consolidation theory examples. example 1 the following results were obtained from an oedometer test...
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Consolidation Theory
Examples
Example 1
The Following results were obtained from an oedometer test on a specimen of saturated clay.
A layer of this clay 8m thick lies below a 4m depth of sand, the water table being at the surface. For both soils, the saturated unit wt=21kN/m3
A 4m depth of fill of unit weight 21kN/m3 is placed on the sand over an extensive area.
Pressure(kN/m2)2754107214429
Void ratio1.2431.2171.1441.0680.994
Continued… Example1
Determine the final settlement due to consolidation of the clay.
Solution: The settlement of the clay
layer
He
eeS
o
oc
1
1
Appropriate values of e are obtained from the following figure:
The clay layer will be divided into four sub-layer, hence H = 2000mm
Continued… Example1
To get the settlement:
Continued… Example1
Example 2
A soil profile is shown in the Fig. If a uniformly distributed load is applied at the ground surface, what is the settlement of the clay layer caused by primary consolidation if:
A) The clay is normally consolidated B) The preconsolidation pressure=190kN/m2
C) The preconsolidation pressure=170kN/m2
Cs=1/6 Cc
Continued… Example 2
Continued… Example 2
A) The average effective stress at the middle of the clay layer is
2
)()('
/14.79)81.919(2
4)81.918(414*2
)(2
4)(42
mkN
wclaysatwsandsatdryo
mmmS
So
LLC
But
e
HCS
c
c
o
o
o
cc
213213.0)14.79
10014.79log(
8.01
4*27.0
27.0)1040(009.0)10(009.0
)log(1 '
'
Continued… Example 2
B) The average effective stress at the middle of the clay layer is
'''
2'
2''
..
/190
/14.17910014.79
co
c
o
Because
mkN
mkN
mmmS
CC
But
e
HCS
c
cs
o
o
o
sc
63036.0)14.79
10014.79log(
8.01
4*045.0
045.06
27.0
6
)log(1 '
'
Continued… Example 2
C) The average effective stress at the middle of the clay layer is
''''
2'
2''
..
/170
/14.17910014.79
oco
c
o
Because
mkN
mkN
mmm
S
e
HC
e
HCS
c
o
o
o
c
o
c
o
sc
8.460468.0
)170
14.179log(
8.01
4*27.0)
14.79
170log(
8.01
4*045.0
)log(1
)log(1 '
'
'
'
In an oedometer test, a specimen of saturated clay 19mm thick reaches 50% consolidation in 20 min.
How long would it take a layer of this clay 5m thick to reach the same degree of consolidation under the same stress and drainage conditions?
How long would it take the layer to reach 30% consolidation?
Example 3
Continued… Example 3
)*
()( 2d
vv
H
tcfTfU
2
*
d
vv
H
tcT
22
21
2
1
d
d
H
H
t
t
year
H
Htt
d
d
63.25.9
2500*
365*24*60
20
*
2
2
21
22
12
Continued… Example 3
years
tt
UT
UFor
v
95.0
36.0*63.25.0
3.0*
4
60.0,
2
2
5030
2
Example 4
The time rate settlement data shown below is for the increment from 20 to 40 kPa from the test. The initial sample height is 2.54cm, and there are porous stones on the top and at the bottom of the sample.
Determine Cv by the log time- fitting procedure.
Continued… Example 4
Elapsed time (min)
Dial reading (mm)
04.041
0.13.927
0.253.879
0.53.830
13.757
23.650
43.495
83.282
153.035
302.766
602.55
1202.423
2402.276
5052.184
14852.040
Continued… Example 4
Continued… Example 4
yearm
t
HTC
mmH
H
t
dvv
d
d
/597.1
10
365*1440*0388.3
6.9
2.12*196.0
2.12
4.242
)04.2041.4(4.252
min6.9
2
6
2
50
2
50
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