Soil Compressibility

Table of contents

Soil compressibility governs how much a foundation will settle under load.

When a load is applied to soil, the soil skeleton compresses, water is expelled, and settlement occurs. The amount and rate of settlement depend on:

  • Soil type: granular vs. cohesive
  • Stress history: normally consolidated vs. overconsolidated
  • Loading magnitude: net stress increase
  • Drainage conditions: drainage path length and permeability

Types of Settlement

Settlement Type Soil Type Time Frame Mechanism
Immediate (elastic) settlement All soils During construction Elastic deformation of soil skeleton without volume change
Primary consolidation Fine-grained soils Months to years Gradual expulsion of pore water under increased load
Secondary compression (creep) All soils, especially organic Years to decades Viscous adjustment of soil skeleton under constant effective stress

The Consolidation Process

Terzaghi's Theory of One-Dimensional Consolidation

Terzaghi's theory describes the time-dependent settlement of saturated clay layers under increased load.

Fundamental equation:

$$ c_v \frac{\partial^2 u}{\partial z^2} = \frac{\partial u}{\partial t} $$

Where:

  • $c_v$ = coefficient of consolidation
  • $u$ = excess pore water pressure
  • $z$ = depth
  • $t$ = time

The Consolidation Analogy

    Load (Δσ)
       ↓
  ┌───────────┐
  │  Sand     │ Drainage layer
  ├───────────┤
  │  Clay     │ ← Excess pore pressure = Δσ at t=0
  │  Layer    │
  │           │ → Pore pressure dissipates over time
  ├───────────┤
  │  Sand     │ Drainage layer
  └───────────┘

At t=0: All load is carried by pore water ($u = \Delta\sigma$)
At t=∞: All load is carried by soil skeleton ($\sigma' = \sigma_0 + \Delta\sigma$)

The Oedometer Test (AS 1289.6.3.1)

The one-dimensional consolidation test (oedometer test) determines:

Parameter Symbol Determination Method
Compression index $C_c$ Slope of virgin compression line
Swell index $C_s$ Slope of unloading/reloading line
Preconsolidation pressure $\sigma'_p$ Casagrande construction
Coefficient of consolidation $c_v$ Casagrande log-time or Taylor √t method
Coefficient of volume compressibility $m_v$ $m_v = \Delta\epsilon / \Delta\sigma'$
Coefficient of permeability $k$ $k = c_v m_v \gamma_w$

The e-log σ' Curve

The void ratio vs. log effective stress plot is the primary output of an oedometer test:

e
│
│  ┌───────────────────┐
│  │  Recompression    │ ── Swelling line (C_s)
│  │                   │
│  │● σ'_p (Preconsolidation Pressure)
│  │                   │
│  │  Virgin Compression Line (C_c)
│  │                   │
│  └───────────────────┘
│   
└─────────────────────────── log σ'

Key Parameters from the e-log σ' Curve

Compression Index ($C_c$):

$$ C_c = \frac{e_1 - e_2}{\log(\sigma'_2/\sigma'_1)} $$

Correlations for $C_c$:

Formula Source Applicability
$C_c = 0.009(LL - 10)$ Terzaghi & Peck Inorganic clays
$C_c = 0.007(LL - 7)$ Skempton Remoulded clays
$C_c = 0.01(w_n - 5)$ CL clays
$C_c = 0.0046LL + 0.014$ General
$C_c = 1.15(e_0 - 0.27)$ All clays

Swell Index ($C_s$):

$$ C_s \approx 0.1\ \text{to}\ 0.2 \times C_c $$

Preconsolidation Pressure

The Casagrande construction determines $\sigma'_p$:

  1. Identify the point of maximum curvature on the e-log σ' curve
  2. Draw the tangent and horizontal lines at this point
  3. Bisect the angle between them
  4. Extend the straight-line portion of the virgin compression curve
  5. The intersection of the bisector and the virgin line = $\sigma'_p$

Overconsolidation Ratio (OCR)

$$ OCR = \frac{\sigma'_p}{\sigma'_{v0}} $$
OCR Stress History
OCR = 1 Normally consolidated (NC)
OCR > 1 Overconsolidated (OC)
OCR < 1 Underconsolidated (unlikely unless fill recently placed)
OCR Degree of Overconsolidation
1 Normally consolidated
1–2 Slightly overconsolidated
2–4 Moderately overconsolidated
4–10 Highly overconsolidated
> 10 Heavily overconsolidated

Settlement Calculations

Primary Consolidation Settlement

For normally consolidated clay ($\sigma'_{v0} + \Delta\sigma' \leq \sigma'_p$):

$$ S_c = \frac{C_c H}{1+e_0} \log\left(\frac{\sigma'_{v0} + \Delta\sigma'}{\sigma'_{v0}}\right) $$

For overconsolidated clay ($\sigma'_{v0} + \Delta\sigma' > \sigma'_p$):

$$ S_c = \frac{C_s H}{1+e_0} \log\left(\frac{\sigma'_p}{\sigma'_{v0}}\right) + \frac{C_c H}{1+e_0} \log\left(\frac{\sigma'_{v0} + \Delta\sigma'}{\sigma'_p}\right) $$

For overconsolidated clay ($\sigma'_{v0} + \Delta\sigma' \leq \sigma'_p$):

$$ S_c = \frac{C_s H}{1+e_0} \log\left(\frac{\sigma'_{v0} + \Delta\sigma'}{\sigma'_{v0}}\right) $$

Where:

  • $H$ = thickness of clay layer
  • $e_0$ = initial void ratio
  • $\sigma'_{v0}$ = initial effective overburden stress
  • $\Delta\sigma'$ = stress increase at mid-depth of layer

Immediate Settlement (Elastic)

For flexible foundations on clay:

$$ S_i = \frac{qB(1-\mu^2)}{E_s} I_f $$

Where:

  • $q$ = applied pressure
  • $B$ = foundation width
  • $\mu$ = Poisson's ratio (0.3–0.5 for clay)
  • $E_s$ = Young's modulus
  • $I_f$ = influence factor (depends on shape and rigidity)

Typical elastic modulus values:

Soil Type $E_s$ (MPa)
Very soft clay 2–15
Soft clay 5–25
Medium clay 15–40
Stiff clay 30–100
Very stiff clay 80–200
Loose sand 10–25
Medium dense sand 25–50
Dense sand 50–80
Dense sand and gravel 80–200

Secondary Compression (Creep)

$$ S_s = \frac{C_\alpha H}{1+e_p} \log\left(\frac{t_2}{t_1}\right) $$

Where:

  • $C_\alpha$ = coefficient of secondary compression
  • $e_p$ = void ratio at end of primary consolidation
  • $t_1$ = time at end of primary consolidation
  • $t_2$ = design life
Soil Type $C_\alpha / C_c$
Clays 0.04 ± 0.01
Silts 0.03 ± 0.01
Organic clays 0.05 ± 0.02
Peat 0.06–0.10

Rate of Consolidation

Degree of Consolidation

$$ U = \frac{S_t}{S_c} \times 100\% $$

Where $S_t$ = settlement at time $t$, $S_c$ = final primary consolidation settlement.

Time Factor

$$ T_v = \frac{c_v t}{d^2} $$

Where:

  • $d$ = **drainage path length** (half the layer thickness for double drainage; full thickness for single drainage)
  • $c_v$ = coefficient of consolidation

Relationship between $U$ and $T_v$:

$U$ (%) $T_v$
0 0
10 0.008
20 0.031
30 0.071
40 0.126
50 0.197
60 0.287
70 0.403
80 0.567
90 0.848
95 1.163
99 1.815

Approximate formulas:

$$ T_v = \frac{\pi}{4}\left(\frac{U}{100}\right)^2 \quad \text{for} \ U < 60\% $$ $$ T_v = 1.781 - 0.933\log(100 - U) \quad \text{for} \ U > 60\% $$

Time to Reach a Given Settlement

$$ t = \frac{T_v d^2}{c_v} $$

Coefficient of Consolidation Determination

Casagrande's Log-Time Method

Plot dial reading vs. log time:

  1. Identify the theoretical zero point (final parabolic portion)
  2. Identify the theoretical 100% primary consolidation (intersection of tangents)
  3. The time at 50% consolidation ($t_{50}$) gives $c_v$:
$$ c_v = \frac{0.197 d^2}{t_{50}} $$

Taylor's √t Method

Plot dial reading vs. square root of time:

  1. Extend the initial straight-line portion
  2. Draw a second line at 1.15 times the slope
  3. The intersection with the curve gives $\sqrt{t_{90}}$
  4. Calculate $c_v$:
$$ c_v = \frac{0.848 d^2}{t_{90}} $$

Stress Distribution in the Ground

Boussinesq's Solution

The vertical stress increase at depth $z$ below a point load $Q$:

$$ \Delta\sigma_z = \frac{3Q}{2\pi z^2} \left[ \frac{1}{1+(r/z)^2} \right]^{5/2} $$

Approximate Methods

2:1 Method (simplified):

$$ \Delta\sigma_z = \frac{Q}{(B+z)(L+z)} $$

Where $B$ = foundation width, $L$ = foundation length

Influence Factors for Common Shapes

Shape Stress at depth $z$ below centre
Square ($B \times B$) $\Delta\sigma_z = q \times I$
Strip (width $B$, infinite length) $\Delta\sigma_z = \frac{q}{\pi}(\alpha + \sin\alpha\cos(\alpha+2\beta))$
Circular (diameter $B$) $\Delta\sigma_z = q\left[1 - \frac{1}{(1+(B/2z)^2)^{3/2}}\right]$

Practical Considerations

Allowable Settlements

Structure Type Maximum Total Settlement Maximum Differential Settlement
Isolated footings — clay 65 mm 1:300
Isolated footings — sand 50 mm 1:300
Raft foundations 50–75 mm 1:500
Frame structures 50 mm 1:300
Steel structures 100 mm 1:200
Bridges 50 mm 1:300
Machinery foundations 25 mm 1:1000
Warehouses 150 mm 1:150

Australian Standards

Standard Relevance
AS 2870 Residential slabs (differential settlement limits)
AS 2159 Piling (settlement of piled foundations)
AS 5100 Bridge design (foundation settlement limits)
AS 4678 Earth retaining structures (settlement behind walls)

9.3 Methods to Reduce Settlement

Method How It Works Applicable Soil
Preloading Apply surcharge before construction Soft clays, silts
Vertical drains (PVDs) Shorten drainage path Soft clays
Deep compaction Increase density Granular soils
Stone columns Reinforce + drain Soft clays, silts
Deep mixing Improve soil stiffness Soft clays, organic
Grouting Fill voids, improve stiffness All soils
Pile foundations Transfer load to competent strata All weak soils

Worked Example

Problem: A 3 m × 3 m square footing carries a load of 900 kN on a 5 m thick clay layer. The clay has $e_0 = 1.0$, $LL = 45\%$, $\sigma'_p = 150$ kPa, $\sigma'_{v0} = 80$ kPa, $C_c = 0.32$, $C_s = 0.05$, $c_v = 5$ m²/year. Water table at surface.

Step 1 — Stress increase at mid-depth (2.5 m):

Using 2:1 method:

$\Delta\sigma = \frac{900}{(3+2.5)(3+2.5)} = \frac{900}{30.25} = 29.8$ kPa

Step 2 — Check OCR at mid-depth:

$\sigma'_{v0} = 80$ kPa, $\sigma'_p = 150$ kPa $OCR = 150/80 = 1.88$ → Overconsolidated

Step 3 — Settlement calculation:

$\sigma'_{v0} + \Delta\sigma = 80 + 29.8 = 109.8$ kPa < $\sigma'_p = 150$ kPa

Use OC equation:

$$ S_c = \frac{0.05 \times 5}{1+1.0} \log\left(\frac{109.8}{80}\right) = 0.125 \times \log(1.373) = 0.125 \times 0.138 = 0.017\ \text{m} = 17\ \text{mm} $$

Step 4 — Time to 90% consolidation:

$d = 2.5$ m (double drainage, clay on sand) $T_{90} = 0.848$ $$ t = \frac{0.848 \times 2.5^2}{5} = 1.06\ \text{years} $$