# Slope Stability Analysis Causes of Slope Failures

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```					      Slope Stability Analysis                       Causes of Slope Failures
• Infinite Slopes                              •   Gravitational Force
– Constant slope of infinite extent          •   Seepage Pressures
• Mountain face
•   Undercutting by Rivers
• Finite Slopes
•   Sudden Drop of Water Levels
– Slopes with heights approaching critical
values                                     •   Earthquakes
• Embankments, earth dams

1
Factor of Safety
• Factor of Safety w.r.t. Strength
– FSs = f / d
 f = c’ + ’ tan ’
 d = cd’ + ’ tan d’

- FSs = [c’ + ’ tan ’ ] / [cd’ + ’ tan d’]

Factor of Safety                                   Factor of Safety
• Factor of safety w.r.t. Cohesion                 • When FSs = 1, the slope is in the state of
– FSc’ = c’ / cd’                                  impending failure
• Factor of Safety w.r.t. Friction
– FS’ = t ’ / t d’
tan    tan                                Generally, FSs ~ 1 5 i acceptable f
• G      ll        1.5 is       t bl for
design – up to 1.75 for earth dams

Finite Slopes
• Culman Method – Planar failure surface
– Fairly good results for near vertical slopes             Finite Slopes
only
with
• Swedish Method – Circularly cylindrical
sliding surface                                       Planar Failure Surfaces
– Good for analysis of embankments and their             Cullman Method
foundations
• Method of Slices

2
cr = (4c’/) [sincos’/(1-cos(-’)]

Finite Slopes
                                                     with

cr                     Circularly Cylindrical
cr = (+’d) / 2
Failure Surfaces

Types of Slope Failures                               Types of Slope Failures
O                                                  O

Toe Circle                                     Slope Circle

Firm Base

Firm Base

Types of Slope Failures                               Types of Slope Failures
O
O

Shallow Slope Failure
X

Midpoint Circle

X

Base Failure

Firm Base                                      Firm Base

3
Homogeneous Clays
Critical Circles (=0)
(
 = 0, (Undrained)
O                                                                                                       •   FSs = 1

•   cd/Hc = m (Stability Number)
Radius = r                                                                                  Md = W1l1 - W2l2
                                                          MR = cdr2
•   m = fn {slope angle, depth to firm base)
H
•     > 53o
W1                                                     cd=[W1l1-W2l2] / r2
l2       l1
– All circles are toe circles
cm
FSs = cu / cd                     – Center of circle found using
W2                                                                                                              • Fig 13.7 in text

cm

Critical Circles (=0)
(
• For  < 53o
• Critical circle may be a toe, slope or
midpoint circle depending on depth to firm
layer
• Depth = DH; x = nH
• Fig 10.10 text
o
Figure 13.7 Location of center of critical circles for  > 53
• Max m = 0.181

Example Analysis
• 60o cut in soft clay
– cu=800 psf; =122 pcf

depth,
• Determine maximum cut depth Hc
• Determine FSs if cut is 25 ft

o
For  < 53 :
Above region=mid-point circle
Within region=toe circle
Below region=slope circle

4
Example Analysis                                Types of Slope Failures
O
• Hc = ?
•  = 60o > 53o, so critical circle is toe circle

Toe Circle

Firm Base

Example Analysis
•   Hc = ?
•    = 60o > 53o, so critical circle is toe circle
•   From Fig 13.6
•   m=0.192

Example Analysis                                    Example Analysis
•   Hc = ?                                            •   FSs = ? for H = 25 ft
•    = 60o > 53o, so critical circle is toe circle   •    m = cm / H
•   From Fig 13.6, m=0.195                            •   cm = m  H
•   m = c u /  Hc                                    •   cm = 0.192*122*25 = 585.6 = 586 psf
•   Hc = cu / m = 800psf / [122pcf x 0.192]         •   FS= cu/cm = 800/586 = 1.36
•   Hc = 34 ft

5

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