hemispherical and slope.ppt

8/10/2019 Hemispherical and Slope.ppt

http://slidepdf.com/reader/full/hemispherical-and-slopeppt 1/41

Lecture 5: Hemispherical

Projection & Slopes

D A Cameron

Rock & Soil Mechanics



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



Meridional stereographic

projection net for

Structural Geology

N



EXAMPLE

dip direction, = 135

dip angle, = 50

denoted as 135/050

Plot also 000/090 and 090/000



N

Dip Direction

= 135°



N

Tracing

paper with

central

drawing pin



• Rotate the paper until the line

marking the dip direction

corresponds with the equatorial

position (90)

Step 1



50 40

POLE

GREATCIRCLE

90



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



N

Rotate back

to the North

position



Add in

000/090



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



Plane A,

060/030

Plane B,

340/075



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



intersection line

plane 1plane 2

The intersecting planes

line of intersection



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



35



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



60



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?



Typical Slope Instability

(a) no particular concentration of poles

- circular failure (e.g. waste rock/ fractured slate)

- similar to soil (use Bishop’s method)



Slope Instability

(b) single concentration of poles above cut

slope - plane failure

slope

Discontinuity – strike parallel to that

for the slope



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



The last condition

20 Strike of

discontinuity

< 20°



(c) double concentration of poles = intersecting

joints - wedge failure most common

Slope Instability

slope

discontinuities



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



intersection line

plane 2

plane 1

intersection line,

I12

Slope Great

Circle

Dip of intersection > friction angle

“friction circle”

UNSAFE slope!



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



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



Wedges intersecting slopes

Great circle of

slope surface

intersection lines of

planar discontinuities

with the slope



(d) single concentration of poles below slope

- toppling failure in hard rock

Slope Instability

slope

discontinuity



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



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



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



2.5 m

U2

U1



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

hemispherical and slope.ppt

Documents