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

Given w = 100 mm t = 3 mm μ= 0.25.

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

Step I Tensions in band
The maximum tension in the band is P_{1} .

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

\frac{P_{1}}{P_{2}}=e^{\mu \theta}=e^{\left\{\frac{(0.25 \times 240) \pi}{180}\right\}}=2.85 .

\therefore \quad P_{2}=\frac{P_{1}}{2.85}=\frac{15000}{2.85}=5263 N .              (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,

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

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

M_{t}=\left(P_{1}-P_{2}\right) R=(15000-5263)(0.3) .

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

\text { Since }\left(\frac{a}{b}\right)=\frac{200}{50}=4 \text { and } e^{\mu \theta}=2.85 .

\therefore \quad\left(\frac{a}{b}\right)>e^{\mu \theta} .

The brake is not self-locking.