Change in Compressive Force on Aqueduct Pillars as a Barge Passes: 04-BS-7, May 2016, Q10

Question 10: Change in Compressive Force on Aqueduct Pillars as a Barge Passes (5 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
Barge length × beam	15 m × 3 m
Barge draught	1.2 m
Canal width × depth	5 m × 2 m

Find. The change in compressive force transmitted to the aqueduct pillars while the barge is overhead.

Approach. Apply Archimedes' principle: a freely floating body displaces a volume of water whose weight exactly equals its own weight, so the barge does not add net weight to the water-filled trough it floats in — compute the barge's weight (via its displaced volume) only to show what it would have been if this were NOT true, then explain why the compressive force is unchanged.

Weight the barge appears to add (displaced volume). $$\forall_{disp} = L\,B\,T = 15(3)(1.2) = 54\ \text{m}^3, \qquad W_{barge}=\rho_w g\,\forall_{disp} = 1000(9.81)(54) = \boxed{529.7\ \text{kN}}$$
Apply Archimedes' principle. Because the barge floats freely (does not touch the canal bed or walls), the upward buoyant force on it — supplied entirely by the surrounding water — equals its own weight. This means the barge locally displaces (removes) exactly 54 m³ of canal water while adding exactly the SAME weight (529.7 kN) back as its own hull and cargo; the total weight of (water + barge) sitting in the aqueduct trough is unchanged.
Conclusion. $$\Delta F_{pillars} = \boxed{0\ \text{(no change)}}$$ The aqueduct's compressive load depends only on the total weight of everything the trough carries, and a freely floating object never changes that total — it simply trades its own weight for an equal weight of displaced water.

Quantity	Result
Barge weight (= weight of water displaced)	529.7 kN
Change in compressive force on pillars	0 kN — no change

Topic: Archimedes' principle applied to a structural load question.

Key relations: buoyant force F_B = ρg∀_displaced; for a freely floating body, F_B = W_body.

Why this works. The pillars only "feel" the total weight resting in the trough (structure + water + anything floating in it); a floating barge is entirely supported by buoyancy from the SAME water the pillars already carry, so it cannot add to that total — the classic "does a fish tank get heavier when the fish swims to the top" reasoning.

Common pitfall. Computing the barge's weight (or the extra water displaced sideways/upward in the trough) and mistakenly adding it as extra pillar load; the whole point of Archimedes' principle here is that this addition is exactly cancelled by the water the barge locally displaces.

Source: White, Fluid Mechanics, Ch. 2 (buoyancy, Archimedes' principle).

Question 10 of 13: Change in Compressive Force on Aqueduct Pillars as a Barge Passes

Question 10: Change in Compressive Force on Aqueduct Pillars as a Barge Passes (5 marks)