Very Good Lecture - Strength, Retaining Wall & Slope Stability

ENGR-627 Performance Evaluation of Constructed Facilities, Lecture # 4

Performance Evaluation of Constructed Facilities Fall 2004 Prof. Mesut Pervizpour Office: KH #203 Ph: x4046

1 Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Soil Strength

2 Dr. Mesut Pervizpour

ENGR-627 Fall 2004

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Soil Strength Shear Strength of Soil (τ): Internal resistance of soil / unit area. MOHR-COULOMB Failure Criteria: Theory of rupture for materials Æ failure under combined σ and τ Æ any stress state that combined effect reaches the failure plane Along the failure plane τf = f(σ) Failure envelope is a curved line Æ approximated by linear relationship Mohr-Coulomb failure criteria: τf = c + σ tanφ In terms of effective parameters: τf = c’ + σ’ tanφ’

τ

Mohr’s failure envelope

Cohesion

φ: internal friction angle Mohr-Coulomb failure criteria

c σ 3 Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Soil Strength Inclination of the Plane of Failure Caused by Shear: Failure

when shear stress on a plane reaches τf (line) determine inclination (θ) of failure plane with major & minor principal plane h τ

Æ Æ σ1

A

B

σ3

F D E

θ σ1

σ3

g f

C

c

φ

e O σ3

σ1 > σ3

fgh Æ failure plane s = c + σ tanφ ab Æ major principal plane ad Æ failure plane Æ θ to 2θ angled Angle bad = 2θ = 90 + φ Î θ = 45 + φ/2

τf = c + σ tanφ

d

2θ a

σ1

b σ

σ1 = σ3 tan2(45+φ/2) + 2c tan(45+φ/2) Similarly for effective parameters. Shear failure for saturated soils: τf’ = c’ + σ’ tanφ’

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Soil Strength Shear Strength Parameters in Laboratory: Unconfined Compression Test of Saturated Clay: Æ A type of unconsolidated-undrained triaxial test Æ For clayey samples (Cohesive) Æ σ3 = 0 (confining pressure) Æ Axial load (σ1) applied to fail the sample (relatively rapid) Æ At failure σ3f = 0 and σ1f = major principal stress Æ Therefore undrained shear strength is independent of confining pressure τf = σ1 / 2 = qu / 2 = Cu or Su

σ1

τ

qu: unconfined compressive strength, cu (Su): undrained shear strength

σ1 Cu or Su

φ=0 Total stress Mohr’s Circle at failure

σ3

σ1 = qu

σ 5

Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Soil Strength Direct Shear Test (stress or strain controlled): Specimen is square or circular Box splits horizontally in halves Normal force is applied on top shear box Shear forces is applied to move one half of the box relative to the other (to fail specimen) Stress Controlled: Shear force applied in equal increments until failure Failure plane is predetermined (horizontal) Horizontal deformation & ∆H is measured under each load.

Loading plate

Shear Force

Normal force

Sample

τ

Porous Stone

Shear Stress

Strain Controlled: Constant rate of shear displacement Restraining shear force is measured Volume change (∆H) (Advantage: gives ultimate & residual shear strength)

Peak shear strength

Dense sand Loose sand

Ultimate shear strength

τf

Expansion

τ ∆H

Shear Box Dr. Mesut Pervizpour

τf Shear Displacement Dense sand Shear Displacement Loose sand

Compression

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Soil Strength Direct Shear Test (continued): Repeat Direct Shear under several normal stresses. Plot the normal stress vs. shear stress values.

τf

τf = σ tan φ c = 0 for dry sand and σ = σ’ φ = tan-1(τf / σ)

Dry sand φ σ

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ENGR-627 Fall 2004

Soil Strength Drained Direct Shear Test on Saturated Sand & Clay: Test conducted on saturated sample at slow rate of loading Æ allowing excess pore water to dissipate. For sand (k is high Æ pwp dissipates quickly) Therefore φ under drained conditions ~ same For clay (k is low Æ under load consolidation takes time, therefore load needs to be applied very slow).

τf

General Comments on Direct Shear Test:

OC clay

φ’ c’

τf = c’ + σ’ tan φ’ NC clay, c=0

τf = σ’ tan φ’

Failure is not along the weakest plane (forced at horizontal plane) Represents angle of friction between soil and foundation material:

τf = ca + σ’ tan δ Ca: adhesion

φ’ σ

δ: angle of friction between soil and foundation material 8

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ENGR-627 Fall 2004

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Soil Strength Triaxial Shear Test: Reliable method for determination of shear strength parameters.

cap

σ1 membrane

σ3

σ3 Porous stone

Axial stress (deviator stress) is applied to cause failure (shear) by vertical loading. Load vs. deformation readings are recorded. Three general types of triaxial test are: 1. Consolidated – drained test (CD) 2. Consolidated – undrained test (CU) 3. Unconsolidated – undrained test (UU)

σ1 σ3: confining pressure applied all around sample (air/water/glycerine)

∆σd σ3 σ3

σ3 Porous stone

σ3 σ = σ3 + ∆σd ∆σd 1 9 Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Soil Strength Triaxial Shear Test: Consolidated-drained test: Specimen is subjected to confining stress σ3 all around. As a result the pwp of the sample increases by uc. If the valve is opened at this point the uc will dissipate and sample will consolidate (∆V decreases under σ3) σ3 σ3

σ3

B=

uc

σ3

Skempton’s pwp parameter (B~1.0 for saturated soils)

σ3 ∆σd σ3 σ3

σ3 ud = 0 σ3 ∆σd

End of consolidation stage uc = 0. Application of deviator stress (∆σd): For drained test ∆σd is increased slowly, while the drainage valve is kept open, & any excess pwp generated by ∆σd is allowed to dissipate. (∆V can be measured by measuring amount outflow-water, since S=100%) CD test Æ excess pwp completely dissipated Æ σ3 = σ3’ 10 Dr. Mesut Pervizpour

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Soil Strength Triaxial Shear Test: Consolidated-drained test (Continued): At failure (Axial stress) Æ σ1 = σ1’ = σ3 + (∆σd)f σ1’ Æ major principal stress at failure σ3’ Æ minor principal stress at failure Conduct other triaxial (CD) tests under different σ3 (confining) pressure and obtain the corresponding σ1’ at failure and plot the Mohr’s circle for each test. σ1

τ

τf =

θ = 45 + φ / 2

σ3

φ

Total and Effective Stress Failure Envelope

B

σ3

for OC clays

σ’ t

’ anφ

A

σ1 φ1 c

σ3 = σ3’

O

2θ

σ1 = σ1’

σ

2θ

(∆σd)f (∆σd)f 11 Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Soil Strength Triaxial Shear Test: Consolidated-undrained test (CU): Consolidation of S=100% sample under σ3 (confining stress) & allow uc to dissipate. Drainage valve is closed after complete consolidation (uc = 0) Deviator stress (∆σd) is applied and increased to failure. ∆ud is developed (due to no drainage). σ3 σ3

σ3

σ3 ∆σd σ3 σ3

σ3 ∆ud ≠ 0 σ3 ∆σd

End of consolidation stage uc = 0 (and close valves).

A=

∆ud ∆σ d

Skempton’s pwp parameter

Loose sand & NC clay Æ Dense sand & OC clay Æ

∆ud increases with strain ∆ud increases with strain up to a certain point and drops & becomes negative (due to dilatation of soil) 12

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Soil Strength Triaxial Shear Test: Consolidated-undrained test (Continued): Total and Effective principal stresses are not the same. At failure measure (∆σd)f and (∆ud)f Major principal stress at failure is obtained as: Total: σ3 + (∆σd)f = σ1 Effective: σ1 - (∆ud)f = σ1’ Minor principal stress at failure is obtained as: Total: σ3 Effective: σ3 - (∆ud)f = σ3’

τ

Mohr’s Circle for CU Test:

= τf φ’

Note: σ1 - σ3 = σ1’ - σ3'

n ta σ’

φ’ Effective Stress

Failure Envelope

τf =

nφ c u σ ta Total Stress

Failure Envelope

B

A O

σ3’ (∆ud)f

σ

σ1

σ1’

σ3

φcu

(∆ud)f 13 Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Soil Strength Triaxial Shear Test: Consolidated-undrained test (Continued): For OC Clay: φcu

τ

τf =

for OC clays

τ f = c cu

+ σ ta

φ1cu

A

A = Af =

σ3’

(∆ud ) f (∆σ d ) f

σ3

nφ c u

n φ 1cu

ccu O

σ ta

σ1’

B

σ

σ1

0.5 Æ 1 for NC clay -0.5 Æ 0 for OC clay

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Soil Strength Triaxial Shear Test: Unonsolidated-undrained test (UU): Drainage in both stages is not allowed. Therefore application of

σ3

And application of

∆σd Æ

u = uc + ∆ud

Æ

uc = B σ3 ∆ud = Ặ ∆σd

u = B σ3 + Ặ ∆σd = B σ3 + Ặ (σ1 - σ3)

Æ

It can be seen that tests conducted with different σ3 results in the same (∆σd)f, resulting in mohr’s circle with same radius. τ

Effective

φ φ = 0 Failure envelope

Cu σ3’

σ3

σ1’

σ3

σ1 σ1

σ

σ1’ = [σ3 + (∆σd)f] – (∆ud)f = σ1 - (∆ud)f σ3’ = σ3 - (∆ud)f Example: σ3 ↑ by ∆σ3 ⇒ ∆uc = ∆σ3 σ3’ = σ 3 + ∆σ3 - ∆uc = σ3

Æ (∆σd)f will be the same. Dr. Mesut Pervizpour

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Soil Strength General Comments on Triaxial Tests: Failure plane not predetermined Field strength Æ function of rate of application of load and drainage Granular soil Æ drained shear strength parameters NC Clay

Æ Under footing Æ Undrained conditions

Excavation in OC Clay

Æ Drained case (more critical)

Control of stress states are possible in Triaxial test

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Soil Strength Undrained Cohesion of NC and OC Deposits: NC clay Æ undrained shear strength cu or Su increase with effective overburden pressure cu / σ’ = 0.11 + 0.0037 (PI)

Skempton (1957) Ladd for OC clas (1977)

{PI: in %}

(cu/σ’)OC / (cu/σ’)NC = (OCR)0.8

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ENGR-627 Fall 2004

Soil Stresses At A Point Due to Poisson’s effect Æ lateral flow (creep) εx = µ εz

0.0 ≤ µ ≤ 0.5

K Æ Ratio of lateral to vertical stress:

K = σh / σv

Kf Æ Maximum strength failure line K0 < 1

NC soils

K0 < 1

Slightly OC soils Æ OCR < 3

K0 > 1

Highly OC soils Æ OCR > 3

z h

x

σv = γt h σh

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General Comments

CD Æ Long-term Stability (earth embankments & cut slopes) CU Æ Soil initially fully consolidated, then rapid loading (slopes in earth dams after rapid drawdown) UU Æ End of construction stability of saturated clays, load rapidly & no drainage (Bearing capacity on soft clays)

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Slope Stability

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Slope Stability Slope Stability: The engineering assessment Of the stability of natural and man-made Slopes as influenced by natural or induced Changes to their environment. Studied by analytical (closed-form) or numerical (approximate) methods. Both methods are simplification of actual Geological, mechanical and other aspects.

The stability of a slope depends on its ability to sustain the effects of load increases or environmental changes. Pre-failure analysis: to assess safety of slope and its intended performance. Post-failure analysis: study of failure and processes causing it. 21 Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Slope Stability Slope Stability analysis (continued): Determination of shear stress developed on the most likely rupture surface and comparing to shear strength of soil. Likely rupture surface: is the critical surface with minimum factor of safety.

Steepened Slope to Wall To increase space

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Slope Stability The effective evaluation of slope stability requires: •

Site characterization (geological – hydrological conditions)

•

Groundwater conditions (pore pressure model)

•

Geotechnical parameters (strength, deformation, drainage)

•

Mechanisms of movement ( kinematics – potential failure modes)

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Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Landslide Components

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Landslide Components

Varnes (1978), Morgenstern (1985) 25 Dr. Mesut Pervizpour

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Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Rotational Slides

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Slope Stability Components of Slopes Facing

Crest

Toe Slope angle Foundation

Reinforcement Reinforced fill

Retained Fill Foundation

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Slope Stability Possible Failure Modes of Slopes

Local failure

Surficial failure

Slope failure Global failure

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Slope Stability Typical Surfical Failure: •

Shallow failure surface up to 1.2 m (4ft)

•

Failure mechanisms: –

Poor compaction

–

Low overburden stress

–

Loss of cohesion

–

Saturation

–

Seepage forces

Original ground surface

Slip Surface

Slide Mass

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ENGR-627 Fall 2004

Slope Stability Analytical Solutions – Limit Equilibrium: •

Widely applied analytical technique, where force (or moment) equilibrium

•

The analyses is based on material strength, rather than stress-strain

conditions are determined based on statics. relationships. •

A “Factor of Safety”, is defined as a tool of evaluating the slope stability with limit equilibrium approach.

FS =

resisting forces shear strength of material = driving forces shear stress required for equilibrium

Where FS > 1.0 represents a stable slope and FS < 1.0 stands for failure. Required values: Limit Equilibrium: FS = 1.0 Under Static Loads: FS ≥ 1.3 – 1.5 Under Seismic Loads: FS ≥ 1.1 30 Dr. Mesut Pervizpour

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Slope Stability Limit Equilibrium: Overall measure of the amount by which the strength of the soil would have to fall short of the values described by c and φ in order for the slope to fail.

FS =

FS =

resisting forces shear strength of material = driving forces shear stress required for equilibrium c + σ tan φ

τ eq

=

τf τd

τf : Average Shear strength of soil τd : Shear stress developed on potential surface

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ENGR-627 Fall 2004

Slope Stability Limit Equilibrium (continued): Fundamentals of limit equilibrium method (Morgenstern, 1995): • Slip mechanism results in slope failure • Resisting forces required to equilibriate disturbing mechanisms are found from static solution • The shear resistance required for equilibrium is compared with available shear strength in terms of Factor of Safety •The mechanism corresponding to the lowest FS is found by iteration

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Slope Stability Stability of Infinite Slopes without Seepage (Surficial slope stability): Soil Shear Strength: τf = c’ + σ’ tanφ’ Pore water pressure: u=0 Failing along AB at a depth H Static equilibrium of forces on the block. Assume F on ab and cd are equal. Along line AB: Developed resistance: τf = cd’ + σ’ tanφd’ = cd’ + γ H cos2β tanφd’ Driving force due to weight: τd = γ H cosβsinβ

2c tan φ + γ H sin 2β tan β

For c = 0:

FS =

tan φ tan β

FS = 1 Æ H = Hcr

d

L a

β

Factor of Safety:

FS =

Forces: Na = γ L H cosβ Ta = γ L H sinβ σ‘ = γ L H cos β / (L/cosβ) = γ H cos2β τ= γ L H sinβ / (L/cosβ) = γ H cosβsinβ Nr = γ L H cosβ Tr = γ L H sinβ

Na

F

W β

B

Ta

F H b

β

c

Tr

A

R

β Nr 33

Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Slope Stability Stability of Infinite Slopes with Seepage (Surficial slope stability): Soil Shear Strength: τf = c’ + σ’ tanφ’ Forces: GWT at surface, pore pressure u=γwh= γwHcos2β Na = γsat L H cosβ Failing along AB at a depth H Ta = γsat L H sinβ σ = γsat L H cos β / (L/cosβ) = γsat H cos2β Static equilibrium of forces on the block. τ= γsat L H sinβ / (L/cosβ) = γsat H cosβsinβ Assume F on ab and cd are equal. Nr = γsat L H cosβ Along line AB: Tr = γsat L H sinβ Developed resistance: h= Hcos2β τf = cd’ + σ’ tanφd’ = cd’ + (σ-u) tanφd’ d = cd’ + (γsat - γw) H cos2β tanφd’ L Driving force due to weight: E a PAG τd = γ H cosβsinβ SEE Factor of Safety: W β F Na 2 c' γ ' tan φ ' β FS = +

γ sat H sin 2β

B

γ sat tan β

H

For c = 0:

FS =

γ ' tan φ ' γ sat tan β

FS = 1 Æ H = Hcr

Ta

F

b

β A

c

Equipotential line

Tr R

β Nr 34

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Slope Stability Slope Stability with Plane Surface: AC Æ Trial failure place B

Na

C

W θ Ta

H Tr A

β θ

For c = 0:

Factor of Safety:

FS =

2 c sin β + H γ sin (β − θ ) cos θ tan φ Hγ sin (β − θ )sin θ

FS =

tan φ tan β 35

Dr. Mesut Pervizpour

ENGR-627 Fall 2004

Slope Stability Modes of Failure of Finite Slopes:

Shallow slope failure

Base failure

Slope failure

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Slope Stability Circular surface – Slip circle analysis (φ = 0): Circular slip surfaces are found to be the most critical in slopes with homogeneous soil. There are two analytical, statically determinate, methods used for FS: the circular arc (φ=0) and the friction circle method.

FS = Circular failure surface in φ=0 soil is defined by its undrained strength, cu.

FS =

M r cu LR resisting moment = = Md Wx driving moment

cu R 2θ Mr = M d W1l1 − W2l2

W1 W2

l2

l1

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Slope Stability Circular surface – Friction circle (φ, c soil): Trial circle through toe. The friction circle method attempts to satisfy the requirement of complete equilibrium by assuming that the direction of the resultant of the normal and frictional component of strength mobilized along the failure surface corresponds to a line that forms a tangent to the friction circle with radius: Procedure (Abramson et al 1996 more detailed) C parallel to ab P passes through intersection W-C P makes φm with line through center of friction circle, & tangent to FC U often taken 0 Force polygon Æ determine C Critical circle Æ developed cohesion is maximum For FS = 1, the critical height: C’ / (γ Hcr) = f(α, β, θ, φ’) = m (stability No.)

Rf = R sinφm

β P φm

φ > 3 deg Æ critical circles all toe circles 38 Dr. Mesut Pervizpour

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Slope Stability Method of Slices (limit equilibrium): Soil divided to vertical slices, width of each can vary. The previous methods do not depend on the distribution of the effective normal stresses along the failure surface. The contribution is accounted for by dividing the failing slope mass into smaller slices and treating each individual slice as a unique sliding block. Non-circular:

Circular:

The discretization of the slip surface to elements results in two force components acting on each: Normal and Shear forces. The other unknown is the location of line of action of the normal force for each element. However the equilibrium conditions: ΣFx=0, ΣFy=0, ΣM=0 No. of unknowns = No. of slices * 3 Therefore assumptions should be made.

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Slope Stability Circular surface (Bishop method): Soil divided to vertical slices, width of each can vary. Can be applied to layered soil, with different properties. Find minimum FS by several trials. ΣM0 = 0 n

FS =

∑ (c' ∆l i =1

i

+ Wi cos α i tan φ ')

n

∑ (W sin α ) i =1

i

i

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Slope Stability Search for Minimum Factor of Safety: Minimum FS values for the failure surface for every center is obtained, and recorded by the center of rotation, the contours indicate the location of the center with minimum overall FS.

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Slope Stability Slope Stability with Seepage (u ≠ 0): Obtain the average pwp at the bottom of the slice using the phreatic line. Total pwp for the slice is un ∆Ln

Phreatic surface h z H

Seepage

β FS modified (from Bishop method) for pore pressure: n

FS =

∑ [c' ∆l + (W − u ∆l )cos α i =1

i

i

i

i

i

tan φ ']

n

∑ (W sin α ) i =1

i

i

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Lateral Earth Pressure

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Lateral Earth Pressure Lateral Earth Pressure Coefficient:

H

σz’ σx’

P=(1/2)K γ H2 1/3 H

K=σx’/σz’

σx’ = Kσz’= KγH

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Lateral Earth Pressure Lateral Earth Pressure Coefficient at Rest: Relationship between σz’ and σx’ at a given depth (at rest means no shear). Ko : Coefficient of earth pressure at rest, Ko

= σx’ / σz’

Rigid Wall No movement

H

σz’ σx’

P=(1/2)K γ H2 1/3 H

K=σx’/σz’

σx’ = Kσz’= KγH

For coarse-grained soils: (ok for loose sand) Ko = 1 - sinφ’ For fine grained NC soils: Ko = m - sinφ’ m: 1 for NC cohesionless or cohesive m: 0.95 OCR > 2 Massarch (1979) Ko = 0.44 + 0.42 (PI% / 100) For OC clays: Ko = Ko(NC) (OCR)(1/2) Or Ko = (1 - sinφ’) OCRsinφ’

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Lateral Earth Pressure Coefficient of Active Lateral Earth Pressure: Wall moves away from the soil (pushed out).

Movement

H

σz’ σx’

Ka=σx’/σz’

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Lateral Earth Pressure Wall Movement Required to Reach the Active Condition: Soil Type

Horizontal movement required to reach the active state

Dense sand

0.001 H

Loose sand

0.004 H

Stiff clay

0.010 H

Soft clay

0.020 H

(From CGS, 1992)

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Lateral Earth Pressure Coefficient of Passive Lateral Earth Pressure: Wall moves towards the soil (pressed in).

Movement

H

σz’ σx’

Kp=σx’/σz’

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Lateral Earth Pressure Wall Movement Required to Reach the Passive Condition: Soil Type

Horizontal movement required to reach the passive state

Dense sand

0.020 H

Loose sand

0.060 H

Stiff clay

0.020 H

Soft clay

0.040 H

(From CGS, 1992)

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Lateral Earth Pressure In Summary: 1. 2. 3.

If the wall moves away from the fill (soil) pressure will decrease and reach to active state. (σh = Ka σv) If the wall moves towards the fill (soil) pressure will increase and reach to passive case. (σh = Kp σv) More deformation is generally required to achieve passive case than the active case.

Kp

Ko Ka Movement away From backfill

Movement towards backfill 50 Dr. Mesut Pervizpour

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Lateral Earth Pressure Classical Lateral Earth Pressure Theories: •

Coulomb’s Earth Pressure Theory (1776)

•

Rankine’s Earth Pressure Theory (1857)

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Lateral Earth Pressure Rankine’s Earth Pressure Theory: Assumptions: • The soil is homogeneous and isotropic • Frictionless wall • Failure surfaces are planar • The ground surface is planar • The wall is infinitely long (plane strain condition) • At the active or passive state (plastic equilibrium, every point in soil about to fail) • The resultant on the back of the wall is at angle parallel to ground surface

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Lateral Earth Pressure Rankine’s Earth Pressure Theory: Attainment of Rankine’s Active State

Attainment of Rankine’s Passive State

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Lateral Earth Pressure Rankine’s Earth Pressure Theory – Force Diagram: β

C

W

β P A

θ

T

Rankine’s Earth Pressure Theory Force Equilibrium N

P β

N

T

W θ

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Lateral Earth Pressure Rankine’s Theory – Critical Angle of Failure Plane: Critical angle of failure plane: The angle (θ) when the thrust (P) reaches the maximum value for the condition or the minimum value for the passive condition At the active state: θcritical = 45o + φ / 2 At the passive state: θcritical = 45o - φ / 2

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Lateral Earth Pressure Rankine’s Theory – Earth Pressure Distribution (c’=0): β

H

P = (1/2) K γ H2 β

PH = P cosβ = (1/2) K γ H2 cosβ

H/3 β

σ = K σz = K γ H

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Lateral Earth Pressure Rankine’s Theory – Coefficient of Active Earth Pressure: For β ≤ φ’:

For β = φ’:

Ka =

cos β − cos 2 β − cos 2 φ ' cos β + cos 2 β − cos 2 φ '

K a = tan 2  45 o − φ '  2 

Rankine’s Theory – Coefficient of Passive Earth Pressure:

For β ≤ φ’:

For β = φ’:

Kp =

cos β + cos 2 β − cos 2 φ ' cos β − cos 2 β − cos 2 φ '

K p = tan 2  45 o + φ '  2 

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