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Unformatted text preview: SOLUTION (19.1)
Known: The angle of wrap on the motor pulley is 1600, the slack—side tension is known. The belt friction factor is known. Centrifugal force is to be assumed
negligible.
Find: Determine the torque capacity of the motor pulley. Schematic and Given Data: Slack Side
P2 = N Assumptions:
1. The coefﬁcient of friction is constant throughout the angle of wrap.
2. The capacity of the belt drive is determined by the capacity of the small pulley. 3. The belt withstands the load.
4. Centrifugal force is negligible. Analysis: 1. For (i) = 1600 = 2.79 rad, P1/P2 = e“? = e(03)(279) = 2.31 2. For «1; = 1600, T =(P1— P2) r = (2.31 P2  P2) r =1.31P2 r 3. T: 1.31 P2r= 1.31(40 N)(50 mm) =2620 Nomm I Comment: If the initial angle of wrap had been 1500 rather than 160°, the torque capacity would be 2388 Nomm; i.e., the less angle of wrap on the driving pulley the
lower the torque capacity of the pulley. 19—1 SOLUTION (19.2) Known: The angle of wrap on a pulley increases from 1600 to 2000 without change in
slack—side tension. Find: Determine the percentage increase in torque capacity of the pulley. Schematic and Given Data:
Slack Side Slack Side Assumptions: 1 . The coefﬁcient of friction is constant throughout the angle of wrap. 2. The capacity of the belt drive is determined by the capacity of the small pulley.
3. The belt withstands the load. 4. Centrifugal force is negligible. Analysis: 1. For i) = 1600 = 2.79 rad, P1/P2 = em) = e(0.3)(2.79) = 2,31 2. For (1) = 2000 = 3.49 rad, P1/P2 = ef‘l’ = e(03)(349) = 2.85 3. For (I) = 160°, T = (P1 — P2) r = (2.31 P2 — P2) r = 1.31 P2 r
4. For (1) = 200°, T = (P1 — P2) r = (2.85 P2  P2) r = 1.85 P2 r 5. Increased torque capacity = = 0.41 = 41% I 192 Comment: If the initial angle of wrap had been 1500 rather than 1600, then the
increase in torque capacity would have been 55% rather than 41%. SOLUTION (19.3)
Known: The parameters c, r1, r2, and 0c are known for two pulleys and a belt. Find: Develop an equation for the belt length, L, as a function of c, r1, IQ, and or. Schematic and Given Data:
 Parallel Lines Assumptions:
1. The effect of gravity on the belt shape is negligible.
2. The belt has sufﬁcient tension to prevent sagging of the top and bottom belt strands. Analysis:
1. L=2ll +12+l3 11 = 0 cos 0:
12 = 2M4 lsggoza)
13= ml—lgggoz“) 180 + 20:) + 2m1(180  2a) 2. L=ZCCOSOL+27CI 360 360 Thus, L = 20 cos (X+—9% [r2{90 + or) + r1(90  00] I 193 SOLUTION (19.4)
Known: Figure Pl9.4 suggests an approximation length ABCD that will be equal to
half the length of the belt. Find: From the half length approximation, develop an equation relating center
distance, 0, belt length, L, and pulley radii, r1, and r2. Schematic and Given Data: Assumptions: 1. The effect of gravity on the belt shape is negligible. 2. The belt has sufﬁcient tension to prevent sagging of the top and bottom belt
strands. Analysis:
1. L/2=BC+AB+CD 2. BC = 1/ c2 + (rzrl)2 AB=%7U1 CD = %TCI2
3. %=VC2+(r2r1)z+—é—Ttr1+lm‘2 Thus, c2 = 51—2 — 2[—I:1r(r1 + r2)T+z1?1t2(r1 + 12)2  (r2 — r1)2
[L2  mm, + r2) + 1:2(1r1 + r92] — (r2  r1)2 _1_
4
ﬁ—[L «(n + r2>]2  (r1  r2)2 or c2 or c2 19—4 SOLUTION (19.5)
Known: The angles of wrap for two pulleys and a belt are given. Each pulley has a known radius. Find: Derive an equation relating 0t, r1, I2, and 0. Schematic and Given Data: Parallel Lines Assumptions:
1. The effect of gravity on the belt shape is negligible.
2. The belt has sufﬁcient tension to prevent sagging of the top and bottom belt strands.
1'2  I} C I Analysis: From inspection of the drawing, sin 05 = Comment: For r1 = 7 mm, r2 = 17 mm and c = 34 mm as shown in the ﬁgure,
on = sin'1 (10/34) = 17.10. The angle of wrap on the smaller pulley is 145.80. SOLUTION (19.6)
Known: A motor of given horsepower and speed drives an input pulley of known diameter and angle of wrap. The size 5 V belts have a known unit weight and angle [3.
The maximum belt tension is 1501b and the coefﬁcient of friction is 0.20. Find: Determine the number of belts required. 19—5 Schematic and Given Data: Multiple V—belt, 13 = 18 0, size 5V
Unit weight = 0.012 lb/in. Power input = 25 hp Pmax = P1 = 150 lb Number of belts = ? Assumptions: 1. The maximum tension in the belt is limited to 150 lb.
2. The coefﬁcient of friction will be at least 0.20.
3. Power is shared equally by each belt. Analysis:
1. We ﬁrst calculate terms in Eq. (19.3a): with Eq. (19.2), Pc = mV2 where V = 3.7 (n) 1773? = 339 ill/sec 0012 2:
Pc— 386 (339) 3.57 lb e<f¢/ sin B) = e((O.2)(2.88)/ sin 5) = 6.45
2. Substituting in Eq. (19.3a) and solving for P2: 150— 3.57 _ _ _
————P2_3.57 _ 6.45 or 146.4 _ 6.45 P2 23.0 Hence, P2 = 26.3 lb
3. From Eq. (18.24), T = (P1  P2) 1' = (150 — 26.3) = 229 lb in. 1750 (229)
5252 (12) 196 4. From Eq. (1.3), w per bolt = = 6.36 hp/belt 5. For 25 hp, £6— = 3.93, and 4 belts are required. I Comment: If a 30 hp motor was used then 5 belts would be required. As more and
more belts are needed however, the effects of misalignment of the shafts (and
consequent unequal sharing of the load) becomes important. SOLUTION (19.7) Known: A pulley of given diameter drives a ﬂat drum of given diameter. The center
distance is known. Find: Determine the sliplimited capacity of the pulley relative to that of the drum. Schematic and Given Data: Assumptions:
1. The slack side of the belt does not sag.
2. The friction coefﬁcient is uniform and equal to 0.25 for both the pulley and drum. Analysis:  =T2'l‘1z6020. =147o
1. FromProb. 19.2, smoc C 120 ,0: 9. 2. e1 (pulley) = 180°  2(19.47°) = 141.060 = 2.462 rad (b2 (drum) = 1800 + 2(19.47°) = 218.940 = 3.821 rad
3. The drum torque capacity: 197 191/132 = em = e<025) (3821) = 2.6, hence, P1 = 2.60 P2
Torque = (P1 — P2) r = 1.60 P2 (60) = 96 P2 4. The pulley torque capacity:
131/132 2 e(f¢/sin B) = e(0.25) (2.462/sin 18°) = 7.33 Torque = (P1  P2) r = 6.33 P2 (20) = 126.6 P2 5. The pulley has = 32% more capacity than the drum. I Comment: The above calculations used f = 0.25. SOLUTION (19.8) Known: A pulley of given rotational speed, radius, and angle of wrap drives a Vbelt
with known friction, weight, and maximum tension. Find: Determine the maximum power transmitted by the pulley. Schematic and Given Data: n = 4000 rpm
r = 100 mm
pulley radius q)=170O \ Belt Maximum
Tension = 1300 N Belt Unit
Weight: 1.75 N/m Assumptions:
1. The friction coefﬁcient is uniform throughout the contact area.
2. The effect of centrifugal force is important. Analysis:
, 1.75 N/m 4000 2  2 _ _______ _. _
1. From Eq. (19.2), Fc _ m v _(9.81m/Sz)( 60 0.21: m/s ) _ 313 N
2. From Eq. (19.3), ____1%00 :9) g3 = e(02/Sin 18°><170w180> = 6.82. Hence, P2 = 458 N 2 _
3. T =(P1 P2) r = (1300 — 458) (0.100) = 84 Nm 4000 (84) 4. From Eq. (1.2), W: = 35.18 kW I 9549 19—8 SOLUTION (19.9) Known: A pulley of given rotational speed, radius, and angle of wrap drives a V—belt
with known friction, weight, and maximum tension but (a) two V—belts are needed, (b)
a single V—belt with twice the maximum tensile capacity is used. Find: Determine the power transmitted by the pulley for cases (a) and (b). Schematic and Given Data: Case (a) Two Vbelts, Case (b) One V—belt with
twice the cross sectional area n = 4000 rpm
r = 100 mm
pulley radius ¢=170° \ Belt Maximum
Tension = 1300 N Belt Unit
Weight = 1.75 N/m Assumptions: 1. If two belts are used, each will have the same capacity and the load will be shared
equally between them. 2. If a belt with twice the cross section is used, there will be no loss in the tensile
strength capacity because of a size effect. Analysis:
1. For case (a) each of the two identical belts could transmit 35 kW, hence total
power capacity is 70 kW. I 2. For case (b) doubling the section would double m', giving PC = 626 N.
Doubling PC and P1 with no change in eWSin B would double P2. Thus, power
capacity is doubled in cases (a) and (b). I SOLUTION (19.10)
Known: A pulley of given rotational speed, diameter, and angle of wrap drives a V—
belt with known friction, weight, and maximum tension. Find: Determine the maximum power transmitted by the smaller pulley. 199 Schematic and Given Data: 11 = 3500 rpm
(11 = 6 in.
pulley diameter
4) = 1700 Driver pulley
Driven pulley \ Belt Maximum
Tension = 250 lb Belt Unit
Weight = 0.012 lb/in. Assumptions: 1. The friction coefﬁcient is uniform throughout the contact area.
2. The effect of centrifugal force is important. Analysis: 1. F E _ 19.2 , p ._._ IV2 __: 0.012 lb/in. 3500 .  . 2 2
mm q ( ) C m ((32.2 ft/sz)(12in./ft))( 60 6“ "1/5)
= 37.5 lb Pl'PC
PzPc 2. From Eq. (19.3), = ef¢ where f = i— for the Vbelt
sinB 250 — 37.5 = e(0.2/sir118°)(1701t/180) = 632
P2  37.5 P2 = 68.7 lb
3. T = (P1 — P2) r = (250 68.7) (3): 544 lb in. 4. From Eq. (1.3) W = “T lb'ft = = 30.2 hp I 5252 5252 SOLUTION (19.11) Known: A V—belt pulley of given diameter and rotational speed drives a second pulley at known rotational speed. The V—belt has a given unit weight and coefﬁcient of
friction. 1910 Find: (a) Determine values of P1 and P2. (b) Determine the loads applied by the belt to each shaft. (c) Determine initial belt tension when the drive is not operating. (d) Determine values of P1 and P2 when the drive is operating at normal speed but
transmitting only 6 kW. Schematic and Given Data: Single Vbelt, = 18 ° Unit weight = .2 N/m Power transmitted = 12 kW Initial belt tension just adequate to prevent slippage Driving pulley _ n = 1750 rpm DnVeD Puney 1:020 n= 1050 rpm
f = 0.20 Assumption: Initial belt tension is marginally adequate to prevent slippage. Analysis:
(a) P1 and P2
1. The larger pulley diameter = 180 = 300 mm 2. From Prob. 19.5, sin a = I25“ = 2496393 : 01 = 8.630 3. For smaller pulley, o = 1800  201 : (1) = 162.70 = 2.84 rad
4. From Eq. (192): PC: mo)er 2
PC = (Engzlngd. (.090 m)2 = 61.0 N
. S 5. From Eq. (19.3): 9% = e<<020/ sin 18°)<284>) = 6.285 ~—(1)
2 _ . 6. From Eq. (1.2): T = 9549 W = W = 65.48 N m
n 1750 1911 T = (P1 — P2) r : 65.48 = (P1 — P2) 0.090; hence, P1  P2 = 727.54 N ———(2)
Simultaneous solution of equation (1) and equation (2) gives P2 = 198.7, P1 =
926.2. Rounding off: P1 = 926 N, P2 = 199 N II 00\] (b) Shaft loads 1. Driving shaft torque,
T = (926.2 N —l98.7 N) (0.090 m) = 65.5 N In I
(which checks with calculated torque of 65.48 N m) 2. Driven shaft torque: T = (926.2 N ~198.7 N) (0.150 m)
T = 109.2 N m I 3. Radial load = 4/ (P1 + P2 cos 2002 + (F2 sin 2002 / 2 o 2
= (926.2 +1987 C0817.260) + (198.7 Sin 17.26) = 1117.6 N 4. Radial load applied to each shaft = 1118 N I
5. Assume P1 + P2 remains = 926.2 + 198.7 = 1124.9 N (c) Initial belt tension (no rotation) lel2£ziﬁ=5625N I (d) E; and P; for 6 kW 1. The torque is reduced to 65248 = 32.74 N m Hence, (P1 a P2) (0.090 m) = 32.74 N m; P1  P2 = 363.78 N
2. Assuming P1 + P2 remains = 1124.9, P1 = 744 N, P2 = 381 N II SOLUTION (19.12)
Known: A V—belt pulley of known diameter and rotational speed transmits a given
power to a driven pulley of known diameter. This is accomplished with a belt of speciﬁed unit weight, angle and coefﬁcient of friction. Find: (a) Determine values of P1 and P2. (b) Determine the loads applied by the belt to each shaft. (c) Determine values of P1 and P2 if the power is reduced to a known value. 19—12 Schematic and Given Data: Case (a) and (b) Power = 12 hp,
Case (0) Power = 3 hp SingleVbelt, 5: 18° Unit weight = 0.012 lb/in. Driven pulley
12 in.dja.
n = ? rpm Driving pulley f = 0.20
6 in.dia. n = 1750 rpm f = 0.20 Assumption: The initial belt tension is just adequate to prevent slippage (given). Analysis:
(a) P_ and P_2 for 12 hp 1. From Prob. 19.4, sin CL = r232 = §%. HCHCG, 01 = 8.630 2. For the smaller pulley, q) = 1800 — 20L 2 162.70 = 2.84 rad
3. From Eq. (19.2), PC = m, (oer ‘2
pC = (0.012 1b/in.)(1750 X 2" 20 (3 in.)2 = 9.401b 386 in./s2 60 S .1
P1— 9.40 . o
= (0.20/ 18 : ..___..
4. From Eq. (19.321), PT 940 e sm 6.285 (1)
. From Eq. (1.3), r = 525,12 W = 12% = 36.0 lb ft = 432.21b in.
6. T = (P1  P2) r; 432 = (P1 — P2) (3), hence P1 — P2 = 144.05 lb (2) 7. Simultaneous solution of equation (1) and equation (2) gives:
P2 = 36.671b (37 lb) and P1 = 180.72 lb (181 lb) . (b) Shaft loads
1. Driving shaft torque = T = (180.72  36.67) (3/12) = 36 lb ft
2. Driven shaft torque = T = (180.72 — 36.67) (6/12) = 72 lb ft 1913 3. Radial load = «/ (P1 + P2 cos 2002 + (P2 sin 2002 o 2 o 2
= ’V (180.72 + 36.67 cos 17.26 ) + (36.67 sin 17.26 ) = 2161b (applied to each shaft) I (c) P_ and P2 for 3 hp 1. Torque is reduced to 4352 = 216.1 lb in. Hence, (P1  P2) (3) = 216.]; P1 — P2 = 72.0 lb 2. With the assumption that average tension remains unchanged; i.e., no adjustments
are made, we have P1 + P2 = 180.72 + 36.67 = 217.39 lb 3. Simultaneous solution gives: P1 = 144.7 lb, P2 = 72.7 lb, rounding off, P1 = 145
lb, P2 = 73 1b I SOLUTION (19.13D)
Known: A web site address is given as http://www.grainger.com. Find: Select an A—type Vbelt with a length of 32 in. List the manufacturer,
description, and price. Analysis: A product search of the web site gives: Item #: 1A095 Mfg. Name: Browning Description: A Type VBelt—32", 12" Top Width—5/16" Thick—RMA #A3O
Price: $8.22 SOLUTION (19.14D)
Known: A web site address is given as http://www.grainger.c0m Find: Select an ANSI No. 40 standard single riveted steel roller chain. List the
manufacturer, description, and price. Analysis: A product search of the web site produced: Item #: 2W093 Mfg. Name: U.S. Tsubaki Description: ChainANSI#40—10' Long, Standard Single Riveted RollerSteel
Price: $22.16 SOLUTION (19.15) Known: A ﬂuid coupling connects an electric motor to a driven machine. The motor
and coupling curves are given. Find: (a) Estimate the speed of the machine input shaft and the percent of motor
output power converted into heat in the coupling. (b) Estimate the slip during
overload operation. (c) Determine the motor rpm at overload when the machine stops
and the percent of motor output power converted into heat for this condition. 19—14 Schematic and Given Data: Fluid Coupling 
performance curves 
Fig. 19.11 Electric Motor
n = 1780 rpm
55% rated power Driven Machine Assumptions: 1. The ﬂuid coupling is the type shown in Fig. 19.10. 2. The motor and coupling performance curves are given in Fig. 19.11. 3. The rated motor output corresponds to 1750 rpm, 100% rated torque. Analysis: (a) The 1780 motor rpm corresponds to approximately 2% slip. Machine input rpm
is 1780 (0.98) = 1744 rpm. I
As assumed, 100% of motor torque reaches the machine, hence 2% of the motor
power is converted to heat. I (b) From Fig. 19.11, at 100% rated torque, slip = 3.5%. I (c) From Fig. 19.11, at maximum coupling torque, the stall speed is approximately 775 rpm.
At this point, the motor delivers 168% of the rated torque at 775/ 1750 = 44% rated speed. Hence power is (1.68) (0.44) = (0.744) times rated power, or
approximately 74% of rated power is converted into heat. I SOLUTION (19.16)
Known: A ﬂuid couplingconnects an electric motor to a driven machine. The motor and coupling curves are given. Find: Determine the factor by which the ﬂuid coupling diameter could be reduced if
the (i) power required is reduced by half (ii) the rotational speed is doubled. 1915 Schematic and Given Data: Fluid Coupling —
Electric Motor 10 kW size
n = 1750 rpm
Power = 10 kW Driven Machine Fluid Coupling 
Electric Motor 5 kW size
n = 1750 rpm
Power = 5 kW Driven Machine Fluid Coupling Electric Motor
11 = 3500 rpm
Power = 5 kW Driven Machine Assumption: The ﬂuid couplings are geometrically similar (given).
Analysis: 1. From Eq. (19.7), T cc mZDS. Hence, power oc (1)3135. 2. If power is halved, D5 can be halved, giving DEW = 0.5 DEM . 5 I
3 . Dnew = Dold = Dold
pOWel' 4. If power is halved and a) is doubled, then with D cc 3 (D
5 Q;
Dnew = Dold 23 = D and I)new = 0.574 Dold 7 19—16 SOLUTION (19.17) Known: A hydrodynamic torque converter is used to provide a known torque
multiplication of the known torque of a motor. Find: Estimate the torque applied to the oneway clutch. Schematic and Given Data: Hydrodynamic Torque
Convener — provides a
2.4 torque multiplication Driving Motor
Torque = 100 N In Driven Machine Assumption: The hydrodynamic torque converter is like the one represented in Fig.
1 9. 1 3 . Analysis:
1. From Eq. (19.9), Ti + To + T; = 0; where Tr is the reaction torque applied by the one—way clutch.
2. With To = —2.4 Ti; we have Ti + (2.4 Ti) + T; = O. 3. Solving for Tr gives Tr = 1.4 Ti. I 1917 ...
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This note was uploaded on 09/29/2009 for the course ME 530.230 taught by Professor Katz during the Spring '09 term at Johns Hopkins.
 Spring '09
 KATZ

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