ENGR-627 Performance Evaluation of Constructed Facilities, Lecture # 4
Performance Evaluation of Constructed Facilities
Fall 2004
Prof. Mesut Pervizpour
Office: KH #203
Ph: x4046
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
Soil Strength
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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
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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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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
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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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ENGR-627 Fall 2004
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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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Dr. Mesut Pervizpour
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
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Dr. Mesut Pervizpour
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
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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’
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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
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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)
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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
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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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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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ENGR-627 Fall 2004
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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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Dr. Mesut Pervizpour
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
Slope Stability
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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.
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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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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
Dr. Mesut Pervizpour
ENGR-627 Fall 2004
Landslide Components
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Landslide Components
Varnes (1978), Morgenstern (1985)
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
Slope Stability
Possible Failure Modes of Slopes
Local
failure
Surficial failure
Slope
failure
Global failure
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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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Dr. Mesut Pervizpour
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
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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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Dr. Mesut Pervizpour
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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
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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
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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
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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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
21
Lateral Earth Pressure
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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
22
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
23
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
24
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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Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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
ENGR-627 Fall 2004
25
Lateral Earth Pressure
Classical Lateral Earth Pressure Theories:
•
Coulomb’s Earth Pressure Theory (1776)
•
Rankine’s Earth Pressure Theory (1857)
51
Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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
52
Dr. Mesut Pervizpour
ENGR-627 Fall 2004
26
Lateral Earth Pressure
Rankine’s Earth Pressure Theory:
Attainment of Rankine’s Active State
Attainment of Rankine’s Passive State
53
Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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
54
Dr. Mesut Pervizpour
ENGR-627 Fall 2004
27
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
55
Dr. Mesut Pervizpour
ENGR-627 Fall 2004
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
56
Dr. Mesut Pervizpour
ENGR-627 Fall 2004
28
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

57
Dr. Mesut Pervizpour
ENGR-627 Fall 2004
29