Course
Year
Version
: S0484/Foundation Engineering
: 2007
: 1/0
Session 13 – 14
SHEET PILE STRUCTURES
SHEET PILE STRUCTURES
Topic:
• General
• Types of Sheet Pile
• Lateral Pressure Diagram
• Cantilever Sheet Pile
SHEET PILE
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.
SHEET PILE TYPES
CANTILEVER SHEET PILE
SHEET PILE TYPES
ANCHORED SHEET PILE
Free Earth Support
SHEET PILE TYPES
ANCHORED SHEET PILE
Fixed Earth Support
SHEET PILE TYPES
ANCHORED SHEET PILE
anchor plate or beam
SHEET PILE TYPES
ANCHORED SHEET PILE
tie back
vertical anchor pile
SHEET PILE TYPES
ANCHORED SHEET PILE
anchor beam with batter piles
LATERAL EARTH PRESSURE DIAGRAM
LATERAL EARTH PRESSURE DIAGRAM
LATERAL EARTH PRESSURE DIAGRAM
Fixed Earth Support
LATERAL EARTH PRESSURE DIAGRAM
Free Earth Support
LATERAL EARTH PRESSURE DIAGRAM
Free Earth Support
CALCULATION STEPS
CANTILEVER SHEET PILE - SAND
CALCULATION STEPS
CANTILEVER SHEET PILE - 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 the resultant of the area ACDE (P)
5. Determine the z (the center of pressure for the area ACDE)
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
p5
A1 
 '.K p  K a 
8P
A2 
 '.K p  K a 
A3 
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
L44  A1 L34  A2 L24  A3 L4  A4  0
9. Calculate p4
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

 

 z' 
M max  P z  z '  1 . '.(z ' ) 2 ( K p  K a )  
2
3
with
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
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)  9.763 kPa
p2   .L1   '.L2 K a  (15.9)( 2)  (19.33  9.81)30.307
p2  18.53 kPa
• Step 3 (Calculate L3)
L3 
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)
z

1 
2
3
3



 2  
9
.
763
0
.
66

3


29
.
289
0
.
66


13
.
151
0
.
66


6
.
115

0
,
66






  


58.32 
3
2
3






 3  

z  2.23 m
• 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
EXAMPLE
CANTILEVER SHEET PILE - SAND
• Step 7 (calculate A1 – A4)
A1 
p5
214.66

 7.66
 '.K p  K a  (9.52)(2.943)
8P
(8)(58.32)
A2 

 16.65
 '.K p  K a  (9.52)(2.943)
A3 
A3

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 
(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  7.66L34  16.65L24  151.39L4  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)  349.14
kPa
• Step 10 (calculate p3)
p3   ' ( K p  K a ) L4  (9.52)( 2.943)( 4.8)  134.48 kPa
EXAMPLE
CANTILEVER SHEET PILE - SAND
• Step 11 (Calculate L5)
p3 .L4  2 P (134.48)( 4.8)  2(58.32)
L5 

 1.09 m
p3  p4
134.48  349.14
• 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
EXAMPLE
CANTILEVER SHEET PILE - SAND
EXAMPLE
CANTILEVER SHEET PILE - SAND
Dimension of Sheet Pile
z' 
(K p
2P

 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
CALCULATION STEPS
CANTILEVER SHEET PILE - CLAY
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)
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 


P1. P1  12.c.z1
0
 .L1   '.L2   2.c
5. Calculate L4
D4.c   .L1   '.L2   P1
L4 
4.c
6. Calculate p6 and p7
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
CALCULATION STEPS
CANTILEVER SHEET PILE - CLAY
8. Calculate the maximum bending moment
M max


p6 .( z ' ) 2
 P1 z1  z ' 
2
with
P1
z' 
p6
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
Determine the penetration depth (D) and dimension of sheet pile
EXAMPLE
CANTILEVER SHEET PILE - CLAY
• Step 1 (Determine ka and kp)

32 


K a  tan 2  45    tan 2  45    0.307
2
2



0


K p  tan 2  45    tan 2  45    1.00
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
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
• Step 5 (calculate L4)
D4.c   .L1   '.L2   P1
L 
4
52.252.2  1247 1.78
0
15.92  19.33  9.813  247 
D = 2.13 m
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
EXAMPLE
CANTILEVER SHEET PILE - CLAY
• Calculation of moment
z' 
M max
M max

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