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Chapter 16
16-1
(a) ¸1 = 0°, ¸2 = 120°, ¸a = 90°, sin ¸a = 1, a = 5in
120°
0.28pa(1.5)(6)
Eq. (16-2): Mf = sin ¸(6 - 5 cos ¸) d¸
1
0°
= 17.96pa lbf · in
120°
pa(1.5)(6)(5)
Eq. (16-3): MN = sin2 ¸ d¸ = 56.87pa lbf · in
1
0°
c = 2(5 cos 30ć%) = 8.66 in
56.87pa - 17.96pa
Eq. (16-4): F = = 4.49pa
8.66
pa = F/4.49 = 500/4.49 = 111.4 psi for cw rotation
56.87pa + 17.96pa
Eq. (16-7): 500 =
8.66
pa = 57.9 psi for ccw rotation
A maximum pressure of 111.4 psioccurs on the RH shoe for cw rotation. Ans.
(b) RH shoe:
0.28(111.4)(1.5)(6)2(cos 0ć% - cos 120ć%)
Eq. (16-6): TR = = 2530 lbf · in Ans.
1
LH shoe:
0.28(57.9)(1.5)(6)2(cos 0ć% - cos 120ć%)
Eq. (16-6): TL = = 1310 lbf · in Ans.
1
Ttotal = 2530 + 1310 = 3840 lbf · in Ans.
Fx Force vectors not to scale
(c)
Fx
F
30
F
Fy
30
y
Fy
y
Secondary
Primary Rx
shoe
shoe
Ry
R
Rx
R
x
Ry
x
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Chapter 16 397
RH shoe: Fx = 500 sin 30° = 250 lbf, Fy = 500 cos 30° = 433 lbf
ć%
120 2Ä„/3 rad
1 ¸ 1
Eqs. (16-8): A = sin2 ¸ = 0.375, B = - sin 2¸ = 1.264
2 2 4
0ć% 0
111.4(1.5)(6)
Eqs. (16-9): Rx = [0.375 - 0.28(1.264)] - 250 =-229 lbf
1
111.4(1.5)(6)
Ry = [1.264 + 0.28(0.375)] - 433 = 940 lbf
1
R = [(-229)2 + (940)2]1/2 = 967 lbf Ans.
LH shoe: Fx = 250 lbf, Fy = 433 lbf
57.9(1.5)(6)
Eqs. (16-10): Rx = [0.375 + 0.28(1.264)] - 250 = 130 lbf
1
57.9(1.5)(6)
Ry = [1.264 - 0.28(0.375)] - 433 = 171 lbf
1
R = [(130)2 + (171)2]1/2 = 215 lbf Ans.
16-2 ¸1 = 15°, ¸2 = 105°, ¸a = 90°, sin ¸a = 1, a = 5in
105°
0.28pa(1.5)(6)
Eq. (16-2): Mf = sin ¸(6 - 5 cos ¸) d¸ = 13.06pa
1
15°
105°
pa(1.5)(6)(5)
Eq. (16-3): MN = sin2 ¸ d¸ = 46.59pa
1
15°
c = 2(5 cos 30°) = 8.66 in
46.59pa - 13.06pa
Eq. (16-4): F = = 3.872pa
8.66
RH shoe:
pa = 500/3.872 = 129.1 psi on RH shoe for cw rotation Ans.
0.28(129.1)(1.5)(62)(cos 15° - cos 105°)
Eq. (16-6): TR = = 2391 lbf · in
1
LH shoe:
46.59pa + 13.06pa
500 = Ò! pa = 72.59 psi on LH shoe for ccw rotation Ans.
8.66
0.28(72.59)(1.5)(62)(cos 15° - cos 105°)
TL = = 1344 lbf · in
1
Ttotal = 2391 + 1344 = 3735 lbf · in Ans.
Comparing this result with that of Prob. 16-1, a 2.7% reduction in torque is achieved by
using 25% less braking material.
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16-3 Given: ¸1 = 0°, ¸2 = 120°, ¸a = 90°, sin ¸a = 1, a = R = 90 mm, f = 0.30,
F = 1000 N = 1kN, r = 280/2 = 140 mm, counter-clockwise rotation.
LH shoe:
f pabr a
Mf = r(1 - cos ¸2) - sin2 ¸2
sin ¸a 2
0.30pa(0.030)(0.140) 0.090
= 0.140(1 - cos 120ć%) - sin2 120°
1 2
= 0.000 222pa N · m
pabra ¸2 1
MN = - sin 2¸2
sin ¸a 2 4
pa(0.030)(0.140)(0.090) 120° Ä„ 1
= - sin 2(120°)
1 2 180 4
= 4.777(10-4) pa N · m
180ć% - ¸2
c = 2r cos = 2(0.090) cos 30ć% = 0.155 88 m
2
4.777(10-4) - 2.22(10-4)
F = 1 = pa = 1.64(10-3) pa
0.155 88
pa = 1/1.64(10-3) = 610 kPa
f pabr2(cos ¸1 - cos ¸2)
TL =
sin ¸a
0.30(610)(103)(0.030)(0.1402)
= [1 - (-0.5)]
1
= 161.4N· m Ans.
RH shoe:
Mf = 2.22(10-4) pa N · m
MN = 4.77(10-4) pa N · m
c = 0.155 88 m
4.77(10-4) + 2.22(10-4)
F = 1 = pa = 4.49(10-3) pa
0.155 88
1
pa = = 222.8kPa Ans.
4.49(10-3)
TR = (222.8/610)(161.4) = 59.0N· m Ans.
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Chapter 16 399
16-4
(a) Given: ¸1 = 10°, ¸2 = 75°, ¸a = 75°, pa = 106 Pa, f = 0.24,
b = 0.075 m (shoe width), a = 0.150 m, r = 0.200 m, d = 0.050 m, c = 0.165 m.
Some of the terms needed are evaluated as:
¸2
¸2 ¸2
¸2
1
A = r sin ¸ d¸ - a sin ¸ cos ¸ d¸ = r -cos ¸ - a sin2 ¸
¸1
2
¸1 ¸1
¸1
75°
75°
1
= 200 -cos ¸ - 150 sin2 ¸ = 77.5mm
10°
2
10°
75Ä„/180 rad
¸2
¸ 1
B = sin2 ¸ d¸ = - sin 2¸ = 0.528
2 4
¸1
10Ä„/180 rad
¸2
C = sin ¸ cos ¸ d¸ = 0.4514
¸1
Now converting to pascals and meters, we have from Eq. (16-2),
f pabr 0.24[(10)6](0.075)(0.200)
Mf = A = (0.0775) = 289 N · m
sin ¸a sin 75°
From Eq. (16-3),
pabra [(10)6](0.075)(0.200)(0.150)
MN = B = (0.528) = 1230 N · m
sin ¸a sin 75°
Finally, using Eq. (16-4), we have
MN - Mf 1230 - 289
F = = = 5.70 kN Ans.
c 165
(b) Use Eq. (16-6) for the primary shoe.
f pabr2(cos ¸1 - cos ¸2)
T =
sin ¸a
0.24[(10)6](0.075)(0.200)2(cos 10° - cos 75°)
= = 541 N · m
sin 75°
For the secondary shoe, we must first find pa.
Substituting
1230 289
MN = pa and Mf = pa into Eq. (16-7),
106 106
(1230/106) pa + (289/106) pa
5.70 = , solving gives pa = 619(10)3 Pa
165
Then
0.24[0.619(10)6](0.075)(0.200)2(cos 10° - cos 75°)
T = = 335 N · m
sin 75°
so the braking capacity is Ttotal = 2(541) + 2(335) = 1750 N · m Ans.
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(c) Primary shoes:
pabr
Rx = (C - f B) - Fx
sin ¸a
(106)(0.075)(0.200)
= [0.4514 - 0.24(0.528)](10)-3 - 5.70 =-0.658 kN
sin 75°
pabr
Ry = (B + f C) - Fy
sin ¸a
(106)(0.075)(0.200)
= [0.528 + 0.24(0.4514)](10)-3 - 0 = 9.88 kN
sin 75°
Secondary shoes:
pabr
Rx = (C + f B) - Fx
sin ¸a
[0.619(10)6](0.075)(0.200)
= [0.4514 + 0.24(0.528)](10)-3 - 5.70
sin 75°
=-0.143 kN
pabr
Ry = (B - f C) - Fy
sin ¸a
[0.619(10)6](0.075)(0.200)
= [0.528 - 0.24(0.4514)](10)-3 - 0
sin 75°
= 4.03 kN
y
Note from figure that +y for secondary shoe is opposite to
R
+y for primary shoe.
RV
Combining horizontal and vertical components, x
x
RH
RH =-0.658 - 0.143 =-0.801 kN
RV = 9.88 - 4.03 = 5.85 kN
y
R = (0.801)2 + (5.85)2
= 5.90 kN Ans.
16-5 Preliminaries: ¸1 = 45° - tan-1(150/200) = 8.13°, ¸2 = 98.13°
¸a = 90°, a = [(150)2 + (200)2]1/2 = 250 mm
98.13°
1
Eq. (16-8): A = sin2 ¸ = 0.480
8.13°
2
¸2
98.13°
Let C = sin ¸ d¸ =- cos ¸ = 1.1314
8.13°
¸1
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Chapter 16 401
Eq. (16-2):
f pabr 0.25pa(0.030)(0.150)
Mf = (rC - aA) = [0.15(1.1314) - 0.25(0.48)]
sin ¸a sin 90°
= 5.59(10-5) pa N · m
98.13Ä„/180 rad
¸ 1
Eq. (16-8): B = - sin 2¸ = 0.925
2 4
8.13Ä„/180 rad
pabra pa(0.030)(0.150)(0.250)
Eq. (16-3): MN = B = (0.925)
sin ¸a 1
= 1.0406(10-3) pa N · m
Using F = (MN - Mf )/c, we obtain
104.06 - 5.59
400 = pa or pa = 203 kPa Ans.
0.5(105)
f pabr2C 0.25(203)(103)(0.030)(0.150)2
T = = (1.1314)
sin ¸a 1
= 38.76 N · m Ans.
Ć
16-6 For +3Ãf :
f = fÅ» + 3Ãf = 0.25 + 3(0.025) = 0.325
Ć
0.325
Mf = 5.59(10-5) pa = 7.267(10-5) pa
0.25
Eq. (16-4):
104.06 - 7.267
400 = pa
105(0.500)
pa = 207 kPa
207 0.325
T = 38.75 = 51.4N· m Ans.
203 0.25
Ć
Similarly, for -3Ãf :
f = fÅ» - 3Ãf = 0.25 - 3(0.025) = 0.175
Ć
Mf = 3.913(10-5) pa
pa = 200 kPa
T = 26.7N· m Ans.
16-7 Preliminaries: ¸2 = 180° - 30° - tan-1(3/12) = 136°, ¸1 = 20° - tan-1(3/12) = 6°,
¸a = 90ć%, a = [(3)2 + (12)2]1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in.
136ć%
0.30(150)(2)(10)
Eq. (16-2): Mf = sin ¸(10 - 12.37 cos ¸) d¸
sin 90°
6°
= 12 800 lbf · in
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136°
150(2)(10)(12.37)
Eq. (16-3): MN = sin2 ¸ d¸ = 53 300 lbf · in
sin 90°
6°
LH shoe:
cL = 12 + 12 + 4 = 28 in
Now note that Mf is cw and MN is ccw. Thus,
53 300 - 12 800
FL = = 1446 lbf
28
FL 1446 lbf
4"
14
Fact 361 lbf
16"
0.30(150)(2)(10)2(cos 6° - cos 136°)
Eq. (16-6): TL = = 15 420 lbf · in
sin 90°
RH shoe:
pa pa
MN = 53 300 = 355.3pa, Mf = 12 800 = 85.3pa
150 150
On this shoe, both MN and Mf are ccw.
Also cR = (24 - 2 tan 14°) cos 14° = 22.8in
Fact = FL sin 14° = 361 lbf Ans.
FR = FL/ cos 14° = 1491 lbf
355.3 + 85.3
Thus 1491 = pa Ò! pa = 77.2 psi
22.8
0.30(77.2)(2)(10)2(cos 6° - cos 136°)
Then TR = = 7940 lbf · in
sin 90°
Ttotal = 15 420 + 7940 = 23 400 lbf · in Ans.
16-8
¸2
Mf = 2 ( f dN)(a cos ¸ - r) where dN = pbr d¸
0
¸2
= 2 f pbr (a cos ¸ - r) d¸ = 0
0
From which
¸2 ¸2
a cos ¸ d¸ = r d¸
0 0
r¸2 r(60°)(Ä„/180)
a = = = 1.209r
sin ¸2 sin 60°
Eq. (16-15)
4r sin 60°
a = = 1.170r
2(60)(Ä„/180) + sin[2(60)]
1491 lbf
F
R
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Chapter 16 403
16-9
(a) Counter-clockwise rotation, ¸2 = Ä„/4 rad, r = 13.5/2 = 6.75 in
4r sin ¸2 4(6.75) sin(Ä„/4)
a = = = 7.426 in
2¸2 + sin 2¸2 2Ä„/4 + sin(2Ä„/4)
e = 2(7.426) = 14.85 in Ans.
(b)
Fy
Fx
P Actuation
lever
Rx 3" 6.375"
Ry
Ä… = tan-1(3/14.85) = 11.4°
0.428P
2.125P
MR = 0 = 3Fx - 6.375P
tie rod
2.125P
0.428P
Fx = 2.125P
Fx = 0 =-Fx + Rx
0.428P
Rx = Fx = 2.125P
2.125P y
P
F = Fx tan 11.4ć% = 0.428P
2.125P
y
1.428P
Fy =-P - F + Ry
Ry = P + 0.428P = 1.428P
Fy Left shoe lever.
Fx
MR = 0 = 7.78Sx - 15.28Fx
15.28"
15.28
Sx = (2.125P) = 4.174P
Sx
7.78
Sy
Sy = f Sx = 0.30(4.174P)
7.78"
Rx
= 1.252P
y
Ry
Fy = 0 = Ry + Sy + F
y
Ry =-F - Sy
=-0.428P - 1.252P
=-1.68P
Fx = 0 = Rx - Sx + Fx
Rx = Sx - Fx
= 4.174P - 2.125P
= 2.049P
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0.428P 1.428P
2.125P
2.125P
1.252P
4.174P
Ans.
4.174P
1.252P
2.049P 2.049P
1.68P 2.68P
Left shoe lever Right shoe lever
(c) The direction of brake pulley rotation affects the sense of Sy, which has no effect on
the brake shoe lever moment and hence, no effect on Sx or the brake torque.
The brake shoe levers carry identical bending moments but the left lever carries a
tension while the right carries compression (column loading). The right lever is de-
signed and used as a left lever, producing interchangeable levers (identical levers). But
do not infer from these identical loadings.
16-10 r = 13.5/2 = 6.75 in, b = 7.5in, ¸2 = 45°
From Table 16-3 for a rigid, molded nonasbestos use a conservative estimate of
pa = 100 psi, f = 0.31.
In Eq. (16-16):
2¸2 + sin 2¸2 = 2(Ä„/4) + sin 2(45°) = 2.571
From Prob. 16-9 solution,
pabr
N = Sx = 4.174P = (2.571) = 1.285pabr
2
1.285
P = (100)(7.5)(6.75) = 1560 lbf Ans.
4.174
Applying Eq. (16-18) for two shoes,
T = 2af N = 2(7.426)(0.31)(4.174)(1560)
= 29 980 lbf · in Ans.
16-11 From Eq. (16-22),
pabD 90(4)(14)
P1 = = = 2520 lbf Ans.
2 2
f Ć = 0.25(Ä„)(270°/180°) = 1.178
Eq. (16-19): P2 = P1 exp(- f Ć) = 2520 exp(-1.178) = 776 lbf Ans.
( P1 - P2)D (2520 - 776)14
T = = = 12 200 lbf · in Ans.
2 2
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16-12 Given: D = 300 mm, f = 0.28, b = 80 mm, Ć = 270°, P1 = 7600 N.
f Ć = 0.28(Ą)(270ć%/180ć%) = 1.319
P2 = P1 exp(- f Ć) = 7600 exp(-1.319) = 2032 N
2P1 2(7600)
pa = = = 0.6333 N/mm2 or 633 kPa Ans.
bD 80(300)
D 300
T = ( P1 - P2) = (7600 - 2032)
2 2
= 835 200 N · mm or 835.2 N · m Ans.
16-13
125
200
P1 P2
P2 F
P1 125
275
125
Ä… = cos-1 = 51.32°
200
Ć = 270° - 51.32° = 218.7°
Ä„
f Ć = 0.30(218.7) = 1.145
180°
(125 + 275)F (125 + 275)400
P2 = = = 1280 N Ans.
125 125
P1 = P2 exp( f Ć) = 1280 exp(1.145) = 4022 N
D 250
T = ( P1 - P2) = (4022 - 1280)
2 2
= 342 750 N · mm or 343 N · m Ans.
16-14
(a)
D = 16", b = 3"
n = 200 rev/min
P1
f = 0.20, pa = 70 psi
P2
P1 P
P2
pabD 70(3)(16)
Eq. (16-22): P1 = = = 1680 lbf
2 2
f Ć = 0.20(3Ą/2) = 0.942
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Eq. (16-14): P2 = P1 exp(- f Ć) = 1680 exp(-0.942) = 655 lbf
D 16
T = ( P1 - P2) = (1680 - 655)
2 2
= 8200 lbf · in Ans.
Tn 8200(200)
H = = = 26.0hp Ans.
63 025 63 025
3P1 3(1680)
P = = = 504 lbf Ans.
10 10
(b)
Force of belt on the drum:
R = (16802 + 6552)1/2 = 1803 lbf
1680 lbf
655 lbf
Force of shaft on the drum: 1680 and 655 lbf
1680 lbf
655 lbf
TP1 = 1680(8) = 13 440 lbf · in
13,440 lbf" in 5240 lbf" in
TP2 = 655(8) = 5240 lbf · in
1803 lbf
Net torque on drum due to brake band:
1803 lbf
T = TP1 - TP2
= 13 440 - 5240
1680 lbf
= 8200 lbf · in
655 lbf
8200 lbf" in
The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted with
the drum at center span, the bearing radial load is 1803/2 = 901 lbf.
(c) Eq. (16-22):
2P
p =
bD
2P1 2(1680)
p|¸=0° = = = 70 psi Ans.
3(16) 3(16)
As it should be
2P2 2(655)
p|¸=270° = = = 27.3 psi Ans.
3(16) 3(16)
16-15 Given: Ć =270°, b =2.125 in, f =0.20, T =150 lbf · ft, D =8.25 in, c2 = 2.25 in
Notice that the pivoting rocker is not located on the vertical centerline of the drum.
(a) To have the band tighten for ccw rotation, it is necessary to have c1 < c2 . When fric-
tion is fully developed,
P1
= exp( f Ć) = exp[0.2(3Ą/2)] = 2.566
P2
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If friction is not fully developed
P1/P2 d" exp( f Ć)
To help visualize what is going on let s add a force W parallel to P1, at a lever arm of
c3. Now sum moments about the rocker pivot.
M = 0 = c3W + c1 P1 - c2 P2
From which
c2 P2 - c1 P1
W =
c3
The device is self locking for ccw rotation if W is no longer needed, that is, W d" 0.
It follows from the equation above
P1 c2
e"
P2 c1
When friction is fully developed
2.566 = 2.25/c1
2.25
c1 = = 0.877 in
2.566
When P1/P2 is less than 2.566, friction is not fully developed. Suppose P1/P2 = 2.25,
then
2.25
c1 = = 1in
2.25
We don t want to be at the point of slip, and we need the band to tighten.
c2
d" c1 d" c2
P1/P2
When the developed friction is very small, P1/P2 1 and c1 c2 Ans.
(b) Rocker has c1 = 1 in
P1 c2 2.25
= = = 2.25
P2 c1 1
ln( P1/P2) ln 2.25
f = = = 0.172
Ć 3Ą/2
Friction is not fully developed, no slip.
D P1 D
T = ( P1 - P2) = P2 - 1
2 P2 2
Solve for P2
2T 2(150)(12)
P2 = = = 349 lbf
[( P1/P2) - 1]D (2.25 - 1)(8.25)
P1 = 2.25P2 = 2.25(349) = 785 lbf
2P1 2(785)
p = = = 89.6 psi Ans.
bD 2.125(8.25)
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(c) The torque ratio is 150(12)/100 or 18-fold.
349
P2 = = 19.4 lbf
18
P1 = 2.25P2 = 2.25(19.4) = 43.6 lbf
89.6
p = = 4.98 psi Ans.
18
Comment:
As the torque opposed by the locked brake increases, P2 and P1 increase (although
ratio is still 2.25), then p follows. The brake can self-destruct. Protection could be
provided by a shear key.
16-16
Ä„pad
(a) From Eq. (16-23), since F = (D - d)
2
then
2F
pa =
Ä„d(D - d)
and it follows that
2(5000)
pa =
Ä„(225)(300 - 225)
= 0.189 N/mm2 or 189 000 N/m2 or 189 kPa Ans.
Ff 5000(0.25)
T = (D + d) = (300 + 225)
4 4
= 164 043 N · mm or 164 N · m Ans.
(b) From Eq. (16-26),
Ä„pa
F = (D2 - d2)
4
4F 4(5000)
pa = =
Ä„(D2 - d2) Ä„(3002 - 2252)
= 0.162 N/mm2 = 162 kPa Ans.
From Eq. (16-27),
Ä„ Ä„
T = f pa(D3 - d3) = (0.25)(162)(103)(3003 - 2253)(10-3)3
12 12
= 166 N · m Ans.
16-17
(a) Eq. (16-23):
Ä„pad Ä„(120)(4)
F = (D - d) = (6.5 - 4) = 1885 lbf Ans.
2 2
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Chapter 16 409
Eq. (16-24):
Ä„ f pad Ä„(0.24)(120)(4)
T = (D2 - d2)N = (6.52 - 42)(6)
8 8
= 7125 lbf · in Ans.
Ä„(0.24)(120d)
(b) T = (6.52 - d2)(6)
8
d, in T, lbf · in
2 5191
3 6769
4 7125 Ans.
5 5853
6 2545
(c) The torque-diameter curve exhibits a stationary point maximum in the range of
diameter d. The clutch has nearly optimal proportions.
16-18
(a)
Ä„ f pad(D2 - d2)N
T = = CD2d - Cd3
8
Differentiating with respect to d and equating to zero gives
dT
= CD2 - 3Cd2 = 0
dd
D
d* = " Ans.
3
d2T
=-6 Cd
dd2
which is negative for all positive d. We have a stationary point maximum.
6.5
(b) d* = " = 3.75 in Ans.
3
"
Ä„(0.24)(120) 6.5/ 3
T * = [6.52 - (6.52/3)](6) = 7173 lbf · in
8
(c) The table indicates a maximum within the range:
3 d" d d" 5in
d
(d) Consider: 0.45 d" d" 0.80
D
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Multiply through by D
0.45D d" d d" 0.80D
0.45(6.5) d" d d" 0.80(6.5)
2.925 d" d d" 5.2in
"
d 1
= d"/D = " = 0.577
D
3
which lies within the common range of clutches.
Yes. Ans.
16-19 Given: d = 0.306 m, l = 0.060 m, T = 0.200 kN · m, D = 0.330 m, f = 0.26.
12
12
60
165 Ä… = tan-1 = 11.31°
60
153
Not to scale
C
L
Uniform wear
Eq. (16-45):
Ä„(0.26)(0.306) pa
0.200 = (0.3302 - 0.3062) = 0.002 432pa
8 sin 11.31°
0.200
pa = = 82.2kPa Ans.
0.002 432
Eq. (16-44):
Ä„pad Ä„(82.2)(0.306)
F = (D - d) = (0.330 - 0.306) = 0.949 kN Ans.
2 2
Uniform pressure
Eq. (16-48):
Ä„(0.26) pa
0.200 = (0.3303 - 0.3063) = 0.002 53pa
12 sin 11.31°
0.200
pa = = 79.1kPa Ans.
0.002 53
Eq. (16-47):
Ä„ pa Ä„(79.1)
F = (D2 - d2) = (0.3302 - 0.3062) = 0.948 kN Ans.
4 4
16-20 Uniform wear
1
2
Eq. (16-34): T = (¸2 - ¸1) f pari ro - ri2
2
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Chapter 16 411
Eq. (16-33): F = (¸2 - ¸1) pari(ro - ri)
2
(1/2)(¸2 - ¸1) f pari ro - ri2
T
Thus, =
f F D f (¸2 - ¸1) pari(ro - ri)(D)
ro + ri D/2 + d/2 1 d
= = = 1 + O.K. Ans.
2D 2D 4 D
Uniform pressure
1
3
Eq. (16-38): T = (¸2 - ¸1) f pa ro - ri3
3
Eq. (16-37):
1
2
F = (¸2 - ¸1) pa ro - ri2
2
3
(1/3)(¸2 - ¸1) f pa ro - ri3
T 2 (D/2)3 - (d/2)3
= =
2
f F D 3 [(D/2)2 - (d/2)2D]
(1/2) f (¸2 - ¸1) pa ro - ri2 D
2(D/2)3(1 - (d/D)3) 1 1 - (d/D)3
= = O.K. Ans.
3(D/2)2[1 - (d/D)2]D 3 1 - (d/D)2
16-21 É = 2Ä„n/60 = 2Ä„ 500/60 = 52.4 rad/s
H 2(103)
T = = = 38.2 N · m
É 52.4
Key:
T 38.2
F = = = 3.18 kN
r 12
Average shear stress in key is
3.18(103)
Ä = = 13.2MPa Ans.
6(40)
Average bearing stress is
F 3.18(103)
Ãb =- =- =-26.5MPa Ans.
Ab 3(40)
Let one jaw carry the entire load.
1 26 45
rav = + = 17.75 mm
2 2 2
T 38.2
F = = = 2.15 kN
rav 17.75
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The bearing and shear stress estimates are
-2.15(103)
Ãb = =-22.6MPa Ans.
10(22.5 - 13)
2.15(103)
Ä = = 0.869 MPa Ans.
10[0.25Ä„(17.75)2]
16-22 É1 = 2Ä„n/60 = 2Ä„(1800)/60 = 188.5 rad/s
É2 = 0
From Eq. (16-51),
I1I2 Tt1 320(8.3)
= = = 14.09 N · m · s2
I1 + I2 É1 - É2 188.5 - 0
Eq. (16-52):
188.52
E = 14.09 (10-3) = 250 kJ
2
Eq. (16-55):
E 250(103)
T = = = 27.8ć%C Ans.
Cpm 500(18)
16-23
n1 + n2 260 + 240
n = = = 250 rev/min
2 2
260 - 240
Cs = = 0.08 Ans.
250
É = 2Ä„(250)/60 = 26.18 rad/s
E2 - E1 5000(12)
I = = = 1094 lbf · in · s2
CsÉ2 0.08(26.18)2
m W
2 2
Ix = do + di2 = do + di2
8 8g
8gI 8(386)(1094)
W = = = 502 lbf
2
602 + 562
do + di2
w = 0.260 lbf/in3 for cast iron
W 502
V = = = 1931 in3
w 0.260
Ä„t Ä„t
2
Also, V = do - di2 = 602 - 562 = 364t in3
4 4
Equating the expressions for volume and solving for t,
1931
t = = 5.3in Ans.
364
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Chapter 16 413
16-24 (a) The useful work performed in one revolution of the crank shaft is
U = 35(2000)(8)(0.15) = 84(103) in · lbf
Accounting for friction, the total work done in one revolution is
U = 84(103)/(1 - 0.16) = 100(103) in · lbf
Since 15% of the crank shaft stroke is 7.5% of a crank shaft revolution, the energy
fluctuation is
E2 - E1 = 84(103) - 100(103)(0.075) = 76.5(103) in · lbf Ans.
(b) For the flywheel
n = 6(90) = 540 rev/min
2Ä„n 2Ä„(540)
É = = = 56.5 rad/s
60 60
Since Cs = 0.10
E2 - E1 76.5(103)
I = = = 239.6 lbf · in · s2
CsÉ2 0.10(56.5)2
Assuming all the mass is concentrated at the effective diameter, d,
md2
I =
4
4gI 4(386)(239.6)
W = = = 161 lbf Ans.
d2 482
16-25 Use Ex. 16-6 and Table 16-6 data for one cylinder of a 3-cylinder engine.
Cs = 0.30
n = 2400 rev/min or 251 rad/s
3(3368)
Tm = = 804 in · lbf Ans.
4Ä„
E2 - E1 = 3(3531) = 10 590 in · lbf
E2 - E1 10 590
I = = = 0.560 in · lbf · s2 Ans.
CsÉ2 0.30(2512)
16-26 (a)
(1)
F21 rG
rP
T2 T2
(T2)1 =-F21rP =- rP = Ans.
F12
rG -n
T1 T2
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(2) Equivalent energy
rG
rP
2 2
(1/2)I2É2 = (1/2)(I2)1 w1
2
É2 I2
(I2)1 = I2 = Ans.
IL
2
É1 n2
2 2 2
IG rG mG rG rG
(3) = = = n4
IP rP m rP rP
P
IG n4IP
From (2) (I2)1 = = = n2IP Ans.
n2 n2
IL
(b) Ie = IM + IP + n2IP + Ans.
n2
100
(c) Ie = 10 + 1 + 102(1) +
102
= 10 + 1 + 100 + 1 = 112
reflected load inertia
reflected gear inertia Ans.
pinion inertia
armature inertia
16-27 (a) Reflect IL , IG2 to the center shaft
n
IP IG1
IL
IP m2IP
IM
m2
Reflect the center shaft to the motor shaft
IP
IP m2IP IL m2
IM n2IP
n2
IP m2 IL
Ie = IM + IP + n2IP + + IP + Ans.
n2 n2 m2n2
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Chapter 16 415
IP R2IP IL
(b) For R = constant = nm, Ie = IM + IP + n2IP + + + Ans.
n2 n4 R2
" Ie 2(1) 4(102)(1)
(c) For R = 10, = 0 + 0 + 2n(1) - - + 0 = 0
"n n3 n5
n6 - n2 - 200 = 0
From which
n* = 2.430 Ans.
10
m* = = 4.115 Ans.
2.430
Notice that n*and m* are independent of IL .
16-28 From Prob. 16-27,
IP R2IP IL
Ie = IM + IP + n2IP + + +
n2 n4 R2
1 100(1) 100
= 10 + 1 + n2(1) + + +
n2 n4 102
1 100
= 10 + 1 + n2 + + + 1
n2 n4
nIe
1.00 114.00
Ie
1.50 34.40
2.00 22.50
100
2.43 20.90
3.00 22.30
4.00 28.50
5.00 37.20
20.9
6.00 48.10
n
7.00 61.10 0 1 2 4 6 8 10
2.43
8.00 76.00
9.00 93.00
10.00 112.02
Optimizing the partitioning of a double reduction lowered the gear-train inertia to
20.9/112 = 0.187, or to 19% of that of a single reduction. This includes the two addi-
tional gears.
16-29 Figure 16-29 applies,
t2 = 10 s, t1 = 0.5s
t2 - t1 10 - 0.5
= = 19
t1 0.5
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The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is
1300(12)
TL = = 1560 lbf · in
10
The rated motor torque Tr is
63 025(3)
Tr = = 168.07 lbf · in
1125
For Eqs. (16-65):
2Ä„
Ér = (1125) = 117.81 rad/s
60
2Ä„
És = (1200) = 125.66 rad/s
60
-Tr 168.07
a = =- =-21.41
És - Ér 125.66 - 117.81
TrÉs 168.07(125.66)
b = =
És - Ér 125.66 - 117.81
= 2690.4 lbf · in
The linear portion of the squirrel-cage motor characteristic can now be expressed as
TM =-21.41É + 2690.4 lbf · in
Eq. (16-68):
19
1560 - 168.07
T2 = 168.07
1560 - T2
One root is 168.07 which is for infinite time. The root for 10 s is wanted. Use a successive
substitution method
T2 New T2
0.00 19.30
19.30 24.40
24.40 26.00
26.00 26.50
26.50 26.67
Continue until convergence.
T2 = 26.771
Eq. (16-69):
-21.41(10 - 0.5)
I = = 110.72 in · lbf · s/rad
ln(26.771/168.07)
T - b
É =
a
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Chapter 16 417
T2 - b 26.771 - 2690.4
Émax = = = 124.41 rad/s Ans.
a -21.41
Émin = 117.81 rad/s Ans.
124.41 + 117.81
É = = 121.11 rad/s
Å»
2
Émax - Émin 124.41 - 117.81
Cs = = = 0.0545 Ans.
(Émax + Émin)/2 (124.41 + 117.81)/2
1 1
2
E1 = I Ér = (110.72)(117.81)2 = 768 352 in · lbf
2 2
1 1
2
E2 = I É2 = (110.72)(124.41)2 = 856 854 in · lbf
2 2
E = E1 - E2 = 768 352 - 856 854 =-88 502 in · lbf
Eq. (16-64):
E = Cs I É2 = 0.0545(110.72)(121.11)2
Å»
= 88 508 in · lbf, close enough Ans.
During the punch
63 025H
T =
n
TLÉ(60/2Ä„) 1560(121.11)(60/2Ä„)
Å»
H = = = 28.6hp
63 025 63 025
The gear train has to be sized for 28.6 hp under shock conditions since the flywheel is on
the motor shaft. From Table A-18,
m W
2 2
I = do + di2 = do + di2
8 8g
8gI 8(386)(110.72)
W = =
2 2
do + di2 do + di2
If a mean diameter of the flywheel rim of 30 in is acceptable, try a rim thickness of 4 in
di = 30 - (4/2) = 28 in
do = 30 + (4/2) = 32 in
8(386)(110.72)
W = = 189.1 lbf
322 + 282
Rim volume V is given by
Ä„l Ä„l
2
V = do - di2 = (322 - 282) = 188.5l
4 4
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where l is the rim width as shown in Table A-18. The specific weight of cast iron is
Å‚ = 0.260 lbf · in3 , therefore the volume of cast iron is
W 189.1
V = = = 727.3in3
Å‚ 0.260
Thus
188.5 l = 727.3
727.3
l = = 3.86 in wide
188.5
Proportions can be varied.
16-30 Prob. 16-29 solution has I for the motor shaft flywheel as
I = 110.72 in · lbf · s2/rad
A flywheel located on the crank shaft needs an inertia of 102I (Prob. 16-26, rule 2)
I = 102(110.72) = 11 072 in · lbf · s2/rad
A 100-fold inertia increase. On the other hand, the gear train has to transmit 3 hp under
shock conditions.
Stating the problem is most of the solution. Satisfy yourself that on the crankshaft:
TL = 1300(12) = 15 600 lbf · in
Tr = 10(168.07) = 1680.7 lbf · in
Ér = 117.81/10 = 11.781 rad/s
És = 125.66/10 = 12.566 rad/s
a =-21.41(100) =-2141
b = 2690.35(10) = 26903.5
TM =-2141Éc + 26 903.5 lbf · in
19
15 600 - 1680.5
T2 = 1680.6
15 600 - T2
The root is 10(26.67) = 266.7 lbf · in
É = 121.11/10 = 12.111 rad/s
Å»
Cs = 0.0549 (same)
Émax = 121.11/10 = 12.111 rad/s Ans.
Émin = 117.81/10 = 11.781 rad/s Ans.
E1, E2, E and peak power are the same.
From Table A-18
8gI 8(386)(11 072)
W = =
2 2
do + di2 do + di2
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Chapter 16 419
Scaling will affect do and di, but the gear ratio changed I. Scale up the flywheel in the
Prob. 16-29 solution by a factor of 2.5. Thickness becomes 4(2.5) = 10 in.
Å»
d = 30(2.5) = 75 in
do = 75 + (10/2) = 80 in
di = 75 - (10/2) = 70 in
8(386)(11 072)
W = = 3026 lbf
802 + 702
3026
v = = 11 638 in3
0.26
Ä„
V = l(802 - 702) = 1178 l
4
11 638
l = = 9.88 in
1178
Proportions can be varied. The weight has increased 3026/189.1 or about 16-fold while
the moment of inertia I increased 100-fold. The gear train transmits a steady 3 hp. But the
motor armature has its inertia magnified 100-fold, and during the punch there are decel-
eration stresses in the train. With no motor armature information, we cannot comment.
16-31 This can be the basis for a class discussion.
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