Geotechnical Engineering

DHARM

COMPRESSIBILITY AND CONSOLIDATION OF SOILS 245

Clay A Clay B

e 0 = 0.572 e 0 = 0.612

e 1 = 0.505 e 1 = 0.597

(^) σ 0 = 120 kN/m^2 σ 0 = 120 kN/m^2
σ 1 = 180 kN/m^2 σ 1 = 180 kN/m^2
avA = ∆
∆
e
σ
=0 067
60
.
m^2 /kN m
e
vB==
∆
∆σ
0 015
60
.
m^2 /kN
mvA=+
0 067
60

. /(1 0 572. ) mvB=+0 015
60
. /(1 0 612. )

= 7.10 × 10–4 m^2 /kN = 1.55 × 10–4 m^2 /kN

HA/HB = 1.5 and tt (^50) BA/ 50 = 3
T 50 = Cvt 50 /H^2
∴ T 50 =
Ct
H
Ct
H
v
A
v
B
.. (^50) ABB
2
50
= 2
C
C
t
t
H
H
v
v
A
B
A
B
B
A
=^50
50
2
. 2 = 3 × (1.5) (^2) = 6.75
But Cv = k/mvγw
or k = Cvmv.γw
∴ kA/kB =
cm
cm
vv
vv
AA
BB
.
.

..
.

=××

×

−

(^675) −
710 10
155 10
4
4 = 30.92 ≈^31.
Example 7.8: A saturated soil has a compression index of 0.25. Its void ratio at a stress of
10 kN/m^2 is 2.02 and its permeability is 3.4 × 10–7 mm/s. Compute:
(i) Change in void ratio if the stress is increased to 19 kN/m^2 ;
(ii) Settlement in (i) if the soil stratum is 5 m thick; and
(iii) Time required for 40% consolidation if drainage is one-way.
(S.V.U.—B.Tech., (Part-time)—Apr., 1982)
Compression index, Cc = 0.25
e 0 = 2.02 σ 0 = 10 kN/m^2 k = 3.4 × 10–7 mm/s
(^) σ 1 = 19 kN/m^2
(i) Cc =
∆e
log ( 10 σσ 1 / 0 )∴ 0.25 =
∆e
log ( 10 19 10/ )
∴ ∆e = 0.25 log 10 (1.9) ≈ 0.07
or Void ratio at a stress of 19 kN/m^2 = 2.02 – 0.07 = 1.95
av = ∆∆e/ σ = 0.07/9 = 0.00778 m^2 /kN
mv = av/(1 + e 0 ) = 0.00778/(1 + 2.02) = 2.575 × 10–3 m^2 /kN
(ii) Thickness of soil stratum, H = 5 m.
Settlement, S =
HC
e
HC
e

. cc
()

log

.

()

log

110 10

0

00

10 1

+ 0

F +

HG

I

KJ

=

+

F

HG

I

KJ

σσ

σ

σ

σ

∆