Diaphragm Sheetpile

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Construction

Grab for cohesionless soil

Hydrofraise for very cohesive soils 2

Construction

• Vibratory hammer • Hammer 3

Similar method of design • Definition of the dimension • Embedment • Type of sheetpile or thickness of the slurry wall

• Design of the anchor system if required • Number • Length • Kind

• Global stability analysis • Analysis of hydraulic flow under the wall

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• Rigid plastic method

• Elastic plastic method
Modulus of subgrade reaction method

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• Bidimensional analysis • Rigid plastic behavior • Stresses normal to the wall (Rankine theory) • Strains in the wall remain small

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Deflection y

« Rotation point »

O

Active earth pressure

Passive earth pressure 7

Maximum active earth pressure

z H

Maximum passive earth pressure

A f0 0.2f0

Max counter passive earth pressure

O Resulting diagram of earth pressure

Max counter active earth pressure

Simplified model

D

FC

Final model

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• Hypotheses • Rotation around O • Balance equations of the system (unknown : f0, Fc) Active pressure : a ( z ) = K a .γz Passive pressure : b( z ) = − K p .γ ( z − H ) For point A :

r r ∑F = 0 r r ∑ M /o = 0

a ( z ) + b( z ) = 0 ⇒ z A =

K pH K p − Ka

⇒ D = 1.2 f 0 +

Ka H K p − Ka

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• Hypotheses Point B : 0 ≤ y ( B ) ≤

T k

T

d

B H

Point O : end restraint

y (O ) = 0 y′(O) = 0 Unknowns :

T , f0 , Fc

A D

FC O


we state : y ( B ) = 0

r r ∑F = 0 r r ∑ M /o = 0

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• Solvation of the problem
Iterative method


Blum method (equivalent beam) Hyp : M f ( A) = 0

y′(O) = 0 rejected B

T

A

R

A

FC

R

O 11

• In case of several rows of anchors
displacements restrained


plastic equilibrium is no more correct


Elastic plastic method

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z≤H σ H′ (z ) p p = K p .σ v′ ( z )

p i = K 0 .σ v′ ( z )

pi

p a = K a .σ v′ ( z )

y

yp

ya

• Elastic domain :

: at rest pressure

ya ≤ y ≤ y p

p ( y ) = pi + K h ( y ). y

Kh

: subgrade reaction modulus

• Plastic domain :

y ≤ ya

y ≥ yp

p ( y ) = pa

ya ≈ 0.0005 H

p( y) = p p

y p ≈ 0.005H

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p out

z≥H

K h out

p a out

p p out

y

p a in

K h in

p p in

p in p = pout − pin 3

y

p a out − p p in

1

2

pp out − pa in

1 : K h in 2: K h in + K h out 3 : K h out


5 linear portions

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d2y EI 2 = − M dz

Bending moment for beam theory

∂M ∂2M T =− ⇒ dT = pdz = − 2 dz ∂z ∂z d4y EI 4 = p ( y, z ) dz

with

Shear force

p( y, z ) = pi ( z ) + K h ( y, z ). y

solvation with numerical technics

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loading stage p

Assumption :

new state
M’

M moves to M’ to an « elastic » state

∆σ v

Case 1 :

K p ∆σ v

K 0∆σ v

K a∆σ v

M : initial state

y Case 2 :

new state
M’ K 0∆σ v

M : initial state

K 0 ∆σ v

new state
M’

Case 3 :

K 0∆σ v

M : initial state 16

unloading stage p

Assumption :

K p ∆σ v

M does not change Case 1 : ∆σ v

Soil

K a∆σ v

M =M’

y

removed

Case 2 : M=M’

K 0 ∆σ v

M

Case 3 : M’

exception !!! 17

Displacements y(z) Anchor

Reality : arching between anchor and foundation

Assumption for reaction modulus method

Assumptions : • No shear component on the wall • Soil : independent layers Correct estimate of pressure on the wall Bad estimate of wall displacements

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• From pressumeter tests • Correlations with site experiments


industrial software

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