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Settlement of footings

Intermediate Geotechnical and Foundation Engineering Reviewed 24 Oct 2026 2 min read

Every footing settles. The question is whether settlement, and particularly the difference in settlement between neighbouring footings, stays within what the structure can tolerate. Excess total settlement breaks pipes and services; differential settlement tilts walls, cracks brickwork and frames, and jams doors. In many designs on clay, settlement rather than bearing capacity decides the footing size.

Settlement has three parts. Immediate (elastic) settlement happens as load is applied, from soil distortion without volume change; it dominates in sands and is estimated from elastic theory or from SPT and CPT correlations. Consolidation settlement occurs in saturated clays as water drains out over time, and is computed from oedometer data. Secondary compression continues slowly after that, mainly in organic soils.

Diagram: Indiacon original, not to scale.

For sands, empirical methods based on SPT N values or cone resistance are widely used, because undisturbed samples cannot be taken. Plate load tests can be scaled to footing size, but the plate stresses only shallow soil, so a weak layer at depth may be missed. For clays, divide the compressible zone into layers, find the added stress at the middle of each using a stress distribution method, compute each layer's settlement and add. Codes give permissible total and differential settlements and angular distortions depending on soil type and structure type; framed buildings on sand typically tolerate less total settlement than those on clay.

Tip

Compare neighbouring footings, not just one; differential settlement is what cracks the frame.

Key formulas
  • Elastic: Si = q B (1 − ν²) If / Es
  • Consolidation (NC): Sc = Cc H / (1 + e₀) log(σ′f / σ′₀)
  • Total: S = Si + Sc + Ss
  • Angular distortion: β = δ differential / spacing
Common mistake

Using one plate load test result for a wide raft without checking deeper layers.

Solved example

Problem. A 2.5 m square footing applies q = 160 kPa on a sand with Es = 30 MPa and ν = 0.3. Take influence factor If = 0.95 (flexible centre, approximate). Find the immediate settlement.

  1. Si = 160 × 2.5 × (1 − 0.09) × 0.95 / 30 000 m.
  2. = 160 × 2.5 × 0.91 × 0.95 / 30 000.
  3. = 345.8 / 30 000 = 0.0115 m.

Answer. About 12 mm

Quick quiz

1Which part of settlement dominates in sand?

Immediate (elastic) settlement.

2Why can a plate load test mislead?

It stresses only shallow soil, missing weak deep layers that a large footing would load.

3What is angular distortion?

Differential settlement divided by the distance between the two points.

Code note
  • IS 8009 (Part 1) (India): Calculation of settlement of shallow foundations.

Clauses and documents to read

  • IS 8009 (Part 1):1976 (settlement of shallow foundations)
  • IS 1904:1986 (permissible settlements, see its settlement provisions)
  • IS 1888:1982 (plate load test)

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.

  • B. M. Das, Principles of Foundation Engineering, Cengage, latest edition.
  • J. E. Bowles, Foundation Analysis and Design, McGraw Hill, 5th edition.
  • M. J. Tomlinson, J. Woodward, Pile Design and Construction Practice, CRC Press, latest edition.

Free and official sources

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