h cos2β. Seepage. Flow. Failure Surface. Slope Surface h β h. W. N'+U. Pore Water Force. U = γ w bh cosβ b. T β β γ Ï Î² γ γ cos sin h. 'tan ) ( cos ) - (h + c'. = FS.
SOILS AND FOUNDATIONS Lesson 06 Chapter 6 – Slope Stability
Testing
Theory
Experience
Topics g Topic
1 (Section 6.0 – 6.8)
- Stability analysis of slopes g Topic
2 (Section 6.9)
- Improving the stability of embankments
SLOPE STABILITY
Lesson 06 - Topic 1 Stability analysis of slopes Section 6.0 – 6.8
Learning Outcomes g At
the end of this session, the participant will be able to:
- Recall modes of slope failure - Explain effects of water on slope stability - Discuss slope stability circular and block -
analyses Compute safety factor by chart solution
Stability Problems g Shallow
translational failure (Infinite Slope) g Circular
Failure
Embankment Fill
Firm Soil
Stability Problems g Sliding
block failure
g Lateral
squeeze
- Lesson 7
Effect of Water on Slope Stability g Frictional
Soils
- Below Water Table, Buoyancy Reduces Shearing Resistance g Clays
- Cohesive Strength Decreases as Water Content Increases Cohesive Strength
Water Content
Effect of Water on Slope Stability (Cont’d) g Fills
on Clays and Silts
- Soil Consolidates as Water is Squeezed Out Factor of Safety Increases With Time g Cuts
in Clay
- Soil Absorbs Water When Overburden Pressure Removed - Factor of Safety Decreases With Time
Effect of Water on Slope Stability (Cont’d) g Shales,
Claystones, Siltstones, Etc.
- Weak Rock Materials “Slake” When Exposed to Water - Embankments Undergo Internal Settlement or Failure
Design Factor of Safety g Minimum g Use
FS = 1.25 for highway side slopes
FS = 1.3 to 1.5 for critical slopes such as end slopes under abutments, slopes containing footings, major retaining structures
Design Factor of Safety g Selection
of FS depends on:
- Method of stability analysis - Method used to determine shear strength - Degree of confidence in reliability of subsurface data
- Consequences of failure - Criticality of the application
Infinite Slope Analysis g Slope
that extends for a relatively long distance and has consistent subsurface profile can be considered as infinite slope g Failure plane parallel to slope surface
Embankment Fill
Firm Soil
Infinite Slope Analysis in Dry Sands β b Slope Surface
W=γbh S
N = W cos β
W h
N
N
T S N Failure Surface
g FS
W T = W sin β β
S = N tan φ ForceφPolygon S N tan tan φ FS = = = T W sin β tan β
is independent of slope height h g FS is a function of only φ and β
c-φ soils with Water β
Slope Surface
Seepage Flow
h
b
W
h
h cos2β T N'+U Pore Water Force U = γwbh cosβ Failure Surface
c' + h ( γsat - γ w ) cos 2 ( β ) tan φ' FS = γsat h sin β cos β
For c' = 0
γ' tan φ' FS = γsat tan β
Circular Arc Failure
Resi stan ce
F orc
e
Circular Arc Failure
Resisting Moment FS = Overturnin g Moment
FS =
Total Shear Strength × LS Weight Force × L W
Simple Rule-of-Thumb for FS g Only
for preliminary “guestimate” for FS 6c FS ≅ γ Fill × H Fill
c = cohesion of foundation soil γFill = unit weight of fill HFill = Height of fill g No
water
What is the FS for following case?
(6)(1100 psf) FS = = 1.69 (130 pcf)(30 ft)
Circular Arc Stability Analysis (ORDINARY METHOD OF SLICES) le Circ
R s u i Rad
O
Fill
Firm Soft Firm
Slip Surface
R
Step by Step Procedure 1. 2. 3.
Draw cross-section to natural scale Select failure surface Divide the failure mass into 10-15 slices using suggestions on Page 6-14
Forces on a Slice Without Water
With Water
Failure Mass Divided into Slices O α= +
2:1 R 9
−7 °
−15°
4°
3°
+3
+4
1°
4 32 +5
1
4°
+16°
+9 °
°
5 +5
°
1 1
− 24
6
8
10
+1°
−3
2°
+ 25
° −4 9 ° 53 − 4 − 15 14 13 12 2°
16
7
60°
R
Note that slices 1 through 9 have positive α angles and contribute to the driving force. Slices 10 through 16 have negative α angles and reduce the net driving force.
Step by Step Procedure 4. Compute total weight ( WT ) of each slice 5. Compute frictional resisting force for each slice N Tanφ - ul 6. Compute cohesive resisting force for each slice Cl 7. Compute tangential driving force (T) for each slice 8. Sum resisting and driving forces for ALL slices and compute FS ∑ Resisting Forces ∑ N tan φ + ∑ c 1 FS = = ∑T ∑ Driving Forces
Example of One Slice w/o Water Assume: g γ total = 120 pcf, slice height = 10’ slice width = 10’, φ = 25°, α = 20°, l =11’, C = 200 psf. g Find:
Resisting and Driving Forces
Compute Slice Weight and Normal Force W T = γ total x slice area (x 1’ thick) = 120 pcf x 10’ x 10’ = 12,000 lbs N
= WT cos α - ul = 12,000 lbs x cos 20° = 11,276 lbs
Compute Resisting and Driving Forces N Tan φ = 11276 x Tan 25° = 5,258 lbs Cl = 200 psf x 11’ x 1’ = 2,200 lbs T = Wt Sin α = 12,000 lbs x Sin 20° = 4,104 lbs
Group Exercise g Assuming
the water is 5’ above the slice base, which of the force components change in this exercise?
Solution g The
water will affect the normal force, N
N = WT Cos α - ul = 12,000 lbs x Cos 20° - 5 x 62.4 x 11 = 11,276 lbs – 3,432 lbs = 7,844 lbs (N=11,276 lbs for original water level)
Tabular Form for Calculations g Figure
6-11 g Figure 6-12
Recommended Stability Methods g Limit
equilibrium methods
- Summation of moments, vertical and horizontal forces g Ordinary
Method of Slices (OMS) ignores both shear and normal interslice forces and considers only moment equilibrium
Recommended Stability Methods g Variations
of OMS are Bishop method, Simplified Janbu method, Spencer method, etc.
g Bishop
method
- Also known as Simplified Bishop method - Includes interslice normal forces - Neglects interslice shear forces - Satisfies only moment equilibrium
Recommended Stability Methods g Simplified
Janbu method
- Includes interslice normal forces - Neglects interslice shear forces - Satisfies only horizontal force equilibrium g Spencer
method
- Includes both normal and shear interslice forces - Considers moment equilibrium - More accurate than other methods
Recommended Stability Methods g OMS
is conservative and gives unrealistically lower FS than Bishop or other refined methods g For purely cohesive soils, OMS and Bishop method give identical results g For frictional soils, Bishop method should be used as a minimum g Recommendation: Use Bishop, Simplified Janbu or Spencer
Slope Stability Guidelines for Design g Table
6-1
g Computer
analysis is now-a-days commonly performed by use of slope stability software
- XSTABL, UTEXAS, ReSSA, SLOPE/W, etc
Remarks on Safety Factor g Minimum
FS = 1.25 using OMS
g Use
FS = 1.3 to 1.5 for critical slopes such as end slopes under abutments, slopes containing footings, major retaining structures
g Use
FS = 1.5 for cut slopes in fine-grained soils which can lose strength with time
Critical Failure Surface
Critical Failure Surface g Check
multiple failure surfaces and compare the lowest safety factors g Search all areas of slope to find the lowest safety factor g Be careful of “secondary” features such as thin weak layers g Evaluate all loading and unloading conditions, e.g., rapid drawdown g Use stability charts to develop a “feel” for the safety factor
Stability Charts g Assumptions
- Two-dimensional limit equilibrium analysis - Simple homogeneous slopes - Circular slip surfaces only g Useful
for preliminary analysis prior to computer analysis to develop a “feel” for safety factor
Taylor’s Stability Charts g Stability
Number
c′ Ns = Fc γ H g In
terms of Fc
See Figure 6-15
c′ Fc = N s γH g FS
= Fc = Fφ Slope Angle, β
Taylor’s Stability Charts for φ′=0 conditions and β 53º, use Figure 6-14
β = 53º
Determination of Factor of Safety c′ tan φ ′ τ′ = + σ′ FS FS FS
τ ′d = c ′d + σ ′ tan φ ′d τ′ τ ′d = ; FS
φ′ φ ′d = FS
c′ ; c ′d = FS
tan φ ′ tan φ ′d = FS
Example 6-1 g 30-ft
high slope g Slope angle, β = 30º g Total unit weight, γ = 120 pcf g Effective cohesion, c′ = 500 psf g Effective friction angle, φ′=20º g Determine
the Factor of Safety, FS
Example Computation g Assume
FS = 1.6 g FS = Fc = Fφ φ′ 20 o φ ′d = = = 12 . 5 o FS 1 .6
φ′d=12.5º and β = 30º, the stability factor, Ns, is 0.06. Thus, 500 psf 0 . 06 = ( 1 . 6 ) ( 120 pcf ) ( H )
g For
500 psf H = = 43 . 4 ft > 30 ft ( 1 . 6 ) ( 120 pcf ) ( 0 . 06 )
Taylor’s Stability Charts
0.075 0.06
Slope Angle, β
Example Computation g Since
43.4 ft > 30 ft, the actual FS is higher than 1.6. g Assume FS=1.9 g FS
= Fc = Fφ= 1.9 =>
φ′ 20o = = 10.5o φ′d = FS 1.9
=> Ns ≈ 0.075
500 psf H = = 29 . 2 ft ( 1 . 9 ) ( 120 pcf ) ( 0 . 075 ) g Computed
H is close to actual height of 30 ft
- Therefore, FS ≈ 1.9
Janbu’s Stability Charts g Account
for:
- Surcharge loading at top of slope - Submergence - Tension cracks - Seepage g Section
6.6.3
Sliding Block Failure Types 1
Fill Shallow Weak Soil Layer Firm Soil
2
Fill
Thin Seam Weak Clay Firm Soil Firm Soil
3
Fill
Lens of Silt or Sand w/o Frictional Resistance Impermeable Clay Clay Clay
After Slide CL SR 42 Oregon Fill
18’ 12’
Sandstone Silty Clay
24’
Active Wedge
Sliding Block Analysis
Fill
Central Block
Passive Wedge
Pa Pp
Sand Soft Clay Seam
cL
Sand
L
Pa = Active Driving Force = ½ γ H2Ka Pp = Passive Resisting Force = ½ γ H2Kp cL = Resisting Force Due To Clay Cohesion
Resisting Forces Pp + cL = FS = Driving Forces Pa
Example 6-3 the Safety Factor for the 20′ high embankment by the simple sliding block method using Rankine pressure coefficients, for the slope shown below
g Find
2
20′
10′
1’
γT = 110 pcf φ = 30°
1
γT = 110 pcf φ = 30° Soft Clay C = 400psf Firm Material
Example 6-3 g Add
solution
Student Exercise 2 g Using
a Rankine sliding block analysis, determine the safety factor against sliding for the embankment and assumed failure surface shown 2 30′
Sand Fill γ = 120 pcf φ = 30°
45° - φ/2 30°
10′ 5′ 16′
1 OGS
OGS
Sand γ = 120 pcf φ = 30° Soft Clay
Sand γ ′ = 60 pcf φ = 30°
45° + φ/2 C = 250 psf
Solution K a = tan 2 ( 45° − φ ) = tan 2 ( 45° − 30° ) = 0.33 2 2 Kp = tan 2 ( 45° − φ ) = tan 2 ( 45° + 30° ) = 3.0 2 2 ( per ft.) Pa = 1 γ H 2 K a = 1 (0.120kcf )( 40ft )2 (0.33)(1ft ) = 32 Kips → 2 2 Pp = 1 γ H 2 K p = 1 (0.120kcf )(10ft )2 (3.0)(1ft ) = 18Kips ← 2 2 cL = (0.250ksf )(60ft )(1ft ) = 15Kips ←
Σ Resisting Forces Pp + cL 18Kips + 15Kips FS = = = Pa 32 Kips Σ Driving Forces
F.S. = 1.03 – TOO LOW!!
Student Exercise g Same
as previous exercise except that water table rises of 10 ft to OGS
Solution Pa1 = γ1H1K a1 = (0.120kcf )(30' )(0.33) = 1.2ksf ( per foot ) PaFill = (1.2 Ksf )(30' )( 1 )(1' ) = 18Kips → 2 Pa 2 = 1.2ksf + (0.060kcf )(10' )(0.33) = 1.4ksf ( per foot ) (1.2ksf + 1.4ksf ) PaSand = (10' )(1' ) = 13Kips → 2 PaTotal = 18Kips + 13Kips = 31Kips → Pp = 1 γ b H 2 K p = 1 (0.060kcf )(10' )2 (3) = 9 Kips ←
