1.
Settlement and Consolidation
CHAPTER 4

2.
§4 Settlement and Consolidation
§4.1 General
§4.2 Oedometer test
§4.3 Preconsolidation pressure
§4.4 Consolidation settlement
§ 4.5 Terzaghi’s theory
§ 4.6 Degree of consolidation

3.
§4.1 General
§4 Settlement and Consolidation
 compressibility
–volume changes in a soil when
subjected to pressure –giving
AMOUNTS of settlement
 consolidation
-rate of volume change with time –
giving TIME to produce an
amount of settlement required

4.
§4.2 Oedometer test
§4 Settlement and Consolidation
P1
s1
e1
e0
P
t
e s
t
P2
s2
e2
P3
s3
e3

5.
e
H
e
H


 1
1
1
0
0
)
1
( 0
0
0 e
H
s
e
e 


curve
curve
curve

6.
Compression coefficient a
p1 p2
e1
e2
M1
M2
e0
e
p
curve e-p
△p
△e
p
e
a
d
d


1
2
2
1
p
p
e
e
p
e
a





 ＝
Evaluation of compression with a1-2
 a1-2＜0.1MPa-1 , Low compressibility
 0.1MPa-1≤a1-2＜0.5MPa-1, Middle compressibility
 a1-2≥0.5MPa-1, High compressibility
1
2
2
1
＝
p
p
e
e
p
e
a
slope







7.
e - lgσ′Curve
compression index Cc
p
C
dp
de
a c
v 


1
2
2
1
1
2
2
1
lg
)
(
lg
lg
p
p
e
e
p
p
e
e
Cc






p
dp
de
p
p
e
e
C
c
c
lg
lg
2
1





8.
Preconsolidation pressure-the maximum effective vertical stress that has acted on the
clay in the past
p
s
OCR



OCR=1： lack consolidation
OCR>1： normal consolidation
OCR<1： over consolidation
How to obtain the preconsolidation pressure:
1 Produce back the straight-line part (BC) .
2 Determine the point (D) of maximum curvature on the
recompression part (AB) of the curve.
3 Draw the tangent to the curve at D and bisect the angle
between the tangent and the horizontal through D.
4 The vertical through intersection point of the bisector and
CB produced gives the approximate value of the
preconsolidation pressure.
§4.3 Preconsolidation pressure
§4 Settlement and Consolidation

9.
s
c
d S
S
S
S 


 coefficient of volume compressibility or the compression index
 Consider a layer of saturated clay of thickness H.
 an elemental layer of thickness dz at depth z.
1
2
2
1
p
p
e
e
p
e
a





 ＝
1
1
1
2
1
)
(
1
H
E
p
H
p
p
e
a
s
s





1
1
2
1
2
1
1
H
e
e
e
H
H
s





§4.4 Consolidation settlement
§4 Settlement and Consolidation

10.
Curve of gravity stress and additional
stress at the central of base
Determine the calculation depth
Determine the layer
The settlement of each layer
The whole settlement
i
i
i
i
i h
e
e
e
s
1
2
1
1







n
i
i
s
s
1
d
depth
σc
σz
z
i-1
1 2
3 4
5 6
1 2
aip0
ai-1p0
z
i
Ai
Ai-1

11.
Example 1

12.
The assumptions made in the theory are:
1 The soil is homogeneous and fully saturated.
2 There is a unique relationship, independent of time, between void ratio and
effective stress.
3 The solid particles and water are incompressible.
4 Compression and flow are one-dimensional (vertical).
5 Strains are small.
6 Darcy’s law is valid at all hydraulic gradients.
7 The coefficient of permeability and volume compressibility remain constant.
§4.5 Terzaghis theory of one-dimensional consolidation
§4 Settlement and Consolidation

13.
  2
1
2
w
k 1 e
u u
t a z

 

  
The total stress increment
soil skeleton
increasing effective stress
the excess pore water pressure decreases


a
e
k
cv
)
1
( 1


)
4
/
exp(
2
sin
1
4 2
2
1
2
, v
m
z
t
z T
m
H
m
m
u 




 


t
H
c
T v
v 2


14.
c
ct
s
s
U 
the consolidation settlement at time t being given
 by the product of U and
 the final settlement.











S
S
H
e
a
dz
e
a
dz
dz
U t
z
t
z
z
t
z
t
1
1
,
,
1
1




§4.6 Degree of consolidation
§4 Settlement and Consolidation

15.
Determine the degree of consolidation
Time factor
permeable
impermeable
degree
of
consolidation
Curve 1
Curve 2
Curve 3
P

16.
 H=10m ; e1=0.8; a=0.00025kPa-1; k=0.02m/year
？ ① settlement after one year St
② time(t) when Uz=0.75
③ if the bottom layer is permeable，time(t) when Uz=0.75
Example 2
157kPa
235kPa
H
p
clay
impermeable
permeable

17.
Solution
 1. When t=1 year
mm
H
e
a
S z 273
1 1


 
year
m
a
e
k
c
w
v /
4
.
14
)
1
( 2
1




144
.
0
2

 t
H
c
T v
v
5
.
1
157
235


a
From figure Ut=0.45
mm
S
U
S z
t 123


 2. If Uz=0.75 From Uz=0.75，a＝1.5 then Tv＝0.47
year
26
.
3
2


v
v
c
H
T
t
 3. If open layer, Uz=0.75
From Uz=0.75,a＝1,H5m then Tv＝0.49
year
85
.
0
2


v
v
c
H
T
t