Question 6 of 6: Single Short-Shoe External Drum Brake
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — December 2014 — 07-Mec-B1, Advanced Machine Design. Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; answer any THREE of Part II (Problems 3–6). All six problems are solved here as a complete study resource.
Given. A pivoted short shoe pressing on the outside of a rotating drum; the friction force line of action passes a vertical distance $c=r-e$ from the pivot $O_1$.
Given data
Drum width / radius
$w=40$ mm, $r=35$ mm
Lever geometry
$a=110$ mm, $b=70$ mm, $e=25$ mm
Wrap (half-angle to edge)
$\theta=40^\circ$
Max lining pressure
$p_{\max}=1.3$ MPa
Friction coefficient
$\mu=0.3$
Find. The torque capacity, the actuating force $F_a$, and the value of $c$ that makes the brake self-locking.
Short-shoe external brake: normal force on the vertical centre-line (arm $b$ to $O_1$), friction force horizontal at the top contact (arm $c=r-e$ to $O_1$).
Approach. Treat the short shoe as a point contact where the whole normal force acts on the drum’s vertical centre-line. Size the normal force from the projected contact area at the maximum lining pressure, get the torque from the friction force at the drum radius, take moments about the pivot for the actuating force, and set that force to zero for the self-locking condition.
Normal force from projected area. For a short shoe the resultant normal force is $p_{\max}$ times the projected contact area, $A_p=w\cdot 2r\sin(\theta/2)$: $$N=p_{\max}\,w\,(2r\sin\tfrac{\theta}{2})=(1.3\times10^6)(0.040)(2\cdot0.035\sin20^\circ)=1.245\times10^{3}\ \text{N}.$$
Torque capacity. The friction force $\mu N$ acts at the drum radius: $$\boxed{\;T=\mu N r=(0.3)(1245)(0.035)=13.1\ \text{N}\cdot\text{m}.\;}$$
Actuating force (moments about $O_1$). The normal reaction has arm $b$; the friction force is horizontal at the top contact, a vertical distance $c=r-e=35-25=10$ mm from the pivot, and for this rotation sense it assists the actuation (self-energizing). Thus $$F_a=\frac{N(b-\mu c)}{a}=\frac{1245\,[0.070-(0.3)(0.010)]}{0.110}=758\ \text{N}.$$
Self-locking condition. The brake self-locks when no actuating force is needed, i.e. $F_a\le0\Rightarrow b-\mu c\le0$: $$c\ge\frac{b}{\mu}=\frac{70}{0.3}=233\ \text{mm}.$$ With the actual arm only $c=10$ mm the brake is well clear of self-locking (as it should be for a controllable brake).
