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1

8 Sheet Pile Walls

Types of sheet piles:

1. Cantilever Sheet Piles.

2. Anchored Sheet Piles ( Free earth support – Fixed Earth Support)

1- Cantilever sheet pile:

.مىغزست في انتزبت انزمهيت أو انطيىيت بذون وجود أي دعم خارجي Sheet pile في هذا انىوع تكون انـ

Example 1) (Cantilever sheet pile immersed in sandy soil)

For the shown sheet pile:

Calculate the theoretical depth of embedment theoD and actual depth actD

Find the section modulus S, if MPaall 172 .

o

3

32

/9.15:

mKNSand

3/33.19: mKNSand sat

2

1- Find Pa kk , For each layer:

25.3307.0

1

2

3245tan

307.02

3245tan

2

2

p

a

k

k

2- Find the values of lateral pressure and draw pressure diagram:

0.00zAt

00.02 aaa kcHqk

m2zAt 2/77.99.152307.0 mkNa

m5zAt just above

( (دائما عىذوجود طبقت تزبت جذيذة وأخذ قبم وبعذ، في هذا انسؤال نم تختهف طبقاث انتزبت ونكه سىعمم هذي انحساباث نهتوضيح فقط )

)Active and Passive دائما عىذ خط انحفز وأخذ قبم وبعذ ووحسب انضغطيه )

2

2

/546.18

00.0

/546.1881.933.193307.09.152307.0

mkN

mkN

p

a

m5zAt just after

2

2

/546.18

00.02

/546.1881.933.193307.09.152307.0

mkN

kcHqk

mkN

ppp

a

Now draw the pressure diagram

To find the distance L3:

3L5zAt

33

p

33

33

925.2546.1898.30

point At this

98.30052.925.3

925.2546.1852.9307.0546.18

LL

LL

LL

a

p

a

mL 66.03

9.77

18.546

21502.28 4 L

3

4L5.66zAt

21502.28

94.306.216052.952.966.052.939.15225.3

922.293.152.9307.0307.052.966.0

4

44

44

L

LL

LL

ap

p

a

Find the length of line CD:

4

4

1.2866.0

546.18LCD

L

CD

Find the forces from the pressure diagram:

2

444

554544

4

3

2

1

05.141.282

1

5.10706.2821502.281.282

1

/12.6545.1866.02

1

/164.13377.9545.182

1

/31.29377.9

/77.9277.92

1

LLLA

LLLLLLA

mKNP

mKNP

mKNP

mKNP

ACD

CBE

4

For equilibrium:

5.10706.28

364.5805.14

00.005.145.10706.2812.6164.1331.2977.9

00.0

4

2

45

2

4554

L

LL

LLLL

Fx

305.14

35.10706.28

44.012.666.1164.1316.231.293267.477.9

00.0

42

45

554

4444

LL

LLLL

LLLL

M base

3

4

2

5

2

544 683.483.35353.9364.58130 LLLLL

By trial and error:

4L 5L Check the relation

4.8m 0.18m OK (right part semi equal left part)

mD

DD

mD

actual

theoactual

theo

55.646.52.1

3.12.1

46.58.466.0

3- Find the maximum moment on the sheet pile:

Take section at distance x below the distance L3:

0.00Vmoment maximumFor

05.14364.58

: Vsection at the force

05.141.282

1

1.2866.0

546.18

2

2

XV

Shear

XXXP

XLX

Lo

o

2.04mX

3

04.205.1448.212.6

7.3164.132.431.29367.677.9

3

maxM

mKNM .43.209max

4- Calculate the section modulus S:

mmS /1021.110172

1043.209 3396

length of the wall

5

Example 2) (Cantilever sheet pile immersed in clay)

For the shown sheet pile:


Find maxM


32

3045tan

12

045tan

3

1

2

3045tan

12

045tan

2

2

2

1

2

2

2

1

p

p

a

a

k

k

k

k


0.00zAt

2/2012022012 mkNkcHqk aaa

m2zAt Just before:

2/201202202201 mkNa

m2zAt Just after:

2/203

10220220

3

1mkNa

2/20 mkNq

32 /20 /20: mKNmkNCClay

3/18

30:

mKN

Sand o

3

2

/19

/50:

mKN

mkNCClay

6

m5zAt Just before:

00.0

/383

10218320220

3

1 2

p

a mkN

m5zAt Just after:

2

2

2

/8614100

/100150201

/141502183202201

mkN

mkN

mkN

ap

p

a


D5zAt

2

2

/314

/19214150219183220201

100191502191

mkN

mDkND

DD

ap

p

a

20

20

38 86

314

7


DA

LLA

mkNP

mkNP

mkNP

mkNP

ABDE

CDF

86

200863142

1

/273182

1

/60320

/101202

1

/101202

1

44

4

3

2

1

For equilibrium:

08620027601010

00.0

4

DL

FX

435.043.04 DL

00.079.3023.2

00.02

863

2001275.16033.310667.410

00.0

2

44

DD

DD

LLDDDD

M base

theoactual

theo

DD

mD

6.14.1

2.3

3- Find the maximum moment on the sheet pile:

Take section at distance X below the dredge line

mmkNM

mXXdX

dM

MFor

XXM

XXX

XXXM

/.147

011.1087860

:

66.1038743

286127

5.16033.31067.410

max

max

2

8

(Anchored Sheet pile immersed in sandy soil)

Free Earth support

ثم يتولد أسفل خط الحفز ضغط من نوع Active في هذا النوع يكون سلوك الضغط على الجدار أعلى خط الحفز

.على يسار الجدار Passive معاكس

Example 3) For the shown sheet pile:


Calculate the anchor force F.


3

3

1

2

3045tan 2

p

a

k

k


0.00zAt

00.0a

m05.3zAt

2/267.161605.33

1mkNa

m15.9zAt just above

2/97.3569.91.61605.33

1mkNa

m15.9zAt just after

o

mkNSand

30

/16: 3

o

sat mkNSand

30

/5.19: 3

9

2/97.3569.91.61605.33

1mkNa


Find the point of zero pressure:

3L15.9zAt

33

p

33

33

23.397.3507.29

point At this

07.29069.93

23.397.3569.93

197.35

LL

LL

LL

a

p

a

mL 39.13

4L54.10zAt

4p

44

44

84.25

07.294.4069.9339.169.93

23.34.4069.93

139.169.9

3

197.35

L

LL

LL

a

p

a


2

4445

4

3

2

1

92.1284.252

1

/2539.197.352

1

/1.601.67.192

1

/23.991.627.16

/807.2405.327.162

1

LLLP

mkNP

mkNP

mkNP

mkNP

For Equilibrium:

00.03

201.992.12083.825

5867.51.6057.423.99503.0807.24

00.0

4

2

4

LL

M anchor

53.116515.13 2

4

3

4 LL

mL 7.24

16.27

35.97

10

mKNF /115

theoactual

theo

DD

mD

4.13.1

1.439.17.2

27.292.12251.6023.99807.24

00.0

F

FX

(Anchored Sheet pile immersed in CLAY)

Free Earth support Example 4) For the shown sheet pile:

Calculate the theoretical depth of embedment theoD



69.32

3545tan

12

045tan

271.02

3545tan

12

045tan

2

2

2

1

2

2

2

1

p

p

a

a

k

k

k

k


0.00zAt

00.0a

ft8.01zAt

ksfpsfa 316.03161088.10271.0

opcfSand 35 108:

o

sat pcfSand 35 2.127:

psfcpcfClay sat 850 2.127:

11

ft4.23zAt just above

00.0

695.0695271.0024.622.1276.211088.10271.0

p

a ksfpsf

ft4.23zAt Just after:

ksfpsf

ksfpsf

ksfpsf

ap

p

a

834.08348661700

7.117001850201

866.0866185024.622.1276.211088.101


D 4.23zAt

ksfpsfDD

DD

DD

ap

p

a

834.08348.648668.641700

8.641700185028.641

8.64866185028.648.646.211088.101

Find the forces from pressure diagram:

316.0

695.0 0.834

12

DP

ftkP

ftkP

ftkP

834.0

/0932.46.21379.02

1

/8256.66.21316.0

/7064.18.10316.02

1

4

3

2

1

For Equilibrium:

ftD

DD

M base

6.7

0272

834.08.19093.42.16825.68.1706.1

00.0

ftkF

F

FX

/3.6

6.7834.0093.4825.6706.1

00.0

13

Extra problem For the shown sheet pile:

Calculate the theoretical depth of embedment theoD


Answers

D = 7ft

F = 3 k/ft

opcfSand 32 115:



sat=122.4pcf

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