Solution Manual for Mechanical Vibrations, 6th Edition

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SOLUTIONSMANUALMECHANICALVIBRATIONSSIXTHEDITIONSINGIRESUS.RAOUniversityofMiamiJanuary2016

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DownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com

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ToLordSriVenkateswara

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ContentsPreface1.FundamentalsofVibration1-1-1-992.FreeVibrationofSingle-Degree-of-FreedomSystems2-1-2-1843.HarmonicallyExcitedVibration3-1-3-1384.VibrationUnderGeneralForcingConditions4-1-4-1465.Two-Degree-of-FreedomSystems5-1-5-1336.Multidegree-of-FreedomSystems6-1-6-1527.DeterminationofNaturalFrequenciesandModeShapes~~7-1-7-718.ContinuousSystems8-1-8-839.VibrationControl9-1-9-8410.VibrationMeasurementandApplications10-1-10-1811.NumericalIntegrationMethodsinVibrationAnalysis~~11-1-11-5012.FiniteElementMethod12-1-12-9613.NonlinearVibration13-1-13-5214.RandomVibration14-1-14-28

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ERRATA(FORTEXTBOOK)MechanicalVibrationsSixthEdition,SingiresuS.Rao(January10,2016)1.Page238Problems2.55-2.58line3changek=10,000tok=10,000N/m2.Page239Problem2.65Replace(a)bythefollowing:(a)Findwhetherthetrunkofthetreewillbuckleundertheweightofthecrownofthetreeassumingthattheweightofthecrownactsasanaxialloadatthetopofthetrunk.Assumetheheightofthetrunkas/=I=10m.3.Page253Problem2.121Addthefollowinglineafterpart(d):Assumem=10kgandk=10,000N/m.4.Page253Problem2.122‘Addthefollowinglineafterpart(d):Assumem=10kgandk=10,000N/m.5.Page254Problem2.123Addthefollowinglineafterpart(d):Assumem=10kgandk=10,000N/m.6.Page368Problem3.87line3modifytheendofthesentenceasfollows:Replace“respectively.”by“respectively,and&=m/M.”

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Chapter1%.>)..SFundamentalsofVibrationaJLee@=[Heed|Head,andneck-.uppertorso,uppertorsoAhhipsandlegsspinal5=fegsfipsdpespinalPEERSSe7|columnnn(ilegs#~--7shoulderP9legs<:T.(i)Gin)SeatRestraint\WindshieldoN\5)Instrument3panel.ICHE;—OTR|StEWNtoutImpactforceee1-1

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unbalanced>©)unbalancedforces1forces3,E)1!Massofframe&aMassofframe,Treciprocatingengine+reciprocatingx.Tengine&!og-pdamping&fadndatienblockx”stiffnessofxy:boltsand-elasticpad«damping|blockstiffnessxpoodamping©--dlstiffnessofsoil(@)onedegreeofiFreedommodel(b)Twodegreeoffreedommodel)-od|massofbodymassofbodySuspensionofwheelsdampingrearWheelsof30°suspension©wheels(i)777(i1-2

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[[emassofpassengers%plusseatsE)ha<hizdamping©stiffnessofseatsaa=o-bod.dampingRo_—-ydamping2stiffnessstiffnessof\.yz"ofrearsuspensionFront"wsuspensionwheelswheelsdepiaNE:-7damping&stiffnessrissa*00»ofrearwheels77’7777(iii)®)RadiatorsheetmetalKNCenteNe¥.7BumperDriveline©)[Shockabsorber53Ll))]Plow1-3

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=®EINSRNSULr.SNz1k,Kp24g0~egTXX¥Eleg3Ky+2kyKy+keky~e%FXxwvAtxK--ep,Keg+*ykgksKymKe(BgrKe).tksksrzhblksrkvakgkspalakaksPAKatesKy+Kak+2KK3Kg+2KKsKgpity+ikgKsthikJ1-4

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FoSE&JA’x|€TJLre©ooxz@)i+LFFiFab){tefyFromFig.(@),x=PIE(x2—X01)9,+42ICRRI(10122,+4;#verticalforceeppilibrivmfromFrg.u(b):F=F+F,(2>Momenteguilibriumaboutc’(Fig.hd):Fp2,=F,2,>SolutionofEgs.(2)and(3):f=FlapoFA4)gL,+L,22+[P}Displacementsofsprings*,amdtk;aregivenbyooBo_FlooR_Fh*1,(4+12)27Th,(G12)DisplocementofforceFcanbefoundusingEps-CsDinEg.(1:¢REPA1.SETHEE1+,1,(1,+12)Lita4p(L442)-_F(2+1;Ly(6)ISRAELTheepuivalntAp2ingconstandoftheAyslominthe1-5

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-gdirectionofx,Kg,“givenbyEp.(6):ike===Lr)wkbikeD):NTEPaeRce1-6

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<3>3Epuivalenceofpotentialenergiesgives&z2fTEEAC0)L100)+3kalods)=£fey=p4p3=Forseriesspringsi»tyandki:«0kaLESEyy=—————————KipsTRkeks123Kd+Kpfea+KakUsingenergyepuivalence,’2z23keyot=hkkyothKusO+3ksOF)+ike(eR)~Heg=ky+Kizz+R*%g+R"kefk,®=K|®2%3)2CTerty+R(st®s)ee@1)Forsimplysupportedbeam,12mSorloadotmiddle,+3i54Ged«=ee_Wbosro(27)PapT=—3-=12-36x10’NfmwhereT=(L2)(er)5mt“on._om_59.84-78=originaldeflection===200xto=—396-8447xl0™Whenspring&isadded,Keg,=&+Kg;mg)4mg@)Newdeflection=Z%_22ge=ZF=)dkeyTF©”TsC4=k+HkiSok=3%=37.08x10N/m!(b)Newdeflection=+.3;k=2mLgJ?75,k=%,z12:36x10N/m=k+Kk,©)Newdeflection=TZ4mflection==24;key®$F_4,%383Cik=dk=402x10Nm=+k,Ef1-7

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>i»)ForbarwithlengthL,YoungsmodulusEandE)cross-sectionA,Theaxialstiffness(x)isgivenby&=AEa]af®=7whencross-sectionissolidcircularwithdiameterd,area=A=Td,(2)whencross—sectionissguarewithsided.ove=Ap=4(3)Whencross~sectioniShollowcircularwithmeandio.dandwaltthicknesstzo-1d,area=TTdt=Td(61d)=oT4%(CY)}Forspecifiedvalueofk=%&,cross-sectionareasicedis:fereguirisA=KL=c(constant(CD)Weightofbar:withsolidcircularsection:cw=rdL=clkwithd=RES©>withhollowciceularsection:Wy,=omdL=orm(HAE)L=okcb(€))withspuaresection:=okwy=dL=%C|=4w,=2732Wi8))™Ed+Theshaftwiththehollowciceularcross-sectioncorrespondsfominimumweight.eyerm1-8

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>>@Stiffressofocantileverbeamunderobending3wvforceoffreeend=5<p)JLForoApresfledvalueof*k=%,I=zl-=Cc=constant(©)3ForaSolidcircularsectionwithdiameterd5CLh_64REE3)I===c=4d=ord==>(weightofbeam=Wy=wd*f—-wk6c4%Jets=3.5449{JC“)forahollowcircularsectionwithmeandiameterdandwollthicknesst=o:1d,weightofbeam(wp)is:wozI(do—d)k=ptsfey-(4-07)=Li(444)=wdtl=WHEdD—1=owREE-igoLeGDForovAguassecionwithside4d.weightofthebeam(wy)4:2wz=d*4=[REE=4.5135Lic")BycomponingEs:WY,(8)amd(6,Theminimumweight1-9

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z3SpringSorceisgivenbyF=Soox+40x?(DD)2staticequilibriumoftherubbermounting(x*)under+theweightoftheeledmiomicinstrumentisgivenbyE=200=8a0x*+40x?462*®4900x"~200=o(2Therootsofthecubiceppation(2)cambefordFromMATLABofx*=02492»—orlz4éLE447734@ThusHesalieguilibriampositionoftherubbermountingisgivenbytherealrootofEg-@):x*=02432in4>@)EguivalentlinearAleringconstantofvubbermountingokitsshoeeguilibriumposition,usingCACkeg=|=860+126x*=goo+1200(0-24%2)=got4521lb/in52>bYDeflectionofrubbeyrmountingcorrespondingfotheequivalentLinearApringconstantis:BEPEAAT.©)k=keTwor4s2tSE1-10

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zg@)F(x)=zoox+50x?410x?a)adLwhentheSpringundergoesasteadydeflectionof:x*=0'5induringtheoperationoftheengine,theforceexertedonthepringconbefoundosg=200(e'5)+So(05)+10(o-5)°=13.75Lb(2dEquivalentLinearSpringconstantatifAteadlydeflectionisgivenbyEy-G7):Keg=a8=200+100230egd=x=x¥2=200+loo(0:8)+30(o'5)=253.75(b/inJ1-11

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