Question 6 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, May 2014 — 07-Mec-B1 Advanced Machine Design. Open book, 3 hours, 100 marks. Part I (Problems 1–2) compulsory; answer three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Reference texts: R. G. Budynas & J. K. Nisbett, Shigley’s Mechanical Engineering Design, 10th ed. (shafts & fatigue §6–7, journal bearings §12, clutches & brakes §16); R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design (lubrication, brakes); R. C. Hibbeler, Mechanics of Materials (beam deflection, impact).
Problem 6: Double Short-Shoe External Drum Brake (20 marks)
Given. Two pivoted external short shoes act on a common drum; each is analysed as a concentrated (short-shoe) contact at the drum crown.
Given data
Quantity
Value
Drum width $w$ / radius $r$
60 mm / 40 mm
Lever geometry $a$, $b$, $e$
90 mm, 80 mm, 30 mm
Shoe arc $\theta$
$30^\circ$
Max lining pressure $p_{\max}$
1.3 MPa
Friction coefficient $\mu$
0.3
Find. The brake torque capacity, the actuating force $F_a$, and the value of the friction moment-arm $c$ that would make a shoe self-locking.
Free body of the self-energising (upper) shoe: normal force $N$, friction $\mu N$ (arm $r-e$ about pivot $O_1$), and $F_a$ at arm $a$.
Approach. Get the short-shoe normal force from the projected contact area at $p_{\max}$, sum both shoes for the torque, then take moments about each pivot to find $F_a$; self-locking follows from the condition that makes $F_a$ vanish.
Normal force per shoe. For a short shoe the contact is treated as concentrated over the projected width $2r\sin(\theta/2)$:
$$N = p_{\max}\,w\,\big(2r\sin\tfrac{\theta}{2}\big) = (1.3\times10^6)(0.060)(2\!\cdot\!0.040\sin15^\circ) = 1615\ \text{N}.$$
Torque capacity. Each shoe contributes $\mu N r = 0.3(1615)(0.040) = 19.4\ \text{N}\!\cdot\!\text{m}$; with both shoes at $p_{\max}$,
$$\boxed{\,T = 2\mu N r = 38.8\ \text{N}\!\cdot\!\text{m}\,}$$
Friction moment arm. The friction force acts tangent to the drum at the crown, a vertical distance $r-e = 40-30 = 10\ \text{mm}$ from the pivot; this is the effective arm $c$ in the moment balance.
Actuating force (moments about the pivot). For the self-energising shoe the friction moment assists application, $F_a\,a = N\,b - \mu N(r-e)$:
$$F_{a,\text{self}} = \frac{N\,[\,b-\mu(r-e)\,]}{a} = \frac{1615(0.080-0.3\!\cdot\!0.010)}{0.090} = 1382\ \text{N}.$$
For the de-energising shoe the friction opposes, giving $F_{a,\text{de}} = N[b+\mu(r-e)]/a = 1489\ \text{N}$.
Self-locking condition. The brake self-locks when the required $F_a \to 0$, i.e. when the friction moment alone balances $N b$: $b - \mu c = 0$, hence
$$\boxed{\,c = \frac{b}{\mu} = \frac{80}{0.3} = 266.7\ \text{mm}\,}$$
The present geometry gives an arm of only $r-e = 10\ \text{mm} \ll 266.7\ \text{mm}$, so this brake is far from self-locking — it is fully controllable, which is the desired behaviour.
Check / assumptions: The figure’s dimension $c$ is interpreted as the vertical moment arm of the friction force about the pivot; in the given design that arm equals $r-e=10$ mm. Both shoes are taken to reach $p_{\max}$ (so each develops the same $N$) for the torque capacity; because one shoe is self-energising and one de-energising, their actuating forces differ slightly as shown.
