ch10 - SOLUTION(10.4 Known A double-threaded Acme stub...

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Unformatted text preview: SOLUTION (10.4) Known: A double-threaded Acme stub screw of known major diameter is used in a jack having a plain thrust collar of known mean diameter. Coefficients of running friction are estimated as 0.10 for the collar and 0.11 for the screw. Find: (a) Determine the pitch, lead, thread depth, mean pitch diameter, and helix angle of the screw. (b) Estimate the starting torque for raising and for lowering a 5000 lb load. (0) If the screw is lifting a 50001bload at the rate of 4 ftfmin, determine the screw rpm. Also determine the efficiency of the jack under this steady-state condition. (cl) Determine if the screw will overhaul if a ball thrust bearin g (of negligible friction) were used in place of the plain thrust collar. Schematic and Given Data: Double-threaded Acme stub screw (1 a 2 in. dc = 2.75 in. f6 = 0.10 Assumptions: 1. The starting friction is about 13 higher than running friction. 2. The screw is not exposed to vibration. Analysis: 1. From Table 10.3, there are 4 threads per inch. p = U4 = 0.25 in. Because of the double-threaded screw, L = 2p = 0.50 in. From Fig. 10.4 (0). Threaded depth = 0.3 p = 0.075 in. dm = ti — 0.3p = 1.925 in. From Eq. (10.1), A. = tan—1 L =tarr1 ( 0'5 )= 4.73° “din 1.9253 2. For starting, increase the coefficients of friction by 113: f}; = 0.133,f = 0.147 From Eq. (10.6), . c an =tan‘l (tan 0; cos )1) = tan‘l (tan 14.5 cos 4.73 ) = 14.45° From Eq. (10.4), T— de mam —Loosun +chc 2 IfidmCOSCIn+fL+ 2 _.3500(1 925) 0.1473(1925) + 0. 5003 14. 45° +3500(1).133x175) 2 Qflmmfiohfififlsfigfi 3 T = 799.9 + 640.1 = 1440 lb in. to raise the load I From Eq. ( 10.5), ___ 3500(1.925) 0.1471(1.92_5)+ 0.5005 14.45" + 3500(0.133)(2.75) 2 fi(1.925)oos 14.45°-0.147(0.5) 2 T = 230.0 + 640.1 = 8?0.1 lb 111. to lower the load I 4(12)in.lmin__ 01—511m— 96er ' 4. From Eq. (10.4), Withfc = 0.1,f = 0.11 T _ 3500(1.925) 0.113(1.925)+0.5003 141.45” + 3500(0.I)(2.?5) _ 2 “(1.923003 14.450 - 0.11(0.5) 2 T = 667.1 + 481.3 = 1148.4 lb in. 5. From Eq. (1.3), - 4 Wout- (33332) = O. 424 hp Therefore, efficienCy- - “Iii—154:0 24: -24% I 6. From Eq. (10.7), the screw is self-locking if L cos an _ 0.5(cos 14. 45°) faw— 310.925) =0‘08 Thus, if f = 0.11, the screw is self-locking and not overhauling. SOLUTION (10.9) Known: A jack uses a single square-thread screw to raise a lmown load. The major diameter and pitch of the screw and the thrust collar mean diameter are known. Running friction coefficients are estimated. Find: (a) Determine the thread depth and helix angle. (b) Estimate the starting torque for raising and lowering the load. (c) Estimate the efficiency of the jack for raising the load. (cl) Estimate the power required to drive the screw at a constant 1 revolution per second. Schematic and Given Data: Single square-thread screw (1 = 36 mm p=6mm do = 80 mm f = 0.15 i"c =0.12 Assumption: The starting friction is about 1:3 higher than running friction. Analysis: 1. From Fig. 10.4(c), Thread depth = pl2 = 62’2 = 3 mm I 2. From Eq. (10.1), n rtdm Wheredm d 2 33mm 6 35(33) a=tan-1 =3.31° I 3. For starting, increase the coefficients of friction by H3. then f: 0.20,fc = 0.16 SOLUTION (10.7) Known: A square-threaded, single thread power screw is used to raise a known load. The screw has a mean diameter of 1 in. and four threads per inch. The collar mean diameter is 1.75 in. The coefficient of friction is estimated as 0.1 for both the thread and the collar. Find: (a) Determine the major diameter of the screw. (b) Estimate the screw torque required to raise the load. (c) If collar friction is eliminated, determine the minimum value of thread coefficient of friction needed to prevent the screw from overhauling. Schematic and Given Data: Assumption: The screw is not exposed to vibration. Analysis: 1. From Fig. 10.4(8), d=dm+g=1+u2i=rtzsm l 2. From Eq. (10.43). T = wan, [mam + L]+ wt. dc 2 «am-art: 2 (0.1m1)+0.2§ +(i3750](0i)(1.75) 10(1) - (0.1)(025) 2 T = 1245 lb in. + 1203 lb in. = 2448 lb in. I 3. From Eq. (10.751), the screw is self-locking if L — ”-25 = 0.03 f * Mahdi f a 0.08 I _(13750x1) ‘ 2 Therefore, the minimum value of thread coefficient of friction needed to prevent the screw from overhauling is 0.08. From Eq. (10.4a), [mm + L)+ ch c1c mam -.fL 2 Wd T=—m 2 (50,000)(0. 033) jth + W) 2 __ 35(0. 033) — (0.20)(0.006) 2 T = 215 +320 T = 535 N-m to raise the load I From Eq. (10.521), 2 de [fndm — L]+ ch dc 2 ndm +_f L 2 _ (so,ooo)(o. 033) 0.20::(0. 033) — 0- 006 + (50,000)(g16)(o.0§bml ‘ 2 “(0.033) + (0.20)(0.006) 2 T = 116 + 320 T = 436 N-m to lower the load I From Eq. (10.4a), withf = 0.15,fc = 0.12, 31(0033) — (0. 15)(0.006) (50,000)(0.033) r = —---—— (50.000)(0. 12)(0.080) 2 + _____mfi___u_ 2 T: 173+240=4l3 N'm Work input to the screw during one revolution = ZJET = 2313(413) = 2595 N'm Work output during one revolution = W' p = (50,000)(0.006) = 300 N°rn 10. Check: Torque during load raising with f = f c = 0 0 + 0.006 1:01.033) — 0 = (50,000)(0.033) T 2 +0 T = 47.8 N'm Twi zero ri ion Efficiency = ...C__.l_*}____L‘E_} _. 47mg Tiaclud) - 7113' = 11.6% 11. Check (partial): Torque during load raising if collar friction is eliminated = 173 N'm Efficiency (screw only) = % = 28% 12. From Eq. (1.2), — —_—"T — ——(6O)(4l3) = 2.6 kW _ 9549 _ 9549 SOLUTION (10.10) Known: An ordinary C-clamp uses a 1l2 in. Acme thread and a collar of SIS in. mean diameter. Find: Estimate the force required at the end of a S-in. handle to develop a 200 lb clamping force. ' Schematic and Given Data: Assumptions: 1. Coefficients of running friction are estimated as 0.15 for both the collar and the screw. 2. The screw has a single thread. Analysis: 1. From section 10.3.1, and considering that service conditions may be conducive to relatively high friction, estimatef =f¢ as 0.15 (for running friction). 2. From Table 10.3, p = 0.1 in., and with a single thread, L = 0.1 in. 3. From Fig. 10.4(a), dm=d — g = 0.5 — 0.05 = 0.45 in. 01:14.5“J 4. FromEq.(10.1). _ -1_L_= -l 0-1 _ o l—tan J'tdm tan 35(0.45)"4‘05 5. From Eq. (10.6), on = tan"1 (tan or cos 3..) = tan'1 (tan 14.5o cos 4.05“) = 14.470 (Note: with it u 4°, it is obvious that an 2: a and well within the accuracy of assumed friction coefficients) From Eq. (10.4), T = de (fmdm + Lcosan) + chdc 2 admcos un—fL 2 ___ (200)(0.45_) ( (0. 15)u(0.45) + 0.1(cos 14. 47°) + _{_2_0(_})(0.15}(0.6252 2 Ic(0.45)(cos 14. 47°) _ (0.15)(o. 1) 2 T = 10.27 + 9.37 = 19.64 lb in. Use T =s 20 lb in. At the end of a 5-in. handle, the clamping force required as % = 4 lb I SOL TION( .11) . Known: Two identical 3 in. major diameter screws (single threaded) with modified square threads are used to raise and lower a SCI-ton sluice gate of a dam. An estimated friction coefficient is only 0.1 for the screw. Because of gate friction, each screw must provide a lifting force of 26 tons. Find: Detennine the power required to drive each screw when the gate is being raised at the rate of 3 fti’min. Also calculate the corresponding rntatin g speed of the screws. Schematic and Given Data: inn-inn diameter power screw Illlllllflllflll- f=0.l g “ltd-lidsfl'fifii‘m '_ . . recast-time - 50 ton sluice gate .2- =I I: I: I——r.l I=I I=l I: I: I: D :1 I: =- I: El El l:l :- l=l :- =- m :3 I... H '"L'fig-rlrr‘r Hag '5 I &%T;HT2&$2£% . 15 'L 'l' at: a; c e a -: :2a..a..i it Assumption: Collar friction can be neglected. Analysis: 1. From Table 10.3, p = L = 1:11.75) = 0.571 in. 2. From Fig. 10.4(d), a... =d _ g: 3 , ___0.5£7l = 2.71 in. and a = 2.50 3. Since cos OL = 0.999, use Eq. (10.4a): T = de (fmdm + L)+ ch dB 2 m1... —-.fL 2 = (52,000)(2.71) ( (0.1)n(2.71) +0571 ) + 0 2 :c(2.71) — (0.1)(0571) T = 11.851 lb in. = 988 lb ft (during load raising) To raise the gate 36 inJrnin with L = 0.571 in. requires 3610.571 = 63.05 as 63 rpm From Fig. (1.3). - 63. W: (—33%: 11.9hp say 12 hp Therefore, 12 hp are required to drive each screw. Check: Work output per gate — 52,000 lb (3 ftfmin) = 156,000 lb flfmin = 4.73 hp — 4;: -1M11_= 0 A43“ :rcdm tan :c(2.71) 3'84 From Fig. 10.8, Efficiency .. 40 % Thus, WWW .. % = 11.83 hp ...
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