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Unformatted text preview: SOLUTION (10.4) Known: A doublethreaded 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 steadystate 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: Doublethreaded 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 doublethreaded 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 selflocking if L cos an _ 0.5(cos 14. 45°) faw— 310.925) =0‘08 Thus, if f = 0.11, the screw is selflocking and not overhauling. SOLUTION (10.9) Known: A jack uses a single squarethread 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 squarethread 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=tan1 =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 squarethreaded, 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 «amart: 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 selflocking 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 Nm 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 Nm 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 Cclamp uses a 1l2 in. Acme thread and a collar of SIS in. mean
diameter. Find: Estimate the force required at the end of a Sin. 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 01 _ 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 5in. 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 SCIton 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: inninn diameter
power screw Illlllllﬂllﬂll f=0.l g “ltdlidsﬂ'ﬁﬁi‘m '_ . .
recasttime  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'ﬁgrlrr‘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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 Spring '08
 ZHOU

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