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ME 452: Machine Design II
Solution to Homework Set 13
Problem 1
(Problem 161, page 851).
We will solve the problem for a clockwise rotation of the drum.
For a clockwise rotation of the drum,
the right shoe is selfenergizing, because for the right shoe the friction force acting on the shoe causes a
clockwise moment about its hinge pin, which is in the same direction as the applied force, F.
In this
way, the friction helps apply the brake (see Figure 167, page 814 and Example 162, page 817).
(a)
The moment of the friction force about the hinge pin, see Equation (162), page 815, can be written
as
2
1
sin (
cos )
sin
a
f
a
fp br
M
ra
d
θ
θθ
=−
∫
The moment of the friction force about the hinge pin, by carrying out the above integration, see
Equation (168), page 816 and Example (162), page 817, can be written as
()
22
21
2
1
cos
cos
sin
sin
sin
2
a
f
a
fp br
a
Mr
⎡⎤
−
−
−
⎢⎥
⎣⎦
The angles
1
and
2
, see Figure for Problem (161), page 851 and Figure (167), page 814, are given as
12
0
and
120
=°
=
°
The angle at which the maximum pressure on the shoe occurs, see Figure 166(b), page 814 and
Example (162), page 817, is given as
90
a
=
°
The distance
a
, see Figure (167), page 814 and Figure for Problem (161), page 851, is given as
5 in
aR
=
=
The distance
c
, see Figure (167), page 814 Figure for Problem (161), page 851, is given as
2c
o
s
3
0
cR
=
°
Therefore, the distance
c
is given as
25c
o
s
3
0
i
n
c
=××
°
=
8
.
6
6
Therefore, the moment of the friction force about the hinge pin is given as
0.28
1.5 6
5
6
cos120
cos0
sin 120
sin 0
17.96
inlb
sin90
2
a
f
a
p
M
p
×××
=
× − ×
°−
° −
° =
°
The moment of the normal force about the hinge pin, see Equation (163), page 815, can be written as
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2
1
2
sin
sin
a
N
a
pbra
M
d
θ
=
∫
The moment of the normal force about the hinge pin, by carrying out the above integration, see Equation
(168), page 816 and Example (162), page 817, can be written as
()
21
2
1
11
sin2
sin
2
4
a
N
a
M
θθ
⎡⎤
=−
−
−
⎢⎥
⎣⎦
Therefore, the moment of the normal force about the hinge pin is given as
1.5 6 5
1 120
0
1
sin 240
sin0
56.87
inlb
sin90
2
180
180
4
a
N
a
p
Mp
ππ
××
×
×
×
⎛⎞
=×
−
−
°
−
°
=
⎜⎟
°
⎝⎠
For the selfenergizing (right) shoe, the actuating force, see Equation (164), page 815, can be written as
Nf
M
M
F
c
−
=
The actuating force is specified to be F = 500 lb.
Therefore the actuating force is given as
( )
56.87
17.96
500
4.49
8.66
aa
right
right
a
right
pp
p
−
==
Therefore, the maximum pressure at the selfenergizing (right) shoe, is given as
500
111.4 psi
4.49
a
right
p
(b)
The braking torque provided by the right shoe, see Equation (166), page 815, can be written as
( )
2
12
(cos
cos
)
sin
a
right
right
a
fp
b
r
T
−
=
Therefore, the braking torque provided by the right shoe is given as
2
0.28 111.4 1.5 (6) (cos0
cos120 )
2526.6 inlb
right
T
×
°
−
°
°
The left shoe is selfdeenergizing.
Therefore, for the left shoe, the actuation force, see Equation (167),
page 816, can be written as
M
M
F
c
+
=
Therefore the actuation force is given as
( )
56.87
17.96
500
8.64
8.66
left
left
a
left
p
+
3
Therefore, the maximum pressure at the left shoe is given as
()
500
57.9 psi
8.64
a
left
p
==
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This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue University.
 Fall '08
 Staff
 Machine Design

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