Question 5 of 6: Single short-shoe drum brake — torque, actuating force and the self-locking condition (20 marks)

Question 5: Single short-shoe drum brake — torque, actuating force and the self-locking condition (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. An external shoe pressed onto the outside of a drum by a pivoted lever, with all lever geometry measured from the pivot $O_1$:

Find. The torque capacity $T$, the actuating force $F_a$ required to develop it, and the value of the friction moment arm $c$ that would make the brake self-locking.

Approach. Treat the short shoe as carrying a uniform pressure, so the resultant normal force is the peak pressure acting on the projected contact area; get the torque from the friction force at the drum radius; then take moments about the pivot, in which the friction force is self-energising, and set that moment expression to zero to find the self-locking geometry.

1. Justify the short-shoe (uniform-pressure) model. The pressure on a pivoted shoe actually varies as $\sin\phi$ around the arc, but for an arc of roughly $45^\circ$ or less the variation is small enough that the resultant is well approximated by taking the peak pressure over the projected area. At $\theta=40^\circ$ that approximation is comfortably valid, and the “single short-shoe” wording sanctions it. The projected width is the chord subtended by the arc, not the arc length itself: $$L_{\text{proj}}=2r\sin\frac{\theta}{2}=2\left(35\right)\sin20^\circ=2\left(35\right)\left(0.3420\right)=23.94\ \text{mm}$$
2. Resultant normal force at the allowable pressure. Loading the lining to its limit, $$F_n=p_{\max}\,w\,L_{\text{proj}}=1.3\left(40\right)\left(23.94\right)=\boxed{\ 1245\ \text{N}\ }$$ Using the arc length $r\theta=24.4$ mm instead of the chord would overstate $F_n$ by 2 % here; the error grows with $\theta$, so the habit of taking the chord matters.
3. Torque capacity. The friction force acts tangentially at the drum surface, so its moment about the drum axis is $$T=\mu F_n r=0.3\left(1245\right)\left(35\right)=1.307\times10^{4}\ \text{N}\cdot\text{mm}$$ $$\boxed{\ T=13.1\ \text{N}\cdot\text{m}\ }$$ That is the torque this single shoe can absorb without exceeding the lining pressure limit — modest, as one would expect from a 70 mm drum with a $40^\circ$ shoe, and the reason real brakes use two shoes, a wider arc or a much larger drum.
4. Identify the friction moment arm about the pivot. The friction force acts along the tangent at the contact, that is along a horizontal line at height $r$ above the drum axis; the pivot $O_1$ sits at height $e$ above the same axis. The perpendicular distance between them is therefore the dimension $c$ on the figure: $$c=r-e=35-25=10\ \text{mm}$$ This is the geometric heart of the problem, and it is worth stating explicitly: $c$ and $e$ are measured from opposite ends of the drum radius and must sum to $r$.
5. Moment equilibrium of the lever about the pivot. The drum turns so that its surface at the contact sweeps towards the actuating end, and the friction force on the shoe therefore points that way too. Its moment about $O_1$ helps press the shoe on, which is what “self-energising” means. Summing moments about $O_1$: $$F_a\,a=F_n\,b-\mu F_n\,c\qquad\Longrightarrow\qquad F_a=\frac{F_n\left(b-\mu c\right)}{a}$$ $$F_a=\frac{1245\left(70-0.3\times10\right)}{110}=\frac{1245\left(67.0\right)}{110}=\boxed{\ 758\ \text{N}\ }$$ The friction moment is recovering $\mu c/b=4.3$ % of the actuation effort. If the drum ran the other way the friction moment would oppose engagement instead (de-energising), and the same torque would demand $F_a=F_n\left(b+\mu c\right)/a=826$ N — 9 % more. A brake whose duty is bidirectional must be sized on the de-energising direction.
6. Self-locking condition. The brake is self-locking when the friction moment alone is enough to hold the shoe on, so that no actuating force is needed and the brake grabs. Setting $F_a=0$: $$b-\mu c=0\qquad\Longrightarrow\qquad \boxed{\ c\ \ge\ \frac{b}{\mu}=\frac{70}{0.3}=233\ \text{mm}\ }$$ Confirming: at $c=233.3$ mm, $F_a=1245\left(70-0.3\times233.3\right)/110=0$ exactly.
7. Interpret that result physically. Since $c=r-e$, demanding $c=233$ mm on a 35 mm-radius drum requires $$e=r-c=35-233=-198\ \text{mm}$$ that is, the pivot would have to sit 198 mm below the drum axis instead of 25 mm above it — nearly six drum radii away, an arrangement that cannot be built around this drum. So the honest engineering answer is twofold: mathematically self-locking needs $c\ge b/\mu=233$ mm; practically, with $b=70$ mm and $\mu=0.3$ fixed, this brake cannot be made to self-lock, and at the actual $c=10$ mm it sits at only 4 % of the self-locking threshold, i.e. very comfortably controllable. Self-locking is normally a fault to be avoided in a service brake (it makes the torque uncontrollable and the release unpredictable), and is sought only in back-stops and hoist holding brakes.