BE Civil Engineering (IOE, TU) Engineering Geology II (IOE, CE 601) Question Paper 2079 Nepal

(a) Objectives and sequence of dam site investigation

Objectives:

• Determine the depth, type and engineering properties of foundation rock/soil.
• Locate weak zones, shear/fault zones, karst cavities and the depth to sound bedrock for the dam seat.
• Assess permeability for seepage/grouting design and abutment stability.
• Provide data for bearing capacity, settlement and excavation quantities.

Typical sequence: desk study (maps, air photos, seismicity records) → reconnaissance/geological mapping → geophysical surveys (indirect) → exploratory drilling, pitting and adits (direct) → in-situ and laboratory testing → monitoring/instrumentation.

Direct methods: trial pits, trenches, exploratory drifts/adits, rotary core drilling, in-situ tests (plate load, permeability/Lugeon tests).

Indirect (geophysical) methods: seismic refraction, electrical resistivity, gravity and ground-penetrating radar.

(b) Core Recovery and RQD

Core Recovery (CR):

$$CR = \frac{\text{Total length of core recovered}}{\text{Total length of core run}}\times100 = \frac{155}{200}\times100 = \mathbf{77.5\%}$$

RQD counts only intact sound pieces $\ge 10\,\text{cm}$. From the list, pieces $\ge 10$ cm are: $24,\ 11,\ 18,\ 32,\ 14,\ 21,\ 13$.

Sum of these pieces:

$$24+11+18+32+14+21+13 = 133\,\text{cm}$$ $$RQD = \frac{\sum(\text{pieces}\ge10\,\text{cm})}{\text{Total core run length}}\times100 = \frac{133}{200}\times100 = \mathbf{66.5\%}$$

Classification (Deere's RQD chart):

With $RQD = 66.5\%$, the rock-mass quality is FAIR.

(a) The RMR system

Bieniawski's Geomechanics Classification (RMR) rates a rock mass on five basic parameters, each assigned a rating; their sum is the basic RMR (0–100):

1. Uniaxial compressive strength of intact rock.
2. Rock Quality Designation (RQD).
3. Spacing of discontinuities.
4. Condition of discontinuities (roughness, aperture, infill, weathering, persistence).
5. Groundwater conditions.

A sixth adjustment is then applied for the orientation of discontinuities relative to the engineering work (tunnel/foundation/slope). Favourable orientation gives small/zero deduction; unfavourable gives a larger negative correction. The adjusted RMR governs support design, stand-up time and shear-strength estimates.

(b) Calculation

Basic RMR = sum of the five basic ratings:

$$RMR_{basic} = 12 + 13 + 10 + 25 + 10 = \mathbf{70}$$

Adjusted RMR = basic RMR + orientation adjustment $(-5)$:

$$RMR_{adj} = 70 + (-5) = \mathbf{65}$$

Classification (RMR 61–80 = Class II = Good rock):

With $RMR_{adj}=65$ the rock mass is Class II – Good rock.

Engineering guidance for Class II:

• Stand-up time: of the order of 6 months for a ~8 m span (good, mostly self-supporting).
• Cohesion of rock mass: $c \approx 300\text{–}400\,\text{kPa}$.
• Friction angle: $\phi \approx 35^\circ\text{–}45^\circ$.
• Support: generally locally bolted crown with occasional mesh/shotcrete; light support.

(a) Landslide classification

By type of movement (after Varnes): falls, topples, slides (rotational / translational), lateral spreads, flows (debris flow, earthflow, creep), and complex/compound movements.

By type of material: rock, debris (coarse + soil mix), and earth (predominantly fine soil). Combining the two gives names such as rock fall, debris flow, earth slide.

Causal factors (any four): increase in pore-water pressure (heavy/prolonged rainfall), toe erosion or undercutting (river/road cut), seismic shaking, slope over-steepening or surcharge loading, weathering and reduction of shear strength, unfavourable discontinuity orientation, deforestation.

(b) Factor of Safety – planar sliding (dry)

For a dry plane the limit-equilibrium FoS is:

$$FoS = \frac{c\,A + W\cos\psi_p \tan\phi}{W\sin\psi_p}$$

Resisting force components:

• Cohesive resistance: $c\,A = 25\,\text{kPa}\times 26\,\text{m} = 650\,\text{kN/m}$
• Normal force on plane: $N = W\cos\psi_p = 4500\times\cos35^\circ = 4500\times0.8192 = 3686.3\,\text{kN}$
• Frictional resistance: $N\tan\phi = 3686.3\times\tan30^\circ = 3686.3\times0.5774 = 2128.4\,\text{kN}$

Total resisting force:

$$R = 650 + 2128.4 = 2778.4\,\text{kN}$$

Driving (shear) force:

$$T = W\sin\psi_p = 4500\times\sin35^\circ = 4500\times0.5736 = 2581.2\,\text{kN}$$

Factor of Safety:

$$FoS = \frac{2778.4}{2581.2} = \mathbf{1.076}$$

Comment: $FoS \approx 1.08$ is only marginally above unity. The slope is technically stable but unsafe — well below the usual design requirement of $FoS \ge 1.5$ for permanent rock slopes. Any water pressure on the joint or seismic loading would drop it below 1.0, so stabilisation (rock bolts/anchors, drainage, re-grading) is required.

(a) Geological problems in tunnelling

1. Squeezing / swelling ground — weak, overstressed rock or clay (e.g. phyllite, montmorillonite shale) deforms inward with time; needs heavy yielding/flexible support, invert closure.
2. Running / flowing ground — cohesionless sand or crushed fault gouge below water table flows into the heading; needs forepoling, spiling, grouting or shields.
3. High water inflow — permeable, jointed or karstic rock gives sudden inrush; needs probe drilling ahead, pre-grouting, drainage.
4. Fault and shear zones — sheared, low-strength material with possible loose blocks; demands close ring support and careful blasting.
5. Rock burst / high in-situ stress — brittle strong rock under deep cover fails violently; controlled by stress-relief, bolting, sequencing.
6. Gas and high temperature — methane/CO₂ or geothermal heat in deep tunnels; needs ventilation and monitoring.

Each problem dictates the excavation method (drill-blast vs TBM), round length, and the type and timing of support.

(b) Barton's Q-system calculations

(i) Maximum unsupported (safe) span:

$$\text{Max span} = 2\,(ESR)\,Q^{0.4} = 2\times1.6\times(4.5)^{0.4}$$

$Q^{0.4} = 4.5^{0.4}$. Compute: $\ln4.5 = 1.5041$, $\times0.4 = 0.6016$, $e^{0.6016}=1.825$.

$$\text{Max span} = 2\times1.6\times1.825 = \mathbf{5.84\,\text{m}}$$

(ii) Permanent roof support pressure (Barton):

$$P_{roof} = \frac{2}{J_r}\,Q^{-1/3} = \frac{2}{1.5}\times(4.5)^{-1/3}$$

$4.5^{1/3}$: $\ln4.5=1.5041$, $/3=0.5014$, $e^{0.5014}=1.651$, so $4.5^{-1/3}=1/1.651=0.6057$.

$$P_{roof} = 1.3333\times0.6057 = \mathbf{0.808\,\text{kg/cm}^2}\ (\approx 79\,\text{kPa})$$

(iii) Self-supporting check: the required span is $B=6\,\text{m}$ but the maximum unsupported span is only $5.84\,\text{m}$. Since $6\,\text{m} > 5.84\,\text{m}$, the tunnel is NOT self-supporting — permanent support (systematic bolting with ~0.8 kg/cm² capacity, e.g. bolts plus shotcrete) is required for the crown.

(a) Seismotectonic setting of Nepal

Nepal lies on the Himalayan collision belt where the Indian Plate underthrusts the Eurasian Plate at ~4–5 cm/yr along the Main Himalayan Thrust (MHT), the basal décollement. Surface expressions, from south to north, are the:

• Main Frontal Thrust (MFT) — youngest, active frontal fault.
• Main Boundary Thrust (MBT) — separates Siwaliks from Lesser Himalaya.
• Main Central Thrust (MCT) — separates Lesser from Higher Himalaya.

Continuous strain accumulation along the locked MHT and its periodic release in great earthquakes (e.g. 1934 Bihar–Nepal, 2015 Gorkha $M_w$ 7.8) put Nepal in a high seismic-hazard zone. Young, weak, fractured Himalayan rocks and steep terrain amplify shaking and trigger co-seismic landslides.

(b) Distance and magnitude

(i) S–P relation. Travel times for the same path of length $D$:

$$t_p = \frac{D}{V_p},\qquad t_s = \frac{D}{V_s}$$ $$t_{SP} = t_s - t_p = D\left(\frac{1}{V_s} - \frac{1}{V_p}\right)$$

Solving for $D$:

$$D = \frac{t_{SP}}{\dfrac{1}{V_s} - \dfrac{1}{V_p}} = \frac{t_{SP}\,V_p V_s}{V_p - V_s}$$

Substitute $t_{SP}=12.5\,\text{s}$, $V_p=6.0$, $V_s=3.5\,\text{km/s}$:

$$D = \frac{12.5\times6.0\times3.5}{6.0-3.5} = \frac{12.5\times21}{2.5} = \frac{262.5}{2.5} = \mathbf{105\,\text{km}}$$

(ii) Richter local magnitude. With trace amplitude in mm converted from the Wood–Anderson reference:

$$M_L = \log_{10}A - \log_{10}A_0$$

Here $A = 8\,\text{mm}$, so $\log_{10}A = \log_{10}8 = 0.903$, and $\log_{10}A_0 = -2.8$:

$$M_L = 0.903 - (-2.8) = 0.903 + 2.8 = \mathbf{3.70}$$

So the epicentral distance is about 105 km and the local magnitude is about $M_L \approx 3.7$ (a minor earthquake).

(a) Darcy's law and definitions

Darcy's law: the specific discharge (Darcy velocity) through a saturated porous medium is proportional to the hydraulic gradient:

$$q = K\,i,\qquad Q = K\,i\,A$$

where $K$ = hydraulic conductivity, $i = dh/dl$ = hydraulic gradient, $A$ = cross-sectional area.

• Hydraulic conductivity ($K$): the ease with which water flows through the medium, i.e. discharge per unit area per unit gradient (units m/day). It depends on both the fluid and the medium.
• Transmissivity ($T$): the rate at which water is transmitted through the full saturated thickness of an aquifer per unit width per unit gradient; $T = K\,b$ (units m²/day).

(b) Calculations

Transmissivity:

$$T = K\,b = 25\,\text{m/day}\times18\,\text{m} = \mathbf{450\,\text{m}^2/\text{day}}$$

Discharge per metre width (take width $w = 1\,\text{m}$, so flow area $A = b\times w = 18\,\text{m}^2$):

$$Q = K\,i\,A = 25\times0.004\times18 = \mathbf{1.8\,\text{m}^3/\text{day per m}}$$

Equivalently $Q = T\,i = 450\times0.004 = 1.8\,\text{m}^3/\text{day per m width}$, confirming the result.

Geological considerations for a dam and reservoir site

1. Foundation suitability

• The foundation must safely carry the dam load without excessive settlement, sliding or crushing. Sound, unweathered, low-permeability rock (e.g. fresh granite, massive limestone with no karst) is ideal.
• Avoid thick weathered mantles, soluble rocks (gypsum, cavernous limestone), and zones of weak shale/clay that can fail in shear.

2. Reservoir watertightness (leakage)

• The reservoir rim and floor must retain water. Continuous permeable beds, open joints, faults, or buried valleys can leak water to adjacent valleys.
• Karstic terrain and dipping permeable strata daylighting downstream are particularly hazardous.

3. Abutment stability

• Valley sides must resist sliding under reservoir loading and rapid drawdown.
• Discontinuities dipping out of the slope toward the valley, or toward the downstream face, threaten abutment/slope failure.

4. Other factors: seismicity and active faults across the site, reservoir-rim landslide potential (e.g. Vajont-type), silting from upstream erosion, and availability of construction materials.

Methods to control foundation seepage

• Grout curtain: pressure grouting (cement/chemical) along a line near the heel to form a low-permeability barrier; verified by Lugeon tests.
• Cut-off trench / concrete cut-off wall keyed into impervious strata.
• Consolidation grouting of the near-surface foundation to seal open joints.
• Upstream impervious blanket to lengthen the seepage path.
• Drainage gallery and relief wells downstream of the curtain to relieve uplift pressure and collect residual seepage.

(a) Definitions

• Dip ($\psi$): the maximum angle of inclination of a planar discontinuity measured from the horizontal, in the vertical plane perpendicular to the strike.
• Dip direction: the compass bearing (azimuth) of the horizontal projection of the line of steepest descent (true dip line); it is always $90^\circ$ from strike.
• Apparent dip: the inclination of the plane measured in any vertical section not perpendicular to strike; it is always smaller than the true dip (equal only when the section is along the true-dip direction).

(b) Apparent dip calculation

The apparent dip $\alpha$ in a vertical section making angle $\beta$ with the strike is:

$$\tan\alpha = \tan\psi\,\sin\beta$$

Given true dip $\psi = 40^\circ$ and $\beta = 50^\circ$:

$$\tan\alpha = \tan40^\circ\times\sin50^\circ = 0.8391\times0.7660 = 0.6428$$ $$\alpha = \tan^{-1}(0.6428) = \mathbf{32.7^\circ}$$

As expected the apparent dip ($32.7^\circ$) is less than the true dip ($40^\circ$).

Landslide investigation

Investigation precedes any control: geological/geomorphological mapping of the slide, identifying scarp, flanks and toe; subsurface exploration (boreholes, test pits) to locate the slip surface; monitoring of movement (inclinometers, extensometers, survey pins) and pore pressure (piezometers); and assessment of triggering factors (rainfall, toe erosion, seismicity).

Mitigation / control measures

(i) Surface and subsurface drainage (most cost-effective — water is the main trigger)

• Surface: shaping and sealing the slope, lined catch/interceptor drains, chutes and cascades to carry runoff away quickly; sealing tension cracks.
• Subsurface: horizontal sub-drains (drainage holes), drainage galleries/adits, relief wells and trench/blanket drains to lower the water table and reduce pore pressure on the slip surface.

(ii) Slope geometry modification

• Unloading the head (excavating/benching the upper part to reduce driving weight).
• Loading the toe (counterweight berm) to increase resisting moment.
• Flattening (re-grading) the overall slope angle and benching to break long slopes.

(iii) Structural / retaining measures and bioengineering

• Structural: retaining walls, gabion walls, breast/toe walls, rock/soil anchors and ground nails, rock bolts, dowels, and dentition; check dams in gullies; rockfall nets/barriers.
• Bioengineering: turfing, brush layering, live check dams, fascines, jute/coir geotextiles and plantation of deep-rooted vegetation to bind topsoil and reduce surface erosion.

Often a combination is used — drainage first, then geometry correction, then structural/bio reinforcement.

(a) Seismic refraction — advantage and limitation

Advantage: rapid, relatively inexpensive coverage of large areas; gives depth to bedrock/water table and a measure of rock quality (rippability) from layer velocities.

Limitation: it requires velocity to increase with depth — a low-velocity (softer) layer beneath a faster layer (velocity inversion) is not detected ("hidden layer" problem); also a thin intermediate layer may be missed and depths can be ambiguous.

(b) Depth to the refractor

For a single horizontal refractor, the intercept time relates to top-layer thickness $h_1$ by:

$$t_i = \frac{2h_1\sqrt{V_2^2 - V_1^2}}{V_1 V_2}\quad\Rightarrow\quad h_1 = \frac{t_i\,V_1 V_2}{2\sqrt{V_2^2 - V_1^2}}$$

Compute the root term:

$$V_2^2 - V_1^2 = 2000^2 - 600^2 = 4\,000\,000 - 360\,000 = 3\,640\,000$$ $$\sqrt{3\,640\,000} = 1907.9\,\text{m/s}$$

Substitute:

$$h_1 = \frac{0.020\times600\times2000}{2\times1907.9} = \frac{0.020\times1\,200\,000}{3815.8} = \frac{24\,000}{3815.8}$$ $$h_1 = \mathbf{6.29\,\text{m}}$$

The top (slower) layer is about 6.3 m thick before the high-velocity refractor (bedrock).

(a) The Q-system

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

The three quotients (each a ratio) represent:

• $\dfrac{RQD}{J_n}$ — relative block size (degree of jointing / fragmentation of the rock mass).
• $\dfrac{J_r}{J_a}$ — inter-block shear strength (roughness vs. alteration/infill of the joints).
• $\dfrac{J_w}{SRF}$ — active stress (water pressure reduction vs. stress reduction factor / loosening).

(b) Q calculation

$$Q = \frac{75}{6}\times\frac{2}{1}\times\frac{0.66}{2.5}$$

Step by step:

• $\dfrac{75}{6} = 12.5$
• $\dfrac{2}{1} = 2.0$
• $\dfrac{0.66}{2.5} = 0.264$

$$Q = 12.5\times2.0\times0.264 = 25.0\times0.264 = \mathbf{6.60}$$

Category (Barton scale):

With $Q = 6.60$, the rock mass is classified as FAIR.