S2-Geo-3-Rock Slope stability.pdf

7/29/2019 S2-Geo-3-Rock Slope stability.pdf

1/30

Rock Slope StabilityAnalysis Method


2/30

Stability Analysis Method

Stability analysis define the geometry of blocks or of the system ofblocks 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.


3/30

Static and Dynamic Equilibrium Method

Rigid block instability modes on inclined plane producesindividual block sliding and toppling

Static equilibrium problem variablesDynamic equilibrium problem variables


4/30

Static equilibrium analysis

Dynamic equilibrium analysis

Stability Chart


5/30

Safety Factor and Limit Equilibrium Method

Wsina

WcosaWa

R

Assuming the shear stress t

on sliding surface defined by

Coulomb criterion

t tan c

at tancosA

Wc

at tancosWcAAR Limit block equilibrium condition

aa tancossin WcAW If c = 0 a

)(mobilized

mobitable)(maximum

t

tF


6/30

Safety Factor and Limit Equilibrium Method

Safety factor : a number for which the available shear strength

parameters (c- ) 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 method

Solution 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

t

t

F


7/30

Effect of Water Pressure in Rock Discontinuities

Water filling discontinuities involves a lowering of stability conditions for

natural or artificial slopes

Subvertical

discontinuity plane

VU

VVp

0nV

A

UW

a

cos

VW

UWcAF

a

a

sin

tancos


8/30

A. Rock Slope Stability

Rock slide occurs at the weakness plane (discontinuities plane), cracks or

shear zone

Rock discontinuity plane with the angle ofa :Rock weight = W

Resistance force = FR

Driving force = FD

W

N T

A a

Discontinuities

plane

A1

WN = Wcosa

T = Wsina


9/30

Consider 1 m the slide

T = tangential component (W) at discontinuity plane

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

a = slope inclination

= internal rock friction angle

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

W

N

T

A a

A1

WN = Wcosa

T = Wsina asinWTFD CNFR tan

CWFR a tancos


10/30

Stable condition : safety factor : (SF)

Hence :

If C = 0 very weak discontinuity plane

So or

W

N

T

A a

A1

WN = Wcosa

T = Wsina

mAACc

0,11

RD FF

Cohesion / unit area:

a

a

sin

tancos

W

CW

F

FSF

D

R

a

a

sin

tancos

W

W

F

FSF

D

R

a

tan

tanSF


11/30

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

a

a

sin

tancos

W

UW

SF


12/30

B. Rock block slide on discontinuity plane

a

W

Wcosa Wsina

a

h

b B

x

zH

a

wz

U

Bedrock

Hard rock wz

E

A

B C

K

J

T


13/30

a

W

Wcosa Wsina

a

h

b B

x

z

H

a

wz

U

Batuan

keras

Batuan

keraswz

L

E

A

B C

K

J

max

min

e

CL

Neff

a

W

Wcosa Wsina

a

h

b B

x

z

H

a

wz

U

Batuan

keras

Batuan

keraswz

L

E

A

B C

K

J

max

min

e

CLCL

Neff

asin

zhL

L

eaveragehext

61

)1( mLNeff

averageh

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

so:

Where

effective normal stress component

T

CUNCNT eff tantan

UNNeff


14/30

a

W

Wcosa Wsina

a

h

b B

x

z

H

a

wz

U

Batuan

keras

Batuan

keraswz

L

E

A

B C

K

J

max

min

e

CL

Neff

a

W

Wcosa Wsina

a

h

b B

x

z

H

a

wz

U

Batuan

keras

Batuan

keraswz

L

E

A

B C

K

J

max

min

e

CLCL

Neff

T

acosWN

Normal component of rock

block (W) sliding on the

shear plane

zxhbhBbW 2

1

a

tan

1

tan

1

2

1

2

h

z

W

The weight of the sliding rock

mass block


15/30

a

W

Wcosa Wsina

a

h

b B

x

z

H

a

wz

U

Batuan

keras

Batuan

keraswz

L

E

A

B C

K

J

max

min

e

CL

Neff

a

W

Wcosa Wsina

a

h

b B

x

z

H

a

wz

U

Batuan

keras

Batuan

keraswz

L

E

A

B C

K

J

max

min

e

CLCL

Neff

T

Note:

W = weight of sliding rock mass

block

a = inclination of the discontinuity

plane to horizontal plane

u = uplift pressure

a

sin

21 zhzUw

= rock internal friction

C = c.A = c.L (1 m the canvas)

A = total shear plane area (1 m the canvas)

L = length of discontinuity

= unit weight of sliding rock material

z = depth of the crack

= slope inclination


16/30

a

W

Wcosa Wsina

a

h

b B

x

z

H

a

wz

U

Batuan

keras

Batuan

keraswz

L

E

A

B C

K

J

max

min

e

CL

Neff

a

W

Wcosa Wsina

a

h

b B

x

z

H

a

wz

U

Batuan

keras

Batuan

keraswz

L

E

A

B C

K

J

max

min

e

CLCL

Neff

T

Resultant of Shear Force Resultant (FR):

a tansinHTFR

a tansintan HCNeff

CHUN a tansin

CHUW aa tansincos

FH=NH tan

H

NH

a

Resultant of driving force (FD):

aa cossin HWFD Safety factor (SF):

aa

aa

cossin

tansincos

HW

CUHWSF


17/30

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

0 and c = 0 dry rock or wet

0 and c 0 dry rock or wet

aa

aa

cossin

tansincos

HW

CUHWSF


18/30

When earthquake occurs

WF1

ag

WamF

1

m = rock mass which is sliding

g = gravity acceleration

a = earthquake acceleration = kg.g

kg = earthquake coefficientF1 = kg.W

Safety factor (SF):

aaa

aaa

coscossin

tansinsincos

1

1

HFW

CUHFWSF

D

R

F

FSF


19/30

C. Analysis of Plane Slides (Goodman, 1980)

A simple formulation of conditions forlimiting 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 offield 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


20/30

H

ZZw

a

Tension crack

1. Tension crack delimits

the top of the slide

H

a

Zw

Z

2. Tension crack intercepts

the slope face


21/30

Z = vertical distance from the crest of the slope to the bottom of the crack

If 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 slope If 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:

a = the dip of the sliding surface

cj and j = shear strength intercept (cohesion) and friction angle of the sliding

surfaceW = the weight of the potentially sliding wedge

A = length (are per unit width) of the sliding surface

U = resultant of water pressure along the sliding surface

V = resultant of water pressure along tension crack

jj VUWAcVW aaaa tansincoscossin


22/30

H

ZZw

a

H

a

Zw

Z

asin

ZHA

AZU ww 212

21

ww ZV

a cotcot12

1

2

2

HZHW

1tancotcot12

12

2

aa H

Z

HW


23/30

Solve the above equation with the known geometry and presumed water

conditions at the time of failure to yield a value forcj, since this quantityis 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 chart

for design, in which H is plotted against cos a.

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 aaaa tansincoscossin

aa

aaaa

tancossin

costansintantancossincos

Fb

AcFVUFa j

a cot1

2

12

2

H

ZHa 2

2

1Hb


24/30

From Hoek and Bray (1977):

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

A reduction in j reduces the factor of safety (FS) ofhigh slopes

more than low slopes.

Filling a tension crack with water reduces the stability ofall heights

and angles of slopes.

Drainage is frequently found to be effective in stabilizing rock

slopes that exhibit tension cracks and other signs in incipient

movement.


25/30

8 m

5 m

3 m

45o

A

B

C

Tension

crack

51Example

The cohesion along the sliding surface is 80 kN/m

2

and internalfriction angle = 35o. Unit volume weigth of the rock = 24 kN/m3.

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


26/30

8 m

5 m

3 m

45o

A

B

C

51

wz

a=21.04

WcosaWsina

Bedrock

wzT

WH

Hcosa

Hsina

00.4

cossin

tansincos

aa

aa

HW

CUHWSF

Without earthquake force:

a

sin

21 zhz

Uw

asin

38

L

2

2

1 zH w

BALcC

5 5


27/30

8 m

5 m

3 m

45o

A

B

C

51

wz

a=21.04

WcosaWsina

Bedrock

wz

WH

Hcosa

Hsina

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

aaa

aaa

coscossin

tansinsincos

1

1

HFW

CUHFWSF

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

Considering earthquake force:

ag

WamF

1

F1

F1cosaF1sina


28/30

Two blocks sliding

on a rock slope

Vertical joint

131

1113cos

sin

WR

23322

2233212

cossin

tansincos

RW

RWF

W1

= force exceeding the mobilitable resistant force

1 : (i = 1,2,3) = friction angle values on Plane 1,2,3

c is assumed to be 0

Safety factor of Block 2 (F2) =

overall safety factor


29/30

Analisis pada bidang longsor datar

B. Lereng terbatas (f in i te slope)

B.1. Analisis bidang longsor datar (Culmans method)

aa

sinsinsin

HW2

21

aaaa

a

sinsin

sincossinH/

sin/H

Na 21

1

W

P

Na

Nr

Ta

Tr

a

H

A

BC

a

aa

at sinsinsinsinH/

sin/H

Ta2

21

1


30/30

W

P

Na

Nr

Ta

Tr

a

H

A

BC

Shear resistant (td) at AB:

ddd c t tan

Critical condition F=1

t = td

aaa

sin

tancossinsinH/c dd 21

0

a

dc

a2

dc

4

1 H

cossin

cosc

d

dd

Critical condition

F=1 cd = c ; d =

cos

cossincHc

1

4

S2-Geo-3-Rock Slope stability.pdf

Documents

Transcript of S2-Geo-3-Rock Slope stability.pdf