Introduction to Vibration in Engineering [3 ed.]

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IN VIBR ENGIN

THIRDEDITION

BYSUNG LEE AND WERELEY NORMAN MARYLAND UNIVERSITY OF

cogne ay SAN DIEGO

Bassim and Publisher CEO Hamadeh, Carrie Revisions Carer and Author Baarns, Manager, Kaela Editor Martin, Project Production Editor Rees, Jeanine Senior Villavicencio, Emely Designer Graphic Alexa Lucido, Licensing Manager Director ofMarketing Natalie Piccotti, Kassie Senior Vice ofEditorial Graves,President Director of Academic Jamie Giganti, Publishing © 2022 Inc. Nopart Allrights reserved. of this be may publication reprinted, reproduced, Copyright by Cognella, or orhereafter or now or in means, form other known utilized any any transmitted, electronic, mechanical, invented, by orinany written information retrieval without the and system permission including photocopying, microfilming, recording, Inc. For ofCognella, audio and other forms of any translations, inquiries permissions, regarding foreign rights, rights, at contact the reproduction, please Cognella Licensing Department [email protected]. or names or areused Notice: be Trademark Product t rademarks trademarks and for identimay corporate registered only to intent fication and without explanation infringe. 2020 © Cover LP/Jobalou. iStockphoto image copyright inthe States Printed United ofAmerica.

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TABLE OF CONTENTS PREFACE 1 CHAPTER

—_ OF RESPONSE SDOF DYNAMIC SYSTEMS

1.1Equation ofMotion DW 1.2Free Vibration ofUndamped SDOF FF Systems 1.3Free Vibration NS ofDamped SDOF Systems 1.4Forced Vibration NYO 1.4.1 ASDOF toaStep Load wo System Subjected 1.4.2 AStep Load Duration ofFinite 1.4.3 Impulse Loading 1.4.4 Convolution AYN Integral 1.5Numerical Methods toDetermine CO Response System 1Problem NO wo Sets Chapter 2 CHAPTER

DO

OF SDOF NO Se) STEADY-STATE RESPONSE SYSTEMS 2.1Response to ofDamped SDOF Systems O Harmonic Ww Loading 2.2Force tothe Transmitted Base DD 2.3Rotating wihMass Unbalance WO Systems 2.4SDOF GO with Base Motion Ff Systems 2Problem Sets Fr FF Chapter

W

CHAPTER 3

OF SYSTEMS 47 RESPONSE DYNAMIC MDOF

48 3.1.Equation viaNewton's ofMotion Second Law 3.2.Natural and Modes ofUndamped Frequencies 49 MDOF Systems 3.3Response 61 ofUndamped MDOF Systems 3.4.Additional 75 Topics 3.4.1 75 ofMDOF Steady-State Response Systems 3.4.2 76 Free Vibration ofDamped Systems 3.4.3 inTime 79 Numerical Integration 3Problem 80 Sets Chapter 4 CHAPTER

ENERGY APPROACH 85 4.1.Kinetic and Potential Energy Energy 86 ofDynamic Systems 4.2Equation via the ofMotion Lagrange Equation89 4.3Damping 94 Effect 96 Problem Sets Chapter 4 CHAPTER 5 STABILITY 101 DYNAMIC 102 5.1.Follower Force Problem 5.1.1 102 ofMotion Equation 5.1.2 106 Behavior Static O7 5.1.3 Behavior Dynamic 110 5.2Dynamic Aeroelasticity AModel 5.2.1 110 Wing

117 5.2.2 Flutter Analysis 5.2.3 120 Phenomena Other Dynamic 121 5.2.4 Real and Other Surfaces Wings Aerodynamic 5Problem 124 Sets Chapter CHAPTER 6

127 ABSORBER VIBRATION aMass a 128 6.1Absorber with and Spring 6.2Steady-State 130 ofMDOF Response Systems aMass, aSpring, aDamper 6.3Absorber 131 with and 134 6Problem Sets Chapter CHAPTER 7 VIBRATION OF SLENDER BODIES 137 7.1Longitudinal 38 Vibration ofaSlender Body 7.2Torsional 45 Vibration ofaSlender Body 7.3Vibration 46 ofaTaut String 7.4Bending 49 Vibration ofaSlender Body 7.5Assumed Mode Method forApproximate Solution58 7Problem 63 Sets Chapter CHAPTER 8 FINITE ELEMENT MODELING 167 8.1Longitudinal 168 Vibration ofaSlender Body 8.1.1 169 Model ofaSDOF Construction 8.1.2 Slender Flying Body Undergoing ATwo-DOF Motion: Model 170 Longitudinal

8.2Finite Element of Modeling L ongitudinal a Vibration of Slender Body 8.3Torsional Vibration ofaSlender Body a 8.4Bending Vibration of Slender Body forBeam 8.4.1 Formulation Element Bending nofMotion 8.4.2 Model forthe Global Equatio 8.4.3 Alternate Finite Element Formulation f£ orBending Vibration 8.5Bending ation Vibr under Axial Force 8Problem Sets Chapter CHAPTER 9

FORCE MODAL AND INPUT IDENTIFICATION 9.1.Modal ication Identif 9.2.Input Force Determination 9.2.1 Numerical Construction of Function Impulse Response 9.2.2 Force Determination Input Using Function Impulse Response 9Problem Sets Chapter CHAPTER 10

EL FLUTTER WING AND PAN

10.1 teSpan ofFini Wings 10.1.1 Kinetic Strain Energy, Energy, Incremental and Work Done 10.1.2 Finite Element Formulation forWing Flutter 10.1.8 Assun ned Mode Formulation Flutter229 forWing

10.2 230 Panel Flutter 10.2.1 Finite Element Formulation forPanel Flutter231 10.2.2 Assumed Mode Formulation Flutter234 forPanel 10 236 Problem Sets Chapter

APPENDICES A.1: Nomenclature A.2: List ofFigures A.3: References

239 239 242 244

PREFACE

asatextbook are iswritten This book for and students who undergraduate first-year graduate careers sector. inaerospace inprofessional We have the ofteachinterested had primarily privilege acourse a inaerospace in The fornumber ofyears. vibration, ing acourse including aeroelasticity, challenge on a is vibrations ofthis that of t he textbooks focus from many type existing developing different from that ofanaerospace perspective engineer. orstructure can mass An be vibration with distributed and aerospace system complex, undergoing a or a a structure contains For stiffness. skin stressed multitude wing example, surrounding seem are to fuselage ofsapars and ribs that the that Yet,physical defy simple analyses. insights developed a a from careful of of freedom described system (SDOF) simple, study single degree by ordinary amass a in differential and and and is, spring equation—that damper—are profound, possibly can a cases, For limited describe the vibration ofsuch with systems wing. bending multi-degrees amode asa an notion offreedom of of the f reedom described system (MDOF), single degree by ismanifested inthe ofthe normal differential mode concept equations. equation ordinary a a solid istodevelop and of t his book foundation of Therefore, key objective understanding can as more to SDOF be u sed and MDOF that blocks tackle difficult systems general building canbe tointroduce is Asecond of this book the which systems. equation Lagrange approach, goal toconstruct of For structures, motion for used vibrations aerospace systems. equations complex or Interactions often involve forces. the flexible among structure, aerodynamic motion-dependent can to inertia loads and lead instabilities. models motion-dependent dynamic Accordingly, simple are two to with discuss the offreedom introduced of which degrees subject instability, dynamic canoccur as a a oran in for such slender under follower forceaircraft systems wing flight. astructural can to body is Another of this book and illustrate the tools which key body goal develop by

INENGINEERING INTRODUCTION TO VIBRATION

an via into is t he element This be transformed finite method oformotion offinite offreedom. equation degrees a as motions done slender Combined torsional, using by body examples. undergoing longitudinal, bending are to contents and also introduced modaland within forces, input experimental analytical approaches identify context the ofthe time domain method. an most istoprovide But t his textbook the student. the of book for purpose important perhaps approachable are is For each illustrated short the and Moreover, concepts covered, topic explanations. examples goal through or toshow to to howcompute whether and illustrate how numerical, aerospace solutions, complex analytical can are a set in in structural vibrations be reduced using of tools thateasily software programmed asMATLAB.thecomplexity to such seeks models, aerospace compute results, engineer Ultimately, develop analytical to structure. assessments, make and then decide what be d one the of t he should aerospace improve performance toenable isonpractical the the solutions the focus of with of Therefore, computing application goal techniques, iteration. aerospace engineering design serves asthe oursenior-level one-semester inthis inAerospace The material book textbook for class Vibrations. an on version both We have andhonors ofthis class based these materials each concurrently taught regular can to the be omitted the and of t he Some of class. year. advanced topics easilyslowpacepressure content. tothe As We and have overall thank wourith ofthis students contributed any many kind, project colleagues our our allof students andgraduate who have taken the students and assistants, class, postdoctora teaching we to various In Drs. all of whom have contributed the of this book. thank Soonwook aspects staff, particular, was Dr. instruKwon inenhancing Garatsa and for their the III, manuscript. John Draper Ray Draper help sets. in We Dr. the solution for the Becnel for mental manual also thank Andrew initiating problem providing on contents valuable the of t his book. feedbacks

a

and

W. Lee Prof. Sung Norman Prof. M.Wereley ofAerospace Department Engineering of University Maryland

Park, College Maryland

DYNAMIC RESPONSE OF SDOF SYSTEMS canbe as motion m odeled offreedom systems Many degree physical undergoing dynamic single a mass, a a a is linear A with and of f reedom system (SDOF) spring, (DOF) damper. degree aa oran a motion Once that describes the of system. displacement displacement angular asacollection a a onecanconstruct anphysical ismodeled ofamass, and system equation spring, damper, oranenergy The motion via m otion ofmasotion N ewton's second law o f apequation approach. an one can to which mathematical differential pears equation (ODE) apply techniques ordinary todetermine the response. system wewill an construct Inthis SDOF ofmotion for systems equation chapter, applying a by in We Newton's second law om f otion. will then consider “free― which undervibration, system no starts motion vibrational with free of)externally load. The free vibration with goes (or applied a at start terms in motion. the“initial conditions― of and the ofthe displacement velocity specified occurs a at a We will that free vibration the findasmass, constant, and determined system frequency amplitude by and and the initial conditions. The parameters constant) (such spring damping on then vibration under discussion free vibration will be forollowed the offorced by externally study as we anumerical loads such function load will introduce step applied impulse. Subsequently, canbe todetermine intime method that used for numerical of ofmotion the equation integration to to in of of The be SDOF loads. c overed this response any systems types topics subjected applied aresummarized asfollows: chapter 1,1Equation ofmotion 1.2Free vibration SDOF ofundamped systems 1.3. Free SDOF vibration ofdamped systems 1.4Forced vibration: function load and step impulse to 1.5Numerical methods determine response system wewill both Inthis SIsystem and the the the ofunits and chapter chapters, adopt following units in units in States. The SIsystem of still used the United basicthe system engineering English are or for to meter for and mass, time relevant for second (sec (m) s) dynamics kilogram (kg) length are units inthe mass, while the basic for foot for and system engineering (ft) English slug length or a as newton x time. isdefined Inthe SIasystem 1 N=1 1m/sec’. second for force of (N) (sec s) kg as1lb is =1 Inthe force of defined systempound ft/sec’. engineering (Ib) English slugx1

INENGINEERING INTRODUCTION TO VIBRATION

1.1 Equation ofMotion

amass, a a canbe The ofmotion foraSDOF with and derived system equation spring, using dashpot damper Newton's second law. 1.1.1Spring-mass-damper system Example

FO) y(t), 1.1 Figure Mass-spring-dashpot damper system. mass crepresent constant constant m,spring mass, InFigure concentrated and the effective 1.1, k, damping orDOF isthe and ofthe while ofthe and system, system, stiffness, y properties displacement energy-dissipating are as Fis the load. and acceleration ofthe follows: system applied Velocity expressed 2 dy dy y —: yswey —dtvelocity, deacceleration mass a asshown To Newton's second isolate the and draw below. law, (FBD) apply free-body diagram j

y

=

=

k 1.2 Figure Free-body diagram. toNewton's isthe isthe Inthe and force. second FBD, force, restoring spring restoring ky damping According cy

law, =F my—ky-cy or

my+cy +ky=F 1.1.2Effect ofgravity Example

1.3 SDOF under effect. system gravity Figure

(1.1.1)

DYNAMIC RESPONSE OF SDOF SYSTEMS is ofmotion Equation (1.1.2) my +cy+ky=mg+F a constant ofthe For where and measured from the gisthe yisthe gravity displacement spring, original length mass atstatic under gravity, equilibrium mg (1.1.3) ky, as ycanbe Displacement expressed (1.1.4) 9 yuy,t+ static where from the ofthe measured _y;: spring displacement original length static measured the from position equilibrium 4:displacement into equation equation (1.1.4) (1.1.2), Placing F (1.1.5) mg M(IAY) +N+R(y, =0 and that equation ¥,= j, (1.1.3) noting Using + =F (1.1.6) my jtky term ismeasured static when the the the from position, gravity Accordingly, displacement equilibrium disapmotion. from the of pears equation =

+I, +H) + =

aninstrument mass minstalled nose cone 1.1.3Consider inthe of Example package arocket acushion aspring isrepresented of with vibration. The cushion kand against by adamper c.The a isfired rest rocket withgiven acceleration For this from d. vertically one can use a a a as 1.4. case, model withmoving baseshown inFigure of SDOF system

amoving 1.4 SDOF with base. system Figure mass a todraw account into the the and forces (instrument) spring Isolating diagram, taking free-body damping on mass, forces the and Newtons second law, acting applying

od-k(y-d)-c( my =—ky yd)

kd-cj

+

+

=

(1.1.7)

INENGINEERING INTRODUCTION TO VIBRATION

d, y,y— Introducing p-Hp m(jp+ —ky d) + =

(1.1.8) (1.1.9) myn Cj+kyy=—md aSDOF with inwhich F=—md under load the represents system y, (1.1.9) Equation displacement applied to isfixed end the base. bottom ofthe spring =

1.2 Free Vibration ofUndamped SDOF Systems

a

noexternal to isthen (F0), For with the free vibration. When vibrates force said system system undergo starts motion initial free the the with initial and initial called vibration, system given displacement velocity, conditions. an free Let's consider vibration of undamped system. (c0) =

=

nodamping. 1.5 with SDOF system Figure

ofmotion: my Equation +ky=0 Initial conditions: initial displacement: y(0)=y, initial jy velocity: (0)

(1.2.1)

=

Note: a nd values. represent ,initial 1)yThe given ¥, tothe strain inthe initial stored energy spring 2) displacement corresponds t kyp mass to T he initial the initial kinetic the of energy 3) velocity corresponds tm, no +strain isconserved is total vibration. There With kinetic free energy energy (= damping, energy) during strain the the ofenergy between kinetic and vibration. energy energy exchange during asecond-order a is differential with equation Equation (1.2.1) (in t ime), (ODE) ordinary homogeneous solution ofthe form: following

(= (=) ) :

:

Act equation equation (1.2.2) (1.2.1), Placing =

y

to

+k) =0 Ae― (mp?

(1.2.2)

DYNAMIC RESPONSE OF SDOF SYSTEMS nonzero For nontrivial y(i.e., A), +k=0

mp’

(1.2.3) (1.2.4)

ft

@,m Introducing =

canbe as (1.2.3) Equation expressed (1.2.5) p +@.=0 From equation (1.2.5), > and p,=-i@ (1.2.6) p,=i@, p=tia, asalinear corto twosolutions isthen The combination solution of t he equation (1.2.1) general expressed asfollows: top,and p,inequation (1.2.6) n

responding +Ase +A,e =A yHAe Ot Recall the Euler's formula: following =cosB+isinB e―

=cosB—isinB e-― canbe as equation equation (1.2.8), (1.2.7) Using expressed sing,t =C, cos@,t+C, y(t) are new set From where and equation (1.2.9), C, C,a ofconstants.

(1.2.7)

(1.2.8)

cos@,t (1.2.10) sin@,t+C,@, j(t)=—-C,, aredetermined asfollows: Constants and the initial conditions C, C, by applying t 0, Attime initial from y,,and equation (1.2.9), displacement: y(0)= = (1.2.11) Ci is(0) Initial from j, and equation (1.2.10), velocity J =

=

‘

,

(1.2.12) C0,Jo7G canbe asa intime, function such that sinusoidal (1.2.9) equation Alternately, expressed single (1.2.13) y(t) =Csin(@,t+) C;amplitude where @: phase angle inonesecond, covered called natural @,,: (rad/sec) angle frequency =

=< a

INENGINEERING INTRODUCTION TO VIBRATION

arerelated asfollows. oneof tothe and i nitial conditions the trigonometric phase Using angle Amplitude canbe as formulas, equation (1.2.13) expressed cos + @, (1.2.14) tcos@ C(sin tsing) y(t) @, with equation equation (1.2.9), Comparing (1.2.14) =

=Csing C, =Ccoso C, From and also from the sketch, equation (1.2.15)

(1.2.15)

|| ES

“

2

|

XoJ CcC°+C =— > =tan? tang 2 i C,(20) oO 2 2

2

(1.2.16)

Note:

free vibration with natural SDOF system 1)Anundamped undergoes frequency

rad/sec ol. T: and in

2) period > o,T=2n o, orHertzisdefined 3)Natural (Hz) asfollows: cycles frequency fin

pa2t n

f 1_% T 274 =

=

(1.2.17) (1.2.18

DYNAMIC RESPONSE OF SDOF SYSTEMS aspacecraft onthe 1.2.1Consider For vertical the spacemotion, Example g round. asa mass k. is modeled with M and T he the craft SDOF system spring spring represents vertical the stiffness ofthe Under the ofthe gear. spring l anding d eadweight s pacecraft, cm. 7 .62 displaces vertically by inthe Determine natural vibration vertical the offree direction. (a) frequency isdoubled. Determine the natural when the stiffness ofthe spring (b) frequency Solution: (a) > Mg -t M static J

ky o,=M=| =

:

aie

81=11.35 28 Hz f 2n=1.806 Vu 0.0762rad/sec,

V 4 =|

=

ot constant mass, For @, (b) Vk Vk fr = 2.554 Hz f=2x1.806 ~

1.3Free Vibration ofDamped SDOF Systems

a a a consists that ofamass, Consider and system spring, damper.

Adamped 1.6 SDOF Figure system. an crepresents constant The law with mechanism. Newton's second apdashpot damping energy-dissipating tothe mass: isolated plied

my F—ky—cy =

my+cyt+ky=F ¢.k F j+—jt—y=— .

mom

m

nowintroduce anondimensional ratio Let's Casfollows: damping

(1.3.1)

INENGINEERING INTRODUCTION TO VIBRATION

~=2¢0, m

Cc

Then, aan=—— 2mo,

(1.3.2

canbe as rewritten The last ofequation (1.3.1) F y=— m §+2¢@,j+@, where

For vibration free (F 0), of m otion: equation

(1.3.3)

=

initial conditions:Â¥+2c,j+@7y=0 ypand y(0) (0)5, =

(1.3.4) (1.3.5)

=

Note:

out Inthe vibration will ofdamping, free die presence eventually. to ishomogeneous, isofthe Asolution which form: equation (1.3.4), following

Ae― y(t) Then, y(t) ¥(t)= pAc", p’Ae™ = into and equations (1.3.6) (1.3.7) equation (1.3.4), Placing =0 p+ 2c, Ae― (p+ @ ; nonzero For nontrivial

(1.3.6) (1.3.7)

=

(1.3.8)

y(i.e.,A), =0 p+2¢o,p+@,

(1.3.9)

from Then, equation (1.3.9),

=0 (p+¢o,)° -(c@,)° +@, -1) (p+¢o,)° =@,(¢° p+so, =o, Jo? -1 -1 p=-co, +0, Jo

(1.3.10)

onthe onecanconsider cases: ofthe the three ratio, Depending magnitude damping following

DYNAMIC RESPONSE OF SDOF SYSTEMS case 1)Overdamped >1) (¢

pp =-S0, +@,/97-1

(1.3.11) P,=-S0,-@,/5°-1 are tothe of motion in(1.3.4) isexpressed For real numbers. The solution and >1,both p, p, ¢ negative equation asfollows: (1.3.12) y(t) ce" +c,¢°" =

aredetermined astime tozero Note a nd the initial with Constants from conditions. that, c, c, y elapses, drops nooscillation. case (¢ 2) damped C ritically For ¢=1, =1) =

=—®,

PiP2 asfollows: tothe of in(1.3.4) isexpressed The motion solution equation +e, ce" te" y(t)= time Note Constants initial a nd from the conditions. c, c, determined that, are

as

oscillation.

no

case

— =

0

Israric

―

(1.4.13)

INENGINEERING INTRODUCTION TO VIBRATION canthen as be (1.4.12) Equation expressed Y=Ysraric(1— cos, (1.4.14) t) onecan see From that under the condition equation (1.4.14), dynamic rat (1.4.15) [ane 2a under Also, condition, dynamic panne (1.4.16) 2%) force) R(max (2sraric) max(spring rayne) |y| So, overload (1.4.17) 2F, F, dynamic F, This illustrates the ofdynamic effect. importance simple example aspacecraft aspecified 1.4.1 to orientation C onsider rotating angular using Example a anapplied is at r est. control moment. For under spacecraft initially rigid spacecraft The moment 7, applied 16=T as moment moment isexpressed Iisthe where control ofinertia ofthe The spacecraft. T=-K(0-0,)—CO is where final Kand the ofmotion Carecontrol desired and Then, gains. equation 6,isthe angle, 16+C0+K0@=K@6, canbe as tothe The solution above ofmotion equation expressed +C, cos@,t A(t) sin@,t)+6, *"(C, one c anfind the initial that

= =

=

—

=

=

=e

conditions, Applying

|2] â„¢Csin(a,t+ 6) +0,

C= 0, C,=-0,,

Alternatively, 0 =e

where

=

DYNAMIC RESPONSE OF SDOF SYSTEMS

1.4.2AStep Load Duration ofFinite F(t)

Fo tp 1.8 duration. Astep load finite Figure of

° OS¢tSt , F(t)=F, Vv (1.4.18) 0$tSt,, 1.4.1 For ofSection still holds. equation (1.4.12) t t>¢,, force when for the free vibration with the system >t». Applied disappears Accordingly, undergoes at t= “initialâ conditions tp. att=tp, From ofSection 1.4.1, equation (1.4.12) cost) (1.4.19) F, sin@,t» H(tp)= 4 is For free the ofmotion vibration, equation undamped (1.4.20) myt+ky=0 to is that the solution the above Recall equation homogeneous (1.4.21) sin@,t y=C,cos@,t+C, ll

vl6»)=-P(1.

.

and

(1.4.22) onecanshow inequation that the initial conditions free t=¢, (for vibration) given (1.4.19), Applying C,= -1) (1.4.23) ESiN C, tp sin@,t+C, cos@,t) jy=@,(-C, at =

(0s, 7p

INENGINEERING INTRODUCTION TO VIBRATION

1.4.3Impulse Loading over case atrest, — € isinitially isapplied inwhich 0o=tan 1-R*

(2.1.25)

mo) ese]

ratio R=—: where Q frequency a, into and equations (2.1.14) (2.1.15) equation (2.1.20), Placing

32

(2.1.23)

(2.1.24)

STEADY-STATE RESPONSE OF SDOF SYSTEMS asfollows: Insummary, y,canbe expressed YpYsin(Qr-)

(2.1.26)

=

are asfollows: where and @ Y, amplitude phase expressed angle

1 F, + okVa-R?Y (2gRy 1 2¢R prt TR

(2.1.27)

_.

(2.1.28)

arealso ingraphical inFigure 2.1. shown form They Note: resonance at R=1. For small ¢, occurs 1) static ofF,. load under 2) Ysrarre displacement Yomagnification From factor 3)B,/k (MF). equation (2.1.27),

rs =

?

:

——__1_ = +(2gR)’ V(Q-R’)’ Consider

(2.1.29)

MF

cases asfollows: 4) three special R30. 0:The isstatic. Then MF>1and response ¢—(Gj) and (i)R=1.Then 2 MF=—== 2¢ Reo.Then > (iii) MF> 0 and 7. Resonance: a occurs tothe QQ isclose Forlightly when the natural freresponse system, damped frequency large forcing resonance. are resonance a is This The the called natural also called frequencies. quency @,. phenomenon frequencies resonance. For avoid system (1) safety, canbe a to Resonance For used determine the natural of this, system. @, (2) experimentally frequency resonance occurs. resonance occurs at at F sin Qwith t variable Q apply and look for If then Q, Q,which ~

=

@,Q,. ~

Q,,

33

INENGINEERING INTRODUCTION TO VIBRATION 5 4

: —

----

—

\"

—--

(a) 6=0.05 (b) €=0.20 © t =0.0 (d) C=1.00 }

m2 @

(a) 6=0.05 =0.20 (b) € (©) 6=0.50 (d) F=1.00 1.5 2 2.5 3 —

----

—

—--

2.1 factor and angle. Figure Magnification phase

Note:

Ttcanbe shown for that F=F, cosQt, ¥p Y, cos(Qt—-) isgiven inequation inequation where and @ Y, (2.1.27) (2.1.28). =

34

(2.1.30) (2.1.31)

STEADY-STATE RESPONSE OF SDOF SYSTEMS

Alternative Approach: now a asa cos We Qtand sinQt consider function variable such may F, F, expressed complex forcing combining

that

(2.1.32) F=F,(cosQt+isinQt) Fie aSDOF toaforce inequation isthen For the ofmotion system (2.1.32), equation subjected (2.1.33) my+cyt+ky Fe =

=

or

j+02y=— the solutions mRe 54260,

as particular Setting iQe

(2.1.34)

_

Iya¥e

(2.1.36)

Ye=-Q’Ye™ i,=iNYe™, yp(iQ)

(2.1.36)

Then

=

into and equations equation (2.1.35) (2.1.34), (2.1.36) Substituting

-O’ Rei 4 1 —Q? (@; +i2¢@,Q) =— Ye From the above Fee Ye+2¢@, )+@7 (i2Ye™ Yem =

—2

i

O24:

_

pio

(2.1.37)

equation iQe E0im eee’ YX ;

_

,

_

—Q? m(@, +i2¢@,Q) ei Fie ; » @ F, k Q 1—-| +i]2g (1-R? ov― mo| +i2gR) 28) , | — Qo, _

,

_

i

—

note We that

1-R? Ae® +i2¢R A(cos@+isin@) =

=

(2.1.39) 35

INENGINEERING INTRODUCTION TO VIBRATION

From the above, equation

Acos0=1—R’, AsinO=2¢R

(2.1.40)

the above from relations, Then,

and=F = >0=tan* +QgR)

(RF

A=

+

R2

Accordingly, 20 20 eit

YeE k —

—

ei) -&

E

=

Yeit)

(2.1.41) (2.1.42)

Ja-R?P+(2gRPe® JQ-R?Y +(2gR)!? Owith

¢, Replacing 2¢R ypYeY, tang (2.1.48) [cos(Qt-@)+isin(Qt—O)], 1-R* inequations which solutions the ofthe and with combination agrees (2.1.26) (2.1.31). =

=

=

2.2 Force tothe Transmitted Base

on rigid 2.2 SDOF foundation. system Figure a : force tothe transmitted base E,

(2.2.1) (2.2.2) and (2.2.3) =Y,Qe0s(Qt-$) into and equations equation (2.2.2) (2.2.3) (2.2.1), Placing (2.2.4) F,=kY, sin(Qt—)+cY,Qeos(Qe—-) F,=ky+oy F is For F, sinQt, recall that the response steady-state y=Yj sin(Qe-) =

36

STEADY-STATE RESPONSE OF SDOF SYSTEMS tothe tothe the force transmitted base The represents response. (2.2.4) Equation steady-state corresponding can a asfollows: in sine time side of be function of equation (2.2.4) expressed right-hand single (2.2.5) E, =Fsin(Qt-@+/)

where

P=J(kY,Y +(cY,Q)

(2.2.6) (2.2.7)

—

y=tan™ 2¢R 0 tan From and equations (2.2.5) (2.2.6), =

py

(2.2.8) =kY, kY, (2¢R)’ +(c¥,Q) 1+ |F,|= F=,/(kY,)

max

=

anondimensional measure onemay tothe As offorce transmitted introduce force TR, base, transmissibility as defined follows: ree (2.2.9)

me F

into equation equation (2.2.8) (2.2.9), Placing

(2.2.10) ( 2gRY FG5Ry 1+ =MFJ1+(2gR) k uf

“e

TR=0

where

=

-o

+(2gR)° V(l-R?)’ tt

MF=

Accordingly, 1+(2gRY (1-R°) +(2gR) R= isshown ingraphical inFigure which 2.3. form [ra Note: For R=1, S 1) TR=1 forR= 2) V2

(2.2.11)

_

Trost

(2.2.12)

37

INENGINEERING INTRODUCTION TO VIBRATION as increases. todecrease > TR increases ForRV2, TR. decrease ¢ So, 3) damping

2

— \

\

----

\\

Ls

(a) €=0.10 (b) €=0.25 (©) 6=0.50 1.00 (d) C=

—

—

||

—--

eo 0.5

0-4 0

Waa ==~~~4

2

3

4

Q/a, versus 2.3 Force ratio. Figure ransmissibility frequency

2.3.Rotating with Mass Unbalance Systems Various SDOF with unbalance be modeled sinusoidal load systems system rotating subjected transmission tail attached power shafts, system engine wing Examples fuselage, helicopter a attached tail automobile aircraft fans and machines. boom, vertical wheels, laundry with ForSDOFrotating unbalance be motion, mass

are toa a e

can

toa

asa

toa

an

or

a

or

rotor

mass can a mass m, system by single undergoing represented asshown axis inFigure 2.4. offset distance from the ofrotation

mass 2.4 with unbalance. system Figure Rotating

38

STEADY-STATE RESPONSE OF SDOF SYSTEMS canbe asfollows: via law ofmotion derived Newton's second Equation 2 =—6j t Qt) (=m, 94m, =

(yes

mass. is where the total m

sin 9+ —cy— Qt) ky (m=, )¥+m, m,e(-Q― sin Qt Qt my toy kym,eQ? t+ EF, sin =

(2.3.1)

=

=

(2.3.2) (2.3.3)

where F,m,eQ? The side ofequation the vertical ofthe force. The represents component right centrifugal steady-state to(2.3.2) =

is response equation (2.3.2) corresponding y=Y, sin(Qr—-)

where

1 meQ? Y,=

_

Moe

R?

(2.3.4) (2.3.5) ~

+(2cr))™ (2cR) f(1-R?)’ RJR?) + a

2¢R 9 =tanI-R

(2.3.6)

a gas onanaircraft rearfuselage. 2.3.1Consider turbine mounted engine Example mount The Nand the natural ofthe and 5,300 e ngine weighs frequency assembly engine is56.75 is is The turbine out b alance and the u nbalance of gas rad/sec. represented mass is 38.82 N-mm where is the u nbalanced and the distance. o ffset by e mg m, e maximum tothe at the transmitted the Determine force aircraft when turbine operates 0.1, Assume rom. 6,500 ¢ =

=

Solution: 1 myeQ? Hany ke [Gee _

0

39

INENGINEERING INTRODUCTION TO VIBRATION

82107 em,38.8210" kg-m 9.81=3.95710― =

2

=1.74x10° N/m k=m@? 5675) =

680. 34 6500X2X7 Q= 680.34 R= =11.988 rad/sec, s ec 60 56.75 =

m Y,7.36810 Accordingly, tothe Maximum =33.3N force transmitted fuselage kY,/1+(2¢R)’ =

=

2.4 SDOF with Base Motion Systems

a

aninstrument on Consider mounted vehicle. package flight

amoving 2.5 with base. SDOF system Figure

ofthe package ofthe mounting damping constant of k:springthe mounting instrument ofthe y:displacement d:displacement base ofthe mass over the and the forces it, acting Isolating considering m:mass instrument c: constant

(2.4.1) my =—k(y—d)—c(y—d)

40

STEADY-STATE RESPONSE OF SDOF SYSTEMS

Relative displacement: the relative displacement: Introducing n= y~4

equation (2.4.1), —Chp =—kyp m(Fp +d) =—md Mn +H thyy

to

isoscillating Ifthe with Qand Asuch base that frequency amplitude d=AsinQt

Then, d=—AQ?sinQt

and becomes equation (2.4.4) t =MAQ’ sin Qt sin Qe Mp +CVq hyp where F,mAQ? canbe as (2.4.7) Equation expressed sin Qt Fn +260,AQ’ or to The is response equation (2.4.7) (2.4.8) steady-state corresponding =

=

F,

jp+yp O? =

YqY,sin(Qt—6) =

(2.4.2)

(2.4.3) (2.4.4) (2.4.5) (2.4.6) (2.4.7) (2.4.8) (2.49)

where y,

,

=(2)

mag

.

YOR +(26RY J-R?Y+(2gRP k=?)

ran __-1_25R Oe TR tothe Force transmitted base:

E, =hyg tGp

|F,|=F =kY,J1+(2¢R)

max

2.4.10 ea)

(2.4.11)

(2.4.12) (2.4.13) 4]

INENGINEERING INTRODUCTION TO VIBRATION

Maximum absolute acceleration: From equation (2.4.1)

mj=-K(y—d) 0 cA)“hy,

and

(2.4.14) (2.4.15)

j= m axl ¥F

maxm| From

equation (2.4.13)

| V1+(2sRy “bY, A Q? Also,max|d |= max

(2.4.16) (2.4.17)

equation Then, using (2.4.10)

1+(2¢RY | ¥| max|d| +(2gR)’ —J(1-R?)?

max

_

tothe for isidentical in(2.2.12). TR (2.4.18) equation Equation Absolute displacement: From equation (2.4.1)

myt+cyt+ky= kd+ced =kAsinQt+cAQcos myt+cyt+ky Qt for

(2.4.18)

(2.4.19) (2.4.20)

response ‘Then, steady-state

1 AQ 1 kA (01-0) sin(Qt— Ok +(26RP mn

IG—R?)co

")

14|E

cos(Qt— 1-9) (2.4.21)

2°

A or

42

(2.4.22) sin(Qt-O+y) (1-R’)’+(2gR)

STEADY-STATE RESPONSE OF SDOF SYSTEMS

1+(2gRY sin(Qt-9+7) +(26R) Ja-R?)? max |y| J1+(2cRY +(2¢R)° =

(2.4.28)

(2.4.24) on flight Nistobemounted 2.4.1Aninstrument 890 package weighing Example a Anelastic is used to the v ibration t he base. vehicle. from of mounting protect package a H z. is 1 8 The base withfrequency of vibrating maximum is the the acceleration allowed forthe Under condition, s teady-state p ackage maximum acceleration base. Determine the bound tothe ofthe ofthe effec10% upper tive constant t he of spring mounting.

max|d|/(1—R?Y

Solution: 11

43

INENGINEERING INTRODUCTION TO VIBRATION 5 5pct 11 11 @? by 211w@? < Beto11na?

Q?

>

m

Accordingly,

k + =0 =0 M(ti@)’ ge Ke“ (K-@°M) ge" (3.2.4) From equation (3.2.4), (3.2.5) (K-@’M)9=0 must isahomogeneous For be nontrivial satisfied: , the (3.2.5) equation. equation Equation algebraic following det =

2

:

(K-@°M)=0 where det stands for the determinant.

(3.2.6)

arecalled Kand For holds for values of These values M, @*. equation given (3.2.6) eigenvalues specific specific todetermine onecanuse isa For @, Note c alled that 9, equation equation (3.2.5) eigenvector. (3.2.5) given an aconstant In are @ isdetermined within vibration, equation. eigenvector homogeneous Accordingly, multiple. are natural natural modes. called and called eigenvectors undamped frequencies inthe 3.2.1Let's consider the shown sketch. Example system

nodamping. 3.4 with Two-DOF Figure system

50

DYNAMIC RESPONSE OF MDOF SYSTEMS isasfollows: The ofmotion equation

O + k,-k, L F, M, OoM,on 4 -k,k+k, , hb E, or Mq+Kq=F For free vibration (F= 0) Mg+Kq=0 _

(3.2.7) (3.2.8)

,

case: the consider Now, following

=m,M, M, =2m,k, =k, =k,

Then,

3.2.10) 3.2.11) 3.2.12)

asfollows: To determine the and thus natural let's (a) frequencies, proceed eigenvalues

For free vibration,

(K-@’M)g=0 as

(3.2.13)

abeds SPs}

(3.2.14)

M

as Kandshown inequations isexpressed With and equation (3.2.13) (3.2.11) (3.2.12),

or

(3.2.15)

where

(3.2.16) 5]

INENGINEERING INTRODUCTION TO VIBRATION or

by ile

|o 2-27 ¢,[{ 9 For nontrivial -l |=0 det1-p -1 2-25

~~

(1-p)(2-25)-1=0 — 2p°-4p+1=0 weobtain From equation (3.2.19),

(3.2.19)

-1

,

|}

(3.2.18)

_

2

4

p=it or

=1--—=0.293, =1+ B, ~1.707 1+ p,

P=1B= 50293, From equation (3.2.16),

a1.

(3.2.20) 2,

(3.2.21)

and

k m =,mF-92938 O, k 1.707 k > k , =5, @,=,|1.707

[0.293

m _, O, =

=

m

m

m

(3.2.22) (3.2.23)

asfollows: To thus the natural modes let's determine the and offree vibration, (b) eigenvectors proceed into p,inequation equation (3.2.20) (3.2.17), Placing

V2, i-f-y) 2.08 2=,| {f{3h a

-1

52

Q-

_

%

0

2.

DYNAMIC RESPONSE OF MDOF SYSTEMS 1

{¢}-{o} qh: nae -l1a >,

(3.2.25)

canbe asfollows: intwoequations Equation (3.2.25) expressed

=0 -$,+V29, V2, is Wethat

(3.2.26a) (3.2.26b) see are to two the identical So, equation (3.2.26a) equation (3.2.26b). equations linearly dependent, one twounknowns. isonly and there for equation From equation (3.2.26), (3.2.27) =1 =0> + =0 -$,

5h a" 1 d,

1 Inmatrix Ifwechoose then =1, @, @, form =

o-{’ |-|| fre} ofa) {2-3} [oa V2 ; Peal | |-{ yeh

(3.2.28)

to Note an vector constant iscalled itcorresponds The above indicates that that any eigenvector. p, Subscript“1― a tothe isalso ofthe above solution eigenvector equation. multiple .

into p,inequation (3.2.20) equation (3.2.17), Similarly, placing

3

[OJ18

“1 4

%

so “!

(3.2.30)

canbe asfollows: intwoequations Equation (3.2.30) expressed 1

=0 -$,-V20,

(3.2.31a) (3.2.31b) ~$, 20, are note to two isidentical We that the So, equation (3.2.31a) equation (3.2.31b). equations linearly dependent, twounknowns. is oneequation and there for

-

> h=0 =0

only

53

INENGINEERING INTRODUCTION TO VIBRATION

From equation (3.2.31), 1

= 1 Inmatrix Ifwechoose form =1, $, then@, yr"

he

=

9%, 1

2-|is} o-{

(3.2.32)

an vector to Note constant isalso itcorresponds The above “2― indicates that that any eigenvector. p, Subscript a to isalso above solution ofthe eigenvector equation (3.2.30). multiple .

Summarizing:

The mode: first =

1 the the natural first first natural mode @, 2 1 1 frequency,

fo293= ;-||-|

Q,

:

/a

3.5a 1. Mode Figure

The second mode: 1 : the natural natural second second mode @, Q,=, : the 2 2 frequency,

1.707% ;||-|| =

1/3

2. 3.5b Mode Figure

54

DYNAMIC RESPONSE OF MDOF SYSTEMS asystem masses aspring. 3.2.2Consider connected oftwo Example by

aspring. masses Two 3.6 connected Figure by

Note:

to isallowed The translation. system undergo rigid body ofmotion: Equation

(3.2.33) or

(3.2.34) Mg+Kq=F For free vibration, (3.2.35) Mg+Kq=0 M : cases: mass ratio —+=c =m, Let's consider the M, following 2 Determination natural ofthe and thus the (a) eigenvalues frequencies: .

For free vibration,

(K-@°M)9=0 be

(3.2.36) Equation expressed can

as

(3.2.36) (3.2.37)

or

(3.2.38) 55

INENGINEERING INTRODUCTION TO VIBRATION

where or

2

a.1-3/4, 1]|_Jo

p= -1 1-p 1

% For nontrivial ,

>,

[> [o

l1-c ? -1l -l1 il-p

or

is] a (1-cp)(1—-

p)-1=0 > 1-(1+c)p+@p"-1=0 plp—(c+1)J=0 weobtain From equation (3.2.42), >

c 5_k

f,=——

=P

__k and0,>=~p,—=0 m > @,=0 k k > o,=ct1k 0, moeom cm Determination ofthe and thus the natural modes offree vibration: eigenvectors (b)

ctl =3,

(3.2.39)

(3.2.4 i

(3.2.41) (3.2.42)

(8.2.48)

(3.2.44) (3.2.45) (3.2.46)

into p,inequation equation (3.2.43) (3.2.40), Placing

56

Ie ||-|

t E

(3.2.47)

DYNAMIC RESPONSE OF MDOF SYSTEMS canbe asfollows: intwoequations (3.2.47) Equation expressed =9 (3.2.48a) 9-9, =0 (3.2.48b) —o, +, see one to twounknowns. isidentical isonly We that there for So, equation equation (3.2.48b). equation (3.2.48a) From (3.2.49) equation @, (3.2.48), =@, w e =1.Inmatrix If choose then form: =1, @, @, Ja JAE (3.2.50) vector note constant of isthe is We The above column first that the above any eigenvector. eigenvector multiple a to also solution equation (3.2.47). into (3.2.40), p,inequation equation (3.2.43) Similarly, placing 1 1-(c+1) Jo % (3.2.51) 4 |

{|-{1|

{ef-{eh 5alt ~$,-*9,=0 -c

-l

4c d,0

ve

canbe asfollows: intwoequations Equation (3.2.52) expressed =0 —, —,

(3.2.53a) c (3.2.53b) one note to two isidentical isonly We that there for So, equation equation equation (3.2.53a) (3.2.53b). unknowns. From (3.2.54) 6,=—c@, equation (3.2.53), =—c Inmatrix then Ifwechoose form: =1, @, @, I ~e (3.2.55) >,3 .

0-1"

-|

57

INENGINEERING INTRODUCTION TO VIBRATION

).

vector to We note constant isthe The above column that second any p, eigenvector multiple (corresponding a to is ofthe above also solution equation eigenvector (3.2.52).

Summarizing:

The mode: first

-|

0-0, 9,

(3.2.56)

1for 3.7a Mode translation. Figure rigid body

Note: isarigid isnostrain inthe there Mode mode and stored energy spring. body 1 The second mode: +1

o,=25 c

(3.2.57)

motion. 3.7b Mode Figure 2forelastic

Free Vibration ofUndamped MDOF Summary: Systems For Ndegrees free with offreedom, vibration (F0)ofasystem (3.2.58) Mg+Kq=0 asolution vector. isan NX1 ishomogeneous DOF The above and admits ofthe equation where following q form: (3.2.59) =

q=9e" 58

DYNAMIC RESPONSE OF MDOF SYSTEMS noenergy or For p=+i@ with loss and response gain, oscillatory q=

ge"

into equation equation (3.2.58), (3.2.60) Placing

=0 (K-@°M)ge*

(3.2.60) (3.2.61)

From equation (3.2.61),

(K-@’M)9=0

(3.2.62) mustbe isa homogeneous For the nontrivial ®, (3.2.62) equation. equation Equation algebraic following satisfied: det (3.2.63) (K-@°M)=0 arecalled For a @ Ndegrees values of with offreedom, that system equation (3.2.63) Specific satisfy eigenvalues. are roots Neigenvalues there counting multiple separately. as @ @, isexpressed For each equation (3.2.62) =0 (3.2.64) (K-@7M)q, onecanuse todetermine For the The mathematical @,, @,. given equation (3.2.64) eigenvector corresponding exercise iscalled ofdetermining and eigenvectors analysis. eigenvalue eigenvalues From equation (3.2.64) (3.2.65) Kg, =@/M@, =

Note: i. natural number ofmode ; frequency 1)@ :natural to mode @, @, corresponding are For N Nnatural Nnatural with there and modes. DOFs, asystem 2)Forconvenience, frequencies are an in of from order @,arranged @such , that increasing starting 3)O,| J} 14834x10~ 1.4834x10~ —1.8648 107i J} x 2.4002 —3.0173x 2.4002 107 107 107i +3.0173x107i 107.9337 X107i +7.9337X107i ‘ 5.9734x10" 3 5.9734x 4.9033 +.4.9033 3.691810" 107i 3.691810 * 107i _

_

"|

—

_}

_}

—

78

DYNAMIC RESPONSE OF MDOF SYSTEMS are are toeach toeach We that other while other. observe and and @; @, @, @, conjugate conjugate can as 2X1 These also be follows: eigenvectors expressed -0.61803 » e» 9, ~0.61803 1 e's = 1

| QO | whereand =

0,=—0,

1 1 oi um > 0,= 0.61803 eo, 0.61803 arescaled vector toshow entries ineach Note the where that the column that with agree 0, = —6,. they case. zero for the damping eigenvectors =

inTime 3.4.3Numerical Integration aMDOF Consider model with the ofmotion, equation (3.4.16) Mq+Cq+Kq=F we can we can c onvert intime,work For numerical with the the above above integration equation. directly Alternatively, a as anew v to inthe section variable such that second-order first-order equation equation previous by introducing (3.4.17) q=v into equation equation (3.4.17) (3.4.16), Placing (3.4.18) Mv+Cv+Kq=F or (3.4.19) v=M'(F—Cv—Kq) canbe a asfollows: into matrix and combined equation Equations (3.4.17) (3.4.19) single v q 3.4.20) we canbe as (3.4.20) Equation expressed x (x,t) 3.4.21) =f

{:|-|c-4)| ("| _

3.4.22) 79

INENGINEERING INTRODUCTION TO VIBRATION

| |

(3.4.23) f(x,t)= M(E—Ce—Kq) wecanusethe in For numerical of t he Euler method f irst-order described integration (in t ime) equations, oramore as a ccurate 1 such the method. method fourth-order (RK4) Runge-Kutta Chapter anexercise inExample 3.3.3 inpart As solve the reader the with the ofmotion and may equation (a) problem a to in in time the initial conditions confirm that method for numerical solutions part (d) integration produces tothose in the identical given example.

3Problem Sets Chapter 3.1 the matrix, the For two-DOF stiffness and natural modes follows: matrix, system, a

mass

as given

are

wf2}efi} wef] wens Sf

B. Find the numerical values ofAand (a) Find the natural ofthe system. (b) frequencies to motion? Is the allowed (c) system undergo rigid body

athree-DOF 3.2 Consider shown below. system

mass toderive motion. Draw for each the of equation (a) free-body diagram m a otion inmatrix the ofmvector. theassmatrix, the the form, matrix, equation Express (b) Identify damping stiffness and the load matrix,

inProblem 3.3 For the shown 3.2, system

80

DYNAMIC RESPONSE OF MDOF SYSTEMS no outan todetermine Assume natural the and (a) (c= ). C arry analysis damping eigenvalue frequencies to 0 in is 1. Scale each the natural modes. such that the entry eigenvector (in equal largest m agnitude) Sketch the modes. (b) 1and 2. Check the between mode mode (c) orthogonality

acrude onateststand. inthe two-DOF The shown sketch model 3.4 ofarocket may system represent

M,=+M, =2M, M: total mass

=k k,=k, =k, toconstruct a mass for law the of Draw each and Newton's second equation (a) free-body apply diagram motion. motion inmatrix the form. equation Express Determine the natural model. and ofthe @, @, (b) frequencies Determine the that the natural modes and of model. Scale the modes such the @, @, (c) (in m agnilargest to in is 1. each mode entry tude) equal Confirm the of natural modes. (d) orthogonality acrude onalaunch Shown below model ofarocket 3.5 may represent pad. mass, 0.25M, M,0.5M,M:total 2

M,2

=

8]

INENGINEERING INTRODUCTION TO VIBRATION mass matrix. Construct the (a) matrix. Construct the stiffness (b) Determine For the natural and the natural modes. the natural scale such that the modes, (c) frequencies to in 1. isequal the each Sketch modes, (in entry eigenvector

largest magnitude)

Note:p=@ 2M Introduce acrude a sled onafrictionless inthe 3.6 The three-DOF shown sketch model for track system represents jet vibration. undergoing longitudinal mass matrix. the Construct (a) matrix. Construct the stiffness (b) Determine natural the and the natural modes. Sketch the modes. (c) frequencies Confirm ofthe modes. (d) orthogonality Note: For ineach Introduce the natural such that the scale @? entry modes, (in p eigenvector largest magnitude) to is 1. _

=

—

.

equal

F393

mass, M, 0.2M, M,0.6M,M:total =

=

a sled onahorizontal at 3.7 inthe shown Consider the model of track The sled, jet previous jet problem. initially a an a to blast delivered atthe constant rest, issubjected tail followed thrust by explosive by loading charge, the ofPapplied after blast immediately loading.

Note:

on isrepresented inthe For the blast effect ofthe form of M, impulse simplicity, I, mass tofuel isassumed isneglected. The loss burn track and the due frictionless, tothe Determine the initial conditions (a) corresponding tothe impulse. Derive the for mode. determine the initial conditions. Also, &, equation (b) corresponding rigid body Determine ©, (t). toanelastic Derive the mode. initial conditions. for determine the Also, Of, equation (c) corresponding Determine (t). toanelastic Derive the for mode. determine the initial conditions. Also, a,corresponding (d ~Determine (t). .

@, equation a,

82

DYNAMIC RESPONSE OF MDOF SYSTEMS vector Determine the nodal q. to (e) displacement in in(e). Determine the force each the (f) spring displacement corresponding atwo-DOF mass asfollows: matrix 3.8 matrix stiffness Consider with the and the system given

tothe was was atrest. Itturnsoutthat force which the acceleration of recorded system applied initially can as motion the be ensuing expressed

A

A(cos@,t+cos@,t) A(cos@,t—cos@,t)

4 0.707 Gn constant. Aisa the Determine where given following: =

=

mass. each ofterms and (a) velocity displacement on mass eachin ofA. force acting (b) alinearasshown two 3.9 Consider the connected sketch. springinthe pendulums by

canbe motion Show small rotational the vibration of for free that, equations (a) assuming expressed angles, as inmatrix form .

ci 1 cil 0 5a 1+6 m20101 a> +mgL -1 146qr 0 mg

-|I

kh show natural that the _

are out to (b) frequencies Carry eigenvalue analysis

[7 -/£

o,

,

=

1+26)

83

INENGINEERING INTRODUCTION TO VIBRATION

natural Find the modes. (c) Derive the for aas nd a, @,. equation (d) are Determine Initial conditions follows: the (e) given q,(0)=A,, 4,(0)=0, 4,(0)=0, 4,(0)=0, to for initial conditions the and Of, @,. equations corresponding Determine and (t). (f) @,(t) ar, Determine and q,(¢) (g) q,(t). not to@,. 1and isvery Describe and 6« but the behavior of4,(t) when thus close, @, q,(t) (h) equal, inExample 3.10 3.4.1 =1.6 the with Ns/m. C,C, Repeat problem =

84

APPROACH ENERGY

wehave Newton's motoconstruct Inthe the second law of previous equation chapters, applied wewill analternate tion isenergy Inthis ofasystem. that This introduce based. chapter, approach more a is convenient the when withsystem called method, equation Lagrange approach, dealing a it We number when involves with ofdegrees offreedom, especially angular displacements. large a not to inthis will attempt introduce formal ofthe equation proof Lagrange approach chapter. we a to to it its will few demonstrate that indeed Instead, apply example problems application can in masotion results ofare thatbe obtained the ofNewton's second law. equations by application follows: covered topics Specific 4.1Kinetic and ofdynamic energy energy systems potential via 4.2Equation ofmotion the equation 4.3Damping effect Lagrange wewill construct InSection the kinetic ofsimple and 4.1, energy energy systems, potential we a or the will that with rotational effect of observe for Here, system neglecting damping. canbe afunction aswell asangular of the kinetic of energy freedom, angular angular degree displacement wewill in Section introduce the without and 4.2, equation Lagrange proof velocity. Subsequently, to to it several itsvalidity InSection then and usefulness. 4.3, appreciate problems apply example wewill can as an effect ofdamping be eitherexternally discuss how the incorporated applied or a inlater force function. The will be u sed equation Lagrange introducing dissipation approach toexamine construction the ofdynamic and the ofequation ofmotion systems chapters stability via the element method. finite

85

INENGINEERING INTRODUCTION TO VIBRATION

4.1 Kinetic and Potential Energy Energy

ofDynamic Systems

4.1.1ASDOF system. Example

4.1 SDOF system. Figure Undamped

sma okt

T The kinetic energy: V=U= The energy: potential isthe strain where energy 4.1.2Asimple Example pendulum A:fixed pivot point L:length ofthe string =

U

(4.1.1) (4.1.2)

4.2 Figure Simple pendulum. Assumptions: islight The and assumed massless. 1) string mass. mass aconcentrated at isassumed The the ofthe end string 2) The kinetic energy:

1 +9° Taine 5mx +9") x and mass canbe as interms The ofthe ofangle @ ycoordinates expressed

x=Lsin@, y=LcosO

time derivatives ofequation (4.1.4), Taking dxdO =

=(Lcos0)0 40 dydy =(-L sin0)0

(4.1.4)

_

=

86

(4.1.3)

(4.1.5)

ENERGY APPROACH into equation (4.1.5) equation (4.1.3), Placing

5m(t8y m6? =116°

(4.1.6)

T=

or

Tat 2

2

where

: mass moment inertia I=mL’ of

asa isused The Ifthe energy: reference, pivot point potential height V=—mgL cosO

(4.1.7) (4.1.8)

anaxis. 4.1.3The about kinetic ofarigid energy rotating Example body K,

are The equations Lagrange

_ aod) 00°00! _, 4(oT)_ oT dt\do} aob v ob oTOV

“(5) ―

For M,=2M,, M,=M, T= For K,=K,=K 1 1 V=U=—K6@’+—K(o-0) 2 7 (g-@) into and equations equation (5.1.11) (5.1.12) (5.1.10), Placing

5 +5 MQL'6 OMoL')e +M,V'6

(5.1.10) (5.1.11)

(5.1.12)

M,L’6+2K0—Ko=F, 3M,V°6+ (5.1.13) M,U’6+M,L’6+K(-6)=E, Inmatrix form: (5.1.14)

INENGINEERING INTRODUCTION TO VIBRATION

Pasfollows: To the consider incremental the thrust find for and let's the work done expressions F, F,, by dW —(P cos) =—(Psin@)dx, dy, + dOLsingd@)(6.1.15) =—(Psin@)(Lcos0d0 Lcos$d)—(Pcos)(—L

sin —

or

+F,do dW =—PL(sin@cos0—cossin@)d0 F.d0 where —PL 0—cos@sin@) F, =—PL(sin@cos sin(@—@) =

=

F,=0 on Note that F,isdependent angular displacement. For small angles,

F=-PL(9-6)

Inmatrix form:

(5.1.18)

ol] tafe’ pL Fe f-x[: beef?

(5.1.19)

into equation equation (5.1.19) (5.1.14), Placing

6

or

or

m3 14ii¢ “L1 ML 1 6 ¢ 14} ““L1 1

(5.1.20) (5.1.21)

MU14} ° ¢

(5.1.22)

thrust parameter p=nondimensional be

(5.1.23)

Mg+Kjq=0

(5.1.24)

|

1

with

:

can as Equation (5.1.22) expressed symbolically

DYNAMIC STABILITY

where 1 ;massmatrix :

S 7 |

Kf= 2

P : effective matrix stiffness

on and isdependent Note that P NONSYMMETRIC. K,y can anondimensional asfollows: We time further the above equation simplify by introducing Ksuch that: Divide equation (5.1.22) by

oar Gee deemed os

ascaled asfollows: Introduce time“ tT― defined

T=at

Then,

dtdtdtdt 6-0 at dOdo 2" _

(5.1.26) (6.1.27) (5.1.28

_

.

g6=—= de dt

Also,

? dod 2dt de! into and equations equation (5.1.28) (5.1.29) (5.1.25), Placing do 2 3 14|) de° G2 [| 2-P-1+P@(_Jo K 1114]do -1 1 d 0 Mol dt x,

—FP_

GP

5

| Wechoose such that can

a

2

Mtr

a’ K =1 or

K

(5.1.29

>

|}

(6.1.30)

INENGINEERING INTRODUCTION TO VIBRATION

Then,

ao ])

9 0 3.1 dr? -14+P + 2-P 1 14]do -1 1 o 0

a

with

|J =

(5.1.32) (5.1.33)

canbe as (5.1.32) Equation symbolically expressed iq=0 Mg+K

where

(5.1.34)

1 M 3 11

(5.1.35)

g=dt― dt―

(5.1.36)

-|#9do

(5.1.37) (5.1.38) canuse toexamine We the for P. equation (5.1.32) given stability

5.1.2Static Behavior to isreduced For static the ofmotion condition, equation

[Pr s+}

For nontrivial solution (i.e., non-straight configuration), -1+P-0 det2-P -1 1

(5.1.39)

(5.1.40)

DYNAMIC STABILITY matrix is determinant above the ofthe However,

2-P+P-1=140

(5.1.41)

Ht

ve

to is the and solution equation (5.1.39)

notbuckleand remains itcanbe ifthe the does unstable However, So, system statically dynamically straight. is thrust level

high enough. 5.1.3Dynamic Behavior ofmotion: Equation Mg+Kjq=0 to isofthe The form solution equation (5.1.43) q= into equation equation (5.1.44) (5.1.43), Placing

(5.1.43) (6.1.44) (6.1.45)

Ge"

(Ky +4°M)=0 (5.1.46) det(Ky +4°M)=0 The which the solutions ofequation be (k=1,2), (5.1.46), expressed A, eigenvalues For nontrivial ®,

are

can

as

(5.1.47) >0) (@, A, Re(A,)+ilm(A,) = 0, +i, where 0,=Re(A,) (5.1.48) Then, Ttisin@,T) 0% OHME oFlOOMF (5.1.49) (cosey,. time where ofamplitude with e°" change isin oscillation with @,T @, (cos “frequency― @,T) >0for criteria: If0, =Re(A,) the ofvibration excited initial disturbance will any k, Stability amplitude by to intime. isthen The be said unstable. system grow dynamically =

=

X

~

~

=

INENGINEERING INTRODUCTION TO VIBRATION

Kand Mas Expressing mM, My ky, ky M= KyRyRog My,My as canbe written Equation (5.1.46) =

(5.1.50)

|

AA‘ +BA’>+C=0

(5.1.51)

where

A=My My, Bak mythm, km, km, C= Ryko —kyokp, From equation (5.1.51),

Mm —M

—

(5.1.52)

—

VB’ Weobserve follows: -4AC R -B+ 2A

(5.1.53) can as For the for small and and all and thus 0 A? system 6, A=+i@,, P, (1) +i@, Re(A,)=0 isdynamically stable. we roots =0andobserve As —4.AC double for Awith B* @,@,. Pincreases, (2) one As —4AC t he further and the real of of becomes and the increases, P P

A

Frequency

oS |

a

>P

5.2 Real and model under oftwo-DOF Figure part plots frequency afollower force.

DYNAMIC STABILITY

Note:

cancheck vector remains via We control. how the behaves ifthe thrust line vertical thrust body always

@®

kr

ex

(x2; y2) M) y P nodirectional 5.3 Two-DOF model under force with Figure applied change. v

canconsider We the incremental work done the thrust by

— +F,do

dW=—Pdy, =—PL(—sin@d0 sinodb)=F,d0 toshow that

F,PLsing FPLsin@, =

=

canthen as small the ofmotion be equation expressed Assuming angles,

(5.1.54) (5.1.55) (5.1.56)

(5.1.57)

P*-3P+1=0

the above equation, Solving

345 P= 2

(5.1.58) (5.1.59)

INENGINEERING INTRODUCTION TO VIBRATION twovalues tothe The critical load smaller ofthe the (i.e., load). corresponds buckling (6.1.61) P, 03800 wecandetermine into the above value the the of equation eigenvector (i.e., mode) (5.1.57), Placing buckling asfollows: P, 6 0.6180 (5.1.62)

25 {:

{e+ |

onecan outfree todetermine P(< Note that for vibration the natural and carry given P ,), analysis frequencies as as not is it the natural modes, The stable buckle does system long statically.

5.2 Dynamic Aeroelasticity

to a will first the We two-DOF of consider aircraft model flutter importance wing using appreciate simple inertia motion motion We between torsional will then other and ofthe introduce wing. coupling bending briefly as such aeroelastic flutter and stall flutter. phenomena panel

5.2.1AModel Wing asshown canrepresent two inthe Amodel with sketch essential characteristics of freedom the of wing degrees a the behavior ofreal wing.

5.4 model Two-DOF wing. Figure orupstream U:flight speed velocity vertical q,: displacement rotation gy: angle constant linear stiffness ofthe spring wing representing k,: bending

110

DYNAMIC STABILITY constant torsional torsional stiffness ofthe spring representing wing k,: e:distance center and axis between the the elastic (a.c.) (e.a.) aerodynamic

Note: axis center. isthe The elastic locus ofshear (1) use as to axisthe torsion. Wethe elastic reference from (2) structurally decouple bending are static the from q, measured (3)qa, nd equilibrium.

The ofMotion: Equation canbe asfollows: via For the the ofmotion the two-DOF derived model, equation equation Lagrange applying Strain energy: 1 1 Vea 2 Kinetic energy:

ka toha

(5.2.1)

a2

5.5 model vertical and rotational motion. Two-DOF Figure wing undergoing axis coordinate the chord from the elastic starting &: along mass m: per chord length mass oflength dT: kinetic of small energy dé aT2 —€4, +(E4, c0sq,)' sing,)"] 1,, + sin― cosqn qs (mdb (cos* qn)] 54rd. [gi 5°42 5 1,, + cosq, 26414 dr 5°43] =Fly. + T= 4,4, cosq, —2€

==(ihd€){(4, .

=

_

.

_

.

.

—

.

=

=

(6.2.2a)

linds) 5)ai €7q)\mnd& (5.2.2b) far fndé-4u4, |Eide Jere 54 54, ~

1

..

L2y

cosq, +1

4

INENGINEERING INTRODUCTION TO VIBRATION outthe tothe the from integrations Carrying leading edge trailing edge, 1 1 + cosqy ~Sudics

T=SMai Tada totalofthe wing frag of about the elastic Sue Side J

(5.2.3)

where M=

mass

: :

mass moment static

QA :mass moment inertia

J

.

.

axis

.

.

the elastic I,= md& of about termin inertia motion The second between the vertical and represents equation (5.2.3) coupling (bending) ameasure c.m. e.a. ameasure motion. is i nertia the t he also rotational of offset from theand of (torsional) S, c.m. noinertia e.a.,=0and is Ifthe coincides with the there coupling. coupling. For +1and S, small rotation, cosq, +1 T 1 Sadie (5.2.4) =a axis

glade Mai

Incremental work done: + dWFaq, E,dq, moment: force and Aerodynamic =

(6.2.5) 92 Meas €.a.

onthe 5.6 moment model liftand wing. Figure Aerodynamic acting : moment tothe axis with elastic respect M,, aerodynamic

M.,M,,Lecosq, M,,+Le moment to can

(5.2.6)

(37)

(5.2.7a)

=

+

=

Note for small the due be rotation, that, drag neglected. the equation, Lagrange Applying aT, + F, ‘ > =L Mg, -Sy4,+ kg, dt\

04,oq, 94,

DYNAMIC STABILITY

d oT+ov =F, -S,4,+1,4)4 kody dt\04,oq, 99, where F=L,FE, =M.,,

(27) >

=M.,

Adopting quasi-steady aerodynamic theory,

L=qS—~GQy qoC, Ja asfollows: isexpressed where the effective ofattack, 01,,, angle

=a-t So,BF ay

aC ft4 L=gqS— U

[«

pressure q:dynamic area S: ofthe model wing planform U:flight speed ofthe model liftslope wing

(5.2.76 (5.2.8 ) (5.2.9) (6.2.10)

S,

anairfoil term inequation section To the second let’s consider with velocappreciate (5.2.9), moving upward asshown ityq, below:

n u————> e.a.

Udt

Nan

5.7 ineffective due tovertical motion ofthe AOA wing. Figure Change overtime asseen anobserver inthe isshown The ofanairparticle with the dt, airfoil, moving displacement by sketch: inangle Decrease ofattack:

a fit

ao= t

(6.2.11)

INENGINEERING INTRODUCTION TO VIBRATION

from Also, aerodynamics,

“4 dC =qSc M,, de (5.2.12 ) c:the where chord length. on to isdependent isconstant shows that while with respect M,, M,, (5.2.12) Equation angular velocity, angle ofattack. Then, ac ac4 (6.2.13)

=GD

we

(1-2)

e4S +08 M,, =eL+M,, Fo ava Inequations and the and 4,) represent (5.2.10) (5.2.13), velocity-dependent (4, “aerodynam damping.â =

terms

are Insummary, two-DOF the ofmotion ofthe model equations wing

Mg, Sad hg, Lb + 1,4, M., SoG hog +

=

=

-

where

(5.2.14) (6.2.15)

oC q L=qS—+|q,-4 U

[« Fo [a i)va at Sella GSS 95-2 Ja, eosin RoaC,

(6.2.16) dC4 eqS +qSc (5.2.17) M,, =eL+M,, into and and and equations equations (5.2.16) (5.2.17) (5.2.14) (5.2.15) rearranging, Placing oc,1, (5.2.18) Mi+S kia,+ aC IG, aCy,+(Ryp)eorion (5.2.19) Sadi €4S t)Ut 12

aryadC

4

=

.

.

=

).

.

=

.

—

where

€95 0a (Reg vertical vibrationdriven rotational represents =

—

(5.2.20)

a and Equation (5.2.18) (torsional) displacement by (bending) a rotational acceleration while vibration driven vertical and acceleration, represents equation (5.2.19) by velocity. Static Case: Wing Divergence to case, static For and reduce equations (5.2.18) (5.2.19)

114

DYNAMIC STABILITY

ka dC

(5.2.21) (6.2.22) (5.2.23)

So 9 (Reponion where do dC, effective torsional spring Ro gS (Rp)eosion =

=

.

=

=

;

Fo (64

;

constant

constant torsional stiffness ofthe spring representing wing k,torsional : constant torsional —egS [64aerodynamic spring asthe onion ishigh airspeed increases. that Ifthe Note CeCreases pressure (thus dynamic q) speed enough, (k,-) 0 in torsion. is and the loses the torsional the will This fail stiffness, wing wing statically phenomenon at isdetermined The called from the=0condition, ie., pressure q,,wing wing divergence. dynamic divergence :

Se (keg)osion =

k,,

~egySE=0

ky

(5.2.24)

Then,

kL ID=— eS 0a

JE

and the U,is wing speed divergence

(6.2.25)

2 =,,—2 (6.2.26) U, p air where: density Note: isastatic 1) W ing phenomenon, divergence ebetween can a.c. e.a. most to i ncrease the the the The T o decrease offset distance and q,,we try important 2)single e. is for the offset distance parameter wing divergence wecan toincrease i ncrease in T o torsional this also the stiffness. lead However, q,, may trytoincrease 3) weight. Case: Inertia and Flutter Dynamic Coupling What ifc.m.coincides with (A) =0)? e.a.(S, asfollows: =0and the the reads ofmotion equation S, ignoring Setting aerodynamic damping, oC, Mg, 48 k,q, Jabh 9 (6.2.28) Tidy dension (keg Io +

+

=

=

INENGINEERING INTRODUCTION TO VIBRATION >9, intorsion vibration with the natural free of represents equation (5.2.28) frequency (ky)onion Bap Iension

For

|

torsion

a

aSDOFof toaforce toq, qsubjected , represents system (5.2.27) Equation proportional .

FDeorsion Deending [ki M =

notgrow time isnoflutter. Inequations The ofq,and with there and and q,do (5.2.27) amplitudes are not in structural and the ofdampHowever, presence present. (5.2.28), damping aerodynamic damping out, motion DOF vibration thus torsional free (with diesand the vertical(with q,)also ing, ) DOFq, out. dies nowa a with cm.behind > the section For Consider with acceleration. (B) wing moving wing 0), upward e.a.(S, c.m. amoment rotates inertia section axis. force induces that the clockwise around the elastic This acting through increases AOA and the which the promotes lift, upward motion—destabilizing.

“1 |

v

behind elastic axis. 5.8a ofmass center Figure Wing a c.m.ahead rotates section For with the This ofe.a.(S, acceleration counterclockwise. wing 0) isthe ofdynamic called flutter, 2)Inertia (with wing S, coupling instability

DYNAMIC STABILITY c.m. or occurrence tothe toprevent isused Mass the closer forward elastic of wing axis) balancing (moving delay or mass areasfollows: via inertia flutter the of Examples eliminating reducing coupling, can rotor tothe balancing blades: lead be close 1)H elicopter placed weights edge. or ortabs: oraleading to amass arod can horn be of Control surfaces used ahead rudder, (aileron, 2) elevator) place the axis. hinge

5.2.2Flutter Analysis are: The ofmotion ofthe two-DOF model equations wing + =L Ma, -S,4, kg, + M., Ly, —SyGi kgs =

(5.2.29) (5.2.30)

where

aC q,-2 4 L=qS—4| D0 Uac

[« }¢4S>~ (5.2.29)(«.-4) expressed

(5.2.31) (6.2.32)

Erdeh opts] GS || ||

we

4

ac4s gSes M,,=eL+M,, canbe as inmatrix and form (5.2.30) Equations M -S,4 k,0 “a{_ Lb | |} =

or

=F Mg+Kq

where

h Ce

M M_-S,

SaI,

+

(6.2.34)

(5.2.35) (5.2.36)

INENGINEERING INTRODUCTION TO VIBRATION

(5.2.37)

(5.2.38) as vector Inmatrix Fcanbe load the form, expressed 1 0 ac, ac, F= L " 0a Lh 4S , eo Oe OL or 0a —

[of{*| =

4Sac, ~

F=4K,q-4qC,q

where

(5.2.39)

(5.2.40) (5.2.41) (5.2.42)

onq, vector Note Fisdependent that the load qand q. is the ofmotion Then, equation

Mg+Kq=F=4K,q—-4qC,q

Mgq+qC,q+(K—qK,)q=0

Mg+Cq+K,q=0

or

where

: matrix C=C, aerodynamic damping : effective matrix stiffness =K-«K, K, K:structural matrix stiffness : matrix stiffness —qK,, (nonsymmetric) aerodynamic

118

(5.2.48)

DYNAMIC STABILITY

Note that a k, —qS

K, 0 k,-¢Se 24 oC, Ja =

(5.2.44)

afree orsudden togust The initial above describes vibration disturbances due equation homogeneous following a at certain notdie outwith control the offree vibration does time, the If, condition, input. wing amplitude flight to issaid influtter. be occurs canbe asfollows: atwhich The determined flutter pressure dynamic to the ofmotion equation Neglecting damping simplifies

Mg+K,,q=0 to isofthe solution form

equation (5.2.45) A q=e" into equation (5.2.46) equation (5.2.45),

Placing

(5.2.45) (5.2.46)

(Ky +2°M)p=0 For nontrivial ®, det(Ky +2°M)=0 from whichfind A=

(6.2.47)

where+

(5.2.49)

(6.2.48)

A,, A,. eigenvalues canbe the used for the follower force discussed equation (5.2.48) procedure problem Following previously, as expressed +C=0 AA‘ BA’ onecan

A=MI,-S. B= oCoC

rahe fas (i

—qSe 0a da― aC, C=k, —qSe 0a From equation (5.2.49), A> ~B+VB*-4AC 2A _

(5.2.50)

INENGINEERING INTRODUCTION TO VIBRATION canthen asfollows: We observe Oe

qF

>4

a

ae

i

g

5.9 and Real part plots. Figure frequency >0and For for small and 0) (@, J,=Re(A,) +ilm(A,)=6,

where 0,=Re(A,)

Then,

+isin oH 0% HM! oFOHO @,t) (cosen.t =

=

—

(5.2.62)

time where ofamplitude with e°" change with oscillation @, (cos@,t£isin @,t) frequency criteria: Flutter out.

by flutter. experiences ~

~

DYNAMIC STABILITY

two The lowest each other and qy-coalescence. @, @, approach frequencies no at This is Under the of q,. assumption @, @, frequency damping, Flutter with included: (2) analysis damping wecan asfollows: convert case, Inthis intofirst-order equation equation (5.2.54) (in t ime) v such anew variable Introduce that =

a

a=v

into equation (5.2.63) equation (5.2.54), Placing

Mv+Cv+K,,q=0 and be combined follows: (5.2.63) (5.2.64) Equations can

as

(6.2.63) (5.2.64)

Ewis. }{3} “He

(6.2.65)

I Bz-—Az=0

canbe as isanidentity matrix. where Equation (5.2.65) expressed

where Z=

v

loa

(5.2.66)

(5.2.67) (5.2.68)

(5.2.69) to isafirst-order isofthe Asolution equation. equation (5.2.66) (5.2.66) Equation homogeneous following form: Z=Ge (5.2.70)

INENGINEERING INTRODUCTION TO VIBRATION

into equation (5.2.70) equation (5.2.66), Placing > Bode“—Age“=0, (A—AB)ge*=0 (5.2.71) — (A-AB)9=0 or

Ag= ABO

(6.2.72)

Note: an isastandard Por Aand for B, given equation (5.2.72) equation (1) eigenvalue analysis. are at to # With but other. close each q,, (2) damping, @, @, they

5Problem Sets Chapter the launch vehicle 5.1 two-DOF ofslender flexible Consider model crude model and introduced asa as T time

teststand.scaled vertical Using defined a

a

ona

wasderived asshown inequation two-DOF the model ofmotion for the equation (5.1.34). are asthe asfollows: The initial disturbances initial conditions given expressed =0.1 =0, 0,6(0) =0, (0) #(0) (0) over100 to @ units inscaled time Use RK4 the method and for the @ P. plot following time? =1.8.Does the with grow P (a) amplitude P= time? Does the with grow (b)2,3. amplitude to inproblem Dtohe 5.2 ofthe 5.1. described system investigate following dynamic stability versus Plot real the the of part P. (a) eigenvalues versus two P. Plot (b) frequencies LVstructure Determine the becomes unstable. (c) P,atwhich dynamically =

DYNAMIC STABILITY

a

on a 5.3 literature Conduct search oflaunch vehicles and submit report “pogo instability― three-page typed

that includes the following: ispogo itoccur? What does (a) instability? Why Historical (b) to examples. How pogo prevent (c) instability. List ofreferences.

(d)

asubsonic asfollows: 5.4 Consider model with properties wing given M=8,7 5.4 84 slug, slug-ft, slug-ft’ S, I, x X 1.310° 6.310! lb/ft, Ib-ft/rad k, k, e=0.19 ¢=5.0 fe, ft, 150 fe,Ol=1872 =

=

=

= $=

“

the and do Neglect aerodynamic damping following: 1: Case sealevel at Determine Assume and conditions. q,, (a) wing divergence. U,, versus atthe onset Determine Plot the offlutter. pressure pressure q q,, (b) dynamic dynamic Re(A,). s ea at Determine the flutter level. speed U,, versus Do Plot and the coalesce? q @, @,. (c) frequencies e 0.38 onthe eon it 2:Repeat 1with Case Case C omment Does ft. effect of and flutter. wing divergence as as the affect flutter wellthe wing speed? speed significantly divergence onthe on 1with 3:Repeat 1.5 Case Case Comment effect of and flutter. wing slug-ft. divergence S, S, onthe on 1with 4:Repeat =—1.5 effect ofS, flutter. Case Case Comment and wing slug-ft. divergence S, are as 0, inProblem =0 For the with initial 5.5 the conditions 5.4, q, wing properties given given 4, q,=0, to Use RK4 the and the method do 4,0.1. following: Forq=0.97q,, and q, q,. (a) plot For and q q, q,. (b) plot 1.034,, moment x + and torsional with 1.3 do the 10°(1 Replace following: F(q,) 4q3)q, For0.974, and q, q,. (c) plot alimit oscillation. For q=1.034q,, q,and q,toobserve (d) plot cycle =

=

=

=

=

=

q=

INENGINEERING INTRODUCTION TO VIBRATION

a a a

on a attransonic search 5.6 Conduct literature flutter and submit report. wing regime three-page on nose a 5.7 Conduct literature ofthe and submit survey report. gear “shimmy― landing three-page aliterature on“chatter― nose a ofthe submit 5.8 and Conduct survey gear report. landing three-page on canbe a Mathieu into 5.9 Conduct literature excitation― of that m odeled survey s ystems “parametric a and submit report. equation three-page

ABSORBER VIBRATION toa atorclose toanatural Adynamic is could be that system subject frequency frequency forcing wecanalter amass a away case, in Inthis the such ofthe and system. system spring by adding are that the natural the ofthe from m odified system separated frequencies adequately forcing a mass in constitute level v ibration. The reduced of added and spring resulting greatly frequency, a“vibration canalso tofurtherthe A Inthis absorber.― be a dded vibration level. mitigate damper wewill can to examine how vibration absorber be introduced the suppress chapter, steady-state or we use a a will For vibration ofthe with system. system primary primary, simplicity, original, as an are as to offreedom be follows: discussed topics single degree example. Specific amass 6.1 Absorber with and spring 6.2 MDOF of response systems Steady-state amass, a a 6.3 Absorber with and spring, damper

a a

INENGINEERING INTRODUCTION TO VIBRATION

aMass aSpring 6.1 Absorber with and

an asinusoidal Q SDOF Consider the of under offrequency response system steady-state undamped loading asshown is below. for simplicity. Damping neglected

Qr sin q,Fy, Fo =

ky 6.1 under SDOF sinusoidal system Figure loading. is The ofmotion ofthe system equation sinQe (6.1.1) mq, Fy +k,q, resonance occurs. tothe Qisclose To that the natural and /m, @, Suppose frequency frequency /k, forcing we can a as to the and the vibration, suppress system system spring modify by mass(m,) (k,) adding original twonatural inthe inasystem sketch. This will with from the shown modification result away placed frequencies =

=

forcing frequency.

a

mass 6.2 with and absorber spring. Figure System nowatwo-DOF aredifferent twonatural that We have with from the Q. system frequency frequencies forcing a nswer to at is But select and To let's the what the best look steady-state of way m, k,? this response question, the modified system. is The the ofmotion for system equation

BERS Hh fis (ae

Jo 0 k, || =k, state solution ‘The ofthe above be O m, %

or can as particular equation expressed steady

L

A

|.

(6.1.3)

VIBRATION ABSORBER into and equation (6.1.3) equation (6.1.2) rearranging, Placing A,_|% +k, -k, k, -Q’m, 0 A, k, Q?n, -k, wecanfind From equation (6.1.4), (k, -Q?m (k, +k,—Q'm,)F, (ky —Q?m,)-k; A k,F, (k, +k, )(k, —Q'm, -Q?m,)k arechosen Ifk, and such that m, (k, -Q’m,)=0 Then

| ||-{

:

°

(6.1.4) (6.1.5) (6.1.6)

F, =0, A, A,=

and

=e =9, F, ,

qi q,=-—*sinQe

k,

(6.1.7)

mass notvibrate. From the ofthe does system equation (6.1.6), Accordingly, original

ke

=Q? (6.1.8) my mass and atuned issatisfied, constitute the ifequation The added vibraSo, q,=O. (6.1.8) spring (m,) (k,) can a tion is is It absorber. be shown forsystem with but small ifequation q,still that, q,#0, (6.1.8) damping, satished. mass ratios: the modified with the 6.1.1Consider system Example following

wecandetermine resonance as From natural free vibration the (or analysis, frequencies frequencies) =1.1705,fk, Case @, @, (a) =0.8543,/k,/m,, /m, Case @,0.8011 @, (b) =1.2483,/k, /m,, /m, a mass iswider. Case with natural ratio, So, separation (b), frequency higher a athigh is vibration of t he tail of 6.1.2 v ertical AOA jet B ending Example fighter asa m ass a constant to modeled of m, and harmonic SDOF system spring k,, subject =

Jk,

129

INENGINEERING INTRODUCTION TO VIBRATION

@due tofluctuation ofgiven ofthe flow. loading amplitude s eparated frequency F,and Itturns out that the is close to the natural ofthe verymass first frequency forcing frequency an itisdecided toretrofit constant to and absorber of m,and Due tail, spring k,. maximum isspecithe allowable ofthe absorber space limitation, displacement spring fied. the effect ofdamping and dothe Neglect following: mass onthe constant maximum Determine the the absorber and absorber based allowable spring (a) displacement. 48a > fy From (6.1.7), where value. 2 (q,) k, (q,) specified equation (42 ax From m, 2 equation (6.1.8), =o tothe Describe how would the absorber vertical tail. you system (b) implement

-2 7

max

=

max

6.2Steady-State ofMDOF Response Systems

aMDOF toharmonic sin is F F, Qt. Consider with The ofmotion system equation subject damping loading then =

sinQe Mg+Cq+Kq =F, For response, steady-state BeosQt q=AsinQt+ Then,

(6.2.1)

(6.2.2)

A= Bsn) =—-Q*(A Qt+ g sinBoos Qt) into and equations (6.2.2) (6.2.3) equation (6.2.1), Placing + + cos + +

O(A cosGat-

-Q?M(A Bsin sinQt cosQt and Collecting

(6.23)

sin sin sinQtBos Qt— QtBcosMt) Qt (6.2.4) F, Mt) QC(A Qt) K(A =

terms,

(6.2.5) (K Q’M)A-QCB-=F, |+cosQ¢| from (K-Q’M)B which, sin Qe

|=0

(K-Q°M)A-QCB=F, (6.2.6) (K-Q°M)B+QCA =0

VIBRATION ABSORBER canbe a asfollows: two into The above combined equations equation single —~Q?QC fo FE K-Q*M Al_} (6.2.7) |(B 0 QC K-Q"M canbe anN-DOF canbe For A B. For the above s olved and for Q, system, given equation equation (6.2.2) as inanexpanded form expressed q A, B, _

B, op)A,psinQtt+) ¢cosQt =)

°

*

(6.2.8)

IN Ay By cosQt q,A, sinQt+ =

or

B,

(6.2.9)

sin(Qt—-@,) =A) +B; =|q,| sin(Qe-9,) where q,

(6.2.10)

ofqi (6.2.11) +B’: =A? q| amplitude aMass, aSpring, aDamper 6.3.Absorber with and

6.3 with absorber with Figure System damping. mass toa isattached vibration and with absorber ofmass m,, c, m,, system spring primary spring k,, damper a to sin Qt. and force of c, Then, F F, subject k,,damper m, -c : O ° cto, k, +k,—k,E fn “kook

A

=

Je-| ‘| w-| e-|‘ 0 m,

cy Cy

2

2

0

(6.3.1)

INENGINEERING INTRODUCTION TO VIBRATION .

mass ratio m.

»|

:

With B=—+, m1 Also define M=

For See Section y=1(Note: 6.1),

and

"2k, =

+ k,+k, k, Bk, -Bk, -k, a -Bh, Bh, Accordingly, 0 & 148 Q (K -p 1 0 |1 1+B -Ba] 4

K-|}|-B)|4|| iF o f aw) *] al, | -|[ | OB (K-9°M)=k, _

_

(63.3) (6.3.4)

k, Bk, =

(6.3.2a, b)

=

7;}

(6.3.5)

ant)

_

and finallyM4

B-Ol -B BA-2")

where o=Q

(6.3.6) (6.3.7)

=

.

wecan asfollows: For the Qm C atrix, proceed

26,9 fA

25 p= mo =

VIBRATION ABSORBER

[Bm [tm, hig "Jo shay | ne=o] fpas[ (268m) 7 (asin) lala|

= 26 26 with c, ac, Then, + Co CC, 1 cy Cy -l -1 (2¢BQ -1 =(2¢p k Q @+1 =k]a@+1 O21|=k, m,

(6.3.8) (6.3.9)

=

=

—

_

1 1

=

Finally, a=

(6.3.10) canbe to tosolve inthe section Aand and introduced for equation previous Equations (6.3.6) (6.3.10) (6.2.7) we can a as B.For introduce nondimensional convenience,

displacement : Ls ofq, amplitude AK

—_

,

canbe 6.3.1Suppose is of t he small and p rimary system Example damping n eglected. versus For for t he Q axis) (horizontal axis) following parameters: plot q,(vertical y=10, =0.01 and for¢ (a) B=0.1 B=0.05 and for ¢=0.01 ¢=0.1 (b) B=0.1 100 (a) —p=01,0.01 0.01 --B0.05, C= 80

+

=

mass 6.4 Nondimensional ratio. ofthe fordifferent Figure displacement primary system

INENGINEERING INTRODUCTION TO VIBRATION a mass ratio ratio For how the the of observe and 0 .01, you may response system given Baffects damping two

of natural ofthe modified system. separation frequencies

(b)

100

—B=0.1,6=0.01 --B=0.1,0=0.1 B is

80 |

IF 5 60

4

40 20 0+ 0

5

=

0.5

1

1.5

2

Q 6.5 ratio. Nondimensional ofthe for different primary system Figure displacement damping a mass ratio ratio For of0.1, observe how the affects the response. you may system given damping

6Problem Sets Chapter 6.1 For the in 6.3.1, determine nondimensional absorberof

mass 2and the problem Example displacement on case. force the absorber acting spring k,foreach mass mass toa i s with 6.2vibration absorber of and attached m,, c, m,, system primary spring k, damper canbe toaforce A sin is For the small of Qt. and system, spring F, F, areintroduced primary k, subject damping neglected. to The nondimensional the ofmotion. parameters equation following simplify =

m,

=

ofq, lal amplitude al mRk,

Q __ m,

and c, 2¢Bym, =

:

(®)

VIBRATION ABSORBER

For that y#1,wecanshow

K=k, and B=By’ where “

7 K(2abr0| ; , Mam |@,Q following B (a) plot OC=

1

and

onwhat comment For the parameters and observe: 0.1, you y=0.95 versusfor and ¢=0.01 ¢=0.1. versus onwhat comment For =1.05 for the and observe: Q you =0.1, parameters y % (b) 8 plot following and ¢=0.01 ¢=0.1. can an For ofy? and find value =0.01, you ¢ optimum (c)B=0.1 =

mass and constitute anabsorber inthe 6.3 For the three-DOF shown added system sketch, m, system springk, tosuppress excessive vibration ofthe system. primary ttohe ofmotion the ofthe Set two-DOF and determine up response system primary (a) steady-state aequation sin force Qt, Plot nondimensional and with F,F, subject displacement amplitudes (g,q,)respect toQ, tothe isclose Icturnsoutthat Q natural first ofthe Consider the condition system. (b) primary frequency mass to zero for vibration of m,select k,. of t he To the three-DOF with the vibration absorber for different response system operatinvestigate (c) to with nondimensional and Q. Q, 4, q,)respect ing Compare speed plot displacement amplitudes (%, with the responses. system primary to of For the three-DOF with force system, respect spring (nondimensionalized (d) plot amplitude F,) to each with with the the forces of respect system. spring Compare springprimary sin Qt q2, FyFo =

.

=

INENGINEERING INTRODUCTION TO VIBRATION

1 1 m=7 Mo

1

k,=K,, m,=M,, m=Mo» =ZK, Q Q= la.) = la,| L=(41) z (41) |as| VK./M, (9) static ofq,under F, displacement (4, Yt inthe For 6.4 the shown and m,isconnected m,via system sketch, m, '

°

sui

:

sac

on lever resting rigid twofulcrums. iseand the The the left the the distance between fulcrum and fulcrum distance between right mass

tomasses

a

isd.Rotation isassumed the left fulcrum lever small. ofthe m,and qv Fy

mass

state atwhich tomass inChapter is Review 2.7 and the condition the force transmitted 2, m, (a) problem tozero sin Ot. can for equal F, F, or to in Describe how the the sketch be vibration ofaircraft. system wing (b) applied mitigate fuselage =

OF VIBRATION SLENDER BODIES weconsider Inthis vibration ofslender bodies torsional and chapter undergoing are taut longitudinal, m otion We vibration. also of of consider lateral vibration strings. Equations bending afreeof cutoutof For small the slender constructed bodies using body length infinitesimally body. one can exact natural and natural ofuniform find solutions for modes cross-section, frequencies means. via analytical note inthe We that ofnatural modes observed for discrete systems previous orthogonality are also holds for slender bodies, mathematical among chapter Noteworthy equivalence longitudiare taut nal torsional vibration vibration of and distinct. vibration, strings although they physically an are section in The last of t his deals with method which assumed modes approximate chapter tofind via introduced solutions the approximate equation approach. Lagrange exact isimportant solutions for vibration of s lender bodies for practical Finding analytical an asthe to benchmark solutions solution such method, approximate applications. They provide orthe inthis which will assumed mode method introduced finite element method be chapter, are as inthe inthis discussed follows: covered topics following chapter. Specific chapter aslender 7.1Longitudinal vibration of body aslender 7.2Torsional vibration of body 7.3Vibration ofatautstring 74Bending ofaslender vibration body 7.5Assumed for solution mode method approximate

INENGINEERING INTRODUCTION TO VIBRATION

7.1 Longitudinal Vibration ofaSlender Body aslender asshown inthe Consider vibration direction below. body undergoing longitudinal

7.1 vibration. Slender Figure body undergoing longitudinal unit force per f(x,t): : applied atxlength L axial force P(t) applied area cross-sectional atx A(x): :axial cross-section ofthe located u(x,t) displacement : mass unit per m(x)= pA(x) : mass volume p per length cutsat afree ataninstant create intime To ofmotion, let’s derive equation imaginary body by introducing xand x+dx. f(xt)dx =

[| | |

— Flt) x

+dest) ——> Fx

x+dx

7.2 motion. ofslender Figure Free-body diagram body undergoing longitudinal onthe atx. isaxial inertia InFigure force surface located the 7.2, effect, F(x,t) acting Including

>(Axial

0 (7.1.1) F(x forces) +dx,t)—F(x,t)+ f(x,t)dx —(mdx)ii =

where

=

Sed (Es F=0,,A=Ee,A=EA2

+x dbut)= F(x Flat) into equation equation (7.1.2) (7.1.1), Substituting

(7.1.2)

mi o>oe mii=0

(7.1.3)

ox

(7.1.4)

ox ox as toaxial isrelated Axial force displacement

VIBRATION OF SLENDER BODIES stress, isaxial isaxial strain into Eisthe where and modulus. O.,, €,, equation (7.1.4) Young's Substituting

equation (7.1.3),

A, =iB, to is The solution equation (7.1.12) general (x)= ce+c,e%* =c,e* +c,e° or +Dcos Bsin G(x) Bx Bx

(7.1.17) (7.1.18) (7.1.19)

=

constants. Band Dareunknown where 7.1.1Aslender cross-section isfixed atthe end and at ofuniform left free Example body

the end. Determine the natural and modes oflongitudinal free vibration. frequencies right case are For this the conditions boundary at x=0, (7.1.20) w=O077=0 at x=L, (7.1.21) ox=03ox onefinds atx 0, atx L, D=0. From the that the condition condition Applying boundary boundary

F=EAst cos BB BL=0 BL=0 The aboveholds for

equation

pi=(n-2)x o,=B, [24[:>FA } Bx n>} Bsin{ np, a shape in(

=

(7.1.22) (7.1.23)

n isthe where Then from the natural ofthe number. mode integer equation (7.1.16), (=1,2,----:) frequency n is mode number

>O, 2 mL

m

From withD=0, equation (7.1.19) Bsn B(x) 2)L = where =

6 .

=sin|

=

n—

|—

2)L isthe modeofthe n-th mode. 140

(7.1.24) (7.1.25) (7.1.26) A.

VIBRATION OF SLENDER BODIES

7.1.2Aslender cross-section isfixed atboth ends. Determine ofuniform Example body the natural and modes vibration. oflongitudinal free frequencies are x L. atx=0 inequation The D=0 conditions and for and #(x)=0 7(x) (7.1.19), boundary Accordingly, =

sinBL=07 BL=na is Then from natural equation (7.1.16), frequency Oo, =n

EA

ann isthe where mode and

(7.1.27) 7.1.28 (7.1.28)

n number, integer (=1,2,-----)

Bsinâ„¢'* Bo,

Bsin Bx U(x) =

=

=

sin

where @,= isthe n-th ofthe mode. mode shape

ofnatural modes: Orthogonality s ¥ canbe r and as For mode mode equation (7.1.11) (rs), expressed

+2 (ea) +2 (eat )@,integrating equation (7.1.31) Multiplying by length, |odx=0 J|mor9, EAS 2 | +2 @, equation (7.1.32) Multiplying by length mo, ox 0 mo", ox 0

and

x=L x=0

(7.1.29) (7.1.30)

(7.1.31) (7.1.32)

the

over

9

(7.1.33)

over the and integrating

x=L

Ox ox J +2

no

(cjos=0 more aS®)

(7.1.34)

INENGINEERING INTRODUCTION TO VIBRATION canbe asfollows: term inequation the second transformed parts, (7.1.33) Integrating by

“db (7.1.3 Aw. ea) [oza%.)― ose [oa{ ja lien right drops 2. Jie (a%®. J Jos eat Similarly, a, (7.1.37) a0, a6, ant Jz [es J Jos. Equations (7.1.36) (7.1.37). (7.1.33) right (7.1.34) (7.1.38) wo stg Jno ease (7.1.39) @? nb, ‘J x=0

orfree onthe hand term For the side outand fixed first boundaries,

2

ay

x=L

(7.1.36)

x=L

a

note tothat isidentical We that the hand side of ofequation equation can n ow as w ritten and be

dx=

xe

ody fesces (@?-@?) |

x=0 ax= from equation equation (7.1.38) (7.1.39), Subtracting =

x=L

mo,,dx=0 for 0,#@, Accordingly, xaL =0

J

dx=0 10,9, and from Also, equation (7.1.33) equation (7.1.37) To

2 99, Ox dx=0 EA ax96 d sin@,t a

(sino 7.2 Torsional Vibration ofSlender =

=

+:

s=l

Body Consider slender undertorques shown below. a

a

as applied body tr

—>

x=0

7.3 torsional motion. Slender undergoing Figure body : twist (x,t) angle : unit per torque fr(T(t) : applied atx Llength x,t) applied : Torque mass moment unit per I(x) ofinertia length ona areshown Attime t,torques dx below. segment acting =

Tht)

fr(out)dx

[|

x

+dx, T(x 0)

xtdx

7.4 intorsion. ofaslender Figure Free-body diagram body onthe x. at istorque cross-section inertia InFigure located the 7.4, effect, position T(x,t) acting Including

=0 (7.2.1) =T(x+dx,t)—T(x,t)+ (torque) f;(x,t)dx —(Idx)o(x,t) >

where

T(oe+dx,t) oe de =T(xt)+ Ox

(7.2.2)

INENGINEERING INTRODUCTION TO VIBRATION

into equation equation (7.2.2) (7.2.1), Substituting

16=0 es, Jie=0 (as 18

(7.2.3)

Also

r=c)%e

(7.2.4) constant isthe isthe torsion cross-section. into where shear modulus and of t he J equation (7.2.4) Placing G(7.2.3) equation oO Ié-—| (7.2.5 ) ° Jogfr axG]— areasfollows: inFigure For the slender shown conditions 7.3, body boundary at x=0, Geometric condition: (7.2.6) @=0 (1) boundary at x=L, T= 0g condition: x =T (7.2.7) (2) Torque boundary are to isidentical We observe that vibration for if and @,GJ,I equation (7.2.5) equation (7.1.5) longitudinal f, m we u,EA, Pand with and observe between conditions, T comparing boundary equivalence replaced f.Also, atx L. to is oftorsional vibration identical that for Ox

(=]

,

|=

=

..

I>―

applied analysis mathematically Accordingly, longitudina vibration. =

7.3Vibration ofaTaut String tension inFigure Consider vibration of string fixed both ends and under shown 7.5, given which violin may represent astring piano. transverse

ataut

ona

at

as

ora

zw

x=L

x=0

7.5 Taut inthe Figure string displaced configuration. unit force per applied length p.(x,t): transverse

w(x,t):displacement a transverse toconsider isnecessary in it For vibration ofataut under axial tensile force, string free-body diagram on a to transverse create tension inorder For the the effect of let's free this, capture position displaced displacement. cuts atx and in x+dx the initial and consider ofthe free body by introducing straight configuration equilibrium as assume transverse inthe inFigure issmall. 7.6 We that shown the displacement displaced body configuration 146 .

VIBRATION OF SLENDER BODIES

+dx, F(x t)

——-

z

7.6 inthe ofataut Figure Free-body diagram string displaced configuration. onthe atx cross-section axial force located F(x,t): acting atx cross-section rotational ofthe located @(x,t): angle z inthe inertia 0,Then From the the effect, direction) (forces free-body diagram, including

¥ + =0 (Fsin@), p,dx—(mdx)w —(Fsin@),,,, where =

+2 (sind

etd (Fsin@) )ds (Fsin6), into equation equation (7.3.2) (7.3.1), Substituting =

—mw dn. > On. —mw=0 =0 |dx ——(Fsin@)+ -—(Fsin®)+ ax ax

p,

p,

P

.

inequation @ sin@ for small @ (7.3.3), Setting p,—mib=0 0 0x into equation (7.3.4), Substituting 0 F—|+ dwPe-m ox ox =0 Finequation T with tensile force (7.3.5) Replacing ~-2|0 (dw m™ oxp2― ox P, constant is For the ofmotion tension, equation 2 ox p,

(7.3.1) (7.3.2)

(7.3.3)

=

-

2 (r6)+ ( 5.

(7.3.4)

=

(} [To i

|

mio =

(73.5

,

(7.3.6 (7.3.7)

INENGINEERING INTRODUCTION TO VIBRATION aslender tothe of isidentical motion We the above vibration observe that for of equation equation longitudinal wealso section EA with uniform ifw,Tand with and observed u,EA p,arereplaced previous body f.Inthe

the mathematical between torsional vibration and vibration. These mathematical equivalence longitudinal aresummarized asfollows: equivalences String TorsionalTaut

Longitudinal ) u

EA

fel I

m

Sr taut

f

w

T m

Pz

a string 7.3.1Consider atboth ends and under tension 7.The fixed Example string now as is given initial and released. follows displacement S2 =2L for2Ox ov, that (7.4.2) Noting V,(x+dx)=V,(x)+ 5x dx, becomes equation (7.4.1) p,—mid > ++p,-—mw=0 (7.4.3) 2+ av. ax |dx=0 ax av, moment Bis the about effect ofrotary inertia, point equilibrium Neglecting (7.4.4) M,(x+dx)—M, (x) pd 0 .

Z

.

.

= V.(a)dx+

VIBRATION OF SLENDER BODIES

where

aM p, p, (7.4.5) —mib ax M,(x+dx)=M,(x)+—>—dx, inequation into the first equation (7.4.5) (7.4.4), Introducing .

=

=0 aM, v, P(x) - (dx)°

(=

2

\

(7.4.6)

termonthe termand canbe as isahigher Inthe the left above second order equation, involving dropped p, 0.Accordingly, dx—

OM |dx=0 —-V, ax from which OM tay ax and equation (7.4.3) (7.4.8), Combining

(7.4.7) (7.4.8)

aM, ax+p.—mib=0

(7.4.9)

7

zero tothe transverse strain isassumed Bernoulli-Euler beam shear and thus theory, According bending

(7.4.10) g--2― isthe inthe plane. One rotational also where ofthe defined clockwise cross-section, may positive angle @ recall that x z -

7.4.11 (7.4.11)

M, =-El,

into py

equation (7.4.11) equation (7.4.9), Placing a owPe Oe Ox?

(

eecen ?

7.412

_

abeam atx 0and atx Lasshown inFigure For fixed free conditions conditions: 7.7, Boundary boundary areasfollows: w=0 at x=0, and * 0 (7.4.13) (1) atx=L, (2) =

0 aw = M, =-El, >=, 9 ew|r? aM, ae Onal

=

(7.4.14 )

INENGINEERING INTRODUCTION TO VIBRATION

Free vibration: is The ofmotion for free vibration equation

2 2. mir canbe as tothe solution The above equation expressed

a1, Zo 2{

w(x)" into

w=

equation equation (7.4.16) (7.4.15), Placing

(7.4.15)

(7.4.16)

2— a ow—mow ||e"+tiwt =0 a2 > EI,

(7.4.17)

a ow 5, Pra

(7.4.18)

itfollows from which that

(21, )-no's= _

0

abeam i. For isafourth The above for ofuniform order differential equation equation homogeneous ordinary m constant cross-section with and becomes equation (7.4.18) EI,,

Ow =0 —mo’w Ox EI,

tothe isofthe The solution above form: equation homogeneous following

= ceâ„¢* into equation equation (7.4.20) (7.4.19), =0

Placing

\ce** (A* 0"

where _

2 me

EI, For nontrivial solution,

(7.4.19) (7.4.20) (7.4.21) (7.4.22)

(7.4.23) +0°)(A’-0°)=0 which=0-9(A' from Mi -0'

=0 A’-o’ V+0°=0,

(7.4.24)

VIBRATION OF SLENDER BODIES or

NV =-0’, A,,=t0 V=0'°>1,,=Hi0,

Then

W=ce°*+c,e +c,e°* +c,¢ * that Recalling

7.4.25) 7.4.26)

coshox=sinhox= 7.4.27) canbe as The ofequation hand side (7.4.26) right rearranged 0x+Feos 0x ox+Dcosh ox+Esin i Csinh (7.4.28) D. with unknown coefficients C,D,Eand 7.4.1Free cross-section vibration ofacantilevered beam ofuniform Example abeam areasfollows: atx L, atx 0and cross-section For ofuniform free fixed conditions boundary ow ow=0 90=-—=0>— At x=0, w=0>#=0, (7.4.29) ox ax Atx=L, =

=

=

aw=0>a M,=5+ -EI, a »owo>2® 0 _oM, "ax dx Ox? ax’

57=0

(7.4.30 )

From equation (7.4.28),

Ow x 7

.

.

ox+Dsinh ox+ Ecos Ox—Fsin (Ceosh ox) Ow0° ox+Dcosh ox Esin 0x—Fcos (Csinh 6x) — rrr 0’ ox+Dsinh ox— ox+ Ecos Fsin aw, (Ccosh ox) es

3, 7

4/..

;

2

;

.

(7.4.31) (7.4.32) (7.4.33)

INENGINEERING INTRODUCTION TO VIBRATION atx =0, the conditions Applying boundary

=0—> D+F=0>5F=-D ow0>C+E=0 E=-C

ox atx Land and the conditions equations (7.4.34) (7.4.35). Applying boundary 2— 0x =0-— +Deosh =0 OL OL— Esin OLFcos OL Csinh

(7.4.34) (7.4.35)

=

or

“

—

oL+sin oL+cos C(sinh OL)+ D(cosh oL)=0

(7.4.36)

ow Ecos oLoL=0 OL +Fsin OL+Dsinh x 0 Ceosh

or

.

Fer

;

oL+cos oL—sin C(cosh OL)+ oL)=0 D(sinh can as Inmatrix form and be equations (7.4.36) (7.4.37) expressed

sinh

oL

(7.4.37)

oL+sin OLcoshoL+cos C 0 D 0 coshhoL+cosOL sinhoL-sinoL must matrix For nontrivial above be solution of t he the determinant of t he coefficient equation, homogeneous tozero. {_}

(7.4.38 _

equal Accordingly, =0 OL-cos +2cosh (7.4.39) (sinh —(sin (cosh OL+(cos OL) OL)’ OL)’ OL)’ [ ] that Noting =1 =1, (7.4.40) (cosh —(sinh (cos +(sin OL) OLY OL)’ OLY to issimplified Equation (7.4.39) +1=0 OL-cos OL cosh (7.4.41) as tothe The solution be above may equation expressed -—

o,L=a,

(7.4.42)

VIBRATION OF SLENDER BODIES n a one isa via where m ode number. method for integer (=1,2,----) equation (7.4.41), example, Solving graphical canshow canthen etc. 7.8851 that be a, a, a, 1.8751, 4.6968, equation equation (7.4.22), (7.4.42) Using as =

=

=

expressed

aL

F

y

and

=

2

=4,

(7.4.43)

EI,

(7.4.44)

O,4,4a are From the the three natural above lowest equation, frequencies EI 22.06 EI EI @,3.516 @, mL mL@,=62,17 mL =

z ,

=

m »

t

(7.4.45)

canbe as 6=0,,, with and Now, (7.4.34 using equations )and (7.4.35), equation (7.4.28) expressed Ww o,,x—cos (7.4.46) =C(sinh o,,x—sin D(cosh 0,x)+ 0x) and

ce O,x—cos o,x—sin 0,,x)+(cosh Ox) D(sinh

gle

can as o=0,,, With be equation (7.4.37) expressed

C(cosh 6,L+cos D(sinh o,L—sin o,L)=0 6,L)+ Then sinho,L—sino,L C__ D_~ o,L cosho,L+cos and

(7.4.47) (7.4.48) (7.4.49)

sinh 0,L—sin 0,L O,,.x—cos (sinh o,,x—sin (x)D cosh @, 0,,x)+(cosh 0,,x) (7.4.50) 06,,L+cos o,,L isthe mode ofthe n-th mode. shape

ae

INENGINEERING INTRODUCTION TO VIBRATION

ofnatural modes: Orthogonality r and s # canbe as Ingeneral, for mode mode equation (rs), (7.4.18) expressed 2

2

2(21, Zs.) =0 mo;, 0 %%, mo;9, =

a.

2,ze.) mo,’d, &.)Ieee \.

(7.4.51) (7.4.52)

over the and equation integrating (7.4.51) Multiplying by@, length,

dx=0

over and the equation (7.4.52) by@, integrating Multiplying length, x=L 2

2

(7.4.53)

ve (22, Jem, 58 Sex se 58 (#1 n= i (#1, [eg fa, a= J

x=0

dx=0

(7.4.54)

terms inequations twice the first and such parts and, (7.4.54) (7.4.53) Integrating by boundary considering as one can free for show conditions and that clamped, hinged simplicity, 0 0 ao, 0° (7.4.55)

from equation equation (7.4.54) (7.4.53), Subtracting mo,¢,dx=0 (@?-@?) | r#s for Accordingly, x=L

x=0

x=L

(7.4.56)

J

(7.4.57)

era fa, Jes[

(7.4.58)

dx=0 10,9, and Tol i 52%03> 0°, 5F0°¢, 9hk=O x=0

VIBRATION OF SLENDER BODIES via vibration modal response analysis: Bending onemay asacombination as For ofbending vibration ofnatural modes response, express displacement analysis + (1), + +O(ty %,4, (7.4.59) w( x,t)(0), are torepresent isthe where number of n atural modes chosen and unknown modal @, time-dependent N into coefficients. equation equation (7.4.59) (7.4.12), Placing =

r=N

ot,OF, Yo =

r=1

r=N

o

@, m6, “ equation @,integrating ra Multiplying by }: length .

+

x

p.

the above natural mode and

(7.4.60)

the

over

=

das (2.3 Je [Lae hiPon 58 Jmordx J02[n, | 4Jo

(7.4.61)

tothe inequations the ofnatural left ofthe and hand side modes (7.4.57) (7.4.58), According orthogonality is above and thus equation simplified (7.4.62) ja,+}Ay Py a= Op.dx or

x=L

x=L

x=L

x=0

x=0

x=0

7.4.63 s=1,2,-,N +K,a, =F, Md, wherex=L 7.4.64 M,= x=0 m@?dx x=L 2 x=L 2 7.4.65 K,= x=0 x=0 =0° mp?dx=0M, x=L 7.4.66 B= x=0 Op.dx canbe NSDOF SDOF Initial conditions for the determined represents systems. systems (7.4.63) Equation as the ofnatural modes follows: using orthogonality

J [¢, | “ < |

(2, jis

y Joma,|

o,mw(x,t)dx (pdx a(t) o° mdx =

=

(7.4.67)

INENGINEERING INTRODUCTION TO VIBRATION

which x=L from 1

1x=L

(7.4.68) t)dx a,,(t)= @muw(x,t)dx, a(t) @,miv(x, Accordingly, 1xsL 1x=L (7.4.69) a,(0)= g,mw(x,0)dx, COE ave

J mw! MM. =

TE Oemelos Obtx 7.5 Assumed Mode Method for Solution J

J

Approximate via Forslender solution the ofnon-uniform find cross-section,

onecan an approximation body Rayleigh-Ri anassumedfield inwhich is into method introduced the for the energy expressions displacement Lagrange one a asshown in For consider slender vibration may equation approach. example, body undergoing bending 7.7 the ofSection (see 7.4), Figure beginning a

Kinetic energy:

x

x+dx

aslender onemay a inbending To the kinetic consider derive for first energy vibration, expression body body asshown inthe rotation the oflength dx sketch. the contribution from around the axis, y segment Neglecting is kinetic dTthe shaded energy portion 2 T1 (7.5.1) entire isthen The kinetic for the energy body =

of

(mdx)w

2 w'mds T=+ f Xo

(7.5.2)

Strain energy: aslender canbe as inbending, strain For the energy body expressed x=L

2 2

U=5 a") da

(7.5.3)

VIBRATION OF SLENDER BODIES

where

dz: area ofinertia dy =fe moment Incremental work done loads: by applied inFigure For the shown loads 7.7, applied x=L

(7.5.4)

A

(7.5.5) | = rotation dw that infinitesimal for incremental and 656 dw and 60 Note represent .-, dWx=0Swp.dx+(M,60) +(Pdw),_,

0). ( 0) ( > displacement aninfinitesimal state in i s in while dx space. segment changes length Assumed displacement: onemay assume was For solution, approximate

+ (7.5.6) fo(x)a, fi(x)a, f,(x)a, isanassumed isatimethat where function satisfies conditions and a, geometric ( k=1,2,---++) f,(x) boundary a x one assume at 0, may For unknown coefficient. forcantilevered beam fixed example, dependent x x 3 +

w=

=

2

w=a(2) +a,(4) te beasts

(7.5.7)

where

x =—,

sar

anexercise, acantilever two-term As consider the for beam with approximation w=as +a, +as° =a,s Inmatrix form ° “

H|p»: Jia-| | w=|:

w=| where °° N=[ or

a4

“

i Jess

7.5.8 (7.5.8) (7.5.9) 5,

(7.5.10) (7.5.11)

(7.5.12)

INENGINEERING INTRODUCTION TO VIBRATION mass matrix: Construction ofthe From and equations (7.5.12) (7.5.10)

w=Na, w=a'N'

Then =1 sel x=L 1 1 1 T=2Xow°mdx=— ms 5=0 2 a"N'NamLds=—a"

fsNa

| | J = =+4"Ma 2

where s=1 : mass matrix M=L s=0 matrix: Construction ofthe stiffness

Nds JmN'

(2) 2% + \¢ +6035) ( 2a, =r =F]4,

(7.5.16)

I[sm

1

1 2 6s a

B=|| 2 6s

or

(7.5.14)

(7.5.15)

ow2 10w ax\dx) ds\ds dxJdx Lds’ ax’ where

(7.5.13)

2 1 ror aw 4,4, 1} « ve

1

f=peB =| Then, equations (7.5.16) (7.5.18) equation (7.5.3), placing ira

(7.5.17) (7.5.18)

and into

1 avee Ly (Bw), ds PEL, re B’BaL

v=] art) {2 J 7

-

s=l 1 E ds|ja=—a'Ka , =-al 1B'B 2 > Ds

where Eo ds: matrix K=— stiffness

(7.5.19)

-

1,B'B

(7.5.20)

VIBRATION OF SLENDER BODIES vector: Construction ofthe load vector isconstructed From The load equation equation using (7.5.12) (7.5.5). s2 s;

|

5w=|

Sa, 5a, =da'N' unit is Incremental work done load per p,, applied by length, aol dx Sa'N'p,Lds=6a" ds x=0 Swp, p, J J JN' sal

ss

E Jeo A=LJN"p.ds =

s=0

s=0

(7.5.21) (7.5.22)

where

s=l

s=0

(7.5.23)

Atx=L,s=1and

P(Sa, +6a,) (Péw) _, Also, ds 1 owdw Q@=de Oe De s=1 Then and x=L, at =-M, +36a,) (M,60)_, =

=F

(7.5.24) (7.5.25)

(2a5+30,5")

§ (28% Accordingly, A,+P=$a"E Sa, 5a, 2a | A,+P-3—+ éw=

(7.5.26) (7.5.27)

where

+P-2——4 : vector

F= A,

load

A,+P-3—+

(7.5.28)

INENGINEERING INTRODUCTION TO VIBRATION

ofmotion: Equation asetof the equations, Lagrange Applying d{aT)\) dTo0U SS |-S4+S=k dt\da, da, )da, d{oT) dToU + ~~ F, dt\04, 0a, a, ) asfollows: inthe ofmotion results equation Ma+Ka=F For vibration free (F0), #|

.

=

a=@er― isthe where ofa Then

part time-independent ®

(7.5.29)

(7.5.30) (7.5.31)

.

(7.5.32) (K@°M)9=0—> Kg=a@’Mo a n out todetermine natural and After carrying (thusfrequencies) eigenvectors eigenvalue analysis are as the natural modes determined follows:eigenvalues tia =we —,tiat

w=Na=N@e~ (7.5.33) and w=NO (7.5.34) onecanuse to isthe ofw.For the k-th mode with and part @, @,, equation (7.5.34) time-independent determine natural mode #,. the abeam two-term in For ofuniform results cross-section, approximation EI 34.807 EI Q,3.5327 @, (7.5.35) mL mL —, * we twonatural in exact inequation with values the the equation (7.5.35) (7.4.45), frequencies Comparing tothe exact isquite isnot that natural observe the first natural close value while the second frequency frequency asaccurate. more toadd terms inthe itisnecessary For assumed mode. accuracy, analysis improved Eigenvalue as two follows: also eigenvectors produces 1 : -0.3837 9 —0.8221 (7.5.36 1 onecanobtain as twonatural Then from the modes equation (7.5.34), (7.5.37) #,=N@,, #,=N@, _—

=

=

=

,

=

|

VIBRATION OF SLENDER BODIES canbe to at s=1 maximum isequal For t he natural such value each of modes scaled that the unity plotting, x and with exact inequation the natural mode (orL) compared (7.4.50), =

7Problem Sets Chapter 7.1slender cross-section isfixed0and linearof ofuniform connected

to constant at spring body are as A a k Find the natural and modes oflongitudinal vibration. The conditions frequencies boundary follows: atx =

x =L.

Ou—ku at x=L EA ox

*(} (=e *)body subjected applied suddenly analysis. x=0,

at u=0

mass atx =0 at L.Aconcentrated cross-section isfixed 7.2 Anow slender of u niform and free ofM body atx= L.Find is added The the natural and modes oflongitudinal vibration. frequencies boundary x=

areasfollows: conditions

at x=0 u=0

0x Miiatx=L a)

atx =0and at x L.The atrest, cross-section isfixed is 7.3 Aslender of u niform free initially body, x toanaxial at constant. P Land force held Determine the response using the modal asfollows atboth 7.4 tension Atautstring Tisgiven initial fixed ends and under and then displacement =

=

released. L O S for 2. 2 L

w(x,0)=— xN,Âand matrices. For To with modal ®,Q,arerectangular identification, proceed pseudoinve method. with D!, equation (9.1.24) Postmultiplying or

D.D! =®,A.D? with DI, equation (9.1.27) Postmultiplying

(9.1.29)

D.D! =6,A,D! (9.1.30) canthen isamatrix sso,2). Itcanbe shown that where of R R®, e T = 1,2 (k (9.1.30) Equation @&, as be expressed =R®A,D! RD,D! D.D! (9.1.31) with equation (9.1.31) Postmultiplying (D,D7)'®,, =

=

> =R® =©, )'®, D,D/(D.D! D,D!(D,D/)'® The last ofthe above be

(9.1.32)

can as equation expressed

=, (9.1.33) A,®, (9.1.34) A,=D.D!(D.D!)" measured data. and be be constructed from With equations equation (9.1.25) (9.1.28), (9.1.33) written where can

can

as

o,& AJo,& OS ®, ®, ®, fe}

9

(9.1.35)

AND MODAL INPUT FORCE IDENTIFICATION

column for k, equation (9.1.22), Recalling At =e2A, ®,, ®,, Aa ®,, ®,, A

(9.1.36)

Then Ok 2A,At Ok & =e* Ail, ®,, ®,, canbe as The above equation expressed 5

A,X=AX A,X=AIX > where Ae2hat =

9.1.37)

9.1.38) 9.1.39)

X= ®,,

9.1.40) ®,, are isa standard for After and equation Equation (9.1.38) eigenvectors analysis. eigenvalue eigenvalues canbe and determined from and determined, @, equations (9.1.39), (9.1.40), (9.1.11) A,, ,(=®,,) (9.1.12).

9.1.1 1kg,=k the system below with M,= Example Consider two-DOF shown k, 400 0.8 and Note that this isiden ical toM,= the N/m Ns/m. system e xample problem C,= C,= insection 3.4.2 3. 2

ofChapter

9.1 Two-DOF model. Figure tomass Determine isapplied =1Ns RK the method of the first q,,q,using M,. I, (a) covering Animpulse s ec. are 10 Determine Time also and that measured data. the q,, g, pretend they Apply experimentally tocarry out todetermine Domain Method and modes. analysis frequencies eigenvalue corresponding own time 0.015 for shifting. Select yourAt(say sec)

INENGINEERING INTRODUCTION TO VIBRATION

outan matrix RK4 the method the from the and output A, using carrying Constructing eigenvalue asfollows: the equation using (9.1.38) analysis produces eigenvalues —0.1529+12.3600i, =—0.1529—12.3600i,

A, A, =-1.0478—32.3526i =-1.0478+32,3526i, A, A, wecan as two From these eigenvalues, identify frequencies =32.3526 @,12.3600, @, canalso asfollows: tothe 2X1 The modes be eigenvectors expressed corresponding 1 1 =

=

| oi

‘| oi7"0, 0.6180 0, 0.6180 where and 8,=—0, =

=

o-|Jer o.-| \e |

|

4 —0.6180—0.6180 1 1 ;

where 0,=-0,. wemay a random measurement tothe errors, noise via To simulate accelerations the add calculated (b) we a to noises noise RK For method. MATLAB. random forsignal Then, may generate using example, as ratio TDM of of 150, (SNR) application produces eigenvalues 0.1766 =—-0.1766 +12.3577i, —12.3577i, A, A, =-1.0596 =-1.0596 32.3483: +32.3483i, A, A, wecan as two From these eigenvalues, identify frequencies =32.3483 =12.3577, @, @, canalso asfollows: tothe 21eigenvectors The modes be expressed corresponding 1 6,_ 1 0, 9 0.6183 0.6183 where and 8,=—0, in -0.6161 ea e°, 93=06161 QO, 1 1 = where 0,=—0,. asslender wenote continuous For structural such and that bodies, shells, solids, systems equation plates, a can at in in section holds discrete The method described this be for then system. points (9.1.1) displacements once areavailable. to at extended modal of s uch accelerations measured discrete systems points analysis easily canalso to The strain via be method modified modal identification data measured =—

—

_

_}

sensors. strain

214

\| \| _

_}

easily

using

piezoelectri

AND MODAL INPUT FORCE IDENTIFICATION

9.2 Input Force Determination

measure forces ona to cases, itisnotpossible Invarious Insuch system. input acting practical directly dynamic or we use to the data acceleration determine characteristics situations, maymeasured system output strain) (e.g,, an are constitutes inverse the unknown force. This force and input mathematically problem. Examples dynamic inhypervelocity wind tunnel control and determination surface experiments including deployment unsteady can inhighspeed time-invariant For load determination tunnels. linear relation wind the systems, input-output canthen anumerical to via inthe time We be t he convolution method domain expressed develop integrals. construct the force deconvolution. a input using toa For time-invariant linear the convolution systems, output input F(t) d(¢) integral relating single single canbe T=t as expressed

| J

(9.2.1) d(t)= b(t—t)F(t)de T=0 att=0. toanunit is is The where function the above response response h(t) impulse impulse applied integral can as commutative andbe alternatively expressed F(t—t)h(t)dr d(t) (9.2.2) T=0 wehave anSDOFwith Insection 1.4.4 the convolution ofChapter for introduced 1, system integral applied as as the force the and and hold for and However, any output. input equations (9.2.1) (9.2.2) input displacement can a a at to at strain For of be t he measured location due force output systems. pair example, applied d(¢) dynamic can nowusethe twoconvolution atwo-step to We another location ofastructural method body. develop integrals isavailable via Atime for force when measured domain method output input experiments. constructing developed by is in review inAppendix For Lee the subsections. the references A.3. and summarized details listed Draper following T=t

=

9.2.1Numerical ofImpulse Function Construction Response inthe istoconstruct For The first the function method hammer response two-step this, step (IRF). impulse can as can to construction be the known measured which be used of forces used force for generate output input can use a to We then determine unknown IRF. IRF theameasurement the constructed and measured force. outputcanbe input as atadiscrete in time For taken Â),equation ¢, (9.2.2) point(t= expressed T=, (9.2.3) d(t,)= t=0F(,-t)b(t)de canbe successive time isaninteger The above with which ofsamsegment integration At, segmented multiple time such that segment pling (9.2.4)

J

INENGINEERING INTRODUCTION TO VIBRATION canthen as The side hand ofequation be (9.2.3) expressed right

JFGf“P(g, de 4f“P(, -1)b(t)de

+ d(t,) ab(t)dt4DW) (9.25) nowassume andtobe overeach ina interms We n odal values functions of may segment F(t) h(t) polynomial manner tothe similar element such finite formulation that d=AbAy, (9.2.6) vector to a we construct For A ofd(t,). the dis column where hammer forces, may given input corresponding we vector at inequation intime, and withcan dofthe measured then, may output attempt (9.2.6) points sampling weintroduce tofind inversion. A. A hby be i ll-conditioned for H owever, proper inverting Accordingly, on can vector to constraints inverse We the the the determine the h improve of problem. then IRF conditioning via the least method. square pseudoinverse 9.2.2Input Force Determination Function Impulse Response Using wecannowreturn to iscommutative IRF Wich the that determined, equation (9.2.1). equation (9.2.1) Noting can as to roles and the ofF(t) and be discretized equation equation (9.2.2) reversing h(t), (9.2.1) symbolically =

d=BFAr, F

(9.2.7)

vector at vector isacolumn matrix Band where of With of the d measured output ). points given F(t, sampling wemay tofind vector intime, is matrix B inversion. F, the for ill-conditioned H owever, attempt input again wefirst tothe vector to constraints F introduce assumed the conditionproper input improve Accordingly, can vector inverse We then the least ofthe determine the forcevia square input ing problem. pseudoinver method. m= 1kg, anSDOF mass 9.2.1Consider with constant Example = =system spring c

k 1000 0.63245 The isinitially atrest. constant system N/m, damping Ns/m. an asfollows. construct To the consider force IRE (a) applied expressed 100 6=1.575x10" FO)= 5

4x10°(t-5)) N, s pee find numericalmethod such the method. that RK4 and Pretend

a as integration using displacement are atasamthe obtained numerical measured values integration displacements by experimentally recover Construct IRF and then the Hz(ie., off,=12,000 the At, applied pling frequency =1/f.). force. to is Now N.Pretend determine the force F(t)=10sin(200t) (b) displacement applied corresponding use in unknown thenthe the IRF and determined and constructed part (a) numerically displacement torecover the force. applied

AND MODAL INPUT FORCE IDENTIFICATION

Solution: tothe insections 9.2.1. IRF The below the constructed described compares (a) procedure figure according no exact 9.2.2 We and with the IRF from observe that withdismatch, equation (1.4.35), they closely in cernible the differences plot. -

0

0.05

0.1

Time (s)

Analytical Constructed

0.15

0.2

vs,exact 9.2 constructed IRF IRF. Figure Numerically

Time (ms) vs.applied. 9.3 Pulse force: reconstructed Figure

inthis section in 9.3 shows the force reconstructed the described input Figure procedure following with the actual force. comparison applied

INENGINEERING INTRODUCTION TO VIBRATION

incomparison The shows reconstructed with the actual below the force force of input (b) applied figure sinusoidal shape. 10

° (N)

Force

Time (ms)vs. 9.4 reconstructed Sinusoidal force: Figure applied. an an toa inthis section The relates method described response output given input impulse through canbe extended to function. the method and (SIMO) Output Input-Multi easily Single Accordingly, one one use Forinput Multi andn systems. outputs, mayequation (MIMO) (9.2.6) Output Input-Multi ntimes nIRFs, asetof togenerate

9Problem Sets Chapter in 9.1.1. 9.1 Consider theDOF described system

Example tomass Determine isnowapplied An RK the the ofI, =1Ns method 4,,4, M,. using covering (a) impulse s ec. 10 Also determine and them. first ,4, plot are g, in Domain that obtained measured data. Time Pretend the part 4,,4, (b) (a) experimentally Apply to out to and Select Method determine modes. your carryeigenvalue frequencies analysis corresponding ownAt 0.015 time in for shifting. the results with the 3, Compare (saysec) example problem Chapter section 3.4.2, areadded measurement tothe errors, noises To simulate MATLAB accelerarandom using (c) generated a to tions inpart noise ratio calculated forsignal of150. part (a). (b) (SNR) Repeat inExample inChapter 9.2 3.3.3 3ismodified The two-DOF described such that system by adding damping is matrix the damping two

_

08 400

AND MODAL INPUT FORCE IDENTIFICATION tomass For=1.0, isapplied Animpulse Ns 41/42 RK the ofI,=1 determine method B (a) M,. using s ec. 10 Determine them. the first also and covering 4,4, plot tocarry out to in the accelerations determined the TDM part (a), (b) Using apply analysis eigenvalue ownAt 0.01for time determine and modes. Select your sec) frequencies (say corresponding shifting, to noises in MATLAB with of the accelerations SNR 1 50 Add random calculated using (c) generated are to out that measured the TDMcarryeigenvalue data, part (a). Pretending they experimentally apply to determine and modes. corresponding analysis frequencies =0.05 9.3 9.2 =0.005. Problem for and the and modes with the values shown Repeat 8 Compare B frequencies on comment inExample 3.3.3 and what observe. you on atwo-DOF reconstruction inAppendix 9.4 the listed and consider Review A.3 references force input are as at rest inthe isinitially The shown sketch. and follows. system parameters system system N/m, c,=5Ns/m, c,=10Ns/m. k,=1000N/m, k,=2000

m,=m,=1kg,

asfollows: tomass canbe Ahammer force M, described (a) applied

100

§=1.575x10"s ( 4x10°(t-6))N, ew FQ)=5 Determine RK4 numerical such the method method. Pretend using integration displacements a

as

are isunknown numerical that the force and the determined integration applied displacements by a at Construct Hz(ie., values measured off, =10,000 experimental frequency At, sampling 1/f,). recover IRF =4-. the and then the hammer force, using R, R, applied to is Now N.Pretend the force determine F(t)=10sin(60zt) (b) displacements corresponding applied use to in IRF unknown and the constructed andthe determined part (a) numerically displacement recover Use the 4. force. R,R, applied set5.4. inFigure inProblem 9.5 two-DOF 5.4 the the Consider model shown with properties wing given ashort as duration force described Suppose now

=

=

=

=

\"

(-0.55 ‘

J bp

a TU F(t)= atthe isapplied axis. with and elastic 6=0.0126s, a=2X107,b=1/15 vertically Determine RK4 U0).construct the method for wind-off condition 4,and 4,via (i.e, (a) are twoIRFs in that Pretend d etermined obtained data and q,toand 4, part (a) (b) experimentally two Use the offreedom. =5. R, corresponding degrees =7. Reconstruct Use the and with the actual force compare force R, (c) F(t). applied =

AND PANEL WING FLUTTER a wasused todescribe In is with which two-DOF model 5, flutter, wing wing Chapter simple a moto motion between torsional dueinertia and wing dynamic instability coupling bending mass was asthe to most center tion. axisidentified The relative elastic ofwing importance wing onset for ofthe flutter. parameter important wing we weassume a Inthis For will consider the flutter ofwings of f inite span. chapter simplicity, asacantilevered cross-section. box ofuniform the wing wing body Treating undergoing bending we strain increand torsional motion, the for the kinetic and energy, energy, expressions develop we to For mental work flutter both the dueaerodynamic loads. analysis, consider assumed mode For formulation and the finite element formulation. the assumed mode the modes for formulation, are over construct to rotation entire torsional and introduced the span wing displacement bending onset m otion. is via For the of The of flutter then determined equation wing analysis. eigenvalue rotation finite element the assumed for and the assumed for formulation, displacement bending are toconstruct arethen ineach torsion matrices. the relevant introduced element element They to toconstruct canbe motion then assembled the of which for u sed equation eigenvalue analysis atflutter onset. the determine pressure dynamic wewill can insupersonic A thin thin of flow. investigate panel Subsequently, dynamics panels oneside we toa isexposed For the when the flutter of flow. experience supersonic panel simplicity, assume isoffinite inthe itisinfinitely size while will that the direction ofsupersonic flow panel we to construct in For the flow direction normal the direction. the ofmotion, first equation long strain the for the kinetic incremental work ofaerodynamic loads and energy, energy, expressions associated with deflection. The finite element formulation and the assumed mode method panel can to inthe otion will result which be determine the ofm used for equation eigenvalue analysis to onset.

condition the flutter corresponding flight

22)

INENGINEERING INTRODUCTION TO VIBRATION

10.1. ofFinite Wings Span

wewill nowa rootto Lmeasured For 10,1 consider of from tip. wing wing wing simplicity, Figure straight length a X in is axis. shows which theaxis the elastic wing planform along

x=0

ee

axis elastic

10.1 Figure Wing planform. aninfinitesimal 10.2 shows ofwing and torsional with the span Figure displacement undergoing bending mass center the center axis elastic and the of (c.m.) the chord. (a.c.), (e.a.), placed aerodynamic along

w(x)

—— x+dx 10.2 oflength dxinthe direction. segment spanwise Figure Wing

Inthe figure,

chord length

c:

e:offset distance between the

center axis and the elastic aerodynamic section wing displacement bending totorsion twist section due @: wing angle 10.1.1 Kinetic and Incremental Work Done Strain Energy, Energy, w:

Strain energy: asthe are torsion axis the elastic reference ofthe and axis, wing Choosing bending structurally decoupled.

Accordingly,

―

vials) otfo(2)

AND PANEL FLUTTER WING

Kinetic energy: section motion motion inwhich 10.3 shows vertical and rotational wing Figure undergoing axis the coordinate the chord from elastic starting &: along mass area mm: per wing planform

10.3 section motion motion. vertical and rotational undergoing Figure Wing can as Kinetic dT then be per span energy wing expressed aT

J [(0-Gbc089)+(G $)] = (iid) (rndé)[w? 21d coso-+E74"(cos? J +si n ― => 10.1.2) i?2:cos +26" J J | Jemdé pind -+ =5i 6° the from the the indg—iidc Einde+ integrations Carrying leading edge trailing edge, —

out

to

(10.1.3)

iii® +5 =< 3,106 1,6" |mnd&:

aT where

cos

in=

mass span per wing mass moment static span per wing Syede: mass moment ofinertia per span wing &mdé: I, termin motion The second “inertia between the the and represents equation (10.1.3) coupling― bending measure a c.m. e.a. ameasure is inertia If torsional and ofthenooffset from theand thus of coupling. the motion, S, c.m. e.a., =0 is inertia For the there and coincides with and small the kinetic rotation, 1, cos S, coupling. is ofthe energy per span wing dT2 (10.1.4) 2 rootto entire For the from tip, xe wing wing length x=L x=L x=L Las 1 iardx T a indx (10.1.5) woS, I,

J J

=

=

~

w'm—w5, +26°T, Jats ohJats|e

=

_

x=

INENGINEERING INTRODUCTION TO VIBRATION

Incremental work done loads: by aerodynamic InFigure 10.4, lift per span i: : wing moment center about the per span m,, aerodynamic aerodynamic ai

x+dx onwing 10.4 moment liftand ofinfinitesimal span acting Figure Aerodynamic length. to tolift moment is incremental work due and 10.4, Figure Referring x=L x=L x=L x=L Swidx+ Swidx+ (10.1.6) +el)dx x=0 x=0 x=0 x=0 5W,,, Sprindx= A

"

A

J J f | Spin, where axis tel: about the elastic per span =

moment m,, =m,, aerodynamic for quasi-steady aerodynamic theory simplicity, Adopting

2-2 Cr,

ass) and

(10.1.7)

(10.1.8) m,.dx q(cdx)c leg t) isthe where and The above show that lift qisthe pressure upstream U,, equations dynamic velocity, aerodynamic are on moment and dependent deformation. wing cannowbe orthe motion via The c onstructed the assumed mode of for flutter formulation equation analysis to istobe isidentical It element the mode finite formulation. n oted that for the assumed formulation procedure a inthe that for element finite element formulation. single 10.1.2 Finite Element Formulation for Flutter Wing — wemay an assume islinear iscubic twist For finite element that and for bending canbe asfollows: displacement modeling, angle twist For element. the convenience, assumed angle expressed =

AND PANEL FLUTTER WING

pena o=(-9ot9,=L0 sh ―

(10.1.9)

g=N,d=d'N;

(10.1.10)

| N,=| 6,ow,66, , Ww,

(10.1.11) (10.1.12)

0 1s 00

or

where

0 0 1l-s0 0 :

a-| | Similarly, displacement expressed bending the assumed

be

can

as

w=|N,|

(10.1.13)

w=N,d=d'N? where

(10.1.14)

N,0 N,N,0

or

N.=|N,|

(10.1.15) N,0 N,N,0 are in 8. Recall that N,,N,,N,,N,, given Chapter mass Element matrix: anelement, canbe into the kinetic assumed fields for energy equation (10.1.5) Introducing displacement as expressed (10.1.16) 2 T=1d'm'd

INENGINEERING INTRODUCTION TO VIBRATION

where mi s=0ANN, s=0 mass isthe matrix. element matrix: Element stiffness Note that

(10.1.17) T,NIN, (NIN, +NjN,,)ds 5,

if atl]af

=

s=0

1 1 ow1ew10N,d)=—B,d=—d'B? i Os? P5a ( ) ig P Ox― (10.1.18) 1 loiror 19N,d)=-B,d=-d'B a¢_10¢_ axlds735 1° (10.1.19) =

=

w

=

=

where

a

w

w

Nod) =7 8s a

(10.1.20)

anelement canthen as to strain be the for energy equation (10.1.1), According expressed 1

2 U,=—d'k'd

(10.1.21)

where s=l

s=l 1 ELBIB 1 yow

GyBzB, J J k= dss isthe matrix. as element stiffness ow

s=0

s=0

(10.1.22)

vector: Element load From equation (10.1.6),

=6d" dw, +el)dx=5d"F(10.1.23) [Niidx+6d" [Nj(sh,

vector is where the element load F= det + dx (10.1.24) twist into the assumed and and equations (10.1.7) (10.1.8) Introducing angle displacement ‘! 1N.d x Idx (cd: Nd @ (10.1.25) x ) qe

el) (1i1,, fNi fNj

"

dC,U.*

=

ri

.

m,.dx q(cdx)c aC,, oa =

3

(10.1.26)

AND PANEL FLUTTER WING

into itcanbe shown that and equations (10.1.25) (10.1.26) equation (10.1.24), Substituting Fi

=qKid—qCid where

(10.1.27)

PUceds | NENode STING c PSE aC, N'N,d Nt aC, NG pat! JeSNG 150 iF if Equation assembling equation a=

s=l1 a1 KL ac i

1i"

s=l1

+

ac N

7

T

2 â„¢,xT

xa N

in"

T

(10.1.28) (10.1.29) (10.1.30)

ofmotion: is After contributions from all the ofmotion elements,

(10.1.31) Mgq+Kq=F where symbolically F beexpressed (10.1.2) F=4K,q~-qC,q islinearly into Note that and q q, q. equation (10.1.32) equation (10.1.31), Placing dependent F (10.1.33) Mg+Kq qK,q-qC,q can

as

on

=

or

(10.1.34) matrix where (10.1.35) C=qC,: aerodynamic damping and : effective matrix stiffness (10.1.36) =K~-dK, Ky matrix matrix isnonsymmetric. Kissymmetric The structural stiffness while stiffness —qK,, aerodynamic afree isasecond describes vibration initial order which equation (10.1.34) Equation homogeneous following a or to to control the reduces disturbances duegust sudden ofmotion input. equation Neglecting damping, (10.1.37) Mg+K,q=0 to isofthe The solution form equation (10.1.37) a=oe" (10.1,38) into equation equation (10.1.38) (10.1.37), Placing (K,+4’M)@=0 (10.1.39)

Mg+Cq+K,,q=0

INENGINEERING INTRODUCTION TO VIBRATION canbe as the above equation expressed Alternatively,

(10.1.40)

K,9=AMg where

(10.1.41) A=-2’. is standard The the for pressure (10.1.40) equation Equation eigenvalue analysis. dynamic corresponding a

canthen asdescribed onset inSection 5.2.4 offlutter be determined of

5. Chapter

to

Wing divergence: wemay to set 0and0in case, static For Then the reduces q q equation equation (10.1.34). (10.1.42) q=0—(K-qK,)q=0 Ky (10.1.4) Kq=4K,q to at then via We consider the above determine the may pressure equation wing using dynamic divergence we not that delete allbending affect does However, AOA, may noting eigenvalue analysis. displacement bending DOF before eigenvalue analysis. conducting asubsonic 10.1.1 the Consider aircraft ofrectangular with folwing Example planform properties: lowing =

=

L=61m,c=1.8m,e=0.1c

I,

149 8.9 kg/m, kg-m’/m, kg-m/m Sy N-m? N-m’, 9.91x10° GJ1.03x10° z EI, im38.2 =

=

80aLog|

=

=

=

-(2) L

afinite the elements and do element model with four ofequal using Neglect aerodynamic damping following length. arange vs.q. out 0 Determine for ofdynamic pressure q.Plot q;at (a) Re(A) Carry analyses eigenvalue occurs. s ea at which the flutter level. vs. flutterDetermine U,, speed twolowest @ qfor from the ofA, and the part imaginary (b) Identify frequencies plot frequencies frequencies. areshown inFigures 10.5 10.6. The results and =

AND PANEL FLUTTER WING an °

© &

AN

RMN bow 0

0.5

15

1

2

q(N/m?)

25

104

x

vs.dynamic 10.5 Real ofeigenvalue pressure. part Figure tothe q, =19320 and km/h. results, N/m’ U,639.4 analysis According eigenvalue =

90 85 bss

80 75+

aiaieiee|

wy

oS |

70

~~

+

65+ 60 554+ 50 \ 450 0.5 F

+

|

NI

t

=

aan |

i

1 4

15

L

2

2.5

(N/m) x1ot vs.dynamic 10.6 pressure. Figure Frequency

10.1.3 Assumed Mode Formulation Flutter for Wing

weintroduce construct via To the ofmotion the assumed for assumed mode modes formulation, equation wing we assume rotation. For and torsional that may bending displacement example,

(10.1.44)

INENGINEERING INTRODUCTION TO VIBRATION s x/L. inequation Note where the the conditions that assumed modes (10.1.44) geometric satisfy boundary are inmatrix Written and thusadmissible. form, =

@=N,q=q'N;

where

000ss° 8° stS| N,=|0 a,a,a,b, b,bb, , can q' | =|4, b, as the be assumed Similarly, bending displacement expressed

w=N,q=q'Ny where

(10.1.45) (10.1.46) (10.1.47) (10.1.48)

=|s*° N, equationwing, equation resulting

8 00000 (10.1.49) onecanfollow a inthe section With these assumed the described for element modes, previous procedure single a toconstruct to in isinform the ofmotion for the The identical that w

equation can same the be the of described (10.1.34). dynamic stability winginvestigated procedure following Accordingly, inthe section. previous atwo-DOF oneintroduced canbe a tothe inChapter Note 5 similar that model c onstructed for wing wing atthe to via The offinite the assumed mode formulation. reader end ofthis span tryaproblem may chapter the appreciate procedure.

10.2 Panel Flutter

a thin onone the in to to For flow due pressure side, supersonic panel exposed panel change aerodynamic can as

deflection be expressed

1 M?-2 pu? PaJVmM2—-1\ 0x ot

(%|

M?-1U,, where inaerodynamic p,ichange pressure a ir p: density air upstream U,,: speed M:Mach number

(10.2.1)

ofthe displacement panel bending coordinate with U,, parallel a asshown inFigure For inthe 10.7. consider ydirection simplicity, simply-supported panel infinitely long us a to u nit in This consider slice of width the direction. y permitsonly w:

x:

230

AND PANEL FLUTTER WING

x=0

x=L

onone side. flow

Asimply-supported 10.7 tosupersonic Figure panel exposed canthen as strain The kinetic and the ofthe be energy energy panel expressed

et2JW'mdx Xo

,

ow4* ore u=— er|—"]

[] by?

24, [are

(10.2.2) 10.2.3 oe)

mismass area tothose unit These similar for beam here However, per expressions multiplied by bending. a or turns out unit it unit width forthin of that for materials and, width, isotropic panel plate

[,

=———~

(10.2.4)

12(1-v") isthe ratio. is where histhe and Poisson's Incremental work done pressure thickness, panel by change V 6W= =1- dx =L

x

(10.2.5) dxdy Owp, Jou, J via The be assumed mode ofmotion for flutter constructed the formulation equation panel analysis isidentical The for the finite element formulation. the assumed mode formulation that for A

x=0

cannow

procedure inthe element finite element formulation.

or

to

single

a

10.2.1 Finite Element Formulation forPanel Flutter mass matrix matrix: Element and stiffness mand mass areidentical constant tothose EIYthe matrix matrix element and the stiffness for beam Assuming in element the previous presented chapters. vector to Element load due pressure aerodynamic change: anelement, as is For the assumed displacement expressed

w=Nd=d'N’

(10.2.6)

INENGINEERING INTRODUCTION TO VIBRATION

d

are inSection is 8.The where Nand incremental work the 8.4.3 ofChapter done pressure given by change then s=1 x=Xx, dx= F (10.2.7) Sup, where

p,lds=(8d)' J J(5d)" NT

bv, dw, 60, 66, (dd)" | and -|

(10.2.8)

s=l

N*p,lds [ into observe that element load F'is equation equation (10.2.1) (10.2.9), F=

s=0

Placing as deformation

vector

onecan

(10.2.9)

on dependent

1dw Nd puUz_M’=2 10.2.10) ( ot U,, Pa

â)€œ JN )

p=- 7 Vie where

a

°

1 owdwds ldw_101a Nd)=N)d=-Ad l dxds dxIds =

=

with

=

Tas! Tas!

(10.2. 2.

ON, ON, Os ON: dsON, OsOsds into and equations equation (10.2.6) (10.2.12) (10.2.11), Placing

A=2n)=|

(10.2.12)

a

F =-

M’-21 atlFAds naa aU,

cad F

we

-1U.

r 2 N’ |dpu, Ads >

5

Vp|

canbe as The above equation expressed symbolically

=qKid—qCid where

5PU! N'Ads

a=

Ki =-

s=l

all"

k

r |d (10.2.13) N’Nads 2,

(10.2.14)

(10.2.15) (10.2.16)

AND PANEL FLUTTER WING 2

s=1

Coat cette

(10.2.17)

Ttcanbe shown that

Ki 2}

;

1 i. i. 41 2 1 2 «10 10 10Lh 60 boOIe) — oO ai) aD 5 erc)

2 2,45

(10.2.18)

JRorie

entire For the x=Lstructure,

(10.2.19) Sup, dx=6q"F J over canbe as vector After all the load elements, expressed assembly global x=0

F=4K,q—qC,q ofmotion: Equation structure, entire For the Mg+Kq=F into equation (10.2.20) equation (10.2.21), Placing

Mg+Cq+Kq=0

(10.2.21) (10.2.22) (10.2.23)

Mg+Kyq=0

(10.2.24)

Mq+Kq 4K,q—-4C,4 =

or

(10.2.20)

matrix where C=qC,; aerodynamic damping : effective matrix stiffness =K-«K, Ky matrix K:structural stiffness matrix stiffness non-symmetric —qK_,: aerodynamic The above describes caused initial disturbances. ofthe response equation homogeneous dynamic panel by a can via Panel be the described flutter, procedure investigated eigenvalue analysis dynamic instability, following inthe section. previous is the ofmotion equation Neglecting damping,

INENGINEERING INTRODUCTION TO VIBRATION tothe isofthe The solution above form equation q=Ge"

into equation equation (10.2.25) (10.2.24), Substituting

+2°M)o=0 (Ky the above be equation Alternatively, expressed can

as

(10.2.25) (10.2.26)

AM@

Kp= (10.2.27) isastandard where for A=—A’, equation (10.2.27) Equation eigenvalue analysis. afinite canbe as For element with elements model ofequal equation (10.2.26) length, expressed

(10.2.28) +V’M)o=0 K and non-dimensional where matrices and ,

f

M

are

Aaa|@L' EI, Also, _

(10.2.29)

-K-K, K, where

(10.2.30)

Us TreEn

|al

(10.2.31)

DONG where

(10.2.32)

z=

1

10.2.2 Assumed Mode Formulation forPanel Flutter we consider as For assumed may the example, displacement expressed N w=

k=1

=s'—s\" N, s

(10.2.33) canbe withx/L. The assumed mode satisfies the conditions. Equation geometric (10.2.32) boundary as inmatrix form expressed (10.2.34) w=Nq=q'N* =

234

AND PANEL FLUTTER WING

100150200250300350400

q

vs.9 10.8 Real for the Figure part panel

vs.gfor 10.9 the Figure panel Frequency

INENGINEERING INTRODUCTION TO VIBRATION

For N=5,

N,NN , N5| , N=|N, (10.2.35) &&4 (10.2.36) q=|4 45\ section in With these the for element the follow assumed described modes, procedure single previous onecan

a

to isinaform the The the ofmotion for identical equation equation equation panel. resulting canbe same the of the described (10.2.23). procedure panel investigated Accordingly, dynamic stability following in section. the previous the via assumed method 10.2.1. out flutter mode Example Carry analysis panel with N 5, areshown inFigures 10.8 The results 10.9 and toconstruct

for the the

=

10 Problem Sets Chapter 10.1 in For the subsonic aircraft of described 10.1,

wing rectangular neglect aerodynami planform six and t he element model with eExample lements do finite ofequal using damping following length. matrices matrix. Construct the element structural stiffness and the structural stiffness (a) global mass mass matrices matrix. Construct the element and the b) global matrices. the element stiffness Constructaerodynamic (c) matrix. Construct the aerodynamic stiffness (d) a occurs. out global atwhich o vs.q.Determine P lot for of flutter range q . q,, (e) Carry analyses eigenvalue atsealevel. Determine the flutter U,, speed v s. twolowest [email protected]: the Plot (f) frequencies plot only frequencies. at Determine the pressure q,, wing (g) dynamic divergence. 10.2 cross-section, ofwings ofuniform Consider divergence twoelements is Show model with the for the ofequal for that, equation (a) length, eigenvalue analysis

a

[2ah4a

“71 80 JGbe The smallestleads The pressure qp. -1 1 q3 24,1 2 q3

v TE ac,

__

can as to at wing eigenvalue dynamic divergence expressed

_

GJ Lee 0a

me

oc, a

4 2.60. isthe For where smallest the two-element confirm that model, eigenvalue. =

AND PANEL FLUTTER WING aconverges to2.47. toshow with number ofelements that Repeat (b) increasing analysis eigenvalues athin in ononeside asdescribed insection 10.3 5.2. For Consider flow the supersonic panel simplicity, panel m to constant isassumed in be Assume the direction. and y infinitely aerodynamic long Neglect Introduce El,. damping. 2

3

(4 }

4

q=—Pus VieEI, [me, 7-6, +10, matrix element stiffness variable element matrix, structural and =1(

_

_

Tan y

mass express (a) Introducing 6=10, matrix interms element stiffness ofw,,6, and mass w,, aerodynamic is The with three elements ofequal size. the Construct modeled 6.. matrix, (b) panel global global matrix. structural stiffness and stiffness matrix, global aerodynamic asair increases. vs. out to Plot q. (c)Carry investigate 6, speed eigenvalue analysis dynamic stability vs. Do occurs. atwhich Determine flutter Plot the t he normalized q, J. @, frequencies frequencies atflutter? coalesce a whether the Check experiences (d) panel divergence. amodel 10.4 Problem 10.3 with five ofequal Repeatusing elements length. canbe a tothe inChapter Asimple two-DOF 10.5 5 model similar model introduced c onstructed for wing as wand via offinite the assumed mode method with span wing expressed @ w(t)filxdqr(t), O26) (*1) f(>)a(¢) are to Inthe First the chosen the behavior of the above and assumed modesrepresent equation f, f, must atthe root.wing, ofathe assumed the modes conditions ll, geometric wing boundary Obviously, onthe an wemay issatisfy As the of choice of the assumed modes. accuracy approximation example, dependent assume that (*2) f,=1-cosf,=sin twist inequation For and and q,isthe q,isthe wing (*2), wing tip tip f, f, d isplacement angle. in that assumed modes the conditions. Confirm the equation (*2) geometric (a) boundary satisfy use todetermine in in For the the assumed modes the described 10.1, (b) wing equation (*2) Example occurs. at which the flutterNeglect pressure dynamic aerodynamic damping. occurs. atwhich in For the determine also the pressure part (c) wing (b), wing dynamic divergence inthis For the described the assumed mode method with functions offifth 10.6 try wing chapter, polynomial s infor torsion. order both and bending vs. occurs. sealevel o at atthe D etermine Determine Plot q. the which flutter flutter gq, (a) speed U, condition. twolowest wvs.q. Note: Plot the (b) frequencies plot only frequencies. at Determine the pressure q,, wing (c) dynamicdivergence. =

=

=

INENGINEERING INTRODUCTION TO VIBRATION

10.7 10.6 Problem the assumed with modes functions ofsixth order. Repeat using polynomial anassumed a inthis 10.8 For the described mode with function ofseventh try panel chapter, polynomial s in order for bending displacement. twolowest Plot real ofeigenvalues the Note: and part (a) plot only frequencies. frequencies. at Determine the condition (b) flight divergence.

APPENDICES 2 A.1: Nomenclature Chapter issinusoiofapplied which force, F,; amplitude 1 intime dal Chapter SDOF Q:frequency ofa system ofapplied force SDOF k:spring ofa solution response system y,: particular steady-state ofSDOF of response system m:mass

constant c: a constant F;forcetoaSDOF system

or

Y,: amplitude damping steady-state MF: factor applied magnification t:time @: phase angle and acceleration TR: force y,ÂÂ¥: ¥, displacement, velocity,mg: transmissibility mass ofaSDOF unbalanced system mass at e:offset initial initial and distance of u nbalanced from the Yo, Jo: displacement velocity start axis rotation the ofvibration of a constant d: ofmoving base g:gravity displacement toa a mass of SDOF relative Yp: displacement ofa moving base o,=â„¢natural (in rad/sec) frequency

(E:

SDOF system 3 Chapter or vector inHertz ofdegrees offreedom ofa perq:Column cycles f= natural frequency MDOF second system matrix T:period M:mass ofMDOF system matrix MDOF of C: C: system amplitude damping aMDOF matrix K: s tiffness of system @: phase angle aMDOF vector ratio F: load of ¢: system applied damping or I:impulse natural of : eigenvector, mode, independent time @: rotational angle

2,

aa

INENGINEERING INTRODUCTION TO VIBRATION @: naturalinrad/sec

moment about the M,,: frequency aerodynamic aerodynam center in coefficient the modal of ,;time-dependent analysis moment axis forced vibration about the elastic M,,: aerodynamic a: of attack angle area 4 S: wing Chapter planform c:chord T:kinetic energy length V:potential and energy, gravity potential including dC : lift strain energy slope 5 U:strain energy L:length moment per x,y:Cartesian inaplane 0Cy coordinate wing a“+:slope angular velocity

or

I:massmoment inertia of mass M:concentrated

k:linearconstant constant torsional

spring k,: D: spring function

6 Chapter mass, constant, and ofthe spring m,,k,,¢;: damping system primary mass, constant, and ofthe spring m,,k,,¢: damping vibration absorber Q:frequency ofapplied load ratio mM, B=—mass

dissipation 5 Chapter masses concentrated M,,M,,M,: constants torsional K,,K,,K: spring rotational 0, @ : angles L:length of massless bar rigid P:thrust ratio work dW: incremental done infinitesimally by applied frequency load .

==:

nondimensional thrust P: parameter matrix effective stiffness

7: time nondimensional scaled K U:flight upstream speed velocity T:

or

=:

nondimensional ofapplied load frequency

vertical q,: displacement ratio rotation absorber q,: ¢: angle damping constant linear stiffness spring representing k,: bending 7 ofthe wing Chapter or constant torsional torsional axial ofaslender spring representing u(x,t): k,: displacement longitudinal stiffness ofthe wing body e:distance center m:mass unit between the the and per aerodynamic length axis L:length elastic of the slender body mass area M:total A: the c ross-sectional of the slender of wing mass moment static axis unit inbody about the elastic of force the per Ff: S,: wing applied length longitudina mass moment i nertia of about the elastic direction I,:w ing onthe F;axial axis cross-section force acting natural ofthe n-th mode q:dynamic pressure @,: frequency L:lift n-th natural mode ,(x): 240

APPENDICES

twist modal r-th mode torsional coefficient for the ,: g: angle a or a taut transverse inslender tension in T:tororque ofbeam string w(x,t): body displacement bending area moment inertia kinetic of for beam energy bending I: in cross-section inbending twist of cross-section torsion of the @: rotational t he @: angle angle moment transverse inertia unit I:mass of per per length applied length p.(x,t): transverse unit incremental Sw: torque per Sr: applied displacement infinitesimally transverse length in w(x,t): displacement bending onthe transverse unit cross-section F; force a xial force per acting applied p.(x,t): length :area moment inertia Q.: of for beam rotational velocity bending I, cross-section in of t he @: rotational angle bending onthe moment cross-section 9 vector acting Chapter on u: cross-section shear the force V,: acting displacement U:strain energy Q,: eigenvector M:massmatrix A,: eigenvalue vector ii:a matrix K:stiffness cceleration vector F:load d:output transverse incremental h:impulse function Sw: response infinitesimally inbending displacement F: cross- input rotation time incremental of 66: segment At,: sampling infinitesimally section inbendingangleAt,: time for function segment response impulse time ratios segment R,,R,: 8 Chapter or 10 axial ofaslender u(x,t): Chapter displacement longitudinal m: mass

M;:

per span wing body mass moment unit static per span per wing length S,: mass moment L:length inertia ofthe slender of per span wper ing body area I: A:cross-sectional of the slender 1: lift span wing body moment incremental axial about the du: per span infinitesimally displacement m,,: aerodynamic aerodycenter namic incremental work done OW: infinitesimally by applied moment axis force at the about elastic per m,,: aerodynamic i 4i: nodal node span displacement orreaction atnode i inaerodynamic force for Pa: pressure R,: applied change a ir element nodal coordinates X,,X, p: density or air I:element upstream U_;: speed speed length flight vector M: DOF Mach number q:global mass matrix M:global matrix K:global stiffness vector F:global load m:mass

+

24]

INENGINEERING INTRODUCTION TO VIBRATION

4 A.2: List of Chapter F igures 1 4.1Undamped SDOF system. Chapter Figure

1.1Mass-spring-dashpot 4.2Simple system. Figure damper Figure pendulum. anaxis. 1.2 4.3. about rotating Free-body Figure body Figure diagram. Rigid masses a two 1.3. effect. 4.4 of connected SDOF under system gravity string. Figure Figure System by a 1.4 with base. SDOF 4.5 Torsional system movingFigure spring. Figure no 1.5 SDOF 4.6 with torsional with system Figure Figure Rotating damping. body no spring. 4.7Two-DOF 1.6Adamped SDOF with system. system Figure Figure damping, constant toahorizon1.7 step intime.Figure 4.8Simple load of F, F(t) Figure pendulum subjected 1.8A offinite tal force. load duration. step Figure vs. A 1.9Nonlinear force 4.9SDOF with system spring Figure Figure damping. vs.timedisplacement asan 1.10 4.10 SDOF treated with system Figure Figure Displacement damping vs. 1.11 time external force. two Figure Velocity vs. time 4.11 1.12 Two-DOF Acceleration withdampers, Figure Figure system 2 5 Chapter Chapter 2.1. with follower force. 5.1 factor and Two-DOF model Figure Figure Magnification phase angle. a on two2.2SDOF foundation. and of 5.2 Real system part Figure Figure rigid frequency plots versus a ratio. DOF 2.3Force model under follower force. Figure transmissibility frequency a mass 2.4 unbalance. under 5.3Two-DOF with model force system Figure Rotating Figure applied a nodirectional 2.5 SDOF with base. with moving 5.4Two-DOF Figuresystem change. model wing, Figure verti3 5.5 Two-DOF model wing Figure Chapter undergoing motion. 3.1Atwo-DOF cal rotational and system. Figure mass moment on 1. 3.2Free-body for 5.6 lift a nd acting Figure Figure diagram Aerodynamic mass 3.3 2. the for model wing. Figure Free-body diagram no tovertical in 3.4Two-DOF 5.7 AOA with effective due system Figure Figure damping. Change 1. motion 3.5a ofthe Mode wing, Figure mass center axis. 2. of elastic 5.8a 3.5b Mode behind Figure Figure Wing masses a mass center axis. ofelastic 3.6Two connected of ahead 5.8b spring, Figure Figure Wing by 3.7a translation. and 5.9Real Mode for part Figure Figure plots. rigid body frequency 1 motion. elastic Mode for 3.7b Figure 2 3.8Decoupled 6 SDOF systems. Chapter Figure 3.9Iwo-DOF 6.1SDOF under loads. under sinusoidal system system Figure applied Figure loading, mass masses a t wo 6.2 A 3.10system of with absorber connected and spring. Figure System Figure by loads. 6.3 with absorber with under spring Figure System applied damping. masses 6.4 3.11 Two connected Nondimensional ofthe springs Figure Figure by displacement vs. mass t 3.12 =1.0 ratio. forB for different system primary Figure q,/A 3.13 6.5 forB=1.0 Nondimensional ofthe priFigure q,/Avs.t Figure displacement vs.tfor=0.05 ratio. 3.14 for different mary system B Figure q,/A damping vs.tfor 3.15 =0.05 Figure q,/A B =

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APPENDICES 7 8.9Slender Chapter Figure body undergoing longitudinal 7.1 motion motion Slender and torsional Figure body undergoing longitudinal 8,10 vibration Cantilevered slender under Figure body a 7.2Free-body ofslender loads body bending Figure diagram motion 8.11 Two-node element Figure undergoing longitudinal bending motion notations 8,12 New 7.3Slender-body torsional element DOFs for Figure Figure undergoing a ona 7.4Free-body in Figure 8,13 element ofslender Loads acting Figure body diagram bending torsion 8,14 for Figure bending Three-element model in 7.5 vibration Taut the string Figure displaced analysis element with 8.15 Two-node Figure configuration bending a taut inthe 7.6Free-body DOFs of string alternate Figure diagram configuration 7.7displaced 9 Cantilevered slender under Chapter Figure body 9.1Two-DOF load a model vs.exact Figure bending in Figure 7.8Free-body 9.2Numerically IRF ofslender constructed Figure body diagram IRF bending vs. 9.3 Pulse reconstructed force: Figure applied vs. 8 9.4 Sinusoidal force: reconstructed Chapter Figure applied 8,1 Slender Figure undergoing longitudinal body 10 vibration Chapter a 8.2Two-DOF 10.1 model ofslender Figure body Figure Wing planform 10.2 inthe oflength dx vibration segment Figure Wing undergoing longitudinal 8.3Spring ofthe two-DOF model spanwise direction Figure analogy section 8.4 Three-element model for 10,3 vertical Figure Figure Wing longitudinal undergoing motion motion vibration rotational and on moment 10.4 8.5Two-node element for liftand acting Figure Figure longitudinal Aerodynamic vibration the ofinfinitesimal span wing vs. length 8.6 the 10.5 Three-element model withleft endFigure Real ofeigenvalue part Figure dynamic fixed pressure vs. 8.7Natural 10.6 modes obtained the threepressure Figure Figure Frequency by dynamic to A element 10.7 model Figure panel exposed simply-supported onone 8.8Slender torsional flow side supersonic Figure body undergoing vs. motion 10.8 Real the for part q panel Figure vs 10.9 the for Figure Frequency panel

q

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INENGINEERING INTRODUCTION TO VIBRATION

T.1965. Donald A.3: ReferencesGreenwood, Principles Dynamics. of Prentice-Hall. Cliffs, NJ: Englewood

tothe H.Norman. 1958. AnIntroduction S.R. Least “Double 1986, Abramson, Ibrahim, Squares Approach in Press York: for New U se Ronald Structural Modal Identification,†J., Dynamics of Airplanes. AIAA No. vol. 499-503. 24, 3, pp. Company. N.1999. E.C. S.R. and “A Method Maher Structural 1977, Bismarck-Nasr, Ibrahim, Mikulcik, Dynamics inAeronautical AIAA Direct Parameters Education for Identification ofVibration Engineering. Free Series. from Shock and Vibration Bulletin, Response,’ 183-198. Holt and vol. 4,pp. L., 47, Halfman, pt. Ashley, Bisplinghoff, L.Raymond 2008. 1955. Daniel Robert 3"ed. Inman, Vibration, J. Engineering Aeroelasticity. Addison-Wesley. and Prentice-Hall. S.W. “Smooth 2019, III, Lee, Cliffs, J .W. NJ: Draper Englewood et Functions and Construction ofImpulse A., Brincker, R., M.R., Response Ashory, Malekjafarian, a Domain Time Deconvolution Loads “Identification ofclosely modes al., using using Applied spaced retime S ound and vol. Ibrahim domain method: 443, Method,― V ibration, J. experimental 430-443. the International Modal pp. sults,’ Proceedings of Analysis 2013. S.W. E.C. and Feb IT, Lee, Marineau, 2018, 11-14, J.W,, Draper Conference, “Numerical ofImpulse Construction 1985. Leonard. Elements Vibration Meirovitch, Response of Functions New and York: McGraw-Hill. 2" Sound e d. Reconstruction,’ Input Signal Analysis, J. I.2003. Passive New vol. Isolation. 259-271. and Vibration 432, Rivlin, Vibration, pp. Eugene 16.91 ASME Press. MIT 1974. Structural York: Courses, John. Dugundji, 16.92 Marie D,1997. and Advanced William and Thomson, T., Dahleh, Dynamics Aeroelasticity. I. 1997. in Theory Hubert The Revolution Vibration with eEnglewood d. 5* Flomenhoft, Applications, of FL: Cliffs, Structural Palm Beach Prentice-Hall. Gardens, NJ: Dynamics. Press.

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