Measurement, analysis and interpretation of structural geological data

Rock Mass Rating (RMR), also called the Geomechanics Classification, was proposed by Bieniawski (1973, revised 1989). It gives a single rating from 0 to 100 that describes rock mass quality for tunnels, slopes and foundations, and guides support design.

Parameters and ratings

$$RMR = R_1 + R_2 + R_3 + R_4 + R_5 + R_6$$

Example ratings: UCS above 250 MPa = 15, RQD 90 to 100 % = 20, spacing above 2 m = 20, completely dry = 15.

Condition of discontinuities (maximum 30)

Rock mass classes

Use: the class gives stand-up time (e.g. 20 years for 15 m span in class I, 30 minutes for 1 m in class V), shear strength parameters and recommended excavation and support.

Rock Quality Designation (RQD), proposed by Deere (1967), is the percentage of sound core pieces longer than 10 cm in the total length of a core run. It is an index of the degree of jointing and fracturing of the rock mass.

From drill core

$$RQD = \frac{\sum \text{length of core pieces} \geq 10\ \text{cm}}{\text{total length of core run}} \times 100\ \%$$

Only natural, hard and sound pieces are counted; breaks caused by drilling or handling are ignored. Example: in a 150 cm run the pieces of 12, 25, 8, 30, 15 and 20 cm are counted as 12 + 25 + 30 + 15 + 20 = 102 cm, so RQD = 102/150 × 100 = 68 % (fair).

From outcrops (no core)

1. Volumetric joint count (Palmstrom):

$$RQD = 115 - 3.3\,J_v \quad (J_v \geq 4.5),\qquad RQD = 100\ \text{for}\ J_v < 4.5$$

where $J_v$ is the number of joints per cubic metre (sum of joint frequencies of all sets per metre). 2. Scanline (Priest and Hudson): count discontinuities per metre $\lambda$ along a scanline, then $RQD = 100\,e^{-0.1\lambda}(0.1\lambda + 1)$.

RQD classes

Importance

• It is a quick, cheap index of rock quality from boreholes.
• It is one of the parameters of the RMR system (rating 3 to 20) and of the Q-system ($RQD/J_n$ gives block size).
• It helps in choosing tunnel support, estimating rock mass strength and deformability, and deciding grouting and excavation methods.
• Limitation: it ignores joint orientation, condition and filling.

The Q-system was developed by Barton, Lien and Lunde (1974) at the Norwegian Geotechnical Institute from many tunnel case histories. It gives the rock tunnelling quality index Q, used to classify rock mass and to design tunnel support (shotcrete, rock bolts, spans).

Calculation of Q

$$Q = \frac{RQD}{J_n} \times \frac{J_r}{J_a} \times \frac{J_w}{SRF}$$

The three quotients mean: $RQD/J_n$ = relative block size, $J_r/J_a$ = inter-block shear strength, $J_w/SRF$ = active stress. Q varies from 0.001 to 1000.

Example: $RQD = 75,\ J_n = 9,\ J_r = 3,\ J_a = 1,\ J_w = 1,\ SRF = 1$ gives $Q = (75/9)(3/1)(1/1) = 25$, i.e. good rock.

Rock mass classes

The support is then chosen from the Q chart using the equivalent dimension $D_e = \text{span or height} / ESR$, where ESR is the excavation support ratio.

Intact rock is the unfractured block of rock between discontinuities, small enough to be tested as a laboratory core specimen. Rock mass is the in-situ rock together with all its discontinuities (joints, bedding, faults, shear zones), i.e. the intact rock blocks plus the structural planes between them.

Class II is good rock with RMR 61 to 80 (cohesion 300 to 400 kPa, friction angle 35° to 45°, stand-up time about 1 year for a 10 m span). Guidelines for a 10 m span tunnel:

• Excavation: full face, advance of 1.0 to 1.5 m; complete support about 20 m from the face.
• Rock bolts: locally in the crown, 3 m long, spaced 2.5 m, with occasional wire mesh.
• Shotcrete: 50 mm in the crown where required.
• Steel sets: not required.

The rock is mostly self-supporting, so only local support for loose blocks is needed.

Fair rock is Class III with RMR 41 to 60 (cohesion 200 to 300 kPa, friction angle 25° to 35°, stand-up time about 1 week for a 5 m span). The guideline (Bieniawski, 1989, for a 10 m span horse-shoe tunnel, drill and blast) is:

Excavation

• Top heading and bench method.
• Advance of 1.5 to 3 m in the top heading.
• Support is started after each blast, and the support is completed within 10 m of the face.

Support

   Top heading   ____________
                /  bolts 4 m \
   Bench       |  mesh+shotcr |
               |______________|

The bolts and shotcrete hold the loosened blocks together and let the rock mass carry itself. Local support is increased where water or weak seams occur.

Class IV is poor rock with RMR 21 to 40 (cohesion 100 to 200 kPa, friction angle 15° to 25°, stand-up time about 10 hours for a 2.5 m span).

Excavation method

• Top heading and bench.
• Advance of 1.0 to 1.5 m in the top heading.
• Install support concurrently with excavation, and complete it within 10 m of the face.

Support system (10 m span tunnel)

        ___________________
      /  shotcrete 100-150  \
     |  bolts 4-5 m @1-1.5 m |
     |   light-medium ribs   |
     |_______________________|

Because the rock stands only a few hours, short rounds and immediate shotcrete are needed, and the invert may need support if the floor is weak.

The Geomechanics Classification is the Rock Mass Rating (RMR) system of Bieniawski (1973, updated 1989). It classifies a rock mass by adding ratings of six parameters, so that engineers can estimate stand-up time, shear strength and the support needed.

Parameters

1. Uniaxial compressive strength of intact rock (0 to 15)
2. RQD (3 to 20)
3. Spacing of discontinuities (5 to 20)
4. Condition of discontinuities: persistence, aperture, roughness, infilling, weathering (0 to 30)
5. Groundwater (0 to 15)
6. Adjustment for orientation of discontinuities (0 to -12)

$RMR = \sum$ ratings (0 to 100).

Classes

Each class has cohesion (above 400 to below 100 kPa) and friction angle (above 45° to below 15°) and a recommended excavation and support (bolts, shotcrete, steel ribs).

Other rock mass classification systems

Stereographic projection is a method of representing three-dimensional orientations (planes and lines) on a two-dimensional circle. A reference sphere is cut by the plane or line through its centre, and the point or trace where it meets the lower hemisphere is projected onto the horizontal equatorial plane. A plane appears as a great circle, its normal as a pole, and a line as a point. The equal-angle (Wulff) and equal-area (Schmidt) nets are used.

Uses in engineering geology

1. Slope stability: kinematic analysis of plane, wedge and toppling failure; selecting safe cut-slope angle and direction (Markland test, friction circle).
2. Joint analysis: plotting poles and contouring them to find joint sets and their mean orientation (rose and contour diagrams).
3. Structural geology: finding true dip from two apparent dips, apparent dip in any direction, angle between planes or lines, and intersection of two planes (line of intersection and its plunge).
4. Folds and faults: locating fold axis (π and β diagrams), axial plane and plunge; analysing fault slip direction.
5. Tunnels and underground openings: deciding tunnel axis relative to joint sets, and identifying unstable roof wedges.
6. Dams, bridges and foundations: checking stability of abutments and foundation rock against sliding along discontinuities.
7. Rock excavation and quarrying: planning orientation of faces and blasting.
8. Rotation of structures: restoring beds to original position (tilt correction) to interpret the geological history.

Properties of rock mass

1. Physical: density, porosity, water content, permeability.
2. Mechanical (strength and deformation): compressive, tensile and shear strength, modulus of deformation, Poisson's ratio, cohesion and friction angle.
3. Lithological/petrographic: rock type, mineral composition, grain size, texture, weathering and alteration.
4. Structural: discontinuities, folds, faults and their geometry.
5. Hydrogeological: groundwater level, seepage, secondary permeability.
6. In-situ stress: magnitude and direction of stresses.
7. Time-dependent behaviour: creep, swelling, slaking, stand-up time.

Properties of discontinuities (ISRM) used for classification

The RMR system is one of the most widely used tools for design of underground openings (tunnels, caverns, mines, shafts) because it turns field observations into numbers that can be used directly in design.

Role / importance

1. Quality assessment: divides the rock into five classes (very good to very poor), so the tunnel can be divided into sections of similar behaviour (geotechnical zoning).
2. Stand-up time: gives the unsupported span and time the rock will stand, which decides how soon support must be installed.
3. Support design: the class gives guidelines for rock bolts (length, spacing), shotcrete thickness, steel ribs and wire mesh.
4. Excavation method: suggests full-face, top heading and bench, or multiple drifts and the length of advance per round.
5. Strength parameters: cohesion and friction angle for each class, and the modulus of deformation $E = 2\,RMR - 100$ (GPa, for $RMR > 50$) for numerical models.
6. Orientation adjustment: the tunnel axis direction relative to joints (favourable to very unfavourable) can reduce RMR by up to 12, which helps choose the best alignment.
7. Cost estimate and planning: quantities of support and construction time can be forecast; it is also used at the face for quick checks during construction (updating the design).
8. Communication: a simple common language between geologists, designers and contractors.

It is used together with the Q-system for cross-checking, and the empirical relation $RMR = 9\ln Q + 44$ links the two.

Plane failure is the sliding of a rock block along a single plane of weakness (joint, bedding or foliation) that daylights in the slope face. According to Hoek and Bray, it occurs when all these conditions are satisfied:

1. Strike: the plane strikes parallel, or nearly parallel (within about $\pm 20^\circ$), to the slope face. That is, the dip directions of the plane and the slope are within $20^\circ$.
2. Daylighting: the plane dips less steeply than the slope face, $\psi_p < \psi_f$, so that the plane is exposed in the face.
3. Friction: the dip of the plane is greater than its friction angle, $\psi_p > \phi$, so that the block can slide under gravity.
4. Release surfaces: the sliding block must be free to move laterally, for example by lateral joints with negligible resistance or by the slope face ending (or a tension crack at the top).

   crest   tension crack
      \     |
       \    |__
   face \     \ \   sliding plane
         \     \ \ (dip psi_p)
          \_____\_\___

Stereonet test: the pole of the plane falls inside the daylight envelope (between the slope great circle and the friction circle), i.e. the great circle of the plane lies between the slope circle and the friction circle (the shaded zone). If $\psi_p \leq \phi$ the slope is stable.

Method: the top of the limestone is at 200 m, 220 m and 240 m depth in A, B and C. All three boreholes are vertical and the bed is 60 m thick in each (260 - 200 = 280 - 220 = 300 - 240 = 60 m), so 60 m is the vertical thickness. First find the attitude of the bed (three-point problem), then convert vertical thickness to true thickness.

Step 1: coordinates (B at origin, x east, y north)

$$\begin{aligned} A &: x = 600\sin 28^\circ = 281.68\ \text{m},\ y = 600\cos 28^\circ = 529.77\ \text{m} \\ C &: x = 400\sin 190^\circ = -69.46\ \text{m},\ y = 400\cos 190^\circ = -393.92\ \text{m} \end{aligned}$$

Step 2: plane of the top of the bed

Depth to the top: $d = 220 + a x + b y$. Substituting A and C:

$$\begin{aligned} 200 &= 220 + 281.68a + 529.77b \\ 240 &= 220 - 69.46a - 393.92b \end{aligned}$$

Solving gives $a = 0.0366$ and $b = -0.0572$ (metres of depth per metre).

Step 3: dip and dip direction

$$\tan\delta = \sqrt{a^2+b^2} = 0.0680 \Rightarrow \delta = 3.89^\circ$$

The bed becomes deeper towards the direction $(a,b)$, so the dip direction is $\arctan(a/b)$ measured from north = $147.38^\circ$ (toward SSE). The strike is perpendicular to this: $57.38^\circ$, i.e. N57.4°E.

Attitude of the bed: N57.4°E / 3.9° towards SSE (dip direction 147°).

Step 4: true thickness

$$t = t_v\cos\delta = 60\cos 3.89^\circ = 59.86\ \text{m}$$

(The bed is almost flat, so the true thickness is only slightly less than 60 m.)

Answer: true thickness of the limestone = 59.86 m (about 59.9 m); bed dips 3.9° towards SSE.

Poor rock is Class IV (RMR 21 to 40, cohesion 100 to 200 kPa, friction angle 15° to 25°, stand-up time about 10 hours for 2.5 m span).

Support system (10 m span tunnel)

• Rock bolts: systematic, 4 to 5 m long, spaced 1 to 1.5 m in crown and walls, with wire mesh.
• Shotcrete: 100 to 150 mm in the crown and 100 mm in the sides.
• Steel ribs: light to medium ribs at 1.5 m spacing where required.

Excavation method

• Top heading and bench.
• Advance of 1.0 to 1.5 m in the top heading.
• Support installed together with excavation and completed within 10 m of the face.

Short rounds and quick support are needed because the rock deteriorates fast after excavation.

A bed is a plane, so its level changes uniformly in plan: $z = z_B + a x + b y$, where $a$ and $b$ are the rates of change of level per metre east and north. This is the analytical form of the three-point (strike-line) method.

Data and coordinates (B at origin, x east, y north)

A = (0, 900), C = (-800, 0), B = (0, 0). Levels of the top: A = -350 m, B = -310 m, C = -390 m. The layer is 60 m thick in each hole (-350 to -410, -310 to -370, -390 to -450), so $t_v = 60$ m.

Gradients of the top surface

$$\begin{aligned} b &= \frac{z_A - z_B}{900} = \frac{-350 - (-310)}{900} = -0.0444 \\ a &= \frac{z_B - z_C}{800} = \frac{-310 - (-390)}{800} = 0.1000 \end{aligned}$$

The level falls northward and rises eastward, so the bed descends towards the NW/WNW.

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.1094 \Rightarrow \delta = 6.25^\circ$$

The bed descends toward the direction $\arctan(-a/-b)$ = 294.0° (WNW), so the strike is $\ 294.0^\circ - 90^\circ = 204.0^\circ$, i.e. N24.0°E.

True thickness

All three boreholes show the same vertical thickness, 60 m, so

$$t = t_v\cos\delta = 60\cos 6.25^\circ = 59.64\ \text{m}$$

Answer: attitude of the layer = strike N24.0°E, dip 6.2° towards WNW (dip direction 294°); true thickness = 59.6 m.

A bed is a plane, so its level changes uniformly in plan: $z = z_B + a x + b y$, where $a$ and $b$ are the rates of change of level per metre east and north. This is the analytical form of the three-point (strike-line) method.

Data and coordinates (B at origin)

A = (0, 700), C = (-600, 0). Levels of the top: A = -410 m, B = -380 m, C = -430 m. The layer is 20 m thick in each hole (-410/-430, -380/-400, -430/-450), so $t_v = 20$ m.

Gradients of the top surface

$$\begin{aligned} b &= \frac{-410 - (-380)}{700} = -0.0429 \\ a &= \frac{-380 - (-430)}{600} = 0.0833 \end{aligned}$$

The level falls northward and rises eastward, so the layer dips towards the NW.

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.0937 \Rightarrow \delta = 5.35^\circ$$

The bed descends toward the direction $\arctan(-a/-b)$ = 297.2° (WNW), so the strike is $\ 297.2^\circ - 90^\circ = 207.2^\circ$, i.e. N27.2°E.

True thickness

All three boreholes show the same vertical thickness, 20 m, so

$$t = t_v\cos\delta = 20\cos 5.35^\circ = 19.91\ \text{m}$$

Answer: attitude of the layer = strike N27.2°E, dip 5.4° towards WNW (dip direction 297°); true thickness = 19.9 m.

In the RMR system each rock mass class is given shear strength parameters of the rock mass, cohesion $c$ and angle of internal friction $\phi$. They decrease as the class goes from very good (I) to very poor (V):

These values are used in the Mohr-Coulomb relation $\tau = c + \sigma\tan\phi$ for stability and support design.

According to the RMR system (Bieniawski, 1989) the classes needing steel ribs together with shotcrete are the two weakest:

Class V definitely needs steel ribs and thick shotcrete, applied immediately after blasting. For Class IV ribs are used where required (weak zones, high water or squeezing), with thick shotcrete. Classes I to III do not need steel sets.

The support type for an underground opening is selected from the quality of the rock mass, the size of the opening and the expected loads, using classification systems and site conditions.

Steps of selection

1. Investigate and classify: map the rock, estimate RQD, joint sets, strength and groundwater, and determine RMR and/or Q for each tunnel section.
2. Find the span and use: the larger the span (or equivalent dimension $D_e = \text{span}/ESR$), the heavier the support. A permanent hydropower tunnel needs more safety than a temporary adit.
3. Use the charts and tables: RMR class guidelines (bolts, shotcrete, ribs) or the Barton Q-chart.
4. Check special conditions: high stress, squeezing or swelling ground, running water, fault zones, joint orientation relative to the axis.

Typical selection (RMR)

Justification

• Good rock carries itself, so only local support is economical.
• Poor and very poor rock loosens rapidly and needs immediate and stiff support (shotcrete and ribs) to prevent collapse.
• Bolts reinforce the rock so it supports itself, shotcrete seals the surface, and ribs carry the loose rock load.
• The final choice is checked and adjusted by observation and monitoring during construction.

Wedge failure is the sliding of a wedge-shaped block formed by two intersecting discontinuity planes along their line of intersection. The conditions (Hoek and Bray) are:

1. Two planes (joints, bedding or foliation) intersect each other and also the slope face, forming a wedge with its apex at the bottom.
2. Daylighting: the line of intersection must plunge out of the slope face; the plunge direction lies within the slope face, i.e. the plunge of the line is less than the apparent dip of the slope face in that direction, $\psi_i < \psi_{f}$.
3. Friction: the plunge of the line of intersection is greater than the friction angle, $\psi_i > \phi$. Hence the condition is

$$\psi_f > \psi_i > \phi$$

4. Dip directions: the line's plunge direction must be roughly toward the slope face (within about $90^\circ$ of the slope dip direction).

        \  J1     J2  /
         \  \    /  /
          \  \  /  /
           \  \/  /  <- wedge slides along
            \ line of intersection

Stereonet test: the point where the two great circles cross (the intersection) must lie in the region between the slope great circle and the friction circle.

Toppling failure is the forward rotation (overturning) of columns or slabs of rock about a fixed base, occurring in steep slopes with closely spaced discontinuities that dip steeply into the slope.

Conditions (Goodman and Bray)

1. Discontinuities dip into the hill: the dip direction of the discontinuity set is opposite to the dip direction of the slope face, $\alpha_p \approx \alpha_f \pm 180^\circ$ (within about $10^\circ$ to $30^\circ$); the strike is nearly parallel to the slope face.
2. Steep dip: the planes dip steeply, with

$$\psi_p \geq (90^\circ - \psi_f) + \phi$$

so that interlayer slip occurs and the layers can flex and rotate. 3. Slender columns: the slabs have a thickness small compared with height, with a continuous set of cross joints that form the base of the columns. 4. Steep slope face with the toe free to move, and the centre of gravity of the block falling outside its base.

        \  |  |  |  |
         \ |  |  |  |  columns tilt
          \|  |  |  |  out of slope
           \__|__|__|__

Stereonet test: the poles of the planes plot inside the toppling zone, bounded by the great circle of the slope rotated by $(90^\circ-\phi)$, within the small circle of dip $\geq 90^\circ - \psi_f + \phi$ and at the opposite side to the slope dip direction.

Kinematic analysis checks whether the geometry of the discontinuities permits a block to move out of a rock slope, without considering forces. It is done on a stereonet for each failure mode (Markland test and others).

If a mode passes its test, the slope is kinematically unstable for that mode and a stability calculation (factor of safety) is then required. If none passes, the slope is kinematically stable.

The factor of safety (FS) of a slope is the ratio of the forces (or shear strength) resisting failure to the forces (or shear stress) driving failure along the potential failure surface.

$$FS = \frac{\text{resisting force (shear strength)}}{\text{driving force (shear stress)}}$$

For plane failure of a block of weight $W$ on a plane of dip $\psi_p$, area $A$, cohesion $c$ and friction angle $\phi$ (dry):

$$FS = \frac{cA + W\cos\psi_p\tan\phi}{W\sin\psi_p}$$

• $FS > 1$: stable; $FS = 1$: limiting equilibrium; $FS < 1$: failure.
• Design values are usually $FS \geq 1.3$ to $1.5$ for permanent slopes and about $1.2$ to $1.3$ for temporary slopes. Water and earthquake loads reduce the factor of safety.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data and assumptions

Slope face N40°W/55° dips south-west, so it is taken as 55/230 (dip/dip direction). Joint set 1 N50°W/37° dips SW: 37/220. Joint set 2 N70°E/67° dips SE: 67/160. $\phi = 28^\circ$.

Individual planes

Intersection of the joint sets

Result

• J1 strikes almost parallel to the face (10° difference), daylights ($37^\circ < 55^\circ$) and dips more than $\phi$ ($37^\circ > 28^\circ$), so plane failure along J1 is possible.
• J2 is steeper than the face ($67^\circ > 55^\circ$), so it alone cannot slide.
• J1 and J2 form a wedge whose line plunges $36.4^\circ$ towards $232^\circ$, within the face (apparent dip of face $55.0^\circ$) and above $\phi$, so wedge failure is also possible.

Conclusion: the slope is unstable (plane failure along J1 and wedge failure along J1-J2). To stabilise it, flatten the cut to below about $35^\circ$ (less than the intersection plunge $36.4^\circ$ and the J1 dip $37^\circ$), or use rock bolts, drainage and benching.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data and assumptions

Hill slope N41°W/56° dipping south-west = 56/229; discontinuity N50°W/37° dipping SW = 37/220; $\phi = 28^\circ$ (dip directions are assumed on the same side, which is the critical case).

Possible failure mode

• Dip directions differ by only $9^\circ$ (< 20°).
• Joint dip $37^\circ <$ slope $56^\circ$, so the joint daylights in the natural slope.
• Joint dip $37^\circ > \phi = 28^\circ$.

All plane-failure conditions are satisfied, so the possible mode is plane failure along the discontinuity.

Design of the cut slope

Plane failure is avoided if the cut slope does not expose the plane, i.e. the cut angle is not steeper than the dip of the discontinuity:

$$\psi_{cut} \leq \psi_p = 37^\circ$$

On the stereonet the great circle of the cut slope must plot outside (on the hill side of) the great circle of the joint, with its dip direction kept parallel to the joint.

Answer: cut the slope at about 35° (not more than 37°) with dip direction parallel to the joint (about N41°W strike). If a steeper cut (up to 56°) is needed for the road, support it with rock bolts or shotcrete and drain the slope. Dip of plane (37°) is above the friction angle (28°), so the sliding risk is real if the cut is steeper than 37°.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data and assumptions

Natural slope N48°E/52° dipping SE = 52/138; discontinuity N60°E/75° dipping SE = 75/150; $\phi = 30^\circ$.

Possible failure mode

• Dip directions differ by $12^\circ$ (< 20°), so the plane is nearly parallel to the slope.
• The plane dips more steeply ($75^\circ$) than the natural slope ($52^\circ$), so it does not daylight in the natural slope and the natural slope is stable.
• If the road cut is made steeper than $75^\circ$ the plane will daylight, and since $75^\circ > \phi = 30^\circ$ it will slide: plane failure is the possible mode.

Design of the cut slope

Keep the cut angle below the dip of the discontinuity:

$$\psi_{cut} < \psi_p = 75^\circ$$

Answer: the cut slope may be made up to a maximum of about 70° (always less than 75°), with the face striking parallel to the joint. A steeper cut will open the joint plane and cause plane failure. Use a bench or rock bolts if a cut steeper than 75° is unavoidable.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data and assumptions

Rock slope N45°W/70° dipping SW = 70/225; discontinuity N55°W/50° dipping SW = 50/215; $\phi = 28^\circ$.

Possible failure mode

• Dip directions differ by $10^\circ$ (< 20°).
• $\psi_p = 50^\circ < \psi_f = 70^\circ$: the plane daylights.
• $\psi_p = 50^\circ > \phi = 28^\circ$: sliding is possible.

So the existing slope is liable to plane failure along the discontinuity.

Design

The plane is not exposed in the cut face if the cut angle does not exceed the dip of the plane:

$$\psi_{cut} \leq 50^\circ$$

Answer: design the road cut at not more than 50° (say 45° for a margin of safety), striking parallel to the joint plane (N55°W). The existing 70° face should be cut back by about 20° or reinforced with bolts and drainage.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip direction/dip)

Hill slope HS = 170°/50°, bedding plane BP = 160°/42°, $\phi = 26^\circ$. In dip/dip direction form: HS = 50/170, BP = 42/160.

Result

• Dip directions differ by only $10^\circ$ (within $\pm 20^\circ$).
• $\psi_p = 42^\circ < \psi_f = 50^\circ$, so the bedding plane daylights in the face.
• $\psi_p = 42^\circ > \phi = 26^\circ$.

On the stereonet the great circle of BP lies between the slope great circle and the friction circle (the pole falls in the daylight envelope). Possible mode of failure: plane failure (sliding along the bedding plane). It can be controlled by reducing the slope below $42^\circ$ or by bolting and drainage.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip/dip direction)

HS 40/050; J1 46/040; J2 28/200; J3 70/103; J4 39/348; $\phi = 25^\circ$.

Single planes

Intersections

(The other 2 intersections plunge into the hill, away from the face, and cannot slide.)

Result

• J1 dips almost with the slope but is steeper ($46^\circ > 40^\circ$), so it does not daylight.
• J2 dips back into the hill at $200^\circ$ (far from the face) and is not toppling-type since it is gentle.
• J3 and J4 individually do not daylight in a sliding sense.
• The intersection J3 x J4 plunges $32.4^\circ$ towards $026^\circ$, which is between $\phi = 25^\circ$ and the face apparent dip $37.5^\circ$ and points out of the face.

Possible mode of failure: wedge failure formed by joints J3 and J4, sliding along their line of intersection. (Plane and toppling failure are not kinematically possible.)

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip direction/dip)

Hill slope 175°/52°, bedding plane 150°/48°, $\phi = 38^\circ$. In dip/dip direction: HS 52/175, BP 48/150.

Stability analysis

• Dip direction of BP is $25^\circ$ from the slope direction: just outside the usual $\pm 20^\circ$ limit.
• $\psi_p = 48^\circ < \psi_f = 52^\circ$ and the apparent dip of the face in the BP direction is $49.2^\circ$, so the bedding plane just daylights.
• $\psi_p = 48^\circ > \phi = 38^\circ$, so the block can slide if it daylights.

So the great circle of BP lies close to the edge of the safe zone: the slope is marginally stable, with a potential for plane (bedding) failure which becomes real if the toe is cut, or if water pressure or weathering lowers the friction.

Suggestions

1. Cut back the slope to $\leq 45^\circ$ (less than the bedding dip) so that bedding does not daylight.
2. Alternatively install rock bolts or anchors across the bedding and shotcrete the face.
3. Provide surface and sub-surface drainage (weep holes) to prevent water pressure.
4. Avoid undercutting the toe; provide a catch ditch or retaining wall.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip/dip direction)

HS 65/202; F 27/350; J1 81/180; J2 69/078; J3 73/021; J4 50/293. The friction angle is not given, so $\phi = 30^\circ$ is assumed (typical for rock joints).

Single planes

Intersections

(The other 8 intersections plunge into the hill, away from the face, and cannot slide.)

Result

• J3 (73/021) dips into the hill almost exactly opposite to the face (face dips 202°, opposite = 022°). Its dip $73^\circ$ is more than $90^\circ - 65^\circ + 30^\circ = 55^\circ$, so toppling failure is possible.
• J1 strikes nearly parallel to the face but dips $81^\circ$, steeper than the face, so it does not daylight.
• F (27°) dips into the hill and is below $\phi$; J2 and J4 do not daylight individually.
• The intersection J1 x J4 plunges about 45.2° towards 261°, between $\phi$ and the face apparent dip, so wedge failure is also possible.

Conclusion: the slope is unstable. The main modes are toppling along J3 and wedge sliding along the J1-J4 intersection. Reduce the slope angle, remove the potential wedge, and use bolts or anchors and drainage.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip/dip direction)

HS 65/202; F 27/350; J1 81/180; J2 69/078; J3 73/021; J4 52/293. The friction angle is not given, so $\phi = 30^\circ$ is assumed (typical for rock joints). (The last digit of F in the scan is unclear; $27^\circ$ is used. As F dips into the hill at less than $\phi$ it does not affect the result.)

Single planes

Intersections

(The other 8 intersections plunge into the hill, away from the face, and cannot slide.)

Result

• J3 (73/021) dips into the hill almost exactly opposite to the face (face dips 202°, opposite = 022°). Its dip $73^\circ$ is more than $90^\circ - 65^\circ + 30^\circ = 55^\circ$, so toppling failure is possible.
• J1 strikes nearly parallel to the face but dips $81^\circ$, steeper than the face, so it does not daylight.
• F (27°) dips into the hill and is below $\phi$; J2 and J4 do not daylight individually.
• The intersection J1 x J4 plunges about 47.1° towards 261°, between $\phi$ and the face apparent dip, so wedge failure is also possible.

Conclusion: the slope is unstable. The main modes are toppling along J3 and wedge sliding along the J1-J4 intersection. Reduce the slope angle, remove the potential wedge, and use bolts or anchors and drainage.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip/dip direction)

H (hill slope) 49/249; F 56/243; J1 58/071; J2 30/175; J3 83/286; $\phi = 25^\circ$.

Single planes

Intersections

(The other 3 intersections plunge into the hill, away from the face, and cannot slide.)

Result

• F has almost the same dip direction as the slope but is steeper ($56^\circ > 49^\circ$), so it does not daylight: no plane failure.
• J1 dips into the hill (071 against 249 - 180 = 069). The toppling limit is $90^\circ - 49^\circ + 25^\circ = 66^\circ$, but $\psi = 58^\circ <66^\circ$, so toppling is not expected.
• J3 is very steep and J2 is gentle and dips at right angles to the face: neither slides alone.
• The intersection J2 x J3 plunges $27.7^\circ$ towards $200^\circ$; this is greater than $\phi = 25^\circ$ and smaller than the apparent dip of the face ($36.9^\circ$), so it plots between the friction circle and the slope great circle.

Conclusion: the slope is stable against plane and toppling failure, but wedge failure along the intersection of J2 and J3 is possible, and should be treated by flattening the face to below about $27^\circ$ in that direction or by bolting.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip direction/dip)

Hill slope 138°/45°; J1 234°/38°; J2 098°/58°; J3 315°/60°; $\phi = 25^\circ$. In dip/dip direction: HS 45/138, J1 38/234, J2 58/098, J3 60/315.

Single planes

Intersections

(The other 2 intersections plunge into the hill, away from the face, and cannot slide.)

Interpretation

• J1 dips toward 234°, $96^\circ$ away from the face direction: it runs across the slope and cannot slide out.
• J2 is steeper ($58^\circ$) than the face ($45^\circ$): it does not daylight.
• J3 dips into the hill (315° against 138° + 180° = 318°) at $60^\circ$; toppling needs $\psi \geq 90^\circ - 45^\circ + 25^\circ = 70^\circ$, so it is below the limit.
• The only intersection that points out of the face is J1 x J2, plunging $21.3^\circ$ towards 174°. Its plunge is less than $\phi = 25^\circ$, so the wedge cannot slide.

Conclusion: the rock slope is kinematically stable against plane, wedge and toppling failure and the canal alignment is acceptable. Provide drainage and lining to prevent seepage water from raising pore pressure, and check local loosening.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip direction/dip)

HS 310/43; BP 306/31; J1 350/40; J2 288/37; J3 237/32. In dip/dip direction: HS 43/310, BP 31/306, J1 40/350, J2 37/288, J3 32/237. The friction angle is not stated, so $\phi = 30^\circ$ is assumed.

Single planes

Intersections

Result

• BP dips almost in the same direction as the face (4° difference), daylights ($31^\circ < 43^\circ$) and is just above $\phi$ ($31^\circ > 30^\circ$): plane failure along the bedding is marginally possible.
• J1 (dip dir. 40° off) and J2 (22° off) each daylight, but they are outside the $\pm 20^\circ$ range for plane failure. J3 dips across the slope at 32° (73° off).
• The J1 x J2 wedge (plunge $34.1^\circ$ towards 314°) and BP x J1 wedge (plunge $31.0^\circ$ towards 306°) lie between the friction circle and the slope great circle: wedge failure is possible.

Conclusion: the slope is marginally unstable, with bedding-plane sliding and wedge failure along J1-J2 as the possible modes. If the true friction angle is $35^\circ$ or more, all of them become stable. Remedies: reduce the slope angle to less than about $31^\circ$, or use rock bolts, benches and drainage.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip/dip direction)

HS 75/230; BP 68/045; J1 49/215; J2 55/265; J3 65/199; $\phi = 32^\circ$.

Single planes

Intersections

(The other 3 intersections plunge into the hill, away from the face, and cannot slide.)

Result

• J1 dips $49^\circ$ at 215°, only $15^\circ$ from the face direction. $32^\circ < 49^\circ < 75^\circ$: plane failure along J1.
• BP dips $68^\circ$ at 045°, opposite to the face (230° - 180° = 050°) and steeper than $90^\circ - 75^\circ + 32^\circ = 47^\circ$: toppling failure.
• J1 x J2, J2 x J3 (and marginally J1 x J3) intersect with plunges above $32^\circ$ but below the face apparent dip: wedge failure is also possible because the face is very steep.

Conclusion: the main modes are plane failure along J1 and toppling of bedding slabs (BP), with secondary wedge sliding. The slope needs to be flattened to less than about $45^\circ$ (J1 dip is $49^\circ$) and reinforced.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip direction/dip)

HS 110/40; B (bedding) 130/20; J1 100/40; J2 200/50; $\phi = 32^\circ$. In dip/dip direction: HS 40/110, B 20/130, J1 40/100, J2 50/200.

Failure checks for the existing 40° face

• B: dip direction within 20° of the slope, but $20^\circ < \phi = 32^\circ$, so it is stable.
• J1: dip direction differs by $10^\circ$ and dip $40^\circ$ equals the face angle: a steeper cut would daylight J1, and as $40^\circ > \phi$ it would slide (plane failure).
• J2: dips at $90^\circ$ from the face direction and cannot slide on its own.
• J1 x J2 intersection plunges $32.1^\circ$ towards 142°, just above $\phi$. It lies in the face if the apparent dip of the face towards 142° is greater than $32.1^\circ$, i.e. the cut angle is greater than $36.4^\circ$ (from $\tan\psi\cos 31.7^\circ = \tan 32.1^\circ$).
• B x J1 plunges $15.4^\circ$ and B x J2 $20.0^\circ$ (both below $\phi$): safe.

Design

Answer: the cut slope should be limited to about 35° (maximum 36°, below both the J1 dip of 40° and the wedge limit of 36.5°). Then plane, wedge and toppling failure are all kinematically avoided.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data (dip direction/dip)

NS (natural slope) 320/70; F 155/68; J1 240/80; J2 310/35; J3 010/85; $\phi = 30^\circ$. In dip/dip direction: NS 70/320, F 68/155, J1 80/240, J2 35/310, J3 85/010.

Single planes

Intersections

(The other 2 intersections plunge into the hill, away from the face, and cannot slide.)

Result

• J2 dips only $10^\circ$ from the face direction at $35^\circ$: $30^\circ < 35^\circ < 70^\circ$, so plane failure is possible.
• F dips into the hill at 155° (opposite direction of the face is 140°, a 15° difference) and is steep ($68^\circ \geq 90^\circ - 70^\circ + 30^\circ = 50^\circ$): toppling of slabs is possible.
• J1 and J3 are steeper than the usual daylight range and do not slide alone.
• J1 x J2 (plunge $34.3^\circ$/323°) and J2 x J3 (plunge $32.0^\circ$/283°) are between $\phi$ and the face apparent dip: wedge failure is possible.

Conclusion: the slope is unstable; plane failure on J2, wedge failure (J1-J2, J2-J3) and toppling on F can occur. Flatten the slope to less than $35^\circ$ (the J2 dip), install bolts and anchors, and provide drainage.

Method (lower-hemisphere stereonet, equal-angle): plot the great circle of the slope face and of each discontinuity, draw the friction circle of radius $\phi$, and mark intersections of great circles. Then apply the kinematic tests below.

Data and assumptions

All planes strike as given and are taken to dip on the side that appears in the figure: Hill slope N45°E/70° SE = 70/135; Bedding N55°E/45° SE = 45/145; J1 N47°W/31° SW = 31/223; J2 N77°E/36° SE = 36/167; J3 N20°E/10° SE = 10/110. The friction angle is not stated, so $\phi = 30^\circ$ is assumed.

Single planes

Intersections

Result

• Bedding dips $10^\circ$ off the face direction at $45^\circ$: $30^\circ < 45^\circ < 70^\circ$, so it daylights and can slide: plane failure.
• J3 dips only $10^\circ$, below $\phi$: stable. J1 dips across the slope (88° from the face direction).
• The Bedding x J2 intersection plunges $32.6^\circ$ towards 195°, in the face (apparent dip $53.8^\circ$) and above $\phi$: wedge failure is also possible.

Possible mode of failure: plane failure along the bedding planes (main), with wedge failure formed by bedding and J2. Reduce the cut angle below $45^\circ$ and provide bolting and drainage.

Method: the bed is a plane; its depth varies linearly with position. Take N as origin, x east, y north.

Data

• M is west of N: M = (-360, 0), depth 160 m
• N = (0, 0), depth 60 m
• P is north of N: P = (0, 360), depth 240 m
• O is north of M: O = (-360, 360)

Depth gradients

$$\begin{aligned} \text{east-west:}\quad & a = \frac{60 - 160}{360} = -0.2778\ \text{m/m (depth decreases eastward, M to N)} \\ \text{north-south:}\quad & b = \frac{240 - 60}{360} = 0.5\ \text{m/m (depth increases northward)} \end{aligned}$$

Depth $d = 60 - 0.2778\,x + 0.5\,y$ (m) - here $x,y$ in metres. Check M: $60 + 0.2778 \times 360 = 160$ m. Check P: $60 + 0.5 \times 360 = 240$ m.

Attitude

$$\tan\delta = \sqrt{0.2778^2 + 0.5^2} = 0.572 \Rightarrow \delta = 29.8^\circ$$

The bed deepens toward the direction $(-0.2778\ \text{east},\ +0.5\ \text{north})$, i.e. $\arctan(0.2778/0.5) = 29.05^\circ$ west of north: dip direction $= 330.95^\circ$ (N29°W). The strike is perpendicular: $240.95^\circ$, i.e. N61°E (to S61°W).

Attitude of the quartzite: strike N61°E, dip 29.8° towards N29°W (NNW).

Depth at O

$$d_O = 60 - 0.2778(-360) + 0.5(360) = 60 + 100 + 180 = 340\ \text{m}$$

Check using the parallelogram rule: $d_O = d_M + d_P - d_N = 160 + 240 - 60 = 340$ m.

Answer: attitude N61°E/29.8° NNW (dip direction 331°); borehole O meets the quartzite at 340 m depth.

Principle: for a plane bed and a square (or any parallelogram) layout of vertical boreholes, the depth at the fourth corner is found from the other three corners:

$$d_P = d_{\text{adjacent 1}} + d_{\text{adjacent 2}} - d_{\text{opposite}}$$

because the depth changes uniformly along each side, so the depth at P is the sum of the two gradients along the sides from the opposite corner.

The depths of the other three boreholes are not given in the question. The companion problem of the same square layout (side 300 m) is used: SW corner A = 15 m, SE corner B = 45 m, NW corner C = 60 m, with P at the NE corner.

Calculation

• Gradient along the south side (A to B): $\dfrac{45 - 15}{300} = 0.10$ m/m
• Gradient along the west side (A to C): $\dfrac{60 - 15}{300} = 0.15$ m/m
• Depth at P (NE corner): $d_P = 15 + 0.10 \times 300 + 0.15 \times 300 = 90$ m

Check: $d_P = d_B + d_C - d_A = 45 + 60 - 15 = 90$ m.

Attitude (for completeness)

$\tan\delta = \sqrt{0.10^2 + 0.15^2} = 0.1803 \Rightarrow \delta = 10.2^\circ$, dip direction $= \arctan(0.10/0.15) = 33.7^\circ$ (NE), strike $303.7^\circ$ (N56°W).

Answer: with the above data, the coal seam is met at P at 90 m depth. For any other data use $d_P = d_B + d_C - d_A$ (the sum of the two adjacent corners minus the opposite corner).

Method: the coal seam is a plane. Take A (SW corner) as origin, x east, y north. B (SE) = (300, 0), C (NW) = (0, 300).

Data

Depth gradients

$$\begin{aligned} a &= \frac{45 - 15}{300} = 0.10\ \text{(depth increases eastward)} \\ b &= \frac{60 - 15}{300} = 0.15\ \text{(depth increases northward)} \end{aligned}$$

Depth $d = 15 + 0.10\,x + 0.15\,y$.

Dip and dip direction

$$\tan\delta = \sqrt{0.10^2 + 0.15^2} = 0.1803 \Rightarrow \delta = 10.22^\circ$$ $$\text{dip direction} = \arctan\left(\frac{0.10}{0.15}\right) = 33.69^\circ\ \text{from north (NNE)}$$

The strike is perpendicular: $33.69^\circ - 90^\circ = -56.31^\circ$, i.e. N56.3°W (= 303.69°).

Answer: attitude of the coal seam = strike N56.3°W, dip 10.2° towards N33.7°E (NNE).

A bed is a plane, so its depth changes uniformly in plan: $d = d_A + a x + b y$, where $a$ and $b$ are the rates of change of level per metre east and north. This is the analytical form of the three-point (strike-line) method.

Layout (plan, north up)

   B(50) ------- A(200)      B is west of A
     |             |         C is south of B
     |             |         D is south of A
   C(100) ------ D(?)

Origin at A; x east, y north: A = (0, 0), B = (-900, 0), C = (-900, -900), D = (0, -900). Depths: A = 200 m, B = 50 m, C = 100 m.

Gradients of depth

$$\begin{aligned} a &= \frac{d_A - d_B}{900} = \frac{200 - 50}{900} = 0.1667\ \text{(deeper eastward)} \\ b &= \frac{d_B - d_C}{900} = \frac{50 - 100}{900} = -0.0556\ \text{(deeper southward)} \end{aligned}$$

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.1757 \Rightarrow \delta = 9.96^\circ$$

The bed deepens toward the direction $\arctan(a/b)$ = 108.4° (ESE), so the strike is $\ 108.4^\circ - 90^\circ = 18.4^\circ$, i.e. N18.4°E.

Depth at D

$$d_D = d_A + b(-900) = 200 + (-0.0556)(-900) = 250\ \text{m}$$

Check with the parallelogram rule: $d_D = d_A + d_C - d_B = 200 + 100 - 50 = 250$ m.

Answer: limestone attitude = N18.4°E / 10.0° towards ESE (dip direction 108°); depth at D = 250 m.

A bed is a plane, so its level changes uniformly in plan: $z = z_B + a x + b y$, where $a$ and $b$ are the rates of change of level per metre east and north. This is the analytical form of the three-point (strike-line) method.

Coordinates (B at origin)

$$\begin{aligned} A &: x = 900\sin 315^\circ = -636.40\ \text{m},\ y = 900\cos 315^\circ = 636.40\ \text{m} \\ C &: x = 700\sin 160^\circ = 239.41\ \text{m},\ y = 700\cos 160^\circ = -657.78\ \text{m} \end{aligned}$$

Levels of the top: A = -350 m, B = -310 m, C = -390 m. The bed is 60 m thick in each hole, $t_v = 60$ m.

Plane of the top surface

$z = -310 + a x + b y$:

$$\begin{aligned} -350 &= -310 - 636.40a + 636.40b \\ -390 &= -310 + 239.41a - 657.78b \end{aligned}$$

Solving: $a = 0.2900$, $b = 0.2272$.

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.3684 \Rightarrow \delta = 20.23^\circ$$

The bed descends toward the direction $\arctan(-a/-b)$ = 231.9° (SW), so the strike is $\ 231.9^\circ - 90^\circ = 141.9^\circ$, i.e. N38.1°W.

True thickness

All three boreholes show the same vertical thickness, 60 m, so

$$t = t_v\cos\delta = 60\cos 20.23^\circ = 56.30\ \text{m}$$

Answer: attitude of the sandstone = strike N38.1°W, dip 20.2° towards SW (dip direction 232°); true thickness = 56.3 m.