Course
Year
: S0705 – Soil Mechanic
: 2008
TOPIC 6
LATERAL EARTH PRESSURE
CONTENT
•
•
RETAINING EARTH WALL (SESSION 21-22 : F2F)
RANKINE METHOD (SESSION 21-22 : F2F)
– ACTIVE LATERAL PRESSURE
– PASSIVE LATERAL PRESSURE
•
COULOMB METHOD (SESSION 23-24 : OFC)
– ACTIVE LATERAL PRESSURE
– PASSIVE LATERAL PRESSURE
•
•
Bina Nusantara
LATERAL PRESSURE DUE TO EXTERNAL LOAD
(SESSION 23-24 : OFC)
DYNAMIC EARTH PRESSURE (SESSION 23-24 : OFC)
SESSION 21-22
RETAINING EARTH WALL
RANKINE METHOD
Bina Nusantara
RETAINING EARTH WALL
Defined as a wall that is built to resist the lateral pressure
of soil – especially a wall built to prevent the advance of a
mass of earth/soil
Bina Nusantara
RETAINING EARTH WALL
Bina Nusantara
RETAINING EARTH WALL
Bina Nusantara
RETAINING EARTH WALL
Bina Nusantara
RETAINING EARTH WALL
Bina Nusantara
RETAINING EARTH WALL
Bina Nusantara
RETAINING EARTH WALL
Bina Nusantara
EARTH LATERAL PRESSURE
• Defined as soil stress/pressure at horizontal direction
and a function of vertical stress
• Cause by self weight of soil and or external load
• 3 conditions :
– Lateral Pressure at Rest
– Active Lateral Pressure
– Passive Lateral Pressure
Bina Nusantara
EARTH LATERAL PRESSURE
Lateral Pressure at rest
Case I
Bina Nusantara
EARTH LATERAL PRESSURE
Active Lateral Pressure
Case II
Bina Nusantara
EARTH LATERAL PRESSURE
Passive Lateral Pressure
Case III
Bina Nusantara
EARTH LATERAL PRESSURE
q
Jaky, Broker and Ireland  Ko = M – sin ’
Sand, Normally consolidated clay  M = 1
Clay with OCR > 2  M = 0.95
v =  . z + q
z
Broker and Ireland
Ko = 0.40 + 0.007 PI , 0  PI  40
Ko = 0.64 + 0.001 PI , 40  PI  80
v
h
h
K
v
At rest, K = Ko
Bina Nusantara
Sherif and Ishibashi  Ko =  +  (OCR – 1)
 = 0.54 + 0.00444 (LL – 20)
 = 0.09 + 0.00111 (LL – 20)
LL > 110%   = 1.0 ;  = 0.19
RANKINE METHOD
ACTIVE LATERAL PRESSURE
a = v . tan2(45-/2) – 2c . tan (45-/2)
a = v . Ka – 2cKa
Bina Nusantara
Ka = tan2 (45 - /2)
1 = 3 . tan2 (45+/2)+2c.tan (45+/2)
RANKINE METHOD
PASSIVE LATERAL PRESSURE
Bina Nusantara
RANKINE METHOD
PASSIVE LATERAL PRESSURE
p= v . tan2(45+/2) + 2c . tan (45+/2)
Bina Nusantara
RANKINE METHOD
PASSIVE LATERAL PRESSURE
Kp = tan2 (45 + /2)
h = v . Kp + 2cKp
Bina Nusantara
EXAMPLE
q = 20 kN/m2
h1 = 2 m
Sheet
Pile
1 = 15 kN/m3
1 = 10 o
c1 = 0 kN/m2
2 = 15
h3 = 4 m
kN/m3
h2 = 8 m
2 = 15 o
c2 = 0 kN/m2
Questions:
1. Determine the active and passive lateral pressure of sheet pile structure
2. Determine the total lateral pressure
Bina Nusantara
SOLUTION
Coefficient of Lateral Pressure :
Active ; ka =
tan2(45-1/2)
= 0.704
q = 20 kN/m2
2m
Pa1
Passive ; kp = tan2(45+2/2) = 1.698
8m
4m
Pw2
Pp1
Pa2
Pq1
Pw1
Active Lateral Pressure
Pa1 = ka . 1 . h1 – 2 . c . ka = 0.704 . 15 . 2 – 2 . 0 . 0.704 = 21.12 kN/m2
Bina Nusantara
Pa2 = ka . (1 . h1 + 1’ . h2) – 2 . c . ka = 49.28 kN/m2
SOLUTION
Coefficient of Lateral Pressure :
Active ; ka =
tan2(45-1/2)
= 0.704
q = 20 kN/m2
2m
Pa1
Passive ; kp = tan2(45+2/2) = 1.698
8m
4m
Pw2
Pp1
Pa2
Active Lateral Pressure
Pq1 = ka . q = 0.704 . 20 = 14.08 kN/m2
Bina Nusantara
Pw1 = kw . w . h2 = 1 . 10 . 8 = 80 kN/m2
Pq1
Pw1
SOLUTION
Coefficient of Lateral Pressure :
Active ; ka =
tan2(45-1/2)
= 0.704
q = 20 kN/m2
2m
Pa1
Passive ; kp = tan2(45+2/2) = 1.698
8m
4m
Pw2
Pp1
Pa2
Pq1
Pw1
PASSIVE LATERAL PRESSURE
Pp1 = kp . 2’ . h3 + 2 . c . kp = 1.698 . 5 . 4 + 2 . 0 . 1.698 = 33.96 kN/m2
Bina Nusantara
Pw2 = kw . w . h3 = 1 . 10 . 4 = 40 kN/m2
SOLUTION
Coefficient of Lateral Pressure :
Active ; ka =
tan2(45-1/2)
= 0.704
q = 20 kN/m2
2m
Pa1
Passive ; kp = tan2(45+2/2) = 1.698
Pa
4m
8m
za
Pp
zp
Pw2
Pp1
Pa2
Pq1
Pw1
ACTIVE LATERAL FORCE
Pa = 0.5 . Pa1 . h1 + (Pa1+Pa2)/2 . H2 + Pq1 . (h1+h2) + 0.5 . Pw1 . h2 = 763.52 kN/m
Bina Nusantara
za = 3.56 m
SOLUTION
Coefficient of Lateral Pressure :
Active ; ka =
tan2(45-1/2)
= 0.704
q = 20 kN/m2
2m
Pa1
Passive ; kp = tan2(45+2/2) = 1.698
Pa
4m
8m
za
Pp
zp
Pw2
Pp1
Pa2
Pq1
PASSIVE LATERAL FORCE
Pp = 0.5 . Pp1 . h3 + 0.5 . Pw2 . h3 = 147.92 kN/m
Bina Nusantara
zp = 4/3 m
Pw1
RANKINE EARTH PRESSURE FOR INCLINED BACKFILL
Ka  cos 
cos   cos 2   cos 2 
cos   cos 2   cos 2 
Pa 
Kp  cos 
1
2
. .H 2 .Ka
cos   cos 2   cos 2 
cos   cos 2   cos 2 
Pp  1 2 . .H 2 .Kp
Bina Nusantara
SESSION 23-24
COULOMB METHOD
LATERAL PRESSURE DUE TO EXTERNAL
LOAD
DYNAMIC EARTH PRESSURE
Bina Nusantara
COULOMB METHOD
ACTIVE LATERAL PRESSURE
Assumption:
-Fill material is granular
-Friction of wall and fill
material is considered
-The failure surface in
the soil mass would be a
plane (BC1, BC2 …)
Pa = ½ Ka .  . H2
Ka 
Bina Nusantara
sin 2 (  )

sin(    ). sin(    ) 
2
sin . sin   1 

sin(



).
sin(



)


2
EXAMPLE
Consider the retaining wall shown in the following figure. Given
- H = 4.6 m
-  = 16.5 kN/m3
-  = 30 o
-  = 2/3 
-c=0
-=0
-  = 90 o
Calculate the Coulomb’s active force
per unit length of the wall

Bina Nusantara
SOLUTION
Ka 
sin 2 (  )

sin(    ). sin(    ) 
2
sin . sin   1 

sin(



).
sin(



)


Ka = 0.297
Pa  1
2
.  . H 2 . Ka
Pa = 51.85 kN/m
Bina Nusantara
2
COULOMB METHOD
PASSIVE LATERAL PRESSURE
Kp 
sin2 (  )

sin(   ). sin(   ) 
sin2 . sin   1 

sin(   ). sin(  ) 

Pp = ½ Kp .  . H2
Bina Nusantara
2
COULOMB METHOD WITH A SURCHARGE ON THE BACKFILL
Bina Nusantara
LATERAL EARTH PRESSURE DUE TO SURCHARGE
2q
a 2b

.
nH a 2  b 2


2
a > 0,4
4q
a 2b

.
nH a 2  b 2


2
a  0,4
q
0,203b
 .
H 0,16  b 2

Bina Nusantara

2
LATERAL EARTH PRESSURE DUE TO SURCHARGE

q
  sin . cos 2
H
P
q
H2  1 
90
 b' 
1  tan 1 

H
 a'b' 
 2  tan 1 

 H 
H  2  1   R  Q   57,30a' H
z
2H 2  1 
2
Bina Nusantara
R  a' b' 90   2 
2
Q  b'2 90  1 
LATERAL EARTH PRESSURE FOR EARTHQUAKE
CONDITIONS
Pae 
1
2
1  k v K ae
.

.
H
2
 kh 
'  tan 1 

1  k v 
K ae 
Bina Nusantara
sin 2     '

sin    sin  '   
2
cos '. sin . sin  '  1 





sin





'
sin





2
LATERAL EARTH PRESSURE FOR EARTHQUAKE
CONDITIONS
Bina Nusantara
LATERAL EARTH PRESSURE FOR EARTHQUAKE
CONDITIONS
z
0.6 H Pae    H  Pa
3
Pae
Pae  Pae  Pa
Bina Nusantara
LATERAL EARTH PRESSURE FOR EARTHQUAKE
CONDITIONS
Ppe  1 2 . .H 2 1  k v K pe
K pe 
Bina Nusantara
sin 2   '  

sin    sin    ' 
cos '. sin 2 . sin    '901 





sin





'
sin





2
LATERAL EARTH PRESSURE FOR EARTHQUAKE
CONDITIONS
Bina Nusantara
EXAMPLE
Refer to the following figure. For kv = 0 and kh = 0.3,
determine :
a. Pae
b. The location of the resultant, z, from the bottom of the
wall
 = 35 o
 = 18 kN/m3
5m
Bina Nusantara
 = 17.5 o
SOLUTION
Part a.
Pae 
1
2
1  k v K ae
.

.
H
2
Kae = 0.47
Pae 
1
2
. 18 . 52 1  0 . 0.47
Pae  105.75 kN / m
Bina Nusantara
SOLUTION
• Part b.
Where :
Pa  1
2
.  . H 2 . Ka
Ka = 0.25
 = 90o
 = 17.5o
 = 0o
Pa = 56.25 kN/m
Pae = Pae – Pa = 105.75 – 56.25 = 49.5 kN/m
z
Bina Nusantara
0.6H Pae    H  Pa
3
Pae
 2.29 m