7.4 Karst Topography, Rock Slopes & Rockfall Hazards

7.4 Karst Topography, Rock Slopes & Rockfall Hazards

Geotechnical engineering in rock formations presents distinct hazards associated with karst terrain dissolution, rock slope instability, and rockfall hazards. Unlike soils, rock masses are strongly anisotropic, dominated by geological discontinuities (joints, bedding planes, faults). PE Civil examinees must be proficient in sinkhole genetic mechanisms, geophysical exploration, kinematic stereonet analysis, limit equilibrium stability equations for jointed rock blocks, and kinetic energy rockfall protection systems.

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1. Karst Topography & Sinkhole Mechanics

Chemical Dissolution Reactions

Karst topography develops through chemical weathering of soluble bedrock, primarily carbonate rocks (limestone, $\text{CaCO}_3$; dolomite, $\text{CaMg(CO}_3)_2$) and evaporites (gypsum, $\text{CaSO}_4\cdot 2\text{H}_2\text{O}$; rock salt, $\text{NaCl}$). Carbonate dissolution is driven by carbonic acid formed when rainwater absorbs atmospheric and soil $\text{CO}_2$:

$$\text{CaCO}_3 + \text{H}_2\text{O} + \text{CO}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{HCO}_3^-$$

This dissolution expands rock joints into underground conduits, caves, and pinnacled bedrock profiles.

Genetic Sinkhole Classifications

1. Cover-Collapse Sinkhole: The most catastrophic hazard. Cohesive overburden clay bridges across a bedrock void. Percolating water erodes the roof internally, forming a dome-shaped cavity that migrates upward until the soil roof collapses abruptly into the bedrock cavern.
2. Cover-Subsidence Sinkhole: Occurs in non-cohesive sandy overburden. Sand continuously pipes downward into small bedrock fissures, producing gradual ground surface depressions.
3. Solution Sinkhole: Bedrock exposed directly at the ground surface dissolves uniformly along joints, creating shallow bowl-shaped depressions.

Geophysical Void Exploration & Remediation

• Electrical Resistivity Tomography (ERT): Identifies air-filled cavities (very high resistivity) or clay-filled voids (very low resistivity).
• Ground Penetrating Radar (GPR): Maps shallow soil piping and shallow void roofs ($< 5 - 10\text{ m}$ depth).
• Microgravity Surveys: Detects localized subsurface mass deficits caused by deep caverns.
• Remediation Techniques: Low-slump compaction grouting ($50\text{ to }100\text{ mm}$ slump under high pressure) to densify soil, cap grouting across bedrock openings, and high-capacity micropiles drilled into sound bedrock.

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2. Rock Slope Failure Modes & Kinematic Criteria

Stability of rock slopes is dictated by joint orientation relative to the slope face. Kinematic analysis uses stereographic projections (stereonets) to determine feasible failure modes:

                      ROCK SLOPE FAILURE MODES
 ┌──────────────────┬──────────────────┬──────────────────┬──────────────────┐
 │  Planar Failure  │  Wedge Failure   │ Toppling Failure │ Circular Failure │
 ├──────────────────┼──────────────────┼──────────────────┼──────────────────┤
 │ Single continuous│ Intersection line│ Steep joints dip │ Heavily jointed  │
 │ joint plane dips │ of 2 joint planes│ INTO slope face; │ rock mass or soil│
 │ OUT of slope face│ plunges OUT face │ layers tilt out  │ like weak rock   │
 └──────────────────┴──────────────────┴──────────────────┴──────────────────┘

Kinematic Conditions for Failure

• Planar Failure Kinematic Criteria:
  1. The strike of the planar joint must be within $\pm 20^\circ$ of the slope face strike.
  2. The joint dip ($\psi_p$) must be less than the slope face dip ($\psi_f$) (joint daylighting): $\psi_p < \psi_f$.
  3. The joint dip must exceed the friction angle of the joint surface ($\phi'$): $\psi_p > \phi'$.
  $$\psi_f > \psi_p > \phi'$$
• Wedge Failure Kinematic Criteria: The plunge of the intersection line ($\psi_i$) of two joint planes must daylight on the slope face and exceed the joint friction angle: $$\psi_f > \psi_i > \phi'$$
• Toppling Failure Kinematic Criteria: Joints dip steeply into the slope face. Toppling initiates when the joint dip ($\psi_j$) satisfies: $$\psi_j \ge (90^\circ - \psi_f) + \phi'$$

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3. Limit Equilibrium Analysis & Joint Mechanics

Barton-Bandis Joint Shear Strength Model

For un-grouted rock discontinuities, shear strength is highly non-linear, modeled via the Barton-Bandis Equation:

$$\tau = \sigma'_n \cdot \tan \left[ JRC \cdot \log_{10}\left( \frac{JCS}{\sigma'_n} \right) + \phi_b \right]$$

where $JRC$ is Joint Roughness Coefficient ($0\text{ to }20$), $JCS$ is Joint Wall Compressive Strength, $\sigma'_n$ is effective normal stress, and $\phi_b$ is basic friction angle.

Factor of Safety Equation for Planar Rock Slope

For a rock block resting on a planar slip surface of length $A$ with joint cohesion $c'$, friction angle $\phi'$, joint uplift water force $U$, tension crack water force $V$, and tensioned rockbolt capacity $T$ inclined at angle $\theta$ below horizontal:

$$FS = \frac{c' A + \left[ W \cos\psi_p - U - V \sin\psi_p + T \sin(\theta + \psi_p) \right] \tan\phi'}{W \sin\psi_p + V \cos\psi_p - T \cos(\theta + \psi_p)}$$

                            V (Tension Crack Water)
                            │
                            ▼
                   ┌──────────────┐
                   │   ROCK       │
                   │   BLOCK      │ ══► T (Rockbolt Tension)
                   │   (Weight W) │
                   └──────────────┘
         ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  <-- Joint Plane (Dip ψp)
                   ▲
                   │
                   U (Uplift Water Pressure)

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4. Rockfall Mechanics & Hazard Mitigation

Kinetic Energy Dynamics

Rockfall hazard intensity is dictated by total kinetic energy ($E_k$), combining translational and rotational components:

$$E_k = E_{trans} + E_{rot} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$

where $m$ is rock mass, $v$ is translational velocity, $I$ is mass moment of inertia, and $\omega$ is rotational velocity.

Protection Mitigation Systems

1. High-Capacity Flexible Catch Fences: Dynamic net barriers rated from $500\text{ to }10,000\text{ kJ}$ impact capacity featuring energy-dissipating friction rings.
2. Draped High-Tensile Steel Wire Mesh: Wire mesh (e.g., Tecco mesh) anchored at the crest, controlling rockfall trajectories and guiding detached rocks safely into a ditch.
3. Catchment Ditches & Berms (Ritchie Criteria): Ditch depth ($D_{ditch}$) and width ($W_{ditch}$) sized as a function of slope height ($H_{slope}$) and slope angle to contain bouncing rocks.
4. Pattern Rock Bolting & Shotcrete: Active fully-grouted rock bolts preventing initial block detachment.

5. Comprehensive Worked Engineering Example

Problem Statement

A highway rock cut is excavated in sandstone at a slope face angle $\psi_f = 60^\circ$. Geological mapping identifies a continuous bedding plane discontinuity dipping out of the slope face at $\psi_p = 36^\circ$. A vertical tension crack is present at the slope crest. Geotechnical data per unit width ($1.0\text{ m}$) are as follows:

• Rock block weight $W = 900.0\text{ kN/m}$
• Joint plane base area $A = 14.0\text{ m}^2/\text{m}$
• Joint cohesion $c' = 25.0\text{ kPa}$
• Joint friction angle $\phi' = 30.0^\circ$
• Calculated joint uplift water force $U = 180.0\text{ kN/m}$
• Calculated tension crack water force $V = 50.0\text{ kN/m}$

Required:

1. Calculate the unreinforced Factor of Safety ($FS_{unreinforced}$) against planar sliding along the joint.
2. Determine if the slope meets the required design stability standard of $FS \ge 1.50$.
3. Calculate the required rockbolt tension capacity ($T$) per meter of wall width to achieve $FS = 1.50$, assuming pattern rock bolts are installed at an inclination $\theta = 14^\circ$ below horizontal.

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Step-by-Step Calculation Solution

Step 1: Unreinforced Factor of Safety Calculation

Compute statutory geometric trigonometric values:

• $\sin(36^\circ) = 0.58779$
• $\cos(36^\circ) = 0.80902$
• $\tan(30^\circ) = 0.57735$

Calculate total driving force ($D$) down the slip plane:

$$D = W \sin\psi_p + V \cos\psi_p$$ $$D = (900.0 \times 0.58779) + (50.0 \times 0.80902) = 529.01 + 40.45 = 569.46\text{ kN/m}$$

Calculate net effective normal force ($N'$) acting perpendicular to the slip plane:

$$N' = W \cos\psi_p - U - V \sin\psi_p$$ $$N' = (900.0 \times 0.80902) - 180.0 - (50.0 \times 0.58779)$$ $$N' = 728.12 - 180.0 - 29.39 = 518.73\text{ kN/m}$$

Calculate available shear resistance ($R$) along the slip plane:

$$R = c' A + N' \tan\phi'$$ $$R = (25.0\text{ kPa} \times 14.0\text{ m}^2/\text{m}) + (518.73\text{ kN/m} \times 0.57735)$$ $$R = 350.0 + 299.49 = 649.49\text{ kN/m}$$

Calculate Unreinforced Factor of Safety ($FS_{unreinforced}$):

$$FS_{unreinforced} = \frac{R}{D} = \frac{649.49\text{ kN/m}}{569.46\text{ kN/m}} = 1.140$$

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Step 2: Evaluation Against Design Standard

Since $FS_{unreinforced} = 1.140 < 1.50$, the unreinforced rock cut is unstable and presents an unacceptable hazard for highway operations. Structural rockbolt reinforcement is required.

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Step 3: Required Rockbolt Tension Capacity ($T$) for $FS = 1.50$

Rockbolts are inclined at $\theta = 14^\circ$ below horizontal. Angle between rockbolt force $T$ and failure plane: $\theta + \psi_p = 14^\circ + 36^\circ = 50^\circ$.

• $\sin(50^\circ) = 0.76604$
• $\cos(50^\circ) = 0.64279$

Set up the reinforced Factor of Safety equation with $FS = 1.50$:

$$FS = 1.50 = \frac{R_{unreinforced} + T \sin(\theta + \psi_p) \tan\phi'}{D_{unreinforced} - T \cos(\theta + \psi_p)}$$ $$1.50 = \frac{649.49 + T (0.76604 \times 0.57735)}{569.46 - T (0.64279)}$$ $$1.50 = \frac{649.49 + 0.44227 T}{569.46 - 0.64279 T}$$

Multiply both sides by $(569.46 - 0.64279 T)$:

$$1.50 \times (569.46 - 0.64279 T) = 649.49 + 0.44227 T$$ $$854.19 - 0.96419 T = 649.49 + 0.44227 T$$

Rearrange terms to isolate $T$:

$$854.19 - 649.49 = 0.44227 T + 0.96419 T$$ $$204.70 = 1.40646 T$$ $$T = \frac{204.70}{1.40646} = 145.54\text{ kN/m}$$

Engineering Conclusion: Installing tensioned rock bolts providing a minimum structural capacity $T = 146\text{ kN/m}$ inclined at $14^\circ$ below horizontal increases the planar rock slope Factor of Safety to the target design standard of $FS = 1.50$.