1

Rock Slope Stability Analysis Method

(Teuku Faisal Fathani)

Stability Analysis MethodStability analysis define the geometry of blocks or of the system of blocks isolated by discontinuity planes and exposes on the examined excavation face or the natural slope

Sliding Analysis by using Static or Dynamic Equilibrium Method

Static limit equilibrium method :

(1) Examines kinematics possibility of sliding or topping of each block which has a face exposed on the slope

(2) Only examines the incipience motion and does not consider the subsequent behavior of whole system of the blocks.

Dynamic equilibrium method :

(1) Simulate the behavior of a blocky system a more realistic hypothesis by referring to the examined physical phenomenon;

(2) For problem consists in computation of the block motion or when the block are subjected to cyclic stresses or pulsing loads.

2

Static and Dynamic Equilibrium MethodRigid block instability modes on inclined plane produces individual block sliding and toppling

Static equilibrium problem variables Dynamic equilibrium problem variables

Static equilibrium analysis

Dynamic equilibrium analysis

Stability Chart

3

Safety Factor and Limit Equilibrium Method

Wsinα

WcosαWα

R

Assuming the shear stress τon sliding surface defined byCoulomb criterion

φστ tan+= c

φατ tancosA

Wc +=

φατ tancosWcAAR +=⋅=Limit block equilibrium condition

φαα tancossin WcAW += If c = 0 φα =

)(mobilized mobitable) (maximum

ττ

=F

Safety Factor and Limit Equilibrium Method

Safety factor : a number for which the available shear strength parameters (c - φ) must be divided to reach limit equilibrium condition

Principal hypothesis :

- Failure surface simple or composite shear failure surface- Sliding mass single or more intact stiff blocks which can move without significant strain or failure of block rock matrix

Limit equilibrium method = overall analysis methodSolution is given for a system of blocks or for a single block by means of overall safety factor constant on the whole examined surface

)(mobilized

e)mobilitabl (maximum τ

τ=F

4

Effect of Water Pressure in Rock Discontinuities Water filling discontinuities involves a lowering of stability conditions for natural or artificial slopes

Subverticaldiscontinuity plane

VU

VVp =

0=nV

( )A

UW −=

ασ cos

( )VWUWcAF

+−+

=α

φαsin

tancos

A. Rock Slope Stability• Rock slide occurs at the weakness plane (discontinuities plane), cracks or

shear zone

Rock discontinuity plane with the angle of α :

Rock weight = W

Resistance force = FR

Driving force = FD

WN

T

A α

Discontinuities plane

A1

C N tan φ

WN = Wcosα

T = Wsinα

5

• Consider 1 m ⊥ the slide

T = tangential component (W) at discontinuity plane

W = rock weight slides on the slide plabe A-A1

α = slope inclination

φ = internal rock friction angle

C = c (A-A1) ( resistant due to cohesion along the shear zone)

WN

T

A α

A1

C N tan φ

WN = Wcosα

T = WsinααsinWTFD ==

CNFR += φtan

CWFR +⋅= φα tancos

Stable condition : safety factor : (SF)

Hence :

If C = 0 very weak discontinuity plane

So or

WN

T

A α

A1

C N tan φ

WN = Wcosα

T = Wsinα

( )( )mAACc

0,11−=

RD FF =

Cohesion / unit area:

αφα

sintancos

WCW

FFSF

D

R +⋅==

αφα

sintancos

WW

FFSF

D

R ⋅==

αφ

tantan

=SF

6

For non cohesive material SF is not dependence on the height of the slope and the shape of rock mass.

When the rain occurs and the water infiltrates into discontinuity plane pore water pressure (u)

Uplift pressure works : SF U should be considered

( )α

φαsin

tancosW

UWSF ⋅−=

B. Rock block slide on discontinuity plane

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Bedrock

Hard rock γwz

LE

A

B C

K

J

σmax

σmin

e

CL

Neff

T

7

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Batuankeras

Batuankeras γwz

LE

A

B C

K

J

σmax

σmin

e

CL

Neff

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Batuankeras

Batuankeras γwz

LE

A

B C

K

J

σmax

σmin

e

CLCL

Neff

αsinzhL −

=

⎟⎠⎞

⎜⎝⎛ ±=

Le

averagehext61σσ

)1( mLNeff

averageh ⋅=σ

Shear resistant (T) occurs at discontinuity plane along AK (=L),so:

Where effective normal stress component

T

( ) CUNCNT eff +−=+⋅= φφ tantan

UNNeff −=

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Batuankeras

Batuankeras γwz

LE

A

B C

K

J

σmax

σmin

e

CL

Neff

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Batuankeras

Batuankeras γwz

LE

A

B C

K

J

σmax

σmin

e

CLCL

Neff

T

αcosWN =

Normal component of rock block (W) sliding on the shear plane

( ) ( ) ( )[ ]zxhbhBbW ⋅−⋅−+= γ21

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−⎟⎠⎞

⎜⎝⎛−

=θα

γtan

1tan

1

21

2

hz

W

The weight of the sliding rock mass block

8

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Batuankeras

Batuankeras γwz

LE

A

B C

K

J

σmax

σmin

e

CL

Neff

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Batuankeras

Batuankeras γwz

LE

A

B C

K

J

σmax

σmin

e

CLCL

Neff

T

Note:

W = weight of sliding rock mass block

α = inclination of the discontinuity plane to horizontal plane

u = uplift pressure

( )α

γ

sin2

1 zhzU w −⋅

=

φ = rock internal frictionC = c.A = c.L (1 m ⊥ the canvas)A = total shear plane area (1 m ⊥ the canvas)L = length of discontinuityγ = unit weight of sliding rock materialz = depth of the crackθ = slope inclination

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Batuankeras

Batuankeras γwz

LE

A

B C

K

J

σmax

σmin

e

CL

Neff

α

W

Wcosα Wsinα

a

h

θ

b Bx

zH

α

γwz

U

Batuankeras

Batuankeras γwz

LE

A

B C

K

J

σmax

σmin

e

CLCL

Neff

T

Resultant of Shear Force Resultant (FR):

( ) φα tansinHTFR −=

φαφ tansintan HCNeff −+⋅=

( )[ ] CHUN +−−= φα tansin

[ ] CHUW +−−= φαα tansincos

FH=NH tanφ

H NH

α

Resultant of driving force (FD):

αα cossin HWFD +=

Safety factor (SF):

( )αα

φααcossin

tansincosHW

CUHWSF+

+−−=

9

Safety factor (SF):

SF = 1,0 stable condition, when it is going to slide

SF < 1,0 not stable

SF > 1,0 stable

In general SF ≥ 1,5 long time

SF ≥ 1,3 short time (earthquake load)

Special condition:

H = 0, U = 0 dry rock or good drainage

H ≠ 0, U = 0 water fill the crack only

φ ≠ 0 and c = 0 dry rock or wet

φ ≈ 0 and c ≠ 0 dry rock or wet

( )αα

φααcossin

tansincosHW

CUHWSF+

+−−=

When earthquake occurs

WF1

ag

WamF ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛=⋅=1

m = rock mass which is slidingg = gravity accelerationa = earthquake acceleration = kg.gkg = earthquake coefficientF1 = kg.W

Safety factor (SF):

( )ααα

φαααcoscossin

tansinsincos

1

1

HFWCUHFWSF

+++−−−

=

D

R

FFSF =

10

C. Analysis of Plane Slides (Goodman, 1980)

A simple formulation of conditions for limiting equilibrium of a plane slide provides useful in back calculating actual failure cases

Important step in attempting to design a new excavation in a rock mass

Rework field data using an appropriate model rather than to attempt a program of field tests.

Two cases of plane failure:

1. Tension crack delimits the top of the slide at a point beyond the crest of the slope

2. Tension crack intercepts the slope face

H

ZZw

α

θ

Tension crack

1. Tension crack delimits the top of the slide

H

α

θZw

Z

2. Tension crack intercepts the slope face

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Z = vertical distance from the crest of the slope to the bottom of the crackIf the tension crack is filled with water to depth Zw, it can be assumed that water seeps along the sliding surface losing head linearly between the tension crack and the toe of the slopeIf the slide mass behaves like a rigid body, the condition for limiting equilibrium is reached when the shear force directed down the sliding surface equals the shear strength along the sliding surface Failure occurs when:

α = the dip of the sliding surfacecj and φj = shear strength intercept (cohesion) and friction angle of the sliding

surfaceW = the weight of the potentially sliding wedgeA = length (are per unit width) of the sliding surfaceU = resultant of water pressure along the sliding surfaceV = resultant of water pressure along tension crack

( ) jj VUWAcVW φαααα tansincoscossin −−+=+

H

ZZw

α

θ

H

α

θZw

Z

( )αsinZHA −

=

AZU ww ⋅⋅= γ21

2

21

ww ZV ⋅= γ

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+= θαγ cotcot1

21 2

2

HZHW

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡−⋅⎟

⎠⎞

⎜⎝⎛ −= 1tancotcot1

21 2

2 θααγHZHW

12

Solve the above equation with the known geometry and presumed water conditions at the time of failure to yield a value for cj, since this quantity is hard to measure in the laboratory.

When the distribution of values for cj has been determined in this way from case histories, that equation can be used to generate a slope chartfor design, in which H is plotted against cos α.

Multiplying the Factor of Safety (F) to the left side of the equation.

In which the tension crack is assumed to intercept the slope crest

( ) jj VUWAcVW φαααα tansincoscossin −−+=+

( ) ( )[ ]( )φαα

αφαφφααθ

tancossincostansintantancossin

cos−

−−++−=

FbAcFVUFa j

αγ cot121 2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+=

HZHa 2

21 Hb γ=

From Hoek and Bray (1977):

A reduction in cj affects steep slopes more than flat slopes.

A reduction in φj reduces the factor of safety (FS) of high slopesmore than low slopes.

Filling a tension crack with water reduces the stability of all heights and angles of slopes.

Drainage is frequently found to be effective in stabilizing rockslopes that exhibit tension cracks and other signs in incipient movement.

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8 m

5 m

3 m

45o

A

B

C

Tension crack

51Example

The cohesion along the sliding surface is 80 kN/m2 and internal friction angle = 35o. Unit volume weigth of the rock = 24 kN/m2.

Due to SNI-1726-2002, the slope located at Seismic zone 3 with peak bedrock acceleration: a = 0,15g.

Calculate the safety factor (FS)

Bedrock

8 m

5 m

3 m

45o

A

B

C

51

γwz

α=21.04°

Wcosα Wsinα

U

Bedrock

γwz

Neff

T

WH

Hcosα

Hsinα

( ) 00.4cossin

tansincos=

++−−

=αα

φααHW

CUHWSF

Without earthquake force:

( )α

γ

sin2

1 zhzU w −⋅

=

αsin38 −

=L

2

21 zH w ⋅= γ

BALcC −⋅=

14

8 m

5 m

3 m

45o

A

B

C

51

γwz

α=21.04°

WcosαWsinα

U

Bedrock

γwz

Neff

T

WH

Hcosα

Hsinα

If the site located at seismic zone 3 a = 0.15g F = 2.89

( )ααα

φαααcoscossin

tansinsincos

1

1

HFWCUHFWSF

+++−−−

=

If the site located at seismic zone 5 a = 0.25g F = 2.43

Considering earthquake force:

ag

WamF ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛=⋅=1

F1

F1cosα

F1sinα

Two blocks sliding on a rock slope

Vertical joint

( )( )131

1113 cos

sinβφφ

φβ−+

−=WR

( )[ ]( )23322

2233212 cossin

tansincosβφβ

φβφβ−+

−+=

RWRWF

W1

= force exceeding the mobilitable resistant force

φ1 : (i = 1,2,3) = friction angle values on Plane 1,2,3c is assumed to be 0

Safety factor of Block 2 (F2) =

overall safety factor

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Analisis pada bidang longsor datarB. Lereng terbatas (finite slope)

B.1. Analisis bidang longsor datar (Culman’s method)

( )⎟⎟⎠

⎞⎜⎜⎝

⎛αβα−β

γ=sinsin

sinHW 221

( )( )( ) ( )

αβα−βααγ

=α

=σsinsin

sincossinH/sin/HNa 21

1

W

P

Na

Nr

Ta

Tr

αβ

H

L

A

BC

γ

( )( )( ) ( )

αβα−βαγ

=α

=τsinsin

sinsinH/sin/HTa

2211

W

P

Na

Nr

Ta

Tr

αβ

H

L

A

BC

γ

Shear resistant (τd) at AB:

ddd c φσ+=τ tan

Critical condition F=1 τ = τd

( ) ( )( )⎟⎟⎠

⎞⎜⎜⎝

⎛β

φα−αα−βγ=

sintancossinsinH/c d

d 21

0=δαδ dc

⎟⎠⎞

⎜⎝⎛ φ+β

=α2

dc

( )4

1 Hcossin

coscd

dd

γ⎟⎟⎠

⎞⎜⎜⎝

⎛φβφ−β−

=

Critical conditionF=1 cd = c ; φd = φ

( )⎟⎟⎠⎞

⎜⎜⎝

⎛φ−β−φβ

γ=

coscossincHc 1

4