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

) increasing the density (

增加密度

)

process of 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 grained soils ( 细粒土)

faster rate

than fine 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

Loading ( 荷载 ) at time = T ) 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 v y 简单化 v x v z v z (water flow)

( 竖向荷载 ) Vertical settlement ( 竖向沉降 ) 土样本 Oedometer ( 固结仪 ) 侧向应变为零 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 )  h f s ’ f

2. 1-D compression curve (

单向压缩曲线

)

Unloading and Reloading ( 卸载与再加载 ) Re-compression curve ( 回弹曲线) ) Loading ( 加载 Compression curve ( 压缩曲线)   v   h H 0 where  v is change in vertical strain ( 向应变增量 ), 竖  h is the change in height for each loading increment ( 每次荷载增 量 的 高 度 改 变 ) and H 0 is the initial specimen height ( 样本初始高度)

(1)Coefficient of volume compressibility [

体积压缩系数

] (m v )

No lateral strain (侧向应变为零)   v   h H 0   e 1  e 0 m v    vol  s ' v   1 1  e 0  e 1 s ' v 1  e 0  s ' v 0 e 0 e 1 where  vol 应变增量 ), is volumetric strain ( 体积 s ’ v is change in vertical effective stress ( 应力增量 ) and e 0 initial void ratio ( 初始孔隙比 ) is s ’ v0 s ’ v1

Example (

例子

)

m v   1 1  e 0  e 1 s ' v 1  e 0  s ' v 0 m v   1 1  1 .

2  0 .

8 200  1 .

2  100  1 .

82  10  3 m 2 / kN

How about

a 100-200

(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:

E s

 s  '

z z

 1 

e

0

a v

 1

m v

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: E  1  z  s z    s x    s y   s z  z  1  2  2 1    E s  1  2  2 1   soil is not ideal elastic material,above equation is only an approximate formula.

(3) e——log s ’ curve e 0 e 1 e 2 e 3 s ’ v0 s ’ v1 1). Compression Index [ 压缩指数 ] (C c ) C c  log   e 1 s ' v 1   e o log  s ' vo 2). Expansion Index [ 回弹指数 ] (C e ) C e  log   e 3 s ' v 1   e 2 log  s ' vo

4.3 Calculation formulae of soil compression with zero lateral strain

土的压缩量计算



H



e

1 1  

e e

1 2

H

1  

e

1 

e

1

H

1 

H

 1

a v



e

1

pH

1 

H



m v pH

1 

H



p E s H

1

4.4 Settlement Calculation

沉降计算

Settlement found in soil due to loading may be divided into three categories: S t  S i  S c  S s where S t S c is total settlement ( settlement ( 瞬时沉降 is primary consolidation ( 有效应力改变 ] 沉降 ) – [time, and 时间 ] S s 总沉降 ), 主固结沉降 S i is immediate ) – [elastic deformation, 弹性变形 ] , ) – [change of is secondary consolidation ( s ’, 次固结

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, S i = 0 Similar to rubber ( 橡皮 )

 ) Based on elastic theory ( 弹性理论 S i  1   2 E  b  p  I 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

)

.

the It is calculated from following two methods: 1.

2.

e : s ’ curve e : log s ’ curve  h f 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 e 1i \e 2i 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 ), stress ( 现今有效应力 ) and s ’ vmax s ’ v is current effective 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/m 3 and unit weight of clay (  clay ) is 17 kN/m 3 . 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 ) Clay ( 粘土 ) 10 m 6 m Dense sand ( 密砂 ) ) Sand ( Clay ( 粘土 ) 砂土 15 m 6 m Dense sand ( 密砂 ) Pressure (kN/m 2 ) 0 25 50 100 200 400 Height (mm) 19.0

18.8

18.3

17.6

16.1

14.5

After reclamation ( 填 海后 )

(1) Consider clay deposit as a single layer and in the middle of clay layer: 5m ) Sand ( 砂土 15 m Clay ( 粘土 ) 6 m Dense sand ( 密砂 ) Before reclamation ( 填 海前 ) s ' v 0  ( 17  9 .

8 )  3  21 .

6 kN/ m 2 After reclamation ( 填 海后 ) s ' v 1  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

m v   1 1  e 0  s ' e 1 v 1  e  s ' 0 v 0 Pressure, s ’ v (kN/m 2 ) 0 25 50 100 200 400 Height, H (mm) 19.0

18.8

18.3

17.6

16.1

14.5

&  H H 0   e 1  e 0  H (mm) -0.2

-0.5

-0.7

-1.5

-1.6

 H/H 0 (H 0 = 19 mm) 0.011

0.026

0.037

0.079

0.084

m v s ’ v (kN/m 2 ) 25 25 50 100 200   1 H 0   H  s ' v m v (m 2 /kN) 4.2 x 10 -4 1.05 x 10 -3 7.4 x 10 -4 7.9 x 10 -4 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 15 m 6 m Clay ( 粘土 ) Dense sand ( 密砂 ) Layer 1 Depth of middle of layer (m) 1.5

2 4.5

s ’ v0 (kN/m 2 ) s ’ v1 (kN/m 2 ) 7.2*1.5 = 10.8

7.2*4.5 = 32.4

20*5+10.2*10 +10.8 = 212.8

20*5+10.2*10 +32.4 = 234.4

H (m) 3 3

Example ( 例子 ) Pressure, s ’ v (kN/m 2 ) Height, H (mm)  H (mm)  H/H 0 (H 0 = 19 mm) s ’ v (kN/m 2 ) m v (m 2 /kN) 0 25 50 100 200 400 19.0

18.8

18.3

17.6

16.1

14.5

-0.2

-0.5

-0.7

-1.5

-1.6

0.011

0.026

0.037

0.079

0.084

25 25 50 100 200 4.2 x 10 -4 1.05 x 10 -3 7.4 x 10 -4 7.9 x 10 -4 4.2 x 10 -4 S 1   [ 4 .

2 0 .

 461 10 m  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 S 2   [ 1 .

05  10  3 0 .

448 m  ( 50  32 .

4 )  7 .

4  10  4  50  7.9

 10 -4  100  4 .

2  10  4  ( 234 .

4  200 )]  3 S  S 1  S 2  0 .

461  0 .

447  0.908

m

NC soil ( 正常固结土 ) e 0 e 1 e 2 e 3 s ’ 0 Consider a soil layer of thickness H: S    e 1 1   e 0 e 0  H  H 1  e 0 C c  log   s ' 1 s ' 0   s ’ 1 C c  log   e 1 s ' 1   e 0 log  s ' 0 S  1  H e 0 C c  log   s ' 1 s ' 0  

OC soil ( 超固结土 ) (i) If final stress ( s ’ 1 )  pre-consolidation pressure ( s ’ c ) S    e 1 1   e 0 e 0  H  1  H e 0 C e  log   s ' 1 s ' 0   S  1  H e 0 C e  log   s ' 1 s ' 0  

OC soil ( 超固结土 ) (ii) If final stress ( s ’ 1 ) > pre-consolidation pressure ( s ’ c ) 

e’



e’’

 e   e '   e ''      C e  log   s ' c s ' 0    C c  log   s ' 1 s ' c      S    1  e e 0 H  H 1  e 0     C e  log   s ' c s ' 0    C c  log   s ' 1 s ' c     

An oil tank of 10 m diameter which carries an uniform pressure of 100 kN/m 2 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/m 3 , G s = 2.7, s ’ c = 76 kPa, C c = 0.2 and C e = 0.05

10 m Oil Tank 100 kN/m 2 2 m Clay ( 粘土 ) Dense sand ( 密砂 )

Consider the clay deposit as single layer and the stresses in the middle of the clay layer are Before building the oil tank s ' v 0  19  1  19 kN/ m 2 e 0  G  s  sat /   sat w /   1 w  2 .

7  19 / 9 .

8 19 / 9 .

8  1  0 .

81 After building the oil tank p = 100 kN/m 2 , r 0 = 5 m and z = 1 m  K r = 0.992 [from 土力学 表 2-5 p.55] s ' v 1  19  100  0 .

992  118 .

2 kN/ m 2

s ' v 0  19 kN/ m 2 s ' v 1  118 .

2 kN/ m 2 s ' c  76 kN/ m 2 e 0  0 .

81 S  H 1  e 0     C e  log   s ' c s ' v 0    C c  log   s ' v 1 s ' c      S  1  2 0 .

81  0.174

m    0 .

05   log  76 19  0 .

2   log  118 .

2 76  

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 ) [m v ] 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