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Q: The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the hinge pins A and B
supports a pair of shoes. The actuating mechanism is to be arranged to produce the same force F on each shoe. The
face width of the shoes is 75 mm The material used permits a coefficient of friction of 0.26 and a maximum pressure of
1,300 kPa.
Q: The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the hinge pins A and B
supports a pair of shoes. The actuating mechanism is to be arranged to produce the same force F on each shoe. The
face width of the shoes is 75 mm. The material used permits a coefficient of friction of 0.24 and a maximum pressure of
1000 kPa. (a) Determine the...
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Q: The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the hinge pins A and B
supports a pair of shoes. The actuating mechanism is to be arranged to produce the same force F on each shoe. The
face width of the shoes is 75 mm The material used permits a coefficient of friction of 0.26 and a maximum pressure of
1,300 kPa.
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The Figure Shows A 400-mm Diameter Brake Drum With... | Chegg.com
Q: The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the hinge pins A and B
actuating mechanism is to be arranged to produce the same force F on each shoe. The
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face width of the shoes is 75 mm. The material used permits a coefficient of friction of 0.24 and a maximum pressure of
1000 kPa. (a) Determine the...
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The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of
the hinge pins A and B supports a pair of shoes. The actuating mechanism is to be arranged to
produce the same force F on each shoe. The face width of the shoes is 75 mm. The material
used permits a coefficient of friction of 0.24 and a maximum pressure of 1000 kPa
.
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Mechanical...
Fundamental Fundamental
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9th Edition
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(a) Determine the actuating force.
(b) Estimate the brake capacity.
(c) Noting that rotation may be in either direction, estimate the hinge-pin reactions.
Step-by-step solution
Step 1 of 11
Given:
Diameter of the brake drum,
Face width of the shoe,
Coefficient of the friction,
Maximum pressure,
And
Comment
Step 2 of 11
a) The moment of frictional forces,
Where
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Step 3 of 11
The moment of normal forces,
The actuating force is given by
The actuating force,
Comment
Step 4 of 11
b) The torque applied by the primary shoe is given by
For the secondary shoe,
The moment of frictional forces,
The moment of normal forces,
Actuating force,
Comment
Step 5 of 11
The torque applied by the secondary shoe is given by
The total braking capacity
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Brake capacity,
Comment
Step 6 of 11
c) For primary shoes
Reactions on hinge pin are given by
Where
And
Comment
Step 7 of 11
Horizontal reaction,
Horizontal reaction on primary shoe,
Comment
Step 8 of 11
Vertical reaction,
Vertical reaction on primary shoe,
Comment
Step 9 of 11
For secondary shoes:
Reactions on hinge pin are given by
Horizontal reaction,
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Horizontal reaction on secondary shoe,
Comment
Step 10 of 11
Vertical reaction,
Vertical reaction on secondary shoe,
Comment
Step 11 of 11
Net horizontal reaction,
Net vertical reaction,
Resultant reaction,
Resultant reaction,
Comment
27
2
Recommended solutions for you in Chapter 16
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Chapter 16,Home
Problem 9P
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The shoes on the brake
depicted in the figure
subtend a 90 arc on the
drum of this external
pivoted-shoe brake. The
actuation...
The block-type hand
brake shown in the figure
has a face width of 1.25
in and a mean coefficient
of friction of 0.25. For
an...
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CHAPTER 16
Q. 16.4
Shigley’s Mechanical Engineering Design – Solution Manual [EXP-48072] (https://holooly.com/sources/shigleys-mechanicalengineering-design-solution-manual-exp-48072/)
The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the hinge pins A and B
supports a pair of shoes. The actuating mechanism is to be arranged to produce the same force F on each shoe. The face
width of the shoes is 75 mm. The material used permits a coefficient of friction of 0.24 and a maximum pressure of 1000 kPa.
(a) Determine the actuating force.
(b) Estimate the brake capacity.
(c) Noting that rotation may be in either direction, estimate the hinge-pin reactions.
Step-by-Step
Report Solution (https://holooly.com/report-a-problem/)
Verified Solution 
(a) Given: θ1
= 10∘ , θ2 = 75∘ , θa = 75∘ , pa = 106 P a, 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:
A = [r ∫
θ2
​
​
θ1
sin θdθ − a ∫
θ2
θ1
​
​
θ2
1
sin θ cos θdθ] = r[− cos θ]θθ21 − a [ sin2 θ]
2
θ1
​
​
​
​
​
​
​
​
​
75∘
∘
1
2
= 200[− cos θ]75
= 77.5 mm
10∘ − 150 [ sin θ ]
2
10∘
​
B=∫
C=∫
θ2
sin2 θdθ = [
​
​
θ1
θ2
​
​
θ
1
− sin 2θ]
= 0.528
2
4
​
​
​
​
sin θ cos θdθ = 0.4514
​
θ1
​
​
Now converting to Pascals and meters, we have from Eq. (16-2),
Mf =
​
f pa br
sin θa A
​
​
=
0.24(106 )(0.075)(0.200)
(0.0775)
sin 75∘
= 289 N ⋅ m
106 (0.075)(0.200)(0.150)
(0.528)
sin 75∘
= 1230 N ⋅ m
​
​
From Eq. (16-3),
MN =
​
pa bra
sin θa B
​
​
=
​
​
Finally, using Eq. (16-4), we have
F =
MN −Mf
c
​
​
​
=
1230−289
165
​
= 5.70 kN
(b) Use Eq. (16-6) for the primary shoe.
f pa br2 (cos θ1 − cos θ2 )
sin θa
0.24 (106 ) (0.075)(0.200)2 (cos 10∘ − cos 75∘ )
=
= 541 N ⋅ m
sin 75∘
For the secondary shoe, we must first find pa . Substituting
https://holooly.com/solutions/the-figure-shows-a-400-mm-diameter-brake-drum-with-four-internally-expanding-shoes-each-of-the-hinge-pins-a-and-b-su… 1/3
T =
​
​
​
​
​
​
5/23/22, 4:01 PM
The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the hinge pins A and B supports a…
For the secondary shoe, we must first find pa . Substituting
​
MN =
(https://holooly.com)
1230
289
​
5.70 =
106
​
pa and Mf =
​
​
106
​
(1230/106 )pa +(289/106 )pa
​
165
pa
​
​

into Eq. (16 – 7),
​
pa = 619 (103 ) P a
, solving gives
​
Then
T =
0.24[619(103 )]0.075(0.2002 )(cos 10∘ −cos 75∘ )
sin 75∘
so the braking capacity is Ttotal
​
= 335 N ⋅ m
= 2(541) + 2(335) = 1750 N ⋅ m
​
(c) Primary shoes:
pa br
(C − f B) − Fx
sin θa
6
10 (0.075)0.200
=
[0.4514 − 0.24(0.528)] (10−3 ) − 5.70 = −0.658 kN
sin 75∘
Rx =
​
​
​
​
​
pa br
(B + f C) − Fy
sin θa
6
10 (0.075)0.200
=
[0.528 + 0.24(0.4514)] (10−3 ) − 0 = 9.88 kN
sin 75∘
Ry =
​
​
​
​
​
Secondary shoes:
pa br
(C + f B) − Fx
sin θa
0.619 (106 ) 0.075(0.200)
=
[0.4514 + 0.24(0.528)] (10−3 ) − 5.70
sin 75∘
= −0.143 kN
Rx =
​
​
​
​
​
pa br
(B − f C) − Fy
sin θa
0.619 (106 ) 0.075(0.200)
=
[0.528 − 0.24(0.4514)] (10−3 ) − 0
sin 75∘
= 4.03 kN
Ry =
​
​
​
​
​
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
________________________________________________________________________________________________
____________________________
Eq. (16-2): Mf
Eq. (16-3):
Eq. (16-4):
Eq. (16-6):
f pa br
sin θa
= ∫ f dN (r − a cos θ) =
​
MN = ∫ dN (a sin θ) =
​
F =
MN −Mf
c
​
θ
​
​
pa bra θ2
sin θa ∫θ1
​
​
​
​
​
​
∫θ12 sin θ(r − a cos θ)dθ
​
​
​
sin2 θdθ
​
​
​
f pa br2 θ2
∫ sin θdθ
sin θa θ1
2
f pa br (cos θ1 − cos θ2 )
=
sin θa
T = ∫ f rdN =
​
​
​
​
​
​
​
​
​
​
​
​
Eq. (16-7):
F =
MN +Mf
c
​
​
​
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