Extra Credit for Mc Gran n Test 2
2
For the Assigned problem 6 separate bearings must be selected. The bearings will be for Connections A,B,C,D,E,F. connections E and B have thrust loads on them, this thrust bearing in addition to the radial load requires a value F e to be calculated. This value equates the value of the thrust load and axial load damage to an equivalent radial load only (what bearings are rated for) each bearing supports a shaft, and each shaft is held by only two bearings. First we will analyze bearing E, because it has an axial as well as radial loading. Bearing E: The axial Loading =500 lb The radial loading is found by taking the components of the two reaction forces. = 2631 lbs. From this we can use the equation we calculate an this will be used later in the calculations, for this analysis we first We get an initial c 10 value of 20069.1 lbf. which is 89.3076 kNs from this we must select a X 2 and a Y 2 to calculate an initial F e =(.56)(1)(2631)+(1.63)(500)=2288.36 lbf. which is 10.1 kN we use this value to recalculate a c 10 value and get 77.042, from this we are able to pick a value from the timken catalog to select a bearing that would support this load, the bearing has the e=.33 and Y=1.79 from this we get a F e =10.539kN, this checks out with the bearing 2557025520 for the 10,000 hour life expectancy of the bearing for bearing E. Next we will calculate the loading at bearing F again the first step is to calculate the radial load this radial loading can be put into the c 10 equation with the same x d =48, this calculation yields a value of 9249.92 lbf which is 41.1621 kN, The analysis can stop here because their is no axial component, for this radial loading the bearing 306W works. Similarly we will do the analysis for the next shaft which travels at 240 rev/min which yields the x D =144
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In fulfillment of an ME degree at Binghamton
Transcript of Extra Credit for Mc Gran n Test 2
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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
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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
