ENVIRONMENTAL GEOTECHNICS

Slope Stability Exercises

Prof. Ing. Marco Favaretti

University of PadovaDepartment of Civil, Environmental and Architectural EngineeringVia Ognissanti, 39 – Padova (Italy)

phone: +39.049.827.7901e-mail: marco.favaretti@unipd.itwebsite: www.marcofavaretti.net

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2

EXAMPLE 01

Embankment made of cohesive soil cu = 20 kPaφu = 0°γ = 19 kN/m3

Question: determine FoS for the trial circle

The weight of the sliding sector is 346 kN, acting at an eccentricity of 5 m from the centre of rotation.

Disturbing moment = 346 x 5 = 1730 kN mRestoring moment = cu r2 θ = 20⋅92⋅70/180⋅π = 1980 kN m F = 1980/1730 = 1.14

Area of the removed portion = 1.5 ⋅3 = 4.5 m2

Weight of the removed portion = 4.5⋅19 = 85.5 kNEccentricity from O = 5.7 mRelief of disturbing moment = 5.7 ⋅85.5 = 488 kN F = 1980/(1730-488) = 1.6

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Embankmentc’ = 20 kPaφ’ = 20°γ = 19.3 kN/m3

Foundationc’ = 75 kPaφ’ = 7°γ= 19.3 kN/m3

EXAMPLE 02

determine FoS for the trial circles

Classic case of an embankment resting on a stiff layer.

• CIRCLE 01: tangent to the lower layer• CIRCLE 02: crossing the foundation soil

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Embankment

c’ = 20 kPaφ’ = 20°γ = 19.3 kN/m3

EXAMPLE 02 - A

Slice n. Area (m2)

Weight (kN)

N(kN)

T (kN)

1 3.7 71 71 -7

2 8.7 168 163 42

3 11.6 224 191 116

4 7.7 148 104 106

ΣN = 529 ΣT = 257

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EXAMPLE 02 - A

85.1257

192284F

kN 284180767.1020rc

kN 192364.0529tanN

=+

=

=π⋅⋅⋅=θ⋅⋅

=⋅=φ⋅∑

59.1257

192217F

kN 217180587.1020rc

kN 192364.0529tanN

=+

=

=π⋅⋅⋅=θ⋅⋅

=⋅=φ⋅∑

°θ

=⋅=

φ

+°⋅γ

=

58 become

m 96.243.13.19

402

45tanc2h

cracks with

c

6

Embankmentc’ = 20 kPaφ’ = 20°γ = 19.3 kN/m3

Foundationc’ = 75 kPaφ’ = 7°γ= 19.3 kN/m3

EXAMPLE 02 - B

Slice n. Area (m2)

Weight (kN)

N(kN)

T (kN)

1 3.7 71 61 -36

2 9.7 187 184 -33

3 16.6 320 316 52

4 19.2 370 322 186

5 14.3 276 162 224

ΣT = 393

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EXAMPLE 02 - B

2.95 becomes F cracks with

39.3393

1691163F

kN 11631804515.920

1808515.975rc

kN 169123.0883364.0162tanN

kN 883 layer lower kN 162 layer upper N

=+

=

=π⋅⋅⋅+π⋅⋅⋅=θ⋅⋅

=⋅+⋅=φ⋅

==

∑

∑

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EXAMPLE 03

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EXAMPLE 03

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EXAMPLE 03

An embankment has a slope of 1V:2H. The soil properties are:c’ = 25 kPa φ’ = 20° γ = 16 kN/m3 H = 31 mUsing Taylor’s charts determine the F value for the slope

From the charts:φ’ = 20° β = 26.6° stability number = 0.017

Step 1: Fφ = 1

1F 96.243.8

25F

kPa 43.8017.03116c 017.0H

c

c

mm

=>>==

=⋅⋅==⋅γ

φ

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EXAMPLE 03

Step 2: Fφ = 1.5

From the charts:

φ’m = artg (tan20°/1.5) = 13.5° β = 26.6° stability number = 0.047

5.1F 07.13.23

25F

kPa 3.23047.03116c 047.0H

c

c

mm

=<<==

=⋅⋅==⋅γ

φ

Step 3: Fφ = 1.35

φ’m = artg (tan20°/1.35) = 15° β = 26.6° stability number = 0.037

35.1F 37.13.18

25F

kPa 3.18037.03116c 047.0H

c

c

mm

=≈==

=⋅⋅==⋅γ

φ

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EXAMPLE 04

Embankmentc’ = 12 kPaφ’ = 20°γ= 19.2 kN/m3

R = 9.15 mθ = 89°

θ

n. hw(m)

u(kPa)

z(m) ru

1 0.654 6.42 0.95 0.352

2 1.958 19.21 2.44 0.41

3 2.44 29.90 3.32 0.376

4 2.02 19.82 3.50 0.295

5 0.246 2.41 1.74 0.072

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EXAMPLE 04

33.19.207

170.6106.5F

kN 6.1701808915.912R'cl'c

89

=+

=

=π⋅⋅⋅=θ⋅⋅=⋅

°=θ

Conventional method

Rigorous method

Trial 1: Fφ = 1.5 43.19.207

297.5F ==

42.19.207

295.8F ==Trial 2: Fφ = 1.43

α⋅φ

+

α=

α

Ftan'tan1

secm1

Conventional method = Fellenius MethodRigorous method = Simplified Bishop Method

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EXAMPLE 04

Fellenius

Simplified Bishop

28.2

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EXAMPLE 05

Conventional method = Fellenius MethodRigorous method = Simplified Bishop Method

Embankmentc’ = 12 kPaφ’ = 20°γ= 19.2 kN/m3

R = 9.15 mθ = 89°

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EXAMPLE 05

Fellenius

Simplified Bishop

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EXAMPLE 05

Fellenius

Simplified Bishop

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EXAMPLE 06-A

φ’ = 30°γ = 18.0 kN/m3

β= ???1. Dry2. Filtrazione // pendioF = 1.25

462.025.1

5774.0tan

tantanF

==β

βφ

=

( )

°=β

=⋅⋅−

=β

βφ

⋅γγ

=

12

21.01825.1

5774.081.918tan

tantan'F

sat

1 2

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EXAMPLE 06-B

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EXAMPLE 06-B