公 上游久通大￥ Example-Solution Shaft reactions in x and y directions are: Fxb3=-(Ft23+Fr43)=-(-0.546+0.199)=0.347kN Fyb3=-(Fr23+Ft43)=-0.199-0.546)=0.347kN The resultant shaft reaction is F63=V0.347+0.3472=0.491kW 国上清通大学 Forces on gears Forces on helical gears 11
11 Example – Solution Shaft reactions in x and y directions are: Fxb3 = - (Ft 23 + Fr 43) = -(-0.546 + 0.199) = 0.347 kN Fyb3 = - (Fr 23 + Ft 43) = -(0.199-0.546) = 0.347 kN Fb 0.347 0.347 0.491kN 2 2 3 The resultant shaft reaction is Forces on gears Forces on helical gears
公 上游久通大学 Forces on gears Forces on helical gears 针齿圆挂齿轮传动受力分析 B-B. B--B. 国上清通大学 Forces on gears Tangential component: F,=2T1/d1 Radial component: F=F tga cos B Axial component: F。=F,gB 1 f Total normal force: F=F I cos an cos B B-Helix angle a-Normal pressure angle 法向压力角 a-Transverse pressure angle端面压力角 12
12 Forces on gears Forces on helical gears Tangential component: 1 1 Ft 2T / d Fr Ft tg n / cos Fa Ft tg / cos cos Fn Ft n αn—Normal pressure angle 法向压力角 αt —Transverse pressure angle 端面压力角 β—Helix angle Radial component: Axial component: Total normal force: Forces on gears
图 上酒短大峰 Forces on gears f=-f2F1=-f2F1=-f2 directions:FF same as spur gears Fa-depends on directions of rotating and helix angle n2 2 《分 《心 1 n 图上涤夫廷大学 Forces on gears ▣Forces on bevel gears c Pitch angle tany=NG tanT =NP Cone distance A。 Uniform Pitch angle clearance -Pitch diameter D Back-cone radius. 13
13 2 1 Ft1 Ft 2 Fr 1 Fr 2 directions：Ft 、Fr same as spur gears Fa — depends on directions of rotating and helix angle Fa1 Fa 2 n2 n1 Fr2 Fr1 Ft2 Ft1 n1 n2 Fa1 Fa2 Forces on gears Forces on bevel gears Forces on gears