Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 16-Mec-B1 Advanced Machine Design, December 2017. Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; only three of the four Part II problems (Problems 3–6) are required. All six problems are solved as a study resource. “State all assumptions clearly… assume any missing data and properly state it” is an explicit exam instruction.
Figure 5.1 — Single short-shoe external brake. Shoe pivots at $O_1$; actuating force $F_a$ acts at lever distance $a$; $b$ is the pivot-to-drum-axis horizontal offset; $e$ the pivot height above the axis; $r$ the drum radius; $2\theta$ the wrap angle.
Given. Drum width $w=40$ mm, $a=110$ mm, $b=70$ mm, $e=25$ mm, $r=35$ mm, half-wrap $\theta=40^\circ$, $p_{max}=1.3$ MPa, $\mu=0.3$. Short-shoe idealisation (uniform pressure, single resultant normal force).
Find. Torque capacity $T$, required actuating force $F_a$, and the value of the pivot arm $c$ that makes the brake self-locking.
Approach. For a short shoe the pressure is taken uniform, so the resultant normal force acts on the projected contact area at the drum surface; friction torque is $\mu N r$. A moment balance of $N$, $\mu N$ and $F_a$ about the pivot gives $F_a$; self-locking occurs when the friction moment alone can hold the shoe, i.e. the required $F_a\le0$.
Normal force from the projected area. A short shoe of wrap $2\theta$ presents a projected width $2r\sin\theta$ to the pressure, so $$N=p_{max}\,w\,(2r\sin\theta)=1.3\times10^{6}(0.040)(2\times0.035\sin20^\circ)=\boxed{1245\text{ N}}.$$
Torque capacity. $$T=\mu N r=0.3(1245)(0.035)=\boxed{13.1\text{ N}\cdot\text{m}}.$$
Actuating force (moment about $O_1$). The normal force acts at arm $b$ and the friction force $\mu N$ at arm $c=r-e=35-25=10$ mm (the friction force is self-energizing for the counter-clockwise drum). Balancing moments, $$F_a\,a=N\,b-\mu N\,c\ \Rightarrow\ F_a=\frac{N(b-\mu c)}{a}=\frac{1245\,(70-0.3\times10)}{110}=\boxed{758\text{ N}}.$$
Self-locking condition. The brake self-locks when the friction moment $\mu N c$ equals or exceeds the normal-force moment $N b$, so that $F_a\le0$: $$\mu N c\ge N b\ \Rightarrow\ c\ge\frac{b}{\mu}=\frac{70}{0.3}=\boxed{233\text{ mm}}.$$ At the design pivot ($c=10$ mm) the brake is far from self-locking (as required for a controllable brake); it would only self-lock if the pivot were placed an impractically large $c\ge233$ mm below the axis, confirming a stable, non-grabbing design.
