BE Civil Engineering (IOE, TU) Engineering Geology II (IOE, CE 601) Question Paper 2076 Nepal

(a) Objectives and staged programme (4 marks)

Objectives of dam site investigation:

• Determine the depth, type and engineering properties of foundation rock and overburden.
• Locate and characterise geological discontinuities (faults, shear zones, bedding, joints) and weak phyllite bands that may control sliding/seepage.
• Assess bearing capacity, deformability and shear strength of the foundation.
• Evaluate permeability and groundwater conditions for seepage and grouting design.
• Assess reservoir water-tightness, slope stability of the rim and availability of construction materials.

Staged programme:

1. Desk study / reconnaissance — review of regional geological maps, topographic maps, aerial photos/satellite imagery, seismic records; identification of regional structures and hazards.
2. Surface geological mapping — engineering geological mapping of the dam axis and reservoir, discontinuity surveys (orientation, spacing, persistence, infilling), mapping of weathering and landslides.
3. Sub-surface exploration — core drilling along the dam axis and abutments, exploratory adits/trenches, geophysical surveys (seismic refraction, ERT), in-situ tests (Lugeon water-pressure tests for permeability, plate load, dilatometer), and laboratory tests on cores.

(b) TCR and RQD computation (6 marks)

Run (core barrel) length, $L = 2.0\,\text{m} = 200\,\text{cm}$.

Sum of all recovered intact core pieces:

$$\sum l_i = 0.05+0.12+0.22+0.31+0.08+0.45+0.18+0.15 = 1.56\,\text{m}$$

Total Core Recovery:

$$TCR = \frac{\text{total length of core recovered}}{\text{length of core run}}\times100 = \frac{1.56}{2.0}\times100 = \mathbf{78\%}$$

RQD counts only sound pieces $\geq 0.10\,\text{m}$ (10 cm). Qualifying pieces:

$$0.12,\ 0.22,\ 0.31,\ 0.45,\ 0.18,\ 0.15\ \text{(the }0.05\text{ and }0.08\,\text{m pieces are excluded)}$$ $$\sum l_{i\geq 0.1} = 0.12+0.22+0.31+0.45+0.18+0.15 = 1.43\,\text{m}$$ $$RQD = \frac{\sum l_{i\geq 0.1\,\text{m}}}{L}\times100 = \frac{1.43}{2.0}\times100 = \mathbf{71.5\%}$$

Classification (Deere): $50\% < RQD \leq 75\%$ → "Fair" rock-mass quality. The relatively low RQD compared to TCR reflects the soft, closely-fractured phyllite bands within the quartzite, which would govern foundation treatment and grouting.

RMR system (3 marks)

The RMR (Geomechanics Classification, Bieniawski 1989) rates a rock mass by summing ratings of five parameters: (1) uniaxial compressive strength of intact rock, (2) RQD, (3) spacing of discontinuities, (4) condition of discontinuities, and (5) groundwater conditions. The sum (max 100) gives the basic RMR; a sixth adjustment for discontinuity orientation relative to the engineering work is then applied to give the adjusted RMR. The result places the rock mass into one of five classes (I very good → V very poor), each linked to cohesion, friction angle, stand-up time and support recommendations.

Computation (5 marks)

Basic RMR:

$$RMR_{basic} = 12+13+10+25+7 = \mathbf{67}$$

Orientation adjustment (favourable for tunnels) $= -2$:

$$RMR_{adj} = 67 + (-2) = \mathbf{65}$$

Class and support (2 marks)

$RMR_{adj} = 65$ lies in $61$–$80$ → Class II, "Good rock".

• Cohesion $\approx 300$–$400\,\text{kPa}$, friction angle $\approx 35^\circ$–$45^\circ$.
• Stand-up time guidance: roughly 1 week for a 8 m span (Class II) i.e. an unsupported span/time well within working limits.
• Support: systematic rockbolts 4 m long spaced 1.5–2 m in crown and walls with occasional wire mesh, and 50 mm shotcrete in crown where required; full-face or top-heading-and-bench excavation is feasible.

(a) Derivation (3 marks)

Resolve the block weight $W$ along and normal to the sliding plane dipping at $\psi_p$:

• Driving (downslope) force: $T = W\sin\psi_p$
• Normal force on the plane: $N = W\cos\psi_p$

Resisting force = cohesive + frictional shear strength along plane area $A$:

$$R = cA + N\tan\phi = cA + W\cos\psi_p\tan\phi$$

Factor of safety (dry):

$$FoS = \frac{\text{resisting}}{\text{driving}} = \frac{cA + W\cos\psi_p\tan\phi}{W\sin\psi_p}$$

(b) Computation (4 marks)

Driving force:

$$W\sin\psi_p = 4500\times\sin35^\circ = 4500\times0.5736 = 2581.1\,\text{kN}$$

Normal force:

$$W\cos\psi_p = 4500\times\cos35^\circ = 4500\times0.8192 = 3686.3\,\text{kN}$$

Cohesive resistance ($c = 25\,\text{kPa} = 25\,\text{kN/m}^2$, $A = 34.9\,\text{m}^2$):

$$cA = 25\times34.9 = 872.5\,\text{kN}$$

Frictional resistance:

$$W\cos\psi_p\tan\phi = 3686.3\times\tan30^\circ = 3686.3\times0.5774 = 2128.5\,\text{kN}$$

Total resisting force:

$$R = 872.5 + 2128.5 = 3001.0\,\text{kN}$$ $$FoS = \frac{3001.0}{2581.1} = \mathbf{1.16}$$

Comment: $FoS \approx 1.16 > 1$, so the dry slope is marginally stable but with a low margin of safety; it is below the usual design requirement of $\approx 1.3$–$1.5$ and would be considered unsafe for permanent works.

(c) Effect of water (1 mark)

Water in a tension crack and along the plane produces an uplift water force $U$ that reduces the effective normal stress (and a crack-thrust force $V$). The frictional resistance becomes $(W\cos\psi_p - U)\tan\phi$ while driving force increases by $V\,$. Both effects reduce the FoS, easily dropping it below 1 and triggering failure — hence drainage is the key stabilisation measure.

(a) Geology for an arch dam (4 marks)

An arch dam transfers most of the reservoir thrust horizontally into the valley walls, so abutment strength is critical.

Favourable conditions:

• Strong, massive, low-deformability rock (e.g. granite, hard limestone, gneiss) in both abutments capable of resisting concentrated arch thrust.
• Narrow, steep-sided V- or U-shaped valley giving a small crest length-to-height ratio.
• Discontinuities oriented so that they do not form wedges dipping out of the abutment toward the downstream/valley; bedding/joints dipping into the abutment are favourable.
• Watertight reservoir basin (low-permeability rock, no karst, no leaky cols).

Unfavourable conditions:

• Weak, weathered, schistose or highly jointed abutment rock; presence of faults or shear zones cutting the abutments.
• Discontinuities or thrust planes dipping downstream/out of the slope (potential abutment sliding).
• Soluble (karstic) or highly permeable rock causing reservoir leakage.

(b) Lugeon value (4 marks)

Test data: section length $L = 3.0\,\text{m}$, test pressure $P = 1.0\,\text{MPa}$ (= reference pressure, so no pressure correction needed), water take $Q = 24\,\text{L/min}$.

Specific water take per metre at the test pressure:

$$q = \frac{Q}{L} = \frac{24\,\text{L/min}}{3.0\,\text{m}} = 8.0\ \text{L/min/m}$$

Since the test pressure equals the reference 10 bar (1 MPa), the Lugeon value equals this specific take:

$$Lu = q\times\frac{P_{ref}}{P} = 8.0\times\frac{10}{10} = \mathbf{8\ \text{Lugeon}}$$

Equivalent hydraulic conductivity:

$$k \approx 8 \times 1.3\times10^{-7} = 1.04\times10^{-6}\,\text{m/s}$$

Interpretation: A value of about 8 Lugeon indicates a moderately permeable rock mass (typically 1–5 Lu = low, 5–15 Lu = moderate, >15 Lu = high). For a dam foundation a tightness criterion of $\leq 1$–$3$ Lugeon is usually required, so consolidation/curtain grouting is needed to reduce seepage and uplift to acceptable levels.

(a) Seismotectonics of Nepal (4 marks)

The Himalaya results from the continued northward under-thrusting of the Indian Plate beneath the Eurasian Plate at $\sim$4–5 cm/yr. This convergence is accommodated along major north-dipping thrust faults that bound the morphotectonic zones:

• MFT (Main Frontal Thrust / HFT) — the southernmost, active thrust separating the Sub-Himalaya (Siwaliks) from the Indo-Gangetic plain; it is the surface expression of the locked thrust.
• MBT (Main Boundary Thrust) — separates the Sub-Himalaya from the Lesser Himalaya.
• MCT (Main Central Thrust) — separates the Lesser Himalaya from the Higher Himalaya crystallines.

These splay upward from a common basal décollement, the Main Himalayan Thrust (MHT), which is locked in its shallow part where strain accumulates and is released as great earthquakes. The strain build-up in segments that have not ruptured recently constitutes a seismic gap (e.g. the western Nepal gap), where a large earthquake is overdue. Repeated great events (1934 Bihar–Nepal $M\approx8.0$, 2015 Gorkha $M_w\,7.8$) confirm high hazard.

(b) Magnitude calculations (4 marks)

(i) Amplitude ratio. Richter magnitude is $M=\log_{10}A + (\text{const})$, so a unit increase in $M$ means a 10× larger recorded amplitude:

$$\frac{A_Y}{A_X} = 10^{(M_Y-M_X)} = 10^{(8.0-6.0)} = 10^{2} = \mathbf{100}$$

Ground-motion amplitude of Y is 100 times that of X.

(ii) Energy ratio. Using $\log_{10}E = 4.8 + 1.5M$:

$$\frac{E_Y}{E_X} = 10^{1.5(M_Y-M_X)} = 10^{1.5\times2.0} = 10^{3.0} = \mathbf{1000}$$

Earthquake Y releases 1000 times more energy than X.

(Check, absolute values: $\log E_X = 4.8+1.5(6.0)=13.8 \Rightarrow E_X=10^{13.8}\,\text{J}$; $\log E_Y = 4.8+1.5(8.0)=16.8 \Rightarrow E_Y=10^{16.8}\,\text{J}$; ratio $=10^{3.0}=1000$.)

Implication: Because energy scales so steeply with magnitude, design ground motions and structural detailing must be governed by the rare large ($M\approx8$) event, not by frequent moderate shocks; site-specific seismic hazard analysis and ductile detailing are essential for Himalayan dams.

Varnes classification by type of movement (3 marks)

Causative factors (2 marks)

Preparatory (make slope susceptible): steep slope geometry, weak/weathered rock, adverse discontinuity orientation, loss of vegetation, weathering, removal of toe support by river erosion or excavation.

Triggering (initiate movement): intense/prolonged rainfall and rise in pore-water pressure, earthquakes, undercutting by streams, and human activities (poor road cutting, loading the crest, blocked drainage).

Mitigation for Nepal hill roads (1 mark)

• Surface and sub-surface drainage (catch drains, horizontal drains) to reduce pore pressure — the single most cost-effective measure.
• Slope support / bioengineering — retaining/gabion walls, rock bolts, and vegetative bioengineering (plantation, brush layering) to stabilise cut slopes.

(a) Geological problems in tunnelling (3 marks)

• Squeezing ground — slow plastic inward movement under high overburden stress in weak/ductile rock (phyllite, shale, fault gouge); $\sigma_{cm}/\sigma_v$ low.
• Swelling ground — volume increase on water access in clay-rich or anhydrite/montmorillonite rock, producing high invert/crown pressures.
• Water inrush — sudden high-pressure inflow when intersecting permeable, jointed or karstic zones, faults or aquifers; hampers excavation.
• Overbreak — excess excavation beyond the design profile along unfavourable joints/blasting damage.
• Rock burst — sudden violent failure (spalling) of strong, brittle rock under high stress at depth.

(b) Terzaghi rock load (3 marks)

For moderately blocky and seamy rock, upper bound:

$$H_p = 0.35(B + H_t) = 0.35\,(6 + 6) = 0.35\times12 = \mathbf{4.2\ \text{m}}$$

Support (vertical) pressure on crown:

$$p_v = \gamma\,H_p = 25\,\text{kN/m}^3 \times 4.2\,\text{m} = \mathbf{105\ \text{kN/m}^2\ (\approx 0.105\ \text{MPa})}$$

This pressure is used to size the steel sets / shotcrete-bolt support in the crown.

(a) Definitions (3 marks)

• Aquifer — a saturated permeable formation that stores and readily transmits usable quantities of groundwater. Example: sandstone, gravel.
• Aquitard — a poorly permeable bed that transmits water very slowly (leaky); can leak between aquifers. Example: silty clay, sandy shale.
• Aquiclude — a saturated but essentially impermeable bed that stores water yet does not transmit significant amounts. Example: clay.
• Aquifuge — an impermeable formation that neither stores nor transmits water. Example: massive unfractured granite.

(b) Darcy's law computation (3 marks)

Hydraulic gradient:

$$i = \frac{\Delta h}{L} = \frac{145 - 140}{500} = \frac{5}{500} = 0.01$$

Darcy (specific discharge) velocity:

$$v = k\,i = (8\times10^{-5}\,\text{m/s})(0.01) = 8\times10^{-7}\,\text{m/s}$$ $$\boxed{v = 8\times10^{-7}\,\text{m/s}}$$

Discharge per metre width (aquifer thickness $b = 12\,\text{m}$, width $w = 1\,\text{m}$, area $A = 12\,\text{m}^2$):

$$Q = v\,A = (8\times10^{-7})(12) = 9.6\times10^{-6}\,\text{m}^3/\text{s per metre width}$$

Converting to litres/day for feel:

$$Q = 9.6\times10^{-6}\times86400 = 0.829\,\text{m}^3/\text{day} \approx \mathbf{829\ \text{litres/day per metre width}}$$

Q-system (2 marks)

Barton's Q (NGI, 1974) characterises rock-mass quality for tunnelling from six parameters combined as three quotients:

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

where $\dfrac{RQD}{J_n}$ ≈ relative block size, $\dfrac{J_r}{J_a}$ ≈ inter-block shear strength (roughness/alteration), and $\dfrac{J_w}{SRF}$ ≈ active stress (water and stress reduction). $Q$ ranges from 0.001 (exceptionally poor) to 1000 (exceptionally good) and is used to estimate support and equivalent dimension.

Computation (3 marks)

$$Q = \frac{70}{6}\times\frac{2}{1}\times\frac{1}{2.5}$$

Step by step:

$$\frac{70}{6} = 11.667,\qquad \frac{2}{1} = 2,\qquad \frac{1}{2.5} = 0.4$$ $$Q = 11.667 \times 2 \times 0.4 = 11.667 \times 0.8 = 9.33$$ $$\boxed{Q \approx 9.3}$$

Classification: $Q$ between $4$ and $10$ → "Fair" rock mass. Light, systematic support (rockbolts with spot shotcrete) would typically be adequate for a normal-span tunnel.

(a) Discontinuity properties (2.5 marks)

• Orientation (dip/dip-direction) — controls whether wedges/planes daylight; primary input to kinematic analysis.
• Spacing — governs block size and rock-mass strength/permeability.
• Persistence (continuity) — length over which a discontinuity acts; long persistent joints form through-going failure surfaces.
• Roughness — contributes to shear strength (dilatancy, JRC).
• Aperture / infilling — weak clay infill lowers shear strength and raises permeability.
• Seepage / water condition — water pressure reduces effective stress and stability.

(b) Kinematic check for plane sliding (2.5 marks)

Conditions for plane failure:

1. Joint dips in roughly the same direction as the slope face (within $\pm20^\circ$): slope faces $180^\circ$, joint dips towards $175^\circ$; difference $=5^\circ < 20^\circ$ → satisfied.
2. The joint daylights in the face, i.e. joint dip $<$ slope-face dip: $40^\circ < 70^\circ$ → satisfied (the plane emerges in the cut face).
3. Joint dip exceeds the friction angle $\psi_p > \phi$: $40^\circ > 30^\circ$ → satisfied (driving shear overcomes friction).

All three conditions are met, so plane sliding is kinematically possible and the slope is at risk of plane failure along this joint set; a detailed limit-equilibrium (FoS) analysis and support/drainage would be required.

Answer any four; each is worth 2 marks.

(i) Weathering grades (Grade I–VI)

• I Fresh, II Slightly weathered, III Moderately weathered, IV Highly weathered, V Completely weathered, VI Residual soil.
• As grade increases, strength, RQD and seismic velocity fall while permeability and compressibility rise. Grades I–II are good foundation rock; grades V–VI behave as soil and must be excavated or specially treated. Mapping weathering grade guides founding depth and slope angles.

(ii) Seismic refraction

• A surface energy source generates P-waves; geophones record first-arrival travel times along a spread. Plotting time vs distance gives layer velocities and, from cross-over distances/intercept times, the depth to refractors (e.g. bedrock).
• Layer velocity increases with density/induration, so it distinguishes overburden ($\sim$300–800 m/s) from rock ($>$2000 m/s). Rippability is judged from the bedrock P-velocity: low velocity → rippable, high velocity → blasting required.

(iii) Schmidt rebound hammer

• A spring-loaded mass impacts the rock surface and the rebound number $R$ is read. $R$ correlates (with rock density) to uniaxial compressive strength via published charts; it is a quick, non-destructive index of surface hardness used for field strength estimates and discontinuity-wall (JCS) strength in the JRC–JCS shear model. Several readings are averaged and the lowest discarded.

(iv) Reservoir-induced seismicity (RIS)

• Filling a large, deep reservoir can trigger earthquakes by (a) increasing the vertical load and (b) raising pore-water pressure along pre-existing faults, which reduces effective normal stress and promotes slip. RIS is most likely where active/critically-stressed faults underlie deep reservoirs; it requires fault mapping, micro-seismic monitoring and controlled impounding (rate of filling).

(v) Drainage gallery / uplift relief

• Uplift (pore-water) pressure acts on the base of a gravity dam, reducing effective weight and stability. A drainage gallery with a line of vertical drain holes just downstream of the grout curtain intercepts seepage and relieves base pressure, lowering the uplift diagram and improving sliding/overturning safety. It also allows inspection, foundation grouting and instrumentation.