Primary Consolidation Settlement

s_{c} = \frac{C _{c} H}{1 + e _{0}} lo g \frac{σ _{2}^{'}}{σ _{1}^{'}}

One-dimensional primary settlement of normally consolidated clay. Cc is the compression index (slope of e–log σ' in the virgin range), H the layer thickness, e0 the initial void ratio. If the clay is overconsolidated and σ'_2 stays below σ'_p, replace Cc by the recompression index Cr. Time rate uses Tv = cv t / H_dr².

$s_{c}$ (m): Primary consolidation settlement
$C_{c}$: Compression index
$H$ (m): Layer thickness
$e_{0}$: Initial void ratio
$σ_{1}^{'}, σ_{2}^{'}$ (kPa): Initial and final effective stress

Step-by-step solved example

NC clay: Cc = 0.35, H = 3.0 m, e0 = 0.90, σ'1 = 80 kPa, σ'2 = 160 kPa. Find sc.

1-D settlement of a clay layer: s = Cc H/(1+e0) log(σ2'/σ1').

1. Stress ratio
σ'2/σ'1 = 2.00, log10(2) = 0.3010.
$lo g (2) = 0.301$
2. Settlement
sc = 0.35×3.0×0.3010 / 1.90 = 1.05×0.3010/1.90 = 0.166 m (166 mm).
$s_{c} = 0.166 m$

Answer: 0.166 m (166 mm)

10 practice questions

0/10 correct

1.If σ'2 = 2 σ'1, log(σ'2/σ'1) equals

2.Doubling H, all else equal, makes sc

3.For OC clay with σ'2 < σ'_p use

4.Time factor Tv = cv t / H_dr². Double drainage means H_dr is

5.Cc = 0.40, H = 4 m, e0 = 1.0, σ'2/σ'1 = 2. sc =

6.Secondary compression continues

7.U = 50% consolidation for a clay layer occurs at Tv ≈

8.Sands settle mainly by

9.mv (coefficient of volume compressibility) gives sc =

10.Raising e0 (same Cc, H, stress ratio) makes sc