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Rock Slope StabilityAnalysis Method
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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.
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Static and Dynamic Equilibrium Method
Rigid block instability modes on inclined plane producesindividual block sliding and toppling
Static equilibrium problem variablesDynamic equilibrium problem variables
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Static equilibrium analysis
Dynamic equilibrium analysis
Stability Chart
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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/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
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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
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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
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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
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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
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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