A differential band brake is shown in Fig. 12.24(a). The width and the thickness of the steel band are 100 mm and 3 mm respectively and the maximum tensile stress in the band is 50 N/mm². The coefficient of friction between the friction lining and the brake drum is 0.25. Calculate:
(i) the tensions in the band;
(ii) the actuating force; and
(iii) the torque capacity of the brake.
Find out whether the brake is self-locking.

Question

Accepted Answer

Given w = 100 mm t = 3 mm μ= 0.25.

[latex] \sigma_{t}=50 N / mm ^{2} \quad R=300 mm =0.3 m [/latex].

Step I Tensions in band
The maximum tension in the band is [latex] P_{1} [/latex].

[latex] P_{1}=\sigma_{t} w t=50(100) 3=15000 N [/latex].

[latex] \frac{P_{1}}{P_{2}}=e^{\mu heta}=e^{\left\{\frac{(0.25 imes 240) \pi}{180} ight\}}=2.85 [/latex].

[latex] herefore \quad P_{2}=\frac{P_{1}}{2.85}=\frac{15000}{2.85}=5263 N [/latex]. (i).

Step II Actuating force
The free-body diagram of forces acting on the band and the actuating lever is shown in Fig. 12.24(b).
Taking moment of forces about the fulcrum,

[latex] P_{1}(50)+P(950)-P_{2}(200)=0 [/latex].

(15 000)(50) + P(950) – (5263)(200) = 0
or P = 318.5 N (ii)
Step III Torque capacity of brake

[latex] M_{t}=\left(P_{1}-P_{2} ight) R=(15000-5263)(0.3) [/latex].

= 2921.1 N-m (iii)
Step IV Self-locking property

[latex] ext { Since }\left(\frac{a}{b} ight)=\frac{200}{50}=4 ext { and } e^{\mu heta}=2.85 [/latex].

[latex] herefore \quad\left(\frac{a}{b} ight)>e^{\mu heta} [/latex].

The brake is not self-locking.

Question 12.10: A differential band brake is shown in Fig. 12.24(a). The wid...

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