1×(3×8×300×5)=3.15×1099911210577.121015.3?????iNN7)查教材10-23图得:K1??=0.92K2??=0.94齿轮的接触疲劳强度极限:取失效概率为1%,安全系数S=1,应用公式(10-14)得:[H?]1=SKHHN1lim1?=0.92×650=598MPaKt=1.31T=51350N?mm0.35R???Hlim1?650Mpa?Hlim2?550MpaEZ=189.8ZE=2.5K1??=0.92K2??=0.9410[H?]2=SKHHN2lim2?=0.94×550=517MPa取[H?]2和[H?]1中的较小者作为该齿轮副的接触疲劳许用应力,即[H?]=[H?]2=517MPa4.1.2.2设计计算1)试算小齿轮的分度圆直径,带入??H?中的较小值得1td??mm85.77235.05.0135.0513503.145178.1895.232???????????????2)调整小齿轮分度圆直径1圆周速度V????mmddRtm23.645.035.0185.775.0111???????????100060V11ndm?4.91m/s2当量齿轮的齿宽系数d?mmudbitR46.302/1)25/50(85.7735.02/122????????474.023.64/46.30/1d???mdb?3)计算载荷系数HK1由表10-2查得使用系数AK=12根据V=4.91m/s,7级精度查图表(图10-8)得动载系数vK=1.083直齿锥齿轮精度较低,取齿间载荷分配系数1??HK根据大齿轮两端支撑,小齿轮悬臂布置用插值法查表10-4得?HK=1.375由此,得到实际载荷系数485.11375.108.11????????HHVAHKKKKK[H?]=517MPaV=4.91m/sb=30.46mm474.0d??
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