Ejercicio resuelto de un pórtico, por el método de pendiente - deflexión. Cortesía del catedrático de la UC, Yupanqui Vargas José Luis.Descripción completa
Full description
guidelines for constructing pipeline in steep slopes. Also how to prevent pipeline damage as a result of landslip.
Slope Deflection for beams and frames by Hibbeler.Full description
Rock Slope Stability Analysis Method
Stability Analysis Method
Stability 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.
Static and Dynamic Equilibrium Method
Rigid block instability modes on inclined plane individual block sliding and toppling
Static equilibrium problem variables
produces
Dynamic equilibrium problem variables
Static equilibrium analysis
Stability Chart
Dynamic equilibrium analysis
Safety Factor and Limit Equilibrium Method
Wsin
Assuming the shear stress t on sliding surface defined by Coulomb criterion
t c tan R W
Wcos
t c
W cos a
R t A cA W cos a tan
Limit block equilibrium condition
W sin a cA W cos a tan F
t (maximum mobitable) t (mobilized )
A
tan
If c = 0
a
Safety Factor and Limit Equilibrium Method F
t (maximum mobilitabl e) t (mobilized )
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
Effect of Water Pressure in Rock Discontinuities
Water filling discontinuities involves a lowering of stability conditions for natural or artificial slopes
Subvertical discontinuity plane
V U
V p
V
V n
0
F
W cos a U A
cA W cos a U tan W sin a V
A. Rock Slope Stability • Rock slide occurs at the weakness plane (discontinuities plane), cracks or shear zone A1 W
N = Wcos W
Discontinuities plane
N
T = Wsin
T
A
Rock discontinuity plane with the angle of : Rock weight = W Resistance force = FR Driving force = F
• Consider 1 m the slide A1 W
N = Wcos W T = Wsin
N
A
F D
T W sin a
F R
N tan C
F R
W cos a tan C
T
T = tangential component (W) at discontinuity plane W = rock weight slides on the slide plabe A-A 1 a = slope inclination = internal rock friction angle
C = c (A-A 1) ( resistant due to cohesion along the shear zone)
A1 W
N = Wcos W
Cohesion / unit area:
N
T = Wsin
T
c A
Stable condition : F D
SF
Hence :
If C = 0 So
F R F D
F R
C
A A1 1,0m
safety factor : (SF)
W cos a tan C W sin a
very weak discontinuity plane
SF
F R F D
W cos a tan W sin a
or
SF
tan
tan a
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
SF
U should be considered
W cos a U tan W sin a
B. Rock block slide on discontinuity plane B
b B
W Wcos
h
Wsin
H K
Hard rock
U
a E
Bedrock
C z
wz wz
T
A
x
J
B
b
x
J
B
C
z W
H K
Wcos
h
Wsin
z w
Batuan keras
T
w
N e f f A
z
L
h z sin a
6e ext h average1 L
U
Batuan keras
a
L
x m a
e
h average
E
C L
N eff L (1m)
n m i
Shear resistant (T) occurs at discontinuity plane along AK (=L), so:
T N eff tan C N U tan C
Where N eff N U
effective normal stress component
B
b
x
J
B
C
N W cos a
z H
W
K Wcos
h
Batuan keras
Wsin
T
w
N e f f A
z
w
z
Normal component of rock block (W) sliding on the shear plane
U
Batuan keras
a
L
e
x m a
E
C L
n m i
The weight of the sliding rock mass block
W
1 2
b B h b h x z
z 2 1 1 h 1 W 2 tana tan
B
b
x
J
B
C
z H
W
K Wcos
h
Batuan keras
Wsin
T
w
N e f f A
z
w
Note: W = weight of sliding rock mass block a = inclination of the discontinuity plane to horizontal plane
z
U
u = uplift pressure
Batuan keras
a
L
e
x m a
E
C L
n m i
U
1 z h z 2 w sin a
= 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
B
b
x
J
B
C
Resultant of Shear Force Resultant (F R):
z W
H K
Wcos
h
Batuan keras
Wsin
T
F R
z
w
N e f f A
w
z
N eff tan C H sin a tan
U
N U H sin a tan C
Batuan keras
a
T H sin a tan
L
a x m
e
E
C L
W cosa U H sin a tan C
n m i
Resultant of driving force (F D): H
NH
F D
W sin a H cos a
Safety factor (SF): FH=NH tan
SF
W cos a H sin a U tan C W sin a H cos a
Safety factor (SF):
SF
W cos a H sin a U tan C W sin a H cos a
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
When earthquake occurs
F1
W F 1 m a a g W
m = rock mass which is sliding g = gravity acceleration a = earthquake acceleration = k g.g kg = earthquake coefficient F1 = kg.W
Safety factor (SF):
SF
F R F D
SF
W cos a F 1 sin a H sin a U tan C W sin a F 1 cos a H cos a
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
Tension crack
Z
Zw H
1. Tension crack delimits the top of the slide
Z Zw H
2. Tension crack intercepts the slope face
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 Z w, 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:
W sin a V cos a c j A W cos a U V sin a tan j a = the dip of the sliding surface
c j and j = shear strength intercept (cohesion) and friction angle of the sliding surface W = 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
A Zw
Z
H
H Z sin a
U 1 w Z w A 2 V
1 Z 2 2 w w
2 Z 2 W H 1 cot a cot 2 H
1
Z Zw H 2 Z 2 W H 1 cot a cot a tan 1 2 H
1
W sin a V cos a c j A W cos a U V sin a tan j
Solve the above equation with the known geometry and presumed water conditions at the time of failure to yield a value for c j, since this quantity is hard to measure in the laboratory.
When the distribution of values for c j 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.
a F sin a cos a tan U tan V sin a tan F cos a c j A cos b F sin a cos a tan In which the tension crack is assumed to intercept the slope crest 2 1 Z 2 a H 1 cot a H 2
b
1 2
H 2
From Hoek and Bray (1977):
A reduction in c j affects steep slopes more than flat slopes.
A reduction in j reduces the factor of safety (FS) of high slopes more 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 rock slopes that exhibit tension cracks and other signs in incipient movement.
5
5m
Example
1 C 3m
8m
B
Bedrock
Tension crack
A 45o
The cohesion along the sliding surface is 80 kN/m 2 and internal friction angle = 35 o. Unit volume weigth of the rock = 24 kN/m 3.
Due to SNI-1726-2002, the slope located at Seismic zone 3 with peak bedrock acceleration: a = 0,15g.
Calculate the safety factor (FS)
5
5m
1 C Hcos
W Wcos
8m
wz
Wsin
B
T
H
3m
Hsin wz
H 1 w z 2 2 A
=21.04 45o
Bedrock
L
Without earthquake force:
SF
W cos a H sin a U tan C W sin a H cos a
U
4.00
1 z h z 2 w sin a 83 sin a
C c L A B
5
5m
1 C
F1cos
Wcos
8m
Hcos
F1 F1sin
W Wsin
H
wz
B
3m
Hsin wz
A
=21.04 45o
W F 1 m a a g
Bedrock
Considering earthquake force:
SF
W cos a F 1 sin a H sin a U tan C W sin a F 1 cos a H cos a
If the site located at seismic zone 3
a = 0.15g
F = 2.89
If the site located at seismic zone 5
a = 0.25g
F = 2.43
Two blocks sliding on a rock slope W1
Vertical joint
R3
W 1
sin 1 1 cos 1 3 1
= 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 (F 2) =
W 1 cos 2 R3 sin 3 2 tan 2 F 2 W 2 sin 2 R3 cos 3 2
overall safety factor
Analisis pada bidang longsor datar B. Lereng terbatas (f i n i t e s l o p e ) B.1. Analisis bidang longsor datar (Culman’s method)
