Geotechnical Engineering

DHARM

COMPRESSIBILITY AND CONSOLIDATION OF SOILS 247

For the clay stratum:

w = 50%

G = 2.65

Since it is saturated, e = w.G = 0.50 × 2.65 = 1.325

This is the initial void ratio, e 0.

γsat =

()

()

(.. )

(.)

.

.

Ge

e w

+

+

≈ +

+

×=×

1

2 65 1 325

1 1 325

10 3 975 10

2 325

γ kN/m^33 kN/m

= 17.1 kN/m^3
γ = (γsat – γw) = 7.1 kN/m^3
Initial effective overburden pressure at the middle of the clay layer:
σ 0 = (0.4 × 18.5 + 1.6 × 9.0 + 2 × 7.1) t/m^2
= 36 kN/m^2
Let us assume that the applied surface pressure of 4.0 t/m^2 gets transmitted to the
middle of the clay layer undiminished.
∴ ∆σ = 40 kN/m^2
The compression index, Cc may be taken as:
Cc = 0.009 (wL – 10) ...(Eq. 7.8)
∴ Cc = 0.009 (65 – 10) = 0.495
The consolidation settlement, S, is given by:

S =

HC

e

. c
()

log

1 0 10

0

+ 0

F +

HG

I

KJ

σσ

σ

∆

=

400 0 495

1 1 325

36 40

(^1036)
×

. +
(.)

log

()

cm

≈ 27.64 cm.

(b) Thickness of the laboratory sample = 25 mm.

Since it is two-way drainage with porous discs on either side, the drainage path,

H = 25/2 = 12.5 mm.

Time for 90% primary compression, t 90 = 81 minutes.

Time factor, T 90 , for U = 90% is known to be 0.848.

(Alternatively, T = – 0.9332 log 10 (1 – U) – 0.0851

= – 0.9332 log 10 0.10 – 0.0851 = 0.9332 – 0.0851 = 0.8481

∴ T 90 =

Ct

H

v 90

2

Coefficient of consolidation

Cv =

TH

t

90

2

90

0 848 1 25^2

81

. .(.)
= × cm^2 /min.

=

0 848 1 25

81 60

.(.)×^2

× cm

(^2) /s
= 2.726 × 10–4 cm^2 /s.