Lecture 5: Hemispherical Projection & SlopesD A CameronRock & Soil Mechanics

Why are these plots needed?to provide a simple visual reference of the various joint sets seen in rock mass exposuresto evaluate the potential for instability of engineering works in these massese.g. dip angle and dip direction of slopes can be compared with prevailing joint sets

Meridional stereographic projection net for Structural Geology

EXAMPLEdip direction, = 135dip angle, = 50denoted as 135/050

Plot also 000/090 and 090/000

Dip Direction = 135

Tracing paper with central drawing pin

Step 1Rotate the paper until the line marking the dip direction corresponds with the equatorial position (90)

Steps 2 and 3Measure 50 ( = the dip direction, ) from the outer circle RHS and trace the great circle for the plane as shownMeasure (90 - ) or 40 from the outer circle, but this time from the LHS to locate the POLE of the great circle or plane

POLEGREAT CIRCLE

The Normal or PoleThe normal to a plane is an imaginary line drawn perpendicular to the planeThe downward direction of this line is plotted on the projection The point representing the normal on the stereonet is referred to as the pole of the plane

Normal to the planeThe trend n and plunge n of the normal to a plane are given by:n = w 180n = 90 - w

Rotate back to the North position

Add in 000/090

IntersectionsTwo planes A and B have orientations A:060/030 and B:340/075 These planes intersect on the stereonet at the point A:B- this point represents the line of intersection of the discontinuities represented by the planes

Plunge of intersection lineRotate tracing until intersection point lies on the E-W lineRead off the number of degrees from the perimeter to the intersection point= the plunge of the intersection line

intersection lineplane 1plane 2

Plunge of intersection lineRotate tracing until intersection point lies on the E-W lineRead off the number of degrees from the perimeter to the intersection point= the plunge of the intersection line

Dip direction of intersection lineRotate tracing back to the datumMark off dip direction as indicatedThe intersection point can be designated as 060/035

60

SLOPE STABILITYStereonet information can be used to indicate the likely instabilityPlots of the poles of discontinuity planesContours to indicate high concentrations in areas of the netprevailing discontinuitiesPosition 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 Bishops method)

Slope Instability(b) single concentration of poles above cut slope - plane failure slope

Conditions for planar failureThe plunge of the slope > dip of the discontinuityDiscontinuity daylights on the slope faceDiscontinuity has a dip angle > for the jointmechanically possibleDip direction of the discontinuity and slope lie within 20

The last condition 20

Slope Instability(c) double concentration of poles = intersecting joints - wedge failure most commonslope

Conditions for wedge failureThe plunge of the slope > dip of the Intersection line Intersection line daylights on the slope faceIntersection line has a dip angle > for the jointsmechanically possible

intersection lineplane 2plane 1Dip of intersection > friction angleUNSAFE slope!

The Friction Circle Outer radius = 1, represents = 0Radius of friction circle = (90 - )/90 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

(d) single concentration of poles below slope - toppling failure in hard rockSlope Instability

Mechanics of a Planar FailureQn 8.1, Priest (1993) = 27 kN/m3. Inclined tension crack. Silt filled joint: cj = 10 kPa, j = 32

6 mArea of wedge = 44.43 m2 W = 1200 kNpwp at point D = 2x9.8 = 19.6 kPa U1 = 22.66 kN and U2 = 101.83 kN

FN = Wcos30 - U1sin60 - U2= 917.8 kN

Sliding resistance = (10.38x10 kPa) + 917.8(tan32)= 677.3 kNSliding force =Wsin30 +U1cos60= 611.3 kN

ANSWER: FoS = 1.11

SUMMARY Key PointsGreat Circles & polesstrikes, dip directions & dip anglesLines of intersectionConditions for slope instability4 potential typesFriction circle applicationcohesionless jointsAnalysis of planar failures