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2026 Statistics
Key Facts: HKIE Professional Assessment — Geotechnical Discipline Exam
Portfolio
Assessment Mode
HKIE PA Regulations
2 Hours
Written Essay Duration
HKIE PA Regulations
HK$ 3,100
Total Application & Fee
HKIE Fee Schedule 2026
4 Areas
HKIE Competence Standard
HKIE Competence Standards
45 Mins
Interview Duration
HKIE PA Regulations
RPE Geotechnical
Target Qualification
HKIE Membership Regulations
The HKIE Professional Assessment (Geotechnical Discipline) evaluates candidates for Corporate Membership (MHKIE) through portfolio review, interview, and a 2-hour technical essay. This 100-question study bank offers an English-language MCQ adaptation covering soil mechanics, rock weathering (Geoguide 3), slope stability (Geoguide 1 & GCO Pub 1/90), soil nails (Geoguide 7), pile foundations, retaining walls, ELS deep excavation, seepage flow, and HKIE ethics to support candidate technical preparation.
Sample HKIE Professional Assessment — Geotechnical Discipline Practice Questions
Try these sample questions to test your HKIE Professional Assessment — Geotechnical Discipline exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A 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?
A.104.95 kPa
B.154.00 kPa
C.49.05 kPa
D.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.
2According 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?
A.93.23 kPa
B.124.47 kPa
C.81.23 kPa
D.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.
3Under Geoguide 3 (Guide to Soil and Rock Description), how is Grade III rock material classified within the rock weathering classification scheme for Hong Kong rocks?
A.Moderately Weathered Rock, where discoloration indicates weathering of rock material and along joints, but less than half of the rock material is decomposed or disintegrated.
B.Completely Weathered Rock, where all rock material is decomposed to soil state but original rock texture and structure are preserved.
C.Slightly Weathered Rock, where discoloration is present along major joints only and rock material shows no sign of weakening.
D.Highly Weathered Rock, where more than half of the rock material is decomposed or disintegrated to a soil state.
Explanation: Geoguide 3 classifies rock weathering into six grades: Grade I (Fresh), Grade II (Slightly Weathered), Grade III (Moderately Weathered), Grade IV (Highly Weathered), Grade V (Completely Weathered), and Grade VI (Residual Soil). Grade III represents Moderately Weathered Rock, in which discoloration is widespread and less than half of the rock material is decomposed/disintegrated.
4A 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?
A.285.2 mm
B.570.4 mm
C.190.1 mm
D.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.
5A 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?
A.1.36 years
B.5.43 years
C.2.71 years
D.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.
6In rock core logging per Geoguide 3, how is Rock Quality Designation (RQD) defined and what RQD range corresponds to 'Good' rock quality?
A.Percentage of intact core pieces ≥ 100 mm in length over total run length; 75% to 90% corresponds to Good quality.
B.Percentage of intact core pieces ≥ 50 mm in length over total run length; 50% to 75% corresponds to Good quality.
C.Ratio of total core recovery (TCR) to solid core recovery (SCR); > 90% corresponds to Good quality.
D.Percentage of intact core pieces ≥ 100 mm in length over total run length; > 90% corresponds to Good quality.
Explanation: RQD is defined as the sum of length of intact hard core pieces equal to or greater than 100 mm expressed as a percentage of total core run length. RQD classification: < 25% (Very Poor), 25-50% (Poor), 50-75% (Fair), 75-90% (Good), and > 90% (Excellent).
7What 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?
A.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.
B.In a CD test, consolidation is prevented during the initial stress application phase, whereas in a CU test, full consolidation is permitted.
C.In a CD test, total stress parameters c_u and ϕ_u are directly obtained, whereas CU tests only measure hydraulic conductivity.
D.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'.
8A 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?
A.High plasticity clay (CH), because its Plasticity Index (PI = 30%) lies above the A-line (PI_A = 25.55%).
B.High plasticity silt (MH), because its Plasticity Index (PI = 30%) lies below the A-line.
C.Low plasticity clay (CL), because its Plasticity Index is below 35%.
D.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).
9In Hong Kong geology, what is a key distinguishing mineralogical and mechanical characteristic of Completely Weathered Volcanics (CWV) compared to Completely Weathered Granite (CWG)?
A.CWV typically has a finer matrix, higher clay fraction, higher plasticity, and lower permeability than CWG.
B.CWV consists predominantly of coarse quartz sand grains with negligible silt and clay content.
C.CWV exhibits significantly higher permeability and lower effective friction angle than CWG.
D.CWV contains abundant micro-granitic quartz gravels causing higher field permeability than CWG.
Explanation: Completely Weathered Volcanics (CWV) derived from fine-grained tuffs have a finer soil particle grading, higher silt/clay content, higher plasticity, and lower permeability compared to Completely Weathered Granite (CWG), which is quartz-rich and well-graded gravelly sand.
10During a ground investigation in Hong Kong, a Standard Penetration Test (SPT) is conducted in submerged fine sand below the water table, yielding a field N-value of 22. If N > 15 in submerged fine sand, what dilatancy correction is recommended per Terzaghi & Peck?
A.N_corrected = 15 + 0.5 × (N - 15) = 18.5 ≈ 18 or 19
B.N_corrected = N × 1.5 = 33
C.N_corrected = N - 15 = 7
D.N_corrected = 15 + 1.5 × (N - 15) = 25.5
Explanation: When SPT is performed in submerged fine or silty sand with N > 15, pore pressure build-up during dynamic driving increases resistance. Terzaghi & Peck's dilatancy correction is N' = 15 + 0.5 × (N - 15). For N = 22, N' = 15 + 0.5 × (22 - 15) = 15 + 3.5 = 18.5.
About the HKIE Professional Assessment — Geotechnical Discipline Practice Questions
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