General limit equilibrium method for the estimation of 2D slope stability

C. Grandas and A. Niemunis

February 5, 2016

Contents1 Introduction, definition of safety factor 1

1.1 Key problem: overly restrictive assumptions on ∆Ei . . . . . . 2

2 Simplified methods without balance of internal forces 32.1 Simplified Fellenius . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Simplified Bishop . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Generalized limit equilibrium (GLE) methods 4

4 Fellenius method (FEL) 4

5 Slices with water 5

6 Simplified Fellenius with Mathematica (dry version) 66.1 Target function and graphics . . . . . . . . . . . . . . . . . . . 66.2 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76.3 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76.4 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

7 Multi-method solution with Fortran 87.1 Files and modules . . . . . . . . . . . . . . . . . . . . . . . . . 97.2 Preparation of input . . . . . . . . . . . . . . . . . . . . . . . . 107.3 Output and graphics . . . . . . . . . . . . . . . . . . . . . . . . 117.4 To be modified . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

A Notation 12

B Why logarithmic spiral 14

C Intersection between the spiral and a line 15

D On the uniqueness of the solution 15

La Conchita 1995, Photo: T.B.Edil

1 Introduction, definition of safety factor

A computer program for slope stability analysis is presented. The failure mechanism is expected in the form of a rigidblock with rotational sliding along a circular or spiral slip line. The program determines the safety factor F using themethods of slices. Various positions of the slip line are considered, among which the minimum safety factor is searchedfor.The soil strength τ is described by the friction angle ϕ and the cohesion c. According to the strength by Krey andTiedemann we have ϕ = ϕ′, c = c′ or ϕ = ϕs and c = 0 for overconsolidated or normally consolidated soils, respectively.For dense soils with dilatancy, ψ > 0, or for loose soils with contractancy, ψ < 0, the slip line may have the form of anevolving or convolving logarithmic spiral, respectively. The centre c and the radius of the spiral are usually found bytrial and error minimizing the safety factor. For a given geometry, both nominal cohesion c and nominal friction tanϕare scaled by the same, usually unknown, safety factor F

τm = (c+ σ tanϕ)/F tm = mobilized (1)

until the equilibrium of weights, reactions, and friction forces is established. Note that the mobilized strength τm iscomputed with the effective normal stress σ. The same factor F applies to all slices and the slope is safe for F > 1. Thesliding block of soil, Fig. 1, is divided into n vertical slices of thickness b each, interacting with each other with lateral

1

forces E (not necessarily horizontal). The inclination of lateral forces has to be assumed. It cannot be calculated fromthe mobilized strength because there is no sliding on the vertical contacts between slices.The sliding is directed to the right and the slices are numbered i = 1, 2, . . . n from the left. A slice i has two verticalcontacts numbered i − 1 on the left and i on the right. The numbers of contacts k are used to index the lateral forcesk = 0, 1, 2, . . . n. The driving forces are mainly self weights Wi and external loads. The resistance forces are calculatedfrom the reactions of the subsoil Qi = Ni + Tm

i and from cohesion forces Ci, both applied at the bottom surface of theslices. The bottom slide line is approximated by straight sections, Fig. 1, but all slices are subjected to a single rigidrotation, i.e. there are no velocity jumps across the vertical interfaces1. First, for the sake of simplicity, we consider adry slope (i.e. pore pressure is zero everywhere and, therefore, total and effective stresses are identical).

1.1 Key problem: overly restrictive assumptions on ∆Ei

The method of slices uses the equilibrium of forces in each slice i

Wi + ∆Ei + Qi + Ci = 0 for i = 1, 2, . . . n (2)

wherein ∆Ei = Ei−1−Ei results from forces Ei−1 and −Ei acting on the i-th slice from the left and from the right. Theequilibrium of moments is disregarded for individual slices. The most common solution methods are• The simplified Fellenius method: assumes that ∆Ei are parallel2 to the bottom surfaces of the respective (i-th) slices• The simplified Bishop method: assumes that ∆Ei are horizontalThese methods assume that the inclinations of the resulting forces are known a priori, Fig. 1. Given the directions∆ei of ∆Ei and the directions qi of Qi found using tanϕ/F with a temporarily guessed value of safety factor F , sayF = 1.5, the only unknowns in (2) are the values ∆Ei and Qi. One may solve the equilibrium equations (2) independentlyfor individual slices obtaining and ∆Ei and Qi. Unfortunately the solutions ∆Ei = ∆Ei∆ei are usually incompatible.Having all ∆Ei from (2) we may find all inter-slice forces Ei. Note that E0 = 0 on the left side of the first slice. Wemay find E1 = E0 + ∆E1 and next we find E2 = E1 + ∆E2 etc. Summing up,

E0 = 0 Ej =j∑

i=1∆Ei and En = 0 (3)

because there is no force acting on the last (n-th) slice from the right (cf. Fig. 1 with n = 5). In order to satisfy (3)we may try to modify F iteratively. Even then, however, the system (2) with (3) cannot be satisfied adjusting just oneparameter F . Indeed, we have 2n + 2 equation: n pairs from (2) and one pair from (3) and only 2n + 1 unknowns: Qi,

3

3ΔE −E

2 E

2 E3ΔE

3 −E

2 E

4 E 4 − E3 − E

2 − E

1 − E

3 E

1 E1

2

3

45

1

2

3

4

WW

W

WW E

E1

2

3

1

2

34

3

E 4

E 1

2

4

5

ϕ

C

C

Q

Q

1

C1

Q1

2

C2

Q 2

3

3

3 4

4

4

Q 5

C5

5

cv

v −ψ

/F

Figure 1: A logarithmic slip line with contractancy ψ < 0: numbering of slices i = 1, 2, . . . , n and their interfacesi = 1, 2, . . . , n− 1 (left). Sign convention of the inter-slice forces Ei (middle). Formally we may use E0 = En = 0. Theassumption about the direction of ∆Ei = Ei−1 −Ei is made in the Fellenius’ method.

∆Ei and F . Therefore the assumption of all directions of all ∆Ei to be prescribed is overly restrictive. In order to handlethis problem the simplified methods propose to replace two equations (3) by a single equation of static moments withrespect to the centre of the sliding circle (or spiral).Another solution could be a slight increase of inclinations of the unit vector ∆ei parallel to ∆E. This can be done by anunknown parameter λ common for all slices. Using a double iteration we may expect to find such F and λ that (2) and(3) are satisfied.

1Differently to the KEM with translation mechanism.2This assumption is not explicitly stated in the original book by Fellenius (1927) [?] or in its latter paper [?] from 1936, but it is attributed

to Fellenius in the geotechnical literature.

2

2 Simplified methods without balance of internal forces

The proposed approximate solutions by Fellenius and by Bishop may slightly violate the equilibrium equations (3), i.e.∑ni=1 ∆E 6= 0. These methods determine forces Qi, ∆Ei and F (together 2n+ 1 unknowns) using:

• 2n equations (2) of equilibrium of individual slices• a single equilibrium (4) of moments with respect to c

∑i

(rWi ×Wi + rQ

i ×Qi + rQi ×Ci) = 0 . (4)

instead of two scalar equations (3). Vector rWi describes the position of Wi (from c to the gravity centre of the slice) and

vector rQi describes the position of Qi and Ci (from c to the middle of the bottom of the slice).

Note that ∆Ei have been omitted in (4). This equation is written as if the forces ∆Ei were in equilibrium, i.e. as if∑ni=1 ∆E = 0 and

∑ni=1 r∆E × ∆E = 0 . Apart from replacing of (3) by (4), the essential assumption of simplified

methods is the omission of ∆Ei in (4).

2.1 Simplified Fellenius

In the simplified method by Fellenius (4) can be directly solved for F . For this purpose we observe that ∆Ei = Ei−1−Eiis assumed parallel to the bottom of the slice and hence it has no influence on the normal force Ni at the bottom. Thisforce can be easily found from the projection

Ni = ni(ni ·Wi) (5)

wherein ni is a unit vector perpendicular to the bottom, Fig. 8. Weight Wi contains all external (also inclined) loads onthe slice. Using the Coulomb criterion without cohesion we obtain

Qi = Ni + 1F

ti‖Ni‖ tanϕ , (6)

wherein ti is a unit vector parallel to the bottom of slice i and directed against the movement of the slope, Fig. 8.Substituting (6) into (4) we obtain∑

i

(rWi ×Wi + rQ

i ×Ni + 1F

[rQ

i × (ti‖Ni‖ tanϕ) + rQi ×Ci)

]= 0 (7)

which is a scalar equation which can be easily solved for F using

F =

∑i

[rQ

i × (ti‖Ni‖ tanϕ) + rQi ×Ci)

]∑

i

[rQ

i ×Ni + (rWi ×Wi)

] (8)

For circular sliding line with radius r = const we have additional simplifications

rQi ×Ni = 0, rQ

i × ti = r and rQi ×Ci = r‖Ci‖ (9)

2.2 Simplified Bishop

In the simplified Bishop’s method F is calculated iteratively from (2).• Assume a temporary value F for the first linearization, e.g. F = 1.5• Using F find inclinations qi

• solve the linearized system (2) for the values ∆Ei and Qi[∆eix qix

∆eiy qiy

]·

{∆Ei

Qi

}= −

{Wix + Cix

Wiy + Ciy

}for i = 1, 2, . . . n (10)

using the horizontal direction ∆ei and qi calculated with F .

3

• Substitute the results ∆Ei and Qi into (4) treating F as unknown. Find a new F that satisfies (4).

Having 2n+ 1 unknowns and 2n+ 1 equations in (2) and (4) the problem can be solved and a rough estimation of safetyfactor F is obtained.

Summing up:• The essential assumption is the usage of (4) with (2) instead of satisfying (3) and (2)• Internal forces ∆Ei do not enter (4) as if

∑i ∆Ei = 0 was satisfied.

• For a given F we can solve equation pairs for individual slices independently.• Owing to the smart choice of direction of ∆Ei parallel to the bottom of each slice, no iterative solution is needed in

the simplified Fellenius’ method. The major advantage is that the normal contact forces Ni do not depend on F , Fig.2 and that the tangential compontnts of Q are linearly dependent on F , see also (7).

Remark:Assumption ∆Ei parallel to the bottom surface (simplified Fellenius) pertains to the effective tractions only. It may beused for calculations with buoyant densities and seepage forces. In calculation with total densities and pore pressuresrequires consideration of horizontal pore water pressures Fig. 2.

3 Generalized limit equilibrium (GLE) methods

The GLE solution is based on equilibrium of forces in individual slices (with slightly modified assumption about in-clinations of ∆ei) and on two consistency equations (3) for the internal forces, i.e.

∑ni=1 ∆Ei = 0. The methods

introduce an unknown parameter λ ≈ 1, which modifies (rotates) all inclinations of interslice forces from {∆Ei1,∆Ei2}→to {∆Ei1, λ∆Ei2}→ or the inclinations of {Ei1, Ei2}→ to {Ei1, λEi2}→. Let us introduce an unknown parameter λ ≈ 1,which modifies (rotates) all inclinations of interslice forces from {∆Ei1,∆Ei2}→ to {∆Ei1, λ∆Ei2}→. Now, apart fromforces Qi and ∆Ei in each slice we have additionally two global variables F and λ. They will be found iteratively from (3)and (2). With 2n+ 2 unknowns λ, F, λ,Qi and ∆Ei the system of equations (3) and (2) is not overconstrained anymore:we have n+ 2 unknown and n+ 2 equilibrium conditions.We may solve the system iteratively:• Assume initial values of F and λ, e.g. F = 1.5, λ = 1 (first linearization)• Solve the linearized system (2) for ∆Ei and Qi in individual slices i[

∆eix qix

∆eiy qiy

]·

{∆Ei

Qi

}= −

{Wix + Cix

Wiy + Ciy

}for i = 1, 2, . . . n (11)

using the directions ∆ei and ∆qi (calculated with assumed F and λ).• Correct F and λ minimizing the discrepancy in (3). The Newton method for corrections of F and λ can be used. F

depends mainly on the horizontal component of∑

∆E and λ depending on the vertical component of∑

∆E. Theiteration is finished when (3) is satisfied that is when

∑∆E = 0.

• For∑

∆E 6= 0 repeat solution of the linear system (2 ) with corrected F and λ

The solution is not unique and may depend on the initial values F and λ and on the correction algorithm. We prefersmall inclinations of ∆Ei and hence small values of λ.

Unfortunately one cannot guarantee the uniqueness of the solution for F and λ because the system of equations isnonlinear, see Appendix D.

4 Fellenius method (FEL)

Alternatively we may replace the assumptions about the directions of ∆Ei by assumptions of directions of Ei. Given alldirections of Ei we write the equilibrium condition (2) of the i-th slice in an equivalent form

Wi + Ei−1 −Ei + Qi + Ci = 0 for i = 1, 2, . . . n with E0 = En = 0 (12)

As before, the directions of Qi can be found using the inclination tanϕ/F to the normal of the bottom. The system (12)has 2n equations and we have 2n unknowns: n skalars Qi, n − 1 skalars Ei and F . The system can be solved withoutadditional simplifications but the equations are nonlinear. We may solve the system iteratively:

4

• Assume F = 1 values (first linearization)• solve the linearized system (12 ) for Qi and Ei with E0 = 0 but without(!) assumption En = 0.

[ei−1x qix −eix

ei−1y qiy −eiy

]·

Ei−1

Qi

Ei

= −{

Wix + Cix

Wiy + Ciy

}for i = 1, 2, . . . n (13)

For each slice we have two equations and three unknowns so we cannot solve the slices independently. However, thefirst slice i = 1 has no lateral force E0 = 0 on the left and hence Ei−1 = 0. Using this observation we can solve thefirst slice obtaining Q1 and E1. Given E1 we can be solve the second slice etc. Of course, such sequential solution is notmandatory and one can alternatively construct the following global system (here for 3 slices) and solve it collectively.

q1x −e1x

q1y −e1y

e1x q2x −e2x

e1y q2y −e2y

e2x q3x −e3x

e2y q3y −e3y

·

Q1

E1

Q2

E2

Q3

E3

(14)

• correct F as long as En 6= 0.• repeat solution of the linear system (12 ) with corrected F

Unfortunately, the solution is not unique and may depend on the initial value of F and on the correction algorithm.One should therefore check, whether the final solution contains compression forces only. The solution assures the globalequilibria of forces and moments automatically. A graphic solution analogous to this method is proposed in the originalpaper by Fellenius [?]. There are no recommendations on the particular inclinations of Ei, however.

5 Slices with water

Two equivalent methods can be used to calculate water in the slope

• TOT : use total weights and pore pressures around each slice• EFF : use buoyant weights and seepage forces

Let us consider a single slice partly or totaly under water. The usual index denoting the number of the slice is omittedfor simplicity. The mobilized strength along the slip line is obtained from the effective stress

τm = c+ (σtot − u) tanϕF

(15)

wherein u is the water pressure at the bottom and σtot is the normal component of the total traction ttoti = σtot

ij nj there.The equilibrium condition can be expressed in two ways:

TOT: using the total stress from total weight Wtot = γbh~g with water pressure force U = n u b/ cosβ at the bottom actingalong vector n normal to the bottom surface. Moreover we consider horizontal forces UL and UR from water pressureon interfaces between slices.

EFF: using effective stress obtained with self weight W = Wtot + B reduced by the vertical buoyancy force B = −~g γwbzand supplemented with the seepage force J. We denote B = ‖B‖.

These methods can be shown to be equivalent with or without seepage forces if

U + UL + UR = B + J (16)

holds. Discrepancies in safety factor F between TOT and EFF could however arise for UR +UL 6= 0, Fig. 2 (no seepage),if assumptions about the inclinations of inter-slice forces applied on one hand to the total forces ∆Etot and on the otherhand to forces ∆E′ obtained from effective tractions only. In order to preserve the consistency between the TOF andEFF solutions we assume that inclinations of inter-slice forces pertain always to inter-slice forces from effective stresses.Therefore, beside U, also the horizontal pore water pressures UL and UR need to be considered in the calculation as anexternal load acting on a slice. Fig. 2 illustrates the problem.

5

NTm

C

U

UL

B

b

zh

UR

β FelleniusN=(Wtot-B)cos β=W cos β

Tm

W=Wtot+B

N

ΔE

C

U

UL

BUR

βϕ

Tm

Wtot

Wtot

N

ΔE

ΔE

C

βϕ

U

UL

BUR

Tm

Wtot

N

ΔEC

βϕ

Tm

W=Wtot+B

N

ΔEC

βϕ

Bishop (N tan ϕ+C)sin β+(N+U)cos β=Wtot

Figure 2: In the simplified Fellenius method the resulting effective(!) inter-slice force ∆E is inclined parallel to thebottom. Without seepage (a horizontal water table at z) the difference U −B is horizontal and not parallel to bottomso TOT and EFF will give different safety factors if UL + UR is disregarded in TOT. The simplified Bishop’s methodassumes horizontal ∆E and therefore, at the absence of seepage, UL +UR is included in ∆E. Is not necessary to calculateUL + UR as load in TOT in problems without seepage if the Bishop’s method is used.

6 Simplified Fellenius with Mathematica (dry version)

The method of slices can be easily programmed with Mathematica . The main part of the script computes the targetfunction, that is the safety factor by Fellenius. Note that many global variables are used in the script. The mainprocedures are: 3

The user is supposed to input the profile of the slope and the material parameters. Note that the dilatancy angle ´ \psi´and not only tanψ is needed as input. The points of geometric intersection of the slope defined via ´ points´ with the log-spiral slide line is found in a primitive algorithm with a limitation one intersection pro segment. Therefore an additionalpoint in the middle of the incline slope is introduced in the examples.

6.1 Target function and graphics

Target function is the safety factor F as a function of radius and centre of the spiral. Just a single set of materialparameters is used so a prior averaging is necessary for layered soils.getGraphics [ ] := Module [{ g0 , g1 , g2 , g2a , g3 , g4 , g5 , k r e s k i } ,g0 = Graphics [ Point [ c0 ] ] ;g1 = Parametr icPlot [ s p i r a l ,{ t , Pi , 2 Pi } ] ;g2 = Parametr icPlot [ parametrizedSegments , {u , 0 , 1 } ] ;g2a = Graphics [ Point [ s l opePo in t s ] ] ;k r e s k i= Line [#]& /@ ({ topPoints , bottomPoints } // Transpose ) ;g3 = Graphics [ Point [ { interS l i ceXX , inte rS l i c eYY }// Transpose ] ] ;g4 = Graphics [ Point [ { interS l i ceXX , i n t e r S l i c e Z Z }// Transpose ] ] ;g5 = Graphics [ k r e s k i ] ;{g0 , g1 , g2 , g2a , g3 , g4 , g5}]

Fe l l en iusTargetFunct ion [ r 0 ?NumericQ , c0x ?NumericQ , c0y ?NumericQ ] := Module [{ segment , i n t e r s e c t i o n s ,m, ru le ,so lu , tSo lu t i ons , he ights , widths , tmiddles , weights , dSpi ra ldt , t angen t i a l s , normals , r e s i s t a n c e s , r ad i i ,

resistantMoment , activeMoment , activeMomentPlus , rotate90CCW , t0 , u0 , d i s t , iSegment , i S l i c e } ,s p i r a l = {c0x , c0y} + r0∗ Exp [ tanPsi ∗ t ]{Cos [ t ] , Sin [ t ]} ;i n t e r s e c t i o n s = {} ;Do [ segment = parametrizedSegments [ [ i ] ] ;

d i s t = ( segment−s p i r a l ) . ( segment−s p i r a l ) //N;{m, r u l e } = FindMinimum [{ d i s t , 0<= u<= 1 && 2 < t < 7 } ,{{ u , 0 . 5} ,{ t , 4 . 5}} ] //N;I f [ Abs [m] < 10ˆ−5 , AppendTo [ i n t e r s e c t i o n s , segment / . r u l e ] ] ;

,{ i , nSegments } ] ;I f [ Length [ i n t e r s e c t i o n s ] != 2 , Return [ 20 ] ] ;

in te rS l i c eXX = i n t e r s e c t i o n s [ [ 1 , 1 ] ] + Range [ 0 , n S l i c e s ] ( i n t e r s e c t i o n s [ [ 2 , 1 ] ] − i n t e r s e c t i o n s [ [ 1 , 1 ] ] ) / n S l i c e s ;in te rS l i c eYY= i n t e r s e c t i o n s [ [ 1 , 2 ] ] + Range [ 0 , n S l i c e s ] ( i n t e r s e c t i o n s [ [ 2 , 2 ] ] − i n t e r s e c t i o n s [ [ 1 , 2 ] ] ) / n S l i c e s ;Do [ so lu = Solve [ parametrizedSegments [ [ iSegment ,1] ]== inte rS l i c eXX [ [ i S l i c e ] ] , u ] [ [ 1 ] ] ; u0 = u / . so lu ;

I f [0<= u0 <= 1 , inte rS l i c eYY [ [ i S l i c e ] ] = ( parametrizedSegments [ [ iSegment , 2 ] ] / . s o lu ) ] ;,{ i S l i c e , 2 , n S l i c e s } , { iSegment , 1 , nSegments} ] ;i n t e r S l i c e Z Z= i n t e r s e c t i o n s [ [ 1 , 2 ] ] + Range [ 0 , n S l i c e s ] ( i n t e r s e c t i o n s [ [ 2 , 2 ] ] − i n t e r s e c t i o n s [ [ 1 , 2 ] ] ) / n S l i c e s ;t S o l u t i o n s = {} ;Do [ so lu = FindRoot [ s p i r a l [ [ 1 ] ] == interS l i c eXX [ [ i S l i c e ] ] ,{ t , 4 . 5 } ] ; t0 =t / . so lu ;

I f [ 3 <= t0 <= 6 , i n t e r S l i c e Z Z [ [ i S l i c e ] ] = ( s p i r a l [ [ 2 ] ] / . s o lu ) ] ; AppendTo [ tSo lu t i ons , t0 ] ;,{ i S l i c e , 1 , n S l i c e s+1 } ] ;

3 For Mathematica insiders: In this listing produced with listings.sty a problem with Mathematica symbols in the last line of themodule getGraphics[ ] becomes evident. Only varepsilon ( ε = Esc ce Esc ) and not epsilon (ε Esc e Esc) can be use in Mathematica scripts.One may copy paste all Greek letters from PDF back to Mathematica only via Sumatra and not via Adobe Acrobat

6

topPoints = { interS l i ceXX , inte rS l i c eYY }// Transpose ;bottomPoints = { interS l i ceXX , i n t e r S l i c e Z Z }// Transpose ;

he i gh t s = topPoints [ [ All , 2 ] ] − bottomPoints [ [ All , 2 ] ] ;widths = Drop [ topPoints [ [ All , 1 ] ] ,1]− Drop [ bottomPoints [ [ All , 1 ] ] ,−1] ;weights =spec i f i cWe igh t ∗( he i gh t s [ [ # ] ] + he i gh t s [ [ #+1 ] ] ) ∗widths [ [ # ] ] / 2 & /@ Range [ n S l i c e s ] ;tmiddles = ( Drop [ tSo lu t i ons , 1 ] + Drop [ tSo lu t i ons ,−1] ) / 2 ;dSp i ra ld t= D[ s p i r a l , t ] // S imp l i f y ; rotate90CCW= {{0 ,−1} ,{1 ,0}} ;t a n g e n t i a l s = Normalize [#] & /@ ({ ( dSp i ra ld t [ [ 1 ] ] / . t −> tmiddles ) , ( dSp i ra ld t [ [ 2 ] ] / . t −> tmiddles )}// Transpose ) ;normals = rotate90CCW . # & /@ t a n g e n t i a l s ;r e s i s t a n c e s =( normals [ [ All , 2 ] ] ∗ weights ∗ tanPhi + cohes ion ∗widths / t a n g e n t i a l s [ [ All , 1 ] ] ) ;r a d i i = r0 Exp [ Tan [ Psi ] ∗ t ] / . t−> tmiddles ;resistantMoment = Plus @@ ( r e s i s t a n c e s ∗ r a d i i ∗Cos [ Psi ] ) ;activeMoment = Plus @@ ( ( c0x − ( s p i r a l [ [ 1 ] ] / . t−> tmiddles ) )∗weights ) ;activeMomentPlus = Plus @@ ( r a d i i ∗ ( normals [ [ All , 2 ] ] ∗ weights ) ∗Sin [ Psi ] ) ; (∗ vector ∗ vector = vectorin Mma ∗)I f [ activeMoment < 0 , Return [ 20 ] ] ;

resistantMoment /( activeMoment + activeMomentPlus ) (∗ output F e l l e n i u s s a f e t y f a c t o r ∗)]

Important global variables (some are initialized within the target function = ugly programming):´ t´ = θ for parametric description of spiral;´c0x,c0y,r0 ´ = c, r0 geometry of the spiral´ cohesion tanPhi, Psi, tanPsi´ soil parameters´ segments, nSegments ´ = list of coordinate pairs of points on the ground surface and length of this list´ ParametrizedSegments[[i]] ´ segments of the ground surface each parametrized with u ∈ (0, 1)´ spiral = {c0x , c0y} + r0 * Exp[ tanPsi *t ]{ Cos [ t ] , Sin [ t ]} ; ´ logarithmic spiral parametrized with ´ t´ = θ in radians´ interSliceXX ´ = x of lines between slices of two ´ intersections ´.´ interSliceYY ´ y-coordinates of intersections the vertical lines between slices and the ´ ParametrizedSegments[[i]] ´´ interSliceZZ ´ y-coordinates of intersections the vertical lines between slices and the ´ spiral ´´ topPoints, bottomPoints ´ x, y pairs of ntersections the vertical lines between slices and ground surface or spiral.

Important local variables of ´FelleniusTargetFunctions ´´ kreski, g..´ = graphics;´ intersections ´ = x, y coords of two points of intersection of ´ spiral´ with ´ParametrizedSegments ´. One intersection persegment is allowed. Artificial points on the slope (in polygon ´ segments´ of the ground surface) are sometimes necessary.´ heights, widths, weights ´ dimensions hi, wi and weights W i (scalar) of slices´ tmiddles ´ θi-s at the middle of slice´ dSpiraldt ´ analytical expression for derivative of the log spiral with respect to θ´ tangentials ´ unit vectors ti parallel to bottom´ normals ´ unit vectors ni ⊥ to bottom directed upwards´ resistances ´ Ri = (n2W tanϕ+ cw/t1) cosψ, note that force (n2W tanϕ+ cw/t1) is not perpendicular to radius vector´ resistantMoment ´

∑riRi

´ activeMoment ´ moment∑W ixi

b from self weight W i where xib is x coordinate of the middle of the bottom

´ activeMomentPlus ´ additional moment∑N i sinψri from reactions N i = ‖Ni‖ = Wn2 at the bottom. The normal forces

Ni increase the active moment if ψ > 0, see (7)´ radii´ length of radius ri from centre to the middle of the bottom of a slice (for the moments of resulting forces)The condition ´ _?NumberQ´ is imposed on all arguments of ´FelleniusTargetFunction´. It is important when ´FelleniusTargetFunction´is invoked within Mathematica ´ FindMinimum´ function because ´ FindMinimum´ tries to analyse its target function analyticallyto find possible simplifications. The conditions ´ NumberQ´ prevent these default attempts to simplify the target function.

6.2 Example 1

A computation with a single spiral slip linecohes ion= 10 ; tanPhi =Tan [20 \ [ Degree ] ] ; s p e c i f i cWe igh t = 20 ; Psi = −5 \ [ Degree ] ; tanPsi = Tan [ Psi ] ;s l opePo in t s= { {−10, 10 } , {10 , 10} , {20 , 5} , {30 , 0} , {90 , 0}} ;segments = { Drop [ s lopePo ints , −1] , Drop [ s lopePo ints , 1 ]} // Transpose ;nSegments = Length [ segments ] ; n S l i c e s = 10 ;parametrizedSegments =segments [ [ # ,1 ] ]+ u∗( segments [ [ #, 2 ] ] −segments [ [# , 1 ] ] )& /@ Range [ nSegments ] ;

r0 = 25 ; {c0x , c0y} = {25 , 13} ; sa f e tyF= Fe l l en iusTargetFunct ion [ r0 , c0x , c0y ] ;c0 = {c0x , c0y } ; {g0 , g1 , g2 , g2a , g3 , g4 , g5} = getGraphics [ ] ;Show [{ g0 , g1 , g2 , g2a , g3 , g4 , g5 , Graphics [ Text [ sa fe tyF ] ] } ]

Note that the target function calculates not only the safety factor but also some data for graphics. They are accessiblevia global variables.

6.3 Example 2

A computation with grid values of the centre point. For simplicity r in not varied but assumed that the slip lines passesthrough the foot of the slope.

7

1.35723

Figure 3: Graphic output obtained directly from Mathematica for a single geometry of example 1.

sa f e tyFs = {} ; z0 = {8 , 10} ;Do [ r = z0 − {x0 , y0} ; r0 = Norm [ r ] / Exp [ (2 Pi + Arg [ r . { 1 . 0 , I } ] )∗ tanPsi ] // N ;

AppendTo [ sa fetyFs , { r0 ,{ x0 , y0 } , Fe l l en iusTargetFunct ion [ r0 , x0 , y0 ]} // Flatten ] ;, { x0 , 15 , 32 , 1 } , {y0 , 10 , 24 , 1 } ]

bes t = sa f e tyFs [ [ Ordering [ sa f e tyFs [ [ All , 4 ] ] , 1 ] ] ] [ [ 1 ] ] ;r0 = best [ [ 1 ] ] ; c0= best [ [ 2 ; ; 3 ] ] ; {c0x , c0y} = c0 ; sa fe tyF=A´Fel len iusTargetFunct ion [ r0 , c0x , c0y ] ;{g0 , g1 , g2 , g2a , g3 , g4 , g5} = getGraphics [ ] ;g6= ListContourPlot [ s a f e tyFs [ [ All , 2 ; ; 4 ] ] , Contours −> 10 , ContourLabels−>True , ColorFunction−>”DarkRainbow” ] ;{g0 , g1 , g2 , g2a , g3 , g4 , g5} = getGraphics [ ] ;Show [{ g0 , g1 , g2 , g2a , g3 , g4 , g5 , Graphics [ Text [ sa fe tyF ] ] , g6 } ]

1.27718

1.5

1.7

1.8

1.8

2

2

2.2

2.2

2.3

2.3

2.5

2.7

2.8

3

Figure 4: Graphic output obtained directly from Mathematica shows slices and isolines of safety factor. The minimumFs = 1.277 has been found for r = 25.69, and c = {26, 16}

6.4 Example 3

A computation with optimization of the centre point and radius (three parameter optimization) returns F = 1.268.r =. ; x =. ; y =. ;

FindMinimum [ Fe l l en iusTargetFunct ion [ r , x , y ] , {{ r , 11} , {x , 15} , {y , 15} } ] // Quiet

(∗ r e tu rns {1 .26836 , { r −> 26 .3169 , x −> 26 .0285 , y −> 16.6038}} and hence ∗)r0 = 26.316850710464983 ‘ ;{c0x , c0y} = {26.028520001942308 ‘ , 16 .60380876119284 ‘} ;sa f e tyF = 1.2683579792702355 ‘ ;{g0 , g1 , g2 , g2a , g3 , g4 , g5} = getGraphics [ ] ;Show [{ g0 , g1 , g2 , g2a , g3 , g4 , g5 , Graphics [ Text [ sa fe tyF ] ] } ]

(∗ r e tu rns {1 .26836 , { r −> 26 .3169 , x −> 26 .0285 , y −> 16.6038}} ∗)

The minimum safety is similar with the worst one obtrained from the calculations on grid in example 2. The slip line isindeed passing through the foot of the slope as assumed in example 2.

7 Multi-method solution with Fortran

The methods for the estimation of the slope stability described in Sec. 3 (i.e. General Equilibrium Method (GLE)and Fellenius (FEL)) have been implemented in a Fortran 90 program. The program computes factor of safety F of ahomogeneous soil slope. The landslide has the profile of a logarithmic spiral. To determine the forces along the slip surface,

8

1.26836

Figure 5: Graphic output obtained Mathematica for optimal geometry

the sliding block is first divided into vertical slices. The user can choose between analysis with pore water pressures andtotal weights (TOT) and analysis with seepage forces and weights modified by buoyancy and seepage forces (EFF). Thereare two versions of the program: with and without optimization of the geometry to find the minimum factor of safety.This section refers to the version without optimization. Both versions read the same input file. Therefore, some optionsare read from the input file but will have no effect. The factor of safety F computed by the program corresponds to theinitial slip surface given in the input file, and not to the global minimum factor of safety.

The forces acting on each slice are written by the program in a text file and plotted in a Postscript file as force polygon.In addition, the program draws the geometry of the slip surface and the slope into a second Postscript file.

7.1 Files and modules

The program is divided into seven files: six containing modules and one containing the main program, see Table 1.

File Module Purpose, ContentspiralSlope.f90 spiralSlope Main program, target function to

compute the factor of safetygeneralTools.f90 generalTools subroutines for outer product, num-

ber comparison, etc.geometricTools.f90 geometricTools subroutines for segments intersec-

tion, spiral generation, slices con-struction

globalBlock.f90 globlaBlock declaration of global variablesshared among modules

inData.f90 inData definitions of the structures layer,VectorOfX, controls, and slice

NLSolver.f90 NLSolver subroutines for solving the nonlinearsystem of equations

psTools.f90 psTools subroutines for graphical output inform of Postscript files

Table 1: Fortran files and modules of the program

To compile the executable file spiralSolpe.exe, the following command line can be used:

gfortran generalTools.f90 inData.f90 globalBlock.f90 geometricTools.f90 psTools.f90NLSolver.f90 spiralSolpe.f90 -o spiralSolpe.exe

The fortran files must appear in the above specific order, because all modules on which the currently compiled file dependsmust be already exist. The executable file spiralSolpe.exe reads the input text file input.txt. The output of theprogram consists of three ouput files: a text file output.txt with the magnitude of the forces acting on each slice, andtwo postscript files (with file extension .ps) containing graphics with the geometry of the slope and the force polygon ofall slices. The postscript files can be visualized with the program Ghostview4.

4http://pages.cs.wisc.edu/ ghost/

9

7.2 Preparation of input

The text file input.txt contains the geometry of the slope, the position ground water level and the geometry of the slipsurface, the material properties of the soil, and options to control the calculation. The name of the input file must beinput.txt. No blank lines are allowed. Some input data are not being used in the current version of the program.

The input file has the following structure:

1. Problem name: a sequence of characters without blanks. It will appear in the output files.

2. Control parameters: a sequence of characters with the format ’AAA.BBB.CCC.DDD’ (the dots and the uppercase areobligatory). The first three characters AAA can be set to GLE to choose the General Limit Equilibrium method orFEL to choose Fellenius method. The characters BBB (5th to 7th) can be either EFF to make the analysis in termsof effective stresses, buoyant weight and seepage forces or TOT to perform the analysis in terms of effective stressesbut with total weight (γr) and with pore water pressures around . The characters CCC (9th to 11th) can be eitherOPT to optimize the geometry of the initial slip surface to find a global minimum factor of safety or NOP to omitoptimization and compute the factor of safety for the slip surface given in this input file only. The characters DDD(13th to 15th) can be either SEQ to solve the global system of equation sequentially or GLO, to solve the globalsystem of equations directly for all slices at once.

3. Number of layers, number of slices. This version of the program allows for one soil layer only.

4. r0, ∆r, xc, yc, ∆xc, ∆yc, ψ, F0, λ0. This line contains the geometry of the slip surface (radius r0, center pointxc, yc and dilatancy/contractancy angle ψ). The parameters ∆r, ∆xc, and ∆yc are the initial variation of the slipsurface geometry used in the optimization algorithm (with the option OPT). The initial estimation of safety factorF0 and of the inclination increase λ0 are required by both solution methods.

5. npGS: Number of points defining the ground/slope surface

6. npGS input lines containing the x and y coordinates of the next point on the ground surface

7. List of soil parameters: c, ϕ, γd, and γr: cohesion c, friction angle ϕ, dry γd and saturated γr unit weight.

8. npGW : Number of points defining the ground water surface (phreatic line)

9. npGW input lines containing the x and y coordinates of the next point on the water surface

10. List of water parameters: cw, ϕw, γdw, and γw: cw = 0, friction angle ϕw = 0, dry γd = 0 and saturated γr unitweight of water. The values cw, ϕw, γdw are ignored by the program.

11. npE: number of points defining the inclination of the interslice force Ei or ∆Ei

12. npE input lines containing: xE , Eix, Eiy where, xE is the x−coordinate at which Ei or ∆Ei is defined, and Eix

and Eiy are the components x and y−components of the direction vector Ei or ∆Ei

An example of the input file for an slope on clay (c = 10 kPa, ϕ = 20◦, γd = 20 kN/m3, γr = 20 kN/m3, ψ = −5◦ ) withan inclination of 26.5◦ is shown below.

slopeTest ! TitleFEL.TOT.NOP.SEQ ! Method: 1) GLE or FEL 2) EFF or TOT 3) OPT or NOP 4) GLO or SEQ1, 10 ! number of layers, number of slices25 , 1, 25.0,13.0, 1.0, 1.0, -5, 2.0, 1.0 ! r, dr, xc,yc, dxc,dyc, psi, F, lambda5 ! number of points of the first layer-10.0, 10.0 ! x and y coordinates of the first point...10.0, 10.020.0, 5.030.0, 0.090.0, 0.010, 20, 20, 20 ! c, phi, gammad, gammar5 ! number of points of the water layer-10.0, 8.0 ! x and y coordinates of the first point10.0, 7.020.0, 2.030.0, 2.090.0, 2.00, 0, 0, 10 ! (c, phi, gammad,gammar) of water ! formally needed0 ! number of points defining external loads (not implemented yet)3 ! number of points defining the inclination of the interslice force Ei or dEi-5, 1, -0.1 ! x coordinate, x and y components of the direction vector Ei or dEi20, 1, -0.180, 1, -0.1

10

Figure 6: Output file Geometry.ps from the input given in Section 7.2. The ground surface, the slip surface and theslices are shown

Factor of safety = 1.4742990

100

W1-Er1C1Q1

W2

-Er2C2

Q2

W3

-Er3C3

Q3

W4

-Er4C4

Q4

W5

-Er5C5

Q5

W6

-Er6C6

Q6

W7

-Er7C7

Q7

W8

-Er8C8

Q8

W9-Er9C9

Q9

W10-Er10C10Q10

Figure 7: Output file Forces.ps from the input given in Section 7.2. All interslice forces Er are identically inclined.

7.3 Output and graphics

The program generates three files: one text file (output.txt) and two graphic (Postscript) files (Forces.ps Fig. 6 andGeometry.ps Fig. 7).

output.txt contains: the name of the problem, the factor of safety, the slip surface and the list of the forces acting oneach slice. The input file from Section 7.2 generates the following output.txt:

TITLE: slopeTestFACTOR OF SAFETY = 1.474299SLIP SURFACE: xc = 25.000 yc = 13.000 r0 = 25.000THE FORCES ACTING ON EACH SLICE ARE:SLICE W C Q El Er

1 164.12 44.789 179.34 0.0000 122.872 385.84 28.016 417.38 122.87 343.443 461.48 23.057 467.81 343.44 506.374 478.66 20.605 472.12 506.37 587.735 462.44 19.298 457.29 587.73 589.236 418.73 18.647 426.83 589.23 517.987 349.95 18.532 378.45 517.98 392.548 256.32 18.934 306.40 392.54 239.269 150.45 19.984 212.95 239.26 94.92710 48.886 22.144 103.61 94.927 0.40917E-10

The postscript files obtained for the input given Section 7.2 are shown in Figs. 6 and 7.

7.4 To be modified

The task is to implement two effects caused by an earthquake

11

1. increase of horizontal acceleration to µhg

2. accumulation of pore water pressure and decrease of the effective normal stress due to self weight of soil by factorµu

For vertical gravity, g = g {0,−1}, we may easily implement the first effect using {WixWiy} ={µh,−1

}* s%weight in (11)

or in (13). For this purpose we need to modify type(slice) adding a variable for µh.The second effect can be implemented increasing the gradient γw of pore water pressure. For slices totally under waterwe find the vertical gradient of pore pressure γw as follows

γw = γw + µu(γr − γw) and γ′ = γr − γw, (17)

wherein γr is the unit weight of saturated soil and γ′ is the modified ”buoyant unit weight”. In the EFF method suchγ′ may be taken to calculate the buoyant weight of the slice and j = −iγw is the seepage force, wherein the hydraulicgradient i can be assumed parallel to the water table in the slice and directed upstream.In the TOT method we may calculate the resultant forces of linearly distributed pressures along all four walls of the slicewith the geometry of the slice and with γw.For slices with part A1 above water and A2 under water the situation is more complicated. We need to define the verticalpore pressure gradient as

γw = γw + µu

[A1

A2γd + γr − γw

]and γ′ = γr − γw (18)

Note that the weight of the slice is W = A2γ′ +A1γd and γ′ can be negative! In the special case of (18) with µu = 1 we

obtain

W = A2γ′ +A1γd = A2γr −A2γw +A1γd (19)

= A2γr −A2

[γw + A1

A2γd + γr − γw

]+A1γd = A2γr − [���A2γw +A1γd +A2γr����−A2γw] +A1γd = 0 (20)

Hence, despite being partly flooded only, fully liquefied slices are weightless in the EFF method. The seepage force canbe taken as A2iγw.In the TOT method we may calculate the resultant forces of linearly distributed pore pressures along floaded parts ofthree walls of the slice using γw from (18). The total weight is W = A2γr + A1γd and the total resultant of the porepressures acting on A2 can be calculated by Gauß divergency theorem

Fi = −∫

S

γwx2δijnjdS = −∫

V

γwx2,jδijdV = −A2γwδ2i (21)

The vertical component of this force is simply −A2γw and hence the total weight with pore pressure results for µu = 1 in

A2γr +A1γd −A2γw = A2γr +A1γd −A2(��γw + A1

A2γd + γr���−γw) = 0 (22)

as expected.

A Notation

Sliding movement is always to the right (along positive x axis) and the numbers of slices increases to the right. Onlyone set of soil parameters 5is possible, hence one cannot calculate layered soils. All forces are written in bold face. Theircomponents are positive if they point along increasing x or y axes.

• a acceleration vector due to earthquake (largests value in unfavourable direction)• β inclination of the bottom line, e.g. t = {cosβ, sin β} and n = {sin β, cosβ}• n toal number of slices• n− 1 toal number of interfaces• b width of a slice• z underwater height of a slice

5With increasing number of layers with low friction angles the limit state method becomes inaccurate [?].

12

m

spiezometric line

slip surface

ground surface

c

x

y

b

θθl

θr

r

r0

t

n

stsn

w

Figure 8: Logarithmic spiral slip surface and the unit normal and tangent vectors. The unit tensors n and t are thenormal and tangential directions with respect to the slip surface. n is directed inwards the block. t is directed with theexpected movement, see Section B

• h total height of a slice• c, ϕ strength parameters of a slice• σ, τ effective stress components normal and tangential to the bottom of a slice• u pore water pressure at the bottom• ~t = (t)→ = t/‖ t ‖ normalization• ‖ t ‖ norm• t × t cross product• ei = ~Ei the direction of the inter-slice force across the i-th slice is denoted by• ∆ei = (∆Ei)→ the direction ∆Ei = Ei−1 −Ei

• g, g gravitation as a vector and as a value• r position vector• c centre of rotation• n vector normal to the bottom pointing upwards, if circle n = {sin β, cosβ}• t vector tangential to the bottom pointing against the movement, if circle t = {cosβ, sin β}• Qi = qiQi (no sum) resistance force at the bottom of i-th slice and its norm• Ei = eiEi (no sum) inter-slice force exerted by slice i on slice i+ 1• ∆Ei = ∆ei∆Ei (no sum) resulting force from from both interfaces in slice i• F global safety factor• λ global increase of inclination of ∆Ei

• Wi = Wi~g weight (reduced by buoyancy) of slice i as a vector and its value• Bi = −Bi~g buoyancy force in slice i• C = −tcb/ cosβ• Ni = nini ·Q (no sum) normal component of Qi

• Tm = titi ·Q (no sum) tangential component of Qi

• U = nu b/ cosβ• URUL

• A = sns(z − h)γw = Asn resulting force from water pressure above slice (only for z > h)• Qi = aWi/g Horizontal seismic load• m centre of mass of a slice. The weight W and the seismic force Q are applied there.• b the middle of the bottom of the slice, Force Q = N + Tm is applied there• s the mid-point of the top of the slice. The force A = of water pressure is applied there• sn the unit vector normal to the top surface, pointing inward the slice , Fig. 8

13

• st the unit vector tangential to the top surface, Fig. 8

B Why logarithmic spiral

The logarithmic spiral is used to describe the slip surface. The logarithmic spiral is more suitable for the description ofthe slip line than the circle because one can consider dilatancy (evolving spiral) or contractancy (convolving spiral). Botheffects may accompany the shearing in soil. The circular slip line can be obtained from the spiral as a special case.Expressions for the normal and tangent vectors to the slip surface must be provided in the calculation of the normal Nand the tangential T forces. Consider the logarithmic spiral given in Figure 9. In polar coordinates, the equation of thecurve is

r = r0 exp(kθ). (23)It can be shown that k = tanψ is related to the dilatancy angle ψ . The cartesian coordinates of a point P on the curve

x

y

rn

t

m

θr0

P

Figure 9: Logarithmic spiral and the vectors t,n,m

is given by the vector x = {x, y} withx = r0 exp(kθ) cos θ, (24)y = r0 exp(kθ) sin θ. (25)

A tangent vector x′ = ∂x∂θ

at point P can be written as

x′ = {∂x/∂θ, ∂y/∂θ} , (26)wherein

∂x

∂θ= r0 exp(kθ) (k cos θ − sin θ) , (27)

∂y

∂θ= r0 exp(kθ) (k sin θ + cos θ) . (28)

The unit tangent vector t is

t = x′

‖x′‖ = 1√1 + k2

{k cos θ − sin θ, k sin θ + cos θ} . (29)

The unit normal vector n is

n = t′

‖t′‖ = 1√1 + k2

{− cos θ − k sin θ, k cos θ − sin θ} . (30)

Let us define a unit vector m perpendicular to the radius rm = {− sin θ, cos θ} . (31)

t ·m = 1√1 + k2

= cosϑ. (32)

If k and 1 are the sides of a right-angled triangle and ψ is the angle between them, then its hypotenuse is√

1 + k2, andk = tanψ (33)

The spiral can be discretized with i = 1, 3, . . . nS points pi and saved as ´ spi ´ . Intersection with a line from b to ecan be checked using ´ segments_intersection(p1,p2,b,e) ´ in which points ´ p1,p2´ run along the spiral. Intersection with a linefrom b to e can be checked using ´ forall(i=1,ns-1) ´ in which points ´ p1,p2´ run along the spiral.

14

C Intersection between the spiral and a line

A considerable computational effort in GLE optimization is dedicated to geometrical problems. Many problems can bereduced down to finding an intersection between a spiral and a straight line. The equation of the spiral{

x = xc + r0 exp kθ cos θ,y = yc + r0 exp kθ sin θ (34)

and the equation of a segment between points 1 and 2

y = − y2 − y1

x2 − x1(x2 − x) + y2 (35)

must be simultaneously satisfied. Combining equations (34), (34) and (35) we get

g(θ) = yc + r0 exp kθ sin θ + y2 − y1

x2 − x1(x2 − xc − r0 exp kθ cos θ)− y2 = 0. (36)

Each θ that satisfies g(θ) = 0 correspond to an an intersection point between the line and the spiral.

D On the uniqueness of the solution

The system of 2n equations

Wi + Ei−1 −Ei + Qi + Ci = 0 for i = 1, 2, . . . n with E0 = En = 0 (37)

must be solved for the 2n scalar unknowns Ei (with i = 1, 2, ..., n − 1), Qi (with i = 1, 2, . . . n) and F . Let us define avector di perpendicular to qi (where qi is the direction of Qi, i.e. Qi = Qiqi), so that qi×di = {0, 0, 1} and qi ·di = 0.Multiplying equation (37) by di and qi allow us to write two scalar equations for each slice i

(Wi + Ei−1 − Eiei + Ci) · di = 0 (38)(Wi + Ei−1 − Eiei +Qiqi + Ci) · qi = 0 (39)

from which the scalars Ei and Qi can be found as functions of F . This equations can be solved sequentially (say, startingfrom i = 1 and ending at i = n). Doing so, it should be noticed that En = 0, so Qn can be found directly from (39) andequation (38) becomes

R(F ) = (Wi + Ei−1 − 0 + Ci) · di = 0 (40)

which is an scalar equation with only F as unknown. Now the problem reduces to find the F that satisfies R(F ) = 0.Unfortunately, the solution of R(F ) = 0 is not unique. In deed, by inspection of the equation it can be seen, that thenumber of solutions of R(F ) = 0 is n+ 1. The Fig. 11 shows the function R(F ) for an example with n = 3 (see Fig. 10)and n = 10.

Figure 10: Example problem with c = 10 kPa, ϕ = 20◦, ψ = −5◦, and γ = 20 kN/m3. The arrows indicate the assumeddirection of the interslice forces Ei.

In the example, the equation R(F ) = 0 for n = 3 has 3 solutions F1 = −0.196268, F2 = −1.05662∗10−18, F3 = 0.128997,and F4 = 1.82077, see Fig. 12. For n = 10, the equation R(F ) = 0 has 11 solutions: F1 = −0.62324, F2 = −0.296338,F3 = −0.163666, F4 = −0.0753517, F5 = −0.00544314, F6 = 8.01121 · 10−18, F7 = 0.0557533, F8 = 0.114171, F9 =0.175706, F10 = 0.250541, and F11 = 1.50682.The problem now is to identify the “correct” solution among the n+1 solutions. For this purpose we must apply additionalcriteria to reject unrealistic solutions. For example, solutions with F < 0 are not acceptable as they would lead to Ci

15

-2 -1 1 2 3

-6000

-4000

-2000

2000

4000

6000R

F

n=3

-2 -1 1 2 3

-6000

-4000

-2000

2000

4000

6000R

F

n=10

Figure 11: Graphic representation of equation (40) (i.e. function R(F ) as function of F ) for an example with n = 3slices (left) and n = 10 slices (right).

F1= -0.196268W1

E1

C1

Q1

W2

E2C2

Q2

W3C3

Q3

W1

E1

C1

Q1

W2

E2

Q2

W3

C3 Q3

F3=0.128997

C2Q1

W1E1W2E2

Q2

W3

C3Q3

F2= -1.05662 . 10-18

W1

E1C1

Q1

W2

E2C2

Q2

W3

C3

Q3

F4=1.82077

Figure 12: Polygon of forces according to the solutions F1, F2, F3, F4 for the example with n = 3 slices.

forces acting in the same direction of the slope failure and not as “resistant” forces. In addition, solutions that lead totension in forces Ei or Qi are also not allowed.The Fig. 12 shows the polygon of forces for each of the four solutions obtained for the example with n = 3 slices. Ofcourse, each solution Fi produces a closed polygon of forces, i.e. all forces in all slices are in equilibrium. The first twosolutions F1 and F2 are negative and they lead to unrealistic cohesion forces. F3 is positive but produces tension forcesEi. It only remains F4 as an acceptable solution.

16