S e c .1 . 4
23
L4odeling and EnergyMethods In termsof g = 9.31m/s2,thisbecomes Sximum
a c c e t e r a t i o: n
':::"
t )
T ' : ' s o - 7 . 6 8e ' s 9 . 8 1m / s '
Example1.3.3 Computethe form of the responseof an underdampedsystemusingthe Cartesianform of the solutiongivenin Window 1.5. + y) : sinxcosy * cosxsiny.Applyingthis Solution Frombasictrigonometrysin(x to equation(1.36)with x = a7t and y : $ yields x(t) : 4"-t'" sin(or7t+ 0) : r-{''(ArsinoT/ * A2cosuat) where,41: cosSandA2= sin0,asindicatedinWindowl.5.Evaluatingtheinitialconditions yields x ( 0 ) : x o : e o ( A r s i n 0+ A 2 c o s 0 ) Solvingyields,42 : xo.Next differentiate.r(r) to get .i:
-(6,s-(','(Arsin<,rrl + lrcosorrt) + aoe-L',,t(Arcostodt - ,42sinorTl)
Applying the initial velocityconditionyields 'uo= i(0) : -La,(ArsinO + x6cos0)+ ao(A1cos0- rosin0) Solvingthis last expressionyields A^1
u0 I
(anxs
-
(D4
Thus the free responsein Cartesianform becomes
,1ry: .{'.'("#*
*'o.or.ur,) sino,,/
tr 1.4 MODELINGAND ENERGYMETHODS Modeling is the art or processof writing down an equation,or systemof equations, to describethe motion of a physicaldevice.For example,equation(1.2)was-gbtained by modelingthe spring-masssystemof Figure1".4.8ysummingthe forcesactingon the massin the x directionand employingthe experimentalevidenceof the mathematicalmodel of the force in a springgivenby Figure1.3,equation(1..2)canbe obtained.The success of this modelis determinedby how well the solutionof equation (1.2)predictsthe observedbehaviorof the system.This comparisonbetweenthe vibrationresponseof a deviceandthe responsepredictedby the analyticalmodelis discussedin Section1.6.Themajorityof this book is devotedto the analysisof vibration
24
Introduction to Vibrationand the FreeResponse
Chao.1
models. However, two methods of modeling-Newton's law and energy methodsare presented in this section. More comprehensive treatments of modeling can be found in Doebelin (1980),Shames(1980,1989),and Cannon (1967),ryxample. The force summation method is used in the previous sectionsVd should be familiar to the reader from introductory dynamics (see,e.g.,Shames,1980).Newton's law of moticn (called Newton's second /aw) states that the rate of change of the absolute momentum of the mass center is proportional to the net applied force vector and acts in a direction of the net force. For systemswith constant mass (such as those considered here) moving in only one direction, the rate of change of momentum becomes the scalar relation -dl m, x l . \ : m x dt
which is often calledthe inertia force.The physicaldeviceof interestis examinedby noting the forcesactingon the centerof mass.The forcesare then summed(asvectors) to producea dynamicequationfollowingNewton'slaw.For motion in the x direction only,this becomesthe scalarequation
2f,,: ^*
(1.4e)
where f, denotes the ith force acting on the mass r.txin the x direction and the summation is over the number of such forces. In the first three chapters, only singledegree-of-freedomsystemsmoving in one direction are considered;thus Newton's law takes on a scalar nature. In more practical problems with many degrees of freedom, energy considerationscan be combined with the concepts of virtual work to produce Lagrange's equations,as discussedin Section 4.7.Lagrange's equations also provide an energy-basedalternative to summing forces to derive equations of motion. For bodies that are free to rotate about a fixed axis, the sum of the torques about the rotation axis through the center of mass of the object must equal the rate of change of angular momentum of the mass.This is expressedas
) ruo,: "'g
(1.s0)
where M6i are the torquesactingon the objectthroughpoint 0, -/ is the moment of inertia (sometimesdenoted1s)aboutthe rotationaxis,and 0 is the angleof rotation. This is discussed in more detail in Example1.5.1. If the forcesor torquesactingon an objector mechanicalpart are difficult to determine,an energyapproachmay be more efficient.In this method the differential equationof motion is establishedby usingthe principleof energyconservation. This principleis equivalentto Newton'slaw for conservativesystemsand statesthat the sum of the potentialenergyand kinetic energyof a particleremainsconstantat eachinstantof time throughoutthe particle'smotion. IntegratingNewton'slaw
Sec.
'1.4
25
Modelingand EnergyMethods
(F : mi) over an increment of displacementand identifying the work done in a conservative field as the change in potential energy yields
(1 . s 1 )
Ur-Ur:Tz-Tt
where U1and {./2representthe particle'spotential energyat the times /r and /t, respectively,and T1 and T2representthe particle'skinetic energyat timest, andt2,reto yield spectively. Equation(1.51)canbe rearranged (7.52)
T+U:constant
where T andU denotethe total kinetic and potentialenergy,respectively. For periodic motion, if rt is chosento be the time at which the moving mass passesthrough its staticequilibriumposition,Ul canbe set to zero at that time, and if 12is chosenasthe time at which the massundergoesits maximumdisplacementso that its velocityis zero (f, : O),equation(1.51)yields
(1.s3)
Tr: Uz
Since tlre reference potential energy Uliszero, U2in equation (1.53) is the maximum value of potential energy in the system. Because the energy in this system is conserved, Z2 must also be a maximum value so that equation (1.53) yields Z-u" :
(1.54)
U-u*
for conservative systemsundergoing periodic motion. Since energy is a scalar quantity, using the conservation of energy yields a possibility of obtaining the equation of motion of a system without using vectors. Equations (1.52), (1.53), and (1.54) are three statementsof the conservation of energy. Each of these can be used to determine the equation of motion of a spring-mass syslem.As an illustration, consider the energy of the spring-mass system of Figure 1.1-4,hanging in a gravitational field of strength g. The effect of adding the mass /11to the masslessspring of stiffnessk is to stretch the spring from its rest position at 0 to the static equilibrium position A. The total potential energy of the spring-mass system is the sum of the potential energy of the spring (or strain energy; see,e.g.,Shames,1989) and the gravitational potential energy.The potential energy of the spring is given by
u,p,ing:lt1n+x12
lo V
(b)
(1.ss)
system Figure1,14 (a)A spring-mass hangingin a gravitationalfield. Here A is the staticequilibriumpositionand x is the displacementfrom equilibrium. (b) The free-bodydiagramfor static equilibrium.
andtheFreeResponse Introduction to Vibration
25
Chap.1
The gravitationalpotentialenergyis Ugro, :
-mgx
(1.56)
wherethe minussignindicatesthat the massis locatedbelow the rcferencepoint xs The kinetic energyof the systemis (r.s7) 7 : !m*2 into equation(1.52)yields Substitutingtheseenergyexpressions - mgx I )t<11,+ r)2 : constant i**' Differentiatingthis expressionwith respectto time yields *(mi + kx) + *(kL - ms) :0
(1.s8) (1.se)
Sincethe staticforce balanceon the massfrom Figure 1.14(b)yields the fact that kL, : mg, equation(1.59)becomes (1.60)
i(mi+kx):g
The velocity i cannot be zero for all time; otherwise, x(t) : constant and no vibration would be possible.Hence equation (1.60) yields the standard equation of motion
(1.61)
mi * kx:0
This procedure is called the energy method of obtaining the equation of motion. The energy method can also be used to obtain the frequency of vibration directly for conservative systems that are oscillatory. The maximum value of sine (and cosine) is 1. Hence, from equations (1.3) and (1.a), the maximum displacement is ,4 and the maximum velocity is o,A. Substitution of these maximum values into the expression for U-u* and Z*u* and using the energy equation (1.54) yields
lm(.,A)2: it ,4 Solving this for on yields the standard natural frequency relation ,^ :
(r.62) {t
/^ .
Example1.4.1 systemhitting a bump.Calculatethe FigureL.15is a crudemodelof a vehiclesuspension naturalfrequencyof oscillationusingthe energymethod.Assumethat no energyis lost duringthe contact.
Figure 1,15 Simple model of an automobile suspension system.The rotation of the wheel relative to the horizontal as it hits a bump is given by 0. It is assumed that the wheel roils without slipping as the car hits the bump.
S e c .1 . 4
andEnergyMethods Modeling
27
Solution From introductory dynamics,the rotationalkinetic energyof the wheel is T*, : lJ02 where-/ is the massmomentof inertiaof the wheeland 0 : 0(l) is the angle of rotation of the wheel.This assumes that the wheelmovesrelativeto the surfacewithout slippingasit climbsthe bump (so that no energyis lost at contact).The translational kinetic energyof the wheelis T7 : lm*2. The rotation 0 and the translationx are relatedby x : r0. Thus i : rd and T,o.: |J *2112.At maximumenergyx : A and* : a,,A,so that T_ ' max
1 ., = 1 + l lrz)af,* * l J . , f'^^^ i^i'*^- ,7 t(^
and U.u*: t1t<*2^-:ltct Using conservationof energyin the form of equation(1.54)yieldsZ.u* = f/-.^, or 1(
/\,
;lm+-l/U.J;:;K z \ r-/
1, /-
Solvingthis last expressionfor o,, vields
the desired frequency of oscillation of the suspensionsystem. The denominator in the frequency expression derived in this example is called the effectivenass because the term (m + J lr2) has the same effect on the natural frequency as does a mass cf valu e (m + J lr2). !
Example 1.4.2 Determine the equation of motion of the simple pendulum shown in Window 1.1 using the energy method. Solution Several assumptions must first be made to ensure simple behavior (a more complicated version is considered in Example 1.4.6).The size of the mass,z, is assumed to be much smaller than the length of the pendulum, /. Furthermore, the mass of the pendulum arm is considered to be negligible compared to the mass rz. With these assumptions the mass moment of inertia about point 0 is Jo : mlz The angular displacement 0(r) is measuled from the static equilibrium or rest position of the pendulum. The kinetic energy of the system is 7:)Jsa2:)m1262 The potential energy of the system is
U:mgl(I-cos0)
to Vibrationand the FreeResponse lntroduction
28
Chap'1
since I ( 1 - cos 0 ) is the geometric change in elevation of the pendulum mass'Substitu(1.52) and tion of these expressionsfor the kinetic and potential energy into equation differentiating yields
) *o lt i- m t ' { +
m s t ( t- c o s o l :] s
or m l 2 e 6+ r z g l ( s i n 0 ):00 Factoringout 0 yields o(mtzo+ ngl sinO)= 6 Since6 (r) cannotbe zerofor all time,this becomes ml20+mglsin0:0 or
0+ ]sin0 t
-n
in Section1.10.However,here,since This is a nonlinearequationin 0 and is discussed linear equationof motion for the penthe angles, small sin 0 can be approximatedby 0 for dulum becomes q
e + ;o =0 I for initial condito an oscillationwith naturalfrequency. , .: filt This corresponds = 0. sin 0 approximation by the as defined small, 0 remains tions suchthat
Example1.4.3 Determinethe equationof motion of the shaftand disk illustratedin Window 1.1using v ethod. t h ee n e r g .m Solution The shaft and disk of Window 1.1 are modeled as a rod stiffnessin twisting, resulting in torsional motion. The shaft, or rod, exhibits a tolque in twisting proportional tolhe angle of twist 0(t).The potential energy associatedwith the torsional spring stiffne.s is U - jke2, where the stiffness coefficient k is determined much like tire metnoO used to determine the spring stiffness in translation, as discussed in Section 1.1.The angle 0(r) is measuredfrom the static equilibrium, or rest, position.The : I le2. rnis kinetic energy associatedwith the disk of mass moment of inertia "r is T assumes that the inertia of the rod is much smaller than that of the disk and can be neglected. Substitution of these expressionsfor the kinetic and potential energy into equa(1.52) and differentiating yields tion d dt
(lto' + it s'): (rd + to)e: o
so that the equation of motion becomes (because 0 + 0)
,16+ko:0
Sec.1.4
29
Modelingand EnergyMethods
This is the equationof motion fo1lqrsionalvibrationof a disk on a shaft.The natural frequencyof vibrationis o, : V klJ . I ExamPle1.4.4 Model the massof the springof the systemshownin Figure1.16and determinethe effect of includingthe massof the springon the valueof the naturalfrequency'
Figure 1.16 Spring-masssystemwith a springof nonnegligiblemass,m'.
Solution One approachto consideringthe massof the springin analyzingthe system vibration responseis to calculatethe kineticenergyof the spring.Considerthe kinetic energyof the elementdy of the spring.If n" is the total massof the spring,\nt"[) dy is 't)d,maybe approxithe massof the elementzly.The velocityof this element,denoted mated by assumingthat the velocityof the tip, i(r), varieslinearly over the length of the spring: oo, :
i rt'l
The total kinetic energy of the spring is the kinetic energy of the element dy integtated over the length of the spring: T _ spfmg
I ft m,[r.l',
,J, t l7r)o' rlm'\., 2\3 t
From the form of this expression,the effective massof the spring is m,f3, or one-third of that of the spring.Following the energymethod,the maximum kinetic energy of the svstemis thus
r^*::(* * )),;* Equatingthis to the maximumpotentialenergy,|k'4,yields the fact that the natural frequencyof the systemis
to Vibrationand the FreeResponse lntroduction
30
Chap'1
the natural Thus includingthe effectsof the massof the springin the systemdecreases n, the mass system than the much smaller is frequency.Note that if the massof spring is negligible' frequency natural the on effectof the springmass
Example1.4.5 cxhibit vibration.Calculatethe naturalfrequenaswell assolidsysterrrs, Fluid systems, cy of oicillationof the fluiciin the U tubemanometerillustratedin FigureL.17usingthe energymethod.
^y: weight density (volume) A : cross-sectionalarea / : Iength of fluid
FigureL.17 U-tube manometer consistingof a fluid movingin a tube
Solution The fluid hasweightdensity1 (i.e.,the specificweight).The restoringforce of c.g.)] is providedby gravity.Thepotentialenergyof the fluid [(weight)(displacement is potential energy in is 0.5("y,4x)xin eachcolumn,so that the total change U : Uz - Ur : Lt Ar, - (-lt at ) : 1 Axz The changein kineticenergyis
r:r
1 AI'v
s
Ah
( i ' - 0 ) : t IT r -
Equatingthe changein potentialenergyto the changein kinetic energyyields IA\.) ;;
*:lAxz
1
y sin(onl + $) and evaluating this Assuming an oscillating motion of the form x(t) : expression for maximum velocity and position yields 1I
: Yz i'ru2'x2 of vibration.Solvingfor ornyields whereX is usedhereto denotetheamplituAe
tr ,, : v7
fi
I
Sec.1.4
I I
Modeling andEnergyMethods
3l
which is the naturalfrequencyof oscillationof the fluid in the tube.Note that it depends only on the accelerationdue to gravityandthe lengthof the fluid.Vibrationof fluidsinsidemechanicalcontainers(calledsloshing)occursin gastanksin both automobilesand airplanesand forms an importantapplicationof vibrationanalysis. ! Example 1.4.6 Considerthe compoundpendulumof Figure1.18pinned at point 0.A compoundpendulum is a pendulumwith a significantmassmomentof inertiaresultingfrom a distribution of massaboutits length.In the figure G is the centerof mass,O is the pivot point, and 0(t) is the angulardisplacement of the centerlineof the pendulumof massrn and moment / measuredabout the z-axisat point O. Point C is the centerof percussion, which is definedasthe distance46alongthe centerlinesuchthat a simplependulum(a massless rod pivotedat zerowith massm at its tip asin Example1.4.2)of radius46has the sameperiod.Hence J Qn=-
mr
where r is the distancefrom the pivot point to the centerof massNote that the pivot point O and the centerof percussionC can be interchangedto producea pendulumwith the samefrequency. The radiusof gyration,ku,is the radiusof a ring that hasthe sameresistanceto angularaccelerationas the rigid body does.Theradiusof gyrationand center of percussionare relatedby eor : k2o Considerthe equationof motionof the compoundpendulum.Taking momentsaboutits pivot point O yields 2Mo : /0 (r) : -mgr sinl(t)
lE ia x
(b)
Figure 1.18 (a) Compoundpendulum pivoted to swingabout point O under the influenceof gravity.(b) Free-body diagram.
andthe FreeResponse to Vibration ntroduction
32
Chap'1
For small 0(/) this nonlinear equation becomes
" 1 6 1 r+; m g r \ ( t ) = 0 The naturalfrequencyof oscillationbecomes
u , , = !t^8' "/ in termsof the centerof percussionas This frequencycanbe expressed
a"- : r E
Yqo
be seenby exwhich is just the frequencyof a simplependulumof length4o'This can 1.19(a)or reFigure of pendulum (massless) the simple aminingihe forcesacting-on 1"4'2' Example in method energy the by obtained result ihe calling SummingmomentsaboutO Yields ml26: -mgl sin0 or after approximatingsin 0 with 0' o
e + ;I o : 0 : This yieldsthe simplependulumfrequencyof orn 'fglt 'which is equivalent : qs' to that obtainedpreviouslyfor the compoundpendulumusing/ NextconsidertheuniformlyshapedcompoundpendulumofFigurel.l9(b)of gyration' length/. Here it is desiredto calculatethe centerof percussionand radiusof .I.hemassmomentofinertiaaboutpointoisl,sothatsummingmoments about O vields
"r0:
-^s
f,"ine
G = centerof mass
Figure 1.19 (a) SimPlePendulum consistingof a masslessrod pivoted at point O with a massattachedto its tip. (b) Compoundpendulum consistingof a shaft with a center of massat point G. Here /6 is the pin reaction force.
S e c .1 . 4
33
Modelingand EnergyMethods
since the massis assumedto be evenly distributed and the center of massis at r : I p.The m o m e n t o f i n e r t i a f o r a s l e n d e r r o d a b o u tO i s J : \ m l z ; h e n c e t h e e q u a t i o n o fm o t i o n i s n112,
;u
"
l -
ms10:0
where sin 0 hasagainbeenapproximatedby 0 assumingsmallmotion.This becomes
ii+--"0:o 2l so that the naturalfrequencyis
tri '": lr7 The centerof percussionbecomes I.,
4o= mrJ and the radius of gyration becomes
ko: lw
I V3
These positions are marked on Figure 1.19(b). The center of percussion and pivot point play a significant role in designing an automobile. The axle of the front wheels of an automobile is considered as the pivot point of a compound pendulum parallel to the road. If the back wheels hit a bump, the frequency of oscillation of the center of percussion will annoy passengers.Hence automobiles are designed such that the center of percussion falls over the axle and suspension system.away from passengers.
u
So far, three basic systemshave been modeled: rectilinear or translational motion of a spring-mass system,torsional motion of a disk-shaft system,and the pendulum motion of a suspended mass system. Each of these motions commonly experiences energy dissipation of some form.The viscous damping model of Section 1.3 developed for translational motion can be applied directly to both torsional and pendulum motion. In the caseof torsional motion of the shaft, the energy dissipation is assumed to come from heating of the material and/or air resistance.Sometimes,as in the caseof using the rod and disk to model an automobile crankshaft,or camshaft, the damping is assumed to come from the oil that surrounds the disk and shaft, or bearings that support the shaft. The damping mechanism for the pendulum comes largely from friction in the joint, or point of attachment of the pendulum arm to ground, and a little from the mass pushing air out of the way as the pendulum swings.The energy dissipation or damping associatedwith the spring material in rectilinear motion and the rod material in torsional motion is much larger than that associatedwith the pendulum.
34
Introduction to Vibration andthe FreeResponse
Chap.1
TABLE 1.1 A COMPARISON OF RECTILINEAR AND ROTATIONAL SYSTEMSAND A SUMMARYOF UNITS
Rectilinear x (m) Spring force Dampingforce Inertia force Equationof motion Stiffnessunits Dampingunits Inertia units Force/torque
kx c* mi mi*c*+kx:0 N/m N's/m, kgls kg N : kg.m/s2
Torsional/pendulum 0 (rad)
ko CH
JO
t6+cg+to=o N.m/rad m.N.s/rad kg.m2/rad N.m : kg.m2ls2
In all three cases,the dampingis modeledas proportionalto velocity (i.e., -c* or : -c0 The equationsof motion are then of the form indicatedin f, : f,. ). Table1.1.Both of theseequationscan be expressed as a dampedlinear oscillator given in the form of equation (1.48).Hence each of these three systemsis characterizedby a natural frequencyand a dampingratio. Each of thesethree systems has a solutionbasedon the nature of the dampingratio (, as discussedin Section1.3.
1.5 STIFFNESS The stiffnessin a spring,introducedin Section1.1,canbe relatedmore directlyto material and geometricpropertiesof the spring.This sectionintroducesthe relationshipsbetweenstiffness,elasticmodulus,and geometryof varioustypes of springs, and illustratesvarioussituationsthat canlead to simpleharmonicmotion.A springlike behaviorresultsfrom a varietyof configurations, includinglongitudinalmotion (vibration in the directionof the length),transversemotion (vibration perpendicular to the length),and torsionalmotion (vibrationrotatingaroundthe length).Consideragainthe stiffnessof the springintroducedin Section1.1.A springis generally madeof an elasticmaterial.For a slenderelasticmaterialof length /, cross-sectional areaA, and elasticmodulusE (or Young'smodulus),the stiffnessof the bar for vibration alongits lengthis givenby .EA I
(1.63)
This describesthe springconstantfor the vibration problem illustratedin Figure 1.20, wherethe massof the rod is ignored(or very smallrelativeto the mass,??in the figure).The modulusE hasthe unitsof pascal(denotedPa),which are N/m2.The modulus for severalcommonmaterialsis sivenin Thble1.2.
Sec.1.5
35
Stiffness
E = elastic modulus ,4 : cross-sectionalarea / : length of bar x(t) = deflect'on
Figure 1.20 Stiffnessassociatedwith the longitudinal vibration of a sler.der prismaticbar.
TABLE 1.2 PHYSICALCONSTANTSFOR SOME COMMONMATERIALS
Material
Young'smodulus, E(N/m'z)
Density, (kgl-')
Shearmodulus, G(N/m'z)
Steel Aluminum Brass Copper Concrete Rubber Plywood
2.0 x 7 . 1x 10.0x 6.0 x 3.8 x 2.3 x 5.4 x
7.8 x 2.7 x 8.5 x 2.4 x 1.3 x 1 . 1x 6.0 x
g.0 x 2.67 x 3.68 x 2.22 x
1011 1010 1010 1010 10e 1,0e 10e
103 1,03 103 I03 103 103 102
1010 1010 1010 r0r0
8.21x 108
Next considera twistingmotion with a similar rod of circularcrosssectionas a polar momentof inertia,"Ip, illustratedin F'igure1,.2l.Inthiscasethe rod possesses the caseof a wire or shaft of 1.2). For (see Table G of rigidity, (shear) modulus and : N/m2.The torsionalstiffunits has rigidity of modulus naa d', J p diameter llZ.ihe nessis (recallWindow 1. 1) GJ, , (1,.64) I which is usedto describethe vibrationproblemillustratedin Figure1.21,wherethe massof the shaftis ignored.In the figure,0(/) representsthe angularpositionof the shaftrelative to its equilibriumposition.Thedisk of radiusr and rotationalmoment of inertia ,I will vibrate aroundthe equilibriumposition0(0) with stiffnessGJP/1.
k-
e(0)
Figure 1.21
GJ" :
=stiffnessofrod
J : miss moment of inertia of the disk G = shear modulus of rigidity of the rod Jp : polar moment of inertia of the rod / = length of rod 0 : angular displacement
Stiffness associated with the torsional vibration of a shaft
36
Introduction to Vibrationand the FreeResponse
Chap.1
Example1.5.1 Calculatethe naturalfrequencyof oscillationof the torsionalsystemgivenin Figure1.21. Solution Usingthe noment equation(1.50),the equationof motionfor thissystemis
16@: -ko?) This may be wrilten as t-
t i l r y+ ) o ( r ) : o This agreeswith the resultobtainedusingthe energymethodas indicatedin Example 1.4.3.This indicatesan oscillatorymotion with frequency
li IGi ',:vi:ln Supposethat the shaftis madeof steeland is 2 m long with a diameterof 0.5 cm.If the disk hasmassmomentof inertia : 0.5kg.m2 and consideringthat the shearmodu"/ lus of steelis G : 8 x 1010 N/m2,the frequencycan be calculatedby t ox* z m ) 4 / s z ] , _ *Glp _(B x1010N/m2)[1r'(0.5 = k-
(r):
J
t.t
(2 m)(0.5kg.-r)
:4.9 (rarl2/s2) Thus the natural frequency is orn: 2.2 rad/s.
n Consider the helical spring of Figure I.22.In this figure the deflection of the spring is along the axis of the coil. The stiffnessis actually dependent on the "twist', of the metal rod forming this spring.The stiffnessis a function of the shear modulus G.
l,^ {
r/ : 2R : n : x(4 :
diameter of spring material diameter of turns number of turns deflect on
l_*,,, ,
tJO'
64nRr
Figure 1.22 Stiffness associatedwith a helical spring.
S e c .1 . 5
37
Stiffness
x(o)
T
L"
x(t)
E : elasticmodulus I : lengthof beam areaaboutthe neutralaxis 1 : momentof inertiaof cross-sectional Figure 1.23 Beam stiffnessassociatedwith the transversevibration of the tip of a beam (Blevins,1987).
the diameterof the rod, the diameterof the coils,and the numberof coils.The stiffnesshasthe value , ''
Gd4
(1.6s)
64nR3
The helical-shapedspringis very common.Someexamplesare the springinside a retractablebalfpoinfpen and the spring containedin the front suspensionof an automobile. Next considerthe transversevibration of the end of a "leaf " springillustratedin of an automobile Figure1.23.Thistypeof springbehavioris similarto the rearsuspension beam,E is the of the length the / is aswell as the wings of someaircraft.In the figure, inertia of the of (area) moment the elastic(Young's)hodulus of the beam,and 1 is frequency with oscillate will beam area.The massr??at the tip of the cross-sectional @r:
EEI t- \,1 mF
(1.66)
in the direction perpendicular to the length of the beam x(t)' Example1.5.2 Consideran airplanewing with a fuel pod mounte
r#r
Vertical wing vibration Figure 1.2 Simplevibration model of an airplanewing with a fuel pod mounted at its extremity.
-T-"(o) I V r(t)
38
Introductionto Vibrationand the Free Response
Chap. 1
change in the natural frequency of vibration of the wing, modeled as in Figure 1.24,as the airplane usesup the fuel in the wing pod. The estimated physical parameters of the b e a m a r eI - 5 . 2 x 1 0 5 m a ,E - - 6 . 9 x 1 0 e N / m 2 , a n d l : 2 m . Solution The natural frequency of the vibration of the wing modeled as a simple massless beam with a tip mass is given by equation (1.66).The natural frequency when the fuei pod is fuil is
. u r ' r: V r
EEI
ffi
:ll'6rad/s
^p:l
which is about 1.8Hz (.L8cyclesper second). The naturalfrequencyfor the wing when the fuel pod is emptybecomes
loTs2x ro'r : [ 5 r a d / s o e _ p r :ItEr V ; r , - Vfijxest ro(af or 18.5 Hz. Hence the natural frequency of the airplane wing changes by a factor of 10 (i.e., becomes 10 times larger) when the fuel pod is empty. Such a drastic change mav cause changesin handling and performance characteristicsof the aircraft. !
If the springof Figure1.23is coiledin a planeas illustratedin Figure 1,.25,the stiffnessof the springis greatlyaffectedand becomes , E: I K
t
( r .61)
Severalother springarrangements andtheir associated stiffnessvaluesare listed in Thble 1.3.Textson solid mechanics, suchas Shames(1989),shouldbe consulted for further details.
,--
EI I
;-->, ,
xltl
/ : total length of spring E : elastic modulus of spring 1 = moment of inertia of cross-sectionalarea
Figure 1,25 Stiffness associated with a coiled spring.
S e c 1. . 5
Stiffness
39
TABLE1.3 SAMPLE SPRING CONSTANTS Axial stiffnessof a taperedbar of length/, modulusE, and end diameters d1 and d, Torsional stiffnesson a hollow uniform shaft of shear modulus G, length/, insidediameterdt, and outsidediameterd2
n Ed,d" 4I - dl\ , - - nc(d1
"-
3a
Tiansversestiffnessof a pinned-pinnedbeamof modulus,E, area moment of inertia 1, and length/ for a load appliedat point a from its end
. '-
3EII a 2 U- d 2
Tiansversestiffnessof a clamped-clampedbeam of modulus,E, area moment of inertia /, and length/ for a load appliedat its cenier
, ''
I92EI 13
Example1.5.3 As anotherexampleof vibrationinvolvingfluids,considerthe rollingvibrationof a ship in water.Figure1.26illustratesa schematicof a ship rolling in water.Computethe natural frequencyof the ship asit rolls back and forth aboutthe axisthroughM.
17: buoyant force
G: center of gravity M: metacenter B: center of buoyance
Figure 1.26 Dynamicsof a ship in rolling water.
Introduction to Vibration andthe FreeResponse
40
Chap.1
In the figure,G denotesthe centerof gravity,B denotesthe centerof buoyancy, M is the point of intersectionof the buoyantforce beforeand after the roll (calledthe metacenter), andft is the iengthof GM.The perpendicularline from the centerof gravity to the iine of actionof the buoyantforceis markedby the point Z.Here W denotes the weightof the ship,"/denotesthe massmomentof the shipaboutthe roll axis,and 0(t) denotesthe angleof roll. Solution SummingmomentsaboutM yields: /6(r) : -wGZ : -whsing(r) Again for smallenoughvaluesof 0,this nonlinearequationcanbe approximatedby
/ 6 ( / )+ w h o ( t ) : 0 Thus the natural frequencv of the svstem tt * ,r:\/
,
All of the springtypesmentionedarerepresented schematically asindicatedin Figure1.1.If more than one springis presentin a givendevice,the resultingstiffness of the combinedspringcanbe calculatedby two simplerules,asgivenin Figurel.27. Theserulescanbe derivedby consideringthe equivalentforcesin the system.
Springs in series
k2
K1
, -" *, o
ffi abc
7 llk,+Llk,
Springs in parallel k1
+-i
,,qnn!b t. L2
*
k o o = k t +k z
Figure 1.27 Rules for calculatingthe equivalentstiffnessesfor parallel and seriesconnectionsof springs.
Example1.5.4 Considerthe spring-massarrangementof FigureI.28(a) and calculatethe naturalfrequencyof the system. Solution To tind the equivalentsinglestiffnessrepresentationof the five-springsystem givenin Figure1.28(a),the two simplerulesof Figure!.27 areapplied.First,the parallel arrangementof k1 and k2 is replacedby the singlespring,as indicatedat the top of
4l
Stiffness
S e c .1 . 5
':{-
(b)
(a)
-3-
+
+
+
l^l
-T-
+
(d)
(c.)
systemto an equivalentsingle-spring-mass Figure1..28 Reductionof a five-springone-mass svstem.
Figure 1.28(b).Next,the seriesarrangementof k3 and kais replacedwith a singlespring of stiffness 1
uh+ w asindicatedin the bottom left sideof Figure1.28(b).Thesetwo parallelspringson the bottom of Figure1.2S(b)are next combinedusirrgthe parallelspringformulato yield a singlespringof stiffness
ktu
1
tlt,+w
asindicatedin Figure1.28(c).Thefinal stepis to realizethat the springactingat the top ofFigure 1.28(c)and that at the bottom both attachthe massto groundand henceact in parallel.Thesetwo springsthen combineto yield the singlestiffness k-k,*k)+k.
-
:kt-tk2+ks+
I,-
1 . -,,
l/kt + t/K4
ktko _(kt+ b+ k4
t < r + t < r ) ( k t +k o )+ k t t c o h+
k4
asindicatedsymbolicallyin Figure 1.28(d).Hence the natural frequencyof this systemis -'-!
m(t
Note that even though the systemof Figure 1.28containsfive springs,it consistsof only one massmoving in only one (rectilinear)direction and henceis a single-degree-offreedomsystem.
42
lntroduction to Vibrationand the FreeResponse
Chap.1
Springs are usually manufactured in only certain increments of stiffness values depending on such things as the number of turns, material, and so on (recall Figure 1.22).Becausemass production (and large sales)brings down the price of a product, the designer is often faced with a limited choice of spring constants when designing a system.It may thus be cheaper to use several "off-the-shelf" springs to create the stiffness value necessary than to order a special spring with specific stiffness.The rules of combining parallel and series springs given in Figure 1..27can then be used to obtain the desired, or acceptable,stiffness and natural frequency Example 1.5.5 Considerthe systemof Figure1.28(a)with kt : 0. Comparethe stiffnessand frequency of a 10-kg massconnectedto ground,first by two parallel springs (k. : ko : 0, kr : 1000 N/m, and kz : 3000 lf/-), then by two series springs (tc, : tc, : O, k: : 1000N/m. andko : 3000N/m). Solulion First consider the case of two paraliel springs so that fu : ko : 0, k1 : 1000N/m, andk, : 3000N/m.Then the equivalentstiffnessis givenby Figure1.27 to be the simplesum or k"o : 1000N/m + 3000N/m : 4000N/m and the correspondingfrequencyis *
: 20 rad/s
pdiltel
In the caseof a seriesconnection(kr: kr: 0), ttre two springs(t, : 1OOO N7m k+ : 3000N/m) combineaccordingto Figure1.27to yield t- _
1
_3000
" " u - V l o o 0 +t l l o o o - 3 + l
3000 4
750N/m
Thecorresponding naturalfrequency becomes : osc.i.s
r.t^" : ,ry50 V
rO,
8'66rad/s
Note that usingtwo identicalsetsof springsconnectedto the samemassin the two different waysproducesdrasticallydifferentequivalentstiffnessand resultingfrequency. A seriesconnectiondecreases the equivalentstiffness,while a parallel connectionincreasesthe equivalentstiffness.This is importantin designingsystems.
Example 1.5.5illustrates that fixed valuesof spring constantscan be used in various combinations to produce a desired value of stiffnessand corresponding frequency.It is interesting to note that an identical set ofphysical devicescan be used to create a system with drastically different frequencies simply by changing the physical arrangement of the components.This is similar to the choice of resistorsin an electric circuit. The formulas of this section are intended to be aids in designing vibration systems. In addition to understandingthat effect of stiffnesson the dynamics-that ig on the natural frequency-it is important not to forget static analysiswhen using springs.
Sec.1.6
Measurement
43
In particular, the static deflection of each spring system needs to be checked to make sure that the dynamic analysis is correctlf interpreted. Recall from the discussionof Figure 1.14 that the static deflection has the value
A:ry where m is themasssupportedby a spnngJf rtiffn"r, k in a gravitationalfield providing accelerationof gravity g. Static deflectionis often ignored in introduciory treatmentsbut is.usedextensivelyin springdesignand is essentialin nonlinearanalysis.Staticdeflectionis denotedby a variety of symbols.The symbols6, a, 6", and xq are all usedin vibration publicationsto denotethe deflectionof a springcausedby the weight of the massattachedto it. 1.6 MEASUREMENT Measurements associated with vibrationareusedfor severalpurposes. First,the quantities requiredto analyzethe vibratingmotion of a systemall requiremeasurement. The mathematicalmodelsproposedin previoussectionsall require knowledgeof the mass,damping,and stiffnesscoefficientsof the deviceunder itudy.Thesecoefficientscan be measuredin a variety of ways,as discussedin this section.Vibration measurements are alsousedto verify andimproveanalyticalmodels.Thisis discussed somedetail in Chapter7. Other usesof vibrationtestingtechniquesincludereliai1 bility and durability studies,searchingfor damage,and testingfor acceptabilityin terms of vibration parameters. Thesetopicsare alsodiscussed briefly in -haptei Z. In many cases,the massof an objector deviceis simplydeterminedby usinga scale.In fact,massis a relativelyeasyquantityto measure. However,the massmoment of inertia may require a dynamicmeasurement. A method of measuringthe mass moment of inertia of an irregularlyshapedobjectis to placethe objecton the platform of the apparatusof Figure1..29andmeasurethe periodof osciliationof the system, z. By usingthe methodsof Sectionr.4,it canbe shownthat the momentof
Suspension wires of length/
Disk of known moment 16, mass rno and radius rg
Figure 1,29 Tiifilar suspensionsystem for measuringthe momentof inertia of irregularly shapedobjects.
44
Introduction toVibration andthe FreeResponse
Chap.1
inertia of an object,l (about a verticalaxis),placedon the disk of Figure 1.29rvith its masscenteralignedverticallywith that of the disk,is givenby gTzrf,(ms+ m)
Io
It-
412l
(1.68)
Here m is the massof the part beingmeasured, ms is the massof the disk,rs the radius of the disk,/ the length of the wires,1sthe moment of inertia of the disk, and g the accelerationdue to gravity. The stiffnessof a simplespringsystemcanbe measuredassuggested in Section 1.1The elasticmodulus,E, of an objectcanbe measuredin a similarfashionby performinga tensiletest (see,e.g.,Shames, 1989).In this method,a tensiletestmachineis employedthat usesstraingages(discussed in Chapter7) to measurethe strain,e,in the testspecimenaswell asthe stress, o, bothin the axialdirectionof the specimen.This producesa curvesuchasthe oneshownin Figure1.30.The slopeof the curvein the linear regiondefinesthe Young'smodulus;or elasticmodulus,for the testmaterial.The relationshipthat the extensionis proportionalto the forceis known asFtrooke's law.
€ Strain
Figure 1.30 Stress-strain curve of a test specimen for determining the elastic modulus.
The elastic modulus can also be measuredby using some of the formulas given in Section 1.5 and measurement of the vibratory response of a structure or part. For insiance, consider thc cantilevered arrangement of Figure I.23.If tlggrass at the tip is given a snrall deflection, it witl oscillate with frequency a, : V k/m. If o is measured, the modulus can be determined from equation (1.66), as illustrated in the following example. Example1.6.L Considera steelbeam configurationas shownin Figure L23.-the beam has a length / : 1 m and momentof inertia1 : 10-ema,witha massrrz: 6 kg attachedto the tip. If the massis given a smallinitial deflectionin the transversedirectionand oscillates with a periodof T : 0.62s,calculatethe elasticmodulusof steel. Sofution SinceZ : 2rf otn,equation (1.66)yields T :Ztr
mt-
3EI
S e c .1 . 6
Measurement
45
Solvingfor -Eyields E:
4r2ml3 _
3T2I
4 T ' z ( 6k g ) ( 1 m ) 3
3(0.62s)2(1oi m4)
: 205 x 10eN/m2
The period 7, and hencethe frequencyc'ln,can be measuredwith a stopwatch for vibrationsthat are large enoughand last long enoughto see.However,many vibrationsof interesthavevery smallamplitudesand happenvery quickly.Henceseveral very sophisticateddevicesfor measuringtime and frequenryhavebeendeveloped and are discussedin Chapter7. The dampingcoefficientor, alternatively,the dampingratio is the most difficult quantity to determine.Both massand stiffnesscan be determinedby static tests;however,damping requiresa dynamic test to measure.A record of the displacementresponseof an underdampedsystemcanbe usedto determinethe damping ratio. One approachis to note that the decayenvelope,denotedby the dashed line in Figure 1.31,for an underdampedsystemis Ae-t','. The measuredpoints r(0), x(rr),x(t.r),x(tr),andso on canthenbe curvefit to A, As-tu^\,Ae-('ntz, n"-tu"tt, and so on.This will yield a valuefor the coefficient(a,.If. m and /care known,( and c can be determinedfrom (con. This approachalsoleadsto the conceptof logarithmicdecrement, denotedby 6 and definedby E:ln
x:(:t \ x(t + T)
(1.6e)
whereZ is the period of oscillation.Substitutionof the analyticalform of the underdampedresponsegivenby equation(1.36)yields E:ln
^r-La"t sin(oar + g)
4"-La"o+r) sin(orar+ oaT + g)
(i.70)
Displacement(mm)
1.0 0.5
Time (s)
Figure 1.31 Underdampedresponse usedto measuredamping.
andtheFreeResponse to Vibration Introduction
46
Chap.1
sin(oar + $),andtheexpression Sinceo2? :2n,thedenominatorbecomes e-{a"(J+r) for the decrement reduces to
(1.71)
E:lne(',r:L@,7 The period Z in this case is the damped period \2n /<,:a)so that
')6 - : 'itr,. - - 4----:+ ( ,,ft for ( yields Solvingthisexpression 5-
2rL
(1.72)
{1=
(1..73)
\/4;';E
which determinesthe dampingratio giventhe valueof the logarithmicdecrement. Thus if the valueof x(r) is measuredoff the plot of Figure 1.3I at any two sucpeaks,sayx(rr) andx(rr),equation(1.69)canbe usedto producea measured cessive valueof b, and equation(1,73)canbe usedto determinethe dampingratio.The formula for the dampingratio,equations(1.29)and (1.30),and knowledgeof m and k subsequentlyyield the value of the dampingcoefficientc. Note that peak measurementscanbe usedover anyintegermultiple of the period (seeProblem66) to increase takenat adjacentpeaks. the accuracyover measurements The computationin Problem1.66yields
E:!,n( ,'i')=,) n \x(t + nT) / where n is any integer number of successive(positive) peaks.While this does tend to increase the accuracy of computing 6, the majority of damping measurements performed today are based on modal analysismethods, as discussedin Chapter 7. Example 1.6.2 The free responseof the systemof Figure1.9with a massof 2 kg is recordedto be of the form givenin Figure 1.31.A staticdeflectiontest is performedand the stiffnessis deat tr and f2 are measuredto be 9 and terminedto be 1.5 x 103N/m.The displacements 1 mm, respectively.Calculatethe dampingcoefficient. Solution From the definition of the logarithmicdecrement 6:lnl
fr(r,).1 . Isn*l
.-: l:lnl l:2.1e12 Lx(I2)J LImml
From equation(1.73),
2.t972 5
{G,itIefi
= 0.33 or 33%
Also,
,,, : 2fk^:
2@
: 1.095x 1o2kgls
S e c .1 . 6
47
Measurement and from equation(1.30)the dampingcoefficientbecomes : ) 3 6 . 1 5k g / s c : c,,L: (t.ols x 1c'z)(0.33
Example1.6.3 Mass and stiffnessare usuallymeasuredin a straightforwardmanneras suggestedin that precludeusingthesesimple Section1.6.However,there are certaincircumstances methods.In thesecasesa measurement of the frequencyof oscillationboth beforeand after a known amountof massis addedcanbe usedto determinethe massand stiffness of the original system.Supposethen that the frequencyof the systemin Figure1,.32(a) is measuredtobe?radf s andthe frequencyof FigureI.32(b)with an addedmassof 1 kg is known to be I radls.Calculatem andk.
Figure 1.32 Using added mass and measured frequencies to detertnine n and k.
(4.)
Solution
From the definition of natural frequency .. @1
--
k m
and
- . @0 -
4m: k
and
m+1,:k
--
I L
I
r
--
k ^+I
Solving for m and k yields
or
*:tut
and r=lNz-
This formulationcan alsobe usedto determinechangesin massof a system.As an example,the frequencyof oscillationof a hospitalpatientin bedcanbe usedto monitor the changein the patient'sweight(mass)without havingto movethe patientfrom is consideredto be the changein massof the original the bed.In this casethe mass1116 system.If the original massand frequencyare known, measurementof the frequency
Introduction to Vibrationand the FreeResponse
48
Chap.1
canbe usedto determinethe changein massne. Giventhat the originalweightis 120lb, the originalfrequencyis 100.4Hz,and the frequencyof the patient-bedsystemchanges to 100Hz, determinethe changein the patient'sweight. From the two frequencyrelations a 2 1 m: k and a$(m+ *o):
k
Thus, olrru : ,?,(m + no). Solving for the change in mass trs yields
mo:ml
l'l
,\ (06
.\
Ll /
Multiplying by g and convertingthe frequencyto hertz yields
t) w,:w\fi,_ or
'o!.4 ry)' - r' )I wo-- l2orb ' " ' " 1I \r t o o H z) : 0.96lb Since the frequency decreased,the patient gained almost a pound. An increase in tiequency would indicate a loss of weight.
n
Measurement of m, c, k. <,s,,and( is used to verify the mathematicalmodel of a system and for a variety of other reasons.Measurement of vibrating systemsforms an important aspectof the activity in industry related to vibration technology. Chapter 7 is specifically devoted to measurement, and comments on vibration measurements are mentioned throughout the remaining chapters.
1.7 DESIGN CONSIDERATIONS
Design in vibration refers to adjusting the physicalparameters of a device to cause its vibration responseto meet a specified shape or performance criterion. For instance,consider the responseof the single-degree-of-freedomsystemof Figure 1 .9. The shape of the responseis somewhat determined by the value of the damping ratio in the sensethat the responseis either overdamped, underdamped, or critically damped (( > 1, ( < 1, L -- l,respectively).The damping ratio in turn depends on the values of m, c, and k. A designer may choose these values to produce the desired response.
Sec.1.7
DesignConsiderations
49
Section1.5on stiffnessconsiderations is actuallyan introductionto designas well.The formulasgiven there for stiffnessin terms of modulusand geometricdimensionscan be usedto designa systemthat has a givennatural frequency.Example I.5.2 pointsout one of the importantproblemsin design,that often the properties that we are interestedin designingfor (frequencyin this case)are very sensitiveto operationalchanges. In Example1,.5.2,the frequencychangesa greatdealasthe zrirplaneusesup fuel. Another important issuein designoften focuseson usingdevicesthat are already available.For example,the rulesgivenin Figure I.27 aredesignrulesfor producinga desiredvalue of springconstantfrom a setof "available"springsby placing them in certaincombinations,asillustratedin Example 1.5.5.Often designwork in engineeringinvolvesusingavailableproductsto produceconfigurations(or designs) that suit your particularapplication.In the caseof springstiffness, springsare usually massproduced,and henceinexpensive, in only certaindiscretevaluesof stiffness. The formulas given for parallel and seriesconnectionsof springsare then usedto producethe desiredstiffness.If costis not a restriction,then formulassuchasgiven inTable 1.3 may be usedto designa singlespringto fit the statedstiffnessrequirements.Of course,designinga spring-masssystemto havea desirednaturalfrequency may not produce a systemwith an acceptablestaticdeflection.Thus,the design processbecomescomplicated.Design is one of the most activeand excitingdisciplines in engineeringbecauseit often involvescompromiseand choicewith many acceptablesolutions. Unfortunately,the valuesof m, c,andk haveotherconstraints. In particular,the sizeand materialof which the deviceis madedeterminestheseparameters. Hencethe designprocedurebecomesa compromise.For example,for geometricreasons,the massof a devicemay be limited to be between2 and 3 kg, and for staticdisplacement conditions, the stiffnessmaybe requiredto be greaterthan200N/m. In thiscase, the natural frequencymust be in the interval 8.16rad/s = 0), S 10 rad/s
(r.14)
This severely limits the design of the vibration response,4s illustrated in the following example. Example 1.7.1 Considerthe systemof Figure1.9with massand stiffnesspropertiesassummarizedby inequality(1.74).Supposethat the systemis subjectto an initial velocitythat is always lessthan 300mm/s,and to an initial displacement of zero (i.e.,x6 : 0, ?ro- 300mm/s). For this rangeof massand stiffness, choosea valueof the dampingcoefficientsuchthat the amplitudeof vibrationis alwayslessthan 25 mm. Solution This is a design-oriented example,and henceastypicalof designcalculations there is not a nice,cleanformulato follow.Rather,the solutionmust be obtainedusing
50
Introduction to Vibration andthe FreeResponse
Chap.1
theory andparameterstudies. First,note that for zeroinitial displacement, the response may be written from equation(1.38)as x(/)
uo A7
e-(',tsin(orat)
Also note that the amplitudeof this periodicfunctionis 1)^
t'-"' Thus,for smallo2 the amplitudeis largerthanfor largeror7.Hencefor the rangeof frequenciesof interest,it appearsthat the worstcase(largestamplitude)will occurfor the smallestvalueof the frequency(o, : 8.16rad/s).Also,the amplitudeincreases with oo so that usingoo : 300mm/s will ensurethat amplitudeis a large aspossible.Now,o6and r,rnare fixed,so it remainsto investigatehow the maximumvalue of .r(t) variesas the dampingratio is varied.One approachis to computethe amplitudeof the responseat the first peak.From Figure1.10the largestamplitudeoccursat the first time the derivativeof -r(r) iszero.Takingthederivativeof x(l) andsettingitequaltozero yieldsthe expressionfor the time to the first peak: s)de*Lantcos(o;t)
-
(ans-|'^t sin(co,r) :
0
Solvingthis for I and denotingthis valueof time by T- yields
= Itun-,ryL--L) r^: -t-ra"-'13) @7
\Eo"/
64
\
(
/
The valueof the amplitudeof the first (and largest)peak is calculatedby substituting the valueof T- into .x(t),resultingin A^(()=x\T^):
-(2\\ -'lft / ( il - r - - - r - t u " - ( v ' " - /( s' )i n \ t " - ' \ - - a - / / ,,\/t-(2
Simplifyingyields
A.(L) : ?t;1|4'^"
-'l''4-\ \s )
For fixed initial velocity(the largestpossible)and frequency(the lowestpossible),this value of ,4-(() determinesthe largestvaluethat the highestpeak will haveas ( varies. The exactvalue of ( that will keepthis peak,and hencethe response,at or below 25 mm canbedeterminedbynumericallysolvingA.(():0.025(m)foravalueof(.Thisyields ( : 0.281.Usingthe upperlimit of the massvalues(rn : 3 kg) then yieldsthe valueof the required damping coefficient: c :Zma,(:
2 ( 3 X 8 . 1 6 X 0 . 2 8:1 ) 1 4 . 1 , 5 k g / s
For this valueof the dampingthe responseis neverlarger than 25 mm. Note that if there is no damping, the same initial conditions produce a response of amplitude A : u s f a n : 3 7m m . n
Sec. 1.7
DesignConsiderations
5t
As another example of design,consider the problem of choosing a spring that will result in a spring-mass system having a desired or specified frequency.The for_ mulas of Section 1.5 provide a means of clesigninga rpting to have a specified stiffness in terms of the properties of the spring material (modulus) and its geometry. The following example illustrates this. Example 1.7.2 Considerdesigninga helicalspringsuchthat when attachedto a 10-kgmass,the resulting spring-masssystemhasa naruralfrequencyof 10 rad/s (about 1.6Hz). Solution From the definition of the natural frequency,the springis required to have a stiffnessof k:
l o 2 , m : ( 1 0 ) ' z ( 1 0=) 1 0 3 N / m
The stiffnessof a helicalspringis givenby equation(i.65) to be k : r G N / *'=
Gdo. 64nR'
or
Gd: 6.4 x I0o: nR'
This expressionprovidesthe startingpoint for a design.Thechoicesavaiiableare the type of materialto be used(hencevariousvaluesof G); the diameterof the material.d: the radiusof the coils,R; and the numberof turns,n. The choicesof G and d are,ofcourse, restrictedby availablematerials,n is restrictedto be an integer,and R may have re_ strictionsdictatedby the sizerequirementsof the device.Here it is assumedthat steel of l.-cmdiameteris available.The shearmodulusof steelis about G : 9.273x 1010 N/m2 so that the stiffnessformula becomes 6.4 x 704N/m :
( s . z t t x I o r o N / m 2 ) ( l om - r) 4 nR3
nR3 = I.292 x l0 2 If the coil radiusis chosento be 10crn,this yieldsthe fact that the numberof tumsshouldbe l'29]<-lo-2 mj : r 2 . 9o r 1 3 , = IU"M'
Thus, if 13 turns of l-cm-diametersteelare coiled at a radiusof 10 cm, the resultins springwill havethe desiredstiffnessand the 10-kgmasswill oscillateat approximatel! 10 radls.
tr In Example1'.7.2,several variableswerechosento producea desireddesign.In eachcasethe designvariables(suchasd, R,etc.)are sub.iect to constraints. Suchcon_ straintsare consideredformally in Chapter5. Other aspectsof vibrationdesignare presentedthroughoutthe text asappropriate.There are no set rulesto follow in designwork. However,someorganizedapproachesto designare presentedin chapter 5.
to Vibrationand the FreeResponse lntroduction
52
'l Chap.
The following example illustrates another difficulty in design, by examining what happens when operating conditions are changed after the design is over. Example1.7.3 systemof a smallsportscar, As a lastexampie,considermodelingtheverticalsuspension systemof the form as a single-degree-of-freedom mi*c*+kx=0 wherem is the massof the automobileandc andk arethe equivalentdampingandstiffness systems. The car deflectsthe suspensionsystemunder of the four-shock-absorber-spring is chosen(designed)to be criticaliy damped.If the its own weight0.05m. The suspension carhasa massof 1361kg (massof a PorscheBoxster),calculatethe equivalentdampingand a full gastank,andluggage stiffnesscoefficientsof the suspension system.If fwo passengers, totaling 290kg arein the car,how doesthis affectthe effectivedampingratio? Solution The massis 1361ks and the naturalfrequencvis
so that k:
136ltL'f,
an amountA, calledthe staticdeflection,by the At restthe car'sspringsarecompressed weightof the car.Hence,from a forcebalanceat staticequilibrium,mE : kL',sothat
k ---g A
and i kit:
'":\;/
: \ -i r8/\ ' '
= \ /0q.. 08 \5' '/
:r4racrls
I h e s t i l l n e s so f t h e s u s p e n s i o sn y s r e mi s r h u s : 2 . 6 6 8x 1 0 sN / m k - 1361(14)2 For critical damping ( : I or c : c,,and equation (1.30) becomes c : 2man : 2(1361)(14) : 3.81 x 104kgls Now if the passengers and luggage are added to the car, the mass increases to 136l + 290 : 165l kg. Since the stiffnessand damping coefficient remain the same,the new static deflection becomes
mF I : -: k frequency
1651(9.8) - = 0.06m 2.668x It'
Sec. 1.8
53
Stability Fromequations(L29) and (I.30)the dampingratio becomes _c
3.81x 104: Zma,
-:
3 . 8 1x 1 0 0 ilL,
2(1.65r)(r2.7)
is no longercriticallydampedandwill fuel,andluggage Thusthecarwith passengers, exhibitsomeoscillatorv motionin theverticaldirection.
Note that this illustratesa difficulty in designproblems,in the sensethat the car cannot be exactlycritically dampedfor all passengersituations.Inthis case,if critical dampingis desirable,it really cannotbe achieved.Designsthat changedramaticallywhen one parameterchangesa smallamountare saidnot to be robust.This is discussedin greaterdetail in Chapter5.
1.8 STABILITY the physicalparametersm, c,andk are allconsideredto be In the precedingsections, positive in equation (1,.25).Thisallows the treatmentof the solutionsof equation (1,.25)to be classifiedinto three groups:overdamped,underdamped,or critically Thesefour sodamped.The casewith c : 0 providesa fourth class,calledundamped. grow with time and their amthat they do not well behaved in the sense lutions are all plitudesare finite.There are manysituations,however,in which the coefficientsare not positive,and in thesecasesthe motion is not well behaved.This situationrefers to the stabiliryof solutionsof a system. Recallingthat the solution of the undampedcase(c : 0) is of the form (co,r + +), it is easyto seethat the undampedresponseis bounded.That is,if ,4 sin lr(t)l denotesthe absolutevalueof x, then
+ o)l : e: l'(r)l= Alsin(o,r
Lr,T*o * ,1
(1..7s)
for every value of r. Thus lr(r)l is alwayslessthan somefinite number for all time and for all finite choicesof initial conditions.In this casethe responseis well behaved and said to be stable(sometimescalled marginallystable).If, on the other hand,the valueof k in equation (1,.2) is negativeandm is positive,the solutionsare of the form x(t) : ,4 sinh a,t I B cosh ornt
(1.76)
which increaseswithout bound as t does.In this caselx(t)l no longerremainsfinite 1.33illustratesa stablereand suchsolutionsare calleddivergentor unstable.Figure sponseand Figure 1.34illustratesan unstable,or divergent,response.
Introduction to Vibrationand the FreeResponse
54
Chap.
(mm) Displacement
1.0 0.5 Time (s)
0.0 10
'\/
).
-0.5 - 1.0 Figure1.33 Exampleof a stableresponse. (mm) Displacernent
Time(s)
Figure 1.34 Example of an unstable,or divergent,response.
Consider the responseof the damped systemof equation (1.25)with positive coefficients.As illustrated in Figures 1.10,1.11,1.12,and 1.13,it is clear that x(t) approaches zero as I becomes large becauseof the exponential-decay terms. Such solutions are called asymptotically stable.Again, if c or k is negative (with m positive), the motion grows without bound and becomes unstable as in the undamped case.In the damped case,however, the motion may be unstable in one of two ways. Similar to overdamped solutions and underdamped solutions, the motion may grow without bound and not oscillate, or it may grow without bound and oscillate. The nonoscillatory case is called divergent instability and the oscillatory case is called flutter instability, or sometimes just flutter. Flutter instability is sketched in Figure 1.35. Displacement(mm) 8 o
4 2 0
-6 -8
Time (s) 46810t21416 Figure 1.35 Exampleof flutter instability.
S e c .1 . 8
Stability
55
The trend of growing without bound for large r contirrues in Figures 1.34 and 1.35, even though the figure stops.These types of instability occur in a variety of situations, often called self-excitedvibratiorzs,and require some source of energy.The following example illustrates such instabilities.
Example1.8.1 Considerthe invertedpendulumconnectedto two equalsprings,shownin Figure1.36. Assumethat the springsare undeflectedwhen in the verticalpositionand that the massm of the ball at the end of the pendulumrod is substantiallylargerthan the mass of the rod itself,so that the rod is consideredto be massless. If the rod is of length/ and the springsare attachedat the point I p,the equationof motion becomes ml26 +
/kt2
\t''u/
\
c o s O- r n g l s i n 0: 0
(r.77)
This is obtainedby summingmomentsabout the point of attachmentof the pendulum to ground(hencethe reactionforceat the pin doesnot enterinto the equation).For valu e so f 0 l e s st h a n a b o u tr f 2 0 . s i n 0 a n d c o s 0c a n b e a p p r o x i m a t ebdy s i n 0 = 0 a n d cos0 = 1.Applying this approximationto equation(1.77)yields k12 ml26 + - 0 mgl\ :0 z which upon rearrangingbecomes 2mt6\) + (kt - zms)o(t) : o where0 is now restrictedto be small(smallerthanr 120).Ifft, /, and m are allsuchthat the coefficientof 0, calledthe effectivestiffness, is negative,that is,if kl-Zmg<0 the pendulummotion will be unstableby divergence, asillustratedin Figure1.34.
Figure 1.36 Inverted pendulum oscillator and its free-body diagram. Heref force at the pin, and the pendulum is of length /.
is the total reaction
to Vibrationand the FreeResponse Introduction
56
Chap.1
ExampleL.8.2 The vibrationof an aircraftwing canbe crudelymodeledas mi*ci+kx:1* where m, c and k are the mass,damping,and stiffnessvaluesof the wing,respectively, modeledasa single-degree-of-freedom system,andwhere1i is an approximatemodel of the aerodynamicforceson the wing (ry > 0 for high speed).Rearrangingthis expressionyields mi+(c-"1,)t+kx:0 -
I f l a n d c a r e s u c h t h a t c 1 > 0 , t h e s y s t e m i s a s y m p t o t i c a l l y s t a b l e . H o w e1vi se r , i f h fe f o r m s u c h t h a t c- 1 < 0 , t h e n ( : ( c - 1 ) p m a ^ < 0 a n d t h e s o l u t i o n s a r et o + s) xv ) = Ae-tu''sin(urrl where-(or,/ > 0 for all r > 0. Suchsolutionsincreaseexponentiallywith time,asindicatedin Figure1.35.This is an exampleof flutter instabilityand self-excitedoscillation.
n
OFTHETIMERESPONSE 1.9 NUMERICALSIMULATION So far the vibration problems examined have all been cast as linear differential equations that have solutions that can be determined analytically. These solutions are often plotted versustime in order to visualizethe physicalvibration and obtain an idea of the nature of the response.However, there are many more complex systems that cannot be solved analytically.The pendulum equation given in Example 1.4.2is such a system.In order to solve ihe pendulum equation analytically,the approximation of sin ( 0 ) : 0 was used.This allowed the analytical determination of an approximate solution that holds only for those initial conditions for which 0 remains less than about 10 degrees.For caseswith larger initial conditions, a numerical integration routine may be used to compute and plot a solution to the equation of motion. The free responseof any single-degree-of-freedomsysten may be easily computed by simple numerical means such as Euler's method or Runge-Kutta methods. This section examines the use of these common numerical methods for solving vibration problems that are difficult to solve in closedform. Runge-Kutta schemescan be found on calculators and in most common mathematical software packages such as Mathematica, Mathcad, Maple, and Mant-as, or they may be programmed in more traditional languages,such as FORTRAN, or into spreadsheets.This section reviews the use of numerical methods for solving differential equations and then applies these methods to the solution of severalvibration problems consideredin the previous sections.These techniques are then used in the following section to analyze the response of nonlinear systems.Appendix F introduces the use of Mathematica, Mathcad, and Mert-ae for numerical integration and plotting. Many modern curriculums introduce these methods and codes early in the engineering curriculum, in which casethis section can be skipped, or used as a quick review
Sec.1.9
Numerical Simulation of theTimeResponse
57
There are many schemesfor numericallysolvingordinary differentialequations,such as thoseof vibration analysis. Two numericalsolution schemesare presentedhere.The basisof numericalsolutionsof ordinarydifferentialequationsis to essentiallyundo calculusby representingeachderivativeby a smallbut finite difference(recall the definition of a derivativefrom calculusgivenin Window 1.6).A numericalsolutionof an ordinary differentialequationis a procedurefor constructing approximatevalues:x1,xz,. . . , x,, of the solutionx(r) at the discretevaluesof time: to 4 tt 1tz"'. < 1,. Effectively,a numericalprocedureproducesa list of discrete valuesxt : xlt,) that approximatethe solutionrather than a continuousfunction x(t), which is the exactsolution.Theinitial conditionsof the vibrationproblemof interestform the startingpoint. For a single-degree-of-freedom systemof the form :0 : : * mi ci * kx r(0) xe .i(0) uo (1.78) the initial valuesxoand oeform the first two pointsof the numericalsolution.Let Ty be the total lengthof time over which the solutionis of interest(i.e.,the equationis tobesolvedforvaluesof lbetweenr : 0andt : T).ThetimeintervalT, - }isthen dividedup into n intervals(sothat L,t : Trln).Thbnequarion(1.78)is ialculatedat thevaluesof /0 : 0,/r : At,tr: 2L,t,...,tn: nLt : Tytoproduce an approximate representation, or simulation,of the solution. Window 1.6 The definition of a derivative of x(r) at t : t,is dx\t) . dt
:llm A1+0
x(t,*r) * x(t,)
Lt
where ti+t : tr * Ar and x(r) is continuous.
The concept of a numerical solution is easiestto grasp by first examining the numerical solution of a first-order scalar differential equation. To this end consider the first-order differential eouation *(t) : ax(t)
x(0) : xe
(r.7e)
The Euler method proceeds from the definition of the slope form of the derivative given in Window 1.6,before the limit is taken: xt+r - xi
:o*,
N
(1.80)
where rr denotes x(ti), xi+t denotes x(t,*r),and Ar indicates the time interval between t, and t,*, (i.e., Ar : ti+t - r;). This expressioncan be manipulated to yield xi+l : x, I
Ltlaxi)
(1.81)
This formula computes the discrete value of the response x,n, from the previous value x;, the time step A/, and the system's parameter a.This numerical solution is called
to Vibrationand the FreeResponse lntroduction
58
Chap.1
an Euler or tangentline method.The following example illustrates the use of the Euler formula for computing a solution using vtbl-Z. Example1.9.1 -3x, t(0) = 1 for varUse the Euler formulato computethe numericalsolutionof i : ious time incrementsin the time interval0 to 4, and comparethe resultsto the exact solution. Solution First,the exactsolutioncanbe obtainedby directintegrationor by assuming a solutionof the form x(t) : Arxt.Substitutionof this assumedform into the equation * : -3x yields , tr'er': -3Ae\', or \ = -3, so that the solution is of the form x(,t) : 4r-zt.Applying the initial conditions;r(0) = 1 yieldsA : L.Hencethe analytical solutionis simplyx(t) : ,-' ' by equation Next considera numericalsolutionusingthe Euler methodsuggested 3 , s o t h a t x ; n i : + A / ( 3 x ; ) . S u p p o sethatavery l r (1.81).Inthiscasetheconstanta: : fiom the interval over is formed and the solution (i.e., 0.5) Al is taken step crudetime / : 0 to r : 4.Then Table1.4illustratesthe valuesobtainedfrom equation(1.81): xo:1
: -0.5 xr = ro + (o.s)(-3)(.ro) :0.25 xz: -05 - (1.s)(-0.s)
forms the column marked "Euler." The column marked "exact" is the value of e-'' at the indicated elapsedtin.refor a given index. Note that while the Euler approxlmate gets close to the correct final value, the value oscillates around zero while the exact value does not.This points out a possible source of error in a numerical solution. On the other hand, if Ar is taken to be very small, the difference between the solution obtained by
OFTHE EXACTSOLUTIONOF * : -3x' x(0) : 1 TABLE 1.4 COMPARISON TO THE SOLUTIONOBTAINEDBY THE EULERMETHODWITH LARGETIME f : OTO 4 STEP(Af : 0.5) FORTHE INTERVAL
Index
00 1 2 3 4 5 6 7 8
Elapsed time
0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000
Exact
1.0000 0.2231 0.0498 0.0111 0.0025 0.0006 0.0001 0.0000 0.0000
Euler
1.0000 -0.5000 0.2500 -0.1250 0.062s -0.0312 0.0156 -0.0078 0.0039
Absolute error
0 -0.7231 0.2002 -0.1361 0.0600 -0.0318 0.0155 -0.0078 0.0039
S E c1 . 9
NumericalSimulationof the Time Resoonse
59
A r : 0 . 1s ,Ar:0.5s
Time (s)
0.5
1
1.5
Figure1.37 Plotofx(1,)versust,fori: .r(0) : 1.
2
2.s
3
3.5
4
-3rusingA/ - 0.linequation(1.81)with
the Euler equationandthe exactsolutionbecomeshard to see,asFigure1.37illustrates. Figure 1.37is a plot of x(r) obtainedvia the Euler formula for At : 0.1.Note that it looksvery muchlike the exactsolutione-3'. ! It is important to note from the example that two sourcesof error are present in computing the solution of a differential equation using a numerical scheme such as the Euler method. The first is called the /o rmula error, which is the difference between the exact solution and the solution obtained by the Euler approximation.This is the error indicated in the last column of Table 1.4. Note that this error accumuIates as the index increasesbecausethe value at each discrete time is determined by the previous value, which is already in error. This can be somewhat controlled by the time step and the nature of the formula. The other source of error is the round-off eruor due to machine arithmetic. This is,of course,controlled by the computer and its architecture. Both sourcesof error can be significant.The successfuluse of a numerical method requires an awarenessof both sourcesof errors in interpreting the results of a computer simulation of the solution of any vibration problem. The Euler method can be improved upon by decreasing the step size,as Example 1.9.1 illustrates. Alternatively, a more accurate procedure can be used to improve the accuracy (smaller formula error) without decreasing the step size Lt. Several methods exist (such as the improved Euler method and various Taylor series methods) and are discussed in Boyce and DiPrima (1997), for instance. Only the Runge-Kutta method js discussedand used here. The Runge-Kutta method was developed by two different researchers from about 1895 to 1901 (C. Runge and M. W. Kutta). The Runge-Kutta formulas (there
to Vibrationand the FreeResponse lntroduction
60
Chap'1
are several) involve a weighted average of values of the right-hand side of the differential equation taken at different points between the time intervals /, and l' + A/. The derivations of various Runge-Kutta formulas are tedious but straightforward and are not presented here (see Boyce and DiPrima,1991). One useful formulation c a n b e s t a t e d f o r t h e f i r s t - o r d e r p r o b l et :m f ( x , r ) , x ( 0 ) : x n , w h e r e / i s a n y s c a l a r function (linear or nonlinear) as
xn+t:r,, *
+
(k,,t* zk,z+ 2k,3+ k,4)
(1.82)
where
k6 : f \x,,t,)
k,,r:f (r, * I0,r,,,* +) * +) * I o,r,t" k,.:f(,,r, k,,q: f(xu + Ltkn,t, + Lt) The sum in parenthesesin equation (1.82) representsthe average of six numbers each of which looks like a slope at a different time; for instance,the term ft,1 is the slope of the function at the "left" end of the time interval. Such formulas can be errhancedby treating At as a variable, A/r. At each time step /,, the value of Ar, is adjusted based on how rapidly the solution x(r) is changing. If the solution is not changing very rapidly, a large value of A/r is allowed without increasing the formula error. On the other hand, if x(r) is changing rapidly, a small Ar, must be chosen to keep the formula error small. Such step sizescan be chosen automatically as part of the computer code for implementing the numerical solution. The Runge-Kutta and Euler forrnulas just listed can be applied to vibration problems by noting that the most general (damped) vibration problem can be put i n t o a f i r s t - o r d c rf o t m . Returning to a damped systemof the form mi(t) + ci(t) + kx(t) :0
x(0) - x0,
t(0) : i0
(1'83)
the Euler method of equation (1.81) can be applied by writing this expression as two first-order equations.To this end, divide equation (1.83)by the massm and define two new variablesby x1 : x(t) and 12 : t(/).Then differentiatethe definition of 11(r), rearrange equation (1.83), and replace x and its derivative with .rr and "t2to get i1Q) : x2(t)
*r(r):-k*'U)-**'(t)
(1.84)
subject to the initial conditions r,(0) : ro and xz(O) : i6. The nvo coupled firstorder differential equations given in (1.8a) may be written as a single expression by
Sec.1.9
Numerical of theTimeResponse Simulation
6l
using a vector and matrix form determinedby first defining the vector 2 x 1 x(/) and the 2 x 2matrix Aby
A:l
t ko c1l -;) I L-*
x(/) : [;;[;]]
: x(o)
[;;lS]]
(18s)
The matrix ,4 definedin this way is calledthe statematrix and the vectorx is called the statevector.The position xt and the velocity x2 lre called the statevariables. Using thesedefinitions(seeAppendix C), the rules of vector differentiation(element by element)andmultiplicationof a matrix timesa vector,equations(1.84)may be written as *(r) : Ax(t)
(1.86)
subject to the initial condition x(0). Now the Euler method of numerical solution given in equation (1.81) can be applied directly to this vector-matrix formulation of equation (1.86), by simply calling the scalar x, the vector x, and replacing the scalar a with the matrix,4 to produce
*(r,*,):x(r,) + A,tAx(r,)
(1.87)
This, along with the initial condition x(0), defines the Euler formula for integrating the general single-degree-of-freedomvibration problem described in equation (1.82) for computing and plotting the time response. As suggested,the Euler-formula method can be greatly improved by using a Runge-Kutta program. For instance,Merree has two different Runge-Kutta-based simuiations:ode23and ode45.Theseare automaticstep-sizeintegrationmethods (i.e., A/ is chosen automatically). The Engineering Vibration Toolbox has one fixed-step Runge-Kutta-based method, vtbl_3, for comparison.The M-file ode23usesa simple second-and third-order pair of formulas for medium accuracyand ode45usesa fourthand fifth-order pair for greater accuracy.Each of these corresponds to a formulation similar to that expressedin equations (1.82) with more terms and a variable step size Ar. In general, the Runge-Kutta simulations are of a higher quality than those obtained by the Euler method. Example 1.9.2 Use the ode45functionto simulatethe responseto 3; + i + 2x : 0 subjectto the initial conditionsx(0) : 0, t(0) : 0.25over the time interval0 < t < 20. Solution The first stepis to write the equationof motion in first-orderform.This yields xt=
Next an M-file is created to store the equations of motion.An M-file is created by choosing a name, say sdof . m,and entering
62
Introduction to Vibration and the FreeResoonse
C h a p .1
!
^5
0.2 0.1
-.!_
0>,
E>
0 -0.1 *0.2 Ilme
Figure1.38 Plot of the displacement x(t) of thesingle-degree-of-freedom systemof Example1.9.2(solidline) and the corresponding velocityi(t) (dashedline). function xdot = sdof(t,x) xdot = zeros(2,I); xdot(1) = x(2); x d o t ( 2 ) = - ( 2/ 1 1 " * r L ) - ( t / 3 ) ' , x ( 2 ) ; Next, go to the command mode and enter t0 =0;tf = 20; x0 = [0 0.25]; ' [ t , x ] = o d e 4 5( s d o f ' , [ t 0 t f ] , x 0 ) ; pl ot (t, x) The first line establishesthe initial (t0) and final times (tf).The second line creates the vector containing the initial conditions x0. The third line creates the two vectors t, containing the time history and x containing the response at each time increment in r, by calling ode45 applied to the equations set up in sdof.The fourth line plots the vector x versus the vector t. This is illustrated in Fieure 1.38.
The preceding example may also be solved using Mathematica,
Mathcad, and
Maple by writing a FORTRAN routine or by using any number of other computer codes or programmable calculators.The following example illustrates the commands required to produce the result of Example 1.9.2using Mathematica and again using Mathcad.These approachesare then used in the next section to examine the response of certain nonlinear vibration Droblems. Example 1.9.3 SolveExample1.9.2usingthe Mathematicaprogram. Solution The Mathematicaprogramusesan iterativemethodto computethe solution and acceptsthe second-order form of the equationof motion.The text after the prompt
S e c .1. 9
NumericalSimulationof the Time Response
63
"Inl1l:="is typed by the userand returnsthe solutionstoredin the variablex[t]. In the argumentof the NDMathematicahasseveralequalsignsfor differentpurposes. to be solved,followedby the equation the differential types in function the user Solve and the nameof the independent initial conditions,the nameof the variable(response), variablefollowedby the intervalover which the solutionis sought.NDSoI','ecomputes the solutionand storesit asan interpolatingfunction;hencethe codereturnsfollorving the output prompt Out [1] =.The plot commandrequiresthe nameof the interpolating function returnedby NDSolve,x[t] in this case,the independentvariable,t, and the rangeof valuesfor the independentvariable. In [1-]: = N D S o l v e [ { x "[ t ] + ( 1 , / 3 ) ' ! x ' [ t ] + ( 2 / 3 ) * x [ t ] = - 0 , x ' [ 0 ] = = 0 . 2 5 , x [ 0 ] = = 9 1 , x , { t , 0 ,20ll P l o t [ E v a Iu a t e [ x [ t ] / . % 1 ,{ t , 0 , 2 0 } l 0 u t [ 1 ] = { { x - > I n t e r p o l a t ' i n g F u n c toi nt { t 0 . , 2 0 .} } , + l } }
0.15 0.1
-0.1
Out[2]= -Craphics-
u Example 1.9.4 SolveExample1.9.2usingthe Mathcadprogram. Solution The Mathcadprogramusesa fixed time stepRunge-Kuttasolutionandreturns the solution as a matrix with the first column consistingof the time step,the secondcolumn containingthe response,and the third column containingthe velocity response. First type in the initial
condition vectoi-:
[01
L 0 . 2 Is
64
Introduction to Vibration andthe FreeResoonse Then type in the systemin first-order
Chao.1
form:
r-l ,=t| Y_r+r,)_ D(t,y) 1r,_I) L \3 S o l v eu s i n g Runge-Kutta: Z := rkfixed(y,0,20,1000,D) N a m et h e t i m e v e c t o r f r o m t h e R u n g e - K u t t am a t r i x s o l u t i o n : +
.-
7<0>
N a m et h e d i s p l a c e m e n t v e c t o r f r o m t h e R u n g e - K u t t a m a t r i x s o l u t i o n : x i= Z'1' N a m et h e v e l o c i t y
v e c t o r f r o m t h e R u n g e - K u t t am a t r i x s o l u t i o n : dxdt := Z'2'
Pl ot the sol ut'i ons.
X
d-xdt
D The useof thesecomputationa!programsto simulatethe responseof a vibrating systemis fairly straightforward. Further informationon usingeachof theseprogramscan be found in AppendixF or by consultingmanualsor any one of numerous bookswritten on usingthesecodesto soivevariousmath and engineeringproblems. You are encouragedto reproduceExample I.9.4 and then repeat the problem for variousdifferentvaluesof the initial conditionsand coefficients. In this way,you can
S:c.1.10
Coulomb Friction andthePendulum
65
build some intuition and understanding of vibration phenomena and how to design a svstem to produce a desired response.
..10
C O U L O M BF R I C T I O N A N D T H EP E N D U L U M In the previoussections,all of the systemsconsideredare linear (oi linearized)and havesolutionsthat can be obtainedby analyticalmeans.In this sectiontwo systems are analyzedthat arenonlinearand do not havesimpleanalyticalsolutions. The first is a spring-masssystemwith slidingfriction (coulomb damping),and the secondis the full nonlinearpendulumequation.In eachcasea solutionis obtainedby usingthe numericalintegrationtechniques introducedin Section1.9.Theabilityto computethe solutionto generalnonlinearsystemsusingthesenumericaltechniquesallowsus to considervibration under more sophisticated effects. Nonlinearvibrationproblemsare much more complexthan linear systems. Their numericalsolutions, however,are oftenfairly straightforward. Severalnewphenomenaresultwhen nonlineartermsare considered. Most notably,the ideaof a single equilibrium point of a linear systemis lost.In the caseof coulomb damping,a continuousregion of equilibrium positionsexists.In the caseof the nonlinearpendulum,an infinite numberof equilibriumpointsresult.This singlefact greatlyComplicatesthe analysis, measurement, and designof vibratingsystems. A commondampingmechanismoccurringin machinesis causedby slidingfriction or dry friction and is calledCoulombdamping.Coulombdampingis characterized bv the relation
f , : F , ( * 1:
{lJ;:3}
wheref, is the dissipationforce,l/ is the normalforce (seeany introductoryphysics text), and p is the coefficientof slidingfriction (or kinetic friction). Figure 1.39is a schematicof a massru slidingon a surfaceand connectedto a springof stiffnessk.
Figure 1.39 Spring-masssystemsliding on a surfacewith kinetic coefficientof friction u.
66
Introductionto Vibrationand the Free Response
Chap. 1
TABLE1,5 APPROXIMATE COEFFICIENTS OF FRICTION FORVARIOUSOBJECTSSLIDINGTOGETHER
Material
Kinetic
Static
Metal on metal (lubricated) Wood on wood Steel on steel (unlubricated) Rubber on steel
0.07 0.2 0.3 1.0
0.09 0.25 0.75
F-
1an
r(r)
/.:
pN : pmq
(b)
(4.)
Figure 1.40 Free-body diagram of the forces acting on the sliding block system of Figure 1.39: (a) mass moving to the left (t < 0) (b) mass moving to the right ( i > 0 ) . F r o m t h e y d i r e c t i o n ,N - m g .
The frictionalforce f, always opposesthe direction of motion causing a system with coulomb friction to be nonlinear. Table 1.5 lists some measured values of the coefficient of kinetic friction for several different sliding objects.Summing forces in part (a) of Figure 1.40in the x direction yields that (note that the mass changesdirection when the velocity passesthrough zero)
mIIkx:Wmg
for i<0
(1.88)
Here the sum of the forces in the vertical direction yields the fact that the normal force ,4y'is just the weight, mg, where g is the accelerationdue to gravity (not the case if rz is on an inclined plane). In a similar fashion, summing forces in part (b) of Figure 1.40yields
mi*kx:_pmg
for i>0
(1.8e)
Since the sign of i determines the direction in which the opposing frictional force acts, equations (1.88) and (1.89) can be written as the single equation mi * pmg sgn(i) + kx :0
(1.e0)
wheresgn(r) denotesthe signumfunction,definedto havethe value 1 for r > 0, -1 for t { 0, and 0 for t : 0. This equationcannotbe solveddirectly using methods suchasthe variationof parametersor the methodof undeterminedcoefficients. This is becauseequation(1.90)is a nonlineardifferentialequation.Rather,equation(1.90) can be solvedby breakingthe time intervalsup into segmentscorrespondingto the
Sec.1.'10
67
CoulombFrictionand the Pendulum
: 0). Alchangesin directionof motion (i.e.,at thosetime intervalsseparatedby i ternatively,equation(1.90)canbe solvednumerically. To solveequation(1.90),first assumethat the initial conditionsaresuchthat the to an initial massis movingto the left, asindicatedin Figure1.40(a).Thiscorresponds : < themass,sequilibrium 0andx(0) xo,wherexgistotherightof conditionofi(0) positionand is large enoughso that the restoringforce,kxs,overcomesthe friction iorce.Recall that the staticfriction coefficientis alwaysgreaterthan the coefficientof dynamicfriction.Becauseof friction,the equilibriumpositionis not a singleposition, :0This but rather there are many positionsfor which r(/) < pmclk andi(r) systems. existenceof multiple equilibria is also a characteristicof nonlinear With xs to the right of any equilibrium the massis moving to the left, the friction forceis to the right,and equation(1.88)holds.Equation(1.88)hasa solutionof the form
x ( / ): , 4 1 c o s, o n*t B l s i n r o. r y/
(1.e1)
where .^ : f kJ^and A1 and 81 are constantsto be determined by the initial conditions. To that end, applying the initial conditions yields Lmg
x ( 0 ): A ' + f = x s i(0):onB1
(1,.92)
(1.e3)
:0
(1.91).Thus constantsinequation HenceBr : 0 andAy- -rs ILmglkspecifiesthe as it moves the left, to (at moves and rs) rest from starts the mass when x\t):
/ \"0-
tt^g \ L /cos(l)nt
wmg
(1..e4)
This motion continuesuntil the first time *(t) :0. This happenswhen the derivative of equation(1.94)is zero,or when
t(/) :-'"('o
-"f)sinorn/1 :o
(1.es)
*(t) :0 and the mass starts to move to the right provided Thus when t1 : 11f11.,, that the spring force,kx,is large enough to overcome the maximum frictional force p,mg.Henceequation (1.89) now describesthe motion. Solving equation (1.89) yields
* B2sino,t ,r(t): A2cos.;.nt ry
(1.e6)
for r f a, < t < /r, where /2is the secondtime that i becomeszero.T'heinitial conditionsfor equation(1.96)are calculatedfrom the previoussolutiongivenby equation (L.94)au1: /n\ xl -l
\ o",/
: (ro* flcosn
. / " r \ = -"(to xt-l \ to",/
-
flsint:
pmg k
o
:
2wmg- X g
(r.e7) (1.e8)
68
Introduction to Vibrationand the FreeResponse
Chap.1
4pN\
- - l
KI
\'oo":(4#1)
/ - \/' Y o -
2pN\ t /
Figure 1.41 Free response of a svstemwith Coulomb friction present.
From equation (1.96) and its derivatives it follows that Az:
xo -
3pmg
(1.ee)
Bz: 0
The solutionfor the secondintervalof time then becomes
x ( r ):
(
3w*S \
\"-;/coso,r
_
p-g
n
T
*.,.il
Zn
(1.100)
This procedure is repeateduntil the motion stops.Themotion will stop when the velocity is zero (* : 0) and the spring force (ftr) is insufficient to overcome the maximum frictional force ( pmg ) . The responseis plotted in Figure 1.41, Several things can be noted about the free responsewith Coulomb friction versus the free responsewith viscousdarnping.First, with Coulomb damping the amplitude decayslinearly with slope 2p"mga,
(1.101)
rather than exponentially as does a viscously damped system. Second, the motion under Coulomb friction comes to a complete stop, at a potentially different equilibrium position than when initially at rest,whereasa viscorrslydamped system oscillutes around a single equilibrium, r : 0, with infinitesimally small amplitude. Finally, the frequency of oscillation of a system with coulomb damping is the same as the undamped frequency,whereas viscous damping alters the frequency of oscillation. ExampleL.10.1 The responseof a massoscillatingon a surfaceis measuredto be of the form indicated in Figure1.41.The initial positionis measuredto be 30 mm from its zerorest position, and the final positionis measuredto be 3.5mm from its zero rest positionafter four cyclesof oscillationin 1 s.Determinethe coefficientof friction.
10
69
andthePendulum CoulombFriction
rad/s,sincefour cycleswere Solution First the frequencyof motion ts 4 Hz, or 25.1"3 completedin 1,s.The slopeof the line of decreasingpeaksis -{tt
+
1a
-26.5mm/s
T:
(:l.l0l). Therefore.from expression -2t"g -2wg a: -21"ry'8" : -26'5 mm/s : ' 1tk
7t
a'.
T0"
Solvingfor p yields n (25.13rud/ s)(-26.5mm/s) (-2X9.81 x 103mm/s2)
= 0.107
is probablyverysmoothor lubricated thatthesurface Thissmallvaluefor p indicates The responseof the systemof equation(1.90)can alsobe obtainedby the numericalintegrationtechniquesof the previoussection,which is substantiallyeasier thanthe precedingconstructionof the solution.For example,vtbl-5 usesa fixed step equationof moRunge-Kuttamethodto integrateequation(1.90).Thesecond-order tion canbe reformulatedinto two first-orderequationssomewhatlike equation(1.86) and integratedby the Euler method of equation (1.87),or standardRunge-Kutta methodsmay be employedasdescribedin Appendix F.Figure1.42illustratesthe re-
Time (s) Figure 1.42 Free response(displacementversustime) of a systemsubject to coulomb friction with two differentinitial positions(5 m, solid line;and 4.5m, dashed line) for zero initial velocity (m : 1000kg, p : 0.3,k : 5000N/m)'
Introduction to Vibration andtheFreeResponse
Chap.1
sponseof a systemsubjectto Coulomb friction for two different initial conditions usingMathcad'sfixed time stepRunge-Kuttaroutine.Note in particularthat the system comesto rest at a differentvalueof 11dependingon the initial conditions.Such a systemhasthe samefrequencyyet couldcometo restanyvherein the regionbounded by the two vertical lines (x : +p,mg lk). The responsewill come to rest at the first time the velocityis zero and the displacement is within this region. Comparingthe responseof a linear spring-mass systemwith viscousdamping (saythe underdarnpedresponseof Figure1.10)to the responseof a spring-masssystem with Coulomb dampinggivenpreviously,an obviousand signiticantdifference is the rest position.'fhesemultiple rest positionsconstitutea major feature of nonlinear systems;the existenceof more than one equilibriumposition. The equilibriumpoint of a system,or setof governingequations, may be defined bestby first placingthe equationof motion in statespaceaswasdone in the previoussectionfor the purposeof numericalintegration.A generalsingle-degree-of-freedom svstemmav be written as
i ( r ) + f ( * @ . * ( r ) ): 0
(1.102)
wherethe function/ cantake on anyform,linearor nonlinear.For example,for a linearspring-masssystemthefunction/isjust cic(t)+ kx(t),whichisalinf(*,i): ear function of the statevariablesof positionand velocity.For a nonlinearsystem/ will be somenonlinearfunction of the statevariables.For instance,the pendulum equationderivedand discussed in Example 1.4.2canbe written in the form of equat i o n ( 1 . 1 0 2 ) b y d e f i n i nt g o "bfe / ( 0 , 0 ) : ( S l t ) s i n ( 0 ) , w h e r e 0 i s n o w t h e d i s p l a c e ment variable. The generalstatespacemodel of equation(1.102)is written following the procedurein the previoussectionby definingthe two statevariablesx, : x(r) and x2: *U). Then equation(1.102)can be written asthe first-orderpair ir(t) : xr(t) *r(t) : -f(*,, *r)
(1.103)
This state space form of the equation is used both for numerical integration (as before for the Coulomb friction problem) as well as for formally defining an equilibrium position by defining the state vector, x, used in equation (1.86) and a nonlinear vector function, F as
F: I xzu) ,l L-I\xr, xz))
(1.104)
Equations(1.103)may now be written in the simpleform * : F(x)
( 1.10s)
An equilibrium point of this system,denotedx", is definedto be any value of x for whichF(x) is identicallyzero(calledzerophasevelocity).Thusthe equilibriumpoint is any vector of constantsthat satisfiesthe relations
F(x"): o
(1.106)
Sec.1.10
71
andthePendulum Friction Coulomb
by For the Coulombfriction case,the solutionto this expressionis a regiondefined xz:0 - wmg k:
wmg k
positionlies withThis describesthe conditionthat the velocity(xt) is zero and the conditions,the initial the on Depending friction. of force in the regiondefinedby the the equilibUsually, region. this in somewhere x, of value response"willend up ai a illustrates. example following as the rium valuesare setsof numbers Example1.10.2 * x- Bzxr:0' calculatetheequilibriumpositionforthenonlinearsystemdefinedbyi : form,lettingx1 x asbefore, or in stateequation *t=
xz
*r:
x t ( B z x -l t )
correspondto Solution Theseequationsrepresentthe vibrationof a "soft spring"and = x - x' f6' 1.4.2'wheresinx of Example an approximationoithe pendutumproblem The equationsfor the equilibrium position are xz: 0
t , ( 9 ' * ? - 1 ): o to the three There are three solutionsto this set of algebraicequations,corresponding equilibrium positionsof the soft spring'They are
l-,1-l t-+l *"= tol Lol Lol'L o-l !
in The next exampleconsidersthe full nonlinearpendulumequationillustrated point pivot its Figure 1.43.physicaliythe pendulummay swingall the way around Unstable equilibrium
l 6 + 4| s i n o = o i
lStable | equilibrium
,--) l__--/ t^
= 1T.Jlr, )rr. i xl tx2:v (a)
(b)
(c)
/ and tip mass/r? Figure 1.43 (a) A pendulum consistingof a masslessrod of length position' equilibrium up (c) The straight iti,fr" toulgiti Oownequilibriumposition'
to Vibration andthe FreeResponse lntroduction
72
Chap.'l
and has equiiibrium positions in both the straight-up and straight-down positions as i l l u s t r a t e di n F i g u r e 1 . 4 3 ( b )a n d 1 . 4 3 ( c ) . Example1.L0.3 Calculatethe equilibriumpositionsof the pendulumof Figure1.43. form is givenby Solution The pendulumequationin state-space xt: rz .6,/\ xz: -
o
7sm\jr1l
so that the vectorequationF(t) : 0 yieldsthe followingequilibriumsolutions: x z : 0 a n dx r = 0 , r , 2 r , 3 n , 4 r , 5 r . . . sincesin(x1) is zero for any multiple of n. Note that there are an infinite number of equiiibriumpositions,or vectorsx,. Theseare all eitherthe up positioncorresponding to evenmulto the odd valuesof n [Figure1.43(c)],or the downpositioncorresponding tiplesof tr [Figure1.43(b)].Thesepositionsform two distincttypesof behavior.The responsefor initial conditionsnearthe evenvaluesof n is a stableoscillationaroundthe to initial conditionsnear down position,just asin the linearizedcase,whilethe response even valuesof T movesawayfrom the equilibriumposition(calledunstable)and the without bound. valueof the responseincreases I The stability of equilibrium of a nonlinear vibration problem is of major importance and is based on the definitions given in Section 1.8.However, in the Iinear case there is only one equilibrium value and every solution is either stable or unstable.In this case the stability condition is said to be a global condition.In the nonlinear case there is more than one equilibrium point, and the concept of stability is tied to each particular equilibrium point and is therefore referred to as local stability. As in the example of the nonlinear pendulum equation, some equilibrium points are stable and some are not. Furthermore, the stability of the responseof a nonlinear system depends on the initial conditions.In the linear case,the initial conditions have no influence on the stability, and the system parameters and form of the equation completely determine the stability of the response.To see this, look again at the pendulum of Figure 1.43.If the initial position and velocity are near the origin, the system response will be stable and oscillate around the equilibrium point at zero. On the other hand, if the same pendulum (i.e., same /) is given initial conditions near the equilibrium point at 0 : Tr rad, the response will grow without bound. Hence, 0 : tr rad is an unstable equilibrium. Even though nonlinear systems have multiple equilibrium and more exotic behavior, their response may still be simulated using the numerical-integration methods of the previous section.This is illustrated for the pendulum in the following example, which compares the response to various initial conditions of both the nonlinear pendulum equation and its corresponding linearization treated in Example 1,4.2.
Sec.1.10
CoulombFrictionand the Pendulum
Example1.10.4 Comparethe responses of the nonlinbarand linear pendulumequationsusingnumericalintegrationandthe value(gl {) : (0.1)'z, for (a) the initialconditionsxo = 0.1rad and oo : 0.1radT's, and(b) the initial conditionsx0 : 1 rad and oo = L rad/s,by plottingthe responses. Here x and r,'are usedto denote0 and its derivative,respectively, in order to accommodatenotationavailablein computercodes. Solution Dependingon which programis usedto integratethe solutionnumerically, the equationsmustfirst be put into first-orderform andthen eitherEuler integrationor Runge-Kutta routine may be irrrplementedand the solutionsplotted.Integrationsin MerLee, Mathematica,and Mathcadare presented.More detailscan be found in Appendix F. Note that the responseto the linear systemis fairly closeto that of the full nonlinearsystemin case(a) with slightlydifferentfrequency, while case(b) with larger initial conditionsis drasticallydifferent.The Mathcadsolutionfollows. Finst enter the initial ve:= 0.1
c o n d i t . i o n sf o r e a c h r e s p o n s e : xo:= 0.1
v1o:= 0.1-
v1o:= 0.1
N e x t d e f i n e t h e f i ' e q u e n c ya n d t h e n u m b e ro f a n d s i z e o f t h e t i m e s t e p s : o:=
0.1
N := 2000
i:=0..N
. 6'n ot=r+
The nonlinear Euler integration is vi.l+xi [ * r - ' 1 . _ '[I L u ' ' ' _ . J L - r ' . s i n ( x i ) . A + v 1_ l The "linearEuler integration is
f * 1 , , r 1 .' -_ t L v 1 , - ,I
v 1 ; . A + x 1 -I1
L - . t ( A ) . x 1 r+ v 1 1 J
The plot of these two solutions yields
74
Introduction to Vibrationand the FreeResoonse Here the da s o ' lu t i o n s a equilibrium
ne i s t h e l i n e a r s o l u t i o n . N e x t c o m p u t et h e s e ing initial conditions close to unstable
T
[''.'-]
Lvr+r l
Chao.1
v6 := 0.1
A+x1
I
s i n ( x i )' A + v 1 _ l
x1s := T
vlo
[xr1+rl f v l 1+ 1 J
:=
._t '-
0.t"
v 1 -.iA + x l 1 I . . x 1 r+ v 1 1I L-(,)- ( A ) |
?
Xt
Iir
The Marlen code for running the solutions (using Runge-Kutta this time) and plotting is obtained by first creating the appropriate M-files (named 1 i n*pend-dot. m and NL-pend-dot. m,defining the linear and nonlinear pendulum equations, respectively): function xdot = lin_pend_dot(t,x) n / od e f i n e t h e n a t u r a l f n e q u e n c y omega= 0.1; xdot(1,1) = x(2); x d o t ( 2 , 1 ) = - o m e g a , r 2 * x ( 1 ;) function xdot = NL_pend_dot(t,x) o m e g a= 0 . 1 ; % define the natural frequency xdot(1,1) = x(2); x d o t ( 2 , 1 ) = - o m e 9 a , r 2 * s i n ( x ( 1 );) In the command mode type the following: 'linear % Overplot & non'linear simulations of the free % response of a pendulum. x0 = 0.1; v0 = 0.1; ti = 0; tf = 200; %
.l
i near
ltime-lin,sol-lin]=eis451'1in-pend-dot', % nonl i near [ t ' i m e - N L , s o l - N L ] = o d e 4 5 ( ' N L - p e n d - d o t[' t, i
[ti
tf], [x0 v0]) ;
tf], [x0 v0]) ;
Sec.1.'l0
CoulombFrictionand the Pendulum % overplot displacements f i g ur e plot(time-l i n,sol-l i n(: ,1) , '-') hol d p l o t ( t i m e - N L , s o l - N L (: , 1 ) , ' - ' ) ') x l a b e l( ' t i m e ( s ) y l a b e ' l( ' t h e t a ' ) ' v s . n o n l i n e a r p e n d u l u mw i t h x 0 = title(['Linear = ' num2str(v0)l) num2str(x0)' and v0 l e g e n d ( ' I i n e a r ' , ' n o n l i n e a r ')
...
Here the plots have been suppressedas they are similar to those from the Mathcad solution. Next consider the Mathematica code to solve the same problem. First we load the Add-on package that will enable us to add a legend to our plot. In [55] := <
. In[3] := o=0.1;
The following cell solves the linear differential equation, then the non-lir.reardif' ferential equation and then produces a plot containing both responses. I n t 8 5 l : = x l i n = N D S ov' le [ { x ] " t t l + t o 2 " " x l[ t ] = = 9 , x 1 [ 0 ] = = . 1 , x ' l ' [ 0 ] = = ' 1 ] ' xl [t] , {t,0, 200}l x n o n l i n = N D S o l v e [ { x n ] " [ t ] + r o z ' r $ - i n t x n l [ t ] l = = 0 , x n 1[ 0 ] = = . 1 , x n l ' [ 0 1 = = ' 1 1 ' xnl [t] , {t,0, 200}l P'lot[{Evaluate[xl [t]/.xl inl , Eva'luatelxnl [t],/.xnon] inl] ' {t'0'200}, P l o t S t y ' l e - + D a s h in g [ { } ] , D a s h i n g [ { . 0 1 , . 0 I } ] , P l o t l a b e l - - +" L i n e a r a n d Nonlinear Response, Stable Equ'ilibrium"' A x e s L a b e l - + { " t i m e , s " , " " } , P l o t L e g e n d - - +{ " L i n e a r " , " N o n -L i n e a r " } , L e g e n d P o s ' i t i o n - - +{ 1 , 0 } , L e g e n d S i z e - +{ . 7 , . 3 } l ; O u t [ 8 5 ] = { { x l i t J - - +I n t e r p o l a t ' i n g F u n c t io n [ { { 0 . , 2 0 0 .} } , + ] t t l } } O u t [ 8 6 ] = { { x n ] [ t ] - + I n t e n p o l a t i n g F u n c t io n t { { 0 . , 2 0 0 . } i , < > l [ t ] ] ] StableEquilibrium Linear and NonlinearResponse,
Linear
\
Non-Linear
76
Introduction to Vibrationand the FreeResponse
Chao.1
In[88]:= Clear[xlin, xnonlin, xl, xn]l x ' l i n = N D S o l v e [ { x l ' ' [ t ] 4 1 o 2 ' : x 1t t l = = 0 , x l [ 0 ] = = l r , x l ' [ 0 ] = = . 1 ] , x 1 [ t ] , {t, 0, 200}l xnonlin=NDSolve[{xnl"[t]+
40
-
Linear
-'-----'
Nonlinear
time,s
Note from the plots in Example 1.10.4that even in the casewhere the initial conditions are small, the linear responseis not exactly the same as the full nonlinear system response. Ilowever, if the initial velocity is changed to zero, the solutions are very similar. This is illustrated in Figure 1.44. In summary, nonlinear systemshave severalinteresting aspectsthat linear systems do not. In particular, nonlinear systemshave multiple equilibrium positions rather than just one, as in the linear case.Some of these extra equilibrium points may be unstable and some stable.The stability of a response depends on the initial conditions, which can send the solution to different equilibrium positions and hence different types of response.Thus the behavior of the response depends on the initial conditions, not just the parameters and form of the equation, as is the casefor the linear system.This is illustrated in Example 1.10.4. Even though the responseof a nonlinear systemis much more complicated and closed-form analytical solutions are not always available,the responsecan be simulated using numerical integration. In modeling real systems,the nonlinearity is always present. Whether or not it is important to include the nonlinear part of the
77
Problems
a
Y.
8
x1, -0.1
-0.2 Figure 1.44 The response of both the linear (dashed line) and nonlinear system (solid line) of Example 1.10.4with the initial velocities set to zero'
modelin computingthe responsedependson the initial conditions.If the initial conditions are suchthat the system'snonlinearitycomesinto play,then thesetermsshould Somewhatthe same be included.Otherwisea linearresponseis perfectlyacceptable. can be said for includingdampingin a systemmodel.Which effectsto include and which not to includewhen modelingand analyzingavibrating systemform one of the important aspectsof engineeringpractice.
PROB LEMS Thoseproblemsmarkedwith an asteriskareintendedto be solvedusingcomputationalsoftware or the Merlan Toolbox. Section1.1 loadedwith massand the corresponding(static) 1.1. The springof Figure1.2is successively displacementis recordedas follows.Plot the data and calculatethe spring'sstiffness. Note that the datacontainsomeerror.Also calculatethe standarddeviation. m(kd
10
1.1
12
L3
14
15
16
x(m)
1..I4
1.25
I.37
i.48
1.59
I.7r
I.82
L.2. Derive the solution of mi * kx : 0 and plot the result for at leasttwo periods for the casewith @n: 2rad/s, xs : 1 mm, and oe : \6 m-/s. 1.3. Solvemi * kx : 0fork : 4Nlm,m : 1kg,x6: 1 mm,andoo: 0.Plotthesolution. 1.4. The amplitudeof vibrationof an undampedsystemis measuredto be 1 mm.The phase shift from r : 0 is measuredtobe? rad and the frequencyis found to be 5 rad/s.Calculate the initial conditionsthat causedthis vibration to occur.
78
Introduction to Vibrationand the FreeResponse
Chap.1
1.5. An undampedsystemvibrateswith a frequencyof 10 Hz and amplitude1 mm. Calculate the maximumamplitudeof the system'svelocityand acceleration. L.6. Showby calculationthat -4sin(<,.r,t + $) can be represented as -Bsinco,r* Ccosto,t and calctilateC and B in terms of A and S. L.7. Using the solutionof equation(1.2)in the form x ( t ) : B s i n r o n*/ C c o s o , / calculatethe valuesof B and C in termsof the initial conditionsrs and o6. 1.8' Using Figure 1.6,verify that equation(1.10)satisfiesthe initial-velocitycondition. 1.9. (a) A 0.5-kgmassis attachedto a linearspringof stiffness0.1N/m. Determinethe natural frequencyof the systemin hertz.(b) Repeatthis calculationfor a massof 50 kg and a stiffnessof 10 N/m and compareyour resultto that of part (a). 1.10. Derive the solutionof the single-degree-of-freedom systemof Egure 1.4by writing Newton's law,ma : -kx,in differentialform usinga dx = u du andintegratingtwice. 1.11. Determinethe naturalfrequencyof the two systemsillustratedin FigureP1.11.
(b)
(a)
FigureP1.11 * 1 . 1 2 .P l o t t h e s o l u t i o n g i v e n b y e q u a t i o n ( 1 . 1 0 ) f o r t h e c=a s1e0k0 0 N / ma n d m : 1 0 k g f o r two completeperiodsfor eachof the following setsof initial conditions:(a) xo : 0, 'u6: 1 m/q(b) to = 0.01m,uo = 0,and(c) ro : 0.01m,tre: 1 m/s. *1.13. Make a three-dimensional surfaceplot of the amplitudeA of an undampedoscillator given by equation (1.9) versusx0 and or for the rangeof initial conditionsgiven by -1 < rn < 0.1mand-1 < ?.)0 < 1m/sforasystemwithnaturalfrequencyof 10rad/s. 1.14. Use a free-bodydiagramof the pendulumof Window 1.1and derive the equationof motion givenin the windowby usingthe approximationsin(0) : 0 for small0. 1,.15. A pendulum haslength of 250 mm.What is the system'snatural frequency inHertz? 1.16. The pendulumin Window 1.1is requiredto oscillateonceeverysecond.What length shouldit be? L.17. The approximationof sin0 : 0 is reasonable for 0 lessthan10'.If a pendulumof length 0.5m hasan initial positionof 0(0) : 0, what is the maximumvalueof the initial angular velocity that can be given to the pendulumwithout violatingthis small-angleapproximation?(Be sureto work in radians.) Section 1.2 *1.1.8.Plotthesolutionofalinearspring-masssystemwithfrequencycon = 2radfs,xo: 1mm, and oo : 2.34mm/s,for at leasttwo periods.
Chap.1
Problems
79
*L.L9. Computethe naturalfrequencyandplot the solutionof a spring-masssystemwith mass of 1 kg andstiffnessof 4 N/m andinitial conditionsof xe : 1 mm and uo : 0 mm/s.for at leasltwo periods. 1.20. To designa linear spring-masssystemit is often a matter of choosinga springconstant suchthat the resultingnaturalfrequencyhasa specifiedvalue.Supposethat the massof a systemis 4 kg and the stiffnessis 100N/m. How much must the springstiffnessbe changedin order to increasethe naturalfrequencyby 10%? 1.21. Referringto Figure1.7,if the maximumpeakvelocityof a vibratingsystemis 200mm/s is 5000mm/s2,whatwiil the peak at 4Hz andthe maximumallowablepeakacceleration be? displacement and accelerationin Figure 1.7haveslopesof 1.22. Show that linesof constantdisplacement *1 and -1, respectively. If rms valuesinsteadof peak valuesare used,how doesthis affect the slope? 1.23. An automobile is modeled as a 1000-kgmass supported by a spring of stiffness ft : 400,000N/m. When it oscillatesit doesso with a maximum deflectionof 10 cm. the massincreasesto as much as 1300kg. Calculatethe When loadedwith passengers, velocityamplitude,andaccelerationamplitudeif the maximumdechangein frequency, flection remains10 cm. 1.24. A machineoscillatesin simpleharmonicmotion and appearsto be well modeledby oscillation.Its accelerationis measuredto an undampedsingle-degree-of-freedom have an amplitudeof 10,000mm/sz at 8 Hz. What is the machine'smaximumdisplacement? 1.25. A simpleundampedspring-masssystemis setinto motion from restby givingit an initial velocityof L00mm/s.It oscillateswith a maximumamplitudeof L0 mm.What is its natural frequency? of maximumamplitude5 cm 1.26. An automobileexhibitsa verticaloscillatingdisplacement that the automobile Assuming of 2000 cm/s'. acceleration maximum and a measured systernin the verticaldirection,calculate canbe modeledasa single-degree-of-freedom the naturalfrequencyof the automobile. Section1.3 1 . 2 7 . S o l v ei + 4 i * , r = 0 f o r x n : 1 m m . u n : 0 m m / s . S k e t c h y o u r r e s u l t s a n d d e t e r r ' , . r i n e which root dominates. 1 , . 2 8S. o l v ei + 2 t * 2 x : O f o r x s : O m m , o o : 1 m m / s a n d s k e t c h t h e r e s p o n s e . Y o u m a y wish to sketchx(t) : e"' andx(l) : -t-' ttttr. 1.29. Derive the form of tr1,and \2 givenby equation(1.31)from equation(i.28) and the definition of the dampingratio. 1.30. Use the Euler formulasto deriveequation(1.36)from equation(1.35)andto determine the relationshipslistedin Window 1.4. 1.31. Using equation(1.35)asthe form of the solutionof the underdampedsystem,calculate the valuesof the constantsa1and a2in termsof the initial conditionsxe and z'0. 1.32. Calculatethe constants-4 and S in termsof the initial conditionsand thusverify equation (1.38)for the underdampedcase. 1.33. Calculatethe constantsa1and a2in termsof the initial conditionsand thusverify equations (1.42)and (1.43)for the overdampedcase.
80
to Vibrationand the FreeResponse Introduction
Chap.1
L.34. Calculatethe constantsa1and a2in termsof the initial conditionsancithusverify equation (1.46)for the criticallydamDedcase. 1.35. Using the definitionof the dampingratio and the undampednaturalfrequency,derive equation(1.48)from (1.47). 1.36. For a dampedsystem,m,c,andt areknowntobe nr = 1 kg,c : Zkg/s,k = 10N/m. or critically underdamped, Is the systemoverdamped, Calculatethe valuesof ( andcu,,. damped? *1.37. Plot x(r) for a dampedsystemof naturalfrequencyan = 2 rad/s and initial conditions xo : 1 mm, : 0, for the followingvaluesof the dampingratio:( : 0.01,( : 0.2, ( : 0 . 6 , ( : "o 0 . 1 , L- 0 . 4 , a n d (: 9 . 3 . *1.38. Plot the responsex(r) of an underdamped andoo : 0 systemwith <,r , : 2radf s'( : 0.1", : : : mm, and 10 5 mm, ro mm, xs xs L displacements: initial for the following xo : 100mm' landus:0forx(t)andsketchtheresPonse. 1 . 3 9 .S o l v ei - i + x : 0 w i t h x 6 : systemhasmassof 100kg, stiffnessof 3000N/m, and damping 1.40. A spring-mass-damper coefficientof 300kg/s.Calculatethe undampednaturalfrequenry,the dampingratio,and Does the solutionoscillate? the dampednaturalfrequency. systemhasmassof 150kg, stiffnessof 1500N/m and damping L.41. A spring-mass-damper coefficientof 200 kg/s. Calculatethe undampednaturalfrequency,the dampingratio Is the systemoverdamped,underdamped,or critically and the dampednatural freqrrency. damped?Does the solutionosciilate? *1.42. "the systemof Problem 1.40is given a zero initial velocityand an initial displacement of 0.1 m. Calculatethe form of the responseand plot it for as long as it takes to die out. *L.43. The systemof Problem 1.41is given an initial velocityof 10 mm/s and an initial displacementof -5 mm.Calculatethe form of the responseandplot it for aslong asit takes to die out. FIow long doesit take to die out? *1.44. Choosethe dampingcoefficientof a spring-mass-damper systemwith massof 150kg and stiffnessof 2000N/m suchthat its responsewill die out after about2 s,givena zero initial positionand an initial velocityof 10 mm/s. 1.45. Derive the equationof motion of the systemin FigureP1.45and discussthe effectof gravity on the naturalfrequencyand the dampingratio.
t, FigureP1'45 1.46. Derive the equationof motion of the systemin FigureP1.46and discussthe effectof gravity on the natural frequencyand the dampingratio.You may haveto make someapproximationsof the cosine.Assumethe bearingsprovidea viscousdampingforce only
/-hrn
I
Problems
8l
in the vertical direction. (From A Diaz-Jimenez, South African M echanical Engine er, Yol. 26, pp. 65-69, I97 6.)
FigureP1.46
Section1.4 1.47. Calculatethe frequencyof the compoundpendulumof Figure 1.19(b)if a massm. is addedto the tip, by usingthe energymethod. 1.48. Calculatethe total energyin a dampedsystemwith frequency2 rad/s anddampingratio ( : 0.01with mass10kg for the case"r0= 0.1andoo : 0.Plot the total energyversustime. 1.49. Use the energymethodto calculatethe equationof motion and naturalfrequencyof an airplane'ssteering-gear mechanismfor the nosewheelof its landinggear.The mechanism is modeledasthe single-degree-of-freedom systemillustratedin FigureP1.49.
(Steering wheel)
(Tire-wheel assembly)
FigureP1.49 Single-degree-of-freedom modelof a steeringmechanism.
The steering-wheel and tire assembly are modeled as being fixed at ground for this calculation. The steering-rod gear system is modeled as a linear spring-and-masssystem (^, kt) oscillating in the -r direction. The shaft-gear mechanism is modeled as the disk of inertia .I and torsional stiffnessk1.The gear / turns through the angle 0 such that the disk does not slip on the mass.Obtain an equation in the linear motion.r.
Introductionto Vibrationand the Free Response
82
Chap. 1
system 1.50. A control pedal of an aircraftcan be modeledas the single-degree-of-freedom shaftand the pedalas a lumpedmass of FigureP1.50.Considerthe leveras a massless at the end of the shaft.Use the energymethodto determinethe equationof motion in Assumethe springto be unstretched 0 andcalculatethe naturalfrequencyof the system. atO:0.
Figure P1.50 Model of a foot pedal usedto operatean aircraft control surface.
1.51. To savespace,two largepipesare shippedone stackedinsidethe other asindicatedin FigureP1.51.Calculatethe naturalfrequencyof vibrationof the smallerpipe (of radius R1)rolling back and forth insidethe largerpipe (of radiusR). Use the energymethod and assumethat the insidepipe rolls without slippingand hasa massof rn.
Large pipe
Tiuck bed (u) Figure P1.51 pipe.
(b) (a) Pipes stacked in a truck bed. (b) Vibration model of the inside
1.52. Considerthe exampleof a simplependulumgivenin Examplel.4.Z.Thependulummotion is observedto decaywith a dampingratio of ( : 0.001.Determinea dampingcoefficientand add a viscousdampingterm to the pendulumequation.
Llh:n
1
Problems
83
L.53. Determinea dampingcoefficientfor the disk-rod systemof Example1.4.3.Assuming that the dampingis due to the materialpropertiesof the rod, determinec for the rod if it is observedto havea dampingratio of ( : 0.01. 1.54. The rod and diskof Window L.Larein torsionalvibration.Calculatethe dampednaturai frequencyif "r : 1000m2.kg.c : 20 N.m.s/rad, and ft : 400N.m/rad. 1.55. Considerthe systemof Figure1.15,whichcould alsorepresenta simplemodelof an aircraft landingsystem.Supposethat a viscousdamperis addedto the systemof damping coefficientc.What is the dampednaturalfrequency? 1..56.ConsiderProblem1.55with k = 400,000 N. n,ffi = 1500kg,"/ : 100m2.kg, r : 25 cm. = and c 8000N'm's. Calculatethe darnpingratio and the dampednaturalfrequency. How much effectdoesthe rotationalinertiahaveon the undamDednaturalfrequencv? Section 1.5 1.57. A helicopterlanding gear consistsof a metal framework rather than the coil springbasedsuspensionsystemusedin a fixed-wingaircraft.The vibration of the frame in the vertical directioncan be modeledby a springmade of a slenderbar as illustrated in Figure 1.20, where the helicopter is modeled as ground. Here / : 0.4 m, E : 20 x 1010 N/mz, and m : 100kg.Calculatethe cross-sectional areathat shouldbe . usedif the natural frequencyis to be fi : 500 Hz. 1.58. The frequencyof oscillationof a personon a divingboardcanbe modeledasthe transversevibration of a beam as indicatedin Figure 1.23.Let mbe the massof the diver (rn : 100kg) and / - 1 m. If the diver wishesto oscillateat 3 Hz, what value of EI shouldthe diving-boardmaterialhave? 1.59. Considerthe springsystemof Figurel.28.Let kt : ks * kz : 100N/m, ft: : 50 N/m. and ko : 1 N/m. What is the equivalentstiff'ness? 1.60. Springsare availablein stiffnessvaluesof 10,100,and i000 N/m. Designa springsystem usingthesevaluesonly,so that a 100-kgmassis connectedto groundwith frequency of about 1.5rad/s. 1.61. Calculatethe naturalfrequencyofthe systemin Figure1.28(a)it ky : k2: 0. Choosenz and nonzerovaluesof kt, ko,andft5so that the naturalfrequencyisJ, : 1ggttr. *1.62. Example1.4.4examinesthe effectof the massof a springon the naturalfrequencyof a simplespring-masssystem.Use the relationshipderivedthereand plot the naturalfrequencyversusthe percentthat the springmassis of the oscillatingmass.[Jseyour plot to commenton circumstances when it is no lonser reasohableto neslectthe massof the spring. 1.63. Calculatethe naturalfrequencyand dampingratio for the systemin FigureP1.63given thevaluesm: l0 kg,c = 100kg/s,k1: 4000N/m,kz: 200N/m,andk. = 1000N/m. Assumethat no friction actson the rollers.Is the systemoverdamped, criticallydamped, or underdamped?
Figure P1.63
to Vibrationand the FreeResponse lntroduction
84
Chap' 1
1.64. RepeatProblem1.63for the systemof FigureP1'64'
Figure P1.64 (E : 2 x 1011N/m2) and sizes the 1.65. A manufacturer makes a leaf spring from a steel Later, to save weight, the spring is frequency. .p""iti. u has the device spring so that '7.L x ttito N7m2).assuming that the massoJ the spring is much made of aluminum (E : if the frequency insmaller then that oi th" device the spring is attached to, determine creasesor ciecteasesand by how much'
Section 7.6 1.66. Show that the logarithmic decremetrt is equal to
a:Ltn! fr
xn
where x, is the amplitude of vibration after n cycleshave elapsed' 1.67. Derive equation (1.68) for the trifilar suspensionsystem' unknown modulus. An exper1.6g. A prototype composite material is formed and hence has beam of length 1 m and a cantilevered imlnt is performed consisting of forming it into , is given an initial system The end. its at : attached ,nos 10 ma with u d-kg moment 1 modulus E' the s' Calculate 0.5 period of a with to oscillate found displacement and : 400,000 N/m is obof k L.69. The free response of a 1000-kg automobile with stiffness as a single-deautornobile the Modeling 1.31. Figure givenin form served to be of the coefficient gree-of-freedom oscillation in the vertical clirection,determine the damping 12' at cm and0'22 if the displa.ement at /, is measuredto be 2 crn its damping ratio' 1.70. A pendulum decays from 10 crr to 1 cm over one period. Determine ratio ( is often approx1.71. The reiationship between the log decrement E and the darnping good approximation imated as 6 : 2rr(. For what values of ( would you consider this a to equation (1.72)? mass of the system ls meaI.72. Adamped system is modeled as illustrated in Figure 1.9.The N/m. It is observed that 5000 be to sured to be 5 kg and its spring constant is measured value after five cycles' initial its of 0.25 to decavs during free vibration the amplitude c' coefficient, damping viscous Calculate the Section 1.7 (see also Problems 1.60 and 1'85 through 1'90) placed in parallel with the 1.73. Choose a dashpot's visccus damping value such that when s. spring of Example 1.7.2reduces the frequency of oscillation to 9 rad,i
^2n
I
Problems
85
: 0 and?ro: 10mm/s.Determjnem, c,andft suchthat 1.74. For an underdampedsystem,.ro the amplitudeis lessthan L mm. 1.75. RepeatProblem1.74if the massis restrictedto lie between10 kg < m < 1.5kg. 1.76. Use the formula for the torsionalstiffnessof a shaft from equation(1.6a)to designa 1-m shaftwith torsionalstiffnessof 105N. m/rad.. 1.77. RepeatExample1.7.2usingaluminum.What differencedo you note? 1.78. Tiy to designa bar (seeFigure1.20)that hasthe samestiffnessas the springof Example 1.7.2.Note that the bar mustremainat least10 timesaslong asit is wide in order to be modeledby the formulaof Figure1..20. 1.79. RepeatProblem1.78usingplastic(E:1.40 x 10eN/m2) andrubber(E'= 7 x 106N/m2). Are any of thesefeasible? Section 1.8 (seealso Problem 1.39) 1.80. Considerthe invertedpendulumof Figure1.36as discussed in Example1.8.1.Assume that a dashpot(of dampingrate c) alsoactson the pendulumparallelto the two springs. How doesthis affectthe stabilitypropertiesof the pendulum? 1.81. Replacethe massless rod of the invertedpendulunof Figure 1.36with a solid-object compoundpendulumofFigure f.i9(b). Calculatethe equationsof vibrationand discnss valuesof the parameterfor whichthe systemis stable. 1.82. Considerthe disk of FigureP1.82connectedto two springs.Use the energymethod to calculatethe system'snaturalfrequencyof oscillationfor smallangles0(r).
Figure P1.82 Vibration model of a rolling disk mountedagainsttwo springs, attachedat point s.
Section7.9 *1.83. ReproduceFigure1.37for the varioustime stepsindicated. *1.84. use numerical integrationto solvethe system of Example 1.7.3with m : 1.36rkg, k : 2-688x 105N/m, c : 3.81 x 103kgls subjectto th; initial conditionsx(0) = b and o(0) : 0.01m/s. Compareyour resultusingnumericalintegrationto just plotting the analyticalsolution (usingthe appropriateformula from Section1.3)by plotting both on the samegraph. *1.85. Consideragainthe dampedsystemof Problem 1.84and designa dampersuchthat the oscillationdiesout after 2 seconds. There are at leasttwo waysto do this.Here it is intendedto solvefor the responsenumerically,following Examples!.9.2,1.9.3,or 1.9.4, usingdifferentvaluesof the dampingparameterc until the desiredresponseis achieved. *L.86. Consideragainthe dampedsystemof Example7.9.2 anddesigna dampersuchthat the oscillation dies out after 25 seconds.There are at least two ways to do this. Here it is
to Vibrationand the FreeResponse lntroduction
86
Chap.1
followingExamplest.9.2'1'.9.3,ot1.9.4, intendedto solvefor the responsenumerically, is achieved.Is usingdifferentvaluesof the dampingparameterc until the desiredresponse or criticallydamped? underdamped, your resultoverdamped, *1.87. RepeatProblem1.85for the initial conditions'r(0) : 0.1m and ?r(0): 0.01mm/s' *1.88. A springand damperare attachedto a massof 100kg in the arrangementgivenin Figure L.9.Thesystemis giventhe initial conditionsx(0) : 0.1m and?r(0)= 1 mm/s.Designthe springand damper(i.e.,chooseft and c) suchthat the systemwill cometc rest Tiy to keep c as small as in 2 secondsand not oscillatemore than two completecycles. possible.Also compute(. *1.89. RepeatExampleI.7.lby usingthe numericalapproachof the previousfive problems. *1.90. RepeatExample1.7.1for the initial conditionsx(0) : 0.01m and r;(0) = I mm/s.
Sectionl.I0 1.91. A 2-kg massconnectedto a springof stiffness103N/m hasa dry-slidingfriction force 20 cm in 15 cycles.How long (,f'r) of f N.As the massoscillates, its amplitudedecreases doesthis take? x 103N/mwithafriction m:Skgandk:9 1 . 9 2 . C o n s i d e r t h e s y s t e mFoifg u r e l . 4 0 w i t h force of magnitude6 N. If the initial amplitudeis 4 cm, determinethe amplitudeone cyclelater aswell as the dampedfrequency. *1.93. Compute and plot the responseof the systemof Figure P1.93for the casewhere t o = 0 . 1r l , o 0 : 0 . 1m / s ,$ : 0 . 0 5 , m : 2 5 0 k g , 0 : Z } " , a n d f t = 3 0 0 0 N / m ' H o w l o n g doesit take for the vibrationto die out?
FigureP1.93
*L.94. Computeand plot the responseof a systemwith Coulombdampingof equation(1.88) forthe casewherero : 0.5ltr,oo: 0m/s, p : 0.1,m: 100kg,and t : 1500N/m' How long doesit take for the vibrationto die out? *1.95. A massmovesin a fluid againstslidingfriction asillustratedin FigureP1.95.Model the dampingforce asa slowfluid (i.e.,linearviscousdamping)plus Coulombfriction because of ttre sliding,with the followingparameters:m : 250kg, p = 0.01,c : 25 kg/s, and k : 3000N/m. (a) Computeandplot the responseto the initial conditionsxo : 0.1m, oo : 0.1 m/s (b) Computeand plot the responseto the initial conditionsro : 0.1 m, a6 : I mf s.How long doesit take for the vibrationto die out in eachcase?
Chap.1
87
Problems Fluid
ffi
kt2
m
AAf,\AA
vv\y\yv\v kt2
FigureP1.95
*1.96. ConsiderthesystemofProbleml.g5(a)andcomputeanewdampingccefficient,c,that will causethe vibrationto die out after one oscillation. L.97. Computethe equilibriumpositionsof i + 62,,x* $x2 : 0. How many are there? 1.98. Computethe equilibriumpositionsof i + (t)2,r- Fzxt + 1x5 = 0.How many are there? *1.99. Considerthe pendulumof Example1.10.3with a lengthof i. m and initial conditionsof 0o: rll1rad and do = 0. Comparethe differencebetweenthe responseof the linear versionof the pendulumequation[i.e.,with sin(0) : 0] and the responseof the nonlinear versionof the pendulumequationby plottingthe responseof both for four periods' *1.100.RepeatProblem1.99if the initial displacementis 8s : T f 2 rad. 1.101..If the pendulumof Example1.10.3is givenan initial conditionnearthe equilibriumposition of 0o = n rad and d o : 0 doesit oscillatearoundthis equilibrium? *1.102. Calculate the responseof the systemof Problem 1.98 for the initial conditions of = g' xo : 0.0LfiI,oo : 0 m/s,and a naturalfrequencyof 3 rad/s and for F : fOO,l = 0) *1.103.RepeatProblem1.L02andplottheresponseofthelinearversionofthesystem(p versions nonlinear on the sameplot to comparethe differencebetweenthe linear and of this equationof motion.
M n r L n g @E N G I N E E R I N G V I B R A T I O N T O O L B O X Dr. JosephC. Slater of Wright State Llniversityhas authored a Mnrlas Toolbox keyed to this text.The EngineeringVibrationToolbox(EVT) is organizedby chapter and may be usedto solvethe ToolboxProblemsfound at the end of eachchapter. In addition,the EVT may be usedto solvethosehomeworkproblemssuggested for computerusagein Sections1.9and 1.L0,ratherthan usingM.cTI-aedirectly.MarLABand the EVT are interactiveand are intendedto assistin analysis,parametric studiesand design,as well as in solvinghomeworkproblems.The EngineeringVibrationToolbox is licensedfree of chargefor educationaluse.For professionaluse, usersshouldcontactthe EngineeringVibrationToolbox author directly. The EVT is updatedand improvedregularlyand can be downloadedfor free. To download,update,or obtain informationon usageor current revision,go to the EngineeringVibration Toolboxhomepageat edu/-vtoolbox http://www.cs.wright. earlier versionsof Menan, aswell run on This siteincludeslinks to editionsthat who use the EVT is maininstructors as the most recentversion.An e-mail list of updates.The EVT is delatest tained so userscanreceivee-mailnotificationof the (including MacintoshandVMS) signedto run on any platform supportedby Merlen versionof Merthe current with and is regularlyupdatedto maintaincompatibility home page as the on LAB.A brief introduction to Merles and UNIX is available after vtoolbox help type to get startedand well. Pleaseread the file Readme.txt
88
to Vibrationand the FreeResponse lntroduction
Chap' 1
curinstallation to obtain an overview. Once installed, typing vtbud will display the revithe current download to and attempt your installation of status rent revision sion status from the anonymous FTP site. Updates can then be downloaded incrementally as desired. Please seeAppendix G for further information.
TOOLBOX PROBLEMS : 1 mm] the values T81.1. Fix [your choiceor usethe valuesfrom Example1.3.1with x(0) of the initial veof values range for a x(t) plot the responses and .r(0) c,k, and of m, locity i(0) to seehow the responsedependson the initial velocity.Rememberto use numberswith consistentunits. x(l) for a range TBl.2. Usingthe valuesfrom ProblemTBi..1andi(0) : 0, plot the response ofvalues ofx(0) to seehow the responsedependson the initial displacement' TB1.3. ReproduceFigures1.10,1'11,andl'.72. . T81.4. ConsidersolvingProblemL.32andcomparethe time for eachresponseto reachand staybelow0.01mm. T81.5. SolveProblems1.88,1.89,and 1.90usingthe Engineeringvibration Toolbox. T8,1.6. SolveProblems1.93,1.94,and 1.95usingthe EngineeringVibrationToolbox.
Response to Hormonic Excitotion Thlschcpter focuseson the most fundomentolconcept in vibroiion onolysis, Thisconcept isthot of resononce. Resoncnceoccurswhen o periodicexternolforceisopplied to o systemhovingo nolurol frequencyequol to tfre driving frequency. Thisoflen hcppenswhen the excitoiionforce isderivedfrom some rototingpcrt sucl-ros the helicopiershovyn on the top left.The rototingblode couseso hormonic force to be opplied to the body of the helicopter. lf the frequencyof the blode rotction correspondsto the noturolfrequencyof the body, resononcewilloccur os describedin section2,L Resononcewillccuse lorgedeflectionswhich moy exceed the elosiiclimitsond cousethe structureto foil,An exomplefomilior to mostisthe resononcecousedby on out of boloncetireon o cor (bottom photo).Thespeed of tire rotctionconespondsto the driving frequency. At o certoinspeed,the out-of-boloncetire couses resononce, which mcy be felt os shckingof the steering-wheel column.lf ihe cor isdrivenslower, or foster,the frequencymovesowoy fromthe resoncnceconditionof ihe drivingfrequencybeingexocfly equolto the noturolfrequencvond the shokingslops.
89
