上海交通大学：《Design & Manufacturing II and Project》课程教学资源（作业）01_Power Screw Solutions

ME 357 Power Screw Homework Solutions 1.11-1 a)and b) c)Make a scale drawing of a)and b).Draw a section 50mm long,mark out all the dimensions. Laj Thread Depth 2.5mm Width 2.5mm d.-25-2*1.25=22.5mm d,=25-2*2.5=20mm Lead=Pitch=5mm (b) --5rm:- Depth 2.5mm Width at pitch line 2.5mm d =22.5mm d,=20mm Lead=Pitch=5mm

ME 357 Power Screw Homework Solutions 1. 11-1 a) and b) c) Make a scale drawing of a) and b). Draw a section 50mm long, mark out all the dimensions

2.11-6 Sol. (a)Revolution speed of motor 1720r/min Gears ratio i=75:1 Revolution speed of threads 1720/75=23r/min Pitch of screw p=1=0.5 in Speed of press head 0.5*23=11.5in/min (b)press rated load 50001b=2F dn=D-2=3-0.25=2.75im f=0.05 f=0.06 de=5in Power of motor H= Tn 63000×0.95 1=+人-a号2 2 =225016x2.75m05in+元x0.05x275 Sinxsec14.5+2501bx0.06×5im, 2 π×2.75im-0.05x0.5mxsc145+ 2 =1505.21b-in H=1505.21b-imx23pm） 63000×0.95 =0.578hp

2. 11-6 Sol. (a) Revolution speed of motor 1720r/min Gears ratio i =75:1 Revolution speed of threads 1720/75=23r/min Pitch of screw p=l=0.5 in Speed of press head 0.5*23=11.5 in/min (b) press rated load 5000lb=2F 3 0.25 2.75 2 m p d  D     in 0.05 0.06 5 c c f f d in    Power of motor 63000 0.95 Tn H   sec 2( ) 2( ( ) ) 2 sec 2 2500 2.75 0.5 0.05 2.75 sec14.5 2500 0.06 5 2[ ( ) ] 2 2.75 0.05 0.5 sec14.5 2 1505.2 m m c c screw bearing m Fd l fd Ff d T T T d fl lb in in in lb in in in lb in                              (1505.2 23 ) 63000 0.95 0.578 lb in rpm H hp     

3.11-8 5 Known:Acme thread ~6 7 f=015,/,=0.15.d=i6m,2=29 Question:FindFfor applied force of6b,2.75 in.from the scre centerie. SoL P=L=1/6=0.1667” seca= 1=1.033 cosa d=d-05*p=1 =0.5417 812 { =F0,541701667+x*0.15*0.54171.033 =0.0696F 20.5417-0.15*0.1667*1.033 0.15*7 T=44=F6=0.0328F) 2 ∴.Ta=T+T-0.102F T=F●r=6b*2.75=16.51b-m 0.102F=16.5bim F-161.8b

3. 11-8 f f F

4.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 diameted-卫_`_⊥` =0.5417° 2812 1 Lead angle:元=tan-'(）=tan(6 πd （x03417=5597 Comparing tan Acosa and frictional coefficient f to check self-locking? tan Acosa=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 lb*in pitch=lead=1/6 in 1 e=F 161.81b× 6-=26% 2πT2π×16.5lb·im

4. 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      