Civil Engineering Reference
In-Depth Information
Worked examples
Example 15.1: Interpretation of oedometer test results
The first two columns of
Table 15.1 contain data from a single increment of an oedometer test in which the total
vertical stress was raised from
0 the sample was
20 mm thick and it was allowed to drain from the top and from the bottom.
For two-way drainage the drainage path is
H
σ
=
90 kPa to
σ
=
300 kPa. At
t
=
10mm. The degree of consoli-
dation
U
t
is given by Eq. (15.24), taking the final settlement as
=
ρ
∞
=
1.920mm
corresponding to
t
=
24 h.
√
t
method
.
Figure 15.12(a) shows
U
t
plotted against
√
t
. Scaling from the
(a)
diagram,
√
t
1
=
4.6 and hence
t
1
=
21.2min. From Eq. (15.36),
10
−
3
)
2
3
H
2
4
t
1
=
3
×
(10
×
1.9m
2
/year
=
×
×
×
=
c
v
60
24
365
4
×
21.2
(b) Log
t
method. Figure 15.12(b) shows
U
t
plotted against log
t
. From the figure,
log
t
50
=
0.70 and
t
50
=
5.01min. From Eq. (15.38),
0.196
H
2
t
50
10
−
3
)
2
0.196
×
(10
×
2.1m
2
/year
c
v
=
=
×
60
×
24
×
365
=
5.01
The mean value for the coefficient of consolidation from the two methods is
c
v
2m
2
/year.
=
σ
During the increment the vertical effective stress changes from
=
90 kPa at the
σ
=
start to
300 kPa at the end. The vertical strain is
ε
=
1.920/20
=
0.096 and
z
from Eq. (8.9) the coefficient of compressibility is
=
ε
0.096
300
z
σ
z
=
10
−
4
m
2
/kN
m
v
90
=
4.6
×
−
Table 15.1
Results of an oedometer test - Example 15.1
√
t
(
min
1/2
)
Time (min)
Settlement
p
t
(mm)
U
t
log t
0
0
0
0
-
0.25
0.206
0.107
0.5
-0.602
1
0.414
0.216
1
0
2.25
0.624
0.325
1.5
0.352
4
0.829
0.432
2
0.602
9
1.233
0.642
3
0.954
16
1.497
0.780
4
1.204
25
1.685
0.878
5
1.398
36
1.807
0.941
6
1.556
49
1.872
0.975
7
1.690
24 h
1.920
1.000
-
-
