BE Civil Engineering (IOE, TU) Engineering Geology II (IOE, CE 601) Question Paper 2078 Nepal

(a) RMR computation (Bieniawski 1989)

We assign the five parameter ratings (R1–R5), then apply the joint-orientation adjustment.

Basic RMR:

$$RMR_{basic} = R1+R2+R3+R4+R5 = 7+13+10+25+10 = 65$$

Orientation adjustment (R6): For a tunnel driven perpendicular to strike, against the dip, with dip 35° (range 20°–45°), the orientation is rated Fair, giving an adjustment of −5.

$$RMR_{adjusted} = 65 + (-5) = \mathbf{60}$$

(b) Classification and support

RMR = 60 falls in the range 61–80? No — 60 lies in the band 41–60 → Class III, "Fair rock."

• Stand-up time: approximately 1 week for a 5 m unsupported span (Class III).
• Support concept (for ~6 m span, Class III): systematic rock bolts 4 m long spaced 1.5–2 m in crown and walls, with wire mesh; 50–100 mm shotcrete in crown and 30 mm on walls. Excavation by top heading and bench, 1.5–3 m advance, support commenced 10 m from face.

Final answers: Basic RMR = 65; Adjusted RMR = 60 → Class III (Fair rock).

Selection and construction of a concrete gravity dam site — geological considerations

(i) Favourable site requirements (≈3 marks)

• A narrow gorge with widening upstream (bottle-neck) minimises dam length and maximises storage.
• Strong, sound, relatively impervious, homogeneous bedrock at shallow depth on both abutments and in the river bed, capable of taking high foundation stresses without excessive settlement.
• Abutments steep enough to seat the dam but stable against sliding.
• Watertight reservoir rim and availability of construction materials (aggregate, impervious fill) nearby.

(ii) Influence of geological structure on the foundation (≈4 marks)

• Rock type: Fresh igneous and metamorphic rocks (granite, gneiss, quartzite) make excellent foundations. Shale, clay-bearing or soluble rocks (limestone, gypsum) are problematic due to low strength, swelling or solution cavities.
• Attitude of beds: Horizontal or upstream-dipping beds are favourable; beds dipping downstream form potential sliding planes and increase leakage paths.
• Faults and shear zones: Crushed/gouge-filled zones reduce bearing capacity, create leakage and differential settlement, and may be seismically active. They must be excavated and dental-concreted or treated.
• Joints: Open or persistent joints increase permeability and reduce shear strength; they govern grout-curtain design.

(iii) Reservoir leakage — investigation and treatment (≈3 marks)

• Causes: permeable strata, solution channels in carbonates, buried valleys, faults connecting the reservoir to adjacent low ground, and continuous pervious beds dipping out of the basin.
• Investigation: geological mapping of the rim, drilling with water-pressure (Lugeon) tests, geophysical surveys, dye/tracer tests and piezometric monitoring.
• Treatment: consolidation grouting of foundation, a deep grout curtain along the dam axis to cut seepage, drainage galleries with relief wells to reduce uplift, and upstream impervious blanketing of leaky reaches.

A well-investigated site reconciles strength (bearing), stability (sliding) and watertightness (leakage and uplift).

Planar (plane) failure — limit equilibrium

The factor of safety is the ratio of resisting force to driving force along the plane:

$$FS = \frac{cA + (W\cos\psi_p - U)\tan\phi}{W\sin\psi_p + V}$$

Useful values: $\cos 30° = 0.8660$, $\sin 30° = 0.5000$, $\tan 32° = 0.6249$.

(a) Dry state ($U = 0,\ V = 0$):

Normal force $N = W\cos\psi_p = 2700 \times 0.8660 = 2338.3\ \text{kN}$

Resisting force $R = cA + N\tan\phi = (25 \times 18) + 2338.3 \times 0.6249$ $R = 450 + 1461.2 = 1911.2\ \text{kN}$

Driving force $D = W\sin\psi_p = 2700 \times 0.5000 = 1350.0\ \text{kN}$

$$FS_{dry} = \frac{1911.2}{1350.0} = \mathbf{1.42}$$

(b) With water ($U = 160$ kN, $V = 90$ kN):

Effective normal force $N' = W\cos\psi_p - U = 2338.3 - 160 = 2178.3\ \text{kN}$

Resisting force $R = cA + N'\tan\phi = 450 + 2178.3 \times 0.6249 = 450 + 1361.2 = 1811.2\ \text{kN}$

Driving force $D = W\sin\psi_p + V = 1350.0 + 90 = 1440.0\ \text{kN}$

$$FS_{wet} = \frac{1811.2}{1440.0} = \mathbf{1.26}$$

(c) Comment and remedy (2 marks)

Water reduces FS from 1.42 to 1.26 (a ~11 % drop) by adding uplift (less normal stress, less friction) and an extra down-slope thrust. The slope remains nominally stable (FS > 1) but the margin is small and could fall below 1 under heavier recharge or seismic loading.

Remedial measure: install sub-horizontal drainage holes to relieve water pressure (restoring FS toward the dry value); alternatively rock-bolt/anchor the block, or unload the head/regrade the slope.

(a) Geological problems in tunnelling (5 marks — 1¼ each)

(b) Terzaghi rock load (3 marks)

Rock-load height:

$$H_p = 0.35\,(B + H_t) = 0.35 \times (8 + 6) = 0.35 \times 14 = 4.9\ \text{m}$$

Vertical support (rock-load) pressure on the roof:

$$p_v = \gamma\,H_p = 25 \times 4.9 = \mathbf{122.5\ \text{kN/m}^2\ (kPa)}$$

Total vertical load per metre run of tunnel:

$$P = p_v \times B = 122.5 \times 8 = \mathbf{980\ \text{kN/m}}$$

(a) Seismotectonics of the Nepal Himalaya (5 marks)

• The Himalaya formed by the continued continental collision of the Indian plate underthrusting beneath the Eurasian plate at ~4–5 cm/yr. This convergence is accommodated mainly along a gently north-dipping detachment, the Main Himalayan Thrust (MHT), on which great earthquakes nucleate.
• Strain accumulates as the locked MHT stores elastic energy between great earthquakes; its periodic release makes Nepal one of the most seismically active regions on Earth.
• Major thrust systems (south to north):
  • MFT – Main Frontal Thrust: youngest, southernmost thrust at the Siwalik–Gangetic plain front; presently most active surface break.
  • MBT – Main Boundary Thrust: separates the Siwaliks from the Lesser Himalaya.
  • MCT – Main Central Thrust: separates the Lesser Himalaya from the Higher Himalayan crystallines.
• Seismic gap: a segment of a fault that has not ruptured for a long time and has therefore accumulated large strain, implying high probability of a future great earthquake. The Western Nepal seismic gap (between the 1505 and 1934 ruptures) is a prominent example; the 2015 Gorkha event only partly filled the central gap.

(b) Energy ratio (3 marks)

For two magnitudes the energy ratio is:

$$\frac{E_1}{E_2} = 10^{\,1.5\,(M_{s1}-M_{s2})}$$

Difference in magnitude: $\Delta M = 7.8 - 6.0 = 1.8$

$$\frac{E_1}{E_2} = 10^{\,1.5 \times 1.8} = 10^{2.7}$$

$10^{2.7} = 10^{0.7}\times10^{2} = 5.0119 \times 100 \approx 501$

The Gorkha main shock released about 500 times more energy than the $M_s = 6.0$ aftershock.

(a) Darcy's law (3 marks)

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

$$q = K\,i,\qquad i = \frac{\Delta h}{L}$$

Hydraulic gradient:

$$i = \frac{3.0}{500} = 6.0 \times 10^{-3}$$

Darcy velocity:

$$q = K i = (2.5 \times 10^{-4}) \times (6.0 \times 10^{-3}) = 1.5 \times 10^{-6}\ \text{m/s}$$

Darcy velocity $q = 1.5 \times 10^{-6}\ \text{m/s}$.

(b) Flow rate through the strip (3 marks)

Cross-sectional area = thickness × width:

$$A = 12\ \text{m} \times 1000\ \text{m} = 12{,}000\ \text{m}^2$$

Discharge:

$$Q = qA = (1.5 \times 10^{-6}) \times 12{,}000 = 1.8 \times 10^{-2}\ \text{m}^3/\text{s}$$

Converting: $Q = 0.018\ \text{m}^3/\text{s} \times 86400 \approx \mathbf{1555\ \text{m}^3/\text{day}}$.

Flow rate $Q = 0.018\ \text{m}^3/\text{s} \approx 1.56 \times 10^{3}\ \text{m}^3/\text{day}.$

(a) Subsurface investigation methods (3 marks)

• Direct (intrusive) methods: trial pits and trenches, auger/wash/percussion/rotary core drilling, exploratory adits and shafts — give samples and direct logging.
• Indirect (geophysical) methods: seismic refraction/reflection, electrical resistivity, ground-penetrating radar, gravity and magnetic surveys — infer subsurface conditions without excavation.
• In-situ tests: standard penetration test (SPT), pressuremeter, plate load, permeability (Lugeon) tests — measure properties in place.

(b) RQD computation (3 marks)

RQD counts only sound core pieces ≥ 10 cm measured along the core axis.

Pieces ≥ 10 cm: 11, 14, 22, 31, 18, 26, 15, 12 cm. (Excluded: 8, 9, 5 cm and the fragments.)

Sum of qualifying lengths:

$$11+14+22+31+18+26+15+12 = 149\ \text{cm}$$

Total core-run length $= 3.0\ \text{m} = 300\ \text{cm}$.

$$RQD = \frac{\sum (\text{pieces} \ge 10\ \text{cm})}{\text{core run length}}\times100 = \frac{149}{300}\times100 = \mathbf{49.7\,\% \approx 50\,\%}$$

Classification: RQD ≈ 50 % lies in the 25–50 % band → "Poor" rock-mass quality (upper boundary of Poor / lower boundary of Fair).

(a) Classification by type of movement (3 marks)

(Any three with correct example earn full marks.)

(b) Causative vs triggering factors (2 marks)

• Causative (preparatory) factors are the inherent, long-acting conditions that make a slope susceptible: steep slope geometry, weak/weathered and sheared rock, unfavourable joint/bedding orientation, deforestation, and adverse groundwater regime.
• Triggering factors are the short-term events that initiate failure on an already-susceptible slope: intense monsoon rainfall, earthquakes (e.g. 2015 Gorkha), toe erosion by rivers, and undercutting by road construction.

In the Nepal hills the same slope is prepared by fragile Lesser-Himalayan geology and steep terrain and is triggered mainly by monsoon cloudbursts and seismic shaking.

(a) Barton Q-system (2 marks)

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

• $\dfrac{RQD}{J_n}$ — block size (degree of jointing).
• $\dfrac{J_r}{J_a}$ — inter-block shear strength (roughness vs alteration of joints).
• $\dfrac{J_w}{SRF}$ — active stress (water pressure and in-situ stress effects).

(b) Q value (3 marks)

$$Q = \frac{70}{9}\times\frac{1.5}{1.0}\times\frac{0.66}{2.5}$$

Step 1: $\dfrac{70}{9} = 7.778$

Step 2: $\dfrac{1.5}{1.0} = 1.5$

Step 3: $\dfrac{0.66}{2.5} = 0.264$

$$Q = 7.778 \times 1.5 \times 0.264 = 7.778 \times 0.396 = \mathbf{3.08}$$

Classification: $Q = 3.08$ lies in the range 1–4 → "Poor" rock mass.

Thiem equation for confined steady-state flow (6 marks)

$$Q = \frac{2\pi T (s_1 - s_2)}{\ln(r_2/r_1)}\quad\Rightarrow\quad T = \frac{Q\,\ln(r_2/r_1)}{2\pi (s_1 - s_2)}$$

(In a confined aquifer the drawdown at the nearer well is larger; $s_1 - s_2$ is the head difference between the two radii.)

Step 1 — radius ratio and its log:

$$\frac{r_2}{r_1} = \frac{120}{20} = 6,\qquad \ln 6 = 1.7918$$

Step 2 — drawdown difference:

$$s_1 - s_2 = 2.4 - 1.0 = 1.4\ \text{m}$$

Step 3 — substitute:

$$T = \frac{0.030 \times 1.7918}{2\pi \times 1.4} = \frac{0.053754}{8.7965}$$ $$T = 6.11 \times 10^{-3}\ \text{m}^2/\text{s}$$

Transmissivity $T \approx 6.1 \times 10^{-3}\ \text{m}^2/\text{s}$ (≈ 528 m²/day).

(a) Kinematic analysis principle (3 marks)

Kinematic analysis uses the stereographic projection of discontinuity planes (as great circles or poles) together with the slope face and the friction cone to test whether a block is free to move — ignoring the magnitude of forces, only the geometry. The dominant failure modes (planar, wedge, toppling) are identified from the relative orientation of joints and the slope face.

Conditions for planar sliding:

1. The discontinuity must dip in the same general direction as the slope face (within about ±20° of the dip direction of the slope).
2. The discontinuity must daylight in the slope face, i.e. its dip $\psi_p$ is less than the slope face dip $\psi_f$ ($\psi_p < \psi_f$).
3. The discontinuity dip must exceed the friction angle ($\psi_p > \phi$).
4. Release surfaces (lateral limits) must be present.

(b) Feasibility test (3 marks)

Given: slope face $\psi_f = 60°$ towards $180°$; joint $\psi_p = 40°$ towards $190°$; $\phi = 30°$.

Check each condition:

1. Direction: joint dip direction 190° vs slope dip direction 180°; difference $= 10° \le 20°$ → satisfied (roughly parallel).
2. Daylighting: $\psi_p = 40° < \psi_f = 60°$ → satisfied (joint daylights in the face).
3. Friction: $\psi_p = 40° > \phi = 30°$ → satisfied (driving exceeds frictional resistance).

All three primary conditions are met:

Planar sliding IS kinematically feasible on this joint set. The slope should be analysed by limit equilibrium and provided with drainage/support.