What it does
When you load saturated clay, water has to squeeze out of its tiny pores before the soil can compress, so the ground settles slowly, over months or years. This page answers both questions engineers ask: how much will it settle, and how long will it take?
- How much: for each layer, the page finds the effective stress at mid-depth before (σ′0) and after (σ′f) loading, compares it with the preconsolidation pressure σ′p, and uses the compression index Cc (normally consolidated), the recompression index Cr (over-consolidated) or both (over-consolidated clay pushed past σ′p).
- How long: Terzaghi’s theory gives the average degree of consolidation U from the time factor Tv = cv·t/Hdr². The page sums the series exactly, finds the time to any U, and draws the settlement against time and the isochrones (how the excess pore pressure dies away through the clay).
- cv from a lab test: from Casagrande’s t50 or Taylor’s t90 on an oedometer curve.
How to use it
- Paste or open the layers from the top down: name (optional), thickness, bulk unit weight γ, initial void ratio e0, Cc, Cr and σ′p. Put Cc = Cr = 0 for a sand layer that only adds weight, and σ′p = 0 for normally consolidated clay.
- Set the water table depth (the page uses γ − 9.81 below it) and, for thick layers, split each layer into sublayers.
- Choose the load: a wide fill adds the same stress at every depth; a footing spreads by the 2:1 rule.
- Give cv and the drainage: sand above and below the clay means it drains both ways, so the drainage path Hdr is half the clay thickness.
Worked example
All numbers are made up. Under 2 m of sand fill lie Clay A (3 m, e0 = 1.10, Cc = 0.45, Cr = 0.06, σ′p = 80 kPa) and Clay B (4 m, e0 = 0.85, Cc = 0.30, Cr = 0.04, σ′p = 95 kPa), with the water table at 1.5 m. A wide fill adds 60 kPa. At the middle of Clay A, σ′0 = 42.6 kPa rises to 102.6 kPa, passing σ′p, so both Cr and Cc apply: 93.0 mm. Clay B adds 102.3 mm. The page shows Settlement 195 mm; 90 % after 4.16 years for cv = 2.5 m²/year with drainage top and bottom (Hdr = 3.5 m). Half the settlement happens in the first 11.6 months.
In the lab box, a 19 mm sample drained both ways with t50 = 4.5 minutes gives cv = 2.08 m²/year.
Good to know
- Over-consolidated clay settles far less until the load passes σ′p. Getting σ′p right matters more than anything else.
- Real clay often has sand seams that shorten the drainage path a lot, so field consolidation is often faster than lab cv predicts.
- Secondary compression (creep) continues after primary consolidation ends and is not included here.
- For study and planning, not a substitute for professional judgement.
FAQ
Why mid-layer stresses? The log of stress is curved, so one point per thick layer slightly misjudges the settlement. Splitting into sublayers fixes that; the table still lists the mid-layer stresses.
What if σ′p is below σ′0? The clay is under-consolidated (still settling under its own weight). The page treats it as normally consolidated and says so, but such a site needs special care.
Why does U start like a square root? Early on, U ≈ √(4Tv/π), so settlement plotted against log time has the S-shape Casagrande used to find t50.
Where the numbers come from
- Primary consolidation settlement S = Cc·H/(1 + e0)·log(σ′f/σ′0), with Cr below the preconsolidation pressure: one-dimensional consolidation theory, general soil mechanics.
- Terzaghi's one-dimensional consolidation theory: U = 1 − Σ 2/M² exp(−M²Tv), M = π(2m + 1)/2, Tv = cv·t/Hdr², and the isochrone series; Casagrande's log-time and Taylor's root-time methods for cv (Tv at 50 % and 90 % computed from the series).
- The 2:1 stress spread under a footing: a common approximate rule in soil mechanics texts.
- Example values are made up by fibub for illustration.
Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.
Last updated 2026-11-09.