Settlement

Settlement is the downward movement of foundations as soil compresses. Learn immediate (elastic) settlement, primary consolidation (compression index, preconsolidation, OCR), time rate with c_v and T_v, secondary compression, stress distribution and allowable settlements.

The Leaning Tower of Pisa is the world's most famous settlement problem: soft clay under one side compressed more than the other. Excess or uneven settlement cracks walls, jams doors, breaks pipes and tilts structures.

Clays settle slowly as water squeezes out (consolidation) — sometimes over decades — while sands settle quickly.

An analogy

Pressing on a wet sponge: it compresses gradually as water escapes; the thicker the sponge and the smaller its pores, the longer it takes.

Components of settlement

Immediate (elastic distortion, fast), primary consolidation (drainage of pore water, slow in clays), and secondary compression (creep, continues after primary). Total settlement is their sum; differential settlement between points causes damage.

Consolidation and stress history

An oedometer test gives the e–log σ′ curve. Normally consolidated (NC) clay is at its highest past stress and compresses along C_c; overconsolidated (OC) clay has been more heavily loaded before and compresses less (C_r) until the preconsolidation pressure σ′_p is exceeded. OCR = σ′_p/σ′₀.

Time rate

Terzaghi's 1D consolidation theory: time ∝ H_dr² / c_v. Double drainage (sand above and below) halves H_dr and cuts time by four. Prefabricated vertical drains shorten drainage paths to accelerate consolidation under preloads.

Stress distribution and limits

Load spreads with depth (Boussinesq, Westergaard, 2:1 method), so compressible layers far below are less stressed. Typical limits: 25 mm (isolated footings on sand), 40–65 mm (rafts on clay), angular distortion ≤ 1/500 to avoid cracking. Monitoring with settlement plates and piezometers validates predictions.

Formulas

Worked example: Consolidation settlement

Given: NC clay 4 m thick, C_c = 0.3, e₀ = 0.9; σ′₀ = 80 kPa at mid-layer, Δσ = 60 kPa

Find: S_c.

Answer: ≈ 153 mm

Worked example: Time to 90 %

Given: Double drainage, H = 4 m → H_dr = 2 m; c_v = 2 m²/year

Find: t₉₀.

Answer: ≈ 1.7 years

Common mistakes

Applications

Key takeaways

Going further