22-Mec-A4 Design and Manufacture of Machine Elements · Undated paper

Question 5 of 6: External symmetric pivoted-shoe brake

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

Notes on this paper

Paper format. National Examinations, May 2019 — 16-Mec-A4 Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator. Six questions in two parts: Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element analysis). The rubric asks for two questions from each part; all six are solved here. All questions carry equal value (25 %).

Reference texts. S. Kalpakjian and S. Schmid, Manufacturing Engineering and Technology, 7th ed. (Part A: Ch. 16 sheet-metal forming, Ch. 26 grinding); M. Groover, Fundamentals of Modern Manufacturing, 6th ed. (Ch. 20, 25); R. Budynas and K. Nisbett, Shigley's Mechanical Engineering Design, 10th ed. (Ch. 3 stress, Ch. 6 fatigue, Ch. 7 shafts and keys, Ch. 16 brakes); R. Hibbeler, Mechanics of Materials, 10th ed. (Ch. 6, 7, 9 transverse shear and stress transformation).

Question 5: External symmetric 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. A symmetric two-shoe external brake. Each lining subtends $2\theta_a = 90^\circ$ on a drum of 13.5 in diameter ($r = 6.75\ \text{in}$); $\mu = 0.30$; the drum turns counterclockwise. The shoe pins sit at the drum-centre height, 7.78 in above the shoe-lever base pivots, and the link pins are 15.26 in above those pivots. The handle lever is pinned to the top of the right-hand shoe lever, takes the link 3 in above that fulcrum, and carries $P$ at a horizontal arm of $6\tfrac38 = 6.375\ \text{in}$.

Find. (a) the pin spacing $e$; (b) the force in every member as a multiple of $P$, with free-body diagrams; (c) whether reversing the drum changes the braking torque.

Q5 — external symmetric pivoted-shoe brake. Each shoe pin sits at the friction-neutral offset a = 7.426 in from the drum centre, so the dimension e = 2a = 14.85 in.

Check: the 13.5 in dimension is read as a diameter, not a radius. Its dimension line runs through the drum centre with an arrowhead on each rim, which is the diameter convention, and the reading is confirmed independently by the geometry: with $r = 6.75\ \text{in}$ the friction-neutral pin offset computed in step 1 gives $e = 14.85\ \text{in}$, which matches the drawn spacing of the base pivots to within 3 % when scaled against the 15.26 in lever height. Reading 13.5 in as a radius would demand a 27 in drum inside a 15 in-tall frame — geometrically impossible — and would double every force in the answer.

Approach. The whole question turns on part (a): the dimension $e$ is not a free choice but the value that puts each shoe pin at the point where the friction traction exerts no moment about the pin. Once that offset is known, the shoe carries no friction moment, the lever chain becomes a sequence of single-unknown moment equations, and the braking torque collapses to the compact identity $T = \mu N a$ per shoe. Part (c) then answers itself.

Locate the pin so the friction moment vanishes. With pressure distributed as $p = p_a\sin\theta$ over the lining and $\theta$ measured from the line joining the drum centre to the pin, the moment of the friction traction about the pin is $$M_F = \mu p_a b r \left[\, r\!\int \sin\theta\,d\theta \;-\; a\!\int \sin\theta\cos\theta\,d\theta \,\right]$$ Setting $M_F = 0$ and evaluating the integrals over a lining symmetric about $\theta = 90^\circ$ gives the standard result $$a = \frac{4r\sin\theta_a}{2\theta_a + \sin 2\theta_a}$$ with $\theta_a$ the half-angle of the lining. Here $\theta_a = 45^\circ = \pi/4$, so $$a = \frac{4(6.75)\sin 45^\circ}{\tfrac{\pi}{2} + \sin 90^\circ} = \frac{19.092}{2.5708} = 7.426\ \text{in}$$
Convert the offset into the drawn dimension. The two pins straddle the drum centre symmetrically, and $e$ is measured between them, so $$e = 2a = 2(7.426) = \boxed{e = 14.85\ \text{in} \approx 14\tfrac{7}{8}\ \text{in}}$$ Note $a > r$, as it must be: the pin of an external shoe lies outside the drum surface.
(b) Handle lever. Take moments about the handle fulcrum — the pin joining the handle to the top of the right-hand shoe lever. The actuation force $P$ acts downward at a horizontal arm of 6.375 in; the link, which is horizontal, attaches 3 in above the fulcrum, so its vertical arm is 3 in: $$F_{\text{link}}(3) = P(6.375) \;\Longrightarrow\; \boxed{F_{\text{link}} = 2.125\,P}$$ The fulcrum pin itself carries the vector closure of the two, $\sqrt{(2.125P)^2 + P^2} = 2.348P$, directed up and to the left on the handle.
Shoe levers. Each shoe lever is a vertical bar pinned to the frame at its base. It is loaded by the horizontal link force at 15.26 in and by the shoe reaction at 7.78 in. Because the pin was placed at the friction-neutral point, the resultant the shoe applies to its lever passes through that pin, and the friction component of it is vertical — acting along a line through both the shoe pin and the base pivot directly below it. A force whose line of action passes through the moment centre has zero arm, so friction drops out of the lever equation entirely: $$N(7.78) = F_{\text{link}}(15.26)\;\Longrightarrow\; N = 2.125P\left(\frac{15.26}{7.78}\right) = \boxed{N = 4.168\,P}$$ The base pivot carries $N$ horizontally plus the friction force $\mu N = 1.250P$ vertically, i.e. $4.352P$ at $16.7^\circ$ to the horizontal. By symmetry the left-hand shoe lever is the mirror image, with the identical normal force.
Braking torque. With the pin at the friction-neutral offset, the friction moment about the pin is zero, which means the entire friction force acts as though it were applied at the radius $a$ — the pin offset is the effective friction radius. Hence for each shoe $$T_{\text{shoe}} = \mu N a = (0.30)(4.168P)(7.426) = 9.29P$$ and for the pair $$\boxed{T = 2\mu N a = 18.57\,P \ \ [\text{lb}\cdot\text{in per lb of } P]}$$ As a worked illustration, a hand force of $P = 100\ \text{lb}$ produces $T = 1857\ \text{lb}\cdot\text{in} = 154.8\ \text{lb}\cdot\text{ft}$, and the corresponding maximum lining pressure follows from $N = p_a b r \int\sin\theta\,d\theta$ once a lining width $b$ is chosen.
(c) Effect of reversing the drum. No — the braking torque is unaffected by the direction of rotation. Look at where rotation could possibly enter: only through the moment of the friction traction about the shoe pin, whose sign flips with the drum. But that moment is identically zero by the construction of step 1, and zero does not change sign. The normal force therefore remains $N = 4.168P$ and the torque remains $18.57P$ in either direction. In the language of brake design, each shoe is neither self-energising nor de-energising: the pin was deliberately placed at the point that removes the friction feedback.
Why a designer pays for that. A short-shoe or badly pinned brake gains torque in one direction and loses it in the other, sometimes by 40 % or more, and if $\mu$ rises when hot the self-energising shoe can grab or lock. This design trades peak torque for predictability: the same pedal effort gives the same retardation whichever way the drum turns, which is exactly what is wanted on a hoist or winch that must hold a load while lowering and while raising. Placing the pin at $a = 4r\sin\theta_a/(2\theta_a+\sin2\theta_a)$ is the whole design idea of the pivoted shoe.

Q5 — final results (per unit actuation force $P$)
Quantity	Symbol	Result
Lining half-angle	$\theta_a$	45°
Friction-neutral pin offset	$a$	7.426 in
(a) Pin spacing	$e = 2a$	14.85 in ($\approx 14\tfrac78$ in)
(b) Link force	$F_{\text{link}}$	2.125 P
(b) Normal force per shoe	$N$	4.168 P
(b) Friction force per shoe	$\mu N$	1.250 P
Braking torque, one shoe	$T_{\text{shoe}}$	9.29 P lb·in
Total braking torque	$T$	18.57 P lb·in (1857 lb·in at P = 100 lb)
(c) Sensitivity to rotation direction	—	None — neither self- nor de-energising

Topic. Long-shoe external drum brakes with pivoted (symmetrically hinged) shoes.

Key relations. Pressure distribution $p = p_a\sin\theta$; friction-neutral pin offset $a = 4r\sin\theta_a/(2\theta_a + \sin 2\theta_a)$; normal-force moment $M_N = p_a b r a\!\int\!\sin^2\theta\,d\theta$; friction moment $M_F = \mu p_a b r[r\!\int\!\sin\theta\,d\theta - a\!\int\!\sin\theta\cos\theta\,d\theta]$; torque per shoe $T = \mu N a$ when $M_F = 0$.

Why this works. The sinusoidal pressure distribution follows from assuming the lining deflects about the pin, so the local normal deflection — and hence the pressure — is proportional to the distance from the pin line. Once that shape is accepted, the position of the pin is the only free geometric parameter, and there is exactly one location at which the resultant friction traction passes through it. Placing the pin there removes the friction feedback that makes ordinary shoes direction-sensitive, and it also lets the torque be written without ever evaluating $p_a$ or the lining width, because $\mu N a$ already contains the integral.

Common pitfall. Measuring $\theta$ from the vertical or from the lining's own axis instead of from the drum-centre-to-pin line, which corrupts every integral. A second and very common error is to assume that a two-shoe brake must have one self-energising shoe and therefore that direction matters; here the pivot geometry is chosen precisely to make it not matter, and saying so is the point of part (c).

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §16-3 (external contracting rim clutches and brakes) and §16-4 (the symmetrically pivoted shoe).

← Question 4 Next: Question 6 →