FortheAssignedproblem6separatebearingsmustbeselected.ThebearingswillbeforConnectionsA,B,C,D,E,F.connectionsEandBhavethrustloadsonthem,thisthrustbearinginadditiontotheradialloadrequiresavalueFetobecalculated.Thisvalueequatesthevalueofthethrustloadandaxialloaddamagetoanequivalentradialloadonly(whatbearingsareratedfor)eachbearingsupportsashaft,andeachshaftisheldbyonlytwobearings.FirstwewillanalyzebearingE,becauseithasanaxialaswellasradialloading. BearingE:TheaxialLoading=500lbTheradialloadingisfoundbytakingthecomponentsofthetworeactionforces.

=2631lbs.Fromthiswecanusetheequation

wecalculatean thiswillbeusedlaterinthecalculations,forthisanalysiswefirstWegetaninitialc10valueof20069.1lbf.whichis89.3076kNsfromthiswemustselectaX2andaY2tocalculateaninitialFe=(.56)(1)(2631)+(1.63)(500)=2288.36lbf.whichis10.1kNweusethisvaluetorecalculateac10valueandget77.042,fromthisweareabletopickavaluefromthetimkencatalogtoselectabearingthatwouldsupportthisload,thebearinghasthee=.33andY=1.79fromthiswegetaFe=10.539kN,thischecksoutwiththebearing2557025520forthe10,000hourlifeexpectancyofthebearingforbearingE. NextwewillcalculatetheloadingatbearingFagainthefirststepistocalculatetheradialloadthisradialloadingcanbeputintothec10equationwiththesamexd=48,thiscalculationyieldsavalueof9249.92lbfwhichis41.1621kN,Theanalysiscanstopherebecausetheirisnoaxialcomponent,forthisradialloadingthebearing306Wworks.Similarlywewilldotheanalysisforthenextshaftwhichtravelsat240rev/minwhichyieldsthexD=144

ForbearingCwewillagaincalculatecreateavectorwiththetwocomponentvectors.Thisvalueisthenpluggedintothec10equationwhichyields

avalueof146kN,sincethereisnoradialforcethisistheforcewewilluse.Thebearing314WworksforthisloadForbearingDthecalculationsaresimilaragain,thexDstillisequalto144,wegetavectorfromthecomponents thisthenisisconvertedto46.0299kN,forthisradialload,sincethereisagainnoaxialload,thevalueof46.0299kNistheonewewilluse.forthisloadwewilluse307Wbearing. nextwewillmovetotheshaftthathasbearingsAandbearingsBthexDvaluebecomes720becauseofanewRPM.firstwewilldobearingAbecausethereisnoaxialload,weget,onceagaintheradialloadbysummingthesquaresofthecomponentstoget

fromthisvalueandthec10equationweget6088.32lb.whichis27.093kNthisradialloadingisthenusedtofindabearing,Abearing306wasusedtosupportthisload. thefinalbearingtobeanalyzedwastheBbearing.forthisbearingtheaxialthrustis500lbandtheradialforcefromthetwocomponentsisfoundtobe

fromthisradialloadingweusethec10tofind121.981kNfromthisvalueweselectataperbearingthatwillsupportthisloadandreadofftheeandYvalues.e=.4andY=1.49weusethesenewvaluesandtheFeformulatogetanewvalueofFe=6.58336kNthisispluggedbackintothec10equationtogetamoreaccuratevalueof113.1kNfromthisabearingof38723820wasfoundtobesufficienttohandleboththeaxialandradialloadings.

Bearing c10 Fe Timkenpart#

A 27.093kN 263.519lbs 306W

B 113.1kN 1479.41lbs 38723820

C 146kN 2434.18lbs 314W

D 46.0299kN 765.57lbs 307W

E 80.3kN 2568.36lbs 2557025520

F 41.1621kN 987.377lbs 306W