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