This document provides calculations for determining braking forces and torques in a drum brake system. It gives equations and applies them to example problems, providing values for forces, pressures, and torques. In one example, a maximum pressure of 734.5 kPa is found on the right shoe for clockwise rotation. The total braking torque is calculated to be 422 N-m. In another example, a braking capacity of 1750 N-m is determined.

1. Chapter 16
16-1
Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN,  1 = 0,
 2 = 120, and  a = 90.
From which, sin a = sin90 = 1.
Eq. (16-2):
0.28 pa (0.040)(0.150) 120
0 sin  (0.150  0.125 cos  ) d
1
 2.993 10 4  pa N · m
Mf 
Eq. (16-3):
pa (0.040)(0.150)(0.125) 120 2
4
0 sin  d  9.478 10  pa N · m
1
MN 
c = 2(0.125 cos 30) = 0.2165 m
Eq. (16-4):
F 
9.478 10 4  pa  2.993 10 4  pa
0.2165
 2.995 10 3  pa
p a = F/ [2.995(103)] = 2200/ [2.995(103)]
= 734.5(103) Pa for cw rotation
Eq. (16-7):
2200 
9.478 10 4  pa  2.993 10 4  pa
0.2165
p a = 381.9(103) Pa for ccw rotation
A maximum pressure of 734.5 kPa occurs on the RH shoe for cw rotation.
(b) RH shoe:
Eq. (16-6):
0.28(734.5)103 (0.040)0.1502 (cos 0o  cos120o )
TR 
 277.6 N · m
1
LH shoe:
381.9
TL  277.6
 144.4 N · m Ans.
734.5
T total = 277.6 + 144.4 = 422 N · m Ans.
Ans.
Ans.
Chapter 16, Page 1/27

2. (c)
RH shoe:
F x = 2200 sin 30° = 1100 N, F y = 2200 cos 30° = 1905 N
Eqs. (16-8):
1

A   sin 2  
2

0o
Eqs. (16-9):
Rx 
120o
Ry 
 0.375,
734.5 103  0.040(0.150)
1
3
734.5 10  0.04(0.150)
R  [ 1007 
LH shoe:
2 / 3 rad
 1

B    sin 2 
2 4

0
2
 1.264
[0.375  0.28(1.264)]  1100  1007 N
[1.264  0.28(0.375)]  1905  4128 N
1
 41282 ]1/ 2  4249 N Ans.
F x = 1100 N, F y = 1905 N
Eqs. (16-10): Rx 
381.9 103  0.040(0.150)
[0.375  0.28(1.264)]  1100  570 N
1
381.9 103  0.040(0.150)
[1.264  0.28(0.375)]  1905  751 N
Ry 
1
1/ 2
R   597 2  7512   959 N Ans.
______________________________________________________________________________
16-2
Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN,  1 = 15,
 2 = 105, and  a = 90.
From which, sin a = sin90 = 1.
Eq. (16-2):
0.28 pa (0.040)(0.150) 105
4
Mf 
15 sin  (0.150  0.125 cos  ) d  2.177 10  pa
1
Chapter 16, Page 2/27

3. MN 
Eq. (16-3):
pa (0.040)(0.150)(0.125) 105 2
4
15 sin  d  7.765 10  pa
1
c = 2(0.125) cos 30° = 0.2165 m
F 
Eq. (16-4):
7.765 10 4  pa  2.177 10 4  pa
0.2165
 2.58110 3  pa
p a = 2200/ [2.581(10 3)] = 852.4 (103) Pa
= 852.4 kPa on RH shoe for cw rotation Ans.
RH shoe:
TR 
Eq. (16-6):
LH shoe:
2200 
0.28(852.4)103 (0.040)(0.1502 )(cos15  cos105)
 263 N · m
1
7.765 10 4  pa  2.177 10 4  pa
pa  479.110
3

0.2165
Pa  479.1 kPa on LH shoe for ccw rotation
Ans.
0.28(479.1)103(0.040)(0.150 2 )(cos15  cos105)
 148 N · m
1
 263  148  411 N · m Ans.
TL 
Ttotal
Comparing this result with that of Prob. 16-1, a 2.6% reduction in torque is obtained by
using 25% less braking material.
______________________________________________________________________________
16-3
Given:  1 = 0°,  2 = 120°,  a = 90°, sin  a = 1, a = R = 3.5 in, b = 1.25 in, f = 0.30,
F = 225 lbf, r = 11/2 = 5.5 in, counter-clockwise rotation.
LH shoe:
Eq. (16-2), with  1 = 0:

f pabr 2
f p br 
a 2 
Mf 
 sin   r  a cos   d  sin a a  r (1  cos  2 )  2 sin  2 
sin  a 1


0.30 pa (1.25)5.5 
3.5 2

o
5.5(1  cos120 )  2 sin 120
1


 14.31 pa lbf · in
Eq. (16-3), with  1 = 0:

pabra 2 2
pabra  
1

MN 
 sin  d  sin a  22  4 sin 22 
sin  a 1




pa (1.25)5.5(3.5) 120    1
 2  180   4 sin 2(120) 
1




 30.41 pa lbf · in

Chapter 16, Page 3/27

4.  180o   2 
o
c  2r cos 
  2(5.5) cos 30  9.526 in
2


30.41 pa  14.31 pa
 1.690 pa
F  225 
9.526
pa  225 / 1.690  133.1 psi
Eq. (16-6):
f pabr 2 (cos 1  cos  2 ) 0.30(133.1)1.25(5.52 )

[1  (0.5)]
sin  a
1
 2265 lbf · in  2.265 kip · in Ans.
TL 
RH shoe:
30.41 pa  14.31 pa
 4.694 pa
9.526
pa  225 / 4.694  47.93 psi
47.93
TR 
2265  816 lbf ·in  0.816 kip·in
133.1
F  225 
T total = 2.27 + 0.82 = 3.09 kip  in
Ans.
______________________________________________________________________________
16-4
(a) Given:  1 = 10°,  2 = 75°,  a = 75°, p a = 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 here:

2
 r 2 sin  d  a 2 sin  cos  d   r   cos  2  a  1 sin 2  
A 
1
1
2

 1




1
75
1

 200   cos  10  150  sin 2    77.5 mm
2
10
75 /180 rad
2
 1

2
B   sin  d    sin 2 
 0.528
1
2 4
10 /180 rad
75
C 
2

sin  cos  d  0.4514
1
Now converting to Pascals and meters, we have from Eq. (16-2),
0.24 106  (0.075)(0.200)
f pabr
Mf 
A
(0.0775)  289 N · m
sin  a
sin 75
Chapter 16, Page 4/27

5. From Eq. (16-3),
MN
pabra
106 (0.075)(0.200)(0.150)

B 
(0.528)  1230 N · m
sin  a
sin 75
Finally, using Eq. (16-4), we have
F 
MN  M f
c

1230  289
 5.70 kN
165
Ans.
(b) Use Eq. (16-6) for the primary shoe.
T 

fpabr 2 (cos 1  cos  2 )
sin  a
0.24 106  (0.075)(0.200)2 (cos 10  cos 75)
sin 75
 541 N · m
For the secondary shoe, we must first find p a . Substituting
1230
289
pa and M f  6 pa into Eq. (16 - 7),
6
10
10
(1230 / 106 ) pa  (289 / 106 ) pa
5.70 
, solving gives pa  619 103  Pa
165
MN 
Then
T 
0.24 619 103   0.075  0.2002   cos 10  cos 75 


sin 75
 335 N · m
so the braking capacity is T total = 2(541) + 2(335) = 1750 N · m
Ans.
(c) Primary shoes:
pabr
 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
( B  f C )  Fy
Ry 
sin  a
106 (0.075)0.200

[0.528  0.24(0.4514)] 10 3   0  9.88 kN
sin 75
Rx 
Chapter 16, Page 5/27

6. Secondary shoes:
Rx 

pabr
(C  f B)  Fx
sin  a
0.619 106  0.075(0.200)
sin 75
 0.143 kN
p br
Ry  a ( B  f C )  Fy
sin  a

0.619 106  0.075(0.200)
 4.03 kN
sin 75
[0.4514  0.24(0.528)] 10 3   5.70
[0.528  0.24(0.4514)] 10 3   0
Note from figure that +y for secondary shoe is opposite to
+y for primary shoe.
Combining horizontal and vertical components,
RH  0.658  0.143  0.801 kN
RV  9.88  4.03  5.85 kN
R  (0.801) 2  5.852
 5.90 kN Ans.
______________________________________________________________________________
16-5
Given: Face width b = 1.25 in, F = 90 lbf, f = 0.25.
Preliminaries:  1 = 45°  tan1(6/8) = 8.13°,  2 = 98.13°,  a = 90°,
a = (62 + 82)1/2 = 10 in
Eq. (16-2):

f pabr 2
0.25 pa (1.25)6
Mf 
 sin   r  a cos   d 
sin  a 1
1
98.13

sin   6  10 cos   d
8.13
 3.728 pa lbf · in
Eq. (16-3):

MN 
pabra 2 2
p (1.25)6(10)
 sin  d  a 1
sin  a 1
98.13

sin 2  d
8.13
 69.405 pa lbf · in
Eq. (16-4): Using Fc = M N  M f , we obtain
90(20)  (69.405  3.728) pa

pa  27.4 psi
Ans.
Chapter 16, Page 6/27

7. Eq. (16-6):
fp br 2  cos 1  cos  2  0.25(27.4)1.25  6   cos8.13  cos 98.13 
T  a

sin  a
1
 348.7 lbf · in Ans.
______________________________________________________________________________
2
16-6
ˆ
For 3 f :
ˆ
f  f  3 f  0.25  3(0.025)  0.325
From Prob. 16-5, with f = 0.25, M f = 3.728 p a . Thus, M f = (0.325/0.25) 3.728 p a =
4.846 p a . From Prob. 16-5, M N = 69.405 p a .
Eq. (16-4): Using Fc = M N  M f , we obtain
90(20)  (69.405  4.846) pa

pa  27.88 psi
Ans.
From Prob. 16-5, p a = 27.4 psi and T = 348.7 lbf⋅in. Thus,
 0.325   27.88 
T 

 348.7  461.3 lbf ·in
 0.25   27.4 
Ans.
ˆ
Similarly, for 3 f :
ˆ
f  f  3 f  0.25  3(0.025)  0.175
M f  (0.175 / 0.25) 3.728 pa  2.610 pa
90(20) = (69.405  2.610) p a
 p a = 26.95 psi
 0.175  26.95 
T 

 348.7  240.1 lbf · in Ans.
 0.25  27.4 
______________________________________________________________________________
16-7
Preliminaries:  2 = 180°  30°  tan1(3/12) = 136°,  1 = 20°  tan1(3/12) = 6°,
 a = 90, sin a = 1, a = (32 + 122)1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in, p a = 150 psi.
Eq. (16-2):
Mf 
0.30(150)(2)(10) 136o
6 sin  (10  12.37 cos  ) d  12 800 lbf · in
sin 90
Eq. (16-3):
MN 
150(2)(10)(12.37) 136 2
6 sin  d  53 300 lbf · in
sin 90
LH shoe:
c L = 12 + 12 + 4 = 28 in
Chapter 16, Page 7/27

8. Now note that M f is cw and M N is ccw. Thus,
FL 
Eq. (16-6):
53 300  12 800
 1446 lbf
28
TL 
0.30(150)(2)(10) 2 (cos 6  cos136)
 15 420 lbf · in
sin 90
RH shoe:
M N  53 300
pa
 355.3 pa ,
150
M f  12 800
pa
 85.3 pa
150
On this shoe, both M N and M f are ccw. Also,
c R = (24  2 tan 14°) cos 14° = 22.8 in
Fact  FL sin14  361 lbf Ans.
FR  FL / cos14  1491 lbf
Thus,
1491 
Then,
TR 
355.3  85.3
pa  pa  77.2 psi
22.8
0.30(77.2)(2)(10) 2 (cos 6  cos136)
 7940 lbf · in
sin 90
T total = 15 420 + 7940 = 23 400 lbf · in Ans.
______________________________________________________________________________
16-8
2
M f  2 ( fdN )(a cos   r )
0
where dN  pbr d
2
 2 fpbr  (a cos   r ) d  0
0
From which
2
2
0
0
a cos  d  r  d
r 2
r (60)( / 180)
a 

 1.209r
sin  2
sin 60
Ans.
Chapter 16, Page 8/27

9. Eq. (16-15):
a 
4r sin 60
 1.170r
2(60)( / 180)  sin[2(60)]
Ans.
a differs with a ¢ by 100(1.170 1.209)/1.209 =  3.23 %
Ans.
______________________________________________________________________________
16-9
(a) Counter-clockwise rotation,  2 =  / 4 rad, r = 13.5/2 = 6.75 in
Eq. (16-15):
4r sin  2
4(6.75) sin( / 4)

 7.426 in
a 
2 2  sin 2 2
2 / 4  sin(2 / 4)
e  2a  2(7.426)  14.85 in
Ans.
(b)
 = tan1(3/14.85) = 11.4°
M
F
x
R
 0  3F x  6.375P
 0  F x  R x
 F x  2.125P
 R x  F x  2.125P
F y  F x tan11.4o  0.428P
 Fy   P  F y  R y
R y  P  0.428P  1.428P
Left shoe lever.
 M R  0  7.78S x  15.28F x
15.28
Sx 
(2.125P)  4.174P
7.78
S y  f S x  0.30(4.174 P)  1.252P
 Fy  0  R y  S y  F y
R y   F y  S y  0.428P  1.252 P  1.68P
 Fx  0  R x  S x  F x
R x  S x  F x  4.174P  2.125P  2.049P
Chapter 16, Page 9/27

10. (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
designed 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 = 6 in,
 2 = 45° =  / 4 rad.
From Table 16-3 for a rigid, molded non-asbestos lining use a conservative estimate of
p a = 100 psi, f = 0.33.
Equation (16-16) gives the horizontal brake hinge pin reaction which corresponds to Sx in
Prob. 16-9. Thus,
p br
100(6)6.75
N  S x  a  2 2  sin 2 2  
2  / 4   sin  2  45  


2
2
 5206 lbf


which, from Prob. 6-9 is 4.174 P. Therefore,
4.174 P = 5206
 P = 1250 lbf = 1.25 kip
Ans.
Applying Eq. (16-18) for two shoes, where from Prob. 16-9, a = 7.426 in
T  2a f N  2(7.426)0.33(5206)
 25 520 lbf · in  25.52 kip · in Ans.
______________________________________________________________________________
16-11 Given: D = 350 mm, b = 100 mm, p a = 620 kPa, f = 0.30,  = 270.
Chapter 16, Page 10/27

11. Eq. (16-22):
pabD
620(0.100)0.350

 10.85 kN
2
2
P 
1
Ans.
f   0.30(270)( / 180)  1.414
Eq. (16-19):
P 2 = P 1 exp( f  ) = 10.85 exp( 1.414) = 2.64 kN
Ans.
T  ( P  P2 )( D / 2)  (10.85  2.64)(0.350 / 2)  1.437 kN · m Ans.
1
______________________________________________________________________________
16-12 Given: D = 12 in,
Eq. (16-22):
f = 0.28, b = 3.25 in,  = 270°, P 1 = 1800 lbf.
pa 
2P
2(1800)
1

 92.3 psi
bD 3.25(12)
Ans.
f   0.28(270o )( / 180o )  1.319
P2  P exp( f  )  1800 exp(1.319)  481 lbf
1
T  ( P  P2 )( D / 2)  (1800  481)(12 / 2)
1
 7910 lbf · in  7.91 kip · in Ans.
______________________________________________________________________________
16-13
 M O = 0 = 100 P 2  325 F  P 2 = 325(300)/100 = 975 N Ans.
 100 
  cos 1 
  51.32
 160 
  270  51.32  218.7
f   0.30(218.7)  / 180   1.145
P  P2 exp( f  )  975exp(1.145)  3064 N
Ans.
1
T   P  P2  ( D / 2)  (3064  975)(200 / 2)
1
 209 103  N · mm  209 N · m
Ans.
______________________________________________________________________________
Chapter 16, Page 11/27

12. 16-14 (a) D = 16 in, b = 3 in
n = 200 rev/min
f = 0.20, p a = 70 psi
Eq. (16-22):
P 
1
Eq. (16-14):
pabD
70(3)(16)
 1680 lbf

2
2
f   0.20(3 / 2)  0.942
P2  P exp( f  )  1680 exp(0.942)  655 lbf
1
D
16
 (1680  655)
2
2
 8200 lbf · in Ans.
Tn
8200(200)
H 

 26.0 hp
63 025
63 025
3P
3(1680)
 504 lbf Ans.
P  1 
10
10
T  ( P  P2 )
1
Ans.
(b) Force of belt on the drum:
R = (16802 + 6552)1/2 = 1803 lbf
Force of shaft on the drum: 1680 and 655 lbf
TP1  1680(8)  13 440 lbf · in
TP2  655(8)  5240 lbf · in
Net torque on drum due to brake band:
T  TP1  TP2
 13 440  5240
 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.
Chapter 16, Page 12/27

13. (c) Eq. (16-21):
2P
bD
2P
2(1680)
1


 70 psi
3(16)
3(16)
p 
p
  0
Ans.
2P2
2(655)

 27.3 psi Ans.
3(16)
3(16)
______________________________________________________________________________
p
  270

16-15 Given:  = 270°, b = 2.125 in, f = 0.20, T =150 lbf · ft, D = 8.25 in, c 2 = 2.25 in (see
figure). 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 c 1 < c 2 . When
friction is fully developed,
P / P2  exp( f  )  exp[0.2(3 / 2)]  2.566
1
If friction is not fully developed,
P 1 /P 2 ≤ exp( f  )
To help visualize what is going on let’s add a force W parallel to P 1 , at a lever arm of
c 3 . Now sum moments about the rocker pivot.
M
 0  c3W  c1P  c2 P2
1
From which
c2 P2  c1P
1
c3
The device is self locking for ccw rotation if W is no longer needed, that is, W ≤ 0.
It follows from the equation above
W 
P
c
1
 2
P2
c1
When friction is fully developed
2.566  2.25 / c1
2.25
c1 
 0.877 in
2.566
When P 1 /P 2 is less than 2.566, friction is not fully developed. Suppose P 1 /P 2 = 2.25,
Chapter 16, Page 13/27

14. then
c1 
2.25
 1 in
2.25
We don’t want to be at the point of slip, and we need the band to tighten.
c2
 c1  c2
P / P2
1
When the developed friction is very small, P 1 /P 2 → 1 and c 1 → c 2
Ans.
(b) Rocker has c 1 = 1 in
P
c
2.25
1
 2 
 2.25
P2
c1
1
ln( P / P2 ) ln 2.25
1
f 

 0.172

3 / 2
Friction is not fully developed, no slip.
T  ( P  P2 )
1
P
D
D
 P2  1  1
2
 P2
2
Solve for P 2
2T
2(150)(12)

 349 lbf
[( P / P2 )  1]D
(2.25  1)(8.25)
1
P  2.25P2  2.25(349)  785 lbf
1
2P
2(785)
p  1 
 89.6 psi Ans.
2.125(8.25)
bD
P2 
(c) The torque ratio is 150(12)/100 or 18-fold.
349
P2 
 19.4 lbf
18
P  2.25P2  2.25(19.4)  43.6 lbf
1
89.6
p 
 4.98 psi Ans.
18
Comment:
As the torque opposed by the locked brake increases, P 2 and P 1 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 Given: OD = 250 mm, ID = 175 mm, f = 0.30, F = 4 kN.
Chapter 16, Page 14/27

15. (a) From Eq. (16-23),
2  4000 
2F
pa 

 0.194 N/mm 2  194 kPa
Ans.
 d ( D  d )  (175)(250  175)
Eq. (16-25):
Ff
4000(0.30)
T 
(D  d ) 
(250  175)10  3  127.5 N · m Ans.
4
4
(b) From Eq. (16-26),
4F
4(4000)

 0.159 N/mm 2  159 kPa
2
2
2
2
 ( D  d )  (250  175 )
pa 
Eq. (16-27):
T 

f pa ( D3  d 3 ) 

(0.30)159 103  2503  1753 10 3 
Ans.
3
12
12
 128 N · m Ans.
______________________________________________________________________________
16-17 Given: OD = 6.5 in, ID = 4 in, f = 0.24, p a = 120 psi.
(a) Eq. (16-23):
 pa d
 (120)(4)
F 
(D  d ) 
(6.5  4)  1885 lbf Ans.
2
2
Eq. (16-24) with N sliding planes:
 fpa d 2
 (0.24)(120)(4)
T 
(D  d 2 ) N 
(6.52  42 )(6)
8
8
 7125 lbf · in Ans.
(b)
T 
 (0.24)(120d )
8
(6.52  d 2 )(6)
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) Eq. (16-24) with N sliding planes:
Chapter 16, Page 15/27

16.  f pa d ( D 2  d 2 ) N
T 
8

 f pa N
8
D d  d 
2
3
Differentiating with respect to d and equating to zero gives
dT
 f pa N 2

 D  3d 2   0
8
dd
D
d* 
Ans.
3
 f pa N
3 f pa N
d 2T
 6
d 
d
2
8
4
dd
which is negative for all positive d. We have a stationary point maximum.
(b)
d* 
6.5
 3.75 in
3
Eq. (16-24):
T* 

Ans.

 (0.24)(120) 6.5 / 3  2
2
6.5  6.5 / 3  (6)  7173 lbf · in
8





(c) The table indicates a maximum within the range: 3 ≤ d ≤ 5 in
d
(d) Consider:
 0.80
0.45 
D
Multiply through by D,
0.45D  d  0.80D
0.45(6.5)  d  0.80(6.5)
2.925  d  5.2 in
*
1
d
 0.577
   d * /D 
3
D
which lies within the common range of clutches.
Yes. Ans.
______________________________________________________________________________
16-19 Given: d = 11 in,
l = 2.25 in, T = 1800 lbf · in,
D = 12 in,
f = 0.28.
 0.5 
  tan 1 
  12.53
 2.25 
Chapter 16, Page 16/27

17. Uniform wear
Eq. (16-45):
 f pa d 2
D  d2
8sin 
 (0.28) pa (11) 2
1800 
12  112   128.2 pa
8sin12.53
1800
 14.04 psi Ans.
pa 
128.2
T 
Eq. (16-44):
 pa d
F 
2
(D  d ) 
 (14.04)11
2
(12  11)  243 lbf
Ans.
Uniform pressure
Eq. (16-48):
 f pa
 D3  d 3 
12 sin 
 (0.28) pa
1800 
123  113   134.1pa
12sin12.53
1800
pa 
Ans.
 13.42 psi
134.1
T 
Eq. (16-47):
 pa
 (13.42)
122  112   242 lbf Ans.
4
4
______________________________________________________________________________
F 
(D 2  d 2 ) 
16-20 Uniform wear
Eq. (16-34):
Eq. (16-33):
Thus,
1
( 2  1) f pa ri  ro2  ri 2 
2
F = ( 2   1 ) p a r i (r o  r i )
T 
(1 / 2)( 2  1) f pa ri  ro2  ri 2 
T

f ( 2  1) pa ri (ro  ri )( D)
f FD
r  ri
D / 2  d / 2 1
d
 o

 1   O.K .
D
2D
2D
4
Ans.
Uniform pressure
Eq. (16-38):
T 
1
( 2  1) f pa  ro3  ri3 
3
Chapter 16, Page 17/27

18. F 
Eq. (16-37):
1
( 2  1) pa  ro2  ri 2 
2
Thus,
(1 / 3)( 2  1) f pa  ro3  ri3 
T
2  ( D / 2)3  (d / 2)3 



 

2
2
2
2
f FD (1 / 2) f ( 2  1) pa  ro  ri  D
3   ( D / 2)  (d / 2) D  


3
2( D / 2)3 1  (d / D)3 

  1  1  (d / D)  O.K . Ans.



3( D / 2)2 1  (d / D) 2  D 3 1  (d / D) 2 


______________________________________________________________________________
16-21
  2 n / 60  2 500 / 60  52.4 rad/s
T 
H


2(103 )
 38.2 N· m
52.4
Key:
T
38.2

 3.18 kN
r
12
Average shear stress in key is
3.18(103 )
 
 13.2 MPa Ans.
6(40)
Average bearing stress is
F
3.18(103 )
b  

 26.5 MPa
Ab
3(40)
Let one jaw carry the entire load.
F 
Ans.
1  26 45 


  17.75 mm
2 2
2
38.2
T
F 

 2.15 kN
rav 17.75
rav 
The bearing and shear stress estimates are
b 
2.15 103 
 22.6 MPa Ans.
10(22.5  13)
2.15(103 )
 0.869 MPa Ans.
 
10 0.25 (17.75) 2 


______________________________________________________________________________
Chapter 16, Page 18/27

19. 16-22
1  2 n / 60  2 (1600) / 60  167.6 rad/s
2  0
From Eq. (16-51),
I1I 2
Tt1
2800(8)


 133.7 lbf · in · s 2
I1  I 2 1  2 167.6  0
Eq. (16-52):
E 
I1 I 2
133.7
2
(167.6  0) 2  1.877 106  lbf  in
1  2  
2  I1  I 2 
2
H = E / 9336 = 1.877(106) / 9336 = 201 Btu
In Btu, Eq. (16-53):
Eq. (16-54):
H
201

 41.9F Ans.
C pW
0.12(40)
______________________________________________________________________________
T 
16-23
n1  n2
260  240

 250 rev/min
2
2
C s = ( 2   1 ) /  = (n 2  n 1 ) / n = (260  240) / 250 = 0.08
n
Eq. (16-62):
Ans.
 = 2 (250) / 60 = 26.18 rad/s
From Eq. (16-64):
I 
6.75 103 
E2  E1

 123.1 N · m · s 2
2
2
Cs
0.08(26.18)
I 
m 2
 do  di2 
8

m
8I
8(123.1)

 233.9 kg
2
d  di
1.52  1.42
Table A-5, cast iron unit weight = 70.6 kN/m3
Volume:
2
o
  = 70.6(103) / 9.81 = 7197 kg / m3.
V = m /  = 233.9 / 7197 = 0.0325 m3
V   t  d o2  d i2  / 4   t 1.52  1.4 2  / 4  0.2278t
Equating the expressions for volume and solving for t,
0.0325
t 
 0.143 m  143 mm
0.2278
Ans.
Chapter 16, Page 19/27

20. ______________________________________________________________________________
16-24 (a) The useful work performed in one revolution of the crank shaft is
U = 320 (103) 200 (103) 0.15 = 9.6 (103) J
Accounting for friction, the total work done in one revolution is
U = 9.6(103) / (1  0.20) = 12.0(103) J
Since 15% of the crank shaft stroke accounts for 7.5% of a crank shaft revolution, the
energy fluctuation is
E 2  E 1 = 9.6(103)  12.0(103)(0.075) = 8.70(103) J
(b) For the flywheel,
Since
Eq. (16-64):
Ans.
n  6(90)  540 rev/min
2 n
2 (540)

 56.5 rad/s
 
60
60
C s = 0.10
E E
8.70(103 )
I  2 21 
 27.25 N · m · s 2
2
Cs
0.10(56.5)
Assuming all the mass is concentrated at the effective diameter, d,
md 2
I  mr 
4
4I
4(27.25)
 75.7 kg Ans.
m 2 
d
1.22
______________________________________________________________________________
2
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 lbf · in
Ans.
4
E2  E1  3(3531)  10 590 in · lbf
E E
10 590
 0.560 in · lbf · s 2 Ans.
I  2 21 
2
0.30(251 )
Cs
______________________________________________________________________________
Chapter 16, Page 20/27

21. 16-26 (a)
(1)
(T2 )1   F21rP  
T2
T
rP  2
rG
n
Ans.
Equivalent energy
(2)
2
(1 / 2) I 22  (1 / 2)( I 2 )1 12 
I
22
( I 2 )1  2 I 2  2
1
n2
2
(3)
2
Ans.
2
I G  rG   mG   rG   rG 
           n4
I P  rP   mP   rP   rP 
From (2)
( I 2 )1 
(b) I e  I M  I P  n 2 I P 
IL
n2
IG
n4 I
 2P  n2I P
n2
n
Ans.
Ans.
______________________________________________________________________________
16-27 (a) Reflect I L , I G2 to the center shaft
Chapter 16, Page 21/27

22. Reflect the center shaft to the motor shaft
I e  I M  I P  n2I P 
I P m2
I
 2 I P  2L 2
2
n
n
mn
(b) For R = constant = nm, I e  I M
(c) For R = 10,
Ans.
IP
R2I P
I
 I P  n I P  2  4  L2
n
n
R
2
Ans.
I e
2(1) 4(102 )(1)
 0  0  2n(1)  3 
00
n
n5
n
n6  n2  200 = 0
From which
n*  2.430 Ans.
10
m* 
 4.115
2.430
Ans.
Notice that n*and m* are independent of I L .
______________________________________________________________________________
16-28 From Prob. 16-27,
IP
R2I
I
 4 P  L2
2
n
n
R
1 100(1) 100
 10  1  n 2 (1)  2 
 2
n
n4
10
1 100
 12  n 2  2  4
n
n
Ie  I M  I P  n2I P 
Chapter 16, Page 22/27

23. 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 additional
gears.
______________________________________________________________________________
16-29 Figure 16-29 applies,
t2  10 s, t1  0.5 s
t2 - t1 10  0.5

 19
t1
0.5
The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is
TL 
1300(12)
 1560 lbf · in
10
The rated motor torque T r is
Tr 
63 025(3)
 168.07 lbf · in
1125
For Eqs. (16-65):
2
(1125)  117.81 rad/s
60
2
(1200)  125.66 rad/s
s 
60
168.07
Tr
a 

 21.41 lbf  in  s/rad
125.66  117.81
 s  r
168.07(125.66)
Trs
b

 2690.4 lbf · in
s  r 125.66  117.81
r 
Chapter 16, Page 23/27

24. The linear portion of the squirrel-cage motor characteristic can now be expressed as
Eq. (16-68):
T M = 21.41 + 2690.4 lbf · in
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 desired. Use a
successive substitution method
T2
0.00
19.30
24.40
26.00
26.50
New T 2
19.30
24.40
26.00
26.50
26.67
Continue until convergence to
T 2 = 26.771 lbf ⋅ in
Eq. (16-69):
a  t2  t1 
21.41(10  0.5)

 110.72 lbf · in · s 2
ln T2 / Tr  ln(26.771 / 168.07)
T b
 
a
T  b 26.771  2690.4
max  2

 124.41 rad/s Ans.
a
21.41
min  117.81 rad/s Ans.
124.41  117.81
 
 121.11 rad/s
2
124.41  117.81
max  min
Cs 

 0.0545 Ans.
(max  min ) / 2 (124.41  117.81) / 2
1
1
E1  I r2  (110.72)(117.81) 2  768 352 in · lbf
2
2
1 2 1
E2  I 2  (110.72)(124.41) 2  856 854 in · lbf
2
2
E  E2  E1  856 854  768 352  88 502 in · lbf
I 
Eq. (16-64):
E  Cs I  2  0.0545(110.72)(121.11) 2
 88 508 in · lbf, close enough Ans.
Chapter 16, Page 24/27

25. During the punch
63 025H
n
TL (60 / 2 ) 1560(121.11)(60 / 2 )
H 
 28.6 hp

63 025
63 025
T 
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 2
W 2
d o  di2 
d o  di2
8
8g
8gI
8(386)(110.72)
W  2

2
d o  di
d o2  di2
I 

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
d o  30  (4 / 2)  32 in
8(386)(110.72)
 189.1 lbf
W 
322  282
Rim volume V is given by
V 
l
d
4
2
o
 di2  
l
4
(322  282 )  188.5l
where l is the rim width as shown in Table A-18. The specific weight of cast iron is
 = 0.260 lbf / in3, therefore the volume of cast iron is
V 
W


189.1
 727.3 in 3
0.260
Equating the volumes,
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
Chapter 16, Page 25/27

26. I = 110.72 lbf · in · s2
A flywheel located on the crank shaft needs an inertia of 102 I (Prob. 16-26, rule 2)
I = 102(110.72) = 11 072 lbf · in · s2
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
r  117.81 / 10  11.781 rad/s
s  125.66 / 10  12.566 rad/s
a  21.41(100)  2141 lbf · in · s/rad
b  2690.35(10)  26903.5 lbf · in
TM  2141c  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
  121.11 / 10  12.111 rad/s
Cs  0.0549 (same)
max  121.11 / 10  12.111 rad/s Ans.
min  117.81 / 10  11.781 rad/s Ans.
E 1 , E 2 , E and peak power are the same. From Table A-18
6
8gI
8(386)(11 072) 34.19 10 
W  2


d o  di2
d o2  di2
d o2  di2
Scaling will affect d o and d i , 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.
d  30(2.5)  75 in
do  75  (10 / 2)  80 in
di  75  (10 / 2)  70 in
Chapter 16, Page 26/27

27. W 
34.19 106 
 3026 lbf
802  702
W
3026
V 

 11 638 in 3
0.260

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
deceleration stresses in the train. With no motor armature information, we cannot
comment.
______________________________________________________________________________
16-31 This can be the basis for a class discussion.
Chapter 16, Page 27/27