22-Mec-A4 Design and Manufacture of Machine Elements · December 2014

Question 6 of 6: Pivoted long-shoe drum brake

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator permitted. Six questions divided into Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine elements); candidates answer two from Part A and two from Part B, and all questions carry equal value (25 % each). All six questions are worked here.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology — Ch. 10–12 (casting processes and design), Ch. 1 and 3 (structure, cold work, annealing), Ch. 21–22 (machining and tool life).
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design — Ch. 3 (stress, Mohr's circle, beam deflection), Ch. 4 (statically indeterminate shafts), Ch. 16 (clutches and brakes).
R. C. Hibbeler, Mechanics of Materials — Ch. 6–9 (bending, transverse shear, combined loading, stress transformation).
W. D. Callister and D. G. Rethwisch, Materials Science and Engineering — Ch. 7 (dislocations, strengthening, recovery and recrystallization).

it reports a single 5,200 N load and a dimension chain of 750 + 450 + 450 = 1,650 mm that "contradicts" the 2,400 mm overall. Q5 is solved against the drawing, not the caption. Likewise in Q4 the 3,000 lb load is read from the drawing as a horizontal force applied through the bolted plate on the neutral axis, and the section as a 8 in deep I-beam (¹⁄₂ + 3¹⁄₂ + 3¹⁄₂ + ¹⁄₂).

Question 6: Pivoted long-shoe drum brake (25 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.

Quantity	Symbol	Value
Drum radius	$r$	175 mm
Lining width	$b$	62.5 mm
Permissible maximum lining pressure	$p_a$	0.70 MPa
Coefficient of friction	$\mu$	0.25
Hinge pin, relative to the drum centre	—	225 mm left, 150 mm above
Arm of the actuating force about the pin	$c$	225 + 275 = 500 mm
Lining extent	—	45$^\circ$ either side of the vertical centreline
Drum speed	$n$	500 rev/min

Find. (a) the actuating force $P$ that brings the peak lining pressure to 0.70 MPa, and (b) the friction power the brake absorbs at 500 rev/min.

Pivoted long-shoe brake geometry.

Approach. Adopt the standard long-shoe model in which lining pressure varies as $\sin\theta$ with $\theta$ measured from the line joining the drum centre to the hinge pin; integrate the normal and friction tractions to get their moments about the pin, and take moments to find $P$. The braking torque follows from the friction traction about the drum axis.

Locate the hinge pin and set up $\theta$. The pin lies 225 mm to the left of and 150 mm above the drum centre, so $$a = \sqrt{225^{2} + 150^{2}} = 270.4\ \text{mm},$$ and the drum-centre-to-pin line makes $\arctan(150/225) = 33.69^\circ$ above the horizontal, i.e. a bearing of $146.31^\circ$. The lining runs from $135^\circ$ to $45^\circ$ measured from the same origin, so in Shigley's $\theta$ (measured from the pin line towards the lining) $$\theta_1 = 146.31^\circ - 135^\circ = 11.31^\circ, \qquad \theta_2 = 146.31^\circ - 45^\circ = 101.31^\circ ,$$ which correctly span the $90^\circ$ of lining.
Pressure distribution. A rigid shoe pivoting about the pin presses the lining into the drum by an amount proportional to $\sin\theta$, so $$p = p_a \frac{\sin\theta}{\sin\theta_a} .$$ Since $\theta_2 = 101.31^\circ$ is greater than $90^\circ$, the sine reaches its maximum inside the lining, and therefore $\theta_a = 90^\circ$ with $\sin\theta_a = 1$. The peak pressure of 0.70 MPa occurs at $\theta = 90^\circ$, i.e. $56.31^\circ$ from the vertical on the leading side.

Lining pressure distribution; the peak falls inside the lining, so theta_a = 90 degrees.

Moment of the normal forces about the pin. The normal traction on an element $r\,b\,d\theta$ has moment arm $a\sin\theta$ about the pin, so $$M_N = \frac{p_a b r a}{\sin\theta_a}\int_{\theta_1}^{\theta_2}\sin^{2}\theta\,d\theta .$$ Evaluating the integral, $\int \sin^2\theta\,d\theta = [\theta/2 - \sin 2\theta/4] = 0.9777$, hence $$M_N = 0.70(62.5)(175)(270.4)(0.9777) = 2.024\times10^{6}\ \text{N}\cdot\text{mm}.$$
Moment of the friction forces about the pin. The friction traction is tangential, with moment arm $(r - a\cos\theta)$ about the pin: $$M_F = \frac{\mu p_a b r}{\sin\theta_a} \int_{\theta_1}^{\theta_2}\sin\theta\,(r - a\cos\theta)\,d\theta .$$ With $\int\sin\theta\,d\theta = 1.1764$ and $\int\sin\theta\cos\theta\,d\theta = 0.4615$, $$M_F = 0.25(0.70)(62.5)(175)\left[175(1.1764) - 270.4(0.4615)\right] = 1914(81.06) = 1.552\times10^{5}\ \text{N}\cdot\text{mm}.$$
Establish the sense of the friction moment. The drum turns clockwise, so at the lining the drum surface sweeps from the left-hand end towards the right-hand end — away from the pin side. Taking moments about the pin shows the friction traction turns the shoe into the drum, in the same sense as the applied force. The shoe is therefore self-energizing, and the friction moment is subtracted from the required actuating moment.
(a) Actuating force. Moments about the hinge pin give $P c = M_N - M_F$, so $$P = \frac{M_N - M_F}{c} = \frac{2.024\times10^{6} - 1.552\times10^{5}}{500} = \boxed{3{,}738\ \text{N} \approx 3.74\ \text{kN}} .$$ Self-energizing action is worth having: without it the same peak pressure would require $(M_N + M_F)/c = 4{,}359$ N, so friction is doing about 14 % of the work of applying the brake.
Braking torque. Every friction element acts tangentially at radius $r$, so its moment about the drum axis is simply $\mu\,dN\,r$: $$T = \frac{\mu p_a b r^{2}}{\sin\theta_a}\int_{\theta_1}^{\theta_2}\sin\theta\,d\theta = 0.25(0.70)(62.5)(175)^{2}(1.1764) = 3.940\times10^{5}\ \text{N}\cdot\text{mm},$$ that is $\boxed{394.0\ \text{N}\cdot\text{m}}$. As a check, the resultant normal force on the lining is $N = p_a b r \int \sin\theta\,d\theta = 9{,}007$ N, and $\mu N r = 0.25(9{,}007)(175) = 3.940\times10^{5}$ N·mm, which agrees.
(b) Friction power. At 500 rev/min the angular speed is $$\omega = \frac{2\pi (500)}{60} = 52.36\ \text{rad/s},$$ so the power dissipated at the lining is $$\dot{W} = T\omega = 394.0 (52.36) = \boxed{20{,}630\ \text{W} \approx 20.6\ \text{kW}} .$$

Twenty kilowatts is a substantial thermal load for a lining of this size. The contact area is $r(\pi/2)b = 17{,}180$ mm², so the mean lining pressure is $9{,}007/17{,}180 = 0.524$ MPa (against the 0.70 MPa peak), and the power density is about 1.2 W/mm². A brake asked to absorb this continuously would need forced cooling or a much larger drum; as a stopping brake acting intermittently it is acceptable, with the drum's thermal mass absorbing the energy of each stop.

Check: lining extent and pin position. The two $45^\circ$ angles on the drawing are read as being measured from the vertical centreline through the drum centre, giving a lining that subtends $90^\circ$ symmetrically about the top of the drum; the 150 mm dimension is read as the height of the hinge pin above the drum centreline, and the 225 mm as its horizontal offset. Because $\theta_2 > 90^\circ$, $\theta_a = 90^\circ$ has been used rather than $\theta_a = \theta_2$; had the lining been shorter, so that $\theta_2 < 90^\circ$, the peak pressure would occur at the trailing edge and $\sin\theta_a = \sin\theta_2$ would appear in every expression.

Quantity	Symbol	Value
Pin offset from the drum centre	$a$	270.4 mm
Lining limits (from the pin line)	$\theta_1$, $\theta_2$	11.31$^\circ$, 101.31$^\circ$
Moment of normal forces about the pin	$M_N$	2.024 × 10⁶ N·mm
Moment of friction forces about the pin	$M_F$	1.552 × 10⁵ N·mm
(a) Actuating force	$P$	3,738 N (3.74 kN)
Braking torque	$T$	394.0 N·m
Drum angular speed	$\omega$	52.36 rad/s
(b) Friction power absorbed	$\dot{W}$	20.6 kW

Topic. External-contracting long-shoe drum brakes: pressure distribution, self-energizing action, actuating force and heat dissipation.

Key relations. $p = p_a \sin\theta/\sin\theta_a$; $M_N = (p_a b r a/\sin\theta_a)\int\sin^2\theta\,d\theta$; $M_F = (\mu p_a b r/\sin\theta_a)\int \sin\theta\,(r - a\cos\theta)\,d\theta$; $Pc = M_N \mp M_F$ (minus for self-energizing, plus for self-deenergizing); $T = (\mu p_a b r^2/\sin\theta_a)\int\sin\theta\,d\theta$; $\dot{W} = T\omega$. If $\theta_2 > 90^\circ$ then $\theta_a = 90^\circ$, otherwise $\theta_a = \theta_2$.

Why this works. The $\sin\theta$ pressure law is not an assumption about the lining material but a kinematic consequence: a rigid shoe rotating by a small angle about the pin displaces each point of the lining by an amount proportional to its perpendicular distance from the pin line, which varies as $\sin\theta$. With a linear-elastic lining, pressure follows displacement. The self-energizing behaviour then falls out of the geometry: the friction traction acts tangentially, and because the pin is offset from the drum centre this tangential force has a moment about the pin that can either help or oppose the actuating force, depending on the direction of rotation. Reversing the drum turns the same brake into a self-deenergizing one needing 4,359 N instead of 3,738 N — which is why brakes required to work equally in both directions are designed with opposed shoe pairs.

Common pitfall. Setting $\theta_a = \theta_2$ without checking whether $\theta_2$ exceeds $90^\circ$; here it does, so the peak pressure lies inside the lining and $\sin\theta_a = 1$. Almost as common is measuring $\theta$ from the vertical or from the horizontal rather than from the drum-centre-to-pin line, which corrupts both integrals. A third is to add the friction moment when the rotation makes the shoe self-energizing, which overstates $P$ by about 17 % here. Finally, the torque integral uses $\int\sin\theta\,d\theta$, not $\int\sin^2\theta\,d\theta$ — the moment arm about the drum axis is the constant $r$, not $a\sin\theta$.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, Ch. 16.2–16.3 (internal and external expanding/contracting rim brakes; self-energizing action) and Ch. 16.8 (energy considerations and temperature rise).

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