Slope Stability

CE 303 GEOTECHNICAL ENGINEERING - II

. .

ARTH –

TABILITY OF

SLOPES by Dr. T. T. Venkata Venkata Bhar Bharat, at, Ph.D. Ph.D.  Assistant Professor Professor Department of Civil Engineering ,

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METHODS OF STABILITY ANALYSIS 

Limit Equilibrium Method (choice of analysis here!) 

based on equilibrium of forces

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requires knowledge of statics

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soil is considered to be on the verge of failure

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based on equilibrium of stresses

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requires numerical methods

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generally, analysis is done using software packages such as as Plaxis, Plaxis, Geostui Geostuido do (Slope/w (Slope/w)) etc.

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STABILITY OF INFINITE SLOPE 

Infinite slopes have dimensions that extend over reat distances as compared to their depth

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The assumption of an infinite length simplifies the analysis considerably.

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 repr  repres ese ent ntat atiive se sect ctio ion n o in init inite e s ope is considered in the figure. , mechanism should be postulated first.

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It is reasonabl reasonable e to assume assume that that failure failure occurs occurs on on a plane parallel to slope.

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STABILITY OF INFINITE SLOPE …  A slice of soil is considered considered between the the surface of the slo e and the ass assume med d sli lane as shown in figure in the previous slide.

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Draw free-body diagram of forces acting on this slice and then formulate equilibrium equations.

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STABILITY OF INFINITE SLOPE … 

The factor of safety (F ) of a slope is defined as the ratio ratio of of the the avai availab lable le shear shear stre stren n th of the soil soil τ  to the minimum shear strength to maintain stability (mobilized strength, τ m

F  =

τ  f 

where

τ  f

= σ n′ tanφ ′

τ  f

m

=

for Effective Stress  Analysis (ESA) for Total Stress  Analysis (TSA)

u

Case – I: ESA without without the effect of seepage seepage forces F  =

tan tanφ ′

at limitequilibrium

α

= φ ′ 5

STABILITY OF INFINITE SLOPE … Case – II: ESA ESA with the effect effect of of seepage seepage forces forces (J  (J s) 

e us now cons er groun wa er w n es ng mass and assume that the seepage is parallel to the slope. slope. The seepage force is given by

,

 At limit equilibrium,

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STABILITY OF INFINITE SLOPE … Case Case – III: III: TSA  TSA  





e s ear s ress on

e s p p ane or a

s

The factor of safety F for TSA is given by:

 At limit equilibrium, 7

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Critical value of z occurs at:

INFINITE SLOPES – S ALIENT POINTS 

The maximum stable slope in a coarse-grained soiil in th so the e absence of see a e  is  is e ual to the friction angle of the soil.

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The maximum stable slope in a coarse-grained soil, in presence of seepage, seepage, is roughly half of the friction angle of the soil. soil.

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The critical slope angle in fine-grained soils is 45° and the critical depth is equal to the depth of the tension cracks 2s /  .

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Infinite slope mechanism is usually not observed for for fin finee- raine rained d so soil ils. s. For For such such so soil ilss rotat rotatio ional nal failure mechanism is more common.

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INFINITE SLOPE – A N E XAMPLE 

Dry sand is to be dumped from a truck on the side of a roadw adwa . The ro erti rties of the sand are are ’ = 30°, = 17 kN/m3 and sat = 17.5 kN/m3. Determine the maximum slope angle of the sand sa nd in , , seepage and (c) the saturated state if groundwater is present and seepage occurs parallel para llel to the slope towar s t e toe o t e s ope. W at is t e sa e s ope in the dry state for a factor of safety of 1.25? (will be solved in the class) 9

ROTATIONAL SLOPE F AILURE 

Slopes made up of soils have been observed to fail through a rotational a ure mec an sm.

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The failure surface is right) or noncircular (bottom right).

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The analysis also takes into account the presence of a p reat c sur ace w t n t e sliding mass.

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STABILITY A NALYSIS OF A ROTATIONAL F AILURE  A free-body diagram of the assumed circular mechanism would show the wei ht W of the soil within the sliding mass acting at the centre of mass.

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If seepage is present, the seepage forces (Js) would be present.

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The forces resisting the clockwise rotation of the sliding mass are the shear forces mobilized by the soil alon the circular sli surface.

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We must now use statics to determine whether the disturbin moments cr created b  W and J exceed the restoring moment provided by the soil.

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= 0 Anal sis Q (kN) dw

dQ 

F  =

 = F  =

cu La r 

 = F  = Q

Presence of load, Q

cuLa r  w

u a

Wd 

 = F  = w 

Presence of crack

cu La r  Q

w

w 

Presence of load and crack

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FRICTION-CIRCLE METHOD Considered forces: forces: 

Wei Wei ht of so soil il mass mass in failure zone, W 

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Sum of cohesive forces ac ng para e o c or  AB, C m

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The resultant of frictional forces, R

Factor of safety equa on s ase on:

τ m

=

τ  f 

=

c c

+

σ ′ tanφ ′ φ 

such that F = F = F c = F φ 

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FRICTION-CIRCLE METHOD… 

Important relations m



=

c F c

×

or

engt

AB

LC m

=

Arc Arc Le Len n th

 AB

( Chord Length) AB

× r 

tanψ  =

′

tan

F φ 

 Procedure: 

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



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 Assume a failure plane such as ABDA  Obtain the weight of soil mass, W , in the failure zone by graphical techniques Find the direction (parallel to chord AB) and distance of C m from center, O  Assume F φ  and draw friction circle with radius r×Sinψ Find the direction of R (passes through intersection of W and W and C m, and runs tangent to φ-circle)

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Draw force polygon and find the magnitude of C m

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Obtain F c and compare with assumed F φ 

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Change the value of F φ and repeat the procedure till F c = F φ 

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METHOD OF SLICES 

One approach that is commonly used to analyze rotati rotationa onall failu failure re is is to divide divide the slidin slidin mass mass into into an arbitrary number of vertical slices and then sum the forces and moments of each slice.

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METHOD OF SLICES… 

Of course, the larger the number of slices, the bett better er the the accu accura rac c of our our solu soluti tion on..

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However, dividing the sliding mass into a number of vertical slices poses new problems.

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We now have to account for the internal or interfacial forces between two adjacent slices.

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Let’s now attempt to draw a free-body diagram of an arbitrary vertical slice and examine the orces acting on t is s ice.

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FORCES A CTING ON A V  VERTICAL SLICE

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METHOD OF SLICES – K NOWN

UANTITIES

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METHOD OF SLICES – UNKNOWN

UANTITIES

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METHOD OF SLICES… 

If there are n slices, we have to obtain the values of 6n-1 arameters.

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However, we only have 4n number of equations. a

eaves us w

n- un nowns.

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Therefore, the problem is statically .

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For example, if there are 10 slices, we’ll have 6x10= = .

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Therefore, in order to obtain a solution, we have to make certain sim lif in assum tions or use an iterative method

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METHOD OF SLICES… 

Several solution methods have been developed de endin on the assum tions made about the unknown parameters and which equilibrium condition (force, moment or both) have been .

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Tables on the next two pages provide a summary of .

Computer programs (such as SLOPE/W or  XSTABL are available for all the methods listed in the table.

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SWEDISH CIRCLE METHO HOD D 

Forces acting on a slice: 

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Wei ht of soil oil mass Cohesive forces (C ) in the opposite to the direction of pro a e we ge movement Reaction (R) at the base , assuming slippage is imminent

 Assumptions:

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The intersli interslice ce reaction reaction forces forces are equal equal and opposite opposite ear orces at t e nter-s ce are assume to e zero

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SWEDISH CIRCLE METHO HOD D… Factor of safety:

F  =

∑ ⎡⎣cb secα  j =1

+ W j cosα j  tanφ ⎤⎦

n

 j =1



 j

⎣W  j sinα j  ⎦

It may be noted that the tangential component, T  j, and base angle, α  j, may be negative for few slices

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SWEDISH CIRCLE METHO HOD D…  N and  N and T curves: T curves:



∑N = A

N

× γ 

∑T = A × γ  T 

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where A where AN  and A and AT  are areas of N - and T - diagrams diagrams,, respecti respectively vely

RIG IGOR OROU OUS S ME METH THOD ODS S  

Bishop's Simplified an u s

mp

e

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Janbu Janbu's 's Ge Gene neral raliz ized ed

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Spencer



Morgenstern-Price

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General Limit Equilibrium (GLE)

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Lowe-Karafiath 25

BISHOP’S SIMPLIFIED 

The effect of forces acting on the sides of the individual slices are taken into account

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Disregards the shear forces on the inter-slices ( X   X 1 = X 2  = 0)

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Method satisfies moment equilibrium and vertical force equilibrium

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BISHOP’S SIMPLIFIED … 

Factor of Safety: n

F  =

∑m  j =1

α 

⎣c j′bj + (Wj − ubj  ) tanφ ′⎦ n

 j =1

⎣

 j

s nα j  ⎦

where mα  = (1 + tanφ ′ ta tanα F ) cosα  

s appears on o e s es, era ve approac s requ re by assuming the F and finding the value. The assumed value is compared against the computed.

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The process is continues until both the values match

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INTERSLICE FORCES 

Interslice shear forces are required to calculate the normal force at the base of each slice. sli ce.

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 X i) is com The interslice interslice shear shear force force ( X  compu pute ted d as as a percentag percentage e of the intersli interslice ce normal normal force force (Ei) acco accord rdin ing g to the followin em irical e uation ro osed b Morgenstern and Price (1965):

where: percentage (in decimal form) of the function λ = the percentage , f(x) = interslice interslice force function representing representing the the relative direction direction of of the resultan resultantt interslice interslice force

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 ARIOUS INTERSLICE

FORCE FUNCTIONS

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METHODS

OF

SLOPE STABILITY

NALYSIS

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SSUMPTIONS IN

ARIOUS

METHODS

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COMPARISON OF DIFFERENT METHODS

Provided in the “ Additional_read  Additional_read”” folder folder in CE303 dropbox dropbox link

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DESIGN CHARTS 

Slope stability analysis based on design charts is useful 

for preliminary analysis

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for rapid means of checking the results of detailed

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to compare alternates that can later be examined by rigorous analysis to determine the approximate value of the F as it allows some quality control check for the subsequent computer-generated solutions To back-calculate strength values for failed slopes to aid in planning remedial measures 33

DESIGN CHARTS… 

Taylor’s chart (1948)

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Spencer (1967)

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Janbu 1968

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Hunter & Schuster (1968)

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Chen Chen & Gig Giger er (197 (1971) 1)

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O’Connor & Mitchell (1977)

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Cousins (1978)

 

ar es

oares 1984

Barnes (1991) 34

DESIGN CHARTS… 

Taylor’s chart (1948)

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Spencer (1967)

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Janbu 1968

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Hunter & Schuster (1968)

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Chen Chen & Gig Giger er (197 (1971) 1)

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O’Connor & Mitchell (1977)

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Cousins (1978)

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ar es

oares 1984

Barnes (1991) 35

T AYLOR’S CHARTS 1948 

Taylor’s charts provide the stability values in terms of “stabilit number S  ” us usin fric fricti tio on-ci n-circ rclle method Condition:

= c=

φ 

 Analysis by these charts is valid for for simple sections

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In general, 

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failure surface passes through the toe when the slope is steep base failure (failure extends below toe) occurs when either either the the slo slo es are are flatt flatter er or/and or/and firm stratum stratum exists exists below the toe

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T AYLOR’S CHARTS 1948 …

Fig. Conceptual section by Taylor

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next figure

   )    8    4    9    1    (

φ = 0 Analysis:

Stability number

   S    R    A    H    C    S    ’    R    O

In terms of F.S.

F c

=

c cd

=

c Nsγ H

   Y    A

   T 38

c' −φ  ' Analysis:  Analysis: −φ ' 

   )    8    4    9    1    (

In terms of F.S.

   S    R    A    H    C    S    ’    R    O

c

F 

=

c′

c′

cd

Nsγ H

φ ′ φ d 

   Y    A

   T 39

PROBLEM - 1 

Given a soil slope with height, H = H = 12 m, DH  m, DH = = 18 m,  β  , , , 





F of S 

The distance from toe to the point where critical circle appears on the ground F of S , if there are heavy loadings outside the toe.

(solve it during the tutorial class)

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PROBLEM - 2 

Given a soil slope with height, H = H = 12 m,  β = 300, c’ = 24 kPa ’ = 200 and = 19 kN/m3. What is the factor of safety of the slope?

(solve it during the tutorial class)

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SPENCER’S CHARTS 1967 

Based on solutions computed using Spencer’s method which sa satisfies co com lete e uilibrium

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Charts are used to determine the required slope angle for a preselected F of F of S 

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Solutions for three different pore pressure ratios, ru: 0, 0.25, 0.5. 

Pore water pressure ratio (ru) is the ratio of pore water force on a slip surface to the total force due to

 Assumption: firm stratum is at great depth below below the slope

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   )    7    9    1    (   s   r   a    h   c    ’   r   e   c   n   e   p    S

φ d = tan-1(tanφ /F )

43 4 3

PROBLEM - 3 

Given a slope with height H = H = 18 m, c’ = c’ = 9.6 kPa, φ ’  ’  = 300 = 19.6 kN/m3 r = 0.25 0.25 dete determ rmin ine e the the maximum slope angle  β for F of F of S of S of 1.5.

so ve it uring t e tutoria c ass

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PROBLEM FOR A SSIGNMENT SSIGNMENT - 1 

Given a soil slope with height, H = H = 12 m,  β = 300, c’ = 24 kPa ’ = 200 and = 19 kN/m3, find the factor of safety of the slope using the following methods: 

φu = 0 analysis

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Friction-circle Friction-circle method

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Swedish circle method

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Bishop’s simplified method

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