ F ðx; t Þ  W subject to the initial conditions

ð4:145Þ

248

4

Vibration of Continuous Systems

)

Z ðx; 0Þ ¼ f ðxÞ  W ðx; 0Þ Z_ ðx; 0Þ ¼ gðxÞ  W_ ðx; 0Þ

ð4:146Þ

and the boundary conditions a1 Z ð0; t Þ  b1 Z 0 ð0; t Þ ¼ 0

) ð4:147Þ

a2 Z ðl; t Þ þ b2 Z 0 ðl; t Þ ¼ 0

Observe that Z(x, t) satisfies the new inhomogeneous partial differential equation given by Eq. 4.145 subject to the homogeneous boundary conditions given by Eq. 4.147. The method described in this section can be used to solve a large class of vibration problems in which one point on the continuous media is subjected to a specified excitation. This is demonstrated by the following example. Example 4.11

Consider the longitudinal vibration of a bar with length l. One end of the bar is assumed to be fixed, while the other end has a specified sinusoidal motion. If the initial conditions are assumed to be zero, find the series solution of the vibration equation of this system. Solution. The partial differential equation of the system is utt ¼ c2 uxx , 0 < x < l,

t>0

subject to the inhomogeneous boundary conditions uð0; t Þ ¼ 0,

uðl; t Þ ¼ sin ωt

and the initial conditions uðx; 0Þ ¼ 0,

u_ ðx; 0Þ ¼ 0

Observe that the function W ¼ W ðxÞ sin ωt is twice differentiable and satisfies the inhomogeneous boundary conditions. Let ω ¼ kc for some constant k. This choice of the function W satisfies the differential equation as well as the boundary conditions. Clearly, d2 W þ k2 W ¼ 0, dx2

W ð0Þ ¼ 0,

W ðl Þ ¼ 0

(continued)

4.6 Inhomogeneous Boundary Conditions

249

which implies that W ð xÞ ¼

sin kx sin kl

and accordingly W ðx; t Þ ¼ W ðxÞ sin ωt ¼

sin kx sin kct sin kl

The function u(x, t) can then be written as uðx; t Þ ¼ Z ðx; t Þ þ W ðx; t Þ where Z(x, t) is an unknown twice differentiable function. Substituting the preceding equation into the partial differential equation, one obtains the following partial differential equation: Z€  c2 Z 00 ¼ 0 subject to the boundary conditions Z ð0; t Þ ¼ 0,

Z ðl; t Þ ¼ 0

and the initial conditions Z ðx; 0Þ ¼ W ðx; 0Þ ¼ 0 kc sin kx Z_ ðx; 0Þ ¼ W_ ðx; 0Þ ¼ sin kl Note that in this case, the boundary conditions are homogeneous, and also the right-hand side of the resulting partial differential equation is zero because the selected function W(x, t) satisfies the original partial differential equation. The function W(x, t) does not have to satisfy this condition which leads in this example to a simplified equation. The new partial differential equation in Z can be solved by the methods described in the preceding sections. The solution u can then be obtained using Eq. 4.144, and it is left to the reader to verify that uðx; t Þ ¼

1 sin kx 2cω X ð1Þnþ1 nπx nπct sin ωt þ sin sin 2 2 sin kl l n¼1 ω  ðnπc=lÞ l l

250

4.7

4

Vibration of Continuous Systems

Viscoelastic Materials

In the case of undamped free and forced vibrations of continuous systems, the effect of discrete damping elements as well as structural damping is not considered. It is, however, clear at this point that the effect of external damping as the result of discrete damping elements such as dashpots can be introduced to the vibration equations by developing an expression for the virtual work of the damping forces. Internal damping, however, is present in most materials as the result of friction between the particles. It has been observed that the amplitude of free vibration of a solid specimen decreases with time even in the cases in which the specimen is isolated from any source of external damping. Such materials are called viscoelastic. There are various models for describing the behavior of viscoelastic materials. The simplest model is called the Kelvin–Voigt model in which the stress acting on the body is assumed to be proportional to the strain and its time derivative, that is, σ ¼ Eε þ γ ε_

ð4:148Þ

where σ and ε are, respectively, the stress and strain, and E and γ are constants that depend on the properties of the material. In order to demonstrate the use of the simple Kelvin–Voigt model in the vibration analysis of viscoelastic continuous systems, we consider the vibration of a uniform rod of length l and cross-sectional area A. The elastic force acting on the rod can be written as P ¼ σA. The strain displacement relationship for this simple example is given by ε ¼ ∂u/∂x. By using this equation, the stress of Eq. 4.148 can be expressed in terms of the displacement as 2

σ¼E

∂u ∂ u þγ ∂x ∂x∂t

ð4:149Þ

Using this equation, the elastic force P ¼ σA can be written as 2

P ¼ EA

∂u ∂ u þ γA ∂x ∂x∂t

ð4:150Þ

Substituting this expression for P into Eq. 4.2 of Sect. 4.1, one obtains 2

2

∂ u ∂ ∂u ∂ u EA þ γA ρA 2 ¼ ∂t ∂x ∂x ∂x∂t

! þ F ðx; t Þ

ð4:151Þ

where ρ is the mass density and F(x, t) is the external force. If the cross-sectional area is assumed to be constant, Eq. 4.151 reduces to 2

2

3

∂ u ∂ u ∂ u ¼ c2 2 þ b2 2 þ F ðx; t Þ=ρA ∂t 2 ∂x ∂x ∂t

ð4:152Þ

4.7 Viscoelastic Materials

251

pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi where c ¼ E=ρ and b ¼ γ=ρ. If γ ¼ 0, Eq. 4.152 reduces to the same equation obtained for the undamped longitudinal vibrations of rods. If the external forces are absent, Eq. 4.152 reduces to 2

2

3

∂ u ∂ u ∂ u ¼ c 2 2 þ b2 2 2 ∂t ∂x ∂x ∂t

ð4:153Þ

A solution of this equation can be obtained using the separation of variables technique. To this end, we assume a solution of the form u(x, t) ¼ ϕ(x)q(t). Substituting this equation into Eq. 4.153 leads to   ϕ€q ¼ c2 ϕ00 q þ b2 ϕ00 q_ ¼ c2 q þ b2 q_ ϕ00 ð4:154Þ  2  This equation can be rewritten as €q= c q þ b2 q_ ¼ ϕ00 =ϕ. Since the left-hand side of this equation depends on t only, and the right-hand side depends on x only, one must have

c2 q

€q ϕ00 ¼ β2 ¼ 2 ϕ þ b q_

ð4:155Þ

where β2 is a constant. One therefore has €q þ 2ξωq_ þ ω2 q ¼ 0,

ϕ00 þ β2 ϕ ¼ 0

ð4:156Þ

where ω ¼ cβ and ξ ¼ b2β/2c. The solutions of Eq. 4.156 in the case of lightly damped system can be expressed as q ¼ X 1 eξωt sin ðωd t þ ψ 1 Þ



ϕ ¼ X 2 sin ðβx þ ψ 2 Þ

ð4:157Þ

pffiffiffiffiffiffiffiffiffiffiffiffiffi where ωd ¼ ω 1  ξ2 and X1, ψ 1, X2, and ψ 2 are constants which can be determined using the boundary and initial conditions as described in the preceding sections. The use of this procedure leads to the definition of the eigenfunctions ϕj. The general solution can then be written as the sum of the normal modes as uðx; t Þ ¼

1 X

  ϕj ðxÞX 1j eξ j ω j t sin ωd j t þ ψ 1 j

ð4:158Þ

j¼1

where subscript j refers to the jth mode of vibration. Wave Motion One may observe that each term in Eq. 4.158 is the product of two harmonic functions, one of them depends on x only, while the other depends on t only. Using the definition of ϕ(x) given by Eq. 4.157, each term in the series of Eq. 4.158 can be written as

252

4

Vibration of Continuous Systems

    u j ðx; t Þ ¼ X 1 j X 2 j eξ j ω j t sin β j x þ ψ 2 j sin ωd j t þ ψ 1 j

ð4:159Þ

By using the trigonometric identity sin α1 sin α2 ¼ 12 f cos ðα1  α2 Þ  cos ðα1 þ α2 Þg, Eq. 4.159 can be written as      u j ðx; t Þ ¼ A j eξ j ω j t cos β j x  ωd j t þ ψ mj  cos β j x þ ωd j t þ ψ pj ð4:160Þ where A j ¼ X 1 j X 2 j =2, ψ mj ¼ ψ 2 j  ψ 1 j , and ψ pj ¼ ψ 2j + ψ 1j. The terms between the brackets in Eq. 4.160 represent two harmonic waves, traveling in opposite directions with the same speed ωdj/βj. The amplitude of these two waves is multiplied by the exponential decay eξ j ω j t . Consequently, the amplitude of the resulting wave motion will decrease with time as the result of the damping effect. If the damping is equal to zero, the amplitude of the two harmonic waves remains constant. Recall from the analysis presented in Sect. 4.1 that in the case of undamped elastic rod, harmonic waves with differentp frequencies travel with the same phase ffiffiffiffiffiffiffiffi velocity which is constant and equal to c ¼ E=ρ. In this case, the elastic medium is said to be nondispersive, and the group velocity is constant and equal to the phase velocity. This is not, however, the case when viscoelastic materials are considered. The phase velocity of harmonic waves that have different frequencies are not in general equal. Consequently, the group velocity is no longer equal to the phase velocity and it becomes dependent on the wave number.

4.8

Energy Methods

Lagrange’s equation, which utilizes scalar energy quantities, can be used as an alternative to the methods described in the preceding sections for the formulation of the vibration equations of continuous systems. We demonstrate the use of Lagrange’s equation by considering the longitudinal vibration of rods. It was shown, in the case of longitudinal vibration, that the solution can be assumed, Pusing the technique of the separation of variables, in the form uðx; t Þ ¼ 1 j¼1 ϕj ðxÞq j ðt Þ. For simple boundary conditions, the kinetic and strain energies and virtual work are given, respectively, by T¼

1 2

ðl

ρAu_ 2 dx,

U¼

0

1 2

ðl

EAu0 dx, 2

δW ¼

0

ðl

F ðx; t Þδu dx

ð4:161Þ

0

P Substituting for u_ , u0 , and δu from uðx; t Þ ¼ 1 j¼1 ϕj ðxÞq j ðt Þ and using the appropriate orthogonality conditions, the kinetic and strain energies and the virtual work can be written as T¼

1 1X mj q_ 2j , 2 j¼1

U¼

1 1X k j q2j , 2 j¼1

δW ¼

1 X j¼1

Q j δqj

ð4:162Þ

4.8 Energy Methods

253

where mj and kj are, respectively, the modal mass and stillness coefficients, and Qj is the generalized force associated with the jth modal coordinate. By using the expressions for the kinetic and strain energies, the virtual work, and Lagrange’s equation, or the virtual work principle, one can show that the same system of differential equations of motion developed in the preceding sections can be obtained. This system of equations is written as mj €qj þ k j qj ¼ Q j , j ¼ 1,2,3,. . .. Clearly, if the boundary conditions are not simple, the expressions for the kinetic and strain energies may take a different form. For example, if a mass m is attached to the end of the rod, the kinetic energy expression becomes T¼

1 2

ðl

1 ρAu_ 2 ðx; t Þdx þ mu_ 2 ðl; t Þ 2 0

ð4:163Þ

Similarly, if a spring of stiffness k is attached to the end of the rod, the strain energy expression becomes 1 U¼ 2

ðl

1 2 EAu0 ðx; t Þdx þ ku2 ðl; t Þ 2 0

ð4:164Þ

In the case of torsional oscillations, the expressions for the kinetic and strain energies in the case of simple boundary conditions are, respectively, defined as T¼

1 2

ðl

ρI p θ_ 2 ðx; t Þdx,

U¼

0

1 2

ðl

GJθ0 ðx; t Þdx

ð4:165Þ

EIv00 ðx; t Þdx

ð4:166Þ

2

0

and in the case of transverse vibrations T¼

1 2

ðl 0

ρAv_2 ðx; t Þdx,

U¼

1 2

ðl

2

0

Conservation of Energy From the analysis presented in this section, it is clear that regardless of the type of vibration (longitudinal, torsional, or transverse), the kinetic energy and strain energy of the system  can be written, respectively, as  P1  P1 2 2 _ T¼ =2 and U ¼ m k q q j¼1 j j j¼1 j j =2, where mj and kj are, respectively, the modal mass and stiffness coefficients which can be defined using the orthogonality relationships, and qj and q_j are, respectively, the modal coordinates and velocities. It was shown in the case of free undamped vibration that the modal coordinate qj can be expressed in the form qj ¼ B1j sin ωjt + B2j cos ωjt, which can be rewritten in another alternate form as qj ¼ Cj sin (ωjt + ψ j), where B1j, B2j, Cj, and ψ j are constants that can be determined by using  the initial  conditions. The modal velocity q_j can also be written as q_j ¼ ωj C j cos ω j t þ ψ j . Therefore, the kinetic and strain energies can be written in a more explicit form, respectively, as

254

4

T¼

Vibration of Continuous Systems

9 1   1X > ω2j C2j mj cos 2 ωj t þ ψ j > > > = 2 j¼1

1   1X U¼ C 2j k j sin 2 ωj t þ ψ j 2 j¼1

> > > > ;

ð4:167Þ

Recall that k j ¼ ω2j mj , and consequently, the kinetic expressed in P energy can be   1 2 2 terms of the modal stiffness coefficients as T ¼ =2. j¼1 C j k j cos ω j t þ ψ j The sum of the kinetic and strain energies can then be written as T þU ¼

1 1   1X   1X C 2j k j cos 2 ωj t þ ψj þ C2j k j sin 2 ωj t þ ψj 2 j¼1 2 j¼1

¼

1      1X C 2j k j cos 2 ωj t þ ψj þ sin 2 ωj t þ ψj 2 j¼1

¼

1 1 1X 1X C 2j k j ¼ C 2 ω2 m j 2 j¼1 2 j¼1 j j

ð4:168Þ

That is, the sum of the kinetic and strain energies is a constant. Since both the kinetic and strain energies are nonnegative numbers, the strain energy becomes maximum when the kinetic energy is minimum. Similarly, the kinetic energy becomes maximum when the strain energy is minimum. The contribution of each mode to the total energy of the system depends on the initial conditions as represented by the constant Cj, the natural frequency of this mode, and the modal mass and stiffness coefficients. As discussed in Chap. 2, the fact that the energy of the system, during the undamped free vibration, is conserved can be utilized to derive the equations of motion. To this end, one can write " # 1   d d 1X ðT þ U Þ ¼ mj q_2j þ k j q2j ¼ 0 dt dt 2 j¼1

ð4:169Þ

Since mj and kj are constant  taking the derivatives in the preceding P  coefficients, equation, one obtains 1 qj þ k j qj q_ j ¼ 0. If we assume q_ j , j ¼ 1,2,3,. . ., to be j¼1 m j € independent velocities, their coefficients in the preceding equation must be equal to zero. This yields mj €qj þ k j qj ¼ 0, j ¼ 1,2,3,. . ., which is the same system of equations that can be obtained using Lagrange’s equation in the case of undamped free vibration. These are uncoupled differential equations expressed in terms of the modal coordinates. Since there is no coupling between the modes, one may expect that there is no exchange of energy between different modes, and as a consequence, the total energy associated with each mode is conserved. In order to see this, the  associated with an arbitrary mode j can be written as  total energy T j þ Uj ¼ mj q_ 2j þ k j q2j =2, where Tj and Uj are, respectively, the kinetic and strain

4.9 Approximation Methods

255

energies associated with the jth modal coordinate. Since in the undamped free vibration,the modal coordinates and modal velocities functions, one  are harmonic     has T j ¼ ω2j C2j mj cos 2 ω j t þ ψ j =2 and U j ¼ C 2j k j sin 2 ω j t þ ψ j =2, which yields T j þ U j ¼ C 2j k j =2 ¼ C2j ω2j mj =2. That is, the total energy associated with each mode is indeed constant. Owing to the fact that the modal coordinates and velocities are simple harmonics, the total energy associated with a particular mode j can be written as T j þ U j ¼ C 2j k j =2 ¼ U j , or alternatively as T j þ U j ¼ C2j ω2j mj =2 ¼ T j , where T j and U j are the maximum kinetic and strain energies of the mode j. Therefore, the total kinetic energy associated with a mode is constant and is equal to the maximum kinetic energy or the maximum strain energy.

4.9

Approximation Methods

It is clear from the analysis presented thus far that the exact solution of the free and forced vibration of continuous systems is represented by an infinite series expressed in terms of the principal modes of vibration. In many applications, highfrequency modes of vibration may not have a significant effect on the solution of the vibration equations such that the contribution of these high-frequency modes can be neglected and the solution may be represented in terms of a finite number of modes or in terms of assumed polynomials that describe the shape of deformation of the continuous systems. In this section and the following sections, we discuss several approximate methods that can be used to determine the fundamental natural frequencies of the continuous systems. In some of these methods, a finite-dimensional model, which is in a form similar to the mathematical model that governs the free vibration of the multi-degree of freedom system, is first developed. The resulting reduced system of equations can be used to determine a finite set of natural frequencies and mode shapes. We shall present first an approximate method, called Rayleigh’s method, which can be used to determine the fundamental natural frequency of the continuous systems. Rayleigh Quotient Recall from the analysis presented in the preceding section that in the case of free undamped vibration, the energy associated with any mode of vibration is conserved. The sum of the kinetic energy and strain energy of a mode remains constant and equal to the maximum kinetic energy or equal to the maximum strain energy. Consequently, for the jth mode of vibration, the conservation of energy in the case of undamped free vibration leads to T j ¼ U j , j ¼ 1,2,3,. . ., where T j and U j are, respectively, the maximum kinetic and strain energies associated with mode j. By using the expressions for T j and U j given in the preceding section, it is clear that the natural frequency of the jth mode can be defined using the following equation:

256

4

ω2j ¼

U j kj ,  ¼  mj Tj

Vibration of Continuous Systems

j ¼ 1,2,3,. . .

ð4:170Þ

where T j ¼ ω2j Tj . Note that in the case of longitudinal vibration, the preceding equation and Eq. 4.104 yield ðl ω2j ¼

0

2 EAϕ0 j ðxÞdx

ðl 0



l

2 EAϕ0 j ðxÞ

0

ð4:171Þ

ρAϕ2j ðxÞdx

where E, ρ, A, and l are, respectively, the modulus of elasticity, mass density, crosssectional area, and Ðlength of the rod.  ÐIn the case ofsimple end conditions, Eq. 4.171 l l 02 2 reduces to ω j ¼ 0 EAϕ j ðxÞdx = 0 ρAϕ2j ðxÞdx . In the case of transverse vibration, the conservation of energy of mode j leads to ðl ω2j ¼

0

2 EI z ϕ00j dx

þ

h

EI z ϕ00j

ðl 0

0

ϕj 

EI z ϕ00j ϕj0

i

l

0

ð4:172Þ

ρAϕ2j dx

where E, ρ, Iz, and l are, respectively, the modulus of elasticity, mass density, moment of inertia of the cross section, and length the case of simple end Ð of the beam.  In Ðl l 00 2 2 conditions, Eq. 4.172 reduces to ω j ¼ 0 EI z ϕ j dx = 0 ρAϕ2j dx . Note that the numerators in Eqs. 4.171 and 4.172 are proportional to the strain energy of mode j while the denominator is proportional to the kinetic energy of that particular mode. The expressions on the right-hand side of these equations are called Rayleigh quotients. Frequency Approximation In many practical applications, it is difficult to determine the exact mode shapes. Nonetheless, due to the fact that in many of these applications the shape of the first mode is simple, a guess, with reasonable degree of accuracy, can be made for a shape that resembles the first mode. In this case, the Rayleigh quotient can be used to determine an approximate value for the fundamental natural frequency. In order to demonstrate this procedure, let us consider the case of a cantilever beam and choose the following function to approximate the fundamental mode shape  πx ϕ1 ¼ b 1cos 2l

ð4:173Þ

where b is a constant. Since we have simple end conditions in the case of a cantilever beam, the Rayleigh quotient of Eq. 4.172 in the case of simple boundary conditions can be used to obtain an estimate for the fundamental natural

4.9 Approximation Methods

257

frequency ω1. In this case of simple boundary conditions, the use of Eq. 4.173 leads to 2 π2 πx cos dx 2l 4l2 ω21  ω21 ¼ ð0 l ¼ ð 0l  πx2 2 ρAϕ1 ðxÞ dx ρA 1  cos dx 2l 0 0 ðl

EI z ϕ100 ðxÞ dx 2




ðl

EI z

ð4:174Þ

where ω1 is the approximate fundamental natural frequency. If we assume the case of a uniform beam which is made of homogeneous  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi material, one can show that Eq. 4.174 leads to ω1 ¼ ω1  3:667=l2 EI z =ρA.p The exact fundamental natural   ffiffiffiffiffiffiffiffiffiffiffiffiffiffi frequency of a cantilever beam is ω1 ¼ 3:5156=l2 EI z =ρA. This equation shows that the error in the solution obtained by using the approximate Rayleigh method is less than 5%. The assumed mode of Eq. 4.173 satisfies all the geometric boundary conditions at the clamped end but satisfies only one of the natural boundary conditions at the free end (ϕ00 (l, t) ¼ 0). In Rayleigh’s method, the assumed functions must satisfy only the geometric boundary conditions; it is not necessary that these functions satisfy the natural boundary conditions. The error in the fundamental natural frequency obtained using the Rayleigh quotient becomes smaller if the integrals  00 2   Ðl Ðl  2 00 2 2 0 EI z ϕ1  ϕ1 dx and 0 ρA ϕ1  ϕ1 dx are small. The use of Rayleigh quotient to predict the fundamental natural frequency always leads to an   approximate eigenvalue which is higher than the exact one, that is, ω1  ω1 is always greater than or equal to zero. In order to prove this result, let the assumed eigenfunction of the beam have an arbitrary shape that can be represented as a linear combination of the exact eigenfunctions, that is, 1 P ϕ1 ¼ α1 ϕ1 þ α2 ϕ2 þ    ¼ αj ϕj , where αj, j ¼ 1,2,3,. . ., are constants. In this j¼1

case, Rayleigh quotient for the beam can be written as ðl R¼

0

ðl

1 X

!2 αj ϕ00j

2 EI z dx EI z ϕ100 ðxÞdx 0 j¼1 ¼ ð ðl !2 1 l X 2 ρAϕ dx ρA αj ϕj dx 0

0

ð4:175Þ

j¼1

which upon using the orthogonality conditions leads to 1 P

R¼

j¼1 1 P j¼1

1 P

α2j k j ¼ α2j mj

α2j ω2j mj

j¼1 1 P

j¼1

ð4:176Þ α2j mj

Since ω1 < ω2 <   , and the modal mass coefficients are always positive, one has

258

4 1 P

R¼

α2j ω2j mj

j¼1 1 P

j¼1

ω21 

α2j mj

1 P

j¼1 1 P

j¼1

Vibration of Continuous Systems

α2j mj

α2j mj

¼ ω21

ð4:177Þ

This equation indicates that the exact fundamental natural frequency is indeed the minimum of the Rayleigh quotient, and any assumed shape for the fundamental mode will lead to an approximate value for the natural frequency that is higher than the exact one. This fact can be demonstrated by using another example. We consider again the case of a cantilever beam and assume the fundamental mode to be the static deflection of the beam. In this case, the function ϕ1 can be chosen as  3 2 2 ϕ1 ¼ b x4  4lx þ 6l x , where ρA/24EIz. It follows  b is a constantdefined as b ¼ that ϕ01 ¼ b 4x3  12lx2 þ 12l2 x and ϕ001 ¼ 12b x2  2lx þ l2 . Substituting ϕ1 and ϕ001 into the Rayleigh quotient in the case of simple boundary conditions, ffiffiffiffiffiffiffiffiffiffiffiffiffiffi  of Eq. p4.172 =ρA . Comparing this estimated value with one obtains ω1  ω1 ¼ 3:53=l2 EIp   z ffiffiffiffiffiffiffiffiffiffiffiffiffiffi the exact solution, ω1 ¼ 3:5156=l2 EI z =ρA, it is clear that the error in the estimated fundamental natural frequency, using the static deflection as the assumed mode, does not exceed 0.5%. Effect of the Rotary Inertia The Rayleigh quotient can be used to examine the effect of the rotary inertia on the fundamental frequency as well as higher natural frequencies. To this end, we consider the same example as the one discussed in Dym and Shames (1973). We have previously shown that the exact eigenfunctions of a simply supported beam are given by ϕj ¼ sin jπx/l, j ¼ 1, 2,. . . . In this case, one has 9 jπx l > > dx ¼ > > > l 2 > 0 0 >
2 ð l ðl = 2> jπ jπx ð jπ Þ 2 dx ¼ ϕ0j ðxÞdx ¼ cos 2 l l 2l > 0 0 > >
4 ð l ðl > 4 > jπ ðjπ Þ > > 00 2 2 jπx ; dx ¼ 3 > ϕj ðxÞdx ¼ sin l l 2l 0 0 ðl

ϕ2j ðxÞdx ¼

ðl

sin2

ð4:178Þ

Rayleigh quotient that accounts for the effect of the rotary inertia can be written for mode j as ðl

ω2j

EI z ϕ00j dx 0 ¼ ðl ðl 2 2 ρAϕj dx þ ρI z ϕ0j dx 0

2

ð4:179Þ

0

Assuming the case of a uniform beam with constant modulus of elasticity and  using the identities of Eq. 4.178, one obtains ω2j ¼ EI z ðjπ Þ4 =2l3 =ððρAl=2Þþ   ρI z ðjπ Þ2 =2l , which can be written as

4.9 Approximation Methods

259

ω2j ¼ h

EI z ðjπ Þ4 =ρAl4 i   1 þ j2 π 2 =12 ðh=lÞ2

ð4:180Þ

where h is the height of the cross section. The exactpnatural ffiffiffiffiffiffiffiffiffiffiffiffiffiffifrequency ωj, neglecting the effect of the rotary inertia, is ω j ¼ ðjπ=lÞ2 EI z =ρA. The fundamental natural frequency obtained by including the effect of the rotary inertia is given by Eq. 4.180 with j ¼ 1 as ω21 ¼ h

EI z π 4 =ρAl4 1 þ ðπ 2 =12Þðh=lÞ2

i

ð4:181Þ

Clearly, in the case of a long thin beam ((h/l)  1), the effect of the rotary inertia on the fundamental mode of vibration is not significant as compared to the frequencies of higher modes of vibration. Rayleigh–Ritz Method In the Rayleigh method discussed in this section, the fundamental natural frequency of the system was predicted by assuming an approximate shape for the first mode. The Rayleigh–Ritz method can be considered as an extension of the Rayleigh method. It allows, not only to obtain a more accurate value of the fundamental natural frequency, but also to determine the higher frequencies and the associated mode shapes. In the Rayleigh–Ritz method, the shape of deformation of the continuous system is approximated using a series of trial shape functions that must satisfy the geometric boundary conditions of the problem. The shape of deformation of the continuous system can be written as wðxÞ ¼ c1 ϕ1 ðxÞ þ c2 ϕ2 ðxÞ þ    þ cn ϕn ðxÞ ¼

n X

cj ϕj ðxÞ

ð4:182Þ

j¼1

where ϕ1 , ϕ2 , . . . , and ϕn are the trial shape functions which can be the eigenfunctions, a set of assumed mode shapes, or a set of polynomials, and c1, c2, . . ., cn are constant coefficients called the Ritz coefficients. In Eq. 4.182, the functions ϕ1 , ϕ2 , . . . , and ϕn are assumed to be known, while the coefficients c1, c2, . . ., and cn are adjusted by minimizing the Rayleigh quotient with respect to each of these coefficients. This procedure leads to a homogeneous system of n algebraic equations. For this system of equations to have a nontrivial solution, the determinant of the coefficient matrix must be equal to zero. This defines an equation of order n in ω2, and the roots of this equation define the approximate natural frequencies, ω1, ω2, . . ., and ωn. These approximate natural frequencies can be used to determine a set of approximate mode shapes for the system. By using this procedure, the continuous system which has an infinite number of degrees of freedom is represented by a model which has n degrees of freedom. The accuracy of the finite-dimensional model in representing the actual infinite-dimensional model depends on the choice of the trial shape functions.

260

4

Vibration of Continuous Systems

We have shown previously that Rayleigh quotient can be written as ω2 ¼ R ¼ U  =T , where, by using the expression of Eq. 4.182, T and U can be written as 1 T ¼ 2

n X n X

mij ci cj ,

i¼1 j¼1

U ¼

n X n 1X k ij ci cj 2 i¼1 j¼1

ð4:183Þ

where mij and kij are mass and stiffness coefficients that depend on the shape functions. For example, in the case of longitudinal vibration of a rod with simple Ðl boundary conditions, the mass and stiffness coefficients are mij ¼ 0 ρAϕi ϕj dx and Ðl kij ¼ 0 EAϕ0i ϕ0j dx, respectively, whereas in the case of transverse vibration of a beam with simple boundary conditions, the mass and stiffness coefficients are Ðl Ðl mij ¼ 0 ρAϕi ϕj dx and kij ¼ 0 EI z ϕ00i ϕ00j dx, respectively. Note that if the shape functions are chosen such that they satisfy some orthogonality relationships, the mass coefficients mij or the stiffness coefficients kij can be equal to zero if i 6¼ j. By using matrix notation, Eq. 4.183 can be written as 1 T ¼ cT Mc, 2

1 U  ¼ cT Kc 2

ð4:184Þ

where M and K are, respectively, the system mass and stiffness matrices, and c ¼ ½ c1 c2    cn T is the vector of Ritz coefficients. Substituting Eq. 4.184 into the Rayleigh quotient ω2 ¼ R ¼ U  =T , one obtains U  1 cT Kc R ¼  ¼ 12 T T 2 c Mc

ð4:185Þ

In order to minimize Rayleigh quotient with respect to the Ritz coefficients, we differentiate R with respect to these coefficients and set the result of the differentiation equal to zero. This leads to   ∂R T ð∂U  =∂cÞ  U  ∂T =∂c ¼ ¼0  2 ∂c T

ð4:186Þ

  That is, T ð∂U  =∂cÞ  U  ∂T =∂c ¼ 0, which upon dividing by T and using ω2 ¼ R ¼ U  =T , one obtains ∂U  ∂T  ω2 ¼0 ∂c ∂c   Because (∂U/∂c) ¼ cTK and ∂T =∂c ¼ cT M, Eq. 4.187 leads to 

K  ω2 M c ¼ 0

ð4:187Þ

ð4:188Þ

4.9 Approximation Methods

261

This is a system of n algebraic homogeneous equations that has a nontrivial solution if the determinant of the coefficient matrix is equal to zero. That is, jK  ω2Mj ¼ 0. This leads to a nonlinear equation of order n in ω2. The roots of this equation define the system natural frequencies ω21 , ω22 , . . . , ω2n . Associated with each natural frequency, ωj, there is a vector of Ritz coefficients cjhwhich can be i determined to within an arbitrary constant by solving the equation K  ω2j M cj ¼ 0, which can be written more explicitly as 2

k11  ω2j m11





k12  ω2j m12



   6 6 k  ω2 m 2  ω m k 6 21 21 22 22 j j 6 4 ⋮ ⋮    kn1  ω2j mn1 kn2  ωj2 mn2 2 3 0 6 7 607 7 ¼6 6 ⋮ 7, j ¼ 1,2,3, . . . , n 4 5 0

  ⋱ 

  32 3 c1 k 1n  ω2j m1n   76 7 6c 7 k 2n  ω2j m2n 7 76 2 7 76 ⋮ 7 ⋮   54 5 k nn  ω2j mnn cn j

ð4:189Þ

The approximate mode shape associated with the frequency ωj can be obtained by using Eq. 4.182 as w j ðxÞ ¼ c1 j ϕ1 ðxÞ þ c2 j ϕ2 ðxÞ þ    þ cnj ϕn ðxÞ n X ¼ ckj ϕk ðxÞ, j ¼ 1,2,3,. . . , n

ð4:190Þ

k¼1

Once the natural frequencies and mode shapes are determined, the continuous system which has an infinite number of degrees of freedom can be represented by an equivalent multi-degree of freedom system. In this case, the vibration of the continuousP system can be described using the separation of variables as wðx; t Þ ¼ 1 j¼1 w j ðxÞq j ðt Þ, where qj(t), j ¼ 1,2,3,. . ., n are the time-dependent coefficients. Illustrative Example It can be shown that if the assumed shape functions happen to be the exact eigenfunctions, the Rayleigh–Ritz method yields the exact natural frequencies and the exact mode shapes. In order to demonstrate this fact, consider the case of a simply supported beam with a uniform cross section and constant modulus of elasticity. We choose the shape functions to be the exact eigenfunctions and consider the case in which the number of Ritz coefficients is equal to three, that is, ϕj ðxÞ ¼ sin ðjπx=lÞ, j ¼ 1,2,3. The mass coefficients mij and the stiffness coefficients kij are

262

4

ðl

Vibration of Continuous Systems



if i ¼ j if i ¼ 6 j

m=2 0 0 ( ðl ðiπ Þ4 EI z =2l3 kij ¼ EI z ϕ00i ϕ00j dx ¼ 0 0

mij ¼

ρAϕi ϕj dx ¼

if i ¼ j if i ¼ 6 j

where m ¼ ρAl is the total mass of the beam. The matrices M and K are given by 2 1 m4 0 M¼ 2 0

0 1 0

3 0 0 5, 1

2 1 0 EI z π 4 4 K¼ 0 16 3 2l 0 0

3 0 0 5 81

Substituting the matrices M and K into jK  βMj ¼ 0, one obtains

1β

0

0

0 16  β 0

0

0

¼ 0 81  β

where β ¼ ω2(ml3/EIzπ 4). The preceding equation leads to ð1  βÞð16  βÞð81  βÞ ¼ 0 It is clear that the roots of this equation are βj ¼ j4, which define the three fundamental natural frequencies of the beam as rffiffiffiffiffiffiffi EI z ω j ¼ ðjπ Þ , j ¼ 1,2,3 ml3 2

These are the same as the exact natural frequencies of the simply supported beam. The Ritz coefficients associated with each natural frequency can be obtained by using Eq. 4.189 which can be written for this system as 2

1  βj 4 0 0

0 16  β j 0

32 3 2 3 0 c1 j 0 0 54 c2 j 5 ¼ 4 0 5 81  β j c3 j 0

The solution of this equation for j ¼ 1,2,3 is 2

3 c11 c1 ¼ 4 0 5, 0 Therefore, the first mode shape is

2

3 0 c2 ¼ 4 c22 5, 0

2

3 0 c3 ¼ 4 0 5 c33

4.9 Approximation Methods

263

w1 ¼ c11 ϕ1 ðxÞ þ c21 ϕ2 ðxÞ þ c31 ϕ3 ðxÞ πx ¼ c11 ϕ1 ðxÞ ¼ c11 sin l Similarly, the second and third mode shapes are, respectively, w2 ðxÞ ¼ c22 sin

2πx , l

w3 ðxÞ ¼ c33 sin

3πx l

which are the exact eigenfunctions. Example 4.12

Consider the case of a cantilever beam which has a uniform cross-sectional area and a constant modulus of elasticity. Assume that the shape of deformation of the beam is in the form wðxÞ ¼ c1 ϕ1 ðxÞ þ c2 ϕ2 ðxÞ where ϕ1 ðxÞ ¼ ξ2 and ϕ2 ðxÞ ¼ ξ3 , where ξ ¼ x/l and l is the length of the beam. The assumed shape functions ϕ1 ðxÞ and ϕ2 ðxÞ satisfy the geometric boundary conditions, that is, the deflection and slope are equal to zero at the fixed end (x ¼ 0). Using the assumed shape functions ϕ1 ðxÞ and ϕ2 ðxÞ, the mass coefficients are m11 ¼

ðl 0

ρAϕ12 ðxÞdx ¼

m12 ¼ m21 ¼

ðl

ðl

ðl

ρAξ4 dx ¼

0

ρAϕ1 ðxÞϕ2 ðxÞdx ¼

0

ð1

ð1

mξ4 dξ ¼

0

ð1

mξ5 dξ ¼

0

m 5

m 6

m m22 ¼ ρAϕ22 ðxÞdx ¼ mξ6 dξ ¼ 7 0 0 where m is the total mass of the beam. Similarly, the stiffness coefficients are k11 ¼

ðl 0

#2 2 1 ∂ ϕ1 ðxÞ ¼ EI z 2 dx l ∂ξ2 0 ðl

"

 2 2 4EI z EI z l 2 dξ ¼ 3 l l 0 " # ðl ðl 2 2 1 ∂ ϕ1 ðxÞ ∂ ϕ2 ðxÞ 00 00 ¼ k21 ¼ EI z ϕ1 ðxÞϕ2 ðxÞdx ¼ EI z 4 dx l ∂ξ2 ∂ξ2 0 0   ð1 12ξ 6EI z ¼ EI z l 4 dξ ¼ 3 l l 0 ¼

k12

2 EI z ϕ100 ðxÞdx

ð1

(continued)

264

4

k22 ¼

ðl 0

2 EI z ϕ002 ðxÞdx

# 2 1 ∂ ϕ2 ð xÞ ¼ EI z 2 dx l ∂ξ2 0 ðl

"

2 6ξ2 12EI z ¼ EI z l 2 dξ ¼ 3 l l 0 ð1



Vibration of Continuous Systems

Therefore, the matrices M and K can be defined as 2

1 65 M ¼ m6 41 6

3 1 67 7, 15 7

K¼

 EI z 4 l3 6

6 12



The characteristic equation is

4  β=5

6  β=6

6  β=6

¼0 12  β=7

where β ¼ ω2(ml3/EIz). One, therefore, has the frequency equation


 β β β 2 4 ¼0 12   6 5 7 6 which yields the quadratic equation β2  1224β þ 15120 ¼ 0 This equation has the roots β1 ¼ 12:4802,

β2 ¼ 1211:51981

From which the first two fundamental frequencies can be determined as rffiffiffiffiffiffiffi EI z ω2 ¼ 34:8069 ml3 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi The exact fundamental natural frequency is 3:5156 EI z =ml3 . The error in the frequency estimated using the Rayleigh–Ritz method is less than 0.5%. The exact value of the ffi second natural frequency of the cantilever beam is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffi EI z , ω1 ¼ 3:53273 ml3

22:0336 EI z =ml3 . Therefore, the error in the second natural frequency is about 58%. A significant improvement in the second natural frequency can be obtained by increasing the number of shape functions.

4.10

4.10

Galerkin’s Method

265

Galerkin’s Method

Galerkin’s method is a technique used for obtaining an approximate solution for both linear and nonlinear partial differential equations. In order to apply Galerkin’s method, an approximate solution for the displacement field is first assumed. Since this approximate solution is not, in general, the same as the exact solution, substitution of the assumed solution into the partial differential equations and the boundary conditions leads to some error called a residual. In Galerkin’s method, the criterion for selecting the assumed solution is to make the error small. In order to demonstrate the use of Galerkin’s method, we write the partial differential equations of the systems discussed in this chapter in the following general form: Dðwðx; t ÞÞ  F ðx; t Þ ¼ 0

ð4:191Þ

where F(x, t) is an arbitrary forcing function and D(w(x, t)) is a function that depends on the type of the problem considered. For example, in the case of longitudinal vibration of rods, the function D(w(x, t)) is given by 2

Dðwðx; t ÞÞ ¼

2

∂ u ∂ u  c2 2 2 ∂t ∂x

ð4:192Þ

where in this case w(x, t) ¼ u(x, t) is the longitudinal displacement of the rod and D is the differential operator defined as 2

D¼

2

∂ ∂  c2 2 ∂t 2 ∂x

ð4:193Þ

pffiffiffiffiffiffiffiffi and c ¼ E=ρ. Similar definitions can be made in the case of the torsional oscillation of shafts. In the case of transverse vibrations of beams, the function D(w(x, t)) can be recognized as 2

Dðwðx; t ÞÞ ¼

4

∂ v ∂ v þ c2 4 2 ∂t ∂x

where w(x, t) ¼ v(x, t) is the transverse displacement of the beam, c ¼ D is a differential operator that can be defined as 2

D¼

ð4:194Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi EI z =ρA, and

4

∂ ∂ þ c2 4 ∂t 2 ∂x

ð4:195Þ

In Galerkin’s method, the unknown exact solution w(x, t) is approximated by the ðx; t Þ which can be written as function w ðx; t Þ ¼ w

n X j¼1

ϕj ðxÞqj ðt Þ

ð4:196Þ

266

4

Vibration of Continuous Systems

where ϕj ðxÞ are assumed shape functions and qj(t) are unknown coordinates that depend on time. The assumed functions ϕj ðxÞ are chosen to satisfy the boundary conditions of the problem. Due to the fact that the assumed approximate solution is not the same as the unknown exact solution, Eq. 4.196 will not satisfy the partial differential equation of Eq. 4.191, that is, Dðwðx; t ÞÞ  F ðx; t Þ 6¼ 0. Therefore, one has ðx; t ÞÞ  F ðx; t Þ ¼ Re Dðw

ð4:197Þ

where Re is the residual or the error that results from the use of the approximate solution. By taking a virtual change P n in the assumed approximate solution ðx; t Þ ¼ j¼1 of Eq. 4.196, one obtains δw ϕj ðxÞδqj ðt Þ. Multiplying both sides of Eq. 4.197 by the virtual change and integrating over the solution domain yields n ð l X

ðx; t ÞÞ  F ðx; t Þ  Re ϕj ðxÞdx δqj ðt Þ ¼ 0 ½Dðw

ð4:198Þ

0

j¼1

Since the coordinates qj(t) are assumed to be linearly independent, the preceding equation yields ðl

½Dðwðx; t ÞÞ  F ðx; t Þ ϕj ðxÞdx ¼

0

ðl

Re ϕj ðxÞdx,

j ¼ 1,2,. . . , n

ð4:199Þ

0

In Galerkin’s method, the assumed shape functions ϕj ðxÞ are chosen such that ðl

Re ϕj ðxÞ dx  0,

j ¼ 1,2,. . . , n

ð4:200Þ

0

Consequently, Eq. 4.199 yields n differential equations which can be written as ðl

ðx; t ÞÞ  F ðx; t Þ ϕj ðxÞdx ¼ 0, ½Dðw

j ¼ 1,2,. . . , n

ð4:201Þ

0

It is clear that in Galerkin’s method the assumed shape functions ϕj ðxÞ can be considered as weighting functions selected such that the error Re over the solution domain is small. Consequently, Galerkin’s method can be considered as a special case of the weighted residual techniques. In the weighted residual techniques there are broad choices of the weighting functions or error distribution principles, whereas in Galerkin’s method the weighting functions are selected to satisfy the criterion of Eq. 4.200. Example 4.13

In order to demonstrate the use of Galerkin’s method, we consider the longitudinal forced vibration of a prismatic rod. The governing partial differential equation of the system is (continued)

4.10

Galerkin’s Method

267

2

ρA

2

∂ u ∂ u  EA 2 ¼ F ðx; t Þ ∂t 2 ∂x

We assume a solution in the form uðx; t Þ ¼ ϕ1 ðxÞq1 ðt Þ þ ϕ2 ðxÞq2 ðt Þ where ϕ1 ðxÞ ¼ 1  ξ,

ϕ2 ð xÞ ¼ ξ

in which ξ ¼ x/l. It follows that δ uðx; t Þ ¼ ϕ1 δq1 þ ϕ2 δq2 Multiplying both sides of the partial differential equation by δ u and integrating over the volume, we obtain ðl " 0

# 2 2 ∂ u ∂ u ρA 2  EA 2  F ðx; t Þ δ u dx ¼ 0 ∂t ∂x

The first term in this equation represents the virtual work of the inertia forces and can be written as ðl

2

∂ u ρA 2 δ u dx ¼ ∂t 0

ðl

  ρA ϕ1 €q1 þ ϕ2 €q2 ϕ1 δq1 þ ϕ2 δq2 dx

0

which can be written using matrix notation as ðl 0

where q ¼ ½ q1

2

ρA

∂ u €T Mδq δ u dx ¼ q ∂t 2

T

q2 , and M is the mass matrix defined as  2 ϕ1 M ¼ ρA ϕ1 ϕ2 0 ðl

ϕ1 ϕ2 ϕ22



2

1 63 dx ¼ m6 41 6

3 1 67 7 15 3

where m ¼ ρAl is the mass of the rod. The virtual work of the elastic forces can be integrated by parts to yield

l ð l 2 ∂ u ∂ u

∂ u∂  u þ EA δ u dx  EA 2 δ u dx ¼ EA δ ∂x 0 ∂x ∂x ∂x 0 0 ðl

(continued)

268

4

Vibration of Continuous Systems

While this equation automatically accounts for any set of boundary conditions, in this example, for simplicity, we consider simple boundary conditions. Consequently

l ∂ u

u ¼0 EA δ ∂x 0 One, therefore, has 

ðl

2

EA 0

∂ u δ u dx ¼ ∂x2 ¼

ðl EA 0

ðl

0

∂ u ∂  δ u dx ∂x ∂x

  EA ϕ10 q1 þ ϕ02 q2 ϕ10 δq1 þ ϕ02 δq2 dx

¼ qT Kδq where the stiffness matrix K is defined as K¼

"

ðl EA 0

ϕ10 2

ϕ10 ϕ02

ϕ10 ϕ02

ϕ02 2

#

" # EA 1 1 dx ¼ l 1 1

The virtual work of the external forces can be written as ðl

F ðx; t Þδ u dx ¼

0

ðl

 F ðx; t Þ ϕ1 δq1 þ ϕ2 δq2 dx

0

¼ Q1 δq1 þ Q2 δq2 in which Q1 ¼

ðl

F ðx; t Þ ϕ1 ðxÞdx,

0

Q2 ¼

ðl

F ðx; t Þ ϕ2 ðxÞdx

0

One can then write the virtual work of the external forces as ðl

F ðx; t Þδ u dx ¼ QT δq

0

where Q ¼ ½ Q1

Q2 T . It follows that  T € M þ qT K  QT δq ¼ 0 q (continued)

4.11

Assumed-Modes Method

269

Since the components of the vector q are assumed to be linearly independent, the preceding equation leads to M€ q þ Kq ¼ Q, which can be written more explicitly as 2

1 63 m6 41 6

4.11

3 1 " # " #" # " # €q1 1 1 q1 Q1 EA 67 7 þ ¼ 5 1 l 1 1 €q2 q2 Q2 3

Assumed-Modes Method

The assumed-modes method is closely related to the Rayleigh–Ritz and Galerkin methods discussed in the preceding sections. In fact, it leads to the same dynamic formulation as Galerkin’s method if the same assumed displacement field is used. In the assumed-modes method, the shape of deformation of the continuous system is approximated using a set of assumed shape functions. This approach can be used in the dynamic analysis of structures with complex geometrical shapes and complex boundary conditions. For such systems, it is difficult to determine the exact eigenfunctions. If, however, the deflected shape of the structures resembles the shape of deformation of some of the simple systems as the ones discussed in this chapter, the eigenfunctions of the simple systems can be assumed as the shape functions of the more complex structure. Another alternative is to use experimental testing to determine the shape of deformation by measuring the displacements at selected nodal points. The shape of deformation between these selected nodal points can be defined using polynomial functions. As was shown in this chapter, the vibration of continuous systems can be represented in terms of an infinite number of vibration modes. Each of these modes can be excited independently by the use of appropriate initial conditions or forcing functions. Therefore, the determination of the shape of the fundamental modes of vibration by experimental measurements is possible. In fact, the use of modal testing to determine the mode shapes and natural frequencies of continuous systems is an integral part in the design of many mechanical systems. In addition to the mode shapes and natural frequencies, modal testing also provides the modal stiffness, mass, and damping coefficients associated with these modes. By using these coefficients, a finite-dimensional model can be developed for the continuous systems. If the shape functions that define the deflected shape of the continuous system can be obtained experimentally or by inspection, an approximation for the P ndisplacement field can be assumed in terms of these shape functions as wðx; t Þ  j¼1 ϕj ðxÞqj ðt Þ, where w(x, t) ¼ u(x, t) in the case of longitudinal vibration, w(x, t) ¼ θ(x, t) in the case of torsional vibration, and w(x, t) ¼ v(x, t) in the case of transverse vibration, ϕj ðxÞ may be the jth eigenfunction or an assumed polynomial, qj is a time-dependent coordinate, and n is the number of elastic degrees of freedom used in the assumed

270

4

Vibration of Continuous Systems

solution. By using a finite number of terms in the solution, the kinetic energy and strain energy can be written as T¼

n X n 1X m jk q_ j q_k , 2 j¼1 k¼1

U¼

n X n 1X k jk qj qk 2 j¼1 k¼1

ð4:202Þ

The equations of vibration can be developed using these energy expressions and Lagrange’s equation. This procedure leads to the following n ordinary differential equations of motion: n  X

 m jk €qk þ k jk qk ¼ Q j , j ¼ 1,2,3,. . . , n

ð4:203Þ

k¼1

These equations can be written compactly in a matrix form as M€ q þ Kq ¼ Q, where M and K are, respectively, the n n mass and stiffness matrices, q ¼ ½ q1 q2 . . . qn T is the vector of generalized coordinates, and Q ¼ ½ Q1 Q2 . . . Qn T is the vector of generalized forces. The preceding matrix equation is in a form similar to the matrix equation of the multi-degree of freedom systems, and therefore, the techniques discussed in the preceding chapter can be applied to obtain a solution for this matrix equation, which represents an equivalent finitedimensional model for the infinite degrees of freedom continuous system. An equivalent single degree of freedom model can be obtained if n is equal to 1. The use of this coordinate reduction procedure is demonstrated by the following example. Example 4.14

If the transverse vibration of a beam with one end fixed and the other free is approximated by vðx; t Þ ¼ ϕ1 ðxÞq1 ðt Þ þ ϕ2 ðxÞq2 ðt Þ where ϕ1 ðxÞ ¼ 3ξ2  2ξ3 ,

  ϕ 2 ð xÞ ¼ l ξ 3  ξ 2

in which l is the length of the beam and ξ ¼ x/l, obtain the matrix equation of motion of this system. Solution. The displacement and velocity of the beam can be expressed as 

vðx; t Þ ¼ ϕ1 ðxÞ

ϕ2 ð xÞ

" # q1 ðt Þ

 v_ ðx; t Þ ¼ ϕ1 ðxÞ ϕ2 ðxÞ

q2 ðt Þ " # q_1 ðt Þ q_2 ðt Þ (continued)

4.11

Assumed-Modes Method

271

Also 00

v ðx; t Þ ¼



ϕ001 ðxÞ

ϕ002 ðxÞ



6 ¼ 2 ð1  2ξÞ l

" # q1 ð t Þ q2 ð t Þ

2 ð3ξ  1Þ l

"

q1 ð t Þ

#

q2 ð t Þ

Therefore, the kinetic energy and strain energy are given by T¼

1 2

ðl

1 ρAv_ 2 dx ¼ q_ T Mq_ , 2 0

U¼

1 2

ðl

1 2 EIv00 dx ¼ qT Kq 2 0

where q ¼ [q1 q2]T and M and K are the mass and stiffness matrices defined as M¼

ðl

" ρA

0

# ϕ1  ϕ2

"

ϕ1

ðl

11l=210

13=35

¼ ρAl

ϕ2 dx ¼

" ρA

0

#

ϕ12

ϕ1 ϕ2

ϕ2 ϕ1

ϕ22

11l=210 l2 =105 ð l " 00 2 ð l " 00 # ϕ1 ϕ1  00 00 K ¼ EI 00 ϕ1 ϕ2 dx ¼ EI ϕ2 0 0 ϕ002 ϕ001 " # EI 12=l2 6=l ¼ l 6=l 4

ϕ100 ϕ002

# dx

#

ϕ200 2

dx

The virtual work of the external force is δW ¼

ðl

F ðx; t Þδv dx ¼

0

ðl 0

 F ðx; t Þ ϕ1

ϕ2

  δq1 dx ¼ ½ Q1 δq2

 Q2

δq1 δq2



where Q1 ¼

ðl

F ðx; t Þ ϕ1 dx,

0

Q2 ¼

ðl

F ðx; t Þ ϕ2 dx

0

Therefore, the matrix equation for the equivalent two degree of freedom system can be written as " ρAl

13=35

11l=210

11l=210

l2 =105

#"

€q1 €q2

#

" EI 12=l2 þ l 6=l

6=l 4

#"

q1 q2

#

" ¼

Q1 Q2

#

272

4

Vibration of Continuous Systems

Problems 4.1. Determine the equation of motion, boundary conditions, and frequency equation of the longitudinal vibration of the system shown in Fig. P4.1, assuming that the rod has a uniform cross-sectional area.

Fig. P4.1

4.2. The system shown in Fig. P4.2 consists of a uniform rod with a spring attached to its end. Derive the partial differential equation of motion and determine the boundary conditions and frequency equation of the longitudinal vibration.

Fig. P4.2

4.3. Obtain the equation of motion, boundary conditions, and frequency equation of the system shown in Fig. P4.3.

Problems

273

Fig. P4.3

4.4. Determine the equation of motion, boundary conditions, and frequency equation of longitudinal vibration of a uniform rod with a mass m attached to each end. Check the fundamental frequency by reducing the uniform rod to a spring with end masses. 4.5. Derive the equation of motion of longitudinal vibration of the system shown in Fig. P4.4. Obtain the boundary conditions and the frequency equation for the case in which k1 ¼ k2 ¼ k.

Fig. P4.4

4.6. Determine the equation of motion, boundary conditions, and natural frequencies of a torsional system which consists of a uniform shaft of mass moment of inertia I and a disk having mass moment of inertia I1 attached to each end of the shaft. 4.7. Determine the equation of torsional oscillation of a uniform shaft with one end fixed and the other end attached to a disk with inertia I1. Obtain the boundary conditions and the frequency equations. 4.8. Obtain the equation of torsional oscillation of a uniform shaft with a torsional spring of stiffness k attached to each end.

274

4

Vibration of Continuous Systems

4.9. Derive the partial differential equation of the torsional oscillations of a uniform shaft clamped at the middle and free at the two ends. The shaft has length l, modulus of rigidity G, mass density ρ, and cross-sectional area A. 4.10. For the system shown in Fig. P4.5, obtain the partial differential equation of the longitudinal vibration, the boundary conditions, and the frequency equation.

Fig. P4.5

Check the results by letting the two rods have the same dimensions and by comparing with the results obtained by solving problem 1. 4.11. Determine the equation of transverse vibration, boundary conditions, and frequency equation of a uniform beam of length l clamped at one end and pinned at the other end. 4.12. Obtain the frequency equation of the transverse vibration of the beam shown in Fig. P4.6.

Fig. P4.6

4.13. A uniform rod which is fixed at one end and free at the other end has the initial conditions u(x, 0) ¼ A0 sin (πx/2l) and u_ ðx; 0Þ ¼ 0, where A0 is a constant. Obtain the general solution of the longitudinal vibration. 4.14. Repeat Problem 13 for the initial conditions u(x, 0) ¼ 0 and u_ ðx; 0Þ ¼ V 0 sin ðπx=2lÞ, where V0 is a constant. 4.15. For the system of Problem 13, examine the validity of using the initial conditions u(x, 0) ¼ A0 cos (πx/2l) and u_ ðx; 0Þ ¼ 0, where A0 is a constant. What is the expected solution? Comment on the results.

Problems

275

4.16. A uniform shaft which is fixed at one end and free at the other end has the initial conditions θ(x, 0) ¼ A0 sin (nπx/2l) and θ_ ðx; 0Þ ¼ 0, where n is a fixed odd number and A0 is a constant. Obtain the solution of the free torsional vibration of this system. 4.17. Repeat Problem 13 in the case where the initial conditions are given by u(x, 0) ¼ A0x and u_ ðx; 0Þ ¼ 0, where A0 is a constant. 4.18. Repeat Problem 16 by considering the initial conditions θ(x, 0) ¼ A0(x + x2) and θ_ ðx; 0Þ ¼ 0, where A0 is a constant. 4.19. For a uniform rod with both ends free, obtain the solution for the longitudinal vibration if the initial conditions are given by u(x, 0) ¼ A0 cos(nπx/l ) and u_ ðx; 0Þ ¼ 0, where n is an integer and A0 is a constant. 4.20. Repeat Problem 19 using the initial conditions u(x, 0) ¼ A0x2 and u_ ðx; 0Þ ¼ 0. Comment on the validity of using these initial conditions, where A0 is a constant. 4.21. A uniform rod with one end fixed and the other end free is subjected to a distributed axial load in the form F(x, t) ¼ x sin 5t. Determine the response of the system as the result of application of this forcing function. 4.22. Determine the response of the system shown in Fig. P4.7 to the concentrated harmonic forcing function F(t).

Fig. P4.7

4.23. The system shown in Fig. P4.8 is subject to a concentrated axial force at the mass attached to the end of the rod. Determine the response of this system to this end excitation.

Fig. P4.8

276

4

Vibration of Continuous Systems

4.24. Determine the response of the system shown in Fig. P4.9 to the harmonic motion of the support.

Fig. P4.9

4.25. Determine the response of the torsional system shown in Fig. P4.10 to the harmonic torque excitation.

Fig. P4.10

4.26. Repeat Problem 22 if the forcing function F(t) is the periodic function shown in Fig. P4.11.

Fig. P4.11

Problems

277

4.27. Determine the response of the beam shown in Fig. P4.12 to the forcing function F(t) ¼ F0 sin ωft, where F0 and ωf are constants.

Fig. P4.12

4.28. Repeat Problem 27 if F(t) is given by the periodic function defined in Problem 26. 4.29. The system shown in Fig. P4.13 consists of a uniform beam with one end pinned and the other end free. Determine the response of this system to the harmonic support excitation.

Fig. P4.13

4.30. Repeat Problem 22 if the force F(t) is given by the rectangular pulse shown in Fig. P4.14.

Fig. P4.14

278

4

Vibration of Continuous Systems

4.31. If the longitudinal displacement of the system of Problem 22 is represented by u (x, t) ¼ ϕ1(x)q1 + ϕ2(x)q2, where ϕ1(x) ¼ sin(πx/2l) and ϕ2(x) ¼ sin(3πx/2l), and q1 and q2 are time-dependent coordinates, obtain the matrix equation that governs the forced vibration of this system. 4.32. Repeat Problem 31 if ϕ1(x) ¼ 0 and ϕ2(x) ¼ x. 4.33. Obtain the solution of the longitudinal vibration of a uniform rod with one end fixed and the other end with a condition u(l, t) ¼ d cos ωt, where d and ω are constants. Assume zero initial conditions. 4.34. By using the Rayleigh–Ritz method, determine the natural frequencies and mode shapes of a beam with free ends. Assume that the modulus of elasticity and the cross-sectional area of the beam are constant. Use the following assumed function vðxÞ ¼ c1 þ c2 x þ c3 x2 þ c4 x3 Compare the results obtained using the Rayleigh–Ritz method and the exact solution. 4.35. Determine the natural frequencies and mode shapes of the system shown in Fig. P4.2 using the Rayleigh–Ritz method. Assume that the trial functions are the exact eigenfunctions of a rod with one end fixed and the other end free. Consider the two cases in which n ¼ 2,3. Plot the approximate mode shapes and compare the results with the exact ones. 4.36. Repeat Problem 35 assuming that the trial function is uðxÞ ¼ Comment on the results.

Pn j¼1

c jx j.

4.37. Using Galerkin’s method, obtain a three degree of freedom mathematical model that represents the system of Problem 34. Use the mode shapes obtained in Problem 34 as the assumed shape functions. 4.38. Use Galerkin’s method to obtain a two and three degree of freedom mathematical model for the system of Problem 35. Use the mode shapes obtained in Problem 35 as the assumed shape functions.

5

The Finite Element Method

The approximate methods presented at the end of the preceding chapter for the solution of the vibration problems of continuous systems are based on the assumption that the shape of the deformation of the continuous system can be described by a set of assumed functions. By using this approach, the vibration of the continuous system which has an infinite number of degrees of freedom is described by a finite number of ordinary differential equations. This approach, however, can be used in the case of structural elements with simple geometrical shapes such as rods, beams, and plates. In large-scale systems with complex geometrical shapes, difficulties may be encountered in defining the assumed shape functions. In order to overcome these problems, the finite element (FE) method has been widely used in the dynamic analysis of large-scale structural systems. The FE method is a numerical approach that can be used to obtain approximate solutions to a large class of engineering problems. In particular, the FE method is well suited for problems with complex geometries. In the FE method, the structure is discretized to small regions called elements which are rigidly interconnected at selected nodal points. The deformation within each element can then be described by interpolating polynomials. The coefficients of these polynomials are defined in terms of physical coordinates called the element nodal coordinates that describe the displacements and slopes of selected nodal points on the element. Therefore, the displacement of the element can be expressed using the separation of variables as the product of space-dependent functions and time-dependent nodal coordinates. By using the connectivity between elements, the assumed displacement field can be written in terms of the element shape function and the nodal coordinates of the structure. Using the assumed displacement field, the kinetic and strain energies of each element can be developed, thus defining the FE mass and stiffness matrices. The energy expressions of the structure can be obtained by summing the energy expressions of its elements. This leads to the definition of the structure mass and stiffness matrices.

# Springer Nature Switzerland AG 2019 A. Shabana, Vibration of Discrete and Continuous Systems, Mechanical Engineering Series, https://doi.org/10.1007/978-3-030-04348-3_5

279

280

5

The Finite Element Method

Fig. 5.1 Finite element discretization

Fig. 5.2 Finite elements: (a) truss element; (b) triangular element; (c) beam element; (d) rectangular element

Figure 5.1 shows a continuous system or a structure which is divided to a finite number of elements which collectively approximate the physical description of the system. These elements are assumed to be rigidly connected at selected nodal points whose displacements and slopes are the elastic coordinates of the system. Figure 5.2 shows some typical finite elements which are commonly used in the dynamic analysis of the structural systems. Different numbers of nodes can be used for different types of elements. The nodal coordinates of these elements are also shown in Fig. 5.2.

5.1 Assumed Displacement Field

281

In this chapter, we discuss the formulation of the mass and stiffness matrices of the continuous systems using the FE method. In Sects. 5.1, 5.2, and 5.3, the general procedure for defining the assumed displacement field in terms of the element shape functions and the nodal coordinates of the continuous system is described. In Sect. 5.4, the formulation of the mass matrix is developed using the kinetic energy of the elements. In Sect. 5.5, the element strain energy is used to define the stiffness matrix of the structure. The formulation of the structure equations of motion is presented in Sect. 5.6, while the convergence of the FE solution is examined in Sect. 5.7. Sections 5.8 and 5.9 are devoted to the analysis of higher-order and spatial elements, respectively. The large rotation and large deformation problem is explained in Sect. 5.10, which provides a brief introduction to the FE absolute nodal coordinate formulation (ANCF). In Sect. 5.11, sub-structuring techniques used to obtain reduced-order models are discussed. Examples are presented throughout the chapter in order to demonstrate the use of the formulations presented for different types of elements.

5.1

Assumed Displacement Field

Each finite element represents a continuous system which has an infinite number of degrees of freedom. However, by choosing the size of the element to be small, the deformation of the element can be approximated by using relatively low-order polynomials. The coefficients of these polynomials in the dynamic case are time dependent. Therefore, in a selected coordinate system of element j, the displacement field is defined as u j ¼ S1j a j ,

j ¼ 1,2,. . . ,ne

ð5:1Þ

where S1j is a space-dependent matrix, a j is the vector of the time-dependent coefficients of the polynomials, and ne is the total number of elements used to discretize the continuous system. The displacement u j can be a scalar, two-dimensional or three-dimensional vector depending on the type of element used. The time-dependent coefficients in Eq. 5.1 lack an obvious physical meaning. In the FE method, these coefficients are expressed in terms of the time-dependent element nodal coordinates. This relationship can be written as a j ¼ S2j q j ,

j ¼ 1,2,. . . ,ne

ð5:2Þ

where S2j is a constant matrix which can be obtained by defining the displacements and slopes at the nodal points, and q j is the vector of the element nodal coordinates. By substituting Eq. 5.2 into Eq. 5.1, the displacement field of the element can be expressed in terms of the element nodal coordinates as u j ¼ S1j S2j q j , j ¼ 1,2,. . . ,ne , which can be written as

282

5

The Finite Element Method

u j ¼ S j q j , j ¼ 1,2,. . . ,ne

ð5:3Þ

where S j is the space-dependent element shape function matrix defined as S j ¼ S1j S2j , j ¼ 1,2,. . . ,ne . The procedure described in this section for writing the displacement field in terms of the space-dependent element shape function and the time-dependent nodal coordinates is a general procedure which can be used for many elements. The use of this procedure is demonstrated by the following examples. Truss Element The truss element shown in Fig. 5.2a is assumed to carry only tension or compression loads. Therefore, only axial displacement is allowed. The displacement field of this element is expressed using a first-order polynomial as u j ¼ a1j þ a2j x

ð5:4Þ

where a1j and a2j are time-dependent coefficients. Equation 5.4 can be written in the form of Eq. 5.1 as " u j ¼ ½ 1 x

a1j

# ð5:5Þ

a2j

where the space-dependent matrix S1j and the time-dependent vector a j are  T recognized as S1j ¼ ½ 1 x  and a j ¼ a1j a2j . As shown in Fig. 5.2(a), this element has two nodal points. Each nodal point has one degree of freedom which represents the axial displacement of this node. One can, therefore, write the following conditions:   u j x ¼ l j ¼ u2j

u j ðx ¼ 0Þ ¼ u1j ,

ð5:6Þ

where u1j and u2j are the element nodal coordinates and l j is the length of the element. Substituting the conditions of Eq. 5.6 into Eq. 5.5 yields " u j ð x ¼ 0Þ ¼ ½ 1

0

a1j

#

a2j

¼

u1j ,

" #     a1j j j u x¼l ¼ 1 l ¼ u2j a2j j

9 > > > > > = > > > > > ;

ð5:7Þ

These two equations can be written in one matrix equation as "

1

0

1

lj

#"

a1j a2j

#

" ¼

u1j u2j

# ð5:8Þ

The solution of this equation defines the coefficients a1j and a2j in terms of the nodal coordinates u1j and u2j as

5.1 Assumed Displacement Field

"

a1j

283

#

a2j

" 1 lj ¼ j l 1

0

#"

u1j

# ð5:9Þ

u2j

1

where S2j of Eq. 5.2 can be recognized as S2j

  1 lj 0 ¼ j l 1 1

ð5:10Þ

The element shape function matrix S j of Eq. 5.3 can then be defined as 2

1 S j ¼ S1j S2j ¼ ½ 1 x 4 1 lj ¼ ½ ð1  ξ Þ ξ 

3   0 x x 5 1 j j 1 ¼ l l j l

ð5:11Þ

where ξ is the dimensionless parameter ξ ¼ x/l j. One can then write the displacement of the element in terms of the element nodal coordinates as " u ¼ S q ¼ ½ð1  ξÞ ξ j

j

j

u1j

# ð5:12Þ

u2j

This equation can also be written as u j ¼ ϕ1j ðξÞu1j þ ϕ2j ðξÞu2j

ð5:13Þ

where ϕ1j ðξÞ ¼ 1  ξ and ϕ2j ðξÞ ¼ ξ. Note that when x ¼ 0, that is, ξ ¼ 0, ϕ1j ð0Þ ¼ 1, and ϕ2j ð0Þ ¼ 0, while when x ¼ l j, that is, ξ ¼ 1, ϕ1j ð1Þ ¼ 0 and ϕ2j ð1Þ ¼ 1. That is, the element of the shape function ϕi j takes the value one at node i and the value zero at the other node. Triangular Element Several FE formulations for the two-dimensional triangular elements can be found in the literature. In this section, we consider the simplest one which is called the constant-strain triangular element. The element is shown in Fig. 5.2b and has three nodal coordinates. Each node has two degrees of freedom which represent the horizontal and vertical displacements of the node as shown in Fig. 5.2b. Therefore, the element has six nodal coordinates given by h q j ¼ u1j

v1j

u2j

v2j

u3j

v3j

iT

ð5:14Þ

The displacement field is approximated by the following polynomials: u j ¼ a1j þ a2j x þ a3j y v j ¼ a4j þ a5j x þ a6j y

) ð5:15Þ

284

5

The Finite Element Method

This element in which the displacement field is defined by the preceding equation is called constant strain since 9 ∂v j j> > ¼ a6 > ¼ = ∂y > ∂u j ∂v j > > þ ¼ a3j þ a5j γ xyj ¼ ; ∂y ∂x

εxj

∂u j ¼ a2j , ¼ ∂x

εyj

ð5:16Þ

That is, the normal strains εxj and εyj and the shear strain γ xyj are the same at every point on the element. Let (x1, y1), (x2, y2), and (x3, y3) be the coordinates of the nodal points of the element. It is clear that )

ukj ¼ a1j þ a2j xk þ a3j yk

ð5:17Þ

vkj ¼ a4j þ a5j xk þ a6j yk , k ¼ 1,2,3

where ukj and vkj are the nodal coordinates of the jth node. By using a similar procedure to the one described in the case of the truss element, Eq. 5.17 can be used to write the coefficients a1j , a2j ,. . . , a6j in terms of the nodal coordinates u1j , v1j , u2j , v2j , u3j , v3j. This leads to the following displacement field: 3 u1j 6v j 7 1 7 6 6 7 0 6 u2j 7 6 7 N 3 6 v2j 7 6 j7 4 u3 5 2

 uj ¼

uj vj



 ¼

N1 0

0 N1

N2 0

0 N2

N3 0

ð5:18Þ

v3j

where the scalars N1, N2, and N3 are defined by 9 1 > ½ x y  x y þ x ð y  y Þ þ y ð x  x Þ  2 3 3 2 3 2 > > 2 3 > 2A j > > = 1 N2 ¼ j ½x3 y1  x1 y3 þ xðy3  y1 Þ þ yðx1  x3 Þ > 2A > > > > 1 > ; N3 ¼ ½ x y  x y þ x ð y  y Þ þ y ð x  x Þ  1 2 2 1 2 1 1 2 j 2A N1 ¼

ð5:19Þ

in which A j is the area of the element defined by the determinant.

1

1 j A ¼

1 2

1

x1 x2 x3

y1

y2

y3

ð5:20Þ

5.1 Assumed Displacement Field

285

Consequently, the matrix S j of Eq. 5.3 can be recognized as  Sj ¼

0 N1

N1 0

N2 0

0 N2

N3 0

0 N3

 ð5:21Þ

Observe that  N k ðxi ; yi Þ ¼

0

if k 6¼ i

1

if k ¼ i

ð5:22Þ

Beam Element The beam element shown in Fig. 5.2c has two nodal points. Each node has three nodal coordinates that represent the longitudinal and transverse displacements and the slope. These nodal coordinates are  q j ¼ u1j

v1j

θ1j

u2j

v2j

θ2j

T

ð5:23Þ

Since we selected six nodal coordinates, our displacement field must contain six coefficients. The longitudinal displacement is approximated by a first-degree polynomial, while the transverse displacement is approximated by a cubic polynomial. The displacement field is, therefore, given for element j by )

u j ¼ a1j þ a2j x v j ¼ a3j þ a4j x þ a5j x2 þ a6j x3

ð5:24Þ

Because the axial displacement is described by a first-order polynomial, the axial strain is constant within the element and ∂2u j/∂x2 ¼ 0. On the other hand, the transverse displacement is described using a cubic polynomial and consequently the shear force is assumed to be constant within the element. At the nodal points, one has 9 u2j ¼ a1j þ a2j l j , v1j ¼ a3j > > = j j j j j  j 2 j  j 3 v 2 ¼ a3 þ a4 l þ a5 l þ a6 l > > j j j j j j j  j 2 ; θ1 ¼ a4 , θ2 ¼ a4 þ 2a5 l þ 3a6 l u1j ¼ a1j ,

ð5:25Þ

where l j is the length of the element. The preceding equations can be used to define the displacement field in terms of the nodal coordinates. This leads to 

uj u ¼ vj j

 ¼ S jq j

ð5:26Þ

where q j is the vector of the element nodal coordinates defined by Eq. 5.23, and S j is the space-dependent element shape function matrix defined by

286

5

2 x 6 1  lj 6 j S ¼6 6 4 0

The Finite Element Method

0

0

 2  3 x x 13 j þ2 j l l

 2  3 # x x x l j2 j þ j l l l 3

0

0

x lj

"

j

0

7 "  2 # 7 3 7 x x 5 lj  lj lj

 2  3 x x 3 j 2 j l l

ð5:27Þ The shape function matrix of the truss element can be obtained from the shape function matrix of the beam element as a special case in which the transverse deformation is neglected. Rectangular Element The rectangular element shown in Fig. 5.2d has four nodes. Each node has two degrees of freedom that represent the horizontal and vertical displacements. The nodal coordinates of the element are defined by the vector q j as  q j ¼ u1j

v1j

u2j

v2j

u3j

v3j

u4j

v4j

T

ð5:28Þ

The assumed displacement field of this rectangular element can be written as u j ¼ a1j þ a2j x þ a3j y þ a4j xy

) ð5:29Þ

v j ¼ a5j þ a6j x þ a7j y þ a8j xy

Following the procedure described in the case of the truss, triangular, and beam elements, one can show that the shape function matrix of the four-node rectangular element is given by " S ¼ j

N1

0

N2

0

N3

0

N4

0

0

N1

0

N2

0

N3

0

N4

# ð5:30Þ

where 1 ðb  xÞðc  yÞ, 4bc 1 N3 ¼ ðb þ xÞðc þ yÞ, 4bc

N1 ¼

9 1 > ðb þ xÞðc  yÞ > = 4bc > 1 > N4 ¼ ðb  xÞðc þ yÞ ; 4bc N2 ¼

Observe again that Nk(xi, yi) has the same property defined by Eq. 5.22.

ð5:31Þ

5.2 Comments on the Element Shape Functions

5.2

287

Comments on the Element Shape Functions

The assumed displacement field of the element is represented by simple functions expressed in the form of polynomials. The use of these simple functions guarantees that the displacement field within the element is continuous. Furthermore, the assumed displacement field must include low-order terms if the approximate solution is to be close enough to the exact solution. For example, consider the assumed displacement field of the beam element given by Eq. 5.24. If the beam is subjected to a static axial end load F, the exact solution is u j ¼ (F/(E jA j))x and v j ¼ 0, where E j and A j are, respectively, the modulus of elasticity and the cross-sectional area of the element j. Omitting the term a2j x in the assumed axial displacement of the first equation in Eq. 5.24 deletes the term that contains the exact answer. It is, therefore, important that the order of the polynomial is selected in such a manner that the approximate solution converges to the exact solution. This property is called the completeness. The completeness requirement guarantees that the approximate solution converges to the exact solution as the number of elements increases. The order of the polynomials in the assumed displacement field can be increased by increasing the number of nodes of the element or/and by increasing the number of coordinates of the nodes by considering higher derivatives of the displacements. It is clear that as a part of the completeness requirements, the assumed displacement field of the element must be able to represent the state of constant strain. The rigidbody motion without strain is a special case of the state of constant strain. Therefore, rigid-body modes must be present in the assumed displacement field. In this section, this property for the elements discussed in the preceding section is examined. The procedure for doing this is to assume that the element undergoes an arbitrary rigidbody motion. The values of the nodal coordinates as the result of this motion can be determined and substituted into the displacement field of the element in order to check whether this assumed displacement field can represent this rigid-body motion. Truss Element If the truss element whose assumed displacement field is defined by Eq. 5.12 undergoes a rigid-body translation dx in the direction of the x-coordinate, the values of the nodal coordinates as the result of this rigid-body motion become  j T  T u1 u2j ¼ d x l j þ dx . Substituting this vector into Eq. 5.12, the displacement of an arbitrary point on the element as the result of this rigid-body motion is given by " u ¼ ½ ð1  ξÞ j

ξ "

¼ ½ ð1  ξÞ

ξ

u1j

#

u2j dx l j þ dx

ð5:32Þ

# ¼ ξl j þ d x

which indicates that each point on the element indeed translates by the amount dx. Consequently, the truss-element shape function as defined by Eq. 5.11 can be used to describe an arbitrary rigid-body translation along the x-coordinate of the element.

288

5

The Finite Element Method

Rectangular Element While the truss element is used to describe only axial displacements, the linear rectangular element is used to describe an arbitrary planar displacement. An arbitrary rigid-body motion is described by a translation and a rotation. If the linear rectangular element undergoes an arbitrary rigid-body translation dx in the horizontal direction, the element nodal coordinates as the result of this rigid-body displacement become q j ¼ ½ b þ dx

c

b þ dx

c

b þ dx

c

b þ d x c T

ð5:33Þ

Substituting this vector into the displacement field presented in the preceding section, one obtains the displacement of an arbitrary point on the element as the result of this rigid-body motion as 2

" u ¼ j

uj vj

#

" ¼

N1

0

N2

0

N3

0

N4

0

N1

0

N2

0

N3

0

b þ dx

3

6 c 7 7 6 7 6 6 b þ dx 7 7 6 # 7 0 6 6 c 7 7 6 7 6 N 4 6 b þ dx 7 7 6 7 6 c 7 6 7 6 4 b þ dx 5

ð5:34Þ

c " ¼

ðN 1 þ N 2 þ N 3  N 4 Þb þ ðN 1 þ N 2 þ N 3 þ N 4 Þdx

#

ðN 1  N 2 þ N 3 þ N 4 Þc

One can show that N1 + N2 + N3 + N4 ¼ 1, N1  N2 + N3 + N4 ¼ y/c, and N1 + N2 + N3  N4 ¼ x/b and accordingly u j ¼ ½ x þ dx

y T ¼

    x d þ x y 0

ð5:35Þ

That is, the displacement of an arbitrary point on the element as the result of this rigidbody translation is dx. It is, therefore, clear that the element shape function as defined in the preceding section can be used to describe an arbitrary rigid-body translation in the x-direction. By using a similar procedure, one can also show that this element shape function can also describe an arbitrary rigid-body translation in the y-direction. If the element undergoes a rigid-body rotation α, the value of the nodal coordinates as the result of this rigid-body rotation is  q j ¼ ðb cos α þ c sin αÞ

ðb sin α  c cos αÞ ðb cos α þ c sin αÞ

ðb sin α  c cos αÞ ðb cos α  c sin αÞ ðb sin α þ c cos αÞ  ðb cos α  c sin αÞ ðb sin α þ c cos αÞ T

ð5:36Þ

5.2 Comments on the Element Shape Functions

289

Substituting this vector in the assumed displacement field of the bilinear rectangular element defined in the preceding section, one obtains " u ¼ j

uj

#

" ¼

ðN 1 þ N 2 þ N 3  N 4 Þb cos α þ ðN 1 þ N 2  N 3  N 4 Þc sin α

#

ðN 1 þ N 2 þ N 3  N 4 Þb sin α þ ðN 1  N 2 þ N 3 þ N 4 Þc cos α # " #" # x cos α  y sin α cos α  sin α x ¼ ¼ x sin α þ y cos α sin α cos α y vj

"

ð5:37Þ which indicates that the element shape function of the bilinear rectangular element can describe an arbitrary rigid-body rotation. This rotation can be finite or infinitesimal. By using similar procedures, one can also show that the shape function of the constant strain triangular element can describe an arbitrary rigid-body motion. Beam Element Following the procedure described in this section, it can be shown that the shape function of the beam element presented in the preceding section can describe an arbitrary rigid-body translation. In order to examine the ability of this shape function in describing an arbitrary rigid-body rotation, we consider the case in which the element rotates a rigid-body rotation defined by the angle α. In this case, one must have 

uj vj



 ¼

cos α

 sin α

  x

sin α

cos α

0

 ¼

x cos α



x sin α

ð5:38Þ

where x is the coordinate of an arbitrary point on the element. In this case, θ j ¼ ∂v j/∂x ¼ sin α, and accordingly the vector of nodal coordinates as the result of this  T rigid-body rotation is given by q j ¼ 0 0 sin α l j cosα l j sin α sinα . Substituting this vector into the displacement field of the beam element as defined by Eqs. 5.26 and 5.27, one obtains 

uj vj



 ¼

x cos α x sin α



 ¼

cos α

 sin α

  x

sin α

cos α

0

ð5:39Þ

In order to arrive at this result, it is assumed that the slope θ j ¼ ∂v j/∂x ¼ sin α instead of the angle α or tan α. Clearly, sinα  tan α  α if the rigid-body rotation α is infinitesimal. In the classical FE literature, the slopes at the nodes are assumed to represent infinitesimal rotations in order to allow treating the element coordinates using vector algebra. Transforming the nodal coordinates between different frames is then possible. The use of infinitesimal rotations as nodal coordinates, however, leads to difficulties in the large rotation and deformation analysis of structural systems as discussed in Sect. 5.10 of this chapter.

290

5.3

5

The Finite Element Method

Connectivity Between Elements

In the preceding sections, the displacement field of the element was defined in the element coordinate system. In the FE approach, a structural component is discretized using a number of elements, each of which has its own coordinate system. Because of geometrical considerations imposed by the shape of the structural components, the axes of the element coordinate system may not be parallel to the axes of the global coordinate system. The connectivity conditions between elements, however, require that when two elements are rigidly connected at a nodal point, the coordinates of this point as defined by the nodal coordinates of one element must be the same as defined by the coordinates of the other element. Therefore, before applying the connectivity conditions, the nodal coordinates of the elements must be defined in one global system. Coordinate Transformation As shown in Fig. 5.3, let β j be the angle which the X j-axis of the element j coordinate system makes with the X-axis of the global coordinate system. Let C j be the transformation matrix from the element coordinate system to the global coordinate system. The transformation matrix C j is given by " C ¼ j

cos β j

 sin β j

sin β j

cos β j

# ð5:40Þ

Therefore, the displacement field u j of Eq. 5.3 can be written in the global coordinate system as ugj ¼ C j u j ¼ C j S j q j

ð5:41Þ

In this equation, the vector of nodal coordinates q j is defined in the jth element coordinate system. This vector can be expressed in terms of the global coordinates as j q j qj ¼ C g

Fig. 5.3 Connectivity conditions

ð5:42Þ

5.3 Connectivity Between Elements

291

where qgj is the vector of nodal coordinates defined in the global coordinate system,  j is an orthogonal transformation matrix. In the case of the beam element, for and C  j is given by example, the matrix C " j ¼ C

#

C1jT

0

0

C1jT

ð5:43Þ

where C1j is the matrix 2 C1j

cos β j

6 ¼ 4 sin β j

 sin β j cos β

0

0

3

7 05 ¼

j

0

"

1

Cj 0T

0 1

# ð5:44Þ

Substituting Eq. 5.42 into Eq. 5.41, the global displacement vector can be expressed j q j , which can be in terms of the global element nodal coordinates as ugj ¼ C j S j C g written as ugj ¼ Sgj qgj

ð5:45Þ

j . where Sgj ¼ C j S j C Connectivity Conditions Having defined the element nodal coordinates in the global system, one can proceed a step further to write the displacement vector in terms of the total vector of nodal coordinates of the structure. To this end, we define the total vector of nodal coordinates of the structure as qg ¼ ½ q1

q2

   q n T

ð5:46Þ

where n is the total number of nodal coordinates of the structure. One can then write the vector of nodal coordinates of the element in terms of the structure nodal coordinates as qgj ¼ B j qg

ð5:47Þ

where B j is a Boolean matrix whose elements are either zeros or ones. For example, we consider the beam shown in Fig. 5.4. The beam is divided into two finite beam elements. For these two elements, the nodal coordinates are  q1g ¼ u11  q2g ¼ u21

T 9 =  2 T;

v11

θ11

u12

v12

θ12

v21

θ21

u22

v22

θ2

ð5:48Þ

Since the structure has three nodes, the total vector of the structure nodal coordinates is given by

292

5

The Finite Element Method

Fig. 5.4 Connectivity between two beam elements

qg ¼ ½ u1

v1

θ1

u2

v2

θ2

u3

v3

θ 3 T

ð5:49Þ

The vector q1g of the nodal coordinates of the first element can be written as q1g ¼ B1 qg , where B1 is the matrix 2

1 60 6 60 B1 ¼ 6 60 6 40 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

0 0 0 0 0 0

0 0 0 0 0 0

3 0 07 7 07 7 07 7 05 0

ð5:50Þ

0 0 0 0 1 0

3 0 07 7 07 7 07 7 05 1

ð5:51Þ

Similarly, for the second element q2g ¼ B2 qg , where 2

0 60 6 60 2 B ¼6 60 6 40 0

0 0 0 0 0 0

0 0 0 0 0 0

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

That is, the structure of the matrices B1 and B2 defines the second node as a common node for the two elements in this example. In general, the matrix B j has the number of rows equal to the number of element coordinates and the number of columns equal to the number of nodal coordinates of the structure. The global displacement vector ugj of element j can now be expressed in terms of the total vector of the nodal coordinates of the structure. To this end, we substitute Eq. 5.47 into Eq. 5.45 to obtain ugj ¼ Sgj B j qg ,

j ¼ 1,2,. . . ,ne

ð5:52Þ

This equation will be used in the following sections to develop the element mass and stiffness matrices. The structure mass and stiffness matrices can then be obtained by assembling the element matrices.

5.3 Connectivity Between Elements

293

Kinematic Constraints In many structural applications, some of the nodal coordinates may be specified. For example, one end of a beam or a truss in a structural system may be fixed. Such conditions can be used to reduce the number of the elastic degrees of freedom of the structure. In these cases, the vector of  T nodal coordinates qg can be written as qg ¼ qfT qsT , where qs is the vector of specified nodal coordinates, and qf is the vector of free nodal coordinates or the system degrees of freedom. If the specified nodes are fixed, one can write the vector qg in terms of the system degrees of freedom as qg ¼ Bc qf

ð5:53Þ

where Bc is an appropriate matrix whose entries are either zeros or ones. If no nodal coordinates are fixed, Bc is the identity matrix. Substituting Eq. 5.53 into Eq. 5.52 yields ugj ¼ Sgj B j Bc qf

ð5:54Þ

The use of the development presented in this section is demonstrated by the following examples. Example 5.1

For the structural system shown in Fig. 5.5, obtain the connectivity matrices. Assume that the system is divided into two beam elements.

Fig. 5.5 Structural system

Solution. As shown in the figure, the structure has three nodal points. The nodal coordinates of the structure defined in the global coordinate system are qg ¼ ½ u1

v1

θ1

u2

v2

θ2

u3

v3

θ 3 T (continued)

294

5

The Finite Element Method

The coordinate systems of the elements are also shown in the figure. For the  first element, it is clear that β1 ¼ 90 and, accordingly, the matrix C j of Eq. 5.40 is given by " C ¼ 1

cos β1 sin β1

 sin β1 cos β1

#

" ¼

cos 90 sin 90



 sin 90



cos 90



#





0 1 ¼ 1 0



For the second element β2 ¼ 0 and, accordingly, the matrix C2 is the identity matrix, that is, C2 ¼ I. Therefore, the matrices C1j , j ¼ 1,2, of Eq. 5.44 are given by 2

0 C11 ¼ 4 1 0

3 1 0 0 0 5, 0 1

2

1 0 C21 ¼ 4 0 1 0 0

3 0 05 1

One can then write the nodal coordinates of the two elements as j q j , j ¼ 1,2, where C j is the matrix qj ¼ C g " j ¼ C

C1jT

0

0

C1jT

#

That is, the nodal coordinates of the element defined in the global system are  T q1g ¼ v11 u11 θ11 v12 u12 θ12  T q2g ¼ u21 v21 θ21 u22 v22 θ22 Since these two vectors are defined in the global coordinate system, one also has q1g ¼ ½ u1

v1

θ1

u2

v2

θ2 T

q2g ¼ ½ u2

v2

θ2

u3

v3

θ3 T

That is, the connectivity matrices B1 and B2 are defined for this system as 2

1 60 6 60 1 B ¼6 60 6 40 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

0 0 0 0 0 0

0 0 0 0 0 0

3 0 07 7 07 7 07 7 05 0 (continued)

5.3 Connectivity Between Elements

2

0 60 6 60 B2 ¼ 6 60 6 40 0

295

0 0 0 0 0 0

0 0 0 0 0 0

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

3 0 07 7 07 7 07 7 05 1

Furthermore, since in this example node 1 is fixed, one has u1 ¼ v 1 ¼ θ 1 ¼ 0 and the degrees of freedom of the system are given by qf ¼ ½ u2

θ2

v2

u3

v3

θ 3 T

The vector of nodal coordinates of the structure can be expressed in terms of these degrees of freedom as 3 2 0 u1 6 v1 7 6 0 6 7 6 6 θ1 7 6 0 6 7 6 6 u2 7 6 1 6 7 6 6 v2 7 ¼ 6 0 6 7 6 6 θ2 7 6 0 6 7 6 6 u3 7 6 0 6 7 6 4 v3 5 4 0 0 θ3 2

0 0 0 0 1 0 0 0 0

0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 1 0 0

3 0 2 3 07 7 u2 07 7 6 v2 7 6 7 07 76 θ 2 7 6 7 07 7 6 u3 7 6 7 07 7 4 v3 5 07 7 θ 05 3 1

0 0 0 0 0 0 0 1 0

where the matrix Bc in this example can be recognized as 2

0 60 6 60 6 61 6 Bc ¼ 6 60 60 6 60 6 40 0

0 0 0 0 1 0 0 0 0

0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 1 0

3 0 07 7 07 7 07 7 07 7 07 7 07 7 05 1

1 , B1, Bc, and the vector qf defined in this example, By using the matrices C1, C the displacement field of the first element can be defined by using Eqs. 5.45 and 5.54 as (continued)

296

5

The Finite Element Method

1 B1 Bc qf u1 ¼ C1 S1 C Similarly, the displacement field of the second element is 2 B2 Bc qf u2 ¼ C2 S2 C

5.4

Formulation of the Mass Matrix

In this section, the element mass matrix is formulated using the kinetic energy expression. Two approaches are commonly used in the FE formulation of the mass matrix. The first approach is the lumped-mass approach in which the element mass matrix is obtained by lumping the element mass at the nodal points. One way of using the lumped-mass approach is to describe the inertia of the element by concentrated masses which have zero moment of inertia about their centers. This approach leads to a singular mass matrix in problems involving beams since there is no inertia assigned to the rotational degrees of freedom. Clearly, this approach has the drawback that the kinetic energy becomes a positive-semidefinite quadratic form instead of positive definite. Another way of using the lumped-massapproach is to describe the inertia of the element using small rigid bodies attached to the nodal points of the element. The process of doing this, however, is arbitrary. The second approach which is used in the FE formulation of the mass matrix is the consistent-mass formulation. In this approach, the assumed displacement field of the element is used to define the element kinetic energy. Consequently, the resulting mass matrix is often nondiagonal and positive definite. This is the approach discussed in this section for the formulation of the element mass matrix. Element Mass Matrix The kinetic energy of the element j is defined as Tj ¼

1 2

ð V

j

_ gj dV j , ρ j u_ jT g u

j ¼ 1,2,. . . ,ne

ð5:55Þ

where V j and ρ j are, respectively, the volume and mass density of the element j and u_ gj is the global velocity vector which is defined using Eq. 5.54 as u_ gj ¼ Sgj B j Bc q_ f , where B j is the Boolean matrix that defines the connectivity of the element, Bc is the matrix of the kinematic constraints, and Sgj is the shape function matrix of Eq. 5.45. Substituting u_ gj into Eq. 5.55, one obtains 1 T j ¼ q_ Tf BcT B j T 2 which can also be written as

ð Vj

 ρ j Sgj T Sgj dV j B j Bc q_ f

ð5:56Þ

5.4 Formulation of the Mass Matrix

297

1 T j ¼ q_ Tf Mfj q_ f , 2

j ¼ 1,2,. . . ,ne

ð5:57Þ

where Mfj is recognized as the jth element mass matrix and it is defined as ð j

Mf ¼

BcT B j T

Vj

ρ

j

Sgj T Sgj dV j

 B j Bc

j ¼ 1,2,. . .ne

ð5:58Þ

This matrix has dimension equal to the number of degrees of freedom of the structure. Using the definition of the matrix Sgj and the orthogonality of the transformation matrix C j, the mass matrix Mfj of Eq. 5.58 can be written as j T M j C j B j Bc , Mfj ¼ BcT B j T C

j ¼ 1,2,. . . ,ne

ð5:59Þ

where M j is the element mass matrix defined in the element coordinate system. This element mass matrix is given by ð Mj ¼

Vj

ρ j S j T S j dV j ,

j ¼ 1,2,. . . ,ne

ð5:60Þ

The mass matrix M j is symmetric and has dimension equal to the number of the element nodal coordinates. The global element mass matrix Mfj , on the other hand, has a dimension equal to the number of free coordinates of the structure. Structure Mass Matrix The mass matrix of the structure can be defined by sumPe j ming the kinetic energies of its elements as T ¼ nj¼1 T , where T is the kinetic hP i j ne _ f , which energy of the structure. Using Eq. 5.57, one has T ¼ ð1=2Þq_ Tf j¼1 Mf q can be rewritten as T ¼ ð1=2Þq_ Tf Mf q_ f , where Mf is recognized as the structure mass matrix and is defined as Mf ¼

ne X

Mfj ¼

j¼1

ne X

j T M j C j B j Bc BcT B j T C

ð5:61Þ

j¼1

where the element mass matrix M j is defined by Eq. 5.60. Example 5.2

For the truss element with the shape function defined by Eq. 5.11, the element mass matrix M j defined in the element coordinate system is given by  mj 2 ρ S S dV ¼ M ¼ 6 1 Vj ð

j

j

jT

j

j

1 2



(continued)

298

5

The Finite Element Method

Fig. 5.6 Two-truss member

where m j is the mass of the element j. The system shown in Fig. 5.6 consists of two truss elements which have equal masses. The transformation matrices C j  j for these elements are the identity matrices. The connectivity matrices and C 1 B and B2are 

 0 0 , 1 0

1 B ¼ 0 1



0 1 B ¼ 0 0 2

0 1



Therefore, the matrices M1f and M2f are 1T M1 C 1 B1 ¼ B1T M1 B1 M1f ¼ B1T C 2 3 " #" 1 0 2 1 1 0 m1 6 7 ¼ 40 15 6 1 2 0 1 0 0

0

¼

0

2T M2 C 2 B2 ¼ B2T M2 B2 M2f ¼ B2T C 2 3 #" 0 0 " 2 2 1 0 1 m 6 7 ¼ 41 05 6 1 2 0 0 0 1

0 1

2

#

m1 6 41 2 6 0 0 2

# ¼

2 1

2

0 0

m 6 40 2 6 0 1

0

3

7 05 0 0

3

7 15 2

If we assume that m1 ¼ m2 ¼ m, the structure mass matrix Mf can then be defined as Mf ¼

2 X

Mfj ¼ M1f þ M2f

j¼1

2

2 1

m6 ¼ 41 2 6 0 0

0

3

2

0

7 m6 05 þ 40 6 0 0

0 2 1

0

3

2

2

7 m6 15 ¼ 41 6 2 0

1

0

3

4

7 15

1

2

Example 5.3

For the beam element with the shape function defined by Eq. 5.27, the element mass matrix Mj is given by (continued)

5.4 Formulation of the Mass Matrix

299

ð Mj ¼

Vj

ρ j S j T S j dV j

2

3j

1=3

6 6 0 6 6 0 6 ¼ m j6 6 1=6 6 6 4 0 0

13=35

7 7 7 7 7 7 7 7 7 5

Symmetric

11l=210

l2 =105

0

0

1=3

9=70

13l=420

0

13=35

13=420

l =140

0

11l=210

2

l2 =105

where m j and l j are, respectively, the mass and length of the element and superscript j on the major bracket implies that all quantities inside the bracket are superscripted with j. If we consider the structural system of Example 5.1, for the first element, one can show by direct matrix multiplication that 2

0 60 6 6 1 B1 Bc ¼ 6 0 C 60 6 41 0

0 0 0 1 0 0

0 0 0 0 0 1

0 0 0 0 0 0

0 0 0 0 0 0

3 0 07 7 07 7 07 7 05 0

1 , B1, and Bc are evaluated in Example 5.1. If we assume where the matrices C that the two elements have the same mass m and equal length l, one can then write the mass matrix of the first element using Eq. 5.59 as 1T M1 C 1 B1 Bc M1f ¼ BcT B1T C 2 13=35 0 11l=210 0 6 0 1=3 0 0 6 6 2 6 11l=210 0 l =105 0 6 ¼ m6 6 0 0 0 0 6 6 0 0 0 0 4 0

0

0

0

0

0

3

0

7 07 7 07 7 7 07 7 7 05

0

0

0 0 0

For the second element, one has 2

1 60 6 60 2 2  C B Bc ¼ 6 60 6 40 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

3 0 07 7 07 7 07 7 05 1 (continued)

300

5

The Finite Element Method

Therefore, the mass matrix M2f of the second element is 2T M2 C 2 B2 Bc M2f ¼ BcT B2T C 2 1=3 0 0 6 13=35 11l=210 6 0 6 6 0 11l=210 l2 =105 6 ¼ m6 6 1=6 0 0 6 6 9=70 13l=420 4 0 0

1=6

0

0

9=70

0

13l=420

1=3

0

0

13=35

0

11l=210

13=420 l =140 2

3

0

7 13=420 7 7 l2 =140 7 7 7 7 0 7 7 11l=210 5 l2 =105

The composite mass matrix of the structure can then be obtained as Mf ¼

2 X

Mfj ¼ M1f þ M2f

j¼1

2

74=105 6 0 6 6 6 11l=210 ¼ m6 6 1=6 6 6 4 0 0

5.5

0 74=35

11l=210 11l=210

1=6 0

0 9=70

11l=210 0

2l2 =105 0

0 1=3

13l=420 0

9=70

13l=420

0

13=35

13=420

l =140

0

11l=210

2

3 0 13=420 7 7 7 l2 =140 7 7 7 0 7 7 11l=210 5 l2 =105

Formulation of the Stiffness Matrix

There are several techniques for developing the FE stiffness matrix. In the analysis presented in this section, the strain energy is used to formulate the stiffness matrices. To this end, some basic concepts and definitions which are important in the analysis of continuous systems are briefly discussed. Stress–Strain Relationships For an element j, the strain energy is given by

Uj ¼

1 2

ð Vj

σ j T ε j dV j

ð5:62Þ

where V j is the element volume and σ j and ε j are, respectively, the vectors of stress and strain. The stress and strain vectors are in general six-dimensional vectors which can be written as

5.5 Formulation of the Stiffness Matrix

σ j ¼ ½ σ1 σ2 σ3

σ4 σ5

ε j ¼ ½ ε1 ε2 ε3

ε4 ε5

301

 T 9 σ 6 T ¼ σ x σ y σ z σ xy σ yz σ zx =  T ; ð5:63Þ ε6 T ¼ εx εy εz εxy εyz εzx

For an isotropic material, the stress components can be expressed in terms of the strain components as 9 E ½ð1  2vÞεx þ vεt  > > > > ð1 þ vÞð1  2vÞ > > >  > E > ð1  2vÞεy þ vεt = σy ¼ ð1 þ vÞð1  2vÞ > > > E > ½ð1  2vÞεz þ vεt  > σz ¼ > > ð1 þ vÞð1  2vÞ > > ; σ xy ¼ Gεxy , σ yz ¼ Gεyz , σ zx ¼ Gεzx σx ¼

ð5:64Þ

where E, G, and v are, respectively, the modulus of elasticity, modulus of rigidity, and Poisson’s ratio, and εt ¼ εx+εy+εz. The stress vector σ j can be written compactly using the constitutive equations as σ j ¼ E jε j

ð5:65Þ

where E j is the matrix of elastic coefficients. The preceding equation is the generalized Hooke’s law. The matrix E j of the elastic coefficients depends on the material properties of the structure. This matrix can be written explicitly in terms of the modulus of elasticity, modulus of rigidity, and Poisson’s ratio as 2 6 6 6 6 E j 6 E ¼ ð1 þ vÞð1  2vÞ 6 6 6 4

1v v v

v

v

1v v v 1v

0

0

0

0 0

0 0

0 0

0

0

0

0

0

ð1  2vÞ=2

0 0

0 0

0 0

0 0

3

ð1  2vÞ=2 0 0 ð1  2vÞ=2

7 7 7 7 7 7 7 7 5

ð5:66Þ The inverse of this matrix is 2

 j 1 E

1 6 v 6 16 6 v ¼ 6 E6 0 6 4 0 0

v 1 v 0 0 0

v v 1 0 0 0

3 0 0 0 7 0 0 0 7 7 0 0 0 7 7 7 2ð 1 þ v Þ 0 0 7 5 0 2ð 1 þ v Þ 0 0 0 2ð 1 þ v Þ

ð5:67Þ

302

5

The Finite Element Method

Accordingly, the strain components can be expressed in terms of the stress components as 9   1 > > σx  v σy þ σz > > E > > > >  1 > > = εy ¼ σ y  vðσ x þ σ z Þ E >   1 > > εz ¼ σ z  v σ x þ σ y > > > E > > σ xy σ yz σ zx > > ; εxy ¼ , εyz ¼ , εzx ¼ G G G εx ¼

ð5:68Þ

where the modulus of rigidity G can be expressed in terms of the modulus elasticity E and Poisson’s ratio v as G ¼ E/(2(1 + v)). For plane strain, εz ¼ εxz ¼ εyz ¼ 0, the stress–strain relationships reduce in this special case to 9   E > > ð1  vÞεx þ vεy > > ð1 þ vÞð1  2vÞ > > > =   E ð1  vÞεy þ vεx σy ¼ ð1 þ vÞð1  2vÞ > > > >   > vE > εx þ εy , σ xy ¼ Gεxy > σz ¼ ; ð1 þ vÞð1  2vÞ σx ¼

ð5:69Þ

which can be written in a matrix form as 2 3 σx 1v 6 v 6 σy 7 E 6 6 7¼ 4 σ z 5 ð1 þ vÞð1  2vÞ 4 v σ xy 0 2

v 1v v 0

3 2 3 0 εx 7 0 74 εy 5 5 0 εxy ð1  2vÞ=2

ð5:70Þ

In this case of plane strain, one has σ z ¼ v(σ x + σ y). The inverse relationships are given by   9 1 þ v > σx  v σx þ σy > εx ¼ > > E > >   = 1 þ v ð5:71Þ σy  v σx þ σy εy ¼ > E > > > > 2ð 1 þ v Þ > ; σ xy εxy ¼ E which can be written in a matrix form as 2 3 εx 1  v v ð 1 þ v Þ 4 v 1  v 4 εy 5 ¼ E εxy 0 0 2

32 3 σx 0 0 54 σ y 5 σ xy 2

ð5:72Þ

5.5 Formulation of the Stiffness Matrix

303

Another special case is the case of plane stress. In this case, σ z ¼ σ yz ¼ σ zx ¼ 0. By using a similar procedure to the case of plane strain, the constitutive relationships in the case of plane stress can be obtained. Strain-Displacement Relationships For a linear elastic material, the straindisplacement relationships are 9 > > > =

∂u ∂v ∂w εx ¼ , εy ¼ , εz ¼ , ∂x ∂y ∂z ∂u ∂v ∂v ∂w þ , εyz ¼ þ , εxy ¼ ∂y ∂x ∂z ∂y

∂w ∂u > > > εxz ¼ þ ; ∂x ∂z

ð5:73Þ

where u, v, and w are the translational displacements of an arbitrary point on the elastic medium in the x, y, and z directions, respectively. That is, u j ¼ ½ u v w T. The strain-displacement relationships can be written compactly as ε j ¼ D ju j

ð5:74Þ

where D j is a differential operator defined as 2

∂=∂x 6 0 6 6 0 j D ¼6 6 ∂=∂y 6 4 0 ∂=∂z

0 ∂=∂y 0 ∂=∂x ∂=∂z 0

3 0 0 7 7 ∂=∂z 7 7 0 7 7 ∂=∂y 5 ∂=∂x

ð5:75Þ

Element Stiffness Matrix The constitutive equations and the strain-displacement relationships can be used to express the strain vector in terms of the derivatives of the displacement. Substituting Eq. 5.74 into Eq. 5.65 yields σ j ¼ E jD ju j, which can be written in terms of the nodal coordinates of the structure as  j B j Bc qf σ j ¼ E jD jS jC

ð5:76Þ

Similarly, the strain ε j can be written as j B j Bc qf ε j ¼ D jS jC

ð5:77Þ

Substituting Eqs. 5.76 and 5.77 into the strain energy of Eq. 5.62, one obtains Uj ¼

1 2

ð Vj

  jT D j S j T E j D j S j C j B j Bc qf dV j , qfT BcT B j T C

ð5:78Þ

304

5

The Finite Element Method

where the symmetry of the matrix of elastic coefficients has been utilized. Equation 5.78 can be written in a simple form as U j ¼ ð1=2ÞqfT Kfj qf , where Kfj is the global element stiffness matrix defined as j

Kf ¼

 jT BcT B j T C

ð



j

DS

Vj

 j T



 j B j Bc E D S dV C j

ð5:79Þ

j ¼ 1,2,. . . ,ne

ð5:80Þ

j

j

j

The global stiffness matrix Kfj can be written as j T K j C  j B j Bc , Kfj ¼ BcT B j T C

where K j is the element stiffness matrix defined in the element coordinate system and is given by ð



K ¼ j

V

T D j S j E j D j S j dV j ,

j ¼ 1,2,. . . ,ne

j

ð5:81Þ

Illustrative Example In order to illustrate the procedure discussed in this section for formulating the element stiffness matrix, we consider the four-node, eight degree of freedom planar rectangular element. The shape function of this element is defined in Sect. 5.1 as " Sj ¼

N1

0

N2

0

N3

0

N4

0

0

N1

0

N2

0

N3

0

N4

#

where 1 ðb  xÞðc  yÞ, 4bc 1 N3 ¼ ðb þ xÞðc þ yÞ, 4bc N1 ¼

1 ðb þ xÞðc  yÞ, 4bc 1 N4 ¼ ðb  xÞðc þ yÞ 4bc N2 ¼

where 2b and 2c are the dimensions of the element as shown in Fig. 5.2d. In the case of two-dimensional displacement, the differential operator D j is given as 3

2

∂ 6 ∂x 6 6 6 j D ¼ 60 6 6 4∂ ∂y Therefore,

0

7 7 ∂7 7 7 ∂y 7 7 ∂5 ∂x

5.5 Formulation of the Stiffness Matrix

2

∂N 1 6 ∂x 6 6 6 j j DS ¼6 0 6 6 4 ∂N 1 ∂y

0 ∂N 1 ∂y ∂N 1 ∂x

305

∂N 2 ∂x

0 ∂N 2 ∂y ∂N 2 ∂x

0 ∂N 2 ∂y

∂N 3 ∂x 0 ∂N 3 ∂y

0 ∂N 3 ∂y ∂N 3 ∂x

3

∂N 4 ∂x

0 7 7 ∂N 4 7 7 7 ∂y 7 7 ∂N 4 5 ∂x

0 ∂N 4 ∂y

in which ∂N 1 ∂x ∂N 2 ∂x ∂N 3 ∂x ∂N 4 ∂x

ð c  yÞ , 4bc ð c  yÞ , ¼ 4bc ð c þ yÞ , ¼ 4bc ð c þ yÞ , ¼ 4bc

∂N 1 ∂y ∂N 2 ∂y ∂N 3 ∂y ∂N 4 ∂y

¼

ð b  xÞ 4bc ð b þ xÞ ¼ 4bc ð b þ xÞ ¼ 4bc ð b  xÞ ¼ 4bc ¼

Therefore, the matrix D jS j is given by 2 1 6 DS ¼ 4 4bc j

 ð c  yÞ

0

ð c  yÞ

0

0

ðb  xÞ

0

ðb þ xÞ

j

ðb  xÞ ð c þ yÞ 0 0

ð b þ xÞ

ðc  yÞ ðb þ xÞ ðc  yÞ 3 ðc þ yÞ 0 7 ð b þ xÞ 0 ð b  xÞ 5 ðc þ y Þ

ð b  xÞ

If we consider the case of plane stress, given by 2 1 E 4v Ej ¼ ð 1  v2 Þ 0

ðc þ yÞ

the matrix of elastic coefficients E j is v 1 0

3 0 5 0 ð1  vÞ=2

where E and v are, respectively, the modulus of elasticity and Poisson’s ratio of the element. Assuming that the element has a constant thickness and constant modulus of elasticity and Poisson’s ratio, the use of Eq. 5.81 leads to Kj ¼

ðc ðb c



T D j S j E j D j S j t dx dy

b

where t is the thickness of the element. Using the matrices D jS j and E j, the stiffness matrix of the four-node rectangular element can be written explicitly as

306

5

The Finite Element Method

3

2

k 11 6 k 21 6 6k 6 31 6k Et 6 41 j K ¼ 6 12ð1  v2 Þ 6 k 51 6 6 k 61 6 4 k 71 k 81

k 22 k 32 k 42 k 52 k 62 k 72 k 82

k 33 k 43 k 53 k 63 k 73 k 83

7 7 7 7 7 7 7 7 7 7 7 5

symmetric k44 k54 k64 k74 k84

k55 k65 k75 k85

k66 k76 k86

k77 k87

k88

in which the stiffness coefficients kij are k11 ¼ k33 ¼ k 55 ¼ k77 ¼ 4r þ 2ð1  vÞr 1 , k22 ¼ k44 ¼ k 66 ¼ k88 ¼ 4r 1 þ 2ð1  vÞr, 3 k21 ¼ k 61 ¼ k52 ¼ k 43 ¼ k83 ¼ k74 ¼ k 65 ¼ k87 ¼ ð1 þ vÞ, 2 k31 ¼ k75 ¼ 4r þ ð1  vÞr 1 , 3 k41 ¼ k 81 ¼ k32 ¼ k 72 ¼ k63 ¼ k54 ¼ k 85 ¼ k76 ¼  ð1  3vÞ, 2 k51 ¼ k73 ¼ 2r  ð1  vÞr 1 , k71 ¼ k53 ¼ 2r  2ð1  vÞr 1 k42 ¼ k86 ¼ 2r 1  2ð1  vÞr, k 62 ¼ k84 ¼ 2r 1  ð1  vÞr k82 ¼ k64 ¼ 4r 1 þ ð1  vÞr where r ¼ c/b. Structure Stiffness Matrix The structure stiffness matrix can be defined by sumPe ming the strain energy expressions of its elements, that is, U ¼ nj¼1 U j, where U is the strain energy of the structure. Using the expression hP ifor the element strain energy j j ne j T T U ¼ ð1=2Þqf Kf qf , one has U ¼ ð1=2Þqf j¼1 Kf qf , which can be written as U ¼ ð1=2ÞqfT Kf qf , where Kf is the stiffness matrix of the structure which is defined Pe Kfj , or in a more explicit form as as Kf ¼ nj¼1 Kf ¼

ne X

j T K j C  j B j Bc BcT B j T C

ð5:82Þ

j¼1

where K j is the jth element stiffness matrix defined in the element coordinate system. Example 5.4

The strain energy for the truss element of Example 5.2 is Ð U j ¼ ð1=2Þ V j σ xj εxj dV j , where σ xj and εxj are, respectively, the stress and (continued)

5.5 Formulation of the Stiffness Matrix

307

strain components. The stress–strain relationship in this simple case is given by σ xj ¼ E j εxj , where E j is Young’s modulus of the element material. The strain displacement relationship is εxj ¼

∂u j ¼ D ju j ∂x

where the differential operator of Eq. 5.74 reduces in this case to D j ¼ ∂/∂x. The displacement u j can be written in terms of the nodal coordinates as u j ¼ S jq j, where S j is the element shape function defined as  S ¼ j

x 1 j l



x lj



That is, D jS j ¼

1 ½1 lj

1

Therefore, the stiffness matrix of Eq. 5.81 can be written as ð

 j j T j j j j D S E D S dV

K ¼ j

V

¼A

j

ðl j

j 0

" # 1 1 j E ½ 1 l j2 1

" E jA j 1 1 dx ¼ j l 1

1

#

1

where A j is the cross-sectional area of the truss element j. The stiffness matrix K j is defined in the element coordinate system. The global matrix can be obtained using a procedure similar to the one described for the mass matrix of the system discussed in Example 5.2.

Example 5.5

Using the assumptions of the elementary beam theory and neglecting rotary inertia and shear deformation, one can show that the strain energy for a beam element j can be written as Uj ¼

1 2

ðl j 0

2



4E j A j ∂u ∂x

j

2 þ E jI j

!2 3 ∂ v 5dx ∂x2 2 j

(continued)

308

5

The Finite Element Method

where E j is the modulus of elasticity, A j is the cross-sectional area, I j is the second moment of area, and u j and v j are, respectively, the axial and transverse displacements. The strain energy can be written in the following form: 1 U ¼ 2

ðl j

j



0

j 0

ðu Þ

  E jA j ðv Þ 0

0 E jI j

j 00



 0 ðu j Þ 00 dx ðv j Þ

where "

ðu j Þ

0

j 00

ðv Þ

#

" ¼

0

Sxj

#

00 Syj

qj

where Sxj and Syj are the rows of the element shape function matrix defined by Eq. 5.20, that is, " S ¼ j

Sxj

#

Syj

One can then write the strain energy as U j ¼ (1/2)q j TK jq j, where the stiffness matrix K j is defined as ðl j h i 0 0 00 00 E j A j Sxj T Sxj þ E j I j Syj T Syj dx K ¼ j

0

Using the shape function of Eq. 5.27, it can be shown that the stiffness matrix of the beam element is given explicitly by 3j

2

A=I 6 0 6 E jI j 6 0 j K ¼ j 6 A=I l 6 6 4 0 0

12=l2 6=l 0 12=l2 6=l

7 7 7 7 7 7 5

Symmetric 4 0 6=l 2

A=I 0 0

12=l2 6=l

4

where superscript j on the major bracket implies that all the variables inside the bracket are superscripted with j. Note that K j is the element stiffness matrix defined in the element coordinate system. For the structural system of Example 5.1, if we assume that the two elements have the same dimensions and material properties and follow a procedure similar to the one used for the mass matrix in Example 5.3, one can show that the stiffness matrices K1f and K2f of the two elements are given, respectively, by (continued)

5.6 Equations of Motion

309

2

12=l2 6 0 6 EI 6 6=l K1f ¼ 6 l 6 6 0 4 0 0

0 A=I 0 0 0 0

6=l 0 4 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

3 0 07 7 07 7 07 7 05 0

and 2

A=I 6 0 6 EI 6 0 K2f ¼ 6 l 6 6 A=I 4 0 0

0 12=l2 6=l 0 12=l2 6=l

0 A=I 6=l 0 4 0 0 A=I 6=l 0 2 0

0 12=l2 6=l 0 12=l2 6=l

3 0 6=l 7 7 2 7 7 0 7 7 6=l 5 4

where E, I, l, and A are, respectively, the modulus of elasticity, second moment of area, length, and area of the elements. The composite stiffness matrix of the structure can then be obtained as P Kf ¼ 2j¼1 Kfj ¼ K1f þ K2f . That is, 2 6 6 EI 6 Kf ¼ 6 l 6 6 4

5.6

A=I þ 12=l2 0 6=l A=I 0 0

 

0  A=I þ 12=l2 6=l 0 12=l2 6=l

6=l 6=l 8 0 6=l 2

A=I 0 0 A=I 0 0

0 12=l2 6=l 0 12=l2 6=l

3 0 6=l 7 7 2 7 7 0 7 7 6=l 5 4

Equations of Motion

In the preceding sections, expressions for the element kinetic and strain energies were obtained and used to identify, respectively, the element mass and stiffness matrices. The structure kinetic and strain energies can be obtained by adding the energy expressions of its elements. This leads to the definition of the structure mass and stiffness matrices. In this section, the energy expressions are used with the virtual work of the external forces to define the equations of motion of the structure. Let F j be the vector of the external forces that acts at a point p j of element j. If F j is defined in the global coordinate system, the virtual work of this force vector is given by δW j ¼ F j T δugj . Using Eq. 5.54, the virtual change δugj can be written as j B j Bc δqf δugj ¼ Sgj B j Bc δqf ¼ C j S j C

ð5:83Þ

310

5

The Finite Element Method

where S j is evaluated at the point of application of the force. Therefore, the virtual j B j Bc δqf , which can be work δW j ¼ F j T δugj can be written as δW j ¼ F j T C j S j C jT j j written as δW ¼ Qf δqf , in which Qf is the generalized force vector given by jT S j T C j T F j , Qfj ¼ BcT B Tj C

j ¼ 1,2,. . . ,ne

ð5:84Þ

The virtual work of the forces acting on the structure is given by δW ¼

ne X j¼1

δW j ¼

ne X

Qfj T δqf ¼ QfT δqf

ð5:85Þ

j¼1

where Qf is the vector of generalized forces associated P e withj the generalized nodal coordinates of the structure and is defined as Qf ¼ nj¼1 Qf . Having defined in the preceding sections the kinetic and strain energies of the structure and having obtained the generalized forces associated with the structure nodal coordinates, one can use Lagrange’s equation to develop the equations of vibration of the €f þ Kf qf ¼ Qf , where Mf and Kf are, respectively, the symmetric structure as Mf q mass and stiffness matrices developed in the preceding sections, qf is the vector of nodal coordinates defined in the global coordinate system, and Qf is the vector of generalized forces. If the system is viscously damped, the equations of motion, €f þ Kf qf ¼ Qf , can be modified in order to include the damping effect. In this Mf q case, the matrix equation of motion takes the form €f þ Cf q_ f þ Kf qf ¼ Qf Mf q

ð5:86Þ

where Cf is the damping matrix. Equation 5.86 is in a form similar to the matrix equation obtained for the multi-degree of freedom systems. Therefore, the solution techniques discussed in Chap. 3 can be used to obtain a solution for the matrix equation of Eq. 5.86. If the number of nodal coordinates of the structure is very large, modal truncation methods can be used in order to reduce the number of degrees of freedom. The use of the procedure described in this section for formulating the equations of motion of structural systems is demonstrated by the following example. Example 5.6

Figure 5.7 shows the structural system discussed in Examples 5.1, 5.3, and 5.5. The force vector F(t) ¼ [Fcos α Fsin α]T acts at point B as shown in the figure. The virtual work of this force is given by δW 2 ¼ FT δu2gB , where u2gB is the position vector of point B defined in the global system. The virtual change δu2gB can be obtained using Eq. 5.83 as 2 B2 Bc δqf δu2gB ¼ C2 S2 C (continued)

5.6 Equations of Motion

311

Fig. 5.7 Forced vibration

2 are identity matrices and, accordingly, δu2 In this example, C2 and C gB reduces to δu2gB ¼ S2 B2 Bc δqf in which the shape function S2 must be evaluated at point B at which x ¼ l. By using the shape function of Eq. 5.27 and substituting x ¼ l, one obtains 

0 S ¼ 0 2

0 0

0 1 0 0

0 1

0 0



Furthermore, from the results of Example 5.5, B2Bc is the identity matrix. Therefore, the virtual work reduces to δW 2 ¼ FT S2 δqf

3 δu2 6 δv2 7 7 6 7 0 0 6 6 δθ2 7 7 6 1 0 6 δu3 7 7 6 4 δv3 5 2

 ¼ ½ F cos α

F sin α 

0 0 0 0

0 0

1 0

3 δu2 6 δv2 7 7 6 6 δθ 7 6 27 0 6 7 6 δu3 7 7 6 4 δv3 5 δθ3 2

¼½0 0

0

F cos α

F sin α

δθ3

2 which can be written compactly as δW 2 ¼ Q2T f δqf , where Qf is the vector of generalized forces resulting from the application of the force F and is given by

(continued)

312

5

Q2f ¼ ½ 0

0

0 F cos α

F sin α

The Finite Element Method

0 T

Since there are no forces acting on the first element, one has Q1f ¼ 0. The generalized forces of the structure can then be obtained as Qf ¼ Q1f þ Q2f ¼ Q2f The mass and stiffness matrices of this structural system were obtained, respectively, in Examples 5.3 and 5.5. Using the results presented in these two examples and Eq. 5.86, the undamped vibration equations of this structural system can be written as €f þ Kf qf ¼ Qf Mf q or in a more explicit form as 2

74=105

0

11l=210 1=6

0

32

0

€u2

3

6 7 13=420 7 76 €v2 7 7 6 7 6€ 7 0 13l=420 l2 =140 7 2l2 =105 76 θ2 7 76 7 76 €u3 7 0 1=3 0 0 76 7 76 7 13l=420 0 13=35 11l=210 54 €v3 5 θ€3 0 13=420 l2 =140 0 11l=210 l2 =105 32 3 2  2 u2 0 6=l A=I 0 0 A=I þ 12=l 7 6   6 6 v2 7 0 12=l2 6=l 7 0 A=I þ 12=l2 6=l 7 76 6 6 76 7 6 θ2 7 6=l 6=l 8 0 6=l 2 7 EI 6 7 76 þ 6 6 76 u 7 l6 A=I 0 0 A=I 0 0 76 3 7 6 76 7 6 7 6 6=l 0 12=l2 6=l 7 0 12=l2 54 v3 5 4

6 0 74=35 6 6 6 11l=210 11l=210 6 m6 6 1=6 0 6 6 4 0 9=70

3

2

0

11l=210

0

6=l

2

9=70

0

6=l

4

θ3

0 6 0 7 7 6 7 6 6 0 7 7 6 ¼6 7 6 F cos α 7 7 6 7 6 4 F sin α 5 0

Remarks The close relationship between the FE method and the classical approximation techniques, discussed at the end of the preceding chapter, should be apparent by now. In fact, the FE method can be regarded as a special case of the Rayleigh–Ritz method. While in both techniques the separation of variables is used to write the

5.7 Convergence of the FE Solution

313

displacement field in terms of a set of assumed shape functions and a set of timedependent coordinates, there are some important differences between the two techniques. For example, in the FE method, the time-dependent coordinates represent physical variables such as the displacements and slopes at selected nodal points. By defining these variables in a global coordinate system, the FE assembly is possible. In the Rayleigh–Ritz method, the coefficients in the assumed displacement field may lack any physical meaning. Furthermore, in the Rayleigh–Ritz method, the assumed displacement field is defined for the whole structure. In many practical applications, especially in structures with complex geometry, it is difficult to assume a suitable displacement field. This basic problem is avoided by using the finite element method, since the assumed displacement field is defined over the domain of a small element. Consequently, the global deflected shape of the structure is defined by simply selecting the appropriate boundary conditions imposed on the element nodal coordinates. By assuming the displacement field for relatively small elements, low-order interpolating functions can be used, leading to a simple formulation for the element mass and stiffness matrices. The element matrices, which are often computer generated, are assembled in order to obtain the mass and stiffness matrices of the structure.

5.7

Convergence of the FE Solution

The FE theory guarantees that the approximate FE solution converges to the exact solution by increasing the number of elements. The use of more elements, however, leads to an increase in the number of degrees of freedom and, accordingly, an increase in the number of differential equations which must be solved. In this section, the accuracy of the FE solution is examined. In particular, the FE solution of the eigenvalue problem is considered, and the obtained results are compared with the exact solution. €f þ Kf qf ¼ 0, where Mf and Kf In the case of free undamped vibration, one has Mf q are, respectively, the mass and stiffness matrices of the structure formulated using the FE method and qf is the vector of free nodal coordinates. As in the case of multi-degree of freedom systems, a solution is assumed in the form qf ¼ A sin (ωt + ϕ), where A is the vector of amplitudes, ω is the frequency, t is time, and ϕ is the phase angle. It follows that [Kf  ω2Mf]A ¼ 0. For this system of equations to have a nontrivial solution, the determinant of the coefficient matrix must be equal to zero, that is, jKf  ω2Mfj ¼ 0. This is the characteristic equation which is of order n in ω2, where n is the number of degrees of freedom of the system. The roots of the characteristic equation define the eigenvalues ω21 , ω22 ,. . .ω2n . The eigenvector or mode shape associated with the natural frequency ωi is obtained using the equation Kf  ω2i Mf Ai ¼ 0. In order to examine the convergence of the FE solution, we consider the simple example of the longitudinal vibration of a prismatic rod with one end fixed. It was shown in the preceding chapter that the exact natural frequencies and mode shapes of pffiffiffiffiffiffiffiffi the rod are, respectively, ωk ¼ ðð2k  1Þπ=2lÞ E=ρ and ϕk ¼ A1k sin (ωkx/c), k ¼ 1,2,3,. . ., where E, ρ, and l are, respectively, the modulus of elasticity, mass density, and length of the rod, A1k is an arbitrary constant, and c is the wave velocity

314

5

The Finite Element Method

pffiffiffiffiffiffiffiffi without dispersion defined as c ¼ E=ρ. Consider the case in which the rod is represented by one truss element. In this case, since one end of the rod (x ¼ 0) is fixed, the assumed displacement field is u(x, t) ¼ ξq(t), where ξ ¼ x/l and q(t) is the axial displacement The Ð l  of the  free Ðend.  kinetic energy of the rod in this 1 case is T ¼ 0 ρAu_ 2 dx =2 ¼ ρAlq_ 2 0 ξ2 dξ =2 ¼ ρAlq_ 2 =6. The strain energy is

Ð  l U ¼ 0 EAðu0 Þ2 dx =2 ¼ EAq2 =2l. Clearly, in this case, the mass and stiffness matrices reduce to scalars since the system has one degree of freedom. The mass Therefore, and stiffness coefficients are, respectively, m11 ¼ ρAl/3 and k11 ¼ EA/l. ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the approximate solution for the first natural frequency is ω ¼ k 1 pffiffiffiffiffiffiffiffi 11 =m11 ¼ pffiffiffiffiffiffiffiffi ð1:73205=lÞ E=ρ. The exact solution is ω1 ¼ ð1:570796=lÞ E=ρ, that is, the error in the FE solution using one element is about 10%, and the estimated solution is higher than the exact solution since the continuous system is represented by one degree of freedom. If the same rod is divided into two equal elements, each has length l/2, the system has two degrees of freedom, and the resulting mass and stiffness matrices of the rod are  ρAl 4 Mf ¼ 12 1

 1 , 2

 2EA 2 Kf ¼ 1 l

1 1



In this case, the characteristic equation is given as

2  4β

1  β

1  β

¼0 1  2β

where β ¼ (ρl2/24E)ω2. The characteristic equation can be expressed in terms of β as (2  4β)(1  2β)  (1 + β)2 ¼ 0, which can be written as 7β2  10β + 1 ¼ 0. The roots of this equation are β1 ¼ 0.108194 and β2 ¼ 1.320377. Since 2 ω ¼ 24Eβi =ρl2 , i ¼ 1,2, thepapproximate natural frequencies are ω1 ¼ ð1:6114=lÞ ffiffiffiffiffiffiffiffi pi ffiffiffiffiffiffiffiffi E=ρ and ω2 ¼ ð5:6292=lÞ E=ρ. Clearly, the use of two elements instead of one element leads to an improvement in the fundamental natural frequency. The error in the case of the two elements is less than 2.6%. The exact value of the second natural pffiffiffiffiffiffiffiffi frequency is ω2 ¼ ð4:71239=lÞ E=ρ. The error in the second natural frequency using two elements is about 19.5%. By increasing the number of elements, a better approximation can be obtained for the second natural frequency. Roughly speaking, a reasonable approximation for a set of low-frequency modes can be achieved by using a number of degrees of freedom equal to twice the number of the desired natural frequencies. Table 5.1 and Fig. 5.8 show the computer results obtained using the FE method. The number of degrees of freedom used in this model is 150 degrees of freedom. In Table 5.1, the obtained FE results for the modal mass and stiffness coefficients as well as the natural frequencies are compared with the exact solution. These results are presented for a rod with length 3.6 m, modulus of elasticity E ¼ 2  1011 N/m2, and mass density ρ ¼ 7870 kg/m3. The rod has a circular cross-sectional area with diameter 0.0185 m. In general, by using a large number of

5.7 Convergence of the FE Solution

315

Table 5.1 Modal coefficients Modal mass coefficients Mode Exact Finite element number solution solution 1 3.808 3.807 2 3.808 3.807 3 3.808 3.806 4 3.808 3.804 5 3.808 3.802 6 3.808 3.799 7 3.808 3.796 8 3.808 3.792 9 3.808 3.788 10 3.808 3.783 11 3.808 3.777 12 3.808 3.771 13 3.808 3.765 14 3.808 3.757 15 3.808 3.750 16 3.808 3.742 17 3.808 3.733 18 3.808 3.724

Modal stiffness coefficients Finite element Exact solution solution 1.84234  107 1.84231  107 1.65811  108 1.65796  108 4.60585  108 4.60476  108 9.02747  108 9.02334  108 1.49230  109 1.49118  109 2.22923  109 2.22675  109 3.11355  109 3.10873  109 4.14527  109 4.13672  109 5.32436  109 5.31032  109 6.65085  109 6.62888  109 8.12472  109 8.09199  109 9.74598  109 9.69896  109 1.15146  1010 1.14489  1010 1.34307  1010 1.33414  1010 1.54941  1010 1.53752  1010 1.77049  1010 1.75498  1010 2.00631  1010 1.98641  1010 2.25687  1010 2.23169  1010

Natural frequencies Exact Finite element solution solution 2199.56 2199.83 6598.70 6599.27 10997.82 10999.40 15396.95 15401.51 19796.10 19804.28 24195.19 24210.35 28594.30 28617.28 32993.47 33028.85 37392.56 37441.68 41791.71 41860.26 46190.83 46286.48 50589.95 50714.75 54989.02 55144.14 59388.29 59590.92 63787.37 64031.66 68186.48 68483.22 72585.61 72946.67 76984.77 77412.68

Fig. 5.8 Mode shapes. (—— Fourier Method, ----- Finite Element Method)

316

5

The Finite Element Method

nodal coordinates, the FE method yields very good approximations for the fundamental natural frequencies. Note also, from the results presented in Fig. 5.8, the good agreement between the approximate mode shapes obtained using the FE method and the exact eigenfunctions obtained by solving the partial differential equation. Semidefinite Systems Systems which admit rigid-body motion are called semidefinite systems. The strain energy of such systems is a positive-semidefinite quadratic form in the displacement coordinates. That is, the strain energy can be zero for nonzero values of the system coordinates. It was shown previously that the element shape functions can describe rigid-body displacements. The shape function of the truss element has one rigid-body mode that describes the rigid-body translation along the axis of the element. The rectangular and triangular element shape functions have three rigid-body modes that describe an arbitrary rigid-body planar motion. While the beam element shape function has three rigid-body modes, it can be used to describe only infinitesimal rigidbody rotations and arbitrary rigid-body translations in two directions. Clearly, if the number of the fixed coordinates of a structure discretized using the FE method is less than the number of the rigid-body modes of the element shape function matrix, the structure will have rigid-body degrees of freedom. Consequently, the strain energy of the system is a positive-semidefinite quadratic form. Consider the case of the four-node, eight degree of freedom bilinear rectangular element shown in Fig. 5.9. The shape function matrix of this element was developed in the first section of this chapter, and its stiffness matrix was developed in the preceding section. The vector of nodal coordinates of this element is  T q j ¼ u1j v1j u2j v2j u3j v3j u4j v4j . As previously mentioned, the shape function matrix of the bilinear rectangular element has three rigid-body modes. It can be shown that possible three rigid-body modes are A1j ¼ ½1 0 1 0 1 0 1 0T A2j ¼ ½0 1 0 1 0 1 0 1T A3j ¼ ½c  b c b  c b  c  bT where 2b and 2c are the dimensions of the element as shown in Fig. 5.9. The vectors A1j and A2j describe rigid-body translations in two orthogonal directions, while the

Fig. 5.9 Rectangular element

5.8 Higher-Order Elements

317

vector A3j describes a rigid-body rotation. Note also that the vectors A1j , A2j , and A3j are orthogonal, that is, AijT Akj ¼ 0 for i 6¼ k. Clearly, a rigid-body motion of the bilinear rectangular element can be expressed as a linear combination of the vectors A1j , A2j , and A3j . By using the stiffness matrix of the bilinear rectangular element developed in the preceding section, one can show that the rigid-body modes A1j , A2j , and A3j have no contribution to the elastic forces. In fact, the use of direct matrix multiplications shows that K j Akj ¼ 0, k ¼ 1,2,3. Consequently, for a more general rigid-body displacement that can be expressed as a linear combination of A1j , A2j , and A3j , the vector of elastic forces is equal to zero. Furthermore, since K j Akj is equal to zero, for k ¼ 1,2,3, one has AkjT K j Akj ¼ 0, k ¼ 1,2,3. That is, the stiffness coefficients associated with the rigid-body modes A1j , A2j , and A3j are equal to zero. Therefore, the strain energy can be zero for nonzero values of the coordinates, and it is indeed a positive-semidefinite quadratic form when the element has one or more of its rigidbody modes. The mass coefficients associated with the rigid-body modes A1j , A2j , and A3j are not equal to zero. By using the shape function of the bilinear rectangular element developed in the first section of this chapter, it can be shown that the element mass matrix is given by 2

4 60 6 62 6 j6 m j 60 M ¼ 36 6 61 60 6 42 0

0 4 0 2 0 1 0 2

2 0 4 0 2 0 1 0

0 2 0 4 0 2 0 1

1 0 2 0 4 0 2 0

0 1 0 2 0 4 0 2

2 0 1 0 2 0 4 0

3 0 27 7 07 7 17 7 07 7 27 7 05 4

where m j is the mass of the element. Observe that A1jT M j A1j ¼ m j , A2jT M j A2j ¼ m j ,  jT j j j 2 2 and A3 M A3 ¼ m b þ c =3. That is, the mass coefficients associated with the rigid-body modes of the element are not equal to zero. Consequently, the kinetic energy of the element is a positive-definite quadratic form in the velocities. In fact, the mass coefficients associated with the rigid-body modes A1j , A2j , and A3j are the exact mass coefficients that are used in the dynamics of rigid bodies. Moreover, the vectors A1j , A2j , and A3j are orthogonal with respect to the mass matrix, that is, AijT M j Akj ¼ 0 if i 6¼ k.

5.8

Higher-Order Elements

Thus far, we have considered simple elements in order to demonstrate the use of the powerful FE technique in the dynamic analysis of structural systems. It was shown that the degree of the polynomial used to describe the element displacement field depends on the number of nodal coordinates of this element. In fact, the number of

318

5

The Finite Element Method

coefficients of the interpolating polynomial must be equal to the number of the degrees of freedom of the element. Therefore, the order of these polynomials can be increased by increasing the number of nodal coordinates. Two approaches can be used to achieve this goal. In the first approach, the number of nodal points of the element is increased using the same type of nodal coordinates. In the second approach, the same number of nodal points is used, and the number of nodal coordinates is increased by using higher derivatives of the displacement variables. This leads to an increase in the number of degrees of freedom of each nodal point. While the use of higher-order elements leads to an increase in the dimensions of the element matrices and consequently larger number of calculations are required, it has the advantage that fewer higher-order elements are needed in order to obtain the same degree of accuracy. Furthermore, the use of higher-order elements produces more accurate results in applications where the gradient of the displacements cannot be approximated by low-order polynomials. Illustrative Example To demonstrate the development of the kinematic and dynamic equations of higher-order elements, we consider the truss element shown in Fig. 5.10. This element has three nodal points, and each node has one degree of freedom that represents the axial displacement at this node. The second node is assumed to be centrally located between the end points of the element. Since this element has three degrees of freedom, a quadratic interpolating polynomial with three coefficients can be used. This polynomial can be written as u j ¼ a1j þ a2j x þa3j x2 , which can also be written as 2  uj ¼ 1

x

a1j

3

6 j 7 7 x2 6 4 a2 5 j a3

ð5:87Þ

where a1j , a2j , and a3j are three coefficients which can be determined using the conditions 9 u j ðx ¼ 0Þ ¼ u1j > > = j j u ðx ¼ l=2Þ ¼ u2 > > ; u j ðx ¼ lÞ ¼ u3j where l is the length of the truss element.

Fig. 5.10 Three-node truss element

ð5:88Þ

5.8 Higher-Order Elements

319

Substituting Eq. 5.88 into Eq. 5.87, one obtains 2

u1j

3

2

1

32 j 3 a1 7 76 j7 2 l =4 56 4 a2 5 l2 a3j

0

6 j7 6 6u 7 ¼ 41 4 25 1 u3j

0

l=2 l

This equation can be solved for the coefficients a1j , a2j , and a3j in terms of the nodal coordinates u1j , u2j , and u3j as 2

a1j

3

2

1

32 j 3 u1 7 76 6 1=l 54 u2j 7 5 j 2=l2 u3

0

6 j7 6 6 a 7 ¼ 4 3=l 4 25 2=l2 a3j

0

4=l 4=l2

Substituting this equation into Eq. 5.87, the displacement field of the element can be written in terms of the nodal coordinates as 2

1

 6 u j ¼ 1 x x2 4 3=l 2=l

2

0

32 j 3 u1 7 76 j7 1=l 56 4 u2 5 2=l2 u3j 0

4=l 4=l2

This equation leads to u j ¼ N 1 u1j þ N 2 u2j þ N 3 u3j

ð5:89Þ

where 9 N 1 ¼ ð1  2ξÞð1  ξÞ > =

N 2 ¼ 4ξð1  ξÞ

ð5:90Þ

> ;

N 3 ¼ ξð1  2ξÞ

in ξ ¼ x/l. The functions Ni, i ¼ 1,2,3 satisfy the properties N 1 þ N 2 þ N 3 ¼ Pwhich 3 N ¼ 1 and Ni ¼ 1 at node i and is equal to zero at all nodes other than i.  i i¼1 Ð l  2 The kinetic energy of the three-node truss element is T j ¼ 0 ρA u_ j dx =2, where ρ and A are, respectively, the mass density and cross-sectional area of the element, and u_ j ¼ N 1 u_ 1j þ N 2 u_ 2j þ N 3 u_ 3j . Substituting into T j yields T j ¼ q_ jT M j q_ j =2,  T where q j ¼ u1j u2j u3j , and M j is the 3  3 element mass matrix defined as

Mj ¼

ðl 0

2

N 21

6 ρA4 N 2 N 1

N1N2 N1N3 N 22

N3N1 N3N2

3

7 N 2 N 3 5dx ¼ m N 23

ð1 0

2

N 21

6 4 N2N1

N1N2 N1N3 N 22

N3N1 N3N2

3

7 N 2 N 3 5dξ N 23

ð5:91Þ

320

5

The Finite Element Method

where m is the total mass of the element. Upon carrying out the integrations in Eq. 5.91, it can be shown that the element mass matrix M j is given by 2 3 4 2 1 m 25 ð5:92aÞ M j ¼ 4 2 16 30 1 2 4 The strain energy of the three-node truss element is given by Ð l  0 2 U j ¼ 0 EA uj dx =2, in which (0 ) denotes differentiation with respect to x and 0

uj ¼ N 01 u1j þ N 02 u2j þ N 03 u3j. Substituting into U j, the strain energy of the three-node truss element can be written as U j ¼ q jT K j q j =2, where K j is the 3  3 element stiffness matrix defined as 2 3 2 3 N 0 21 N 01 N 02 N 01 N 03 7 8 1 ðl 6 7 EA 6 7 0 0 02 0 0 7 ð5:92bÞ K j ¼ EA6 4 N 2 N 1 N 2 N 2 N 3 5dx ¼ 3l 4 8 16 8 5 0 1 8 7 N 03 N 01 N 03 N 02 N 0 23 Consider the case of a rod with one end fixed and the other end free. If we assume that only one three-node truss element is used, the system has two degrees of freedom and the eigenvalue problem is given by KA ¼ ω2MA, where A is the vector of amplitudes, ω is the frequency, and M and K are defined in this case as  m 8 M¼ 15 1

 1 , 2

 EA 16 K¼ 3l 8

8 7



Therefore, the characteristic equation of the system can be written as



16  8β 8  β



8  β 7  2β ¼ 0 in which β ¼ (ml/5EA)ω2. It follows that (16  8β)(7  2β)  (8 + β)2 ¼ 0, which can be written as 15β2  104β + 48 ¼ 0. This quadratic equation has two roots β1 ¼ 0.497193 and 2 ¼ 6.43614, which define the first two natural frequencies pβffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi of the system as ωi ¼ 5EAβi =ml, i ¼ 1,2. That is, sffiffiffiffi rffiffiffiffiffiffi EA 1:57669 E ω1 ¼ 1:57669 ¼ ml l ρ sffiffiffiffi rffiffiffiffiffiffi EA 5:67280 E ω2 ¼ 5:67280 ¼ ml l ρ Comparing these results with the results obtained in the preceding section using the two-node truss element, it is clear that the three-node truss element gives better accuracy for the fundamental natural frequency. This is an expected result since higher-order interpolating polynomials and more degrees of freedom are used in the case of the three-node truss element as compared to the two-node truss element.

5.9 Spatial Elements

321

Fig. 5.11 Higher-order triangular element

While only the three-node truss element is discussed in this section as an example, other higher-order elements can be developed using a similar procedure. For example, in the case of a four-node truss element, a cubic interpolating polynomial with four coefficients must be used. The element in this case has four degrees of freedom and, consequently, the resulting mass and stiffness matrices are 4  4. Other examples of higher-order elements are the six-node and the ten-node triangular elements shown in Fig. 5.11. The six-node triangular element has 12 degrees of freedom. The displacement field of this element is approximated using a quadratic polynomial, that is u j ¼ a1j þ a2j x þ a3j y þ a4j x2 þ a5j xy þ a6j y2

)

v j ¼ a7j þ a8j x þ a9j y þ a10j x2 þ a11j xy þ a12j y2

ð5:93Þ

The ten-node triangular element has 18 degrees of freedom. The displacement field within this element is approximated using cubic polynomials. In this case, the displacements u j and v j are defined by )

u j ¼ a1j þ a2j x þ a3j y þ a4j x2 þ a5j xy þ a6j y2 þ a7j x3 þ a8j x2 y þ a9j xy2 þ a10j y3 v j ¼ a11j þ a12j x þ a13j y þ a14j x2 þ a15j xy þ a16j y2 þ a17j x3 þ a18j x2 y þ a19j xy2 þ a20j y3

ð5:94Þ An internal node (node 10) is introduced for this element in order to have complete cubic polynomials.

5.9

Spatial Elements

The procedure for developing the kinematic and dynamic equations for spatial elements is very similar to that of the planar elements. There are slight differences in the analytical development of the spatial elements as compared to the planar elements. While in planar elements, the displacement of a nodal point is at most two-dimensional, axial, and transverse displacements, in spatial elements the displacement of a nodal point can be three-dimensional. Furthermore, bending and extension of the threedimensional element in more than one plane (as well as torsion) are possible. Therefore, in general, using the same order of approximation, three-dimensional elements have a larger number of nodal coordinates as compared to planar elements.

322

5

The Finite Element Method

Fig. 5.12 Spatial transformation

Another difference between the spatial and planar elements lies in the transformation of coordinates from the element coordinate system to the global coordinate system. In planar elements, the transformation depends only on one independent angle. This transformation is given by Eq. 5.40. In the spatial analysis, however, three independent orientational coordinates must be used. Figure 5.12 shows the coordinate system X jY jZ j of the element j. The orientation of this coordinate system in the XYZ global coordinate system can be described using the direction cosines. Let i j, j j, and k j be unit vectors along the axes X j, Y j, and Z j of the element, and let i, j, and k be unit vectors along the global axes X, Y, and Z, respectively. Let β1j be the angle between X j and X, β2j be the angle between X j and Y, and β3j be the angle between X j and Z. The components of the unit vector i j along the X-, Y-, and Z-axes are given, respectively, by 9 α11 ¼ cos β1j ¼ i j  i > > = α12 ¼ cos β2j ¼ i j  j > > ; α13 ¼ cos β3j ¼ i j  k

ð5:95Þ

where α11, α12, and α13 are the direction cosines of the X j-axis with respect to the X-, Y-, and Z-axes, respectively. In a similar manner, the direction cosines α21, α22, and α23 of the axis Y j and the direction cosines α31, α32, and α33 of the axis Z j can be defined as α21 ¼ j j  i, α22 ¼ j j  j, α23 ¼ j j  k and α31 ¼ k j  i, α32 ¼ k j  j, α33 ¼ k j  k. Since the direction cosines αij represent the components of the unit vectors i j, j j, and k j along the axes X, Y, and Z, one has 9 i j ¼ α11 i þ α12 j þ α13 k = > j j ¼ α21 i þ α22 j þ α23 k > ; k j ¼ α31 i þ α32 j þ α33 k

ð5:96Þ

Consider the vector u whose components in the element coordinate system are  In the global coordinate system the components of the vector denoted as u, v, and w. u are denoted as u, v, and w. The vector u can, therefore, have the following two

5.9 Spatial Elements

323

 j or u ¼ ui + vj + wk. Using these different representations: u ¼ ui j þ vj j þ wk equations and Eq. 5.96, one obtains    i u ¼ α11 u þ α21 v þ α31 w   þ α12 u þ α22 v þ α32 w j   þ α13 u þ α23 v þ α33 w k By comparing this equation with u ¼ ui + vj + wk, one concludes  u ¼ α11 u þ α21 v þ α31 w  v ¼ α12 u þ α22 v þ α32 w  w ¼ α13 u þ α23 v þ α33 w That is, the relationship between the global and local components can be written in the following matrix form 2 3 2 u α11 6 7 6 4 v 5 ¼ 4 α12

α21 α22

α13

α23

w

32 3 u α31 76 7 α32 54 v 5  w α33

ð5:97Þ

from which the transformation matrix Cj can be recognized as 2

α11 6 j C ¼ 4 α12

α21 α22

3 α31 7 α32 5

α13

α23

α33

ð5:98Þ

This transformation matrix is expressed in terms of nine direction cosines αij, i, j ¼ 1,2,3. These nine components, however, are not independent. They are related by the six algebraic equations αk1 αl1 þ αk2 αl2 þ αk3 αl3 ¼ δkl ,

k,l ¼ 1,2,3

ð5:99Þ

where δkl is the Kronecker delta, that is,  δkl ¼

1 if k ¼ 1 0 if k 6¼ 1

ð5:100Þ

Therefore, there are only three independent direction cosines in the transformation matrix of Eq. 5.98. This transformation matrix can be used to define the nodal coordinates of the three-dimensional element in the global coordinate system. In the remainder of this section, we discuss the assumed displacement field of some of the three-dimensional elements. The development of the mass and stiffness matrices of these elements is left as an exercise, since a similar procedure to the one previously discussed in this chapter can be used. The mass matrix can be defined using the kinetic energy expression, while the stiffness matrix can be defined using the strain energy expression and the general elasticity relationships presented in this

324

5

The Finite Element Method

Fig. 5.13 Beam element

chapter. In the case of beam and plate elements, the classical beam and plate theories may be used in order to simplify the elasticity equations. Beam Element Figure 5.13 shows an example of a three-dimensional beam element. This element has two nodal points, one located at each end. Each node has six degrees of freedom ui j , vi j , wi j , θxij , θyij , θzij , i ¼ 1,2, where ui j , vi j , and wi j are the translational displacements of the node i, and θxij , θyij , and θzij are its rotations about the three perpendicular axes of the element. Therefore, the element has 12 degrees of freedom which are defined as h q j ¼ u1j

v1j

w1j

θxj1

θyj1

θzj1

u2j

v2j

w2j

θxj2

θyj2

θzj2

iT

ð5:101Þ

The shape function matrix S j of this element is 2

S jT

1ξ   6 ξ  ξ2 η   6 ξ  ξ2

6 6 6 6 6 6 0 6 6 6 ð1  4ξ þ 3ξÞlζ 6  6 6 1 þ 4ξ  3ξ2 lη 6 ¼6 ξ 6   6 6 6 ξ þ ξ2 η 6 6 6ξ þ ξ2 ζ 6 6 6 0 6   6 4 2ξ þ 3ξ2 lζ   2ξ  3ξ2 lη

0

0

3

7 7 7 0 1  3ξ2 þ 2ξ3 7 7 7 ð1  ξÞlζ ð1  ξÞlη 7   7 2 3 7 0 ξ þ 2ξ  ξ l 7   7 7 ξ  2ξ2 þ ξ3 l 0 7 7 0 0 7 7 2 3 7 3ξ  2ξ 0 7 7 2 3 0 3ξ  2ξ 7 7 7 lξζ lξη 7 7 0 ξ2  ξ3 5  2  3 ξ þ ξ l 0 1  3ξ2 þ 2ξ3

0

where ξ ¼ x/l, η ¼ y/l, and ζ ¼ z/l, and l is the length of the beam element.

ð5:102Þ

5.9 Spatial Elements

325

Fig. 5.14 Solid element

Solid Element Figure 5.14 shows the eight-node solid or brick element. This element has 24 degrees of freedom which represent the displacements of the nodes along three perpendicular axes. The shape function of the brick element may be derived using the interpolating polynomials u j ¼ a1j þ a2j x þ a3j y þ a4j z þ a5j xy þ a6j yz þ a7j zx þ a8j xyz v j ¼ a9j þ a10j x þ a11j y þ a12j z þ a13j xy þ a14j yz þ a15j zx þ a16j xyz w ¼ j

a17j

þ

a18j x

þ

a19j y

þ

a20j z

þ

a21j xy

þ

a22j yz

þ

a23j zx

þ

9 > > = > >

; a24j xyz

ð5:103Þ

The 24 nodal coordinates are  q j ¼ u1j v1j w1j u2j v2j w2j u3j v3j w3j u4j v4j w4j u5j v5j w5j ð5:104Þ  u6j v6j w6j u7j v7j w7j u8j v8j w8j T By using Eqs. 5.103 and 5.104, one can show that the shape function matrix of the solid element is given by 2

N1 S ¼4 0 0 j

where

0 N1 0

0 0 N1

N2 0 0

0 N2 0

0 0 N2

. . . N8 ... 0 ... 0

0 N8 0

3 0 0 5 N8

ð5:105Þ

326

5

1 N 1 ¼ ð1  ξÞð1  ηÞð1  ζ Þ, 8 1 N 2 ¼ ð1  ξÞð1 þ ηÞð1  ζ Þ, 8 1 N 3 ¼ ð1  ξÞð1 þ ηÞð1 þ ζ Þ, 8 1 N 4 ¼ ð1  ξÞð1  ηÞð1 þ ζ Þ, 8

The Finite Element Method

9 1 > N 5 ¼ ð1 þ ξÞð1  ηÞð1  ζ Þ > > > 8 > > > > 1 > N 6 ¼ ð1 þ ξÞð1 þ ηÞð1  ζ Þ > = 8 > 1 > N 7 ¼ ð1 þ ξÞð1 þ ηÞð1 þ ζ Þ > > > 8 > > > > 1 > N 8 ¼ ð1 þ ξÞð1  ηÞð1 þ ζ Þ ; 8

ð5:106Þ

in which ξ ¼ x/a, η ¼ y/b, and ζ ¼ z/c, where 2a, 2b, and 2c are, respectively, the dimensions of the solid element in the X-, Y-, and Z-directions. Tetrahedral Element A typical constant strain tetrahedral element is shown in Fig. 5.15. The element has straight sides with four nodes, one at each corner. Each node has three nodal coordinates that represent the translational displacements of the nodes along three perpendicular axes. Therefore, the element has 12 degrees of freedom. The displacement field of the element may be described using the following interpolating polynomials: u j ¼ a1j þ a2j x þ a3j y þ a4j z v j ¼ a5j þ a6j x þ a7j y þ a8j z wj ¼

a9j

þ

a10j x

þ

a11j y

þ

9 > > = > >

; a12j z

ð5:107Þ

The vector of nodal coordinates of the tetrahedral element is  T q j ¼ u1j v1j w1j u2j v2j w2j u3j v3j w3j u4j v4j w4j

ð5:108Þ

By using these nodal coordinates and the assumed displacement field of Eq. 5.107, it can be shown that the element shape function matrix is

Fig. 5.15 Four-node tetrahedral element

5.9 Spatial Elements

2

N1 Sj ¼ 4 0 0

0 N1 0

327

0 0 N1

N2 0 0

0 N2 0

0 0 N2

N3 0 0

0 N3 0

0 0 N3

3 0 0 5 ð5:109Þ N4

N4 0 0 N4 0 0

where the scalars Ni, i ¼ 1,2,3,4 are 9 N 1 ðx; y; zÞ ¼ C11 þ C 21 x þ C31 y þ C 41 z > > > > N 2 ðx; y; zÞ ¼ C12 þ C 22 x þ C32 y þ C 42 z =

ð5:110Þ

N 3 ðx; y; zÞ ¼ C13 þ C 23 x þ C33 y þ C 43 z > > > > ; N 4 ðx; y; zÞ ¼ C14 þ C 24 x þ C34 y þ C 44 z in which

x2 1

C 11 ¼

x3 6V j

x 4

x1 1

C 13 ¼

x2 6V j

x 4

1 1

C 21 ¼  j 1 6V

1

1 1

C 23 ¼  j 1 6V

1

1 1

C 31 ¼

1 6V j

1

1 1

C 33 ¼

1 6V j

1

1 1

C 41 ¼  j 1 6V

1

1 1

C 43 ¼  j 1 6V

1

y2 y3 y4 y1 y2 y4 y2 y3 y4 y1 y2 y4 x2 x3 x4 x1 x2 x4 x2 x3 x4 x1 x2 x4

z2



z3 ,

z4

z1



z2 ,

z4

z2



z3 ,

z4

z1



z2 ,

z4

z2



z3 ,

z4

z1



z2 ,

z4

y2



y3 ,

y4

y1



y2 ,

y

4



x1 1

C12 ¼  j x3 6V

x 4

x1 1

C14 ¼  j x2 6V

x 3

1 1

C 22 ¼

1 6V j

1

1 1

C 24 ¼

1 6V j

1

1 1

C 32 ¼  j 1 6V

1

1 1

C 34 ¼  j 1 6V

1

1 1

C 42 ¼

1 6V j

1

1 1

C 44 ¼

1 6V j

1

and V j is the volume of the tetrahedral element defined as

y1 y3 y4 y1 y2 y3 y1 y3 y4 y1 y2 y3 x1 x3 x4 x1 x2 x3 x1 x3 x4 x1 x2 x3

z1



z3

z4

z1



z2

z3

z1



z3

z4

z1



z2

z3

z1



z3

z4

z1



z2

z3

y1



y3

y4

y1



y2

y

3

328

5



1

1 1 j V ¼

6 1

1

x1 x2 x3 x4

y1 y2 y3 y4

z1

z2

: z3



z4

The Finite Element Method

ð5:111Þ

Plate Element Figure 5.16 shows a plate element that has four nodal points, one at each corner. Each nodal point has five degrees of freedom, three displacements along three perpendicular directions, and two rotations about the X j-and Y j-axes. This plate element, therefore, has 20 nodal coordinates which can be written in vector form as  q j ¼ u1j v1j w1j θxj1 θyj1 u2j v2j w2j θxj2 θyj2 ð5:112Þ  u3j v3j w3j θxj3 θyj3 u4j v4j w4j θxj4 θyj4 T The extension of the plate (membrane effect) may be described using the interpolating polynomials of the planar bilinear rectangular element, that is, ) u j ¼ a1j þ a2j x þ a3j y þ a4j xy ð5:113Þ v j ¼ a5j þ a6j x þ a7j y þ a8j xy By using these interpolating polynomials and the shape functions of plate bending presented by Prezemieniecki (1968), a plate element shape function matrix that accounts for the membrane and bending effect can be defined as 2

S jT

ð1  ξÞð1  ηÞ 6 0 6 6 0 6 6 6 0 6 6 0 6 6 ð1  ξÞη 6 6 0 6 6 6 0 6 6 0 6 6 0 6 ¼6 6 ξη 6 6 0 6 6 0 6 6 0 6 6 6 0 6 6 ξð1  ηÞ 6 6 0 6 6 0 6 6 4 0 0

0 ð1  ξÞð1  ηÞ 0 0 0 0 ð1  ξÞη 0 0 0 0 ξη 0 0 0 0 ξð1  η Þ 0 0 0

3 0 7 0 7 2 27 ð1 þ 2ξÞð1  ξÞ ð1 þ 2ηÞðηÞ 7 7 ð1 þ 2ξÞð1  ξÞ2 ð1  η2 Þηb 7 7 ξað1  ξÞ2 ð1 þ 2ηÞð1  ηÞ2 7 7 7 0 7 7 0 7 7 ð1 þ 2ξÞð1  ξÞ2 ð3  2ηÞη2 7 7 ð1 þ 2ξÞð1  ξÞ2 ð1  ηÞη2 b 7 7 7 ξð1  ξÞ2 ð3  2ηÞη2 a 7 7 7 0 7 7 0 7 2 2 7 ð3  2ξÞξ ð3  2ηÞη 7 7 2 2 ð3  2ξÞξ ð1  ηÞη b 7 7 2 2 7 ð1  ξÞξ ð3  2ηÞη a 7 7 0 7 7 0 7 7 2 ð3  2ξÞξ2 ð1 þ 2ηÞð1  ηÞ 7 7 5 ð3  2ξÞξ2 ð1  ηÞ2 ηb ð1  ξÞξ2 ð1 þ 2ηÞð1  ηÞ2 a ð5:114Þ

5.10

Large Rotations and Deformations

329

Fig. 5.16 Plate element

where ξ ¼ x/a and η ¼ y/b, and a and b are, respectively, the dimensions of the plate element along the X j and Y j element axes as shown in Fig. 5.16.

5.10

Large Rotations and Deformations

The conventional finite elements discussed so far in this chapter either employ a low order of interpolation that does not ensure continuity of the rotations, strains, and stresses at the nodal points or employ rotations as element degrees of freedom. These elements have geometry and linearization issues that have been discussed in the literature. Rectangular, triangular, solid, and tetrahedral elements, in particular, are referred to as isoparametric elements, since both the position and the deformation of these elements can be described using the same shape function matrix. This is mainly due to the fact that the shape function matrix and the nodal coordinates of these isoparametric elements can describe an arbitrary rigid-body displacement as demonstrated in Sect. 5.2 of this chapter. In the classical FE literature, beams and plates are not considered as isoparametric elements because the displacement gradients at the nodes are used to describe infinitesimal rotations. Such a linearization of the slopes leads to incorrect rigid-body equations of motion when the beams and plates rotate as rigid bodies (Shabana 1996A). Such a difficulty can be circumvented if no infinitesimal or finite rotations are used as nodal coordinates. Absolute Nodal Coordinate Formulation In this section, a simple procedure referred to as the absolute nodal coordinate formulation (ANCF) is discussed. This formulation, in which no infinitesimal or finite rotations are used as nodal coordinates, can be used in many large deformation applications. In this formulation, the nodal coordinates are defined in the inertial frame in terms of absolute nodal positions and position gradients, and, as a consequence, no coordinate

330

5

The Finite Element Method

transformations are required to account for the large displacements. ANCF finite elements lead to a constant mass matrix, and as a consequence, an efficient procedure can be used for solving the nodal accelerations. In the absolute nodal coordinate formulation, the beam and plate elements can be treated as isoparametric elements since their shape functions and nodal coordinates can describe an arbitrary rigidbody displacement. Furthermore, no distinction is made between plate and shell elements since shell geometries can be systematically described using ANCF plate elements by a proper selection of the nodal coordinates in the reference stress-free configuration (Pappalardo et al. 2017, 2018). The position gradients at the nodes can be determined in the undeformed reference configuration using simple rigid-body kinematics. In this section, we briefly describe the absolute nodal coordinate formulation, which can be used in the large rotation and deformation analysis of beam, plate, and shell structure applications, such as the vibrations of cables and flexible space antennas. For simplicity, in the development presented in this section, we drop the superscript j, which refers to the element number. ANCF Element Equations In the absolute nodal coordinate formulation (Shabana 2018A), the coordinates of the element are defined in an inertial coordinate system, and as such no time-varying coordinate transformation is required for the elements. Furthermore, no infinitesimal or finite rotations are used as nodal coordinates, and instead slopes that are obtained as the derivatives of the position field are used. In this case, the FE matrix equation of motion can be written in the simple form M€e þ Ke ¼ Q, where M is the element mass matrix, K is the element stiffness matrix, e is the vector of the element nodal coordinates, and Q is the vector of generalized nodal forces. In the ANCF equations, the element mass matrix is constant, and it is the same matrix that appears in linear structural dynamics. The element stiffness matrix, on the other hand, is a nonlinear function of the nodal coordinates even in the case of small deformations. Slopes and Rigid-Body Kinematics In order to demonstrate the use of the absolute nodal coordinate formulation in the analysis of rigid-body motion, we consider the uniform slender beam shown in Fig. 5.17. The beam has length l, cross-sectional area A, mass density ρ, volume V, and mass m. The coordinate system of this beam element is assumed to be initially attached to its left end, which is defined by point O as shown

Fig. 5.17 Absolute nodal coordinate formulation

5.10

Large Rotations and Deformations

331

in the figure. Even though, in the absolute nodal coordinate formulation, the displacements in the orthogonal directions can be interpolated using the same polynomials, we use in this section, for the purpose of demonstration, the shape function matrix of a planar beam element that was defined in Sect. 5.1 of this chapter as " S¼

1ξ

0

0

1  3ξ2 þ 2ξ3

ξ

0

0

3ξ2  2ξ3

0

  l ξ  2ξ2 þ ξ3

0

#

  l ξ3  ξ2 ð5:115Þ

where ξ ¼ x/l. The vector of nodal coordinates associated with the shape function of Eq. 5.115 is e ¼ [e1 e2 e3 e4 e5 e6]T, where e1 and e2 are the translational coordinates at the node at O, e4 and e5 are the translational coordinates of the node at B, and e3 and e6 are the slopes at the two nodes. An arbitrary rigid-body displacement of the beam is defined by the translation R ¼ [Rx Ry]T of the reference point O and a rigid-body rotation θ. As the result of this arbitrary rigidbody displacement, the vector of nodal coordinates e can be defined in the global coordinate system as  e ¼ Rx Ry sin θ

Rx þ l cos θ

Ry þ l sin θ

sin θ

T

ð5:116Þ

The slopes in the preceding equation are defined using simple rigid-body kinematics, since in the case of a rigid-body motion, the location of an arbitrary point on the element as the result of the translation R and the rotation θ can be defined as  r¼

rx ry



 ¼

Rx þ x cos θ Ry þ x sin θ



It follows that ∂ry/∂x ¼ sin θ. It can be demonstrated that the element shape function matrix of Eq. 5.115 and the vector of element nodal coordinates can describe exactly the rigid-body motion of the element if the slopes are used instead of the infinitesimal rotations. This is clear from the following equation: 

Rx þ x cos θ Se ¼ Ry þ x sin θ

 ð5:117Þ

which is the same as the vector r previously determined using simple rigid-body kinematics. The preceding equations clearly demonstrate that the element shape function matrix and the nodal coordinates can describe an arbitrary rigid-body displacement provided that the coordinates are defined in the inertial frame and the slopes are defined in terms of trigonometric functions. Therefore, the conventional FE shape function matrix can describe exactly a rigid-body displacement. Using the slopes as nodal coordinates, the element kinetic energy takes the simple form T ¼ e_ T M_e=2, whereÐM is the constant mass matrix that was previously defined in this chapter as M ¼ ρST SdV. V

332

5

The Finite Element Method

Strain Energy While in this section only the case of small deformations, often used in the linear vibration theory, is considered for simplicity, ANCF element can be used to solve large deformation problems. One needs only to change the form of the stiffness matrix, which is nonlinear even in the case of small deformations. If point O on the beam element is selected as the reference point, the components of the relative displacement of an arbitrary point with respect to point O can be defined in the inertial coordinate system as  u¼

ux uy



 ¼

ðS1  S1O Þe ðS2  S2O Þe

 ð5:118Þ

where S1 and S2 are the rows of the element shape function matrix, and S1O and S2O are the rows of the shape function matrix defined at the reference point O. In order to define the longitudinal and transverse displacements of the beam, the unit vector i along a selected beam axis is first defined as  T rB  r O i ¼ ix iy ¼ jrB  rO j

ð5:119Þ

where rO and rB are the global position vectors of the two end points of the beam. A unit vector j perpendicular to i can be obtained as j ¼ [jx jy]T ¼ k  i, where k is a unit vector along the Z axis. Therefore, the longitudinal and transverse deformations of the beam can be defined as  ud ¼

ul ut



 ¼

   ux i x þ uy i y  x uT i  x ¼ ux jx þ uy jy uT j

ð5:120Þ

If a linear elastic model is assumed, a simple expression for the strain energy U can be written as !2 1  2 2 1 @ ∂ul ∂ ut A dx EA þ EI U¼ 2 0 ∂x ∂x2 ðl

0

ð5:121Þ

where E is the modulus of elasticity and I is the second moment of area. It can be shown that the use of ANCF elements produces zero deformation in the case of an arbitrary rigid-body motion. Using the simple definition of the strain energy presented in the preceding equation, one can show that the stiffness matrix in the case of a linear elastic vibration model is nonlinear. In the case of large deformation analysis, another expression for the strain energy that is based on the nonlinear strain-displacement relationships must be used. Since the governing equations in the case of small and large deformation analyses are nonlinear and must be solved numerically, there is little to be gained by using the small strain assumptions.

5.11

Sub-structuring Techniques

333

Fig. 5.18 Crude oil sloshing in railroad vehicle braking scenario (Grossi and Shabana 2018A, B). (1-tank, 2-stub sill, 3-bolster, 4-wheelset, 5-frame, 6-equalizer, 7-rail, 8-coupler, 9-crude oil)

Application to Modeling Complex Nonlinear Systems ANCF elements can be used to model the oscillations of complex physics and engineering systems. A complex vibration problem which can be studied using ANCF elements is liquid sloshing in moving tank cars or containers as in the case of transporting hazardous materials (HAZMAT) by rail. If the frequency of the liquid sloshing loads is close to one of the fundamental natural frequencies of the vehicle, the vehicle structural components can fail, resulting in deadly and environmentally damaging accidents. As discussed in Chap. 1, the two basic elements which define a typical vibration problem are the inertia and the restoring force. Liquid sloshing is not a typical vibration problem because fluids do not have restoring forces; the fluid dynamic behavior is governed by viscous forces defined using Navier–Stokes equations. Wei et al. (2015) developed an ANCF liquid sloshing formulation applicable to Newtonian fluids. This approach was validated for fluid modeling applications against experimental and numerical results (Grossi and Shabana 2018B). However, most hazardous materials, such as crude oil, exhibit non-Newtonian behavior which cannot be properly analyzed using linear constitutive equations. An algorithm that allows capturing the sloshing effects of highly viscous non-Newtonian crude oils was proposed (Grossi and Shabana 2018A). The effect of crude oil sloshing loads on railroad vehicle dynamics was investigated in different motion scenarios including a braking scenario as the one shown in Fig. 5.18. The simulation results show that crude oil sloshing plays an important role in defining vehicle stability, coupler forces, as well as loads on bogies and rails. Figure 5.19 compares the coupler force results obtained in the case of transporting heavy, medium, and light crude oils using virtual prototyping techniques.

5.11

Sub-structuring Techniques

Developing flexible body models for mechanical and aerospace structures requires the formulation of the equations of motion of the deformable components that have infinite number of degrees of freedom. As discussed in this book, infinite-dimension

334

Fig. 5.19 Coupler forces in the cases of heavy (

5

), medium (

The Finite Element Method

), and light (

) crude oils

spaces are not suitable for computational algorithms, and therefore, the FE approximation methods are often used. In the FE methods, as discussed in this chapter, the bodies or structures are divided into small regions, called elements, connected at nodal points, each of which has a finite number of nodal degrees of freedom. For large-scale systems, such as vehicles and airplanes, the resulting FE models can have millions of degrees of freedom. In the early 1960s and 1970s of the past century, the computer power was very limited, in both speed and array space. This motivated researchers to propose sub-structuring techniques that allow dividing the structure into sub-structures, reduce the sub-structure dimension, and reassemble the sub-structures to obtain a reduced-order model that can be solved efficiently with reasonable array space requirements. For example, an airplane model can be divided into four separate sub-structures: the fuselage sub-structure, two wing sub-structures, and tail sub-structure. Different computer-aided engineering (CAE) groups can work independently to develop different FE meshes for the sub-structures and use sub-structuring techniques to reduce the number of degrees of freedom of each sub-structure. The reduced-order model of the airplane can then be obtained by assembling the reduced-order models of the fuselage, wings, and tail at nodes called the interface nodes. Craig–Bampton Sub-structuring Technique The need for such computational efficiency and collaboration between research and design groups within an industry on modeling complex systems was one of the main reasons that motivated the development of sub-structuring techniques. Because the mesh of each sub-structure can be developed independently and coordinate reduction techniques can be applied in order to reduce the dimensionality of the sub-structure model, a

5.11

Sub-structuring Techniques

335

reduced-order FE mesh of the structure can be obtained by assembling its reducedorder sub-structure models at the interface nodes. A widely used sub-structuring technique is the Craig–Bampton (CB) method (Craig Jr. and Bampton 1968), in which a coordinate partitioning is employed to separate the sub-structure elastic coordinates into boundary and internal coordinates. Using static condensation, constraints are imposed on selected boundary nodes in order to determine a set of modes called static correction modes. By applying constraints on the boundary nodes, an eigenvalue problem associated with the internal degrees of freedom can be identified to define a set of fixed-interface modes. The combination of the static correction and the fixed-interface modes defines the CB transformation. Craig and Bampton (1968), in their original paper, explained clearly the goal of their sub-structuring approach, which is to obtain a reduced-order model by reducing the dimensions of the sub-structures. Craig and Bampton demonstrated the use of their sub-structuring method to obtain accurate solutions, and such a sub-structuring technique does not alter the problem to be solved and the boundary conditions or leads to a different or improved solution. They showed that their approach, if correctly applied, leads to a solution that converges to the unique solution of the original problem (O’Shea et al. 2018). Coordinate Reduction As discussed in this book, when coordinate reduction techniques are used, the resulting coordinate transformation matrix is non-square. Nonetheless, the accuracy of the approximate solution, regardless of the coordinate reduction technique used, should be measured against the solution obtained when the square transformation is used. The reduced-order model solution is judged accurate if it converges to the solution obtained without coordinate reduction. Consider, for example, the original problem defined by the mathematical model M€ qf þ Kqf ¼ Q. In sub-structuring methods, the coordinate vector qf is written in terms of another coordinate vector p as qf ¼ Bp, where B is a coordinate transformation matrix. Substituting the transformation qf ¼ Bp into the equation qf þ Kq by the  T M€  fT ¼ Q and pre-multiplying T € þ B KB p ¼ B Q. In the case of a transpose of the matrix B leads to B MB p square transformation matrix B, one can also pre-multiply by the inverse of B because linear combinations of the equations of motion produce valid equations (O’Shea et al. 2018). In this case, the similarity transformation B1MB should not alter or improve the solution of the original problem. techniques are used, B is no longer a  If sub-structuring    € þ BT KB p ¼ BT Q is used and leads to a square matrix, in this case BT MB p reduced-order model. Nonetheless, the solution resulting from using sub-structuring techniques is judged accurate if it converges to the solution obtained using the square transformation. Craig–Bampton Transformation As an example of sub-structuring methods, the Craig–Bampton (CB) method is considered in this section. As in other sub-structuring methods, the CB sub-structuring technique allows dividing large and complex structures into smaller sub-structures, reduce the dimension of each sub-structure, and assemble the reduced-order sub-structure models to obtain a lower-order structure model. The free vibration of the structure is governed by the

336

5

The Finite Element Method

equation M€ qf þ Kqf ¼ 0. The vector of coordinates qf is portioned into boundary  T coordinates (qf)b and internal coordinates (qf)i such that qf ¼ ðqf ÞbT ðqf ÞiT and rewrite the equation M€ qf þ Kqf ¼ 0 using this coordinate partitioning as "

ðMÞbb ðMÞib

#"   # " €f b q ðKÞbb   þ ðMÞii ðKÞib €f i q

ðMÞbi

ðKÞbi

#"

ðKÞii

ðqf Þb

#

ðqf Þi

¼

" # 0 0

ð5:122Þ

By neglecting the effect of the inertia forces, the second matrix equation in the preceding equation leads to (K)ib(qf)b + (K)ii(qf)i ¼ 0. This equation, upon assuming that (K)ii is nonsingular and upon employing the technique of static condensation, can be used to write the coordinates of the internal nodes in terms of the coordinates of the boundary nodes as  ðqf Þi ¼ ðKÞ1 ii ðKÞib ðqf Þb ¼ HCB ðqf Þb

ð5:123Þ

 CB ¼ ðKÞ1 ðKÞ . The preceding equation can be used to write In this equation, H ib ii all the coordinates in terms of the boundary coordinates as qf ¼ HCB(qf)b, where the matrix HCB is " HCB 

I  HCB

#

" ¼

#

I ðKÞ1 ii ðKÞib

ð5:124Þ

The columns of this transformation, which has a number of rows equal to the number of coordinates qf and a number of columns equal to the number of boundary coordinates (qf)b, are called the static correction modes. These static correction modes represent the change in the internal node coordinates as the result of a unit displacement in the boundary node coordinates. In order to develop the Craig–Bampton transformation, the second equation in Eq. 5.122 is used again by considering the free vibration of the internal nodes with respect  to the boundary nodes. In this case, one has the equation €f i þ ðKÞii ðqf Þi ¼ 0, which can be used to define an eigenvalue problem. ðMÞii q This eigenvalue problem defines a number of eigenvectors, called fixed-interface modes, equal to the number of the coordinates of the internal nodes. The fixedinterface modes define the columns of a matrix ΦCB , which when combined with the static correction modes defines the Craig–Bampton transformation " qf ¼

ðqf Þb ðqf Þi

"

# ¼

I

0

HCB

ΦCB

#"

pb pf

# ¼ ΦCB p

ð5:125Þ

In this equation, pb ¼ (qf)b, pf is the vector of fixed-interface modal coordinates, and ΦCB is the Craig–Bampton transformation defined as " ΦCB ¼

I

0

HCB

ΦCB

# ð5:126Þ

Problems

337

In most applications, the insignificant fixed-interface modes are eliminated in order to reduce the number of coordinates. In this case, the Craig–Bampton transformation ΦCB is not a square matrix. Substituting Eq. 5.125 into the equation M€ qf þ Kqf ¼ 0 that defines the free vibration of the structure and pre-multiplying by the transpose of the CB transformation ΦCB, one obtains  T   T  €f þ ΦCB KΦCB pf ¼ 0 ð5:127Þ ΦCB MΦCB p  T  Because  Tin the CB  procedure, the resulting mass and stiffness matrices ΦCB MΦCB and ΦCB KΦCB are not diagonal, some commercial software use the preceding equation to solve for another eigenvalue problem in order to define diagonal modal mass and stiffness matrices (O’Shea et al. 2018). Problems 5.1. The beam element shown in Fig. P5.1 has two nodal points. The coordinates of each node are given by vi and θi, i ¼ 1,2. The element has length l. If the displacement is described using the polynomial v ¼ a1 + a2x + a3x2 + a4x3, obtain the element shape function matrix.

Fig. P5.1

5.2. For the structure shown in Fig. P5.2, obtain the displacement field of each element in terms of the structure nodal coordinates. Assume that the structure is discretized using the truss elements shown in the figure.

Fig. P5.2

5.3. Obtain the displacement field of each truss element of the structure shown in Fig. P5.3 in terms of the nodal coordinates of the structure. Identify the element transformation and connectivity matrices.

338

5

The Finite Element Method

Fig. P5.3

5.4. Using the shape function obtained in Problem 1 for the beam elements, obtain the displacement field of each element of the structure shown in Fig. P5.4 in terms of the structure nodal coordinates. Identify the element transformation and connectivity matrices. Assume that the length of element j is l j.

Fig. P5.4

5.5. Repeat Problem 4 using the shape function matrix of the beam element given in Sect. 5.1 of this chapter. 5.6. Using the beam element of Problem 1, obtain the displacement field of each element expressed in terms of the nodal coordinates of the structure shown in Fig. P5.5. Identify the element transformation and connectivity matrices.

Fig. P5.5

Problems

339

5.7. Repeat Problem 6 using the shape function of the beam element given in Sect. 5.1 of this chapter. 5.8. Obtain the displacement field of the elements of the structure shown in Fig. P5.6 by using the element defined in Problem 1. Assume that the elements have equal lengths. Identify the element transformation matrices as well as the connectivity matrices.

Fig. P5.6

5.9. Repeat Problem 8 using the finite beam element given in Sect. 5.1 of this chapter. 5.10. By using the FE method, obtain the mass and stiffness matrices of the structure of Problem 2. 5.11. Derive the mass and stiffness matrices of the structure of Problem 3. 5.12. Obtain the mass and stiffness matrices of the structure of Problem 4. 5.13. Develop the mass and stiffness matrices of the structure of Problem 5. 5.14. For the structure given in Problem 6, define the structure mass and stiffness matrices. 5.15. For the structure of Problem 7, obtain the structure mass and stiffness matrices. 5.16. Using beam elements, obtain the mass and stiffness matrices of the structure of Problem 8.

340

5

The Finite Element Method

5.17. Define the mass and stiffness matrices of the structure of Problem 9. 5.18. Obtain the generalized forces associated with the structure generalized coordinates as the result of the application of the force F shown in Fig. P5.7. Assume that the structure is discretized using two equal beam elements and the element shape functions are the same as obtained in Problem 1.

Fig. P5.7

5.19. Repeat Problem 18 using the shape function of the beam element given in Sect. 5.1 of this chapter. 5.20. The structure shown in Fig. P5.8 is discretized into three equal truss elements. Obtain the generalized forces associated with the structure generalized coordinates as the result of the application of the force F.

Fig. P5.8

5.21. The structure shown in Fig. P5.9 is discretized using two-dimensional beam elements. The length, mass, and moment of area of element j are denoted, respectively, by l j, m j, and I j. Obtain the generalized forces associated with the structure generalized coordinates as the result of application of the forces F1 and F2. Use the beam element defined in Problem 1.

Problems

341

Fig. P5.9

5.22. Repeat Problem 21 using the two-dimensional beam element defined in Sect. 5.1 of this chapter. 5.23. Using the shape function of the beam element defined in Sect. 5.1 of this chapter, obtain the generalized forces associated with the structure generalized coordinates as the result of applying the forces F1 and F2 on the structure shown in Fig. P5.10.

Fig. P5.10

5.24. Repeat Problem 23 using the beam element defined in Problem 1. 5.25. Derive the matrix differential equations of free vibration of the structure of Problem 2. 5.26. Obtain the differential equations of free vibration of the structure of Problem 6. 5.27. Obtain the equations of free vibration of the system of Problem 7. 5.28. Obtain the equations of motion of the system of Problem 20. 5.29. Derive the differential equations of motion of the system of Problem 21.

342

5

The Finite Element Method

5.30. Derive the differential equations of motion of the system of Problem 22. 5.31. Identify a set of rigid-body modes for the two-dimensional beam element and discuss the orthogonality of these modes with respect to the mass and stiffness matrices. 5.32. Identify a set of rigid-body modes for the two-dimensional triangular element and discuss the orthogonality of these modes with respect to the element mass and stiffness matrices. 5.33. Determine the element shape function matrix of the six-node triangular element. 5.34. Determine the element shape function matrix of the ten-node triangular element. 5.35. Develop the mass and stiffness matrices of the three-dimensional beam element. 5.36. Obtain the mass and stiffness matrices of the solid element. 5.37. Obtain the mass and stiffness matrices of the tetrahedral element. 5.38. Identify a set of rigid-body modes for the three-dimensional beam element and discuss the orthogonality of these modes with respect to the element mass and stiffness matrices. 5.39. Identify a set of rigid-body modes for the solid element and discuss the orthogonality of these modes with respect to the element mass and stiffness matrices.

6

Methods for the Eigenvalue Analysis

In addition to the matrix-iteration method discussed in Chap. 3, there are several other computer methods that are widely used for solving the eigenvalue problem of vibration systems. Among these methods are the Jacobi method and the QR method. In these methods, which are based on the similarity transformation, a series of transformations that convert a given matrix to a diagonal matrix which has the same eigenvalues as the original matrix are used. Not every matrix, however, is similar to a diagonal matrix, and therefore we find it appropriate to devote several sections of this chapter to discuss the similarity transformation before we briefly discuss the computer methods used for solving the eigenvalue problem of vibration systems. Several definitions will be used repeatedly throughout the development presented in this chapter. Some of these definitions are summarized below. Monic Polynomials A monic polynomial is a nonzero polynomial with a leading coefficient equal to one. An example of such a polynomial in λ is f(λ) ¼ λ8  5λ3 + λ2 + 8. The coefficient of the term that has the highest power in λ in this nonzero polynomial is one, and therefore the preceding polynomial is a monic polynomial. The definition of the monic polynomial will be used repeatedly in this chapter, and for this reason we will slightly change the way we write our characteristic matrix for a matrix B to (λI  B) instead of (B  λI). Quotient and Remainder If f(λ) and g(λ) are two polynomials such that g(λ) 6¼ 0, then there exist unique polynomials q(λ) and r(λ) such that f ðλÞ ¼ qðλÞgðλÞ þ r ðλÞ

ð6:1Þ

with r(λ) ¼ 0 or the degree of r(λ) is less than the degree of g(λ). The polynomial q(λ) is called the quotient and r(λ) is called the remainder. If r(λ) ¼ 0, g(λ) is said to be a divisor or a factor of f(λ). We also say that f(λ) is divisible by g(λ). For example, the polynomial f(λ) ¼ λ2 + 6λ + 5 can be written as f(λ) ¼ (λ + 5)(λ + 1). It follows that # Springer Nature Switzerland AG 2019 A. Shabana, Vibration of Discrete and Continuous Systems, Mechanical Engineering Series, https://doi.org/10.1007/978-3-030-04348-3_6

343

344

6

Methods for the Eigenvalue Analysis

the polynomials (λ + 5) and (λ + 1) are divisors or factors of f(λ). If, on the other hand, we consider the polynomial g(λ) ¼ (λ + 2), then f(λ) can be written as f(λ) ¼ (λ + 4)g(λ)  3, where, in this case, the remainder r(λ) ¼ 3, which is a polynomial of a degree less than the degree of g(λ). Greatest Common Divisor A monic polynomial g(λ) is said to be the greatest common divisor of the nonzero polynomials f1(λ) and f2(λ) if the following two conditions are satisfied: 1. g(λ) divides both f1(λ) and f2(λ). 2. If h(λ) divides both f1(λ) and f2(λ), then h(λ) divides g(λ). Least Common Multiple The least common multiple of two polynomials f1(λ) and f2(λ) is a polynomial g(λ) such that: 1. f1(λ) and f2(λ) are factors (divisors) of g(λ). 2. If h(λ) is a polynomial such that f1(λ) and f2(λ) are divisors of h(λ), then g(λ) is a divisor of h(λ).

6.1

Similarity Transformation

Two square matrices B and D are said to be similar if there exists a nonsingular matrix Φ such that B ¼ Φ1 DΦ

ð6:2Þ

It is clear that if B is similar to D, then D is similar to B since D ¼ ΦBΦ1. Furthermore, if B is similar to D and D is similar to E, then B is similar to E, for 1 1 B ¼ Φ1 1 DΦ1 and D ¼ Φ2 EΦ2 imply that B ¼ ( Φ2Φ1) E( Φ2Φ1). It can also be verified that
m Φ1 Bm Φ ¼ Φ1 BΦ

ð6:3Þ

Example 6.1

Let  B¼

1 0

 1 , 1

 Φ¼

1 0

2 1



It is clear that (continued)

6.1 Similarity Transformation

345

Bn ¼



 1 n , 0 1





Φ1 ¼



2 1

1 0



Since 1 2 Φ BΦ ¼ 0 1 1

1 0

1 1

1 ¼ 0

1 1



  2 1 ¼ 1 0

1 0

1 1



one has


1

Φ BΦ

n



n



1 n ¼ 0 1



Characteristic Polynomial Similar matrices have the same characteristic polynomials. In order to demonstrate this, let B and D be two similar matrices such that B ¼ Φ1DΦ. The characteristic polynomial of B is jλI  Bj ¼ 0, where λ is the eigenvalue of the matrix B. It follows from the similarity transformation that       λI  B ¼ λI  Φ1 DΦ ¼ Φ1 ðλI  DÞΦ     ¼ Φ1 λI  DΦ Since jΦ1jjΦj ¼ 1, one has jλI  Bj ¼ jλI  Dj

ð6:4Þ

which implies that similar matrices have the same characteristic polynomial and, consequently, have the same eigenvalues. It also follows that similar matrices have the same trace and the same determinant. The equality of the characteristic polynomial is a necessary, but not a sufficient, condition for the similarity of two matrices. For example, the two matrices 

1 B¼ 0

 0 , 1



1 D¼ 0

1 1



have the same characteristic polynomial (1  λ)2, but B and D are not similar, since for any nonsingular matrix Φ, one has Φ1 BΦ ¼ Φ1 Φ ¼ I Note that because B and D have the same characteristic polynomial, they have the same eigenvalues λ1 ¼ λ2 ¼ λ ¼ 1. Associated with these repeated eigenvalues, the matrix B has the two independent eigenvectors

346

6

  1 A1 ¼ , 0

Methods for the Eigenvalue Analysis

  0 A2 ¼ 1

This is because the characteristic matrix (λI  B) has rank zero, and, as a consequence, there are two independent solutions for the homogeneous system of algebraic equations (λI  B)A ¼ 0. On the other hand, the matrix D does not have two independent eigenvectors associated with the repeated roots λ1 and λ2 since the characteristic matrix (λI  D) has rank one, and, therefore, there is only one independent solution for the system of homogeneous algebraic equations (λI  D)A ¼ 0. The independent eigenvector of the matrix D takes the form A ¼ [1 0]T. Recall that the number of dependent variables in a system of homogeneous algebraic equations is equal to the number of independent equations, which is the same as the rank of the coefficient matrix in this system of equations. It follows that the number of independent solutions is the same as the degree of singularity of the coefficient matrix and is equal to the dimension n minus the rank r of the matrix. In general, a matrix that has s repeated eigenvalues has s linearly independent eigenvectors associated with these repeated eigenvalues if the matrix is similar to a diagonal matrix. If a matrix is not similar to a diagonal matrix, the s repeated eigenvalues do not correspond to s linearly independent eigenvectors, but they do correspond to a set of generalized eigenvectors, which will be introduced in later sections. As demonstrated by the example of the two matrices B and D previously presented in this section, having the same characteristic polynomial is not sufficient for the similarity of two matrices. In fact, one should not be able to find a nonsingular matrix Φ such that Φ1DΦ is equal to a diagonal matrix.

6.2

Polynomial Matrices

A polynomial in an arbitrary square matrix B can be written as f ðBÞ ¼ α0 I þ α1 B þ α2 B2 þ    þ αn Bn

ð6:5Þ

where α0, α1, . . ., and αn are scalar coefficients. It can be shown that f1(B) and f2(B) are two polynomials in the arbitrary matrix B such that f 1 ðBÞ ¼ α0 I þ α1 B þ α2 B2 þ    þ αn Bn f 2 ðBÞ ¼ β0 I þ β1 B þ β2 B2 þ    þ βm Bm

) ð6:6Þ

then f 1 ðBÞ þ f 2 ðBÞ ¼ f 2 ðBÞ þ f 1 ðBÞ f 1 ðBÞf 2 ðBÞ ¼ f 2 ðBÞ þ f 1 ðBÞ

) ð6:7Þ

6.2 Polynomial Matrices

347

which imply that operations with polynomial matrices follow the same rules as scalar polynomials. This fact can be used to obtain matrix identities, as demonstrated by the following example. Example 6.2

It is known from scalar polynomials that λ2  1 ¼ ðλ  1Þðλ þ 1Þ
 λ3 þ 1 ¼ ðλ þ 1Þ λ2  λ þ 1 It is follows that for any arbitrary square matrix B, one has B2  I ¼ ðB  IÞðB þ IÞ
 B3 þ I ¼ ðB þ IÞ B2  B þ I

Cayley–Hamilton Theorem The Cayley–Hamilton theorem, which we present here without proof, states that every symmetric matrix satisfies its characteristic polynomial. Example 6.3

The matrix 2

4 1 B ¼ 41 0 2 0

3 2 05 0

has the characteristic polynomial jλI  Bj ¼ λ3  4λ2  5λ ¼ 0 The matrices B2 and B3 are 2

21 B2 ¼ 4 4 8

4 1 2

3 8 2 5, 4

2

104 B3 ¼ 4 21 42

21 4 8

Using the characteristic polynomial, it can be shown that B3  4B2  5B ¼ 0

3 42 8 5 16

348

6

Methods for the Eigenvalue Analysis

The Cayley–Hamilton theorem can also be used to define the inverse of nonsingular square matrices. As an example, we consider the matrix 2

3 0 25 1

1 5 B ¼ 4 1 1 2 2

which has the characteristic polynomial λ3  λ2 + 30 ¼ 0. It follows from the Cayley–Hamilton theorem that B3  B2 + 30I ¼ 0, or I ¼ {B3  B2}/30. Premultiplying this equation by B1, one obtains 2

B1

1 66 6  61 1  2 B B ¼6 ¼ 66 30 6 4 0

3 1 1 6 3 7 7 1 1 7 7 30 15 7 7 2 1 5 5 5

Minimal Polynomial The minimal polynomial of an arbitrary square matrix B is the nonzero monic polynomial of least degree that has B as a zero. Every matrix has a unique minimal polynomial, for if f1(B) and f2(B) are two different minimal polynomials of B, then f1(B)  f2(B) would be a nonzero polynomial of lower degree that has B as a zero. If this polynomial is divided by the leading coefficient, one obtains a monic minimal polynomial with lower degree. This contradicts the fact that f1(B) and f2(B) are minimal polynomials of B. Therefore, B has a unique minimal polynomial. Every polynomial that has the matrix B as a zero is divisible without remainder by the minimal polynomial of the matrix B. In order to prove this fact, suppose, on the contrary, that g(B) ¼ 0 is a polynomial in B that is not divisible by the minimal polynomial f(B) of the matrix B. It follows that gðBÞ ¼ f ðBÞhðBÞ þ r ðBÞ

ð6:8Þ

where h(B) is the quotient and r(B) is the remainder that has a degree less than the degree of f(B). Since f(B) ¼ 0 and g(B) ¼ 0, one also has r(B) ¼ 0. This contradicts the assumption that f(B) is the minimal polynomial of the matrix B. Consequently, the minimal polynomial of a matrix B is a divisor of any polynomial that has B as its zero. In particular, the minimal polynomial is a divisor of the characteristic polynomial of the matrix. It was previously pointed out (Eq. 6.3) that if B is an arbitrary square matrix, and Φ is any nonsingular matrix, then Φ1BmΦ ¼ (Φ1BΦ)m for any integer m. It follows from this identity that
 Φ1 f ðBÞΦ ¼ f Φ1 BΦ

ð6:9Þ

6.2 Polynomial Matrices

349

This identity can be used to show that similar matrices have the same minimal polynomial. In order to prove this assertion, suppose that B and D are two similar matrices such that B ¼ Φ1DΦ. If f(D) ¼ 0, then
 f ðBÞ ¼ f Φ1 DΦ ¼ Φ1 f ðDÞΦ ¼ 0

ð6:10Þ

which implies that the set of polynomials having B as a zero is the same as the set of polynomials having D as a zero, and hence the minimal polynomials of two similar matrices are identical. As in the case of the characteristic polynomial, the equality of the minimal polynomials is a necessary, but not sufficient, condition for the similarity of two matrices, as demonstrated by the following example. Example 6.4

Consider the two matrices 2 1 B ¼ 40 0

3 0 0 5, 2

0 2 0

2

1 0 D ¼ 40 1 0 0

3 0 25 2

The characteristic polynomials of these two matrices are jλI  Bj ¼ ðλ  1Þðλ  2Þ2 jλI  Dj ¼ ðλ  1Þ2 ðλ  2Þ Since these two characteristic polynomials are not the same, B is not similar to D. It can be shown, however, that B and D have the same minimal polynomials: gðBÞ ¼ ðB  IÞðB  2IÞ ¼ 0 gðDÞ ¼ ðD  IÞðD  2IÞ ¼ 0 These are the monic polynomials of least degree that have B and D as zeros. If B is the block diagonal matrix 2 6 6 B¼6 4

3

B1 B2 0

0 7 7 7 ⋱ 5

ð6:11Þ

Bs

then the characteristic polynomial of B is the product of the characteristic polynomials of its diagonal blocks, that is, jλI  Bj ¼ jλI  B1 jjλI  B2 j  jλI  Bs j

ð6:12Þ

The minimal polynomial of a block diagonal matrix is the least common multiple of the minimal polynomials of its diagonal blocks.

350

6.3

6

Methods for the Eigenvalue Analysis

Equivalence of the Characteristic Matrices

It can be shown that the eigenvectors associated with distinct eigenvalues of a matrix are linearly independent, and hence a matrix that has distinct eigenvalues is similar to a diagonal matrix. For example, if B is an n  n matrix that has the distinct eigenvalues λ1, λ2, . . ., λn, then the eigenvectors A1, A2, . . ., An associated with these eigenvalues are linearly independent. Furthermore, since BAi ¼ λiAi, one has BΦ ¼ ΦΛ

ð6:13Þ

where Φ is the modal matrix, and Λ is a diagonal matrix that has the eigenvalues as the diagonal elements. The matrices Φ and Λ are Φ ¼ ½A1 A2 . . . An 

ð6:14Þ

and 2 6 6 6 Λ¼6 6 4

3

λ1 λ1 0

0 7 7 7 7 ⋱ 7 5

ð6:15Þ

λn

Premultiplying both sides of Eq. 6.13 by Φ1, one gets Φ1 BΦ ¼ Λ

ð6:16Þ

which implies that if B has linearly independent eigenvectors, then B is similar to a diagonal matrix whose diagonal elements are the eigenvalues of B. Furthermore, the matrix of the similarity transformation Φ is the modal matrix whose columns are the linearly independent eigenvectors of B. Condition for Similarity In the analysis presented in Chap. 3, it was pointed out that the number of independent eigenvectors associated with repeated roots depends on the rank of the characteristic matrix. In fact, two matrices B and D are similar if their characteristic matrices (λI  B) and (λI  D) are equivalent, that is, if (λI  B) can be obtained from (λI  D) by a series of elementary operations. Similarity implies the equivalence of the characteristic matrices since, if B is similar to D, then there exists a nonsingular matrix Φ such that D ¼ Φ1BΦ. It follows that ðλI  DÞ ¼ Φ1 ðλI  BÞΦ

ð6:17Þ

which shows that the characteristic matrices of the two similar matrices B and D are equivalent. Conversely, if the characteristic matrices (λI  B) and (λI  D) are equivalent, then there exist two nonsingular matrices V and W such that (λI  D) ¼ V(λI  B)W. By comparing the coefficients of λ and the constant terms,

6.3 Equivalence of the Characteristic Matrices

351

we conclude that VW ¼ I and D ¼ VBW, which imply that D ¼ W1BW, proving that B and D are similar if their characteristic matrices are equivalent. Invariant Factors A characteristic matrix can be reduced by a series of elementary operations to the diagonal form 2 6 6 4

f 1 ðλÞ

3 f 2 ðλÞ 0

⋱

0 7 7 5 f n ðλÞ

ð6:18Þ

This diagonal matrix is called a canonical diagonal λ-matrix if each diagonal element fi(λ) is a divisor of the next element fi+1(λ) and if all nonzero diagonal elements are monic polynomials. Therefore, an example of a canonical diagonal matrix takes the form 2 6 6 6 6 6 6 6 6 6 6 6 6 4

3

1 ⋱

7 7 7 7 7 7 7 7 7 7 7 7 5

0 1 f 1 ðλÞ

⋱

f k ðλÞ 0

0

⋱

ð6:19Þ

0 where f1(λ), . . ., fk(λ) are monic polynomials such that fi(λ) is a divisor of fi+1(λ). In addition to having the same characteristic and minimal polynomial, similar matrices have the same canonical diagonal matrix; that is, their characteristic matrices can be reduced by a series of elementary operations to the same diagonal form. Therefore, the diagonal elements of the canonical diagonal λ-matrix, which are called the invariant factors, are unique. It follows that similar matrices have the same invariant factors. Example 6.5

The matrices 2

4 B ¼ 41 2

3 1 2 0 0 5, 0 0

2

3 4 D ¼ 4 1 1 2 2

3 6 1 5 2 (continued)

352

6

Methods for the Eigenvalue Analysis

are similar since there exists a nonsingular matrix 2

3 0 1 5 1

1 1 0

1 V ¼ 40 0 such that D ¼ VBV1, where 2

V1

1 ¼ 40 0

3 1 1 1 15 0 1

The matrices B and D have the same eigenvalues: λ1 ¼ 1,

λ2 ¼ 5,

λ3 ¼ 0

The eigenvectors of B associated with these eigenvalues are 2

AB1

3 1 ¼ 4 1 5, 2

AB2

2 3 5 ¼ 4 1 5, 2

2

AB3

3 0 ¼ 4 25 1

The eigenvectors of D are 2

AD1

3 2 ¼ 4 1 5, 2

2

AD2

3 4 ¼ 4 1 5, 2

2

AD3

3 2 ¼ 4 35 1

Note that because of the similarity of B and D, ADi ¼ VABi The characteristic matrices of B and D are 2

λ4

6 ðλI  BÞ ¼ 4 1 2 6 ðλI  DÞ ¼ 4

2

1

2

3

λ

7 0 5

0

λ

λ3

4

1

λþ1

2

2

6

3

7 1 5

λ2

Define the two matrices (continued)

6.3 Equivalence of the Characteristic Matrices

353

2

ΦB ¼ ½AB1 AB2

ΦD ¼ ½AD1 AD2

3 0 25 1 3 4 2 1 35

1 4 AB3  ¼ 1 2 2 2 AD3  ¼ 4 1 2

5 1 2

2

1

It can be shown that 2

3 λþ1 0 0  BÞΦB ¼ 4 0 λ  5 05 0 0 λ 2 3 λþ1 0 0 4 0 λ  5 05 Φ1 D ðλI  DÞΦD ¼ 0 0 λ Φ1 B ðλI

In this case, the canonical diagonal λ-matrix, which can be obtained by elementary operations, is 2 3 1 0 0 40 1 5 0 0 0 λ ð λ þ 1Þ ð λ  5Þ Note that the matrix ΦB and its inverse, which reduce the characteristic matrix to a diagonal form, are the products of elementary matrices. These matrices can be written as ΦB ¼ ðE4 E3 E2 E1 Þ1

Φ1 B

2

1 66 6 61 ¼ ðE4 E3 E2 E1 Þ ¼ 6 66 6 4 0

3 1  37 7 1 7 7 15 7 7 15  5

1 6 1 30 2 5



where E1, E2, E3, and E4 are the elementary matrices 2

1 6 E1 ¼ 4 1

0 1

3 0 7 0 5,

2

0

1

1 6 6 E3 ¼ 6 0 4 0



2

1 6 1 2  5

3 0

7 7 0 7, 5 1

2

1 0 6 E2 ¼ 4 0 1

3 0 7 25

0 0 1 2 3 1 0 0 1 6 7 05 E4 ¼ 4 0 30 0 0 1 (continued)

354

6

Methods for the Eigenvalue Analysis

The inverses of the elementary matrices are 2

1

0

0

3

2

1

0

0

3

6 7 6 7 E1 E1 1 ¼ 4 1 1 0 5, 2 ¼ 4 0 1 2 5 2 0 1 0 0 1 2 3 1 2 3 0 1 1 0 0 6 7 6 6 7 6 7 E1 E1 05 3 ¼ 6 0 1 0 7, 4 ¼ 4 0 30 4 5 2 0 0 1 0 0 5 Similar comments apply to the matrix ΦD. Each characteristic matrix can be reduced to a canonical diagonal λ-matrix whose diagonal elements are the invariant factors. It can be shown that the kth diagonal element (invariant factor) of the canonical diagonal matrix, which is a monic polynomial, is the greatest common divisor of all minors of order k divided by the greatest common divisors of all minors of order k  1. For example, using the matrix B of the preceding example, one can show that the minors of order 2 of the characteristic matrix of B are M 11 ¼ λ2 ,

M 12 ¼ λ,

M 13 ¼ 2λ

M 21 ¼ λ,

M 22 ¼ λ  4λ  4,

M 23 ¼ 2

M 31 ¼ 2λ,

M 32 ¼ 2,

M 33 ¼ λ2  4λ  1

2

The polynomial with a leading coefficient one, which represents the greatest common divisor of all the minors of the matrix (λI  B), is 1. This is also the greatest common divisor of the minors of order 2 of the matrix (λI  D). It can be shown that equivalent matrices have the same greatest common divisors of the kth-order minors.

6.4

Jordan Matrices

While similar matrices have equivalent characteristic matrices that can be reduced by a series of elementary operations to a unique diagonal form, not every matrix is similar to a diagonal matrix. A matrix B is similar to a diagonal matrix Λ if and only if B has a complete set of linearly independent eigenvectors. For, if B has linearly independent eigenvectors, the matrix Φ whose columns are these independent eigenvectors is nonsingular. In this case, the eigenvalue problem leads to BΦ ¼ ΦΛ, where Λ is a diagonal matrix whose diagonal elements are the eigenvalues of B. The preceding equation demonstrates that B and Λ are similar. Conversely, if B is similar to a diagonal matrix, then B has a complete set of linearly independent eigenvectors since V1BV ¼ Λ implies that BV ¼ VΛ, which proves that V is the matrix of the

6.4 Jordan Matrices

355

linearly independent eigenvectors which can be determined to within an arbitrary constant. The question of the linear independence of the eigenvectors arises when a matrix has repeated eigenvalues, as is demonstrated by the following example. Example 6.6

The matrix 2

1 6 5 B¼6 4 2 0

2 10 2 0

1 7 2 0

3 0 07 7 05 1

has the characteristic equation jλI  Bj ¼ ðλ  2Þ3 ðλ  1Þ ¼ 0 which defines the four eigenvalues λ1 ¼ λ2 ¼ λ3 ¼ 2, λ4 ¼ 1 The matrix B has three repeated eigenvalues, λ1, λ2, and λ3, and one distinct eigenvalue, λ4. The matrix (λ4I  B) has rank 3, and as such the system of homogeneous algebraic equations ðλ4 I  BÞA4 ¼ 0 has one independent nontrivial solution: 2 3 0 607 7 A4 ¼ 6 405 1 For the repeated roots, one can show that the matrix (λ1I  B) has rank 2, and therefore, the system of homogeneous algebraic equations ðλi I  BÞAi ¼ 0,

i ¼ 1,2,3

defines the following two independent eigenvectors: 2 3 1 657 7 A1 ¼ 6 4 2 5, 0

2 3 0 627 7 A2 ¼ 6 415 0 (continued)

356

6

Methods for the Eigenvalue Analysis

In this case, the matrix B is not similar to a diagonal matrix. One, however, can show that the matrix B is similar to the matrix D defined as 2

2 60 D¼6 40 0

1 2 0 0

3 0 07 7 05 1

0 0 2 0

and B ¼ VDV1 where 2

1 65 V¼6 42 0

0 1 0 0

3 0 07 7, 05 1

0 2 1 0

2

V1

1 6 1 ¼6 4 2 0

0 0 1 2 0 1 0 0

3 0 07 7 05 1

The preceding example demonstrates that not every matrix is similar to a diagonal matrix. In particular, matrices that do not have a complete set of independent eigenvectors are not similar to diagonal matrices, but instead are similar to matrices that belong to a class called Jordan matrices. A matrix of order m in the form 2 6 6 6 6 6 Gb ¼ 6 6 6 6 6 4

γ

1 γ

3 1 γ 0

1   

0    γ

7 7 7 7 7 7 7 7 7 7 15 γ

ð6:20Þ

is called a Jordan block. The characteristic polynomial of this matrix is (γ  λ)m, and as such γ, which has multiplicity m, is the only eigenvalue of Gb. Since the minimal polynomial of Gb divides the characteristic polynomial, the minimal polynomial must be in the form (γ  λ)k, where k  m. It can be shown, however, that k ¼ m, and the minimal polynomial of a Jordan matrix is identical to the characteristic polynomial (γ  λ)m. That is, ðγI  Gb Þk 6¼ 0 ¼0 For example, if Gb, is the 3  3 matrix, then

for k < m for k ¼ m

ð6:21Þ

6.4 Jordan Matrices

357

2

3 0 1 5 0

1 0 0

0 ðγI  Gb Þ ¼ 4 0 0 which shows that 2

0 ðγI  Gb Þ ¼ 4 0 0

3 1 0 5, 0

0 0 0

2

2

0 ðγI  Gb Þ ¼ 4 0 0 3

0 0 0

3 0 05 0

Clearly, (γI  Gb) is a nilpotent matrix of degree 3. It follows that the characteristic matrix of a Jordan block of order m is a nilpotent matrix of order m, and, as a consequence, the minimal polynomial of Gb is (γ  λ)m. We also observe that the rank of the matrix (λiI  Gb) is m  1, and, as a consequence, the system of algebraic equations (λiI  Gb)Ai ¼ 0 has m  1 independent algebraic equations that define one independent nontrivial solution regardless of the order m of the matrix. A Jordan matrix consists of Jordan blocks that may have different characteristic values. Jordan matrices then assume the following form: 2 6 6 6 G¼6 6 6 4

3

Gb1

Gb2



0

7 7 7 7 7 7 5

0 



ð6:22Þ

Gbs

where each Jordan block Gbi has a number of repeated eigenvalues equal to its dimension. An example of a Jordan matrix is the matrix D of the preceding example: 2

Gb1 D¼4 0 0

0 Gb2 0

3

2

2 0 60 0 5¼6 40 Gb3 0

1 2 0 0

0 0 2 0

3 0 07 7 05 1

where 

Gb1

2 ¼ 0

 1 , Gb2 ¼ 2, 2

Gb3 ¼ 1

As demonstrated in the preceding example, the eigenvalues of the matrix D are λ1 ¼ λ2 ¼ λ3 ¼ 2, λ4 ¼ 1

358

6

Methods for the Eigenvalue Analysis

Associated with these four eigenvalues are only three independent eigenvectors, since the rank of the matrix (λ1I  D) is 2, and, as a consequence, the system (λ1I  D)A1 ¼ 0 has only two independent nontrivial solutions associated with the three repeated eigenvalues λ1, λ2, and λ3. Using a similar procedure, it can be shown that each Jordan matrix has a number of linearly independent eigenvectors equal to the number of its Jordan blocks. Since the condition of similarity of two matrices implies the equivalence of their characteristic matrices, similar matrices have the same number of linearly independent eigenvectors. It follows that if a matrix B is similar to a Jordan matrix whose Jordan blocks are not all scalars, the matrix B does not have a complete set of eigenvectors. An example of such matrices was presented in the preceding example. It is clear from the discussion presented thus far that each square matrix is similar to a Jordan matrix since diagonal matrices are special cases of Jordan matrices in which the Jordan blocks are scalars. It is also clear that two Jordan matrices are similar if and only if they consist of the same Jordan blocks and differ only in the distribution of these blocks on the main diagonal. For example, 2

3 60 G1 ¼ 6 40 0

1 3 0 0

0 1 3 0

3 0 07 7, 05 5

2

5 60 G2 ¼ 6 40 0

0 3 0 0

0 1 3 0

3 0 07 7 15 3

are similar, but these two matrices are not similar to 2

3 60 G3 ¼ 6 40 0

0 3 0 0

0 1 3 0

3 0 07 7 05 5

It can be shown that G1 and G2 have only two independent eigenvectors: one associated with the three repeated eigenvalues λ1 ¼ λ2 ¼ λ3 ¼ 3 and one associated with the distinct eigenvalue λ4 ¼ 5. The matrix G3, on the other hand, has three linearly independent eigenvectors: two associated with the repeated eigenvalues λ1 ¼ λ2 ¼ λ3 ¼ 3 and one associated with the distinct eigenvalue λ4 ¼ 5. As another example, we consider the two matrices 2

3 60 G1 ¼ 6 40 0

1 3 0 0

0 1 3 0

3 0 07 7, 15 3

2

3 60 G2 ¼ 6 40 0

1 3 0 0

0 0 3 0

3 0 07 7 15 3

These two matrices have the same repeated eigenvalues, but they are not similar since G1 consists of one Jordan block while G2 consists of two Jordan blocks. As a consequence, G1 has only one independent eigenvector, while G2 has two independent eigenvectors.

6.5 Elementary Divisors

6.5

359

Elementary Divisors

Previously, we defined the kth invariant factor of a matrix as the greatest common divisor of all minors of order k of the characteristic matrix divided by the greatest common divisor of all minors of order k  1. It was also shown that similar matrices have equivalent characteristic matrices and, as a consequence, have identical invariant factors. It can, therefore, be concluded that Jordan matrices which are not similar do not have identical invariant factors. In order to demonstrate this fact, we consider the two matrices  G1 ¼

 2 0 , 0 2

 G2 ¼

2 0

1 2



Their characteristic matrices are 

 0 , λ2

λ2 ðλI  G1 Þ ¼ 0



λ2 ðλI  G2 Þ ¼ 0

1 λ2



The greatest common divisors of the minors of order 1 and 2 of (λI  G1) are, respectively, (λ  2) and (λ  2)2. Therefore, the matrix of invariant factors of G1 is 

λ2 0

0 λ2



The greatest common divisors of the minors of order 1 and 2 of (λI  G2) are 1 and (λ  2)2, respectively. It follows that the matrix of invariant factors of G2 is 

1 0 0 ð λ  2Þ 2



Using a similar procedure and the induction principle, it can be shown that the canonical diagonal matrix of an n-order Jordan block in the form of Eq. 6.20 is given by 2 6 6 Gλ ¼ 6 6 4

3

1 1 0

⋱

7 7 7 7 5

0 1

ð6:23Þ

ðγ  λÞn

Let fi(λ) be the ith invariant factor of an arbitrary matrix that, in general, can be written as f i ðλÞ ¼ es11 es22   eskk

ð6:24Þ

360

6

Methods for the Eigenvalue Analysis

where e1(λ), e2(λ), . . ., ek(λ) are distinct irreducible monic polynomials. The elements es11 , es22 , . . . , eskk are called the elementary divisors of the invariant factor fi(λ). The elementary divisors of an arbitrary matrix are the elementary divisors of all of the nonconstant invariant factors of this matrix. Example 6.7

Consider a matrix that has the invariant factors 1,1, λ, λ2 , λ2 ðλ þ 2Þ, λ2 ðλ þ 2Þ2 The elementary divisor of the first nonconstant invariant factor is λ, and of the second invariant factor it is λ2. The elementary divisors of the third nonconstant invariant factor are λ2 and (λ + 2), and of the fourth invariant factor they are λ2 and (λ + 2)2. Hence, the elementary divisors of the matrix are λ, λ2 , λ2 , λ2 , ðλ þ 2Þ, ðλ þ 2Þ2 Since similar matrices have the same invariant factors, it follows that similar matrices have identical elementary divisors. The elementary divisors can be used to determine whether or not a matrix is similar to a diagonal matrix. Using the definitions of the invariant factors and the elementary divisors, it is easy to verify that the elementary divisors of a diagonal matrix are all of the first degree. Hence, a matrix is similar to a diagonal matrix if and only if all of its elementary divisors are of the first degree. Note that a Jordan block has only one elementary divisor, (γ  λ)n, as demonstrated by Eq. 6.23, where n is the dimension of the Jordan block. The elementary divisors of a Jordan matrix consist of the elementary divisors of its blocks. Recall that a diagonal matrix can be considered as a special case of a Jordan matrix in which the Jordan blocks are of order 1. Example 6.8

The matrix 2

3 1 B ¼ 4 7 2 2 1

3 3 95 4

has the characteristic matrix 2

λ3 ðλI  BÞ ¼ 4 7 2

1 λþ2 1

3 3 9 5 λ4 (continued)

6.6 Generalized Eigenvectors

361

The invariant factors of the matrix B are 1, 1, (λ  1)(λ  2)2. Consequently, the elementary divisors of the matrix B are (λ  1), (λ  2)2. The matrix B then is similar to the Jordan matrix 2

1 G ¼ 40 0

6.6

3 0 0 2 15 0 2

Generalized Eigenvectors

The procedure described in the preceding sections provides a systematic way of determining whether or not a matrix is similar to a diagonal matrix. First, we reduce, by a series of elementary operations, the characteristic matrix to a diagonal canonical form, thus defining the invariant factors. These invariant factors also can be determined using the greatest common divisors of the minors of the characteristic matrix. The invariant factors then can be used to define the elementary divisors. If all of the elementary divisors are polynomials of the first degree, the matrix is similar to a diagonal matrix. If some or all of the elementary divisors are not polynomials of the first degree, the matrix is not similar to a diagonal matrix, but it is similar to a Jordan matrix. Recall also that a matrix which has a complete set of linearly independent eigenvectors is similar to a diagonal matrix regardless of the number of repeated eigenvalues. In this case, the matrix of the similarity transformation is the matrix whose columns are the independent eigenvectors. The matrix of the eigenvectors can also be defined as the product of a set of elementary matrices which reduce a given matrix and its characteristic matrix to a diagonal form. If a matrix B does not have a complete set of eigenvectors, a question arises as to which Jordan matrix is similar to B and what is the matrix of the similarity transformation. In order to briefly discuss this question, we consider the matrix 2

1 6 5 B¼6 4 2 0

1 7 2 0

2 10 2 0

3 0 07 7 05 1

which has the eigenvalues λ1 ¼ λ2 ¼ λ3 ¼ 2 and λ4 ¼ 1. There are two independent eigenvectors associated with the triple eigenvalue λ ¼ 2, and as such the matrix B is not similar to a diagonal matrix, but similar to the Jordan matrix

362

6

2

2 60 G¼6 40 0

1 2 0 0

Methods for the Eigenvalue Analysis

3 0 07 7 05 1

0 0 2 0

The independent eigenvectors associated with the eigenvalues of the matrix B are 2 3 1 657 7 A1 ¼ 6 4 2 5, 0

2 3 0 627 7 A3 ¼ 6 4 1 5, 0

2 3 0 607 7 A4 ¼ 6 405 1

These three eigenvectors represent the first, third, and fourth columns of the matrix of similarity transformation V ¼ ½ A1

A2

A3

A4 

where A2, the missing eigenvector, is called the generalized eigenvector. In order to illustrate the procedure for determining the generalized eigenvector, we write V1BV ¼ G. It follows from this equation that BV ¼ VG, or 2

B½ A1

A2

A3

A4  ¼ ½ A1

A2

A3

λ1 60 A4 6 40 0

1 λ2 0 0

0 0 λ3 0

3 0 07 7 05 λ4

ð6:25Þ

This matrix equation yields BA1 ¼ λ1 A1 , BA3 ¼ λ3 A3 ,

BA2 ¼ λ2 A2 þ A1 BA4 ¼ λ4 A4

ð6:26Þ

Therefore, the generalized eigenvector A2 must satisfy the equation BA2 ¼ λ2 A2 þ A1

ð6:27Þ

Using the matrix B and the results previously obtained for A1, one can show, as demonstrated later in this section, that the generalized eigenvector A2 is A2 ¼ ½ 0 1 0 0 T , and the matrix of the similarity transformation is 2

V ¼ ½ A1

A2

A3

1 65 A4  ¼ 6 42 0

0 1 0 0

0 2 1 0

3 0 07 7 05 1

ð6:28Þ

Note that A1 and A2 are two linearly independent vectors. Using a similar procedure, one can show that if λ1 ¼ λ2 ¼    ¼ λr of a given n-dimensional matrix B have only

6.6 Generalized Eigenvectors

363

one independent eigenvector, the r linearly independent generalized eigenvectors must satisfy the following relationships: BA1 ¼ λ1 A1 ,

BA2 ¼ λ2 A2 þ A1 ,   , BAr ¼ λr Ar þ Ar1

ð6:29Þ

It follows that the independent vectors that form the columns of the matrix of the similarity transformation can be obtained by using either the equation BAi ¼ λi Ai

ð6:30Þ

BAi ¼ λi Ai þ Ai1

ð6:31Þ

or the equation

Since (λiI  B) is a singular matrix, Eq. 6.31 is a system of nonhomogeneous algebraic equations which can be solved for the unknown generalized eigenvector Ai. In order to demonstrate the procedure for solving this system, we consider the equation for the generalized eigenvector A2 of the matrix B considered earlier in this section. The equation ðB  λ2 IÞA2 ¼ A1 yields 2

1 6 5 6 6 4 2 0

1 2 5 10 2

4

0

0

32 3 2 3 A21 1 0 6 7 6 7 0 76 A22 7 6 5 7 7 76 7¼6 7 0 54 A23 5 4 2 5 1

A24

0

The coefficient matrix in this system has rank 2 since the second and the third rows are multiples of the first row. The application of the Gaussian elimination procedure shows that the above system reduces to 2

1

6 0 6 6 4 0 0

1 2 0

0

0 0

0 0

32

2 3 1 7 6 6 7 0 76 A22 7 6 0 7 7 76 7¼6 7 0 54 A23 5 4 0 5 1 0 A24 0

A21

3

This system leads to the following two independent algebraic equations: A21 þ A22  2A23 ¼ 1 A24 ¼ 0 Since the number of these equations is less than the number of unknowns, there is an infinite number of solutions that can be obtained by partitioning the variables as dependent and independent. The number of dependent variables is equal to the

364

6

Methods for the Eigenvalue Analysis

number of equations, while the remaining are the free or the independent variables. If we select A22 and A24 as the dependent variables, the preceding equations can be written as A22 ¼ 1 þ A21 þ 2A23 A24 ¼ 0 It follows that 2

3 2 3 2 3 2 3 0 1 A21 0 6 A22 7 6 1 7 6 1 7 627 7 6 7 6 7 6 7 A2 ¼ 6 4 A23 5 ¼ 4 0 5 þ 4 0 5A21 þ 4 1 5A23 0 0 0 A24

ð6:32Þ

The first vector on the right-hand side of this equation is the particular solution, while the sum of the last two vectors is the homogeneous solution. Since A21 and A23 can be given arbitrary values, there is an infinite number of solutions for A2. One choice of A2 is A2 ¼ ½ 0 1 0 0 T , which was presented previously, is obtained by assuming that A21 ¼ A23 ¼ 0. This vector is linearly independent of A1, which is obtained from the homogeneous part of Eq. 6.32 by selecting A21 ¼ 1 and A23 ¼ 2. Another choice for A21 and A23 will lead to a different generalized eigenvector A2. For instance, if A21 ¼ 1 and A23 ¼ 0 are chosen, one obtains A2 ¼ ½ 1 2 0 0 T . In this case, we obtain a different similarity transformation matrix 2

V1 ¼ ½ A1

A2

A3

1 65 6 A4  ¼ 6 42 0

1 2 0 0

1 2 1 2 2 3 0 0

3 0 07 7 7 05 1

0 2 1 0

3 0 07 7 7 05 1

The inverse of V1 is 2

V1 1

2 6 1 6 ¼6 4 4 0

One can show that 2

2 60 6 V1 1 BV1 ¼ 6 40 0

1 2 0 0

0 0 2 0

3 0 07 7 7 05 1

which demonstrates that the matrix of similarity transformation is not unique since the eigenvectors and the generalized eigenvectors are not unique.

6.7 Jacobi Method

6.7

365

Jacobi Method

In the remainder of this chapter, we discuss some numerical techniques for solving the eigenvalue problem. We start with the Jacobi method, which can be used to determine the eigenvalues and eigenvectors of symmetric matrices only. Recall that if B is a symmetric matrix, then there exists an orthogonal matrix Φ such that ΦT BΦ ¼ D

ð6:33Þ

where D is a diagonal matrix whose diagonal elements are the eigenvalues of B. In the Jacobi method, the matrix Φ is determined as the product of a series of orthogonal transformation matrices called Jacobi rotation matrices. The Jacobi rotation matrix Ri at step i is just a plane rotation matrix which takes the form 3

2

1 0 60 1 6 6 6 6 6 6 Ri ¼ 6 6 6 6 6 6 6 4

⋱

cos θi ⋮ 0 ⋮ sin θi

 ⋱

0  1



0 

sin θi ⋮ 0 ⋮ cos θi

7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

⋱

ð6:34Þ

1 In this matrix, all of the elements other than those appearing in two columns, say k and l, are the same as the elements of the identity matrix. The sine and cosine functions appear in the elements of the kth and lth rows and columns. The corresponding elements in the product S ¼ RiT BRi

ð6:35Þ

are 9 skk ¼ bkk cos 2 θi  2bkl sin θi cos θi þ bll sin 2 θi = skl ¼ slk ¼ ðbll  bkk Þ sin θi cos θi þ bkl ðcos 2 θi  sin 2 θi Þ ; sll ¼ bkk sin 2 θi  2bkl sin θi cos θi þ bll cos 2 θi

ð6:36Þ

where bij are the elements of the matrix B. If the off-diagonal elements skl and slk of the matrix S are to be zero, one must select θi such that θi ¼

1 2bkl tan 1 2 bkk  bll

ð6:37Þ

It follows that, with a proper selection of the angle θi, each step of the Jacobi method can be used to annihilate two off-diagonal elements. While the next transformation may introduce nonzero elements in the positions of the elements

366

6

Methods for the Eigenvalue Analysis

previously set to zeros, the successive applications of the Jacobi rotation matrices lead to the required diagonal matrix. The product of the Jacobi rotation matrices defines the orthogonal matrix Φ, whose columns are the eigenvectors of B, as Φ ¼ R1 R2 . . . Rn

ð6:38Þ

The Jacobi method becomes inefficient as the order of the matrix increases, and as such, this method is recommended only for moderate-size matrices with order about 10. Example 6.9

In order to demonstrate the procedure for applying the Jacobi method, we consider the symmetric matrix B: 2

4 1 B ¼ 41 0 2 0

3 2 05 0

In order to annihilate b12 and b21, we use Eq. 6.37 to define θ1 as θ1 ¼

1 2ð 1Þ  tan 1 ¼ 13:283 2 40

The Jacobi rotation matrix R1 is 2

cos θ1 R1 ¼ 4 sin θ1 0

sin θ1 cos θ1 0

3 2 0 0:9732 0 5 ¼ 4 0:2298 1 0

0:2298 0:9732 0

3 0 05 1

It follows that 2

R1T BR1

4:2358 0:0 ¼ 4 0:0 0:2361 1:9464 0:4596

3 1:9464 0:4596 5 0:0

In order to set to zero the elements in positions (1,3) and (3,1), we define θ2 as θ2 ¼

1 2ð1:9464Þ  tan 1 ¼ 21:2919 2 4:2358  0:0

which defines R2 as 2

cos θ2 R2 ¼ 4 0 sin θ2

3 2 0 sin θ2 0:9317 1 0 5 ¼ 40 0:3631 0 cos θ2

0 1 0

3 0:3631 5 0 0:9317

The second step in the Jacobi method leads to (continued)

6.8 Householder Transformation

367

2

4:9939

6 R2T R1T BR1 R2 ¼ 4 0:1669 0:0

0:1669 0:2361 0:4282

0:0

3

7 0:4282 5 0:75843

Note that the previously set zero elements in positions (1,2) and (2,1) are no longer zeros after the second step. Nonetheless, by continuing this process, a diagonal matrix can be obtained, and the diagonal elements of this matrix are the eigenvalues of B. The exact eigenvalues of the matrix B are 1, 5, and 0.

Generalized Eigenvalue Problem The Jacobi method discussed in this section can be used in the case of symmetric matrices. In the case of the generalized eigenvalue problem, one has KA ¼ λMA

ð6:39Þ

where M and K are, respectively, the symmetric mass and stiffness matrices. Premultiplying the preceding equation by M1, one obtains


 M1 K A ¼ λA

ð6:40Þ

where M1K is not necessarily a symmetric matrix. A symmetric eigenvalue problem, however, can be recovered by using Cholesky decomposition M ¼ LLT, where L is a lower-triangular matrix. Premultiplying Eq. 6.39 by L1, one obtains
 B LT A ¼ λLT A

ð6:41Þ

where B ¼ L1K(L1)T. The matrix B is symmetric and has the same eigenvalues as the matrix M1K, and its eigenvectors are LTA.

6.8

Householder Transformation

Another method for determining the eigenvalues and eigenvectors of a symmetric matrix is to use a sequence of Householder transformations that reduce the symmetric matrix to a simpler tri-diagonal form. The eigenvalues and eigenvectors of the tri-diagonal matrix can be calculated more efficiently. The resulting eigenvalues are the same as those of the original symmetric matrix, and the product of the resulting Householder transformations defines an orthogonal matrix which relates the eigenvectors of the tri-diagonal matrix and the original symmetric matrix. A Householder transformation or an elementary reflector associated with a unit vector is v1 defined as H ¼ I  2v1 v1T

ð6:42Þ

where I is an identity matrix. The matrix H is symmetric and also orthogonal since

368

6

Methods for the Eigenvalue Analysis



 HT H ¼ I  2v1 v1T I  2v1 v1T ¼ I

ð6:43Þ

It follows that H ¼ HT ¼ H1. It is also clear that if v ¼ jvj v1, then Hv ¼ v

ð6:44Þ

Furthermore, if u is the column vector u ¼ ½ 1 0    pffiffiffiffiffiffiffiffi β ¼ jbj ¼ bT b, then ( ) 2vvT Hb ¼ I  b ðjvjÞ2

0 T , and v ¼ b + βu, where

ð6:45Þ

Using the definitions of v and β, one can show that 3 β 6 0 7 7 6 7 6 Hb ¼ βu ¼ 6 0 7 7 6 4 ⋮ 5 2

ð6:46Þ

0 This equation implies that when the Householder transformation constructed using the vector v is multiplied by the vector b, the result is a vector whose only nonzero element is the first element. Using this fact, a matrix can be transformed to an uppertriangular form by successively applying a series of Householder transformations. In order to demonstrate this procedure, we consider the rectangular n  m matrix: 2

b11 6 b21 B¼6 4⋮ bn1

b12 b22 ⋮ bn2

b13 b23 ⋮ bn3

  ⋱ 

3 b1m b2m 7 7 ⋮ 5 bnm

ð6:47Þ

First, we construct the Householder transformation associated with the first column b1 ¼ [b11 b21 . . . bn1]T. This transformation matrix can be written as b1 b T H1 ¼ I  2  T 1 b1 b1

ð6:48Þ

where b1 ¼ b1 þ β1 u1 , in which β1 is the norm of b1, and the vector u1 has the same dimension as b1 and is defined by u1 ¼ ½ 1 0 0 0 T . Using Eq. 6.46, one has 2

β1

3

6 0 7 7 6 7 6 H1 b1 ¼ β1 u1 ¼ 6 0 7 7 6 4 ⋮ 5 0

ð6:49Þ

6.8 Householder Transformation

369

It follows that 2

β1 6 0 6 B 1 ¼ H1 B ¼ 6 6 0 4 ⋮ 0

ðb12 Þ1 ðb22 Þ1 ðb32 Þ1 ⋮ ðbn2 Þ1

ðb13 Þ1 ðb23 Þ1 ðb33 Þ1 ⋮ ðbn3 Þ1

   ⋱ 

3 ðb1m Þ1 ðb2m Þ1 7 7 ðb3m Þ1 7 7 ⋮ 5 ðbnm Þ1

ð6:50Þ

Now we consider the last n  1 elements of the second column of the matrix B1 T which form the vector b2 ¼ ½ ðb22 Þ1 ðb32 Þ1    ðbn2 Þ1  . A Householder transformation matrix H2i can be constructed such that 2

β1

3

6 0 7 7 6 7 6 H2i b2 ¼ β2 u2 ¼ 6 0 7 7 6 4 ⋮ 5

ð6:51Þ

0 where β2 is the norm of the vector b2, and u2 is the (n  1)-dimensional vector defined by u2 ¼ ½ 1 0 0    0 T . Note that, at this point, the Householder transformation is only of order (n  1), and this transformation can be imbedded into the lower right corner of an n  n matrix H2, where  H2 ¼

0 H2i

1 0

 ð6:52Þ

The matrix H2 is orthogonal and symmetric, and when it is multiplied by an arbitrary matrix, it does not change the first row and the first column of that matrix. By premultiplying B1 by H2, one gets 2

β1

6 0 6 6 B2 ¼ H2 B1 ¼ H2 H1 B ¼ 6 6 0 6 4 ⋮ 0

ðb12 Þ1

ðb13 Þ1



β2

ðb23 Þ2



0 ⋮

ðb33 Þ2 ⋮

 ⋱

0

ðbn3 Þ2



ðb1m Þ1

3

ðb2m Þ2 7 7 7 ðb3m Þ2 7 7 7 ⋮ 5 ðbnm Þ2

ð6:53Þ

Next, we consider the vector that consists of the last n  2 elements of the third column of the matrix B2. This vector is b3 ¼ ½ ðb33 Þ2 ðb43 Þ2    ðbn3 Þ2 T . A Householder transformation matrix H3i can be constructed and imbedded into the lower right corner of the n  n matrix H3, where 2

1 H3 ¼ 4 0 0

0 1 0

3 0 0 5 H3i

ð6:54Þ

370

6

Methods for the Eigenvalue Analysis

Using this matrix, one has 2

β1 60 6 6 H3 B2 ¼ H3 H2 H1 B ¼ 6 60 6 4⋮ 0

ðb12 Þ1 β2

ðb13 Þ1 ðb23 Þ2

0 ⋮

β3 ⋮

0

0

3    ðb1m Þ1    ðb2m Þ2 7 7 7    ðb3m Þ3 7 7 7 ⋱ ⋮ 5    ðbnm Þ2

ð6:55Þ

where β3 is the norm of the vector b3. It is clear that, by continuing this process, all of the elements below the diagonal of the matrix B can be set equal to zero. If B is a square nonsingular matrix, the result of the Householder transformations is an uppertriangular matrix. If B is a rectangular matrix, the result of m Householder transformations is 

Bm ¼ Hm Hm1

R1    H1 B ¼ 0

 ð6:56Þ

where R1 is an m  m upper-triangular matrix. If the matrix B is symmetric, a sequence of Householder transformations can be used to obtain a tri-diagonal matrix that is similar to B, as demonstrated by the following example. The product of the resulting sequence of the Householder transformations defines the matrix that defines the relationship between the eigenvectors of B and the eigenvectors of the tri-diagonal matrix. Example 6.10

Use the Householder transformation to determine the eigenvalues and eigenvectors of the symmetric matrix 2

4 1 B ¼ 41 0 2 0

3 2 05 0

Solution. First, we construct the Householder transformation associated with the last n1 elements of the first column. This defines the vector b1 ¼ ½ 1 2 T . Using this vector, we define H1i ¼ I  2v1 v1T where v1 is a unit vector along the vector (continued)

6.8 Householder Transformation

371

v ¼ b1 þ βu1 ¼

      pffiffiffi 1 1 3:236 þ 5 ¼ 2 0 2

It follows that v1 ¼ ½ 0:851

0:526 T

and 

H1i ¼ I 

2v1 v1T

0:448 0:895 ¼ 0:895 0:447



This matrix can be imbedded in a 3  3 Householder transformation matrix: 

1 H1 ¼ 0

0 H1i



2

1 0 ¼ 4 0 0:448 0 0:895

3 0 0:895 5 0:447

The product H1B is 2

4 H1 B ¼ 4 2:238 0

1 0 0

3 2 05 0

Also, 2

4 2:238 H1 BH1 ¼ 4 2:238 0 0 0

3 0 05 0

This is a tri-diagonal matrix that has the eigenvalues λ1 ¼ 1.0, λ2 ¼ 0.0, and λ3 ¼ 5. The corresponding eigenvectors are 2

3 1:0 At1 ¼ 4 2:238 5, 0:0

2

3 0:0 At2 ¼ 4 0:0 5, 1:0

2

3 2:238 At3 ¼ 4 1:0 5 0:0

Since H1 is an orthogonal matrix, B is similar to H1BH1, and as such it has the same eigenvalues. Furthermore, if BA ¼ λA one has (continued)

372

6

Methods for the Eigenvalue Analysis

ðH1 BH1 ÞðH1 AÞ ¼ λðH1 AÞ which demonstrates that if At is an eigenvector of H1BH1, then the eigenvector of B is T A ¼ H1 1 At ¼ H1 At

It follows that 2

3 1:0 A1 ¼ 4 1:0 5, 2:0

6.9

2

3 0:0 A2 ¼ 4 0:895 5, 0:447

2

3 2:238 A3 ¼ 4 0:448 5 0:895

QR Decomposition

As was demonstrated in the preceding section, a sequence of Householder transformations can be used to reduce an arbitrary matrix to an upper-triangular matrix. For example, the first Householder transformation constructed using the first column of an arbitrary matrix B can be used to set the last n1 elements of this column equal to zero. The second Householder transformation sets the last n2 elements of the second column equal to zero, the third transformation sets the last n3 elements of the third column equal to zero, and the procedure continues in this manner to set all of the elements below the diagonal equal to zero. It follows that n1 Householder transformations can be used to systematically reduce an arbitrary n  n matrix to an upper-triangular matrix. Let B be an n  n matrix, and H1, H2, . . ., Hn1 be a sequence of Householder transformations designed to reduce B to an upper-triangular matrix R. One then has Hn1 Hn2   H1 B ¼ R

ð6:57Þ

Since the Householder transformations are symmetric and orthogonal, one has HiT ¼ H1 i ¼ Hi

ð6:58Þ

Using this identity, Eq. 6.57 can be written as B ¼ H1 H2   Hn2 Hn1 R

ð6:59Þ

Since the product of orthogonal matrices defines an orthogonal matrix, the preceding equation can be written as B ¼ QR where Q is an orthogonal matrix defined as

ð6:60Þ

6.9 QR Decomposition

373

Q ¼ H1 H2   Hn2 Hn1

ð6:61Þ

Equation 60 states that an arbitrary square matrix B can be written as the product of an orthogonal matrix Q and an upper-triangular matrix R. Example 6.11

Find the QR decomposition of the matrix 2

0 3 B ¼ 4 2 2 6 2

3 5 3 5 0

Solution. We consider the first column of the matrix B, which we write as b1 ¼ ½ 0 2 6 T . The norm of this vector β 1 ¼ j b1 j ¼

pffiffiffiffiffi 40 ¼ 6:3246

Define v ¼ b1 þ β1 u1 ¼ ½ 6:3246

6:0 T

2:0

A unit vector along v is v1 ¼ ½ 0:7071 0:2236

0:6708 T

The Householder transformation constructed using this vector is 2

0:0 H1 ¼ I  2v1 v1T ¼ 4 0:3162 0:9486

0:3162 0:9 0:3

3 0:9486 5 0:3 0:1

It follows that 2

6:3246 B1 ¼ H1 B ¼ 4 0:0 0:0

2:5296 0:2514 0:0

3 0:9486 1:1190 5 5:6430

The last two elements of the second column of this matrix form the vector b2 ¼ ½ 0:2514 0:0 T . The norm of this vector is β2 ¼ 0:2514 Define a vector (continued)

374

6

Methods for the Eigenvalue Analysis

w ¼ b2 þ β2 u2 ¼ ½ 0:5028 0:0 T 0 T . The Householder transfor-

A unit vector along this vector is w1 ¼ ½ 1 mation constructed using this vector is 

H2i ¼ I 

2w1 w1T

1 ¼ 0

0 1



This matrix can be imbedded in a three-dimensional matrix H2 as 

1 H2 ¼ 0

0 H2i



2

1 ¼ 40 0

0 1 0

3 0 05 1

Using this matrix, one has 2

6:3246 B2 ¼ H2 B1 ¼ H2 H1 B ¼ 4 0 0

2:5296 0:2514 0

3 0:9486 1:1190 5 5:6430

which is an upper-triangular matrix. Therefore, the matrix B can be written as B ¼ QR, where 2

0 6 Q ¼ H1 H2 ¼ 4 0:3162 0:9486 2 6:3246 6 R ¼4 0 0

0:3162 0:9 0:3 2:5296 0:2514 0

3 0:9486 7 0:3 5 0:1

3 0:9486 7 1:1190 5 5:6430

Using the QR decomposition, an iterative procedure can be developed for determining the eigenvalues and eigenvectors of an arbitrary square matrix B. In this procedure, B is first decomposed into its QR factors as B ¼ Q1 R1

ð6:62Þ

Using the Q1 and R1 matrices in this decomposition, a new matrix B1 is defined as B1 ¼ R1 Q1

ð6:63Þ

It is clear from Eq. 6.62 that R1 ¼ Q1T B, and as a consequence B1 ¼ Q1T BQ1

ð6:64Þ

Problems

375

Since Q1 is an orthogonal matrix, Eq. 6.64 demonstrates that B1 is similar to B, and therefore, these two matrices have the same eigenvalues. In the iterative procedure for determining the eigenvalues and eigenvectors of B, the QR factors of B1 are determined as B1 ¼ Q2R2. A matrix similar to B1 can be defined as B2 ¼ R2Q2. This process continues until a matrix Bi that is similar to B is obtained in a diagonal or a Jordan form. The eigenvalues of Bi are the same as the eigenvalues of B, and the eigenvectors of B can be obtained from the eigenvectors of Bi, using the product of the orthogonal matrices Q in the QR factors. Before using the QR decomposition, it is more efficient computationally to reduce B first to a tri-diagonal form using a sequence of Householder transformations. By reducing B to a simple form, fewer QR factorizations are required and the procedure becomes much less expensive. Another method that can be used to increase the speed of convergence of the QR algorithm is to use a shift of the origin of the eigenvalues. We note that if λ is an eigenvalue of B, then λ + c is an eigenvalue of B + cI, and both B and B + cI have the same eigenvectors. This fact can be demonstrated by adding cA to both sides of the equation BA ¼ λA to yield ðB þ cIÞA ¼ ðλ þ cÞA

ð6:65Þ

This equation shows that (λ + c) is an eigenvalue of (B + cI) and A is the associated eigenvector. More discussion on the selection of the constant c and the convergence of the QR method with shift can be found in the text by Atkinson (1978). Problems 6.1. Find the eigenvalues and eigenvectors of the matrix   4 5 B¼ 2 3 Evaluate also the products (Φ1BΦ)10 and Φ1B10Φ, where Φ is the matrix of the eigenvectors of B. 6.2. Find the eigenvalues and eigenvectors of the matrix 2 3 2:0 0:0 0:0 B ¼ 4 0:0 1:2 0:4 5 0:0 0:4 1:8 Evaluate also the products (Φ1BΦ)10, and Φ1B10Φ, where Φ is the matrix of the eigenvectors of B.

376

6

Methods for the Eigenvalue Analysis

6.3. Find the eigenvalues and eigenvectors of the matrix 2 3 1 1 2 B ¼ 4 0 0 25 0 0 2 and also find the matrix products (Φ1BΦ)10 and Φ1B10Φ, where Φ is the matrix of the eigenvectors of B. 6.4. Using the Cayley–Hamilton theorem, show that the matrix   4 5 B¼ 2 3 satisfies the identity B2  B  2I ¼ 0. Use this identity to find the inverse of the matrix B. 6.5. Using the Cayley–Hamilton theorem, show that the matrix 2 3 2:0 0:0 0:0 B ¼ 4 0:0 1:2 0:4 5 0:0 0:4 1:8 satisfies the identity (B  I)(B  2I)2 ¼
0. Show also that  the inverse of B can be expressed in terms of B as B1 ¼ 14 B2  5B þ 8I . 6.6. Using the characteristic polynomials, determine which of the following matrices is not similar to the others: 2 3 2 3 2 3 1 1 2 52 8 16 1 0 0 B1 ¼ 4 0 0 2 5, B2 ¼ 4 8 8 6 5, B3 ¼ 4 0 2 0 5 0 0 2 16 6 17 0 0 2 6.7. Find the minimal polynomials of the matrices 2 3 2 2:0 0:0 0:0 1 1 B ¼ 4 0:0 1:2 0:4 5, B2 ¼ 4 0 0 0:0 0:4 1:8 0 0

3 2 25 2

6.8. What is the minimal polynomial of the following matrix? 2 3 1 0 0 0 60 2 0 07 7 B¼6 40 0 2 05 0 0 0 2 6.9. Using elementary matrix operations, find the transformation that reduces the characteristic matrix of the matrix

Problems

377

2

1 2 B ¼ 40 3 0 0

3 2 25 5

to a diagonal form. Find also the invariant factors of the matrix B. 6.10. Show that the matrix

2

1 1 B ¼ 40 1 0 0

3 0 15 2

is not similar to a diagonal matrix. Determine the invariant factors of this matrix. 6.11. Find the elementary divisors of the two matrices 2

1 B1 ¼ 4 0 0

1 1 0

3 0 0 5, 2

2

1 B2 ¼ 4 0 0

0 1 0

3 0 05 0

6.12. Using the elementary divisors of the matrix 2 3 1 1 0 B ¼ 40 1 15 0 0 2 determine the Jordan matrix that is similar to B. 6.13. Use the elementary divisors to find the Jordan matrix that is similar to the matrix 2

2 B ¼ 4 1 1

1 4 3

3 0 05 1

6.14. Determine the independent or generalized eigenvectors of the matrix 2

2 B ¼ 4 1 1

1 4 3

3 0 05 1

6.15. Using the Jacobi method, find the eigenvalues and eigenvectors of the matrices 2

3 B1 ¼ 4 1 2

3 4 6 1 1 5, 2 2

2

4 B2 ¼ 4 1 2

1 0 0

3 2 05 0

378

6

Methods for the Eigenvalue Analysis

6.16. Use the QR decomposition to find the eigenvalues and eigenvectors of the matrices 2 3 2 3 3 4 6 4 1 2 B1 ¼ 4 1 1 1 5, B2 ¼ 4 1 0 0 5 2 2 2 2 0 0

Appendix A: Linear Algebra

In this appendix, we summarize some important results from vector and matrix algebra that are useful in our development in this book. Most of the vector and matrix properties presented in the following sections are elementary and can be found in standard texts on linear algebra.

A.1 Matrices An m
n matrix A is an ordered rectangular array that has m
n elements. The matrix A can be written in the form 2

a11

6
 6 6 a21 A ¼ aij ¼ 6 6 ⋮ 4 am1

a12



a22



⋮ am2

a1n

3

7 a2n 7 7 7 ⋱ ⋮ 7 5    amn

ðA:1Þ

The matrix A is called an m
n matrix since it has m rows and n columns. The scalar element aij lies in the ith row and jth column of the matrix A. Therefore, the index i, which takes the values 1, 2, . . ., m, denotes the row number, while the index j, which takes the values 1, 2, . . ., n, denotes the column number. A matrix A is said to be square if m ¼ n. An example of a square matrix is 2

3:0

6 A ¼ 4 6:3 9:0

2:0 0:95

3

0:0

7 10:0 5

3:5

1:25

In this example m ¼ n ¼ 3, consequently, A is a 3
3 matrix. The transpose of an m
n matrix A is an n
m matrix denoted as AT and defined as

# Springer Nature Switzerland AG 2019 A. Shabana, Vibration of Discrete and Continuous Systems, Mechanical Engineering Series, https://doi.org/10.1007/978-3-030-04348-3

379

380

Appendix A: Linear Algebra

2

a21 a22 ⋮ a2n

a11 6
 a 12 AT ¼ a ji ¼ 6 4⋮ a1n

  ⋱ 

3 am1 am2 7 7 ⋮ 5 amn

ðA:2Þ

For example, let A be the matrix  A¼

2:0

4:0

7:5

23:5

0:0

8:5

10:0

0:0



The transpose of A is given by 2

2:0 6 4:0 6 A¼6 4 7:5 23:5

3 0:0 8:5 7 7 7 10:0 5 0:0

That is, the transpose of the matrix A is obtained by interchanging the rows and columns. Definitions A square matrix A is said to be symmetric if aij ¼ aji, that is, if the elements on the upper right half can be obtained by flipping the matrix about the diagonal. For example, 2

3:0 6 A ¼ 4 2:0 1:5

3 2:0 1:5 7 0:0 2:3 5 2:3 1:5

is a symmetric matrix. Note that if A is symmetric, then A is the same as its transpose, that is, A ¼ AT. A square matrix is said to be an upper triangular matrix if aij ¼ 0 for i > j. That is, every element below each diagonal element of an upper triangular matrix is zero. An example of an upper triangular matrix is 2

6:0 2:5

6 0:0 8:0 6 A¼6 4 0:0 0:0 0:0 0:0

10:2 11:0 5:5 3:2

6:0 4:0

0:0

2:2

3 7 7 7 5

A square matrix is said to be a lower triangular matrix if aij ¼ 0 for i < j. That is, every element above the diagonal elements of a lower triangular matrix is zero. An example of a lower triangular matrix is

A.2 Matrix Operations

381

2

6:0 6 2:5 6 A¼6 4 10:2

0:0 8:0

0:0 0:0

5:5

3:2

11:0 6:0

3 0:0 0:0 7 7 7 0:0 5

4:0

2:2

The diagonal matrix is a square matrix such that aij ¼ 0 if i 6¼ j; that is, a diagonal matrix has element aij along the diagonal with all other elements equal to zero. For example, 2

5:0

0:0

6 A ¼ 4 0:0 0:0

0:0

3

7 0:0 5 7:0

1:0 0:0

is a diagonal matrix. The null matrix or the zero matrix is defined to be the matrix in which all the elements are equal to zero. The unit matrix or the identity matrix is a diagonal matrix whose diagonal elements are nonzero and equal to one. A skew-symmetric matrix is a matrix such that aij ¼ aji. Note that since aij ¼ aji for all i and j values, the diagonal elements should be equal to zero. An example of a ~ is skew-symmetric matrix A 2

0:0

~ ¼6 A 4 3:0 5:0

3:0

5:0

3

7 2:5 5

0:0 2:5

0:0

~ ~T ¼ A. Observe that A The trace of an n
n square matrix A, denoted by tr A is the sum of the diagonal elements of A. The trace of A can thus be written as tr A ¼

n X

aii

ðA:3Þ

i¼1

Note that the trace of an n
n identity matrix is n, while the trace of a skewsymmetric matrix is zero.

A.2 Matrix Operations In this section, we discuss some of the basic matrix operations which are used throughout this book. Matrix Addition The sum of two matrices A and B, denoted by AþB is given by
 A þ B ¼ aij þ bij

ðA:4Þ

382

Appendix A: Linear Algebra

where bij are the elements of B. In order to add two matrices A and B, it is necessary that A and B have the same dimensions, that is, the same number of rows and the same number of columns. It is clear from Eq. A.4 that matrix addition is commutative, that is, AþB¼BþA

ðA:5Þ

Example A.1

The two matrices A and B are defined as  A¼

3:0 2:0

1:0 0:0

 5:0 , 2:0

 B¼

2:0 3:0

3:0 0:0

6:0 5:0



The sum AþB is given by  AþB¼

     3:0 1:0 5:0 2:0 3:0 6:0 5:0 4:0 1:0 þ ¼ 2:0 0:0 2:0 3:0 0:0 5:0 1:0 0:0 3:0

while A  B is given by 

     3:0 1:0 5:0 2:0 3:0 6:0 1:0 2:0 11:0 AB¼  ¼ 2:0 0:0 2:0 3:0 0:0 5:0 5:0 0:0 7:0

Matrix Multiplication The product of two matrices A and B is another matrix C defined as C ¼ AB

ðA:6Þ

The element cij of the matrix C is defined by multiplying the elements of the ith row in A by the elements of the jth column in B according to the rule cij ¼ ai1 b1 j þ ai2 b2 j þ    þ ain bnj ¼

X

aik bkj

ðA:7Þ

k

Therefore, the number of columns in A must be equal to the number of rows in B. Observe that if A is an m
n matrix and B is an n
p matrix, then C is an m
p matrix. Observe also that, in general, AB 6¼ BA. That is, matrix multiplication is not commutative. Matrix multiplication, however, is distributive, that is, if A and B are m
p matrices and C is a p
n matrix, then ðA þ BÞC ¼ AC þ BC

ðA:8Þ

A.2 Matrix Operations

383

Example A.2

Let 2

0

6 A ¼4 2 3

4

1

3

1

7 1 5,

2

1

2

0

1

5

2

6 B ¼ 40

3

7 05

Then 2

0

6 AB ¼ 4 2 3

4

1

32

0

1

5

2

1

76 1 54 0

2

1

3

2

5

2

5

5

7 6 05 ¼ 45

3

7 45

Observe that the product BA is not defined in this example since the number of columns in B is not equal to the number of rows in A. The associative law is valid for matrix multiplications. That is, if A is an m
p matrix, B is a p
q matrix, and C is a q
n matrix, then (AB)C ¼ A(BC) ¼ ABC. Determinant The determinant of an n scalar defined as 2 a11 6 6 a21 6 jAj ¼ 6 6⋮ 4 an1


n square matrix A, denoted as jAj, is a a12

   a1n

3

⋮

7    a2n 7 7 7 ⋱ ⋮7 5

an2

   ann

a22

ðA:9Þ

In order to be able to evaluate the unique value of the determinant of A, some basic definitions have to be made first. The minor Mij corresponding to the element aij is the determinant formed by deleting the ith row and jth column from the original determinant jAj. The cofactor Cij of the element aij is defined as Cij ¼ ð1Þiþj M ij

ðA:10Þ

Using this definition of the cofactors Cij, which are determinants of order n  1, the value of the determinant in Eq. A.9 can be obtained in terms of the cofactors of the elements of an arbitrary row i as follows: jAj ¼

n X j¼1

aij C ij

ðA:11Þ

384

Appendix A: Linear Algebra

If A is a 2
2 matrix defined as  A¼

 a12 , a22

a11 a21

the cofactors Cij associated with the elements of the first row are C 11 ¼ ð1Þ2 a22 ¼ a22 ,

C12 ¼ ð1Þ3 a21 ¼ a21

ðA:12Þ

According to the definition of Eq. A.11, the determinant of the 2
2 matrix A can be determined using the cofactors of the elements of the first row as jAj ¼ a11 C 11 þ a12 C 12 ¼ a11 a22  a12 a21 If A is a 3
3 matrix defined as 2

a11

6 A ¼ 4 a21 a31

a12 a22 a32

a13

3

7 a23 5, a33

the determinant of A in terms of the cofactors of the first row is given by jAj ¼

3 X

a1 j C1 j ¼ a11 C 11 þ a12 C 12 þ a13 C13

j¼1

where C 11

 a ¼  22 a32

 a23  , a33 

C 12

 a ¼  21 a31

 a23  , a33 

C 13

 a ¼  21 a31

 a22  a32 

That is, the determinant of A is        a22 a23        a12  a21 a23  þ a13  a21 a22  jAj ¼ a11      a32 a33 a31 a33 a31 a32  ¼ a11 ða22 a33  a23 a32 Þ  a12 ða21 a33  a23 a31 Þ þ a13 ða21 a32  a22 a31 Þ ðA:13Þ One can show that the determinant of a square matrix is equal to the determinant of its transpose, that is, |A| ¼ |AT|, and the determinant of a diagonal matrix is equal to the product of the diagonal elements. Furthermore, the interchange of any two columns or rows changes only the sign of the determinant. One can also show that if a matrix has two identical rows or two identical columns, the determinant of this matrix is equal to zero. This can be demonstrated by the example of Eq. A.13, for example, if the second and third rows are identical, a21 ¼ a31, a22 ¼ a32, and

A.2 Matrix Operations

385

a23 ¼ a33. Using these equalities in Eq. A.13, one can show that the determinant of the matrix A in this special case is equal to zero. A matrix whose determinant is equal to zero is said to be a singular matrix. For an arbitrary square matrix, singular or nonsingular, it can be shown that the value of the determinant does not change if any row or column is added to or subtracted from another. Inverse of a Matrix A square matrix A1 that satisfies the relationship A1 A ¼ AA1 ¼ I

ðA:14Þ

where I is the identity matrix, is called the inverse of the matrix A. The inverse of the matrix A is defined as A1 ¼

Ct jAj

ðA:15Þ

where Ct is the adjoint of the matrix A. The adjoint matrix Ct is the transposed matrix of the cofactors Cij of the matrix A.

Example A.3

Determine the inverse of the matrix 2

1 6 A ¼ 40

1 1

3 1 7 15

0

0

1

Solution. The determinant of the matrix A is equal to one, that is, |A| ¼ 1. The cofactors of the elements of the matrix A are C 11 ¼ 1, C 12 ¼ 0, C 22 ¼ 1, C 23 ¼ 0,

C 13 ¼ 0, C 31 ¼ 0,

C21 ¼ 1 C32 ¼ 1

C 33 ¼ 1 The adjoint matrix, which is the transpose of the matrix of the cofactor elements, is given by 2

C 11 Ct ¼ 4 C 12 C 13

C21 C22 C23

3 2 1 1 C 31 1 C 32 5 ¼ 4 0 0 0 C 33

3 0 1 5 1

Therefore, (continued)

386

Appendix A: Linear Algebra

A1

2 1 Ct ¼ 40 ¼ jAj 0

1 1 0

3 0 1 5 1

It follows that 2

1 1 A A ¼ 40 1 0 0 1

32 0 1 1 54 0 1 0

1 1 0

3 2 1 1 15 ¼ 40 1 0

3 0 0 1 0 5 ¼ AA1 0 1

Note that if A is the 2
2 matrix 

a A ¼ 11 a21

 a12 , a22

the inverse of A can be simply written as A1 ¼

 1 a22 jAj a21

a12 a11



where |A| ¼ (a11a12  a12a21). If the determinant of A is equal to zero, the inverse of A does not exist. This is the case of a singular matrix. If A and B are nonsingular square matrices, then ðABÞ1 ¼ B1 A1

ðA:16Þ

It can also be verified that


A1

T


1 ¼ AT

ðA:17Þ

That is, the transpose of the inverse of a matrix is equal to the inverse of its transpose. A square matrix A is said to be orthogonal if AT A ¼ AAT ¼ I

ðA:18Þ

In this case, AT ¼ A1. That is, the inverse of an orthogonal matrix is equal to its transpose. An example of orthogonal matrices is  A¼ For this matrix, one has

cos θ sin θ

sin θ cos θ

 ðA:19Þ

A.3 Vectors

387



  cos θ sin θ cos θ sin θ A A¼ sin θ cos θ sin θ cos θ   cos 2 θ þ sin 2 θ 0 ¼ 0 sin 2 θ þ cos 2 θ T

Using the trigonometric identity cos2 θ + sin2 θ ¼ 1, one obtains ATA ¼ I, and the matrix A defined by Eq. A.19 is indeed an orthogonal matrix.

A.3 Vectors An n-dimensional vector a is an ordered set a ¼ ð a1 ; a2 ; . . . ; an Þ

ðA:20Þ

of n scalars. The scalar ai,i ¼ 1,2,. . .,n, is called the ith component of a. An n-dimensional vector can be considered as an n
1 matrix that consists of only one column. Therefore, the vector a can be written in the following column form 2

3 a1 6 a2 7 7 a¼6 4⋮5 an

ðA:21Þ

The transpose of this column vector defines the n-dimensional row vector aT ¼ [a1 a2 . . . an]. The vector a of Eq. A.21 can also be written as a ¼ ½ a1

...

a2

an  T

ðA:22Þ

By considering the vector as special case of a matrix with only one column or one row, the rules of matrix addition and multiplication apply also to vectors. For example, if a and b are two n-dimensional vectors, defined, respectively, as a ¼ ½ a1

a2

...

an  T ,

b ¼ ½ b1

b2

. . . b n T ,

then aþb is defined as a þ b ¼ ½ a1 þ b1

a2 þ b2

   an þ bn  T

Two vectors a and b are equal if and only if ai ¼ bi for i ¼ 1,2,. . ., n. The product of a vector a and scalar α is the vector αa ¼ ½ αa1

αa2

. . . αan T

ðA:23Þ

388

Appendix A: Linear Algebra

The dot, inner, or scalar product of two vectors a ¼ ½ a1 a2 . . . an T and b ¼ ½ b1 b2 . . . bn T is defined by the following scalar quantity 2

3 b1 6 b2 7 7 ab ¼ aT b ¼ ½ a1 a2 . . . an 6 4⋮5 bn ¼ a1 b1 þ a2 b2 þ    þ a n bn

ðA:24Þ

which can be written as ab ¼ aT b ¼

n X

ðA:25Þ

ai bi

i¼1

It follows that a  b ¼ b  a. The length of a vector a, denoted as jaj, is defined as the square root of the dot product of a with itself, that is, j aj ¼

pffiffiffiffiffiffiffiffi
1=2 aT a ¼ a21 þ a22 þ    þ a2n

ðA:26Þ

The terms modulus, magnitude, norm, and absolute value of a vector are also used to denote the length of a vector. Example A.4

Let a and b be the two vectors a ¼ ½0

1

3

2 T ,

3 T

b ¼ ½ 1

0

2

3 2 T þ ½ 1

0

2 3 T

then a þ b ¼½0 ¼ ½ 1

1

1 5

5

T

The dot product of a and b is 2

1

3

7 6 6 0 7 7 2 6 6 2 7 5 4 3 ¼ 0 þ 0 þ 6 þ 6 ¼ 12

a  b ¼ aT b ¼ ½ 0

1 3

The length of the vectors a and b is defined as (continued)

A.3 Vectors

389

i1=2 pffiffiffiffiffi pffiffiffiffiffiffiffiffi h 2 aT a ¼ ð 0Þ þ ð 1Þ 2 þ ð 3Þ 2 þ ð 2Þ 2 ¼ 14  3:742 h i pffiffiffiffiffiffiffiffi 1=2 pffiffiffiffiffi ¼ 14  3:742 jbj ¼ bT b ¼ ð1Þ2 þ ð0Þ2 þ ð2Þ2 þ ð3Þ2

j aj ¼

Differentiation In many applications in mechanics, scalar and vector functions that depend on one or more variables are encountered. An example of a scalar function that depends on the system velocities and possibly the system coordinates is the kinetic energy. Examples of vector functions are the coordinates, velocities, and accelerations that depend on time. Let us first consider a scalar function f that depends on several variables q1, q2,. . ., qn and the parameter t, that is, f ¼ f ð q 1 ; q2 ; . . . qn ; t Þ

ðA:27Þ

where q1, q2, . . ., qn are functions of t, that is, qi ¼ qi(t). The derivative of f with respect to t is df ∂ f dq1 ∂ f dq2 ∂ f dqn ∂ f ¼ þ þ  þ þ dt ∂q1 dt ∂q2 dt ∂qn dt ∂t

ðA:28Þ

which can be written using vector notation as 3 dq1 6 dt 7 7 6 6 dq2 7 7 ∂f 6 ∂f 6 7  6 dt 7 þ 7 ∂t ∂qn 6 6 ⋮ 7 7 6 4 dq 5 n dt 2

 df ∂f ¼ dt ∂q1

∂f ∂q2

ðA:29Þ

This equation can be written compactly as df ∂ f dq ∂ f ¼ þ dt ∂q dt ∂t

ðA:30Þ

in which ∂f/∂t is the partial derivative of f with respect to t and 9 q2    qn T > =   ∂f ∂f ∂f ∂f > ; ¼ fq ¼  ∂q ∂q1 ∂q2 ∂qn

q ¼ ½ q1

ðA:31Þ

The second equation in this equation defines the partial derivative of a scalar function with respect to a vector as a row vector. Note that if f is not an explicit function of time ∂f/∂t ¼ 0.

390

Appendix A: Linear Algebra

Example A.5

Consider the function f ðq1 ; q2 ; t Þ ¼ q21 þ 3q32  t 2 where q1 and q2 are functions of the parameter t. The total derivative of f with respect to the parameter t is df ∂ f dq1 ∂ f dq2 ∂ f ¼ þ þ dt ∂q1 dt ∂q2 dt ∂t where ∂f ¼ 2q1 , ∂q1

∂f ¼ 9q22 , ∂q2

∂f ¼ 2t ∂t

Hence, df dq dq ¼ 2q1 1 þ 9q22 2  2t dt dt dt 2 3 dq1  6 dt 7 7 ¼ 2q1 9q22 6 4 dq 5  2t 2

dt where ∂f/∂q can be recognized as the row vector  ∂f ¼ f q ¼ 2q1 ∂q

9q22



Consider the case of vector functions that depend on several variables. These vector functions can be written as
9 f 1 ¼ f 1 q1 ; q2 ; . . . ; qn , t >
> > f 2 ¼ f 2 q1 ; q2 ; . . . ; qn , t = > ⋮ > > ; f m ¼ f m ð q1 ; q2 ; . . . ; qn ; t Þ

ðA:32Þ

where qi ¼ qi(t),i ¼ 1,2,. . ., n. Using the procedure previously outlined in this section, the total derivative of an arbitrary function fj can be written as

A.3 Vectors

391

dfj ∂fj dq ∂fj þ ¼ , dt ∂q dt ∂t

j ¼ 1,2, . . . , m

ðA:33Þ

in which ∂fj /∂q is the row vector  ∂fj ∂fj ¼ ∂q ∂q1

∂fj ∂q2



∂fj ∂qn

 ðA:34Þ

Consequently, 3 2 ∂f1 df1 ∂q1 6 dt 7 6 7 6 6 6 df 7 6 7 6 ∂f2 df 6 6 27 6 ¼ 6 dt 7 ¼ 6 ∂q 7 6 1 dt 6 6 ⋮ 7 6 6 ⋮ 7 6 4 df 5 6 4 ∂f m m dt ∂q 2

1

∂f1 ∂q2 ∂f2 ∂q2 ⋮ ∂fm ∂q2

3 ∂f1 2 dq 3 2 ∂f 3 1 1 ∂qn 7 7 6 ∂t 7 dt 76 7 6 7 76 7 6 7 ∂f2 76 dq ∂f 7 7 6 6 2 2 76  7 7 6 ∂qn 7 dt 7 þ 6 ∂t 7 76 7 7 6 6 7 ⋮ 7 6 ⋮ 7 ⋱ ⋮ 76 7 7 6 6 74 5 4 5 ∂fm ∂fm 5 dqn  dt ∂t ∂qn 

ðA:35Þ

where f ¼ ½ f1

f2

   f m T

ðA:36Þ

Equation A.35 can be written compactly as df ∂f dq ∂f ¼ þ dt ∂q dt ∂t

ðA:37Þ

where the m
n matrix ∂f/∂q, the n-dimensional vector dq/dt, and the m-dimensional vector ∂f/∂t can be recognized, respectively, as 2

∂f1 6 ∂q 6 1 6 6 ∂f 2 6 ∂f ¼ fq ¼ 6 6 ∂q1 ∂q 6 6 ⋮ 6 4 ∂f m ∂q1

3 2 3 2 3 ∂f1 ∂f1 ∂f1 dq1   7 ∂q2 ∂qn 7 6 ∂t 7 6 dt 7 6 7 6 7 7 6 7 6 dq2 7 ∂f 2 ∂f 2 7 ∂f 6 7 2 6 7 7 dq ∂f   6 7 6 7 7 ðA:38Þ ∂q2 ∂qn 7 ∂t 7, dt ¼ qt ¼ 6 dt 7, ∂t ¼ f t ¼ 6 6 7 6 ⋮ 7 7 6 7 6 7 ⋮ ⋱ ⋮ 7 6 ⋮ 7 4 dq 5 7 4 ∂f m 5 n ∂f m ∂f m 5   dt ∂t ∂q2 ∂qn

If fj is not an explicit function of the parameter t, then ∂fj/∂t is equal to zero. Note also that the partial derivative of an m-dimensional vector function f with respect to an n-dimensional vector q is the m
n matrix fq defined in the preceding equation.

392

Appendix A: Linear Algebra

Example A.6

Consider the vector function f defined as 2

f1

3

2

6 7 6 f ¼ 4 f2 5 ¼ 4 f3

q21 þ 3q32  t 2

3

7 8q21  3t 5 2 2 2q1  6q1 q2 þ q2

The total derivative of the vector function f is 3 df 1 6 dt 7 2 2q1 7 6 df 6 df 2 7 7¼6 ¼6 16q1 7 4 dt 6 6 dt 7 ð4q1  6q2 Þ 4 df 5 3 dt 2

32 dq 3

9q22

1

2

2t

3

76 dt 7 6 7 0 54 5 þ 4 3 5 dq2 0 ð2q2  6q1 Þ dt

where the matrix fq can be recognized as 2 6 fq ¼ 4

2q1 16q1

9q22 0

ð4q1  6q2 Þ

ð2q2  6q1 Þ

3 7 5

and the vector ft is ∂f ¼ f t ¼ ½ 2t ∂t

3

0 T

In the analysis of mechanical systems, one may also encounter scalar functions in the form Q ¼ qTAq. Following a similar procedure to the one previously outlined in this section, one can show that ∂Q/∂q ¼ qT(AþAT). If A is a symmetric matrix, that is, A ¼ AT, one has ∂Q/∂q ¼ 2qTA. Linear Independence The concepts to be introduced here are of fundamental importance in the development presented in this book. Their use is crucial in formulating many of the dynamic relationships presented in several chapters of this text. The vectors a1, a2, . . ., an are said to be linearly dependent if there exist scalars e1, e2, . . . , en, which are not all zeros, such that e 1 a1 þ e 2 a2 þ    þ e n an ¼ 0

ðA:39Þ

Otherwise, the vectors a1, a2, . . ., an are said to be linearly independent. In the case of linearly independent vectors, none of these vectors can be expressed in terms of

A.3 Vectors

393

the others. On the other hand, if Eq. A.39 holds, and not all the scalars e1, e2, . . ., en are equal to zero, one or more of the vectors a1, a2, . . ., an can be expressed in terms of the other vectors. Equation A.39 can be written in matrix form as 2 ½ a1

a2

e1

3

6e 7 6 27    an  6 7 ¼ 0 4⋮5

ðA:40Þ

en which can be written compactly as Ae ¼ 0, where e ¼ ½ e1 e2    en T and the columns of the coefficient matrix A are the vectors a1, a2, . . ., an, that is, A ¼ ½ a1 a2    an . If the vectors a1, a2, . . ., an are linearly dependent, the system of homogeneous algebraic equations Ae ¼ 0 has a nontrivial solution. On the other hand, if the vectors a1, a2, . . ., an are linearly independent vectors, then A must be a nonsingular matrix since the system of homogeneous algebraic equations Ae ¼ 0 has only the trivial solution e ¼ A10 ¼ 0. Consequently, in the case where the vectors a1, a2, . . ., an are linearly dependent, the square matrix A must be singular. The number of linearly independent columns in a matrix is called the column rank of the matrix, while the number of independent rows is called the row rank of the matrix. It can be shown that for any matrix, the row rank is equal to the column rank and is equal to the rank of the matrix. Therefore, a square matrix which has a full rank is a matrix which has linearly independent rows and linearly independent columns. One concludes, therefore, that a matrix which has a full rank is a nonsingular matrix. Consequently, if a1, a2, . . ., an are n-dimensional linearly independent vectors, any other n-dimensional vector can be expressed as a linear combination of these vectors. For instance, let b be another n-dimensional vector. We show that this vector has a unique representation in terms of the linearly independent vectors a1, a2, . . ., an. To this end, we write b as b ¼ x 1 a1 þ x 2 a2 þ    þ x n an

ðA:41Þ

where x1, x2, . . ., xn are scalars. In order to show that x1, x2, . . ., xn are unique, Eq. A.41 can be written as b ¼ Ax, where A ¼ ½ a1 a2 . . . an  is a square matrix and x ¼ ½ x1 x2    xn T . Since the vectors a1, a2, . . ., an are assumed to be linearly independent, the coefficient matrix A has a full row rank and, therefore, it is nonsingular. The system of algebraic equations Ax ¼ b has a unique solution x which can be written as x ¼ A1 b. That is, an arbitrary n-dimensional vector b has a unique representation in terms of the linearly independent vectors a1, a2, . . ., an. A familiar and important special case is the case of three-dimensional vectors. One can show that the three vectors 2 3 1 a1 ¼ 4 0 5 , 0

2 3 0 a2 ¼ 4 1 5, 0

2 3 0 a3 ¼ 4 0 5 , 1

394

Appendix A: Linear Algebra

are linearly independent. Any other three-dimensional vector b ¼ ½ b1 b2 b3 T can be written in terms of the linearly independent vectors a1,a2, and a3 as b ¼ b1a1 + b2a2 + b3a3, where the coefficients x1, x2, and x3 can be recognized in this special case as x1 ¼ b1, x2 ¼ b2, and x3 ¼ b3. The coefficients x1, x2, and x3 are called the coordinates of the vector b in the basis defined by the vectors a1, a2, and a3. Example A.7

Show that the vectors 2 3 1 6 7 a1 ¼ 4 0 5, 0

2 3 1 6 7 a2 ¼ 4 1 5 , 0

2 3 1 6 7 7 a3 ¼ 6 415 1

are linearly independent. Find also the representation of the vector b ¼ ½ 1 3 0 T in terms of the vectors a1, a2, and a3. Solution. In order to show that the vectors a1, a2, and a3 are linearly independent, we must show that the relationship e 1 a1 þ e 2 a2 þ e 3 a3 ¼ 0 holds only when e1 ¼ e2 ¼ e3 ¼ 0. To this end, we write 2 3 2 3 2 3 1 1 1 6 7 6 7 6 7 e1 4 0 5 þ e2 4 1 5 þ e3 4 1 5 ¼ 0 0 0 1 which leads to e1 þ e2 þ e3 ¼ 0 e2 þ e3 ¼ 0 e3 ¼ 0 Back substitution shows that e3 ¼ e2 ¼ e1 ¼ 0 That is, the vectors a1, a2, and a3 are linearly independent. In order to find the unique representation of the vector b in terms of these linearly independent vectors, we write (continued)

A.4 Eigenvalue Problem

395

b ¼ x 1 a1 þ x 2 a2 þ x 3 a 3 which can be written in a matrix form as b ¼ Ax, where 2

3 1 7 1 5,

1 1 6 A ¼ 40 1 0 0

2

3 1 6 7 b ¼ 4 35

1

0

Therefore, the vector of coordinates x can be obtained as 2

x1

3

2

1 1

6 7 6 x ¼ 4 x2 5 ¼ A1 b ¼ 4 0

1

0

0

x3

1

32

1

3

2

4

3

76 7 6 7 1 54 3 5 ¼ 4 3 5 1

0

0

A.4 Eigenvalue Problem In the analysis of structural systems, we often encounter a system of homogeneous algebraic equations in the form Ay ¼ λy

ðA:42Þ

where A is a square matrix, y is an unknown vector, and λ is an unknown scalar. Equation A.42 can be written in the form ðA  λIÞy ¼ 0

ðA:43Þ

where I is the identity matrix. Equation A.43 represents an algebraic system of homogeneous equations which have a nontrivial solution if and only if the coefficient matrix (A  λI) is singular. That is, the determinant of this matrix is equal to zero. This leads to jA  λIj ¼ 0

ðA:44Þ

This is called the characteristic equation of the matrix A. If A is an n
n matrix, Eq. A.44 is a polynomial of order n in λ. This equation can be written in the following form: an λn þ an1 λn1 þ    þ a0 ¼ 0

ðA:45Þ

where ai, i ¼ 0, 1, 2, . . ., n, are the scalar coefficients of the polynomial. The solution of Eq. A.45 defines n roots λ1, λ2, . . ., λn. These roots are called the characteristic values or the eigenvalues of the matrix A. Associated with each of these eigenvalues,

396

Appendix A: Linear Algebra

there is an eigenvector yi which can be determined to within an arbitrary constant by solving the system of equations ðA  λi IÞyi ¼ 0

ðA:46Þ

If A is a real-symmetric matrix, one can show that the eigenvectors associated with distinctive eigenvalues are orthogonal, that is, yiT y j ¼ 0 if i 6¼ j

) ðA:47Þ

yiT y j 6¼ 0 if i ¼ j

Example A.8

Find the eigenvalues and eigenvectors of the matrix 2

4 A ¼ 41 2

1 0 0

3 2 05 0

Solution. The characteristic equation of this matrix is  4  λ 1   λ jA  λIj ¼  1   2 0

      λ  2 0

¼ ð4  λÞλ2 þ λ þ 4λ ¼ 0 This equation can be rewritten as λ ð λ  5Þ ð λ þ 1Þ ¼ 0 which has the roots λ1 ¼ 0,

λ2 ¼ 5,

λ3 ¼ 1

The ith eigenvector associated with the eigenvalue λi can be obtained using the equation ðA  λi IÞyi ¼ 0 The solution of this equation yields 2

3 0 y1 ¼ 4 2 5, 1

2 3 5 y2 ¼ 4 1 5 , 2

2

3 1 y3 ¼ 4 1 5 2

Problems

397

Problems A.1. Find the sum of the following two matrices 2 3 2 3:0 8:0 20:5 0 A ¼ 4 5:0 11:0 13:0 5, B ¼ 4 17:5 7:0 20:0 0 12:0

3 3:2 0 5:7 0 5 6:8 10:0

Evaluate also the determinant and the trace of A and B. A.2. Find the product AB and BA, where A and B are given in Problem 1. A.3. Find the inverse of the following matrices: 2

3 2 1 1 0 5, 1 1

1 A¼4 2 0

2

0 B ¼ 4 2 6

3 2 2

3 5 3 5 0

A.4. Show that an arbitrary square matrix A can be written as A ¼ A1 þ A2 where A1 is a symmetric matrix and A2 is a skew-symmetric matrix. A.5. Show that the interchange of any two rows or columns of a square matrix changes only the sign of the determinant. A.6. Show that if a matrix has two identical rows or two identical columns, the determinant of this matrix is equal to zero. A.7. Let  A¼

A11 A21

A12 A22



be a nonsingular matrix. If A11 is square and nonsingular, show by direct matrix multiplications that A

1


¼

1 A1 11 þ B1 H B2 1 H B2



B1 H1 H1

where B2 ¼ A21 A1 B1 ¼ A1 11 A12 , 11 H ¼ A22  B2 A12 ¼ A22  A21 B1 ¼ A22  A21 A1 11 A12



398

Appendix A: Linear Algebra

A.8. Let a and b be the two vectors a ¼ ½1

0

5 T ,

3 2

b ¼ ½ 0 1

2

3 8 T

Find aþb, a  b, |a| and |b|. A.9. Find the total derivative of the function f ðq1 ; q2 ; q3 ; t Þ ¼ q1 q3  3q22 þ 5t 5 with respect to the parameter t. Define also the partial derivative of the function f with respect to the vector q(t), where qðt Þ ¼ ½ q1 ðt Þ q2 ðt Þ q3 ðt Þ T . A.10. Find the total derivative of the vector function 3 2 3 2 2 f1 q1 þ 3q22  5q34 þ t 3 5 f ¼ 4f2 5 ¼ 4 q22  q23 f3 q1 q4 þ q2 q3 þ t with respect to the parameter t. Define also the partial derivative of the function f with respect to the vector q ¼ ½ q1 q2 q3 q4 T . A.11. Let Q ¼ qTAq, where A is an n
n square matrix and q is an n-dimensional vector. Show that
 ∂Q ¼ qT A þ A T ∂q A.12. Show that the vectors 2 3 0 a1 ¼ 4 0 5, 1

2 3 0 a2 ¼ 4 1 5 , 1

2 3 1 a3 ¼ 4 1 5 1

are linearly independent. Determine also the coordinates of the vector b ¼ ½ 1 5 3 T in the basis a1, a2, and a3. A.13. Find the rank of the following matrices 2

2 A ¼ 46 4

3 5 1 9 3 5, 0 2

2

3 B ¼ 42 7

5 1 0 1 1 2

3 0 35 9

Problems

399

A.14. Find the eigenvalues and eigenvectors of the following two matrices 2

2 A ¼ 4 1 0

3 1 0 2 1 5, 1 1

2

6 B ¼ 4 2 4

2 2 4

3 0 3 5 3

A.15. Show that if A is a real-symmetric matrix, then the eigenvectors associated with distinctive eigenvalues are orthogonal.

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Index

A Absolute nodal coordinate formulation (ANCF), 98, 281, 329–333 Analysis of higher modes, 169, 232 Angular acceleration, 6, 9, 12, 101 Angular oscillations, 23, 51, 58, 112, 135, 187, 200 Approximation methods, 201, 255–264, 334 Arbitrary forcing function, 1, 39–42, 265 Assumed displacement field, 269, 279, 281, 286, 287, 289, 296, 313, 314, 323, 326 Assumed-modes method, 269–271 B Beam element, 280, 285–287, 289, 291–293, 298, 307, 308, 316, 324, 330, 332, 337–342 Boolean matrix, 291, 296 Boundary conditions, 184, 188, 201, 205, 208–210, 212, 213, 216, 218–220, 226, 227, 229, 235, 238, 240, 241, 243, 245, 247, 248, 253, 256–260, 265, 266, 268, 269, 272–274, 313, 335 Brick element, see Solid element

Concentrated loads, 242 Condition for similarity, 350 Connectivity conditions, 290, 291 Connectivity of the finite elements, 279, 290–296 Conservation of energy, 55, 90, 107, 140–143, 253, 255, 256 Conservation theorems, 55, 88–93 Conservative systems, 55, 88, 89, 93 Consistent-mass formulation, 296 Constant strain triangular element, 283 Constitutive equations, 301, 303, 333 Constrained motion, 7 Continuous systems, 1, 11, 16, 42, 192, 201–271, 279–281, 300, 314 Convergence of the finite-element solution, 281, 313–317 Convolution integral, 40 Coordinate reduction, 270, 334, 335 Coordinate transformation, 125, 290, 329, 330, 335 Coulomb damping, 1, 26, 28 Critical damping coefficient, 18–24, 30, 35, 39, 50 Critically damped systems, 20, 21, 48

C Cayley–Hamilton theorem, 347, 348, 376 Characteristic equation, 17, 18, 20, 21, 119, 121, 137, 138, 159, 165, 184, 186, 189, 206, 264, 313, 314, 320, 355, 395 Characteristic matrix, 155, 159, 343, 346, 350, 351, 353, 354, 357, 360, 361, 376 Characteristic polynomial, 345–349, 356 Characteristic values, 119, 357, 395 Characteristic vector, 119 Coefficient of sliding friction, 26 Completeness, 287

D Damped natural frequency, 22, 38, 39, 48, 147 Damping general viscous, 108, 152 proportional damping, 107, 146, 147, 161 viscous damping, 17, 24–26, 28, 108, 151 Damping factor, 18–24, 31, 33, 35, 38, 39, 48, 147, 148 Damping matrix, 146, 147, 151, 154, 310 Deformation modes, 134, 139, 176, 229, 239 Degree of freedom, 7

# Springer Nature Switzerland AG 2019 A. Shabana, Vibration of Discrete and Continuous Systems, Mechanical Engineering Series, https://doi.org/10.1007/978-3-030-04348-3

405

406 Degrees of freedom, 9, 12, 14, 42, 56, 57, 67, 83, 96, 97, 107, 108, 119, 132, 134, 143, 148, 161–165, 180, 184, 192, 201, 208, 232, 259, 261, 269, 270, 279, 283, 286, 293, 295–297, 310, 313, 314, 316, 318, 320, 321, 324–326, 328, 329, 333–335 Direct numerical integration, see Numerical integration Direction cosines, 322, 323 Dry friction, see Coulomb friction Duhamel integral, 40 E Eigenfunctions, 201, 205, 206, 209, 212, 217, 219, 230, 232, 236, 239–241, 245, 251, 257–259, 261, 263, 269, 278, 316 Eigenvalue analysis, 343–378 Eigenvalues, 42, 43, 119, 124, 125, 133, 138, 142, 145, 151, 153–155, 159, 163, 165–167, 169, 190, 192, 257, 313, 320, 335–337, 343–378, 395, 396, 399 Eigenvectors, 119, 120, 124, 125, 133, 137, 142, 145, 152–154, 156, 159, 165, 190, 313, 336, 345, 346, 350, 352, 354–356, 358, 361–367, 370–372, 374–378, 395, 396, 399 Elastic waves, 204 Element mass matrix, 296–298, 317, 330 Element nodal coordinates, 279, 281–283, 285, 288, 291, 297, 313, 330, 331 Element stiffness matrix, 303, 306, 308, 319, 320, 330 Elementary beam theory, 201, 221, 223, 307 Elementary divisors, 359–361, 377 Energy loss, 25 Energy methods, 252–255 Equations of motion, 1, 9, 13, 42–44, 55, 59, 66, 67, 81, 84, 86, 90, 91, 93, 95, 97, 99, 101–105, 107–112, 114, 115, 117, 118, 123, 131, 143–145, 151, 160, 162–164, 189, 193, 194, 196, 198, 200, 201, 242, 253, 254, 270, 281, 309–313, 329, 333, 335, 341 Equivalent coefficients, 22 Euler–Bernoulli law, 223 Experimental modal analysis, 147, 148 Exponential matrix forms, 153 F Field matrix, 183, 187 Finite element (FE) method, 279–337 Finite elements (FE) beam element, 191, 280, 285, 291

Index plate element, 324, 328–330 rectangular element, 280, 286, 288, 316, 328 solid element, 325 tetrahedral element, 326, 327, 329 triangular element, 280, 283, 289, 316 truss element, 280, 282, 287, 288, 297, 298, 306, 307, 314, 316, 318, 320, 321 Flexibility coefficients, 117 Flexibility formulation, 165, 166, 168, 176, 177 Flexibility matrix, 117, 118 Force transmission, 32, 33 Forced damped vibration, 109 Forced vibration longitudinal and torsional, 201, 221, 232, 245 transverse, 201, 221–232, 236, 237, 253, 256, 265, 269, 270, 274 of undamped systems, 21, 31 Formulation of the mass matrix, 281, 296–300 Formulation of the stiffness matrix, 300 Free damped vibration, 109 Free longitudinal vibration, 202–213 Free nodal coordinates, 293, 313 Free torsional vibration, 214–220, 275 Free transverse vibration, 221–232, 236 Free undamped vibration, 90, 109, 176, 253, 255, 313 Free vibration, 1, 13, 17, 22–24, 38, 40, 90, 98, 100, 107, 116, 118–123, 125, 131, 134, 138, 140, 141, 155, 180, 193, 197, 199, 203, 208, 224, 226, 241, 250, 254, 255, 335–337, 341 Frequency equation, 209, 211–214, 218–220, 227, 229–231, 264, 272–274 G Galerkin’s method, 265–269, 278 General viscous damping, 108, 151–160 Generalized coordinates, 55–61, 63–67, 69–71, 74, 75, 82, 83, 93, 102, 117, 177, 180, 270, 340 Generalized eigenvalue problem, 192, 367 Generalized eigenvectors, 346, 361–364, 377 Generalized forces, 55, 59–65, 67, 68, 70, 78, 80–82, 85, 93, 111, 117, 160, 253, 270, 310, 311, 340 Generalized Hooke’s Law, 301 Generalized inertia forces, 71–73, 78, 135 Geometric boundary conditions, 205, 226, 227, 257, 259, 263 Greatest common divisors, 344, 354, 359, 361 Group velocity, 252

Index H Hamilton’s principle, 55, 84–88, 93, 104, 105 Hamiltonian, 89–92 Higher order elements, 317–321 Holzer’s method, 180, 181, 184 Householder transformations, 367–375 Hysteretic damping, see Structural damping I Impulse response function, 38–40 Inhomogeneous, 201, 247–249 Inhomogeneous boundary conditions, 201, 247–249 Initial conditions, 19–22, 27, 29, 36, 38, 40, 41, 46, 48, 50–52, 119, 121–123, 127, 131–133, 144, 145, 157, 197, 204, 206, 207, 216, 217, 226, 227, 243, 247–249, 251, 253, 254, 269, 274, 275, 278 Invariant factors, 351, 359–361, 377 Inverse of the modal matrix, 127, 128, 131, 132 Isotropic material, 301 J Jacobi method, 343, 365–367, 377 Jordan matrices, 354–359 K Kelvin–Voigt model, 250 Kinematic constraints, 14, 44, 57, 293, 296 Kinetic energy of rigid bodies, 75, 76 of system of particles, 66, 74, 99 Kronecker delta, 323 L Lagrange’s equation, 64, 71, 74, 78, 80–82, 84, 85, 88, 93, 95, 99, 101–103, 109, 111, 114–116, 193, 196, 197, 201, 252–254, 270, 310 vector form of, 78 Lagrangian approach, 55 dynamics, 55–97 Large rotations and deformations, 329–333 Least common multiple, 344, 349 Linear algebra, 120, 379–399 Linear impulse, 37–39 Linear oscillations, 14–16

407 Longitudinal vibrations, 202–214, 216, 217, 233, 234, 240–242, 244, 247, 248, 251, 252, 256, 260, 265, 269, 272–275, 278, 313 Lumped mass approach, 296 Lumped mass systems, 107 M Magnification factor, 30, 31, 34, 35 Mass matrix, 75–77, 79, 95, 97, 124, 125, 128–131, 133, 136, 138, 142, 146, 155, 164, 165, 170, 189, 267, 281, 296–300, 307, 317, 323, 330, 331 Matrix algebra, 379 Matrix iteration methods, 165–179, 200, 343 Method of transfer matrices, 179 Minimal polynomial, 348, 351, 356, 357, 376 Modal coordinates, 107, 123, 125–128, 131–134, 140, 141, 143–149, 151, 158, 160–162, 240, 242–245, 253–255, 336 Modal damping coefficients, 147, 148 Modal damping factor, 147 Modal damping matrix, 146 Modal force coefficients, 232 Modal mass coefficients, 232 Modal mass matrix, 126, 133 Modal matrix inverse of, 128, 131–133 Modal stiffness coefficients, 129, 131, 140, 150, 232, 234, 237, 239, 245, 254 modal matrix, 315 Modal stiffness matrix, 126 Modal transformation, 107, 123, 125–127, 140, 143, 144, 146, 149, 151, 156, 162, 163, 192 Modal truncation, 108, 161, 163, 192, 310 Mode shapes linear independence of, 124, 125 Modulus of elasticity, 191, 202, 203, 209, 212, 227, 233, 236, 239, 256, 258, 261, 263, 278, 287, 301, 305, 308, 309, 313, 314, 332 Modulus of rigidity, 115, 195, 215, 218, 219, 234, 274, 301, 302 Monic polynomials, 343, 344, 348, 349, 351, 354, 360 Multi-degree of freedom systems viscously damped, 148, 155, 157 N Natural boundary conditions, 205, 226, 227, 257

408 Natural frequencies, 18–22, 24, 27, 38, 39, 42, 43, 45, 46, 48–50, 107, 118, 119, 122, 124, 127–129, 137, 138, 141, 142, 148, 150, 159, 161–163, 165, 166, 169, 171, 176, 179–181, 184, 189, 192–194, 196, 197, 200, 206, 208, 209, 211, 213, 216–219, 226–229, 234, 236–239, 241, 254–259, 261, 262, 264, 269, 273, 278, 313–315, 320, 333 Newton–Euler equations, 6–9, 12 Nodal coordinates free, 293 specified, 293 Nodal points, 150, 269, 279–282, 284, 285, 290, 293, 296, 313, 318, 321, 324, 328, 329, 334, 337 Nondispersive medium, 252 Nonlinear equations, 43 Nonlinear systems, 1, 43, 44, 91, 95–97, 164, 189–192, 333 Nonlinear vibration, 96 Nonoscillatory motion, 17–24 Normal modes, 119, 191, 251 Normalized mode shapes, 128 Numerical integration, 97, 163, 164 O Orthogonality conditions, 4, 125, 137, 170–172, 175, 206, 207, 237, 239, 243, 252, 257 Orthogonality of the eigenfunctions, 201, 205, 232–239 Orthogonality of the mode shapes, 123–127, 129–133, 143, 146, 161, 176, 207, 217, 245 Oscillatory motion, 45, 204 Overdamped system, 18 P Parallel axis theorem, 95 Partial differential equations, 42, 201–203, 208, 212, 216, 223, 224, 232, 236, 239, 240, 244, 247–249, 265–267, 272, 274, 316 Phase velocity, 252 Planar kinematics, 2, 4 Planar motion, 2, 9, 12, 101, 316 Plane strain, 302, 303 Plane stress, 303, 305 Plate element, 324, 328–330 Point matrix, 187

Index Poisson’s ratio, 301, 302, 305 Polynomial matrix, 346–349 Positive definite continuous system, 239 Positive semidefinite continuous system, 239 Principal modes, 119, 201, 255 Principle of virtual work in dynamics, 93, 95, 99, 105, 135, 136 Proportional damping, 107, 146, 147, 161 Q QR decomposition, 372–375, 378 Quadratic form indefinite, 77 matrix of, 77 negative definite, 77 negative semidefinite, 77 positive, 77 positive semidefinite, 77 semidefinite, 77 Quotient, 343, 348 R Rayleigh quotient, 141–143, 255–260 Rayleigh’s method, 255, 257, 259 Rayleigh–Ritz method, 259, 278, 312, 313 Rectangular element, 280, 286, 288, 289, 304, 305, 316, 317, 328 Rectilinear mass-spring systems, 181–183, 186, 188 Rectilinear motion, 109 Remainder, 16, 146, 154, 166, 323, 343, 348, 365 Repeated roots, 125, 346, 350, 355 Residual, 265, 266 Resonance, 31, 43, 146, 148–150 Rigid bodies dynamics of, 64, 71 kinematics of, 1–6, 330, 331 Rigid-body modes, 133–140, 176–179, 192, 197, 228, 238, 287, 316, 317, 342 Ritz coefficients, 259–262 Rotary inertia, 223, 227, 229, 230, 258, 307 Runge–Kutta method, 164 S Semidefinite systems, 83, 107, 133, 140, 176, 239, 316 Separation of variables, 201, 203, 208, 209, 212, 216, 224, 233, 236, 240, 244, 251, 252, 261, 279, 312

Index Shape functions, 260, 261, 263, 266, 269, 278, 279, 281–283, 285–289, 296–298, 304, 307, 308, 311, 313, 316, 317, 325, 326, 328, 329, 331, 337–342 Similarity transformation, 335, 343–346, 350, 361–364 Simple end conditions, 233, 236, 237, 239, 256 Single degree of freedom systems, 1, 11, 13–18, 22, 29, 36–42, 45, 56, 61, 67, 85, 90, 98, 100, 107, 118, 125, 127, 133, 143, 144, 147, 148, 151, 158, 160, 163, 164, 199, 207, 208, 240, 241 Solid element, 325, 326, 342 Spatial elements, 281, 321–329 Spatial kinematics, 3, 4 Specified nodal coordinates, 293 Standard eigenvalue problem, 118, 138 State space representation, 154 State transition matrix, 158, 160, 199, 200 State vector, 154, 155, 157, 183, 184, 187 Steady state solution, 29, 30, 149, 198 Stiffness formulation, 165, 166 Stiffness matrix, 82, 83, 116, 117, 119, 124–126, 128–130, 133, 134, 136–138, 146, 165, 168, 176, 177, 181, 189, 192, 268, 281, 300–309, 316, 317, 319, 320, 323, 330, 332 Stodla method, 165, 179 Strain energy, 35, 55, 81, 83–85, 91, 111, 114, 115, 140, 239, 253–256, 270, 271, 281, 300, 303, 306–308, 314, 316, 317, 320, 323, 332 Strain-displacement relationships, 202, 250, 303, 307, 332 Stress–strain relationship, 300, 302, 307 Structural damping, 24, 25, 148, 250 coefficient, 25 Structure mass matrix, 297, 298 Structure stiffness matrix, 306 Sweeping matrix, 171–173, 175 System of particles, 59, 64, 66, 74, 75, 99 T Tetrahedral element, 326, 327, 329, 342 Torsional oscillations, 114–116, 145, 162, 215, 216, 234, 253, 265, 273

409 Torsional system, 115, 116, 121, 162, 167, 172, 178, 180, 181, 186, 188, 196, 234, 276 Torsional vibrations, 195, 201, 214–221, 232, 244, 245, 269, 275 Transfer matrices, 179 Transient solution, 29 Transmissibility, 33–36 Transmitted force, 33 Transverse vibrations, 201, 221–232, 236, 237, 239, 244, 245, 253, 256, 260, 265, 269, 270, 274 Trial shape functions, 259 Triangular element, 280, 283, 289, 316, 321 Truss element, 280, 284, 286–288, 298, 306, 314, 316, 318, 320, 337 U Undamped free vibration, 90, 107, 118–123, 125, 140, 254, 255 Undamped system, 21, 107, 143–146, 162, 198 Undamped vibration, 176, 253, 255, 312, 313 Underdamped systems, 18, 21, 35, 147 Unit impluse, 38 V Vector algebra, 289 Velocity vector, 5, 6, 9, 14, 74, 75, 79, 92, 296 Vibration of continuous systems, 201–271 Virtual displacement, 60, 65, 70, 80, 244 Virtual work, 55, 59–64, 66, 70, 80, 81, 83–87, 93–95, 98, 99, 101, 111, 114, 115, 135, 136, 242, 252, 267, 268, 271, 309–311 Viscoelastic materials, 201, 250–252 Viscously damped systems, 145–150 W Wave equation, 205 Wave motion, 232, 251, 252 Wave number, 205, 252 Wave velocity, 205, 206, 313 Weighted residual techniques, 266 Work per cycle, 33