Question 6 of 6: Short-shoe external drum brake — torque, actuation, self-locking

Question 6: Short-shoe external drum brake — torque, actuation, 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. Drum width $w=40\text{ mm}$, radius $r=35\text{ mm}$; shoe wrap $\theta=40^\circ$; lever geometry $a=110\text{ mm}$ (pivot–to–$F_a$), $b=70\text{ mm}$ (pivot–to–normal-force line), pivot height $e=25\text{ mm}$ above the drum centre; $p_{max}=1.3\text{ MPa}$, $\mu=0.3$; rotation is self-energizing (drum surface drags the shoe toward the pivot).

Find. the braking torque capacity, the actuating force $F_a$, and the value of the friction-arm $c$ that makes the brake self-locking.

Approach. For a short shoe the pressure is taken uniform, giving a single resultant normal force $N$ and friction $\mu N$ acting at the shoe centre (drum top). The torque is $\mu N r$. A moment balance about the pivot $O_1$ gives $F_a$; the friction term subtracts because the rotation is self-energizing, and driving that term to cancel the normal-force moment gives the self-locking condition.

1. Resultant normal force from the lining-pressure limit. With uniform pressure the radial resultant over the $\theta=40^\circ$ arc is $$N = p_{max}\,w\,\big(2r\sin\tfrac{\theta}{2}\big) = 1.3\times10^6\,(0.040)(2\times0.035\sin20^\circ) = 1.24\times10^{3}\text{ N}.$$
2. Torque capacity. The friction force $\mu N$ acts at radius $r$: $$T = \mu N r = 0.3(1245)(0.035).$$ $$\boxed{T = 13.1\text{ N}\cdot\text{m}.}$$
3. Actuating force (moment balance about $O_1$). The normal force acts at horizontal arm $b$; the friction force (horizontal) acts at the vertical arm $c=r-e=35-25=10\text{ mm}$ and, being self-energizing, reduces $F_a$: $$F_a\,a = N\,b - \mu N\,c \;\Rightarrow\; F_a = \frac{N(b-\mu c)}{a} = \frac{1245\,(70 - 0.3\times10)}{110}.$$ $$\boxed{F_a \approx 758\text{ N}.}$$
4. Self-locking condition. The brake grabs with no applied force when the friction moment alone balances the normal-force moment, $F_a\le0$: $$N\,b - \mu N\,c \le 0 \;\Rightarrow\; c \ge \frac{b}{\mu} = \frac{70}{0.3}.$$ $$\boxed{c \ge 233\text{ mm to self-lock.}}$$ The design value $c=10\text{ mm}$ is far below this, so the brake is safely non-self-locking; self-locking would require relocating the pivot about $233-35\approx198\text{ mm}$ below the drum surface.