CivilGeotechnicalhardCourseworkPE / Advanced

Primary Consolidation Settlement of a Clay Layer

A normally consolidated clay layer is $H = 3.0\ \text{m}$ thick.

Initial void ratio $e_0 = 0.80$
Compression index $C_c = 0.30$
Initial effective stress at mid-layer $\sigma'_0 = 100\ \text{kPa}$
Stress increase from new construction $\Delta\sigma = 50\ \text{kPa}$

Find the primary consolidation settlement.

Given

H	3 m	Layer thickness
e0	0.8 -	Initial void ratio
Cc	0.3 -	Compression index
σ'0	100 kPa	Initial effective stress
Δσ	50 kPa	Stress increase

Hint 1

Normally consolidated means the full compression index C_c applies over the whole stress increase.

Hint 2

S_c = (C_c·H/(1+e₀))·log((σ'₀+Δσ)/σ'₀).

Hint 3

That logarithm is base 10. Using ln overestimates settlement by a factor of 2.3.

Worked solution — try the problem first

For a normally consolidated clay:

$S_c = \frac{C_c H}{1 + e_0}\log_{10}\!\left(\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0}\right)$

Coefficient: $\frac{C_c H}{1 + e_0} = \frac{0.30 \times 3.0}{1 + 0.80} = \frac{0.90}{1.80} = 0.500\ \text{m}$

Log term. Note this is $\log_{10}$ , not $\ln$ — a common and costly slip:

$\log_{10}\!\left(\frac{100 + 50}{100}\right) = \log_{10}(1.50) = 0.1761$

Settlement: $S_c = 0.500 \times 0.1761 = 0.0880\ \text{m}$

$\boxed{S_c = 88\ \text{mm}}$

Why normally consolidated matters. If the clay were overconsolidated with a preconsolidation pressure above 150 kPa, the recompression index $C_r$ (typically $C_c/5$ to $C_c/10$ ) would apply instead, and settlement would be several times smaller. Establishing the stress history is the first step in any real settlement analysis.

Concepts:Consolidation settlementCompression indexVoid ratioNormally consolidated clay

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