22-Mec-A4 Design and Manufacture of Machine Elements · May 2015

Question 6 of 6: External Pivoted-Shoe Brake

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

Notes on this paper

Paper format. National Examinations, May 2015 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator permitted. Six questions in two parts: Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element design). The rubric asks for two questions from each part, and all questions carry equal value (25 % each). All six are solved here, because this set is a study resource rather than an examination script.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology, 8th ed. — sheet-metal forming, deep drawing, abrasive machining.
M. P. Groover, Fundamentals of Modern Manufacturing, 7th ed. — stretch forming and the flow-curve treatment of forming loads.
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 11th ed. — bolted and welded joints (Ch. 8, 9) and brakes (Ch. 16).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — net-section bending and bearing stress.

Units. The paper mixes systems deliberately: Q1, Q5 and Q6 are in US customary units (inch, pound, psi) and Q4 is metric. Each question is solved in the units in which it is set, as the exam intends.

Question 6: External Pivoted-Shoe 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 diameter (from the figure)	$2r$	13.5 in, so $r = 6.75$ in
Shoe arc, symmetric about the horizontal centreline	$2\theta_a$	90°, so $\theta_a = 45^\circ$
Coefficient of friction	$\mu$	0.30
Shoe-lever base pivot to top pin	—	15.26 in
Shoe-lever base pivot to shoe pin (drum-centre height)	—	7.78 in
Handle lever: pivot to P	—	6.375 in
Handle lever: pivot to the horizontal link	—	3 in
Drum rotation	—	counterclockwise

Find. (a) the dimension e that makes the shoes true pivoted shoes; (b) the force in every member of the linkage in terms of P, with free-body diagrams; (c) whether reversing the drum changes the braking torque.

Figure 5 — the brake. Each shoe is pinned to its lever at the height of the drum centre; the levers pivot on the base and are pulled together at the top by the handle linkage. The dimension e is measured between the two lever axes, so each shoe pin stands e/2 from the drum centre.

(a) The dimension e

Approach. A shoe is a pivoted shoe when its pin is placed so that the friction tractions distributed over the lining exert no net moment about that pin; imposing that condition on the sinusoidal pressure distribution fixes the pin offset uniquely.

State the pressure distribution. Measuring $\theta$ from the centreline of the lining, a shoe that is symmetric about its centreline and free to rotate on its pin wears and loads as $$p = p_a\cos\theta, \qquad -\theta_a \le \theta \le \theta_a$$ with $p_a$ the maximum pressure, occurring at the centre of the lining.
Impose zero friction moment about the pin. Setting the moment of the friction tractions about a pin located a distance $a$ from the drum centre to zero and integrating over the arc gives Shigley's result $$a = \frac{4r\sin\theta_a}{2\theta_a + \sin 2\theta_a}$$ where $\theta_a$ is in radians in the denominator.
Substitute the geometry. With $r = 6.75$ in and $\theta_a = 45^\circ = 0.7854$ rad, so that $\sin\theta_a = 0.7071$ and $\sin 2\theta_a = \sin 90^\circ = 1$: $$a = \frac{4(6.75)(0.7071)}{2(0.7854) + 1} = \frac{19.092}{2.5708}$$ $$\boxed{a = 7.43\ \text{in from the drum centre}}$$
Convert to the dimensioned quantity. The two levers are vertical and symmetric about the drum centre, and e is dimensioned between them, so $$\boxed{e = 2a = 14.85\ \text{in}}$$

Check: what e measures. On the exam drawing the e dimension runs from the left lever axis to the right lever axis, which is why it is reported as $2a$. The design quantity produced by the analysis is the single pin offset $a = 7.43$ in from the drum centre; a reader whose copy of the figure dimensions e from the drum centre to one lever should use 7.43 in. Note that $a > r$, as it must be — the pin has to lie outside the drum surface.

(b) Free-body diagrams and the force chain

The linkage transmits P through three bodies in series. Because the exam asks for forces "expressed in terms of the actuation force P", each is carried through symbolically.

Figure 6 — free-body diagrams. The handle lever multiplies P by 6.375/3 = 2.125 into the horizontal link; each shoe lever multiplies that again by 15.26/7.78 = 1.962 into the shoe. Reaction components at each pin are shown; friction on the shoe pins acts vertically.

Handle lever. The handle is pinned at the top of the right-hand shoe lever. P acts vertically 6.375 in from that pin, and the horizontal link is attached 3 in above it. Taking moments about the handle pivot, $$F_{link}(3.0) = P(6.375) \;\Longrightarrow\; \boxed{F_{link} = 2.125P}$$ Equilibrium of the handle then requires the pin reaction to be 2.125P horizontal and P vertical.
Left shoe lever. This lever is pinned to the base at the bottom, carries its shoe on a pin 7.78 in above that, and is pulled horizontally at its top pin, 15.26 in above the base, by the link. The shoe reaction on the lever consists of a horizontal normal resultant N through the shoe pin and a vertical friction resultant, also through the pin. Because the lever is vertical, that friction resultant passes straight through the base pivot and has no moment arm. Taking moments about the base pivot, $$N(7.78) = 2.125P(15.26)$$ $$\boxed{N = 4.17P \ \text{per shoe}}$$
Right shoe lever. The handle reacts onto this lever's top pin with 2.125P horizontal (pressing the right shoe inward) together with P downward. The vertical component again acts along the lever axis and produces no moment about the base pivot, so moments about that pivot give the identical result, $N = 4.17P$. The brake is balanced: both shoes are pressed on with the same force, and the drum bearings carry no net side load from the braking action.
Braking torque. For a shoe pinned at exactly the offset $a$ found in part (a), integrating the friction tractions and the normal tractions over the lining and dividing produces the compact identity $$T = \mu N a$$ per shoe — the pin offset itself is the effective friction radius. Hence for the pair, $$T = 2\mu N a = 2(0.30)(4.17P)(7.43)$$ $$\boxed{T = 18.6P \ \ \text{lbf}\!\cdot\!\text{in per lbf of P}}$$ A hand force of 100 lbf therefore develops about 417 lbf of normal force on each shoe and roughly 1860 lbf·in, or 155 lbf·ft, of braking torque.

(c) Does the direction of rotation matter?

No — and that is the entire purpose of the pivoted-shoe arrangement. Reversing the drum reverses the direction of every friction traction on both linings, but it changes neither the magnitude of the actuating force nor the braking torque, for two reasons that reinforce one another.

The first reason is the choice of pin location made in part (a). The pin was placed precisely so that the friction tractions have zero net moment about it. A distribution whose moment about the pin is zero has a moment of zero when it is reversed as well, so the shoe cannot rotate about its pin to dig itself in or push itself off. That is exactly the self-energizing or de-energizing behaviour that makes an ordinary fixed long shoe direction-sensitive, and here it has been designed out. The pressure distribution, and hence the normal resultant N, is unchanged by reversal.

The second reason is the geometry of the levers. The friction resultant is transmitted to the shoe lever as a vertical force at the shoe pin, and the base pivot lies directly beneath that pin. The moment arm of the friction force about the base pivot is therefore zero in either direction, so the relation $N = 4.17P$ contains no friction term at all. Since $T = 2\mu N a$ and neither $\mu$, $N$ nor $a$ depends on the sense of rotation, the braking torque is identical clockwise and counterclockwise.

This directional indifference is bought at a price, and it is worth stating: the brake gives up the force multiplication that a self-energizing shoe provides, so it needs a larger actuating force for a given torque than a leading-shoe design would. It is chosen where predictability matters more than mechanical advantage — hoists, winches and any machine that must brake equally in both directions.

Quantity	Result
(a) Pin offset from the drum centre, a	7.43 in
(a) Dimension e (between lever axes)	14.85 in
(b) Force in the horizontal link	2.125 P
(b) Normal force on each shoe, N	4.17 P
(b) Braking torque, both shoes	18.6 P lbf·in
(b) Torque at P = 100 lbf	1860 lbf·in (155 lbf·ft)
(c) Effect of rotation direction	None — neither self-energizing nor de-energizing

Topic. External pivoted-shoe (self-aligning) drum brakes.

Key relations. Pressure distribution $p = p_a\cos\theta$ about the lining centreline; pin offset for zero friction moment $a = 4r\sin\theta_a/(2\theta_a + \sin 2\theta_a)$; normal resultant $N = \tfrac{1}{2}p_a b r(2\theta_a + \sin 2\theta_a)$; friction torque about the drum axis $T = 2\mu p_a b r^2\sin\theta_a$, which reduces to the identity $T = \mu N a$ for a correctly pivoted shoe.

Why this works. A long shoe cannot have a uniform pressure distribution, because the shoe is rigid and the lining wears in proportion to the local sliding contact; the cosine distribution is the one consistent with a shoe free to rotate about its own pin. Once that distribution is accepted, requiring the friction tractions to exert no moment about the pin gives a single equation for where the pin must go, and the answer depends only on the drum radius and the arc, not on $\mu$ or on the load. The elegant consequence is the identity $T = \mu N a$: the pin offset doubles as the effective radius at which the friction force may be considered to act, which makes the torque calculation as simple as for a short shoe even though the pressure varies along the arc.

Common pitfall. Measuring $\theta$ from the vertical, or from the line joining the drum centre to the pin, instead of from the centreline of the lining — for the symmetric shoe here those coincide, but on brakes with an off-axis pin they do not, and the error propagates into every integral. The second pitfall is assuming that because the drum turns counterclockwise the shoes must be one leading and one trailing; on a properly pivoted shoe neither is, which is exactly what part (c) is testing.

Source. Budynas and Nisbett, Shigley's Mechanical Engineering Design, 11th ed., §16–3 (external contracting rim brakes and the pivoted symmetric shoe).

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