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Lecture 5: Hemispherical
Projection & Slopes
D A Cameron
Rock & Soil Mechanics
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Why are these plots needed?
• to provide a simple visual reference of
the various joint sets seen in rock mass
exposures
• to evaluate the potential for instability of
engineering works in these masses
– e.g. dip angle and dip direction of slopes can
be compared with prevailing joint sets
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Meridional stereographic
projection net for
Structural Geology
N
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EXAMPLE
dip direction, = 135
dip angle, = 50
denoted as 135/050
Plot also 000/090 and 090/000
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N
Dip Direction
= 135°
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N
Tracing
paper with
central
drawing pin
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• Rotate the paper until the line
marking the dip direction
corresponds with the equatorial
position (90)
Step 1
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50 40
POLE
GREATCIRCLE
90
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Normal to the plane
The trend n and plunge n
of the normal to a plane are
given by:
n = w ± 180
n = 90 - w
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N
Rotate back
to the North
position
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Add in
000/090
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Intersections
Two planes A and B have orientations
A:060/030 and B:340/075
These planes intersect on the stereonet atthe point A:B
- this point represents the line of intersection of
the discontinuities represented by the planes
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Plane A,
060/030
Plane B,
340/075
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Plunge of intersection line
• Rotate tracing until intersection point lies
on the E-W line
• Read off the number of degrees from the
perimeter to the intersection point
= the plunge of the intersection line
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intersection line
plane 1plane 2
The intersecting planes
line of intersection
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Plunge of intersection line
• Rotate tracing until intersection point lies on
the E-W line
• Read off the number of degrees from the
perimeter to the intersection point
= the plunge of the intersection line
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35
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Dip direction of intersection line
• Rotate tracing back to the datum
• Mark off dip direction as indicated
• The intersection point can be designated as
060/035
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60
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SLOPE STABILITY
Stereonet information can be used to indicate the
likely instability
• Plots of the poles of discontinuity planes
• Contours to indicate high concentrations in areas of
the net
– prevailing discontinuities
• Position of discontinuities with respect to the Great
Circle for the Slope?
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Typical Slope Instability
(a) no particular concentration of poles
- circular failure (e.g. waste rock/ fractured slate)
- similar to soil (use Bishop’s method)
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Slope Instability
(b) single concentration of poles above cut
slope - plane failure
slope
Discontinuity – strike parallel to that
for the slope
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Conditions for planar failure
• The plunge of the slope > dip of the
discontinuity
• Discontinuity daylights on the slope face• Discontinuity has a dip angle > for the joint
– mechanically possible
• Dip direction of the discontinuity and
slope lie within 20
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The last condition
20 Strike of
discontinuity
< 20°
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(c) double concentration of poles = intersecting
joints - wedge failure most common
Slope Instability
slope
discontinuities
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Conditions for wedge failure
• The plunge of the slope > dip of the
Intersection line
• Intersection line daylights on the slope face
• Intersection line has a dip angle > for the
joints
– mechanically possible
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intersection line
plane 2
plane 1
intersection line,
I12
Slope Great
Circle
Dip of intersection > friction angle
“friction circle”
UNSAFE slope!
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The Friction Circle
Outer radius = 1, represents = 0° Radius of friction circle = (90° - )/90°
0.750.5 = 45
= 22.5
= 30
(0.67)
1.0
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The Friction Circle
• The meridional plot is overlaid by the
friction circle (same diameter )
• The slope is safe if the intersection point,
I12 is outside the friction circle () for the joint
- mechanically impossible to fail
- assumes c = 0 kPa for the joint
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Wedges intersecting slopes
Great circle of
slope surface
intersection lines of
planar discontinuities
with the slope
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(d) single concentration of poles below slope
- toppling failure in hard rock
Slope Instability
slope
discontinuity
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Mechanics of a Planar FailureQn 8.1, Priest (1993)
6 m
2 m
10 m
30
60
60
= 27 kN/m3. Inclined tension crack.
Silt filled joint: c j = 10 kPa, j = 32
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6 m
• Area of wedge = 44.43 m2
W = 1200 kN
• pwp at point D = 2x9.8 = 19.6 kPa
U1 = 22.66 kN
and U2 = 101.83 kN
10 m
U2
U1
W
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FN = Wcos30 - U1sin60 - U2
= 917.8 kN
Sliding resistance = (10.38x10 kPa) + 917.8(tan32)= 677.3 kN
Sliding force =Wsin30 +U1cos60
= 611.3 kN
ANSWER: FoS = 1.11
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2.5 m
U2
U1
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SUMMARY – Key Points
• Great Circles & poles
– strikes, dip directions & dip angles
• Lines of intersection
• Conditions for slope instability
– 4 potential types
• Friction circle application
– cohesionless joints
• Analysis of planar failures