Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 · 07-Mec-B1 (22-Mec-B1) Advanced Machine Design.
Open-book, 3 hours, 100 marks. The paper requires all of Part I (Problems 1 and 2)
plus any three of the four Part II problems (3–6). All six problems are solved in full below.
Reference texts. R. G. Budynas & J. K. Nisbett, Shigley’s
Mechanical Engineering Design, 10th ed. (shafts & critical speed §7; bolted joints
§8; lubrication & journal bearings §12; brakes §16); R. C. Juvinall &
K. M. Marshek, Fundamentals of Machine Component Design (bearings, brakes, impact);
R. C. Hibbeler, Mechanics of Materials (bending, impact loading).
Note on scope. The exam instructs the candidate to attempt Part I
plus three of Part II. This document answers every problem so that the paper functions as a complete
study set; on exam day a candidate would submit Problems 1, 2 and any three of 3–6.
Question 6: Single Short-Shoe Drum Brake (20 marks)
Single short-shoe external brake: actuating force $F_a$ at
arm $a$, pivot $O_1$ at $(b,e)$, friction moment arm $c=r-e$.
Find. The torque capacity, the required actuating force $F_a$, and the value of the
pivot dimension $c$ that would make the brake self-locking.
Approach. With the short-shoe assumption (uniform pressure, resultant normal force at the
shoe centre), get the normal force from the projected contact area, then the friction torque; take moments
about the pivot $O_1$ for $F_a$, and set the coefficient of $F_a$ to zero for the self-locking limit.
Normal force. For a short shoe the resultant normal force is the peak pressure times the
projected contact area, $A_{proj} = w\,(2r\sin\tfrac{\theta}{2})$,
$$N = p_{\max}\,w\,(2r\sin\tfrac{\theta}{2}) = 1.5\times10^{6}(0.040)(2\cdot0.035\sin17.5^{\circ}) = \boxed{1263\ \text{N}}.$$
Torque capacity. The friction force $\mu N$ acts at the drum radius, so
$$T = \mu N r = 0.4(1263)(0.035) = \boxed{17.7\ \text{N}\cdot\text{m}}.$$
Moment balance about the pivot $O_1$. The normal force (arm $b$) opposes actuation
while the self-energizing friction force (vertical arm $c=r-e$) assists it:
$$F_a\,a = N\,b-\mu N\,c \;\Rightarrow\; F_a = \dfrac{N(b-\mu c)}{a},\qquad c = r-e = 35-25 = 10\ \text{mm}.$$
$$F_a = \dfrac{1263\,[0.070-0.4(0.010)]}{0.110} = \boxed{758\ \text{N}}.$$
Self-locking condition. The brake grabs on its own when the friction moment cancels the
normal-force moment, i.e. $b-\mu c \le 0$:
$$c \ge \dfrac{b}{\mu} = \dfrac{70}{0.4} = \boxed{175\ \text{mm}}.$$
With the actual $c = 10$ mm the brake is far from self-locking, so it is controllable and stable, as a
service brake should be.
Check: $\theta$ is read as the total included contact angle
(short-shoe projected width $2r\sin\tfrac{\theta}{2}$), and the friction moment arm is taken as
$c = r-e$ for the drawn pivot located $e$ above the drum axis. The self-energizing sign follows the
CCW drum rotation; a CW drum would make friction oppose actuation and remove the self-locking possibility.
