An internal-expanding brake with four identical shoes is shown in Fig. 12.19. Each hinge pin supports a pair of shoes. The actuating mechanism is designed in such a way that it produces the same force P on each of the four shoes. The face width of the friction lining is 50 mm and the maximum

Question

An internal-expanding brake with four identical shoes is shown in Fig. 12.19. Each hinge pin supports a pair of shoes. The actuating mechanism is designed in such a way that it produces the same force P on each of the four shoes. The face width of the friction lining is 50 mm and the maximum intensity of normal pressure is limited to 1 N/mm². The coefficient of friction is 0.30. Calculate:
(i) the actuating force P; and
(ii) the torque-absorbing capacity of the brake.

Accepted Answer

[latex] ext { Given } w=50 mm \quad \mu=0.3 \quad p_{\max }=1 N / mm ^{2} [/latex].

R = 250 mm h = 200 mm
Step I Actuating force
It is assumed that the maximum normal pressure will occur between the drum and the lining on right-hand shoe of the upper half. For the righthand shoe,

[latex] heta_{1}=15^{\circ} \quad heta_{2}=75^{\circ} \quad \phi_{\max }=75^{\circ} [/latex].

[latex] \sin \phi_{\max .}=0.9659 \quad h=200 mm \quad c=200 mm [/latex].

where h is the distance of the pivot axis from the axis of the brake drum.
From Eq. (12.19),

[latex] M_{f}=\frac{\mu p_{\max } R w\left[4 R\left(\cos heta_{1}-\cos heta_{2} ight)-h\left(\cos 2 heta_{1}-\cos 2 heta_{2} ight) ight]}{4 \sin \phi_{\max .}} [/latex] (12.19).

[latex] M_{f}=\frac{\mu p_{\max } R w\left[4 R\left(\cos heta_{1}-\cos heta_{2} ight)-h\left(\cos 2 heta_{1}-\cos 2 heta_{2} ight) ight]}{4 \sin \phi_{\max }} [/latex].

[latex] =\frac{0.3(1)(250)(50)\left[4(250)\left(\cos 15^{\circ}-\cos 75^{\circ} ight)-200\left(\cos 30^{\circ}-\cos 150^{\circ} ight) ight]}{ 4 (0.9659)} [/latex].

= 350 091.19 N-mm
From Eq. (12.20),

[latex] M_{n}=\frac{p_{\max } R w h\left[2\left( heta_{2}- heta_{1} ight)-\left(\sin 2 heta_{2}-\sin 2 heta_{1} ight) ight]}{4 \sin \phi_{\max }} [/latex] (12.20).

[latex] M_{n}=\frac{p_{\max } R w h\left[2\left( heta_{2}- heta_{1} ight)-\left(\sin 2 heta_{2}-\sin 2 heta_{1} ight) ight]}{4 \sin \phi_{\max }} [/latex]

[latex] =\frac{1(250)(50)(200)\left[2\left(\frac{60 \pi}{180} ight)-\left(\sin 150^{\circ}-\sin 30^{\circ} ight) ight]}{4(0.9659)} [/latex].

= 1 355 209.59 N-mm.

From Eq. (12.22),

[latex] P=\frac{M_{n}-M_{f}}{C} [/latex] (12.22).

[latex] P=\frac{M_{n}-M_{f}}{C} [/latex]

[latex] =\frac{1355209.59-350091.19}{200}=5025.59 N [/latex] (i).

Step II Torque-absorbing capacity
From Eq. (12.21), the torque [latex] \left(M_{t} ight)_{R} [/latex] for the right hand shoe in the upper half is given by

[latex] M_{t}=\frac{\mu R^{2} p_{\max .} w\left(\cos heta_{1}-\cos heta_{2} ight)}{\sin \phi_{\max }} [/latex] (12.21).

[latex] \left(M_{t} ight)_{R}=\frac{\mu R^{2} p_{\max } w\left(\cos heta_{1}-\cos heta_{2} ight)}{\sin \phi_{\max }} [/latex].

[latex] =\frac{0.3(250)^{2}(1)(50)\left(\cos 15^{\circ}-\cos 75^{\circ} ight)}{0.9659} [/latex].

= 686 315.98 N-mm.

The maximum intensity of pressure for the left-hand shoe in the upper half is unknown. For identical shoes, it can be seen from the expressions of [latex] M_{n} ext { and } M_{f} [/latex] that both are proportional to [latex] \left(p_{\max } ight) [/latex].
For the left-hand shoe, the maximum intensity of pressure is taken as [latex] \left(p_{\max }^{\prime} ight) [/latex]. Therefore, for the left-hand shoe,

[latex] M_{f}^{\prime}=\frac{350091.19 p_{\max .}^{\prime}}{p_{\max }}=\frac{(350091.19) p_{\max .}^{\prime}}{(1)} [/latex].

[latex] =350091.19 p_{\max }^{\prime} [/latex].

[latex] M_{n}^{\prime}=\frac{1355209.59 p_{\max }^{\prime}}{p_{\max }}=\frac{(1355209.59) p_{\max }^{\prime}}{(1)} [/latex].

[latex] =1355209.59 p_{\max }^{\prime} [/latex].

Using Eq. (12.23) for the left-hand shoe in the upper half,

[latex] P=\frac{M_{n}+M_{f}}{C} [/latex] (12.23).

[latex] P=\frac{M_{n}^{\prime}+M_{f}^{\prime}}{C} [/latex].

[latex] ext { or } \quad 5025.59=\frac{(350091.19+1355209.59) p_{\max }^{\prime}}{200} [/latex].

[latex] herefore \quad p_{\max .}^{\prime}=0.59 N / mm ^{2} [/latex].

[latex] \left(M_{t} ight)_{L}=686315.98\left(\frac{p_{\max }^{\prime}}{p_{\max }} ight)=686315.98\left(\frac{0.59}{1} ight) [/latex].

[latex] =404926.43 \quad N - mm [/latex].

The total torque-absorbing capacity of the pair of shoes in the upper half of the brake drum is given by,

[latex] \left(M_{t} ight)_{u h}=\left(M_{t} ight)_{R}+\left(M_{t} ight)_{L}=686315.98+404926.43 [/latex].

= 1 091 242.41 N-mm
Since all four shoes are identical, the two pairs are also identical. Therefore,

[latex] M_{t}=\left(M_{t} ight)_{u h}+\left(M_{t} ight)_{f f}=2\left(M_{t} ight)_{u h}=2(1091242.41) [/latex].

= 2 182 484.82 N-mm
or, [latex] M_{t}=2182.48 N - m [/latex] (ii).

Question 12.9: An internal-expanding brake with four identical shoes is sho...

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