22-Mec-A4 Design and Manufacture of Machine Elements · December 2017

Question 4 of 6: Free-Body Diagrams of a Bicycle Under Pedal Load

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

Notes on this paper

Paper format. National Examinations, December 2017 — 16-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, non-communicating calculator permitted. Six questions divided into Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element statics); candidates answer two from each part, and all questions carry equal value (25 %). All six questions are solved here.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology — sheet-metal forming, welding metallurgy, polymer processing.
M. P. Groover, Fundamentals of Modern Manufacturing: Materials, Processes and Systems — drawing, moulding and tooling design rules.
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design — columns, welded joints, static failure theories.
R. C. Hibbeler, Mechanics of Materials — free-body diagrams, buckling.

Note on units and material data. The paper is metric throughout. Where a property is needed but not printed on the exam (the modulus of elasticity of steel, the reduction of area of 1015 steel, the minimum yield strength of E60 filler metal), the standard handbook value is used and flagged at the point of use, as the paper's own Note 1 invites.

Question 4: Free-Body Diagrams of a Bicycle Under Pedal Load (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
Rider weight applied to one pedal	$W$	800 N, vertically down
Pedal load line ahead of the crank-axle centreline	$e$	160 mm
Pedal sprocket (chainring) radius	$R_{1}$	100 mm
Rear sprocket radius	$R_{2}$	40 mm
Wheel radius, front and rear	$R_{w}$	330 mm
Rear contact patch to crank-axle vertical	$a$	400 mm
Crank-axle vertical to front contact patch	$b$	600 mm

Find. The complete set of forces on four successive free bodies — the crank assembly, the rear wheel and sprocket, the front wheel, and the whole bicycle plus rider — each drawn in equilibrium, with every force quantified.

The four free bodies. Each is drawn with every external force it carries; the chain tension of 1280 N is the internal action that links diagram (a) to diagram (b), and the ground forces link (b) and (c) to (d).

Approach. Work outward from the applied load. Take moments about the crank axle to get the chain tension, carry that tension to the rear sprocket and take moments about the rear axle to get the ground traction, then take moments about a contact patch on the whole assembly to split the 800 N between the two wheels. Each free body then closes by force summation.

(a) Moment equilibrium of the crank assembly about the axle. The rider's 800 N acts 160 mm ahead of the axle, and the chain leaves the top of the 100 mm chainring horizontally, so $$\sum M_{\text{axle}} = 0:\qquad W e = F_{c} R_{1}$$ $$F_{c} = \frac{W e}{R_{1}} = \frac{(800)(160)}{100}$$ $$\boxed{F_{c} = 1280\ \text{N}}$$ The applied moment is $We = 128\ \text{N}\cdot\text{m}$, and the chain reacts it at a shorter radius, so the chain tension exceeds the pedal force by the ratio $e/R_{1} = 1.6$.
Close the crank free body by force summation. The only remaining force is the bearing reaction the frame applies to the spindle, which must balance the 800 N downward pedal load and the 1280 N rearward chain pull: $$R_{x} = F_{c} = 1280\ \text{N},\qquad R_{y} = W = 800\ \text{N}$$ $$R = \sqrt{1280^{2} + 800^{2}} = 1509\ \text{N} \quad \text{at } \arctan\!\left(\frac{800}{1280}\right) = 32.0^{\circ}$$ This 1.5 kN is the load the bottom-bracket bearings actually see, which is nearly twice the rider's weight — the point of drawing the free body rather than assuming the bearing carries only the pedal load.
(b) Moment equilibrium of the rear wheel about its axle. The same chain now pulls forward on the top of the 40 mm rear sprocket, and the only other force with a moment arm about the axle is the ground traction acting at the contact patch, $R_{w} = 330\ \text{mm}$ below the axle: $$F_{c} R_{2} = F R_{w}$$ $$F = \frac{F_{c} R_{2}}{R_{w}} = \frac{(1280)(40)}{330}$$ $$\boxed{F = 155.2\ \text{N forward}}$$ The driving torque at the rear wheel is $F_{c}R_{2} = 51.2\ \text{N}\cdot\text{m}$, and dividing by the wheel radius converts it into a tractive force at the road.
(d) Take the whole assembly next, to obtain the wheel loads. It is easier to do (d) before (c), because the front wheel's normal force comes from the overall moment balance. On the whole bicycle and rider the only external vertical load is the rider's 800 N weight, whose line of action passes through the pedal, that is $400 + 160 = 560\ \text{mm}$ ahead of the rear contact patch on a 1000 mm wheelbase. Taking moments about the rear contact patch, $$N_{f}(a+b) = W(a+e)$$ $$N_{f} = \frac{800(560)}{1000}$$ $$\boxed{N_{f} = 448\ \text{N}, \qquad N_{r} = 800 - 448 = 352\ \text{N}}$$ Checking independently about the front contact patch, $N_{r}(1000) = 800(440)$, which gives 352 N again.
(c) The front wheel, and the horizontal closure of the whole machine. Horizontal equilibrium of the whole assembly demands that the 155.2 N of rear traction be balanced by an equal and opposite ground force at the front contact patch. A freely rotating wheel cannot supply one: with no brake, moment equilibrium about the front axle would force its ground friction to zero. The front wheel must therefore be braked, and its free body carries a normal force of 448 N, a rearward ground friction of 155.2 N, and a brake torque from the caliper of $$T_{b} = F R_{w} = (155.2)(0.330) = 51.2\ \text{N}\cdot\text{m}$$ which is, as it must be, exactly the torque the chain is feeding into the rear wheel. The frame reaction at the front axle is $\sqrt{155.2^{2} + 448^{2}} = 474\ \text{N}$.
Complete the axle reactions on the rear wheel. Summing horizontally on free body (b), the frame must hold the axle back against both the chain pull and the traction, $1280 + 155.2 = 1435\ \text{N}$ rearward, while carrying $352\ \text{N}$ downward, a resultant of 1478 N. Summing all four diagrams, every internal action — chain tension, axle reactions, brake torque — appears twice with opposite sign and cancels, leaving diagram (d) with only the 800 N weight and the two ground reactions. That mutual cancellation is the check that the set of free bodies is consistent.

Check: the equilibrium statement requires a restraint. The question asks for the four bodies "in equilibrium," but a bicycle standing on level ground with an unbraked front wheel is not in equilibrium under a pedal load — the rear traction of 155.2 N is unbalanced and the machine accelerates. Equilibrium is recovered only if something supplies an equal rearward force. The solution above takes the natural case of the front brake being applied, which is what a rider standing on the pedal of a stationary bicycle actually does, and which closes every diagram exactly. Two alternatives are equally defensible if stated: the machine is held against a restraint at the frame, in which case the 155.2 N appears there instead of at the front contact patch; or the machine is free and accelerating, in which case diagram (d) carries a d'Alembert inertia force of 155.2 N rearward through the combined centre of mass and the front-wheel ground friction is zero. In every version the magnitudes computed in steps 1 to 4 are unchanged; only the location of the balancing horizontal force moves.

Free body	Quantity	Value
(a) Crank assembly	Applied moment about the axle	128 N·m
	Chain tension, $F_{c}$	1280 N
	Bottom-bracket bearing reaction	1509 N at 32.0 degrees
(b) Rear wheel	Driving torque, $F_{c}R_{2}$	51.2 N·m
	Ground traction, $F$	155.2 N forward
	Axle reaction from the frame	1435 N rearward, 352 N down (1478 N)
(c) Front wheel	Normal force, $N_{f}$	448 N
	Ground friction	155.2 N rearward
	Brake torque, $T_{b}$	51.2 N·m
(d) Whole assembly	Rear wheel load, $N_{r}$	352 N
	Front wheel load, $N_{f}$	448 N
	Net horizontal force	0 (155.2 N forward and rearward cancel)

Topic. Systematic free-body analysis of a machine: successive isolation and the transmission of load through a mechanism.

Key relations. Moment equilibrium $\sum M = 0$ and force equilibrium $\sum F_{x} = \sum F_{y} = 0$ on each isolated body; torque-to-tractive-force conversion $F = T/R_{w}$; static wheel-load split $N_{f} = W x/L$ measured from the opposite contact patch; Newton's third law linking successive diagrams.

Why this works. Machine design begins with knowing the load at every joint, and the only reliable way to get it is to cut the machine apart one body at a time. Each cut exposes an internal force as an external one, and because the chain can only pull along its own line and a pin can only push through its centre, each free body has few enough unknowns to be solved by a single well-chosen moment equation. Choosing the moment centre at the point where the unknowns meet — the crank axle, then the rear axle, then a contact patch — makes each step a one-line calculation. The final consistency check is powerful and free: assemble all four bodies back together and every internal force must cancel in pairs, leaving only the true external loads.

Common pitfall. The first is assuming the bottom-bracket bearing carries only the 800 N pedal load; it carries 1509 N, because the chain reaction is a second external force on that body. The second is measuring the rider's weight from the crank axle rather than from a contact patch when splitting the wheel loads, which loses the 160 mm pedal offset and gives 320 and 480 N instead of 352 and 448 N. The third is drawing a ground friction force on a freely rotating wheel; if there is no brake and no drive, moment equilibrium about the axle sets that friction to zero.

Source. Hibbeler, Mechanics of Materials, Ch. 1 (equilibrium of a deformable body, internal resultants). Budynas and Nisbett, Shigley's Mechanical Engineering Design, Ch. 3 (load and stress analysis, free-body diagrams and force flow through a machine).

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