vnk-12.ppt presentation notes of my class

Notes

1. Consolidation

2. Compressibility
•The volume of soil mass is decreases under
stress, this decrease is known as
compression, and the capacity of soil to
decrease in volume under stress is known as
compressibility.

3. The volume decrease of a soil under stress might be due to
•Compression of the solid grains;
•Compression of pore water or pore air;
•Expulsion of pore water or pore air from the voids, thus
decreasing the void ratio or porosity.
Compressibility

4. Assumptions followed in soil mechanics
 Solid grains are incompressible

 Water is incompressible
Hence the compression of saturated soil can occur only
if pore water is expelled out of voids.

5. Skeletal Material
(incompressible)
Pore water
(incompressible)
Voids
Solid
Initial State
The consolidation process

6. Skeletal Material
(incompressible)
Pore water
(incompressible)
Voids
Solid
Voids
Solid
Initial State Deformed State
The consolidation process
Water
+


7. Deformation of saturated soil occurs by reduction of pore
space & the squeezing out of pore water. The water can
only escape through the pores which for fine-grained soils
are very small
The consolidation process

8. Consolidation -Spring analogy
In this system, the spring represents the soil skeleton, and
the water which fills the container represents the pore water
in the soil.

9. 1.The container is completely filled with water, and the hole is
closed. (Fully saturated soil)
2.A load is applied onto the cover, while the hole is still
unopened. At this stage, only the water resists the applied
load. (Development of excess pore water pressure)
3.As soon as the hole is opened, water starts to drain out
through the hole and the spring shortens. (Drainage of
excess pore water pressure)
4.After some time, the drainage of water no longer occurs.
Now, the spring alone resists the applied load. (Full
dissipation of excess pore water pressure. End of
consolidation)

10. Region of high
excess water
pressure
Region of low
excess water
pressure
Flow
The consolidation process
The consolidation process is the process of the
dissipation of the excess pore pressures that occur on
load application because water cannot freely drain from
the void space.

11. During consolidation…
Due to a surcharge q applied at the GL,
GL
saturated clay
q kPa
A
the stresses and pore pressures are increased at A.

u
’
..and, they vary
with time.

12. 17
During consolidation…
 remains the same (=q) during consolidation.
GL
saturated clay
q kPa
A

u
’
u decreases (due to drainage)

u
’
q
transferring the load from water to the soil.
while ’ increases,

13. Total
Stress
Time
The consolidation process

14. Total
Stress
Time
Time
Excess
Pore
Pressure
The consolidation process

15. Effective
Stress
Time
The consolidation process

16. Effective
Stress
Time
Settlement
Time
The consolidation process

17. One Dimensional Consolidation
~ drainage and deformations are vertical(none laterally)
saturated clay
GL
q kPa
~ a simplification for solving consolidation problems
water squeezed out

18. Terzaghi’s One Dimensional
Consolidation Theory

19. Assumptions
1. The soil is completely saturated (S = 100%).
2. The soil grains and water are virtually incompressible
3. The compression is one-dimensional (u varies with z only).
4. The flow of water in the soil voids is one-dimensional, Darcy’s law being
valid.
5. Certain soil properties such as permeability and modulus of volume change
are constant; these actually vary somewhat with pressure. (k and m
v are
independent of pressure).

20. v
v
z
z
z
z




Plan
Area A
Elevation
v
z
z
Derivation of consolidation governing equation
1. Water flow (due to consolidation)

21. v
v
z
z
z
z




Plan
Area A
Elevation
v
z
z
Rate at which water
leaves the element


v
z
zA
Derivation of consolidation governing equation
1. Water flow (due to consolidation)

22. 

v
t
zA
Rate of volume decrease
Derivation of consolidation governing equation
2. Deformation of soil element (due to change in
effective stress)
Plan
Area A
Elevationz

23. Rate at which water
leaves the element
Rate of volume decrease
of soil element
=


v
z
zA


v
t
zA
Derivation of consolidation governing equation
Assume: Soil particles and water incompressible

24. Rate at which water
leaves the element
Rate of volume decrease
of soil element
=


v
z
zA


v
t
zA




v
z

v
t
(3)Storage Equation
Derivation of consolidation governing equation
Assume: Soil particles and water incompressible

25. v k
h
z
v



Assume Darcy’s law
(4)
Derivation of consolidation governing equation
3. Flow of water (due to consolidation)

26. v k
h
z
v



Assume Darcy’s law
(4)
Derivation of consolidation governing equation
3. Flow of water (due to consolidation)
Note that because only flows due to consolidation are of
interest the head is the excess head, and this is related
to the excess pore pressure by
h
u
w


(5)

27. Elastic response
 
v vem  (7)
Assume soil behaves elastically
Derivation of consolidation governing equation
4. Stress, strain relation for soil

28. Derivation of consolidation governing equation
5. Principle of effective stress
Note that these are changes in stress due to consolidation
(8)

29. 



v
z

v
t
(3)Storage Equation
v k
h
z
v



Darcy’s law (4)
Elastic response
 
v ve
m (7)
+
+
Derivation of consolidation governing equation
5. Principle of effective stress
Note that these are changes in stress due to consolidation
(8)

30. Equation of 1-D Consolidation







z
ku
z
m
u
t t
v
w
v
e
[ ] [ ]  (10)
Derivation of consolidation governing equation

31. (10)
Solution of consolidation equation
3. Homogeneous soil







z
ku
z
m
u
t t
v
w
v
e
[ ] [ ] 
(13)

32. c
v is called the coefficient of consolidation
Solution of consolidation equation

33. c
v is called the coefficient of consolidation
c
v has units L
2
/T and can be estimated from an oedometer test.
The procedure will be explained in the laboratory sessions.
Solution of consolidation equation

34. c
v is called the coefficient of consolidation
c
v has units L
2
/T and can be estimated from an oedometer test.
The procedure will be explained in the laboratory sessions.
The coefficient of volume decrease m
v
can be measured
from the oedometer test.
Solution of consolidation equation

35. c
v is called the coefficient of consolidation
c
v has units L
2
/T and can be estimated from an oedometer test.
The procedure will be explained in the laboratory sessions.
The coefficient of volume decrease m
v
can be measured
from the oedometer test.
The value of k
v
is difficult to measure directly for clays but
can be inferred from the expression for c
v.
Solution of consolidation equation

36. Uniformly distributed surcharge q
2H
Z
Homogeneous Saturated Clay Layer free
to drain at Upper and Lower Boundaries
Solution of consolidation equation for 2 way drainage

37. Governing Equation
c
u
z
u
t
v




2
2
 (14a)
Solution of consolidation equation for 2 way drainage

38. Governing Equation
Boundary Conditions
c
u
z
u
t
v




2
2

u = 0 when z = 2H for t > 0
u = 0 when z = 0 for t > 0
(14a)
(14 b,c)
Solution of consolidation equation for 2 way drainage

39. Governing Equation
Boundary Conditions
Initial Condition
c
u
z
u
t
v




2
2

u = 0 when z = 2H for t > 0
u = 0 when z = 0 for t > 0
u = q when t = 0 for 0 < z < 2H
(14a)
(14 b,c)
(14d)
Solution of consolidation equation for 2 way drainage

40. u q Z
where
and
Z
z
H
T
ct
H
n
n
n
T
v
n
v
v
 
 




2
1
1
2
0
2
2


 

sin()e
(n)
Solution
(15)
Solution of consolidation equation for 2 way drainage

41. Approximate Expressions for Average Degree
of Consolidation

42. T=0.80.5 0.30.20.1
0
1
2
0.0 0.5 1.0
Z=z/H
u/q
Variation of Excess pore pressure with depth
Solution of consolidation equation for 2 way drainage

43. Calculation of settlement
S
v
dz
H

0
2

44. Calculation of settlement
S
v
dz
H
m
ve
udz
H

 


0
2
0
2
( )

45. Calculation of settlement
S
v
dz
H
m
ve
udz
H
fromwhichitcanbeshown
S
S
UT
v
n
T
v
n
e

 










0
2
0
2
12
2
2
0
( )
()
(16c)

46. 10
-3 10
-2
10
-1
1 10
Dimensionless Time T
v
0.00
0.25
0.50
0.75
1.00
U
Relation of degree of
settlement and time

47. Approximate Expressions for Degree of Settlement
U
T
T
U e T
v
v
T
v
v
 
 

4
02
1
8
02
2
2
4



( .)
( .)
/

48. Approximate Expressions for Average Degree
of Consolidation

49. Consolidation of a clay sample with double drainage

51. Gravel
4mClay
Clay
Sand
5m
Impermeable
Clay
Final settlement=100mm
c
v
=0.4m
2
/year
Soil Profile
Final settlement=40mm
c
v
=0.5m
2
/year
Example 1: Calculation of settlement at a given time

52. For the upper layer
Now using Figure 5 with T
v = 0.1
Example 1: Calculation of settlement at a given time
T
v
c
v
t
H
 


2
0 1
2
2
01
.4
.

53. 10
-3 10
-2
10
-1
1 10
Dimensionless Time T
v
0.00
0.25
0.50
0.75
1.00
U
Relation of degree of
settlement and time

54. For the upper layer
Now using Figure 5 with T
v = 0.1
U = 0.36
so
S = 100 0.36 = 36mm
Example 1: Calculation of settlement at a given time
T
v
c
v
t
H
 


2
0 1
2
2
01
.4
.

55. For the lower layer
Now using Figure 5 with T
v = 0.02
Example 1: Calculation of settlement at a given time
T
v
c
v
t
H
 


2
051
5
2
002
.
.

56. 10
-3 10
-2
10
-1
1 10
Dimensionless Time T
v
0.00
0.25
0.50
0.75
1.00
U
Relation of degree of
settlement and time
0.020.05

57. For the lower layer
Now using Figure 5 with T
v = 0.02
U = 0.16
so
S = 40  0.6 = 6.4 mm
Example 1: Calculation of settlement at a given time
T
v
c
v
t
H
 


2
051
5
2
002
.
.

58. Example 2: Scaling
Oedometer U=0.5 after 2 minutes. 2 way drainage, H = 5 mm
Calculate time for U= 0.5 for 10 m thick layer of the same
clay, 1 way drainage

59. Example 2: Scaling
Oedometer U=0.5 after 2 minutes. 2 way drainage, H = 5 mm
Calculate time for U= 0.5 for 10 m thick layer of the same
clay, 1 way drainage
Oedometer
T
ct
H
c
c
v
v v
v
 


2 2
2
0005
80000
.

60. Example 2: Scaling
Oedometer U=0.5 after 2 minutes. 2 way drainage, H = 5 mm
Calculate time for U= 0.5 for 10 m thick layer of the same
clay, 1 way drainage
Oedometer
Soil layer

T
ct
H
c
c
v
v v
v
 


2 2
2
0005
80000
.
T
ct
H
ct ct
v
v v v
 



2 2
10 100

61. Example 2: Scaling
Oedometer U=0.5 after 2 minutes. 2 way drainage, H = 5 mm
Calculate time for U= 0.5 for 10 m thick layer of the same
clay, 1 way drainage
Oedometer
Soil layer
T
v (oedometer) = T
v (soil layer)
hence t = 80000000 mins = 15.2 years
T
ct
H
c
c
v
v v
v
 


2 2
2
0005
80000
.
T
ct
H
ct ct
v
v v v
 



2 2
10 100

62. Consolidation Test
Increment of load
Topcap
porous stone
sample
water
confining
ring
settlement
dial gauge
Oedometer/Consolidometer

63. Test procedure - IS:2720 (Part XV)—1986]:
Seating pressure -5 kN/m
2
Pressure variation
25, 50,100,200,400, 800 and 1600 kN/m
2
.
Consolidation Test
Dial gauge readings at
1/4, 1/2, 1,2,4, 8, 15, 30, 60, 120, 240, 480 and 1440
minutes

64. Presentation and Analysis of Compression Test Data

65. V
o
id

r
a
t
io

e ef1
ef2
Fig.Time-compression curve for successive increments of stress

66. Pressure-void ratio relationship for a clay
(natural scale)

67. Pressure-void ratio relationship for a clay
(Pressure to logarithmic scale)

68. H -e Relation
saturated clay
GL
q kPa
saturated clay
GL
q kPa
H
o
Time = 0
+
e = e
o
H
Time 
e = e
o - e
average vertical strain =
oH
ΔH

69. H -e Relation
Consider an element where V
s
= 1 initially.
e
1
e
o
Time = 0
+
Time = 
average vertical strain =
o
e
e


1

70. H -e Relation
Equating the two expressions for average
vertical strain,
oe
e


1


oH
H
consolidation
settlement
initial thickness of
clay layer
initial void ratio
change in void ratio

71. Coefficient of compressibility
• denoted by a
v
• is the ratio of change in void ratio to the
corresponding chang in stress
volumeoriginal
volumein change

Change void
Change stress



e
a
v
no
units
e
e
0
e
1




 

72. Coefficient of volume compressibility
• denoted by m
v
• is the volumetric strain per unit increase in stress
volumeoriginal
volumein change

)1(
)1(
o
vo
v
e
ae
e
V
V
m











73. log 
v
’
v
o
id

r
a
t
io loading

v’ increases &
e decreases
unloading

v’ decreases &
e increases (swelling)
- from the above data

74. Compression and recompression indices
log 
v
’
v
o
id

r
a
t
io
1
C
c
C
c ~ compression index
C
r ~ recompression index
(or swelling index)
1
C
r
1
C
r

75. Preconsolidation pressure
log 
v’
v
o
id

r
a
t
io

p’
preconsolidation
pressure
is the maximum
vertical effective
stress the soil
element has ever
been subjected to

76. Overconsolidation ratio (OCR)
log 
v
’
v
o
id

r
a
t
io
virgin consolidation line

p
’

vo
’
e
o
original
state
Field

vo
’
'
'
vo
p
OCR




77. •In the natural condition in the field, a soil may be either normally consolidated
or overconsolidated. A soil in the field may become overconsolidated through
several mechanisms, some of which are listed below
• Removal of overburden pressure
• Past structures
• Glaciation
• Deep pumping
• Desiccation due to drying
• Desiccation due to plant lift
• Change in soil structure due to secondary compression
• Change in pH
• Change in temperature
• Salt concentration
• Weathering
• Ion exchange
• Precipitation of cementing agents

78. Cc- Correlations

79. Cc- Correlations

83. Settlement computations
e
o
, 
vo
’, C
c
, C
r
,

p
’, m
v
-oedometer test
=q
q kPa
H
Two different ways to estimate the
consolidation settlement:
(a) using m
v
(b) using e-log 
v’ plot
settlement = m
v
 H
H
e
e
settlement
o



1
next slide

84. Settlement computations
~ computing e using e-log 
v’ plot
'
''
log
vo
vo
cCe



initial

vo
’
e
o

vo
’+ 
e
If the clay is normally consolidated,
the entire loading path is along the VCL.

85. Settlement computations
~ computing e using e-log 
v’ plot
'
''
log
vo
vo
rCe




vo
’
initial
e
o

vo’+ 
If the clay is overconsolidated, and remains so by
the end of consolidation,
e
VC
L
note the use of C
r

86. Settlement computations
~ computing e using e-log 
v’ plot
'
''
log
'
'
log
p
vo
c
vo
p
r
CCe



 


vo
’
initial
e
o

vo
’+ 
If an overconsolidated clay becomes normally
consolidated by the end of consolidation,
VC
L

p
’
e

87. C
v
and Preconsolidation pressure
calculation
•c
v is evaluated from the consolidation test data by the use
of fitting method
(a) The square root of time fitting method
(b) The logarithm of time fitting method

88. The Square Root of Time Fitting Method

89. The Log of Time Fitting Method.

90. Typical Values of Coefficient of Consolidation
•The process of applying one of the fitting methods may be repeated
for different increments of pressure using the time-compression
curves obtained in each case.
•The values of the coefficient of consolidation thus obtained will be
found to be essentially decreasing with increasing effective stress.
•The coefficient of consolidation should be evaluated in the laboratory
for the particular range of stress likely to exist in the field.
•The range of values for C
v
is rather wide 5×10
-4
mm
2
/s to 2×10
-2

mm
2
/s.
•Further, it is also found that the value of C
v decreases as the liquid
limit of the clay increases.

91. Range of Cv values

92. -

93. -

94. Determination of preconsolidation
pressure

95. SECONDARY CONSOLIDATION
•This may constitute a substantial part of total compression
in the case of organic soils, micaceous soils, loosely
deposited clays, etc.
•A possible disintegration of clay particles is also
mentioned as one of the reasons for this phenomenon.
•Secondary compression is usually assumed to be
proportional to the logarithm of time. Hence, the
secondary compression can be identified on a plot of void
ratio versus logarithm of time

96. SECONDARY CONSOLIDATION

97. SECONDARY CONSOLIDATION
•The void ratio, e
f, at the end of primary consolidation
can be found from the intersection of the backward
extension of the secondary line with a tangent drawn to
the curve of primary compression
•The equation for the rate of secondary compression
may be approximated as follows:
Here, t1 is the time required for the primary compression to
be virtually complete, t2 any later time, and is Δe is the
corresponding change in void ratio. α is a coefficient
expressing the rate of secondary compression

98. SECONDARY CONSOLIDATION
Another way of expressing the time–rate of secondary
compression is through the ‘coefficient of secondary
compression’, Cα, in terms of strain or percentage of
settlement as follows:
In other words, Cα may be taken to be the slope of the
straight line representing the secondary compression on a plot
of strain versus logarithm of time.

99. SECONDARY CONSOLIDATION
The relation between α and Cα is
Generally α and Cα increase with increasing stress.
Some common values of Cα are given below:

100. Secondary consolidation
It has been pointed out previously that clays continue to settle under sustained
loading at the end of primary consolidation, and this is due to the continued re-
adjustment of clay particles. Several investigations have been carried out for
qualitative and quantitative evaluation of secondary consolidation. The magnitude
of secondary consolidation is often defined by