Course
Year
: S0825/Foundation Engineering
: 2009
RETAINING EARTH STRUCTURE
Session 11 – 16
RETAINING EARTH STRUCTURE
Topic:
• Lateral Earth Pressure
– General
– Active earth pressure
• Rankine earth pressure
• Coulomb earth pressure
• Lateral earth pressure due to surcharge
– Passive earth pressure
• Rankine earth pressure
• Coulomb earth pressure
– Influence of ground water table
• Sheet Pile Structure
–
–
–
–
Bina Nusantara
General
Types of Sheet Pile
Lateral Pressure Diagram
Cantilever Sheet Pile
LATERAL EARTH PRESSURE
SESSION 11 – 12
Bina Nusantara
GENERAL
• Lateral earth pressure represents pressures that are “to
the side” (horizontal) rather than vertical.
• Caused by soil self weight and or external load
• 3 categories:
– At rest earth pressure
– Active earth pressure
– Passive earth pressure
Bina Nusantara
AT REST EARTH PRESSURE
The at rest pressure develops when the wall experiences no lateral
movement. This typically occurs when the wall is restrained from
movement such as a basement wall that is supported at the bottom by
a slab and at the top by a floor framing system prior to placing soil
backfill against the wall.
Bina Nusantara
ACTIVE EARTH PRESSURE
The active pressure develops when the wall is free to move outward
such as a typical retaining wall and the soil mass stretches sufficiently to
mobilize its shear strength.
Bina Nusantara
PASSIVE EARTH PRESSURE
If the wall moves into the soil, then the soil mass is compressed
sufficiently to mobilize its shear strength and the passive pressure
develops.
Bina Nusantara
AT REST EARTH PRESSURE
q
Jaky, Broker and Ireland  Ko = M – sin ’
Sand, normally consolidated clay  M = 1
v =  . z + q
z
Broker and Ireland
v
h
K
h
v
At rest, K = Ko
Bina Nusantara
Clay with OCR > 2  M = 0.95
Ko = 0.40 + 0.007 PI , 0  PI  40
Ko = 0.64 + 0.001 PI , 40  PI  80
Sherif and Ishibashi  Ko =  +  (OCR – 1)
 = 0.54 + 0.00444 (LL – 20)
 = 0.09 + 0.00111 (LL – 20)
LL > 110%   = 1.0 ;  = 0.19
ACTIVE EARTH PRESSURE
Bina Nusantara
RANKINE ACTIVE EARTH PRESSURE
1 = 3 . tan2 (45+/2)+2c.tan (45+/2)
a = v . tan2(45-/2) – 2c . tan (45-/2)
a = v . Ka – 2cKa
Bina Nusantara
Ka = tan2 (45 - /2)
RANKINE ACTIVE EARTH PRESSURE (INCLINED BACKFILL)
Ka  cos 
Pa 
1
cos   cos 2   cos 2 
cos   cos 2   cos 2 
2
.

.
H
.Ka
2
(for granular soil, c = 0)
For c- soil
 a  zK a  zK a ' cos 
where :

c
1 
2
  cos  sin  
Ka ' 
2
cos


2

cos 2  
 z 

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2


c
c 2

2
2
2
2
4 cos  cos   cos   4  cos   8  cos  sin  cos     1

 
 z 
 z 



COULOMB ACTIVE EARTH PRESSURE
Assumptions:
-Fill material is
granular soil
- Friction of wall and
fill material is
considered
- Soil failure shape is
plane (BC1, BC2 …)
Pa = ½ Ka .  . H2
Ka 
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sin 2 (  )

sin(    ). sin(    ) 
2
sin . sin   1 

sin(



).
sin(



)


2
COULOMB ACTIVE EARTH PRESSURE
(SURCHARGE ON BACKFILL)
Bina Nusantara
RANKINE PASSIVE EARTH PRESSURE
Bina Nusantara
RANKINE PASSIVE EARTH PRESSURE
p= v . tan2(45+/2) + 2c . tan (45+/2)
Bina Nusantara
RANKINE PASSIVE EARTH PRESSURE
Kp = tan2 (45 + /2)
h = v . Kp + 2cKp
Bina Nusantara
COULOMB PASSIVE EARTH PRESSURE
Kp 
sin2 (  )

sin(   ). sin(   ) 
sin2 . sin   1 

sin(   ). sin(  ) 

Pp = ½ Kp .  . H2
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2
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

Bina Nusantara
q 0.203b
.
H 0.16  b2

2
LATERAL EARTH PRESSURE DUE TO SURCHARGE
q
    sin . cos 2 
H
q
P  H2  1 
90
 b' 
1  tan 1 

H
 a'b' 
 2  tan 1 

 H 
R  a' b' 90   2 
2
H 2  2  1   R  Q   57,30a' H
z
2H 2  1 
Bina Nusantara
Q  b'2 90  1 
PURPOSE OF LATERAL EARTH PRESSURE
• STABILITY ANALYSIS GRAVITY WALL AGAINST
– SLIDING
– OVERTURNING
Bina Nusantara
PURPOSE OF LATERAL EARTH PRESSURE
Bina Nusantara
PURPOSE OF LATERAL EARTH PRESSURE
Bina Nusantara
SHEET PILE STRUCTURES
SESSION 13 – 14
Bina Nusantara
SHEET PILE
Bina Nusantara
GENERAL
Connected or semi-connected sheet piles are often used to
build continuous walls to retain the lateral pressure caused
by soil or external load.
In contrast to the construction of other types of retaining
wall, the building of sheet pile walls do not usually require
dewatering the site.
Sheet piles are also used for some temporary structures,
such as braced cut.
Bina Nusantara
SHEET PILE TYPES (CANTILEVER)
Bina Nusantara
SHEET PILE TYPES (ANCHORED)
Bina Nusantara
Free Earth Support
SHEET PILE TYPES (ANCHORED)
Bina Nusantara
Fixed Earth Support
SHEET PILE TYPES (ANCHORED)
anchor plate or beam
Bina Nusantara
SHEET PILE TYPES (ANCHORED)
tie back
Bina Nusantara
vertical anchor pile
SHEET PILE TYPES (ANCHORED)
anchor beam with batter piles
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LATERAL EARTH PRESSURE DIAGRAM
Bina Nusantara
LATERAL EARTH PRESSURE DIAGRAM
Bina Nusantara
LATERAL EARTH PRESSURE DIAGRAM
Fixed Earth Support
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LATERAL EARTH PRESSURE DIAGRAM
Free Earth Support
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LATERAL EARTH PRESSURE DIAGRAM
Free Earth Support
Bina Nusantara
CALCULATION STEPS
CANTILEVER SHEET PILE - SAND
Bina Nusantara
CALCULATION STEPS
CANTILEVER SHEET PILE - SAND
1. Determine the value of Ka and Kp


K a  tan 2  45  
2



K p  tan 2  45  
2

2. Calculate p1and p2 with L1 and L2 are known
p1   .L1.K a
p2   .L1   '.L2 K a
3. Calculate L3
z  L   L3

p2
 ' K p  K a 
4. Calculate the resultant of the area ACDE (P)
5. Determine the z (the center of pressure for the area ACDE)
Bina Nusantara
CALCULATION STEPS
CANTILEVER SHEET PILE - SAND
6. Calculate p5
p5   .L1   '.L2 K p   '.L3 K p  K a 
7. Calculate A1, A2, A3, A4
Bina Nusantara
p5
A1 
 '.K p  K a 
A3 
8P
A2 
 '.K p  K a 
A4 

6 P 2.z. ' K p  K a   p5

 '2 K p  K a 2

P 6.z. p5  4 P
 '2 K p  K a 2

CALCULATION STEPS
CANTILEVER SHEET PILE - SAND
8. Determine L4
9. Calculate p4
L44  A1 L34  A2 L24  A3 L4  A4  0
p4  p5   '.L4 K p  K a 
10. Calculate p3
p3   ' ( K p  K a ) L4
11. Calculate L5
L5 
p3 .L4  2 P
p3  p4
12. Draw the pressure distribution diagram
13. Obtain the theoretical depth ; D = L3 + L4
The actual depth of penetration is increased by about 20% - 30%
14. Calculate the maximum bending moment

 
with
Bina Nusantara

 z' 
M max  P z  z '  1 . '.(z ' ) 2 ( K p  K a )  
2
3
z' 
2P
( K p  K a ). '
EXAMPLE
CANTILEVER SHEET PILE - SAND
L1 = 2 m
GWL
L2 = 3 m
d = 15.9 kN/m3
t = 19.33 kN/m3
 = 32o
c = 0 kPa
D
Bina Nusantara
Determine the penetration depth (D) and dimension of sheet pile
EXAMPLE
CANTILEVER SHEET PILE - SAND
•
Step 1 (determine the value of ka and kp)

32 


K a  tan 2  45    tan 2  45 
  0.307
2
2 



32 


K p  tan 2  45    tan 2  45 
  3.25
2
2 


•
Step 2 (calculate p1 and p2)
p1   .L1.K a  (15.9)( 2)( 0.307)  9.763kPa
p2   .L1   '.L2 K a  (15.9)( 2)  (19.33  9.81)30.307
p2  18.53kPa
•
Step 3 (Calculate L3)
L3 
Bina Nusantara
p2
18.53

 0.66 m
 ' K p  K a  (19.33  9.81)(3.25  0.307)
EXAMPLE
CANTILEVER SHEET PILE - SAND
•
Step 4 (calculate P)
P  0.5. p1.L1  p1.L2  0.5. p2  p1 .L2  0.5. p2 .L3
P  0.5(9.763)( 2)  (9.763)(3)  0.5(18.53  9.763)3  0.5(18.53)(0.66)
P  9.763  29.289  13.151  6.115  58.32 kN/m
•
Step 5 (calculate z)

1 
2
3
3



 2  
z
9.763 0.66  3    29.289 0.66    13.151 0.66    6.115 0,66  
58.32m
3
2
3



 3  

z  2.23
•
Step 6 (calculate p5)
p5   .L1   '.L2 K p   '.L3 K p  K a 
p5  (15.9)( 2)  (19.33  9.81)33.25  (19.33  9.81)(0.66)(3.25  0.307)
p5  214.66 kN/m2
Bina Nusantara
EXAMPLE
CANTILEVER SHEET PILE - SAND
•
Step 7 (calculate A1 – A4)
p5
214.66
A1 

 7.66
 '.K p  K a  (9.52)(2.943)
A3 
A3
8P
(8)(58.32)

 16.65
 '.K p  K a  (9.52)(2.943)
6 P 2.z. ' K p  K a   p5

 '2 K p  K a 2
(6)(58.32)(2)( 2.23)(9.52)( 2.943)  214.66

 151.93
A4 
Bina Nusantara

A2 
(9.52) 2 (2.943) 2


P 6.z. p5  4 P
58.32(6)( 2.23)( 214.66)  (4)(58.32)

 230.72
2
2
2
2
(9.52) (2.943)
 ' K p  K a 
EXAMPLE
CANTILEVER SHEET PILE - SAND
•
Step 8 (determine L4)
L44  7.66 L34  16.65L24  151.39 L4  230.72  0
L4  4.8 m
•
Step 9 (calculate p4)
p4  p5   '.L4 K p  K a 
p4  214.66  (9.52)( 4.8)( 2.943)  349.14kPa
•
Step 10 (calculate p3)
p3   ' ( K p  K a ) L4  (9.52)( 2.943)( 4.8)  134.48 kPa
Bina Nusantara
EXAMPLE
CANTILEVER SHEET PILE - SAND
•
Step 11 (Calculate L5)
p3 .L4  2 P (134.48)( 4.8)  2(58.32)
L5 

 1.09
p3  p4
134.48  349.14
Bina Nusantara
•
Step 12
Draw the pressure distribution diagram
•
Step 13 (the penetration dept of sheet pile)
– Theoretical
= 0.66 + 4.8 = 5.46 m
– Actual
= 1.3 (L3+L4) =7.1 m
m
EXAMPLE
CANTILEVER SHEET PILE - SAND
Bina Nusantara
EXAMPLE
CANTILEVER SHEET PILE - SAND
Dimension of Sheet Pile
z' 
2P

( K p  K a ). '

 
( 2)(58.32)
 2.04 m
9.52( 2.943)

 z' 
M max  P z  z '  1 . '.( z ' ) 2 ( K p  K a )  
2
3
 2.04 
M max  (58.32)( 2.23  2.04)  0.5(9.52)( 2.04) 2 ( 2.943)

3


M max  209.39 kN.m
Bina Nusantara
SHEET PILE STRUCTURE
SESSION 15 – 16
Bina Nusantara
CALCULATION STEPS
CANTILEVER SHEET PILE - CLAY
Bina Nusantara
CALCULATION STEPS
CANTILEVER SHEET PILE - CLAY
1. Determine the value of Ka and Kp

2
K a  tan  45  
2



K p  tan 2  45  
2

In case of saturated soft clay with internal friction angle () = 0, we got
Ka = Kp = 1
2. Calculate p1and p2 with L1 and L2 are known
p1   .L1.K a
p2   .L1   '.L2 K a
3. Calculate the resultant of the area ACDE (P1) and z1 (the center of
pressure for the area ACDE)
Bina Nusantara
CALCULATION STEPS
CANTILEVER SHEET PILE - CLAY
4. Calculate the theoretical penetration depth of sheet pile (D)
D 2 4.c   .L1   '.L2   2.D.P1 

5. Calculate L4
D4.c   .L1   '.L2   P1
4.c
6. Calculate p6 and p7
L4 
p6  4.c   .L1   '.L2 
p7  4.c   .L1   '.L2 
7. Obtain the actual penetration depth of sheet pile
Dactual = (1.4 – 1.6) x Dtheoretical
Bina Nusantara

P1. P1  12.c.z1
0
 .L1   '.L2   2.c
CALCULATION STEPS
CANTILEVER SHEET PILE - CLAY
8. Calculate the maximum bending moment
M max

with
P1
z' 
p6
Bina Nusantara

p6 .( z ' ) 2
 P1 z1  z ' 
2
EXAMPLE
CANTILEVER SHEET PILE - CLAY
L1 = 2 m
sand
GWL
L2 = 3 m
d = 15.9 kN/m3
t = 19.33 kN/m3
 = 32o
c = 0 kPa
Clay
D
cu = 47 kPa
=0o
Bina Nusantara
Determine the penetration depth (D) and dimension of sheet pile
EXAMPLE
CANTILEVER SHEET PILE - CLAY
•
Step 1 (Determine ka and kp)

32 

2
K a  tan  45    tan  45    0.307
2
2



0


K p  tan 2  45    tan 2  45    1.00
2
2


2
•
Step 2 (calculate p1 and p2)
p1   .L1.K a  (15.9)( 2)(0.307)  9.763 kPa
p2   .L1   '.L2 K a  (15.9)( 2)  (19.33  9.81)30.307
•
p2  18.53 kPa
Step 3 (calculate P1 and z1)
1
1
P1  p1L1  p1L2   p2  p1 L2
2
2
P1  9.763  29.289  13.151  52.2 kN/m
Bina Nusantara
2

3
3
9.763 3    29.289   13.151 
3

2
3
z1 
52.2
z1  1.78 m
EXAMPLE
CANTILEVER SHEET PILE - CLAY
•
Step 4 (obtain Dtheoretical)
D 2 4.c   .L1   '.L2   2.D.P1 


P1. P1  12.c.z1
0
 .L1   '.L2   2.c
D 2 4 47   2 15.9  19.33  9.813 2 D52.2  
127.64 D 2  104.4 D  357.15  0
•
Bina Nusantara
52.252.2  12 47 1.78
0
15.92  19.33  9.813  247 
D = 2.13 m
Step 5 (calculate L4)
D4.c   .L1   '.L2   P1
L4 
4.c
2.13447   15.92  19.33  9.813 52.2
L4 
447 
L4 = 2.13 m
EXAMPLE
CANTILEVER SHEET PILE - CLAY
•
Step 6 (calculate p6 and p7)
p6  4.c   .L1   '.L2   127.64 kN/m2
p7  4.c   .L1   '.L2   248.36 kN/m2
•
•
Step 7 (draw the lateral diagram)
Step 8 (Obtain Dactual)
Dactual = 1.5 x Dtheorical = 1.5 x 2.13 = 3.2 m
Bina Nusantara
EXAMPLE
CANTILEVER SHEET PILE - CLAY
• Calculation of moment
z' 
M max
M max
Bina Nusantara

P1
52.2

 0.41 m
p6 127.64

p6 .( z ' ) 2
 P z1  z ' 
2
2
127.640.41
 52.21.78  0.41 
 103.59 kN  m
2
CALCULATION STEPS
ANCHORED SHEET PILE – FREE – SAND
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FREE – SAND
1. Determine the value of Ka and Kp

2
K a  tan  45  
2



K p  tan 2  45  
2

2. Calculate p1and p2 with L1 and L2 are known
p1   .L1.K a
p2   .L1   '.L2 K a
3. Calculate L3
z  L   L3

p2
 ' K p  K a 
4. Calculate P as a resultant of area ACDE
5. Determine the center of pressure for the area ACDE ( z )
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FREE – SAND
6. Calculate L4
L34  1,5L24 l2  L2  L3  


3P L1  L2  L3   z  l1
 ' K p  K a 
Determination of penetration depth of sheet pile (D)
Dtheoretical = L3 + L4
Dactual = (1.3 – 1.4) Dtheoretical
Determination of anchor force
F = P – ½ [’(Kp – Ka)]L42
Bina Nusantara
  0
CALCULATION STEPS
ANCHORED SHEET PILE – FREE – CLAY
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FREE – CLAY
1. Determine the value of Ka and Kp

2
K a  tan  45  
2



K p  tan 2  45  
2

In case of saturated soft clay with internal friction angle () = 0, we got
Ka = Kp = 1
2. Calculate p1and p2 with L1 and L2 are known
p1   .L1.K a
p2   .L1   '.L2 K a
3. Calculate the resultant of the area ACDE (P1) and z1 (the center of
pressure for the area ACDE)
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FREE – CLAY
4. Calculate p6
p6  4c  L1   ' L2 
5. Determination of penetration depth of sheet pile (D)
p6.D2 + 2.p6.D.(L1+L2-l1) – 2.P1.(L1+L2-l1-z1) = 0
6. Determination of anchor force
F = P1 – p6 . D
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FIXED – SAND
J
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FIXED – SAND
1. Determine the value of Ka and Kp


K a  tan 2  45  
2



K p  tan 2  45  
2

2. Calculate p1and p2 with L1 and L2 are known
p1   .L1.K a
p2   .L1   '.L2 K a
3. Calculate L3
p2
z  L  L3 
 ' K p  K a 
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FIXED – SAND
4. determine L5 from
the following curve (L1
and L2 are known)
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FIXED – SAND
5. Calculate the span of the equivalent beam as l2 + L2 + L5 = L’
6. Calculate the total load of the span, W. This is the area of the
pressure diagram between O’ and I
7. Calculate the maximum moment, Mmax, as WL’/8
Bina Nusantara
CALCULATION STEPS
ANCHORED SHEET PILE – FIXED – SAND
(moment of area ACDJI about O’)
8. Calculate P’ by taking the moment about O’, or
P' 
1
L'
9. Determine D
D  L5  1.2
6 P'
K p  K a  '
10. Calculate the anchor force per unit length, F, by taking the moment about l, or
F 
Bina Nusantara
1
L'
(moment of area ACDJI about I)