RETAINING EARTH STRUCTURE Session 11 – 16 Course: S0825/Foundation Engineering Year: 2009

RETAINING EARTH STRUCTURESession 11 – 16

Course : S0825/Foundation EngineeringYear : 2009

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RETAINING EARTH STRUCTURETopic:• 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

– General – Types of Sheet Pile– Lateral Pressure Diagram – Cantilever Sheet Pile

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LATERAL EARTH PRESSURE

SESSION 11 – 12

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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

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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.

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ACTIVE EARTH PRESSUREThe 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.

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PASSIVE EARTH PRESSUREIf the wall moves into the soil, then the soil mass is compressedsufficiently to mobilize its shear strength and the passive pressure develops.

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AT REST EARTH PRESSUREq

v

h

zv = . z + q

v

hK

At rest, K = Ko

Jaky, Broker and Ireland Ko = M – sin ’

Sand, normally consolidated clay M = 1

Clay with OCR > 2 M = 0.95

Sherif and Ishibashi Ko = + (OCR – 1)

= 0.54 + 0.00444 (LL – 20)

= 0.09 + 0.00111 (LL – 20)

LL > 110% = 1.0 ; = 0.19

Broker and IrelandKo = 0.40 + 0.007 PI , 0 PI 40Ko = 0.64 + 0.001 PI , 40 PI 80

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ACTIVE EARTH PRESSURE

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RANKINE ACTIVE EARTH PRESSURE

Ka = tan2 (45 - /2)

1 = 3 . tan2 (45+/2)+2c.tan (45+/2)

a = v . tan2(45-/2) – 2c . tan (45-/2)

a = v . Ka – 2cKa

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RANKINE ACTIVE EARTH PRESSURE (INCLINED BACKFILL)

1cossincos8cos4coscoscos4sincos2cos2cos

1'

:

22

2

22222

z

c

z

c

z

cK

where

a

22

22

coscoscos

coscoscoscosKa

Ka.H..Pa 221

(for granular soil, c = 0)

For c- soil

cos'aaa zKzK

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COULOMB ACTIVE EARTH PRESSUREAssumptions:

-Fill material is granular soil

- Friction of wall and fill material is considered

- Soil failure shape is plane (BC1, BC2 …)

Pa = ½ Ka . . H2

2

2

2

)sin().sin()sin().sin(

1sin.sin

)(sinKa

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COULOMB ACTIVE EARTH PRESSURE (SURCHARGE ON BACKFILL)

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RANKINE PASSIVE EARTH PRESSURE

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p= v . tan2(45+/2) + 2c . tan (45+/2)

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Kp = tan2 (45 + /2)

h = v . Kp + 2cKp

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COULOMB PASSIVE EARTH PRESSURE

2

2

2

)sin().sin()sin().sin(

1sin.sin

)(sinKp

Pp = ½ Kp . . H2

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LATERAL EARTH PRESSURE DUE TO SURCHARGE

222

2

ba

ba.

nH

q2

222

2

ba

ba.

nH

q4

22b16.0

b203.0.

H

q

a > 0,4

a 0,4

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LATERAL EARTH PRESSURE DUE TO SURCHARGE

2cos.sinH

q

12H90

qP

12

122

H2

H'a30,57QRHz

H

'btan 1

1

H

'b'atan 1

2

22 90'b'aR

12 90'bQ

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PURPOSE OF LATERAL EARTH PRESSURE

• STABILITY ANALYSIS GRAVITY WALL AGAINST– SLIDING

– OVERTURNING

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SHEET PILE STRUCTURES

SESSION 13 – 14

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SHEET PILE

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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.

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SHEET PILE TYPES (CANTILEVER)

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SHEET PILE TYPES (ANCHORED)

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anchor plate or beam


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tie back vertical anchor pile


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anchor beam with batter piles


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LATERAL EARTH PRESSURE DIAGRAM

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Fixed Earth Support


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Free Earth Support


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Free Earth Support


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CALCULATION STEPSCANTILEVER SHEET PILE - SAND

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245tan

245tan

2

2

p

a

K

K

ap KK

pLLz

'

23

1. Determine the value of Ka and Kp

2. Calculate p1and p2 with L1 and L2 are known

3. Calculate L3

a

a

KLLp

KLp

212

11

'..

..

4. Calculate the resultant of the area ACDE (P)

5. Determine the z (the center of pressure for the area ACDE)

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6. Calculate p5

app KKLKLLp 3215 '.'..

7. Calculate A1, A2, A3, A4

ap KK

pA

'.

51

ap KK

PA

'.

82

22

53

'

'..26

ap

ap

KK

pKKzPA

22

54

'

4..6

ap KK

PpzPA

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8. Determine L4 0443242

341

44 ALALALAL

ap KKLpp 454 '.

43 )(' LKKp ap 9. Calculate p4

10. Calculate p3

11. Calculate L543

435

2.

pp

PLpL

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

').(

2'

3

')()''.(.2

1' 2max

ap

ap

KK

Pz

with

zKKzzzPM

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EXAMPLECANTILEVER SHEET PILE - SAND

GWL

d = 15.9 kN/m3

t = 19.33 kN/m3

= 32o

c = 0 kPa

L1 = 2 m

L2 = 3 m

D

Determine the penetration depth (D) and dimension of sheet pile

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• Step 1 (determine the value of ka and kp)

• Step 2 (calculate p1 and p2)

• Step 3 (Calculate L3)

25.32

3245tan

245tan

307.02

3245tan

245tan

22

22

p

a

K

K

53.18

307.03)81.933.19()2)(9.15('..

763.9)307.0)(2)(9.15(..

2

212

11

p

KLLp

KLp

a

a

66.0)307.025.3)(81.933.19(

53.18

'2

3

ap KK

pL

kPa

kPa

m

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• Step 4 (calculate P)

• Step 5 (calculate z)

• Step 6 (calculate p5)

32.58115.6151.13289.29763.9

)66.0)(53.18(5.03)763.953.18(5.0)3)(763.9()2)(763.9(5.0

..5.0..5.0...5.0 322122111

P

P

LpLppLpLpP

23.2

3

266,0115.6

3

366.0151.13

2

366.0289.29

3

2366.0763.9

32.58

1

z

z

66.214

)307.025.3)(66.0)(81.933.19(25.33)81.933.19()2)(9.15(

'.'..

5

5

3215

p

p

KKLKLLp app

kN/m

m

kN/m2

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• Step 7 (calculate A1 – A4)

66.7)943.2)(52.9(

66.214

'.5

1

ap KK

pA

65.16)943.2)(52.9(

)32.58)(8(

'.

82

ap KK

PA

93.151

)943.2()52.9(

66.214)943.2)(52.9)(23.2)(2()32.58)(6(

'

'..26

223

22

53

A

KK

pKKzPA

ap

ap

72.230

)943.2()52.9(

)32.58)(4()66.214)(23.2)(6(32.58

'

4..62222

54

ap KK

PpzPA

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• Step 8 (determine L4)



072.23039.15165.1666.7 424

34

44 LLLL

L4 4.8 m

14.349)943.2)(8.4)(52.9(66.214

'.

4

454

p

KKLpp ap

48.134)8.4)(943.2)(52.9()(' 43 LKKp ap

kPa

kPa

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• Step 11 (Calculate L5)

• 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

09.114.34948.134

)32.58(2)8.4)(48.134(2.

43

435

pp

PLpL m

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Dimension of Sheet Pile

39.209

3

04.2)943.2()04.2)(52.9(5.0)04.223.2)(32.58(

3

')()''.(.2

1'

04.2)943.2(52.9

)32.58)(2(

').(

2'

max

2max

2max

M

M

zKKzzzPM

KK

Pz

ap

ap

m

kN.m

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SHEET PILE STRUCTURE

SESSION 15 – 16

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CALCULATION STEPSCANTILEVER SHEET PILE - CLAY

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245tan

245tan

2

2

p

a

K

K



a

a

KLLp

KLp

212

11

'..

..

3. Calculate the resultant of the area ACDE (P1) and z1 (the center of pressure for the area ACDE)

In case of saturated soft clay with internal friction angle () = 0, we got

Ka = Kp = 1

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0

.2'..

..12...2'...4

21

111121

2

cLL

zcPPPDLLcD

4. Calculate the theoretical penetration depth of sheet pile (D)

5. Calculate L4

c

PLLcDL

.4

'...4 1214

6. Calculate p6 and p7

216 '...4 LLcp 217 '...4 LLcp

7. Obtain the actual penetration depth of sheet pile

Dactual = (1.4 – 1.6) x Dtheoretical

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8. Calculate the maximum bending moment

with

zpzzPM

2

)'.('

26

11max

6

1'p

Pz

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EXAMPLECANTILEVER SHEET PILE - CLAY

GWL

d = 15.9 kN/m3

t = 19.33 kN/m3

= 32o

c = 0 kPa

L1 = 2 m

L2 = 3 m

D

Determine the penetration depth (D) and dimension of sheet pile

Clay

sand

cu = 47 kPa

= 0 o

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• Step 1 (Determine ka and kp)


• Step 3 (calculate P1 and z1)

00.12

045tan

245tan

307.02

3245tan

245tan

22

22

p

a

K

K

53.18

307.03)81.933.19()2)(9.15('..

763.9)307.0)(2)(9.15(..

2

212

11

p

KLLp

KLp

a

a

2.52151.13289.29763.92

1

2

1

1

21221111

P

LppLpLpP

kPa

kPa

kN/m 78.1

2.5233

151.1323

289.2932

3763.9

1

1

z

z

m

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• Step 4 (obtain Dtheoretical)

• Step 5 (calculate L4)

0

472381.933.1929.15

78.147122.522.522.522381.933.199.152474

0.2'..

..12...2'...4

2

21

111121

2

DD

cLL

zcPPPDLLcD

015.3574.10464.127 2 DD D = 2.13 m

474

2.52381.933.1929.1547413.2.4

'...4

4

1214

L

c

PLLcDL

L4 = 2.13 m

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• Step 7 (draw the lateral diagram)

• Step 8 (Obtain Dactual)

64.127'...4 216 LLcp

36.248'...4 217 LLcp

kN/m2

kN/m2

Dactual = 1.5 x Dtheorical = 1.5 x 2.13 = 3.2 m

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• Calculation of moment

mkNM

zpzzPM

59.1032

41.064.12741.078.12.52

2

)'.('

2

max

26

1max

mp

Pz 41.0

64.127

2.52'

6

1

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CALCULATION STEPSANCHORED SHEET PILE – FREE – SAND

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245tan

245tan

2

2

p

a

K

K



a

a

KLLp

KLp

212

11

'..

..

3. Calculate L3

ap KK

pLLz

'

23

4. Calculate P as a resultant of area ACDE

5. Determine the center of pressure for the area ACDE ( z )

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0'

35,1 1321

32224

34

ap KK

lzLLLPLLlLL

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

6. Calculate L4

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CALCULATION STEPSANCHORED SHEET PILE – FREE – CLAY

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245tan

245tan

2

2

p

a

K

K



a

a

KLLp

KLp

212

11

'..

..

3. Calculate the resultant of the area ACDE (P1) and z1 (the center of pressure for the area ACDE)

In case of saturated soft clay with internal friction angle () = 0, we got

Ka = Kp = 1

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216 '4 LLcp

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

4. Calculate p6

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CALCULATION STEPSANCHORED SHEET PILE – FIXED – SAND

J

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245tan

245tan

2

2

p

a

K

K



a

a

KLLp

KLp

212

11

'..

..

3. Calculate L3

ap KK

pLLz

'

23

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4. determine L5 from the following curve (L1 and L2 are known)

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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

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'

1'L

P

''6

2.15 ap KK

PLD

'

1

LF

8. Calculate P’ by taking the moment about O’, or

9. Determine D

10. Calculate the anchor force per unit length, F, by taking the moment about l, or

(moment of area ACDJI about O’)

(moment of area ACDJI about I)

Documents

RETAINING EARTH STRUCTURE Session 11 – 16 Course: S0825/Foundation Engineering Year: 2009