CEE 340 · Advanced Foundation Engineering · Settlement
Settlement Computations
Bearing capacity asks "will it collapse?" Settlement asks the question that actually sinks most projects: "will it still be usable?" A foundation can be nowhere near failure and still crack every wall in the building.
ImmediateConsolidationSecondaryTime Rate
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Total vs. Differential Settlement
Every foundation settles. That is not a defect — soil is compressible, and loading it squeezes it. What matters is how much, and far more importantly, how unevenly.
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The Leaning Tower of Pisa settled about 3 m and is world-famous. Mexico City's Palacio de Bellas Artes settled roughly 4 m and is merely an inconvenience — you step down into it. The difference isn't the amount. Pisa's south side settled more than its north side, and that difference is what tilted it. Uniform settlement moves a building; differential settlement breaks it.
Total settlement
St
How far the whole foundation goes down. Matters for utility connections, drainage, and access — a pipe entering the building at a fixed elevation does not settle with it.
Differential settlement
δS
The difference between two points. This is what cracks walls, jams doors, and tilts structures. Usually the governing criterion.
Angular distortion
δS / L
Differential settlement divided by the distance between the points. Roughly 1/500 risks cracking in load-bearing walls; 1/150 risks structural damage.
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Why it matters: Settlement, not bearing capacity, governs the design of most shallow foundations on clay. You will often find a footing that is enormous compared to what strength alone requires — sized entirely to keep settlement tolerable.
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The Three Components
Settlement arrives in three instalments, on three completely different timescales, driven by three different mechanisms.
1 — Immediate, Se
Elastic distortion (minutes)
The soil skeleton deforms in shape without changing volume, like squashing a rubber block. Happens as fast as you apply the load. Dominant in sand and unsaturated soil.
2 — Primary, Sc
Consolidation (months to years)
Water is squeezed out of saturated clay pores and the soil actually loses volume. Slow, because water has to physically travel out. Usually the largest component in clay.
3 — Secondary, Ss
Creep (years to decades)
The soil skeleton keeps slowly rearranging at constant effective stress, long after pore pressure has dissipated. Matters in organic soils and peat.
Core Idea
\( S_t = S_e + S_c + S_s \). In sand, Se is essentially everything and it is over before construction finishes. In clay, Sc dominates and can still be developing a decade later.
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Immediate (Elastic) Settlement
Elastic settlement of a flexible foundation
\[ S_e = q_o B\,\frac{1-\mu_s^2}{E_s}\,I_s I_f \]
qo = net applied pressure, B = width, μs = Poisson's ratio, Es = soil modulus, Is = shape factor, If = depth factor.
Rigid vs. flexible
A flexible foundation (a thin mat, an embankment) settles more at the centre than the edges — it dishes. A rigid foundation cannot dish, so it settles uniformly at roughly 93% of the flexible centre value. Real footings are close to rigid.
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The hard part is Es, not the formula. Soil modulus is not a constant — it depends on stress level, strain level and drainage. It is usually estimated from SPT or CPT correlations, so a settlement prediction is rarely better than the site investigation behind it.
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How Clay Consolidates
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A saturated clay is a sponge full of water inside a piston. Push on the piston and, at first, the water takes the entire load — it cannot escape instantly, and water is nearly incompressible. Over time water leaks out through a small valve; as it leaves, load transfers from the water to the sponge's own fibres, and the sponge compresses. That handover from water to skeleton is consolidation.
The governing principle
Effective stress does the work
\( \sigma = \sigma' + u \). The total stress jumps instantly when you load. At t = 0 the pore pressure u carries all of it. As u dissipates, σ' rises by exactly the same amount — and it is σ' that compresses the soil. Settlement tracks the transfer, not the load.
What we measure
The e–logσ' curve
An oedometer test loads a clay sample in stages and plots void ratio against log of effective stress. It gives a flat recompression line of slope Cs and a steep virgin compression line of slope Cc, meeting at the preconsolidation pressure σ'c.
The e–logσ' Curve
The soil "remembers" the highest stress it has ever carried. Below σ'c it responds stiffly along the flat recompression line; push past σ'c and it falls onto the steep virgin line, where settlement is several times larger for the same load increment.
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Normally vs. Overconsolidated
The single most important question about a clay before you compute anything: has it ever carried more stress than it carries today? The overconsolidation ratio answers it.
Overconsolidation ratio
\[ OCR = \frac{\sigma'_c}{\sigma'_0} \]
OCR = 1 → normally consolidated (NC): today's stress is the highest ever. OCR > 1 → overconsolidated (OC): it has been loaded more before, by ice sheets, by soil since eroded, or by desiccation.
This is where the money is. Cs is typically only 1/5 to 1/10 of Cc. Worked Examples 1 and 2 run the identical clay layer and load through Cases 1 and 2 and get 110 mm versus 21 mm. Loading an OC clay while staying below σ'c is the cheapest settlement control there is — and it is why preloading a site before building works.
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Settlement Calculator
Enter σ'c and the calculator picks the right case automatically — watch it switch from Case 2 to Case 3 as you raise the load past the preconsolidation pressure.
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Time Rate of Consolidation
Knowing the settlement will eventually be 110 mm is only half the answer. A client needs to know whether that happens in six months or sixty years.
Intuition
Water has to travel
Consolidation is limited by how fast pore water can escape. The further it must travel and the tighter the soil, the slower it goes — which is why the drainage path, not the layer thickness, is what enters the equation.
The drainage path Hdr
The detail everyone gets wrong
If the clay layer has sand above and below, water escapes both ways and only travels half the thickness: Hdr = H/2. Drainage on one side only gives Hdr = H. Since time goes as Hdr2, getting this wrong is a factor-of-four error.
Time scales with the square of the drainage path. Halving the distance water must travel cuts the time to a quarter — which is exactly the principle behind vertical sand drains and wick drains: they don't make the soil stronger, they just give water a much shorter route out.
t1 is the time at which primary consolidation ends; ep is the void ratio at that moment.
When to worry
Negligible in most inorganic clays — often under 10% of primary. But in peat and organic soils it can exceed the primary settlement entirely and continue for decades. Note it is logarithmic in time: each additional log cycle adds the same increment, so it never quite stops.
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Worked Examples
Example 1
Normally Consolidated Clay
Problem: A 4 m clay layer has e₀ = 0.90, Cc = 0.32, and σ'₀ = 110 kPa at mid-depth. The clay is normally consolidated. A new foundation raises the stress at mid-depth by Δσ' = 50 kPa. Find the primary consolidation settlement.
1
NC, so σ'c = σ'₀ = 110 kPa and the whole increment is on the virgin line.
Problem: Identical layer and identical load, but the clay is overconsolidated with σ'c = 190 kPa and Cs = 0.06. Find the settlement.
1
Check the case: \( \sigma'_0 + \Delta\sigma' = 160\text{kPa} < \sigma'_c = 190\text{kPa} \) — the load never reaches the preconsolidation pressure, so this is Case 2, entirely on the recompression line.
Now push harder: if instead Δσ' = 120 kPa, then σ'₀+Δσ' = 230 > 190, so Case 3 applies: \( S_c = 0.1263\log\frac{190}{110} + 0.6737\log\frac{230}{190} = 0.030 + 0.056 = 0.086\text{m} \)
Answer: 21 mm — against 110 mm for the same load on the same clay when normally consolidated, a factor of 5. Push past σ'c (step 4) and settlement jumps to 86 mm, because the second half of the load rides the steep virgin line.
Example 3
How Long Will It Take?
Problem: The 4 m clay layer of Example 1 is sandwiched between sand above and below, with cv = 2.5 m²/year. Find the time for 50% and 90% consolidation, and the settlement at each.
1
Drainage path: sand on both faces, so \( H_{dr} = H/2 = 2.0\text{m} \)
2
U = 50% (< 60%): \( T_v = \frac{\pi}{4}(0.50)^2 = 0.1963 \), so \( t = \dfrac{T_v H_{dr}^2}{c_v} = \dfrac{0.1963(2.0)^2}{2.5} = 0.31\text{yr} \) (about 3.8 months)
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U = 90% (> 60%): \( T_v = 1.781 - 0.933\log(10) = 0.848 \), so \( t = \dfrac{0.848(4)}{2.5} = 1.36\text{yr} \)
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Settlements: 50% of 110 mm = 55 mm; 90% of 110 mm = 99 mm.
Answer: 50% (55 mm) in 0.31 yr; 90% (99 mm) in 1.36 yr. Note the last 40% takes four times as long as the first 50% — consolidation has a very long tail. If drainage were single-sided, Hdr would double and every time would be four times longer.
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Quick Reference & Quick Check
Total settlementSt = Se + Sc + Ss
Elastic settlementSe = qoB(1−μ²)IsIf/Es
OCRσ'c / σ'₀
NC settlement(CcH/(1+e₀))log[(σ'₀+Δσ')/σ'₀]
OC below σ'cuse Cs in place of Cc
Time factorTv = cvt/Hdr²
Drainage pathH/2 double-drained, H single
Tv at U=50% / 90%0.197 / 0.848
SecondarySs = C'αH log(t₂/t₁)
Typical Cs/Cc1/5 to 1/10
1. A building settles 80 mm uniformly across its whole footprint. Another settles 25 mm at one corner and 5 mm at the other. Which is more likely to be damaged?
2. Why is consolidation settlement slow while elastic settlement is immediate?
3. A clay has OCR = 3. A modest load is applied that keeps σ'₀+Δσ' below σ'c. Compared with the same clay normally consolidated, settlement will be:
4. A clay layer drained on both faces reaches 90% consolidation in 2 years. If the lower sand layer were absent (single drainage), how long would it take?