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Chapter 16
16-1
(a)
θ 1 =
0°,
θ 2 =
120°,
θ a =
90°, sin
θ a =
1, a
=
5in
Eq. (16-2): M f
=
0
.
28 p a (1
.
5)(6)
120°
sin
θ
(6
−
5 cos
θ
) d
θ
1
0°
=
17
.
96 p a lbf
·
in
Eq. (16-3): M N
=
p a (1
.
5)(6)(5)
1
120°
sin 2
θ
d
θ =
56
.
87 p a lbf
·
in
0°
c
=
2(5 cos 30 ◦ )
=
8
.
66 in
Eq. (16-4):
F
=
56
.
87 p a
−
17
.
96 p a
=
4
.
49 p a
8
.
66
p a =
F
/
4
.
49
=
500
/
4
.
49
=
111
.
4 psi for cw rotation
Eq. (16-7): 500
=
56
.
87 p a
+
17
.
96 p a
8
.
66
9 psi for ccw rotation
A maximum pressure of 111
p a =
57
.
.
4 psioccurs on the RH shoe for cw rotation. Ans.
(b) RH shoe :
Eq. (16-6): T R =
0
.
28(111
.
4)(1
.
5)(6) 2 (cos 0 ◦ −
cos 120 ◦ )
=
2530 lbf
·
in Ans.
1
LH shoe :
Eq. (16-6): T L =
0
.
28(57
.
9)(1
.
5)(6) 2 (cos 0 ◦ −
cos 120 ◦ )
=
1310 lbf
·
in Ans.
1
T total =
2530
+
1310
=
3840 lbf
·
in Ans.
(c)
F x
Force vectors not to scale
F
F x
30
F
F y
30
y
F y
y
R y
Primary
shoe
R x
Secondary
shoe
R x
R
R
x
R y
x
672655307.003.png 672655307.004.png
Chapter 16
397
RH shoe :
F x =
500 sin 30°
=
250 lbf, F y =
500 cos 30°
=
433 lbf
1
120 ◦
0 ◦ =
2 −
2 π/ 3 rad
1
4 sin 2
Eqs. (16-8): A
=
2 sin 2
θ
0
.
375, B
=
θ
=
1
.
264
0
Eqs. (16-9): R x
=
111
.
4(1
.
5)(6)
[0
.
375
−
0
.
28(1
.
264)]
−
250
=−
229 lbf
1
R y
=
111
.
4(1
.
5)(6)
[1
.
264
+
0
.
28(0
.
375)]
−
433
=
940 lbf
1
R
=
[(
−
229) 2
+
(940) 2 ] 1 / 2
=
967 lbf Ans.
LH shoe :
F x =
250 lbf, F y =
433 lbf
Eqs. (16-10): R x
=
57
.
9(1
.
5)(6)
[0
.
375
+
0
.
28(1
.
264)]
−
250
=
130 lbf
1
R y
=
57
.
9(1
.
5)(6)
[1
.
264
−
0
.
28(0
.
375)]
−
433
=
171 lbf
1
R
=
[(130) 2
+
(171) 2 ] 1 / 2
=
215 lbf Ans.
16-2
θ 1 =
15°,
θ 2 =
105°,
θ a =
90°, sin
θ a =
1, a
=
5in
0
.
28 p a (1
.
5)(6)
105°
Eq. (16-2): M f
=
sin
θ
(6
−
5 cos
θ
) d
θ =
13
.
06 p a
1
15°
p a (1
.
5)(6)(5)
1
105°
Eq. (16-3): M N
=
sin 2
θ
d
θ =
46
.
59 p a
15°
c
=
2(5 cos 30°)
=
8
.
66 in
Eq. (16-4):
F
=
46
.
59 p a −
13
.
06 p a
=
3
.
872 p a
8
.
66
RH shoe :
p a =
500
/
3
.
872
=
129
.
1 psi on RH shoe for cw rotation Ans.
Eq. (16-6): T R
=
0
.
28(129
.
1)(1
.
5)(6 2 )(cos 15°
−
cos 105°)
=
2391 lbf
·
in
1
LH shoe :
500
=
46
.
59 p a +
13
.
06 p a
⇒
p a
=
72
.
59 psi on LH shoe for ccw rotation Ans.
8
.
66
T L
=
0
.
28(72
.
59)(1
.
5)(6 2 )(cos 15°
−
cos 105°)
=
1344 lbf
·
in
1
T total =
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.
672655307.005.png
398
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
16-3 Given:
θ 1 =
0°,
θ 2 =
120°,
θ a =
90°, sin
θ a =
1, a
=
R
=
90 mm, f
=
0
.
30,
F
=
1000 N
=
1kN, r
=
280
/
2
=
140 mm, counter-clockwise rotation.
LH shoe :
M f
=
fp a br
sin
r (1
−
cos
θ 2 )
−
2 sin 2
θ 2
θ a
=
0
.
30 p a (0
.
030)(0
.
140)
0
.
140(1
−
cos 120 ◦ )
−
0
.
090
2
sin 2 120°
1
=
0
.
000 222 p a N
·
m
M N
=
p a br a
sin
θ a
θ 2
2
−
1
4 sin 2
θ 2
=
p a (0
.
030)(0
.
140)(0
.
090)
120°
2
180
−
4 sin 2(120°)
1
=
4
.
777(10 − 4 ) p a N
·
m
c
=
2 r cos 180 ◦ − θ 2
2
=
2(0
.
090) cos 30 ◦ =
0
.
155 88 m
p a 4
F
=
1
=
.
777(10 − 4 )
−
2
.
22(10 − 4 )
=
1
.
64(10 − 3 ) p a
0
.
155 88
p a =
1
/
1
.
64(10 − 3 )
=
610 kPa
T L =
fp a br 2 (cos
θ 1 −
cos
θ 2 )
sin
θ a
=
0
.
30(610)(10 3 )(0
.
030)(0
.
140 2 )
[1
−
(
−
0
.
5)]
1
=
161
.
4N
·
m Ans.
RH shoe :
M f
=
2
.
22(10 − 4 ) p a N
·
m
M N
=
4
.
77(10 − 4 ) p a N
·
m
c
=
0
.
155 88 m
p a 4
F
=
1
=
.
77(10 − 4 )
+
2
.
22(10 − 4 )
=
4
.
49(10 − 3 ) p a
0
.
155 88
p a
=
1
=
222
.
8kPa Ans.
4
.
49(10 − 3 )
T R
=
(222
.
8
/
610)(161
.
4)
=
59
.
0N
·
m Ans
.
a
1
672655307.006.png
Chapter 16
399
16-4
(a) Given:
θ 1 =
10°,
θ 2 =
75°,
θ a =
75°, p a =
10 6 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:
r θ 2
θ 1
a 1
θ 2
a θ 2
θ 1
r −
θ θ 2
A
=
sin
θ
d
θ −
sin
θ
cos
θ
d
θ
=
cos
θ 1 −
2 sin 2
θ
θ 1
150 1
75°
10° =
=
200 −
cos
θ 75°
10° −
2 sin 2
θ
77
.
5mm
B =
θ 2
sin 2
θ
d
θ =
2 −
1
4 sin 2
θ
75 π/ 180 rad
10 π/ 180 rad =
0
.
528
θ 1
θ 2
C
=
sin
θ
cos
θ
d
θ =
0
.
4514
θ 1
Now converting to pascals and meters, we have from Eq. (16-2),
M f
=
fp a br
sin
A
=
0
.
24[(10) 6 ](0
.
075)(0
.
200)
(0
.
0775)
=
289 N
·
m
θ a
sin 75°
From Eq. (16-3),
M N
=
p a br a
sin
B
=
[(10) 6 ](0
.
075)(0
.
200)(0
.
150)
(0
.
528)
=
1230 N
·
m
θ a
sin 75°
Finally, using Eq. (16-4), we have
F
=
M N − M f
c
=
1230
−
289
=
5
.
70 kN Ans.
165
(b) Use Eq. (16-6) for the primary shoe.
T
=
fp a br 2 (cos
θ 1 −
cos
θ 2 )
sin
θ 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 p a .
Substituting
M N
=
1230
10 6
p a and M f
=
289
10 6
p a into Eq. (16-7),
5
.
70
=
(1230
/
10 6 ) p a
+
(289
/
10 6 ) p a
, solving gives p a =
619(10) 3 Pa
165
Then
T
=
0
.
24[0
.
619(10) 6 ](0
.
075)(0
.
200) 2 (cos 10°
−
cos 75°)
=
335 N
·
m
sin 75°
so the braking capacity is T total =
2(541)
+
2(335)
=
1750 N
·
m Ans.
672655307.001.png
400
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
(c) Primary shoes :
R x =
p a br
sin
θ a ( C
−
fB )
−
F x
=
(10 6 )(0
.
075)(0
.
200)
[0
.
4514
−
0
.
24(0
.
528)](10) − 3
−
5
.
70
=−
0
.
658 kN
sin 75°
R y
=
p a br
sin
θ a ( B
+
fC )
−
F y
=
(10 6 )(0
.
075)(0
.
200)
[0
.
528
+
0
.
24(0
.
4514)](10) − 3
−
0
=
9
.
88 kN
sin 75°
Secondary shoes :
R x =
p a br
sin
θ a ( C
+
fB )
−
F x
=
[0
.
619(10) 6 ](0
.
075)(0
.
200)
[0
.
4514
+
0
.
24(0
.
528)](10) − 3
−
5
.
70
sin 75°
=−
0
.
143 kN
R y
=
p a br
sin
θ a ( B
−
fC )
−
F y
=
[0
.
619(10) 6 ](0
.
075)(0
.
200)
[0
.
528
−
0
.
24(0
.
4514)](10) − 3
−
0
sin 75°
=
4
.
03 kN
Note from figure that
+
y for secondary shoe is opposite to
y
R
+
y for primary shoe.
R V
Combining horizontal and vertical components,
x
x
R H
R H =−
0
.
658
−
0
.
143
=−
0
.
801 kN
R V
=
(0
.
88
−
4
.
03
=
5
.
85 kN
y
R
=
.
801) 2
+
(5
.
85) 2
=
5
.
90 kN Ans.
16-5 Preliminaries:
θ 1 =
45°
−
tan − 1 (150
/
200)
=
8
.
13°,
θ 2 =
98
.
13°
θ a =
90°, a
=
[(150) 2
+
(200) 2 ] 1 / 2
=
250 mm
Eq. (16-8):
A
=
2 sin 2
θ 98 . 13°
8
=
0
.
480
. 13°
θ 2
θ =− cos
θ 98 . 13°
8
Let
C
=
sin
θ
d
=
1
.
1314
. 13°
θ 1
9
1
672655307.002.png

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