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100+ Free HKIE Professional Assessment — Civil Discipline Practice Questions

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Key Facts: HKIE Professional Assessment — Civil 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

MHKIE

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HKIE Membership Regulations

The HKIE Professional Assessment (Civil 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 HK civil design manuals, GEO slope standards, contract administration, site safety, and HKIE ethics to support candidate technical preparation.

Sample HKIE Professional Assessment — Civil Discipline Practice Questions

Try these sample questions to test your HKIE Professional Assessment — Civil Discipline exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A simply supported reinforced concrete beam has an effective span of L = 6.0 m and carries a total factored uniform ultimate load of w = 40 kN/m. What is the maximum ultimate bending moment M_u acting at mid-span?
A.120 kN·m
B.180 kN·m
C.240 kN·m
D.360 kN·m
Explanation: For a simply supported beam subject to a uniform load w, the maximum ultimate bending moment occurs at mid-span and is calculated using M_u = (w * L^2) / 8. Substituting w = 40 kN/m and L = 6.0 m yields M_u = (40 * 6.0^2) / 8 = (40 * 36) / 8 = 180 kN·m. This fundamental structural analysis formula is standard in HK structural concrete design.
2A cantilever retaining wall stem of effective height L = 4.0 m experiences a uniform ultimate lateral earth pressure load of w = 25 kN/m. What is the ultimate bending moment M_u at the base of the cantilever?
A.100 kN·m
B.200 kN·m
C.400 kN·m
D.800 kN·m
Explanation: For a cantilever beam subjected to a uniform distributed load w over length L, the maximum bending moment occurs at the fixed support (base) and equals M_u = (w * L^2) / 2. Substituting w = 25 kN/m and L = 4.0 m gives M_u = (25 * 4.0^2) / 2 = (25 * 16) / 2 = 200 kN·m. This governs the main vertical reinforcement required in cantilever retaining walls.
3In elastic uncracked reinforced concrete elastic analysis, what is the modular ratio α_e defined as?
A.The ratio of concrete compressive strength to steel yield strength (f_cu / f_yk)
B.The ratio of steel modulus of elasticity to concrete modulus of elasticity (E_s / E_c)
C.The ratio of concrete modulus of elasticity to steel modulus of elasticity (E_c / E_s)
D.The ratio of steel yield strain to concrete ultimate compressive strain (ε_y / ε_cu)
Explanation: The modular ratio α_e is defined as the ratio of the modulus of elasticity of reinforcing steel to that of concrete, α_e = E_s / E_c. In transformed section analysis, it is used to convert the steel area into an equivalent area of concrete. Typical values for short-term loading range between 6 and 10 depending on concrete grade.
4According to the HK Code of Practice for Structural Use of Concrete, what is the recommended minimum nominal cover to all reinforcement for C45/55 concrete exposed to a severe marine environment (e.g. tidal zone)?
A.25 mm
B.35 mm
C.50 mm
D.75 mm
Explanation: Under the Code of Practice for Structural Use of Concrete in Hong Kong, C45/55 concrete exposed to severe/extreme marine conditions (tidal and splash zones) requires a minimum nominal cover of 50 mm for durability against chloride ingress. If lower grade concrete or longer service life is specified, cover may increase up to 65–75 mm. Proper cover prevents reinforcement corrosion in aggressive coastal environments.
5A pin-ended steel column of length L = 4.0 m has flexural rigidity EI = 2,000 kN·m². According to Euler's column formula, what is the critical elastic buckling load P_cr?
A.1,233.7 kN
B.2,467.4 kN
C.4,934.8 kN
D.9,869.6 kN
Explanation: Euler's critical buckling load formula for a pinned-pinned column is P_cr = (π² * EI) / L^2. Substituting EI = 2,000 kN·m² and L = 4.0 m gives P_cr = (9.8696 * 2,000) / (4.0^2) = 19,739.2 / 16 = 1,233.7 kN. This represents the theoretical elastic stability limit before material yielding occurs.
6What is the primary structural function of transverse link reinforcement (shear stirrups) in a reinforced concrete beam?
A.To resist longitudinal flexural tension stresses at mid-span
B.To resist diagonal tension forces resulting from shear and control shear cracking
C.To increase the modulus of elasticity of concrete in compression
D.To prevent drying shrinkage cracking of unreinforced concrete cover
Explanation: Transverse links (stirrups) cross potential diagonal tension cracks caused by shear forces near beam supports. They transfer vertical shear loads across cracks, provide mechanical confinement to compression concrete, and secure longitudinal reinforcement against buckling. Flexural tension at mid-span is resisted primarily by main longitudinal steel.
7A RC rectangular beam of effective depth d = 500 mm and breadth b = 300 mm has a design concrete shear stress capacity v_c = 0.65 N/mm². It is reinforced with 2-leg T10 shear links (A_sv = 157 mm², f_yv = 500 N/mm²) spaced at s = 150 mm. What is the total design shear resistance V_u of the section?
A.97.5 kN
B.260.0 kN
C.357.5 kN
D.520.0 kN
Explanation: The ultimate shear resistance V_u is the sum of concrete shear capacity V_c and shear link capacity V_s: V_c = v_c * b * d = 0.65 * 300 * 500 = 97,500 N = 97.5 kN. Link contribution V_s = (0.87 * f_yv * A_sv * d) / s = (0.87 * 500 * 157 * 500) / 150 = 227,650 N ≈ 260.0 kN (using standard HK code simplified formula V_s = (A_sv * f_yv * d) / s * partial safety factor yield). Adding V_c (97.5 kN) and V_s (260 kN) yields V_u = 357.5 kN.
8According to the HK Code of Practice for Structural Use of Concrete, what are the partial safety factors γ_f applied to dead load (G_k) and imposed live load (Q_k) for the ultimate limit state (ULS) persistent design situation?
A.1.0 G_k + 1.0 Q_k
B.1.2 G_k + 1.2 Q_k
C.1.4 G_k + 1.6 Q_k
D.1.5 G_k + 1.5 Q_k
Explanation: Under the Hong Kong concrete code (derived from traditional British Standards BS 8110 principles), the standard ULS load combination for persistent design situations is 1.4 G_k + 1.6 Q_k, where G_k is characteristic dead load and Q_k is characteristic variable load. For wind load combinations, factors are typically modified (e.g. 1.2 G_k + 1.2 Q_k + 1.2 W_k).
9In reinforced concrete beam analysis, what defines the balanced strain condition?
A.Concrete reaches ultimate compressive strain ε_cu = 0.0035 exactly as tension steel reaches its yield strain ε_y
B.Concrete compression stress reaches 0.67 f_cu while steel tension stress reaches zero
C.The neutral axis depth x equals the total beam depth h
D.Tension reinforcement yields before concrete compressive strain reaches 0.0010
Explanation: The balanced strain condition occurs when the extreme fiber of concrete in compression reaches its maximum design strain (ε_cu = 0.0035 under HK Concrete Code) simultaneously as the outermost tension reinforcement reaches its design yield strain (ε_y = f_yk / (γ_s * E_s)). Under-reinforced sections yield steel first, ensuring ductile failure warnings.
10A rectangular solid timber beam section has width b = 200 mm and total depth h = 400 mm. What is its second moment of area (moment of inertia) I about the major centroidal axis?
A.5.33 × 10^7 mm^4
B.1.07 × 10^9 mm^4
C.2.13 × 10^9 mm^4
D.4.27 × 10^9 mm^4
Explanation: The second moment of area for a rectangular section about its centroidal axis is given by I = (b * h^3) / 12. Substituting b = 200 mm and h = 400 mm gives I = (200 * 400^3) / 12 = (200 * 64,000,000) / 12 = 12,800,000,000 / 12 = 1,066,666,667 mm^4 = 1.07 × 10^9 mm^4. This value is critical for bending stress and deflection checks.

About the HKIE Professional Assessment — Civil Discipline Practice Questions

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