CHAPTER 2
Aheavytableissupportedbyflatsteellegs(Fig.P2.1).Itsnaturalperiodinlateralvibrationis0.5sec. Whena50-lbplateisclampedtoitssurface,thenaturalperiodinlateralvibrationislengthenedto0.75 sec.Whataretheweightandthelateralstiffnessofthetable? Tn
1. Determinetheweightofthetable. Taking
2. Determinethelateralstiffnessofthetable. Substitute for
Problem 2.2
Anelectromagnetweighing400lbandsuspendedbyaspringhavingastiffnessof100lb/in.(Fig.P2.2a) lifts200lbofironscrap(Fig.P2.2b).Determinetheequationdescribingthemotionwhentheelectric currentisturnedoffandthescrapisdropped(Fig.P2.2c).
Solution:
1. Determinethenaturalfrequency.
k = 100lbin. m =− 400 386 lbsecin. 2 ωn k m === 100 400386 982.secrads
2. Determineinitialdeflection. Static deflection due to weight of the iron scrap
u () 0 200 100 2 == in.
3. Determinefreevibration. ututt n ()()coscos(.) ==02982 ω
FigureP2.2Problem
Amass m isatrest,partiallysupportedbyaspringandpartiallybystops(Fig.P2.3).Inthepositionshown,thespringforceis mg/ 2.Attime t = 0thestopsarerotated,suddenlyreleasingthemass. Determinethemotionofthemass.
1.
Problem 2.4
The weight of the wooden block shown in Fig. P2.4 is 10 lb and the spring stiffness is 100 lb/in. A bullet weighing 0.5 lb is fired at a speed of 60 ft/sec into the block and becomes embedded in the block. Determine the resulting motion u(t) of the block.
FigureP2.4
Solution:
u kv 0
m
m ==− 10 386 00259.lbsecin. 2
m0 3 05 386 1310 ==×−.lbsecin. 2
k = 100lbin.
Conservation of momentum implies
mvmmu 00 0 0 =+() ()
& () u mv mm 0 2857 00 0 = + = ftsec=34.29in.sec
After the impact the system properties and initial conditions are
Mass = mm+=− 0 00272.lbsecin. 2
Stiffness = k = 100lbin.
Natural frequency:
ωn k mm = + = 0 6063.secrads
Initial conditions: uu (),0003429().sec == in.
The resulting motion is ut u tt n n () () sin.sin(.)== 0 05656063
in.
Problem 2.5
Amass m 1 hangsfromaspring k andisinstaticequilibrium.Asecondmass m 2 dropsthroughaheight h andsticksto m 1 withoutrebound(Fig.P2.5).Determinethesubsequentmotion u( t)measuredfrom thestaticequilibriumpositionof m 1 and k
Solution:
With u measured from the static equilibrium position of m1 and k, the equation of motion after impact is
() mmukum 12 2 ++= g (a)
The general solution is utAtBtm nnk()cossin=++ ωω 2 g (b)
ωn k mm = + 12 (c)
The initial conditions are u ()00 = h mm m u)0( 21 2 + = (d)
The initial velocity in Eq. (d) was determined by conservation of momentum during impact: mummu 2212 0 () () =+
where uh 2 2 = g
Impose initial conditions to determine A and B: uAm k ()00 2 =⇒=− g (e) & () uBB m mm h n n 0 2 2 12 =⇒= + ω ω g (f)
Substituting Eqs. (e) and (f) in Eq. (b) gives
ThepackagingforaninstrumentcanbemodeledasshowninFig.P2.6,inwhichtheinstrumentof mass m isrestrainedbyspringsoftotalstiffness k insideacontainer; m = 10lb/gand k = 50lb/in. Thecontainerisaccidentallydroppedfromaheightof3ftabovetheground.Assumingthatitdoes notbounceoncontact,determinethemaximumdeformationofthepackagingwithintheboxandthe maximumaccelerationoftheinstrument.
1.
Problem 2.7
Consider a diver weighing 200 lbs at the end of a diving board that cantilevers out 3 ft. The diver oscillates at a frequency of 2 Hz. What is the flexural rigidity EI of the diving board?
Solution:
Given:
m ==− 200 322 6211.lbsecft 2
fn = 2Hz
Determine EI: k EI L EIEI === 33 3339 lbft
f k m EI n =⇒=⇒ 1 2 2 1 25590ππ
EI ==− ().455908827 2 π lbft 2
© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.Problem 2.8
Show that the motion of a critically damped system due to initial displacement u(0) and initial velocity u(0)is
Solution:
Equation of motion:
mucuku &&& ++= 0 (a)
Dividing Eq. (a) through by m gives
uuu nn ++=20 2ζωω (b)
where ζ = 1
Equation (b) thus reads
Evaluate Eq. (f) at t = 0 : & () () uAA n 01 0 12 =−+− ω
∴ =+=+ AuAuu nn 21000 () ()() ω ω (g)
Substituting Eqs. (e) and (g) for A1 and A2 in Eq. (d) gives [] {} t n etuuutun ω ω ++=)0()0()0()( (h)
uuu nn ++=20 2 ωω (c)
Assume a solution of the form utest () = . Substituting this solution into Eq. (c) yields
() sse nn st 2220 ++= ωω
Because e st is never zero, the quantity within parentheses must be zero:
ssnn 2220 ++= ωω
or s nnn n = −±− =− 224 2
22 ωωω ω ()
(double root)
The general solution has the following form:
utAeAtenn tt () =+12 ω ω (d)
where the constants A1 and A2 are to be determined from the initial conditions: u () 0 and () u 0
Evaluate Eq. (d) at t = 0 : )0( )0( 11 uAAu=⇒= (e)
Differentiating Eq. (d) with respect to t gives
() () utAeAte n t n t nn=−+−ωω ω ω 12 1 (f)
Problem 2.9
Show that the motion of an overcritically damped system due to initial displacement u(0) and initial velocity u
Solution:
Equation of motion: mucuku++= 0 (a)
Dividing Eq. (a) through by m gives uuu nn ++=20 2ζωω (b) where ζ > 1.
Assume a solution of the form utest () = . Substituting this solution into Eq. (b) yields 0) 2( 2 2 = ++ st essnn
Because e st is never zero, the quantity within parentheses must be zero:
Differentiating Eq. (c) with respect to t gives
The general solution has the following form:
where the constants A1 and A2 are to be determined from the initial conditions: u () 0 and () u 0
Evaluate
The solution, Eq. (c), now reads: (
Problem 2.10
DerivetheequationforthedisplacementresponseofaviscouslydampedSDFsystemduetoinitialvelocity u(0)forthreecases: (a) underdampedsystems; (b) criticallydampedsystems;and (c) overdampedsystems.Plot u(t) ÷˙ u(0)/ωn against t/Tn for ζ = 0.1,1,and2.
Solution:
Equation of motion: 0 2 2 =++ uuunn ωζω (a)
Assume a solution of the form
u( t ) = e st
Substituting this solution into Eq. (a) yields:
s n s n e st 2 2 2 0 ++ F H I K = ζωω
Because e st is never zero
s n s n 2 2 2 0 ++= ζωω (b)
The roots of this characteristic equation depend on ζ
(a) Underdamped Systems, ζ<1
The two roots of Eq. (b) are si n 12 2 1 , =−±− F H I K ωζζ (c)
Hence the general solution is
u( t ) = A1e s1 t + A2e s2 t which after substituting in Eq. (c) becomes uteAeAenDD titit () =+ ζωω ω 12 ej (d) where ωωζ Dn=− 1 2 (e)
Rewrite Eq. (d) in terms of trigonometric functions:
u( t ) = e ζωntA cos ωDt + B sin ωDt () (f)
Determine A and B from initial conditions u(0) = 0 and () u 0 :
AB u D == 0 0 & ()
©
Substituting A and B into Eq. (f) gives ut u et n
t n n () () sin = F H I K 0 1 1 2 2 ωζ ωζ ζω (g)
(b) Critically Damped Systems, ζ = 1
The roots of the characteristic equation [Eq. (b)] are: s n s n 12 =− =− ω ω (h)
The general solution is u( t ) = A1 e ωnt + A2 te ωnt (i)
Determined from the initial conditions u(0) = 0 and () u 0 : AAu 1 00 = = 2 () (j)
Substituting in Eq. (i) gives ututen t () () = 0 ω (k)
(c) Overdamped Systems, ζ>1
The roots of the characteristic equation [Eq. (b)] are: sn 12 2 1 , =−±− F H I K ωζζ (l)
The general solution is:
utAeAestst () =+12 12 (m) which after substituting Eq. (l) becomes utAee nn tt () = −+− F H I K F H I K 1 11 22 ζζωζζω + A 2 (n)
Determined from the initial conditions u ()00 = and () u 0 : −== AA u n
12 0 2 2 1
& () ωζ (o)
tt () () = F H G I
K J 0 21 2 2211 ζω ωζωζ
(d) Response Plots
Plot Eq. (g) with ζ = 0.1; Eq. (k), which is for ζ = 1; and Eq. (p) with ζ = 2.
Forasystemwithdampingratio ζ ,determinethenumberoffreevibrationcyclesrequiredtoreducethe displacementamplitudeto10%oftheinitialamplitude;theinitialvelocityiszero.
© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2.12
Whatistheratioofsuccessiveamplitudesofvibrationiftheviscousdampingratioisknowntobe
(a) ζ = 0.01, (b) ζ = 0.05,or (c) ζ = 0.25?
Solution:
(a) ζ = 001: u u i i +
(b) ζ = 005 . : u u i i +
(c) ζ = 025.: u u i i +
= 1 1065.
= 1 137
= 1 506 .
Problem 2.13
ThesupportingsystemofthetankofExample2.6isenlargedwiththeobjectiveofincreasingitsseismic resistance.Thelateralstiffnessofthemodifiedsystemisdoublethatoftheoriginalsystem.Ifthedampingcoefficientisunaffected(thismaynotbearealisticassumption),forthemodifiedtankdetermine (a) thenaturalperiodofvibration Tn ,and (b) thedampingratio ζ
Solution:
Given:
w = 20.03 kips (empty); m = 0.0519 kip-sec2/in.
k = 2 (8.2) = 16.4 kips/in.
c = 0.0359 kip-sec/in.
(a) T m nk === 22 00519 164 0353 ππ .sec
(b) ζ === c km 2 00359 216400519 00194 (.)(.) = 194.%
Problem 2.14
The vertical suspension system of an automobile is idealized as a viscously damped SDF system. Under the 3000-lb weight of the car, the suspension system deflects 2 in. The suspension is designed to be critically damped.
(a) Calculate the damping and stiffness coefficients of the suspension.
(b) With four 160-lb passengers in the car, what is the effective damping ratio?
(c) Calculate the natural frequency of damped vibration for case (b).
Solution:
(a) The stiffness coefficient is k = 3000 2 = 1500 lb/in.
(b) With passengers the weight is w = 3640 lb. The damping ratio is
(c) The natural vibration frequency for case (b) is
Problem 2.15
The stiffness and damping properties of a mass–spring–damper system are to be determined by a free vibration test; the mass is given as m = 0.1 lb-sec2/in. In this test, the mass is displaced 1 in. by a hydraulic jack and then suddenly released. At the end of 20 complete cycles, the time is 3 sec and the amplitude is 0.2 in. Determine the stiffness and damping coefficients.
Solution:
1. Determine ζ and ωn.
Therefore the assumption of small damping implicit in the above equation is valid. T
n == 2 015 4189.secrads
2. Determinestiffnesscoefficient.
km n === ω224189011755(.)..lbsin.
3. Determinedampingcoefficient.
cmcrn === 220141898377 ω (.)(.).lbsecin.
cccr ===− ζ 0012883770107 .(.).lbsecin.
Problem 2.16
A machine weighing 250 lbs is mounted on a supporting system consisting of four springs and four dampers. The vertical deflection of the supporting system under the weight of the machine is measured as 0.8 in. The dampers are designed to reduce the amplitude of vertical vibration to one-eighth of the initial amplitude after two complete cycles of free vibration. Find the following properties of the system: (a) undamped natural frequency, (b) damping ratio, and (c) damped natural frequency. Comment on the effect of damping on the natural frequency.
Solution:
(a) k == 250 08 3125.lbsin.
w ===− g386 lbsecin. 2 250 0647.
ωn k m == 2198.secrads
(b) Assuming small damping,
This value of ζmay be too large for small damping assumption; therefore, we use the exact equation:
ζζ 22 00271 =−.() ⇒
ζ ==002670163 ..
(c) ωωζ Dn=−=12169 2 .secrads
Damping decreases the natural frequency.
Problem 2.17
Determinethenaturalvibrationperiodanddampingratioofthealuminumframemodel(Fig.1.1.4a) fromtheaccelerationrecordofitsfreevibrationshowninFig.1.1.4b.
values directly from Fig. 1.1.4b:
Problem 2.18
ShowthatthenaturalvibrationfrequencyofthesysteminFig.E1.6ais ω n = ωn (1 w/wcr )1/2 ,where ω n isthenaturalvibrationfrequencycomputedneglectingtheactionofgravity,and wcr isthebuckling weight.
Solution:
1. Determinebucklingload.
2. Drawfree-bodydiagramandsetupequilibrium equation.
0 θ (a)
Substituting Eq. (b) in Eq. (a) gives wLkwL g 2 0 () θθ+−= (c)
3. Co mputenaturalfrequency.
Problem 2.19
AnimpulsiveforceappliedtotheroofslabofthebuildingofExample2.8givesitaninitialvelocityof 20in./sectotheright.Howfartotherightwilltheslabmove?Whatisthemaximumdisplacementof theslabonitsreturnswingtotheleft?
Solution:
For motion of the building from left to right, the governing equation is mukuF +=− (a) for which the solution is
utAtBtu nnF()cossin=+− 22 ω ω (b)
With initial velocity of () u 0 and initial displacement u ()00 = , the solution of Eq. (b) is ut u tut n nFn () () sin(cos)=+− 0 1 ω ωω (c) () ()cos sin ututut nFnn=− 0 ωωω (d)
At the extreme right, ()ut = 0 ; hence from Eq. (d)
tan () ω ω n nF t u u = 01 (e)
Substituting ωπ n = 4 , uF = 015.in. and & () u 0 = 20in.sec in Eq. (e) gives
tan . ω π n t == 20 4 1 015 1061 or sin.cos. ω ω nn tt == 09956 00938 ;
Substituting in Eq. (c) gives the displacement to the right:
u =+−= 20 4 099560150093811449 π (.).(.).in.
After half a cycle of motion the amplitude decreases by 2201503 uF =×=..in.
Maximum displacement on the return swing is u =−=1449031149 ...in.
© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.AnSDFsystemconsistingofaweight,spring,andfrictiondeviceisshowninFig.P2.20.Thisdevice slipsataforceequalto10%oftheweight,andthenaturalvibrationperiodofthesystemis0.25sec.If thissystemisgivenaninitialdisplacementof2in.andreleased,whatwillbethedisplacementamplitude aftersixcycles?Inhowmanycycleswillthesystemcometorest?
Solution:
Problem 2.20 Given:
The reduction in displacement amplitude per cycle is 4 u
The displacement amplitude after 6 cycles is 2.06(0.244)2.01.4640.536in. −=−=
Motion stops at the end of the half cycle for which the displacement amplitude is less than uF . Displacement amplitude at the end of the 7th cycle is 0.536 – 0.244 = 0.292 in.; at the end of the 8th cycle it is 0.292 – 0.244 = 0.048 in.; which is less than uF . Therefore, the motion stops after 8 cycles.