A block brake with a short shoe is shown in Fig. 12.8. It is to be designed so that the product pv is limited to 2. where, p = normal pressure between friction lining and the brake drum (N/mm^2) v = peripheral velocity of brake drum (m/s) The coefficient of friction between the brake drum and the

Question

A block brake with a short shoe is shown in Fig. 12.8. It is to be designed so that the product pv is limited to 2. where,
p = normal pressure between friction lining and the brake drum (N/mm²)
v = peripheral velocity of brake drum (m/s)
The coefficient of friction between the brake drum and the friction lining is 0.2. The cable drum is connected to the brake drum by means of a pair of spur gears. The brake drum rotates four times as fast as the cable drum. The permissible intensity of pressure on friction lining is 1 N/mm². Calculate:
(i) The magnitude of the brake shoe force (P)
(ii) The area of friction lining
(iii) The uniform velocity at which the mass can be lowered. What happens at higher speeds?

Accepted Answer

[latex] ext { Given }(p v)=2 \quad \mu=0.2 \quad p_{\max }=1 N / mm ^{2} [/latex].

Step I Brake shoe force
The following notations are used in this example:

[latex] n_{1}= ext { speed of rotation of cable drum }( rpm ) [/latex].

[latex] n_{2}= ext { speed of rotation of brake drum (rpm) } [/latex].

[latex] \left(M_{t} ight)_{1}= ext { torque on cable drum }( N - mm ) [/latex].

[latex] \left(M_{t} ight)_{2}= ext { torque on brake drum (N-mm) } [/latex].

[latex] v_{1}= ext { peripheral velocity of cable drum }( m / s ) ext {. } [/latex]

It is also the velocity with which the mass is lowered

[latex] v_{2}= ext { peripheral velocity of brake drum }( m / s ) [/latex].

The brake drum rotates four times as fast as the cable drum.

[latex] n_{2}=4 n_{1} [/latex] (a).

Referring to Fig. 12.8(a),

[latex] \left(M_{t} ight)_{1}=(m g) imes 150=(500 imes 9.81) imes 150 N - mm [/latex] ( b ).

Referring to Fig.12.8(b),

[latex] \left(M_{t} ight)_{2}=\mu P imes 200=(0.2 P) imes 200 N - mm [/latex] (c)

[latex] \frac{2 \pi n_{1}\left(M_{t} ight)_{1}}{60 imes 10^{6}}=\frac{2 \pi n_{2}\left(M_{t} ight)_{2}}{60 imes 10^{6}} [/latex].

[latex] herefore \quad n_{1}\left(M_{t} ight)_{1}=n_{2}\left(M_{t} ight)_{2} [/latex].

[latex] herefore \quad n_{1}(500 imes 9.81) imes 150=4 n_{1}(0.2 P) imes 200 [/latex].

∴ P = 4598.44 N (i)
Step II Area of friction lining

[latex] P=p_{\max .} imes ext { Area of lining } [/latex].

[latex] herefore ext { Area of friction lining }=\frac{P}{p_{\max }}=\frac{4598.44}{1} [/latex]

= 4598.44 mm² (ii)
Step III Uniform velocity at which mass can be lowered

[latex] v_{2} p_{2}=2 \quad herefore v_{2}=\frac{2}{p_{2}}=\frac{2}{1}=2 m / s [/latex] (d).

[latex] \omega_{2}=\frac{v_{2}}{R_{2}}=\frac{2}{0.2}=10 rad / s [/latex].

[latex] ext { and } \quad \omega_{1}=\frac{\omega_{2}}{4}=\frac{10}{4}=2.5 rad / s [/latex].

[latex] v_{1}=\omega_{1} R_{1}=2.5(0.15)=0.375 m / s [/latex].

[latex] ext { or } \quad v_{1}=0.375 imes 60=22.5 m / min [/latex] (iii).

When the coefficient of friction is constant, the rate of heat generated is proportional to the product pv. Therefore, at higher speeds the brake drum will be overheated.

Question 12.6: A block brake with a short shoe is shown in Fig. 12.8. It is...

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