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Chapter 11 of 20

Consolidation & Settlement

In the GSSSB Work Assistant Civil syllabus under Soil Mechanics & Foundation Engineering · 2 parts

Consolidation of Soil · Settlement of Foundations

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📑 Contents (16 sections)

Part 1 of 2

Consolidation of Soil

Last reviewed 16 Sept 2026 · 6 min read

What consolidation is

When load is applied to a saturated clay, water cannot escape quickly because permeability is very low. The load is first carried by excess pore water pressure; as water slowly drains, the load transfers to the soil skeleton, effective stress increases and the soil compresses. This time-dependent volume reduction due to expulsion of pore water is consolidation.

Total settlement of a foundation on clay:

S = S_{i} + S_{c} + S_{s}

$S_{i}$ = immediate (elastic) settlement — distortion at constant volume.
$S_{c}$ = primary consolidation settlement — expulsion of pore water (this chapter).
$S_{s}$ = secondary compression — creep of the skeleton after excess pore pressure has dissipated.

Spring–piston analogy (Terzaghi)

A cylinder full of water with a spring (soil skeleton) and a piston with a small valve (low permeability):

Instant of loading ( $t = 0$ ): water carries the whole load; excess pore pressure $Δ u = Δ σ$ ; spring stress unchanged.
During drainage: water escapes through the valve; load gradually shifts to the spring; $Δ u$ falls, $Δ σ^{'}$ rises.
End of consolidation: $Δ u = 0$ ; the spring carries the full load; $Δ σ^{'} = Δ σ$ .

Oedometer (consolidation) test

An undisturbed specimen (typically 60–75 mm diameter, 20–25 mm thick) is confined laterally in a ring between porous stones (one-dimensional compression, drainage top and bottom). Loads are applied in stages (each usually doubled, held about 24 hours) and dial readings recorded with time. Results give void ratio $e$ at the end of each stage.

Compressibility parameters

FormulaCompressibility

Coefficient of compressibility: $a_{v} = - \frac{Δ e}{Δ σ ^{'}}$ (m²/kN)
Coefficient of volume compressibility: $m_{v} = \frac{a _{v}}{1 + e _{0}} = \frac{Δ H / H}{Δ σ ^{'}}$ (m²/kN)
Compression index (slope of the virgin line on $e$ –log σ′): $C_{c} = \frac{e _{1} - e _{2}}{lo g _{10} ( σ _{2}^{'} / σ _{1}^{'} )}$
Swelling / recompression index $C_{s}$ (or $C_{r}$ ): slope of unloading–reloading line, typically $C_{c} /5$ to $C_{c} /10$ .
Skempton's empirical relations: $C_{c} = 0.009 (w_{L} - 10)$ (undisturbed clays of low–medium sensitivity); $C_{c} = 0.007 (w_{L} - 10)$ (remoulded).

Normally consolidated and over-consolidated clays

Preconsolidation pressure $σ_{p}^{'}$ (or $σ_{c}^{'}$ ) — the maximum effective stress the soil has experienced in its history. Found from the $e$ –log σ′ curve by Casagrande's graphical construction (point of maximum curvature, horizontal line, tangent, bisector, intersection with the extended virgin line).
Normally consolidated (NC): present effective stress $σ_{0}^{'} = σ_{p}^{'}$ — soil is on the virgin line; highly compressible.
Over-consolidated (OC): $σ_{0}^{'} < σ_{p}^{'}$ (past erosion, desiccation, lowered water table, glacial load); much less compressible for stresses below $σ_{p}^{'}$ .
Under-consolidated: still consolidating under its own recent weight ( $σ_{0}^{'}$ less than the final value).

Over-consolidation ratio O C R = \frac{σ _{p}^{'}}{σ _{0}^{'}}

Consolidation settlement

FormulaPrimary consolidation settlement of a layer of thickness H

Normally consolidated:

S_{c} = \frac{C _{c} H}{1 + e _{0}} lo g_{10} \frac{σ _{0}^{'} + Δ σ}{σ _{0}^{'}}

Over-consolidated, final stress ≤ $σ_{p}^{'}$ :

S_{c} = \frac{C _{s} H}{1 + e _{0}} lo g_{10} \frac{σ _{0}^{'} + Δ σ}{σ _{0}^{'}}

Over-consolidated, final stress > $σ_{p}^{'}$ :

S_{c} = \frac{C _{s} H}{1 + e _{0}} lo g_{10} \frac{σ _{p}^{'}}{σ _{0}^{'}} + \frac{C _{c} H}{1 + e _{0}} lo g_{10} \frac{σ _{0}^{'} + Δ σ}{σ _{p}^{'}}

Using $m_{v}$ : $S_{c} = m_{v} Δ σ H$ (for a small stress range).

$σ_{0}^{'}$ and $Δ σ$ are taken at the middle of the layer (thick layers are divided into sub-layers).

Terzaghi's one-dimensional consolidation theory

Assumptions

Soil is homogeneous and fully saturated.
Soil grains and water are incompressible.
Flow and compression are one-dimensional (vertical).
Darcy's law is valid; $k$ and $m_{v}$ remain constant during consolidation.
Strains are small; the load is applied instantaneously and remains constant.
The relation between void ratio and effective stress is linear over the stress increment.
Time lag is due only to low permeability (no secondary compression).

Governing equation

\frac{\partial u}{\partial t} = c_{v} \frac{\partial ^{2} u}{\partial z ^{2}}, c_{v} = \frac{k}{m _{v} γ _{w}}

$c_{v}$ = coefficient of consolidation (m²/s or m²/year).

Time factor and degree of consolidation

T_{v} = \frac{c _{v} t}{H _{d}^{2}}

$H_{d}$ = length of the longest drainage path: $H_{d} = H /2$ for double drainage (pervious layers top and bottom), $H_{d} = H$ for single drainage.

Average degree of consolidation $U$ = settlement at time $t$ / final settlement:

FormulaU–Tv relations (uniform initial excess pore pressure)

T_{v} = \frac{π}{4} U^{2} (U \leq 60%)

T_{v} = 1.781 - 0.933 lo g_{10} (100 - U %) (U > 60%)

$U$	10%	20%	30%	40%	50%	60%	70%	80%	90%
$T_{v}$	0.008	0.031	0.071	0.126	0.197	0.287	0.403	0.567	0.848

Key consequence: for the same degree of consolidation, time $\propto H_{d}^{2}$ . Double drainage makes consolidation four times faster than single drainage for the same layer; halving the drainage path (e.g. sand drains, prefabricated vertical drains) greatly shortens consolidation time.

Determining $c_{v}$ from test data

Casagrande's log-time fitting method: find $t_{50}$ from the dial reading–log time curve (locating 0% and 100% primary consolidation) → $c_{v} = \frac{0.197 H _{d}^{2}}{t _{50}}$ .
Taylor's square-root-of-time method: straight initial portion of dial reading vs $t$ ; a line with abscissae 1.15 times cuts the curve at $t_{90}$ → $c_{v} = \frac{0.848 H _{d}^{2}}{t _{90}}$ .

Also: $k = c_{v} m_{v} γ_{w}$ gives permeability of clays indirectly.

Secondary compression

After primary consolidation, compression continues slowly at constant effective stress due to plastic readjustment (creep) of the soil skeleton — important in organic soils and peats and highly plastic clays.

C_{α} = \frac{Δ e}{lo g _{10} ( t _{2} / t _{1} )}, S_{s} = \frac{C _{α} H}{1 + e _{p}} lo g_{10} \frac{t _{2}}{t _{1}}

Worked examples

Worked ExampleExample 1 — settlement of NC clay

A 4 m thick normally consolidated clay layer has $e_{0}$ = 0.90 and $C_{c}$ = 0.30. At its mid-depth, $σ_{0}^{'}$ = 80 kPa. A building increases the stress there by 40 kPa. Find the consolidation settlement.

Solution. $S_{c} = \frac{0.30 \times 4000}{1.90} lo g_{10} \frac{120}{80} = 631.6 \times 0.1761 = 111 mm$

Worked ExampleExample 2 — time for consolidation

For the layer above, drainage is from both top and bottom; $c_{v} = 1 \times 1 0^{- 7}$ m²/s. Find the time for 50% and 90% consolidation. How would the times change with single drainage?

Solution. $c_{v} = 1 \times 1 0^{- 7} \times 3.1536 \times 1 0^{7} = 3.154$ m²/year; $H_{d} = 2$ m. $t_{50} = \frac{0.197 \times 4}{3.154} = 0.25 year$ ; $t_{90} = \frac{0.848 \times 4}{3.154} = 1.08 years$

Single drainage ( $H_{d}$ = 4 m): times four times longer → 1.0 year and 4.3 years.

Worked ExampleExample 3 — compressibility from oedometer data

Void ratio falls from 0.85 to 0.78 as effective stress rises from 100 to 200 kPa. Find $C_{c}$ , $a_{v}$ and $m_{v}$ .

Solution. $C_{c} = \frac{0.07}{lo g _{10} 2} = 0.233$ ; $a_{v} = \frac{0.07}{100} = 7.0 \times 1 0^{- 4} m^{2} /kN$ ; $m_{v} = \frac{7 \times 1 0 ^{- 4}}{1.85} = 3.78 \times 1 0^{- 4} m^{2} /kN$

Worked ExampleExample 4 — Skempton's relation

An undisturbed clay has $w_{L}$ = 50%. Estimate $C_{c}$ .

Solution. $C_{c} = 0.009 (50 - 10) = 0.36$

Frequently tested points

Consolidation: expulsion of pore water under sustained load in saturated clay; time-dependent.
$m_{v} = a_{v} / (1 + e_{0})$ ; $C_{c}$ from the virgin line; $C_{c} = 0.009 (w_{L} - 10)$ .
NC settlement $\frac{C _{c} H}{1 + e _{0}} lo g \frac{σ _{0}^{'} + Δ σ}{σ _{0}^{'}}$ .
OCR $= σ_{p}^{'} / σ_{0}^{'}$ ; Casagrande's construction gives $σ_{p}^{'}$ .
$c_{v} = k / (m_{v} γ_{w})$ ; $T_{v} = c_{v} t / H_{d}^{2}$ ; $T_{50}$ = 0.197, $T_{90}$ = 0.848.
Double drainage: $H_{d} = H /2$ → 4 times faster.
Log-time method uses $t_{50}$ ; root-time method uses $t_{90}$ .

Common MistakeCommon mistakes

Using full layer thickness as $H_{d}$ for double drainage.
Taking stresses at the top of the clay layer instead of its middle.
Using $C_{c}$ for an over-consolidated clay loaded below its preconsolidation pressure.

Revision SummaryChapter summary

Consolidation transfers load from pore water to skeleton as water drains.
The oedometer test gives $C_{c}$ , $C_{s}$ , $m_{v}$ , $c_{v}$ and preconsolidation pressure.
Settlement depends on stress history (NC vs OC).
Terzaghi's theory relates time to $c_{v}$ , drainage path and degree of consolidation.
Secondary compression continues after primary consolidation, especially in organic soils.

Part 2 of 2

Settlement of Foundations

Last reviewed 16 Sept 2026 · 6 min read

Settlement and why it matters

Settlement is the vertical downward movement of a foundation. Every foundation settles; the design objective is to keep the total and especially the differential settlement within limits that the structure can tolerate without damage to structure, finishes or services.

Components

S = S_{i} + S_{c} + S_{s}

Component	Cause	Dominant in	Time
Immediate (elastic) $S_{i}$	Elastic distortion of soil (clays at constant volume; sands with volume change)	Sands; also clays under undrained loading	Instantaneous (during construction)
Primary consolidation $S_{c}$	Expulsion of pore water	Saturated clays	Months to years
Secondary compression $S_{s}$	Creep of the soil skeleton	Organic soils, highly plastic clays	Long-term

Immediate (elastic) settlement

Based on the theory of elasticity for a flexible or rigid footing on a semi-infinite elastic medium:

FormulaElastic settlement

S_{i} = q B \frac{1 - μ ^{2}}{E _{s}} I_{f}

$q$ = net foundation pressure; $B$ = width (or diameter); $E_{s}$ = modulus of elasticity of soil; $μ$ = Poisson's ratio (≈ 0.5 for saturated clays under undrained loading, 0.25–0.35 for sands); $I_{f}$ = influence factor depending on shape, rigidity and the point considered.

Typical influence factors (surface footing, semi-infinite medium):

Shape	Flexible — centre	Flexible — corner/edge	Flexible — average	Rigid
Circle	1.00	0.64 (edge)	0.85	0.79
Square	1.12	0.56 (corner)	0.95	0.82

A rigid footing settles uniformly, at roughly the average settlement of a flexible one (about 0.8 of the flexible centre value).

Settlement of footings on sand

Undisturbed sampling of sand is difficult, so settlement is estimated from field tests:

Plate load test — relates settlement of a test plate to that of the footing (Terzaghi–Peck for sand):

S_{f} = S_{p} [\frac{B _{f} ( B _{p} + 0.3 )}{B _{p} ( B _{f} + 0.3 )}]^{2} (B in m)

For clay (settlement proportional to width): $S_{f} = S_{p} \frac{B _{f}}{B _{p}}$ .

Standard penetration test (SPT) — IS 8009 and IS 6403 provide charts relating corrected $N$ , footing width and settlement (e.g. pressure causing 25 mm settlement).
Static cone penetration test — De Beer / Schmertmann methods using cone resistance.

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