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1、MeasuretheYoung’sModulusviathedynamicmethod45陈锋1.ThematerialsEnergyconversiondevice,Low-frequencygenerator,OscilloscopeandShieldedcables2.TheprincipleForauniformfinerodwithlengthLmuchbiggerthanitsdiameterd,bothofitsendsdoesaslightlyvibration,themotionofitwillsatisfytheequation:∂4y
2、∂x4+ρSEJ∂2y∂t2=0Where,ρisthedensityoftherod,Sisthecrosssectionoftherod,Jisthemomentofinertia,EistheYoung’sModulusandyisthevibrationdisplacementoftherod.Solvingthisequation,wecangetthat:1Xd4Xdx4=ρSEJ1Td2Tdt2And:yx,t=(B1chKx+B2shKx+B3cosKx+B4sinKx)Acos(ωt+φ)Where:ω=(K4EJρS)1/2Thisis
3、thefrequencyequation.Ifthesuspendinglinesuspendsthenodeofthesample,bothendsoftherodareinthefreesituation.Atthismoment,theboundaryconditionsisthebroadwiseforceandthebendingmomentiszero.F=-∂M∂x=-EJ∂3y∂x3=0M=EJ∂2y∂x2=0So:d3Xdx3x=0=0,d3Xdx3x=l=0,d2Xdx2x=0=0,d2Xdx2x=l=0Eliminatingtheeq
4、uationsandusingthenumericalmethod,wecangettheKandlshouldsatisfyKl=0,4.730,7.853,10.996,14.137…Generally,wedefinethefrequencyasbasisfrequencywhenKl=4.730Whenthesamplevibrateatthebasisfrequency,therearetwonodeswhichisatadistance0.224land0.776ltotheendrespectively.FedtheK1=4.730linto
5、thefrequencyequation,wegetthenaturalvibrationangularfrequencyis:ω=((4.730)4EJρS)1/2AndgettheYoung’sModulus:E=1.9978×10-3ρl4SJω2=7.8870×10-2l3mJf2WherethemomentofinertiaJ=y2ds=πd464So:E=1.6067l3md4f2Where,listhelengthoftherod,disthediameter,misthemass,fistheresonancefrequencyofthes
6、ample.Actually,Ealsorelatetotherateofdiameterandlengthd/l,sotheequationshouldmultiplyamodifyingfactorR.E=1.6067Rl3md4f2Whenl≫d,R≈1.Whenlisn’tmuchbiggerthand,weknowthat:d/l0.010.020.030.040.050.06R1.0011.0021.0051.0081.0141.019Whenthefrequencyoftheforceattachtotheresonancefrequency
7、,theothersuspendinglinewillgetthemaximumamplitude.Therelationshipbetweenthenaturalfrequencyandtheresonancefrequencyis:fr=ωr2π=12πω2-2β2βisthedampingcoefficient.Towardsthegeneralmetalmaterials,themaximumdampingcoefficientisthe1%oftheω,sowecanusefrtocomputeinsteadoff.3.Theprocedure(
8、1)TheYoung’sModulusofcopperandsta
