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HKIE Professional Assessment - Geotechnical Discipline - independent engineering foundations study materials
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HKIE Professional Assessment - Geotechnical Discipline - independent engineering foundations practice questions

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Question 1Geotechnical engineering models and applications

A soil profile consists of a 3 m thick upper layer of dry sand above a saturated sand layer. The total unit weight above the groundwater table is 18 kN/m³ and the saturated unit weight below is 20 kN/m³. The groundwater table is located at 3 m depth. What is the vertical effective stress at a depth of 8 m?

  1. 104.95 kPa Correct answer.
  2. 154.00 kPa
  3. 49.05 kPa
  4. 123.50 kPa

Explanation

At a depth of 8 m, total vertical stress is σ = (3 m × 18 kN/m³) + (5 m × 20 kN/m³) = 54 + 100 = 154 kPa. Pore water pressure is u = (8 m - 3 m) × 9.81 kN/m³ = 49.05 kPa. Vertical effective stress is σ' = σ - u = 154 - 49.05 = 104.95 kPa.

Question 2Geotechnical engineering models and applications

According to the Mohr-Coulomb failure criterion, what is the effective shear strength τ_f of a soil with effective cohesion c' = 12 kPa, effective friction angle ϕ' = 32°, total stress σ = 180 kPa, and pore water pressure u = 50 kPa?

  1. 93.23 kPa Correct answer.
  2. 124.47 kPa
  3. 81.23 kPa
  4. 107.23 kPa

Explanation

The effective stress on the failure plane is σ' = σ - u = 180 - 50 = 130 kPa. Using the Mohr-Coulomb equation τ_f = c' + σ' tan ϕ', we get τ_f = 12 + (130 × tan 32°) = 12 + (130 × 0.62486) = 12 + 81.23 = 93.23 kPa.

Question 3Geotechnical engineering models and applications

A 6.0 m thick normally consolidated clay layer has an initial void ratio e_0 = 0.90, compression index C_c = 0.30, and initial vertical effective stress σ'_{v0} = 100 kPa. An foundation loading induces a vertical stress increment Δσ' = 100 kPa at the mid-depth of the layer. What is the ultimate 1D primary consolidation settlement ΔH of the clay layer?

  1. 285.2 mm Correct answer.
  2. 570.4 mm
  3. 190.1 mm
  4. 360.0 mm

Explanation

Using Terzaghi 1D consolidation settlement formula ΔH = [C_c / (1 + e_0)] × H × log10[(σ'_{v0} + Δσ') / σ'_{v0}], we plug in H = 6000 mm, e_0 = 0.90, C_c = 0.30, σ'_{v0} = 100 kPa, and Δσ' = 100 kPa. ΔH = [0.30 / 1.90] × 6000 × log10(200 / 100) = 0.15789 × 6000 × 0.30103 = 285.2 mm.

Question 4Geotechnical engineering models and applications

A 4.0 m thick marine clay layer with double drainage has a coefficient of consolidation c_v = 2.5 m²/year. For 90% degree of consolidation, the time factor is T_90 = 0.848. How long (in years) will it take for the clay layer to reach 90% consolidation?

  1. 1.36 years Correct answer.
  2. 5.43 years
  3. 2.71 years
  4. 0.68 years

Explanation

For double drainage, the maximum drainage path is d = H / 2 = 4.0 / 2 = 2.0 m. Using the time factor equation T_v = (c_v × t) / d², we solve for t = (T_90 × d²) / c_v = (0.848 × 2.0²) / 2.5 = (0.848 × 4) / 2.5 = 3.392 / 2.5 = 1.3568 years ≈ 1.36 years.

Question 5Geotechnical engineering models and applications

What is the primary difference between a Consolidated Drained (CD) triaxial test and a Consolidated Undrained (CU) triaxial test with pore pressure measurement on saturated clay?

  1. In a CD test, drainage is permitted during shear so excess pore pressure is zero, yielding effective stress parameters c' and ϕ'; in a CU test, drainage is prevented during shear and measured excess pore pressure allows effective stress analysis. Correct answer.
  2. In a CD test, consolidation is prevented during the initial stress application phase, whereas in a CU test, full consolidation is permitted.
  3. In a CD test, total stress parameters c_u and ϕ_u are directly obtained, whereas CU tests only measure hydraulic conductivity.
  4. In a CD test, pore water pressure is artifically pumped into the specimen to simulate artesian conditions.

Explanation

In CD triaxial testing, slow strain rates allow full drainage so zero excess pore water pressure develops (Δu = 0), directly yielding c' and ϕ'. In CU testing, drainage is closed during shearing, but measuring Δu via transducers allows calculation of effective stresses σ' = σ - u to obtain c' and ϕ'.

Question 6Geotechnical engineering models and applications

A fine-grained soil in Hong Kong has a Liquid Limit (LL) of 55% and a Plastic Limit (PL) of 25%. According to the Casagrande plasticity chart (A-line equation PI = 0.73 × (LL - 20)), how is this soil classified?

  1. High plasticity clay (CH), because its Plasticity Index (PI = 30%) lies above the A-line (PI_A = 25.55%). Correct answer.
  2. High plasticity silt (MH), because its Plasticity Index (PI = 30%) lies below the A-line.
  3. Low plasticity clay (CL), because its Plasticity Index is below 35%.
  4. Organic silt of high plasticity (OH), because LL exceeds 50%.

Explanation

Plasticity Index PI = LL - PL = 55 - 25 = 30%. The A-line value at LL = 55% is PI_A = 0.73 × (55 - 20) = 0.73 × 35 = 25.55%. Since PI (30%) > PI_A (25.55%) and LL > 50%, the soil lies above the A-line in the high plasticity zone, classifying it as High Plasticity Clay (CH).

Question 7Geotechnical engineering models and applications

A falling head permeability test is performed on a soil sample of length L = 10 cm and cross-sectional area A = 50 cm². The standpipe area is a = 1.5 cm². The water head drops from h_1 = 60 cm to h_2 = 30 cm in a time duration t = 300 s. What is the coefficient of permeability k of the soil?

  1. 6.93 × 10⁻⁶ m/s Correct answer.
  2. 6.93 × 10⁻⁴ m/s
  3. 1.39 × 10⁻⁵ m/s
  4. 3.47 × 10⁻⁶ m/s

Explanation

Falling head permeability formula: k = (a × L) / (A × t) × ln(h_1 / h_2). Substituting values: k = (1.5 × 10) / (50 × 300) × ln(60 / 30) = (15 / 15000) × ln(2) = 0.001 × 0.69315 = 6.9315 × 10⁻⁴ cm/s. Converting to m/s: k = 6.9315 × 10⁻⁶ m/s.

Question 8Geotechnical engineering models and applications

A point load strength test on a 50 mm diameter Grade II granite core specimen gives an axial point load index I_s(50) = 4.0 MPa. Using the empirical correlation recommended for Hong Kong rocks (UCS ≈ 24 × I_s(50)), what is the estimated Unconfined Compressive Strength (UCS)?

  1. 96 MPa Correct answer.
  2. 48 MPa
  3. 120 MPa
  4. 64 MPa

Explanation

The standard conversion factor for Hong Kong granitic and volcanic rock core samples per Geoguide 3 is UCS = 24 × I_s(50). For I_s(50) = 4.0 MPa, UCS = 24 × 4.0 = 96 MPa, placing it in the High Strength rock category (> 50 MPa).

Question 9Geotechnical engineering models and applications

Under Barton's shear strength criterion for rock joints (τ = σ'_n tan[JRC log10(JCS / σ'_n) + ϕ_r]), what effect does an increase in normal stress σ'_n have on the peak friction angle of a rough, unweathered rock joint?

  1. The effective friction angle decreases non-linearly because joint asperities are sheared off rather than ridden over at higher normal stress. Correct answer.
  2. The effective friction angle increases linearly with normal stress due to strain hardening.
  3. The effective friction angle remains strictly constant regardless of normal stress level.
  4. The effective friction angle drops instantly to zero once normal stress exceeds JCS.

Explanation

Barton's criterion captures the non-linear strength envelope of rock joints. At low normal stresses, sliding over roughness asperities adds dilatancy (JRC log10(JCS / σ'_n)). As normal stress σ'_n increases, asperities are sheared off, reducing dilation and causing the secant friction angle to decrease towards the residual value ϕ_r.

Question 10Geotechnical engineering models and applications

In Critical State Soil Mechanics (Cam-Clay model), what do parameters λ (lambda) and κ (kappa) represent in the e vs ln p' space?

  1. λ is the slope of the Normal Compression Line (NCL) and Critical State Line (CSL), while κ is the slope of the elastic unloading-reloading line. Correct answer.
  2. λ is the slope of the Critical State Line in q-p' space, while κ is the plastic shear strain increment.
  3. λ is the coefficient of volume compressibility m_v, while κ is the secondary compression index C_α.
  4. λ is the Poisson's ratio, while κ is the shear modulus G.

Explanation

In Cam-Clay framework, λ (lambda) defines the slope of the isotropic Normal Compression Line (NCL) and the Critical State Line (CSL) in e - ln p' space (or v - ln p' space). κ (kappa) defines the slope of the elastic recompression/swelling line during unloading and reloading.

Question 11Geotechnical engineering models and applications

An infinite soil slope has a slope angle β = 30° and depth to slip surface z = 3.0 m. The soil properties are c' = 10 kPa, ϕ' = 35°, γ_sat = 19.0 kN/m³, and γ_w = 9.81 kN/m³. Groundwater is coincident with the slope surface (full seepage parallel to slope). What is the Factor of Safety (FoS) of the slope?

  1. 0.99 Correct answer.
  2. 1.41
  3. 1.25
  4. 0.76

Explanation

For infinite slope with groundwater at surface and parallel seepage: FoS = [c' + γ' × z × cos²β × tan ϕ'] / [γ_sat × z × sin β × cos β]. γ' = 19.0 - 9.81 = 9.19 kN/m³. cos 30° = 0.8660, sin 30° = 0.5000, tan 35° = 0.7002. Numerator = 10 + [9.19 × 3 × (0.8660)² × 0.7002] = 10 + [27.57 × 0.750 × 0.7002] = 10 + 14.48 = 24.48 kPa. Denominator = 19.0 × 3 × 0.5000 × 0.8660 = 24.68 kPa. FoS = 24.48 / 24.68 = 0.992 ≈ 0.99.

Question 12Geotechnical engineering models and applications

In a planar rock slope failure analysis, a joint plane dips towards the slope face at an angle θ = 35°. The slope face dips at 55°. What is the kinematic condition required for planar sliding to occur under gravity?

  1. ϕ_r < θ < ψ_f, where ϕ_r is joint friction angle and ψ_f is slope face dip angle (joint daylighting). Correct answer.
  2. θ > ψ_f > ϕ_r
  3. θ < ϕ_r < ψ_f
  4. θ + ϕ_r > 90°

Explanation

Kinematic daylighting condition for planar failure: The joint dip θ must be flatter than slope face dip ψ_f so the joint daylighting exists on the slope face (θ < ψ_f), and joint dip θ must be steeper than joint friction angle ϕ_r (θ > ϕ_r) to overcome friction under gravity.

Question 13Geotechnical engineering models and applications

Under Geoguide 7 (Guide to Soil Nail Design and Construction), how is the allowable pullout resistance Q_p per unit length of a soil nail in soil calculated?

  1. Q_p = (π × D × τ_bond) / γ_m, where D is drillhole diameter, τ_bond is ultimate grout-soil bond stress, and γ_m is partial material factor. Correct answer.
  2. Q_p = A_s × f_y × sin α
  3. Q_p = c' × D² × L
  4. Q_p = γ × z × D × tan ϕ'

Explanation

Soil nail pullout capacity depends on the circumferential surface area of the drillhole grout column. Pullout force per unit length is q_p = π × D × τ_bond. Factored pullout resistance applies partial factors per Geoguide 7: Q_p = (π × D × τ_bond) / γ_m.

Question 14Geotechnical engineering models and applications

What is the function of a flexible rockfall / debris flow barrier installed at the toe of a natural terrain catchment in Hong Kong?

  1. To dissipate kinetic energy of falling boulders or debris flows through controlled dynamic elongation of steel wire ring nets and brake elements. Correct answer.
  2. To act as a rigid water-retaining concrete dam that impounds flash floods permanently.
  3. To increase the shear strength of the upper natural hillslope bedrock.
  4. To prevent groundwater infiltration across the entire hillslope catchment.

Explanation

Flexible debris barriers consist of high-tensile steel wire ring nets, support cables, steel posts, and friction energy dissipators (brakes). They absorb and dissipate dynamic impact energy of boulders and debris flows through large elastic/plastic deformation.

Question 15Geotechnical engineering models and applications

Under Geoguide 7, what structural failure modes must be checked when designing the soil nail head assembly (bearing plate, washer, nut, and concrete pad)?

  1. Flexural bending of bearing plate, punching shear of concrete head pad, and tensile failure of threaded nail bar connection. Correct answer.
  2. Buckling of the soil nail bar over its grouted length in soil.
  3. Dynamic impact resonance caused by traffic vibration.
  4. Thermal expansion fatigue of the galvanized washer plate.

Explanation

Soil nail head design must ensure safe load transfer between slope surface facing/pad and nail bar. Required checks: bearing plate bending strength, concrete facing pad punching shear, and threaded bar connection tension/shear capacity.

Question 16Geotechnical engineering models and applications

What is the fundamental difference in load transfer mechanism between a passive soil nail and an active prestressed ground anchor in slope stabilization?

  1. A soil nail develops resisting tensile forces passively as slope movement occurs; a prestressed ground anchor applies an active pre-load force directly to the slope face during installation. Correct answer.
  2. A soil nail is installed vertically only, whereas a prestressed anchor is installed horizontally.
  3. A soil nail utilizes post-tensioned hydraulic jacks, whereas ground anchors are ungrouted steel rods.
  4. A soil nail is designed solely for water drainage, whereas an anchor provides structural support.

Explanation

Soil nails are passive elements; they carry no initial stress and rely on ground deformation to mobilize bond shear stress and bar tension. Prestressed ground anchors are active; they are post-tensioned against a concrete bearing head to apply immediate compressive force to the soil/rock mass.

Question 17Geotechnical engineering models and applications

When performing limit equilibrium slope stability analysis, what key assumption distinguishes Bishop's Simplified Method from Fellenius' (Ordinary) Method for circular slip surfaces?

  1. Bishop's Simplified Method considers interslice forces assuming they act horizontally (X_i = 0), satisfying vertical force equilibrium and overall moment equilibrium, whereas Fellenius neglects all interslice forces. Correct answer.
  2. Bishop's method assumes planar slip surfaces, whereas Fellenius assumes circular surfaces.
  3. Fellenius' method satisfies all 3 equilibrium equations (vertical, horizontal, moment), whereas Bishop's method satisfies none.
  4. Bishop's method is strictly limited to dry cohesionless sand.

Explanation

Fellenius' Ordinary Method ignores interslice forces entirely, leading to conservative FoS estimates. Bishop's Simplified Method includes interslice normal forces (assuming interslice shear forces X_i = 0), satisfying vertical force equilibrium for each slice and overall moment equilibrium, providing accurate FoS results.

Question 18Geotechnical engineering models and applications

In Quantitative Risk Assessment (QRA) for natural terrain landslide hazards in Hong Kong, how is Individual Risk (IR) mathematically expressed?

  1. IR = P(H) × P(S|H) × P(T|S) × V(L|T), where P(H) is hazard frequency, P(S|H) spatial probability, P(T|S) temporal occupancy, and V vulnerability. Correct answer.
  2. IR = Total construction cost / Total slope area
  3. IR = Factor of Safety × Rainstorm intensity
  4. IR = Slope height / Distance to building

Explanation

Individual Risk (IR) is the annual probability of fatality for a specific individual. It equals the product of: Landslide hazard annual frequency P(H), Spatial probability of debris reaching location P(S|H), Temporal presence of person P(T|S), and Vulnerability V(L|T).

Question 19Geotechnical engineering models and applications

When carrying out back-analysis of a landslide in completely weathered granite to estimate mobilized shear strength parameters at failure, what fundamental assumption is made regarding the Factor of Safety at the moment of failure?

  1. FoS = 1.00 along the observed failure surface under the estimated groundwater condition at peak rainstorm intensity. Correct answer.
  2. FoS = 1.40 assuming standard Geoguide design margin.
  3. FoS = 0.50 because the slope collapsed rapidly.
  4. FoS = 2.00 using total stress parameters.

Explanation

Back-analysis sets FoS = 1.00 along the known rupture surface at the collapse instance. By inputting estimated failure geometry and pore pressure u during the rainstorm, back-calculation yields mobilized c' and ϕ'.

Question 20Geotechnical engineering models and applications

Why can severe rainstorms produce a transient perched groundwater table in Hong Kong soil slopes overlying unweathered bedrock?

  1. Because infiltration rate through highly permeable weathered soil (CWG) exceeds the drainage capacity of underlying low-permeability bedrock (Grade I/II), causing water accumulation above the bedrock interface. Correct answer.
  2. Because groundwater rises directly from subterranean ocean tides during typhoons.
  3. Because vegetation roots pump water upwards into the soil slope.
  4. Because concrete shotcrete facings draw water inside via capillary suction.

Explanation

In Hong Kong terrain, high-permeability completely weathered soil (k ~ 10⁻⁵ m/s) overlies unweathered bedrock (k ~ 10⁻⁸ m/s). Rainwater rapidly infiltrates soil cover but cannot penetrate intact bedrock quickly, perching water above the bedrock boundary and causing pore pressure build-up.

Question 21Geotechnical engineering models and applications

What geotechnical instrument is primarily installed in a borehole to measure lateral subsurface shear deformations and locate the shear failure plane in a hillside slope?

  1. Inclinometer (with grooved casing and biaxial probe) Correct answer.
  2. Vibrating wire piezometer
  3. Standpipe observation well
  4. Extensometer rod

Explanation

Inclinometers measure lateral tilt/deflection profiles along a grooved plastic casing installed in a vertical borehole. Reading the casing over time identifies the depth, magnitude, and rate of subsurface shear displacement along slip surfaces.

Question 22Geotechnical engineering models and applications

A continuous strip footing of width B = 2.0 m is founded at a depth of D_f = 1.5 m in a saturated clay soil under undrained conditions (c_u = 80 kPa, ϕ_u = 0°, γ = 18.0 kN/m³). Using Terzaghi's bearing capacity equation (N_c = 5.7, N_q = 1.0, N_γ = 0), what is the net ultimate bearing capacity q_net,ult?

  1. 456 kPa Correct answer.
  2. 483 kPa
  3. 152 kPa
  4. 300 kPa

Explanation

Terzaghi equation for strip footing: q_ult = c_u N_c + γ D_f N_q = (80 × 5.7) + (18.0 × 1.5 × 1.0) = 456 + 27 = 483 kPa. Net ultimate bearing capacity is q_net,ult = q_ult - γ D_f = 483 - 27 = 456 kPa.

Question 23Geotechnical engineering models and applications

What causes Negative Skin Friction (downdrag) on a pile driven through a newly placed marine fill stratum into underlying bedrock?

  1. Settlement of the consolidating soft marine clay/fill layer relative to the rigid pile shaft, inducing downward shear stress on the pile. Correct answer.
  2. Upward artesian water pressure pushing the pile out of the ground.
  3. High tensile forces caused by thermal expansion of the concrete pile head.
  4. Lateral wind loading on the above-ground superstructure.

Explanation

When soft soil or reclamation fill consolidates and moves downwards relative to a stiff end-bearing pile, downward shaft friction (downdrag) acts on the pile upper length above the neutral plane, adding extra axial drag load to the pile structure.

Question 24Geotechnical engineering models and applications

In dynamic pile driving monitoring using the Hiley dynamic pile formula, what does 'set' (s) represent?

  1. The plastic (permanent) penetration of the pile per hammer blow, measured in mm. Correct answer.
  2. The elastic temporary compression of the pile shaft and ground.
  3. The total height of hammer drop in meters.
  4. The weight of the pile driving hammer in tonnes.

Explanation

Set (s) is defined as the permanent downward penetration of the pile produced by a single hammer strike (commonly expressed as mm per blow or blows per 300 mm). Total movement equals set (s) plus temporary elastic compression (c).

Question 25Geotechnical engineering models and applications

What is the key objective of performing High-Strain Dynamic Pile Testing (PDA test with CAPWAP analysis) on driven steel H-piles in Hong Kong?

  1. To estimate ultimate mobilize shaft friction and end bearing capacity, measure driving stresses, and verify structural pile integrity during driving. Correct answer.
  2. To determine the chemical sulphate content of surrounding groundwater.
  3. To measure the long-term consolidation settlement of adjacent buildings over 10 years.
  4. To measure ground electrical resistivity for cathodic protection.

Explanation

PDA testing uses strain transducers and accelerometers near the pile head attached to a dynamic analyzer. CAPWAP signal matching calculates total mobilized geotechnical capacity (skin friction vs end bearing), maximum driving stresses, and detects pile damage.

Question 26Geotechnical engineering models and applications

In completely weathered granite (CWG) soil, how is the ultimate shaft friction τ_s of a friction pile calculated using the β (beta) method?

  1. τ_s = β × σ'_v = K × σ'_v × tan δ, where K is lateral earth pressure coefficient and δ is interface friction angle. Correct answer.
  2. τ_s = c_u × N_c
  3. τ_s = γ × z × sin ϕ'
  4. τ_s = q_ult / 10

Explanation

The effective stress β-method calculates pile shaft resistance in granular soils (CWG/CWV) as τ_s = β × σ'_v, where β = K × tan δ. K depends on soil stress history and installation method (driven vs bored).

Question 27Geotechnical engineering models and applications

When assessing group settlement of a pile group driven into a thick sand stratum, how does group settlement S_group compare to single pile settlement S_single under identical load per pile?

  1. S_group is greater than S_single because the pressure bulb of the pile group extends significantly deeper into underlying soil layers. Correct answer.
  2. S_group is smaller than S_single due to arching effects between pile caps.
  3. S_group is exactly equal to S_single regardless of pile group footprint area.
  4. S_group is zero because friction piles eliminate all settlement.

Explanation

Overlap of individual pile stress fields creates a much wider and deeper composite pressure bulb for the pile group. Since stress penetrates deeper into compresssible soil strata, group settlement S_group exceeds single pile settlement S_single.

Question 28Geotechnical engineering models and applications

What operational precaution must be taken when installing Continuous Flight Auger (CFA) piles in loose saprolitic soils (CWG) with high groundwater table to prevent ground loss?

  1. Controlling auger penetration rate and rotation speed to avoid over-flighting (drawing surrounding soil into auger flights), and maintaining positive concrete pump pressure during auger withdrawal. Correct answer.
  2. Vibrating the auger string at 50 Hz without pumping concrete.
  3. Pumping high-pressure compressed air down the hollow stem continuously.
  4. Withdrawing the auger completely before commencing concrete pumping.

Explanation

Over-flighting (rotating auger without sufficient penetration rate) mines soil from surrounding ground, causing ground collapse/settlement of nearby structures. Maintaining positive concrete pumping pressure during controlled auger extraction prevents shaft necking and hole collapse.

Question 29Geotechnical engineering models and applications

A Plate Loading Test (PLT) using a 300 mm diameter rigid circular plate is conducted on compacted CDG fill to determine modulus of subgrade reaction K_s300. If K_s300 = 60 MPa/m, how is the estimated subgrade reaction modulus K_sB for a 2.0 m wide prototype footing in sand calculated per Terzaghi?

  1. K_sB = K_s300 × [(B + 0.3) / (2 × B)]² = 60 × [(2.0 + 0.3) / (2 × 2.0)]² = 60 × (2.3 / 4.0)² = 60 × 0.3306 = 19.8 MPa/m Correct answer.
  2. K_sB = K_s300 × (2.0 / 0.3) = 400 MPa/m
  3. K_sB = K_s300 = 60 MPa/m
  4. K_sB = K_s300 / 10 = 6.0 MPa/m

Explanation

Terzaghi's size scaling formula for modulus of subgrade reaction in sandy soils: K_sB = K_s300 × [(B + 0.3) / (2 B)]², where B is footing width in meters. Plugging in B = 2.0 m: K_sB = 60 × [(2.3) / (4.0)]² = 60 × 0.3306 = 19.8 MPa/m.

Question 30Geotechnical engineering models and applications

What is the purpose of post-grouting (shaft/base grouting) performed in large-diameter bored piles in Hong Kong?

  1. To compress soft base debris (sloughing), consolidate surrounding disturbed soil, and increase ultimate shaft friction and end-bearing resistance. Correct answer.
  2. To melt subterranean granite rock beneath the pile tip.
  3. To prevent electrochemical corrosion of reinforcement steel in ocean water.
  4. To reduce concrete curing time from 28 days to 2 hours.

Explanation

High-pressure post-grouting via pre-installed tube-a-manchette (TAM) pipes injects cement grout under 20 to 40 bar pressure at the pile toe or shaft. This pre-compresses soft sediments, fills micro-voids, and substantially enhances skin friction and end bearing.

Question 31Geotechnical engineering models and applications

Why are micropiles (mini-piles, typically 100-300 mm diameter with high-strength steel casing or bar core) widely used for foundation underpinning in congested urban sites in Hong Kong?

  1. They can be installed using light drilling rigs in low headroom and tight access conditions, penetrating boulders and socketing into bedrock with minimal vibration. Correct answer.
  2. They carry higher total vertical load per pile than a 2.5 m diameter bored pile.
  3. They do not require any cement grout or drilling fluids.
  4. They are driven dynamically using heavy 10-tonne drop hammers.

Explanation

Micropiles offer small footprint, low headroom capability, low vibration rotary drilling, and ability to drill through hard granite boulders and obstructions to achieve socket grout bond capacity in bedrock, making them ideal for urban underpinning.

Question 32Geotechnical engineering models and applications

In karst marble geology encountered in Yuen Long and Tung Chung (Hong Kong), what major geotechnical hazard affects pile foundation design?

  1. Underground solution cavities, irregular bedrock profiles, and overhangs, requiring extensive core drilling and deep pile sockets to reach sound marble. Correct answer.
  2. High expansive montmorillonite clay swelling pressures exceeding 10 MPa.
  3. Severe volcanic ash explosions during pile drilling.
  4. Liquefaction of solid unweathered marble bedrock.

Explanation

Karstified marble in NW and N Lantau HK contains dissolution cavities, solution channels, overhangs, and pinnacles. Bored piles or H-piles must drill through cavities and verify sound rock thickness below the pile toe (typically min 5 m or 3 pile diameters of sound rock core without cavities).

Question 33Geotechnical engineering models and applications

According to Broms' method for laterally loaded vertical piles, how does a 'short rigid pile' failure mode differ from a 'long flexible pile' failure mode in sand?

  1. A short rigid pile rotates as a rigid body about a point near the toe without yielding structurally; a long flexible pile forms a plastic bending hinge along the shaft at depth of maximum bending moment. Correct answer.
  2. A short rigid pile fails purely in tension, whereas a long flexible pile fails in end bearing.
  3. A short rigid pile exhibits infinite lateral deflection without soil failure.
  4. A long flexible pile rotates as a rigid body around the pile head.

Explanation

In Broms lateral load theory: Short piles have high structural rigidity relative to soil stiffness, failing by rigid body rotation/translation against ultimate soil pressure. Long piles undergo large bending, forming a yield plastic hinge at depth of maximum bending moment while lower pile shaft remains undisturbed.

Question 34Geotechnical engineering models and applications

What is the primary benefit of designing a Piled Raft Foundation compared to a conventional freestanding pile group for a high-rise tower in Hong Kong?

  1. The raft slab directly transfers a significant portion of structural load to surface soil, while piles control total and differential settlements, reducing required pile length or quantity. Correct answer.
  2. The raft slab eliminates all structural bending moments in the building columns.
  3. Piles carry 100% of the load while the raft carries 0%, acting purely as a waterproof membrane.
  4. Piled rafts eliminate the need for site ground investigation.

Explanation

Piled raft foundations optimize soil-structure interaction. The raft slab mobilizes shallow soil bearing capacity for static loads, while strategically placed piles reduce differential settlement and share heavy column loads, resulting in economical foundation design.

Question 35Geotechnical engineering models and applications

A smooth vertical retaining wall of height H = 6.0 m retains horizontal backfill of dry sand with unit weight γ = 19.0 kN/m³ and effective friction angle ϕ' = 30°. Using Rankine's earth pressure theory, what is the total active lateral force E_a per meter length of wall?

  1. 114.0 kN/m Correct answer.
  2. 342.0 kN/m
  3. 57.0 kN/m
  4. 228.0 kN/m

Explanation

Rankine active coefficient: K_a = (1 - sin 30°) / (1 + sin 30°) = 0.5 / 1.5 = 1/3 = 0.3333. Active earth pressure at wall base: p_a = K_a × γ × H = (1/3) × 19.0 × 6.0 = 38.0 kPa. Total active force per meter length: E_a = 0.5 × p_a × H = 0.5 × 38.0 × 6.0 = 114.0 kN/m.

Question 36Geotechnical engineering models and applications

For the retaining wall in question 056 (γ = 19.0 kN/m³, ϕ' = 30°), what is the total passive force E_p mobilized per meter length over an embedded toe depth D = 2.0 m using Rankine theory?

  1. 114.0 kN/m Correct answer.
  2. 38.0 kN/m
  3. 228.0 kN/m
  4. 57.0 kN/m

Explanation

Rankine passive coefficient: K_p = (1 + sin 30°) / (1 - sin 30°) = 1.5 / 0.5 = 3.0. Passive earth pressure at depth D = 2.0 m: p_p = K_p × γ × D = 3.0 × 19.0 × 2.0 = 114.0 kPa. Total passive force per meter: E_p = 0.5 × p_p × D = 0.5 × 114.0 × 2.0 = 114.0 kN/m.

Question 37Geotechnical engineering models and applications

In the common simplified comparison of smooth vertical walls with level cohesionless backfill, how does the Coulomb sliding-wedge approach extend the elementary Rankine active-pressure model?

  1. Coulomb considers wall friction angle δ and inclined wall back/fill geometry using a planar sliding wedge limit equilibrium, whereas Rankine assumes a smooth vertical wall and horizontal ground. Correct answer.
  2. Coulomb assumes zero wall friction while Rankine assumes infinite wall friction.
  3. Coulomb is applicable only to cohesive clays, whereas Rankine applies only to rocks.
  4. Coulomb computes earth pressure using fluid hydrostatic pressure formulas.

Explanation

Coulomb's trial wedge theory includes interface wall friction δ, backfill inclination β, and wall back batter α. Rankine's theory assumes a smooth vertical wall face (δ = 0) with stress state in full plastic equilibrium throughout the backfill.

Question 38Geotechnical engineering models and applications

In pseudo-static seismic design of retaining walls using Mononobe-Okabe analysis, how are earthquake inertial forces incorporated?

  1. Horizontal and vertical seismic inertia coefficients (k_h and k_v) multiply the soil wedge weight W to create dynamic inertia forces k_h W and k_v W acting on the soil mass. Correct answer.
  2. By doubling the soil effective cohesion c'.
  3. By setting pore water pressure u to zero across the backfill.
  4. By replacing soil shear strength with concrete compressive strength.

Explanation

Mononobe-Okabe extends Coulomb's wedge theory to seismic loading by introducing pseudo-static horizontal force F_h = k_h × W and vertical force F_v = k_v × W, resulting in an inclined effective gravity vector θ = atan[k_h / (1 ∓ k_v)].

Question 39Geotechnical engineering models and applications

A gravity retaining wall has a total vertical dead weight V = 400 kN/m, total horizontal active earth force E_ah = 120 kN/m, and base interface friction angle δ = 28° (tan 28° = 0.5317). Base cohesion c' = 0 and uplift pore pressure U = 50 kN/m. What is the Factor of Safety against sliding FoS_sliding along the base?

  1. 1.55 Correct answer.
  2. 1.77
  3. 1.33
  4. 2.11

Explanation

Effective vertical force on wall base: V' = V - U = 400 - 50 = 350 kN/m. Maximum sliding resistance: R_sliding = V' × tan δ = 350 × tan 28° = 350 × 0.5317 = 186.1 kN/m. Factor of Safety against sliding: FoS_sliding = R_sliding / E_ah = 186.1 / 120 = 1.55.

Question 40Geotechnical engineering models and applications

A retaining wall base has width B = 3.0 m. Total vertical load V = 400 kN/m acts with an eccentricity e = 0.20 m from the base centerline (within the middle third, e ≤ B/6 = 0.50 m). What are the maximum and minimum foundation bearing pressures (q_max and q_min)?

  1. q_max = 186.7 kPa, q_min = 80.0 kPa Correct answer.
  2. q_max = 133.3 kPa, q_min = 133.3 kPa
  3. q_max = 266.7 kPa, q_min = 0 kPa
  4. q_max = 220.0 kPa, q_min = 40.0 kPa

Explanation

Average stress q_avg = V / B = 400 / 3.0 = 133.33 kPa. Using linear trapezoidal pressure distribution formula: q = (V / B) × [1 ± (6 × e / B)]. q_max = 133.33 × [1 + (6 × 0.20 / 3.0)] = 133.33 × (1 + 0.40) = 186.67 kPa ≈ 186.7 kPa. q_min = 133.33 × (1 - 0.40) = 80.00 kPa.

Question 41Geotechnical engineering models and applications

In the design of a Mechanically Stabilized Earth (MSE) / Reinforced Soil Wall using geogrids or metallic strips in Hong Kong, what two internal stability checks are required?

  1. Tensile rupture of the reinforcement and pullout failure of reinforcement from backfill soil behind the potential failure surface. Correct answer.
  2. Overall deep circular slip and foundation bearing capacity failure.
  3. Sliding along wall base and overturning about toe.
  4. Concrete compression crushing and steel rebar buckling.

Explanation

MSE wall stability requires checking both Internal Stability (reinforcement tensile strength against rupture, and pullout resistance of reinforcement length beyond failure surface) and External Stability (sliding, overturning, bearing capacity, overall slip).

Question 42Geotechnical engineering models and applications

For a reinforced concrete cantilever L-shaped retaining wall, where does the maximum bending moment occur in the vertical stem slab?

  1. At the junction between the vertical stem and the base footing slab. Correct answer.
  2. At the top crest of the vertical stem.
  3. At the midpoint of the base heel slab.
  4. At the outer tip of the base toe slab.

Explanation

The vertical stem of a cantilever retaining wall acts as a vertical structural cantilever fixed at its base. Lateral active earth pressure (increasing linearly with depth) creates maximum bending moment at the stem-base junction connection.

Question 43Geotechnical engineering models and applications

What lateral earth pressure coefficient K_0 should be used for designing unyielding rigid basement walls constructed in normally consolidated soil (Jaky's formula)?

  1. K_0 = 1 - sin ϕ' Correct answer.
  2. K_0 = tan²(45° - ϕ'/2)
  3. K_0 = tan²(45° + ϕ'/2)
  4. K_0 = 1 + sin ϕ'

Explanation

Rigid non-yielding structures (such as heavy basement box walls restrained by floor slabs) do not undergo sufficient lateral yielding to mobilize active conditions. At-rest lateral earth pressure coefficient is given by Jaky's equation K_0 = 1 - sin ϕ'.

Question 44Geotechnical engineering models and applications

A cohesive soil backfill has effective cohesion c' = 15 kPa, friction angle ϕ' = 20° (√K_a = 0.7002), and unit weight γ = 18.0 kN/m³. What is the theoretical depth z_c of the tension crack that can develop at the top of the un-waterfilled backfill?

  1. 2.38 m Correct answer.
  2. 1.19 m
  3. 4.76 m
  4. 3.50 m

Explanation

Active earth pressure equation for cohesive soil: p_a = K_a × γ × z - 2 c' √K_a. Setting p_a = 0 gives tension crack depth z_c = (2 c') / (γ √K_a). Substituting values: z_c = (2 × 15) / (18.0 × 0.7002) = 30 / 12.604 = 2.38 m.

Question 45Geotechnical engineering models and applications

A uniform surcharge load q = 20.0 kPa acts on the horizontal ground surface behind a vertical retaining wall. Backfill properties are γ = 18.0 kN/m³ and ϕ' = 30° (K_a = 0.3333). Wall height H = 5.0 m. What is the additional horizontal active force ΔE_aq per meter length of wall due to surcharge q?

  1. 33.33 kN/m acting at H/2 (2.5 m) above base Correct answer.
  2. 66.67 kN/m acting at H/3 (1.67 m) above base
  3. 16.67 kN/m acting at base
  4. 100.0 kN/m acting at top of wall

Explanation

Uniform surcharge q produces a uniform lateral active pressure Δp_a = K_a × q = (1/3) × 20.0 = 6.667 kPa over full height H. Additional active force per meter: ΔE_aq = K_a × q × H = 6.667 × 5.0 = 33.33 kN/m. Because pressure distribution is rectangular, resultant acts at mid-height H/2 = 2.5 m above base.

Question 46Geotechnical engineering models and applications

In deep foundation excavations for high-density urban sites in Hong Kong, why are reinforced concrete Diaphragm Walls often selected over interlocking Sheet Pile walls for deep basements (> 15 m depth)?

  1. Diaphragm walls provide extremely high flexural stiffness (EI) to minimize ground settlement, serve as permanent basement structural walls, and offer high water-tightness in soft marine clay and high water tables. Correct answer.
  2. Sheet piles are more rigid than 1.2 m thick reinforced concrete diaphragm walls.
  3. Diaphragm walls require no heavy machinery or slurry trenching support.
  4. Sheet pile driving causes zero noise and vibration in urban streets.

Explanation

Cast-in-place diaphragm walls (typically 0.8 m to 1.5 m thick) possess high bending rigidity (EI), controlling wall deflections and adjacent ground settlements. They double as permanent basement perimeter walls and provide watertight cutoffs in high water table urban areas.

Question 47Geotechnical engineering models and applications

What is the primary formula for checking Basal Heave Factor of Safety FoS_bh at the base of a deep braced excavation in soft clay (Bjerrum & Eide method)?

  1. FoS_bh = (N_c × c_u) / (γ × H + q), where N_c is stability factor, c_u is undrained shear strength below base, γ is soil unit weight, H is excavation depth, and q is surface surcharge. Correct answer.
  2. FoS_bh = (γ × H) / c_u
  3. FoS_bh = tan ϕ' / tan β
  4. FoS_bh = K_p / K_a

Explanation

Bjerrum & Eide basal heave evaluation models bearing capacity failure of soft clay under total vertical overburden pressure (γ H + q). Ultimate resisting pressure is N_c × c_u, yielding FoS_bh = (N_c × c_u) / (γ H + q). Minimum required FoS is typically 1.50.

Question 48Geotechnical engineering models and applications

How do Peck's (1969) apparent earth pressure envelopes for designing struts in braced cuts differ from classical Rankine active earth pressure distributions?

  1. Peck envelopes are trapezoidal or rectangular envelope distributions enclosing maximum measured strut loads across multi-supported walls, whereas Rankine predicts a linear triangular pressure distribution. Correct answer.
  2. Peck envelopes assume zero soil pressure along the entire wall depth.
  3. Rankine distributions give higher earth pressure at the top of the wall than at the bottom.
  4. Peck envelopes apply only to unstrutted cantilever retaining walls.

Explanation

In multi-strutted deep excavations, soil arching transfers loads to strut support levels. Apparent earth pressure envelopes (Peck 1969) use trapezoidal or rectangular shapes to envelope maximum strut forces, unlike the classical triangular distribution of Rankine.

Question 49Geotechnical engineering models and applications

Why is pre-stressing (pre-loading) of steel struts applied during installation in an Excavation and Lateral Support (ELS) system?

  1. To minimize inward lateral wall movement and adjacent ground settlement before further excavation proceed down to lower levels. Correct answer.
  2. To increase the self-weight of steel struts.
  3. To melt the steel wailing beam to wall connections.
  4. To create negative pore water pressure in adjacent marine clay.

Explanation

Pre-stressing struts with hydraulic jacks (typically to 50% - 80% of design load) eliminates slack in mechanical connections and compresses the wall against backfill before further bulk excavation, minimizing lateral wall deflections and ground settlement.

Question 50Geotechnical engineering models and applications

What is the function of the 'free unbonded length' (free tendon length) of a prestressed ground anchor supporting an ELS wall?

  1. To ensure the anchor bond length is located beyond the active soil failure wedge, preventing anchor load loss as the soil wedge moves. Correct answer.
  2. To allow water to drain freely through the anchor head.
  3. To increase friction between steel strands and surrounding upper soil.
  4. To reduce the yield strength of the high-tensile steel tendon.

Explanation

Prestressed ground anchor tendons have a free unbonded length (sheathed in plastic sleeve) passing through the potential active failure wedge, transferring stress exclusively to the fixed bond length anchored deep into competent rock/soil behind the failure plane.