上海交通大学：《Design and Manufacturing》课程教学资源（作业）06 power screw solution

ME 357 Power Screw Homework Solutions 1.11-8 Known:Acme thread 5 ~6 7 f=0.15,/.=0.15.d,-16m,2a=29 ueio:FindFfor applied fore2om therec Sol. P=L=1/6=0.1667” seca=- 1 -=1.033 cosa dn=d-0.5*p= 51 =0.5417 812 T.-FdL+md seca 2πd.-uL seca =F05410.1667+0*0.15*0.5417*1.033 2π0.5417-0.15*0.1667*1.033 =0.0696F 015*7 I-F44=F,6-0.0328F) 2 2 .Tu=Te+T.=0.102F T=F·r=6b*2.75=16.5b-im 0.102F=16.51b-im F-161.8b

ME 357 Power Screw Homework Solutions 1. 11-8 f f F

2.Compute the lead angle for the screw of Problem 11-8.Is it self-locking? Lead=1/6 in Major diameter 5/8 in Pitch diameter=d-p=51 =0.5417 2812 1 Lead angle:2=tan-()=tan(6 πd, r×0.5417)=5597° Comparing tan cosa and frictional coefficient f to check self-locking? tan1cosa=0.098×cos14.5°=0.0949<f=0.15 It's self-locking Compute the efficiency of the screw of Problem 8-8. F=161.81 Ib T=16.5 Ib*in pitch=lead=1/6 in 1 Fl 161.8b× e= 2πT 6=26% 2π×16.5lb-in 3.solutions: dm=40-3=37mm,l=2(6)=12mm From Eq.(8-1)and Eq.(8-6) 10(37) 「12+π(0.10)(37)1,10(0.15)(60) TR= 2 π(37)-0.10(12)」 2 =38.0+45=83.0N·m Since n =V/I=48/12=4 rev/s ω=2πn=2π(4)=8πrad/s so the power is H=Tw=83.0(8π)=2086WAns

2. Compute the lead angle for the screw of Problem 11-8. Is it self-locking? Lead = 1/6 in Major diameter 5/8 in Pitch diameter= " " 5 1 " 0.5417 2 8 12 p d   Lead angle: " 1 1 " 1 6 tan ( ) tan ( ) 5.597 0.5417 m l d          Comparing tan cos   and frictional coefficient f to check self-locking? tan cos 0.098 cos14.5 0.0949     < f =0.15 It’s self-locking Compute the efficiency of the screw of Problem 8-8. F=161.81 lb T=16.5 lb*in pitch=lead=1/6 in " 1 161.8 6 26% 2 2 16.5 lb Fl e   T lb in      3. solutions: