OutlineFundamental Law of GearingTowards the Involute ProfileThe Involute ProfileProfile of the Involute ProfileInvolute in ActionNomenclaturePitches, Etc.Velocity RatioContact RatioMinimum # of TeethSlide 12Simple Gear TrainsSimple Gear TrainPowerPoint PresentationSlide 16Reverse on a CarActual Manual TransmissionSlide 19Loading of GearsGear vs. IdlerLoading ExampleSlide 23Gear FailureAGMA Gear Stress FormulaBending Strength Geometry FactorTip vs. HPSCT LoadingFinding J for the ExampleDynamic (Velocity) FactorDynamic FactorDetermining Kv for the ExampleLoad Distribution Factor KmKm for the ExampleApplication Factor, KaOther FactorsBack to the ExampleCompared to What??Life Factor KLTemperature & Reliability FactorsFatigue Bending Strength for ExampleSafety FactorSlide 42Geometry Factor IElastic CoefficientSurface-Fatigue StrengthsLife Factor CLHardness Ratio CHSafety Factor for LoadingGear DesignSlide 50Gear Design StrategyOutlineGear TheoryFundamental Law of GearingInvolute profileNomenclatureGear TrainsLoadingStressesFundamental Law of Gearingfunctionally, a gearset is a device to exchange torque for velocityP=Tthe angular velocity ratio of the gears of a gearset must remain constant throughout the meshinoutinoutVrrm pitch circle, pitch diameter d, pitch pointgearpinionWhat gear tooth shape can do this?click here!Towards the Involute ProfileA belt connecting the two cylindersline of action, pressure angle The Involute ProfileProfile of the Involute Profilepressure angle, line of action, length of action, addendumInvolute in Actionvideo from http://www.howstuffworks.comNomenclatureFigure 11-8NdpcPitches, Etc.circular pitch (mm, in.)base pitch (mm, in.)diametral pitch (teeth/in.)module (mm/teeth)Ndpccoscbpp dNpdNdm Velocity Ratiopitches must be equal for mating gears, thereforeinoutinoutVNNrrm Contact Ratioaverage number of teeth in contact at any one time =length of action divided by the base pitch, or, cosZpmdp sincoscos2222CrarrarZgggpppwhere C=center distance=(Ng+Np)*1/pd*1/2Minimum # of Teethminimum # of teeth to avoid undercutting with gear and rack2minsin2NOutlineGear TheoryFundamental Law of GearingInvolute profileNomenclatureGear TrainsLoadingStressesSimple Gear TrainsSimple Gear TrainRBRPPBvNNNNNNm RBRGGPPBlackBlackYYBvNNNNNNNNNNNNm Simple Gear TrainFine for transmitting torque between shafts in close proximitywhen mv does not need to be too largeUse third gear (“idler”) only for directional reasons (not for gear reduction)Reverse on a CarActual Manual TransmissionOutlineGear TheoryFundamental Law of GearingInvolute profileNomenclatureGear TrainsLoadingStressesLoading of GearsWt= tangential (transmitted) loadWr= radial loadW= total load(see board for figure or page 711 in book)pppptdTrTW2tantrWW costWW Because Tp is constant, should Wt be static for a tooth?Gear vs. IdlerLoading ExampleGiven:•pinion is driving gear with 50hp at 1500 rpm•Np=20•mv=2 (a.k.a., mg=2)•pd=4 /in.=20 degreesFind:Loading and effects of inputs on loadingOutlineGear TheoryFundamental Law of GearingInvolute profileNomenclatureGear TrainsLoadingStressesGear FailureFatigue Loading Surface Failurefrom contact b/n of teeth•infinite life not possible•failure is gradualfrom bending of teeth•infinite life possible•failure can be suddenLewis, 1892, first formulation of gear tooth fatigue failureYFpWdtbAGMA Gear Stress Formulamany assumptions: see page 714… …including that the contact ratio 1<mp<2IBsvmadtbKKKKKKJFpWJ KvKmKaKsKBKIBending Strength Geometry Factorfrom tables on pgs. 716-718Inputspinion or gearnumber of teethpressure anglelong-addendum or full-depthtip loading or HPSTCJ KvKmKaKsKBKITip vs. HPSCT LoadingFinding J for the Example•pinion is driving gear with 50hp at 1500 rpm•Np=20, Ng=40•pd=4 /in.=20 degrees•Wt=420 lb.Jp=0.34, Jg=0.38Dynamic (Velocity) FactorJ KvKmKaKsKBKIto account for tooth-tooth impacts and resulting vibration loadsfrom Figure 11-22 or from Equations 11.16-11.18 (pages 718-719)Inputs needed:•Quality Index (Table 11-7)•Pitch-line VelocityVt = (radius)(angular speed in radians)Dynamic FactorJ KvKmKaKsKBKI )(200)()1(565011Q6for 412v3/2SIVAAKUSVAAKBAQBBtvBtvvVt,max=(A+Qv-3)2Determining Kv for the ExampleVt=(1500rev/min)(2.5 in)(1ft/12in)(2 rad/1rev)=1964 ft/min(well below Vt,max)therefore, Qv=10 845.019648.838.838.83)397.01(56500.397 41012397.03/2vKABLoad Distribution Factor KmJ KvKmKaKsKBKIto account for distribution of load across face8/pd < F < 16/pdcan use 12/pd as a starting pointKm for the ExampleF=12/pd=3 in.Km=1.63Application Factor, KaJ KvKmKaKsKBKIto account for non-uniform transmitted loadsfor example, assume that all is uniformOther FactorsJ KvKmKaKsKBKIto account for sizeto account for gear with a rim and spokesto account for extra loading on idlerKsKBKIKs=1 unless teeth are very largeKB=1 for solid gearsKI=1 for non-idlers, KI=1.42 for idler gearsBack to the ExampleIBsvmadtbKKKKKKJFpWpsipsipiniongearbb3177)1)(1)(1()845.0()63.1)(1()34.0)(3()4)(420(2842)1)(1)(1()845.0()63.1)(1()38.0)(3()4)(420(Compared to What??b is great, but what do I compare it to?bfRTLfbSKKKSsimilar concept to Se, but particularized to gearsSfb´ Table 11-20Figure 11-25 for steelsHardnessSfb´Life Factor KLto adjust test data from 1E7 to any number of cyclesTemperature & Reliability FactorsKT=1 if T < 250°Fsee Equation 11.24a if T > 250°FFatigue Bending Strength for ExampleClass 40 Cast Iron450 rpm, 8 hours/day, 10 yearsT = 80°FReliability=99%N=(450 rev/min)(60 min/hr)(8 hr/dy)(250 dy/yr)(10 yrs)=5.4E8 revsKL=1.3558N–0.0178=0.948Sfb = (0.948)13ksi = 12.3 ksiSfb´= 13ksiTable 11-20Safety FactorbfbbSNNbpinion =(12300 psi)/(3177psi)= 3.88Nbgear =(12300 psi)/(2842psi)= 4.33equal to 1same as for bendingGear FailureFatigue Loading Surface Failurefrom contact b/n
View Full Document
Unlocking...