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100+ Free HKIE Professional Assessment — Structural Discipline Practice Questions

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Key Facts: HKIE Professional Assessment — Structural Discipline Exam

Portfolio

Assessment Mode

HKIE PA Regulations

7.5 Hours

Written Exam Duration

HKIE Structural Examination Guidelines

HK$ 4,390

Total Application & Fee

HKIE Fee Schedule 2026

4 Areas

HKIE Competence Standard

HKIE Competence Standards

45 Mins

Interview Duration

HKIE PA Regulations

MHKIE / RSE

Target Qualification

HKIE Membership Regulations

The HKIE Professional Assessment (Structural Discipline) assesses candidates for Corporate Membership (MHKIE Structural) via portfolio review, presentation/interview, and a 2-hour technical write-up. Note: This 100-question study bank serves as an English-language MCQ study adaptation for structural mechanics, reinforced concrete design (HK Code of Practice for Structural Use of Concrete), structural steel design (HK Code of Practice for the Structural Use of Steel), wind code (HK Code of Practice on Wind Effects in Hong Kong), seismic design guidelines, HK Buildings Ordinance (Cap. 123), and HKIE ethics, and does not replace the official portfolio review, interview, or technical write-up.

Sample HKIE Professional Assessment — Structural Discipline Practice Questions

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

1In the matrix stiffness method of structural analysis, what is the axial stiffness term for a 2D prismatic frame element of length L, cross-sectional area A, and modulus of elasticity E?
A.E A / L
B.12 E I / L^3
C.4 E I / L
D.6 E I / L^2
Explanation: The axial stiffness of a prismatic bar element subject to direct tension or compression is EA/L. This term relates axial force to axial displacement along the member's local x-axis. The flexural stiffness terms in the 2D frame element matrix are represented by 12EI/L^3 for shear and 4EI/L or 2EI/L for bending moments.
2What is the degree of kinematic indeterminacy (unconstrained degree of freedom) for a 2D rigid jointed plane frame consisting of 2 spans, 1 storey, with 3 fixed column bases?
A.6
B.9
C.12
D.3
Explanation: The frame has 3 fixed bases (0 degrees of freedom at supports) and 2 rigid beam-column joints. Each unconstrained 2D rigid joint has 3 degrees of freedom (horizontal translation, vertical translation, and rotation). With 2 unconstrained joints, the total kinematic indeterminacy is 2 × 3 = 6.
3A pin-jointed steel truss member of length L = 4.0 m and cross-sectional area A = 2000 mm^2 carries a tensile axial force N = 300 kN under applied loads. If E = 200 GPa, what is the axial elongation of this member?
A.3.00 mm
B.1.50 mm
C.6.00 mm
D.4.50 mm
Explanation: Elongation delta = (N × L) / (A × E). Substituting N = 300 × 10^3 N, L = 4000 mm, A = 2000 mm^2, and E = 200,000 N/mm^2 gives delta = (300,000 × 4000) / (2000 × 200,000) = 1,200,000,000 / 400,000,000 = 3.00 mm.
4In the Moment Distribution Method, a continuous beam joint B connects span AB (length 6 m, flexural rigidity EI) and span BC (length 4 m, flexural rigidity 2EI). If both ends A and C are fixed, what is the distribution factor for member BC at joint B?
A.0.75
B.0.25
C.0.50
D.0.60
Explanation: Rotational stiffness for a member fixed at the far end is K = 4EI/L. For AB, K_AB = 4EI/6 = 0.667EI. For BC, K_BC = 4(2EI)/4 = 2.000EI. Total stiffness at joint B is K_total = 0.667EI + 2.000EI = 2.667EI. Distribution factor DF_BC = K_BC / K_total = 2.000 / 2.667 = 0.75 (or 3/4).
5In the Moment Distribution Method, what is the carry-over factor from a near joint to a far support that is pinned or hinged?
A.0.0
B.0.5
C.1.0
D.-0.5
Explanation: A pinned or hinged end cannot resist any bending moment (moment equals zero). Therefore, the carry-over factor to a pinned far support is 0.0. A carry-over factor of 0.5 applies only when the far support is fixed.
6A single-degree-of-freedom (SDOF) structural frame has a lumped mass m = 10,000 kg and lateral stiffness k = 16 MN/m. What is the natural circular frequency omega_n of the structure?
A.40.0 rad/s
B.4.0 rad/s
C.1600 rad/s
D.25.1 rad/s
Explanation: Natural circular frequency omega_n = sqrt(k / m). Expressing k in N/m: k = 16 × 10^6 N/m, m = 10,000 kg. omega_n = sqrt(16 × 10^6 / 10,000) = sqrt(1600) = 40.0 rad/s.
7For a harmonically excited SDOF structure subjected to force vibration at resonance (frequency ratio r = 1.0), what is the dynamic magnification factor DMF for a damping ratio zeta = 0.05 (5% critical damping)?
A.10.0
B.5.0
C.20.0
D.1.0
Explanation: At resonance (r = 1.0), the dynamic magnification factor formula simplifies to DMF = 1 / (2 × zeta). For zeta = 0.05, DMF = 1 / (2 × 0.05) = 1 / 0.10 = 10.0.
8A simply supported beam of span L = 6.0 m carries a point load P = 120 kN at mid-span. If flexural rigidity EI = 30,000 kNm^2, what is the maximum mid-span elastic deflection?
A.18.0 mm
B.9.0 mm
C.36.0 mm
D.24.0 mm
Explanation: Mid-span deflection delta = (P × L^3) / (48 × E × I). Substituting P = 120 kN, L = 6.0 m, and EI = 30,000 kNm^2 gives delta = (120 × 6^3) / (48 × 30,000) = (120 × 216) / 1,440,000 = 25,920 / 1,440,000 = 0.018 m = 18.0 mm.
9Where is the shear center located for a thin-walled open channel section subject to transverse bending parallel to its web?
A.Outside the section on the symmetry axis behind the web
B.At the centroid of the channel web
C.At the geometric centroid of the overall section
D.At the junction of the top flange and web
Explanation: For an open channel section, shear flows in the top and bottom flanges create horizontal force components acting in opposite directions. To prevent twisting (torsion) during bending parallel to the web, the resultant shear force must pass through the shear center, which lies outside the channel section on its axis of symmetry behind the web.
10A vertical steel column of length L = 6.0 m is fixed at the base and completely free at the top (cantilever column). If E = 200 GPa and minimum moment of inertia I = 4.5 × 10^7 mm^4, what is the Euler elastic buckling load P_cr?
A.616.8 kN
B.2467.4 kN
C.1233.7 kN
D.4934.8 kN
Explanation: For a fixed-free column, effective length L_e = 2.0 L = 2.0 × 6000 mm = 12,000 mm. Euler load P_cr = (pi^2 × E × I) / (L_e^2). P_cr = (pi^2 × 200,000 N/mm^2 × 4.5 × 10^7 mm^4) / (12,000 mm)^2 = (9.8696 × 9.0 × 10^12) / 144,000,000 = 88.826 × 10^12 / 1.44 × 10^8 = 616,850 N = 616.8 kN.

About the HKIE Professional Assessment — Structural Discipline Practice Questions

Verified exam format metadata for HKIE Professional Assessment — Structural Discipline (Hong Kong Institution of Engineers) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.