第六章 土压力 Chapter 6 Earth Pressure
In this lecture, we’ll learn
 Coefficients of earth pressure (土压力系数)
 Relationship between wall movement & earth
pressure (墙体移动与土压力的关系 )
 Earth pressure at-rest (静止土压力 )
 Rankine’s earth pressure theory (朗肯土压力理论 )
6.1 Earth retaining structures (挡土结构物)
Soil nailed wall for a cut slope
(土钉墙)
Bridge abutment
(桥台)
Quay wall for container terminal
(码头的岸墙)
Deep excavation – diaphragm wall
(深基坑 – 地下连续墙)
Deep excavation – sheet-pile wall
(深基坑 – 板式挡土墙)
6.2 Coefficient of Earth Pressure (土压力系数)
horizontal effective stress 'h
K

vertical effective stress
'v
地面
' v    H
竖向有效应力
’v H
'h  K  ' v
M
 is unit weight of soil and
z is depth of soil
’h
土压力
1. Wall Movement (墙体的移动)
Earth Pressure At-Rest (静止土压力)
 when a wall does not move at all, i.e. it is
“at rest” [AA’] (墙体没有位移)
B A C
Backfill
(填土 )
Active Earth Pressure (主动土压力)
 when a wall moves away from backfill
[AA’ to BB’] (墙体离开填土方向)
Passive Earth Pressure (被动土压力)
 when a wall is pushed against backfill
[AA’ to CC’] (墙体向填土方向挤压)
B’ A’ C’
How to evaluate the coefficients of
earth pressure?
怎样计算土压力糸数?
6.3 Earth Pressure at Rest (静止土压力)

Assumptions (假定):
- no ground water table (没有地下水)
- no movement of wall (墙体没有位移)

Lateral earth pressure (土压 力) [´h]
acting on the wall is:
´
'h  K o  ' v  K o    H
where  is unit weight of soil, H is depth
of soil and Ko is coefficient of earth
pressure at rest (静止土压力糸数).
´h
´
v

Jaky (1944) proposed following empirical expression to estimate Ko
for normally consolidated soil (估计正常固结土的Ko的经验公式):
K o  1  sin '
where ´ is effective angle of internal friction (有效內摩擦角)

Lateral earth pressure [´h]
increases linearly with depth
and total lateral force (总合力)
[Po] exerted on the wall is:
1
Po  K o    H 2
2
A
H
Po
H
3
B
  H  Ko
6.4. Rankine’s Earth Pressure Theory (朗肯土压力理论)
Following assumptions (假定) are made to develop Rankine’s
earth pressure theory.
 Wall is rigid, vertical, frictionless and extends to an infinite
depth (墙背竖直和光滑)
 Backfill is a dry, homogeneous, isotropic and cohesionless
soil (填土是无粘性土,干,均质和各向同性)
 Soil surface is horizontal (水平表面) and there are no shear
stresses on horizontal and vertical planes, i.e. horizontal
and vertical stresses are principal stresses (水平和竖直方
向为主应力方向)
1 .Active Failure (主动破坏)




In figure on the right, red circle
represents initial stress state or Ko
condition of element A (右图的红圆
代表土单元A的初始或静止状态)
Wall is moved away from element A
to B (墙体从土单元A方向离开)
库伦强度包线
´h will reduce but ´v will remain
constant (´h减少,但 ´v 不变)
Mohr’s circle will expand until
touching Coulomb failure surface
[represented by blue circle]. This is
called active failure. (莫尔圆增大与
库伦强度包线相切时为主动破坏,
见右图的蓝圆)
´
´vf
´hf

Major principal plane is horizontal surface (水平面为最大主应力面)

Slip planes for active failure are inclined at 45+´/2 with respect to
horizontal (主动破坏面与水平面的夹角为45+´/2)
´
´
A' A
C
´
´
H
´vf
a  45 
B
´
'
2
Mohr-Coulomb failure criteria [莫尔-库伦破坏准则]

’f
’
’3f
f
’
c’
’3f
f
’f
'1f '3f
2
’1f
f
'1f '3f
2
’1f
f
’3f
’1f
’
f  45 
' 
' 


' 3f  '1f  tan 2  45    2c' tan 45  
2
2


'
2

Rearranging the equation in the previous slide, following expression for
active earth pressure coefficient (主动土压力糸数) [Ka] is obtained:
' 
' 


' 3f  '1f  tan 2  45    2c' tan 45  
2
2


´3f = ´hf & ´1f =
´vf
Ka 

c´ = 0
'hf
' 

 tan 2  45   
' vf
2

A' A
H
Pa
H
3
B
  H  Ka
Lateral earth pressure [’h] increases linearly with depth and total active
force (总主动土压力) [Pa] exerted on the wall is:
1
Pa  K a    H 2
2
2 。Passive Failure (被动破坏)


In figure on the right, red circle
represents initial stress state or Ko
condition of element B (右图的红圆
代表土单元B的初始或静止状态)
Wall is is pushed against element B (
墙体向土单元B方向挤压)

´h will increase but ´v will remain
constant (´h增加,但 ´v 不变)

Mohr’s circle will first contract and
then expand until touching Coulomb
failure surface [represented by blue
circle]. This is called passive failure.
(莫尔圆先减小後增大,与库伦强度
包线相切时为被动破坏,见右图的
蓝圆)
´hf
´h
´vf

Major principal plane is vertical surface (竖直面为最大主应力面)

Slip planes for passive failure are inclined at 45-´/2 with respect to
horizontal (被动破坏面与水平面的夹角为45-´/2)
´
´
A
A'
´
´
H
´vf
 p  45 
B
´
'
2
Mohr-Coulomb failure criteria [莫尔-库伦破坏准则]

’f
’
’3f
f
’
c’
’3f
f
’f
'1f '3f
2
’1f
f
'1f '3f
2
’1f
f
’3f
’1f
’
f  45 
' 
' 


'1f  ' 3f  tan 2  45    2c' tan 45  
2
2


'
2

Rearranging the equation in the previous slide, following expression for
passive earth pressure coefficient (被动土压力糸数) [Kp] is obtained:
' 
' 


'1f  ' 3f  tan 2  45    2c' tan 45  
2
2


´3f = ´vf & ´1f =
´hf
Kp 

c´ = 0
'hf
'  1

 tan 2  45    
' vf
2  Ka

A
A'
H
Pp
H
3
B
  H  Kp
Lateral earth pressure [’h] increases linearly with depth and total
passive force (总被动土压力) [Pp] exerted on the wall is:
Pp 
1
Kp    H2
2
3。 Relationship between wall movement & earth pressure
(墙体移动与土压力的关系 )

The lateral earth pressure exerted on the wall depends on the wall
movement (土压力的大小与墙体位移有关)

Only small movement of wall away from soil is required to
mobilise full active earth pressure in soil mass but substantial wall
movement towards soil is required to mobilise full passive earth
pressure in soil mass (相对於主动破坏,墙体需要较大位移才能
达到被动破坏)
Wall movement (墙体移动)
What will be the earth pressures if
ground water table exists behind the
wall and the backfill is a cohesive soil?
当墙背有地下水和填土是粘性土时,怎
样计算土压力糸数?
4 。 Effect of ground water (地下水作用)
 If groundwater is present, pore water pressure must be considered
in addition to lateral earth pressure.
 DO NOT multiply pore water pressure with coefficient of earth
pressure.
GWL
H1
H
H2
 If groundwater table is located at a height H2 from bottom of wall,
hydrostatic pore water pressure at base of wall is wH2.
 Total imposed lateral force (Pw) due to pore water pressure is
½wH22.
 Below groundwater table, submerged unit weight (’) is used to
evaluate lateral earth pressure.
H1
GWL
H
Pa
H2
Pw
H2
3
  H1  K a  'H 2  K a
 w  H2
5 。Effect of Cohesive backfill 粘性填土作用
 If backfill of retaining wall is cohesive soil, cohesion (c) is not
zero.
 At a depth z from top of wall, active (pa) and passive (pp) lateral
pressure are as follows:
p a    z  K a  2c  K a
p p    z  K p  2c  K p
Zo
H
Pp
Pa
1
H  Zo 
3
2c  K a
  H  Ka
2c  K p
  H  Kp
 For active failure, a tensile zone (拉力区) of depth zo exists from
top of wall.
Zo
zo 
H
Pa
1
H  Zo 
3
2c  K a
  H  Ka
2c
  Ka
Questions: What will be the earth pressures if a
wall carries uniform surcharge and the backfill is
not homogeneous?当墙体承受均布荷载和填土
是不均质土 时 , 怎 样 计 算 土 压 力 糸 数 ?
Effect of surcharge and non-homogeneous backfill
q
1 , '1
H1
 2 , '2
H2
H
H
均布荷载
不均质填土
Example 1 (例题一)
Calculate (i) the total active force on a vertical wall 5m high retaining a sand of
unit weight 17 kN/m3 for which ´=35°: the surface of the sand is horizontal
and the water table is below the bottom of the wall, (ii) the the total force on the
wall if the water table rises to a level 2 m below the surface of the sand. The
saturated unit weight of the sand is 17 kN/m3.
一个高 5m 的垂直挡土墙, 墙后填土是砂土: 重度 () =17 kN/m3, 有效内摩
擦角 (´) = 35,填土表面是水平面. 计算墙背上的 (i) 总主动土压力 (ii)总合
力当地下水位低于填土表面 2m. 砂土的饱和重度 (sat) =20 kN/m3
 = 17 kN/m3
5m
´= 35
sat = 20 kN/m3
Example 1 (例题一)
(i) No water table
5m
 = 17 kN/m3
´= 35
sat =20 kN/m3
' 
35 


K a  tan 2  45     tan 2  45  
2
2


 0.27
1
1
K a    H 2   0.27 17  5 2
2
2
 57 .4 kN /m
Pa 
Example 1 (例题一)
(ii) Water table below 2 m of ground surface
2m
5m
'ha2m   K a    H  0.27 17  2  9.2 kN /m 2
'ha5m   0.27  17  2  10.2  3  17.4 kN /m 2
Pa 
 = 17 kN/m3
´= 35
sat= 20 kN/m3
´= 10.2 kN/m3
1
1
 9.2  2  9.2  17 .4  3  49 .1 kN /m
2
2
 w 2 m   0
 w 5m   9.8  3  29 .4 kN /m 2
1
Pw   29 .4  3  44 .1 kN /m
2
P  Pa  Pw  49.1  44.1  93.2 kN /m
Example 2 (例题二)
The soil conditions adjacent to a sheet pile wall are given in the following figure.
Calculate the total active force behind the wall and the total passive force in
front of the wall.
计算板桩墙上总主动土压力和总被动土压力.
 = 18 kN/m3
5m
´= 38
sat = 20 kN/m3
5m
c´= 10 kN/m2
´= 28
A
Example 2 (例题二)
Assume the wall rotates at its bottom (i.e. point A). Thus, active earth pressure
acts behind the wall and passive pressure acts in front of the wall.
假没板桩墙上端向外移动但下端不动,墙背产生主动破坏,墙前产生被动破
坏.
5m
Pa
5m
Pp
A
Example 2 (例题二)
(i) Behind the wall (墙背)
5m
Soil 1
Pa
Pp
5m
Soil 2
' 
38 


K a (soil1)  tan 2  45     tan 2  45    0.24
2
2


' 
28 


K a (soil2)  tan 2  45     tan 2  45    0.36
2
2 


'haSoil15m   0.24 18  5  21 .6 kN /m 2
'haSoil25m   0.36 18  5  32.4 kN /m 2
'haSoil210m   0.36  18  5  10 .2  5  50.8 kN /m 2
1
1
Pa   21 .6  5  32 .4  50 .8  5  262 kN /m
2
2
Example 2 (例题二)
(ii) In front of the wall the wall (墙前)
5m
Pa
Pp
5m
Soil 2
' 
28 


K p  tan 2  45     tan 2  45    2.77
2
2 


'hpSoil25m   2 10  2.77  33.3 kN /m2
'hpSoil210m   2.77 10.2  5  2 10  2.77  174.6 kN /m2
1
Pp  33 .3  174 .6  5  519 .8 kN /m
2
Example 2 (例题二)
The soil conditions adjacent to a sheet pile wall are given in the following figure.
Calculate the total active force behind the wall and the total passive force in
front of the wall.
计算板桩墙上总主动土压力和总被动土压力.
 = 18 kN/m3
5m
´= 38
sat = 20 kN/m3
5m
c´= 10 kN/m2
´= 28
A
Example 2 (例题二)
Assume the wall rotates at its bottom (i.e. point A). Thus, active earth pressure
acts behind the wall and passive pressure acts in front of the wall.
假没板桩墙上端向外移动但下端不动,墙背产生主动破坏,墙前产生被动破
坏.
5m
Pa
Pp
5m
Example 2 (例题二)
(i) Behind the wall (墙背)
5m
Soil 1
Pa
Pp
5m
Soil 2
' 
38 


K a (soil1)  tan 2  45     tan 2  45    0.24
2
2


' 
28 


K a (soil2)  tan 2  45     tan 2  45    0.36
2
2 


'haSoil15m   0.24 18  5  21 .6 kN /m 2
'haSoil25m   0.36 18  5  32.4 kN /m 2
'haSoil210m   0.36  18  5  10 .2  5  50.8 kN /m 2
1
1
Pa   21 .6  5  32 .4  50 .8  5  262 kN /m
2
2
Example 2 (例题二)
(ii) In front of the wall the wall (墙前)
5m
Pa
Pp
5m
Soil 2
' 
28 


K p  tan 2  45     tan 2  45    2.77
2
2 


'hpSoil25m   2 10  2.77  33.3 kN /m2
'hpSoil210m   2.77 10.2  5  2 10  2.77  174.6 kN /m2
1
Pp  33 .3  174 .6  5  519 .8 kN /m
2
6.5 Coulomb’s earth pressure theory
(库伦土压力理论 )

Coulomb (1776) proposed that a condition of limit
equilibrium exists in a soil wedge between a retaining wall
and a trial slip plane. (库伦土压力理论假设一个滑动面,整
个滑动土体处于极限平衡状态)

The force between the wedge and the wall is determined by
considering the equilibrium of forces acting on the wedge. (
利用整个滑动土体上力的平衡条件来确定土压力)

Among the trial slip planes, the critical slip plane is the one
which gives the maximum lateral pressure on the wall. (在假
定滑动面中,临界滑动面产生最大的土压力)
1。 Poncelet’s Assumptions 假设

Poncelet (1840) used Coulomb’s limit
equilibrium approach to obtain the active and
passive earth pressure coefficients for cases
where

backfill is dry, homogenous and
cohesionless soil with an angle of internal
friction  (填土是干,均质和无粘性土)

backfill is sloping at an angle  to the
horizontal (填土表面与水平面夹角为)

wall friction o is present (墙背与填土之
间的摩擦角为o )

wall face inclined at an angle  to the
vertical (墙背面与竖直线的夹角为).


H
2。 Active failure (主动破坏 )

For active failure, the wall moves away from the soil mass (墙体离开
填土方向,产生主动破坏 )

Soil wedge ABC will slide down (土体ABC向下滑动)
C


B


W
H

o
R
Pa


A
3.Active failure (主动破坏 )

The forces acting on soil wedge ABC is under equilibrium: its weight (W),
the reactions on the slip plane AC (R) and the wall AB (Pa) (土体ABC的
重量W,滑动面AC上的反力R与墙背AB上的反力Pa达至静力平衡)

Pa and R act below the normal of AB and BC, respectively. (Pa与R的方向
向上)
C


B


W
H

o
R
Pa


A
Consider the force triangle on the right hand side and from
sine rule (力矢三角形如右图,通过正弦定律可得):
Pa
W

sin    sin(90    o    )
90°-o-
Pa
W
The weight of soil wedge ABC (土体ABC的重量)
W     ABC
1
2 sin(90    )  sin(90    )
   AB 
2
sin(  )
Express Pa as a function of 
1
cos(  )  cos(  )  sin(  )
2
Pa    H 
2
cos2   sin(  )  cos(     o  )
R
-
Differentiate Pa w.r.t.  and its maximum value corresponds to
(将Pa 对求导数,并令其等於零 )
1
Pa (max.)    H 2 
2
Pa
0

cos2 (  )

sin(   o )  sin(  ) 
2
cos   cos(   o )  1 

cos(



)

cos(



)


o
1
Pa (max .)    H 2  K a
2
where
Ka 
cos 2 (  )

sin(    o )  sin(   ) 
cos   cos(   o )  1 

cos(



)

cos(



)


o
2
2
2
Ka 
cos 2 (  )

sin(    o )  sin(   ) 
cos   cos(   o )  1 

cos(



)

cos(



)


o
2
2
If the wall is vertical ( = 0), there is no friction between the wall and the
soil (o = 0) and the surface of the backfill is horizontal ( = 0), the
expression of Ka becomes
Ka 
cos2 
1  sin 
2

1  sin 2 
1  sin 2
1  sin 

2

 tan  45  
1  sin 
2


For passive failure, the wall is pushed against the soil mass (墙体向填
土方向挤压, 产生被动破坏 )

Soil wedge ABC will slide up (土体ABC向上滑动)

Pp and R act above the normal of AB and BC, respectively.
C


B


H
W
R

Pp
o


A
4. Passive failure (被动破坏 )
Consider the force triangle on the right hand side and from sine rule
(力矢三角形如右图,通过正弦定律可得):
90°+o-
Pp
sin   

W
sin(90    o    )
Pp
W
Weight of soil wedge ABC
W
1
sin(90    )  sin(90    )
  H2 
2
cos2   sin(  )
Express Pp as function of 
1
cos(  )  cos(  )  sin(  )
2
Pp    H 
2
cos2   sin(  )  cos(     o  )
R
+
Differentiate Pp w.r.t.  and its maximum value corresponds to
(将Pp 对求导数,并令其等於零 )
1
Pp (max.)    H 2 
2
Pp (max .) 
where
Pp

cos2 (  )

sin(   o )  sin(  ) 
cos   cos(   o )  1 

cos(



)

cos(



)


o
2
1
  H2  Kp
2
Kp 
0
cos2 (  )

sin(   o )  sin(  ) 
2
cos   cos(   o )  1 

cos(



)

cos(



)


o
2
2
Kp 
cos2 (  )

sin(   o )  sin(  ) 
2
cos   cos(   o )  1 

cos(



)

cos(



)


o
2
If the wall is vertical ( = 0), there is no friction between the wall and
the soil (o = 0) and the surface of the backfill is horizontal ( = 0), the
expression of Kp becomes
Kp 
cos2 
1  sin 
2

1  sin 2 
1  sin 2
1  sin 

2

 tan  45  
1  sin 
2

How to evaluate the earth pressures from
Coulomb’s method if the soil surface is uneven, the
wall carries surcharge, ground water table exists
behind the wall, the backfill is a cohesive soil?
怎样利用库伦理论计算土压力当填土表面不是
平面,有荷载,填土是粘性土与墙後有地下水?
5. Comparisons between Rankine’s and Coulomb’s Theories
(朗肯与库伦理论的比较)
1. Rankine’s theory is based on the stress equilibrium of a
soil element. (朗肯理论基于土单元体的应力平衡条
件来建立)
2. Rankine assumed that the wall is vertical, frictionless
and the soil surface is horizontal. (朗肯理论假定墙背
面竖直与光滑,填土面为水平)
3. In general, Rankine’s theory over-estimates the active
pressure but under-estimates the passive pressure. (朗
肯理论高估主动土压力与低估被动土压力)
4. Coulomb’s theory is based on limiting equilibrium of
forces within a soil wedge for a given slip surface. (
库伦理论基于滑动块体的静力平衡条件来建立)
5. Coulomb assumed the failure surface is a plane. (库
伦理论假定滑动面为平面)
6. In general, Coulomb’s theory under-estimates the
active pressure and over-estimates the passive
pressure. (库伦理论低估主动土压力与高估被动土
压力)