GDII Lecture 1 [Compatibility Mode].pdf
-
Upload
lee-tin-yan -
Category
Documents
-
view
220 -
download
0
Transcript of GDII Lecture 1 [Compatibility Mode].pdf
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
1/132
BEng in Civil EngineeringSubject (CSE410)
Geotechnical Design II
岩土工程設計II
Jian-Hua YIN 殷建華Office: ZS909, Tel: 2766-6065Email: [email protected]
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
2/132
Outline of Lectures by JH YIN:Lecture 1: Consolidation of soils (1-D, 2-D and 3-D)
Lecture 2: Pile foundation
Lecture 3: Soil nailed slopeLecture 4: Excavation and soil reinforcement
Lecture 5: Ground modification
Essential References:(1) Lecture notes.
(2) Das, Braja M. (2007). Principles of Foundation Engineering (6
th
edition),Thomson, United States (ISBN 0-534-40752-8) (or more updated version).
(3) Craig, R.F. (2004). Soil Mechanics, 7th edition (6thor 5th edition), Spon
Press, London and New York (ISBN 04-415-32702-2) (or more updated
version).
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
3/132
Review of Soil Mechanics
(a) Terzaghi’s theory of one-dimensional consolidation
A total of 8 assumptions are made in the theory:
1 The soil is homogenous (ok for each layer )
2 The soil is fully saturated (ok for soil underwater )
3 The solid particles and water are incompressible (ok )
4 Compression and flow are one-dimensional (vertical) (ok )
5 Strains are small (ok for most civil engineering problems)6 Darcy’s law is valid at all hydraulic gradients (ok for water and
common civil engineering problems)
7 The permeability k and volume compressibility mv
are constants
through the process (approximate for a layer and small pressure)8 There is a unique relationship, independent of time, between void
ratio and effective stress (approximate for a layer and silty soils, no
good for soft clay)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
4/132
z v
dz z
vvdvv z z z z
z
uk
z
uu z
k
z
hk kiv e
w
w
e s
z z
)(
)headtotalstatic(constant sw
s hu
z
dz z
uk
z
uk dz
z
vvdvv e
w
e
w
z z z z 2
2
The condition of continuity: net water
coming out rate = volume compression rate
t
V dxdydz
z
uk vdxdydvvdxdy e
w
in z out z z
2
2
wv
ve
vee
w
ev
ev
e
w
e
w
ev
e sv
vvvv
m
k c
z
uc
t
u
z
uk
t
um
dxdydz t
umdxdydz
z
uk
t
V dxdydz
z
uk
dxdydz
t
umdxdydz
t
uum
dxdydz t
mdxdydz t
mdxdydz
t t
dxdydz
t
V
;
)]([
)(
2
2
2
2
2
2
2
2
''
cv
is the coefficient
of consolidation
dxdydz V V V
dxdydz V
vv
,
dxdydz V v dxdydz V
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
5/132
2
2
z
u
ct
u ev
e
wvv m
k c
The initial value of excess pore water pressure (initial condition):
The boundary condition of excess pore water pressure:
To obtain an analytical solution using the method of “separation
of variables”:
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
6/132
for ui=constant
)12(
2
m M
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
7/132
for ui=constantTwo sides are
free drainage
One side is impermeable
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
8/132
i
e z
u
uU 1
i
d
ed
i
d
e
H
i
H
ei
H
v
H
ev
H
v
H
ve
f
t
u
dz u
du
dz u
dz u
dz uu
dz m
dz m
dz
dz
s
sU
2
0
21
2
0
0
0
0
'0
'1
0
'0
'
0
1
0 12
1
)(
)(
)(
)exp(2
1 2
0
2 v
m
m
T M M
U
Good approximate solution for the average degree of consolidation U :
for ui=constant:
d
i
d
e
H
i
H
ei
f
t
dz u
dz u
dz u
dz uu
s
sU 2
0
2
0
0
0
1
)(
If ui is not constant:
f t sU s )12(2
m M
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
9/132
848.0%90
196.0%50
v
v
T U
T U
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
10/132
Two sides are free drainage
The bottom side is impermeable
d is the maximum
water drainage
distance
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
11/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
12/132
t B:t A=4:1
d
d
t 50
固結
次固結
蠕變
(creep)
lab EOP o
creept
t C
h
h
,
log
lab EOP t ,
tcoefficiensecondaryis
1log1
1
log 00 e
C
t
e
et
C e
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
13/132
X
1.15X
t 90
固結
次固結
蠕變(creep)
)(1
9
10;
0
0
90
0
100
0
0
0
p s
f
s p
f
s p
f
s
r r r
aa
aar
aa
aar
aa
aa
r
EOP t
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
14/132
Why a clayey soil creeps?Creep is continuing deformation under constant loadCreep occurs under effective stress action
Creep is due to –viscous adsorbed water (double layers) on clay
plates
–viscous re-arrangement and sliding of clay plates –viscous deformation of clay plates (small)
–viscous deformation of clay skeleton (structure/frame
like)
Adsorbed water is NOT free water which
cannot flow freely under gravity.
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
15/132
Under
particle
contact
force
(effective
stress)
Creep movement !
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
16/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
17/132
In-situ value of cv
and c h
Use large samples c h
is horizontal coefficient of consolidationvh
cc )4~2(
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
18/132
Solution:
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
19/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
20/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
21/132
(c) Correction for construction period
• Previous solution/chart is suddenly applied load• Common real load is ramp loading (construction loading
likes this) – linear increase and then constant
• How to find a solution to this or make a correction?
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
22/132
f cc
t
ct corr t
c
f t
t t corr t
c
st
t U t
t swhere
t t st s
t t For
st
U t
swhere
P
P t st s
t t For
)2
()2
(
)
2
()(
:
)2
()2
(
)2
()(
0
,
'
'
,
't P
t
t 2
t
corr t s ,
' P
corr t s ,
t
U from previous
solution/chart
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
23/132
Solution:
u
u
'
'
0' u
2
'
/43.29)81.93(
mkN u
kPa72.1443.2921'
kPa72.14
'
mm H m s vcf 110872.1494.0'
Suddenly applied loading case:
Layer is open, thus, d =4m, for t =5 years
73.0437.04
54.122
U d
t cT vv
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
24/132
Dam
Sand
This sand layer is an “artesian” layer with higher
water pressure above ground surface
This clay layer is a
“aquitard” layer
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
25/132
Suddenly applied loading:
%2626.0)2
42.1
()2(
0621.04
4.1)
2(
6.2811026.0
)2
42.1()
2
42.1(
31.2071.06.28
7.14
7.14
)2
42.1
()42.1(
242.10
2242.1
22
242.1
,
U
t
U
d
ct T
mmmm
sU s
mm
s s
yearst t For
t v
v
f t
t corr t
c
42.1t 71.02
42.1
2
t
6.28
)2
(
t s
31.20
)(,
t s corr t
kPa
kPat P
yearst t For
t
c
7.14'
44.107.14)(
242.10
242.1'
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
26/132
Solution:
C c=0.32
(a) One layer 3'3
,' /2.108.920;/2.98.919 mkN mkN clayw sand sat sand
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
27/132
3m
2'1
2'0 /8.179)203(8.119;/8.119)32.10()62.9()217( mkN mkN
C c=0.32
)/log();/log(
'0
'1
10'0
'110
eeC C ee cc
00 11 eee
ee o
v
mm H
e
C H
e
ee H s cov f 1826000
8.119
8.179log
855.01
32.0log
11'
0
'1
00
855.0)100
8.119(32.088.0)
100log(32.088.0 0
'
og ee
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
28/132
f cc
t c
t corr t c
f t t
t corr t c
st
t U t
t swheret
t st st t For
st
U t
swhere P
P t st st t For
)2()2()2()(:
)2
()2
()2
()(:0
,
'
'
,
Good
equations
to use !
335.0875.06
5.226.1)
2(
;5.22
13
2 22
U
d
t t c
T yearst
t
cv
vc
mm st t U t t st s f cc
t corr t 61182335.0)2
()2
()(,
(b) Two layers
Method 1: Settlement is proportional to the thickness H For Layer H 1of 4.5m:
For Layer H 2 of 1.5m:
mm s f 5.13618265.4
1
mm s f 5.4518265.1
2
mm sU sU d
t t c
T f c
cv
v 6.1125.136825.0825.0622.025.2
5.226.1)
2(
1111221
mm sU sU d
t t c
T f c
cv
v 1.445.4597.097.040.15.1
5.226.1)
2(
2222222
mm s s s ccc 7.1561.445.11221
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
29/132
mm sU sU d
t t c
T f c
cv
v 6.1188.143825.0825.0622.025.2
5.226.1)
2(
1111221
mm sU sU d
t t c
T f c
cv
v 8.380.4097.097.040.15.1
5.226.1)
2(
2222222
)7.156(4.1578.386.11821 mmmm s s s ccc
2'
1
2'
0
/15.172)203(15.112;/15.112)25.22.10()62.9()217( mkN mkN
mm H e
C H
e
ee H s cov f 8.1434500
15.112
15.172log
864.01
32.0log
111'
0
'
1
0
1
0
11
864.0)100
15.112log(32.088.0)
100log(32.088.0 0
'
ee
Method 2: Settlements for H 1 of 4.5m and H 2 of 1.5m:
2'
1
2'
0 /75.202)203(75.142;/75.142)25.52.10()62.9()217( mkN mkN
mm H e
C H eee H s cov f 0.401500
75.14275.202log
831.0132.0log
112'
0
'
1
0
2
0
12
831.0)100
75.142log(32.088.0)
100log(32.088.0 0
'
ee
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
30/132
vU
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
31/132
(d) Vertical drains (垂直排水井)
• 1-D vertical consolidation of soft soils takes a long time:
“primary consolidation” takes 3 years to 100 years and
creep settlements takes even longer time.• This cases problems of continuing excessive
settlement/deformation of ground or super-structures
• How to solve this problem?
• Use vertical drains plus pre-loading (預加荷載) –
commonly used in Hong Kong
• Why?
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
32/132
Pre-loading –
removed afterward
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
33/132
Prefabricated band (vertical) drains (排水板)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
34/132
Drainage channels
Geo-textile
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
35/132
Purpose of Vertical Drains
and Preloading
(a) Vertical Drains 排水板
• Speed up escaping/dissipation of excess porewater/ pressure due to preloading
(b) Preloading預壓• Making the soil over-consolidated and reducing both
post-construction “primary” and creep consolidation
More accurately speaking, reducing instant compressionand creep
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
36/132
Bedrock or soil
Marine Deposits
Water Table
Pre-loading fill
Sand fill
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
37/132
Lo g(Pressure) (or Stress)
Settlement/strain - large
Set t lem ent/st ra in - sm all
Large creep set tlem ent/s tra in
Sm all creep set tlem ent
Soil pressure/stress
befo re co nstruc tio n
Soil pressure/stress
after construction
Pre- loading to here
Illustration of settlement/strain reduction using pre-loading technique:
reduction in post-construction “primary” consolidation (instant)
settlement/strain and reduction in “secondary” consolidation (creep)
settlement/strain
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
38/132
2
2
2
2
2
2
)1
( z
uc
r
u
r r
uc
t
u ev
eeh
e
)1)(1(1)1)(1()1( r vr v U U U U U U
22 4);(;);(
R
t cT T f U
d
t cT T f U hr r r
vvvv
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
39/132
)1)(1(1)1)(1()1( r vr v U U U U U U
d
hr r r
vvvv
r
Rn
R
t cT T f U
d
t cT T f U ;
4);(;);(
22
2R=D is different from other books
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
40/132
vU
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
41/132
Solution:mm H m s v f 162106525.0
' mm s s s remain f montht 137251626
%8585.0162
1375.0
f
year t
s
sU
82.0)(;7.24
)2.0(4
5.09.7
4
?)(2.0;2.02/4.0:
222
r r r h
r
d d
T f U T ntrial nn R
t cT
nnnr Rmmr drainSand
82.00235.01
85.011
1
1117.00235.0
10
5.07.422
v
r vv
vU
U U U
d
t cT
v
r r v
U
U U U U U
1
11)1)(1()1(
?)(20;202/40: nnnrRmmrdrainSand
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
42/132
82.0)(;7.24
)2.0(4
5.07.4
4
?)(2.0;2.02/4.0:
222
r r r
h
r
d d
T f U T ntrial nn R
t cT
nnnr Rmmr drainSand
n Tr
Ur
5 0.988 0.98
7.5 0.439 0.94
10 0.247 0.74
20 0.0618
0.26
0
0.1
0.20.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 2 4 6 8 10 12 14 16 18 20 22
n
U r
)(
2.3564.0
8.1
564.0
8.192.0
;9
patten square
m R
S
mnr R
n
d
How to make correction for construction period?
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
43/132
f cc
t
ct corr t
c
f t
t t corr t
c
st
t U t
t swhere
t t st s
t t For
st
U t
swhere
P
P t st s
t t For
)2
()2
(
)2
()(
:
)
2
()
2
(
)2
()(
0
,
'
'
,
't P
t
t 2
t
corr t s ,
' P
corr t s ,
t
U from previoussolution/chart
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
44/132
有蠕變性質的土的固結沉降計算的簡化計算方法
A simplified method for calculation of
consolidation settlement of clayey soils with
creep
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
45/132
Settlement due to creep (not use of “secondary” consolidation)
C e
C e 01
lab EOP o
creept
t C
h
h
,
log
t coefficienionconsolidat " "secondarycalled soisC
e
C
t
e
et C
e
e
00 1log1
1
log
)log(time
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
46/132
For Hong KongMarine Clays:
C e=0.3% to 1%
C e
Wh l il ?
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
47/132
Why a clayey soil creeps?Creep is continuing deformation under constant loadCreep occurs under effective stress action (discussed
before)
Creep is due to –viscous adsorbed water (double layers) on clay
plates
–viscous re-arrangement and sliding of clay plates
–viscous deformation of clay plates (small)
–viscous deformation of clay skeleton (structure/frame
like)
Adsorbed water is NOT free water which
cannot flow freely under gravity.
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
48/132
• Creep always exists under the action of
effective stresses (loading), independent ofthe excess pore water (or pore pressure).
• Therefore, creep has nothing to do with
the“primary”consolidation.
• And creep exists during and after “primary”
consolidation.• Creep rate depends on stress/strain state:
–Creep rate is large in a normally
consolidated state.
–Creep rate is small in an over-consolidated
state.
Bjerrum’s time line model apparent “pre consolidation
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
49/132
Bjerrum s time line model, apparent pre-consolidation
pressure”, ageing and “delayed compression” (Bjerrum 1967)
A
BC
Path A 路径1:A>B
Path B 路径2:A>C>B
The creep rate at B is dependent on the
stress-strain (or void ratio) state, not on theloading history (or path)B点的蠕变率只与应力-应变状态点有关,与怎样到达这点的历史或路径无关
Vertical effective stress (t/m2)
V o i d r a t i o
( e
)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
50/132
A
B
C
Path 1:A>B
Path 2: A>C>B
The creep rate at Point B is dependent on thestress-strain (or void ratio) state, not on theloading path.
Vertical effective stress (t/m2)
V o i d r a t i o
( e
)
Th t th d f 1D ttl t l l ti f
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
51/132
field EOP field EOP o
e
f v
field EOP f v
secondary primarytotalA
t t for H t
t
e
C S U
t t for S U S S S
,,
,
""""
)log(1
Hypothesis A:This method is not correct since the creep
settlement in the “primary” consolidation is not included. This
method underestimates settlement (低估了沉降).
There are two methods for 1D settlement calculation of
soil with creep: Hypothesis A and Hypothesis B
Equation of Hypothesis A method:
H th i B Thi th d i t i th ttl t
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
52/132
Hypothesis B:This method is correct since the creep settlement
in the “primary” consolidation is included. This method givescorrect settlement.
Equations of Hypothesis B method:
strainvertical pressure porewater excessu
water of weight unit ty permeabilik t z
uk
conditioncontiniuty From
z e
w
z e
w
;
;;
)1(
:
2
2
A constitutive model (stress-strain relation) is needed for t
z
Yin and Graham (1989 1994) developed a one dimensional
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
53/132
Yin and Graham (1989, 1994) developed a one-dimensional
elastic visco-plastic (1D EVP) model:
/
'')
'()(exp
'
zo
z ep
zo z
o z
z z
V
Vt V
totalB z z e
zo
e s z ep
zo z
o
e s z
e s z
z
s
e s z z e s z z z
S get u for and Solve
uuV
Vt
t uu
uuV t
pressure porewater staticknownu
uuuuuuknownis
,,,)2()1(
)2())
()(exp
)()(
1
;;;
'
/
'
''
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
54/132
YIN J-H. and GRAHAM J.(1989). Viscous elastic plastic
modelling of one-dimensional time dependent behaviour of clays.
Canadian Geotechnical Journal, 1989,26:,199 - 209.
YIN J H. and GRAHAM J. (1994). Equivalent times and elasticvisco-plastic modelling of time-dependent stress-strain
behaviour of clays. Canadian Geotechnical Journal, 1994,31: 42
- 52.
YIN J H. and GRAHAM J. (1996). Elastic visco-plastic modelling
of one-dimensional consolidation. Geotechnique, 1996, 46(3):
515 - 527.
Maxwell rheology model:
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
55/132
Maxwell rheology model:
modulus sYoung E
E E
Spring
z e
z z e
z
'
;
:''
t coefficienviscous
dashpot Viscous
z v
z
v
z z
'' ;
:
model plasticviscoelasticanmodel veconstituti A
E t
E
addition strainsconnectionSeries
z z z
z z v
z
e
z z
:
:)(
''
''
Yin and Graham’s (1989, 1994) 1-D Elastic
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
56/132
/
'
'
'
'
)()(exp
zo
z ep
zo z
o z
z z
V
Vt V
Yin and Graham s (1989, 1994) 1 D Elastic
Visco-Plastic (1-D EVP) model:
Linear elastic spring (non-linear for 1-D EVP model)
Linear visco dash-pot (non-linear for 1-D EVP model)
z '
z ' Maxwell’s Rheoloical Model:
''
z z z
E
Non-linear elastic strain rate; non-linear visco-plastic strain rate
Linear elastic strain rate; linear visco-plastic strain rate
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
57/132
Time (min)
ototalB H S /
)(kPaue
ie uu
o H
Hypothesis B (Simplified by Yin – 殷簡化計算方法):
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
58/132
ionconsolidat "primary" of end at settlement final S
ionconsolidat of degreeaverageU
t any for S S U
S S S
f
v
creep f v
creep primarytotalB
0)(
""
Hypothesis B (Simplified by Yin 殷簡化計算方法):
For details of this simplified method, please see:
殷建华(2011). 从本构模型研究到试验和光纤监测技术研发。岩 土 工 程学报,
第33卷 第1期, 1~16。
Yin, Jian-Hua (2011). From constitutive modeling to development of laboratory testingand optical fiber sensor monitoring technologies. Chinese Journal of Geotechnical
Engineering. Vol.33 No.1, 1~16.
: Point to1 Point 4
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
59/132
'
z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'
3 z z
),( 4'
4 z z ),( 5'
5 z z
1 2
3
45
Normal
consolidation
Unload/ reload
Over-consolidation '2
tp
z
6 ),( 6'
6 z z
tp
z
tp
z 5
z
or
Figure 1. Relation and states of log(stress)-
void ratio (or strain) from 1D straining
H σ
σ
log e1
C
H σ
σ
log e1
C
s ' zp
'
z4
o
c
' z1
'
zp
o
e
f
H
e
C s
: Point to1 Point
z
z
o
e f '
1
'
2log
1
2
H
e
C s
: Point to point dation Preconsoli
zp
z
o
c f
zp zp
'
'
4
'
log
1
4),(
0log1
)(56
0log1
)(64
'
6
'
5e
'
4
'
6e
H e
C :sreloading Point to Point
H e
C s
:unloading Point to Point
z
z
o
f
z
z
o
f
C c and C e can be determined from (a) compression with time 24 hours (t 24) of
duration or (b) compression at the end of primary consolidation in lab (t EOP,lab)
which is about a few minutes.
Settlement due to creep (not use of “secondary” consolidation)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
60/132
Settlement due to creep (not use of secondary consolidation)
C e
C e 01
lab EOP o
creept
t C
h
h
,
log
t coefficienionconsolidat " "secondarycalled soisC
e
C
t
e
et C
e
e
00 1log1
1
log
)lg(t
In the simplified Hypothesis B method, the “equivalent time” te
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
61/132
In the simplified Hypothesis B method, the equivalent time t e
is used for creep:
o
e
et
t t C ee
00 log
0
o
e
o
e
o
ee
o
eetp
z
e1V
parameter creepat
timeequivalent t
t
t t C t
t t
V
C
t
t t
e
C
e
ee
000
00
0
logloglog11
If C c and C e can be determined from (a) compression with time 24 hours (t 24) of
duration: t o=24 hours=1day.
If C c and C e can be determined from (b) compression at the end of primary
consolidation in lab (t EOP,lab) which is about a few minutes: t o= t EOP,lab
(a) Final point is at a normally consolidated state :
For example Point 4 (Figure 1)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
62/132
For example Point 4 (Figure 1)
0)log(
1 0
""
e
o
eoe f
v
creep primarytotalB
t for H
t
t t
e
C sU
s s s
oe t t t
:usewhyisThis4.Point back toisaboveThe
1day)t(log1
log1
)1
11log(
1log
1log
1
:coupling pressure porewater noIf
'
'
4
'
1
'
0
'
'
4
'
1
'
""
H e
C H
e
C
H t
e
C H
e
C H
e
C
s s s
zp
z
o
c
z
zp
o
e
e
zp
z
o
c
z
zp
o
e
creep primarytotalB
oe t t t
'
z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'
3 z z
),( 4'
4 z z ),( 5'
5 z z
12
3
45
Normal
consolidation
Unload/ reload
Over-consolidation'2
tp
z
6 ),( 6'
6 z z
tp
z
tp
z 5
z
or
(b) Final point is at an over-consolidated state :
For example Point 2 (Figure 1)'L'2
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
63/132
p ( g )
0)log(1 20
""
e
eo
eoe f
v
creep primarytotalB
t for H t t
t t
e
C sU
s s s
o
ee
zp
z c zp z
t
t t
V
C
V
C 0
'
'
loglog
:)1994,1989GrahamandYin(
"timeequivalent"thetoAccording
e
c
e zp z
eC
cC
zp
z
e zp z C
C
zp
z C
V
C
V
o
e
zp
z
e
c
e
zp z
o
e
t t t
C
C
C
V
t
t t
)(1010
log)(log:abovetheFrom
'
')()log()(
0
'
'
0
'
'
o
C
C
zp
z C
V
oe t t t e
c
e zp z
)(10
'
')(
z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'
3 z z
),( 4'
4 z z ),( 5'
5 z z
1 2
3
45
Normal
consolidation
Unload/ reload
Over-consolidation'2
tp
z
6 ),( 6'
6 z z
tp
z
tp z 5
z
or
C
C
zC
V
ttt ec
e zp z
)(10'
2)( 2
'
z Log
12
Over-consolidation'2
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
64/132
o
C
zp
z oe t t t
ee
)(10'
22
o
C
C
zp
z C
V
ooee
e
t t t t t t t
t
e
c
e zp z
2)(10
:)1b(in
'
'
2)(
2
2
0)log(1 2
2
0
""
e
eo
ee f
v
creep primarytotalB
t for H t t
t t
e
C sU
s s s
)1dayt(log1
)log(1
log1
:coupling pressure porewater noIf
'
1
'
2
2
2
0
'
1
'
2
""
H e
C H
t t
t t
e
C H
e
C
s s s
z
z
o
e
eo
ee
z
z
o
e
creep primarytotalB
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'
3 z z
),( 4'
4 z z ),(
5
'
5 z z
3
45
Normal
consolidation
Unload/ reload
tp z
6 ),( 6'
6 z z
tp
z
tp
z 5
z
or
(c) Final point is at an over-consolidated state from unloading :
For example Point 4 to Point 6 (Figure 1)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
65/132
0)log(1 60
""
e
eo
eoe f
v
creep primarytotalB
t for H t t
t t
e
C sU
s s s
0log1 '
4
'
6
H e
C s
z
z
o
e f
o
C
C
zp
z C
V
oe t t t e
c
e zp z
)(10
'
' 6)(
6
6
o
C
C
zp
z C
V
ooee t t t t t t t e
c
e zp z
2)(10'
'
6)(
6
6
)1c(0)log(
1 6
6
0
""
e
eo
ee f
v
creep primarytotalB
t for H
t t
t t
e
C sU
s s s
'
z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'
3 z z
),( 4'
4 z z ),( 5'
5 z z
12
3
45
Normal
consolidation
Unload/ reload
Over-consolidation'2
tp
z
6 ),( 6'
6 z z
tp
z
tp
z 5
z
or
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
66/132
o
C
C
zp
z C
V
oe
e
eo
eecreep
t t t
t for H t t
t t
e
C s
e
c
e zp z
)(10
0)log(1
'
'
6)(
6
6
6
0
6
(d) Final point is at an over-consolidated state from reloading :
For example Point 6 to Point 5 (Figure 1)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
67/132
0)log(1
50
""
e
eo
eoe f
v
creep primarytotalB
t for H t t
t t
e
C sU
s s s
0log1 '
6
'
5
H e
C s
z
z
o
e f
H
H t
t t
e
C
t
t t
e
C H
t t
t t
e
C s
tp
z
tp
z
o
eoe
o
eoe
eo
eoecreep
)(
)log(1
)log(1
)log(1
5
5
0050
)log(1
);log(1
5
0
5
0 o
eoetp
z
o
eoetp
z t
t t
e
C
t
t t
e
C
'
z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),(3
'
3 z z
),( 4'
4 z z ),( 5'
5 z z
12
3
45
Normal
consolidation
Unload/ reload
Over-consolidation'2
tp
z
6),(
6
'
6 z z
tp
z
tp
z 5
z
or
C
C
z C
V
ttt ec
e zp z
)(10'
5)( 5
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
68/132
o
zp
oe t t t
)(10'5
o
C
C
zp
z C
V
ooee t t t t t t t e
c
e zp z
2)(10'
'
5)(
5
5
)1d(0)log(1 5
5
0
""
eeo
ee
f v
creep primarytotalB
t for H t t
t t
e
C sU
s s s
)1dayt(log1
)log(1
log1
:coupling pressure porewater noIf
'
6
'
5
5
5
0
'
6
'
5
""
H e
C H
t t
t t
e
C H
e
C
s s s
z
z
o
e
eo
ee
z
z
o
e
creep primarytotalB
Extension of the above method:
(a) Have vertical drains
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
69/132
(a) Have vertical drains
The U v in (1a) … (1d) is replaced by the total U :
toncondolidaiof degreeaverageradial
r
r v
r v
U
)U )(1U (11U
)U )(1U (1U)(1
(b) Multi-layers (n layers)Appling superposition principle
n
1ii,
n
1ii,""
n
1ii, creep primarytotalBtotalB s s s s
layers! betweencontinuityconsider toneed:n
1i
i,""
primary s
2
2 ;60.0085.0)1log(933.0;60.04 H
t C T U for U T U for U T vvvvvvvv
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
70/132
2
2
22
2
2
22
2
;;
ln44
3
)ln(
)2
exp(1
e
r r
w
s
w
e
s
h
r r
r t C T
r
r S
r
r n
S n
S n
K
K
n
S
S
n
S n
n
m
m
T U
f cc
t c
t corr t c
f t t
t corr t c
st
t U t
t swheret
t st st t For
st
U t
swhere P
P t st st t For
)2
()2
(),2
()(:
)2()2(,)2()(:0
,
'
'
,
sr 2
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
71/132
rw=rd=radius of well
re=R=radius of column
rs=radius of smear zone
Smear zone: remoulded soil zone
pushed by mandrel
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
72/132
Band drain
F 0
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
73/132
f cc
t
ct corr t
c
f t
t t corr t
c
st
t U t
t swhere
t t st s
t t For
st
U t
swhere
P
P t st s
t t For
)2
()2
(
)2
()(
:
)2
()2
(
)2
()(
0
,
'
'
,
't P
t
t 2
t
corr t s ,
'
P
corr t s ,
t
U from previoussolution/chart
How to get U, Ur , Uv?(a) Use existing solutions, tables, figures,
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
74/132
( ) g , , g ,
equations(b) Use new internet based tools for calculation
of the coefficients of consolidation and
porewater pressure in 1-D and 2-D cases
• Tools are developed by JH Yin et al. at PolyU
• http://www.cse.polyu.edu.hk/~civcal(Reclamation molulus)
• Interactive calculations!
• 1-D (oedometer condition) - ramp loading and varying
with depth• 2-D axi-symmetric - ramp loading & vertical drain
Internet based consolidation analysis
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
75/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
76/132
Example 1 :
H is 10m with double drainage in 1D straining. Other parameters are in the
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
77/132
g g p
following table. Calculate final settlements at different states at 5 years. The load issuddenly applied.
07.0e
C 5.0c
C 018.0e
C
21
1eo
oeV
)/(
488.0
2 yr m
C v
)(1 dayt o
%1.0
001.030kPa
z1
'
z1
%54.0
0054.040kPa
z2
'
z2
%15.1
0115.0
60kPa
zp
'
zp
%68.8
0868.0
120kPa
z4
'
z4
%40.8
0840.0
100kPa
z5
'
z5
35%.7
0735.0
50kPa
z6
'
z6
Table 1 Parameters and stress-strain state of soil
(1a) Settlement calculation for the case of loading and final stress-strain state is
normally consolidated (loading from Point 1 to Point 4)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
78/132
858m.010)60
120log
2
5.0
30
60log
2
07.0(
log1
log1 '
'
4
'
1
'
H
e
C H
e
C s
zp
z
o
c
z
zp
o
e f
0.352614.3/0979.04/T4U
0976.05/488.05d/tCT,5m2/10d
vv
22
vv
0.406m0.2935)0.858(0.3526
)(
creep f vtotalB s sU s
0.2935m10)1
1-53651log(2
0.018
)t
tlog(
1
5yearst
o
eo
0
"
H t
e
C s e"creep
' z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'3 z z
),( 4'
4 z z ),( 5'
5 z z
12
3
45
Normal
consolidation
Unload/ reload
Over-consolidation '2
tp
z
6 ),( 6'6 z z
tp
z
tp
z 5
z
or
(1b) Settlement calculation for the case of loading and final stress-strain state is at an
overconsolidated state (loading from Point 1 to Point 2)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
79/132
0437m.01030
40log
2
07.0log
1 '1
'
2
H e
C s
z
z
o
e f
o
C
C
zp
z C
V
oe t t t e
c
e
zp z
)(10
'
'
2)(
2
2
= )(163541)60
40(10 018.0
5.0
018.0
2)0115.00054.0(
day
0.352614.3/0979.04/T4U
0976.05/488.05d/tCT,5m2/10d
vv
22
vv
0.047m0.090)0.0437(0.3526
)(
creep f vtotalB s sU s
0.090m10)163541
163545365log(
2
0.018
)t
tlog(
1
5yearst
e2o
e2
0
"
H t
t
e
C s e"creep
'
z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'
3 z z
),( 4'
4 z z ),( 5'
5 z z
12
3
45
Normalconsolidation
Unload/ reload
Over-consolidation'2
tp
z
6 ),( 6'
6 z z
tp
z
tp
z 5
z
or
(1c) Settlement calculation for the case of unloading and final stress-strain state is
at an overconsolidated state (unloading from Point 4 to Point 6)
'
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
80/132
133m.01012050log
207.0log
1 '4
6
H e
C s z
z
o
e f
o
C
C
zp
z C
V
oe t t t e
c
e
zp z
)(10
'
'
6)(
6
6
=
)(10226.11)60
05(10 9018.0
5.0
018.0
2)0115.07350.0(
day
0.352614.3/0979.04/T4U
0976.05/488.05d/tCT,5m2/10d
vv
22
vv
0.047m0.00])133.0[(0.3526
)(
creep f vtotalB s sU s
0.00m10)10226.11
10226.15365log(
2
0.018
)t
tlog(
1
5yearst
9
9
e6o
e6
0"
H t
t
e
C s e
"creep
'
z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'
3 z z
),( 4'
4 z z ),( 5'
5 z z
12
3
45
Normal
consolidation
Unload/ reload
Over-consolidation'2
tp
z
6 ),( 6'
6 z z
tp
z
tp
z 5
z
or
(1d) Settlement calculation for the case of reloading and final stress-strain state is at
an overconsolidated state (reloading from Point 6 to Point 5)
'
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
81/132
1054m.01050
100log207.0log
1 '6
5
H e
C s z
z
o
e f
o
C
C
zp
z C
V
oe t t t e
c
e
zp z
)(10
'
'
5)(
5
5
= )(18.761)60
010(10 018.0
5.0018.02)0115.0840.0(
day
0.352614.3/0979.04/T4U
0976.05/488.05d/tCT,5m2/10d
vv
22
vv
0.081m0.1252)1054.0(0.3526
)(
creep f vtotalB s sU s
0.1252m10)18.761
76.185365log(
2
0.018
)
t
tlog(
1
5yearst
e5o
e5
0
"
H
t
t
e
C s e"creep
'
z Log
e
cC
eC ),('
zp zp tp
z 5
),( 2'
2 z z
),( 3'
3 z z
),( 4'
4 z z ),( 5'
5 z z
12
3
45
Normal
consolidation
Unload/ reload
Over-consolidation'2
tp
z
6 ),( 6'
6 z z
tp
z
tp
z 5
z
or
Example 2:
Calculate the curves of settlement and time for the case of loading and final stress-
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
82/132
.infiniteisfieldthein, field EOP t
Here we assume U v=0.987 is the end of primary consolidation in the field.T v=1.668, t EOP,field =3118day.
strain state is normally consolidated (loading from Point 1 to Point 4) usingHypothesis A and simplied Hypothesis B method.
Solution:
field EOP
field EOP o
e f v
field EOP f v
ondary primarytotalA
t t for H t
t
e
C S U
t t for S U
S S S
,
,
,
"sec"""
)log(1
)1a(0)log(1 0
""
e
o
eoe f v
creep primarytotalB
t for H t
t t
e
C sU
s s s
Curves of settlement and time from the two methods are shown in the following figure.
Cv= 0.488 m^2/year
sf= 0.858 m
t=(day) TV= Uv Uv*sf S_creep S_totalB S_"secondary" S_totalA
1 5.34795E-05 0.008254 0.007082 0 0.007082 0 0.007082
2 0.000106959 0.011673 0.010015 0.027093 0.010331 0 0.010015
4 0.000213918 0.016508 0.014164 0.054185 0.015058 0 0.014164
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
83/132
8 0.000427836 0.023346 0.02003 0.081278 0.021928 0 0.02003
16 0.000855671 0.033016 0.028327 0.108371 0.031905 0 0.028327
32 0.001711342 0.046691 0.040061 0.135463 0.046386 0 0.040061
64 0.003422685 0.066031 0.056655 0.162556 0.067388 0 0.056655
128 0.00684537 0.093382 0.080122 0.189649 0.097832 0 0.080122
256 0.01369074 0.132062 0.113309 0.216742 0.141933 0 0.113309
512 0.027381479 0.186764 0.160244 0.243834 0.205783 0 0.160244
1024 0.054762959 0.264124 0.226619 0.270927 0.298177 0 0.226619
1825 0.0976 0.352606 0.302536 0.293514 0.406031 0 0.3025362737.5 0.1464 0.431853 0.37053 0.309362 0.504128 0 0.37053
4106.25 0.2196 0.528909 0.453804 0.32521 0.625811 0 0.453804
6159.375 0.3294 0.640382 0.549448 0.341058 0.767855 0 0.549448
9239.063 0.4941 0.760495 0.652505 0.356907 0.923931 0 0.652505
13858.59 0.74115 0.869826 0.746311 0.372755 1.070543 0 0.746311
20787.89 1.111725 0.94784 0.813247 0.388603 1.18158 0 0.813247
31181.84 1.6675875 0.98677 0.846649 0.404451 1.245749 0 0.846649
46772.75 2.50138125 0.99831 0.85655 0.420299 1.276139 0.015848213 0.87239870159.13 3.752071875 0.999923 0.857934 0.436148 1.294048 0.031696427 0.88963
105238.7 5.628107813 0.999999 0.857999 0.451996 1.309995 0.04754464 0.905544
157858 8.442161719 1 0.858 0.467844 1.325844 0.063392853 0.921393
236787. 1 12. 66324258 1 0.858 0. 483692 1. 341692 0. 079241067 0. 937241
355180.6 18.99486387 1 0.858 0.49954 1.35754 0.09508928 0.953089
532770.9 28.4922958 1 0.858 0.515389 1.373389 0.110937493 0.968937
799156.3 42.7384437 1 0.858 0.531237 1.389237 0.126785707 0.984786
1198735 64.10766555 1 0.858 0.547085 1.405085 0.14263392 1.0006341798102 96. 16149833 1 0.858 0. 562933 1. 420933 0. 158482133 1. 016482
2697153 144.2422475 1 0.858 0.578781 1.436781 0.174330346 1.03233
4045729 216.3633712 1 0.858 0.59463 1.45263 0.19017856 1.048179
6068594 324.5450569 1 0.858 0. 610478 1.468478 0.206026773 1.064027
9102890 486.8175853 1 0.858 0. 626326 1.484326 0.221874986 1.079875
13654335 730.2263779 1 0.858 0.642174 1.500174 0.2377232 1.095723
20481503 1095.339567 1 0. 858 0. 658023 1. 516023 0.253571413 1. 111571
30722255 1643.00935 1 0.858 0.673871 1.531871 0.269419626 1.12742
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
84/132
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
1 10 100 1000 10000 100000 1000000 10000000
T o t a l
s e t t l e m e n t ( m )
Time (day)
Simplified Hypothesis B method (Yin)
Hypothesis A method
Example 3 :
At a site in Hong Kong with a clay layer of 5 m thick. The bottom of the layer is
permeable. The water table in the worst case is at the ground level. A uniform
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
85/132
Solution :
year mC dayt C
kPaemkN C C
voe
zpo sat ec
/1,1%;1
;50;2,/5.16;04.0;4.0
2
'3
vertical pressure by fill is assumed to be applied suddenly in two stages. The firststage with pressure 45 kPa is applied and in 6 months later, the second stage of 155
kPa is applied. Calculate and plot the time-settlement curve up 50 years using Yin’s
simplified method (using the initial effective stress at the middle clay layer) since the
first loading. Calculate the final total settlements at time of 0.5, 10, and 50 years
since the first loading. Soil parameters are:
m5
m5.2)( year t
)(kPa p
kPa45
kPakPa 45155
year 5.0 year 2
year 5.1
2
2 ;60.0085.0)1log(933.0;60.04 H
t C T U for U T U for U T vvvvvvvv
0927.05725.61
log4.050
log04.0
s
50725.6145725.16;725.16)81.95.16(5.2
45:1
1f1
''''
m H
kPakPa pkPa
kPa Loading Stage
f
zp zi z zi
'
z Log
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
86/132
)(
128.00.03770927.0984.0;0824.00.03770927.0632.0(
984.0101;632.0101101
6.1
5.2
101;32.0
5.2
21:10;2
?0.03770927.0(:,5.0:
)t
log(1
0927.0(:,5.00:
4160.00377.00927.0319.0(
00754.0)1
1-5.03651log(
21
0.01)
t
tlog(
1
0.0377m5)1
1-5.03651log(
21
0.01)
t
tlog(
1
)t-t(t5year .0t
6.0319.014.3/08.04;080.05.2
5.01
5.0
00634.0725.16
50log
21
04.0log
1;01854.0
5021725.1621
1111
933.0
6.1085.0
933.0
32.0085.0
933.0
085.0
1
2222
1111
00
1111
1111
0
e0
0
1
0
e0
0
1
0e
1221
'
'
0
1
1
filedataand curve see
mm s sU s
U
H
t C
H
t C T years yearsTime
U s sU siscurve settlement timethe year t timeOther
H t
e
C U s sU siscurve settlement timethe year t timeOther
m s sU s
t
e
C
H t
e
C s
ok U H
t C T
yearstime At
e
C
creep f v B
T
v
vvv
vcreep f v B
evcreep f v B
creep f v B
ecreep
ecreep
vv
v
zi
zpe zp f
v
)
)
)
)()
e
cC
eC
),( ' zp zp 1
2
3
Normal
consolidation
Over-consolidation
1 f s
2 f s
1creep s
2creep s
Stage 1
Stage 2
),( 2'
2 zp zp ),( 1'1 z z
026080007540018540
725.6145725.1645)81.95.16(5.2
155:2
111
'
1
f
z kPanew
kPa Loading Stage
'
z Log
O lid ti
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
87/132
02692.0725.61
336.71log
21
04.002608.0log
1
71.336100.632577.22310725.6150
10
-
1-logloglog
-log1log1-log)11
-log1
log1
log1
log1
log1
;log1
:,
02608.000754.001854.0
'
1
'
2
0
12
1645.004.04.0
2100634.002608.0
0.04-0.4
0.04-
0.04-0.4
0.4
1
'
1
''
2
01
'
1
''
2
1
'
1
0
'
0
'
2
00
1'
1
'
2
0
'
'
2
0
'
'
2
0
'
1
'
2
0
1
'
'
2
0
2'
1
'
2
0
12
2
'
2
111
01
z
zpe z zp
C C
e
C C
C
z C C
C
zp zp
ec
zp z z
ec
e zp
ec
c zp
zp z z e
zpc
zpec
zp z
z
zpe
zp
zpc
zp
zpc zp
z
zpe z
zp
zpc zp zp
z
zpe z zp
zp zp
creep f z
e
C
kPa
C C
e
C C
C
C C
C
e
C
e
C
e
C
e
C
e
C
e
C
eC
eC
e
C
e
C
are New
ec zp z
ec
e
ec
c
)(
(
e
cC
eC
),('
zp zp 1
2
3
Normal
consolidation
Over-consolidation
1 f s
2 f s
1creep s
2creep s
Stage 1
Stage 2
),( 2'
2 zp zp ),( 1'1 z z
N
kPanew
kPa Loading Stage
creep f z
z
02608.000754.001854.0
725.6145725.1645)81.95.16(5.2
155:2
'
111
'
1
'
z Log Over-consolidation
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
88/132
m H t
e
C U s sU s
U H
t C T
yrst yrsand
m H t
e
C U s sU s
ok T U H
t C T
yrst timeloading yearstimetotal At
m
H e
C
e
C H S
kPa
are New
evcreep f v B
T
vv
v
evcreep f v B
vvv
v
zp
z c
z
zpe
f f
z
zp zp
zp zp
v
326.05)
1
3655.9log(
21
001.0326.0981.0)
t
log(
1
326.0(
981.0101101;52.15.2
5.91
)5.9(10
183.05)1
3655.1log(
21
001.0326.05529.0)
tlog(
1326.0(
6.05529.014.3/24.0414.3/4;24.05.2
5.11
)5.1(2
326.0506519.05)06435.000084.0(
5)336.71
725.216log
3
4.0
725.61
336.71log
3
04.0(
)log1
log1
(
725.216155725.61
;02692.0;71.336
:,
00
22222
933.0
52.1085.0
933.0
085.0
2222
00
22222
2222
' 2
'
2
0' 1
'
2
022
'
2
2
'
2
2'
2
)
)
e
cC
eC
),('
zp zp 1
2
3
Normal
consolidation
1 f s
2 f s
1creep s
2creep s
Stage 1
Stage 2
),( 2'
2 zp zp ),( 1'1 z z
'
z Log Over-consolidationHtCUU
iscurve settlement timethe year t timeOther
Summary
e )l (09270(
:,5.00:
:
)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
89/132
e
cC
eC
),('
zp zp 1
2
3
Normal
consolidation
1 f s
2 f s
1creep s
2creep s
Stage 1
Stage 2
m s s s
yrsand yrst At
m s s s
yrst At
H t
e
C U U
s sU s sU s s s
iscurve settlement timethe year t timeOther
H t e
C U s sU s s
B B B
B B B
evv
creep f vcreep f v B B B
evcreep f v B B
454.0326.0128.0
:102
265.0183.00824.0
:2
)
t
log(
1
326.00.03770927.0
((
:,5.0:
)t
log(1
0927.0(
21
21
00
21
22211121
00
1111
))
)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
90/132
0.04
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
91/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
92/132
'
i H
0.269 187.5x0.269
Empirical equations to fit a measured settlement-time curve in the
field and find the final settlement Sf (or )S
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
93/132
S
Hyperbolic equation method:
Find parameter :
Find final settlement:
2
22
1
1
1
)(
)(
t
t s s s s
t
t
s s s s
d d
d d
Empirical equations to fit a measured settlement-time curve in the
field and find the final settlement Sf (or )S
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
94/132
S
Hyperbolic equation method:
Find parameter :
Find final settlement:
)(;1
)(
1
)(:)()(
)(
1
)()()(
d d
d d d t
d d d t d t d
d d t
s s A B
s s s s
Band s s
A xt vs y s s
t Plot
Bx A yt s s s s s s
t
s s
t
s s
t
t
t s s s s
0 025
0.03
0.035
Data
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
95/132
y = 0.0149x + 0.0066
0
0.005
0.01
0.015
0.02
0.025
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
Time (year)
t / ( s t - s d )
Data
Linear (Data)
d t s s
t
);(;1
;)(
1
)(
)(
1
)(
d d
d d
d d d t
s s A B
s s s s
Band s s
A
Bx A yt s s s s s s
t
0066.0)(
d s s A
0149.0)(
1
d s s B
0d s
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
96/132
)1)((
)1)((
)1)((
2
1
2
1
t
d d
t d d
t
d d t
e s s s s
e s s s s
e s s s s
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
97/132
time (years) Set (cm)
Set
(prediction 1) t/(st-sd)
Set (prediction
2)
0 0 0 0
0.226 22.45 22.26519974 0.010067 22.67391697
0.302 26.7 26.7 0.011311 27.207697440 425 30 35 32 22375862 0 014003 32 8629422
t
t s s s s d d t
)(
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
98/132
0.425 30.35 32.22375862 0.014003 32.8629422
0.488 33.34 34.48516325 0.014637 35.18080627
0.533 35.81 35.92092809 0.014884 36.65321111
0.61 39.08 38.09188759 0.015609 38.88074447
0.665 40.54 39.4571216 0.016404 40.28227883
0.754 44.09 41.39960169 0.017101 42.27737095
0.814 45.72 42.55338369 0.017804 43.462939030.89 46.95 43.86525718 0.018956 44.8114395
0.92 47.02 44.34270318 0.019566 45.3023439
0.998 48.02 45.49074971 0.020783 46.4830323
1.06 48.19 46.31805672 0.021996 47.33410735
1.168 48.63 47.60669616 0.024018 48.66017864
1.373 49.72 49.63014421 0.027615 50.7434113
1.402 49.88 49.88 0.028107 51.00073482
1.44 50.1 50.19572156 0.028743 51.325919592 53.70756979 54.94505495
2.5 55.71214989 57.01254276
3 57.13379008 58.47953216
3.5 58.19449265 59.57446809
4 59.01623055 60.42296073
4.5 59.67158239 61.09979633
6 61.02694237 62.5
6.5 61.34850754 62.83228613
7 61.62684428 63.11992786
8 62.08456815 63.59300477
9 62.4453037 63.96588486
10 62.7369245 64.26735219
t t
s s s s
s s
s s
s s s s
t
s s
s s
t
t
t
t
d
d
d d
d
1
1
12
1
12
1
1
12
1
1
1
1
1
)(
1
1
0d s
Dt= 1.1
Dt/t1= 3.642384 t
t s s s s d d t
)(
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
99/132
t1/Dt= 0.274545
(s1-sd)/(s2-s1) 1.151855
(s2-s1)/(s1-sd) 0.868165
alfa= 0.438742s_infinite= 65.48946
A= 0.0066
B= 0.0149alfa= 0.442953
s_infinite= 67.11409 t
t
s s
s s
s s
s s
s s s s
t
s s
s s
t
t
t
t
d
d
d d
d
1
1
12
1
12
1
1
12
1
1
1
1
1
)(
1
1
0d s
0 2 4 6 8 10 12
Tim e (year)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
100/132
0
10
20
30
40
50
60
70
S e t t l e m e n t ( c m )
Data
Prediction 1
Prediction 2
0.035
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
101/132
y = 0.0149x + 0.0066
0
0.005
0.01
0.015
0.02
0.025
0.03
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
Time (year)
t / ( s t - s d )
Data
Linear (Data)
(2-D problem)
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
102/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
103/132
See paper by:
Zhu, G.F. and Yin, J.-H. (2001). Consolidation of soil with vertical and horizontal
drainage under ramp load. Geotechnique, Vol.51, No.2, pp.361-367.
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
104/132
r w=r d=radius of wellr e=R=radius of column
r s=radius of smear zone
Smear zone: remoulded
soil zone pushed by
mandrel
sr 2
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
105/132
Band drain
No covered here
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
106/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
107/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
108/132
See Figures later on this approach
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
109/132
ln3
)ln(
)2
exp(1
2222
SS n K S nn
m
m
T U
h
r r
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
110/132
)!4
:(
;;
ln44
)ln(
22
2222
before R
t C T fromdifferent isthisnote
r
t C T
r
r S
r
r n
S n K nS S n
m
r r
e
r r
w
s
w
e
s
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
111/132
See proof later
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
112/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
113/132
o p3 p
2 p
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
114/132
o
1 p
2
00
0
33
0
22
0
11
00
0
33
0
22
0
110
0
33
0
22
0
11
332211332211321
333222111
332211
);(
)()(
;;
;;
Hp M S
p
pU
p
pU
p
pU U
US S p
pU
p
pU
p
pU Hp M
p
pU
p
pU
p
pU
Hp M U Hp M U Hp M U S U S U S U S S S S
S U S S U S S U S
Hp M S Hp M S Hp M S
v
v
vvvt t t t
t t t
vvv
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
115/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
116/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
117/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
118/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
119/132
Vacuum preloading
m
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
120/132
S S f S S f
m H 29.4
vm
vm
suv
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
121/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
122/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
123/132
24
02.002067.0
scm K K md r
zone smear awiththisdo Re
h s s s /105.0;5.02/
:
6
without
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
124/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
125/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
126/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
127/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
128/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
129/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
130/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
131/132
-
8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf
132/132
Fully coupled consolidation modelling programs in CSE:
ABAQUS (3-D finite element program)
Plaxis (2-D finite element program)