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. 22-Mec-B1 Advanced Machine Design — National Exams, May 2015. Open book, 3 hours, 100 marks. Part I (Problems 1–2) compulsory; answer any 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.) — deflection §4, fatigue §6, shafts §7, bolted joints §8, journal bearings §12, brakes & clutches §16; R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design; R. L. Norton, Machine Design: An Integrated Approach; R. C. Hibbeler, Mechanics of Materials.
Given. Two opposed short shoes press externally on a drum of radius $r=40$ mm, width $w=60$ mm; each lever has actuating arm $a=90$ mm, normal-force arm $b=80$ mm, pivot offset $e=30$ mm; contact half-angle set by $\theta=30^\circ$; $p_{\max}=1.3$ MPa, $\mu=0.3$.
Given data
Quantity
Symbol
Value
Drum radius / width
$r,\,w$
40 mm, 60 mm
Actuating arm / normal arm
$a,\,b$
90 mm, 80 mm
Pivot offset
$e$
30 mm
Shoe subtended angle
$\theta$
30°
Max lining pressure
$p_{\max}$
1.3 MPa
Friction coefficient
$\mu$
0.3
Find. The brake torque capacity, the required actuating force $F_a$ per lever, and the friction-arm value $c$ that makes a shoe self-locking.
Double short-shoe external drum brake: two opposed shoes actuated by $F_a$ at arm $a$; pivots $O_1,O_2$ offset $e$ from the drum axis; friction moment arm $c=r-e$.
Approach. Treat each short shoe as carrying a uniform pressure $p_{\max}$ over its projected pad area to get the normal force, then the friction force $\mu N$ and its drum torque. Sum both shoes for capacity. A moment balance of one lever about its pivot gives the actuating force; self-locking is where that balance drives $F_a$ to zero.
Normal force per shoe. For a short shoe the pressure is taken uniform over the projected area $A=w\,(2r\sin\tfrac{\theta}{2})$:
$$N=p_{\max}\,w\,(2r\sin\tfrac{\theta}{2})=1.3\times10^{6}(0.060)(2\!\times\!0.040\sin15^\circ)=1.62\text{ kN}.$$
Torque per shoe and total capacity. The friction force $\mu N$ acts at the drum radius, and both shoes brake the drum:
$$T_{\text{shoe}}=\mu N r=0.3(1615)(0.040)=19.4\text{ N}\!\cdot\!\text{m},$$
$$\boxed{T=2\,\mu N r=2(19.4)=38.8\text{ N}\!\cdot\!\text{m}.}$$
Friction moment arm. The friction force is tangent to the drum (horizontal at the shoe), a perpendicular distance $c=r-e$ from the pivot:
$$c=r-e=40-30=10\text{ mm}.$$
Actuating force (self-energizing lever). Taking moments of one lever about its pivot, the friction moment aids the applied moment on the self-energizing shoe, so
$$F_a=\frac{N\,(b-\mu c)}{a}=\frac{1615\,[\,0.080-0.3(0.010)\,]}{0.090}=1.38\text{ kN}.$$
$$\boxed{F_a\approx1.38\text{ kN per lever}.}$$
Self-locking condition. The shoe self-locks (grabs with $F_a\to0$) when the friction moment alone balances the normal moment, i.e. $b-\mu c\le0$:
$$\boxed{c\ge\frac{b}{\mu}=\frac{80}{0.3}=267\text{ mm}.}$$
The actual arm $c=10$ mm is far below this, so the brake is not self-locking—it needs the external force $F_a$ to apply, which is the intended, controllable behaviour.
Check. For a genuine double-shoe brake with a single rotation sense, one shoe is self-energizing ($F_a=N(b-\mu c)/a$) and the other de-energizing ($F_a=N(b+\mu c)/a$). The capacity above assumes both shoes reach $p_{\max}$ (the limiting brake torque); the reported $F_a$ is for the self-energizing lever, which reaches $p_{\max}$ at the lower force and therefore governs the pressure limit.
