公 上海久通大学 Forces on gears ▣Forces on spur gears Gear 直齿圆柱齿轮传动受力分析 啮合线 nion (a 图上涤夫廷大学 Forces on gears ▣Forces on spur gears Normal: Fn二 2T1 di 7 Tangential F,=2T Id N Radial F,F tga N Normal F=F,/cos a a) 6
6 Forces on gears Forces on spur gears Normal： Tangential — 1 1 2 b n d T F Ft 2T1 / d1 N Radial — Fr Ft tg N Normal — Fn Ft / cos N Forces on gears Forces on spur gears
公 上酒短大峰 Forces on gears ▣Forces on spur gears =-f2 F=-F2 T. Force Direction: F:Opposite to rotation direction F2:Same to rotation direction F,F2:Pointing to rotating center respectively q) 圆上泽支大学 Forces on gears ▣Forces on spur gears n 7
7 Ft1 Ft 2 Fr 1 Fr 2 Force Direction： Ft1: Opposite to rotation direction Ft2: Same to rotation direction Fr1 、Fr2: Pointing to rotating center respectively Forces on gears Forces on spur gears 2 1 n2 n1 Fr2 Fr1 Ft2 Ft1 n1 n2 Forces on gears Forces on spur gears
图 上海通大学 染 ▣Shown as the fig Direction of tangential forces on gear 2 and 3 ⊙8 Z 80 View direction Right answer 圆上泽支大学 ▣Shown as the fig Direction of tangential nF2 forces on gear 2 and 3 ⊙8 ⑧ View direction The answer 8
8 Shown as the fig Direction of tangential forces on gear 2 and 3 Right answer Ex1 View direction Z1 Z4 Z2 Z3 A BCD Ft1 Ft2 Ft3 Ft4 The answer Ex2 Z4 Z3 Z1 Z2 Ft2 Ft1 Ft4 Ft3 A BCD Shown as the fig Direction of tangential forces on gear 2 and 3 View direction
图 上海通大学 Example1 In the figure below,pinion 2 runs at 1750 rev/min and transmits 2.5 kW to idler gear 3. The teeth are cut on the 20 degree full-depth system and have a module of m=2.5 mm. Draw a free-body diagram of gear 3 and show all the forces which act on it. 50T 30T 201 圈上充通大学 Example -Solution y F43 Fs 3 50T 20° F 30 20T Fb3 20° (a) (b) 9
9 Example1 In the figure below, pinion 2 runs at 1750 rev/min and transmits 2.5 kW to idler gear 3. The teeth are cut on the 20 degree full-depth system and have a module of m = 2.5 mm. Draw a free-body diagram of gear 3 and show all the forces which act on it. Example – Solution
图 上游充通大学 Example-Solution Pitch diameters of gears 2 and 3 are d2=N2m=20*2.5=50mm d3=N3m=50*2.5=125mm Transmitted load is W,=60×103×H=60x10'×25=0.546kN πd2n π×50×1750 Thus tangential force of gear 2 on gear 3 is F23=0.546 kN.Therefore Fr23=F23tan200=0.546*tan20°=0.199kN 国上清通大学 Example -Solution ▣Which implies that F23 F230.546 =0.581kN c0s20°c0s20° Gear 3 is an idler,and thus transmits no power or torque to its shaft. Thus tangential reaction of gear 4 on gear 3 is also equal to W,.Therefore Ft43=0.546kN F43=0.199kN F43=0.581kN 10
10 Example – Solution Pitch diameters of gears 2 and 3 are d2 = N2m = 202.5 = 50 mm d3 = N3m = 502.5 = 125 mm Transmitted load is Thus tangential force of gear 2 on gear 3 is Ft 23 = 0.546 kN. Therefore Fr 23 = Ft 23 tan 20o = 0.546tan 20o = 0.199 kN kN d n W H t 0.546 50 1750 60 10 60 10 2.5 3 2 3 Example – Solution Which implies that kN F F o o t 0.581 cos 20 0.546 cos 20 23 23 Gear 3 is an idler, and thus transmits no power or torque to its shaft. Thus tangential reaction of gear 4 on gear 3 is also equal to Wt . Therefore Ft 43 = 0.546 kN Fr 43 = 0.199 kN F43 = 0.581 kN