A band brake working on the back-stop principle is shown in Fig. 12.25(a). The width of friction lining perpendicular to the axis of the drum is 75 mm. In normal operating condition, the drum rotates in clockwise direction.
Calculate:
(i) the minimum value of coefficient of friction between the lining and the drum so that the brake acts as a back-stop brake.
(ii) If the normal pressure between the drum and the lining is to be limited to 0.3 N/mm², what could be the maximum braking torque?
\text { Given } w=75 mm \quad p_{\max .}=0.3 N / mm ^{2} .
R = 150 mm
Step I Minimum coefficient of friction for back-stop brake
Refer to Fig. 12.25(b). Construct a line OB perpendicular to AC.
\therefore \quad O B \perp A C \quad \text { and } \quad A E \perp O A .
It is a pair of two perpendicular lines.
\text { Since } \angle A O B=30^{\circ} \quad \therefore \angle E A C=30^{\circ} .
A B=O A \sin \left(30^{\circ}\right)=150 \sin \left(30^{\circ}\right)=75 mmDA = DB + BA = FO + AB
= 150 + 75 = 225 mm
AC = AD – CD = 225 – 50 = 175 mm.
E C=A C \sin \left(30^{\circ}\right)=175 \sin \left(30^{\circ}\right)=87.5 mm .
The direction of rotation for back-stop action is anti-clockwise as shown in Fig. 12.25(c). In order to prevent rotation in anti-clockwise direction,
P_{2} \times 87.5<P_{1} \times 50 .
\text { or } \quad \frac{P_{1}}{P_{2}}>\frac{87.5}{50} \quad \therefore \frac{P_{1}}{P_{2}}>1.75 .
\frac{P_{1}}{P_{2}}=e^{\mu \theta}=e^{\left\{\frac{(\mu \times 240) \pi}{180}\right\}}>1.75 .
\therefore \quad \frac{(\mu \times 240) \pi}{180}>\log _{e}(1.75) .
\therefore \quad \mu>0.1336 (i)
Step II Maximum braking torque
From Eq. (12.27),
p_{\max }=\frac{P_{1}}{R w} (12.27).
P_{1}=R w p_{\max }=150(75)(0.3)=3375 N\text { and } \quad \frac{P_{1}}{P_{2}}=1.75 .
\therefore \quad P_{2}=1928.57 N .
∴ braking torque = (3375 – 1928.57) × 150
= 216 964.5 N-mm
∴ = 216.96 N-m (ii).