Transcript Document

§6 Earth Pressure and Retaining Wall
土压力理论与挡土墙设计
•
Introduction/引言
• Rankine’s earth pressure theory / 朗肯土压力理论
• Coulomb’s earth pressure theory / 库仑土压力理论
• Discuss / 讨论
• Retaining wall design / 挡土墙设计
6.1 Introduction/引言
 x  k0   z
k0 

1 
-Coefficient of earth pressure at rest 静止土压力
k0  1  sin  '
E0 
主动土压力
1
2
Bishop , 1958
  H 2  K0
被动土压力
6.2 Rankine Earth Pressure theory /朗肯土压力理论
6.2.1 Limit state/极限状态
-Active limit equilibrium/主动极限平衡
-Passive limit equilibrium/被动极限平衡
 x  K0  z
6.2.2 Rankine active earth pressure/主动土压力
•In Non-Cohesive soils (soil surface horizontal)/地表水平无粘性土
  
 3   1  tg  45  
2

2
where
1    H
  45 

2
  
E  H  tg  45  
2
2

1
Ea 
2
1
2
where
2
  H 2  Ka
  
K a  tg  45   , active
2

2
pressure
coefficient
•In Cohesive soils (soil surface horizontal)/地表水平粘性土

 3   1  tg 2  45  


  
  2c  tg  45  
2
2

  45 


2
Z0 
  
tg  45  , vertical

2

2c
wall
depth
Ea 
1
2
E 
 H  Z 0 
2
1
2

   1
    2c
2
2

 tg  45    H  tg  45    2cHtg  45   
2 2
2
2




2
H  K  2cH  K  
2
where
2c
2


2

K a  tg  45  
2

•For sloping soil surface(β) /地表倾斜
Ea 
1
2
cos   cos   cos 
2
  H  cos 
2
2
cos   cos   cos 
2
2
2
6.2.3 Rankine passive earth pressure/被动土压力
•In Non-Cohesive soils (soil surface horizontal) /地表水平无粘性土
  
 1   3  tg  45     3  K p
2

2
  
K p  tg  45  , passive
2

2
where
Ep 
1
2

H 2  tg 2  45  


1
2
  H  K p
2 2
pressure
coefficient
•In Cohesive soils (soil surface horizontal)/ /地表水平粘性土

 1   3  tg 2  45 
 3  H
where
Ep 
1
2


  

2
c

tg

 45     3  K p  2c  K p
2
2


2

K p  tg  45  , passive
2


H 2  tg 2  45  


pressure
coefficient
   1
2
  2c  H  tg  45    H  K p  2cH 
2
2 2

Kp
6.3 Coulomb’s Earth Pressure theory / 库仑土压力理论
6.3.1Basic concept
-Assuming
-Active earth pressure
E  G
dEa
d

sin 90    


sin        
Ea  f ( )
0
-Passive earth pressure
Ep  G

sin 90    


sin        
E p  f ( )
6.3.2 Active Case/主动情形
• General solution
AB 
H
AD  AB  sin    
cos 

BC  AB 
sin 90

   i
sin 90    i

then, G 
1
H 
2


sin      sin 90    i
 E 
2

sin      sin 90    i
dEa
d
1
2

cos   cos  i 
2
2
Let ,
 Ea 
H 
2

cos   cos  i 
2
1



sin 90    


sin        
0
cos    
2
 H 
2

cos   cos   1 

2
sin    sin   i  

cos    cos  i  
2
cos    
2
where,  
1
Ea 
2

2
cos   cos   1 

sin    sin   i  

cos    cos  i  
2
, active
pressure
H 2  a
If i=0,α=0,δ=0,
Then
Ea 
1
H 
2
Ea 
cos 
2
1
2
where
2
1  sin  2

1
 H 
2
2
  H 2  a

a  tg 2  45 



2
1  sin 
1  sin 

1
2

  H tg  45 
2
2




2
coefficient
• Complex boundary conditions/复杂边界条件
S
1
2
1
1
2
2
H  a 2  tg  H  a 2  tg 
ab 
1
2
a  tg  A0  tg  B0
2
G    A0 tg  B0 
where, A0 
1
B0 
1
Let ,
dEa
d
2
2
( H  a )  tg 
2
2
 Ea  G 
where,
( H  a)
1
a  tg 
2
2

sin 90    

1
ab
2

sin        
   A0tg  B0  
cos   
sin    
     
0

A0 cos    
 sin      sin      cos     cos   
then,   A0tg  B0  

0
2
2




sin



sin




cos



B0
tg   2tg  tg  ctg  tg 
2
A0
ctg  tg   0

B0 


 tg  tg  tg  ctg  tg 
A0 

where,
     
-Coefficient of active earth pressure /主动土压力系数λa
a 
Ea  
a
E a    A0  tg  B0 
h
A0tg  B0
tg  tg
 tg  tg  
 a  tg  tg  
where,
cos   
sin    
cos   
sin    
     

cos   
sin    
A0tg  B0
tg  tg
 a
-Distribution of active earth pressure /土压力的分布
-Example
A retaining wall of embankment is 6m high. And the relative
parameters are c=0,φ=30。,α=14 。 00’,δ=φ/3, as well as shown in
Fig. below. Calculate the active earth pressure and draw the
distribution.
-Solution
①Assuming sliding surface is BC
A0 
B0 
1
2
1
2
1
a  H  2h0 a  H   3  6  2  3.4  3  6  71.1
2
ab  b  k h0 
1
2
H H  2a  2h0 tg 
1
2
        30  10  14  26

tg  tg 
 0.488 


1
2
tg 26

0.188  1.7320.488  0.580  0.488  1.540

66  2  3  2  3.4 tg14  41.212



B0 
tg  ctg  tg    tg 26 
A0 

 1.052   90 , Sign"" is
  46.45

3  4.5  4.5  1.53.4 
chosen

41.212 
 

 ctg 30  tg 26 

71.1 


②Verifying sliding surface location
B ' C  H  a tg  9  10.52  9.468m
B ' E  H  tg  b  K  6  0.249  4.5  1.5  7.496m
B ' F  B ' E  MN  7.496  3.5  10.996m
B' E  B' C  B' F ,
So, The Sliding surface agrees with the assumption.
③Calculate the Coefficient of active earth pressure λa
a  tg  tg  
cos   
sin    

 tg 46.45


 tg14 
sin 46.45
  0.197
 26 
cos 46.45  30





④Calculate the active earth pressure Ea and its distribution
Ea  
A0tg  B0
tg  tg
71.1 tg 46.45  41.212

 a  18 
tg 46.45  tg14


 0.197  148.36 KN / m
 0    h0  a  18  3.4  0.197  12.06 KN / m 2
 a    a  a  18  3.0  0.197  10.64 KN / m 2
 H    H  a  18  6.0  0.197  21.28 KN / m 2
⑤ Draw the distribution of the active earth pressure
6.3.3 Passive Case
Ep 
1
2
H
cos   
2

cos   cos   1 

2
p 
sin      sin   i  

cos    cosi    
cos   

2
cos   cos   1 

sin    sin   i  

cos    cosi    
2
2

1
2
H 2   p
If i=0,α=0,δ=0,
Then
Ep 
or
1
cos 
2
H 
2
2
Ep 
1  sin  2
1
2
  
 H 
 H  tg  45  
2
1  sin 
2
2

1
2
1  sin 
1
H 2   p
6.4 Discuss
6.4.1 Equivalent internal friction angle
tg D  tg 

c
H
 D  arctg  tg 

c 

H 
2
2
  
    2c
E  H tg  45    2cHtg  45   
2
2
2



1
2
E 
'
Let
1
2
2

H 2  th2  45 

D 

2 
  D 
    2c
'
Ea  Ea , tg  45 
  tg  45   
2 
2  H


D 

     2c 
 2arctg tg 45   

2
2  H 
 
2
6.4.2 Several-layer soils
6.4.3 wall back in polygonal line
6.4.4 graphical method
6.4.5 Earth pressure of abutments
E 
1
2
H 2   B  h0 H    l 0