25-Nav-B1 Applied Thermodynamics and Heat Transfer (25-Mec-A1) · December 2016
Question 3 of 6: Double Short-Shoe External Drum Brake
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 98-Mar-B1, December 2016. Open book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; three of the four Part II problems (3–6) are required. All six problems are solved below.
Given. A double (two-shoe) external drum brake; the lining subtends a small arc so the short-shoe assumption (uniform pressure at the mean radius) applies.
Given data
Quantity
Symbol
Value
Drum radius / width
$r,\ w$
$40\ \text{mm},\ 60\ \text{mm}$
Lever geometry
$a,\ b,\ e$
$90,\ 80,\ 30\ \text{mm}$
Lining arc
$\theta$
$25^\circ$
Max lining pressure
$p_{\max}$
$1.3\ \text{MPa}$
Friction coefficient
$\mu$
$0.25$
Find. The torque capacity $T$, the actuating force $F_a$, and the pivot friction-arm value $c$ at which the self-energizing shoe becomes self-locking.
One shoe of the double external brake: normal force $N$ presses the lining onto the drum, friction $\mu N$ gives torque; the lever pivots at distance $a$ from $F_a$. The companion shoe (not shown) is a mirror image.
Approach. Use the short-shoe model: the resultant normal force equals $p_{\max}$ times the projected lining area; friction $\mu N$ at radius $r$ gives each shoe’s torque (doubled for two shoes); a moment balance on the self-energizing lever gives $F_a$; and self-locking occurs when that actuating force drops to zero.
Normal force on one shoe. The projected area of the short shoe is $w\times2r\sin(\theta/2)$, so
$$N=p_{\max}\,w\,\bigl(2r\sin\tfrac{\theta}{2}\bigr)=1.3\times10^{6}\times0.060\times\bigl(2(0.040)\sin12.5^\circ\bigr)=1.35\times10^{3}\ \text{N}.$$
Torque capacity (both shoes). Each shoe contributes $\mu N r$; a double brake has two shoes acting at $p_{\max}$:
$$\boxed{\ T=2\,\mu N r=2(0.25)(1350.6)(0.040)=27.0\ \text{N}\cdot\text{m}\ }$$
Actuating force from the lever moment balance. With the pivot located $e$ above the drum axis, the friction force acts on a moment arm $c=r-e=40-30=10\ \text{mm}$. Taking moments about the pivot for the self-energizing shoe (friction assists the applied force),
$$F_a=\frac{N\,(b-\mu c)}{a}=\frac{1350.6\,(0.080-0.25\times0.010)}{0.090}=1.16\times10^{3}\ \text{N}.$$
Self-locking condition. The shoe self-locks when the friction moment alone can hold it, i.e. when $F_a\le0$, which requires $b-\mu c\le0$:
$$c\ge\frac{b}{\mu}=\frac{0.080}{0.25}=0.320\ \text{m}=320\ \text{mm}.$$
The actual arm ($c=10$ mm) is far below this, so the brake is safely not self-locking — self-locking would need the pivot placed about 320 mm from the drum axis.
