第四章 土的压缩与固结
Chapter 4 Compressibility and Consolidation of Soil
4.1 Compaction Vs Consolidation 压实与固结
空气的体积减小
孔隙水的体积减小
What is consolidation? (什么是固结?)

Consolidation is a time-related (时间性) process of
increasing the density (增加密度) of a saturated soil
by draining some of the water out of the voids.

Coarse-grained soils (粗粒土), such as sands and
gravels, consolidate at a much faster rate than finegrained soils (细粒土) such as silts and clays.
The need of consolidation theory 固结理论的重要性
Most settlement in engineering is because of consolidation

Tilting of Piza Tower is due to
differential settlement (不平均沉
降引起塔的倾斜)

Two important questions for
consolidation problems (两个重
要问题)
1. How much ? (沉降有多少)
2. How long ? (时间有多久)
Piza Tower, Italy
Examples
Initial
Loading
condition
(荷载) (初始状况)
at time = T
0
Earth dam (
土堤)
Excess pore water pressure (
超孔隙水压力)
Static pore water pressure
(静孔隙水压力)
4.2 1-D consolidation test (单向固结试验)
Consolidation is a three-dimensional (三维) process, but the
direction of flow of water is primarily vertical (主要在竖向)
or one-dimensional (1-D).
s’z
s’z
Simplify
vy
简单化
vx
vz
vz (water flow)
(竖向荷载)
Vertical settlement (
竖向沉降)
侧向应变为零
土样本
Oedometer (固结仪)
Load – settlement curve
or 1-D compression curve
(单向压缩曲线)
1. Time – related consolidation (固结与时间关系)
For each load increment (每一个荷载增量)
hf
s’f
2. 1-D compression curve (单向压缩曲线)
Loading (加载
)
Compression curve
(压缩曲线)
h
 v 
H0
Unloading and Reloading (
卸载与再加载)
Re-compression curve
(回弹曲线)
where v is change in vertical strain (竖
向应变增量), h is the change in height
for each loading increment (每次荷载增
量 的 高 度 改 变 ) and H0 is the initial
specimen height (样本初始高度)
(1)Coefficient of volume compressibility [体积压缩系数]
(mv)
No lateral strain (侧向应变为零)
h
e
 v 

H0 1  e0
 vol
e1  e 0
1
mv 


s' v 1  e 0 s' v1 s' v 0
e0
where vol is volumetric strain (体积
应变增量), s’v is change in vertical
effective stress (应力增量) and e0 is
initial void ratio (初始孔隙比)
e1
s’v0 s’v1
Example (例子)
e1  e 0
1
mv 

1  e 0 s' v1 s' v 0
1
0.8  1.2
mv 

1  1.2 200 100
 1.82 103 m 2 / kN
How about a100-200
(2)Other compressibility parameters
其它压缩参数
Low compresibility
Mid compresibility
High compresibility
Oedometric modulus [侧限压缩模量] (Es) is defined as the
ratio of vertical effective stress (’z) over vertical strain (z)
under zero lateral strain condition ,which reflects the
capability of the soil resisting against compressive
deformation,and shown as follows:
s ' z 1  e0
1
Es 


z
av
mv
From generalised Hooke’s law [广义虎克定律], elastic strains
[弹性应变] in x, y and z directions are expressed as follows:
where E is Young’s modulus or elastic modulus [弹性模
量]. E represents the stess to strain ratio without confined latral
strain. In 1-D consolidation, x = y = 0，we get:


s
1
E
sz    sx    s y  z
z
z
2 
2 


2

2

  E s  1 

 1 

 1  
 1  


soil is not ideal elastic material,above equation is
only an approximate formula.
(3) e——log s’ curve
1). Compression Index [压缩指数] (Cc)
 e1  e o 
Cc 
logs' v1  logs' vo
e0
e1
e2
e3
s’v0
s’v1
2). Expansion Index [回弹指数] (Ce)
Ce 
 e 3  e 2 
logs' v1  logs' vo
4.3 Calculation formulae of soil compression
with zero lateral strain
土的压缩量计算
e1  e2
e
H 
H1 
H1
1  e1
1  e1
av
H 
pH1
1  e1
H  mv pH1
p
H 
H1
Es
4.4 Settlement Calculation
沉降计算
Settlement found in soil due to loading may be divided into
three categories:
St  Si  Sc  Ss
where St is total settlement (总沉降), Si is immediate
settlement (瞬时沉降) – [elastic deformation, 弹性变形],
Sc is primary consolidation (主固结沉降) – [change of s’,
有效应力改变] and Ss is secondary consolidation (次固结
沉降) – [time, 时间]
1。 Immediate settlement 瞬时沉降

Immediate after loading (加荷後立即发生)

No volume change (体积不变) because time is too short
to drain away pore-water

Vertical settlement induces lateral expansion (侧向变形)

If no lateral expansion, Si = 0
Similar to rubber (橡皮 )

Based on elastic theory (弹性理论
)
1  2
Si 
bpI
E
where p is the pressure at the bottom of the foundation (基底
压力), b is width of the foundation (基础的宽度) [shorter
side of a rectangle (钜形的短边) or diameter of a circle (圆形
的直径)], E is Young’s modulus (弹性模量),  is Poisson’s
ratio (泊松比) and I is influence factor (影响系数).
2 。Primary consolidation


主固结沉降
Primary consolidation is the
settlement due to the gradual
dissipation of the pore water
from the saturated soils or
gradual increase in effective
stress (有效应力的增加).
It is calculated from the
following two methods:
1. e : s’ curve
2. e : log s’ curve
hf
s’f
* e --s’ curve method
Calculate stress of every layers
Two situation for calculation load
Consolidated under self-weight
No-consolidated under self-weight
Look for e1i\e2i
Calculate settlement of every layers
Calculate final settlement
4.5 Effect of stress history 应力历史
where OCR is overconsolidation ratio (超固结比), s’v is current effective
stress (现今有效应力) and s’vmax is maximum past effective stress (过去最
大有效应力) or pre-consolidation pressure [s’c] (前期固结压力)

OCR = 1 Normally consolidated [NC] soil (正常固结土)

OCR > 1 Overconsolidated [OC] soil (超固结土)
(前期固结压力)
s’v
s’vmax
OC soil (超固结土)
NC soil (正常固结土)
* e-- log s’ curve method
Overconsolidated Soil
In-situ recompression
curve
In-situ compression
curve
Underconsolidated Soil
Example (例子)
The unit weight of sand (sand) is 20 kN/m3 and unit weight of clay (clay) is 17
kN/m3. Calculate the consolidation settlement of the clay deposit after
reclamation. Consolidation test results on a clay sample is shown in the following
table:
Sea level (海平面)
10 m
Clay (粘土
)
6m
Dense sand (密砂)
Sand (砂土
)
Clay (粘土
)
15 m
6m
Dense sand (密砂)
After reclamation (填海后)
Pressure
(kN/m2)
Height
(mm)
0
19.0
25
18.8
50
18.3
100
17.6
200
16.1
400
14.5
(1) Consider clay deposit as a single layer and in the middle of clay layer:
Before reclamation (填海前)
5m
Sand (砂土
)
15 m
s' v0  (17  9.8)  3  21.6 kN/m2
After reclamation (填海后)
Clay (粘土
)
6m
Dense sand (密砂)
s' v1  20  5  (20  9.8) 10  (17  9.8)  3
 223.6 kN/m 2
s' v  223.6  21.6  202 kN/m 2
mv 
e e
1
 1 0
1  e 0 s' v1 s' v 0
Pressure, s’v
(kN/m2)
Height, H
(mm)
0
19.0
25
&
H
e

H 0 1  e0
mv 
1 H

H 0 s ' v
H (mm)
H/H0
(H0 = 19 mm)
s’v (kN/m2)
mv (m2/kN)
18.8
-0.2
0.011
25
4.2 x 10-4
50
18.3
-0.5
0.026
25
1.05 x 10-3
100
17.6
-0.7
0.037
50
7.4 x 10-4
200
16.1
-1.5
0.079
100
7.9 x 10-4
400
14.5
-1.6
0.084
200
4.2 x 10-4
S  [4.2 104  (25  21.6)  1.05 103  25  7.4 104  50  7.9 10-4 100
 4.2 10 4  (223.6  200)]  6
 0.922 m
(2) Divide the clay deposit into two layers
5m
Sand (砂土
)
Layer
Depth of
middle of
layer (m)
s’v0
(kN/m2)
s’v1
(kN/m2)
H
(m)
1
1.5
7.2*1.5 =
10.8
20*5+10.2*10
+10.8 = 212.8
3
2
4.5
7.2*4.5 =
32.4
20*5+10.2*10
+32.4 = 234.4
3
15 m
6m
Clay (粘土
)
Dense sand (密砂)
Example (例子)
Pressure, s’v
(kN/m2)
Height, H
(mm)
H (mm)
H/H0
(H0 = 19 mm)
s’v (kN/m2)
mv (m2/kN)
0
19.0
25
18.8
-0.2
0.011
25
4.2 x 10-4
50
18.3
-0.5
0.026
25
1.05 x 10-3
100
17.6
-0.7
0.037
50
7.4 x 10-4
200
16.1
-1.5
0.079
100
7.9 x 10-4
400
14.5
-1.6
0.084
200
4.2 x 10-4
S1  [4.2 104  (25  10.8)  1.05 103  25  7.4 104  50  7.9 10-4 100 4.2 104  (212.8  200)]  3
 0.461m
S2  [1.05103  (50  32.4)  7.4 104  50  7.9 10-4 100 4.2 104  (234.4  200)]  3
 0.448 m
S  S1  S2  0.461 0.447  0.908m
NC soil (正常固结土)
e0
Cc 
e1
e2
e3
 e1  e 0 
logs'1  logs'0
s’1
s’0
Consider a soil layer of thickness H:

e1  e 0 
S
H
1  e0
 s'1 
H

C c  log

1  e0
s
'
0


 s'1 
H

S
Cc  log

1  e0
 s'0 
OC soil (超固结土)
(i) If final stress (s’1)  pre-consolidation pressure (s’c)

e1  e 0 
S
H
1  e0
 s'1 
H

C e  log

1  e0
s
'
0


 s'1 
H

S
Ce  log

1  e0
 s'0 
OC soil (超固结土)
(ii) If final stress (s’1) > pre-consolidation pressure (s’c)
e’
e’’
e
H
S
H 
1  e0
1  e0
e  e' e' '

 s' c 
 s'1 
  C c  log

  C e  log



 s' 0 
 s'c 


 s'c 
 s'1 
  Cc  log

 Ce  log




 s'0 
 s'c 
An oil tank of 10 m diameter which carries an uniform pressure of 100
kN/m2 is founded on a 2 m thick stiff clay deposit which rests on an
incompressible dense sand layer. Calculate the settlement of the oil tank.
The properties of clay are given as follows:
sat = 19 kN/m3, Gs = 2.7, s’c = 76 kPa, Cc = 0.2 and Ce = 0.05
10 m
Oil Tank
100 kN/m2
Clay (粘土
)
Dense sand (密砂)
2m
Consider the clay deposit as single layer and the stresses in the middle of
the clay layer are
Before building the oil tank
s'v0  191  19 kN/m2
G s   sat /  w 2.7  19 / 9.8
e0 

 0.81
 sat /  w  1
19 / 9.8  1
After building the oil tank
p = 100 kN/m2, r0 = 5 m and z = 1 m  Kr = 0.992 [from 土力学 表2-5 p.55]
s' v1  19  100 0.992  118.2 kN/m2
2
s' v0  19 kN/m 2 s' v1  118.2 kN/m
H
S
1  e0
s'c  76 kN/m 2 e 0  0.81

 s'c 
 s' v1 
  Cc  log

 Ce  log




 s' v0 
 s'c 

2
 76 
 118.2 
S
 0.05 log   0.2  log

1  0.81 
 19 
 76 
 0.174m
4.6 Terzaghi’s one-dimensional consolidation theory
(太沙基-单向固结理论)
1 Assumptions (假设)
1)Soil is homogeneous (均质).
2)Soil is fully saturated (饱和).
3)Solid particles and the pore water are incompressible (土粒与水不能压
缩).
4)Flow of water and compression of soil are one-dimensional (单向).
5)Strains are small (小应变).
6)Darcy’s law is valid at all hydraulic gradients.
7）Coefficient of permeability (渗透糸数) [k] and the coefficient of
volume compressibility ( 体 积 压 缩 数 ) [mv] remain constant
throughout the consolidation process.
8）There is a unique relationship between void ratio and effective stress.
Establish of
equations
Find deformation
of T time
Find the time
needed
for a deformation