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