Slope Stability

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CE 303 G

CE 303 GEOTECHNICALEOTECHNICAL EENGINEERINGNGINEERING -- IIII

.

TTAABBIILLIITTYY OOFF AARRTTHH

–

S

LOPESLOPES by by Dr. T

Dr. T. Ven. Venkatakata BharBharat, Pat, Ph.D.h.D.

Assistant P

Assistant Professorrofessor

Department of Civil

Department of Civil EngineeringEngineering

,

(2)

-M

M

EETTHHOODDSS OOFF SSTTAABBIILLIITTYY AANNAALLYYSSIISS



 Limit Equilibrium MethodLimit Equilibrium Method (choice of analysis here!)(choice of analysis here!) 

 based on equilibrium of forcesbased on equilibrium of forces



 requires knowledge of staticsrequires knowledge of statics



 soil is considered to be on the verge of soil is considered to be on the verge of failurefailure



 based on equilibrium of stressesbased on equilibrium of stresses



 requires numerical methodsrequires numerical methods



 generally, analysis is done generally, analysis is done using software packagesusing software packages such a

(3)

S

TTAABBIILLIITTYY OOFF IINNFFIINNIITTEE SSLLOOPPEE



 Infinite slopesInfinite slopes have dimensionshave dimensions

that

that extend extend over over reat reat distancesdistances as compared to their depth

as compared to their depth



 The assumption of an infiniteThe assumption of an infinite

length simplifies the analysis length simplifies the analysis considerably.

considerably.



 re reprpreseseentntatatiive ve sesectctioion o n o inin ininitite se s oope pe iiss

considered in the figure. considered in the figure.

,, mechanism should be postulated first. mechanism should be postulated first.



 ItIt is reasis reasonablonable to ase to assume thsume that failat failure oure occurs occurs on an a

3

plane parallel to slope. plane parallel to slope.

(4)

S

…



 A slice of soil is consi A slice of soil is considered between tdered between the surface ofhe surface of

tthhe e sslloo e e aannd d tthhe e asasssuumemed d sslli i llaanne e as as sshhoowwn n iinn figure in the previous slide.

figure in the previous slide.



 DrawDraw free-bodyfree-body diagram of forces acting on thisdiagram of forces acting on this

slice and then formulate

slice and then formulate equilibriumequilibrium equations.equations.

4

(5)

S

…



 TheThe factor of safetyfactor of safety ((F F ) of a slope is ) of a slope is defined as thedefined as the

rati

ratio oo of thf the ae avaivailablable le sheshear sar stretrenn th th of of the the soisoill τ τ to the

to the minimum shear strengthminimum shear strength to maintainto maintain stability (mobilized strength,

stability (mobilized strength, τ τ _m_m where where f f

F

==

τ

tan tan f f nn τ

τ

==

σ σ

′′

φ φ

′′

for Effective Stressfor Effective Stress

Analysis (ESA) Analysis (ESA) for Total Stress for Total Stress

Case –

Case – I: ESA I: ESA withowithout the ut the effect effect of seof seepage epage forcesforces

m m

f

f uu

τ

==

_{Analysis (TSA)}_{Analysis (TSA)}

α α φ

==

φ

′′

ta tann F F

==

φ φ

′′

at at limit-equilibrium equilibrium 5 5

(6)

S

…

Case –

Case – II: ESII: ESA with A with the efthe effect ofect of seepf seepage foage forces (rces (J J _ss))



 e e uus s nnoow w ccoonnss eer r ggrroouunn wwaa eer r ww n n e e ss nngg

mass and assume that the

mass and assume that the seepage is parallel toseepage is parallel to the slope

the slope. The seepage force is given by. The seepage force is given by

,,



 At limit equilibrium, At limit equilibrium,

6

(7)

S

…

Ca

Case se –– IIIII: I: TSTSA A



 e e ss eeaar r ss rreesss s oon n e e ss p p pp aanne e oor r aa ss



 The factor of safetyThe factor of safety FF forfor TSA TSA is given by:is given by:



 At limit At limit equilibrium,equilibrium,



 Critical value ofCritical value of zz occurs at:occurs at:

7

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I

NFINITENFINITE

S

LOPESLOPES

–

S

ALIENT ALIENT

P

OINTSOINTS



 TheThe maximum stable slopemaximum stable slope in a coarse-grainedin a coarse-grained

so

soiil l in in ththee aabbsseenncce e oof f sseeee aa ee i is s ee uuaal l tto o tthhee friction angle

friction angle of the soil.of the soil.



 TheThe maximum stable slopemaximum stable slope in a coarse-grainedin a coarse-grained

soil, in

soil, in presence of seepagepresence of seepage, is roughly, is roughly half ofhalf of the friction angle

the friction angle of the soilof the soil..



 TheThe critical slopecritical slope angle in fine-grained soils isangle in fine-grained soils is 45°45°

and the

and the critical depthcritical depth is equal to the depth of is equal to the depth of thethe tension cracks

tension cracks 2s2s / /



 Infinite slope mechanism is usually not observedInfinite slope mechanism is usually not observed

fo

for fr finine-e- rairainened d sosoilils. s. FoFor sr such uch sosoilils s rorotattatioionalnal failure mechanism is more common.

failure mechanism is more common. 88 ..

(9)

I

NFINITENFINITE

S

LOPELOPE

–

A

NN

E

XAMPLE XAMPLE



 Dry sand is to be dumped from a truck on the sideDry sand is to be dumped from a truck on the side

o

of f a a rrooadadwwaa . . TThhe e roro eertrtiiees s oof f tthhe e ssaannd d araree ’ =’ = 30°,

30°, = = 17 17 kN/m3kN/m3 andand _sat_sat = 17.5 kN/m3.= 17.5 kN/m3.

Determine the maximum slope angle of the sa

Determine the maximum slope angle of the sand innd in ,

, ,,

seepage and (c) the saturated state if groundwater seepage and (c) the saturated state if groundwater is present and seepage occurs para

is present and seepage occurs parallel to the slopellel to the slope ttoowwaarr s s tt e e ttooe e o o tt e e ss ooppee. . WW aat t iis s tt e e ssaa e e ss ooppee in the dry state for a factor of safety of

in the dry state for a factor of safety of 1.25?1.25?

(will be solved in the class)

9

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R

OTATIONALOTATIONAL

S

LOPELOPE

F

AILURE AILURE



 Slopes made up ofSlopes made up of

--soils have been observed to soils have been observed to fail through a rotational fail through a rotational

a

a uurre e mmeecc aann ssmm..



 The failure surface isThe failure surface is

right) or noncircular right) or noncircular (bottom right).

(bottom right).



 The analysis also takes intoThe analysis also takes into

account the presence of a account the presence of a p

p rreeaatt cc ssuurr aacce e ww tt n n tt ee sliding mass.

sliding mass.

10

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S

STABILITYTABILITY A A NNAALLYYSSIISS OOFF A A RROTATIONALOTATIONAL FF AILURE AILURE



 A A free-body diagramfree-body diagram of the assumed circularof the assumed circular

mechanism would show the

mechanism would show the wweeii hht t WW of the soilof the soil within the

within the sliding masssliding mass acting at the centre ofacting at the centre of mass.

mass.



 If seepage is present, theIf seepage is present, the seepage forces (Js)seepage forces (Js)

would be present. would be present.



 The forcesThe forces resistingresisting the clockwise rotation of thethe clockwise rotation of the

sliding mass are the

sliding mass are the shear forces mobilized byshear forces mobilized by the soil

the soil aalloon n tthhe e cciirrccuullaar r sslli i ssuurrffaaccee..



 We must now useWe must now use staticsstatics to determine whether theto determine whether the

d

diissttuurrbbiin n mmoommeenntts cs crreeaatteed d bb W W andand JJ exceedexceed the

(12)

=

= 0

0 A

An

na

all ssiiss

Q (kN) Q (kN) d d_Q_Q ddww u u aa F F Wd Wd

==

u u aa c c L L r r F F

=

_F_F

=

c L c u au L r ar _F_F

=

c L c u au L r ar 12 12 Q Q w w w w Q Q w w w w Presence of load, Q

(13)

F

RICTIONRICTION

-C

IIRRCCLLEE MMEETTHHOODD

Considered forces Considered forces::



 WeWeii ht ht of of sosoil il mamass ss inin failure zone,

failure zone, W W



 Sum of cohesive forcesSum of cohesive forces a

acc nng g ppaarraa e e o o cc oorr AB,

AB, C C _m_m



 The resultant ofThe resultant of frictional forces, frictional forces, RR



 Factor of safetyFactor of safety

e eqquuaa oon n s s aasse e oonn:: tan tan f f m m cc τ τ _σ_σ _φ_φ τ τ

=

= ++

′ ′

′′

13 13 cc φ φ such that such that F F == F F _cc == F F _φ_φ

(14)

F

RICTIONRICTION

-C

IIRRCCLLEE MMEETTHHOODD

…



 Important relationsImportant relations

cc ArArc c LeLen n thth tantan

′′

  Procedure Procedure:: or or engtengt m m _AB_AB cc F F

=

= ××

( (

Chord LengthChord Length

))

m m AB AB C C AB AB L L

=

××

r r tantan F F _φ_φ ψ ψ

==



 Assume a failure plane such Assume a failure plane such as ABDA as ABDA 

 Obtain the weight of soil mass,Obtain the weight of soil mass, W W , in the failure zone by, in the failure zone by

graphical techniques graphical techniques



 Find the direction (parallel to chord AB) and distance ofFind the direction (parallel to chord AB) and distance of C C _m_m

from center, O from center, O



 Assume Assume F F _φ_φand draw friction circle with radius r×Sinand draw friction circle with radius r×Sinψψ 

 Find the direction ofFind the direction of RR (passes through intersection of(passes through intersection of W W andand

C

C _m_m, and runs tangent to, and runs tangent to φφ-circle)-circle)



 Draw force polygon and find the magnitude ofDraw force polygon and find the magnitude of C C _m_m 

 ObtainObtain F F _cc and compare with assumedand compare with assumed F F _φ_φ 

 Change the value ofChange the value of F F _φ_φ and repeat the procedure tilland repeat the procedure till F F _cc== F F _φ_φ

14

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M

EETTHHOODD OOFF

S

LICESLICES



 One approach that is commonly used One approach that is commonly used to analyzeto analyze

rot

rotatiationaonal fal failuilure ire is to s to divdivide ide the the slislidin din masmass ints intoo an arbitrary number of

an arbitrary number of vertical slices and thenvertical slices and then sum the forces and moments of each slice. sum the forces and moments of each slice.

15

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M

EETTHHOODD OOFF

S

LICESLICES

…



 Of course, theOf course, the larger the number of slices, thelarger the number of slices, the

be

betttter er ththe ace accucurarac c of of ouour sor solulutitionon..



 However, dividing the sliding mass into a numberHowever, dividing the sliding mass into a number

of vertical slices

of vertical slices poses new problems.poses new problems.



 We now have to account for We now have to account for thethe internal orinternal or

interfacial forces between two adjacent

interfacial forces between two adjacent slices.slices.



 Let’s now attempt to draw aLet’s now attempt to draw a free-body diagramfree-body diagram ofof

an

an arbitrary vertical slice and examine thearbitrary vertical slice and examine the o

orrccees s aaccttiinng g oon n tt iis s ss iiccee..

16

(17)

F

ORCESORCES

A

CCTTIINNGG OONN A A

V

ERTICALERTICAL

S

LICELICE

17

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M

EETTHHOODD OOFF

S

LICESLICES

–

K

NNOOWWNN UUAANNTTIITTIIEESS

18

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M

MEETTHHOODD OOFF SSLICESLICES – – UUNNKKNNOOWWNN UUAANNTTIITTIIEESS

19

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M

EETTHHOODD OOFF

S

LICESLICES

…



 If there areIf there are nn slices, we have to obtain the values slices, we have to obtain the values ofof

6n-1

6n-1 arameters.arameters.



 However, we only haveHowever, we only have 4n4n number of equations.number of equations.

a

a eeaavvees s uus s ww n-n- uunn nnoowwnnss..



 Therefore, the problem isTherefore, the problem is staticallystatically

..



 For example, if there areFor example, if there are 1010 slices, we’ll have 6x10-slices, we’ll have

6x10-=

= == ..



 Therefore, in order to obtain a solution, we have Therefore, in order to obtain a solution, we have toto

make certain

make certain ssiimm lliiff iin n aassssuumm ttiioonnss or use anor use an iterative method

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M

E

T

H

O

D

O

F

S

LICES

…



 Several solution methods have Several solution methods have been developedbeen developed

d

dee eennddiin n oon n tthhee aassssuumm ttiioonns s mmaaddee about theabout the unknown parameters and which equilibrium

unknown parameters and which equilibrium condition (force, moment or both) have been condition (force, moment or both) have been

..



 Tables on the next two pages provide a summary ofTables on the next two pages provide a summary of

..



 Computer programs (such asComputer programs (such as SLOPE/WSLOPE/W oror

XSTABL

XSTABL are available for all the methods listedare available for all the methods listed in the table.

in the table.

21

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S

WEDISHWEDISH

C

IIRRCCLLEE MMEETTHOHODD



 Forces acting on a slice:Forces acting on a slice:



 WWeiei ht ht of of ssoioil l mmaassss



 Cohesive forces (Cohesive forces (C C ) in the) in the opposite to the direction of opposite to the direction of p

prroo aa e e wwee gge e mmoovveemmeenntt



 Reaction (Reaction (RR) at the base) at the base ,, assuming slippage is assuming slippage is imminent imminent   Assumptions: Assumptions: 

 The The inteinterslirslicece reacreaction tion forcforces es are are equaequall and and opposoppositeite

22



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S

WEDISHWEDISH

C

…

Factor of safety: Factor of safety: 1 1

sseec c _j_j _{j j}ccoos s tta_j j ann j j n n cb cb W W F F α α α α φ φ ==

⎡

++

⎤⎤

⎣

⎦⎦

==

∑

1 1 sin sin j j j j j j W W α α ==

⎣

⎦⎦



 It may be noted that the tangential component,It may be noted that the tangential component, T T _j_j,,

and base angle,

and base angle, α α _j_j, may be negative for few slices, may be negative for few slices

23

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S

WEDISHWEDISH

C

…



 N N andand T T curves:curves:

N N N N A

=

= ××

A γ γ

∑

T T T T

=

= ××

AA γ γ

∑

where

where A A_N_Nandand A A_T_Tare areas ofare areas of N N -- aanndd T T -- diagdiagramsrams, , resprespectiectivelyvely

24

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R

IGIGOROROUOUSS MEMETHTHODODSS



 Bishop's SimplifiedBishop's Simplified



 aan un u ss mmpp ee



 JanJanbubu's's GeGeneneralralizizeded



 SpencerSpencer



 Morgenstern-PriceMorgenstern-Price



 General Limit Equilibrium (GLE)General Limit Equilibrium (GLE)



 Lowe-KarafiathLowe-Karafiath

25

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B

ISHOPISHOP

’

SS

S

IMPLIFIEDIMPLIFIED



 The effect of forces acting on theThe effect of forces acting on the

sides of the individual slices are sides of the individual slices are taken into account

taken into account



 Disregards the shear forces onDisregards the shear forces on

the inter-slices (

the inter-slices ( X X ₁₁ == X X ₂₂= 0)= 0)



 Method satisfies momentMethod satisfies moment

equilibrium and vertical force equilibrium and vertical force equilibrium

equilibrium

26

(27)

B

ISHOPISHOP

’

SS

S

IMPLIFIEDIMPLIFIED

…



 Factor of Safety:Factor of Safety:

n n

( (

))

1 1 tan tan j j j j j j j j j j n n c c b b W W uubb m m F F α α φ φ ==

′ ′

+

−−

′′

⎣

⎦⎦

==

∑

1 1 s s nn j j j j j j α α ==

⎣

⎦⎦

where

where m m _α_α

=

= ++

( (

1 1 ttaan tn taφ φ

′′

an n α α F F

))

ccoossα α



 s s aappppeeaarrs s oon n o o e e ss eess, , eerraa vve e aapppprrooaac c s s rreeqquu rree by assuming the

by assuming the F F and finding the value. The assumed valueand finding the value. The assumed value is compared against the computed.

is compared against the computed.



(28)

I

NTERSLICENTERSLICE

F

ORCESORCES



 Interslice shear forces are Interslice shear forces are required to calculate therequired to calculate the

normal force at the base of each sli normal force at the base of each slice.ce.



 The The inteinterslicerslice sheashear for force rce (( X X _ii)) is cis comompuputeted ad as as a

perce

percentagntage of the inte of the interslierslicece normnormal forcal force (e (EE_ii)) acaccocordrdiningg tto o tthhe e ffoolllloowwiin n eemm iirriiccaal l ee uuaattiioon n rroo oosseed d bb

Morgenstern and Price (1965): Morgenstern and Price (1965):

where: where: λ

λ = the = the percentagpercentage (in decimal form) of e (in decimal form) of the functionthe function ,,

f(x)

f(x) = int= intersliceerslice force force function function representinrepresenting tg the he relativerelative direc

direction otion of the resuf the resultanltant intet interslicerslice forceforce

28

(29)

ARIOUS

I

NTERSLICENTERSLICE

F

ORCEORCE

F

UNCTIONSUNCTIONS

29

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M

EETTHHOODDSS OOFF

S

LOPELOPE

S

TTAABBIILLIITTYY NNAALLYYSSIISS

30

(31)

S

SSSUUMMPPTTIIOONNSS IINN AARRIIOOUUSS

M

ETHODSETHODS

31

(32)

C

OOMMPPAARRIISSOONN OOFF DDIIFFFFEERREENNTT MMEETTHHOODDSS

32

32 Provided in the “

(33)

D

ESIGNESIGN

C

HARTSHARTS



 Slope stability analysis based on design charts isSlope stability analysis based on design charts is

useful useful



 for preliminary analysisfor preliminary analysis



 for rapid means of checking the results of detailedfor rapid means of checking the results of detailed



 to compare alternates that can later be examined byto compare alternates that can later be examined by rigorous analysis

rigorous analysis



 to determine the approximate value of theto determine the approximate value of the F F as it allowsas it allows some quality control check for the

some quality control check for the subsequentsubsequent

computer-generated solutions computer-generated solutions



 To back-calculate strength values for failed slopes to aidTo back-calculate strength values for failed slopes to aid in planning remedial measures

in planning remedial measures

33

(34)

D

ESIGNESIGN

C

HARTSHARTS

…



 Taylor’s chart (1948)Taylor’s chart (1948)



 Spencer (1967)Spencer (1967)



 JJaannbbuu 11996688



 Hunter & Schuster (1968)Hunter & Schuster (1968)



 ChChen en & G& Gigigerer (1(197971)1)



 O’Connor & Mitchell (1977)O’Connor & Mitchell (1977)



 Cousins (1978)Cousins (1978)



 aarr ees s ooaarreess 11998844



 Barnes (1991)Barnes (1991)

34

(35)

D

ESIGNESIGN

C

HARTSHARTS

…



 Taylor’s chart (1948)Taylor’s chart (1948)



 Spencer (1967)Spencer (1967)



 JJaannbbuu 11996688



 Hunter & Schuster (1968)Hunter & Schuster (1968)



 ChChen en & G& Gigigerer (1(197971)1)



 O’Connor & Mitchell (1977)O’Connor & Mitchell (1977)



 Cousins (1978)Cousins (1978)



 aarr ees s ooaarreess 11998844



 Barnes (1991)Barnes (1991)

35

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T

AYLOR AYLOR

’

SS CCHHAARRTTSS

1948



 Taylor’s charts provide the stability values in Taylor’s charts provide the stability values in termsterms

o

of f ““ssttaabbiilliit t nnuummbbeerr S S ” u” ussin in frfricictitioon-n-cicircrcllee method

method

Condition: Condition:



 Analysis by these charts is valid fo Analysis by these charts is valid for simple sectionsr simple sections

cc φ φ

=

= ==

  In general,In general, 

 failure surface passes through the toe when the slope isfailure surface passes through the toe when the slope is steep

steep



 base failure (failure extends below toe) occurs whenbase failure (failure extends below toe) occurs when eith

either ter the she slolo es aes are fre flattlatter oer or/and r/and firm firm strastratum tum exisexiststs below the toe

(37)

T

AYLOR AYLOR

’

SS CCHHAARRTTSS

1

9

4

8

8 …

…

Fig. Conceptual section by Taylor Fig. Conceptual section by Taylor

37

(38)

next figure

next figure φ φ = 0 Analysis:= 0 Analysis:

4 4 8 8 ) ) S S ( ( 1 1 9

9 _{Stability number}_{Stability number}

C C H H A A R R In terms of F.S. In terms of F.S. O O R R ’ ’ S S cc d d ss c c cc F F c c N N Hγ γ H

=

= ==

T T A A Y Y 38 38

(39)

c' c' −φ −φ ' ' Analysis Analysis:: 4 4 8 8 ) ) S S ( ( 1 1 9 9 In terms of F.S. In terms of F.S. c c

′ ′

cc

′′

C C H H A A R R F F

==

φ φ

′′

cc d d ss c c N N Hγ γ H O O R R ’ ’ S S φ φ _d_d T T A A Y Y 39 39

(40)

P

ROBLEMROBLEM

-

1



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

,

, , , ,,



 F F ofof S S



 The distance from toe to the point where critical circleThe distance from toe to the point where critical circle appears on the ground

appears on the ground



 F F ofof S S , if there are heavy loadings outside the toe., if there are heavy loadings outside the toe.

(solve it during the tutorial class)

40

(41)

P

ROBLEMROBLEM

-

2



 Given a soil slope with Given a soil slope with height,height, H H = 12 m,= 12 m, β β = 30= 3000, c’, c’

=

= 24 24 kkPa Pa ’ ’ = = 202000 _and_{and =}_{= 19}_{19 kN/m}_kN/m33 _{What is the}_{What is the} ..

factor of safety of the slope? factor of safety of the slope?

(solve it during the tutorial class)

41

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S

PENCERPENCER

’

SS CCHHAARRTTSS

1967



 Based on solutions computed using Spencer’sBased on solutions computed using Spencer’s

m

meetthhood d wwhhiicch sh saattiissffiiees cs coomm lleette e ee uuiilliibbrriiuumm



 Charts are used to determine the Charts are used to determine the required sloperequired slope

angle for a preselected

angle for a preselected F F ofof S S



 Solutions for three different pore pressure ratios,Solutions for three different pore pressure ratios,

rr_u_u: 0, 0.25, 0.5.: 0, 0.25, 0.5. 

 Pore water pressure ratio (rPore water pressure ratio (r_u_u)) is the ratio of poreis the ratio of pore water force on a slip surface to the total force due to water force on a slip surface to the total force due to 

 Assumption: firm stratum is at great depth belo Assumption: firm stratum is at great depth beloww

the slope the slope

42

(43)

7 7 ) ) s s ( ( 1 1 9 9 φ

φ _d_d = tan= tan-1-1_(tan_(tan_φ_φ_//F_F ))

c c h h a a r r n n c c e e r r ’ ’ S S p p e e 4 4 3 3 43 43

(44)

P

ROBLEMROBLEM

-

3



 Given a slope with heightGiven a slope with height H H = 18 m,= 18 m, c’ c’ = 9.6 kPa,= 9.6 kPa, φ φ ’ ’

= 30

= 3000 ₌_{= 19.6}_{19.6 kN/m}_kN/m33 _{rr = 0.}_{= 0.25}_{25 de}_dete_term_rmin_{ine t}_{e the}_he

maximum slope angle

maximum slope angle β β forfor F F ofof S S of 1.5.of 1.5.

s

soo vve e iit t uurriinng g tt e e ttuuttoorriia a cc aassss

44

(45)

P

RROOBBLLEEMM FFOORR

A

SSIGNMENTSSIGNMENT

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1



 Given a soil slope with Given a soil slope with height,height, H H = 12 m,= 12 m, β β = 30= 3000, c’, c’

=

= 24 24 kkPa Pa ’ ’ = = 202000 _and_{and =}_{= 19}_{19 kN/m}_kN/m33 _{find the factor}_{find the factor} ,,

of safety of the slope using the

of safety of the slope using the following methods:following methods: 



φφ

_u_u = 0 analysis= 0 analysis



 Friction-circFriction-circle le methodmethod



 Swedish circle methodSwedish circle method



 Bishop’s simplified methodBishop’s simplified method

45