Course
Year
: S0705 – Soil Mechanic
: 2008
TOPIC 4
DEFORMATION AND SETTLEMENT OF SOIL
CONTENT
•
•
•
Bina Nusantara
SETTLEMENT
CONSOLIDATION
TIME RATE OF CONSOLIDATION
SETTLEMENT
• Definition
The total vertical deformation at the surface resulting from :
– External Load
– Dewatering
• Settlement Components
– Immediate Settlement ; Se
– Primary Consolidation Settlement ; Sc
– Secondary Settlement (Creep) ; Ss
S  Se  Sc  Ss
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SETTLEMENT
• Purpose
– Study the settlement behavior
– Determine the settlement value and time
– Study the settlement influence to the structure stability
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SETTLEMENT INFLUENCE
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IMMEDIATE SETTLEMENT
• Defined as settlement which occurred directly after the application of
a load, without a change in the moisture content.
• Caused by soil elasticity behavior
• The magnitude of the contact settlement will depend on the flexibility
of the foundation and the type of material on which it is resting.
• For clay, the immediate settlement generally very small comparing
to the consolidation settlement, therefore this immediate settlement
mostly ignored.
• Usually considered at sand or sandy soil.
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IMMEDIATE SETTLEMENT
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IMMEDIATE SETTLEMENT
General Equation (Harr, 1966)
•
Flexible Foundation
– At corner
Se 
– At center
Se 



B.q o
1   s2 
Es
B.q o
Se 
1   s2  av
Es

– Average
 1  m 2  1 
1   1  m 2  m 



ln
 m. ln 
 1  m 2  1 
   1  m 2  m 



•

B.q o

1   s2
Es
2
;

m
B
L
Rigid Foundation
Se 


B.qo
1   s2 r
Es
Es = Elasticity modulus of soil
B = Foundation width
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L = Foundation Length
; H=
IMMEDIATE SETTLEMENT
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IMMEDIATE SETTLEMENT
General Equation (Bowles, 1982)
1   s2
S e  q o .B'.
.F1
Es


L'
M
B'


1
1  M2  1 M2  N2
M  M2  1 1  N2 
F1  M . ln
 ln

2
2
2
2
 
M 1 M  N 1
M  M  N  1 


N
H
B'
Es = Elasticity modulus of soil
H = Effective thickness of soil layer, e.g. 2 to 4B under foundation
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L
2
At the center
L' 
At the corner
L'  L
B' 
B
2
B'  B
and F1 is multiplied by 4
and F1 is multiplied by 1
IMMEDIATE SETTLEMENT
• Saturated Clay
S e  A1 .A 2
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qo .B
Es
IMMEDIATE SETTLEMENT
• Sandy soil


z
S e  C1.C2 q  q 
0
Iz
z
Es
where :
– Iz = strain influence factor
– C1 = correction factor of foundation embedded thickness
= 1 – 0.5.[q/(q-q)]
– C2 = correction factor for soil creep = 1 + 0.2 . log(t/0,1)
– t = time in year
– q = the stress at foundation base caused by external load
– q =  . Df
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IMMEDIATE SETTLEMENT
Modulus Young
Circular Foundation or L/B =1
z=0
 Iz = 0.1
z = z1 = 0.5 B  Iz = 0.5
z = z2 = 2B
 Iz = 0.0
Foundation with L/B ≥ 10
z=0
 Iz = 0.2
z = z1 = B  Iz = 0.5
z = z2 = 4B  Iz = 0.0
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EXAMPLE
A shallow foundation on a deposit of sandy soil that is 3m x 3m in plan. The actual variation of the
values of Young’s Modulus with depth determined by using the Standard Penetration numbers
(correlation : Es = 766N) are also shown in the following figure.
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Estimate the immediate
settlement of the foundation
five years after construction
by using the strain influence
factor method.
EXAMPLE
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EXAMPLE
Depth
(m)
z
(m)
Es
(kN/m2)
Iz
(average)
(m3/kN)
0.0 – 1.0
1.0
8000
0.233
0.291 x 10-4
1.0 – 1.5
0.5
10000
0.433
0.217 x 10-4
1.5 – 4.0
2.5
10000
0.361
0.903 x 10-4
4.0 – 6.0
2.0
16000
0.111
0.139 x 10-4
Iz
z
Es

 t 
 5 
C2  1  0.2. log 
  1  0.2. log 
  1.34
 0.1 
 0.1 
 q 
 17.8 x1.5

  1  0.5
  0.9
C1  1  0.5
 160  17.8 x1.5 
qq


2B
S e  C1.C2 . q  q 
0
Iz
.z
Es
S e  (0.9)(1.34)(160  17.8 x1.5)(1.55 x10 4 )
S e  24.8 mm
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1.55 x 10-4
CONSOLIDATION
• When the stress on a saturated clay layer in the field is increased,
the pore water pressure in the clay will increase.
Because the coefficients of permeability of clays are very low, it
will take some time for the excess pore water pressure to
dissipate and the stress increase to be transferred to the soil
skeleton gradually.
• Consolidation is the process of dissipation of excess pore water
pressure in a row of time.
Note:
Dissipation of pore water pressure occurs simultaneously with the squeezing out
of the pore water. Therefore the consolidation time depend on:
The distance of pore water to be squeezed out
The coefficient of permeability of soft soil
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CONSOLIDATION
a
o
Valve
(soil’s perm eab ility)
a a
o
Ho
Spring
i
Si
(soil partic les)
(Ho - Si)
Water filled c ha m ber
(water saturated soil’s p ores)
IDEALISASI
Pressure is b orne
by pore water
UNDRAINED
SETTLEMENT CURVE
La tera l d eform ation
IMMEDIATE
SETTLEMENT (Si)
SETTLEMENT
PRIMARY OR
CONSOLIDATION
SETTLEMENT (Sc )
Water is expelled
a
i
a
c
SECONDARY
SETTLEMENT (Ss)
LOG TIME
HYDROSTATIC
PRESSURE
(Ho - Si - Sc )
Sc
Spring c om pressed
Water p ressure reduc ed
CONSOLIDATION
No water flow
a a
c
s
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All loa d is borne by spring
Hyd rostatic pressure
(zero exc ess p ore water pressure)
DRAINED CREEP
LOAD
(Ho - Si - Sc - Ss)
Ss
CONSOLIDATION
First time, suggested by Terzaghi (1920-1924) with several
assumption :
–
–
–
–
–
–
–
–
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1 dimensional
Saturation is complete
Compressibility of water is negligible
Compressibility of soil grains is negligible (but soil grains rearrange)
Darcy’s Law is valid
Soil deformation is small
Soil permeability is constant
Soil skeleton of each layer is homogeneous, so isotropic linier elastic
constitutive law is valid
CONSOLIDATION
• Consolidation Type
– Normal consolidation
Preconsolidation pressure (Pc) just equals the
existing effective vertical overburden pressure (Po)
– Over consolidation
If the soil whose preconsolidation pressure (Pc) is
greater than the existing overburden pressure
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CONSOLIDATION
• Normal Consolidation
pc  po
OR
pc
1
po
Sc 
p  p
Cc
.Hc . log o
1  eo
po
• Over consolidation
pc  po
OR
po + p < pc
po < pc < po+p
Bina Nusantara
pc
1
po
Sc 
p  p
Cs
.Hc . log o
1  eo
po
Sc 
p
p  p
Cs
Cc
.H c . log c 
.H c . log o
1  eo
po 1  eo
pc
CONSOLIDATION
Where :
–
–
–
–
–
–
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eo = initial void ratio which getting from index test
Cc = compression index from consolidation test
Cs = swelling index from consolidation test
pc = preconsolidation pressure from consolidation test
po =  ’.z
p = the total stress at any depth of the clay layer caused by
external load, which can be determined by using method of
Boussinesq, Westergaard or Newmark
DETERMINATION OF CONSOLIDATION PROPERTIES
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DETERMINATION OF CONSOLIDATION PROPERTIES
Procedures :
1. Determine the point O on the elop p curve that has the sharpest
curvature (that is, the smallest
radius of curvature)
2. Draw a horizontal line OA
3. Draw a line OB that is tangent to
the e-log p curve at O
4. Draw a line OC that bisects the
angle AOB
5. Produce the straight line portion
of the e-log p curve backward to
intersect OC. This is point D. The
pressure that corresponds to the
point p is the preconsolidation
pressure, pc.
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DETERMINATION OF CONSOLIDATION PROPERTIES
e1  e 2
Cc 
 p2 
log 
p 

 1
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DETERMINATION OF CONSOLIDATION PROPERTIES
Cs 
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e 3  e4
 p4 
log 
p 

 3
DETERMINATION OF COMPRESSIVE PARAMETER
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CONSOLIDATION SETTLEMENT
e1  e 2
av  '
p 2  p1'
• Other equation
S c  m v .H c .p
av
mv 
1  eo
Where :
mv = Compression Index
Hc = Thickness of soft soil layer
p = The stress increment due to the external
load
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CONSOLIDATION TIME
Tv .H 2
t
Cv
Where :
t = consolidation time
Tv = consolidation factor depended on consolidation degree (U)
2
U = 0 – 60%
  U% 
Tv  

4  100 
U > 60%
Tv  1,781  0,933. log 100  U%
U = consolidation degree in percent, descript as ratio of
design settlement to total settlement
U
S c ,i
Sc
x100%
Cv = coefficient of consolidation, get from consolidation test
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CONSOLIDATION TIME
Tv .H 2
t
Cv
Where :
H = length of water path
Porous Layer
Porous Layer
Hc
Hc
Impermeable
layer
Porous Layer
H = Hc
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H = 0.5Hc
CONSOLIDATION TIME OF LAYERED SOIL
Hc,1
Cv1
Tv .H c ,1 2
2
t1 
= 5.2 years
Cv 1
Tv .H c , 2 2
2
Hc,2
Hc,3
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Cv2
Cv3
t2 
Cv 2
Tv .H c , 3 
= 3.4 years
2
t3 
Cv 3
Take the longest
time
= 6.1 years
t = 6.1 tahun
CONSOLIDATION TIME OF LAYERED SOIL
-Determine the equivalent of Hc of each layer
Hc,1
Cv1
H c' ,i
Cvi
Cvref
 H c ,i
-Determine the sump of equivalent Hc
Hc,2
Cv2
-Determine the equivalent of Cv
Cvek  Cvref
Hc,3
 H 
.

' 2
c
Hc 2

Cv3
-Determine the consolidation time
t
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
Tv .  H c
Cv ek

2
EXAMPLE
Determine the total consolidation time of 3 layer
of clay, which have different value of coefficient
of consolidation and thickness for 90% degree of
consolidation.
1st Layer : thickness 5 m, Cv = 5 x 10-3 cm2/s
2nd Layer : thickness 3 m, Cv = 6 x 10-3 cm2/s
3rd Layer : thickness 8 m, Cv = 7 x 10-3 cm2/s
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SOLUTION
Layer
Thickness
(Hc)
Cv
(m2/s)
Equivalent Thickness
(Hc’)
1
5m
5 x 10-7
5.00 m
2
3m
6 x 10-7
3.29 m
8m
10-7
3
Total
Cvref is Cv1
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7x
9.47 m
17.76 m
Cvek
(m2/s)
T
(years)
6.16 x 10-7
11.18
EXAMPLE
A laboratory consolidation test on a normally clay showed the following result :
Load, p (kN/m2)
140
212
Void ratio at the end of consolidation, e
0.92
0.86
The specimen thickness was 25.4 mm and drained on both sides. The time
required for the specimen to reach 50% consolidation was 4.5 min.
A similar clay layer in the field, 2.8 m thick and drained on both sides, is subjected
to similar average pressure increase that is po = 140 kN/m2 and po+p = 212
kN/m2. Determine the following :
1. The expected maximum consolidation settlement in the field
2. The length of time it will take for the total settlement in the field to reach 40 mm
3. Repeated no.2 problem in case of drained on one side
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EXAMPLE
• Question no.1
e1  e2
Cc 
 p2
log 
 p
 1
Sc 
Sc 
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



Cc 
0.92  0.86
 0.333
 212 
log 

 140 
p  p
Cc
.Hc . log o
1  eo
po
0.333
212
. 2.8 . log
 87.5mm
1  0,92
140
EXAMPLE
• Question no.2
– Determine the coefficient of consolidation (Cv)
From laboratory testing
H2
Cv  Tv
t
where :
Tv = /4 (U2) = 0.197 (U = 50%)
H = Hc/2 = 12.7 mm
t
= 4.5 min
We got
12.7 2
Cv  0.197
 7.061
4.5
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mm2/min
EXAMPLE
• Question no.2
– Determine field consolidation coefficient
U
S c ,i
Sc
x100% 
40
x100%  45,7%
87,5
– Calculate consolidation time
Tv .H 2
t
Cv
Where :
U = 45.7%
Tv = /4 (U2) = 0.164 (U = 45.7%)
H = Hc/2 = 1.4 m = 1400 mm
Cv = 7.061 mm2/min
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We got
0.164 x1400 2 = 45523 min = 31.6 days
t
7.061
EXAMPLE
• Question no.3
– Calculate consolidation time
Tv .H 2
t
Cv
Dimana :
U = 45.7%
Tv = /4 (U2) = 0.164 (U = 45.7%)
H = Hc = 2.8 m = 2800 mm
Cv = 7.061 mm2/menit
Diperoleh
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0.164 x 28002 = 182093 min = 126.5 days
t
7.061
THE INFLUENCE OF PORE WATER PRESSURE
Two influences of pore water pressure to the settlement
are :
– Initial average overburden pressure (po)  should be in effective
condition (po’)
– External Load
the uplift of water pressure will reduce the increase of vertical
pressure by external load
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SECONDARY CONSOLIDATION (CREEP)
• Occur after primary consolidation process finished
• Defined as an adjustment of soil skeleton after the
excess pore water dissipated.
• Depend on time and will be occurred in a long time
• Difficult to be evaluated
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SECONDARY CONSOLIDATION (CREEP)
t p  t
C
Ss 
.H c . log
1  ep
tp
Where :
C 
e
t
log 2
tp
See the graph
ep = void ratio at the end of primary consolidation
tp = time at the end of primary consolidation
t = time increment
t2 = tp +t
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SECONDARY CONSOLIDATION (CREEP)
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EXAMPLE
A laboratory testing of consolidation for specimen thickness 25.4 mm is carried out to
determine the secondary settlement, with the result as shown in the following table :
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EXAMPLE
Assume the thickness of the compressible layer is 10 m and the
consolidation settlement is 30 cm which occurs after 25 years. The
initial void ratio eo is 2.855, and the initial dial reading is 12.700
mm
Required :
Compute the amount of secondary compression that would occur
from 25 to 50 years after construction. Assume the time rate of
deformation for the load range in the test approximates that occur
in the field.
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EXAMPLE
C = 0.052
ep = 2,372
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EXAMPLE
t p  t
C
Ss 
.H c . log
1  ep
tp
Ss 
0.052
 50 
.10. log  
1  2.372
 25 
Ss = 4.6 cm
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