1. Geotechnical EngineeringGeotechnical Engineering IIGeotechnical EngineeringGeotechnical Engineering -- II
Stress Distribution,Stress Distribution,
Slope StabilitySlope StabilitySlope StabilitySlope Stability
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN
1
2. Module IV
Stress in soil:
Module IV
Stress in soil:
Boussinesque's and Westergaard's equations for vertical loads-
Pressure due to point loads and uniformly distributed loads –
Assumptions and limitations –Assumptions and limitations
Pressure bulb –
Newmark charts and their use –
Line loads and strip loadsp
Stability of slopes:
Stability of finite slope-Stability of finite slope-
Stability of infinite slope-
Stability Number-
Method of slices- The Swedish circle methodMethod of slices The Swedish circle method
The friction circle method
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4. Introduction
• When stresses in soil mass are relatively small, the soil may be assumed
to behave elastically
• Based on the elastic theory, stresses at various points in the soil mass can
be calculated
• Fi di t t i th il i i t t i ti ti th ttl t i• Finding out stresses in the soil is important in estimating the settlement in
soil masses
• Many formulae available, most accepted among them are:y p g
• Boussinesq’s formula
• Westergaard’s formula
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5. Boussinesq’s Formula for Point Loadsq
QPOINT LOAD, Q
Q
z
SEMI INFINITE
MEDIUM
r
P
zσ
Point load acting on surface of semi−infinite mass
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Point load acting on surface of semi infinite mass
6. Assumptions:
• Soil mass is elastic, isotropic, homogeneous and semi-infinite
• The soil is weightless, compared to the applied loads
• The load is a point load acting on the surfacep g
Vertical stress at a point P due to a point load Q is:
( )( )
5 22 2
3 1
2 1
z
Q
z r z
σ
π
=
+
Q
( )( )1 r z+
z(J. Boussinesq, 1885)
r
σ
P
2z B
Q
Iσ = zσ2z B
z
BI Boussinesq stress coefficient
(Boussinesq influence factor)
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(Boussinesq influence factor)
7. Westergaard’s Formula for Point LoadsWestergaard s Formula for Point Loads
Vertical stress at a point P due to a point load Q is:
( )
3 22
1
z
Q
σ =
( )( )
3 22 2
1 2
z
z r zπ + Q
Westergaard (1938)
z
Westergaard (1938)
Q
r
σ
P
2z W
Q
I
z
σ =
zσ
WI Westergaard stress coefficient
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8. Comparison of Boussinesq’s and Westergaard’s Formulae
Comparison of values of IW and IB
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• The value of IW is less than IB by 33%, at r/z = 0
9. Variation of stress with depth under a unit point load
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10. • Boussinesq’s equation assumes isotropic condition; Westergaard’s equation
assumes stratified soil condition
H W d’ i i l fi ld di i i di• Hence Westergaard’s equation is closer to field conditions in sedimentary
deposits
• However engineers prefer Boussinesq’s equation to compute stresses since
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• However, engineers prefer Boussinesq s equation to compute stresses, since
settlements computed from these stresses give conservative values
11. Problem 1. A concentrated load of 200 kN acts at the ground surface. Find the
i l l h i f h l d i di l di f 5 dvertical stress along the axis of the load at a point at a radial distance of 5m and
a depth of 10m by a) Boussinesq’s and b) Westergaard’s Formulae. Neglect the
depth of foundation.
( )( )
5 22 2 2
3 1
2 1
z B
Q Q
I
z z r z
σ
π
= =
+
a) Boussinesq’s Formula
( )( )1 r z+
3 1
I
3 1
0 2733
( )( )
5 222 1
BI
r zπ
=
+ ( )( )
5 222 1 5 10π
=
+
0.2733=
2
2kN
kN/m
200
0.55
10
z BIσ∴ = × =
10
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12. 1Q Q
Iσ = =) d’ l
( )( )
3 22 2 2
1 2
z WI
z z r z
σ
π
= =
+
a) Westergaard’s Formula
1 1
( )( )
3 22
1
1 2
WI
r zπ
=
+ ( )( )
3 22
1
1 2 r zπ
=
+
0.1733=
2kN
kN/m
200
0 346Iσ∴ = =2
kN/m0.346
10
z WIσ∴
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13. Line Loads LINE LOAD, q
y
e oads q
x
R
SEMI INFINITE
MEDIUM Pr
• Boussinesq’s equation provides a basis for finding stress at any point P in an
z
elastic semi-infinite mass
• By applying this principle, the stress at any point P due to a line load of
infinite extent in an elastic semi-infinite mass can be calculated
• State of stress - plain strain condition
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• Strain in y-direction parallel to the line load is assumed to be zero
14. Consider an elementary length dy along the line load of intensity q/unit length.y g y g y q/ g
The load q.dy can be taken as a concentrated (point) load
V ti l t t i t P d t thi i t l d iΔ
( )
Vertical stress at a point P due to this point load is:zσΔ
( )
( )( )
5 22 2
3 1
2 1
z
qdy
z r z
σ
π
Δ =
+ ( )( )
2 2 2
r x y= +
( )
( )
5 22
2 2
3 1
2
z
qdy
z x y
σ
π
Δ =
⎛ ⎞+
z
( )
2
1
x y
z
⎛ ⎞+
+⎜ ⎟
⎜ ⎟
⎝ ⎠
r
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2 2 2 2 2 2
R r z x y z= + = + +
15. ( ) 3
5 2
3 qdy z
σΔ =
( )( )
5 2
2 2 22
z
z x y
σ
π
Δ
+ +
( ) 3
3 d+∞ +∞ 3
d
∞
( )
( )( )
3
5 2
2 2 2
3
2
z z
qdy z
z x y
σ σ
π
+∞ +∞
−∞ −∞
= Δ =
+ +
∫ ∫
( )
3
5 22 2 2
0
3
2
2
z qdy
z x yπ
∞
=
+ +
∫
( )
x and z are constants for a given point P. Hence, y is the only variable.
( )
22
2 1
z
q
σ∴ = Vertical stress at a point P
( )( )2
1z x zπ + due to a line load of intensity q/unit length
z Z
q
I
z
σ = ZI Influence factor
0 637 / 0
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z
=0.637, at x/z = 0
16. Strip Loadsp
Required to find the vertical stress at a point P on a vertical plane passing
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Required to find the vertical stress at a point P on a vertical plane passing
through the centre of the strip.
17. Vertical stress at a point P due to an elementary line load of intensity qdx is:
( )
2
2 1qdx
σΔ =
2
2
B
q dx
σ ∫I t ti th idth f l d
( )( )
22
1
z
z x z
σ
π
Δ
+
( )( )
22
2 1
z
B
z x z
σ
π −
=
+
∫Integrating over the width of load,
2B
( )( )
2
22
0
4
1
B
q dx
z x zπ
=
+
∫
tan
x
z
β=
( )
Let
2
secdx z β∴ =
( )
2 22
2
22
4 sec 4
cosz
q q
d d
θ θ
β
σ β β β= =∫ ∫ ( )sin
q
θ θ+=
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( )
22
0 01 tan
z
π πβ+
∫ ∫ ( )
π
18. Uniform load on a circular areaU o oad o a c cu a a ea
UNIFORM LOAD ON
CIRCULAR AREA,
q/UNIT AREA
SEMI INFINITE
MEDIUM
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19. Required to find the stress at a point P under the load, on the vertical axis
through the centroid of the loaded area
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through the centroid of the loaded area.
20. Vertical stress at a point P due to a point load of intensity qdA is:
( )
( )
5 22 2
3 1
2
z
qdA
σΔ =
( )
( )
3
5 22 2
3
2
qdA z
π
=
( )( )
5 22 22 1
z
z r zπ + ( )2 22 r zπ +
h f d
( ) ( )
3 3
3 3q z q z
∑
Integrating over the ring of radius r,
( )
( )
( )
( )
5 2 5 22 2 2 2
3 3
2
2 2
z
q z q z
dA rdr
r z r z
σ π
π π
Δ = =
+ +
∑
( )
3
5 22 2
3z
z
qrdr
r z
σΔ =
+( )
Integrating the above from r=0 to r=a gives the stress due to the entire load
over the circular area
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over the circular area
21. 2 2 2 d d2 2 2
r z n+ =Let r dr ndn=
0r n z= ⇒ =
2 2
r a n a z= ⇒ = +
2 2
3
5
3
a z
z z
ndn
qzσ σ
+
= Δ =∫ ∫
2 2
3
4
3
a z
dn
qz
+
= ∫5z z
A z
q
n∫ ∫
3 2
1
⎧ ⎫⎡ ⎤⎪ ⎪
4
z
q
n∫
( )
2
1
1
1
q
a z
⎧ ⎫⎡ ⎤⎪ ⎪
= − ⎢ ⎥⎨ ⎬
+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
Bq I=
I Boussinesq influence factor
BI Boussinesq influence factor
for circular load
• The above formula can be used for finding stresses under ring loads also
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• The above formula can be used for finding stresses under ring loads also.
22. Problem 2. A ring footing of external diameter 8m and internal diameter 4m
d h f 2 b l h d f I i l d f i irests at a depth of 2m below the ground surface. It carries a load of intensity
150 kN/m2. Find the vertical stress at a depths of 2m, 4m and 8m along the
axis of the footing below the footing base.
( )
3 2
2
1
1z qσ
⎧ ⎫⎡ ⎤⎪ ⎪
= − ⎢ ⎥⎨ ⎬For circular load, Bq I=
( )
2
1
z q
a z
⎢ ⎥⎨ ⎬
+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
B
For a ring load, ( )z zExt zInt BExt BIntq I Iσ σ σ= − = −
3 2
1
1I
⎡ ⎤
⎢ ⎥
For z = 2m (below foundation) and a = 4m
( )
2
1
1 4 2
BExtI = − ⎢ ⎥
+⎢ ⎥⎣ ⎦
(external radius),
0 911
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0.911=
23. 3 2
1
1I
⎡ ⎤
⎢ ⎥F z 2 d a 2 (i t l di ) 0 646
( )
2
1
1 2 2
BIntI = − ⎢ ⎥
+⎢ ⎥⎣ ⎦
For z = 2m and a = 2m (internal radius), 0.646=
Hence the stress at a depth of 2m below the load (foundation) along its axis,
( )I I
2
kN/m39 7=
( )z BExt BIntq I Iσ = −
( )150 0 911 0 646σ∴ × kN/m39.7=( )150 0.911 0.646zσ∴ = × −
Similarly the stresses at depths of 4m and 8m can be determined as
2 2
kN/m and kN/m respectively.54 30 ,
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24. Pressure distribution diagramsPressure distribution diagrams
( )
5 22 2
3 1
2
z
Q
σ =
• Stress varies along depth as well as horizontal distance
( )( )
2 22 1z r zπ +
• Vertical pressure on a given horizontal plane is same in all directions at equal
radial distance from the axis of loading
• A contour connecting all points below the ground surface having equal
vertical pressure: Isobar
• Isobar is a spatial surface in the form of a bulb• Isobar is a spatial surface in the form of a bulb
• The zone in a loaded soil mass bounded by an isobar of a given vertical
pressure intensity is called a pressure bulb
• Vertical pressure at every point on the surface of pressure bulb is same
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25. Dept. of CE, GCE Kannur Dr.RajeshKN
Isobars under a uniformly distributed circular area
26. Significant depth
• Depth of pressure bulb corresponding to 20% of foundation contact pressure
• i e depth of the pressure isobar that represents a pressure of 0 2q where q is
g p
• i.e., depth of the pressure isobar that represents a pressure of 0.2q, where q is
the vertical pressure applied
• Significant depth is approximately equal to 1.5 times the width of
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g p pp y q
square or circular footings
27. thcantdeptSignific
• Terzaghi suggested (1936) that this depth is significantly responsibleg gg ( ) p g y p
for the settlement of structures
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28. h l l d d f l d l l b• When several loaded footings are placed closely, isobars merge into
one large isobar, with a large significant depth
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29. Problem 3. A single concentrated load 1000 kN acts at the ground surface.
C i b f i l f i i 40 kN/ 2 UConstruct an isobar for a vertical pressure of intensity 40 kN/m2. Use
Boussisesq formula.
3 1Q
Boussinesq’s Formula
( )( )
5 22 2
3 1
2 1
z
Q
z r z
σ
π
=
+ 1mz =?mr =
3 1000 1
40
× 2mz =
( )( )
5 22 2
40
2 1z r zπ
=
+
3mz =
2 5
2
3 1000
1r z
×⎛ ⎞
= −⎜ ⎟
⎝ ⎠
3mz =
3.45mz =
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2
2 40zπ⎜ ⎟
×⎝ ⎠
30. Contact pressureContact pressure
• Pressure at the contact surface between base of foundation and the
d l lunderlying soil mass
• Contact pressure distribution depends on:
• Flexural rigidity of footing
• Elastic properties of the sub grade soil
• If the footing is flexible, contact pressure is uniform, irrespective of the type
of soil
• If the footing is perfectly rigid pressure distribution depends on whether the• If the footing is perfectly rigid, pressure distribution depends on whether the
soil is cohesive or cohesionless
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32. Vertical stress beneath loaded areas of
irregular shape
Newmark’s Influence Charts
Newmark (1942)
• To determine the vertical stress at any point under a uniformly loaded area
of any shapeof any shape
• A chart consists of a number of concentric circles and radial lines
E h it l ti l t t th t f th di• Each area unit causes equal vertical stress at the centre of the diagram
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34. Newmark’s Chart
Plan of a building foundation
(drawn on tracing paper)(drawn on tracing paper)
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35. Construction of Newmark’s Influence Chart
• Let a uniformly loaded circular
area of radius r1 cm be divided into
20 sectors20 sectors
• Let q is the load intensity, and σz
is the vertical stress at a depth z
O
p
below the centre of the area
• Each sector OA1B1 exerts a
pressure of σz/20 at a depth z
below the centre of the area
Construction of
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Newmark’s Chart
36. Hence, from the equation for stress below a circular load,
( )
3 2
2
1
1
20 20 1
z
f
q
i q
r z
σ
⎧ ⎫⎡ ⎤⎪ ⎪
= − =⎢ ⎥⎨ ⎬
+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
( )1
20 20 1 r z+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
3 2
1 1
⎧ ⎫⎡ ⎤⎪ ⎪
( )
2
1
1 1
1
20 1
fi
r z
⎡ ⎤⎪ ⎪
= − ⎢ ⎥⎨ ⎬
+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
Influence value
• Let us fix an arbitrary value of 0.005 for if
⎧ ⎫
( )
3 2
2
1
1
0.005 1
20 1
q
q
r z
⎧ ⎫⎡ ⎤⎪ ⎪
∴ = − ⎢ ⎥⎨ ⎬
+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
( )1⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
• For z = 5 cm (say), 1 1.35 cmr =
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1
37. • Hence, if a circle is drawn with radius 1.35 cm, divided into 20 equal sectors,
h t ill t f 0 005 t d th f 5each sector will exert a pressure of 0.005 q at a depth of 5 cm.
• Let the radius of a second circle be r2 cm (to be determined).
• Let the annular space between two circles be divided into 20 equal parts
(area units), like A1A2B2B1.
h t b h th t th ti l t th t d t h f th• r2 has to be such that the vertical pressure at the centre due to each of these
area units is 0.005 q.
• The total pressure due to area units OA B and A A B B is 2x0 005 q• The total pressure due to area units OA1B1 and A1A2B2B1 is 2x0.005 q
• Therefore, vertical pressure due to OA2B2 is:
3 2
1
2 0 005 1
q
q
⎧ ⎫⎡ ⎤⎪ ⎪
× = ⎢ ⎥⎨ ⎬ forcm 5 cm2r⇒
( )
2
2
2 0.005 1
20 1
q
r z
× = − ⎢ ⎥⎨ ⎬
+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
2 forcm, 5 cm2 zr =⇒ =
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38. • Similarly a 3rd, 4th, 5th etc. circles can also be drawn
• For the 10th circle,
3 2
1
1 10 0 005
q q
q
⎧ ⎫⎡ ⎤⎪ ⎪
− = × =⎢ ⎥⎨ ⎬ r⇒ = ∞
( )
2
10
1 10 0.005
20 201
q
r z
= × =⎢ ⎥⎨ ⎬
+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
10r⇒ = ∞
• Hence the Newmark’s Chart is drawn as in figure.
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39. • In the Chart, each area
unit will exert an equal
vertical stress of 0 005 q at a
5
vertical stress of 0.005.q at a
depth of 5 cm at the centre
of the diagram.
• Hence, N area units of the
load will exert a total
pressure of 0.005.q.N at a
depth of 5 cm at the centre
of the diagram
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40. How to Use Newmark’s Influence Chart
• A plan of the loaded area is
drawn on a tracing paper anddrawn on a tracing paper and
placed on the Chart, in such a
way that the point below which
stress is required coincides withstress is required coincides with
the centre of the Chart.
5
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41. • The scale of the plan is such that the length AB on the Chart
corresponds to the depth at which pressure is requiredcorresponds to the depth at which pressure is required
• i.e., Let the distance AB is 5 cm (depth for which the Chart is drawn)
N if th t i i d t d th f 5 th th f d ti l• Now, if the stress is required at a depth of 5 m, then the foundation plan
should be drawn to a scale of 5 cm = 5 m or 1 cm = 1 m
• Similarly, if the stress is required at a depth of 15 m, then they, q p ,
foundation plan should be drawn to a scale of 5 cm = 15 m or 1 cm=3 m
• The total number of area units (including fractions) covered by the plan of the
l d d i t d Nloaded area is counted as N
• Hence, the stress exerted by the load at the point is 0.005.q.N
• Thus, Newmark’s chart can be used to determine the vertical stress at
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any point under a uniformly loaded area of any shape
43. • Earth slopes can be:
• Natural slopes: those existing in nature, formed by natural causes
• Man-made slopes: Cuts, embankments, earth dams etc.
• Slopes are classified according to their extent as:
• Infinite slopes: Constant slope of infinite extent. e.g., mountains
• Finite slopes: Slopes of limited extent. e.g., cuts, embankments etc.
• Result of failure of a natural or man-made slope can be catastrophic
• Hence a study of slope stability is very importanty p y y p
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44. • Causes of failure of slopes:
• Gravitational forceGravitational force
• Force due to seepage water
• Surface erosion due to flowing water• Surface erosion due to flowing water
• Sudden lowering of water adjacent to a slope
• Earthquake• Earthquake
•
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45. Stability analysis of infinite slopes
• Consider an infinite slope AB, inclined at angle i to the horizontal
• Failure of slope involves a sliding of soil mass along a plane parallel to
y y p
• Failure of slope involves a sliding of soil mass along a plane parallel to
surface, at some depth
• Let CD is such a failure plane (critical surface) at depth z Bp ( ) p
• Consider a portion of soil mass PQRS of
width b along inclined surface
i
P
Q
g
• Weight of this portion is
b
cosb i
A
P
. . .cosW z b iγ=
• Vertical stress on CD is:
zσz
D
. .cosz
W
z i
b
σ γ= =
σR
S
i
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τC
i
46. • Components of vertical stress parallel to CD and perpendicular to CD are:
2
cos cosz i z iσ σ γ= = sin cos .sinz i z i iτ σ γ= =
τ• The tangential component parallel to failure surface CD is the shear
stress causing failure
fτ• This shear stress is resisted by the shear strength of the soil
τ f
F
τ
τ
=• Hence factor of safety against sliding is:
• Shear strength includes cohesion and internal friction
tanf cτ σ φ= +
• Two cases are considered here: cohesionless soil and cohesive soil
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47. Case 1: Cohesionless soilCase 1: Cohesionless soil
• OA is Coulomb’s failure envelope for cohesionless soil defined by the
equation
tanfτ σ φ=
equation
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Failure condition for an infinite slope of cohesionless soil
48. • OB represents stresses on CD for varying depths• OB represents stresses on CD for varying depths
2
cos cosz i z iσ σ γ= =
sin cos .sinz i z i iτ σ γ= =
sin
tan
cos
z i
i
i
τ σ
σ σ
= = A constant, for a given slope i
cosz iσ σ
taniτ σ= Equation of curve OB
• For a given value of normal stress , failure happens when fτ τ>σ
• i.e., failure happens when i φ>
F t f f t i t lidi i i b
tan tanf
F
τ σ φ φ
• Factor of safety against sliding is given by:
tan tan
f
F
i i
φ φ
τ σ
= = =
• If the slope is less than , the slope is stableφ
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• If the slope is larger than , the slope is unstable, for any depthφ
49. Case 2: Cohesive soilCase 2: Cohesive soil
• DA is Coulomb’s failure envelope for cohesive soil defined by the equation
tanf cτ σ φ= +
• If the slope is less than , the slope is stableφ
Failure condition for an infinite
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slope of cohesive soil
50. • If the line OF representing stress on the failure plane meets the
failure envelope at F
i φ>
failure envelope at F
• At F, shear stress is equal to shear strength. i.e., failure happens at F.
• FF2 represents shear stress at some depth z
• C1C2 represents shear stress at some depth less than z. At this
• Hence, for any depth less than z, shear stress is less than shear
strength and the slope is stable e g at C
depth, shear strength is CC2
strength, and the slope is stable. e.g., at C1
Failure condition for an infinite
Dept. of CE, GCE Kannur Dr.RajeshKN
slope of cohesive soil
51. • Hence if , slope is stable only for depths less than z, known asi φ>
critical depth Hc.
• Factor of safety for any depth z less than H can be given as:• Factor of safety for any depth z less than Hc can be given as:
tanf c
F
τ σ φ+
= =
τ τ
2
cos cosi z iσ σ γ= =We have cos cosz i z iσ σ γ
sin cos .sinz i z i iτ σ γ= =
We have,
2
cos tan
cos sin
f c z i
F
z i i
τ γ φ
τ γ
+
∴ = =
cos .sinz i iτ γ
Dept. of CE, GCE Kannur Dr.RajeshKN
52. • At the critical depth H 1Fτ τ= ⇒ =• At the critical depth Hc , 1f Fτ τ= ⇒ =
2
cos tan
i 1cc H iγ φ+ c
H⇒ =i 1
cos .sin
.e., c
cH i i
γ φ
γ
=
( ) 2
tan tan cos
cH
i iγ φ
⇒ =
−
• Thus, for given values of i and ϕ, Hc is proportional to cohesion c.
( ) 2
tan tan cos n
c
c
i i S
H
φ
γ
= − = Stability Number,
a dimensionless quantity
Dept. of CE, GCE Kannur Dr.RajeshKN
53. Stability analysis of finite slopesStab ty a a ys s o te s opes
• Most commonly, failure occurs along curved surfaces
• Analysis based on the assumption that the curved surface is circular in shape• Analysis based on the assumption that the curved surface is circular in shape
• Some methods are:
• Swedish circle method (Method of Slices or Slip circle method)• Swedish circle method (Method of Slices or Slip circle method)
• Modified method of slices
Si lifi d Bi h ’ h d f li• Simplified Bishop’s method of slices
• Friction circle method
• Taylor’s stability numbers
Dept. of CE, GCE Kannur Dr.RajeshKN
54. • Types of failure of a finite slope:
• Slope failure: Slide surface intersects slope at or above toe; slope angle
is high; soil close to the toe has a high strength
• Face failure: slide surface intersects slope above toe
• Toe failure: slide surface intersects at toe
• Base failure: Slide surface passes below the toe; Soil beneath the base is
softer; slope angle is low
Face
failure
Toe
failurefailure failure
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Base
failure
55. The Swedish circle method
• Analysis of purely cohesive soil (ϕ=0 analysis)
Swedish Geotechnical Commission and W. Fellenius
• Analysis of soil possessing both cohesion and friction (c-ϕ analysis)
ϕ=0 analysis
• Assume a number of slip circles and find the factor of safety ofAssume a number of slip circles, and find the factor of safety of
each
• The slip circle with lowest factor of safety is taken as the
critical slip circle
Dept. of CE, GCE Kannur Dr.RajeshKN
56. r
Let AD be a trial slip circle (radius r)p ( )
r
Driving moment w.r.t O is resisted by the resisting moment
Driving moment .DM W x=
Weight of soil of the wedge ABDAW
Perpendicular distance of centre of gravity of the wedgex
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from the centre of rotation O
57. Resisting moment . .R uM c L r=g R u
Unit cohesionuc
C d l th f th liL
2 r
L
π δ
Curved length of the slip arcL
360
L=
. .
.
R u
D
M c L r
F
M W x
= =Hence, factor of safety
u
m
c
F
c
=Alternatively,
where
.
.
m
W x
c
L r
=
The shear resistance mobilised
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Can be obtained by dividing the wedge into several slicesx
58. c-ϕ analysis
• As in ϕ=0 analysis, assume a number of slip circles, and find the
factor of safety of each
Th li i l ith l t f t f f t i t k th iti l li• The slip circle with lowest factor of safety is taken as the critical slip
circle
• Soil above the slip arc divided
into several slices
• Weight of each slice, W, has a
normal component N and a
tangential component T
r
g p
• Normal components will not
cause driving moment
Dept. of CE, GCE Kannur Dr.RajeshKN
59. h lDriving moment on each slice
DM rTΔ = r
Resisting moment on each slice
( )( )tanRM r c L N φΔ = Δ +
LΔ
( )tan tanr c cLM L N Nφφ ++ × ×Δ∑ ∑∑( )tan tanR
D
r c
r
cLM
F
M
L N
T
N
T
φφ ++ ×
=
×
= =
Δ∑
∑∑
∑∑Factor of safety
( )tan
F
c L N U
T
φ′ ′
=
+ −∑
∑
Factor of safety in terms of
effective values of cohesion
and friction
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T∑and friction,
where U is the pore (neutral) pressure
60. Problem X:
( )tan
F
c L N U
T
φ′ ′+ × −
=
∑
∑
Factor of safety w.r.t. shear strength and
cohesion (in terms of effective values of
h d f ) T∑cohesion and friction),
( )20 27 900 216 tan18
F
× + −
1 694
Dept. of CE, GCE Kannur Dr.RajeshKN
( )
450
F = 1.694=
61. Friction circle method
friction circle
r sinϕ
r
ϕ
• A small circle is drawn with O as centre and r sinϕ as radius. This circle is
called the friction circle or ϕ-circle
Dept. of CE, GCE Kannur Dr.RajeshKN
called the friction circle or ϕ circle
62. friction circle
F
r sinϕ
D
r
E
A
∆R
• Any line EF tangential to this circle cuts the failure arc AD at an obliquity ϕy g q y
• Conversely, any vector representing reaction ∆R (resultant friction) at an
obliquity ϕ to an element of the failure arc AD will be tangential to the
f i ti i l
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friction circle
63. • Let cm is the mobilised unit cohesion for equilibrium
• Let a is the position of mobilised cohesive force
, assumed acting parallel to AD
mc L
r
( ) ( )m mc L a c L r=L L
rL
a
L
∴ =
L
L is length of the chord ADPosition of mobilised
cohesive force c L
L
cohesive force mc L
• Vector mc L is drawn parallel to AD, at a distance a from O
Dept. of CE, GCE Kannur Dr.RajeshKN
64. • Vector W is the weight of the wedgeg g
mc L• is the vector of mobilised cohesive force
a
W R
c
b mc L
• Vector R is the total frictional resistance, drawn with its line of action
passing through the intersection of the above 2 vectors, and tangential to the
friction circle
• From the force triangle abc, knowing W, the magnitude of , and
thus can be determined
mc L
c
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thus , can be determined.mc
65. • The Factor of safety F with respect to cohesive strength can beThe Factor of safety Fs with respect to cohesive strength can be
determined as:
s
c
F
c
=
mc
• A number of slip circles are taken, and factor of safety of each is foundp y
in the manner described above.
• The slip circle with lowest factor of safety is taken as the critical slip
circlecircle
Dept. of CE, GCE Kannur Dr.RajeshKN
66. Taylor’s Stability Numberay o s Stab ty Nu be
• Taylor proposed the concept of using an abstract number (Taylor’s
Stability Number) for finding factor of safety of slopesStability Number) for finding factor of safety of slopes
• He studied a large number of slopes with various slope angles and angles
of internal friction…
• …and came up with an expression for Taylor’s Stability Number as:
m
n
c
c c
S
F H Hγ γ
= = , assuming full mobilisation of
internal friction
• Knowing slope angle and angle of internal friction, Taylor’s Stability
Number can be found from the tables and charts presented by him.
• Knowing Taylor’s Stability Number, factor of safety of slope can be found
as:
c
F
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c
n
F
S Hγ
=
67. Problem X: An excavation is made with a vertical face in a clay soil which has
c = 50 kN/m2 bulk unit weight = 18 kN/m3 Determine the maximum depthcu = 50 kN/m2, bulk unit weight = 18 kN/m3. Determine the maximum depth
of excavation so that the excavation is stable. The following are given for ϕ=0
Slope angle 600 750 900Slope angle 600 750 900
Stability number 0.191 0.219 0.261
For the excavation, slope angle = 900 because it has a vertical face
Hence stability number Sn is 0.261, from table.
2
50 kN/m
c
n
c
H
Sγ
∴ = 3
50
18
kN/m
kN/m 0.261
=
×
m10.64=
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68. Problem X: Calculate the factor of safety with respect to cohesion of a clay
slope laid 1 in 1 to a height of 8 m if the angle of internal friction ϕ = 200, c = 30p g g ϕ
kN/m2, bulk unit weight = 19 kN/m3. What will be the critical height of slope
in this soil? The following are given for ϕ = 200
Slope angle 300 450
Stability number 0.0625 0.062
For the given slope, slope angle ( )1 01tan 45
1
−
= =( )1
Hence Stability Number Sn is 0.062, from table.
F t f f t ith
c
n
c
F
S Hγ
=
2
3
30
0.062
kN/m
kN/m19 8 m
=
× ×
3.18=
Factor of safety with
respect to cohesion
c
c
H
S γ
=Critical height
2
3
30
0 062 19
kN/m
kN/m
=
×
m25.5=
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nS γ 0.062 19 kN/m×
69. SummarySummary
Stress in soil:Stress in soil:
Boussinesque's and Westergaard's equations for vertical loads-
Pressure due to point loads and uniformly distributed loads –
Line loads and strip loadsLine loads and strip loads
Assumptions and limitations –
Pressure bulb –
Newmark charts and their use –Newmark charts and their use
Stability of slopes:
Stability of finite slope-Stability of finite slope-
stability of infinite slope-
Stability Number-
Method of slices- The Swedish circle methodMethod of slices The Swedish circle method
The friction circle method
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69
70. Success happens when chance meets a prepared mind.
Dept. of CE, GCE Kannur Dr.RajeshKN