Mechanical Vibrations [3 ed.] 0070616795, 9780070616790

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SPECIAL INDIAN EDITION

SCHAUM S

COUTlines

MECHANICAL VIBRATIONS

About the. Authors S. Graham Kelly is Associate Professor ·of Mechanical Engineering and Assistant Provost at The University of Akron. He has been on the faculty at Akron since 1982, serving previously at the Uriiversity of Notre Dame. He holds a B.S. in Engineering Science and Mechanics and an M.S. and a Ph.D. in Engineering Mechanics from Virginia Tech. He is also the author of Fundamentals of Mechanical Vibrations and the accompanying software VIBES, published by McGraw-Hill. Shashidhar K. Kudari is Assistant Professor of Mechanical Enginee1ing at B.V.B. College of Engineering and Technology, Hubli, India. He has been on the faculty at BVBCET since 1988. He holds a BE in Mechanical Engineering and M.Tech. and Ph.D. in Machine Design from the Indian Institute of Technology, Kharagpur. Besides Mechanical Vibrations, he has taught undergraduate courses in Machine Design, Finite Element Methods, Theory of Elasticity, Fracture Mechanics and Composite Materials. He is an active researcher and has several publications to his credit.

MECHANICAL VIBRATIONS S. Graham Kelly Associate Professor of Mechanical Engineering and Assistant Provost The University of Akron Adapted by

Shashidhar K. Kudari Assistant Professor, Department of Mechanical Engineering, BVB College of Engineering and Technology, Hubli

yi Tata McGraw-Hill Publishing Company Limited NEW DELHI McGraw-Hill Offices New Delhi New York St Louis San Francisco Auckland Bogota Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

Tata McGraw-Hill Special Indian Edition 2007 Published in India by arrangement with the McGraw-Hill Companies, Inc., New York Sales Territories: India, Pakistan, Nepal, Bangladesh, Sri Lanka and Bhutan Third reprint 2009 RLXDCRCFRARRB

Mechanical Vibrations Copyright © 1996, by The McGraw-Hill Companies, Inc. All Rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of The McGraw-Hill Companies, Inc. including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. ISBN-13: 978-0-07-061679-0 ISBN-10: 0-07-061679-5 Information contained in this work has been obtained by Tata McGraw-Hill, from sources believed to be reliable. However, neither Tata McGraw-Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGrawHill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGraw-Hill and its authors are supplying information but are not attempting to render professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at The Composers, 260, C.A. Apt., Paschim Vihar, New Delhi 110 063 and printed at Pushp Print Services B-39/12A, Arjun Mohalla, Maujpur, Shandara, Delhi-110 053 Cover Printer: Pushp Print Services The McGraw-Hill Companies

To my son, Graham S. Graham Kelly

Contents

Preface to the Special Indian Edition Preface

1. MECHANICAL SYSTEM ANALYSIS 1.1 1.2 1.3 1.4 1.5 1.6 1.7

Introduction Degrees of Freedom and Generalised Coordinates Types of Vibration Mechanical System Components Equivalent Systems Analysis Torsional Systems Static Equilibrium Position Solved Problems Practice Problems

2. FREE VIBRATIONS OF 1-DEGREE-OF-FREEDOM SYSTEMS 2.1 2.2 2.3 2.4 2.5 2.6

Introduction Derivation of Differential Equations Standard form of Differential Equations Undamped Response Damped Response Free Vibration Response for Systems Subject to Coulomb Damping Solved Problems Practice Problems

xiii xv

1.1-1.29 1.1 1.1 1.2 1.2 1.3 1.4 1.4 1.4 1.24

2.1-2.35 2.1 2.1 2.2 2.3 2.3 2.6 2.7 2.30

viii

Contents

3. HARMONIC EXCITATION OF 1-DEGREE-OF-FREEDOM SYSTEMS 3.1-3.47 3.1 Introduction 3.2 Derivation of Differential Equations 3.3 Harmonic Excitation 3.4 Undamped System Response 3.5 Damped System Response 3.6 Frequency Squared Excitations 3.7 Harmonic Support Excitation 3.8 Multifrequency Excitations 3.9 General Periodic Excitations: Fourier Series 3.10 Coulomb Damping 3.11 Hysteretic Damping 3.12 Seismic Instruments Solved Problems Practice Problems

4. GENERAL FORCED RESPONSE OF 1-DEGREE-OF-FREEDOM SYSTEMS 4.1 4.2 4.3 4.4 4.5 4.6 4.7

Introduction General Differential Equation Convolution Integral Laplace Transform Solutions Unit Impulse Function and Unit Step Function Numerical Methods Response Spectrum Solved Problems Practice Problems

3.1 3.1 3.2 3.2 3.5 3.6 3.7 3.8 3.9 3.10 3.10 3.11 3.12 3.42

4.1-4.27 4.1 4.1 4.1 4.2 4.3 4.4 4.4 4.5 4.23

5. FREE VIBRATIONS OF MULTI-DEGREE-OF-FREEDOM SYSTEMS

5.1-5.58

5.1 Introduction 5.2 Lagrange's Equations 5.3 Matrix Formulation of Differential Equations for Linear System 5.4 Stiffness Influence Coefficients 5.5 Flexibility Matrix 5.6 Normal Mode Solution 5.7 Mode Shape Orthogonality 5.8 Matrix Iteration 5.9 Other Methods 5.10 Damped System Solved Problems Practice Problems

5.1 5.1 5.2 5.3 5.3 5.3 5.4 5.4 5.4 5.4 5.5 5.48

Contents

6. FORCED VIBRATIONS OF MULTI-DEGREE-OF-FREEDOM SYSTEMS 6.1 Introduction 6.2 General System 6.3 Harmonic Excitation 6.4 Laplace Transform Solutions 6.5 Modal Analysis for Systems with Proportional Damping 6.6 Modal Analysis for Systems with General Damping Solved Problems Practice Problems

7. VIBRATIONS OF CONTINUOUS SYSTEMS 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

ix

6.1-6.21 6.1 6.1 6.1 6.2 6.2 6.3 6.3 6.19

7.1-7.30

Introduction Wave Equation Wave Solution Normal Mode Solution Beam Equation Model Superpostion Rayleigh's Quotient Rayleigh-Ritz Method Solved Problems Practice Problems

7.1 7.1 7.2 7.2 7.3 7.3 7.4 7.5 7.6 7.26

8. VIBRATION CONTROL

8.1-8.31

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

Introduction Vibration Isolation Isolation from Harmonic Excitation Shock Isolation Impulse Isolation Vibration Absorbers Damped Absorbers Houdaille Dampers Whirling Solved Problems Practice Problems

9. FINITE ELEMENT METHOD 9.1 9.2 9.3 9.4 9.5

Introductiion General Method _ Forced Vibratiol Bar Element Beam Element.

8.1 8.1 8.2 8.3 8.3 8.4 8.5 8.6 8.7 8.8 8.27

9.1-9.19 9.1 9.1 9.2 9.3 9.3

Contents

x

Solved Problems Practice Problems

10. NONLINEAR SYSTEMS 10.1 Introduction 10.2 Differences from Linear Systems 10.3 Qualitative Analysis 10.4 Duffing's Equation 10.5 Self-Excited Vibrations Solved Problems Practice Problems

11. COMPUTER APPLICATIONS 11.1 Introduction 11.2 Software Specific to Vibrations Applications 11.3 Spreadsheet Programs 11.4 Electronic Notepads 11.5 Symbolic Processors Solved Problems Practice Problems Reference

APPENDIX SAMPLE MATHCAD SCREENS FOR SELECTED PROBLEMS INDEX

9.4 9.16

10.1-10.13 10.1 10.1 10.2 10.2 10.3 10.4 10.12

11.1-11.31 11.1 11.2 11.2 11.2 11.2 11.3 11.29 11.31

A.1-A.13 1.1-1.4

0

Preface to the Special Indian Edition

This book serves as an introduction to the subject Mechanical Vibrations at undergraduate/graduate level. The primary emphasis of this book is on problem solving. It guides the students on all aspects of mechanical vibration analysis. The coverage of the book is quite broad and each chapter gives a clear idea of vibration analysis of various systems. The need of the present *adaptation was to add some essential topics (very brief theory) and add solved and practice problems. I have added several topics of interest in brief as well as solved and practice problems. Nothing from the original work has been removed. The style of prior edition has been retained. In this edition, every chapter starts with an appropriate introduction to the topic, followed by brief theory. Some typical solved and practice problems are added to the earlier edition to make the book complete and interesting. The major changes made in the course of adaptation are, in Chapter 2, Principle of conservation of energy method and Rayleigh's methods are added to demonstrate the analysis of single degree of freedom without damping. In Chapter 3, a brief theory on seismic instruments has been added for better understanding of vibration measurements. In Chapter 5, besides matrix iteration method, other numerical methods of vibration analysis such as Holzar's method, Stodola methods are included to analyze the multi degree of freedom systems. The approximate methods like Dunkerley's and Rayleigh's methods to estimate the natural frequencies of vibratory systems have also been added. Solved problems are added to demonstrate the usefulness of these methods. It is also demonstrated how to obtain mode shapes by basic principles. Computer codes for Matrix iteration and Holzar's methods, and demonstration of their use are given in Chapter 11.

xii

Preface to the Special Indian Edition

This book can be used as textbook for the undergraduate course on mechanical vibrations. The book has exhaustive solved and practice problems, and is an excellent resource of study material for the mechanical engineering students. The author would like to express gratitude to Ms Shukti Mukherjee and other members of the Tata McGraw-Hill team for their support in bringing out this edition. Shahshidhar K. Kudari

Preface

A student of mechanical vibrations must draw upon knowledge of many areas of engineering science (statics, dynamics, mechanics of materials, and even fluid mechanics) as well as mathematics (calculus, differential equations, and linear algebra). The student must then synthesize this knowledge to formulate the solution of a mechanical vibrations problem. Many mechanical systems required modeling before their vibrations can be analyzed. After appropriate assumptions are made, including the number of degrees of freedom, necessary basic conservation laws are applied to derive governing differential equations. Appropriate mathematical methods are applied to solve the differential equations. Often, the modeling results in a differential equation whose solution is well known, in which case the existing solution is used. If this is the case of the solution must be studied and written in a form which can be used in analysis and design applications. A student of mechanical vibrations must learn how to use existing knowledge to do all of the above. The purpose of this book is to provide a supplement for a student studying mechanical vibrations that will guide the student through all aspects of vibration analysis. Each chapter has a short introdUction of the theory used in the chapter, followed by a large number of solved problems. The solved problems mostly show how the theory is used in design and analysis applications. A few problems in each chapter examine the theory in mode detail. The coverage of the books is quite broad and includes free and forced vibrations of 1-degree-offreedom, and continuous systems. Undamped systems and systems with viscous damping are considered. Systems with Coulomb damping and hysteretic damping are considered for 1-degreeof-freedom systems. There are several chapters of special note. Chapter 8 focuses on design of vibration control devices such as vibration isolators and vibration absorbers. Chapter 9 introduces the finite element method from an analytical viewpoint. The problems in Chapter 9 use the finite element method using only a few elements to analyze the vibrations of bars and beams. Chapter 10

Preface focuses on nonlinear vibrations, mainly discussing the differences between linear and nonlinear systems including self-excited vibrations and chaotic motion. Chapter 11 shows how applications software can be used in vibration analysis and design. The book can be used to supplement a course using any of the popular vibrations textbooks, or can be'used as a textbook in a course where theoretical development is limited. In any case the book is a good source for studying the solutions of vibrations problems. The author would like to thank the staff at McGraw-Hill, especially John Aliano, for making this book possible. He would also like to thank his wife and son, Seala and Graham, for patience during preparation of the manuscript and Gara Alderman and Peggy Duckworth for clerical help. S. GRAHAM KELLY

Chapter One Mechanical System Analysis

1.1 INTRODUCTION It is well known that any thing, which has mass and elasticity, can vibrate with some initial disturbance. Any motion, which repeats itself after an interval of time, is called as vibration. For examples: (i) Spring mass system, (ii) Pendulum, (iii) Gitar spring, etc. The theory of vibration deals with the systematic analysis of vibratory motions of the body and the forces associated with them. The vibration analysis of a system helps to improve the design of a system/component, which enhances the performance of the system or life of the component. Usually, vibratory systems are complex, for the mathematical analysis of such systems it is required to obtain an equivalent system. In this chapter several problems have been solved to demonstrate how to obtain an equivalent system.

1.2 DEGREES OF FREEDOM AND GENERALISED COORDINATES The number of degrees of freedom used in the analysis of a mechanical system is the number of kinematically independent coordinates necessary to completely describe the motion of every particle in the system. Any such set of coordinates is called a set of generalised coordinates. The choice of a set of generalised coordinates is not unique. Kinematic quantities such as displacements, velocities and accelerations are written as function of the generalized coordinates and their time derivatives. A system with a finite number of degrees of freedom is called a discrete system, while a system with an infinite number of degrees of freedom is called a continuous system or a distributed parameter system.

1.2 1

Mechanical Vibrations

1.3 TYPES OF VIBRATION If a system, after an initial disturbance is left to vibrate on its own, the resulting vibration is known as free vibration. The frequency of free vibration is known as natural frequency of vibration, which is an important parameter in vibration analysis. If a system is subjected to an external repeating type of force, the resulting vibration is known as forced vibration. The frequency of forced vibration is known as forced frequency of vibration, which is also an important parameter in vibration analysis. It is known that if forced frequency of vibration of the system matches with natural frequency of vibration of the system, the amplitude of vibration is very high and this condition is referred as resonance. If no energy is lost or dissipated in friction or other resistance during vibration or oscillation, the resulting vibration is known as undamped vibration. This is a hypothetical condition, which is possible only in vacuum; otherwise in any medium there is resistance to vibration. If energy is lost or dissipated due to resistance during vibration, it is known as damped vibration. If the value of excitation acting on the vibratory system is known at any given instant (deterministic), the resulting vibration is known as periodic vibration. If excitation is non-periodic, the resulting vibration is called as Random vibration. If all the basic components of a vibrating system behaves linearly the resulting vibration is known as linear vibration. Principles of superposition holds good in mathematical analysis of such systems. If any basic component behaves non-linearly, the resulting vibration is known as non-linear vibration. Non-linear systems involve complicated analysis due to the non-linearity of the governing differential equations.

1.4 MECHANICAL SYSTEM COMPONENTS The first step in the vibration analysis of a system is to understand the behavior of the system mathematically. A mathematical model should include sufficient details of the system to arrive at some mathematical equations, which can be solved for the analysis. It is well known that all engineering systems are complex, hence, it is difficult to consider every detail of the system for vibration analysis. For such systems it is required to identify some basic elements, which play a vital role in the vibration of the system and can be mathematically modeled. It is identified that a vibratory system can be simplified and idealized into three simple components. A mechanical system comprises inertia components, stiffness components and damping components. The inertia components have kinetic energy when the system is in motion. The kinetic energy of a rigid body undergoing planar motion is 1 2 1— 2 T = — mv + — /co (1.1) 2 2 where v is the velocity of the body's mass center, w is the angular velocity about an axis perpendicular to the plane of motion, m is the body's mass, and I is the mass moment of inertia about an axis parallel to the axis of rotation through the mass center. The mass moment of inertia of some common shapes is given in Table 1.1. A linear stiffness component (a linear spring) has a force displacement relation of the form F = kx (1.2) where F is applied force and x is the component's change in length from its unstretched length. The stiffness k has dimensions of force per length.

Mechanical System Analysis

1.3

A dashpot is a mechanical device that adds viscous damping to a mechanical system. A linear viscous damping component has a force-velocity relation of the form F = cv (1.3) where c is the damping coefficient of dimensions mass per time.

Table. 1.1 The mass moment of inertia of some common shapes S. No 1

Name

Figure I

I II

[..c____,.]

Slender bar

L

2

Circular disc

3

Square plate

4

Sphere

I = 1 inl,2 2 I = —1 mr 2 2

/ =ImL2 6

L

/ = mr 2 0

5

1.5 EQUIVALENT SYSTEMS ANALYSIS All linear 1-degree-of-freedom systems with viscous damping can be modeled by the simple massspring-dashpot system of Fig. 1.1. Let x be the chosen generalized coordinate. The kinetic energy of a linear system can be written in the form

1

T = —m

(1.4)

eq The potential energy of a linear system can be written in the form

V = —1 keqx2

(1.5)

2 The work done by the viscous damping force in a linear system between two arbitrary locations x1 and x2 can be written as x2

W= iceg icbc

(1.6)

x,

Comparing with the kinetic energy, potential energy and work done by the actual system with Eqs. (1.4), (1.5) and (1.6), equivalent mass, equivalent stiffness and equivalent damping of a system can be obtained.

keg

3.- X

Meg Ceq

Fig. 1.1

Mechanical Vibrations

1.4

1.6 TORSIONAL SYSTEMS When an angular coordinate is used as a generalized coordinate for a linear system, the system can be modeled by the equivalent torsional system of Fig. 1.2. The moment applied to a linear torsional spring is proportional to its angular rotation while the moment applied to a linear torsional viscous damper is proportional to its angular velocity. The equivalent system coefficient for a torsional system are determined by calculating the total kinetic energy, potential energy, and work done by viscous damping forces for the original system in terms of the chosen generalized coordinate and setting them equal to 2 1 T= —1 2 eq

V= 1k 0

2

2 CI

(1.7)

(1.8)

92

= — f ct dO

(1.9)

e,

1.7

STATIC EQUILIBRIUM POSITION

Systems, such as the one in Fig. 1.3, have elastic elements that are subject to force when the system is in equilibrium. The resulting deflection in the elastic element is called its static deflection, usually denoted by Ast. The static deflection of an elastic element in a linear system has no effect on the system's equivalent stiffness.

Fig. 1.3

SOLVED PROBLEMS 1.1 Determine the number of degrees of freedom to be used in the vibration analysis of the rigid bar of Fig. 1.4, and specify a set of generalized coordinates that can be used in its vibration analysis.

Since the bar is rigid, the system has only 1-degrees-of-freedom. One possible choice for the generalized coordinate is 0, the angular displacement of the bar measured positive clockwise from the system's equilibrium position. 1.2 Determine the number of degrees of

Fig. 1.4

freedom needed for the analysis of the mechanical system of Fig. 1.5, and specify a set of generalized coordinates that can be used in its vibration analysis.

Mechanical System Analysis

1.5

Fig. 1.5

Let x be the displacement of the mass center of the rigid bar, measured from the system's equilibrium position. Knowledge of x, by itself, is not sufficient to determine the displacement of any other particle on the bar. Thus the system has more than 1-degree-offreedom. Let 0 be the clockwise angular rotation of the bar with respect to the axis of the bar in its equilibrium position. If 0 is small, then the displacement of the right end of the bar is x + (L/2)0. Thus the system has 2-degrees-of-freedom, and x and 0 are a possible set of generalized coordinates, as illustrated in Fig. 1.6.

Fig. 1.7

The system of Fig. 1.7 has 4-degreesof-freedom. A possible set of generalized coordinates are ei, the clockwise angular displacement from equilibrium of the disk whose center is at 01; 02, the clockwise angular displacement from equilibrium of the disk whose center is at 02; x1, the downward displacement of block B; and x2, the downward displacement of block C. Note that the upward displacement of block A is given by r1 01 and hence is not kinematically independent of the motion of the disk. 1.4 A tightly wound helical coil spring is

Fig. 1.6 1.3 Determine the number of degrees of

freedom used in the analysis of the mechanical system of Fig. 1.7. Specify a set of generalized coordinates that can be used in the system's vibration analysis.

made from an 18-mm-diameter bar of 0.2 percent hardened steel (G = 80 x 109 N/m2). The spring has 80 active coils with a coil diameter of 16 cm. What is the change in length of the spring when it hangs vertically with one end fixed and a 200 kg block attached to its other end? The stiffness of a helical coil spring is k — GD4

64Nr3

1.6

Mechanical Vibrations

k-

(80x109 N2 J(0.018m)4 m 64(80)(0.08 m)3

311 =3.20x10 Using Eq. (1.2), the change in length of the spring is: xx=

F

1.6 Determine the torsional stiffness of the shaft in the system of Fig. 1.9.

////1/////////////////

where D is the bar diameter, r is the coil radius, and N is the number of active turns. Substituting known values leads to

1.4 m

ri =15 mm r, = 25mm

mg

=

G=80 x109 m2

(200 kg) (9.81 s2 mJ 311 3.20x10

— 0.613 m

1.5 Determine the longitudinal stiffness of the bar of Fig. 1.8.

r- A, E

m

)1

Fig. 1.8 The longitudinal motion of the block of Fig. 1.8 can be modeled by an undamped system of the form of Fig. 1.1. When a force F is applied to the end of the bar, its change in length is: = FL AE

v=

AE A

which is in the form of Eq. (1.2). Thus _ AE key

If an angular moment M is applied to the end of the shaft, the angle of twist at the end of the shaft is determined from mechanics of materials as: 0=

L

or

Fig. 1.9

L

ML JG

where J is the polar moment of inertia of the shaft's cross section. Thus M = JG0

L and the shaft's equivalent torsional stiffness is M/0 given by: kt —

JG

L For the shaft of Fig. 1.9 J= rc .4_ 7,4) 2 (10 1 = 2 [(0.025 m)4 — (0.015 m)4] = 5.34 x 10-7 m4

1.7

Mechanical System Analysis

From mechanics of materials, the deflection of a beam fixed at z = 0 and pinned at z = L due to a unit concentrated load at z= a, for z < a is

Thus N (5.34x10-7 m4 ) (80 x109 m 2 kt —

1.4 m w(z; a) =

= 3.05 x 104 N-m rad 1.7 A machine whose mass is much larger than the mass of the beam shown in Fig. 1.10 is bolted to the beam. Since the inertia of the beam is small compared to the inertia of the machine, a 1-degree-of-freedom model is used to analyze the vibrations of the machine. The system is modeled by the system of Fig. 1.3 neglecting damping. Determine the equivalent spring stiffness if the machine is bolted to the beam at 1m (b) z = 1.5 m

1 z3 a) 1_ El t12 ( L ) 2 [(,)

+

21

—

z2 a (i 4

a) L

(2

(1.11)

1 (a) For a = 1 m, alL = 3 . The using Eqs. (1.10) and (1.11),

(a) z =

_ Keg

1 _ 81E/ w(a; a) — 11a3 81(210x109 N m2

N E = 210 x109 —

m2

(1.5 x10-5 m4 )

/=1.5x10-9 m4

3m

11(1 m)3 Fig. 1.19

Let w(z; a) be the deflection of the beam at a location z due to a unit concentrated load applied at z = a. From Mechanics of materials, the beam deflection is linear, and thus the deflection due to a concentrated load of magnitude F is given by y(z; a) = Fw(z; a) If the machine is bolted to the beam at z = a, the deflection at this location is y(a; a) = Fw(a; a) which is similar to Eq. (1.2) with 1 (1.10) k— w(a; a)

= 2.32 x107 1T m 1 (b) For a = 1.5 m, alL = — . Then using 2 Eqs. (1.10) and (1.11), keg -

1 96E1 w(a; a) 7a 3 96(210 x109 mN2 (1.5 x10-5 m4) 7 (1.5 m)3

= 1.28 x 10711 m•

Mechanical Vibrations

1.8 A machine of mass In is attached to the midspan of a simply supported beam of length L, elastic modulus E, and cross-sectional moment of inertia I. The mass of the machine is much greater than the mass of the beam, thus the system can be modeled using 1-degreeof-freedom. What is the equivalent stiffness of the beam using the midspan deflection as the generalized coordinate?

is x. The free body diagram of Fig. 1.12 shows that the total force acting on the block is F = k + k2x + k3x + • • • + 11

= Eldx i=1

(1.12)

k1 x k2x

The deflection of a simply supported beam at its midspan due to a concentrated load F applied at the midspan is F 3 —L3 48 El The equivalent stiffness is the reciprocal of the midspan deflection due to a midspan concentrated unit load. Thus 48E1

k

L3

1.9 The springs of Fig. 1.11 are said to be in parallel. Derive an equation for the equivalent stiffness of the parallel combination of springs if the system of Fig. 1.11 is to be modeled by the equivalent system of Fig. 1.1

Fig. 1.12 The system of Fig. 1.1 can be used to model the system of Fig. 1.11 if the force acting on the block of Fig. 1.1 is equal to the force of Eq. (1.12) when the spring has a displacement x. If the spring of Fig. 1.1 has a displacement x, then the force acting on the block of Fig. 1.1 is F = k X' (1.13) Then for the forces from Eqs. (1.12) and (1.13) to be equal: keg = Eki i=1

1—)—x

1.10 The spring in the system of Fig. 1.13

kn

are said to be in series. Derive an equation for the series combination of springs if the system of Fig. 1.13 is to be modeled by the equivalent system of Fig. 1.1.

Fig. 1.11 k1

If the block is subject to an arbitrary displacement x, the change in length of each spring in the parallel combination

k2

k3 •

Fig. 1.13

• •

1.9

Mechanical System Analysis

Let x be the displacement of the block of Fig. 1.13 at an arbitrary instant. Let xi be the change in length of the ith spring from the fixed support. If each spring is assumed massless, then the force developed at each end of the spring has the same magnitude but opposite in direction, as shown in Fig. 1.14. Thus the force is same in each spring:

k

Fig. 1.15

Problem 1.9. The result is shown in Fig. 1.16(a). The springs on the left of the block are in series with one another. The result of Solved Problem 1.10 is used to replace these springs by a spring whose stiffness is calculated as:

= k2x2 = k3x3 = • • • = kffx„ = F (1.14) In addition,

x = x1 + x2 + x3 + ••• + x„=

1 1 + 1 +1+ 1 2 3k 3k k 3k The springs attached to the right of the block are in series and are replaced by a spring of stiffness

i a=i x

(1.15) Solving for xi from Eq. (1.14) and substituting into Eq. (1.15) leads to

2k

1 1

(1.16)

1

3

k 2k

The result is the system of Fig. 1.16(b). When the block has an arbitrary displacement x, the displacements in each of the springs of Fig. 1.16(b) are the same, and the total forces acting on the block is the sum of the forces developed in the springs. Thus these springs behave as if they are in the parallel and can be replaced by a spring of stiffness

Noting that the force acting on the block of the system of Fig. 1.1 for an arbitrary x is kee and equating this to the force from Eq. (1.16) leads to keg —

2k

1

k 2k 7k


xf 10 mm

0.118mm

AA 0.47 mm Hence, 22 cycles will be executed. 2.44 The block of Fig. 2.30 is displaced 25 rum and released. It is observed that the amplitude decreases 1.2 mm each cycle. What is the coefficient of friction between the block and the surface?

Fig. 2.29 The differential equation governing the motion of the system of Fig. 2.29 can be shown to be {-µmg, 5c> 0 mi-F/oc= ping, x < 0

Application of Newton's law to the free body diagrams of Fig. 2.31 leads to ink + kx =

{-µmg cos 0 x

>0

µmg cos() ic
0

Fig. 2.32

X mgb 2.6

9 ce 7 mL2 0.. + 16 48

kr 2 0)„ =

+ (11 a2 _

I +9mr.2 (2 2 2.2 - inL 0 + - kL + k, )0 = 0, 3 9

6kL2 + 9kt con = 2.3

mL2

1 -m+M)1+ci+lcc=0, (3 3k

2.7 2.8 2.9 2.10 2.11 2.12 2.13

2.14

in +3M

2.4 134 me + mgLO =0

2.15

ing ) 9,0 4

8.81 Hz 164.6 rad/s 132.3 rad/s 19.9 rad/s 0.512 m 9.32 cm < D < 13.2 cm / = 10.65 kg-m2 3g (Po _ 1 2/ p

con = \1-

2 m (L T - a)

rad/sec

rad/sec

2.35

Free Vibrations of 1-Degree-of-Freedom Systems

2.16 (on = 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29

4T niL

c = 1.55 \ra c = 1.55 x 103 N-s/m c = 1.91 x 103 N-s/m 1.79 m/s C = 0.591 C= 0.026 30.9 mm 13.4 mm 4.826 N sec/m (i) 1.89, (ii) 6.61 -0.293 m (i) 33.43 N sec/m, (ii) 8.7 kg c = 16.80 N sec/m

2.30 0.0515 s 2.31 0(t) = 0.0895e-6.61' sin (28.9t + 1.35) 2.32 (i) mL20 + cb20 + Ka20 = 0 q[K

(ii) cod -

(iii)

-

L

c

2 b4

2

4 a2 L2 m 2 1

cb2 2 aL Km

2.33 0(t) = 0.144e18 6f - 0.055e-47-6' 2.34 (a) 1.15, (b) 0.866 xo

2.35

xo

1, requiring M < 3.74 leads to 1+1-=11+3.74 1 = 1.126 •r>.\1 M Thus, the allowable ranges of frequencies are

3E1

co < 0.856con = 195.2

3

3(200 x109 N )(1.6 x10-5 M4 )

and

2

(1.6 m)3 = 2.34 x 106N m The system's natural frequency is

I

con

keg

m

2.34 x1061\1 m 45 kg

rad

co > 1.125o = 256.5 rad s

3.5 A thin disk of mass 0.8 kg and radius 60 mm is attached to the end of a 1.2 m steel (G = 80 x 109 Nin12, p = 7500 kg/m3) shaft of diameter 20 mm. The disk is subject to a harmonic torque of amplitude 12.5 N-in at a frequency of 700 rad/s. What is the steady-state amplitude of angular oscillations of the disk? The torsional stiffness of the shaft is

= 228.0 rad In order to limit the steady-state amplitude to 0.2 mm, the allowable value of the magnification factor is M=

moe x , F0

JG lc( L

N-m rad The mass moment of inertia of the shaft is = 1.05 x 103

2

rad (45 kg) (228.0 — ) (0.0002 m)

1 = - p7rLrs4 2

125 N = 3.74 For an undamped system, Eq. (3.12) becomes M-

L c (0.01 m)4 (80 x109 N m2 2 1.2m

1 11 - r21

For r < 1, requiring M < 3.74 leads to 1 \11 - 3.74 = 0.856 r< .

=

j kg 7500 -T) n(1.2 m)(0.01 m)4 2

= 1.41 x 10-4 kg-m2 The inertia effects of the shaft are included in a 1-degree-of-freedom model by ieg

=

d

s

=

4 1 Is Ind 1 . _ 3

3.15

Harmonic Excitation of 1-Degree-of-Freedom Systems

k = 5000 N/m

= 1(0.8 kg)(0.06 m)2 2 +

1

c= 10 N sec/m

W = 40 kgf (1.41 x 10 kg-m2)

= 1.49 x 10-3 kg-m2 The natural frequency of the system is

Given F0 = 10, co = 15 rad/sec

1.05 x 103 N-111 rad 1.49 x10-3 kg-m2

ki eq

m=

xs(t) =

The frequency ratio is rad 700— s =0.834 839.5 rad

Fo lk

4 (1_ r2 )2 +(2c1.)2.

1 1 - 3.28 1- r2 1- (0.834)2

c = cc 2mcon 10 2(4.07)(35.04)

Let 0 be the steady-state amplitude of torsional oscillation. The torsional oscillation equivalent of Eq. (3.10) is

= 0.035 (under damped) Phase lag 0 = tan—1

req co,2 ,0

2 cr

•

(1-r2 )'

To

r=

0—

sin (cot -0)

5000 . - 35.04 rad/s w„ = V idm 7 \I 4.07

which leads to a magnification factor of M=

W = 40 _ 4.07 kg 9.81 g

Steady state response of the system is

= 839.5 rad

co r= COn

Fig. 3.15

MT°

co co„

ieg 0 = tan-1 )1(839.5 A

[2(0.035)(0.428)1 1- (0.428)2

3.28 (12.5 N-m) (1.49 x10-3 kg-m2

15 = 0.428 35.04

2

= 0.0390 rad = 2.24° 3.6 Calculate the steady state response of the mass shown in Fig. 3.15 after one second, if an exiting force F(t) = 10 sin 15t acts on the mass.

0 = 2.10° = 0.036 rad xs(t)

(10/5000) x sin (15t + 0.036) (1- 0.4282)2 +(2(0.035)(0.428))2

xs(1) = 1.527 mm

3.16

Mechanical Vibrations

3.7 A 45 kg machine is mounted on four parallel springs each of stiffness 2 x 105 N/m. When the machine operates at 32 Hz, the machine's steady-state amplitude is measured as 1.5 mm. What is the magnitude of the excitation provided to the machine at this speed? The system's natural frequency is keg co), = \I in 4 2 x105 11) ( rad m =133.3 — 45 kg The system's frequency ratio is

r=

co =.

The resonant frequency The amplitude at resonance The phase angle at resonance Frequency corresponding to peak amplitude (v) Peak amplitude, and (vi) The phase angle corresponding to peak amplitude.

(i) (ii) (iii) (iv)

W= 55 N; m= W/g = 5.606 kg k = 1100 N/m Fo = 9 N c = 77 N sec/m (i) Resonant frequency r = 1; co = co„

( cycle )( rad 32 2ir cycle s 133.3 rad

= 1.51 The magnification factor for an undamped system with a frequency ratio greater than 1 is M=

vibrate by a harmonic force of 9 N. Assuming viscons damping co-efficient c = 77 N sec/m, find:

1 r2 -1

I 1100 - 14 rad/s \ 5.606 (ii) Amplitude at resonance =

V(1-r2 )2 +(2cr)2

1 - 0.781 (1.51)2 -1

c _ c _ 2ffico„ cc

Equation (3.10) is rearranged to solve for the excitation force as F0 -

2 MCO?,

Fo /k

X„s -

77 2(5.606)(14)

x

= 0.490 at resonance r = 1

M (45 kg) (133.3 rad

(0.0015 m) ••• Xres

Folk V(202

Fo lk 24'

0.781 = 1.54 x 103 N 3.8 A weight 55 N suspended by a spring of stiffness 1.1 kN/m is forced to

(9/1100) 2(0.490) = 8.339 x 10-3 m

3.17

Harmonic Excitation of 1-Degree-of-Freedom Systems

(iii) Phase angle at resonance

0 = tan-1

2Cr (1-r2 )

= tan-1.0 = 90° = 1.57 rad (iv) Frequency corresponding to peak amplitude COpeak

rpeak

co,

- .J1-2c2

=

2(0.490)2 COpeak = 10.09 rad/sec (v) Peak amplitude Xpeak -

r

Pea

k

T-

4r-

co + 0),,

0.05 s

And the period of beating is 2r

-1s VD- co„i These equations are rearranged to w + co„ = 807r co - co„= 27r which are solved simultaneously yielding Tb -

rad rad w = 417r- , co„ = 397r

Xo

(1- r2)2 + (2cr)2 10.09 - 0.7207 14 (9/1100)

Xpeak

From Fig. 3.5 it is observed that the period of oscillation is

(1- 0.72072 )2

+ (2 (0.49)(0.7207))2 = 9.57 x 10-3 m (vi) Phase angle corresponding to peak amplitude

0 = tan-

, gr

= tan 1

1- rp2 2(0.49)(0.72)

1- 0.722 = 55.750 = 0.973 rad

3.9 A system that exhibits beating has a period of oscillation of 0.05 s and a beating period of 1.0 s. Determine the system's natural frequency and its excitation frequency if the excitation frequency is greater than the natural frequency.

3.10 Repeat Solved Problem 3.4 as if the beam had a damping ratio of 0.08. From the solution of Solved Problem 3.4, the system's natural frequency is 228.0 rad/s, and the maximum allowable value of the magnification factor is 3.74. Thus in order to limit the magnification to 3.74, 3.74 >

1 (1- r2 )2 + [2(0.08)/12

r4 - 1.9744r2 + 1 > 0.07149 r4 - 1.9744/2 + 0.9285 > 0 The above is quadratic in r2. Application of the quadratic formula leads to positive solutions of r = 0.879 and r = 1.096. The magnification factor is less than 3.74 if r < 0.879; co < 200.4 rad/sec or r > 1.096; co > 249.9 rad/sec. 3.11 A 110 kg machine is mounted on an elastic foundation of stiffness 2 x 106 N/m. When operating at 150 rad/s, the machine is subject to a harmonic force of magnitude 1500 N. The steady-state

3.18

Mechanical Vibrations

amplitude of the machine is measured as 1.9 mm. What is the damping ratio of the foundation? The natural frequency of the system is

3.12 The differential equation governing the motion of the system of Fig. 3.16 is M,-, (m + i2 i + ci+ 5loc = -="- sin cot r

1-

fk-

COn = \ 171

Us ng the given values, determine the steady-state amplitude of the block.

2x106 \

110 kg

Mo sin wt

= 134.8

k

rad

The magnification factor during operation is mC.0„2 X

M

m = 10 kg 1= 0.1 kg-m2 r = 10 cm

F0 A 2 ) (0.0019 m)

(110 kg)(134.8

k = 1.6 x 10511m n,,,,Ns c= 0 4U -

1500 N = 2.53 The frequency ratio for operation at 150 rad/s is 150

co

/• =

(4)"

rad s

Mo = 100— m rad (0= 180

Fig. 3.16 — 1.113

134.8 rad

The system's natural frequency and damping ratio are

Equation (3.12) can be rearranged to solve for the damping ratio as

i1

C— 2 rM 2 .‘t

(1 r2 )2

5k

\in + r 2

which for this problem leads to 5(1.6 x105 1 1 m

1 2(1.113)

C—

•

1 \,1 (2.53)2.

= 0.142

11 (1.113)212

0.1 kg-m2 \ 10 kg + (0.1 m) 2 = 200

rad

Harmonic Excitation of 1-Degree-of-Freedom Systems

as am = 2.(0„ (tit + r 2

2 2(200 rad ) 10 kg + 0.1 kg-m 2 (0.1m) = 0.08 The frequency ratio is

r= CD"

rad

M0 r

=

100 N-m — 1000 N 0.1m

meg = m

=20 kg r`

Thus

X—

ar

F0 M(0.9, 0.08) meg (0,2 , (1000 N)(4.19) (20 kg) (200 rad )2

(1 — r2) + 242 =0

r= V1-2c 2

Substituting the value of r into Eq. (3.12) leads to 1

—

[1— (1— 2c2 )]2 / +4c2 (1-2c2 )]

s = 0.9 200 rad

J[1— (0.9)2 ]2 + [2(0.08)(0.9)]2 = 4.19 The steady-state amplitude is determined using Eq. (3.10) with

and

am = 0

Mmax

The magnification factor for the system is M = M(0.9, 0.08) 1

F0 —

[(1 — r2)2 + (2 cr)21-312 [2(1 — r2)(-2r) + 2(20(20]

N 640 N-s m

180

3.19

1 2a1 — c2 3.14 A 120 kg machine is mounted at the midspan of a 1.5 m-long simply supported beam of elastic modulus E = 200 x 109 N/m2 and cross-section moment:of inertia I = 1.53 x 10-6 m4. An experiment is run on the system during which the machine is subject to a harmonic excitation of magnitude 2000 N at• a variety of .excitation frequencies. The largest steady-state amplitude recorded during the experiment is 2.5 mm. Estimate the damping ratio of the system. The stiffness of the beam is k—

48E1 1,3

5.24 mm 3.13 Derive Eq. (3.14) from Eq. (3.12). For a fixed c, the value of r for which the maximum of M(r, 4-) occurs is obtained by finding the value of r such that dM/dr = 0. To this end,

48(200 x109 N )(1.53 x10-6 m4 m2 (1.5 m)3 = 4.35 X 1061 .

Mechanical Vibrations

3.201

The system's natural frequency is con

Fi c \

4.35 x10611rad m =190.4 — 120 kg The maximum value of the magnification factor is Mmax

2 nia)ti Xmax F0

a frequency of 40 Hz. Use this information to estimate the stiffness and damping ratio of the foundation. Using Eqs. (3.10) and (3.14), the maximum steady-state amplitude is related to the damping ratio by 1

Mmax

4- 41 — c2 moon Xmax

A 2

(120 kg)(190.41

) (0.0025 m)

2000 N

(3.53) F0 Then from Eq. (3.15) the natural frequency and the frequency at which the maximum steady-state amplitude occurs are related by (1)max rm„ = J1-24.2 =

= 5.44

(on

Equation (3.14) can be rearranged as —+

— 0 1 4Af tlax

Mmax

(pn

J1-24-2

which is a quadratic equation C2 whose roots are

which when substituted into Eq. (3.53), leads to

11/2

r

[1 [1+ \ill 2

1 Mmax

Substituting Mmax = 5.44 and noting that use of the positive sign in the ± choice leads to a damping ratio greater than 1/5 leads to C = 0.092. 3.15 An 82 kg machine tool is mountedon an elastic foundation. An experiment is run to determine the stiffness and damping properties of the foundation. When the tool is exited with a harmonic force of magnitude 8000 N at a variety of frequencies, the maximum steadystate amplitude obtained 4.1 m at

ma/max Xmax (1— 2c2 )F0

1 2C J1 — c2

cycle ) (27r rad (82 kg) [(40 s cycle (0.0041m)

2

(1— 2C2 )(8000 N) 1 2 01 _c2

Substituting given and calculated values and rearranging, leads to 28.20C-2(1 — C2) = (1 — 242)2 4-4 — 412 ± 0.03107 = 0 = 0.179, 0.984

3.2-1,

Harmonic Excitation of 1-Degree-of-Freedom Systems

However, since a maximum steady-state amplitude is attained only for C< 1/ , C= 0.179. Eq. (3.15) is used to determine the natural frequency as co

con

V1—C2 140

cycle) r 2ir rad s cycle

1 — r2

tan 0 Zr However, since the frequency ratio is greater than 1, the response leads the excitation, and if the phase angle is taken to be between 0 and 180°, the appropriate value is 0 = 180° — 21° = 159°. Thus —

1— (1.288)2 tan-1 (159°) 2(1.288) = 0.0982

Nil— (0.179)2 = 255.5

Equation (3.13) can be rearranged to solve for C as:

—

rad

from which the foundation's stiffness is calculated: k= moe, = (82 kg) (255.5 rad )

= 5.35 x 1061\1 m 3.16 A 35 kg electric motor that operates at 60 Hz is mounted on an elastic foundation of stiffness 3 x 106 N/m. The phase difference between the excitation and the steady-state response is 21°. What is the damping ratio of the system?

3.17 The machine of mass m, of Fig. 3.17, is mounted on an elastic foundation modeled as a spring of stiffness k in parallel with a viscous damper of damping coefficient c. The machine has an unbalanced component rotating at a constant speed W. The unbalance can be represented by a particle of mass mo, a distance e from the axis of rotation. Derive the differential equation governing the machines displacement, and determine its steadystate amplitude.

The natural frequency and frequency ratio are 3x106 1\1 rad co = I — 292.8 m " m 35 kg ( r=

co

60

cycle ) ( rad 27r s cycle 292.8

= 1.288

rad

Fig. 3.17

Rotating unbalance

Free body diagrams of the machine at an arbitrary instant are shown in Fig. 3.18. Summing forces in the vertical direction = e•st

ett

3.22

Mechanical Vibrations

and nothing that gravity cancels with the static spring force leads to -kx - ci = (m - m0).5

+ m0eco2 sin 0 + m0 .5i (3.54) Since co is constant, (3.55) 0 = an + 00 Substituting Eq. (3.49) into Eq. (3.56) and rearranging leads to mz + cx + kx

3.18 A 65 kg industrial sewing machine has a rotating unbalance of 0.15 kg-m. The machine operates at 125 Hz and is mounted on a foundation of equivalent stiffness 2 x 106 N/m and damping ratio 0.12. What is the machine's steady-state amplitude? The natural frequency and frequency ratio of the system are

= -m0eco2 sin (cot + 00) = m0em2 sin (cot + y/) (3.57)

where yi = 00 TC Thus the response of a system due to a rotating unbalance is that of a system excited by a frequency squared harmonic excitation. The constant of proportionality defined in Eq. (3.16) is A = moe The application of Eq. (3.17) leads to mX

= A(r, I)

moe r2

(3.58)

V(1- r2 )2 + (4'02 where r =

co

con=

con 9

=

m

con =

r=

max

65 kg

rad =175.4—

CO

( cycle ) ( 2 ir rad 125 s cycle 175.4 rad = 4.48 From the results of Solved Problem 3.17, the excitation provided to the machine by the rotating unbalance is a frequency squared harmonic excitation with A = moe, the magnitude of the rotating unbalance. Thus using Eq. (3.58) of Solved Problem 3.17,

(4.48)2

m0eco2

R- (4.48)212 + [2(0.12)(4.48)]2

(m-mo)i

Effective forces

Fig. 3.18

= 1,71

mX = A(4.48, 0.12) m0 e

2mco„

Ex ernal forces

2x106 N m

X

= 1.051(0.15 kg-m) 65 kg = 2.43 mm

= 1.051

3.23

Harmonic Excitation of 1-Degree-of-Freedom Systems

3.19 A spring damper mec nical system supports a rotor of mass 1000 kg. It is known that rotor has unbalance mass ovt) of 1 kg located at 5 cm radius. It is found that resonance occurs at 1500 6, rpm. What is the amplitude of vibration if rotor is made to run at 1000 rpm. The damping factor 4' = 0.2. in =1000 kg mo = 1 kg % co„ = 1500 rpm (resonance) c=0.2 4)_ r‘ t)0 e = 0.05 m co= 1000 rpm co 1000 r= = =0.666 co 1500 (1710

e

2

m Jr

X= NI (1—

r2 )2 -I- (2 Cr)2 (1(0.05)) (0.666)2 1000

(1— 0.6662 )2 + (2(0.2)(0.666))2 2.217 x10-5 0.6169 = 3.593 x 10-5 meters 3.20 An 80 kg reciprocating machine is placed on a thin, massless beam. A frequency sweep is run to determine the magnitude of the machine's rotating unbalance and the beam's equivalent stiffness. As the speed of the machine is increased, the following is noted: (a) The steady-state amplitude of the machine at a speed of 65 rad/s is 7.5 mm. (b) The maximum steady-state amplitude occurs for a speed less then 65 rad/s.

(c) As the speed is greatly increased, the steady-state amplitude approaches 5 mm. Assume the system is undamped. Solved Problem 3.17 illustrates that a machine with a rotating unbalance experiences a frequency squared harmonic excitation with A = moe, the magnitude of the rotating unbalance. Figure 3.8 shows that as the frequency ratio grows large, A 1. Thus from condition (c) (80 kg) (0.005 m)

—

1

moe moe = 0.4 kg-m Since, the maximum steady-state amplitude occurs for a speed less than 65 rad/s, it is probable that 65 rad/s corresponds to a frequency ratio greater than 1. Thus, for an undamped system . with r > 1, A=

r2 r2 _ 1

For w = 65 rad/s, (80 kg) (0.0075 m) A= mX = — 1.5 0.04 kg-m moe Thus 1.5 =

r2 r2 65

con

-3 r = 1.73

rad

1.73

= 37.6

rad

k = m(0,2,= (80 kg) 1r

rac1)2 s)

= 1.13 x 105 -111. M.

3.24

Mechanical Vibrations

3.21 A 500 kg tumbler has an unbalance of 1.26 kg, 50 cm from its axis of rotation. For what stiffnesses of an elastic mounting of damping ratio 0.06 will the tumbler's steady-state amplitude be less than 2 mm at all speeds between 200 and 600 r/min?

In order for r < 0.788 over the entire frequency range, r = 0.788 should correspond to a frequency less than 600 r/min. Thus co

(600

r)(2ir rad )rimin) min) r J 1 60 s 0.788

7'

The results of Solved Problem 3.17 show that a machine with a rotating unbalance is subject to a frequency squared excitation with A= moe. Thus in order for the steady-state amplitude to be less than 2 mm when the tumbler is installed on the mounting, the largest allowable value of A is A /

MiYmax

mo e (500 kg) (0.002 in) — 1.587 (1.26 kg) (0.5 m)

From Eq. (3.19), Amax(C = 0.06) = 8.36 > Aa11. Then from Fig. 3.8, since Aall > 1 and C < 1[5 , there are two values of r such that A(r, 0.06) = 1.587. In order for A < 1.587, the frequency ratio cannot be between these two values, which are obtained by solving 1.587 =

r2 (1 — r2 )2 + (0.1202

Squaring the above equation, multiplying through by the denominator of the righthand side, and rearranging leads to 1.519r4 — 5.001r2 + 2.519 = 0 which is a quadratic equation r2 and can be solved using the quadratic formula. The resulting allowable frequency ranges correspond to r < 0.788 or r > 1.634

= 79.73 rad

kmin = (500 kg) (79.73

rad )2

= 3.18 x 106N m In order for r > 1.634 over the entire frequency range, r = 1.684 should correspond to a frequency greater than 200 r/min. Thus ,r rad (1 min ) (200 r (22 min] r ) 60 s 1.634 = 12.82

rad

— )2 kmax = (500 kg) (12.82rad = 8.21 x 104N Hence, the acceptable mounting stiffnesses are k < 8.21 x 10411

and

k> 3.18 x 10611m 3.22 A 40 kg fan has a rotating unbalance of magnitude 0.1 kg-m. The fan is mounted on the beam of Fig. 3.19. The

Harmonic Excitation of 1-Degree-of-Freedom Systems

beam has been specially treated to add viscous damping. As the speed of the fan is varied, it is noted that its maximum steady-state amplitude is 20.3 mm. What is the fan's steady-state amplitude when it operates at 1000 r/min? N

E= 200 x 109

and the system's natural frequency is co„ 5N 4.51 x10

m

40 kg

r_

•*

co Con

moe = 0.1 kg-m 1.2 m

rad =106.2 —

The frequency ratio is

m2 1=1.3 x 104 m4

3.25

r ) (2g rad) (1 min r ) 60 s (1000 min

ro =1000 r/min

Fig. 3.19

106.2 rad The maximum value of A is Amax —

= 0.986 The steady-state amplitude is calculated by

mXmax mo e (40 kg) (0.0203 m) 0.1 kg-m

= 8.12 The damping ratio is determined using Eq. (3.19) 8.12 -

mo e m

A(0.986, 0.0617)

0.1 kg-m 40 kg

1 2c41-c 2

= 0.0617 The beam's stiffness is k=

X=

3E1 L3

3(200x109 N )(1.3 x10-6 m4 ) m2

(1.2 m)3 = 4.51 x 10511 m

(0.986)2 [1- (0.986)2 ]2 + [2(0.0617)(0.986)]2

= 19.48 mm 3.23 The fan of Solved Problem 3.22 is to operate at 1000 r/min, 1250 r/min, 1500 r/min, 1750 r/min, and 2000 r/min. What is the minimum mass that should be added to the fan such that its steadystate amplitude is less than 10 mm at all operating speeds? Adding mass to the fan decreases the system's natural frequency, thus increasing the frequency ratio at each operating speed. With no additional

3.26

Mechanical Vibrations

mass, r = 0.986 for co = 1000 r/min. Adding mass will probably lead to a frequency ratio greater than 1 for w = 1000 r/min„ Figure 3.8 shows that for r > 1, the steady-state amplitude for a frequency squared excitation decreases with increasing excitation frequency. Thus if X< 10 mm for co= 1000 r/min, then X< 10 mm for all co > 1000 r/min. The desired magnification factor for = 1000 r/min is 2 17

M=

mw „ A

kX

F0

mo eco2

connected to the main body of the helicopter by an elastic structure. T'he natural frequency of the tail section is observed as 135 rad/s. During flight, the rotor operates at 900 r/min. What is the vibration amplitude of the tail section if one of the blades falls off during flight? Assume a damping ratio of 0.05.

(4.51 x105 -1(0.01m) - 4.11 (0.1 kg-m) (104.7 rad )2 Thus 1

4.11 -

V(1- r2 )2 + [2(0.0617)r]2

Fig. 3.20 The total mass of the rotor is m = 4(2.3 kg) + 28.5 kg = 37.7 kg The equivalent stiffness of the tail section is

Solving for r leads to r = 1.096. Thus keq

104.7 rad w"

m=

1.096

m2

=

= 95.5

/110),21

) (135 = (37.7 kg

rad 2 s )

rad = 6.87 x 105 — N

s

4.51 x105 N m

- 49.5 kg

(95.5 rad )2

Thus the minimum mass that should be added to the machine is 9.5 kg. 3.24 The tail rotor section of the helicopter of Fig. 3.20 consists of four blades, each of mass 2.3 kg, and an engine box of mass 28.5 kg. The center of gravity of each blade is 170 mm from the rotational axis. The tail section is

If a blade falls off during flight, the rotor is unbalanced and leads to harmonic excitation of the tail section. The magnitude of the rotating unbalance is moe = (2.3 kg)(0.170 m) = 0.391 kg-m The natural frequency of the rotor after one blade falls off is

6.87 x 105 1T m rad =139.3 37.7 kg - 2.3 kg

3.27

Harmonic Excitation of 1-Degree-of-Freedom Systems

The frequency ratio is r= 0)n

(900 m (2n min) rd)(1 r. m1 r 60s 139.3 rad s = 0.677 The steady-state amplitude is calculated using Eq. (3.17) X=

mo e

A(0.677, 0.05)

These approximately constant values are S = 0.2, CD = 1.0 In this case, show that the amplitude of excitation is proportional to the square of the frequency, and determine the constant of proportionality. Solving for v from Eq. (3.59) and setting S = 0.2

CD —

0.391 kg-m

(0.677)2

35.4 kg

[1— (0.677)2 ]2 \I + [2(0.05)(0.677)]2

coD

(3.61) 0.4n Substituting Eq. (3.54) into Eq. (3.60) with CD = 1.0 leads to v—

F0 1 (wD

= 9.27 mm 3.25 When a circular cylinder of length L and diameter D is placed in a steady flow of mass density p and velocity v, vortices are shed alternately from the upper and lower surfaces of the cylinder, leading to a net harmonic force acting on the cylinder of the form of Eq. (3.5). The frequency at which vortices are shed is related to the Strouhal number (S) by coD 2nv

S—

Ea 2

DL

which leads to F0 = 0.317pD3Ld 3.26 As a publicity stunt, a 120 kg man is camped on the end of the flagpole of Fig. 3.21. What is the amplitude of vortex-induced vibration to which the man is subject in a 5 m/s wind? Assume a damping ratio of 0.02 and the mass density of air as 1.2 kg/m3.

(3.59) D=10 cm

The excitation amplitude is related to the drag coefficient CD by • CD —

21- 0.4/r

E=80x10

(3.60)

pv2 DL

The drag coefficient and Strouhal number vary little with the Reynolds number Re for 1 x 103 < Re < 2' x 105.

Fig. 3.21

9 N

m2

3.28

Mechanical Vibrations

The flagpole is modeled as a cantilever beam of stiffness k= 3E1 L3. 3(80 x109 N m2

4

(0.05 m)4

= 1.67 x 10-5 m

The natural frequency of the man is

-=\

= 8.86

m

120 kg

rad

The vortex shedding frequency is

-

11 1 0.47r v 0.42r (5 s D

= 62.8

1.9x10-3 kg-m 120 kg

V[1- (7.09)2 ]2 + [2(0.02)(7.09)]2

= 9.42 x 10311

con= \I

A X= — (A)(7.09, 0.02)

(7.09)2

(5 m)3

9.42 x 103

Then using Eq. (3.17),

3.27 A 35 kg block is connected to a support through a spring of stiffness 1.4 x 106 N/m in parallel with a dashpot of damping coefficient 1.8 x 103 N-s/m. The support is given a harmonic displacement of amplitude 10 mm at a frequency of 35 Hz. What is the steadystate amplitude of the absolute displacement of the block? The natural frequency, damping ratio, and frequency ratio are

0.1m

1.4 x106

rad

con=

\

35 kg

m = 200 rad

Hence, the frequency ratio is 62.8

rad s

= = A — 7.09 d 0)" 8.86 ra Using the results of Solved Problem 3.25, it is noted that vortex shedding provides a frequency squared excitation with

2mw,, 1.8 x103 N-s m

= 0.129

2(35 kg) (200 rad )

A= 0.317pD3L kg = 0.317 (1.2 — (0.1m)3(5 m) m3 = 1.9 x 10-3 kg-m

135

cycle) ( , rad ) 27r s ) cycle 200

rad

s.

- 1.10

3.29

Harmonic Excitation of 1-Degree-of-Freedom Systems

The amplitude of absolute acceleration is obtained using Eq. (3.27) as X= Y7'(1.10, 0.129) 1+ [2 (0.129)(1.10)]2

= (0.01 m)

— (1.10)2 + [2(0.129)(1.10)12 = 29.4 mm 3.28 For the system of Solved Problem 3.27 determine the steady-state amplitude of the displacement of the block relative to its support. The displacement of the block relative to its support is obtained using Eq. (3.26) Z= YA(1.10, 0.129) (1.10)2

= (0.01 m)

[1— (1.10)2 ]2 + [2(1.10)(0.129)12 ,\1 = 34.3 mm

r=

co (1)n

30 1

cycle) ( rad 27r s cycle 75.6

rad

The amplitude of absolute displacement of the flow measuring device is calculated using Eq. (3.27) X= YT(2.49, 0.08) = (0.0005 m)

1+[2(0.08)(2.49)]2 [1— (2.49)2 ]2 + [2(0.08) (2.49)12

= 1.03 x 10 4 m The acceleration amplitude is A = o)2X ( cycle) ( rad )12 30 27r s cycle [ (1.03 x

3.29 A 35 kg flow monitoring device is placed on a table in a laboratory. A pad of stiffness 2 x 105 N/m and damping ratio 0.08 is placed between the apparatus and the table. The table is bolted to the laboratory floor. Measurements indicate that the floor has a steady-state vibration amplitude of 0.5 mm at a frequency of 30 Hz. What is the amplitude of acceleration of the flow monitoring device? The natural frequency and frequency ratio are

con

ITc \ m 11

2x105 IT m 75.6 rad 35 kg

— 2.49

m)

=3.66 ' s2 3.30 A vehicle has a mass 490 kg and the total spring , constant of its suspension system is 58800 N/m. The profile of the road may be approximated to a sine wave of amplitude 40 mm and wave length 4 m (Fig. 3.22). Determine: (i) Critical speed of the vehicle (ii) The amplitude of the steady state motion of the mass when the vehicle is driven at critical speed and = 0.5 (iii) The amplitude of steady state motion of mass when the vehicle is driven at 57 km/hr and damping factor = 0.5

3.30

Mechanical Vibrations

X= 40 x 10-3

.\/F1

X= 56.57 x 10-3 m Road profile

A,

4m —).-1

co —

27rV A,

22r (57 x1000) 4 3600 ) = 24.87 rad/s co/co„ = 2.27 X= 40 x 10-3

Fig. 3.22 m = 490 kg k = 58,800 N/m co— O IT.T; .

V1+ [2(0.5)(2.27)]2

58,800 — 10.95 rad/sec 490

(1— 2.272)2 + [2(0.5)(2.27)12

(i) V(m/sec) + r(sec) = A

= 20.96 x 10-3 m

1 V7 A

-A (co/27r)

27rV =

co —

2 irV A

Critical speed co= co„ = 10.95 rad/s

(ii)

3.31 .A radio set of 20 kg mass must be isolated from a machine vibrating with an amplitude of 0.05 mm at 500 rpm. The set is mounted on four isolators, each-having a spring scale of 31,400 N/m and damping coefficient 392 N sec/m (i) What is the amplitude of vibration of the radio (ii) What is the dynamic load on each isolator due to vibration

24rV — 10.95 A V= 6.973 m/sec V= 25.106 km/hr (critical speed)

m = 20 kg k = 4 x 31,400 = 125,600 N/m c = 4 x 392 = 1,568 N sec/m y= Y sin cot Y= 0.05 mm

= 1, C= 0.5, Y= 40 x 10-3 m

co=

27r x 500 60

= 52.5 rad/s

co„= .Jk/m

x Y

V1+(2cr)2 (1 — r2 )2 + (2 cr)2

1125,600 — 79.2 rad/s \ 20

3.31

Harmonic Excitation of 1-Degree-of-Freedom Systems

= 0.662 (O

cc

1568 2(20)(79.2)

c

2mcon

= 0.496 (under damped) X Y.

(1_r2)2

Z = YA(2.49, 0.08)

(2.49)2

= (0.0005 m)

V1+ 2cr

—

The elastic mounting is placed between the flow measuring device and the table. Hence, its deflection is the deflection of the flow measuring device relative to the table. The amplitude of the relative displacement is calculated using Eq. (3.26)

[1— (2.49)2 ]2 + [2(0.08)(2.49)]2

(2.cr)2

= 0.069 mm or 6.9 x 10-5 m Z

r2

Y

J (1— r2)2 + (2 cr)2

Z = 0.025 mm or 2.5 x 10-5 m

(ii) Dynamic load Fdy = ZNI(cW)2 + k2 = 3.77 N Dynamic load on each isolator 3.77 — 0.94 N 4 or Maximum dynamic load on isolator = Maximum force transmitted through isolator = Maximum inertia force on radio = mco2X = 20 x (52.5)2 x (6.9 x 10-5) = 3.8 N On each isolator =

L8 4

= 0.95 N

3.32 What is the maximum deflection of the elastic mounting between the flow measuring device and the table of Solved Problem 3.29?

= 5.94 x 10-4 m 3.33 A simplified model of a vehicle suspension system is shown in Fig. 3.23. The body of a 500 kg vehicle is connected to the wheels through a suspension system that is modeled as a spring of stiffness 4 x 105 N/m in parallel with a viscous damper of damping coefficient 3000 N-s/m. the wheels are assumed to be rigid and follow the road contour. The contour of the road traversed by the vehicle is shown in Fig. 3.24. If the vehicle travels at a constant speed of 52 m/s, what is the acceleration amplitude of the vehicle? v m= 500 kg k = 4 x 1 051A m s c = 3000 N— m m v = 52 —

Fig. 3.23

3.32

Mechanical Vibrations

Fig. 3.24 The natural frequency and damping ratio of the system are con= ri' in

4x105 '11500 kg

= 28.3 rad

2mco„

3000 N-s m 2(500 kg)(28.3 rad)

52 m = 130.7 4 J

The amplitude of absolute displacement of the vehicle is calculated using Eq. (3.27) X= YT(4.62, 0.106) = (0.01 m)

1+ [2(0.106)(4.62)12 [1— (4.62)2 ]2 + [2(0.106)(4.62)12

= 0.106

The mathematical description of the road contour is Y(C) = 0.01 sin (0.84) m If the vehicle travels with a constant horizontal velocity, C= vt. Thus the timedependent vertical displacement of the wheel is y(t) = 0.01 sin [0.8 mit] Since the wheel follows the road contour, it acts as a harmonic base displacement for the body of the vehicle. The frequency of the displacement is co = 0.87tv = 0.8

Hence, the frequency ratio is rad 130.7 co = s r= — 4.62 co„ 28.3 rad

rad

= 6.87 x 10-4 m The vehicle's acceleration amplitude is A = co2X = (i30.7 rad )2

(6.87 x 10-4 m)

11.7 1 1` s2 3.34 Let A be the amplitude of the absolute acceleration of the vehicle of Solved Problem 3.33. Show that A — R(r, C) con2 y

1+ (2Cr)2 r2 \I (1— r2 )2 + (402

where Y is the amplitude of the road contour.

3.33

Harmonic Excitation of 1-Degree-of-Freedom Systems

The amplitude of acceleration is to2X where X is the amplitude of absolute displacement of the vehicle. From Eq. (3.27), a)

2 x

0

— T(r, C)

a.) 2 y

co2

A

CO„2 2 Y

co

A con2 y

6, 3

T(r,

T (r,

2

(0,1

3.35 Plot R(r, C) from Solved Problem 3.34 as a function of r for the value of C obtained in Solved Problem 3.33. At what vehicle speeds do the relative maximum and minimum of R occur? The plot of R(r, 0.106) is shown in Fig. 3.25. The values of r for which the maximum and minimum of R(r, C) for a given C occur are obtained by setting' dR2/dµ = 0 where ,u = r2. To this end

5

r

Fig. 3.25

1+ (2Cr)2 (1— r2 )2 + (2Cr)2

4

3

Substituting C = 0.106 and rearranging leads to /13 = — 3.909,u2 — 40.5µ + 44.5 = 0 whose positive roots are = 1.025, 8.190 r = 1.012, 2.862 The vehicle speeds for which the maximum and minimum steady-state amplitudes occur are given by (28.3 rad v—

rco„ 0.87r

=

jr 0.87r

— 11.26r

+ 4c2 µ3

R2 —

—2)µ+1 and using the quotient rule for differentiation,

vmax = 11.26(1.012) = 11.40 m s

µ2

dR2 d (2,1+12 C2 /12)[112 +(4c2 —2)/1+1] —(µ2 + 4C2 µ3)[(2µ,+(4c2 —2)] [/./2 +

- 2)// +1]2

Setting the numerator to zero leads to 4Cµ3 + (320 — 16C2)µ2 +(164'2 — 2)µ + 2 = 0

vmin

11.26(2.862) = 32.2 ± 1. i s

3.36 Determine the form of W(r, such that XI Y = W(r, C) for the system of Fig. 3.26. What is Wmax? Free body diagrams of the block are shown at an arbitrary instant in Fig. 3.27. Summation of forces

IF)e..1F)eff

3.34

Mechanical Vibrations

mo)?2 x

, cwY

— M(r, 4)

mo),2 x 1

2cmco„ coY co, X 24-coY X y

y(t) = Ysin cot

— M(r, 4)

— M(r,

= W(r, 4) = 2CrM(r, 4) 24-r

Fig. 3.26

(1— r2 )2 + (2'r.)2 The values of r for which the Maximum of W is obtained by setting dW2/dr = 0. The quotient rule for differentiation is applied, giving

t

mk

c(x—Y) Fig. 3.27

dW2

leads to —kx —

dr —

mr + cic + kx = cj) = coff cos (cot)

(3.62)

8c2 r [(1 — r2 )2 + (24-r)2 ] — 44-2 r2 [2 (1 — r2)(-2r) + 2(24"r) (24)] [(1 _ 1.2)2 +( go y

Setting dW21dr = 0 leads to 2 — 2r4 = 0-p r= 1 and Wmax = 1

+ 24-con ic + (0,2, x c = — wt cos wt = 2cco„Y cos wt k _ c m 2rnco„ Equation (3.55) is of the form of Eq. (3.4) with the excitation of Eq. (3.5) where where

con =

Fo = ccoY,

3.37 Determine the steady-state amplitude of angular oscillation for the system of Fig. 3.28. Free body diagrams of the system at an arbitrary instant are shown in Fig. 3.29 Summing moments about 0, —14 Le— y)r L)— cLe() 4 4 4 4

ty= ± 7

2 Thus the steady-state amplitude is obtained using Eq. (3.10) as

=

12

nto

+ mLe ) 4 4L

(3.63)

3.35

Harmonic Excitation of 1-Degree-of-Freedom Systems

y(t) = Ysin cot

T k

N k= 2 x 105 m

3L

L

Tt

c = 400N —s

m

Slender bar of mass m

L = 1.2 m m = 10 kg Y = 0.01 m w = 350

rad

Fig. 3.28 48 tnL2 e - -17 c

+

kg 0

3 kLy(t)= —3 kYL sin cot =— 4 4 The natural frequency and damping ratio are kg 16 48

N 3 3 F0= — kLY = (2 x105 — 4 4 (1.2 m)(0.01 m) = 1800 N-m 7 = ,e13 (10 kg)(1.2 m)2 = 2.1 kg-m2 The frequency ratio for the system is 2

meg = 48

_ .127 k

— 7 mL2

Equation (3.63) is of the form of Eq. (3.4) with the excitation of Eq. (3.5) where

7m

27(2 x1051 m .277.8 rad 7(10 kg)

350 rad s =1.26 277.8 rad

co =' r= w,,

The system's magnification factor is cL2 16 2cco — 7 Inv 48

3c

14mco n = 0.0309

M(1.26, 0.0309)

3(400 • m 14(10 kg)(277.7 rad)

1 [1— (1.26)2 ]2 + [2(0.0309) (1.26)]2 = 1.69 The steady-state amplitude is obtained using Eq. (3.10) as meq o. F0

— M(1.26, 0.309)

3.36

Mechanical Vibrations

k

3L \4

0

m-6-2

Y

c —0

Lm-0

4

1 me

4

External forces

12

Effective forces

Fig. 3.29

—

to a support undergoing harmonic motion. The system is undamped with a natural frequency of

F0 M(1.26,0.0309) iiteq (,,

(1800 N-m)(1.69)

4.43 x 105 -111

(2.1 kg-m2) (277.8 rad )2

wn=

= 0.0188 rad = 1.08°

Tn

=\

= 42.1

3.38 Determine the steady-state amplitude for the machine in the system of Fig. 3.30. y(t) = 0.005 sin 35t m

250 kg

rad

and frequency ratio

=

35 rad co _ s — — 0.831 42.1 rad Wu

E = 210 x m

250 kg

2

The steady-state amplitude is

= 4.1 x 10-6 m4

X= YT(0.831, 0) —

1.8 m

1— (0.831)2

Fig. 3.30 The system is modeled as a 250 kg block through a spring of stiffness N 1 3(210x109 m2 k=

3E1

0.005 m

=

(4.1 x10-6 m4 )

L3

(1.8 m)3

= 0.0162 m

3.39 Approximate the steady-state amplitude of the block in the system of Fig. 3.31. k =1x106 LI 100 kg

300 sin 40t

itimmiiim ri µ=0.08

Fig. 3.31 = 4.43 x 105 The natural frequency and frequency ratio for the system of Fig. 3.31 are

3.37

Harmonic Excitation of 1-Degree-of-Freedom Systems

N 125 kg E= 200 x 109— m2

lx105 11 m CAn = r i i n = 100 kg

/=4.5x10-6m4

80 cm

= 31.6

rad

Fig. 3.32

s.

40 rad • 0.) r— = — 1.27 31.6 rad o

The system's equivalent stiffness, natural frequency and frequency ratio are

The system's force ratio is ml 0.08(100 kg) (9.81 J s2 300N

µ mg t— F0 = 0.262 The magnification factor is determined using Eq. (3.43.),

_ (402 (n) (1- r)2

1 M

k—

3E1 L3

7r

(0.8 m)3 N

= 5.27

x 106 m 5.27 x 106 IT m 125 kg

= 250.3 rad

[4(0.262)12

1

=

3(200 X 109 N m2 (4.5x10-6 m4 )

— 1.538

[1— (1.27)2 ]2

r=

0.)

(2000 r )(2x rad) min r ) (1 min 60 s 205.4 rad

The steady-state amplitude is X=

Fo M 2

ni

--n

(300 N) (1.538) =

A 2

d (100 kg) (31.6 ra

= 4.61 mm 3.40 When a free vibration test is run on the system of Fig. 3.32, the ratio of amplitudes on successive cycles is 2.5 to 1. Determine the response of the machine due to a rotating unbalance of magnitude 0.25 kg-m when the machine operates at 2000 r/min and the damping is assumed to be viscous.

= 1.02 The logarithmic decrement for under damped vibrations is 8= ln (2.5) = 0.916 from which the viscous damping ratio is calculated as 6

47c2 + 62 0.916 V47r2 + (0.916)2

— 0.144

Mechanical Vibrations

3.38

Noting from Solved Problem 3.17 that the rotating unbalance provides a frequency squared excitation with A = moe, and using Eq. (3.17), X=

mo e 171

F(N 5000

7 7 7. Fl >

0.02

4 0.06 0.08'0.10 0.12 0,14 0.16 t(s) 0.0L_J

—5000

A(1.02, 0.144)

0.25 kg-m

Fig. 3.33

(1.02)2

125 kg

R— (1.02)212 + [2(0.144)(1.02)]2

2 0.04

= 7.02 mm

-0.02

(— 5000) sin 507r1t dt+

_o 0.04

3.41. Repeat Solved Problem 3.40 if the damping is assumed to be hysteretic.

(5000) sin 50 nit dt 0.02

Equation (3.44) is used to determine the hysteretic damping coefficient from the logarithmic decrement h—

3

= (50(5000) I

In 2.5) (-0.292

The steady-state amplitude is obtained using Eqs. (3.18) and (3.46) and calculated by

—1 5Orti

[ cos Id + cos

0 + cos 2ii — cos ni] —

10,000

[1 — (-1)i]

Thus the Fourier series representation for F(t) is

X=

mo e NI (1—

F(t)= 10

r+h2

000 7C

0.25 kg-m

(1.02)

= 7.06 mm:

20,000 iv

3.42 Determine the Fourier series representation for the periodic excitation of Fig. 3.33. The excitation of Fig. 3.33 is an odd excitation of a period 0.04 s. Thus ai = 0 i = 0, 1, 2, .... The Fourier since coefficients are calculated by =

T J

0

F (t) sin

i=1

/

sin 507rit

125 kg1[1— (1.02)2 ]2 +(0.292)2

b.;

1 .[(1) — (-1)I]

t dt

1 — sin 50 git

3.43 Determine the Fourier series representation for the excitation of Fig. 3.34. The excitation of Fig. 3.34 is an even excitation of period to. Hence, bi = 0, i = 1, 2, The Fourier cosine coefficients are

3.39

Harmonic Excitation of 1-Degree-of-Freedom Systems

(2/3)to

Fo cos

27ri

to 2t0

to

3 3

10

4to 5t0 2t0 7to 8to 9t o t 3 3 3 3

3F0 1 —

3Fo (1 if F (t)dt to 0 to

=

/3

2

To

i 2 it 2

0

cos

9F0

(2/3)10

( 3F° )t dt + to

2

2i2 7r2

Fo dt + to /3

3F0 (1 —11 dt1 to

2

00

to

27ti 2 a;= — F (t) cos — t dt to to 0

to

27ri 1 4 Iri + cos 1 3 2 3

i=1, 2, 4,5, 7, 8,...

9F0 2712

F0

3

4

=—F0 3

[(1/3)in f ( 3F0

y COS

27ri 1 — cos ;2

i=1,2,4,5,7,8 I

27ri

to

3.44 Determine the Fourier series representation for the excitation of Fig. 3.35. t dt +

to

to

F(N),

2000

I

I

1

F(t) is F(t)—

2

to

t dt

Thus the Fourier Series representation for

(2/3)k,

=

27t1

i =3, 6, 8,12,...

0

to

L) cos

to

(213)10

Fig. 3.34 a0 =

t dt +

to

(1/3)t„

I

I

I

I

1

I

I

I

I

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16

Fig. 3.35

t(s)

3.40

Mechanical Vibrations

The excitation of Fig. 3.35 is neither even nor odd and has a period of 0.04 s. The Fourier coefficients are 2 0.04

ao

[o.o 1 f 0

J

ci =

1

0.04

(2000) dt +

Then + bi2

2000

(0) dt

~ci

0.01

4000

= 1000 N

ici at -

2 0.04

sin e 7C — /. + (1 - cos — 1. 2 2 \ sin

[0.01

(2000) cos

J 0

0.04

0.04

4

r

t dt +

sin 2 i 2

and xi = tan-I

1- COS — t

2./ti (0) cos t dt 0.04

0.01

2 Thus the Fourier series representation for F(t) is CO

= (50)(2000)

F(t)= 500 +

1 507ri

(sin

/=1 i sin (507ri + /cat

2

i - sin 0)

2000 . 7ri sin — 7ri 2 0.01 bi -

2

0.04

(2000) sin 27ri t dt + 0.04

(0) sin

t dt

1.8x107

(0

1 ( 507r; )

x(t) 2

-

=

\

m

200 kg

= 300

rad

The system response is obtained using Eq. (3.39) as 7r .

1- cos

4

3.45 A 200 kg press is subject to the timedependent excitation of Solved Problem 3.44 and Fig. 3.35. The machine sits on an elastic foundation of stiffness-1.8 x 107 Nina and damping ratio 0.06. Determine the steady-state response of the machine, and approximate its maximum displacement from equilibrium.

to

o oi

2000

sin (7r

The natural frequency of the system is

0.04

= (50)(2000)

4000 v 1

4000 - 1 1 500 + n i =1 1.8x102

i sin (1r i) Al(ri, 4 sin (507c; + xi - (Mt

Harmonic Excitation of 1-Degree-of-Freedom Systems

Table 3.1 i

(Di

1 2 3 4

I.;

157.1 314.1 471.2 628.3 785.4 942.5 1099.6 1256.6

5

6 7

8

where ri =

co; con

ci

0.523 1.047 1.57 2.094 2.672 3.141 3.665 4.188

900.2 636.6 300 0 179.8 212.0 128.6 0

illi

cillili

1.37 6.31 0.68 0.29 0.17 0.11 0.080 0.06

1233 4017 202.7 0 30.7 23.9 10.3 0

Z_

r2

Y

1

Z

4(1- ri2 )2 + (0.1202

17

Table 3.1 illustrates the evaluation of the response. Then xn,„
to, u(r — to)g(t, dr

—Fo

ro to /2

+Fo

— Fo

= fu(r— to)g(t,

+

u(r — to)g(t, r) dr to

= f g(t, T) dr Fig. 4.4

cis'

General Forced Response of 1-Degree-of-Freedom Systems

Thus

f

4.9

1 to ) + u (t — — 2

u(r— to )g(t,T)dr

0

= ,

(1/2)to

(1 — 2 J—) to

sin oUt — 'r) (Pc — u(t — to)

t < to

1 to

— II-) sin co„ (t — 'Odd to

.

g (t , 1- )ctr t > to

Evaluation of the integrals yields

ro

2 FO

X(t)

mCO„2

=u(t —to )f g(t,r)ds'

x(t) —

1 f 2F0 [ ow; to o + (1 — 2 r-ju i — ) to 2

(t)(

t to

1to sill • (.0„ t CO„

2 u t — Ito )[1 2 t + 2 to co„ to (

4.11 Use the results of Solved Problems 4.9 and 4.10 to develop a unified mathematical expression for the response of an undamped 1-degree-of-freedom system due to the triangular pulse of Fig. 4.3. Substitution of the unified mathematical reprsentation of the triangular pulse developed in Solved Problem 4.9 into the convolution integral, Eq. (4.5) with 4 = 0, leads to

{ti

sines„ (t — — 2 to )1— u (t — to ) l— t + 1to sin co„ ( t — to )1} C to co„ 4.12 Use the convolution integral and unit step functions to develop a unified mathematical expression for the response of an undamped 1-degree-of-freedom system to the excitation of Fig. 4.5. F(t) 2 F0

Fa

— (1—

to JJ

(1— . to )1

sm co„(t — 'r)ctr Using the integral formula of Solved Problem 4.10, Fo

x(t) — 2 u(t)fisin co„ (t —T)dT to "n . 0

to

2 to

Fig. 4.5 The unified mathematical representation of the excitation of Fig. 4.5 is

Mechanical Vibrations

4.10

Substitution into Eq. (4.5) with leads to

F(t) = 2Fo[u(t) — u(t — to)] + Fo[u(t — to)— u(t — 2to)] = 2Fou(t) — Fou(t — to) — Fou(t — 2t0)

x(t) —

Substitution into the convolution integral, Eq. (4.5), with C = 0 leads to R

x(t) —

C=0

1

{2000 5[ u( "r) (10kg)(100 rad ) — u(1- — 0.1)] sin 100(t— -I-) dz.

I

) F0 5[2u(r)—u(r—to {

"n

+20 f S('r — 0.25) sin 100( t —

o

—u(r— 2t0 )]sin co„ (t--c)dr

= 2u(t) f sin 100(t — r)

}

—

F0 2u (t)f sines„ (t — z)d'r

mco,,

—u (t — to ) 1 sin con (t -'NT

. to

1 sin 1000 — dr . o + 0.02 sin 100(t — 0.25)u(t — 0.25) = 0.02u(t)(1 — cos 100t) — 0.02u(t — 0.1)[1 — cos 100t — 10)] + 0 .02u(t — 0.25) sin (100t — 25) — 2u(t — 0.1)

4.14 Use unit step functions to develop an —u(t-2t0 ) f sine„ (t —T)ds24,

Fo {u(t)(1 — cos co„t) x(t) — 2 mw„ — u(t — to)[1 — cos co„(t — to)] — u(t — 2t0)[1 — cos co„(t — 2t0)]}

infinite series representation for the periodic function of Fig. 4.6. The graphical breakdown of the excitation of Fig. 4.6 is shown in Fig. 4.7. The representation of F(t) in terms of unit step functions is F(t) = Fo [u (t) — u(t — I to )]

4.13 Determine a unified mathematical expression for the response of an undamped 1-degree-of-freedom system of con = 100 rad/s and a mass of 10 kg subject to a rectangular pulse of magnitude 2000 N and duration 0.1 s followed by an impulse of magnitude 200 N-s applied at t = 0.25 s. The mathematical representation of the excitation is F(t) = 2000[u(t) — u(t — 0.1)] + 203(t — 0,25)

+F0 [u(t—to )—u(t—

2

t0

+F0 [u(t— 2t0 )—u(t—t0 )]+...+ which can be written as F(t) = Fo

fu (t — ito )

u [t — ( 2 i — 1 ) to 1}

4.11

General Forced Response of 1-Degree-of-Freedom Systems

F(t)

Fo

to 2

0

3t0 2

2to

5o

3to

7t° 2

2

9to

4t

2

Fig. 4.6

Fo

—Fo

to 2

+Fo

3to

to

2

+•.. Fig. 4.7 4.15 Use the convolution integral to determine a mathematical representation for the response of an undamped 1-degreeof-freedom system due to the periodic excitation of Fig. 4.6. Substitution of the mathematical form of the excitation developed in Solved Problem 4.14 into the convolution integral, Eq. (4.5) with C = 0 leads to x(t) _

1

r

ma)„

0

The order of integration and summation can be interchanged assuming the infinite series converges for all t. This leads to x(t) —

Fo

mCD„ 1=1

{t fu(' — ito ) sin con (t — 1-) d T 0 rr 1 — f li[r — — (2 i

Fo

Lfuer—ito—u[s-- 1(2i —1)to

]}

sin co„(t — r) ch.

o

2

— 1) to 1

sin to„ (t — s') ch}

4.12

Mechanical Vibrations

The integrals are evaluated using the integral formula developed in Solved Problem 4.10,

Using linearity of the inverse transform, di" 9v-1{

X(t)

IS1

12

mw„ x(t)

F0 °±

fu(t - it0 ) cos CO„ (t — T)

mw„ i = 0

n

s2 + CO2

The inverse transforms are determined using Table 4.1, leading to

T=t

1 (1 cos ton t)

x(t) -

- u [t -1(2 i - 1) to = 110

4.17 Solve the Solved Problem 4.2 using the Laplace transform method.

cos con (t — z)

The excitation force and its Laplace transform are

F0

,

F(t) = Fo (t — ito)

= Fo

F(s) — s Substitution into Eq. (4.7) with x(0) = 0 and x (0) = 0 leads to

"'"'n i = 0

[1 - cos co„ (t - ito)] 1 t - - (2i -1) to - u[ 2

x(s) =

{1 - cos co„ [t - --1- (2i -1)to 1}} 2 4.16 Solve the Solved Problem 4.1 using the. Laplace transform method. The excitation force and its Laplace transform are F(t) = Fo F(s)=

A partial fraction decomposition leads to i(s) - 1 [1 mco,2, s

X(t)

=

12 (Y-1111

M, CO

s 2 +

S

s + CCO„

1

(s + Cco„ )2 + co,21

A partial fraction decomposition yields Fo (i 2 MCO„ S

s2 + 2Ccon s +w,

Completing the square of the quadratic denominator and using linearity of the inverse transform leads to

MS (s2 + 0)D

.V(s)=

s + 2Ccon

Fo

Substituting into Eq. (4.7) with x(0) = 0 and z (0) = 0 and C = 0 leads to F0

Fo 1 ms (s2 + 2Cco„ s + con2 )

)

con

wd -I (s + cco„)2 + o)/})

4.13

General Forced Response of 1-Degree-of-Freedom Systems

The first shifting theorem of Table 4.2 is used to obtain

Use of linearity of the inverse transform leads to

x(t) =

0(t) -

1 ,[1 - e-C(`).` (Y-1{

M (0;'

2 s 2 S + COd

36 Fo 7mL(a2 + con2 ) (

COn 29-1

wd s2+(1)dJ

d

The transform pairs of Table 4.1 are used to obtain x(t) -

a ` -1 { 1 1+ s+a co„

1

C°11 2

(9(t) -

COn [1- e-C6).1 (cos cod t + —sin cod t )1 0)d

12F0 7 mL(a2 + (e at

4.18 Solve the Solved Problem 4.5 using the Laplace transform method. The differential equation of Solved Problem 4.5 can be written as c 0 ,23 0 = 36F0 e-ca 7 mL =

27 k 7m

Assuming the system is at rest in equilibrium at t = 0 and taking the Laplace transform of the differential equation leads to

Partial fraction decomposition yields 36F0

S a

sin con t - cos co„ i

4.19 Use the Laplace transform method to determine the response of an undamped 1-degree-of-freedom system of natural frequency co„ and mass in, initially at rest in equilibrium and subject to the triangular pulse of Fig. 4.3. Form the results of Solved Problem 4.5, the mathematical expression for the triangular pulse is F(t) = 2 Fo

(t) + (1 - 21 ) t0

to

o - (1 - -1-)u (t - to )1 2t to

and its Laplace transform is obtained using the second shifting theorem and Table 4.1 as Fo F(s) = 2— to

7mL(a2 + co,2,) 1

+ a (1)n

ul —

36F0 1 (s)= 7 mL (s + a)(s + con2 )

j(s)=

2 S S + CO„

Use of Table 4.1 leads to

m0)2,

where

22-1

}

Y-1{S2 + CO„

a-s s2

0.)

(

1

2 -s(i0i2) + 1 e-sto

— - —e 2 2

s s

s2

4.14

Mechanical Vibrations

The second shifting theorem is used to obtain

Substitution into Eq. (4.7) leads to 2F0 1- 2 e-s('0/2) 3c-(s)=

s2 (s2

mto

e-sto

_ F(s)= F0 — s

w2)

A partial fraction decomposition yields Y(s) = m (0,,"to (1 2e-s(V2) e—sto )( 1 2 S

i=o Substitution into Eq. (4.7) with x(0) = 0, 540) = 0 and, C = 0 leads to

1 S

2

eisto - e 2(2;-1)st 0 )

2 CO„

Y(s) -

The system response is obtained by application of the second shifting theorem and the transform pairs of Table 4.1. Thus x(t) =

2 m s(s2 + wn ) 2 2i — 1) sto

I [e-ist° - e =o

1 2F0 { t - con sin con t to

inco2

- 2u

1

Fo

- to )[(t -1to ) 2

FO

1

2 MO),

S

IL =

2

e—ist° — e 2

2) CO„ —1)st °

i 0

1 - — sin co„ (t - to ji + u(t - to) 2 con co„ (t - to )]}

[(t - to ) 0)n

Inversion of the transform is performed using the second shifting theorem: x(t) -

Fo

{[1-cos con (t - it0 )]

"`"'n i = 0 u(t - ito)}

4.20 Use the Laplace transform method to determine the response of an undamped 1-degree-of-freedom system of natural frequency co„ and mass m, initially at rest in equilibrium and subject to the periodic excitation of Fig. 4.6. From Solved Problem 4.14, the mathematical representation of the periodic function of Fig. 4.6 is F(t) =

F0 / [u(t - /to ) - u

- (2i 1) to )1

- {1 - cos co„ [1- 1(2i -1)to ]f 2 u[t - i(2i 1 -1)6) ] 2

4.21 Use the Laplace transform method to determine the response of an underdamped 1-degree-of-freedom system of damping ratio C, natural frequency co„ and mass in, initially at rest in equilibrium and subject to a series of applied impulses, each of magnitude I, beginning at t = 0, and each a time to apart.

General Forced Response of 1-Degree-of-Freedom Systems The mathematical form of the excitation and its Laplace transform are F(t)= IDS (t - ito ) i=o x

F (s)=I 1e'st0 =o

Substitution into Eq. (4.7) with x(0) = 0 and (0) = 0 leads to )7(s) -

1

m s2 + 2 cco s + 0),2,

x

isto

0

1

_ I

2 2 m ( S + COO + COd 1= 0

ist°

Inversion of the above transform is achieved using both shifting theorems: x(t) -

e-co)„( -trio

/

mcod

0

sin cod(t - it0)u(t - it0) 4.22 Let v = x. Rewrite Eq. (4.1) as a system of two first-order ordinary differential equations with t as the independent variable and, x and v as dependent variables.

over which a solution is desired is discretized, and recurrence relations are developed for approximations to the dependent variables at the discrete times. Let t1, t2, ..., be the discrete times at equal intervals At. The Eider method is an implicit method using a first-order Taylor series expansion to approximate the time derivatives of the independent variables. Develop the recurrence relations for a 1-degree-of-freedom system using the Euler method. Let f(t) be a continuously differentiable function of a single variable. Its Taylor series expansion is 1 f(t + At) = f(t) + (At)i(t) + - (At)2 (t) 2 + - (At)3f (t) + • • • + 6 Truncation after the linear term and rearranging leads to f (t + At) - f(t)

+ 0 (At) (4.16) At Define xi = x(i At) and vi = v(i At). Application of Eq. (4.16) to Eqs. (4.14) and (4.15) of Solved Problem 4.22 at t = ti = i At leads to f(t) =

From the definition of v, =v and form Eq. (4.1),

xi (4.14)

At

Meg

4.23 Direct numerical simulation of Eq. (4.1) often involves rewriting Eq. (4.1) as two first order differential equations, as in Solved Problem 4.22. The time interval

- xi + 0 (At) = At

vi +1 -vi

1 v = — F(t) - 2cco„v - con2x (4.15) Equations (4.14) and (4.15) form a set of first-order linear simultaneous equations to solve for x and v.

4.15

+ 0 (At)

1

=

F(4)- 2 cco„ - co„2

Meg

or

xi+1 = Xi + (At)v1 + O(At)

V1+1 = V

A

1

Llt(— F

f\

,., fro — (0,2, Xi )

M eg

+ O(At)

4.16

Mechanical Vibrations

4.24 Illustrate the application of the explicit Euler method to a 1-degree-of-freedom system of mass 10 kg, natural frequency 100 rad/s, and damping ratio 0.1 subject to a constant force of magnitude 100 N. The system is at rest in equilib-

rium at t = 0. Use a time increment of 0.001 s, and compare with the exact solution of Solved Problem 4.2. The calculations are illustrated in Table 4.3

Table 4.3 ti 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0.012 0.013 0.014 0.015 0.016 0.017 0.018 0.019 0.02

x (t )

v(t)

0 0 0.00001 2.98E-05 5.91E-05 9.75E-05 0.000145 0.0002 0.000262 0.000332 0.000407 0.000488 0.000572 0.000661 0.000751 0.000844 0.000937 0.001029 0.001121 0.00121 0.001297

F(tdim

gconvi

10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

0 0.2 0.396 0.58608 0.768398 0.94121 1.102881 1.251906 1.386922 1.506719 1.610251 1.696645 1.765208 1.815434 1.847003 1.859786 1.853844 1.829423 1.786951 1.727035 1.650448

0 0.01 0.0198 0.029304 0.03842 0.04706 0.055144 0.062595 0.069346 0.075336 0.080513 0.084832 0.08826 0.090772 0.09235 0.092989 0.092692 0.091471 0.089348 0.086352 0.082522

4.25 A numerical approximation to an integral 1(t)= ff(r) dr is of the form /(ti) =

oci f(ti) At i= 0

(4.17)

(0,2, x; 0 0 0.1 0.298 0.59104 0.975239 1.445844 1.997284 2.623237 3.316699 4.070058 4.875184 5.723506 6.60611 7.513827 8.437328 9.367221 10.29414 11.20885 12.10233 12.96585

x(t; + A)

v(ti + A) x(t) (exact)

0 0.00001 2.98E-05 5.91E-05 9.75E-05 0.000145 0.0002 0.000262 0.000332 0.000407 0.000488 0.000572 0.000661 0.000751 0.000844 0.000937 0.001029 0.001121 0.0021 0.001297 0.001379

0.01 0.0198 0.029304 0.03842 0.04706 0.055144 0.062595 0.069346 0.75336 0.080513 0.084832 0.08826 0.090772 0.09235 0.092989 0.092692 0.091471 0.089348 0.086352 0.082522 0.077906

0 4.96E-60 1.97E-05 4.38E-05 7.69E-05 0.000118 0.000168 0.000225 0.000288 0.000357 0.000431 0.000509 0.000591 0.000675 0.00076 0.000846 0.000932 0.001017 0.0011 0.001181 0.0011258

where ti, t2, ti are called knots, the intermediate values at which the integrand is evaluated. The values of a; are specific to the numerical method used. Develop a form of Eq. (4.17) that can be used to approximate the convolution integral, Eq. (4.2).

4.17

General Forced Response of 1-Degree-of-Freedom Systems

The extension of Eq. (4.17) to the convolution integral is x(ti) =

cciF(ti)h(ti — ti) At =o

Assuming tk = k At, x(tJ)=

yaiF(i At)h[j — i) At] At i =o

4.26 For the trapezoidal rule, the values of ai in Eq. (4.17) of Solved Problem 4.25 are a; = a./ = 0.5, a2 = a3 = • • • = = 1. Illustrate the use of the trapezoidal rule to approximate the time-dependent response of a system of mass 10 kg, natural frequency 50 rad/s, and damping ratio 0.05 subject to the time-dependent excitation of Fig. 4.8. Use At = 0.01 s, and approximate x(0.01), x(0.02) and x(0.03). F(t)

Then, using the trapezoidal rule x(0.01) = [0.5F(0)h(At) + 0.5F(0.01)h(0)] At = [0.5(0)(9.36 x 10-4) + 0.5(50)(0)](0.04) = 0 x(0.02) = [0.5F(0)h(2 At) + F(0.01)h(At) + 0.5F(0.02)h(0)] At = [0.5(0)(1.56 x 10'-3) + (50)(9.36 x 10-4) + 0.5(100(0)](0.01) = 4.68 x 10-4 m x(0.03) = [0.5F(0)h(3 At) + F(0.01)h(2 At) + F(0.02)h(At) + 0.5F(0.03)h(0)] At = 1.72 x 10-3 m 4.27 Define tj = j At. Show that Eq. (4.5) can be rewritten as 1 e_ x(tk) — "d

k 100 N

sin cod tk =I

0.02

0.04

0.06

t(s)

Fig. 4.8

G2 1 j =1

where Gib = J F(r)eC(1).'cos codr d'c ti_, =tf

Note that h(k AO — 1 e— CW" k At sin (cook At) mood = 2.003 x 10-3 e-0.025k sin (0.4995k) h(0 At) = 0, h(At) = 9.36 x 10-4 h(2 At) = 1.56 x 10-3, h(3 At) = 1.84 x 10-3

k

— cos cod tk

G21

F(r)e4).'sin codr dr

Note that sin cod(t — = sin codt cos cod z — cos cod t sin (Ad Substituting the previous trigonometric identity into Eq. (4.5) and rearranging leads to

Mechanical Vibrations

4.18

x(t) —

e mr m ood

t

sin cod t f F (z)eN,' cos cod zdr o

such that the constant interpolate on the interval from to ti is equal to F(t) evaluated at the midpoint of the interval. Determine the appropriate forms of G1j and G22 for a piecewise constant interpolation of F(t). Let

F1 = F[(ti + t")/2]. Then

— cos cod t f F (z)e0M sin o)d zdi 0

G,i =

Noting that g(T)d"C= f g(l")

+ g(nc12+

f Fi eN,T cos co„ z dz-

Fi (1 — c2

0

0

cod

+ f g(z)d r to -

k

e

I

"

t

sill cod t j +

ri

f g(t)d-t

=

1=Ir,

n 11

sin cod ti _ 1 +

Cod

leads to

COS Codtj _ 1

x(tk) —

r.

mcod k

sin cod tk

I,

k

cos cod tk

and G2i = f F./ eCa'”r sin cod

E f F(T)e4).' cos cod Tthr i=ly,

—

(pd COS COd ti

"I" (IT

•

Fj (1 — C2 ) cod

ri

f FeoeCm^T sin wd -cd-tj=1 5-I

4.28 One method of approximating the convolution integral is to interpolate F(t) piecewise and exactly integrate the interpolation times the trigonometric functions using Eq. (4.17) of Solved Problem 4.27. Suppose F(t) is interpolated by piecewise constants chosen

[

e

Wn

N, tj — cos cod t j + c— oi sin cod ti

— COS (Od ti

0)n 11 +C-

0)d sin (pd ti _1

General Forced Response of 1-Degree-of-Freedom Systems

4.29 Discuss the advantages and disadvantages of using each of the following numerical methods to provide a numerical approximation to the solution of Eq. (4.1): (a) implicit Euler method, (b) fourth-order Runge-Kutta method, (c) numerical integration of the convolution integral using the trapezoidal rule (Solved Problem 4.26) and (d) numerical integration of the convolution integral using piecewise constant interpolate for F(t) (Solved Problem 4.28). (a) The implicit Euler method is only first-order accurate. Its application leads to a pair of recurrence relations from which the approximations are successively determined. Application of the recurrence relations does not require evaluation of the excitation at times other than for which approximations are obtained. (b) The fourth-order Ruge-Kutta method is fourth-order accurate. It is an explicit method in that its recurrence relations are used to determine the approximations successively. However, their application requires evaluation of the excitation at times other than those at which approximations are obtained. (c) The trapezoidal rule is a numerical integration of the convolution integral. It provides a linear interpolation to the integrand between two knots. Its application does not lead to a recursion relation for the approximation at subsequent times. The formula must be repeated for each time the approximation is required, but an efficient algorithm is easy to develop. (d) The excitation - is interpolated by piecewise constants, multiplied by

4.19:

appropriate trigonometric functions, and the approximate integrand exactly integrated. An algorithm using this method to approximate the convolution integral is easy to develop in that the approximation for the response at an arbitrary time can be calculated as the response at the previous time plus the approximation to the integral between the two times. 4.30 To protect a computer during a move, the computer is placed on a cushioned crate. The motion of the computer in the crate can be modeled as a 1-degree-of-freedom mass-spring-viscous damper system. During the loading phase of the move, the crate is subject to the velocity of Fig. 4.9. Determine the displacement of the computer relative to the crate. V

V0

to

443

Fig. 4.9 Let y be the acceleration to which the crate is subjected. The differential equation governing the displacement of the computer relative to its crate is Eq. (3.24), + 2 cco„ + co,2, z = — whose convolution integral solution is z(t) —

1 0d

(r) e-Ca).(t r) sin cod (t — 2) C12 0

4.20

Mechanical Vibrations

The time-dependent velocity and acceleration are v = vo

[u(t) - u (t -

to

+ vo[u(t -

to)]

to) - u(t - 4E0)]

dv [u(t) - u(t - to)] dt to

j,

The response of a 1-degree-of-freedom system due to the rectangular pulse is determined in Solved Problem 4.7 as x(t)

F0 {1 — cos o„ t t < to inc02 cos co„ (t - to ) - cos co„ t t > to

—

+ vo -t- At) - 8(t - to)] to + v0[5(t - to) - 3(t - 4t0)]

The relative displacement is z(t) = —

Vo

1

— [u ( r) — u

Whether the maximum occurs for t < to or for t > to depends upon the system parameters. The maximum response for t < to occurs either at t = to or when cos co„t = -1 (co„t = 7r). Thus, if to < glco„, the maximum is not achieved for t < to. For to > irkon,

— to )]

2F0

Wel o tO +

to

MO)„2

[o(r) — S(r — to) + [o(r — to )

— o(T. —

4 to )]} e-Cc°.(' - T)

Sin COd (t

For to < if/co„, ; flax occurs for t > to at a value of t when dx/dt = 0. To this end, for t > to,

-c) dr

vo {u) [e cod to c0,2,

dx

Fo

dt

mw,,

--

[ co sin co„(t "

. r (con sin cod t

to)

+ con sin co„t] dx

-= 0 dt

+ cod cos cod t - cod ]

sin co„t = sin co„(t - t0)

whose solution is u(t - to)[(

0)

t=

) (t-t

° d - 0 )-F0/ ga) n 84- dt ‘

coscod

(t — t 0)] — di — t 0 (t

443 )

e-N, (I -4'0)

j[ (2n -1)7r1 , n = 1, 2, ... - to + 2 2w„

This leads to to


which is plotted in Fig. 4.10.

Co„

General Forced Response of 1-Degree-of-Freedom Systems 2.5 2.0 1.5 1.0 0.5 0.0 , con t0 271.

1/2

Fig. 4.10 4.32 Devise an algorithm to numerically develop the response spectrum for a 1-degree-of-freedom system with viscous damping subject to an arbitrary excitation. Assume that an integration method such as Euler's method or a fourth-order Runge-Kutta is used to solve Eq. (4.1). 1. Equation (4.1) can be nondimensionalized by introducing X * — minx

F0

CO„ t t*

21c

where F0 is the maximum value of F(t). In terms of these nondimensional variables, Eq. (4.1) becomes * + 4 Irci* + 47c 2 x*= 4g 2 F(t*) F0 (4.18) Note that the natural period in nondimensional time is 1. 2. Define a = coto/(2n), which can be viewed as a nondimensional value of the pulse duration. Let a range from Aa to 2.5 in increments of Aa = 0.1. Equation (4.18) should be solved using, say Runge-Kutta for each value of a. For a particular value of a, a time increment and a

r 4.21

final time for the Runge-Kutta simulation must be chosen. It •is important to set the nondimensional time increment small enough such that enough integration steps are used over the duration of the pulse and over 1 natural period. Also the final time must be chosen large enough such that integration is carried out sufficiently beyond the duration of the pulse and over several natural periods. Note that for a < 1, the duration of the pulse is less than the natural period. Possible choices_ for the time increment and final time are a At* = — t= 2/5 20 f For. a> 1, the natural period is less than the duration of the pulse. Possible choices for the time increment and final times are 1 At* = — t* = 2.5a 20f 3. Numerical simulation of Eq. (4.18) is developed for each value of a, as described in step 2. The maximum value of the nondimensional response xtax is recorded. The response spectrum is a plot of xj*nax versus a. 4.33 The force exerted on a structure due to a shock or impact is often modeled by the excitation of Fig. 4.11. The response spectrum for this type of excitation for

Fo to Fig. 4.11

4.22

Mechanical Vibrations

2.00 1.75 1.50 1.25 Le 1.00 0.75 0.50 0.25 0.00

1

0

mo

2

3

2if

C -0

T =0.1

---- 4.= 0.3

— — - c= 0.2 - 0.5

Fig. 4.12 several damping ratios is shown in Fig. 4.12. What is the maximum displacement on an undamped 1000 kg structure of stiffness 5 x 106 N/m subject to such a blast with Fo = 1500 N and to = 0.05s? The natural frequency of the structure is 5x10611

con

. m rad — \ 1000 kg = 70.7

The value of the nondimensional parameter on the horizontal scale of the response spectrum is rad 70.7 )(0.05 s) co„ to s — =0.563 27r 2r From Fig. 4.12, for C = 0

Ama x — 1.2 Fo from which the maximum displacement is calculated as

xmax — ( kxmax 117° =1.2 Fo ) k

5000 N 5 x10611 m

= 1.2 mm 4.34 For what diameters of the circular bar of the system of Fig. 4.13 will the maximum displacement of the block be less than 18 mm when subjected to a blast modeled by Fig. 4.11 with Fo = 10,000 N and to = 0.1 s? 0.8 m 50kg

F(t)

E-200 /109 N m2 Fig. 4.13 Let k be the stiffness of the system. Since, it is desired to limit x < 18 mm, kxm Fo

k(0.018 m) — 1.8 x 10-6k 10,000 N (4.19)

4.23

General Forced Response of 1-Degree-of-Freedom Systems

Also

0„ to 27r

I k 0.1s *\I 50 kg 27r

= 2.25 x 10-3 j (4.20) Since k is unknown, a trial-and-error procedure using the response spectrum of Fig. 4.12 is used to determine allowable values of k. Suppose k = 5 x 105 N/m, thus using. Eq. (4.20), conto/(27c) = 1.59. From the response spectrum this leads t° kxmax/F0 = 1.7. However, from Eq. (4.20), the maximum allowable value of kxmax/F0 for k = 5 x 105 N/m is 0.9. Inspection of Eqs. (4.19) and (4.20) shows that co„tol(27c) increases slower than the maximum allowable value of kxmax/R0. Thus an increase in k is tried. Note that if k =1 x 106 N/m, then using

Eq. (4.20) leads to co„t01(27r)= 2.25. The response spectrum yields kxmax/F0 = 1.8, which is the same as using Eq. (4.19). Thus if k> 1 x 106 N/m, the maximum displacement of the block is less than 18 mm. The maximum area of the bar is calculated as (1x106 1(0.8m) A—

kL

E

=

200 x109 N m2

= 4 x 10-6 m2 Hence, the minimum diameter is 4i4 D

4(4 x10-6 m2)

,\1

ir

= 2.26 x 10-3 m

PRACTICE PROBLEMS 4.1 Use the convolution integral to develop the response of an undamped 1-degreeof-freedom system of mass in and natural frequency con subject to an excitation of the form F(t) = Fo sin cont. The system is at rest in equilibrium at t = 0. 4.2 Use the convolution integral to develop the response of an undamped 1-degreeof-freedom system subject to an excitation of the form F(t) = F0(1 — e'). The system is at rest in equilibrium at t = 0. 4.3 Use the convolution integral to determine the response of a 1-degree-of-freedom system of mass m, damping ration c, and natural frequency con subject to an excitation of the form F(t) = Fo sin wt for w w,,. The system is at rest in equilibrium at t = 0. 4.4 Use the convolution integral to determine the time-dependent response of the system of Fig. 4.14.

k

k= 2 x 102\j

m

m= 60 kg r = 5 cm

/= 0.3 kg-m2 c = 5000

N-s

Mo = 50 N-m to = 0.05 s

to

Fig. 4.14

4.24 4.5

Mechanical Vibrations

An undamped 1-degree-of-freedom system is subject to the excitation of Fig. 4.15. Use the convolution integral to determine the system response for t > to.

Fo

F Fo to

Fig. 4.17

to Fig. 4.15 4.6

)

Develop a unified mathematical expression for the excitation of Fig. 4.16 using unit step functions.

Fig. 4.16 4.7

Use the convolution integral to develop the response of an undamped 1-degreeof-freedom s.ystem subject to the excitation of Fig. 4.16. The system is at rest in equilibrium at/ = 0. 4.8 Use the convolution integral to develop the response of an undamped 1-degreeof-freedom system due to the excitation of Fig. 4.17. The system is at rest in equilibrium at t = 0. 4.9 Solve Practice Problem 4.1 using the Laplace transform method. 4.10 Solve Practice Problem 4.3 using the Laplace transform method.

4.11 Solve Practice Problem 4.7 using the Laplace transform method. 4.12 Solve Practice Problem 4.8 using the Laplace transform method. 4.13 A 50 kg block is attached to a spring of stiffness 2 x 106 N/m. The block is subject to an impulse of magnitude 25 N-s at t = 0 and an impulse of magnitude 15 N-s at t = 0.1 s. What is the maximum displacement of the block? 4.14 Use the Laplace transform method to determine the response of a 1-degree-of-freedom system with damping ratio 4 and natural frequency con when subject to the periodic excitation of Fig. 4.18. F Fo

to

t

30 o

2t0 I

— 5° 2

—Fo

Fig. 4.18 4.15 Determine an equation define the response spectrum for an undamped 1-degree-offreedom system subject to the excitation of Fig. 4.17.

General Forced Response of 1-Degree-of-Freedom Systems

4.16 A 100-kg machine is mounted on an isolator of stiffness 5 x 105 N/m. What is the machine's maximum displacement when subject to the excitation of Fig. 4.17 with Fo = 1000 N and to = 0.11 s? 4.17 The response spectrum of a 1-degree-offreedom system subject to a sinusoidal

4.25

pulse is shown in Fig. 4.19. A 25 kg block , is attached to a spring of stiffness 5 x 106 N/m and subject to a sinusoidal pulse of magnitude 1250 N and duration 0.02 s. What is the maximum displacement of the block?

2/r 0

— — - c= 0.2 — C =0.5

(=0.1 0.3

Fig. 4.19 4.18 A 200 kg machine is to be placed on a vibration isolator of damping ratio 0.1. For what range of isolator stiffness will the maximum displacement of the machine be less than 2 mm when it is subject to a sinusoidal pulse of magnitude 1000 N and duration 0.04 s? The response spectrum for a sinusoidal pulse is shown in Fig. 4.19. 4.19 A 500 kg machine is attached to an isolator of stiffness 3 x 105 N/m and damping ratio 0.05. During start up it is subject to an excitation of the form of Fig. 4.17 with to = 0.1 s and Fo = 5000 N. Use the trapezoidal rule for numerical integration of the convolution integral to approximate the machine's displacement. Use A = 0.04 s. 4.20 Consider the method for numerical evaluation of the convolution integral

introduced in Solved Problems 4.27 and 4.28. Develop an expression for G1 if F(t) is interpolated by a series of impulses. That is, on the interval from ti to ti=i, F(t) is interpolated by an impulse of magnitude F(t* i ) At applied at tj where ej is the midpoint of the interval. 4.21 Use numerical integration of the convolution integral to approximate the response of a machine of mass 250 kg attached to an isolator of stiffness 2 ' 106 N/m and damping ratio 0.05 when subject to the excitation F(t) = 1000 ere N Use the method of Solved Problem 4.28. 4.22 Repeat Practice Problem 4.21 using the interpolations of Problem 4.55.

4.26

Mechanical Vibrations

ANSWERS TO PRACTICE PROBLEMS 4.1

sin

F0

F0

cont

2mco„

2mw,2,

F0

4.2

mco,2, (a2 ± (0,2 )

cos

[ (02 car i

, + a2 cos

4.3

t

4.7

co t

ce2

m2

Fo 2 MCO„ to

(t- 1 sin co„ t u(t) L l C°11 1

(1 to

wn

co„t + acon sin (Ant]

C°11

- t - 4 to -

1

sin o)„(t - to )

u(t -

to )

sin co„(t - 4 to ) I u (t - 4 to )

con

F0

[(q _ 0)2 )2 + (2caxon )2]

m

( t - 5 to -t0 cos co„ (t - 5 to ) [ 24-wo)„ (cos a), t — cos cot) 1

+ (co,2, - co2 )(sin cot - --(1). -sin co„ tj ] con 4.4 0.0111[1 - c13.9`(0.140 sin 99.03t + cos 99.03t)] t < 0.05s 0.0111e-11%0.140 sin 99.03t + cos 99.03t - 0.280 sin (99.03t - 4.95) - 2.00 cos (99.03t - 4.95)] t > 0.05s 4.5

F0

[

to COS W„ (t — t0)

con 4.8

4.14

4.6 F°

t0

, sin it to )

u (t) +

Fo 1- —t to

1 sin co„ t u (t -

Fo 2 [ (t — mco„ to

F0 m0 2

1- e-Cw,,'

sin co„ ti u(t)

to )

t ju(t- 4 t0) +44-to t +Fo 5-to

1 si con n co„tju(t)

1 - (t - to - —sin co„(t - to )ju (t - to) 1 con 4.13 6.28 mm

7)/ (012i t0

+ 1

sin (t - 5 t0 )) - u (t - 5 to )

u(t-5t0 )

e

ji _ c2

Cco (t — "0) n

sin co„ (t — ito

cos con (t it0 )

} u (t - ito )

General Forced Response of 1-Degree-of-Freedom Systems

4.15

1"max

Fo

=1 +

4.16 2.50 mm 4.17 0.4 mm

1

co„ to

I

.v 2(1— cos (0,, to )

4.18 k > 5 x 105 N/m 4.20 F (t.*i ) Atea)"1; cos cod t;

4.27

Chapter Five Free Vibrations of Multi-Degree-of-Freedom Systems

5.1 INTRODUCTION A system with multi-degree of freedom requires multiple coordinates to describe its motion. These coordinates are called generalized coordinates when they are independent of each other. The number of generalized coordinates of a system is equal in number to the degree of freedom of the system. For example, a two-degree freedom system has two equations of motion, which can be solved to obtain two natural frequencies. For each of the natural frequencies, system has a different displacement configuration known as normal mode of vibration. Different methods to analyze a system with multi-degree freedom are demonstrated in this chapter by solving several problems.

5.2 'LAGRANGEF EQUATIONS Let x1, x2, x3 x„ be a set of coordinates for an n-degree-of-freedom system. The motion of the system is governed by a set of n ordinary differential equtions with the generalized coordinates as the dependent variables and time as the independent variable. One method of deriving the differential equations, referred to as the free body diagram method, involves applying conservation laws to free body diagrams of the system drawn at an arbitrary instant. An energy method provides an alternative to derive the differential equations governing the vibrations of a multi-degree-of-freedom system. Let V(xi, x2, ..., x,,) be the potential energy of the

5.2

Mechanical Vibrations

system at an arbitrary instant. Let T(x1, x2, ..., x,,, xl , i2 , ) be the kinetic energy of the system at the same arbitrary instant. The lagrangian L(x1, x2, ..., x,,, i2 , ..., in ) is defined as =T—V (5.1) The lagrangian is viewed as a function of 2n independent variables, with the time derivatives of the generalized coordinates assumed to be independent of the generalized coordinates. Let Sxl, 8x2, 5xn be variations of the generalized coordinates. The virtual work (5W done by the nonconservative forces in the system due to the variations of the generalized coordinates can be written as SW=

Q1 Sx;

(5.2)

t=i Lagrange's equations are d (a) dt ax1

a - Q1 ax;

i =1, 2 ..., n

(5.3)

Application of Lagrange's equations leads to a set of n independent differential equations.

5.3 MATRIX FORMULATION OF DIFFERENTIAL EQUATIONS FOR LINEAR SYSTEM For a linear system, the potential and kinetic energies have quadratic forms: n n V=

2

EElc..x.xJ• n

T=

(5.4)

1=1 j=1

n (5.5)

2 1=1 j=1

If viscous damping and externally applied forces, independent of the generalized coordinates, are the only nonconservative forces, the virtual work can be expressed as n

n

SW =

c•• 1.1

i=1 j=1

F Sx;

xJ Ox•J• + •

(5.6)

1=1

Application of lagrange's equation to the lagrangian developed using Eqs. (5.4) and (5.5) and the virtual work of Eq. (5.6) leads to Mx + Ci + Kx = F (5.7) where M is the 17 X n mass matrix whose elements are mu , K is the n x n stiffness matrix whose element are C is the n x n viscous damping matrix whose element are ciP x is the n x 1 displacement vector whose elements are x1, and F is the n x 1 force vector whose elements are F1. The matrices are symmetric. For example, mu =

Free Vibrations of Multi-Degree-of-Freedom Systems

5.3

5.4 STIFFNESS INFLUENCE COEFFICIENTS Stiffness influence coeflicents are used to sequentially calculate the columns of the stiffness matrix for a linear system. Imagine the system in static equilibrium with xi = 1 and xi = 0 for i #j. The jth column of the stiffness matrix, the stiffness influence coefficent k,,, are the forces that must be applied to the particles whose displacements are described by the generalized Coordinates to maintain the system in equilibrium in the prescribed position. The forces are assumed positive direction of the generalized coordinates. If x; is an angular coordinate, then ku is an applied moment. Maxwell's reciprocity relation implies that =

5.5 FLEXIBILITY MATRIX The flexibility matrix A is the inverse of the stiffness matrix. Flexibility influence coefficients can be used to sequentially calculate the columns of the flexibility matrix. The jth column of the flexibility matrix is the column of values of the generalized coordinates induced by static application of a unit load to the particle whose displacement is described by xj. If xi is an angular coordinate, then a unit moment is applied. The reciprocity relation implies that the flexibility matrix is symmertic, aji .

The flexibility matrix is easier to calculate than the stiffness matrix for most structural systems that are modeled using a finite number of degrees of freedom. The differential equations govering the motion of a linear n-degree-of-freedom system can be written using the flexibility matrix as AM I + AC X + x = AF

(5.8)

5.6 NORMAL MODE SOLUTION Introduction of the normal mode solution x = Xelm (5.9) into Eq.(5.7) with C = 0 and F = 0 leads to the following matrix eigenvalue-eigenvector problem for the natural frequencies w and their corresponding mode shape vectors X: M-1KX = o.)2X (5.10) The natural frequencies, the square roots of the eigenvalues of M-1K, are obtained by setting det - co2Il = 0 (5.11) or alternately det 1K - co2M1 = 0 (5.12) If the flexibility matrix is known rather than the stiffness matrix, the natural frequencies are the reciprocals of the square roots of the eigenvalues of AM and are calculated from detld AM - II = 0 (5.13) Use of Eq. (5.11) or (5.13) leads to an nth-order algebragic equation in co2 with real coefficent. Since, M and K are symmetric, the n roots are real, yielding the system's n natural frequencies, w2 < • • • < w„. If the system is stable, then K is nonnegative definite and the roots are nonnegative. An unrestrained system has a rigid body mode corresponding to a natural frequency of zero.

5.4

Mechanical Vibrations

5.7 MODE SHAPE ORTHOGONALITY Let Xi and Xi be mode shape vector for n-degree-of-freedom system corresponding to distinct natural frequencies co; and wi,, respectively. These mode shapes satisfy the following orthogonality condition: XT. MXJ = 0

(5.14)

=0

(5.15)

The mode shape vector X,, when determined as an eigenvector of M-1K or AM, is unique only to a multiplicative constant. The nonuniqueness is alleviated by requiring the mode shape to satisfy a normalization condition, usually specified as XT MX; = 1

(5.16)

If Eq. (5.16) is used as a normalization condition, then XT KX; = coi2

(5.17)

5.8 MATRIX ITERATION Numerical procedures are often used to calculate natural frequencies of systems with a large number of degrees of freedom. Matrix iteration is a numerical procedure that allows determination of a system's natural frequencies and mode shapes successively, beginning with the smallest natural frequency. Let u0 be an arbitrary n x 1 vector the sequence of vectors = AMti i_ i,

=

ui

(5.18)

I lk 'max

converges to X1, the mode shape corresponding to the lowest natural frequency. Also, 111;1., the largest 'absolute value of an element of ui, converges to 11(q. Matrix iteration can be used to determine natural frequencies and mode shapes for higher modes by using a trial vector orthogonal, with respect to the mass matrix, to all previously determined mode shapes. This technique is referred as sweeping technique, demonstrated in Solved Problem 5.50.

5.9 OTHER MFTHODS There are other numerical methods such as Holzar's method and Stodola method which are popular to analyze the multi degree of freedom systems. One can also use Dunkerley's and Rayleigh's methods to obtain the natural frequencies approximately. Few problems are solved to show the usefulness of these techniques.

5.10 DAMPED SYSTEM The determination of the free and forced response of a multi-degree-of-freedom system is

Free Vibrations of Multi-Degree-of-Freedom Systems

5.5

significantly more difficult than for an undamped system. A special case, which is relatively easy to handle, is proportional damping and occurs when constants a and 13 exist such that C = aK + PM (5.19) For proportional damping, the normal mode solution Eq. (5.9) is applicable. If col, cot, , co„ are the natural frequencies corresponding to the undamped system, then the values of co that satisfy Eq. (5.9) are ± co; j l - X1 2

o3i = Jo); where

(5.20)

a modal damping ratio is

(acui

c..=

2

(5.21)

co,

For more genral forms of C, it is convenient to rewrite Eq. (5.7) as My +ky = 0 where

M=

[0 C '

K=

(5.22)

[-m 0

1

K

(5.23)

are symmertic 2n x 2n matrices and [ is a 2n x 1 column vector. A solution to Eq. (5.23) is assumed as y= OCT'

(5.24) (5.25)

The values of y are the complex conjugate eigenvalues of 14-11Z, and (I) is a corresponding eigenvector. The eigenvector satisfy the orthogonality relation (13TME(I3 =0,

i#j

(5.26)

SOLVED PROBLEM 5.1 Use the free body diagram method to derive the differential equations governing the motion of the system of. Fig. 5.1 using x1 , x2 and x3 as generalized coordinates. 1-0- x1

k

t

2 IL k 2m -Wi-- 2m iv\AI

i

miummfimmitillitmilmit

m

Fig. 5.1

Free body diagrams of each of the blocks of the system of Fig. 5.1 are shown in Fig. 5.2 at an arbitrary instant. Application of Newton's law to each of the free body diagram leads to -kxi + 2k(x2 - x1) = + 3kvi - 2kx2 = 0 -2k(x2 - x1) + k(x3 - x2) = 2n/i2 --> 2mI2 - 21a1 + 3kx2.- kx3 = 0 -k(x3 -x2) = 2ini3 2mi3 - kx2 + kx3 = 0

5.6

Mechanical Vibrations

Considering x1 < x2 .< x3 2k (x2 — x1 )

k (x3— x2)

Fig. 5.2 5.2 Use the free body diagram method to derive the differential equations govering the motion of the system of Fig. 5.3 using x and 0 as generalized coordinates. 1-4

>I< 4

M(t)

)-1-‹

M(t)

k (x —

mk

4 L k (x + — 0)

4 External forces

Effective forces

Fig. 5.4

Fig. 5.3

5.3 Derive ,the differential equation of motion for small oscillation of the pendulam shown in Fig. 5.5 using Newton's method. Assume rods are rigid and mass of rod is neglected.

Free body diagram of the bar at an arbitrary instant are shown in Fig. 5.4 assuming small O. Summing forces on

the bar(IF)ext = 1F)eff k(

x—— 1 L0)— k(x +1LO)=mX 4 2

Fig. 5.5

m3C + 2kx + 1kL0 = 0

4 Summing moments about the mass center of the bar

(EMG)

= mat 12 = m (2a)2

=(1,m,)eff

M(t) + k(x — 1L0) 1L

4

/c(x +1LO ) 1L = Io 2 2 /e + 1kLx + kL2 0 = M(t) 16 4

1202 mg

Fig. 5.6 Free body diagram Equations of motion ma2 61 + mga01 — ka202 + ka201 = 0

5.7

Free Vibrations of Multi-Degree-of-Freedom Systems

mad + (mg + ka)0i — ka02 = 0 4ma2 62 + 2mga02 + ka202 — ka201 = 0 4ma 0.2 — ka0i + (2mg + ka)02 = 0

5.4 Use Lagrange's equations to derive the differential equations governing the motion of the system of Fig. 5.1 using x1, x2, and x3 as generalized coordinates. Write the differential equation in matrix form. The kinetic energy of the system at an arbitrary instant is 1 . 1 .2 • T = — na + — 2mx 2 + 1 2n/x3 2 l 2 2 The potential energy of the system at an arbitrary instant is 1 2 1 V = — kx + — 2IC(X2 — X1)2 2 2 1 — x2)2 2 The lagrangian is + — k(x3

L=T—V=

1

.2

1

+- 2mx• 22 2

1 •2 — 2mx3 —— 1 loc2 2 2 i 1 — — 4.102 — Xi)2 — 103 — X2)2 2 2 Application of Lagrange's equations lead to

d 2mi2) +[2102 — xi)(1) dt

+ k(x3 — x2) (-1)] = 0

=0

—o

ax3

. —(2mx3 ) + [k(x3 — x2)(1)] = 0 dt

Rearranging and writing in matrix form leads to 0 2in 0

0 0

+

0 0

x2

21)1_ x3

3k

—2k

—2k

3k —k

0

0 —kx 2 = 0 k x3 0 0

x1

5.5 Use Lagrange's equation to derive the differential equations governing the motion of the system of Fig. 5.3 using x and 0 as generalized coordinates. Write the differential equation in matrix form. The kinetic energy of the system at an arbitrary instant is 1

•2

1

T= — mx + — /0 2 2 2

The potential energy of the system at an arbitrary instant is 1

V= —k(x —

d (al, j_ dt ax,

all

d (DL) dt

1L0 )2

lk i

2 4 The lagrangian is

2

x

Le

y

2 2

dt

(mxi) Rif x1 + 2102 — x l)(— 1)] = 0

d (u) dt

aX2

—o

L=2

+ /02 — C (X — 1 LO 2 4 2 — 1-k(x + 1LO)2 2 2

Mechanical Vibrations

5.8

If the variations Sx and SO are introduced, the virtual work done by the external moment is SW = M(t) 80 Application of Lagrange's equations leads to

The potential energy of the system is 2

V= 1 k (3 + 1 x2 ) 2 4 4

kx2

2

The Lagrangian is 2

2

+X

L - 1 mri 2 2 2

d (aL) _ _aL = Q dt a ax

+

+1 2

i

r2 Al)

2

dt

(mx) +[k (x

- Le I(1) 4

+k(x+1L0)(1)]=0 2 d (aLj_aL dt aO ae dt

(io)

[k (X

_n

-1 4

sw= m(t)8( x2 LO)(- 1L)

1 1 = -- M(t)8x1 + — M(t)(5x2

4

M(t) x +1LOX' 2 2 + k( Rearranging and rewriting in matrix form leads to [m 0

o r+ Le _I

2k

+ —1 x2 ) + =- kx22 2 4 4 2 If variations Sr, and 8x2 are introuced, the work done by the external moment is - 'kr

1 4

kL

5 j 1 — kL 16 _4

2

[0]

Application of Lagrange's equations leads to d (aL)_ aL dt a, -i1+ i2)( 1) 4_ d[mi dt 2 ) 2)

+

_[ 0

41 )( L

)1 L

xi+x2P)=—M(t) —L

4 4 4 kr

M(t)

5.6 Use Lagrange's equation to derive the differential equations govering the motion of the system of Fig. 5.3 using x1 and x2 as generalized coordinates. Write the differential equation in matrix form. The kinetic energy of the system at an arbitrary instant is (5c1+ -i2) 1 ( T= m + I 2 2 2

— xi

L

d (at, dt

_ aL

r il +5C )( 1) 4_, Ii2-41)()1

[

dt 111

2

2

) 2)

a

L ) L)

+[k( xi -Fi x2 )(-1- )+47c2 ] 4 4 4

)2

= 1M(t)

5.9

Free Vibrations of Multi-Degree-of-Freedom Systems

Rearranging and writing in matrix form leads to I L2

4 m

I

4

p

m

I

4

L

pJ,

in I —+4 L2

.; .2

The lagrangian is

L=

1

2

5.7 Use Lagrange's equations to derive the differential equations governing the motion of the system of Fig. 5.7 using x1, x2 and 0 as generalized coordinates. Write the differential equations in matrix form.

1 -//VP 2

k(x —IA)2 — —1 3k(2r01 — 2r02)2 2 Application of Lagrange's equations leads to

—

3

9

1 k k — —M(t) 16 16 [xi x2 = 3 17 —M(t) —k k L 16 16 _

1 2 + -1, 6 2-

d (aL) dt 1

n

-

ae1

1 )+ [k(x — r01) (—r) dt (I0 + 3k(2r01 — 2r02)(2r)] = 0

j

d [aL dt a62 d dt

a 02

=o

• (12 02) + 3k(2re1 — 202)(-2r) = 0 d raL)_aL _ 0 dt `ax )

ax

— dt 11 (n1)+ k(x — re1)(1)= 0

Rearranging and writing in matrix form leads to

Fig. 5.7 The kinectic energy of the system at an arbitrary instant is T=

1 2

•2

1 •2 +212 02 +—mx 2

The system's potential energy at an arbitrary instant is V= k(x— r01)2 + 13k(2r01 — 2r02)2

2

2

II

0

0

01

0

12

0

02

0

0

m

13kr2

—12kr2

—kr

61

—12kr2

12kr2

02

—kr

0

0 k

0 0 0

5.8 Use Lagrange's equations to derive the differential equations governing the motion of the system of Fig. 5.8 using 01 and 02 as generalized coordinates.

5.10

Mechanical Vibrations

d 1 — dt 3

2 01 • ) + [mg2 L

sin 01

02 Identical slender bars of length L, mass m

Fig. 5.8 The kinetic energy of the system at an arbitary instant is

2 (1 • ) 1 1 T= + — —mL2 0' 2 2 2 2 12 2 + 1 (1 Le I mL262 m 2 2 2 ) + 2 12 2 The potential energy of the system at an arbitrary intant is L V = —mg— cosOi — mg- COS 02 2 1 + — k(a sin 02 — a sin 002 2 The lagrangian is L

+k(a sin 02 — a sin 01 )(—a cos0i )]= 0 d (aL) dt a 62

DL =0

a 02

L26 2) + [mg -sin0 ( Oin 2 dt 2 + k(a sin 02 — a sin 01) (+a cos 02 ) = 0 Linearizing and rearranging leads to — mL261 +(mg— L + ka2 )0i — ka2 02 = 0 3 2 I me 62

3

ka-2 01 +(mg

2

+ ka 2 )02 = 0

5.9 Use Lagrange's equations to derive the differential equations governing the motion of the system shown in Fig. 5.7 using 01 and 0, as generalized coordinates.

1 1 go + 3 mL262 23m I 2 2 L + mg— cos 01 + mgL' cos 02 2 2 1 — 2 k(a sin 02 — a sin 01)2

Application of Lagrange's equations leads to d (aL) dt ad,

aL =

ao,

1 .2 T = — .11 + — J2 022 2 2 1 V= — k(02— 01) considering 02 > 01 2 The lagrangian is L = T — V Lagrange's eqns are: d [aL, dt aeoi

at, = 0

ao,

(free vibration)

Free Vibrations of Multi-Degree-of-Freedom Systems

Substituting L in the above equation Lagrange's equations are: d (aT)_ aT av =0 dt Oi De; a9;

r

If de and Le are equivalent diameter and length of the shaft, we get

( 11 11 4 L4 d 4 d4 d4 d4 d4 e 1 2 3 4

Le

where i = 1, 2 The above Eq. is more simpler to use. Substituting T and V in the above equation. Equations of motion are

de

14

d1 )

4

de 1

+

) 4

J1 01 + k,(92— 69 (-1) = 0

J2 02 +k,(0 —>

01 ) (1) =

+L4 ( 6

cla

3

+ k,01 — k,02 = 0

-->

o

.12 62 — k201 + lc,02 = 0

5.10 Obtain the equation for torsionally equivalent shaft for the torsional system shown in Fig. 5.10.

4

5.11 For the system shown in Fig. 5.11 obtain the equations of motion by Lagrange's equation. Also obtain a condition for system not to have elastic coupling.

JB

d,

d2

d2

d4

J4 1 ,

L1

A

2 ,

3 ,

4 L4

'2

-

Fig. 5.10

1 -4—a ).
-1

Fig. 5.11

+ 02 + 03 + 04 T J

We have, for torsion — =

GO

TL1 TL2 TL3 TL4 JG1 JG2 JG3 JG4 =

32 T (Li L2 L3 L4 ) rc G di d A 2

d a d4 3

)

1 .2 1 • 2 T = — mx + JO 2

5.12

Mechanical Vibrations

V=

2

ki (x — a0)2

2

k2(x + b0)2

x and 0 are generalized co-ordinates. Lagranges equation d ( DT ) DT _ dt ax

av =0

ax

liaT c ) _oT _av = 0 dt ag ) DO ad

Jo + 2ka2 = 0 If system executes harmonic motion x = Al sin (cot — 0) 0 =asin (cot - 0) By substituting equations of x and 0 in equation of motions, the characteristic Eq. obtained are

—mco2 + 2k = 0 and

—J0)22 + 2ka2 = 0

Equations of motion + (k1 + k2) + (k2b — k la) = 0

2k

(01 = m rad/s

Jo + (kb — kia)x + (k2b2 + kia2)9 = 0 In matrix notation Fm 01{1}

L O .1] +

[(k +k2)

(k2b — ki a)

jx

L(k2 b — ki a) (k2b2 + a2 ) 10 {0} 0

From k matrix it is 'clear that if kia = k2b the equations are free from elastic coupling, and x and 0 are independent. The uncoupled co-ordinates are referred as principal coordinates. 5.12 Solve the Solved Problem 5.11 to obtain equations for fundamental frequencies if a = b and k1 = k2. For given condition x and 0 are independent and the equation of motion written in matrix form

[m o

{l e

2 = 12ka2 (0

[2k 0

0 2

2

{1

{0}

6

0

The uncoupled equations of motion are mx + 2kx = 0

rad/s

5.13 The identical disks of mass m and radius r of Fig. 5.13 roll without slip. Use Lagrange's equations to derive the governing differential equations using x1 and x2 as generalized coordinates. The kinetic energy of the system at an arbitrary instant is T.=

I

.2 , 1 1 mr2 22

- nixl

+

1 2 1 1 2 (i 2 mr mx + 2 2 22

The potential energy of the system at an arbitrary instant is 1 2, 1 2 V= - kx, + — 2kac2 2 2 + 2 2k(2x2— 2x1)2

The Lagrangian is L= 13 .2 4_ 1 3 mir.

22 nal 22 2- 2 1 2 - -2k:C2 - — 2k (2 x2 — 2 xi )2 2 2

Free Vibrations of Multi-Degree-of-Freedom Systems

X2

2k

\\\ \\ kll\\\\\\\\

x1

5.13

2k

k

1111111111111111111 II 111111111111111111111111111111111111111111111111111 1111111111111111111111

No slip

Fig. 5.13 Application of Lagrange's equations leads to +

d•t 2111x1 ) —

The kinetic energy of the system at an arbitary instant is 1 .2 1 .2 2 1 2 2 3 The potential energy of the system at an arbitary instant is 1 .2 T = —2mx + —mx + —mx

2k(2x2

— 2x1)(-2)] = 0 d (3

.

— dt 21"x2 )

1 2 1 V= — kv, + — 2 107— _ X1 )2

[21a2 + 2k(2x2

a

— 2x2)(2)] = 0 Rearranging and writing in matrix form yields 3m 2

0

0

3 —in 2

[•2 1

[ 9k

8k x,

—8k

10ki Lx2

1

b. it, (X3

2 2 If the variations 5x1, Sx2 and 5x3 are introduced when the system is in an arbitary state, the work done by the forces in the viscous dampers is —)C2)

3W= —c.i1 3x1 — c(±3 —x2 ) 8 (x3 — x2) — 2 Ci3 8X3 = — Cx, Sx 1 — (— c5c3 +

[o 01

) 5x2

+ 3 ci3 ) 3.,x3 Application of Lagrange's equations yields

5.14 Use Lagrange's equations to derive the differential equations governing the motion of the system of Fig. 5.14 using x1, x, and x3 as generalized coordinates.

d (az, _

r

X2

Xi

k

k 2m

k ---/VVV —

=

dt

m

ic

X3

2c m

///11/1/1/111 11111/1/1/1/11/1/1111111/11 111111111111111111I111111111111111111111111

Fig. 5.14

5.14

I

Mechanical Vibrations

d (2mx1)

dr

The kinetic energy of the system at an arbitrary instant is

Elul+ k(x2 - x1)(-1)]

=—

1

T=

dL

d dt

axe

= V2

dt

+c3

_

ax3 — Q33

(mx3)+ k(x3—x2)(1)= d2 — 3 d3

Rearranging and writing in matrix form leads to ✓ 0 0 il c 0 0 ,ii 0 m 0 .3t 2 + 0 c —c .i2 0 0 m .ii:3 0 —c 3c X3

2m

[

[2k —k 0 xi

(

—k k x3

1 1

—2mL2 0 2 2 12 The potential energy of the system at an arbitrary instant is 1 2 1 V = — kx + — 2k(x + L 0)2 2 • 2 1 —2mg — cos°

If the variations ox and 30 are introduced at an arbitary instant, the work done by the viscous damping forces is + L O) 45(x + LO) SW= — Ci — = — c(2x + L 0)5x — cL(i + Le)OO

Application of Lagrange's equations leads to d ( dt

0

+ —k, 2k —k x2 = 0

0

.2

mx + 1

+

dt (1762 ) [k()C2— Xi)(1) + k(x3 — x2)(-1)] = — d (aL) dt

—2

0 dt

5.15 Use Lagrange's equations to derive the differential equations governing the motion of the system of Fig. 5‘15 using x and 0 as generalized coordinates.

ax

)

[mx +2m(.i += e)(1)]

+ [kx + 2k(x + L0)(1)] —cL0

=

d aL) dt (ao

n ae —

> x

L 6)(L)+ 1 2 mg O]

d [2m(i + —

m

k

2

dt

2

12

+ 2k(x + LO)(L) + mgL sin 0 /////// //////

/////// I/1/1

= —cL X — cL2 Rearranging and linearizing leads to 3m

Slender bar of mass 2m, length L

mL

"lL 1[1 [ 2c cL

3 3k

2k

Fig 5.15

24

ad o] ir xl

[vcr, 2kL2 + mgLiL0

[01 oj

Free Vibrations of Multi-Degree-of-Freedom Systems

5.16 Use stiffness influence coefficients to determine the stiffness matrix for the system of Fig 5.1.

5.17 Use stiffness influence coefficients to determine the stiffness matrix for the system of Fig. 5.3 using x and 0 as generalized coordinates.

The first column of the stiffness matrix is obtained by setting x1 = 1, x2 = 0 and x3 = 0 and solving for the applied forces as shown in Fig. 5.16a. Summing the forces to zero on each free body diagram leads to k11= 3K, k21 = —2k and k31 = 0. The second column is obtained by setting xi = 0, x2 = 1 and x3 = 0 and solving for the applied forces as shown in Fig. 5.16b. Summing forces to zero on each free body diagram of Fig. 5.16b leads to k12 = —2k, k22 = 3k and k32 = —k. The third column is obtained by setting x1 = 0, x2 = 0 and x3 = 1 and solving for the applied forces as shown in Fig. 5.16c. Summing forces to zero on each free body diagram of Fig. 5.16c leads to k13 = 0, k23 = —k and k33 = k. Hence, the stiffness matrix is 3k —2k 01 K= — 2k 3k —k 0

—k

5.15

The first column of the stiffness matrix is obtained by setting x = 1 and 0 = 0 and solving for the applied forces and moment as shown in Fig. 5.17a. Summing forces to zero leads to k11 = 2k. Summing moments about the mass center to zero 'leads to k21 + kL/4 — kL/2 = 0 —> k21 = kL/4. The second column is obtained by setting x = 0, and 0 = 1 and solving for the applied load and moment as shown in Fig. 5.17b. Summing forces to zero leads to k12 = kL/4 while summing moments about the mass center leads to k22 — kL/2(L/2) — kL/4(L/4) k22 = 5kL2/ 16. Hence, the stiffness matrix is 1 — kL 2k 4 K= 1 — kL

k

_4

5

16

2 kL

2k k11

k21

k31

X1 = 1, X2 = 0, X3 = 0 (a)

2k

k k22

k12

k 32

= 0, X2 = 1, X3 = 0 (b)

k k23

k13 Xi = 0, X2 = 0, X3 = 1 (C)

Fig. 5.16

k33

Mechanical Vibrations

5.16

x= 1

0=0

1MA = 0 = k2i(L) — —3 k( 1 L) k21 = 736 k

k

k11

4

4

(a)

— — k ( 3 L) 4 4

1MB = 0 =

k 16 The second column is obtained by setting x1 = 0 and x2 = 1 and solving for the applied forces of Fig. 5.18b. Summing moments about each end of the bar to zero, -4 k

k22 X = ° 0= 1 kL 2

(b)

Fig. 5.17 5.18 Use stifness influence coefficients to

determine the stiffness matrix for the system of Fig. 5.3. using x1 and x2 as generalized coordinates. The first column of the stiffness matrix is obtained by setting x1 = 1 and x2 = 0 and solving for the applied loads shown in Fig. 5.18a. Summing moments about each end of the bar to zero,

11

=

1 L) 1 4= 0 = k22(L) — k(L) — — 4 4 k22 = 11 76 k 1MB "12(.0 — 4 k(4 L ) kit = 6k

The stiffness matrix is Xi = 1 X2 = 0

9 k 3 k K=

k21

16 3

16 k11

3 k 4

16 k

17

k 16 _

5.19 Use stiffness influence coeffcients to

(a)

determine the stiffness matrix for the system of Fig 5.7 using x, 01 and 02 as generalized coordinates. k22

B k 4 (b)

Fig. 5.18

x1 = 0 x2= 1

The first column of the stiffness matrix is obtained by setting x = 1, 01 = 0 02 = 0 and solving for the forces and moments shown in Fig. 5.19a. Summing forces to zero on the block and moments about the pin supports to zero lead to k11 = k, k21 = —Icr and k31 = 0. The second

Free Vibrations of Multi-Degree-of-Freedom Systems

equation of equilibrium to the free body diagram leads to k12 = —kr, k22 = 13kr2 and k32 = —12kr2. The third column is obtained by setting x = 0, 01 = 0 and 02 = 1 and solving for the forces and moments shown in Fig. 5.19c. Application of the equations of equilibrium to these free body diagrams leads to k13 = 0, k 23 = —12kr2 and k33 = 12kr2. Thus the stiffness matrix is

k

k K = —kr 0

= 1, 02 =0 (b)

El ki2

k23

x = 0, 01 = 0, 02 = 1

(c)

—kr

0

13/0.2

—12kr2

—12kr2

12kr2

5.20 Use stiffness influence coefficients to derive the stiffness matrix for the system of Fig. 5.15 using x and 0 as generalized coordinates and assuming small O.

kr

x =0,

5.17

k13

Fig. 5.19 column of the stiffness matrix is obtained by setting x = 0, 01 = 1 and 02 = 0 and solving for the forces and moments shown in Fig. 5.19b. Application of the

The first column of the stiffness matrix is obtained by setting x = 1 and 0 = 0 and solving for the applied force and moment on the free body diagrams of Fig. 5.20a. Application of the equations of equilibrium to these free body diagrams leads to k11 = 3k and k21 = 2kL. The second column is obtained by setting x = 0 and 0 = 1 and solving for the applied force and moment as shown on the free body diagrams of Fig. 5.20b. Application of the equations of equilibrium to these free body diagrams leads to k12 = 2kL and k22 = 2kL2 + mgL. Thus the stiffness.matrix is [ 3k 2kL K= 2kL 2kL2 + mgLi 5.21 Use flexibility influence coefficients to determine the flexibility matrix for the system of Fig. 5.1 using x1, x2 and x3 as generalized coordinates.

Mechanical Vibrations

5.18

k

2 mg

-4— 2 k k 21 x= 1 , 0 = 0 (a)

k12

k22

x= 0, 0= 1 (b)

Fig. 5.20 The first column of the flexibility matrix is obtained by applying a unit load to the block whose displacement is x1. The resulting displacements of the blocks are the flexibility influence coefficients all, a 21 and a31. Application of the equations of static equilibrium to the free body diagrams of Fig. 5.21a leads to —ka li + 2k(a21 — ai 1) + 1 = 0 —2k(a21 — a il) + k(a31 — a21) = 0 k(a31 — a21) = 0 which are solved simultaneously, yielding 1 1 a31 a ll = , k k

The second column of the flexibility matrix is obtained by applying a unit load to the middle block and applying the equations of equilibrium to the free body diagrams of Fig. 5.21b, leading to 1

3 3 32 2k a 2k The third column of the flexibility matrix is obtained by applying a unit load to the block whose displacement is described by x3 and applying the equations of equilibrium to the free body diagrams of Fig..5.21c, leading to al2

a13

a22

1 3 — , = k 2k

a33

5 2k

Free Vibrations of Multi-Degree-of-Freedom Systems

5.19

2k (a 21 - a11) (a)

k (a 32 - a22)

2k (a22 -a 12) (b)

ka

k (a33 - a23)

2k (a23 - a 13) (c)

Fig. 5.21

Thus the flexibility matrix is 1 1 1 k k k A= 1 3 3 k 2k 2k

1'

3

L

(a)

k (a" + - a21

2

5

k 2k 2k

41 5.22 Use flexibility influence coefficients to determine the flexibility matrix for the system of Fig. 5.3 using x and 0 as generalized coordinates. The first column of the flexibility matrix is obtained by applying a unit load to the mass center of the bar and setting x = a ll and 0 = a21. Application of the equations of equilibrium to the free body diagram of Fig. 5.22a leads to =0=1—

a2i 4 J —k(aii + 2 a21

+12 -

4

a22 kia 12 +

(b)

2

2

Fig. 5.22

= 0 = k(ai

4 a21

4

L )L

— 2 2 The above equations are solved simultaneously, leading to 4 5 , a21 all = 9kL 9k The second column of the flexibility matrix is obtained by applying a unit —

k(ai

5.20

Mechanical Vibrations

clockwise moment to the bar, applying the equations of equilibrium to the free body diagram of Fig. 5.22b, and solving simultaneously for the flexibility influence coefficients, leading to 4212

4

32 a22

9kL

9kL2

k (2ra 21 -

a31)

Thus the flexibility matrix is k (ail -ra 21 )

5 9k

A=

4 9kL

4 9kL 32 9ke

5.23 Use fleicibility influence coefficients to derive the flexibility matrix for the system of Fig. 5.23 using x1, x2 and 0 as generalized coordinates.

Fig. 5.24 EM0 = 0 = k(a11 — ra21)r — k(2ra 21 — a.31)(2r) EF = 0 = k(2ra 21 — a31) When a simultaneous solution of the previous equations is attempted, an inconsistency results (for example, 1 = 0). This implies that the flexibility matrix does not exist. This is because the system is unrestrained and the stiffness matrix is singular. 5.24 For the system shown in the Fig. 5.25

Fig. 5.23 The first column of the flexibility matrix is obtained by applying a unit load to the block whose displacement is described by x1 . Application of the equations of equilibrium to the free body diagrams of Fig. 5.24 leads to IF = 0 = 1 — k(a li — ran )

(i) Derive the equations of motion (ii) Setup the frequency equation and obtain the fundamental natural frequencies (iii) Obtain the modal vectors and modal matrix (iv) Draw mode shapes of the system without considering inertia of wheels and friction between the wheels and the surface.

Free Vibrations of Multi-Degree-of-Freedom Systems

K

tvv\j_

K / m

-1\AAT-

m

A Ai r

Fig. 5.25 Equations of motion Use Lagrangian's method to obtain the equations of motions Kinetic energy, 1 . +1 .2 T= —MX, — 2 MX.,

(5.27)

Considering x2 > xi Potential energy, V=

1

2

1 2 1 9 + — k (x2 — xi ) + — 2 2 2 (5.28)

{xi}={x1} x2 the Lagrange s equations are d dt d (aT dt ax2

[2k — mco2]A 1 sin(cot + 0) — kA2 sin(wt + 0) = 0 (5.35) — kA1sin(wt + 4)) — (2k — mco2)

A2 sin(wt + 0) = 0 (5.36) Since, sin(wt + 0) # 0, the above equations reduces to [2k — mco2]A1 — kA2 = 0 (5.37) — kA1 — (2k — mco2)A2 = 0 (5.38) for nontrivial solution of Al and A2 we have from Eqs. (5.37) and (5.38) 2k — mw2

—k

—k

2k — mco2

(5.39) The above equation is referred as a characteristic determinant. Solving, we get frequency equation of the system ni2(04 — 4kmw2 + 3k2 = 0 (5.40) put w2 = A, in the above equation _r4ki

aT ± av = 0 (5.29) axi

) DT av = 0 (5.30) ax2 ax2

by substituting T and V in above equations and after simplifying we get, equations of motion as m x 1 + Val — kx2 = 0 (5.31) (5.32) m + — 21cc2 = 0 (ii) Fundamental natural frequencies Assume xi = Al sin (cot + 0) (5.33) x2 = A2 sin (cot + 0) (5.34) substitute harmonics x1 and x2 from Eqs. (5.33) and (5.34) in equations of motion, Eqs. (5.31) and (5.32), we get

=0

m

x+

3k2 =0 m2

(5.41)

Eq. (5.41) is quadratic in A, i.e. 2 Ai, A2 = co,cq

A'19 "2 = Cq5a4 4k + I( 4k)2 12k2 m2 /11 \ 2

2 2 C°1 '(°2

2 2 a)i ,a)2

4k I 4k2 — + yyi \ m2 2 k (4 ± 2)— jn

2

5.22

Mechanical Vibrations

(4 — 2)

(0 1 —

2

Ir

(4 + 2) ± 1 , (02 — \

2

rad/s

(5.43)

and , the fundamental second natural frequency of the system is:

2k — mco12

A11

k

obtained from Eq. (5.47)

(5.42) from above equation, the fundamental first natural frequency of the system is col =

A21

k 2k — mcoi2

A21

A11

obtained from Eq. (5.48) The first modal vector, represented by {A}1 is {A}1 =

{ A411

{ Al

(5.44)

(iii) The modal vectors and modal matrix For first Principal mode (Mode-I) of vibration, the system vibrates with first fundamental natural frequency, i.e. co= cop Therefore, substitute o) = col in Eqs. (5.37) and (5.38). Also for vibrations under Mode-I, consider A11-amplitude of first mass, mass (m1) due to frequency col A21-amplitude of second mass, mass (m2) due to frequency oh With the above conditions, the responses of the system of the two masses are x1 = A11 sin(aht + 01) (5.45) x2 = A21 sin(wit + 01) (5.46) Therefore the Eqs. (5.37) and (5.38) can be written as [2k — mo0Ail — kA21 = 0 (5.47) —kA 11 —(2k — mco?)A2i = 0 (5.48) Let

A21 A11

=µl = amplitude ratio

(5.49)

Pi Ai

A21

(02 = \I 3k rad/s in

l l

To obtain pi substitute for oh from Eq. (5.43) in any of the two Eqs. (5.47) or (5.48) given above —

2k— m(klm) —1 k

A21

In Eq. (5.49), for simplicity consider A11=1, therefore the first modal vector is All } {A21

A11 111 A11

(5.50) for second Principal mode (ModeII) of vibration, the system vibrates with second fundamental natural frequency, i.e. co = co2. Therefore, substitute to = co2 in Eqs. (5.37) and (5.38). Also for vibrations under Mode-II, consider Al2-amplitude of first mass, mass (m1) due to frequency cot A22-amplitude of second mass, mass (m2) due to frequency co2

Free Vibrations of Multi-Degree-of-Freedom Systems

Now, for the system, the model matrix, which is comprised of both modal vectors, is [{A} 1 {A} 2 ]

With the above conditions, the responses of the two masses are: xi = A l2 sin(w2t + 02) X2 = A22 sin(co2t + 02)

5.23

(5.51) (5.52)

= [. A11 mi A11

Therefore the Eqs. (5.37) and (5.38), can be written as

Al2 /12 Al2

[2k- mo.)DA 12 - kA22 = 0 (5.53) (5.57) —kA l2 — (2k Let,

A22

— nwo2)A22 =

0 (5.54) (iv) Mode shapes

= /12 = amplitude ratio

42

2( ),

X2

1

112

(2k - nwo22

A22

, K m \

K

k

Al2

obtained from Eq. /12

%\i—

(5.53)

k 2k - mco22

A22 Al2

Mode-I

obtained from Eq. (5.54) The first modal vector, represented by {A}2 is Mode-II

{A}2 =

{Al2 1

Al2

A22 JJJ

P2 Al2 JJJ

(5.55)

To obtain 1.12 substitute for cat from Eq. (5.44) in any of the two equations given above 1 / 2-

A22 Al2

2k-m(3k/ m) =

- 1

In Eq. (6.36), for simplicity consider A11 = 1, therefore the first modal vector is = {Al2} A22

Al2

Mode shapes of the system

Fig. 5.26

5.25 Show how flexibility influence coefficients are used to obtain fundamental natural frequency of a system shown in Fig. 5.27 by Stodola Method. K

2K 2m

P2 Al2

AN\J-

2m

X3 h*-

--(VV\i—

/////////2 //////// (/ /) //9////////)//)//

(5.56)

1

2

Fig. 5.27

3

5.24

Mechanical Vibrations

The flexibility influence co-efficients of the system are

The new deflection vector is xi

1 — a12 a13 — a21 — a31 — 2k all 3 a22 = a23 a32 = 2k

X2

11

= 5 mco2

5

a33

=a) m 2

2k The inertia of forces in mass 1, 2 and 3 are Fl = 2mw2x1 F2 = 2mw2x2 F3 = m(02 x3

2.2

=,

v 2

2.6 The new deflection vector { V}2 # {V} ) Assume { V}2 as a new deflection vector for second interation. II Iteration

}

F1' = 2mw2 x' = 2 mw2

If initial Displacement vector{ x2 x

F2 = 2MCO2 4 = 4.4 mw2

3

F3 = mw2 x3 = 2.6 mw2

= {niassumed as

... Displacements at 1, 2 and 3 are

{1

xr = Fi'aii +F2 ai2 +.F3'ai3

F1 = 2mco2, F2 = 2mco2 and F3 = MCO2 Using inertia forces and influence coefficients, total displacements at point 1, 2 and 3 can be otained as

mw 2 + 4.4mco2 2.6 mco 2 + k 2k 2k

Then

x; = Fia mw2

k = F1a21

_ 9 mw2 2k xi' = F1'azi + F2/a 22 +F3 a23

+ F2a12 + F3a13 +

mw 2

k

mw2 ▪ 1na) 2 5mco2

k

2k

2k

7.8 111w 2 2k

23 mw2 2k

F2a22 + .F2 s 23

— mw 2 + 6mw2 + 3mw2 = 1 1MCO k 2k 2k k

1 3.2 mo)2 2k

2

X; = F1a31 + F2a32 + F3a33 mw2 6mw2 5mw2 13mw2 _ + k 2k + 2k k

"

E,

x3 = J•I'a31 + F; a32 -FF3 a33

_ mw2 + 13.2 mw2 + 13 mw 2. k 2k 2k 28.2 mw2 2k

5.25

Free Vibrations of Multi-Degree-of-Freedom Systems

New displacement vector is

x2

x3

1 2.55

9mco2 — 2k

xl

1 = 10.23

= {V}3

III Iteration F1" = 2mw2 x1" = 2ma)2 = 2m0)2 xz = 5.1

MCO2

F3" = inco2 x3 = 3.13 ind Displacements at 1, 2 and 3 are

inco2

-

k

+ F27 n 12 +

5.1mco2

2k

+

a13 3.13m0)2 2k

10.23mco2 2k = Ft" a21 + F2" a22 + F3" a23 inoi2 + 15.3mw2 9.39m0)2

-

k

2k

2k

26.69mco2 2k , x;" = Fi" a 31 + F2 a32 + F3" a33 mw2 15.3mco2 + 2k k

16.5 mc02 2k

33.8 mw2 2k New deflection vector is Xi

X2

10.23mco2 2k

x3

{V}4

= 1113

MO)

2

2k

co= 0.44Xn rad/s

3.13

{17 }3 # (17)2 Assume II.% as a new deflection vector for third iteration.

xr -= Fin a

m

= x1

1 Mocy vector = 2.60 with acceptable 3.30 error. The isadvantage of this method is that it ca be used only to obtain first fundamental natural freqency. 5.26 Find the lowest natural frequency and modal vector of the system shown in Fig. 5.27 by matrix iteration method. Draw mode shape. For the system, mass matrix M and flexibility matrix A are 2m 0 0 M = 0 2m 0 0 0 in [1 1 11 A= I- 1 3 3 2k 3 5 1 Considering trial vector uo = (111) u t = AMuo =1 2k

2m 0 0 0 2m 0 1 3 5 0 0

1 2 6 3 1 2k 2 6 5 1 2 2 1

1 2.60 = v14 3.30

1 2.5 in 2.2 k 2.6

=

5.26

Mechanical Vibrations

II Iteration 2 2 1 {1 } it2 = AMui 2'k 2 6 3 2.2 2 6 5

2.6

1 4.5in

2.55 = 142 3.13 as u2 # Li' try next iteration:

5.27 Three machines are equally spaced along the span of a simply supported beam of elastic modulus E and mass moment of inertia I. Determine the flexibility matrix for a 3-degree-offreedom model of the system as shown in Fig. 5.29. L

EA---71

III Iteration

>IA

L

7t

>IA

L

L

/

rFrrh

[2 2 1 u3 = AMu2— m 2 6 3 2k 2 6 5

1 2.55}

= 5.12

X2

X3

Fig. 5.29

3.13

1 = 5.12 2.61 = u3 k 3.22 as . u3 is close to u2 the solution is converged with acceptable error. w2

X1

The deflection of a particle at a distance z along the neutral axis of a simply supported beam, measured from the left support, due to a concentrated unit load applied at a distance a from the left support is y(z)

—k

= L3 ri_ L)[L(2)z _( 6 EA L

col = 0.44 Vin/k

for z a. The elements of the third column of the flexibility matrix are the machine displacements induced by a unit concentrated load to a = 3L/4. Then

1 Modal vector = 2.61 3.22 Mode shape, I Mode

L3 [15 z rzy] 24E/ 16 L and the flexibility influence coefficients are y(z)—

2k

E, I

ic

2m -VW- m "MAI/-I\AA//

a13 =

2.61 3.32

a23

a33

Fig. 5.28

I,) 7L3 4 ) = 768E/ Y(

= (L) ._ 11L3 2 ) 768E1 (3 f )_ 3L3 256E/

5.27

Free Vibrations of Multi-Degree-of-Freedom Systems

The second column of the flexibility matrix is determined by placing a unit concentrated load at a = L/2. Then L3 3 z ( z l3 12E/ 4 L `LJ Note that due to reciprocity and symmetry of the beam, only a22 must be calculated. To this end, y(z) —

[

for z a. The second column of the flexibility matrix is obtained by setting a = 2L/3, leading to 2 3 7 (z) [( Z ) y(z) 61L ) 27 EIRL) Then a12=

L) 11L3 ) — 4374 EI

Y( 3

(2 L) _ 16L3 Y 3 4374E1 From reciprocity, a12 = a21, and from symmetry, a11 = a22. Thus L3 [16 11 A— 4374E/ 11 16 a22 =

a22 = Y (IL) 2 48E1 Then from reciprocity, a32 = a23, and from symmetry, a12 = a32. Then from symmetry, a ll = a33, and from reciprocity, a21 =a 12 and a31 = a13. Thus the flexibility matrix is 3 11 7 256 768 768 L3 11 1 11 A= El 768 48 768 11 3 7 768 768 256 5.28 Two degrees of freedom are to be used to model the vibrations of a fixed-fixed beam of length L, elastic modulus E, and cross-sectional moment of inertia I. Determine the flexibility matrix for the model assuming the generalized coordinates are displacements of equally spaced particles along the span of the beam. The deflection of a particle along the neutral axis a distance z from the left support due to a concentrated unit load at a distance a from the left support is 2 2 L 1a a ) (Z ) y(z) = L L) El 2 a)2( +2 )( ] _ 1( _L )3 L L 6

5.29 A machine with a large moment of inertia is placed at the end of a cantilever beam. Because of the large moment of inertia, it is decided to include rotational effects in a model. Thus a 2-degree-of-freedom model with generalized coordinates, x, the displacement of the machine, and 0, the slope of the elastic curve at the end of the beam, are used. Determine the flexibility matrix for this model if the beam is of length L, elastic modulus E, and cross-sectional moment of inertia I. Consider first a concentrated unit load at the end of the beam. From strength of materials, the deflection at the end of the beam is a ll = L31(3E1), and the slope of the beam at its end is a21 = L21(2EI). Then if a unit moment is applied to the end of the beam, the deflection at the end of the beam is a12 = L2/2E1) and the slope of the elastic curve at the end of the beam is a22 = MEI). Hence the flexibility matrix for this model is A_ L — EI

3 2 1

5.28

Mechanical Vibrations

5.30 Determine the flexibility matrix for the 4-degree-of-freedom system of Fig. 5.30.

Fig. 5.30

Let x1, x2 and x3 be the displacements of the particles on the beam, and let x4 be the displacement of the particle attached to the beam through the spring. Note that the flexibility matrix for the beam without the additional mass-spring system is determined in Solved Problem 5.27. The first column of the flexibility matrix is determined by placing a unit load acting on the first particle. Summing forces on the free body diagram of the block, shown in Fig. 5.31, shows that the force in the spring is zero. Thus the deflection of the beam is only due to a unit force applied to the beam, and the flexibility influence coefficients air . i = 1, 2, 3, and j = 1, 2, 3, are calculated as in Solved Problem 5.27. Also a41 = a21, a42 = a22, a43 = a23 Now consider a unit load applied to the hanging block. The force developed in

the spring is 1. Hence the beam is analyzed as if a unit load were applied to the midspan. Also k(a44 — a24) = 1 1 1 a44 k a24 a22 Then using the results of Solved Problem 5.27, 3 11 7 11 128 768 768 768 1 11 11 1 768 48 768 48 L3 A= 7 11 3 11 El 768 768 128 68 11 1 11 1 E/ + 768 48 768 48 kL3 5.31 Find the natural frequencies of a unrestrained system shown in Fig. 5.32.

J1

J3

2

Fig. 5.32

Take

J1 = J2 = =

k,,= k, .Use Lagranges equation to get differential equations of motion

T= V=

i k(a41 - a21 ) = 0

2

(J012

1 b.

2

40.,)

J032 ) 1J1 F

pi l/2 )2

The equations of motion are

J81 + k101 — k,02 = 0 J62 — k,611 + 2102 — 1(103 = 0 Fig. 5.31

J93

kt 02 k103 — 0

Free Vibrations of Multi-Degree-of-Freedom Systems In matrix form el

[J 0 0 0 J 0

k, —k, —k, 2k, 02 + n 0 -k, -3

ooJ

11,1

0 01 1

= 2 =0 II

A2 = 0)12 = ki lJ 12

02

k,

5.29

03

co„ , = Vki lf rad/s 3k = 0)112

0

3

—

J

0 Considering system vibrates with SHM 01 = 01 sin (cot + 0) sin (cot + 0) 02 = 03 = 03 sin (cot + 0) Substituting above equations in differential equation of motion we get

3k,

co,,, =

rad/s

5.32 derive the stiffness matrix for the 3-degree-of-freedom unrestrained torsional system of Fig. 5.33. The torsional stiffnesses of the shafts are

0 (k,—J0)2 ) —k, —k, (2k,— J0)2) =0 —k, 0 —k, (k,—J0)2 )

k

referred as characteristic determinant. solving above determinant, j3(06 — 4 j2k104 + 3 jki2 (02 = 0

JAB AB GAB 1AB

LAB L (0.03 m) i (80 x 109 N m2 2 ll 0.6 in

Put cot = A. 3k, A. (A, — ki1J)(A, --) = 0

nB,

= 1.70 x105 N-in rad

ne2 60 cm

100 cm

r= 30 mm

r= 40 mnn

N G= 80 x109— m2

G=100 x109 -1 Im2

Fig. 5.33

/A = 2.5 kg = m2 /B = 4.5 kg = m2 /c = 2.8 kg = m2

5.30

Mechanical Vibrations

k-BC , —

The quadratic equation is used to solve for

IBC GBC LBC

2 (0.0 4 n)4 (100 X109

N

co2 —

5/cm ± V25 k2 m2 — 4(m2)(5 k2) 2m2

Mz

v3)

1.0 in (0= [( + 4.02 x105 N-111 rad Stiffness influence coefficients are used to show_ —k,fB K=

0

k,AB+k,BC

AB

BC

0

kBC

1.70 105 [—1.70 0

—1.70 0 5.72 —4.02 I —4.02 4.02

5.33 The differential equations governing the motion of a 2-degree-of-freedom system are rm [0

0 1F.:1 1 1-2k —kk! i = [01 + L—k 3k iLx2 LO]

ni

Determine the system's natural frequencies. The natural frequencies are determined using Eq. (5.12): det 1K — co2MI = 0 det

F2k —k1 _

L-k 3k]

0)2

in

0

0

2k — (02 m

—k

—k

3k — a)2 ni

=0

2

col = 1.176

2

]1/2

m

\IT , w2 = 1.902 \I 1±n in

5.34 Determine the natural frequencies of the system of Fig. 5.3 if in = 5 kg, I = 0.5 kg-m2, L = 0.8 m and k = 2 x 109 N/m. Substituting the given values into the mass and stiffness matrices determined using x and 9 as generalized coordinates in solved Problem 5.5 leads to M=

[5 0 1 0 0.5

,K=

[4 x105

4 x104

4x104

4x104

The natural frequencies are calculated using Eq. (5.12) 4x105 4x104

—

2[5

4x104 4x104 4x105 -50)2 4x104 4x104

=0

0 0.5 0

4x104 -0.5(02

=0

(4 x 105 — 5oi)(4 x 104 — 0.5co2) — (4 x 104)2 = 0 2.5w4 -4 x 105co2 + 1.44 x 1010 = 0 1/2

=0

(2k — co2m)(3k — o)2m) — (— k)(—k) = 0 m20)2 — 5km(02 + 5k2 = 0

5)2 4x105+ (4 x10 —4(2.5)(1.44 x1010) 2(2.5) rad rad col = 233.9 7, (02 = 324.5

5.31 -

Free Vibrations of Multi-Degree-of-Freedom Systems

5.35 A 500 kg machine is placed 2_m-from the left support of a 6 m fixed-fixed beam while a 375 kg machine is placed 4 m from the left support. IgnOrilig inertia effects of the beam, determine the natural frequencies of the system if E = 200 x 109 N/m2 and I = 2.3_5x .10-6 m4. The system is modeled using 2degrees-of-freedom with the generalized coordinates as the displacements of the machines. Using the results of Solved Problem 5.28 with the given values substituted, the flexibility matrix for this model is A=

[1.68 x10-6 1.16 x10-6 1.16 x10-6 1.68 x10-6

The approximate mass matrix for the model is 500 0 M = [ 0 375]

The natural frequencies are calculated using Eq. (5.13): detico2AM - II = 0 det 10-6 co2

.68 1.161[500 0 1.16 1.68R 0 375 10 01

8.4 x104 co2 -1 4.35 x 104 0)2 5.8 x 104 0)2

=0

(1.47x10-3 )2 1.47x10I+ -4(2.27 x10-7)(1)

a)1 = 28.3

rad,

1 rad co2 =67.

mb = ML = (7800 kg)(4.36 x 10-3 m2)(6 m) = 204.0 kg The mass added at each node represents the mass of a segment of the beam. The boundary between segments for adjacent nodes is midway between the nodes. The boundary of a segment for a node adjacent to a support is midway between the node and the support. The inertia of particles near the supports is neglected. If included equally with other particles, the inertia of the beam would be overapproximated. Thus for this model, the mass added to each node is m3/3 = 68.0 kg. Thus the mass matrix becomes M=

[568 0 0 443]

A procedure similar to that used in Solved Problem 5.35 is followed, leading to rad

,

(02 = 62.3 rad

=0

6.3 x 104 0)2 -1

2(2.77 x10-7)

The total mass of the beam is

col = 26.4

2.77 x 10-7o14 - 1.47 x 10-30)2 + 1 = 0

co =

5.36 The beam described in Solved Problem 5.35 is made of-a material -of-mass density -7800 kg/th3 and has a crosssectional area of 4.36 x 10-3 m2. Determine the naturatfrequencies-of the - system when inertia effects of the beam are approximated by adding particles of appropriate mass at the nodes.

- 1/2

5.37 Determine the mode shape vectors for -- the system. of Solved Problem 5.34 using_ x and 0 as generalized coordinates. The normal mode solution implied that the ratios of the values of the generalized coordinates are constant for each mode. Let X be the displacement of the mass center of the bar at an arbitrary instant

5.32

Mechanical Vibrations

for either mode, and let 0 be the angular rotation of the bar at this instant. The mode shapes are calculated using the results of Solved Problem 5.34: 4x105 -5w2 4 x 104

4 x104 x 0.5w2 [X el

The two equations represented by the previous matrix system are dependent. From the above equation (4 x 105 — 5co2)X + 4 x 1040 = 0 4x105 -5w2 4 x104

X

Substituting o) = 233.9 rad/s leads to 0 = —3.16X. Substituting CO = 324.5 rad/s leads to 0 = 3.16X. Arbitrarily setting X = 1, the mode shape vectors are X=

where C1, C2, C3 and C4 are constants of integration. The initial conditions are

4x 104

[01 0

0—

x(t) = C1X1 cos col t + C2X1 sin am + C3X 2 cos co,t + C4X2 sin wet

X2 = [ 1 3.16

5.38 Both ends of the bar of Solved Problems 5.34 and 5.37 are given a 1.8 mm displacement from equilibrium, and the bar released from rest. Determine the time history of the resulting motion. Use of the normal mode solution leads to four linearly independent solutions to the homogeneous set of differential equations. The most general solution is a linear combination of all homogeneous solutions. To this end x(t) = C1X1e"°,t + e,2 x1 e-iwit

C3 X2 ei°)21 + X2 Cic°,t Euler's identity is used to replace the complex exponentials by trigonometric functions

x(0) =

[0.0018

[01 ,X(0) = 01 0

whose application lead to x (0) = 0.0018 = C1 + C3 0(0) = 0 = —3.16C1 + 3.16C3 .i(0) = 0 = co1 C2 + co2C4 0 (0) = 0 = —3.16w1C2 + 3.16c02C4 whose solution is C1 = C3 = 0.0009, C2 = C4 = 0, leading to [xi (01 Lx2 (t)

[ 0.0009 —0.00284

cos 233.9 t

[0.0009 0.00284

cos 324.5 t

5.39 Determine the natural frequencies of the system of Fig. 5.1. The natural frequencies are the square roots of the eigenvalues of M-1K. To this end 1

0

0

1

1 K=

m

3 —2

0 — 2

0

0

1 — 2_

0

—2 0

3

—2

0

—1

3 2

_1 2

0 —1 2

2_

0

3 —1 —1 1

Free Vibrations of Multi-Degree-of-Freedom Systems

The eigenvalues are calculated from det I M-1K — AII = 0 3k in

0

in 3 k_

2m _lk 2m

0 A,

_1 k 2m lk_ A, 2m

0.697

5.40 Determine the mode shape vectors for the system of Fig. 5.1 and Solved Problem 5.39. Let Xi = [Xil Xi2 Xi f be the mode shape vector corresponding to coi . The equations from which the mode shape vectors are determined are

—20 0

0

30 — 1 2

Xi'

1 1 2

——

2

(1 2

0 — A,

Substituting calculated values of Ai from Solved Problem 5.39 leads to

co3 = 1.97,1

—20

0

X, =

0

k in

(02 =

20 30—A,

9 — p+ =o, p = A --11-12 2 The roots of the cubic equations are 0.129, 1, 3.870. Thus the natural frequencies are

30 —

be used in determining the mode shapes. To this end, arbitrarily choose Xi2 = 1. Then Xi, =

—163 + 5/32

col = 0.359 .\1; ,

5.33

X,=

1

X2 =

1.37 4

r1 1 -1

—2.298 X3 =

1 —0.1484

5.41 Use a 3-degree-of-freedom model to approximate the lowest natural frequency of a simply supported beam. The inertia of the beam is approximated by placing particles at equally spaced nodes along the length of the beam, as shown in Fig. 5.34. The magnitude of the particle masses are obtained as in Solved Problem 5.36. If in is the mass of the beam. .1 0 0 rn M= 0 1 0 4 0 0 1

Xi2

2i

Xi3

00 0 where 0 = k/m. Since the previous equations are dependent, only two must

Fig. 5.34

The natural frequencies are the reciprocals of the square roots of the eigenvalues of( AM.) To this end, using the flexibility 'matrix of Solved Problem 5.27.

Mechanical Vibrations

6 det I AM — All = 0

9 11 7 100 mL3 det 11 16 11 — A 0 1 0 4(768)E/ 7 11 9 001

k

m

k

U -

= 9P —A 110 70 110 160 — A 110 =0 70 110 90 — A mL3 — 3072 El

The roots of the above equations are f3 = 0.444, 2 and 31.556. The lowest natural frequency is (DI _

The stiffness in the coupling between each car is k. Describe the time hiStory of motion of the three cars after coupling. The differential equations governing the motion of the system is

P3 + 302 — 78fi + 28 = 0, p= A

1

0 0

0 0 Fig. 5.35

=9.866 ,1 EI mL3

5.42 Determine and graphically illustrate the second mode of the beam of Solved Problem 5.41. The mode shape corresponding to A = 20 is determined from [90-20 110 70 IX21 [0 110 160-20 110 X22 = 0 110 90-20 X23 0 70 Arbitrarily setting X21 = 1 and using the first two of the previous equations leads to 11X22 + 7X23 = —7 14X22 + 11X23 = —11 whose solution is X22 = 0, X23 = —1, leading to X2 = [1 0 —1]T. The mode shape is illustrated in Fig. 5.34. 5.43 The coupling of three identical railroad cars of mass in is shown in Fig. 5.35.

in 0 0 Xl —k 0 0 in 0 X2 = —k 2k —k 0 0 m Li3 _ 0 —k k [ 01 0 0 where x1, x2 and x3 are the displacements of the railroad cars. The system's initial conditions are 0 x(0) = 0 0 The natural frequencies are calculated as det I K — co2 M I = 0 k — mco2

—k

0

—k

2k — mco2

—k

0

—k

k — mco2

col = 0, co2=

, co3=

=0

k

m The mode shapes are determined as —1 1 1 X1 = [[1, X2= 11, X3 = 2 1 --1

5.35

Free Vibrations of Multi-Degree-of-Freedom Systems

x3(t) =_ t__v F sin 3 2 k

The general solution is 011 -1 -k

(C3 COS

+ C4 sin

-

Application of the initial conditions leads to MO) = 0 = C + C2 - C5 x2(0) = 0 = C1 + 2C5 x3(0) = 0 = CI - C3 - C5 I 3 k C6 m 4+ \

±2 (0) = 0 = C2 + 243 LC6 (0) =

0

Fmc. _ 4

= C2

V C

=

4

13 ilc n c6

\

Thus x1(t) =

M

V

21v k

C-= 61I

3k

11 17n- sin Fc-t 3+2 k m - v .1Ln- sin .13-It 6 3k m

x2(t) =

3t

-

1.70 x105 - 2.50)2 -1.70 x105 0

-1.70 k 105

0

5.72x105, -4.02 x105 - 4.5w2 4.02 x105 -4.02 x105 -2.8w2

=0 -31.50 6 + 1.067 x lOwl - 6.696 x 1011w2 = 0 rad rad col = 0, co2 = 288.5 w3 = 505.4

whose solution is C1 = C3 = C5 = 0, C2= , 3

in

The natural frequencies are calculated using the stiffness matrix derived in Solved Problem 5.32: det IK co2MI = 0

+ C6 sin \13— k t)

C2 + \

-23k 1 1 -sin ,‘I3— /c

5.44 Determine the natural frequencies Of the torsional system of Fig. 5.33.

ITc —t \ rn

-1 +{ (C5 cos.\13t -1

=v=

6

L /11 sin 13 .-t 3 \l3k in

5.45 Demonstrate orthogonality of-the mode shapes of the system of Solved Problem 5.37. 0 1[ I XTMX - [1 -3.16] [5 1 20 -0.5i 3.16 = [1 -3.16] [ 5 1.58 = (1)(5) + (-3.16)(1.58) --- 0 5.46 Demonstrate orthogonality of the mode shapes of the system of Solved Problem 5.39.

Mechanical Vibrations

5.36

XTMX2 = m[0.697 1 1.347] 1 0 0 1 0 2 0 1 0 0 2 —1

1 = m[0.697

=

1 1.347]

2 —2

5.47 Normalize the mode shape vectors for the system of Solved Problem 5.39. Normalization of a mode shape vector X is achieved by dividing every component of the vector by [XTMX]1/2. To this end, for the mode shape vectors of Solved Problem 5.39, XTMX1 = [0.697 1 1.347] 0 0 0 2m 0 0 0 2m

in'

m[(0.697)(1) + (1)(2) + (1.347) + (-2)] 0

XTMX3 = m[0.697 1 1.347] 1 0 0 —2.298 0 2 0 1 0 0 2_ _-0.1484_ = m[0.697 1 1.347] —2.298 2 —0.2968 = m[(0.697)(-2.298) + (1)(2) + (1.347)(-0.2968)] =0 X73.114X2 = m[-2.298 1 —0.1484] 1 0 0 1 0 2 0 1 0 0 2 —1 1 = ,n[-2.298 1 —0.1484] 2 —2 = m[(-2.298)(1) + (1')(2) + (-0.1484)(-2) =0

= [0.697 1 1.347]

0.697 1 1.347 0.697in 2m 1.694m

(0.697)(0.697m) + (1)(2m) + (1.347)(2.694m) =,6.115m in 0 0 1 1 x72'mx2 = [1 1 -1] 0 2m 0 0 0 tin —1 rn [1

1

—1]

2m —2m

= (1)(m)

+ (1)(2m) + (-1)(-2m) = 5m X3MX3 = [-2.298 1 —0.1484] m 0 0 —2.298 1 0 2m 0 1 0 0 2m —0.1484 = [2.298 1 —0.1484] —2.298m 2m -0.2968m = (-2.298)(-2.298n) + (1)(2m) + (-0.1484)(-0.2968m) = 7.325m

5.37

Free Vibrations of Multi-Degree-of-Freedom Systems

The normalized mode shapes are XI =

0.697

1

[0.2819 0.4044 , \I m 0.5447

= 1

V6.115m 1.347

0.4472 = 0 5m -1 \/ m -0.4472] 1

X2 =

Normalizing the mode shape 302.1VIX2 = [1 -0.529] 1 150][-0.529 -0.529] [140 L 40 150 = 99.66 The normalized mode shape is X2

1

r 1

0.0100

499.66 L-°•529] Ho.530

2.298

X3 =

1

r

5.49 Use matrix interation with the trial vector u0 = [1 0 0]T to approximate the lowest natural frequency and its mode shape for the system of Solved Problem 5.41.

1

,j 7.325 m 0.1484 • -0.8491] = 0.3695 VT" -0.0548 1

5.48 A 2-degree-of-freedom system whose mass matrix is M=

[100 401 40 150]

has a normalized mode shape of X1 = [0.0341 0.0682]T. Determine the normalized mode shape for the second mode. Assume the second mode shape as Imposing orthogonality,

[1 a]T.

XTMX2 = 0 = [0.0341 0.0682] [100 401 [11 L 40 150] Lai [100 + 40a1 40 +150a]

= [0.0341 0.0682]

= (0.0341)(100 + 40a) + (0.0682)(40 + 150a) = 6.138 + 11.594a a = -0.529

Using the matrix AM calculated in Solved Problem 5.41, matrix interation is used as shown: 90 110 7011 u i = AMuo = 110 160 110 0 70 110 90 0 90 110 , 70

0.8182 ill = [1 0.6364

90 110 70 0.8182 = AMU, = 110 160 110 1 70 110 90 0.6364 0.7131 22.820 = 3201 112= 0.7019 22.460 90 110 7010.71311 u3 = A1VIii2 = 110 160 110 1 70 110 90 0.7019

5.38

Mechanical Vibrations

23.330 0.7389 = 31.570 , 113= 1 23.310 0.7389 90 110 7(P 0.7389 tt4 = AMil3 = 110 160 110 1 70 110 90 0.7389

5.50 Use matrix interation to determine the second natural frequency and mode shape of the system of Solved Problem 5.41. Using sweeping technique. If X2 = [A B C]1, then orthogonality requires X2MX1 = 0 = [A B C] rn 4

22.820 0.7073 = 32.220 , 114 = [1 22.820 0.7073

0 0

110 70 0.7073 u5 = AMti4 = 110 160 110 1 _ 70 110 90 0.7073 90

22.320 0.7073 = 31.560 , ti5= 1 22.320 0.7073 90 110 70 0.7071 u6 = AMtis -= 110 160 110 1 70 110 90_10.7071 22.310 0.7071 = 31.564], ii._6 = 1 0.7071 22.310 Hence the interation has covereged to A. = 31.560 and X1 = [0.7071 1 0.70711T which leads to 1 oh= J31.56 0 1 3072 EI

=9.866 •I

EI m.0

0

0 0.7071

4 0

0 0.7071 4

'Li (0.7071A + B + 0.7071C) = 0 4 A = -1.414B - C Orthogonality to the first mode is imposed by defining Q=

90 110 70f -1.414 -1 [0 110 160 110 1 0 70 110 90 0 0 1

0 -1.7260 -20 = 0 0.44600 0 0 1.1020 20 Matrix interation, when used with the matrix Q, imposes orthogonality of the iterate to the first mode, and thus the interation converges to the mode shape for the second mode and yields the second natural frequency. To this end select u0 = [0 0 1]T. Then 0 -1.7260 -20 1[01 tit = Quo = 0 0.44600 ( 0 1.1020 0 20 1 -20 0

20

ill =

-1 0 1

5.39

Free Vibrations of Multi-Degree-of-Freedom Systems

0 -1.7260 -20 0 0.4460 0 u2 = Q111 = 0 1.1020 20

-1 0

5.52 Use matrix iteration to determine the highest natural frequency and its corresponding mode shape for the system of Solved Problem 5.39.

El =39.19 ,/ niL3

5.51 Use the results of Solved Problems 5.49 and 5.50 to determine the highest natural frequency of the system of Solved Problem 5.41.

Matrix iteration converges to the highest natural frequency when the matrix WOK is used in the iteration procedure. Using u0 = [1 0 0]7' and M-1K from Solved Problem 5.39 with 0 = khn, ul = M-IKuo

If X3 = [D E FI T, then orthogonality with X1 requires D = -1.414E - F. Orthogonality with X2 requires

30 -20 0

1

-0

0

-2-0 1

1 0 --0 -0 2 2

XP14X2 = 0 in 4

- 0.3333

-0 ,

0 0

0

0

= [D E F] 0

0

4

0 0

rn 4

0

-1

1 - 1 (D - F) = 0 = 0 -›D=F

4 by noting that AMX3 = X3X3. Thus

-

110 160 110 -1.414 70 110 90 1

0.4460 -0.6240 0.4460

U2 =

0

1

30

[1

El

IJJ niL3

T .0

90 110 70

= 82.99

, \10.446( niL3 3072E1)

1

I2( inL3 3072E1) \

J0.4460 1

Hence it is clear that X2 = [1 0 and

1

1

(03 -

20 0 -20

(02 =

Hence A3 = 0.446 and

M-11(11 1

30 -20 0 1 1 0 -? -0.3333 -0 2 0 1 1 0 --0 -0 2 2 3.6670 -1.50 0.16670

u2 = - 0.4091 0.0455

5.40

Mechanical Vibrations

3.8700 = -1.6850 0.24974)

u3 = 30 -20 0 1 3 1 -0.4091 -0 20 20 0.0455 1 4) -0 1 0 -2 2 3.8180 = -1.6364) , 0.22730

u 7 = WI K116 30 -20 3

-0

1 fi3 = -0.4285 0.0595

0

3.8590 1.6740 = [0.24454)

1 u4 =-0.4338 0.0636

U5 = M 11(1714

=

30 -20 3

-0 0

0 - 1 1 -0.4338 2 0 20 0.0636 --1i4) 0 2

3.8680 = -1.6830 , 0.24870,

u5 =

1 -0.4351 0.0643

U6 = M 11(115

-20 0 1 3 -0.4351 = -4) 20 -0 2 0.0643 1 0 -y0 0 30

0 1 -0.4354 20 20 0.0652 1 1 -0 -4) 2 2

3.8710 = -1.6850 , 0.25030

U4 = M 11(113

30 -20 0 C1 3 1 -0.4295 -0 2 0 -20 0.0595 1 1 0 -0 -0 2 2

1 u6 = -0.4354 0.06452

= -0.4353 0.0647

The interation has converged to [1 -0.4353 0.0647]T and 22,3 = 3.8710, leading to

X3 =

cos = V3.8710 =1.967 \i /1in

5.53 For what values of c will both modes of the system of Fig. 5.36 be under damped? 2k

k

m

m

E

2c m= 36 kg

k=1.3x10m

Fig. 5.36 The differential equations governing the motion of the system of Fig. 5.36 are rm 0 _E r 3c -2cirii L-2 c 2 c x2 J L0

3k

-2k1I-xi 1 [01 = -2x k 2k ]L 2 LO]

[

Free Vibrations of Multi-Degree-of-Freedom Systems

The system has viscous damping which is proportional with a = c/k and fi = 0. The undamped natural frequencies are determined from detIK — co2MI = 0 3k — mco2

—2k

—2k

2k — 2mo.)2

5.41

Consider a semidefinite torsional system shown in Fig. 5.37. The equations of motion are J1 61 + k„(0, — 02) = o J2 62 + kii( 02 —

=0

J363

k col = 0.5177 \17 , co2 = 1.932\1— 11 Then from Eq. (5.21),. = 0, the mode with the highest natural frequency has the highest damping ratio. Thus for < 1, 1 — aco2 C2

a/12 ( 3

- L, t) = 0 ax 2 The displacement must be continuous at the junction between the two bars: u1(L, t) = u2 (L, t) The resultant force due to the normal stress distribution must be the same in each bar at their junction: ui(0, t) = 0,

j

= C3 (cot + tan 10)

2

dX1

dr

(L) —

1 dX2 4 dx

(L)

4 cos 4)(cot 0 + tan = — sin 0 + cos 0 tan 20

7.11

Vibrations of Continuous Systems

7.10 Determine the lowest natural frequency of longitudinal motion for the system of Fig. 7.5.

EA \I P —= CU —tan o)L — E k

E = 200 x109 ±1-m2 kg p = 7800 — m3

2x105 -N-

m

2m

Fig.7.5

Substituting given values leads to 1.5 0 = —tan 0 The smallest solution of the transcendental equation is 0 = 1.907, which leads to col =

The differential equation governing u(x, t), the longitudinal displacement of the bar, is E n2 2 uu ax2 P Since the end at x = 0 is fixed, u (0, = 0 The resultant of the normal stress at the right end of the bar must equal the forces in the spring at any instant, thus att EA— (L, t) = —ku(L, t)

ax Application of the normal mode solution u(x, t) = X(x)e1" to the boundary condition leads to dx

(L) = —kX(L)

Using X(0) = 0 in Eq. (7.5) leads to C1 = 0. Application of the second boundary condition to Eq. (7.5) leads to Eilco.1.11, cos (co \LP- L) E

E

,

L\Ip

1.907 2m{

200 x109 N m2 kg 7800 m2

= 4.83 x 103 rad 7.11 Show that the mode shapes of Solved Problem 7.5 satisfy an orthogonality condition of the form of Eq. (7.6). Let co; and (of be distinct natural frequencies of the system of Solved Problem 7.5 with corresponding mode shapes Xi(x) and Xi(x). These mode shapes and natural frequencies satisfy the following problems: d2

dx2

+

& X. = 0 c2 I

d2 X •

CO 2

X•• = 0

+

VE

L

(7.24)

dX dx (L) = 0

40) = 0,

dx2 =— k sin (co

I

EA — tp = —tan 0, 0 = coL , \IL) kL

A=3 x 10-6 m2

X(0) = 0, EA

E

c2 dX •

Xi(0) = 0,

dx

(L) = 0

(7.25)

7.12

Mechanical Vibrations

Multiplying Eq. (7.24) by Xi and integrating from 0 to L leads to L

2 L d2 X • CO X • (X) f x.wx CI + C3 = 0

p = 7800 A = 2.6 x 10-3 m2 L= 1m / = 4.7 x 10-6 m4

Fig. 7.6

d2 X (0).= 0 - 1.2C1 + A2C3 = 0 dx2

The free transverse vibrations of the beam of Fig. 7.6 are governed by Eq. (7.8). Since the beam is fixed at x = 0,

from which it is determined that C1 = C3 = 0. Application of the boundary conditions at x = L leads to d2 x

dx2

=0

(L) = 0 — A3C2 cos AL

dx3 + A3C4 cosh AL = 0 A nontrivial solution of the above equation exists if and only if the determinant of the coefficient matrix is zero, —sin AL cosh AL + sinh AL cos AL = 0 leading to tan AL = tanh AL The solutions of the previous transcendental equation are used with Eq. (7.12) to determine the system's natural frequencies. The smallest solution = 0, for which a nontrivial mode shape exists.

ax ax

(0, t) = 0

The boundary conditions at x z L are determined by applying Newton's law to the free body diagram of the block, Fig. 7.7: Az - (L, t) = 0, ax2

(L) = 0 --> — A.2C2 sin AL + sinh

d43 X

w(0, t) = 0,

E'l

w (L, t) — m'-' w (L, t) ax3 at2 Application of the normal mode solution w(x, t) = X(x)e'r to the boundary conditions leads to dX X(0) = 0, (0) = 0 dx El — a3w (L.,01

ax3

m

Fig. 7.7

D2 w

ate

(L,

7.15

Vibrations of Continuous Systems

d2 X (L) = 0, EI d 3 X =- CO2X dx2 dx3 Application of the boundary conditions to Eq. (7.11) leads to X(0) = 0 + C3 = 0 dX — (0) = AC2 + AC4 = 0 dx d2

(L) = 0 --> - A2 cos ALCI

dx2

- )2 sin ALC2 + A2 cosh ALC3 + A2 sinh ALC4 = 0 EI d3 X

m dx3 PA m

+

Noting that m _ PAL

10 kg (7800 kg )(2.6 x 10-3 m2 )(1 m) m2

= 0.493 the three lowest solutions of the transcendental equation are 01 = 1.423, 02 = 4.113, 03 = 7.192 The natural frequencies are calculated using Eq. (7.12): EI p AL4

co, =

(L) = - to2X(L)

sin

+ A cos AL) CI

PA — cos AL + A. sin AL) C2

=

02

(200 x109 N )(4.7 x10-6 m4) M2 (7800 kg )(2.6 x10-3m2)(1 M)2 M3

= 215.3 07 rad

+ (?" -! sinh AL + A cosh AL) C3

col = 486.1

PA + (— cosh AL + A sinh AL) C4 = 0

cot = 3.642 x 103 rad

where co2 has been replaced using Eq.(7.12). The previous equations represent a system of four homogeneous linear simultaneous equations for C1, C2, C3 and C4. A nontrivial solution exists if and only if the determinant of the system's coefficient matrix is zero. Setting the determinant to zero and simplifying leads to (1 + cos 0 cosh 0 ) m0 + (cos 0 sinh p AL - cosh 0 sin 0) = 0, 0=

co3 = 1.114 x 104 rad 7.16 Demonstrate that the mode shapes of a fixed-free beam satisfy an orthogonality condition of the form of Eq. (7.6). Let co; and coi be distinct natural frequencies of a fixed-free beam with corresponding mode shapes Xi and Xi. The problems satisfied by these natural frequencies and mode shapes are d4

dx4

PA 2X ; =0 El co

(7.29)

7.16

Mechanical Vibrations

dX;

X;(0) = 0,

d3 X

d3 X

(0) = 0

(L) Xi (L) +

dx3 d2 x. (L) = o, dx2

a' x.

pA EI

dx4

2

A

(7.30)

0).X. = 0

d4 X •

dx + f x; 4 ✓ dx 0

dx3

i

d4

L

(L) = 0

2L

.1 Xi (x)Xi (X) dx

N dXi

Xi(0) = 0,

dx

(0) Xi (0)

dx3

which after application of boundary conditions reduces to

(0) = 0

L d4 x d3 X j

d2X •

(L) = 0

(L) = 0,

dx3

dx2

J

dx4

Multiplying Eq. (7.29) by Xi and integrating from 0 to L leads to d4 X

L

r

J o

x.

p A (1) i dx— dx EI

d3 i

(L) X i (0)

X.(x) X.(x) dx

0

EI

(0)

0

Since w; # cop the orthogonality condition is verified. 7.17 Determine the steady-state amplitude of the end of the shaft of Fig.7.8.

dx3

dx3 dXi dx

d3 Xi

EI

PA (0)2:— 0)7) Xi(x) Xi(x) dx = 0

Using integration by parts four times on the first integral leads to Xi (L)

pA 2 L -

Using Eq. (7.30) in the previous equation and rearranging leads to

Xi(x) Xi(x) dx = 0

J

dx

X

(L)

p, G, J

d 2 X. dx2

To sin rot

(L) L

dX • d2X. + j (0)x2 (0) dx d d 2X j +

(L) dx2 d 2 Xi dx2

()

dX.

Fig. 7.8 The problem governing the motion of the system of Fig.7.8 is

(L)

a2 e ate

dx2

G

dX •

P axe ate 0(0, t) = 0

dx

(0) JG

a0 (L, t)= To sin cot

7.17

Vibrations of Continuous Systems

The steady-state response is obtained by assuming 0(x, t) = Q(x) sin cot which when substituted into the partial differential equation and its boundary conditions leads to the following problem for Q(x): G d2 Q

7.18 The block at the end of the beam of Fig.7.6 is a small reciprocating machine that operates at 100 rad/s. Determine the machine's steady-state amplitude if it has a rotating unbalance of 0.15 kg-m. The mathematical problem governing the response of the system is the same as that of Solved Problem 7.15 except for the second boundary condition at x = L. This boundary condition is determined by applying Newton's law to the free body diagram of the machine, as shown in Fig 7.9, a2 „, EI O3w (L, t) = m " (L, t) at2 ax3 + moed sin cot

+ co2Q = 0

P dx2 Q(0) = 0 JG

dQ dx

(L) = To

The solution of the differential equation is Q(x) = C1 cos [co.\11i) x

. + C2

moew2 sin wt

P x) sin [co 11—

Application of the boundary conditions leads to 1 w

Q(0)

El

= 0 —> C1 = 0

M

(1, t)

dQ JG -- (L) = To

External forces

To coJ pG cos[co.\1 P L

Hence the steady-state amplitude of the end of the shaft is

A steady-state solution is assumed as w(x, t) = Q(x) sin cot which when substituted into the partial differential equation and the boundary conditions leads to El

d4 Q

Q(0) = 0,

coJrpd

(

tan co

co2 pAQ = 0

dx4

Q(L) = C2 sin (co11-19-x)

To

Effective forces

Fig. 7.9

dx

C2 =

(1, t) at 2

ax3

P

L

dQ

— (0) = 0 dx

7.18

Mechanical Vibrations

When numerical values are substituted, the previous two equations become

22 Q (L) = 0, dx2

d3 Q

EI

(L) = - mco2Q(L) + moeco2 dx3 The solution of the differential equation is Q(x) = C1 cos fix + C2 sinflx + C3 cosh fix + C4 sinh fix where i 2 )114 co pA = El 1/4 2 100 rad ) (7800 kg ) s m3 (2.6x10-3m2)

= 0.682

(4.7 x10-6 m4 ) Application of the boundary conditions leads to C1 + C3 = 0 C2 + C4 = 0 -cos ow,- sin /3w2 + cosh fiLC3 + sinh sw4 .0

pA

cos fiL + sin 13 L)C

Pm

+ (

pA

+ (Sin +

pA

The steady-state amplitude of the machine is Q(L) = C1 cos SL, + C2 sin + C3 cosh g+ C4 sinh

7.19 A torque T is applied to the end of the shaft of Solved Problem 7.5 and suddenly removed. Describe the resulting torsional oscillations of the shaft. Removal of the torque induces torsional oscillations of the shaft. The initial angular displacement of a particle along the axis of the shaft is the static displacement due to a torque T applied at the end of the shaft. Thus, the initial conditions are

Tx a0 o) — , — o) (x, = o JG ac The natural frequencies and mode shapes are

sin fiL- cos 13L) C2

=

cosh fiL + sinh 13+3

sinhpL+ cosh /3L) ( PA P

C3 = -1.82 x 10-3, C4 = 2.69 x 10-3

= 5.60 x 10-4 m

(200 x109 N m2

n 111 p

- 0.776C1 - 0.630C2 + 1.242C3 + 0.736C4 = 0 0.891C1 - 0.564C2 + 1.153C3 + 1.489C4 = 5.04 x 10-3 The equations are solved simultaneously, leading to C1 = 1.82 x 10-3' C2 = -2.69 x 10-3,

mo e/3 pA

(2i -1)ir I G 2L 11 p

-1)xxi

Xi(x) = Ci sin l- (2i L 2L

7.19

Vibrations of Continuous Systems

The mode shapes are normalized according to Eq. (7.7) by

2 T f x sin [( 2j —1 ) 7rx] dx

A.=

L JGJ

2L

0

X7(x) dx = 1

2 TL2 C? sine 0

L 7t 2JG(2j —1)2

[ (2i —1)/rxi dx 2L

Thus the torsional response is 0(x, t)=

C,=

L The general solution for the torsional - oscillations is u(x,

=

C, i=1

8TL (-1Y+1 sin[(2i—Dnx i 7r 2 xt 2i-1 2L

sm

(2 i

\

L

sin [

2L

(Ai cos wit + B, sin coit)

Application of the initial velocity condition leads to B, = 0. Application of the initial displacement condition leads to [(2i—l)gx] Tx N, 2 sin A 2L JGV i=1 L Multiplying the previous equation by X.(x) for an arbitrary j and integrating from 0 to L leads to

[(2i—Dir r 2L

p

7.20 A time-dependent torque of the form of Fig. 7.10 is applied to the midspan of a circular shaft of length L, shear modulus G, mass density p, and polar moment of inertia J. The shaft is fixed at one end and free at the other end. Use modal superposition to determine the time-dependent torsional response of the shaft. T(t)

To

)/rxi dx L 2L JG L xsin[(2i

to Fig. 7.10

\ 2 1=1

L Ai sin o

[(2i 2L

j(2j —1)7tx] dx 2L Mode shape orthogonality implies that the only _ nonzero term in the infinite sum corresponds to i = j. Thus sm

The partial differential equation governing the torsional oscillation is

a2 e GJ

i-To [u(t)—u(t — to )](5(x —)

2

axe

a2 6, =

,

aT-

(7.31)

7,20

Mechanical Vibrations

From Solved Problem 7.5 the natural frequencies and normalized mode shapes for the shaft are

=

(2n — nit I G 2L p

con

X„(x)= — sin L

n= 1, 2 ...

[(2n —1)1rxi

2L The excitation is expanded in a series of mode shapes using Eq. (7.14) with Ck =

J

To[tl(t) — u(t — to)]

0

To

— — S111

L pJ

(2k-1)1 [u(t) — u(t —to)] 4 (7.32) 9(x, t) is expanded in a series of mode shapes of the form of Eq. (7.13) Substituting this expansion in Eq. (7.31) using Eq. (7.32) leads to

[(2k — 1) ] [1i(t) — u(t — to)] 4

The solution of the previous equation is obtained using the convolution integral as 12T0 pko)=\

.(2k —1)7r1 si 4 j L pAok2 n[

{(1 — cos a)kt) u(t) — [1 — cos cok (t — t0 )] u(t — t0)}

(5(X — L ) 2

( —1)irxi I 2 • [2k dx \ L 2L

=

fik+qPk

7.21 Use modal superposition to determine the time-dependent response of the system of Fig. 7.11. Fu sin cot

To sin[

GJEpk (t)

,12 v k

+

dX 2

k =1

Ck (t)X k (X) k=1

p, E, A, I

2

2

Fig. 7.11 The natural frequencies and normalized mode shapes for a simply supported beam are 2

= pJ1,fik (t) X k (X)

(o„ = (n7)

EI

pAL4

k =1

However,

d2

Xk

dX 2 Thus

p = cok2 —X G k

I f(i),/fik + p./o)/ pk — Ck )Xk =0

k=1

Multiplying the above equation by Xi(x) for an arbitrary j, integrating from 0 to L, and using mode shape orthogonality lead to

11 nirx X„(x)= — sin .\ L L

The differential equation governing the motion of the beam is EI a4 w +pA aw2 = F0 sin (o)t)u(x — dx4 X (7.33) The excitation is expanded in a series of mode shapes according to Eq. (7.14) with

7.21

Vibrations of Continuous Systems

j

E, A

p,

L 2 kirx Ck = f Fo sin(cot)u(x--Visin—dx x L x

m

Fo e-at

o 1
12 lead to

S

Then the minimum static deflection is

As, =

= 41.9 7 rad

d -

g

2

s2

—

m"(70.25

2 rad ) S

= 1.99 mm 8.3 A 150 kg sewing machine operates at 1200 r/min and has a rotating unbalance of 0.45 kg-m. What is the maximum stiffness of an undamped isolator such that the force transmitted to the machine's foundation is less than 2000 N? The excitation frequency and magnitude are co = (1200

min

rad = 125.7 — s

) (27( rad ) (1 60s r

8.9

Vibration Control

F0 = m0eco2

The required transsibility ratio is 4 In-

125.7 rad )2 = (0.45 kg-m) (

T=

= 7.11 x 103 N The maximum transmissibility ratio such that the amplitude of the transmitted force is less than 2000 N is Fr 2000 N T= = 0.281 Fo 7.11x103 N Application of Eq. (8.6) with C = 0 and noting T < 1 only when r> 1 lead to 1 -> r - 2.134 r2 -1 The minimum natural frequency and maximum stiffness are calculated as 0.281 -

125.7 rad con

r

2.134

- 58.9

s2

Amax

0) 2 Y

- 0.365 10.97 s

2 The minimum frequency ratio is calculated by 0.365 = 1 -> r= 1.93 r2 -1 The maximum natural frequency and isolator stiffness are co = r=

209.4 rad 1.93

k= inco2 = (20 kg)

- 108.5 rad (108.5 rad )2

s

= 2.35 x 105 1\1 rn

rad s

8.5 A 100 kg turbine operates at 2000 r/min. What percent isolation is achieved it the turbine is mounted on four identical springs in parallel, each of stiffness 3 x 105 N/m?

k = mw,2, = (150 kg) (58.9 rad)

= 5.20 x 105 I\1 m 8.4 A 20 kg laboratory experiment is to be mounted to a table that is bolted to the floor in a laboratory, Measurements indicate that due to the operation of a nearby pump that operates at 2000 r/min, the table has a steady-state displacement of 0.25 mm. What is the maximum stiffness of an undamped isolator, placed between the experiment and the table such that the experiment's acceleration amplitude is less than 4 m/s2? The excitation frequency is 2000 r/min = 209.4 rad/s. The magnitude of the table's acceleration is co2y (209.4 rad )2(0.00025) = 10.97 s2

The equivalent stiffness of the parallel spring combination is N )= 1.2x 106 11keq = 4k = 4(3 x 105 m When the turbine is place on the springs, the system's natural frequency is 1.2 x106 -11. d m - 109.5 ra con = .\17 = s 11 100 kg Noting that 2000 r/min = 209.5/rad/s, the frequency ratio is 209.4 rad s r= A - 1.912 109.5 rad '°

8.10

Mechanical Vibrations

The transmissibility ratio is T= 1 r2 —1

1 = 0.376 (1.912)2 —1

and thus the percentage isolation is 100 (1 — T) = 62.4 percent 8.6 What can be done to the turbine of Solved Problem 8.5 to achieve 81 percent isolation if the same mounting system is used?

start-up. Adding damping decreases the maximum start-up amplitude. The addition of viscous damping leads to smaller isolator displacements. 8.8 An isolator of damping ratio C is to be designed to be achieved a transmissibility ratio T < 1 Derive an expression, in terms of C and T, for the smallest frequency ratio to achieve appropriate isolation.

In Solved Problem 8.1 it is shown that 81 percent isolation requires a minimum frequency ratio of 2.50 Thus for the system of Solved Problem 8.5, the maximum natural frequency is

The relation between T, r and C given by Eq. (8.6):

rad

Squaring and rearranging the previous equation leads to

209.4 con --

2.50

— 83.8 rad

Since the same mountings as in Solved Problem 8.5 are to be used, the natural frequency is decreased only if the mass is increased. The required mass is

m.

k CO'?

1.2 x 106 N m = 170.9 kg )2 ( rad

1+ 4C2 r2

T= 1

r4 + (4C 2

- 2)

+ 44-2 T 2 — T2

r2 + 1

T 2 —1 I 2 r` + T2 =0

The preceding equation is quadratic in r2. Use of the quadratic formula leads to r2 = 1 — 2‘2

(T2 —1) T2

83.8

8.7 List one negative and two beneficial effects of adding damping to an isolator. From Fig. 8.3 it is seen that in the range of isolation (r > /) the best isolation is achieved for an undamped isolator. Thus negative effect of adding damping is to require a larger frequency ratio to achieve the same isolation. Since the range of isolation occurs for r > 1, resonance is experienced during

2

T2

Thus 81 percent isolation is achieved if 70.9 kg is added to the turbine.

±

\ [2c2 ('

T2 -1 )

T2

T2

Since T < 1, only the choice of the plus sign leads to a positive r2. Hence the smallest allowable frequency ratio is

1— 2z2

[T2 -1) T2

r=

2

T2-1 [20(7'2 111 T2

T2

Vibration Control

8.11

8.9 Solve the Solved Problem 8.1 as if the isolator had a damping ratio of 0.1. Using the results of Solved Problem 8.8 with C = 0.1 and T = 0.19 leads to

r=

[

1- 2(0.08)2

[ (0.281)2 -11 (0.281)2

2 [ (0.281)2 -11 } 2(0.08)2 1

(0.19)2 -11

1-2(0.1)2

r=

(0.19)2

1/2

(0.281)2 (0.281)2 -1 [ (0.281)2

1/2

0.19)2-11} {(1- 2(0.1)`[ ( (0.19)2 [(0.19)2 -11 (0.19)2 = 2.63 Thus the maximum natural frequency and maximum allowable stiffness are

125.7 rad rad co - 57.7 2.18 r k = m co,2, = (150 kg) (57.7

con= r

. rad)(1 min) 100 r (271 min] r ) 60 s ( rad 2

k = ma)2, = (200 kg) (39.82

rad)2 s)

= 4.99 x 105 11 m

2.63 = 39.82

= 2.18 The maximum natural frequency and maximum isolator stiffness are calculated as

rad ) s)

5N = 3.17 x 10 8.10 Solve the Solved Problem 8.3 as if the isolator had a damping ratio of 0.08. From Solved Problem 8.3 the required transmissibility ratio is 0.281. Thus using the results of Solved Problem 8.8 with = 0.08 and T = 0.281 leads to

8.11 What are the maximum start-up amplitude and the steady-state amplitude of the system of Solved Problem 8.10? Recall from Chap. 3 that the amplitude is related to the magnification factor by F0 X = - M(r, c) F0

1

k

(1 _ r 2)2 + ( g-02

Substituting values calculated in Solved Problems 8.3 and 8.10 leads to a steadystate amplitude of

Mechanical Vibrations

X

3000 N > 7.11 x 103 N

7.11 x 103 N

1 + (2Cr)2 (1— r2 )2 (2Cr)2

4.99 x10514m 1 V[1 — (2.18)2 + [2 (0.08)(2.18)]2 = 3.78 mm The maximum value of the magnification factor is

Equation (8.27) can be rewritten in terms of r by noting k = md/r.The result is 7.11 x103 0.03 >

2

r2

2CV1—C2 Thus the maximum amplitude during start-up is F0 X. = IT M.

7.11 x103 1\1m 4.99x105 11

r

2(0.08)V1 — (0.08)2 = 8.93 cm 8.12 Design an isolator by specifying k and C for the system of Solved Problem 8.3 such that the maximum start-up amplitude is 30 mm and the maximum transmitted force is 3000 N. The isolator is to be designed by specifying k and c. Two constraints must be satisfied. The maximum start-up amplitude is 30 mm, which leads to 7.11 x 103 — N m 2Ck j1— S 2

(8.29) 2a1— C2 Equations (8.28) and (8.29) must be simultaneously satisfied. There are many solutions to Eqs. (8.28) and (8.29) which can be obtained by trial and error. One solution is r = 1.98 and 4 = 0.20, which leads to Xmax = 0.03 m and FT = 2998 N. Thus k= mw

1

0.03 m >

m

(150 kg) 1125.7

1

max

(8.28)

(8.27)

and the maximum transmitted force must be less than 3000 N, which leads to

— 6.05 x 10511

m

8.13 A machine of mass one tonne is acted upon by an external force of 2450 N at a frequency of 1500 r/min. To reduce the effects of vibration, an isolator of rubber having a static reflection of 2 mm under machine load and an estimated damping factor of 0.2 are used. Determine: (i) The force transmitted to the foundation (ii) The amplitude of vibration of the machine (iii) The phase lag of transmitted force with respect to external force na = 1 tonne F0 = 2450 N = 1500 r/min 8s, = 2mm = 0.2

Vibration Control (1) (0, _

g _ I 9.81 ost \ 0.002

= 70.036 rad/s 27rN 2ir x1500 60 60 = 157.07 rad/s

CD —

r=

CO

=

co„

,

•'• Ft —

157.07 = 0.242 70.036

Fo 41+(2,cr.)2

(1- r2 )2 + (4'02

8.13

8.14 A machine of mass 500 kg is acted upon by an external force of 2000 N at a frequency of 1500 r/min. To reduce the effect of vibration, an isolator of rubber having a static reflection of 2mm under machine load and an estimated damping factor C = 0.2 are used. Determine (i) The force transmitted to the foundation (ii) The amplitude of vibration of machine in= 500 kg F0 = 2000 N N= 1500 r/min;

2450V1 + [2(0.2)(2.242)] 2

21rN

co = (1 — 2.2422)2 + [2(0.2)(2.242)] 2 .\11

C= 0.2,

= 797.73 N

x-

Fo lk

60

T—

Ft

—

Fo

— 157.07 rad/s 8st = 2 mm 41+ (2cr) 2

4 0 _ r2 + (2c02

4(1— r 2 )2 + (24'02

co„= \Iklm

co„= Vklm. 500 x 9.81 k = W18st — 41 2x10-3

k = 4.90 x 106 N/m X= 0.1210 mm

= 2.45 x 106 N/m. (iii)

= tan-

2Cr

1— r2

2.45 x106 •

= tan-1

(on

500

[2(0.2)(2.242) 1— 2.2422

= 12.55° a= tan-1 (2 Cr) = 41.86° p= (0— a) = 12.55° — 41.86° = — 29.31°

r—

= 70.02 rad/s

0) 157.07 — — 0.243 70.02 Co„

Ft= Fo

41+ (2cr) 2 V(1— r2 )2 + (2Cr)2

r

8.14

Mechanical Vibrations

Ft = 2000 •

V1+ [2(0.2)(2.243)]2

1 (2.21)2 (6.05 x 105 1

(1 — 2.24322 )2 + [2(0.2)(2.243)]2 = 2000 x 0.325 = 650 N (ii) Amplitude of vibration F0 /lc

X=

r2 )2 + (2Cr)2 2000/2.45 x 106 4.1305 = 0.197 x ie m 8.15 Using the isolator designed in Solved Problem 8.12, what, if any, mass should be added to the machine to limit its steady-state amplitude to 3 mm?

(125.7 rad )2 s) = 187.0 kg Hence 37.0 kg must be added to the machine. 8.16 Solve the Solved Problem 8.4 as if the isolator had a damping ratio of 0.13. The required transmissibility ratio is determined in Solved Problem 8.4 as T = 0.365. Using the equation developed in Solved Problem 8.8 with T = 0.365 and C = 0.13 leads to

1 — 2(0.13)2

r=

(0.365)2

The steady-state amplitude is calculated by X—

F0 k

\ l/2

1 2 — 11}2 1(1— 2(0.13)2[ (0.365) (0.365)2

4(1— r 2 )2 + (2Cr)2

When mass is added to the machine, for a fixed k, the natural frequency is decreased, and hence the frequency ratio is increased. Using the values calculated in Solved Problem 8.12, 0.003 m —

[ (0.365)2 —1]

7.11 x103 N 6.05 x 105 -11 1 (1_ r2 ,2) + [2(0.08) /]2

which is solved for r = 2.21 leading to kr 2 k m = —= 2 2 Can a)

(0.365)2 —11 [

(0.365)2

= 2.011 The maximum natural frequency and maximum isolator stiffness are determined as n,

—

co

209.4 rad =

2.011

— 104.1

rad

k = mco,2, = (20 kg) (104.1 rad )2

= 2.17 x 1051I

Vibration Control

8.17 For the isolator designed in Solved Problem 8.16, what is the steady-state amplitude of the experiment, an what is the maximum deformation is the isolator? Using the theory of Chap. 3 the steadystate amplitude of the experiment is X = YT(r, C.)= YT (2.011, 0.13) = (0.00025 m) (0.365) = 9.13 x 10- 5 m The maximum deformation in the isolator is the same as the steady-state amplitude of the relative displacement between the experiment and the table: Z = YA(2.011, 0.13) = (0.00025)

8.15

In view of the above, if the isolator is designed such that sufficient isolation is achieved at the lowest operating speed, the transmitted force must be checked at the highest operating speed. To this end, at the lowest operating speed, F0 = moed rad)2 = (0.25 kg m) (104.7 — = 2740 N T-

= 0.365 =

(2.011)2 4[1- (2.011)2j2 + [2(0.13)(2.010 = 3.27 x 10-4 m 8.18 A 200 kg turbine operates .at speeds between 1000 and 2000 r/min. The turbine has a rotating unbalance of 0.25 kg-m. What is the required stiffness of an undamped isolator such that the maximum force transmitted to the turbine's foundation in 1000 N? The rotating unbalance provides a frequency squared excitation to the machine of the form F0. = moed Thus the transmitted force is of the form FT = moeco2 T(r, C) As r increases above 42., T (r, decreases, However, since FT is also proportional to cot, the transmitted force decreases with increasing co until a minimum is reached.

1000 N 2740 N 1 r2

---> r= 1.93

104.7 rad

co

s

co"

1.93 = 54.2

rad

Checking at the upper operating speed, 209.4 rad s -3.86 54.2 rad

co r= co„ ,) • FT = 'Neff

1

r2

-1

)2 (0.25 kg-m)(209.411 s) = 789 N (3.86)2 _i Hence the isolator design is acceptable with 2

k = rriCo2 = (200 kg) (54.2 rad ) =.5.88 x 1051\1. m

8.15

Mechanical Vibrations

8.19 Repeat the Solved Problem 8.18 as if the isolator had a damping ratio of 0.1. The solution procedure is as described in Solved Problem 8.18. Setting T(r, 0.1) = 0.365, using the equation derived in Solved Problem 8.8, leads to

1-2(0.1)2 [

r=

(0.365)2 -11

2

N1/2

{2(0.1)2 [ (0.365)2- i ] (0.365)2 (0.365)2 -11 [ (0.365)2 = 1.98 104.7 rad s - 52.9 rad ej" = r - 1.98 s Checking the transmitted force at the upper operating speed, 209.4 rad co„

=

52., rad

=3.95

FT = moeco2T(3.95, 0.1) = (0.25 kg-m) (209.4

1+ [2(0.2)(3.95)12 [1 - (3.95)]2 -I- [2(0.1)(3.95)12 = 9.55 N

= 560 x 10

5N

8.20 Repeat the Solved Problem 8.19 as if the upper operating speed were 2500 r/min.

(0.365)2

r=

Hence the isolator design is acceptable with 2 rad = (200 kg) 1r k=m s)

The transmitted force at the upper operating speed for the turbine with the isolator design of Solved Problem 8.19 is calculated as 261.8 rad s. - 3.95 r= = o.)„ 52., rad s FT = moedT(4.95, 0.1) rad = (0.25 kg-m) 261 8 s ( I \

2

. 1 + [ 2 (0.1) (4.95)12 [1 _ (4.95)2 ]2 + [2(0.1)(4.95)12

= 1025 N Thus this isolator is not acceptable. An isolator cannot be designed such that sufficient isolation is achieved over the entire operating range. 8.21 Measurement indicates that .the peak components of the -table vibration of Solved Problem 8.4 are a 0.25 mm component at 100 rad/s and a 0.4 mm component at 150 rad/s. An available isolator has a stiffness of 8 x 104 N/m and a damping ratio of 0.1 Will the acceleration felt by the apparatus exceed 6 m/s2 when this isolator is installed?

8.17

Vibration Control

Let r = 100/ w,,. Then 150/co„ = 1.5r. An upper bound on the acceleration felt by the apparatus is A

a

2

5_ (0.00025 m) (100 rau ) T(r, 0.1) s 2

+ (0.0004 m) (150

= 2.5

A rad)

T(1.5r, 0.1)

1+ [2(0.1)/12

=

[1- (1.5 02 ]2 + [2 (0.1)(1.5r)] 1+ 0.04r2 r4 -1.96r2 +1 1+ 0.09r2 5.06254 - 4.41 r2 + 1

If the proposed isolator is used, then 8.x 104 11

r

Thus for 81 percent isolation with an isolator of h = 0.2 1+ (0.2)2

0.19 =

1+ [2(0.1)(1.5r)]2

+9

ct)"

1 + h2 r2)2 +h2

T1, (r, h) =

r2

)2 + (0.2)2

\ (1= r2 )2 + [2 (0.1)/12

+9\

= 2.5

The appropriate form • for the transmissibility ratio for a system with hystere damping is

=

=

20 kgm

- 63.2

rad s

100 rad • s - 1.58 . rad

63:2

and the upper bound on the acceleration is calculated as 3.86 m/52. Thus the available isolator is sufficient. 8.22 Repeat the Solved Problem 8.1 as if the isolator damping is assumed to be hysteretic with a hysteretic damping coeffidient of 0.2.

1.04 r4 - 2r2 +1.04

(0.19)2 (r4 - 2r2 + 1.04) = 1.04 which is solved yielding r = 2.52. Hence the maximum allowable natural frequency is 104.7 raa rad s -41.5 s. a)" = = 2.52 from which the maximum isolator stiffness is calculated as k = mco,2, =

(200 kg) (41.6 rad l

E

= 3.45 x 105 Solved Problems 8.23 through 8.27 refer to the following: During testing, a 150 kg model of an automobile is subject to a triangular pulse, whose force and displacement spectra are shown in Fig. 8.11. 8.23 If the model is mounted on an isolator of stiffness 5.4 x 105 N/rn and damping ratio 0.1, what is the maximum transmitted force and maximum model

8.18

Mechanical Vibrations

The value of the parameter on the horizontal scale of the spectra is

Fo to

(60 rad )(0.12 s) s = 1.15 27r 2.7r From the force spectra, FT/Fo = 1.30, hence FT = 1.30Fo = (1.30)(2500 N) = 3250 N From the displacement spectrum, kx-max/F0 = 1.35, hence con to

.6 .2 0.8 0.4 0.0

0

2

3

(onto /2/r

F0

;flax = 1.35 -k

1.35 (2500 N)

c=0.1 - - C=0.3 - c=0.5 (a) 1.6

,-- - - „

1.2

t 1

M 0.8

..---

--- --.:.:-

/,--- ------- '

-----

7

0.4 0.0 0

1

N

m

= 6.25 mm 8.24 What is the maximum stiffness of an isolator such that the maximum transmitted force is less than 2000 N for a pulse of magnitude 2500 N and duration 0.12 s? It is desired to set FT/Fo = 2000/2500 = 0.8. From the force spectra, this corresponds to a horizontal coordinate of 0.3. Hence,

,1( ii

•SC

5.4x105

1

co„ to

• cu,;to

27r

- 0.3,

0.3(27r)

co„ = 0.12 s

c=0.1 - - =0.3 - C=0.5 (b)

Fig. 8.11

2

displacement .for a pulse of magnitude 2500 N and duration 0.12 s? The system's natural frequency. is 5.4 x 105 IT

=\

150 kg

= 15.71 rad s The maximum allowable stiffness is

rad = 60 s.

k = mco2 = (150

kg) (15.71 Eal)

s

= 3.70 x 104 8.25 What is the maximum displacement of the model with the isolator of Solved Problein 8.24 installed?

849

Vibration Control

For a horizontal coordinate of 0.3, kxmax/F0 = 0.8, hence, F0 x max. 0.8 k =

0.8(2500 N) 3.70 x 104

— 0.054 m

m

From Eq (8.11), E(0.14) = 1.39. Notethat the range of the inverse—ignt function is taken from 0 to 7r. Then 1 MV2 — xmax = 1.39 2 FT

8.26 If the model is mounted on an isolator of stiffness 5.4 x 105 N/m and damping ratio 0.14, what is the maximum transmitted force if the pulse has a magnitude of 3000 N and a duration of 0.01 s? The natural frequency for the model on this isolator is 60 rad/s, thus the natural period is 0.105 s. Since the pulse duration is much smaller than the natural period, a short duration pulse assumption is used. The magnitude of the impulse is the total area under the force time plot: o.ois 1 I = J F(t)dt = 2 — (0.005 s) (3000 N) 2 = 15 N-s From Eq. (8.10), Q (0.14) = 0.848, thus FT = 0.848/co„

20 (150 kg) 1m ( — s = 1.39 763 N = 1.37 mm 8.28 The 120 kg hammer of a 300 kg forge hammer is dropped from 1.3 m. Design an isolator for the hammer such that the maximum transmitted force is less than 15,000 N and the maximum displacement is a minimum. The velocity of the hammer upon impact is vh =

2gh = \12 9.81 + r ,1)(1.3 m)

m = 5.05 — The principle of impulse and momentum is used to determine the velocity of the machine induced by the impact:

= 0.848 (15 N-s) (60 1.-al )

8.27 What is the model's maximum displacement for the situation described in Solved Problem 8.26? The velocity induced by the application of the impulse is v=

Ill

=

15 N-s 150 kg

(120 kg) (5.05 ml v„, —

= 7.63 x 102 N

m = U.1 —

sh

(

mhv h

300 kg m = 2.02 —

For a specified transmitted force, the minimum maximum displacement is attained by choosing C = 0.4. Then the required natural frequency is obtained by FT mvco„

— Q(0.4)

8.20

Mechanical Vibrations

Using the minimum allowable stiffness, the required absorber mass is

Fr (Dti =

mvQ(0.4) 15000 N

k2

(300 kg) (2.02 11) (0.88)

07 2

rad = 28.1— s Thus the maximum allowable stiffness is 2 k = 1n (02, = (300 kg) 1\28.1 rad ) = 2.37 X 10511 8.29 A 200 kg machine is attached to a spring of stiffness 4 x 10' N/m. During operation the machine is subjected to a harmonic excitation of magnitude 500 N and frequency 50 rad/s. Design an undamped vibration absorber such that the steady-state amplitude of the primary mass is zero and the steadystate amplitude of the absorber mass is less than 2 mm. The steady-state amplitude of the machine is zero when the absorber is tuned to the excitation frequency. Thus k2 \ I — = co • r2 = 1 --, o)22 = (0 m2 When this occurs, the steady-state amplitude of the absorber mass is given by Eq. (8.17). Thus F0 0.002 m — k2 k2 5N = 2.5 x 10

500 N 0.002 m

2.5 x105 N 2

(50 rad) s )

= 100 kg Thus an absorber of stiffness 2.5 x 105 N/m and mass 100 kg can be used. 8.30 What are the natural frequencies of the system of Solved Problem 8.29 with the absorber in place? The natural frequencies of the 2-degree-of-freedom system with the absorber in place are the values of w such that the denominator of Eq. (8.12) is zero. Thus, noting that the mass ratio is it = 100/200 = 0.5. 2 r2 r2 —1.5 r2 1 = 0 r1

2 2 C°11C°22

1.5 + 0)2 H 2 2 C°11 C°22

_ (1.5 co222

0) 11

(01(0 22

=

0

=0

It is noted that CO22 = 50 rad/s, and

coil=

4x1051± 1 m =1 200 kg ln

Ic

rad s Substitution of these values leads to — 5.75 x 103 co2 + 5 x 106 = 0 which is solved for a)2 using the quadratic equation. Taking the positive square roots of the roots leads to = 44.72

col = 32.698

rad

,

(02 = 68.42 rad

8.21 7

Vibration Control

8.31 A piping system experiences resonance when the pump supplying power to the system operates at 500 r/min. When a 5 kg absorber tuned to 500 r/min is added to the pipe, the system's new natural frequencies are measured as 380 and 624 r/min. What is the natural frequency of the piping system and its equivalent mass? The system has natural frequencies corresponding to values of w that makes the denominator of Eq. (8.12) zero. Using the definitions of Eqs. (8.14) and (8.15), this leads to cds _ (0 2 ± /1 + \ 0)2 co2 L n W II C°22

(8.30)

=

Noting that rad 422 = 50 r/min = 52.4 and applying Eq. (8.30) with co = 380 r/min = 39.8 rad/s leads to 2.51 x 106 + 1.17 x 103 41 - 4.35 x 106 (1 + =0 Application of Eq. (8.30) for co = 624 r/min = 65.3 rad/s leads to 1.82 x 107 - 1.51 x 103 cqi - 1.17 x 107 (1 + it) = 0 Simultaneous solution of the previous two equations leads to 411 = 49.2

rad

,

it = 0.225

Hence the piping system's natural frequency is 49.2 rad/s, and its equivalent mass is m1 =

" = 05. 2k2g5

8.32 Redesign the absorber used in Solved Problem 8.31 such that the system's natural frequencies are less than 350 _ r/min and greater than 650 r/min. Applying Eq. (8.30) of Solved Problem 8.29 with w = 350 r/min = 36.7 rad/s, co = 49.2 rad/s, and 422 = 52.4 rad/s leads to µ = 0.414. Applying Eq. (8.30) of Solved Problem 8.31 with co = 650 r/ min = 68.1 rad/s, with the same values of coil and w22 leads to II= 0.330. Thus in order for the natural frequencies of the system with the absorber added to be less than 350 r/min and greater than 650 r/min requires an absorber mass of atleast m2 = pm, = (0.414) (22.2 kg) = 9.19 kg Then the absorber stiffness is rad )2 K2 = m20)222 = (9.19 kg) 1 52.4 = 2.52 x 1041\-I8.33 A 100 kg machine is placed at the midspan of a simply supported beam of length 3 in, elastic modules 200 x 109 N/m2, and moment of inertia 1.3 x 10-6 m4. During operation the machine is subjected to a harmonic excitation of magnitude 5000 N at speeds between 600 and 700 r/min. Design an undamped vibration absorber such that the machine's steady-state amplitude is less than 3 mm at all operating speeds. The beam's stiffness is k= 48E1 L3 48(200 x 10'

- 22.2 kg

2 M

(1.3 x 10-6 m )

(3 m)3 = 4.62 x 105 11

Mechanical Vibrations

The system's natural frequency is

con =

ki

4.62 x 105 14=

1

100 kg

- 68.0

p = 0.652, rad

2

= 3.014 x 105 11 m

rad con = co22 = 68.0 7

1- (0.923)2

= µm1 = (0.652)(100 kg) = 65.2 kg

k2 = ni2 o42 = (65.2 kg) (68.0 -12:1)

Assume that steady-state vibrations are to be eliminated at this speed, then

Note that in Eq. (8.12), for r < 1, the numerator is positive and the denominator is negative, hence, in order to enforce, Xi. < 3 mm for w = 600 r/min = 62.8 rad/s with ri = r2 = 628/68.0 = 0.923, 5000 N -0.003 m = 4.62 x 105 m

11 / 2

8.34 If an optimally designed damped vibration absorber is used on the system of Solved Problem 8.33 with a mass ratio of 0.25, what is the machine's steady-state amplitude at 600 r/min? The optimum absorber tuning is obtained from Eq. (8.19) as 1 1 = 0.8 1+µ 1+0.25 The optimum damping ratio is calculated. using Eq. (8.20); q=

(0.923)2 (0.923)2 - (0.923)2 - (1 + µ) (0.923)2 + 1 which is solved for it = 0.652. For r2 > 1, both the numerator and denominator of Eq. (8.12) are negative. Hence, for w = 700 r/min =.73.3 rad/s and ri = r2 = 73.3/68.0 = 1.078, 0.003 m -

5000 N 4.62 x 105

m

1- (1.078)2 (1078)2 (1.078)2 - (1.078)2 - (1+ /2)(1.078)2 +1 which is solved for p. = 0.525. Since the mass ratios calculated represent the minimum mass ratios for the amplitude to be less than 3 mm at the limits of the operating range, the larger mass ratio must be chosen. Hence

3(0.25) 1,1 8(1+0.25)

- 0.274

Using Eq. (8.18) with these values and ri = 0.923, F0 = 5000 N and ki = 4.62 x 105 N/m leads to Xi = 2.9 cm. 8.35 . A 300 kg machine is placed at the end of a cantilever beam of length 1.8 m, elastic modulus 200 x 109 N/m2, and moment of inertia 1.8 x 10-5 m4. When the machine operates at 1000 r/min, it has a steady-state amplitude of 0.8 mm. what is the machine's steady-stateamplitude when a 30 kg absorber of damping coefficient 650 N-s/m and stiffness 1.5 x 105 N/m is added to the end of beam?

8.23

Vibration Control

The beam's stiffness is 3E1 k= L3

q=

= 1.85 x 1061\1 m and the system's natural frequency is

= IT = \

300 kg

m

= 78.5

rad

The frequency ratio for co = 1000 r/min = 104.7 rad/s is rad 104.7 co ri = = — 1.33 W1 1 78.5 rad The excitation amplitude is calculated from knowledge of the steady-state amplitude before the absorber is added: Fo = k2X1 ( ri2 —1) = (1.85 x106 m ) (0.0008 m) [(1.33)2 — 1] = 1.14 x 103 N The natural frequency of the absorber is

f°22 =

k2 = m

= 70.7

c°22

78.5

11

3(200 x 109 -1\1) (1.8 x 10-5 m4 ) m2 (1.8 m)3

1.85 x 106

70.7

1.05 x105 i\l m 30 kg

rad

Thus the parameters of the absorber design are m2 30 kg = -0.1, m1 300 kg

=

rad rad

= 0.90

2v m2 k2 650

—

N-s m

—0.153

2\I (1.5 x 105 1(30 1 kg) m Application of Eq. (8.18) with these values leads to X1 = 9.08 x 10 -4 m. 8.36 An engine has a moment of inertia of 3.5 kg-m2 and a natural frequency of 100 Hz. Design a Houdaille damper such that the engine's maximum magnification factor is 4.8. If the optimum damper design is used, then setting the maximum magnification factor to 4.8 and using Eq. (8.24) leads to 4.8= + 2,

it = 0.526

The optimum damping ratio is determined from Eq (8.23) as 1 — 0.360 4 2(1.526) (2.526) The Houdaille damper parameters are determined from Eq. (8.22) as = p.11 = (0.526)(3.5 kg-m2) = 1.84 kg-m2 c = 2c J2col = 2(0.360)(1.84 kg-m2) 100 cycle) ( 2/r rad) s ) cycle ) = 832

N-s

8.24

Mechanical Vibrations

8.37 During operation the engine of Solved Problem 8.36 is subjected to a harmonic torque of magnitude 100 N-m at a frequency of 110 Hz. What is the engine's steady-state amplitude when the Houdaille damper designed in Solved Problem 8.36 is used? The frequency ratio is r -

0) 110Hz = 1.1 oh

100 Hz

The frequency ratio is r= D 27.c rad ) (1 min (1000 : min) ( r 60 s

89.4

rad =1.17

The amplitude of whirling is calculated using Eq. (8.26): (0.012 m) (1.17)2

X=

J[1- (1.17)2 ]2 + [2(0.07) (1.17)]2

Thus from Eq. (8.21),

= 4.07 cm =

41,2 r2 4c2 (r2 gr2 1)2 (r2-1)2 r2

Mo

100 N-m s2 (3.5 kg_m2 )[100 „le rad]

8.39 A engine flywheel has an eccentricity of 1.2 cm and mass of 40 kg. Assuming a damping ratio of 0.05, what is the necessary stiffness of its bearing to limit its whirl amplitude to 1.2 mm at all speeds between 1000 and 2000 r/min? The maximum allowable value of A is X.

4(0.360)2 + (1.1)2 4(0.360)2[(1.1)2 + 0.536(1.1)2 - 112 + [(1.1)2 - 1]2 (1.1)2

1.2 mm - 0.1 e 1.2 cm Then using Eq. (8.26), Amax =

0.1
C4 = du

4

(0) = u2 C3 = U2

u(e) = u3 du

0

c,e c2e

•

=

U4 —7-74-.11.2

-1

2

t2

3 42 -[

2 °)til3 t3

.e 2

43

42

▪ — (

+16, ,3 -

t2

leti4

+ Q3

The above expreision is substituted into Eq. (9.12). The term that leads to m34 in the element mass matrix is 42 V 21 pA(3 — 2 — u3 e2 .e3 0

V V —

2

t3

fli4d4

= 2pAeti3 U4

cie + c4 = u3 a...3 = U4

d4

Solving the last two of the previous equations simultaneously leads to 1 C1 = — (2u1 + tu2 — 2u3 + tu4) .e3

— 2.eu2 — 3u3 + eu4)

C2 = f2

4-3 +5 .4

0

45

t5

2

t6

3 5 2) = 2piteU3 (— — t — 5

11 pii,e 2 u3 1:44 105

d

9.12

Mechanical Vibrations

When the quadratic form of the kinetic energy is expanded, it includes a term 2717341%3 U4 . Thus 11 "134 = T 1 0 fPAe 9.11 Use a one-element finite element model to approximate the lowest natural frequency and mode shape for a uniform fixed-free beam.

= 1.622

— 0.622 "="— LZ L3 which is illustrated in Fig. 9.11

0.8 0.6 u(x) 0.4

Since the slope and deflection at the fixed end is zero, u i = u2 = 0. Thus, a one-element model of a fixed-free beam has 2-degrees-of-freedom. The global mass and stiffness matrices are obtained by simply setting 111 = tr2 = 0. This is accomplished by eliminating the first and second rows and columns of the element mass and stiffness matrices of Eqs. (9.16) and (9.17). Thus pAL[ 156 —22L M—

K=

0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 x

L

Fig 9.11 9.12 Determine the global stiffness matrix for a two-element finite element approximation to the system of Fig. 9.12.

420 —22L 4L2 [ 12.-6L

L3 —6L

4L2

L

I El p AL4

The eigenvector corresponding to this first mode is u = [1 1.378/./]T. Thus from Eq.(9.14) the mode shape approximation is 2 3 X X u(x) = 3 — — 2 — L2 L3 X2 4_ X3 ) L (1.378) L2 L3 L )

p, A, E, I

-4

2 Ui

The natural frequency approximations are the square roots of the eigenvalues of M-11{. The lowest natural frequency is approximated as col = 3.53 \

(U4

(U2

Fig. 9.12 The two-element model of the beam of Fig. 9.12 has 4-degrees-of-freedom with the global definitions of nodal displacements illustrated. Then for element 1 U 1 = 0, u 2 = 0, 113 = U1 , u4 = U2. The contribution to the global stiffness matrix from element 1 is 12 —6,e 0 0 K1

EI —62 412 _ 0 e3 0 0

0 0 0 0 0 0

9.13

Finite Element Method

Now consider element 2: u1 = U1, u2 = U2, 113 = U3, u4 = U4. The stiffness matrix for element 2 must be modified to account for the potential energy developed in the spring. 1 EI u 2 EI u 2 7 33 1 2823 3 Thus the element stiffness matrix for element 2 is 1 2 VS = .iku3

12

62 4f2

6f

K = EI 2

—12

e3

—6f

2e

6e

6f

—12

2e

—6e

12.125 —6f

4e

—6t

L = — (3u2 —9u4 ) 64 Hence the kinetic energy of the block is L T - - m L- (-31(2 — 9 /144 b- 2 64 1 mL2 (9u2 - 54/42 ti4 +81/44) 2 4096 2 Noting that in the global system U1 = u2 and U2 = 114, the global mass matrix becomes M—

p AL [ 4L2 —3L2

420l.—3L2 4L2

Hence the global stiffness matrix is +

24

—12

0

6f

4f2 —6e 2e2 EI 0 L3 —12 —6f 12.125 —6e

K= —

6e

2t2

—6e

4e2

9.13 Determine the mass matrix for a oneelement finite element approximation for the system of Fig. 9.13.

4


k

4 --1

p AL3

9 —27] 81 2(4096) L-27

= p AL3

[ 0.0106 —0.0104 —0.0104 0.0194

9.14 Use a two-element finite element model to approximate the two lowest natural frequencies for the system of Fig. 9.14. (U3

0

0

PAL

1.2 m

m= 2

>I
x

+

= 0.0112

r=

10.7 Determine the nature and stability of the equilibrium positions of the system of Fig. 10.7.

Duffing's equation for free vibrations with /1 = 0 is + x + ex3 = 0

Mechanical Vibrations

10.8

Define v = x. Then

fio + xo = 0 —> xo = A sin (t + 0)

dv Vdx- + x + ex3 = 0

xl + x1 = —A3 sin3 (t + 0) Use of trigonometric identities leads to + xi . -4

1

Integrating with respect to x leads to 2 E 4 1 2 — v +1X + — X =C 2 2 4 Application of initial conditions and solving for v leads to

3 [3 sin (t + 0) 11 + x1 = — L'1— 4 — sin 3 (t + 4))] x1 = 8A3 t cos(t + (p)

V = .\IX20 + — 82 X4 8 X4 0 — X2 — — 2

A3 sin 3 (t + 0) 32 Hence, x(t) = A sin (t + 0) +

Noting that v = dxldt,

—

dx

dt = +

8 X4 X20 2 0 — X2 — — 2 I One-quarter of the period is the time the block returns to x = 0 from its initial position. During this time the velocity is negative. Hence integrating between x0 and 0 leads to 0

dx

T=4 xo _ \x02

_ x2 _

x4

10.9 Use a straightforward perturbation expansion to develop a two-term approximation to the solution of Duffing's equation with F = 0. Assume (10.7) x = x0(t) + £x1(t) + 0(E2) Substituting Eq. (10.7) into the unforced Duffing's equation leads to +

+

• + X0 + exi + • • •

+ e(xo + Ex + • • • + )3 = 0 + x0 + E( + x1 + x03) + 0(E2) = 0 Setting coefficients of like powers of £ to zero independently leads to

3 A3 tCOS(t+4))

[8

—

A3

32

sin 3 (t + 0)] 0(6 2)

10.10 The solution developed in Solved Problem 10.9 is not periodic. Why, and what can be done to correct the situation? In Solved Problem 10.8 it is shown that the natural period of the nonlinear system is dependent on initial conditions. The perturbation solution of Solved Problem 10.9 has no mechanism to allow for this dependence. Indeed, the response is developed at the same period as a linearized system. The situation can be corrected by the introduction of a time scale that is dependent on the amplitude: t = w(1 + EX' + e2A2 + • • • +) The above expression can be introduced before making the straightforward expansion, in which case the method is called the Linstedt-Poincare method. It can also be introduced after the aperiodic straightforward expansion is obtained in an effort to render it periodic. This latter

Nonlinear Systems method is called the method of renormalization. In either case the results are 3 E42 + • • • +) t = 41x = A sin (t + 0) -

A3

£32 sin

3

(t + 0) + • • • + 10.11 The block of the system of Fig. 10.7 is displaced 1.0 mm from equilibrium and released. Determine the period of the resulting motion. Using the nondimensionalization of Solved Problem 10.6, the nondimensional initial conditions are 0.001 m x(0) = 5.10, 1.96 x 10-4 m (0) = 0 Application of the initial conditions to the two-term uniform expansion developed in Solved Problem 10.10 and noting from Solved Problem 10.5 that E = - 0.0384 lead to

o=

2

EA3 sin 37r x(0) = A sin 2 32 2 5.10 = A - 0.001243 -> A = 5.28 The nondimensional frequency is

w = 1 + 3EA2 = 1 + 8 8 (-0.0384)(5.28)2 = 0.599 The dimensional frequency and period are rad ) con = 0.599 (223.6 s rad = 133.9 T = 0.074 s

10.9

10.12 Use a perturbation method to approximate the forced response of the system of Fig. 10.7. To avoid sign confusion, define S = E. Since the damping is small, it is ordered with the nonlinearity. To this end

oc

µ = 0.0112 = 0.292

0.0384 Then the equation becomes 1 + 0.584 Sic + x - 8x3 = 0.510 sin 0.671t A straightforward perturbation solution is assumed as x(t) = x0(t) + oxi(t) Substitution into the governing equation and setting coefficients of like powers of 45 to zero lead to 10 + xo = 0.510 sin 0.67 it xo =

0.510 sin 0.671t 1- (0.671)2

= 0.928 sin 0.671t + xt = - 0.584 + = - 0.364 cos 0.67 1t + 0.799 sin3 0.671t = - 0.364 cos 0.671t + 0.599 sin 0.671t - 0.200 sin 2.103t xi (t)

0.364 cos 0.6711 1- (0.671)2 0.599 sin 0.671t 1- (0.671)2 0.200 sin 2.103t 1- (2.013)2 = - 0.662 cos 0.671t + 1.09 sin 0.671t + 0.0655 sin 2.103t

10.10

Mechanical Vibrations

10.13 Discuss quantitative tools that can be used to determine if the motion of a nonlinear system is chaotic. (a) The trajectory in the phase plane will not repeat itself for chaotic motion. (b) If a spectral analysis of the time history of motion yields a continuous spectrum, the motion is chaotic. (c) If the response is sampled at regular intervals, the sampled response of a chaotic motion will appear to be random. 10.14 The Runge-Kutta method has been used to develop the phase planes for Duffmg's equation for various values of the parameters, as illustrated in Fig. 10.8. Which of these motions appear to be chaotic?

e= 4.500 4= 0.00 A = 3.40 R= 1.30

The motion in Fig. 10.8a appears to be chaotic as there is no discernible pattern to the motion. The motion in Fig.10.8b is not chaotic as it settles down into a steady state after an initial transient period. 1

10.15 The Runge-Kutta method has been used to develop Poincare sections for the solution of Duffing's equation, shown in Fig. 10.9. Poincare sections are samples of the phase plane at regular intervals. Comment on the motion for each Poincare section. (a) Since the Poincare section is a collection of apparently random points, the motion is probably chaotic. (b) Since the Poincare section is a closed curve, the motion is periodic, but the sampling frequency is incommen-

x

e= 1.100E + 00

c=0.10 A =1.30 R= 1.20

(b)

Fig. 10.8 surate with the frequency of motion. (c) Since the Poincare section only consists of three points, the motion is periodic, and the period of motion is three times the sampling period.

Nonlinear Systems

When x is small, the coefficient multiplying ± in van der Pol's equation is negative. Thus energy is being added to the system through self-excitation. This causes the response to grow. However, when x grows above 1, the damping coefficient becomes positive, and energy is dissipated causing the motion to decay. This continual buildup and decay of amplitude through self-excitation lead to the limit cycle. This limit cycle is independent of initial conditions.

y

••

s = 1.000 E—.01 c=0.00

10.11

..• . ::•

A= 1.00

r = 1.05

10.17 Show how the method of averaging, or the Galerkin method, can be used to approximate the amplitude of a limit cycle. Let F(x, i) represent the nonconservative forces in the system. The work done by these forces over 1 cycle of motion is

E= 0.00 c= 0.00

W=

1=0.00

r= too

F(x,

dx = F(x, x) x dt

If the system develops a limit cycle, the total work done by the nonconservative forces over each cycle is zero. Assume the system is nondimensionalized such that its linear period is 27r then

y

/—

F(x,

dt = 0

(10.7)

x

E= 1.000 E c=0.10

(c)

= 1.00 r = 1.05

Fig. 10.9 10.16 Use the van der Pol equation to qualitatively explain the phenomenon of limit cycles.

When the Galerkin method is used, a response such as x(t) = A sin t is assumed and substituted into the work integral Eq. (10.7). The integral is evaluated, yielding an approximation to the limit cycle amplitude A. 10.18 Use the method of averaging to approximate the limit cycle of the system governed by - the nondimensional equation + a(i2 + x2 — 1)± + x = 0

10.12

Mechanical Vibrations

2/1-

Application of the method of Solved Problem 10.17 using x(t) = A sin t leads to _ F(x, ic), a(±2. + x2

a [A2 cos2 t + A2 sine t — 1] (A cos t)2 dt = 0

2n.

a(A2 — 1)A2

cos2 t dt = 0

F(A sin t, A cos t)A cos t dt = 0 ,ca (A2 — 1)A2 = 0 A=1

PRACTICE PROBLEMS 10.1 Develop the general equation for the trajectory in the phase plane for a system governed by + x + Ex cos x = 0 10.2 Develop the general equation for a trajectory in the phase plane for a system governed by the equation 1 + x — ax2 = 0 10.3 Determine the equilibrium points and their type for the system of Practice Problem 10.2. 10.4 Sketch the phase plane for the system of Practice Problem 10.2. 10.5 Determine the equilibrium points and their type for a system governed by 1+ — x + Ex2 = 0 10.6 Determine the equilibrium points and their type for a system governed by + 2ci — x + EX3 = 0 10.7 Derive an integral expression for the period of motion of the nonlinear system governed by

0 + sin 0(1 — cos 0) = 0 subject to 0 = 00 when 6 = 0. 10.8 A 50 kg block is attached to a spring whose force displacement relation is F = 2000x + 6000x3 for x in meters and F in newtons. The block is displaced 25 cm and released. What is the period of the ensuing oscillations? 10.9 Use the perturbation method to obtain a two-term approximation to the response of a system governed by +x+

+

i = \IC - x2 — 2x sin x — 2 cos x

sin wt

10.10 Use Galkerkin's method to approximate the amplitude of the limit cycle of van der Pol's equation, 10.11 Explain the jump phenomenon from Fig. 10.2. 10.12 Discuss how the Fourier transform of a response can be used to determine if the response is chaotic.

ANSWERS TO PRACTICE PROBLEMS 10.1

EX2 = F

10.2 =

x2

3

ax3

10.13

Nonlinear Systems

10.3 x= 0 is a center; x= a is a saddle point. 10.5 x = 0 is a stable focus for C < 1 and a stable node for C > 1; x=-1/E is a saddle point. 10.6 x = 0 is a saddle point; x = ± are stable foci for C < and stable nodes for e> .

10.8 0.907 s 10.9

F

„a

10.7 T= 4 I eo

dO

— —I cos 2 eo + 2 cos eo 2 1 + — cos 20 — 2 cos 0 2

sin cot + e

1- 02 1

(1 1-402 10.10 2

[

1 2 (i Fc02)

cos 2 cot

pcoF

(1—co2 )2

cos wt

Chapter Eleven Computer Applications

11.1 INTRODUCTiON Vibration analysis often requires much mathematical analysis and computation. Digital computation can be used in lieu of manual computation for many of the tedious tasks performed in vibration analysis. Computer algebra can be used to perform tedious mathematical analysis. However, the user must understand the sequence of the steps and how the results are used. The focus of this chapter is the use of applications software for vibration analysis. It is worthwhile to know how to program using a higher-order programming language such as C, PASCAL or FORTRAN, and programs can be written in these languages to solve many vibrations problems. However, much of the analysis used in the preceding chapters can be performed on personal computers using applications software. The finite element method, a powerful method for approximating the solution of continuous vibrations problems when an exact solution is difficult to attain, is illustrated in Chap. 9. However, for the sake of illustration and for brevity, the examples presented here use at most four elements. When more elements are used, digital computation is essential in obtaining a solution. Many difficulties are encountered in the development of a large-scale finite element model. These range from efficient methods of assembly of the global mass and stiffness matrices to solution of the resulting differential equations using modal analysis. Thus large-scale finite element programs have been developed. Some are available for use on the personal computer. However, they often require pre- and post-processor programs and are beyond the scope of this book.

11.2

Mechanical Vibrations

11.2 SOFTWARE SPECIFIC TO VIBRATIONS APPLICATIONS Software written specifically for vibrations applications is available. The programs in the software package VIBES, which accompanies the McGraw-Hill text Fundamentals of Mechanical Vibrations by Kelly, include programs that simulate the free and forced response of 1- and multi-degreeof-freedom systems. VIBES also has programs that numerically integrate the convolution integral, develop force and displacement spectra, perform modal analysis for continuous systems, and aid in the design of vibration isolators and vibration absorbers. Many of the files are executable programs while several require user-provided BASIC subprograms to allow for any type of excitation.

11.3 SPREADSHEET PROGRAMS Spreadsheets allow the development of relationships between variables and parameters in tabular form. Spreadsheets also have graphical capabilities for presentation of results. The columns and graphs in a spreadsheet are automatically updated when the value of a parameter is changed. Thus the spreadsheet is a useful tool in "what-if' situations such as design applications. Examples of popular spreadsheets are Lotus Development Corporation's Lolus 1-2-3, Microsoft's Excel, Borland's Paradox, and WordPerfect's Quatro Pro.

11.4 ELECTRONIC NOTEPADS When using an electronic notepad, the user develops the solution on the computer screen as if she or he were using pen, paper and calculator. Electronic notepads such as MathSoft's Mathcad and The Math Works, Inc.'s, MATLAB provide mechanisms for performing complex sets of calculations. Electronic notepads have built-in algorithms that allow the user to quickly perform complicated calculations. These include numerical integrations and matrix eigenvalue algorithms Electronic notepads also have automatic update, so that when the value of a parameter is changed, all subsequent calculations involving the parameter are recalculated. Electronic notepads also have graphical capabilities and allow for limited symbolic processing.

11.5 SYMBOLIC PROCESSORS Symbolic processors such as MAPLE V, MACSYMA and Mathematica perform symbolic manipulations. Examples of symbolic manipulations include differentiation with respect to a variable, indefinite integration, partial fraction decompositions, and solving equations for solutions in terms of parameters. Computer algebra software can also be used for linear algebra and solutions of differential equations.

11.3

Computer Applications

SOLVED PROBLEMS 11.1 Use VIBES to plot the response of a 1-degree-of-freedom system of mass 100 kg, natural frequency 100 rad/s and damping ratio 0.3 subject to the excitation F(t) = 1000 sin 125t N The VIBES program FORCED is used to develop the response as shown in Fig. 11.1. The excitation is plotted simultaneously with the response for comparison. The plot illustrates the transient response giving way to a steady-state response. The plots also illustrate the difference in period between the excitation and response and the phase difference.

F(t)lk Displacement —4

Fig. 11.1 11.2 Use VIBES to determine approximations to the natural frequencies and mode shapes of a uniform fixed-pinned beam when 4-degrees-of-freedom are used to model the beam.

The 4-degree-of-freedom model is illustrated in Fig. 11.2. The flexibility matrix is obtained using the VIBES program BEAM. Unit values of beam properties are used as input in BEAM. Thus the numerical values obtained in this example must be multiplied by L3/E/ to obtain the elements of the flexibility matrix. The output from BEAM is shown in Fig. 11.3. The VIBES program MITER uses matrix iteration to determine natural frequencies and normalized mode shapes of a multi-degree-of-freedom system. The flexibility matrix obtained from BEAM is used as input as well as the mass matrix 1 0 0 0 5 1 0 0 0 5 pAL M= 1 0 0 0 5 1 0 0 0 5 Again unit values of the properties are used. Hence the numerical values shown in Fig. 11.4 obtained using MITER in this example are nondimensional. The dimensional natural frequency approximations are obtained by multiplying these values by EH pAL4. The mode shape vectors determined using WIER have been normalized with respect to the mass matrix. L L L 5 —4-1-4— 5 >14 5 4.1

L L 5 —04-readlib(Heaviside); proc (x) ... end Excitation: >F := t->F0*exp(-alphal)*(Heaviside(t)-Heaviside(t-1/alpha)); F := t -> FOe'l(Heaviside(t)-Heaviside(t - )) Differential equation: >eq := (D@ @2)(y)(t) + omega' 2*y(t) = F(t); r eq = D(2)(y)(t) + co2y(t) = F0e- at(Heaviside(t)-Heaviside(t

11

Solution to differential equation using laplace transform: >dsolve (eq, y(t), laplace); y(t) = y(0)o)2 cos(wt) y(0)a2 cos(wt) D(y)(0)cosin(e)t) D(y)(0)a2 sin(o)t) %1 %1 %1 %1 FO a sin(cot) FO a cos(cot) FO eaat %1 To1o) %1

+ invlaplace

+ invlaplace

FO laplace(-Heaviside(at -1), t, s + ajs a s, t 2 3 2 2 S + s CC + CO S + CO 0C

FO laplace -Heaviside(at 1), t, s + aja a ( , s, t 3 2 2 2 S + S 0C + (0 S + CO CC

%1 := a2 + o>2 Initial conditions: >subs({y(0) = 0,D(y)(0) = 0), "); Y(t) =

FO a sin(wt) FO cos(cot) FO e-ut ÷ (02 + a2 + (02 (a2 (02)(0 I FO laplace(-Heaviside(at -1), t, s + or)s a , s, t + invlaplace s3 s2a + (02s + (02u

+ invlaplace

FO laplace(-Heaviside( at -1), t, s + a a a ) , s, t s 3 s2a + 2S + (02a

Note the following: (1) laplace(Heaviside(t-1/alpha))=exp(-s/alpha)/s; (2) Replacing s by s+alpha in the above leads to exp(-s/alpha-1)=exp(-1)exp(-s/alpha)/(s+alpha) (3) sA3+alpha*s^2+omega^2*s+omega^2*alpha=(s+alpha)*(s^2+omega^2)

Fig. 11.17

11.16

Mechanical Vibrations

Then, the inverse transforms become:

>invlaplace(exp(-s/alpha)/{(s+alpha)^2*(s^2+omega^2)},s,t); 1

Heaviside (t -

1

2 ae

a %1

a

4.

) «( r -1

co sin(o)(t

k a/

a

a2 (02

a -1))

%1

1 a2 sin(w(t - 1)) a co+o(t - -))] a a 2 %1 %1 %1 := 2 a2 w2 +

a4

>invlaplace(exp(-sialpha)*s/((s+alpha)^2*(s^2+omega^2)),s,t); ) -a(t--el-j e a a2 + (.02

1 ) co2e-a(t-j a2e-a(t Heaviside t+ aJ %1 %1

+2

a

wa sin(e)(t - 1 )) co2 cos(w(t -1)) a2 cos(w(t -1 )) a a a %1 %1 %1

%1 := 2 a2 co2 + w4 + a4

Fig. 11.17 (Continued)

\\.\\\\\\\\\\\

The MAPLE V worksheet follows (Fig. 11.19). The natural frequencies are the square roots of the eigenvalues of M-1K. Computer algebra is useful in this problem as the mass matrix contains an unspecified value. Computer algebra is used to develop the characteristic polynomial of M-1K in terms of this k1

k2 m1

ki =1 x 105 N/m k2 =1.2 x 105 N/m

I-

M2

-/

k3 V\/—

k3 =1.8 x 105 N/m k4 =1.35 x 105 N/m

Fig. 11.18

m

mi =125 kg m2 =150 kg

\\\\\\\\\\\\\

parameter. Note that since f(2, m) is linear in m, that when f(2, m) = 0 is solved for m, only one value of m exists for each X. A plot of this function between = (60 rad/s)2 = 3600 and = (80 rad/s)2 = 6400 reveals that the maximum and minimum of the function over this interval are at the ends of the interval (Fig. 11.20). Thus, in order for any of the natural frequencies to be between 60 and 80 rad/s, 57.19 kg < m < 138.23 kg

11.11 For what values of m; will the system of Fig. 11.18 have a natural frequency between 60 and 80 rad/s? Use MAPLE V to perform the algebra.

Computer Applications

11.1.7

Solution of Solved Problem 11.11 using MAPLE V Definition: >k1 := 1*10^5; k1 := 100000 >k2 := 1.2101,5; k2 := 120000.0 >k3 := 1.8*10^5; k3 := 180000.0 >k4 1.35*10^5; k4 := 135000.00 >ml := 125; ml := 125 >m2 := 150; m2 := 150 >with (linalg); Warning: new definition for norm Warning: new definition for trace [BlockDiagonal, GramSchmidt, JordanBlock, Wronskian, add, addcol, addrow, adj, adjoint, angle, augment, backsub, band, basis, bezout, blockmatrix, charmat, charpoly, col, coldim, colspace, colspan, companion, concat, cond, copyinto, crossprod, curl, definite, delcols, delrows, det, diag, diverge, dotprod, eigenvals, eigenvects, entermatrix, equal, exponential extend, ffgausselim, fibonacci, frobenius, gausselim, gaussjord, genmatrix, grad, hadamard, hermite, hessian, hilbert, htranspose, ihermite, indexfunc, innerprod, intbasis, inverse, ismith, iszero, jacobian, Jordan, kernel, laplacian, leastsqrs, linsolve, matrix minor, minpoly, mulcol, mulrow, multiply, norm, normalize, nullspace, orthog, permanent, pivot, potential, randmatrix, randvector, range,rank, ratform, row, rowdim, rowspace, rowspan, rref, scalarmul, singularvals, smith, stack, submatrix, subvector, sumbasis, swapcol, swaprow, sylvester, toeplitz, trace, transpose, vandermonde, vecpotent, vectdim, vector] Mass matrix: >M := matrix(3,3,[[m1,0,0],[0,m2.40,0,m11); 125 0 0 M := 0 150 0 0 0 m Stiffness matrix: >K := matrix(3,3J[k1+k2,-k2,0],[-k2,k2+k3, k3],[0,-k3,k3+k4]]); 220000.0 -120000.0 0 K := -120000.0 300000.0 -180000.0 0 -180000.0 315000.00 >B := multiply(inverse(M),K); 1760.00000 -960.0000000 0 B := -800.0000000 2000.000000 -1200.000000 1 1 0 -180000.0- 315000.00 m_ Characteristics polynomial of B, roots are eigenvalues, which are squares >charpoly(B,lambda); (13m - 315000.00 12 - 3760.000000 12 m + .9684000000 1 - .4867200000 1012)/m >1 := (lambda,m) > lambda^3*m-315000*Iambda^2-3760*Iambda^2*m+0.9684*10^91ambda+ 0.2752* >10^7*Iambda*m-0.4867*10^12; f := m) X3 m - 315000 12 - 3760 12 m + .9684000000 109 1 + 0.2752000000 107 m 1- .4867000000 1012 >g := lambda > solve(f(lambda,m) = 0,m); g := X solve(f(4, m) = 0,m) >g(lambda); -315000. X2 + .968400000109 1 - .48670000001012 X3 - 3760. X2 + .2752000107 1 >plot(g(lambda),lambda = 3600.,6400,m=0.,200); >g(3600); 138.2327410 >g(6400) 57.18886782 1.

Fig. 11.19

Mechanical Vibrations 200

11.13 Use MAPLE V to develop the Fourier

series representation for the periodic function of Fig. 11.22. The Fourier series representation for F(t) is

150

ao F(t) = — + Dai cos coit + bi sin wet), 2 CO

m 100

27ri T The Fourier coefficients are obtained using MAPLE V (Fig. 11.23). co• =

50

11.14 During operation, a punch press of mass

4000 4500 5000 5500 6000 lambda

Fig. 11.20 11.12 Use the Rayleigh-Ritz method using

trial functions (Mx) = Lax — 2Lx3 + x4, 7 10 02(x) = — ex - - L2x3 + X5 3 to approximate the lowest natural frequency of a simply supported beam with a concentrated mass in at its midspan. Use computer algebra to help with the manipulations. The application of MAPLE V to approximate the lowest natural frequency of the beam follows in Fig. 11.21. The Rayleigh-Ritz method is applied as illustrated in Chap. 7. The lowest natural frequency is obtained as 42E1 = 96 3968 pAL4 + 7875 mL3 1

in is subject to an excitation of the form of Fig. 11.22. Set up a spreadsheet to design a vibration isolator of damping ratio C for the punch press such that the maximum transmitted force is less than Fall. Also use the spreadsheet to determine the maximum displacement of the machine when the isolator is used. Use the spreadsheet to determine the maximum stiffness of an isolator of damping ratio 0.1 for in = 1000 kg, a = 0.2, T = 0.1 s, F0 = 20,000 N and Fall = 3000 N. An alternate representation for the Fourier series is ao + 2

F(t) =

c

=

117 +

DO

sin(wit + x';), i=1

,

=tan

bi

ai

Defining ri = wit co„, an upper bound for the maximum displacement is

xm„
f := omega->det(D);

w2 640 625 Inc° pAL11+ 1024 2079 f :=

det(D)

>f(omega); 16 2 12 362 scs032pA _ 125 EIL15(I)2m + 31 co4 p2A2 L 125 20 + co4pAL19m El L 4257792 7 14553 1688 2095632 >solve(f(omega)=0,omega); 12

.r 1145 L2,/pV A '

12

471714D L2 1)-.1A

, 96

.fira312V3968LpA + 7875 tn.,0 3968L4 pA + 7875 mL3

, 96

-./L3/243968LpA .t + 7875 m.rei 3968L4 pA + 7875 mL3 96

.. ,F 12L3/243968LpA + 7875 m NiEI 39680 pA + 7875 mL3

Fig. 11.21 (Continued)

T

(1 + a)T 2T (1 + a/2)T

Fig. 11.22 Similarly, an upper bound for the transmitted force is Fmax
omega := 1->2*p1*1/T; ni 2— := i T Cosine coefficients: >a_1 := 2 /T*(i-rit(f_1(t)*cos(omega(1)*t),1=0..alpha*T/2) + int(f_2(t)*cos(omega(1)1),t=alpha*T/2..alpha*T) >); a_i := 2

1 cos(2nia)TF 1 T(nisin(nia) a - cos(nia))F (1 T(cos(nia) + nisin(nia)a)F 1 TF JI/T 2 eiza 2 _Rya 2 2 2 2 712 i 2C( TL i la

>simplify("); F(-2cos(nia) +1+ cos(2nia)) TC 22a Sine coefficients: >b_1 := 2/T*(int(f_1(t)*sin(omega(1)*t),t=0..alpha*T/2) + int(f_2(t)*sin(omega(1)1),t=alpha*T/2..alpha*T)) >; b i := 2

1 T(-sin(nia)+ If i cos(nia)a)F 1 sin(2nia)TF 1 T(nicos(rria)a + sin(rcia))F /T + 2 712i2a 2 2 IT 2 i 20: It2 i 2CC

>simplify("); F(2sin(ni a) - sin(2nia)) TC 2 i 2CC

Fig. 11.23 11.15 A simplified model of a vehicle suspension system is shown in Fig. 11.25. The vehicle traverses a road whose contour is approximately sinusoidal, as shown in Fig. 11.26. Develop a spreadsheet program that evaluates the amplitudes of displacement of the cab relative to the wheels, the absolute displacement of the cab, and the absolute acceleration of the cab for vehicle speeds between 0 and 80 m/s.

The frequency of the excitation is =

2 iry

d

The equations for the relative displacement, absolute displacement, and absolute acceleration are, respectively, r2 Z= h V(1_ r 2 4_ (2c02 X = ii

1+(2cr)2 (1—r 2 )2 +(gr)2

A = co2X

11.22

Mechanical Vibrations

Solution of Solved Problem 11.14: Design of a vibration isolator for periodic excitation Parameters mass damping ratio period alpha F_O F_all k

Calculated values 1000 kg 0.2 0.1 s 0.2 20000 N 3000 N 1190875 N/m

omega_n F_max x_max

34.50906 rad/s 3000.001 N 0.000389 m

Summation Frequency Frequency Fourier coefficient index ratio omega j r_i a_i b_i 1 62.83185 1.820735 3130.997 2274.802 2 125.6637 3.64147 1081.733 3329.231 3 188.4956 5.462205 -910.784 2803.105 4 251.3274 7.282941 -1853.58 1346.702 5 314.1593 9.103676 -1621.14 1.99E-13 6 376.9911 10.92441 -823.812 -598.534 7 439.823 12.74515 -167.287 -514.856 8 502.6548 14.56588 67.6083 -208.077 9 565.4867 16.38662 38.65428 -28.084 10 628.3185 18.20735 0 0 11 691.1504 20.02809 25.87601 18.80002 12 753.9822 21.84882 30.04813 92.47864 13 816.8141 23.66956 -48.5033 149.2778 14 879.6459 25.49029 -151.312 109.9349

c_i 3870.125 3500.561 2947.359 2291.147 1621.139 1018.287 541.3516 218.7851 47.77932 0 31.9845 97.2378 156.9599 187.0324

Magnification Transmissibility factor factor M Tj c_i*M j c_i*T j 0.412043116 0.509737846 1594.658 1972.749 0.08099443 0.143102488 283.5259 500.939 0.034580126 0.083091015 101.92 244.899 0.0191855 0.059091949 43.5968 135.3883 0.012201394 0.046075904 19.78015 74.69544 0.008444277 0.037853387 8.598701 38.54563 0.006191213 0.03216465 3.351624 17.41239 0.004733834 0.02798428 1.035692 6.122542 0.0037369 0.024777477 0.178547 1.183851 0.003024914 0.022236972 0 0 0.002498723 0.020173203 0.07992 0.64523 0.002098851 0.018462656 0.204088 1.795268 0.001787859 0.017021293 0.280622 2.671661 0.001541224 0.015789894 0.288259 2.953222

Fig. 11.24

Fig. 11.25
I
1 ✓:= 1.2 r2 := root(M(r, 0) —

r)

(b) There is viscous damping ratio C r := .8 ri := root(M(r, C) — Mmax, r) COI :=Y1 •

rad sec

✓:= 1.2

X < X. for co > co2 Guess for r < 1 Solving for r

rl = 0.879

col = 195.434 .

(On

X < Xma, for co < wt Guess for r > 1 Solving for r

= 1.125

co2 = 256.851 .

(02 := r2, • CO„

rad sec

rad sec

X < X. for co < col Guess for r > 1

— Mmax, r)

r2 := root(M(r,

r2 = 1.096

Solving for r

rad X < Xmax for co > co2 sec Alternate method of solution which can be used if your version of MATHCAD has symbolic capabilities: First load the symbolic processor. The magnification factor M is written symbolically as a function of r and C. The [ctrl] = is used to type the equal sign.

co2 = 250.099 •

(02 = r2 • (On

1

M=

4 (1

r2 )2 + (2 • C • r)2

11 _ 2 . 4-2 ..„ 1_ 4 . 4-2 +4 . 0 + 1 M2 \ \ , 11 _ 2 \

. c2

1_ 4 . c2 +4 . r 4 + 1 ' M2 +\

1- 2.c2 _ \1_4.c2 +4.04 +1 M2

1 _ 2 . c2 1_ 4 . c 2 +4 . 1.4 + 1 7 M2 \

Now select the variable r in the equation for M. Choose "Solve for variable" from the "Symbolic" menu. Four solutions for r in terms of M and C are shown to the left, in an array. Note that the second and fourth solutions are negatives of the first and third solutions and of no interest.

A.7

Appendix

The first and third solutions are set equal to functions of M and C by copying them to the clipboard and pasting them to the right hand side of a function. The third solution is the smaller values of r and set equal to r1 while the first solution is set equal to r2 4_ 4 r 4 + 1 M2

_2

r1(M,

:=

r2(M,

:= \11 — 2 • C2 -1-

VV

4 • C2 + 4 • C4 + 1 M2

The solution to the problem is obtained by evaluating the functions for M = M. and C = 0 and C := 0.08 ri(Mmax, C) = 0.856, := 0, r2(Mmax, c) = 1.125 ri(Mmax, C) = 0.879, := 0.08, r2(Mmax, C) = 1.096 The values obtained above are identical to those previously attained using the root function. The upper bound on the speed for w < co„ and the lower bound on the speed for w > co„ are obtained as before. This method is useful if these speeds are to be determined for a range of C.

Further study (1) Make a plot of the upper bound on the operating range for r < 1 vs C for 0 < < 0.7.

(2) Plot the lower bound on the operating range for r > 1 as function of the length of the beam for 1 in < L < 4 m assuming C = 0.08 and all other parameters as given.

Appendix

A.8

Natural Frequencies and Mode Shapes for a 3 DOF System (Schaum's Mechanical Vibrations, Solved Problems 5.39, 5.40 and 5.47 pp. 5.32-5.33, 5.36-5.37) Statement

Determine the natural frequencies and normalized mode shapes of the following system. 1-o- Xi

Assume System Parameters mass := 1 • kg k :=

k 000

1 • newton

1_.)-X2

X3

2k

000\- 2 m --'000`-2m

m

m

Solution The mass and stiffness matrices for this 3 degree-of-freedom system are

( mass

M :=

O. kg

0 kg

0 • kg 2 mass \ O. kg

0 • kg

0 kg 2. mass)

K := o•

3. k

—2k

—2.k

3.k

newton m

—k

0 newton m —k

The inverse of the mass matrix is calculated as

minv

0 kg2

0 • kg2

0•kg 2

2 • mass2

0 kg2

0 • kg2

0 • kg2

2• mass2

(4 • mass3

(1

Minv

4 mass2 1

0

0 0.5 J:1

0\ 0 •kg-1

0 0.5 /

The natural frequencies are the square roots of the eigenvalues of MinvK. To this end

D := Min, • K,

D=

( 3

—2

—1

1.5

0 —0.5

0\ —0.5 0.5)

(3.871 \

w2 := eigenvalues(D),

w2 = 1 \ 0.129,

• sec-2

• sec-2

Appendix

A.9

Thus the natural frequencies are co2 = Jw21 ,

(01 = Jw22

(A3= W 20

rad rad , sec c°2 1. sec ' The mode shapes are the eigenvectors of Min,K. col = 0.359 •

E := eigenvecs(D),

(03 = 1.967 •

rad sec

( 0.915 0.577 0.383\ 0.577 0.55 E = —0.399 0.059 —0.577 0.742,

The eigenvectors for the modes are u, v, and w respectively where , := E`1' w := u := E1 I 0.577 \

/ 0.383 \

v=

u = 0.55

w=

0.577 —0.577)

0.915 \ —0.399 0.059)

The mode shapes are normalized with respect to the mass matrix = WT • NT W, = VT • Af• V, cl: = uT • M• u, c3 = 1.162 • kg c2 = 1.667 • kg, cl = 1.853 • kg, The normalized mode shapes are U„

:=

1

ii,

:=

1

.V ,

W„:=

W,

C300

0200

0.849\ 0.447 • ke.5' w = —0.37 • kg-°.5 0.055)

I 0.4471

(0.282\

= 0.404 • ke 5 ' 0.545)

1

vn =

The previous solution follows the methods used in Schaum's Outline in Mechanical Vibrations. However, Mathcad has feature which determines eigenvalues and eigenvectors for a generalized eigenvalue problem of the form Kx = Alsolx. The natural frequencies are the square roots of the eigenvalues of the generalized eigenvalue problems and the mode shapes are the normalized eigenvectors. However to use this feature the matrix K must be dimensionless. To this end 0 0\ M1 := 0 2 0 , 0 0 ')2)

I 3

Il

K1 :=

—2 0

—2 3 —1

0 —1 1

Appendix

3.871 $.1,3 := genvals(Kl, M1),

w3=

El := genvecs(K1, M1),

0.915 —0.577 —0.383 El = -0.399 —0.577 —0.55 0.059 0.577 —0.742

1 0.129

Note that the eigenvectors returned using genvecs are not normalized with respect to the mass matrix. Extension Investigate mode shape orthogonality using the normalized eignevectors. u„T • M • u„ = 1, U„T • M • V„ = 0, u„T • M • w„ = 0 v„T • Ill • u„ = 0, v„T • M y„= 1, v„T • M • w„ = 0 w„T • M • u„ = 0, W„T • M • V„ 0, MIT •M ' wn = Further study (1) Show numerically that the mode shapes are also orthogonal with respect to the stiffness matrix (2) Show numerically that un TKun = coi2

A.11

Appendix

Undamped Absorbed Design (Schaum's Mechanical Vibrations, Solved Problems 8.29 and 8.30, p. 8.20)

Statement A machine of mass m is attached to a spring of stiffness k1. During operation the machine is subject to a harmonic excitation of magnitude Fo and frequency w. (a) Determine the stiffness and mass of an undamped absorber of minimum mass such that the steady-state amplitude of the machine is zero and the steady-state amplitude of the absorber mass is less than X2max when the machine operates at co. (b) What are the natural frequencies of the system with the absorber in place? System Parameters m 1 : = 200 • kg,

k 1.• = 4.105

newton m

F0: = 500 • newton

co: = 50 . rad , X2max = 0.002. m sec The natural frequency of the primary system is kl

(O H : = ml

,

col

= 44.721 •

rad sec

Solution (a) The steady-state amplitude of the primary mass is zero when the absorber is tuned to the excitation frequency. When this occurs the steady-state amplitude of the absorber mass is X2 =

Fo k2

which is rearranged to yield Fo k2= X 2

The minimum absorber stiffness such that X2 < X2max is k2 :=

F0 X2max

k2 2.5

105 newton m

In order for X1 = 0, the natural frequency of the absorber 022 = (k2/m2)1/2 must be equal to the excitation frequency co. Thus, since the ratio of k2 to m2 is fixed, the absorber with the minimum possible mass corresponds to using the minimum allowable stiffness. This leads to. 1)72

:=

k2 2 CO

m2 = 100 • kg

A.12

Appendix

(b) The mass ratio is m2

µ = 0.5

The natural frequencies of the absorber are the frequencies such that the denominator of Eq. (8.12) is zero. This leads to the following f(01)

(04

(1

04 + 2

01

w222

—[ (022 = 60, 2 (0121 • 2 0011 The natural frequencies are the values of co such that f(a)) = 0. When the absorber is added, the system has two natural frequencies, 0)1 < con, a)2 > coll

:= 50 • rad sec

guess value

col := root(Aco), a)),

wl = 32.679 .rad sec

:= 80 • rad sec

guess value

012 := root(f(co), co), Extension

a)2 = 68.426 •

rad sec

With the absorber in place, for what, values of co will the steady-state amplitude of the absorber be less than Ximax? Xlmax := 2.10 • m The steady-state amplitude of the absorber in terms of co is expressed by r1 (co) :=

co

r2(w) =

W11 X1 (co :—

C°22

F0

1 r2 (a))2

k2

[4(c0)2 • r2 (co)2 — r2 (c0)2 — (1 + /I) • r1 (0))2 +11

Analysis shows that the denominator above is negative fOr col < a.)< co2. The numerator is positive for co < co22 and negative for co > 0)22. Thus appropriate range of co is such that coa < co < cob where Xi(a)a) = — XAmax and X1(4) = Xlmar TOL := 0.000001

The default value of TOL = 0.001 is not tight enough in this problem. Try using the method below with the default value and notice the difference in answers. The smaller the value of TOL, the more accurate the answer.

Appendix

w:=42 •

rad sec

:= root(Xlmax + X 1 (o0), co), co„ = 40.442 :=

D4 •

A.13

rad sec

rad — sec

rad root(Ximax — X i (o)), w), cob = 61.238 • — sec Further study (1) Study the design of an undamped absorber if the goal is to reduce the steadystate amplitude of the primary mass to 2 mm for all speeds between 35 rad/ sec and 65 rad/sec. (2) Repeat the extension of this problem if the excitation is a frequency squared excitation caused by a rotating imbalance, that is, F0 = 0.2o)2.

Index Accelerometer 3.12 Adams method 4.4 Amplitude 2.3 of acceleration 3.33 of excitation 3.2 Angular oscillation 3.34 Backbone curve 10.3 Bar element 9.3 Beam 1.7 element 9.3 Beating phenomenon 3.4 Boundary conditions 7.1 Cantilever beam 1.11, 3.13, 3.28, 5.27 Chaotic 10.10 Chaotic motion 10.10 Characteristic determinant 5.29 equation 7.3 Circuit analysis 1.1 Continuous system 1.1, 7.1 Convolution integral 4.1 Coulomb damping 2.6, 3.10 Critical speed 3.29 Critically damped 2.4 d'Alembert's principle 2.2 Damped natural frequency 2.4 period 2.5 vibration 1.2 Damping coefficient 1.3 component 1.2 factor 3.8 ratio 2.3

Dashpot 1.3 Degrees of freedom 1.1 Differential equation 2.1 spring element 1.14 Direct numerical simulation 4.15 Discrete system 1.1 Displacement spectrum 8.3 vector 5.2 Distributed parameter system 1.1 Duffing's equation 10.2 Dunkerley's method 5.4, 5.46 Dynamic load 3.30 Effective forces 2.2 Eigenvalue-eigenvector problem 5.3 Eigenvalues 9.2 Eigenvectors 9.2 Elastic coupling 5.11 Element mass matrix 9.2 stiffness matrix 9.2 Energy method 2.1, 2.8 Equivalent damping 1.3 external force 3.2 mass 1.3 stiffness 1.3, 1.4 system 1.1, 1.18, 1.20, 3.2 system method 2.1, 2.7 torsional system 1.4 Euler method 4.15 Finite element method 9.1 Fixed-fixed beam 5.27 Fixed-free bar 9.4

1.2 beam 1.11 Flexibility influence coefficients 5.3 matrix 5.3, 11.3 Force spectrum 8.3 vector 5.2 Forced frequency of vibration 1.2 Forced vibration 1.2, 3.1 Forcing frequency 3.1 Fourier series 3.9, 3.38 Free body diagram method 2.2 Free vibration 1.2 vibration test 2.26, 3.37 Frequency of excitation 3.2 ratio 3.5, 3.7 squared excitation 3.6 Friction force 2.6 Galerkin method 10.11 General forced response 4.1 Generalised coordinate 1.1, 1.3, 5.1 Geometric boundary conditions 7.5, 9.1 Harmonic excitation 3.1, 3.2 torque 3.14 Helical coil spring 1.5 Highest natural frequency 5.39 Holzar's method 5.4, 5.41 Houdaille damper 8.6 Hysteretic damping 3.10 Identity matrix 6.2 Impedance matrix 6.2 Impressed force 3.1 Impulse 4.2 isolation 8.3 Inertia components 1.2 Initial conditions 2.3 Isolators 8.3 Jump phenomenon 10.3 Kinetic energy 1.2, 5.2 Lagrange's equations 2.1, 5.2 Laplace transform 4.2, 6.2

Index Limit cycle 10.3, 10.11 Linear spring 1.2 superposition 3.9 vibration 1.2 Linstedt-Poincare method 10.8 Local coordinate 9.1 Logarithmic decrement 2.4 Lolus 11.2 Longitudinal oscillations 7.2 vibrations 7.7 Lowest natural frequency 5.37 MACSYMA 11.2 Magnification factor 3.5 MAPLE V 11.2 Mass matrix 5.2 moment of inertia 1.2 Mass-spring-dashpot system 1.3 Mathcad 11.2 Mathematica 11.2 Mathematical model 1.2 MATLAB 11.2 Matrix iteration 5.4, 5.25, 5.37, 5.38, 5.39 Maxwell's reciprocity relation 5.3 Method of averaging 10.11 renormalization 10.9 virtual work 3.2 Modal analysis 6.3 matrix 5.22, 6.2 superposition 7.19 superposition method 7.3 vector 5.22, 5.25 Mode shape 5.4, 5,23, 5.25, 5.32, 6.2, 7.2 vectors 5.3 Motion transmissibility 3.8 transmissibility ratio 3.8 transmission 8.1 Multi-degree of freedom 5.1, 6.1 Natural frequency 3.1 of vibration 1.2 Newton's method 2.2, 2.8 Nonlinear system 10.1 Non-linear vibration 1.2

Index Normal mode of vibration 5.1 solution 5.3, 7.2 Normalization condition 5.4 Numerical integration 4.19 methods 4.19 Orthogonality condition 5.4, 7.3 of the mode shapes 5.35 Over damped 2.4 Overshoot 2.22 Paradox 11.2 Parallel combination of springs 1.8 Partial fraction 6.7 Periodic excitation 3.38 Periodic vibration 1.2 Perturbation methods 10.2 Phase 2.3 Phase of the excitation 3.2 Phase plane 10.2 Piecewise constant interpolation 4.18 Planar motion 1.2, 2.2 Poincare sections 10.10 Potential energy 1.3, 5.1 Principal coordinates 5.12, 6.2 Principle of conservation of energy 2.1 of energy method 2.19 Proportional damping 5.5 Quatro Pro 11.2 Random vibration 1.2 Rayleigh's method 2.19, 5.4, 5.47 quotient 7.4, 7.22, 7.23 Rayleigh-Ritz method 7.5, 7.23 Recurrence relations 4.15 Resonance 1.2, 3.1, 3.3 Response spectrum 4.4, 4.20 Rotating unbalance 3.21, 3.22 Runge-Kutta method 4.4, 4.19, 10.10 Saddle point 10.4 Seismic instrument 3.11 Series combination of springs 1.8

Shape function 9.1 Shock isolation 8.3 Simple pendulum 2.8 Simply supported beam 1.8, 3.19, 5.26, 5.33 Spring damper mechanical system 3.23 mass system 2.19 Static deflection 1.4 Steady state amplitude 3.5 response 3.5, 3.15 Stiffness 1.2 components 1.2 influence coefficents 5.3 matrix 5.2 of a helical coil spring 1.5 Stodola method 5.4, 5.23 Subharmonic resonance 10.1 Superharmonic resonance 10.1 Support excitation 3.7 Suspension system 2.22 Sweeping technique 5.4, 5.38 Taylor series 10.4 expansion 4.15 Torsional oscillations 7.2 stiffness 1.6 system 1.4, 5.41 viscous damper 1.13 Torsionally equivalent shaft 5.11 Transcendental equation 7.9, 7.11 Transibility ratio 8.2 Transient response 3.5 Transverse vibrations 7.2 Trapezoidal rule 4.17 Unbalance 3.23 Undamped natural frequency 2.3 vibration 1.2 Under damped 2.4 Unit impulse function 4.3 Unit step function 4.4 Unrestrained system 5.3 torsional system 5.29 van der Pol equation 10.3, 10.11

L4 Vehicle suspension 3.31 Velometer 3.12 Vibration 1.1 absorber 6.6, 8.4 control 8.1 isolator 8.1 Vibrometer 3.12 Virtual work 5.2

Index

Viscous damping 1.3, 2.3 matrix 5.2 Wave equation 7.1 Wave speed 7.1 Whirling 8.7 Work-energy method 2.27 Zero frequency deflection 3.5