22-Mec-A4 Design and Manufacture of Machine Elements · December 2013

Question 8 of 8: Double-block brake — the distance a for equal wear

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

Notes on this paper

Paper format. National Examination, 07-Mec-A4 Design and Manufacture of Machine Elements, December 2013 — 3 hours, open book, any non-communicating calculator permitted. Eight questions on six pages, divided into Part A (manufacturing processes, Q1–Q4) and Part B (machine-element design, Q5–Q8). The rubric asks for three questions from Part A and two from Part B, five questions constituting a complete paper, all of equal value (20 % each). All eight questions are solved here.

Reference texts.

Kalpakjian & Schmid, Manufacturing Engineering and Technology, 7th ed. — sand casting and casting defects (Ch. 10–12), adhesive bonding and joint design (Ch. 32), fusion welding and weld defects (Ch. 30–31), sheet-metal shearing and blanking (Ch. 16).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — eccentrically loaded shear joints (Ch. 8–9), combined stresses and stress transformation (Ch. 3), statically indeterminate shafts and deflection (Ch. 4), brakes and clutches (Ch. 16).
Hibbeler, Mechanics of Materials, 10th ed. — transverse shear ($VQ/Ib$), combined loading and Mohr's circle (Ch. 7–9).
Faires, Design of Machine Elements, 4th ed. — double-block (double-shoe) brakes, self-energizing geometry, equal-wear proportioning.
ASM Handbook Vol. 15 Casting and Vol. 6 Welding, Brazing and Soldering — hot-spot/shrinkage defects; hydrogen-induced cold cracking and preheat practice (see also CSA W59 and CSA W47.1 for Canadian fabrication practice).

Check: Part B is figure-driven. Every dimension used below was read from the printed figures. Two readings are stated explicitly in Given so a grader can substitute a different interpretation without redoing the method: (i) in Figure A the low rivet is taken as lying on the same vertical centreline as the third rivet of the top row (75 + 75 = 150 mm from the left-hand rivet); (ii) in Figure D the dimension $a$ is the horizontal spacing, measured along the operating lever, between the pin taking the upper shoe link and the pin taking the lower shoe link, with the 10 in operating arm measured from the lower-link pin.

Question 8: Double-block brake — the distance a for equal wear (20 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	Value
Drum radius, $r$	6 in
Coefficient of friction, $\mu$	0.3
Shoe-arm hinge to drum centreline	7½ in (shoe contact is on that centreline)
Hinge to link pin (total arm length)	7½ + 8½ = 16 in
Arm centreline above / below the drum centre	8 in, so the friction force acts $8-6 = 2$ in from the arm centreline
Operating arm	10 in from the lower-link pin to the applied force
Drum rotation	counter-clockwise as drawn
Shoes	short shoes — the normal force may be taken as a single resultant at the drum centreline

Find. The distance $a$ between the two link pins on the operating lever that makes the wear of the two shoes equal.

Figure 8.1 — Free bodies of the two shoe arms and of the operating lever. Because the friction force acts on the drum surface, 2 in inboard of each arm centreline, it adds to the applied moment on the upper (self-energizing) arm and subtracts from it on the lower (de-energizing) arm.

Approach. Equal wear on short shoes running on the same drum means equal contact pressure, hence equal normal force. Write the moment equation for each shoe arm about its own hinge to relate its normal force to the link force it receives; then write the moment and force equations for the operating lever to relate the two link forces to $a$; and finally impose $N_1 = N_2$.

State the equal-wear criterion. Archard's wear law gives a wear rate proportional to the product of contact pressure and sliding velocity, $\dot{w} \propto p\,V$. Both shoes bear on the same drum at the same radius, so $V$ is identical, and both have the same lining area, so $$p_1 = p_2 \;\Longleftrightarrow\; \boxed{N_1 = N_2}$$ is the condition to be satisfied. (Equal braking torque from each shoe is the same condition here, since $T_i = \mu N_i r$.)
Upper shoe — the self-energizing arm. Take moments about the fixed hinge, 7½ in to the left of the shoe. The link pulls the arm end down with force $F_1$ at 16 in; the drum pushes the shoe up with $N_1$ at 7½ in; and with the drum running counter-clockwise the drum surface at the top moves to the left, so the friction $\mu N_1$ acts to the left on the shoe, along a line 2 in below the hinge. That friction moment turns the arm the same way as $F_1$ does, so $$7.5\,N_1 - \mu N_1 (2) = 16\,F_1 \;\Rightarrow\; N_1\,(7.5-2\mu)=16F_1,$$ $$N_1 = \frac{16F_1}{7.5-0.6}=\frac{16F_1}{6.9}=2.319\,F_1.$$
Lower shoe — the de-energizing arm. The geometry is mirrored, but the drum surface at the bottom moves to the right, so the friction now acts 2 in on the far side of the hinge and opposes the applied moment: $$7.5\,N_2 + \mu N_2 (2) = 16\,F_2 \;\Rightarrow\; N_2\,(7.5+2\mu)=16F_2,$$ $$N_2 = \frac{16F_2}{7.5+0.6}=\frac{16F_2}{8.1}=1.975\,F_2.$$ This asymmetry — 6.9 against 8.1 — is the whole problem: with the two link forces equal the upper shoe would be pressed 17 % harder and would wear out first.
Impose equal normal forces. Setting $N_1=N_2$, $$\frac{16F_1}{6.9}=\frac{16F_2}{8.1}\;\Rightarrow\; \frac{F_2}{F_1}=\frac{8.1}{6.9}=1.1739.$$ The lower link must therefore be pulled 17.39 % harder than the upper one, and the geometry of the operating lever has to deliver exactly that ratio.
Operating lever. The lever carries three forces: $F_1$ from the upper link at the left-hand pin, $F_2$ from the lower link at the second pin a distance $a$ to the right of it, and the operating force $P$ at the far end, 10 in beyond the second pin. Taking moments about the lower-link pin, $$F_1\,a = P\,(10) \;\Rightarrow\; P = \frac{a F_1}{10},$$ and resolving perpendicular to the lever, $$F_2 = F_1 + P = F_1\left(1+\frac{a}{10}\right).$$
Solve for $a$. Equating the two expressions for the force ratio, $$1+\frac{a}{10} = \frac{8.1}{6.9}=1.1739 \;\Rightarrow\; a = 10\left(\frac{8.1}{6.9}-1\right)=10(0.17391),$$ $$\boxed{a = 1.739\ \text{in} \approx 1\tfrac{3}{4}\ \text{in}}$$
Check the result. With $a=1.739$ in, $P = 0.1739F_1$ and $F_2 = 1.1739F_1$. Then $$N_1 = \frac{16F_1}{6.9}=2.319F_1,\qquad N_2 = \frac{16(1.1739F_1)}{8.1}=2.319F_1,$$ which are indeed equal, and the total braking torque is $$T = \mu\,(N_1+N_2)\,r = 0.3\,(4.638F_1)(6) = 8.35\,F_1\ \text{lb}\cdot\text{in per lb of }F_1.$$

Notice what the answer means physically. Moving the upper link's pin 1.74 in away from the lower link's pin makes the lever act as a small unequal-arm balance, deliberately under-loading the shoe that the friction is already helping. If the drum ran the other way the sense of the self-energizing action would reverse and the pins would have to be swapped; a brake proportioned this way is therefore correct for one direction of rotation only, which is exactly why brakes required to work in both directions are built with symmetric, non-self-energizing geometry instead.

Quantity	Value
Friction moment arm about each hinge	8 − 6 = 2 in
Upper (self-energizing) arm: $N_1(7.5-2\mu)=16F_1$	$N_1 = 16F_1/6.9 = 2.319F_1$
Lower (de-energizing) arm: $N_2(7.5+2\mu)=16F_2$	$N_2 = 16F_2/8.1 = 1.975F_2$
Equal-wear condition	$N_1 = N_2$ ⇒ $F_2/F_1 = 8.1/6.9 = 1.1739$
Lever relation	$F_2 = F_1(1 + a/10)$, $P = aF_1/10$
Distance for equal wear	$a = 1.739$ in ≈ 1¾ in
Resulting normal force on each shoe	2.319 $F_1$
Total braking torque	$T = \mu(N_1+N_2)r = 8.35\,F_1$ lb·in

Topic. Double-block (double-shoe) external brakes: self-energizing and de-energizing shoes, and proportioning the linkage for equal wear.

Key relations. Short-shoe assumption $F_f = \mu N$ acting at the drum surface; moment equation about the shoe pivot $N\,c \mp \mu N\,e = F\,\ell$, where $c$ is the pivot-to-shoe distance, $e$ the offset of the friction line from the pivot and $\ell$ the actuating-force arm; braking torque $T = \mu N r$ per shoe; Archard wear $\dot{w}\propto pV$, so equal wear at equal radius and area means equal $N$.

Why this works. A shoe is self-energizing when the friction force it develops produces a moment about its own pivot that presses it harder onto the drum; the sign of that moment depends on whether the friction line of action passes on one side of the pivot or the other, which is why the 8 in and 6 in dimensions matter so much despite never appearing on their own. Here the friction line lies 2 in from each arm centreline, on the drum side of the upper hinge and the far side of the lower one, so the same $\mu$ produces $-2\mu$ in one denominator and $+2\mu$ in the other. Everything else follows: the two shoes need different actuating forces to be pressed equally hard, the operating lever is the only place where that difference can be arranged, and offsetting the two link pins by $a$ turns the lever into precisely the required unequal-arm balance.

Common pitfall. Using the arm centreline offset (8 in) instead of the difference $8-6 = 2$ in as the friction moment arm, which turns the modest 6.9/8.1 asymmetry into a wildly wrong one. Equally common is getting the sense of rotation backwards, which swaps the self-energizing shoe and yields a negative $a$ — a physically impossible answer for the linkage as drawn, and therefore a useful check. Finally, the short-shoe assumption (a single resultant $N$ at the drum centreline) is what makes this a one-line moment equation; for a long shoe the pressure distribution must be integrated over the wrap angle and the effective friction radius changes.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed., Ch. 16 (§16–2 to §16–4, external contracting rim brakes, self-energizing and self-locking); Faires, Design of Machine Elements, 4th ed., Ch. 18 (double-block brakes and equal-wear proportioning); Hibbeler, Engineering Mechanics: Statics, Ch. 8 for the underlying dry-friction moment analysis.

← Question 7 Back to the paper →

Quantity	Value
Drum radius, \(r\)	6 in
Coefficient of friction, \(\mu\)	0.3
Shoe-arm hinge to drum centreline	7½ in (shoe contact is on that centreline)
Hinge to link pin (total arm length)	7½ + 8½ = 16 in
Arm centreline above / below the drum centre	8 in, so the friction force acts \(8-6 = 2\) in from the arm centreline
Operating arm	10 in from the lower-link pin to the applied force
Drum rotation	counter-clockwise as drawn
Shoes	short shoes — the normal force may be taken as a single resultant at the drum centreline

Quantity	Value
Friction moment arm about each hinge	8 − 6 = 2 in
Upper (self-energizing) arm: \(N_1(7.5-2\mu)=16F_1\)	\(N_1 = 16F_1/6.9 = 2.319F_1\)
Lower (de-energizing) arm: \(N_2(7.5+2\mu)=16F_2\)	\(N_2 = 16F_2/8.1 = 1.975F_2\)
Equal-wear condition	\(N_1 = N_2\) ⇒ \(F_2/F_1 = 8.1/6.9 = 1.1739\)
Lever relation	\(F_2 = F_1(1 + a/10)\), \(P = aF_1/10\)
Distance for equal wear	\(a = 1.739\) in ≈ 1¾ in
Resulting normal force on each shoe	2.319 \(F_1\)
Total braking torque	\(T = \mu(N_1+N_2)r = 8.35\,F_1\) lb·in