材料力学例题及Examples of mechanics of materials and.docx
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材料力学例题及Examples of mechanics of materials and.docx
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材料力学例题及Examplesofmechanicsofmaterialsand
材料力学例题及(Examplesofmechanicsofmaterialsand)
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Examplesofmechanicsofmaterialsandguidanceofproblemsolving
(secondtosixthchapters)
Thesecondchapter,stretching,compressionandshear
Inexample2-1,trytofigureouttheaxisofthestraightrodoffigurea
Solution:
thestraightrodissubjectedtoaxialexternalforceatA,B,CandDpoints
ABaxialforceisfirstsought
Thememberiscutoffatany1-1sectionofthesection
Examineleftsegment(Figure2-5b)
ApositiveaxialforceissetonthecrosssectionN1
ThustheequilibriumequationSX=0isobtained
N16=0
N1=6kN
N1wasoriginallyassumedthataforceiscorrect
Italsoshowsthattheaxialforceispositive
TheaxialforceofanysectionintheABsectionisequalto+6kN
BCaxialforceisagainsought
Thememberiscutoffatany2-2sectionoftheBCsection
Theleftsegmentisstillexamined(Figure2-5c)
TheaxialforceN2isstillsetonthecrosssection
BySX=0
618N2=0
N2=12kN
N2minusthatoriginallyassumedtensioniswrong(forpressure)
ItalsoshowsthattheaxialforceN2isnegative
TheaxialforceofanysectionintheBCsectionisequalto-12kN
Similarly,theaxialforceofanysectionintheCDsectionis-4kN
Drawinternalforcechart
ThehorizontalaxisXrepresentsthesectionpositionoftherod
RepresentstheaxialforceofthecrosssectionwiththeverticalaxisofX
Drawanaxlebarattheselectedscale
AsshowninFigure2-5(d)
Fromthis,itisshownthatthemaximumaxialforceoccursintheBCsegment
Problemsolvinginstruction:
whentheaxialforceiscalculatedbythecrosssectionmethod,thepositiveaxialforceisalwayssetonthesectionoftheincisionN
ThentheaxialforceNisobtainedbySX=0
IftheNiscorrect,thepositiveaxialforce(tension)isindicated
Ifnegative,itmeansnegativeaxialforce(pressure)
Example2-2testthefreesuspensionofthestraightbar(Figure2-6a)bylongitudinaluniformdistributionofloadQ(force/length)causedbystressandlongitudinaldeformation
ThebarlengthL,sectionalareaAandelasticmodulusEareallknown
Solution:
takeanarbitrarycrosssectionM-Matthelowerendoftherodatx
TheaxialforceofthissectionisN(x)=QX
Accordingtothisformula,theaxlecanbemadeasshowninfigure2-6b
ThestressoftheM-Mcrosssectioniss(x)=N(x)/A=qx/A
Obviously
ThesuspensionendhasmaximumaxialforceNmax=QLandmaximumpositivestress
Longitudinaldeformationofrod
Duetounequalaxialforcesoneachcrosssection
Theformula(2-4)cannotbeapplieddirectly
AndshouldstartfromalongsectionofDX
TakethemicrosegmentDXatx
Itslongitudinalelongationcanbewrittenas
Totalelongationofbars
Theelongationattheendofthefixedmemberduetodeadweightisinvestigated
Thebarselfweightisauniformlongitudinaldistributionforce
Atthistime,thedistributionforceofunitrodlengthisq=A*1*g
Here,Gistheunitvolumeofmaterial,thatis,bulkdensity
GetQintotheupperform
HereG=Algistheweightofthewholepole
Theupperformshowsthatthetotalelongationcausedbythedeadweightoftheequalstraightrodisequaltohalftheelongationofalltheweightconcentratedatthelowerend
Problemsolving:
arodwithvariableaxialforce
CalculationofaxialdeformationofbarsbyusingHooke'slaw
Deformationshallbecalculatedinsections
Thenthealgebraisaddedtothefullbar
Whentheaxialforceisacontinuousfunctionisrequiredbyintegralroddeformation
Example2-3.Figure2-7showsthattworoundsectionbarmaterialsarethesame
Strainenergyoftwobariscalculated
Andcomparethesize
Solution:
arod:
Brod:
Ratioofstrainenergyoftwobar:
Problemsolvinginstruction:
ascanbeseenfromthisexample
Underthesameforce
Thestrainenergyofthebarwithsmallrigidityislarge
Example2-4parallelbars1,2and3suspensionrigidbeamAB,asshowninfigure2-8a
LoadonthebeamG
Suchasrod1,2,3ofthecross-sectionalarea,lengthandmodulusofelasticityarethesame
TheyareA,landErespectively
TrytofindtheaxialforceN1,N2andN3ofthethreepoles
-
Solution:
undertheactionofloadG
ThebeamismovedtoAPhiPhiBposition(Figure2-8b)
Thenthereductionamountofrod1isDl1
Theelongationofrod2and3isDl2andDl3
TakethecrossmemberABastheseparatingbody
Asshowninfigure2-8c
ExceptforloadG
ThereareaxialforcesN1,N2,N3,andX
Duetotheassumptionthatthe1rodisshortened
2and3rodelongation
Therefore,N1shouldbesetaspressure
AndN2andN3aresetastension
(1)equilibriumequation
(a)
Thethreeequilibriumequationscontainfourunknownforces
Therefore,itisastaticallyindeterminateproblem
(2)deformationgeometryequationsbythedeformationdiagram2-8bshowsthattheB1B=2C1CPhiPhi
Thatis
or
(b)
(3)physicalequation
(c)
Substituting(c)form(b)
Thenitissolvedinconjunctionwith(a)
Available
Problemsolvinginstruction:
insolvingstaticallyindeterminateproblem:
assumingthattheaxialforceofeachrodistensionorpressure
Onthebasisoftheextensionorshorteningofeachrodinthedeformationdiagram
Thetwomustagree
Bycalculation,theaxialforceofthreebarsispositive
Instructionsareassetoutinthedeformationdiagram
Rod2and3elongate
Therod1isshortened
Examplesandproblemsolvinginstruction
Example2-5showninFigure3-6thescrewsaresubjectedtoaxialtensionF
Therelationbetweentheknownpermittedshearstress[t]andthetensileallowancestress[s]is:
[t]=0.6[s]
Therelationbetweenthepermittedcompressivestress[sbs]andthetensileallowancestress[s]is[sbs]=2[s]
TrysettingupD
D
ThereasonableratiobetweenTandthree
Solution:
(1)tensilestrengthofscrews
(2)extrudingstrengthofnut;
(3)shearstrengthofnut;
Got:
D:
D:
T=1.225:
1:
0.415
Problemsolvinginstruction:
payattentiontotheshearingsurfaceandextrusionsurfaceofthisproblem
Example2-6apalletisrivetedtoanuprightpostwith8rivets
Asshowninfigure3-7a
Therivetspacingisa
F=80kN
DistanceL=3A
Knownrivetdiameterd=20mm
Permittedshearstress=[t]=130MPa
Checktheshearstrengthofrivet
Solution:
thecenterCoftherivetgroupislocatedontheYaxisofthecolumn
WillforceFtothepointCtranslationgetaCpointytoforceFandacoupleFlclockwise
TheforceFcausedbyCisequaltotheshearforceresultingfromtheshearsurfaceofeachrivet
ItsvalueisF/8
AsshowninFigure3-7(c)
TheshearforcealongthenegativeYdirectionofthreerivets,1,2and8,isshowninthefigureF/8
TheFlalsocausedtheshearineveryrivet
ItisassumedthatthesheardirectionisorthogonaltotheconnectionoftherivetcentertotheC
Thesizeisdirectlyproportionaltothelengthoftheconnection
Figure3-7(b)showsthatFlinducedshearrivetrivet;1,3,5,7areQofshear1;2,4,6,8shearareQof2
ThesumofthemomentsoftheshearforceoftherivetsisequaltoFl.C
Thatis
Recycle
Substitutionformula
2ofthetotalshearrivetF/8F/4=Q2=3F/8
Thetotalshearforceofrivet1is
Sorivet1and3aremostdangerous
so
=115MPa<[t]
Problemsolving:
whencalculatingtheshearstrengthoftheconnectingcomponentoftherivetgroup
Tocorrectlyanalyzetheforceofeachrivet
Whentheexternalforcepassesthroughthecenteroftherivetcluster
Canbeapproximatedastheforceofeachrivetisthesame
Whentheexternalforcedoesnotpassthroughtherivetcentroid,therivetforceshouldbeanalyzedaccordingtotheactualforcecondition
Thethirdchapteristheexampleandtheinstructionofproblemsolving
Example3.1.TheRevn=300r/minoftheknowndriveshaft(Figure4-5(a))
ThepowerinputofthedrivingwheelAisP=400kW
TheoutputpowerofthethreedrivenwheelsisPB=120kW
PC=120kW
PD=160kW
TryHuazhoutorquediagram
Solution:
(1)calculatetorqueactingoneachwheelM
BecauseAisthedrivingwheel
Therefore,thesteeringofthemAisconsistentwiththesteeringoftheshaft;thetorqueonthedrivenwheelistheresistancetotherotationoftheshaft
Therefore,thetorqueofthedrivenwheelB,CandDisoppositetothatoftheaxle
(2)seekingthetorqueofeachshaft;
Askfor1-1sectiontorquefirst
Cutfromthissection
Retainrightsegment
Andsetoutonthesection
PositivetorqueMT1(Figure4-5(b))
ByequilibriumconditionSmx=0
Yes
MDmAMT1=0
haveto
HereMT1showsthatthecrosssectionofminustorqueisminus
TorqueacrossallsectionsoftheAandBwheels
Equalto12.74kNm
ItcanbeconcludedthatMT2=8.92kNm
MT3=10kNm
(3)drawatorquediagram
Thepositionofthecrosssectionisexpressedinabscissa
Torqueisexpressedinordinate
Torquechartsofthethree,BC,andCDsectionsoftheAB,,andaxleareselectedatselectedscales
Becausethetorqueisconstantineachsegment
Therefore,thetorquediagramconsistsofthreehorizontallines
AsshowninFigure4-5(c)
Themaximumtorque7.64kNmoccursinthemiddlesection
Problemsolving:
whentheaxlecrosssectiontorque
PositivetorqueisalwayssetatthecrosssectionMT
UseSmx=0forthistorque
IfMTisapositivetorque
Ifthesignisnegativetorque
Ifinthiscase,A,Bonthewheel
ThetorquediagramisshowninFigure4-5(d)
Fromthiswecansee
Themaximumvalueoftheinternalforcecanbereducedbyreasonablyarrangingtheload
Improvethecarryingcapacityofthemember
Example3.2itisknownthatthetransmissionsh
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