BE Civil Engineering (IOE, TU) Engineering Geology II (IOE, CE 601) Question Paper 2080 Nepal

(a) RMR (1989) computation

RMR is the sum of five rating parameters plus an orientation adjustment.

R1 — Strength of intact rock ($\sigma_c = 110$ MPa) For $100\text{–}250$ MPa the rating is 12.

R2 — RQD ($78\%$) For $RQD = 75\text{–}90\%$ the rating is 17.

R3 — Spacing of discontinuities ($0.45$ m) Spacing $0.2\text{–}0.6$ m gives rating 10.

R4 — Condition of discontinuities Using the detailed guidelines: slightly rough (+5... combined), aperture $<1$ mm, slightly weathered walls. The standard aggregate condition rating for slightly rough, separation $<1$ mm, slightly weathered is 25.

R5 — Groundwater (dry / completely dry) Rating 15.

$$RMR_{basic} = 12 + 17 + 10 + 25 + 15 = 79$$

Orientation adjustment (R6): Drive direction with the dominant set dipping $35^\circ$ — a dip of $20^\circ\text{–}45^\circ$ driving with dip is rated fair. For tunnels the adjustment for fair orientation is $-5$.

$$RMR = 79 - 5 = \mathbf{74}$$

Classification: $RMR = 74 \Rightarrow$ Class II — Good rock.

(b) Stand-up time and support

For Class II rock with $RMR \approx 74$, Bieniawski's stand-up time chart indicates roughly 6 months for an ~8–10 m unsupported span, i.e. the 9 m cavern roof will stand for several months unsupported.

Generic support guideline for Class II (good rock):

• Rockbolts: systematic 4 m long bolts, $1.5\text{–}2.0$ m spacing in crown and walls, with occasional wire mesh.
• Shotcrete: 50 mm in crown where required; none on walls if conditions remain dry.
• Steel sets: none.

Comment: Because the rock is competent (Class II, dry, moderate spacing), systematic rock-bolting plus a thin shotcrete layer is adequate. The unfavourable feature is the $35^\circ$ joint set daylighting toward the cavern, which can release wedges in the crown/sidewall; spot bolting keyed to mapped wedges should supplement the systematic pattern. No heavy steel support is justified.

Final answer: $RMR_{basic}=79$, adjusted $RMR=\mathbf{74}$ (Class II, Good rock); systematic bolting + thin shotcrete is sufficient, with extra wedge control for the $35^\circ$ set.

(a) Geological suitability factors

1. Foundation rock

• A gravity dam transmits large compressive and shear loads; the foundation must have adequate bearing capacity, low compressibility and high shear strength. Sound limestone is competent, but the interbedded shale layers are weak, deformable and may form low-friction sliding planes.
• Dip is $25^\circ$ upstream, which is favourable for sliding stability (bedding planes dip into the reservoir, against the downstream driving direction). Had they dipped downstream, base sliding along a shale parting would be a major concern.

2. Abutments

• Must be strong and watertight to anchor thrust. Bedding/joint orientation governs abutment wedge stability; weathered shale on abutments needs excavation to fresh rock.

3. Reservoir watertightness

• Limestone is soluble → solution cavities, sinkholes and karst conduits can drain the reservoir into adjacent valleys. Continuity of shale beds (aquicludes) can help confine seepage.

4. Reservoir-rim & slope stability

• Upstream-dipping beds on the rim are generally stable; check for daylighting joints, old landslides, and rapid-drawdown effects on shale slopes.

(b) Karst investigation and treatment

Investigation:

• Geological/geomorphic mapping of sinkholes, springs, dolines along the rim.
• Exploratory drilling with core recovery + downhole camera; log cavity intervals and lost-circulation zones.
• Water-pressure (Lugeon) tests in boreholes to quantify rock-mass permeability.
• Geophysics: Electrical Resistivity Tomography and/or microgravity to map voids; dye / tracer tests between boreholes and downstream springs to prove leakage paths.

Treatment:

• Consolidation grouting of the shallow foundation and a deep grout curtain under the dam axis and into both abutments to cut the karst conduits.
• Cavity treatment: dental concrete / backfill grouting of large voids; slush grouting of open joints.
• Foundation drainage gallery with drain holes downstream of the curtain to relieve uplift.

Final answer: Site is conditionally suitable — upstream dip favours sliding stability, but karstic limestone demands tracer-confirmed leakage mapping plus a grout curtain and cavity treatment before it is acceptable.

(a) Kinematic feasibility

Planar sliding requires:

1. The joint daylights in the face: $\psi_p < \psi_f \Rightarrow 35^\circ < 60^\circ$ ✓
2. Joint dip exceeds friction angle: $\psi_p > \phi \Rightarrow 35^\circ > 30^\circ$ ✓

Sliding is kinematically feasible.

(b) Factor of safety

Geometry — area of sliding plane (per metre run). For a slope of height $H$ with face angle $\psi_f$ and failure plane $\psi_p$ through the toe (no tension crack), the length of the failure plane is

$$L = \frac{H}{\sin\psi_p} = \frac{18}{\sin 35^\circ} = \frac{18}{0.5736} = 31.38\ \text{m}$$

Weight of the sliding wedge (per metre run). The cross-sectional area of the wedge:

$$A = \frac{1}{2}H^2\left(\cot\psi_p - \cot\psi_f\right)$$ $$\cot 35^\circ = 1.4281,\quad \cot 60^\circ = 0.5774$$ $$A = \tfrac{1}{2}(18)^2(1.4281 - 0.5774) = \tfrac{1}{2}(324)(0.8507) = 137.8\ \text{m}^2$$ $$W = \gamma A = 26 \times 137.8 = 3584\ \text{kN per m run}$$

Resolve weight on the plane.

• Normal component: $N = W\cos\psi_p = 3584 \times \cos 35^\circ = 3584 \times 0.8192 = 2936\ \text{kN}$
• Driving (shear) component: $T = W\sin\psi_p = 3584 \times \sin 35^\circ = 3584 \times 0.5736 = 2056\ \text{kN}$

Resisting force (Mohr–Coulomb, dry, $u=0$):

$$R = cL + N\tan\phi = (25 \times 31.38) + 2936 \times \tan 30^\circ$$ $$R = 784.5 + 2936 \times 0.5774 = 784.5 + 1695.4 = 2479.9\ \text{kN}$$

Factor of safety:

$$FoS = \frac{R}{T} = \frac{2479.9}{2056} = \mathbf{1.21}$$

Interpretation: $FoS \approx 1.21 > 1$, so the dry slope is marginally stable. The margin is small; introducing cleft-water pressure (saturation) would reduce $N$ and add an uplift/driving force, easily dropping $FoS$ below 1. Drainage and/or rock anchors are advisable.

(a) Q-value

Barton's Q-system:

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

Substituting:

$$Q = \frac{55}{9}\times\frac{1.5}{4}\times\frac{0.66}{5}$$

Step by step:

• $\dfrac{55}{9} = 6.111$ (relative block size)
• $\dfrac{1.5}{4} = 0.375$ (inter-block shear strength)
• $\dfrac{0.66}{5} = 0.132$ (active stress)

$$Q = 6.111 \times 0.375 \times 0.132 = 0.3025 \approx \mathbf{0.30}$$

Classification: $Q \approx 0.30$ falls in the range $0.1\text{–}1.0 \Rightarrow$ Very poor rock.

(b) Equivalent dimension and support

$$D_e = \frac{\text{Excavation span or height}}{ESR} = \frac{6.0}{1.0} = \mathbf{6.0\ m}$$

With $Q = 0.30$ and $D_e = 6.0$ m, on the Grimstad–Barton support chart this plots in the region requiring fibre-reinforced shotcrete plus systematic bolting (roughly support category 5–6).

Recommended support:

• Systematic rockbolts ~ 3 m long at 1.3–1.5 m spacing (use the bolt-spacing relation $\approx 2\sqrt{Q}\cdot\ldots$; here close, regular pattern).
• Steel-fibre-reinforced shotcrete (Sfr) 90–120 mm thick in crown and upper walls.
• Because $J_a = 4$ indicates clay-coated joints (low residual friction) and a shear zone is present ($SRF=5$), provide reinforced ribs of shotcrete (RRS) across the shear zone and ensure good drainage to counter the medium water inflow ($J_w=0.66$).

Final answer: $Q = \mathbf{0.30}$ (Very poor), $D_e = \mathbf{6.0\ m}$; support with systematic ~3 m bolts at ~1.4 m spacing and 90–120 mm fibre-reinforced shotcrete, with extra RRS reinforcement across the shear zone.

(a) Tectonic setting of the Himalaya

The Himalaya formed from the continent–continent collision of the Indian Plate with the Eurasian Plate, which began ~50 Ma ago. The Indian Plate continues to under-thrust beneath Tibet at roughly 18–20 mm/yr, and this convergence is accommodated along a system of north-dipping thrust faults that splay from a common basal décollement, the Main Himalayan Thrust (MHT).

From north to south the principal thrust systems are:

• Main Central Thrust (MCT) — separates Higher Himalaya from Lesser Himalaya.
• Main Boundary Thrust (MBT) — separates Lesser Himalaya from the Siwaliks.
• Main Frontal Thrust (MFT) — the active southernmost thrust at the Himalaya–Gangetic plain boundary.

Why large earthquakes occur: the locked portion of the MHT accumulates elastic strain from the steady convergence; periodic sudden rupture releases this strain as great earthquakes (e.g. 1934 Bihar–Nepal $M\sim8.1$, 2015 Gorkha $M_w\,7.8$). Nepal sits directly above this seismogenic interface, hence the high hazard.

(b) Magnitude vs. intensity, and design use

• Magnitude is a single, instrumental measure of the energy released at the source (e.g. moment magnitude $M_w$); one earthquake has one magnitude.
• Intensity describes the severity of shaking and damage at a particular place (e.g. Modified Mercalli scale); it varies with distance, site soil and structure, so one earthquake has many intensity values.

Use in design: A seismic hazard assessment (deterministic and probabilistic) provides the design ground acceleration / response spectrum for the site. Dams are checked against Maximum Credible Earthquake (MCE) and Operating Basis Earthquake (OBE); active-fault mapping ensures structures are not founded across a capable fault; tunnels are detailed for fault-crossing deformation; bridges use seismic-resistant detailing per the relevant code (e.g. NBC/IS provisions).

Final answer: India–Eurasia collision along the MHT (with MCT, MBT, MFT splays) drives Nepal's great earthquakes; magnitude measures source energy, intensity measures local shaking, and seismic hazard (MCE/OBE, design spectra, fault avoidance) is built into all major engineering-geological designs.

(a) RQD computation

RQD counts only sound intact pieces $\geq 10$ cm measured along the core axis:

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

Qualifying pieces ($\geq 10$ cm): 12, 25, 18, 11, 32, 21, 15, 14 cm. (The pieces 8, 9, 7, 6 cm are excluded.)

Sum of qualifying pieces:

$$12 + 25 + 18 + 11 + 32 + 21 + 15 + 14 = 148\ \text{cm} = 1.48\ \text{m}$$

Total run length $= 2.0$ m $= 200$ cm.

$$RQD = \frac{148}{200}\times 100\% = \mathbf{74\%}$$

(b) Classification and limitation

• $RQD = 74\%$ lies in the 60–75% range $\Rightarrow$ "Fair" rock quality (74% is right at the Fair/Good boundary; Good begins at 75%).
• Limitation: RQD is directional (depends on borehole orientation relative to joints) and is insensitive to joint condition — it ignores joint filling, roughness, weathering and aperture, and treats a piece just under 10 cm the same as rubble. Hence it must be supplemented by RMR/Q.

Final answer: $RQD = \mathbf{74\%}$ (Fair rock); RQD ignores joint orientation/condition and the borehole-direction bias.

(a) Lugeon value

Definition: One Lugeon (Lu) is a water take of 1 litre per minute per metre of test section at a reference injection pressure of 10 bar (1 MPa). It corresponds to a rock-mass permeability of roughly $1\times10^{-7}\ \text{m/s}$.

General formula:

$$\text{Lugeon} = \frac{Q}{L}\times\frac{P_{ref}}{P_{test}}$$

where $Q$ = water take (L/min), $L$ = test length (m), $P_{ref}=10$ bar.

Here $Q = 24$ L/min, $L = 3.0$ m, $P_{test} = 10$ bar so $P_{ref}/P_{test} = 1$:

$$\text{Lugeon} = \frac{24}{3.0}\times\frac{10}{10} = 8.0\times 1 = \mathbf{8\ Lu}$$

(b) Interpretation

• A value of 8 Lu indicates moderately permeable / moderately tight rock (typical descriptive bands: $<1$ Lu very tight, $1\text{–}5$ low, $5\text{–}15$ moderate, $>15$ high permeability).
• For a dam foundation, the usual acceptance criterion is about $\leq 3\text{–}5$ Lu beneath the core/curtain. At 8 Lu the section exceeds typical limits, so consolidation/curtain grouting is warranted to reduce seepage and uplift.

Final answer: 8 Lugeon — moderately permeable; grouting is recommended to bring the foundation below the ~3–5 Lu acceptance limit.

(a) Classification of mass movements (Varnes-type)

By type of movement and material (rock / debris / earth):

(b) Causal factors in the Nepal Himalaya + mitigation

Main causes: steep relief/active tectonic uplift, intense monsoon rainfall, weak/sheared rock & thick colluvium, earthquakes, river/toe erosion, and human activity (road cutting, deforestation).

Two with mitigation:

1. Intense rainfall → high pore pressure: install surface and subsurface drainage (catch drains, horizontal drains) to lower the water table.
2. Unplanned road cutting (toe undercutting): use engineered cut slopes with retaining/breast walls and bioengineering (vegetation) to restore support.

Final answer: mass movements are classed by movement (fall, topple, slide, spread, flow, complex) and material (rock/debris/earth); Himalayan landslides are driven mainly by monsoon rain, steep tectonic terrain, weak rock and human cutting, mitigated by drainage and engineered/bioengineered slope support.

Overbreak in tunnelling

Definition: Overbreak is the excavation of rock beyond the designed (pay-line) tunnel perimeter. It increases the actual excavated cross-section, raising concrete/shotcrete lining quantities, mucking volume and cost, and can destabilise the crown.

Geological causes:

• Unfavourable joint/bedding orientation that releases wedges and slabs at the periphery.
• Closely spaced or open discontinuities and intersecting joint sets forming loose blocks.
• Weak, weathered, sheared or blocky ground, and shear/fault zones.
• Stress-induced spalling in highly stressed brittle rock.
• Poor blasting practice (excessive charge) interacting with the jointing — though this is a construction cause amplified by geology.

Control measures (any two):

1. Controlled/smooth-wall blasting (pre-splitting, perimeter holes with reduced, decoupled charges) to give a clean profile.
2. Early support — immediate shotcrete (with steel fibre) and systematic rockbolting/forepoling to confine the rock before blocks loosen.
3. Map and bolt daylighting wedges; adjust round length and excavation sequence in poor ground.

Final answer: Overbreak is excavation past the design line, caused chiefly by adverse jointing, weak/sheared rock and blasting; it is controlled by smooth-wall blasting and prompt shotcrete + bolting.

Hydrogeological units (by ability to store and transmit water)

Confined vs. unconfined aquifers

• Unconfined (water-table) aquifer: bounded above by the water table, open to the atmosphere through permeable overburden; recharged directly by infiltration. The water level in a well equals the water table. Example: shallow river-terrace gravel aquifer.
• Confined (artesian) aquifer: bounded above and below by aquicludes/aquitards and holds water under pressure greater than atmospheric; a well penetrating it rises above the top of the aquifer (potentiometric surface), and may flow (artesian) if the head exceeds ground level. Example: a sandstone bed sandwiched between shale beds.

Final answer: aquifer (stores + transmits) > aquitard (slow) > aquiclude (stores, no transmit) > aquifuge (neither); unconfined aquifers have a free water table while confined aquifers are pressurised between impermeable layers.

Geophysical methods of subsurface exploration

Geophysical methods are indirect, non-destructive techniques that infer subsurface conditions by measuring contrasts in physical properties from the surface. They are fast and economical for covering large areas, and are normally calibrated against a few boreholes.

(i) Seismic refraction

• Principle: An energy source (hammer/explosive) generates elastic P-waves; geophones record first arrivals. Waves travel faster in denser/stiffer media, so they refract along high-velocity layers and return to the surface. From travel-time vs. distance plots the layer velocities and depths are computed (velocity generally increases with depth: e.g. soil ~300–700 m/s, weathered rock ~1500 m/s, fresh rock >3000 m/s).
• Application: Determine depth to bedrock / thickness of overburden, locate the water table, and assess rippability (ease of excavation) of rock for foundations, cuts and dam sites.

(ii) Electrical resistivity

• Principle: Current is injected through two electrodes and the potential difference is measured across two others (e.g. Wenner/Schlumberger array); the apparent resistivity is computed. Different materials have characteristic resistivities (clay/saturated soil = low; dry sand, gravel, fresh rock = high). Varying electrode spacing probes increasing depth.
• Application: Locate the water table and map aquifers, delineate clay vs. gravel layers, find buried channels, cavities/karst and the bedrock surface for site investigation.

Final answer: Geophysical surveys are rapid indirect tools calibrated by boreholes — seismic refraction (wave velocity) gives overburden thickness, bedrock depth and rippability; resistivity (current flow) maps groundwater, soil layering and cavities.