Earth Walls

1

Bridge abutments

Walls to support roadways, either single, or tiered

2

Reinforced slopes

Walls for access ramps 3

Facing : recast concrete panels

high adherence galvanised steel strips

4

A > 1m

A

H

D

β

D > 0.40m L # 0.8H

L

D/H 1/5 1/7 1/10 1/20 18°

27°

34°

β 5

+ differential settlements <1/100 6

• Excessive tensile forces within strips
elongation or breakage of strips • Excessive induced shear stresses along strips ( and within the surrounding backfill)
failure by pullout

7

• Aims of design :
Determining maximum developed tension forces
Location along a probable critical slip surface
Testing the resistance provided by strips (pullout capacity & tensile strength)

8

Tmax T0

τ

True maximum tensile loci

dl T

τ

T+dT

τ =

dT 2b.dl 9

# 1mm for 75 years

T Tmax T0

τ

with T0 ≤ 0.75Tmax

∆H ∆H

Tmax

Tmax = σ H′ .∆H per unit width of wall

σ H′ = K .σ v′ • 0 ≤ z ≤ z0 = 6 m

 z z  K = K 0  1 −  + K a  z0  z0

• z0 ≥ 6 m

K = Ka

K 0 = 1 − sinφ π φ  K a = tan2  −   4 2 10

1 Fa (z) = K a z 2 2

σ v′ (z ) 2e L

z

Equivalent uniform vertical stress (Meyerhoff method) (per unit width of wall)

σ v′ ( z ) =

Rv ( z ) L − 2e

with

e=

M (z ) Rv (z)

M (z ) =

1 K a .γ .z 3 6

Rv (z ) = γ .z.L

11

• Design criterion : elongation or breakage of strips

Tmax

1 ≤ nσ a .b.Ec FSb FSb = 1.5 n : number of strips by unit width of wall σ a : allowed maximum stress in strips

12

• Design criterion : pullout resistance

1 ≤ 2b.n ∫ µ * ( z ).σ v′ ( x)dx La FS PO La L

Tmax

FS PO

: length of the active zone

: 1.35 for reinforced slopes : 1.5 for abutments

• 0 ≤ z ≤ z0 = 6 m • z0 ≥ 6 m

 z  z µ = µ 1 −  + tan φ ′  z0  z0 ′ = 36° µ * = tan φ ′ φmin *

* 0

µ0* = 1.5

µ * : coefficient of apparent friction 13

e

Ex: e=1m

14

Density:

n=5

15

Ex: 9m high

Ribbed strips HA:

φ ' = 34° γ = 20kN / m 3

Ec = 5mm b = 40mm σ a = 240 MPa ∆H = 0.73m

Pre-design: L/H=1 for reinforced slope !!! Long term calculation
strip thickness Ec= 5-1mm = 4mm

Tmax = σ H′ ( z ).∆ H

Tbreak =

1 nσ a .b.Ec FSb

1 2b.n * * = 2b.n ∫ µ ( z ).σ v′ ( x)dx ≈ µ ( z ).σ v′ ( z )( L − La ) FS PO FS PO La L

Tpullout

z

K

σ v′ (z ) σ H′ (z ) Tmax Tbreak

La

µ*

T pullout

0.75m 16

1.5m