Question 3 of 6: Double Short-Shoe External Drum Brake
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, 16-Mec-B1 Advanced Machine Design, May 2018. Open book, 3 hours, 100 marks. Part I (Problems 1 & 2) is compulsory; candidates answer only three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Given. Two opposed short shoes press on a drum of radius 40 mm and width 60 mm; each shoe pivots on a lever with the actuating force applied at the given $a,b$ offsets.
Find. Torque capacity $T$, the actuating force $F_a$, and the friction-arm value $c$ that makes the brake self-locking.
Double short-shoe external brake: two shoes at $p_{\max}$ grip the drum; the actuating force acts through the pivot lever with arms $a$ and $b$.
Approach. For a short shoe the pressure is taken uniform at $p_{\max}$ over the projected pad area; the normal force gives the friction torque per shoe, doubled for the two shoes. A moment balance on the self-energizing lever gives $F_a$, and setting $F_a=0$ gives the self-locking condition.
Normal force per shoe. Short-shoe theory uses the projected pad area $A=w\,(2r\sin\tfrac{\theta}{2})$ at uniform $p_{\max}$:
$$N=p_{\max}\,w\,(2r\sin\tfrac{\theta}{2})=1.3\times10^{6}\,(0.060)(2\times0.040\sin15^\circ)=1615\ \text{N}.$$
Friction torque (both shoes). Each shoe contributes $\mu N r$; with two shoes at $p_{\max}$,
$$T=2\mu N r=2(0.3)(1615)(0.040)=\boxed{38.8\ \text{N}\cdot\text{m}}.$$
Actuating force on the self-energizing shoe. Taking moments of $N$ and the friction force $\mu N$ about the pivot $O_1$, with friction moment arm $c=r-e=40-30=10\ \text{mm}$:
$$F_a=\frac{N\,(b-\mu c)}{a}=\frac{1615\,(0.080-0.3\times0.010)}{0.090}=1382\ \text{N}.$$
Self-locking condition. The brake self-locks when the friction moment alone closes the shoe, i.e. $F_a\le0\Rightarrow \mu c\ge b$:
$$c\ge\frac{b}{\mu}=\frac{80}{0.3}=\boxed{266.7\ \text{mm}}.$$
The actual friction arm here is only $c=r-e=10\ \text{mm}\ll266.7\ \text{mm}$, so the brake is not self-locking and needs the 1382 N actuating force — a safe, controllable design.
