Time rate of consolidation On the basis of Terzaghi's consolidation theory

1. 1
Time rate of consolidation
(Terzaghi’s consolidation theory)

2. Time-rate of consolidation 2
The time required to reduce the excess pore water pressure is
primarily a function of the distance the water must travel to exit
the stressed area.
Karl Terzaghi proposed a theory in 1925 that predicted the time-
rate of consolidation. This theory is still in use today.
In essence, Terzaghi derived the basic differential equation that
described the change of the excess pore water pressure, u, as a
function of time and space.

3. Time-rate of consolidation 3

4. Time-rate of consolidation 4
Terzaghi’s differential equation can be solved
with following boundary condition
The solution is

5. Time-rate of consolidation 5
The solution is
Tv is a nondimensional number. Tv provides a useful expression to
estimate the settlement in the field from the results of a lab
consolidation test.

6. Graphical solution of differential equation 6

7. Analytical solution of 1D-consolidation Eq. 7
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
Relative
vertical
position,
z/H
Relative excess pore pressure, p/po
T
v
=
2.0
(t
=
20
days)
Data
E = 1,000 kPa
= 0.0
k = 0.001 m/day
w = 10 kN/m3
H = 1.0 m
po = 1 kPa
Eoed = 1,000 kPa
cv = 0.1 m2/day
Tv = cv.t/H2
cv = k.Eoed/ w
Eoed = (1- )E/[(1+ )(1-2 )]
H
Free drainage
Impermeable boundary
po

8. 8

9. 9

10. Degree of consolidation 10
The degree of consolidation is

11. Degree of consolidation 11

12. 0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
Ave.
degree
of
consolidation
U
(%)
Time factor Tv
U<=60%
U> 60%
Theoretical
Average degree of consolidation 12

13. 13

14. Average degree of consolidation 14
%
100
4



v
T
U
%
100
10
1 933
.
0
085
.
0
















 
 v
T
U
2
100
%
4







U
Tv

 
%
100
log
933
.
0
781
.
1 U
Tv 


Fro U = 0 to 60%
Fro U = 0 to 60%
For Tv ≤ 0.217 (U ≤ 52.6%)
For Tv > 0.217 (U > 52.6%)

15. Average degree of consolidation 15
Relationship between Tv & average degree of consolidation
for a uniform and triangular distribution of initial excess
porewater pressure

16. Logarithm-of-time Method 16

17. Square-root-of-time Method 17

18. Consolidation Test - data
Specimen data
At the beginning of test
Diameter of specimen, D = 2.50 in.
Initial height of specimen, H0 = 0.780 in.
Mass of specimen ring + specimen = 208.48 g
Mass of specimen ring = 100.50 g
Initial moisture content
mass of wet soil + can = 79.97 g
mass of dry soil + can = 66.82 g
mass of can = 35.60 g
specific gravity = 2.72
At the end of test
mass of entire wet specimen + can = 234.54 g
mass of entire dry specimen plus can = 203.11 g
mass of can = 127.20 g
18

19. Consolidation Test - data 19
Time-versus-deformation data
date time deformation dial reading (in.)
06/08/96 9:15 AM 0.0000
9:15.1 0.0067
9:15.25 0.0069
9:15.5 0.0071
9:16 0.0077
9:17 0.0084
9:19 0.0095
9:23 0.0107
9:30 0.0120
9:45 0.0132
10:15 0.0144
11:15 0.0152
1:15 PM 0.0158
5:15 0.0160
06/09/96 8:15 AM 0.0162
11:15 0.0162
Loading
increment from 0
to 500 lb/ft2)

20. Calculations
specimen parameters
20
Specimen parameters (at the beginning of test)
1. Initial wet unit weight
2
2
in.
91
.
4
4


D
Area

   3
3
cm
76
.
62
in.
83
.
3 

 o
H
Area
Volume
g
98
.
107
50
.
100
48
.
208
specimen
wet
of
Mass 


3
g/cm
721
.
1
76
.
62
98
.
107
density
wet
Initial 


Volume
Mass
   3
lb/ft
107.4
62.4
721
.
1
t
unit weigh
wet
Initial 

2. Initial moisture content
%
10
.
42
100
60
.
35
82
.
66
82
.
66
97
.
79





w

21. 21
100
100
content(%)
moisture
Initial
mass
wet
Initial
mass
dry
Initial 


g
99
.
75
100
100
1
.
42
98
.
107




water
of
Density
water
of
Mass
start
at
water
of
Volume 

w
V
3
cm
99
.
31
1
99
.
75
98
.
107



3
cm
94
.
27
1
72
.
2
99
.
75
solids
of
Vol 




w
s
s
s
G
M
V

s
v V
V
V 
 3
cm
82
.
34
94
.
27
76
.
62 



 s
v V
V
V
%
9
.
91
100
82
.
34
99
.
31
100 




v
w
V
V
S
Estimated from initial wet
mass and initial moisture
content
3. Initial degree of saturation 100


v
w
V
V
S
Calculations
specimen parameters

22. 22
Specimen parameters (at the end of test)
1. Final moisture content
%
4
.
41
100
20
.
127
11
.
203
11
.
203
54
.
234





w
2. Initial void ratio
3
cm
91
.
27
1
72
.
2
20
.
127
11
.
203
solids
of
Vol 





w
s
s
s
G
M
V

Determined at the
end of test
s
v V
V
V 

3
cm
85
.
34
91
.
27
76
.
62 


v
V
1.249
91
.
27
85
.
34
ratio
void
Initial 



s
v
o
V
V
e
Calculations
Initial void ratio

23. 23
Height of solid in specimen
  
cm
881
.
0
54
.
2
91
.
4
91
.
27
2



A
V
H s
s
Note: 2.54 cm = 1 in.
Change in void ratio, De
   046
.
0
881
.
0
54
.
2
0
0158
.
0
reading
dial
initial
100





D

D
s
s H
d
H
H
e
To convert in2 into cm2
To convert in into cm
Void ratio for 500 lb/ft2 loading
203
.
1
046
.
0
249
.
1 


D

 e
e
e o
Calculations
change in void ratio

24. 24
Time (min.)
Deformation
dial
reading
(in.)
< = log scale
0.1 1 10 100 1000 10,000
0.0040
0.0060
0.0080
0.0100
0.0120
0.0140
0.0160
0.0180
0% consolidation, d0 = 0.0058
Dd
Dd
t1
Mean of d0
and d50
50% consolidation, d50 = 0.0108
The deformation at t2
should be greater than ¼,
but less than ½ of total
deformation for the load
increment.
time-deformation
curve from data
points
t50 = 8.2 min
100% consolidation, d100 = 0.0158
t2 = 4t1
Pressure increment from 0 to 500 lb/ft2
Calculations
Time vs deformation – time for 50% consolidation

25. 25
Coefficient of consolidation, cv
50
2
196
.
0
t
H
cv 
where
t50 = time required for 50%
consolidation
H = half the thickness of test
specimen at 50% consolidation
For 500 lb/ft2 loading:
t50 = 8.2 min
 
50
2
1
d
H
H o 

where
Ho = initial height of specimen at
beginning of test
d50 = deformation dial reading at
50% consolidation
  in.
385
.
0
0108
.
0
780
.
0
2
1



   /min
in.
10
54
.
3
2
.
8
385
.
0
196
.
0 2
3
2




v
c
Calculations
Coefficient of consolidation, cv