Short-shoe external drum brake — torque, actuation, self-locking: 25-Nav-B1 Applied Thermodynamics and Heat Transfer (25-Mec-A1), May 2013, Q6

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.

Short-shoe external brake: normal force $N$ (radial, vertical) and friction $\mu N$ (tangential, horizontal) at the shoe centre; the lever pivots at $O_1$, a height $e$ above the drum axis. For the rotation shown the friction moment aids $F_a$ (self-energizing).

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.

Resultant normal force from the lining-pressure limit. With uniform pressure the radial resultant over the $\theta=40^\circ$ arc is $$N = p_{max}\,w\,\big(2r\sin\tfrac{\theta}{2}\big) = 1.3\times10^6\,(0.040)(2\times0.035\sin20^\circ) = 1.24\times10^{3}\text{ N}.$$
Torque capacity. The friction force $\mu N$ acts at radius $r$: $$T = \mu N r = 0.3(1245)(0.035).$$ $$\boxed{T = 13.1\text{ N}\cdot\text{m}.}$$
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 $F_a$: $$F_a\,a = N\,b - \mu N\,c \;\Rightarrow\; F_a = \frac{N(b-\mu c)}{a} = \frac{1245\,(70 - 0.3\times10)}{110}.$$ $$\boxed{F_a \approx 758\text{ N}.}$$
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.

Question 6 — results
Quantity	Value
Resultant normal force $N$	$1.24\text{ kN}$
Torque capacity $T$	$13.1\text{ N}\cdot\text{m}$
Actuating force $F_a$	$758\text{ N}$
Friction arm (design) $c=r-e$	$10\text{ mm}$
Self-locking friction arm	$c \ge 233\text{ mm}$

Check / assumptions: The “short-shoe” idealization treats the lining pressure as uniform and lumps $N$ and $\mu N$ at the shoe centre (drum top); at $\theta=40^\circ$ this is slightly optimistic and a long-shoe (integrated-pressure) analysis would refine $T$ and $F_a$ by a few percent. The friction arm is taken as $c=r-e$ for the pivot placed $e$ above the drum axis; the self-locking answer $c\ge b/\mu$ holds for whatever vertical friction-arm the pivot geometry produces.

Topic: Short-shoe external (rim) drum brake — torque, actuation, and self-energizing / self-locking.

Key relations: $N=p_{max}\,w\,(2r\sin\tfrac{\theta}{2})$; $T=\mu N r$; moment balance $F_a a = N b \mp \mu N c$; self-locking $c\ge b/\mu$.

Why this works: A short shoe subtends a small enough arc that the pressure is nearly uniform, so the contact reduces to one normal and one friction force. Whether friction helps or opposes actuation depends on which side of the pivot its line of action passes; when it helps (self-energizing), a large enough friction moment arm makes the brake grab on its own — useful for back-stops but dangerous for controllable braking.

Common pitfall: Mixing the self-energizing ($-\mu N c$) and de-energizing ($+\mu N c$) signs, or using $\mu N r$ for the actuation arm instead of the vertical distance from the pivot to the friction line.

Source: Shigley, Mechanical Engineering Design, §16.6 (short-shoe drum brakes); Juvinall & Marshek Ch. 18.

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)