Indiacon Civil

Subjects › Fluid Mechanics and Hydraulics › Basics

Buoyancy

Basics Fluid Mechanics and Hydraulics Reviewed 24 Oct 2026 2 min read

Any body wholly or partly immersed in a fluid is pushed upward by a force equal to the weight of fluid it displaces. That is Archimedes' principle, and it explains why ships float, why pontoons and caissons can be towed into place, and why empty underground tanks, basements and pipelines can rise out of the ground when the water table is high. The buoyant force acts through the centre of buoyancy, the centroid of the displaced volume.

Whether a floating body is stable depends on the metacentre. When the body tilts slightly, the centre of buoyancy moves sideways. The metacentre M is where the vertical through the new centre of buoyancy cuts the body's original centre line. If M lies above the centre of gravity G, the buoyant force and weight form a restoring couple and the body is stable. The metacentric height GM = I / V − BG, where I is the second moment of the waterplane area about the tilt axis and V the displaced volume. Wide, low bodies are stable; tall narrow ones are not.

Diagram: Indiacon original, not to scale.

In civil works, flotation checks are essential for empty sumps, manholes, septic tanks, swimming pools, basements and buried pipelines. The structure's weight, plus soil above it where it can be relied on, must exceed the uplift with a margin; otherwise add weight, anchors, tension piles or relief drains. Caissons for well foundations and immersed tunnel elements are floated, positioned and then ballasted to sink in a controlled way.

Key formulas
  • Buoyant force: F_B = γ V_displaced
  • Metacentric height: GM = I / V − BG
  • Stable if GM > 0
  • Uplift check: FS = (W_structure + W_soil) / F_B ≥ required
Solved example

Problem. An empty underground tank is 6 × 4 m in plan and 3 m deep, weighs 520 kN, with water table at ground level and no soil cover. Check flotation with a required factor of 1.1.

  1. Uplift = 9.81 × 6 × 4 × 3 = 706 kN.
  2. FS = 520 / 706 = 0.74 < 1.1: the tank will float when empty.
  3. Extra weight needed = 1.1 × 706 − 520 = 257 kN, for example a heavier base slab with a toe projection carrying soil, or tension anchors.

Answer. Unsafe; about 257 kN of extra resistance needed

Quick quiz

1State Archimedes' principle.

The upward force equals the weight of fluid displaced.

2When is a floating body stable?

When the metacentre lies above its centre of gravity (GM > 0).

3Name one way to prevent a tank floating.

Add weight, a base projection carrying soil, tension piles or anchors, or relief drains.

Code note
  • IS 3370 (series) (India): Concrete structures for storage of liquids.

Clauses and documents to read

  • IS 3370 (Part 1 and Part 2) (concrete liquid-retaining structures)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • P. N. Modi, S. M. Seth, Hydraulics and Fluid Mechanics including Hydraulic Machines, Standard Book House, latest edition.
  • V. T. Chow, Open-Channel Hydraulics, McGraw Hill, 1959.
  • K. Subramanya, Flow in Open Channels, McGraw Hill Education (India), latest edition.

Free and official sources

All references for this subject →

Found an error? Write to indiacongroupofengineers [at] gmail [dot] com. For corrections and feedback. We never ask for payment or personal details.