Example 19.5. A double shoe brake, as shown in Fig. 19.10,
is capable of absorbing a torque of 1400 N-m. The diameter of the
brake drum is 350 mm and the angle of contact for each shoe is 100°.
If the coefficient of friction between the brake drum and lining is
0.4 ; find 1. the spring force necessary to set the brake ; and 2. the
width of the brake shoes, if the bearing pressure on the lining
material is not to exceed 0.3 N/mm2.
Answers
Answer:
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Answer:
b=144.2
Explanation:
=1400N–m=1400×10
3
N–mm;d = 350 mm or r = 175 mm ; 2 \theta=100^{\circ}=100 \times \pi / 180=1.75 rad ;2θ=100
∘
=100×π/180=1.75rad;; \mu=0.4 ; p_{b}=0.3 N / mm ^{2}μ=0.4;p
b
=0.3N/mm
2
1. Spring force necessary to set the brake
Let S = Spring force necessary to set the brake.
R_{ N 1} \text { and } F_{t 1}R
N1
and F
t1
= Normal reaction and the braking force on the right hand side shoe, and
R_{ N 2} \text { and } F_{t 2}R
N2
and F
t2
= Corresponding values on the left hand side shoe.
Since the angle of contact is greater than 60^{\circ}60
∘
, therefore equivalent coefficient of friction,
\mu^{\prime}=\frac{4 \mu \sin \theta}{2 \theta+\sin 2 \theta}=\frac{4 \times 0.4 \times \sin 50^{\circ}}{1.75+\sin 100^{\circ}}=0.45μ
′
=
2θ+sin2θ
4μsinθ
=
1.75+sin100
∘
4×0.4×sin50
∘
=0.45
Taking moments about the fulcrum O_{1},O
1
,, we have
S \times 450=R_{ N 1} \times 200+F_{t 1}(175-40)=\frac{F_{t 1}}{0.45} \times 200+F_{t 1} \times 135=579.4 F_{t 1}S×450=R
N1
×200+F
t1
(175−40)=
0.45
F
t1
×200+F
t1
×135=579.4F
t1
\ldots\left(\text { Substituting } R_{ N 1}=\frac{F_{t 1}}{\mu^{\prime}}\right)…( Substituting R
N1
=
μ
′
F
t1
)
\therefore \quad F_{t 1}=S \times 450 / 579.4=0.776 S∴F
t1
=S×450/579.4=0.776S
Again taking moments about O_{2},O
2
,, we have
S \times 450+F_{t 2}(175-40)=R_{ N 2} \times 200=\frac{F_{t 2}}{0.45} \times 200=444.4 F_{t 2}S×450+F
t2
(175−40)=R
N2
×200=
0.45
F
t2
×200=444.4F
t2
\ldots\left(\text { Substituting } R_{ N 2}=\frac{F_{t 2}}{\mu^{\prime}}\right)…( Substituting R
N2
=
μ
′
F
t2
)
444.4 F_{t 2}-135 F_{t 2}=S \times 450 \quad \text { or } \quad 309.4 F_{t 2}=S \times 450444.4F
t2
−135F
t2
=S×450 or 309.4F
t2
=S×450
\therefore \quad F_{t 2}=S \times 450 / 309.4=1.454 S∴F
t2
=S×450/309.4=1.454S
We know that torque capacity of the brake \left(T_{ B }\right)(T
B
),
1400 \times 10^{3}=\left(F_{t 1}+F_{t 2}\right) r=(0.776 S+1.454 S) 175=390.25 S1400×10
3
=(F
t1
+F
t2
)r=(0.776S+1.454S)175=390.25S
\therefore∴ S=1400 \times 10^{3} / 390.25=3587 NS=1400×10
3
/390.25=3587N
2. Width of the brake shoes
Let b = Width of the brake shoes in mm.
We know that projected bearing area for one shoe,
A_{b}=b(2 r \sin \theta)=b\left(2 \times 175 \sin 50^{\circ}\right)=268 b mm ^{2}A
b
=b(2rsinθ)=b(2×175sin50
∘
)=268bmm
2
Normal force on the right hand side of the shoe,
R_{ N 1}=\frac{F_{t 1}}{\mu^{\prime}}=\frac{0.776 \times S}{0.45}=\frac{0.776 \times 3587}{0.45}=6186 NR
N1
=
μ
′
F
t1
=
0.45
0.776×S
=
0.45
0.776×3587
=6186N
and normal force on the left hand side of the shoe,
R_{ N 2}=\frac{F_{t 2}}{\mu^{\prime}}=\frac{1.454 \times S}{0.45}=\frac{1.454 \times 3587}{0.45}=11590 NR
N2
=
μ
′
F
t2
=
0.45
1.454×S
=
0.45
1.454×3587
=11590N
We see that the maximum normal force is on the left hand side of the shoe. Therefore we shall find the width of the shoe for the maximum normal force i.e. R_{ N 2}R
N2
.
We know that the bearing pressure on the lining material \left(p_{b}\right)(p
b
),
0.3=\frac{R_{ N 2}}{A_{b}}=\frac{11590}{268 b }=\frac{43.25}{ b }0.3=
A
b
R
N2
=
268b
11590
=
b
43.25
\therefore∴ b = 43.25 / 0.3 = 144.2 mm