BE Civil Engineering (IOE, TU) Engineering Geology II (IOE, CE 601) Question Paper 2077 Nepal

RMR-1989 uses five basic parameters plus an orientation adjustment.

(a) Parameter ratings

Basic (unadjusted) RMR $= 7 + 17 + 10 + 25 + 10 = 69$.

Orientation adjustment: for a tunnel with favourable drive direction/dip the adjustment is $-2$.

(b) Total RMR

$$RMR = 69 + (-2) = \mathbf{67}$$

RMR 67 falls in the band 61–80.

• Class No. II
• Description: GOOD ROCK

(c) Stand-up time and support implications

For Class II rock, Bieniawski's chart gives an unsupported stand-up time of about 6 months for a span of roughly 8 m. For the 6 m tunnel span this is comfortably on the safe side, so the rock can stand unsupported for several months.

Guideline cohesion and friction of the rock mass (RMR 67):

• $c \approx 300\ \text{kPa}$ (range 300–400 kPa for Class II)
• $\phi \approx 35^\circ$–$45^\circ$

Support implications: Light support is adequate — systematic rock bolts 3 m long spaced 2.5 m in crown and walls with occasional wire mesh, plus 50 mm shotcrete in the crown where required. No steel sets are needed. Because the rock is Class II (Good), excavation can proceed by full-face or top-heading-and-bench with bolts installed 20 m from the face. Final answer: RMR = 67, Class II, Good rock, light/occasional support.

(a) Foundation and abutment conditions

A concrete gravity dam transmits large vertical and horizontal loads, so the foundation and abutments must be strong, watertight and stable.

Favourable conditions:

• Fresh, massive, high-strength bedrock (e.g. gneiss, quartzite, fresh limestone) with high bearing capacity (> 3–5 MPa).
• Low jointing/fracturing; tight, healed joints; high RQD.
• Beds dipping upstream so that sliding is resisted and the seepage path is lengthened.
• Abutments of sound rock at a suitable valley-wall angle, free of large daylighting wedges.
• Absence of active faults, shear zones and soluble rock.

Unfavourable conditions:

• Weak, weathered or crushed rock; thrust/shear zones (common along the MCT and MBT in Nepal).
• Beds dipping downstream (sliding hazard) or steeply jointed abutments forming removable wedges.
• Soluble limestone with karst, buried channels, or open joints causing leakage.
• Compressible alluvium / colluvium of unknown depth in the river bed.

Investigations: geological mapping, core drilling and permeability (Lugeon) tests, exploratory adits, seismic refraction for overburden depth, in-situ deformability tests (plate-jacking), and laboratory rock tests.

(b) Reservoir-area problems

• Leakage / watertightness: through karst, permeable beds, buried valleys or faults connecting the reservoir to adjacent valleys; assessed by Lugeon tests and treated with grout curtains.
• Siltation: Himalayan rivers carry very high sediment loads; rapid loss of storage. Controlled by catchment management, check dams, desilting basins and sediment-flushing outlets.
• Reservoir-induced seismicity (RIS): loading and pore-pressure changes can trigger movement on critically stressed faults; requires fault mapping and seismic monitoring.
• Reservoir-rim slope stability: rapid drawdown and reservoir loading can trigger landslides into the reservoir (surge-wave hazard).

(c) Two Nepal-specific problems and treatment

1. Crossing of active thrust/shear zones (MCT, MBT): sheared, weak gouge. Treatment: relocate or realign the dam axis; dental concrete and consolidation grouting of the sheared rock; deep cut-off to sound rock.
2. High sediment yield and glacial-lake outburst flood (GLOF) risk: Treatment: under-sluices/flushing gates, GLOF early-warning systems, and adequate spillway capacity with freeboard.

Final answer: A favourable site needs strong, watertight, upstream-dipping rock free of active faults; reservoir leakage, siltation and RIS must be evaluated; Himalayan-specific shear zones and high sediment/GLOF risk demand grouting, realignment and flushing measures.

(a) Kinematic check for plane failure

The three conditions for plane failure are:

1. The plane must dip in the same direction as the slope face (assumed satisfied — dips out of face). ✔
2. The plane must daylight in the face: $\psi_p < \psi_f$ → $35^\circ < 60^\circ$. ✔
3. The plane dip must exceed the friction angle: $\psi_p > \phi$ → $35^\circ > 30^\circ$. ✔

All conditions met, so plane failure is kinematically possible and a factor-of-safety check is required.

(b) Factor of safety

Weight of sliding wedge (per metre run):

$$W = \tfrac{1}{2}\gamma H^2(\cot\psi_p - \cot\psi_f)$$

$\cot 35^\circ = 1.4281,\quad \cot 60^\circ = 0.5774$

$$W = 0.5 \times 26 \times 18^2 \times (1.4281 - 0.5774)$$ $$W = 0.5 \times 26 \times 324 \times 0.8507 = 13\,\times 324 \times 0.8507$$ $$W = 4212 \times 0.8507 = 3583\ \text{kN/m}$$

Area of failure plane (per metre run):

$$A = \frac{H}{\sin\psi_p} = \frac{18}{\sin 35^\circ} = \frac{18}{0.5736} = 31.38\ \text{m}^2$$

Resolve weight on the plane (dry, no water):

• Normal force: $N = W\cos\psi_p = 3583 \times \cos 35^\circ = 3583 \times 0.8192 = 2935.3\ \text{kN/m}$
• Driving (shear) force: $T = W\sin\psi_p = 3583 \times \sin 35^\circ = 3583 \times 0.5736 = 2055.4\ \text{kN/m}$

Resisting force (Mohr–Coulomb): $R = cA + N\tan\phi$

$$R = (25 \times 31.38) + (2935.3 \times \tan 30^\circ)$$ $$R = 784.5 + (2935.3 \times 0.5774) = 784.5 + 1694.8 = 2479.3\ \text{kN/m}$$

Factor of safety:

$$FoS = \frac{R}{T} = \frac{2479.3}{2055.4} = \mathbf{1.21}$$

Interpretation: FoS = 1.21 > 1, so the slope is marginally stable under dry conditions. The margin is small; under saturated (wet) conditions water pressure would reduce $N$ and could bring FoS below 1, so drainage and/or rock bolting is advisable. Final answer: plane failure is kinematically possible; dry FoS ≈ 1.21 (marginally stable).

(a) Rock mass quality Q

Barton's equation:

$$Q = \frac{RQD}{J_n} \times \frac{J_r}{J_a} \times \frac{J_w}{SRF}$$ $$Q = \frac{70}{9} \times \frac{3}{1} \times \frac{1}{2.5}$$

Step by step:

• $\dfrac{70}{9} = 7.778$
• $\dfrac{3}{1} = 3$
• $\dfrac{1}{2.5} = 0.4$

$$Q = 7.778 \times 3 \times 0.4 = 7.778 \times 1.2 = \mathbf{9.33}$$

Q = 9.33 lies in the range 4–10, so the rock mass is classed as FAIR rock.

(b) Equivalent dimension and unsupported span

Equivalent dimension:

$$D_e = \frac{\text{Excavation span}}{ESR} = \frac{10}{1.0} = 10$$

Maximum unsupported span:

$$\text{Span}_{max} = 2\,ESR\,Q^{0.4} = 2 \times 1.0 \times 9.33^{0.4}$$ $$9.33^{0.4} = e^{0.4\ln 9.33} = e^{0.4 \times 2.2333} = e^{0.8933} = 2.443$$ $$\text{Span}_{max} = 2 \times 2.443 = \mathbf{4.89\ m}$$

(c) Support comment

The maximum unsupported span for this rock (≈ 4.9 m) is much less than the 10 m cavern span, so the excavation cannot stand unsupported and permanent support is mandatory.

From the Q support chart, with $Q = 9.33$ and $D_e = 10$, the cavern falls in a region requiring systematic rock bolting plus fibre-reinforced shotcrete. Typical design:

• Systematic grouted rock bolts ~4–5 m long, spacing ~1.5–2.0 m.
• Steel-fibre-reinforced shotcrete 50–90 mm in the crown.
• Permanent roof bolt length guide: $L = 2 + \dfrac{0.15\,B}{ESR} = 2 + \dfrac{0.15 \times 10}{1.0} = 3.5\ \text{m}$ (use ~4 m bolts for the wide cavern).

Final answer: Q = 9.33 (Fair rock); unsupported span ≈ 4.9 m < 10 m, so systematic bolting + fibre-reinforced shotcrete is required.

(a) Plate-tectonic setting and thrust systems

Nepal lies on the Himalayan collision belt, where the Indian plate underthrusts the Eurasian plate at roughly 18–20 mm/yr. The convergence is accommodated along a series of north-dipping thrusts that branch upward from a common basal décollement, the Main Himalayan Thrust (MHT):

• MFT (Main Frontal Thrust): southernmost, youngest active thrust at the foot of the Siwaliks.
• MBT (Main Boundary Thrust): separates the Siwaliks from the Lesser Himalaya.
• MCT (Main Central Thrust): separates the Lesser Himalaya from the Higher Himalaya.

Strain accumulates on the locked portion of the MHT and is released as great earthquakes. The Central/Western Nepal seismic gap is a stretch (roughly between the 1505 and 1934 rupture zones) that has not produced a great earthquake recently and stores enough strain for a future $M>8$ event; the 2015 Gorkha ($M_w$ 7.8) earthquake only partially relieved this strain.

(b) Seismic zonation, magnitude vs intensity

• Seismic zonation divides the country into zones of expected ground shaking (peak ground acceleration / seismic coefficient) used in codes (NBC 105). Higher seismic-coefficient zones demand stronger design.
• Magnitude measures the energy released at the source — a single value per event. Scales: Richter (local magnitude $M_L$) and the moment magnitude $M_w$ (preferred for large events). Magnitude is logarithmic; +1 unit ≈ 32× more energy.
• Intensity measures the effect of shaking at a place — varies with distance, soil and construction. Scale: Modified Mercalli Intensity (MMI, I–XII).

(c) Engineering geological measures

• Microzonation of cities (Kathmandu valley lacustrine soils amplify shaking and are liquefiable).
• Avoid active fault traces, steep colluvial slopes and liquefiable saturated sands for important structures.
• Site-response and liquefaction analysis; ground improvement (compaction, stone columns) where unavoidable.
• Earthquake-resistant design to NBC 105 / IS 1893, ductile detailing, base isolation for critical buildings.
• Slope stabilisation and landslide-hazard mapping along roads (co-seismic landslides were a major cause of 2015 losses).
• Lifeline redundancy, early-warning, and public preparedness.

Final answer: Nepal's seismicity is driven by Indian–Eurasian convergence on the MHT and its splays (MFT, MBT, MCT) with an unfilled central seismic gap; design must use seismic zonation, distinguish magnitude (energy, $M_w$) from intensity (effect, MMI), and apply microzonation, liquefaction control and code-based aseismic design.

(a) Definitions

Darcy's law: the specific discharge through a porous medium is proportional to the hydraulic gradient: $q = K\,i$, where $q$ = Darcy velocity, $K$ = hydraulic conductivity, $i$ = hydraulic gradient.

• Hydraulic gradient $i = \dfrac{\Delta h}{L}$ = head loss per unit length of flow path (dimensionless).
• Transmissivity $T = K\,b$ = the rate of flow through the full saturated thickness $b$ per unit width per unit gradient (units m²/s).

(b) Gradient, Darcy velocity, flow per metre width

Hydraulic gradient:

$$i = \frac{\Delta h}{L} = \frac{84.0 - 80.5}{500} = \frac{3.5}{500} = 0.007$$

Darcy velocity:

$$q = K\,i = (8\times10^{-5}) \times 0.007 = 5.6\times10^{-7}\ \text{m/s}$$

Transmissivity: $T = K b = (8\times10^{-5})\times 12 = 9.6\times10^{-4}\ \text{m}^2/\text{s}$.

Flow per metre width (full 12 m thickness):

$$Q = q \times (b \times 1) = (5.6\times10^{-7}) \times 12 = 6.72\times10^{-6}\ \text{m}^3/\text{s per m}$$

Equivalently $Q = T\,i = (9.6\times10^{-4})\times0.007 = 6.72\times10^{-6}\ \text{m}^3/\text{s per m}$. ✔

(c) Seepage (actual) velocity

$$v_s = \frac{q}{n_e} = \frac{5.6\times10^{-7}}{0.20} = 2.8\times10^{-6}\ \text{m/s}$$

Final answers: $i = 0.007$; Darcy velocity $q = 5.6\times10^{-7}\ \text{m/s}$; flow per metre width $Q = 6.72\times10^{-6}\ \text{m}^3/\text{s}$; seepage velocity $v_s = 2.8\times10^{-6}\ \text{m/s}$.

(a) Classification by type of movement (after Varnes)

(b) Causative factors in Nepalese hills

Preparatory (predisposing): steep slopes, weak/sheared and weathered rock (MCT/MBT zones), unfavourable joint/bedding orientation, deep colluvium, deforestation, high relief.

Triggering: intense monsoon rainfall and pore-pressure rise, earthquakes (co-seismic slides), undercutting by rivers, and human activity (unplanned road cutting, irrigation, loading the slope crest).

(c) Four engineering mitigation measures

1. Drainage — surface drains and sub-surface (horizontal) drains to lower pore-water pressure (the single most effective and economical measure).
2. Slope geometry modification — benching, reducing slope angle, and removing/unloading material from the head.
3. Retaining and reinforcing structures — retaining walls, gabion walls, soil nails, rock bolts and anchors.
4. Surface protection / bioengineering — shotcrete, wire mesh on rock faces, and vegetation/turfing and bamboo-grass planting to bind the soil.

Final answer: Movements are classed as falls, topples, slides, spreads, flows and complex; Nepalese slides arise from steep weathered terrain triggered by monsoon rain and earthquakes, and are mitigated chiefly by drainage, slope regrading, retaining/anchoring structures and bioengineering.

(a) Stages of subsurface investigation

1. Desk study / reconnaissance: existing geological maps, aerial photos, topographic data and a site walk-over to plan the programme.
2. Preliminary (exploratory) investigation: geological mapping, geophysical surveys (seismic refraction, resistivity), trial pits and a few exploratory boreholes to establish ground model.
3. Detailed investigation: systematic core drilling, in-situ tests (SPT, permeability/Lugeon, pressuremeter, plate load), sampling, and laboratory testing.
4. Construction-stage / confirmatory investigation: verification during excavation, instrumentation and monitoring; review of the ground model.

(b) RQD determination and computation

RQD is the percentage of the total core-run length made up of sound (intact) pieces 100 mm or longer, measured along the core axis:

$$RQD = \frac{\sum(\text{length of intact pieces} \ge 100\ \text{mm})}{\text{total length of core run}}\times100\%$$

Qualifying pieces (all ≥ 100 mm): 250 + 180 + 110 + 320 + 150 = 1010 mm. Total run = 2.0 m = 2000 mm.

$$RQD = \frac{1010}{2000}\times100 = \mathbf{50.5\%}$$

RQD 50.5% → rock quality class FAIR (range 50–75%).

(c) Core recovery vs RQD

• Core recovery = (total length of core recovered ÷ length drilled) × 100%. It counts all recovered core, including broken/short fragments.
• RQD counts only intact, sound pieces ≥ 100 mm long, so RQD ≤ core recovery. RQD reflects the degree of jointing/fracturing, whereas core recovery only reflects how much material came up.

Final answer: RQD = 50.5% (Fair rock).

(a) Overbreak

Overbreak is the excavation of rock beyond the designed (pay) tunnel perimeter — i.e. the actual excavated cross-section is larger than the theoretical line. It wastes excavation, increases concrete/shotcrete backfill, and can loosen the surrounding rock.

Geological causes: closely spaced and unfavourably oriented joints/bedding (blocks fall out), shear/fault zones, weathered or weak rock, joint sets forming wedges in the crown, and poor blasting practice in fractured ground.

(b) Tunnelling hazards

(c) Two functions of face mapping

1. Verify and update the ground model — record rock type, joint sets, water and weak zones to confirm or revise the predicted geology ahead of the face.
2. Guide support and excavation decisions — provide the RMR/Q data used to choose support class, round length and probe-drilling ahead for hazards.

Final answer: Overbreak is excavation beyond the design line caused mainly by jointing/shear zones; squeezing, rock burst, running ground and water inflow are recognised by convergence, spalling, sand run-in and water gush respectively; face mapping updates the ground model and guides support.

(a) Definition and purpose

The Lugeon unit is a measure of in-situ rock-mass permeability obtained from a water-pressure test. 1 Lugeon = a water loss of 1 litre per minute per metre of test section at a reference injection pressure of 1 MPa (≈10 bar), sustained for the test period. It is used to assess foundation/abutment watertightness and to design grout curtains for dams.

(b) Computation of Lugeon value

At the reference pressure of 1 MPa the Lugeon value is simply the water take per metre per minute:

$$\text{Lugeon} = \frac{Q}{L}\times\frac{P_{ref}}{P} = \frac{42\ \text{L/min}}{5\ \text{m}}\times\frac{1\ \text{MPa}}{1\ \text{MPa}}$$ $$\text{Lugeon} = \frac{42}{5} = \mathbf{8.4\ Lugeon}$$

(c) Interpretation and treatment

A Lugeon value of about 8 corresponds to moderate permeability / moderately jointed rock:

• < 1 Lu: very tight (negligible seepage)
• 1–5 Lu: low permeability
• 5–15 Lu: moderate — measurable seepage
• 15–50 Lu: high; > 50 Lu: very high (open or karstic)

At 8.4 Lugeon the foundation is not adequately watertight for a dam (typical acceptance is ≤ 1–3 Lu beneath the core/cut-off). Treatment: install a consolidation and curtain grouting programme (cement grout) beneath the dam axis to reduce permeability to the acceptable few-Lugeon range, with check holes to confirm the reduction.

Final answer: 8.4 Lugeon (moderate permeability) — grout curtain treatment required.

(a) Discontinuity and its properties

A discontinuity is any plane or surface of weakness in a rock mass across which the rock has little or no tensile strength — e.g. joints, bedding planes, faults, shear zones and foliation.

Geometric properties of a discontinuity set (any five):

1. Orientation (dip and dip direction / strike)
2. Spacing (perpendicular distance between adjacent discontinuities)
3. Persistence (areal extent / continuity)
4. Roughness of the surface
5. Aperture (opening width) and infilling
6. Number of sets and block size
7. Seepage / water condition

(b) Volumetric joint count and RQD

The volumetric joint count is the sum of the number of joints per metre for each set, i.e. the reciprocal of each spacing summed:

$$J_v = \frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3} = \frac{1}{0.50}+\frac{1}{0.40}+\frac{1}{0.25}$$ $$J_v = 2.0 + 2.5 + 4.0 = \mathbf{8.5\ \text{joints/m}^3}$$

Since $4.5 < 8.5 < 35$, the correlation applies:

$$RQD \approx 115 - 3.3\,J_v = 115 - 3.3\times8.5 = 115 - 28.05 = 86.95 \approx \mathbf{87\%}$$

RQD ≈ 87% → rock quality class GOOD (range 75–90%).

(c) Why discontinuities govern behaviour

Intact rock between joints is usually strong, but a rock mass is an assemblage of blocks separated by weak surfaces. Failure, sliding, deformation and water flow occur preferentially along the discontinuities, which have far lower shear strength and far higher permeability than the intact rock. Hence the number, orientation and condition of discontinuities — not the intact strength — generally control stability of slopes, tunnels and foundations.

Final answer: $J_v = 8.5$ joints/m³; RQD ≈ 87% (Good rock).