1

Design example:

Find the depth of embedment for a cantilever sheet pile on cohesionless soil with φ=33o and density 18kN/m3

Solution:

. Find the depth of embedment with factors of safety on passive forces of 1 and 2.

ka

k

=0.295

p

Taking moment about bottom of wall (at depth D)

=3.392

3

2

2 3 2

3 2

3 2

3

1 1 (10 )3.392 18 (180 18 )(10 ) 0.2952 3 2 310.176 (30 3 )(10 )(10 )0.295

(30 3 )(100 20 )0.2953000 30 600 300 3 60(3 90 900 3000)0.2950.885 26.56 265.2 885

9.291 26.

D DD D D D

D D D DD D D

D D D D DD D D

D D DD

+× × × × = × + + ×

= + + +

= + + +

= + + + + +

= + + +

= + + +

− 256 265.6 885 07.95 8

D DD m m

− − == =

The embedment depth increases if we apply a higher factor of safety on the passive force. For example, for a factor of safety of 2 on the passive forces, the embedment depth is as follows:

D

C=0

φ=330

γ=18kN/m3

18D 18(10+D)

10m

2

3

3 2

1 1 1 (10 )3.392 18 (180 18 )(10 ) 0.2952 2 3 2 35.088 (30 3 )(10 )(10 )0.2954.203 26.56 265.6 885 0

12.66

D DD D D D

D D D DD D D

D m

+× × × × × = × + + ×

= + + +

− − − ==

Further calculations for 8m embedment depth:

Passive force= 21 3.392 18 8 1953.8 /2

kN m× × × =

Active force= 21 0.295 18 18 860.2 /2

kN m× × × =

Increase the embedment by 20%

8x1.2=9.6m

Additional passive force=1836 kN/m

Additional Active force=74.78kN/m

1099 42.5

10m

8m

1.6m

50.98 1196.7

3

Design of anchored sheet pile walls

Design example

Find the depth of embedment for an anchored sheet pile on cohesionless soil with φ 33° and γ 18kN/m3

Passive force=

. Also find the pressure taken by the anchor when its size is 1m and 1.2m. Use free earth support and fixed earth methods

212 pk Dγ

Moment about anchor point= 21 2(9 )2 3pk D Dγ +

Active force= 21 ( )2 ak H Dγ +

Moment about anchor point= 2( (10 ) 1)3aP D+ −

3.392

0.2951810

p

a

kk

Hγ

=

===

2 2

2 3 2

2 3 2

3 2

1 2 1 2 23.392 18 (9 ) 0.295 18 (10 ) ( 10 1)2 3 2 3 3274.75 20.35 (100 20 )(15.02 1.77 )

15.02 15.02 300.4 177 1.77 35.41.77 50.42 477.4 15.02

D D D D

D D D D DD D D D D

D D D

× × × + = × × × + × + −

+ = + + +

= + + + + +

= + + +

3 218.58 224.33 477.4 15.02 0D D D+ − − =

10m

C=0

φ=330

γ=18kN/m3 D

1m

4

D=3.28m 3.30m≈

Passive force= 21 3.39 18 3.3 332.5 /2

kN m× × × =

Active Force= 21 0.295 18 13.3 469.4 /2

kN m× × × =

Anchor should provide a lateral resistance of 496.4-332.5=163.9kN/m.

If anchor tie-rods are placed at 1.5m centre to centre along the length of the wall (perpendicular to plane of analysis) force in each anchor=1.5x164=246kN

1.5

33.65 VpP H k

Hσγγ

′′ = × ×

σv’

1m size anchor:

=Vertical pressure at mid-depth of the anchor

1.518 13.65 18 1 3.39 222.7218 1

P kN× = × × × × = ×

1.2m size anchor: 1.5

3 18 13.65 18 1.2 3.39 292.7718 1.2

P kN× = × × × × = ×

Increase embedment depth by 20% for additional factor of safety.

3.3 1.2 4D m= × ≈

Anchor should be placed at a minimum distance of 7.6+2.94=10.54 m ≅ 11 m behind the sheet pile wall.

π/4-φ/2=28.50

π/4+φ/2=61.50

2.94m

7.6m

4m

1.6m

5

Fixed earth method

3.392

0.2951810

p

a

kk

Hγ

=

===

Depth of contraflexure , x=0.088H

0.063 0.63H m× =

Slope on net pressure line, c= ( ) 18(3.392 0.295) 55.74p ak kγ − = − =

53.1 53.1 0.9555.74

a mc

= = =

T

R 17.89

56.44

53.1

1 c

10

a x

6

The earth pressure diagram is divided into two parts. The top part has a depth of (10+x) m. T represents the tie rod and let R be the total reaction from the bottom part of the earth pressure diagram.

Take moment about T.

1 2 1 29.63 56.44 10.63 ( 10.63 1) 38.5 0.63 9 0.632 3 2 3

R × = × × × × − − × × × + ×

R=122.55m

Tie rod force, T= Pa21 10.295 18 10.63 38.5 0.63 122.5

2 2 × × × − × × −

– R= =165.3kN/m

Solve the equation to find y

' ' 24 3 2 '

2 2

6 48 6 (2 ) 0p pp

p Ryp RR Ry y y y yc pc c c c

+ + − − + − =

y x a= −

' 0.295 18 10 (3.392 0.295) 18 0.95 106.06pp kNm= × × + − × × =

R

Pp’ 1

c

10

a x

y

kp-ka

7

Solving the equation by trial and error,

y=4.31≅4.3m

Total depth of embedment=4.3+0.95=5.25m

Increase embedment depth by 20% for additional factor of safety

5.25x1.2=6.3m