ENGR-627-Lect4

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1Dr. Mesut Pervizpour ENGR-627 Fall 2004Dr. Mesut Pervizpour ENGR-627 Fall 2004

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

Performance Evaluation of Constructed Facilities

Fall 2004

Prof. Mesut PervizpourOffice: KH #203

Ph: x4046


Soil Strength

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Soil StrengthShear Strength of Soil (τ):

Internal resistance of soil / unit area.

Dr. Mesut Pervizpour ENGR-627 Fall 2004Dr. Mesut Pervizpour ENGR-627 Fall 2004

MOHR-COULOMB Failure Criteria:

Theory of rupture for materials failure under combined σ and τ

any stress state that combined effect reaches the failure plane

Along the failure plane τf = f(σ)

Failure envelope is a curved line approximated by linear relationship

Mohr-Coulomb failure criteria:

τf = c + σ tanφ

Cohesionφ: internal friction angle

c

σ

Mohr-Coulomb failure criteria

Mohr’s failureenvelope

τf = c’ + σ’ tanφ’In terms of effective parameters:

τ

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Soil StrengthInclination 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


fgh failure plane s = c + σ tanφab major principal planead failure plane θ to 2θ angledAngle bad = 2θ = 90 + φ

θ = 45 + φ/2

σ1 > σ3

σ3

σ1

σ3

σ1

θC

BA

D E

F

τf = c + σ tanφ

φc

σ

τ

O

h

σ3 σ1

d

2θa

be

g

f

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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5

Soil StrengthShear Strength Parameters in Laboratory:

Unconfined Compression Test of Saturated Clay:A type of unconsolidated-undrained triaxial testFor clayey samples (Cohesive)σ3 = 0 (confining pressure)Axial load (σ1) applied to fail the sample (relatively rapid)At failure σ3f = 0 and σ1f = major principal stressTherefore undrained shear strength is independent of confining pressure

τf = σ1 / 2 = qu / 2 = Cu or Su

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

Total stress Mohr’s Circle at failure

φ = 0Cuor

Su

σ

τ

σ3 σ1 = qu

σ1

σ1


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Soil StrengthDirect Shear Test (stress or strain controlled):

Specimen is square or circularBox splits horizontally in halvesNormal force is applied on top shear boxShear forces is applied to move one half of the box relative to the other (to fail specimen)


Sample

Porous Stone

Normal force

ττ

Loadingplate

Shear Force

Shear Box

Stress Controlled: Shear force applied in equal increments until failureFailure plane is predetermined (horizontal)Horizontal deformation & ∆H is measured under each load.

Strain Controlled: Constant rate of shear displacementRestraining shear force is measuredVolume change (∆H)(Advantage: gives ultimate & residual shear strength)

τf τf

Densesand

Loosesand

Peak shear strength

Ultimate shear strength

She

ar S

tress

Shear Displacement

Densesand

Shear Displacement

Loose sand

∆H

Expansion

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

φ

τf = σ tan φ

σ

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

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Soil Strength

Drained Direct Shear Test on Saturated Sand & Clay:Test conducted on saturated sample at slow rate of loading allowing excess pore water to

dissipate.For sand (k is high pwp dissipates quickly) Therefore φ under drained conditions ~ sameFor clay (k is low under load consolidation takes time, therefore load needs to be applied

very slow).


τf

φ’

τf = c’ + σ’ tan φ’

σ

NC clay,c=0

φ’

c’τf = σ’ tan φ’

OC clay

General Comments on Direct Shear Test:

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

τf = ca + σ’ tan δCa: adhesionδ: angle of friction between soil and

foundation material

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Soil StrengthTriaxial Shear Test:

Reliable method for determination of shear strength parameters.


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

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

σ1

σ3σ3

membranecap

Porousstone

σ1 = σ3 + ∆σd

σ3

σ3σ3

Porousstone

∆σd

σ3∆σd

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Soil StrengthTriaxial 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

σ3σ3 3σcuB = Skempton’s pwp parameter (B~1.0 for saturated soils)

End of consolidation stage uc = 0.

Application of deviator stress (∆σd):For drained test ∆σd is increased slowly, while the drainage valveis kept open, & any excess pwp generated by ∆σd is allowed to dissipate.(∆V can be measured by measuring amount outflow-water, since S=100%)

σ3

σ3

σ3σ3

∆σd

∆σd

ud = 0

CD test excess pwp completely dissipated σ3 = σ3’

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Soil StrengthTriaxial Shear Test: Consolidated-drained test (Continued):

At failure (Axial stress) σ1 = σ1’ = σ3 + (∆σd)fσ1’ major principal stress at failureσ3’ 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.

τ f = σ’ tanφ’

φ

c σ

τ

O σ3 = σ3’2θ

σ1 = σ1’2θ

(∆σd)f

(∆σd)f

A

B

φ1

for OC clays

Total and Effective Stress Failure Envelope

σ1

σ1

σ3σ3

θ = 45 + φ / 2


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Soil StrengthTriaxial 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

σ3σ3

d

duAσ∆

∆= Skempton’s pwp parameter

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

σ3

σ3

σ3σ3

∆σd

∆σd

∆ud ≠ 0Loose sand & NC clay ∆ud increases with strainDense sand & OC clay ∆ud increases with strain up to a certain

point and drops & becomes negative (due to dilatation of soil)

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Soil StrengthTriaxial Shear Test: Consolidated-undrained test (Continued):

Total and Effective principal stresses are not the same.At failure measure (∆σd)f and (∆ud)fMajor principal stress at failure is obtained as:

Total: σ3 + (∆σd)f = σ1Effective: σ1 - (∆ud)f = σ1’

Minor principal stress at failure is obtained as:Total: σ3Effective: σ3 - (∆ud)f = σ3’

τ f = σ tanφcu

φcu

σ

τ

O σ3 σ1

(∆ud)f

AB

φ’Total Stress Failure Envelope


σ1’σ3’

τ f= σ’ tanφ’ Effective Stress

Failure Envelope

(∆ud)f

Note:σ1 - σ3 = σ1’ - σ3'

Mohr’s Circle for CU Test:

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Soil StrengthTriaxial Shear Test: Consolidated-undrained test (Continued):


τf = ccu + σ tanφ1cu

φcu

ccuσ

τ

O σ3 σ1

AB

for OC clays

σ1’σ3’

τ f = σ tanφcu

φ1cu

( )( ) fd

fdf

uAA

σ∆

∆==

For OC Clay:

0.5 1 for NC clay-0.5 0 for OC clay

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Soil StrengthTriaxial Shear Test: Unonsolidated-undrained test (UU):

Drainage in both stages is not allowed.Therefore application of σ3 uc = B σ3

And application of ∆σd ∆ud = Ặ ∆σd

u = uc + ∆ud u = B σ3 + Ặ ∆σd = B σ3 + Ặ (σ1 - σ3)It can be seen that tests conducted with different σ3 results in the same (∆σd)f, resulting in

mohr’s circle with same radius.


Failure envelopeφ = 0

Cu

σ

τ

σ3’ σ1’

φEffective

σ3 σ1

σ3 σ1

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 (∆σd)f will be the same.

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Soil Strength

General Comments on Triaxial Tests:

Failure plane not predetermined

Field strength function of rate of application of load and drainage

Granular soil drained shear strength parameters

NC Clay Under footing Undrained conditions

Excavation in OC Clay 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 undrained shear strength cu or Su increase with effective

overburden pressure

Skempton (1957) cu / σ’ = 0.11 + 0.0037 (PI) {PI: in %}

Ladd for OC clas (1977) (cu/σ’)OC / (cu/σ’)NC = (OCR)0.8


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Soil Stresses At A Point

Due to Poisson’s effect lateral flow (creep)

εx = µ εz 0.0 ≤ µ ≤ 0.5

K Ratio of lateral to vertical stress: K = σh / σv

Kf Maximum strength failure line

K0 < 1 NC soils

K0 < 1 Slightly OC soils OCR < 3

K0 > 1 Highly OC soils OCR > 3


σv = γt h

σh

h x

z

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General Comments

CD Long-term Stability (earth embankments & cut slopes)

CU Soil initially fully consolidated, then rapid loading

(slopes in earth dams after rapid drawdown)

UU End of construction stability of saturated clays, load rapidly & no

drainage (Bearing capacity on soft clays)



Slope Stability

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Slope Stability


Slope Stability: The engineering assessmentOf the stability of natural and man-madeSlopes as influenced by natural or inducedChanges to their environment.Studied by analytical (closed-form) or numerical (approximate) methods.Both methods are simplification of actualGeological, 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.

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Slope Stability

Steepened Slope to Wall


To increase space

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.

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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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Landslide Components


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Landslide Components


Varnes (1978), Morgenstern (1985)

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Rotational Slides


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Slope StabilityComponents of Slopes


CrestFacing

Toe

Slope angleFoundation

Reinforcement

Reinforced fill Retained

Fill

Foundation

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Slope StabilityPossible Failure Modes of Slopes


Surficial failure

Localfailure

Global failure

Slope failure

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Slope StabilityTypical Surfical Failure:


Original ground surface Slip Surface

Slide Mass

• Shallow failure surface up to 1.2 m (4ft)• Failure mechanisms:

– Poor compaction– Low overburden stress– Loss of cohesion– Saturation– Seepage forces

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Slope StabilityAnalytical Solutions – Limit Equilibrium:


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

conditions are determined based on statics.

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

relationships.

• A “Factor of Safety”, is defined as a tool of evaluating the slope stability with

limit equilibrium approach.

mequilibriu for requiredstress shearmaterial of strength shear

forces drivingforces resisting

==FS

Where FS > 1.0 represents a stable slope and FS < 1.0 stands for failure.Required values:

Limit Equilibrium: FS = 1.0Under Static Loads: FS ≥ 1.3 – 1.5Under Seismic Loads: FS ≥ 1.1

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Slope StabilityLimit Equilibrium:


mequilibriu for requiredstress shearmaterial of strength shear

forces drivingforces resisting

==FS

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.

d

f

eq

cFSττ

τφσ

=+

=tan

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

potential surface

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Slope StabilityLimit 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 StabilityStability of Infinite Slopes without Seepage (Surficial slope stability):


Soil Shear Strength: τf = c’ + σ’ tanφ’Pore water pressure: u = 0Failing along AB at a depth HStatic 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

β

β

L

F

FH

W

R β Nr

Na

Ta

Trb

c

a

d

Bβ

A

Forces:Na = γ L H cosβTa = γ L H sinβσ‘ = γ L H cos β / (L/cosβ) = γ H cos2βτ= γ L H sinβ / (L/cosβ) = γ H cosβsinβNr = γ L H cosβTr = γ L H sinβ

Factor of Safety:

βφ

βγ tantan

sin+=

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HcFS

For c = 0:

βφ

tantan

=FS

FS = 1 H = Hcr

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β

β

L

F

FH

W

R β Nr

Na

Ta

Trb

c

a

d

Bβ

A

h= Hcos2β

Equipotentialline

SEEPAGE

Slope StabilityStability of Infinite Slopes with Seepage (Surficial slope stability):


Soil Shear Strength: τf = c’ + σ’ tanφ’GWT at surface, pore pressure u=γwh= γwHcos2βFailing along AB at a depth HStatic equilibrium of forces on the block.Assume F on ab and cd are equal.Along line AB:

Developed resistance:τf = cd’ + σ’ tanφd’ = cd’ + (σ-u) tanφd’

= cd’ + (γsat - γw) H cos2β tanφd’Driving force due to weight:τd = γ H cosβsinβ

Forces:Na = γsat L H cosβTa = γsat L H sinβσ = γsat L H cos β / (L/cosβ) = γsat H cos2βτ= γsat L H sinβ / (L/cosβ) = γsat H cosβsinβNr = γsat L H cosβTr = γsat L H sinβ

Factor of Safety:

βφ

γγ

βγ tan'tan'

sin'

satsat HcFS +=

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For c = 0:

βφ

γγtan

'tan'

sat

FS =

FS = 1 H = Hcr

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Slope StabilitySlope Stability with Plane Surface:


H

W

NaTa

Tr

B

A β θ

C

θ

Factor of Safety:

( )( ) θθβγ

φθθβγβsinsin

tancossinsin−

−+=

HHcFS 2

For c = 0:

βφ

tantan

=FS

AC Trial failure place

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Slope StabilityModes of Failure of Finite Slopes:


Slope failure

Shallow slope failure

Base failure

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Slope StabilityCircular 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.

moment drivingmoment resisting

===WxLRc

MMFS u

d

r

Circular failure surface in φ=0 soil is defined by its undrainedstrength, cu.

W1

W2l2 l1

2211

2

lWlWRc

MMFS u

d

r

−==

θ

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Slope StabilityCircular 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:

Rf = R sinφm

P

β

φm

Procedure (Abramson et al 1996 more detailed)C parallel to abP passes through intersection W-CP makes φm with line through center

of friction circle, & tangent to FCU often taken 0Force polygon determine CCritical circle developed cohesion is

maximumFor FS = 1, the critical height:C’ / (γ Hcr) = f(α, β, θ, φ’) = m (stability No.)

φ > 3 deg critical circles all toe circles

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Slope StabilityMethod 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=0No. of unknowns = No. of slices * 3Therefore assumptions should be made.

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Slope StabilityCircular 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.

( )

( )∑

∑

=

=

+∆= n

iii

n

iiii

W

WlcFS

1

1

α

φα

sin

'tancos'

ΣM0 = 0

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Slope StabilitySearch 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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H

β

h z

Phreaticsurface

Seepage

Slope StabilitySlope 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

( )[ ]

( )∑

∑

=

=

∆−+∆= n

iii

n

iiiiii

W

luWlcFS

1

1

α

φα

sin

'tancos'

FS modified (from Bishop method) for pore pressure:

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Lateral Earth Pressure

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Lateral Earth Pressure Coefficient:


σz’H

σx’

K=σx’/σz’

P=(1/2)K γ H2

σx’ = Kσz’= KγH

1/3 H

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Relationship between σz’ and σx’ at a given depth (at rest means no shear).

Ko : Coefficient of earth pressure at rest, Ko = σx’ / σz’

For coarse-grained soils:Ko = 1 - sinφ’ (ok for loose sand)

For fine grained NC soils:Ko = m - sinφ’

m: 1 for NC cohesionless or cohesivem: 0.95 OCR > 2

Massarch (1979)Ko = 0.44 + 0.42 (PI% / 100)

For OC clays:Ko = Ko(NC) (OCR)(1/2)

OrKo = (1 - sinφ’) OCRsinφ’

σz’Hσx’

K=σx’/σz’

P=(1/2)K γ H2

σx’ = Kσz’= KγH

1/3 H


Lateral Earth Pressure Coefficient at Rest:


Rigid WallNo movement

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Wall moves away from the soil (pushed out).


Coefficient of Active Lateral Earth Pressure:


σz’H

σx’

Ka=σx’/σz’

Movement

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Soil Type


Wall Movement Required to Reach the Active Condition:


Horizontal movement required to reach the active state

Dense sand

Loose sand

Stiff clay

Soft clay

0.001 H

0.004 H

0.010 H

0.020 H

(From CGS, 1992)

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Wall moves towards the soil (pressed in).


Coefficient of Passive Lateral Earth Pressure:


σz’H

σx’

Kp=σx’/σz’

Movement

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Soil Type


Wall Movement Required to Reach the Passive Condition:


Horizontal movement required to reach the passive state

Dense sand

Loose sand

Stiff clay

Soft clay

0.020 H

0.060 H

0.020 H

0.040 H

(From CGS, 1992)

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In Summary:


1. If the wall moves away from the fill (soil) pressure will decrease and reach to active state. (σh = Ka σv)

2. If the wall moves towards the fill (soil) pressure will increase and reach to passive case. (σh = Kp σv)

3. More deformation is generally required to achieve passive case than the active case.

Movement towards backfillMovement awayFrom backfill

Ka

Kp

Ko

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Classical Lateral Earth Pressure Theories:


• Coulomb’s Earth Pressure Theory (1776)

• Rankine’s Earth Pressure Theory (1857)

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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


Rankine’s Earth Pressure Theory:


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53

Lateral Earth PressureRankine’s Earth Pressure Theory:


Attainment of Rankine’s Active State

Attainment of Rankine’s Passive State

54


Rankine’s Earth Pressure Theory – Force Diagram:


W

NTA

β

θ

C

βP

N

WT

Pβ

θ

Rankine’s Earth Pressure Theory -Force Equilibrium

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55


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

56


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


P = (1/2) K γ H2

β

β

H

H/3

β σ = K σz = K γ H

PH = P cosβ = (1/2) K γ H2 cosβ

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57


Rankine’s Theory – Coefficient of Active Earth Pressure:


For β ≤ φ’:'coscoscos'coscoscos

φββ

φββ22

22

−+

−−=aK

For β = φ’:

−= 2452 'tan φo

aK

Rankine’s Theory – Coefficient of Passive Earth Pressure:

For β ≤ φ’:'coscoscos'coscoscos

φββ

φββ22

22

−−

−+=pK

For β = φ’:

+= 2452 'tan φo

pK

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ENGR-627-Lect4