圆锥圆柱齿轮减速器说明书,大学课程设计论文
Vz10.711 28
==0.199 100100
按8级精度查图得动载系数Kv=1.017 齿宽 b= aa=0.35×155=54.25mm
取:b2=55 mm b1 60mm
b55==0.77 d171.736
查图知齿轮相对于轴承非对称布置,两轮均为软齿面,得:K =1.11 查表得:K =1.2
载荷系数K=KAKvK K =1.25×1.017×1.11×1.2=1.6933 计算重合度 a, 以计算z :
da1=d1+2mn=71.736+2×2.5=76.736mm da2=d2+2mn =238.266+2×2.5=243.266mm
t=arctan(tan n/cosβ)
= arctan(tan200/0.9758)=20.4550
db1=d1cos t=71.736×0.94=67.41mm
db2=d2cos t=238.266×0.94=223.896mm
at1=arccos at2=arccos =
dd167.41
= arccos =28.5424
76.736da1
dd2223.896
= arccos =23.0190
243.266da2
1
[z1(tan at1-tan t)+z2(tan at2-tan t)] 2 1
tan20.455 +93× tan23.019 tan20.455 ] =[28× tan28.5424
2
=1.53