Question 4 of 6: Single short-shoe external drum brake — torque, actuating force and self-locking (20 marks)

Question 4: Single short-shoe external drum brake — torque, actuating force and self-locking (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single external shoe pivoted on a lever and pressed against a rotating drum by an actuating force; the figure defines the lever geometry and the wrap angle.

Find. The braking torque capacity, the actuating force $F_a$ required to develop it, and the value of the friction moment arm $c$ at which the brake becomes self-locking.

Approach. With a wrap angle of only 50° the shoe qualifies as "short", so the lining pressure may be taken as uniform at $p_{\max}$ and the resultant normal force treated as a single radial force acting through the drum centre. Get $N$ from the projected lining area, the torque from $\mu N r$, and $F_a$ from moments about the shoe pivot — being careful with the sign of the friction moment, which is what makes the shoe self-energizing and eventually self-locking.

1. Projected lining area. For a short shoe the resultant normal force is found from the chord-projected area, i.e. the area seen looking along the line of the resultant, of width $w$ and chord length $2r\sin(\theta/2)$. $$A_{proj} = w\,\bigl(2r\sin\tfrac{\theta}{2}\bigr) = 40 \times 2(35)\sin 25^\circ = 40 \times 29.58 = 1183\text{ mm}^2$$
2. Resultant normal force at the pressure limit. Set the uniform pressure equal to the allowable lining pressure, which is what fixes the brake's capacity. $$N = p_{\max}\,A_{proj} = (1.3\text{ N/mm}^2)(1183\text{ mm}^2)$$ $$\boxed{\;N = 1538\text{ N}\;}$$
3. Torque capacity. The friction force $\mu N$ acts tangentially at the drum surface, so its moment about the drum axis is the braking torque. $$T = \mu N r = 0.3 \times 1538\text{ N} \times 0.035\text{ m}$$ $$\boxed{\;T = 16.15\text{ N}\!\cdot\!\text{m}\;}$$
4. Identify the friction moment arm about the pivot. The normal force $N$ is radial (vertical, through the drum centre) and acts at horizontal distance $b$ from the pivot, so its moment about $O_1$ is $Nb$. The friction force $\mu N$ is tangential (horizontal) and acts at the drum surface, a height $r$ above the drum axis, while the pivot sits a height $e$ above that same axis. Its moment arm about the pivot is therefore the vertical offset $$c = r - e = 35 - 20 = 15\text{ mm}$$
5. Moment balance on the shoe lever. With the drum rotating in the sense shown, the friction force drags the shoe into the drum, so the friction moment about $O_1$ acts in the same direction as the applied moment and helps to apply the brake — the shoe is self-energizing. Summing moments about $O_1$: $$F_a\,a + \mu N c - N b = 0 \;\Longrightarrow\; F_a = \frac{N\,(b - \mu c)}{a}$$ $$F_a = \frac{1538\,(70 - 0.3 \times 15)}{120} = \frac{1538 \times 65.5}{120}$$ $$\boxed{\;F_a = 840\text{ N}\;}$$ Self-energizing action has saved about 6 % of the actuating effort: without it the required force would be $N b / a = 897$ N.
6. Check the reversed rotation case. If the drum turns the other way the friction moment opposes the applied moment and the shoe becomes de-energizing: $$F_{a,rev} = \frac{N\,(b + \mu c)}{a} = \frac{1538\,(70 + 4.5)}{120} = 955\text{ N}$$ The same brake needs 14 % more actuating force in reverse, which is why single-shoe brakes are direction-sensitive and why paired opposed shoes are used where torque must be equal in both directions.
7. Condition for self-locking. The brake self-locks when the friction moment alone can hold the shoe against the drum, i.e. when the required actuating force falls to zero or below. $$F_a \le 0 \;\Longrightarrow\; b - \mu c \le 0 \;\Longrightarrow\; c \ge \frac{b}{\mu}$$ $$\boxed{\;c \ge \frac{70}{0.3} = 233\text{ mm}\;}$$
8. Interpret the self-locking geometry. The present design has $c = 15$ mm, an order of magnitude below the locking threshold, so the brake is safely non-locking. To reach $c = 233$ mm with $r = 35$ mm the pivot would have to be relocated to $$e = r - c = 35 - 233 = -198\text{ mm}$$ that is, about 198 mm below the drum axis rather than 20 mm above it. Self-locking is normally something to design away from in a service brake, because the drum would seize on the lightest touch of the pedal and the vehicle or hoist would be uncontrollable; it is deliberately exploited only in back-stops and one-way holding devices.