Transcript Consolidation - جامعة فلسطين

Consolidation Theory
Examples
Example 1


The Following results were obtained from an
oedometer test on a specimen of saturated clay.
Pressure(kN/m2)
27
54
107
214
429
Void ratio
1.243
1.217
1.144
1.068
0.994
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.
Continued…
Example1

Determine the final
settlement due to
consolidation of the clay.
 Solution:
 The settlement of the clay
layer
eo  e1
Sc 
H
1  eo
Continued…

Example1
Appropriate values of e
are obtained from the
following figure:
 The clay layer will be
divided into four sublayer, hence H =
2000mm
Continued…

Example1
To get the settlement:
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
4
2
 o '  2 dry  4( sat ( sand )   w )  ( sat ( clay)   w )
4
 2 *14  4(18  9.81)  (19  9.81)  79.14kN / m 2
2
CH
  
S c  c log( o ' )
1  eo
o
'
But
Cc  0.009( LL  10)  0.009(40  10)  0.27
So
Sc 
0.27 * 4
79.14  100
log(
)  0.213m  213m m
1  0.8
79.14
Continued…
Example 2
B) The average effective stress at the middle of the clay layer is
 o '   '  79.14  100  179.14kN / m 2
 c '  190kN / m 2
Because.. o   '   c
'
'
CH
  
S c  s log( o ' )
1  eo
o
'
But
Cc 0.27

 0.045
6
6
0.045* 4
79.14  100
Sc 
log(
)  0.036m  63m m
1  0.8
79.14
Cs 
Continued…
Example 2
C) The average effective stress at the middle of the clay layer is
 o '   '  79.14  100  179.14kN / m 2
 c '  170kN / m 2
Because.. o   c   '   o
'
'
'
Cs H
c
Cc H
 o  
Sc 
log( ' ) 
log(
)
'
1  eo
 o 1  eo
o
'
'
0.045* 4
170
0.27 * 4
179.14
log(
)
log(
)
1  0.8
79.14 1  0.8
170
 0.0468m  46.8m m
Sc 
Example 3

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?
Continued…
Example 3
cv * t
Tv 
2
Hd
U  f (Tv )  f (
2
cv * t
)
2
Hd
t 1 H d1

t2 H d 2 2
2
Hd2
t 2  t1 *
2
H d1
20
25002

*
 2.63year
2
60* 24* 365 9.5
Continued…
Example 3
For, U  0.60
Tv 

4
U
2
2
0 .3
t30  t50 * 2
0.5
 2.63* 0.36
 0.95 years
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)
0
0.1
0.25
0.5
1
2
4
Dial reading
(mm)
4.041
3.927
3.879
3.830
3.757
3.650
3.495
8
3.282
15
3.035
30
2.766
60
2.55
120
2.423
240
2.276
505
2.184
1485
2.040
Continued…
Example 4
Continued…
Example 4
t50  9.6 min
(4.041 2.04)
2 H d  25.4 
 24.4
2
H d  12.2m m
2
Tv H d
0.196*12.2 2
1440* 365
Cv 

 3.0388*
t50
9.6
106
 1.597m 2 / year