Question 5 of 6: Double short-shoe external drum brake — torque, actuating force and self-locking (20 marks)

Question 5: Double 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. Two opposed short shoes acting on the outside of a drum, each on its own pivoted lever.

Find. The braking torque capacity, the actuating force $F_a$ required at each lever, and the value of the friction moment arm $c$ that would make the brake self-locking.

Approach. A short shoe (wrap angle below roughly 45°) is treated as if the whole normal force acted at its centre, so the pressure distribution need not be integrated; the normal force follows from the projected contact area at $p_{\max}$, the torque from the friction force at the drum radius, and the actuating force from a moment balance about the shoe pivot in which the friction force has the arm $c = r - e$.

1. Establish the friction moment arm from the figure. The friction force on the upper shoe acts along the tangent at the top of the drum, a distance $r$ above the drum axis, while the pivot $O_1$ sits a distance $e$ above that axis. The perpendicular distance between them is therefore $$c = r - e = 40 - 30 = 10\ \text{mm}$$ This is exactly the dimension the figure labels $c$, and it is the quantity the last part of the question asks about.
2. Projected contact width of the short shoe. Because the resultant normal force is taken as acting through the shoe centreline, the effective bearing area is the drum width multiplied by the chord subtended by the wrap angle: $$\ell = 2r\sin\frac{\theta}{2}=2\times 40\times\sin 12.5^\circ = 80\times 0.21644 = 17.32\ \text{mm}$$
3. Normal force at the pressure limit. Loading the lining to its allowable pressure, $$N = p_{\max}\,w\,\ell = 1.3\times 60\times 17.32$$ $$\boxed{N = 1351\ \text{N per shoe}}$$
4. Friction force and torque per shoe. The tangential friction force is $\mu N$, acting at the drum radius: $$T_{\text{shoe}} = \mu N r = 0.25\times 1351\times 0.040 = 13.51\ \text{N}\cdot\text{m}$$
5. Total torque capacity. Both shoes bear on the drum, and both are taken to the same limiting pressure, so $$T = 2\mu N r = 2\times 13.51$$ $$\boxed{T = 27.0\ \text{N}\cdot\text{m}}$$
6. Moment balance about the pivot of the self-energising shoe. With the drum turning so that friction drags the upper shoe toward its pivot, the friction moment acts in the same sense as the actuating moment and helps to apply the shoe. Summing moments about $O_1$, $$F_a\,a + \mu N c - N b = 0 \quad\Longrightarrow\quad F_a = \frac{N\left(b-\mu c\right)}{a}$$ $$F_a = \frac{1351\times\left(80-0.25\times 10\right)}{90}=\frac{1351\times 77.5}{90}$$ $$\boxed{F_a = 1163\ \text{N}}$$
7. The opposite (de-energising) lever. For the second shoe the friction moment opposes the actuating moment, so the sign of the friction term flips: $$F_a' = \frac{N\left(b+\mu c\right)}{a}=\frac{1351\times 82.5}{90}=1238\ \text{N}$$ If both levers are driven by a common actuator, that actuator must deliver the larger value, 1238 N, and the self-energising shoe will then run slightly above $p_{\max}$ unless the linkage is proportioned to compensate — a point worth stating in the answer.
8. Condition for self-locking. The brake grabs when the friction moment alone is enough to hold the shoe on, i.e. when the required actuating force falls to zero or below: $$b-\mu c \le 0 \quad\Longrightarrow\quad c \ge \frac{b}{\mu}=\frac{80}{0.25}$$ $$\boxed{c \ge 320\ \text{mm for self-locking}}$$ Since $c = r-e$, this would demand $e = 40-320 = -280$ mm, i.e. the pivot placed 280 mm below the drum axis instead of 30 mm above it. The brake as drawn is therefore very far from self-locking, which is the desired condition for a service brake — the actuating force stays proportional to the applied torque, so the brake is controllable rather than grabbing.