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Soil Mechanics Chapter 4
§4 Compressibility, consolidation and settlement
土的压缩、固结与地基沉降
•Compressibility /土的压缩性
•Consolidation /土的固结
•Settlement /地基沉降
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Soil Mechanics Chapter 4
• Introduction / 概述
Compressibility and consolidation can be distinguished as:
·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
These are distinct from:
1. compaction which is the expulsion of air from a soil by
applying compaction energy.
2. immediate or undrained settlement which is the resultant
deformation of a soil under applied stresses without any
volume change taking place
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Soil Mechanics Chapter 4
4.1 Compressibility/土的压缩性
4.1.1 Consolidation Test and compression characteristics/土的固结试验与压缩特性
1 Consolidation test
Assumption:
• Load distribution-uniform
• Stress distribution(in
different height)-the same
• Lateral deformation-0
• The area of the sample
section-unchangeable
• Solid soil-uncompressible
oedometer
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oedometer
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2 Compression curve/压缩曲线
F  h F (h  s )

1  e0
1  e1

e0  e1
s
h
1  e0
s
e1  e0  (1  e0 )
h
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Soil Mechanics Chapter 4
3 compression coefficient av/压缩系数
e1  e2
de
av 
 ,
p2  p1
dp
1
1
MPa , kPa
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av1-2<0.1Mpa-1,
0.1≤av1-2<0.5Mpa-1,
av1-2 ≥ 0.5Mpa-1,
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Low compressibility /低压缩性土
Middle compressibility/中压缩性土
High compressibility /高压缩性土
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4 compression index Cc/压缩指数
Cc 
e1  e2
(e  e )
 1 2
p2
lg p2  lg p1
lg
p1
e1  e2
de
Cc 

p
dp
lg
lg
pc
p
de C c
av  

dp
p
Soil Mechanics Chapter 4
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Soil Mechanics Chapter 4
Stress History Effect—
p>pc, lack consolidation（欠固结）
P=pc, normal consolidation （正常固结）
P<pc, overconsolidation （超固结）
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Determine pc
Soil Mechanics Chapter 4
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Soil Mechanics Chapter 4
4.1.2 compressible deformation calculation/单向压缩量的计算
•Only compression in vertical，Deformation due to void volume decrease
F  h F (h  s)

1  e0
1  e1

ei 
s
e0  e1
h
1  e0
hi  hs hi
 1
hs
hs
e0  e1
p
av  p
1
s 
 h  mv  p  h 
p  h
1  e0
Es
av 
mv 
av
coefficient of volume compressib ility / 体积压缩系数
1  e0
1  e0
Es 
Compressib le mod ulus
av
/ 压缩模量
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4.1.3 Effective stress plot
The soil is described as normally
consolidated when its state exists on
the steeper line (1 and 4) and the
effective stress can only be increased
with subsequent reduction in volume
because decreasing the stress takes the
soil state away from the normally
consolidated line. This line is often
referred to as the virgin compression
curve or line as any change of effective
stress along it will be for the first and
only time whereas any number of
unload/reload paths could be followed.
The
soil
is
described
as
overconsolidated when it occurs
Soil Mechanics Chapter 4
on the flatter portions( 2 and 3) and
volume changes can increase or
decrease with changes in effective
stress.
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4.1.4 reloading curves The past–
geological and stress history has
brought the soil to its present
condition.
– If a sample is taken from the ground
with no moisture content change and
loaded in a laboratory consolidation
apparatus a reloading curve can be
obtained by plotting the void ratio
produced after the soil has
consolidated to a new equilibrium for
each change in effective stress.
– The shape of the reloading curve (as
well as geological information about
the soil at the site) will help to
determine whether the soil is
normally
consolidated
or
overconsolidated. If the soil is
normally consolidated the reloading
curve will continue on the virgin
compression line from its present
condition and will follow the straight
line on a log σ＇plot.
Soil Mechanics Chapter 4
The reloading curve for an overconsolidated
clay will have two portions, one
commencing from its present condition and
following a flatter path until it reaches the
virgin
compression
line
at
the
preconsolidation pressure, followed by a
steeper line corresponding to the virgin
compression curve
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Soil Mechanics Chapter 4
4.1.5 overconsolidation ratio OCR
•The overconsolidation ratio (OCR) defined as:
p'
OCR  c
p0 '
previous max imum effective stress

present
effectivestress
For a normally consolidated clay the present effective stress is also the
previous maximum so the OCR=1. for a heavily overconsolidated clay
the OCR may be 4 or more therefore this type of soil has been subjected
to a much greater stress in the past
– Compared to its present condition.
– The significance of Pc＇for an overconsolidated clay is that if stresses
are kept below this value then settlements can be expected to be small
but if the applied stresses due to loading exceed this value then large
settlements will occur as consolidation will take place along the virgin
compression line.
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4.1.6 Deformation modulus/变形模量
• Coefficient of earth pressure at rest/ 静止侧压力系数
x
k0 
z
• Poisson’s ratio/泊松比 μ
k0 

1 
k
 0
1  k0
• Modulus of deformation/变形模量E0
 z   ( x   y )
E0 
z
2 2
E0  (1 
)  Es
1 
2
or
2k 0
E0  (1 
)  Es
1  k0
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• Modulus of compression/压缩模量 Es
e0  e1
av  p
s
h 
 h  mv  p  h
1  e0
1  e0
s

h
1  e0

p
1
Es  


 e0  e1
av
mv
1  e0
s
p  h
Es
( MPa, kPa)
Soil Mechanics Chapter 4
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Soil Mechanics Chapter 4
4.2 Consolidation - Terzaghi theory of one-D consolidation
饱和粘性土太沙基单向固结理论
4.2.1 Introduction/引言
The theory considers the rate at which water is squeezed out of
an element of soil and can be used to determine the rates of:
• volume change of the soil with time
• settlements at the surface of the soil with time
• pore pressure dissipation with time
t=0, u=p0, σ’=0
t=t1, u+ σ’=p0
t=∞, u=0, σ’=p0
u=f(t,x,y,z)
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Soil Mechanics Chapter 4
4.2.2 Assumptions/基本假定
1. Compression and flow are one-dimensional, i.e. they are
vertical only. Whereas significant horizontal flow can occur in
layered deposits.
2. Darcy’s law is valid at all hydraulic gradients but deviation
may occur at low hydraulic gradients.
3. k and m v remain constant. However, they both usually
decrease during consolidation.
4. No secondary compression or creep occurs. If this occurs the
void ratio-effective stress relation-ship is not solely dependent
on the consolidation process.
5. The load is applied instantaneously and over the whole of the
soil layer. However, loads are applied over a construction
period and usually do not extend over a wide area in relation to
the thickness of the consolidating deposit.
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Soil Mechanics Chapter 4
4.2.3 Differential equation of consolidation/固结微分方程
V  dxdydt
v
q z  dq z   q z  dq z 
t
h
q z  K  i  dxdy  K dxdy
z
2
v
 h
 K 2 dxdydz
h  u
w
t
z
K  2u
v
  2 dxdydz 
 w z
t
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Soil Mechanics Chapter 4
v vv

,Vv  Vs  e
t t
v e
e V
   Vs  
t t
t 1  e0

K  2u
1 e
 2 

 w Z 1  e0 t
e
u
a
t
t
 2u
a u
 2 
 w Z
1  e0 t
k
k  2u
u
 mv
 w Z 2
t
or
u
K
 2u

 2
t  w  mv Z
u
u
 Cv 2
t
Z
2
where,
K
Cv 
 w  mv
(cm 2 / s)
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Soil Mechanics Chapter 4
4.2.4 Solution of the consolidation equation/固结微分方程求解
The basic differential equation of consolidation (equation 3.1) gives the
relationship between three values.
u
k
 2u
 2u

 Cv
2
t
mv w z
z 2
where:
c v is the coefficient of consolidation
m v is the coefficient of volume compressibility
k is the coefficient of permeability
u  C1 cos AZ  C2 SinAZ   C3e  A c t 
2
v
Initial condition and boundary conditions
t=0, 0≤Z≤2H, u=σz=p0,
0<t< ∞, Z=0,u=0
Z=2H, u=0
t=∞, 0≤Z≤2H, u=0
General solution
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4
1
m  z m2 N
u
Sin
e

 m1,3,5 m
2H
where,

N
 2 Cv
4 H
2
t 
Case: triangular distribution
1
mz m2 N
u  2  2 cos
e
 m1,3,5 m
2H
8p

Soil Mechanics Chapter 4
2
4
Special solution
Tv ,
Cv
Tv  2  t Time factor
H
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Soil Mechanics Chapter 4
4.2.5 Degree of consolidation/固结度
-average degree of consolidation
mv   'h  '
u
Uv 
  1
mv    h 

St
U ,
S
-different cases
• Rectangular Distribution
4 p 2H 
z  N 1
3z 9 N

2H
Sin
e

Sin
e

...

dz

udz
0

2H
3
2H



U v  1  20H
 1
2z
 dz
0
Uv  1
8   N 1 9 N 1 25 N

e

e

e

...

...


2 
9
25

Uv  1
8

2
e
N
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Soil Mechanics Chapter 4
• triangular distribution

U  1

v
0
H
0
Uv  1
H
8p
udz
 1
2
p z dz

H
0
z  z  N 1
3zz 9 N


e  cos
e  ...dz
 cos
2H
9
2H


H p H  Z 
0 H  dz
32   N 1 9 N 1 25 N

e

e

e

...


3
 
9
25

Uv  1
32

3
e
N
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Soil Mechanics Chapter 4
Example 已知:H=10m,大面积荷载p0=0.12Mpa,e0=1.0,av=0.3Mpa-1,K=1.8cm/年
求 单面及双面排水条件下: (1)加荷一年的沉降量
(2)沉降达14cm所需时间
Solution
(1)
 w  9.8 KN
m3
 10 KN
m3
 10 10 3 MN
2
K 1  e0  1.8 10 2 1  1.0
m
Cv 

 12
year
av w
0.3 10  2
S 
av z
0.3  0.12
H
10  0.18m
1  e0
1  1.0
m3
 10 2
MPa
m
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• an half-closed layer/单面排水
Uv  1
 1
8

2
8

e
e
2

2
4

2
4
Tv
0.12
 1
8

2

e
 2  121 


4  102 
 39.7%
 St  U v  S  39.7%  0.18  0.0715m 
• a open layer/双面排水
Uv  1
8

2
e

2
4
0.48
St  0.135m
 75.2%
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Soil Mechanics Chapter 4
2
(2)
St 0.14
8  4 Tv
Uv  
 77.8%  1  2 e
S 0.18

Tv=0.524,
t=4.37(years)
for an half-close layer
H=10m
t=1.09(years)
for a open layer
H=5m
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Soil Mechanics Chapter 4
32
U v1  1  3 e  N

Type 0
Type 1
Type 2
Type 0-1
Type 0-2
p1=αp2
p1
p
p
H
2H
2H
H
U v0  1 
H
H
H
2H
p
p
p1=αp2
p
p2
p2
p2
8   N 1  9 N 1  25 
e  e  e 
2 
9
25

U vo  1 
8

N
e
2
u2  u0  u1
u02  u0  u1
U v 2  2U v 0  U v1
u01  u0  u1
U v 01 
2U v 0  1   U v1
1
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Soil Mechanics Chapter 4
The Degree of Consolidation Type 0-1
p1=αp2
H
+
p1
Type 0
 p1    p2 ,
H
=
H
Δp
Type 1
p2
Type 0-1
p 2  p1  p
 p  p2  p1  p2 (1   )
1
U v 0  p1 H  p  H U v1
2U v 0  (1   )U v1
2
U v 01 

( p  p2 )
(1   )
H 1
2
(  1)
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Soil Mechanics Chapter 4
4.3 Settlement/地基沉降计算
4.3.1 Classification of foundation settlements/地基沉降分类
– Settlements produced by applied stresses originate from:
•
consolidation settlement/主固结沉降, Sc
•
secondary compression次固结沉降, Ss
•
immediate settlement/瞬时沉降, Sd
amounts of each must be determined to give the total settlement,
S=Sd+Sc+Ss
S
Sd
Sc
Ss
t
4.3.2 Principle and procedure of settlement using delamination total Method
分层总和法
-Formula/计算公式:
e0i  e1i
avi  pi
Si 
 hi 
 hi  mvi  pi  hi , S   Si
1  e0i
1  e0i
-Compression layer/压缩层: General,
There is soft layer below/有软卧层
z
 20%
0
z
 10%
0
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Soil Mechanics Chapter 4
-Thickness of every layer/分层厚: <0.4B (B- Width of foundation/基础厚度)
-Additional contact pressure/
附加压力/基地净压力:
P
p0     D
F
-Calculating procedure/计算步骤:
1) To delaminate (interface, water table) /分层（土层交界面/地下水位）
(hi<0.4B)
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2) Calculate Stress due to self weight/
计算地基各层自重应力
3) Calculate stress due to load/
计算地基各层附加应力
4) Determine the compression
thickness/ 确定压缩层厚度
5) Calculate average σ0σz /
计
算地基各层自重应力及附加应力的
平均值
6) Calculate e0i,e1i
根据e~p曲线得到各层e0i,e1i
7) Calculate/计算各层 Si
8) Calculate/计算总沉降 S
Soil Mechanics Chapter 4
• 《建筑地基基础设计规范》计算地基沉降
• 地基沉降计算的理论公式法
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Soil Mechanics Chapter 4
《建筑地基基础设计规范》（GBJ 7-89）所推荐的地基最终沉降量计
算方法是另一种形式的分层总和法。它也采用侧限条件的压缩性指标，并运用了
平均附加应力系数计算，还规定了地基沉降计算深度的标准以及提出了地基的沉
降计算经验系数，使得计算成果接近于实测值。
1、 计算公式
规范推荐的地基最终沉降量s(mm)的计算公式如下:
ψ --沉降计算经验系数，根据地区沉降观测资料及经验确定,一般在0.2-1.4之间。
n --地基沉降计算深度范围内所划分的土层数，其分层厚度取法同前面按分层总和法
Po--对应于荷载标准值时的基础底面附加压力(kPa)
Esi--基础底面下第i层土的压缩模量， (MPa)；
Zi和Zi-1--基础底面至第i层土、第i-1层土底面的距离(m)；
ai和 ai-1--基础底面的计算点至第i层土、第i-1层土底面范围内平均附加应力系数
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Soil Mechanics Chapter 4
2、计算深度zn的确定
《建筑地基基础设计规范》用符号 表示地基沉降计算深度，并规定应满足下列
条件(包括考虑相邻荷载的影响)：
式中Sn为最后一层厚为Dz的土层的压缩量。Dz的取值查表。
按上式所确定的沉降计算深度下如有较软土层时，尚应向下继续计算，直至软
弱土层中所取规定厚度Δz的计算沉降量满足上式为止。
当无相邻荷载影响，基础宽度在1～50m范围内时，基础中点的地基沉降计
算深度，规范规定，也可按下列简化公式计算