100+ Free IESL Part II Civil Practice Questions

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2026 Statistics

Key Facts: IESL Part II Civil Exam

40%

Passing Score

IESL exam regulations

5 Core

Subject Sections

IESL Part II civil syllabus

LKR 5,000+

Historical Fee (per paper)

IESL finance desk

AMIESL

Leads To (Academic Route)

Institution of Engineers, Sri Lanka

Dec 2022

Phased Out Date

IESL official council decision

The IESL Qualifying Examination Part II - Civil Engineering is a landmark credential for engineers in Sri Lanka seeking professional recognition. It comprises comprehensive examinations in structural analysis, hydraulics/water resources, soil mechanics, surveying, and materials. Earning AMIESL eligibility requires passing all core sections with a minimum mark of 40%. Although the traditional qualifying exam was phased out in 2022, understanding these core competencies remains vital for the current IESL academic qualification assessment routes.

Sample IESL Part II Civil Practice Questions

Try these sample questions to test your IESL Part II Civil exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A continuous beam has three spans and is supported by four roller supports. The beam is continuous over all supports. Under what condition can this beam be classified as statically determinate?

A.If the beam has exactly three internal hinges strategically placed in the spans.

B.If the beam has exactly two internal hinges placed at the continuous supports.

C.If the beam carries only symmetrical vertical loads.

D.It can never be statically determinate regardless of internal hinges.

Explanation: For a beam, the degree of static indeterminacy is given by D = R - (3 + C), where R is the number of support reactions and C is the number of equations of condition provided by internal hinges. For a three-span continuous beam on 4 roller supports, with horizontal restraint assumed at one support, there are 4 vertical reactions and 1 horizontal reaction (R = 5). For D = 0, C must equal 5 - 3 = 2. Therefore, introducing exactly two internal hinges placed at the continuous supports (Option 1) provides the necessary 2 equations of condition to make the structure statically determinate and stable. If exactly three hinges are introduced, D becomes negative (D = -1), creating an unstable mechanism.

2A rigid jointed plane frame has 5 members, 4 rigid joints (including supports), and is fixed at its two supports. What is its degree of static indeterminacy?

A.3

B.6

C.9

D.12

Explanation: For a rigid-jointed plane frame, the degree of static indeterminacy (D) can be calculated using the formula D = 3M + R - 3J, where M is the number of members, R is the number of support reactions, and J is the number of joints (including supports). Here, we have M = 5 members. Since both supports are fixed, R = 2 * 3 = 6 reactions. There are J = 4 joints (including the supports and internal rigid joints). Thus, D = 3(5) + 6 - 3(4) = 15 + 6 - 12 = 9. Therefore, the frame is statically indeterminate to the 9th degree.

3Determine the determinacy of a planar pin-jointed truss with 11 members and 7 joints, supported by a pinned support at one end and a roller support at the other.

A.Statically determinate externally and internally

B.Statically indeterminate to the 1st degree internally

C.Statically indeterminate to the 1st degree externally

D.Unstable under general loading

Explanation: For a planar pin-jointed truss, the degree of static indeterminacy (D) is given by D = m + r - 2j, where m is the number of members, r is the number of support reactions, and j is the number of joints. Here, we have m = 11 members and j = 7 joints. The supports are one pinned (2 reactions) and one roller (1 reaction), giving r = 3 support reactions. Evaluating the formula: D = 11 + 3 - 2(7) = 14 - 14 = 0. Since D = 0 and the supports are non-parallel and non-concurrent, the truss is statically determinate both externally and internally.

4A cantilever beam of span L and constant flexural rigidity EI carries a concentrated point load P at its free end. Using the moment-area method, what is the vertical deflection at the free end?

A.PL^3 / (3EI)

B.PL^3 / (8EI)

C.PL^3 / (48EI)

D.5PL^3 / (384EI)

Explanation: Using the second moment-area theorem, the tangential deviation of the free end A from the tangent at the fixed support B is equal to the moment of the M/EI diagram between A and B about A. The bending moment diagram is a triangle with maximum ordinate PL at the fixed support and 0 at the free end. The area of this M/EI diagram is (1/2) * L * (PL/EI) = PL^2 / (2EI). The centroid of this triangular area is located at a distance x = (2/3)L from the free end A. Therefore, the deflection delta = Area * x = [PL^2 / (2EI)] * [(2/3)L] = PL^3 / (3EI).

5A simply supported beam of span L and uniform flexural rigidity EI carries a uniformly distributed load (UDL) of intensity w per unit length over its entire span. What is the maximum vertical deflection of the beam?

A.5wL^4 / (384EI)

B.wL^4 / (384EI)

C.wL^4 / (8EI)

D.wL^3 / (48EI)

Explanation: For a simply supported beam with a uniformly distributed load w over its entire span L, the maximum deflection occurs at the midspan. By standard deflection formulas (which can be derived using double integration, Macaulay's method, or conjugate beam method), the maximum deflection is delta_max = 5wL^4 / (384EI). Option B corresponds to a beam fixed at both ends, while Option C is for a cantilever under UDL.

6A simply supported beam of span L is subjected to a central concentrated point load P. What is the maximum bending moment in the beam, and where does it occur?

A.PL/4 at midspan

B.PL/2 at midspan

C.PL/8 at supports

D.PL/4 at supports

Explanation: For a simply supported beam under a central point load P, the support reactions are each P/2. The bending moment increases linearly from zero at the supports to a maximum value at midspan. The maximum bending moment M_max is calculated at midspan (x = L/2) as M = (P/2) * (L/2) = PL/4.

7A simply supported beam of span L = 6 m is subjected to a point load of 30 kN at a distance of 2 m from the left support. What is the maximum bending moment in the beam?

A.40 kNm

B.45 kNm

C.60 kNm

D.90 kNm

Explanation: For a simply supported beam of span L carrying a point load P at a distance 'a' from the left support and 'b' from the right support, the reactions are R_left = P*b/L and R_right = P*a/L. The maximum bending moment occurs directly under the point load and is equal to P*a*b/L. Here, P = 30 kN, L = 6 m, a = 2 m, and b = 6 - 2 = 4 m. Thus, M_max = 30 * 2 * 4 / 6 = 40 kNm.

8A beam of length L is simply supported at its left end and at a distance of 0.8L from the left end, leaving an overhang of 0.2L at the right. If the beam carries a UDL of intensity w over the entire length, what is the bending moment at the right support?

A.-0.02 wL^2

B.-0.04 wL^2

C.-0.08 wL^2

D.0.02 wL^2

Explanation: The overhang portion has a length of 0.2L. The load on this overhang is a UDL of intensity w. Treating the overhang as a cantilever extending from the right support, the bending moment at the support is due solely to the UDL acting on the overhang. Bending Moment M = -w * (length of overhang)^2 / 2 = -w * (0.2L)^2 / 2 = -w * 0.04L^2 / 2 = -0.02 wL^2. The negative sign indicates hogging moment.

9In a pin-jointed truss, three members meet at a joint which is not subjected to any external load or support reaction. If two of the members are collinear, what can be concluded about the force in the third member?

A.It must be zero.

B.It is equal to the sum of the forces in the collinear members.

C.It is equal to the force in one of the collinear members.

D.It cannot be determined without analyzing the whole truss.

Explanation: By joint equilibrium, if we align the x-axis with the two collinear members, the force equilibrium in the perpendicular y-direction (sum of forces in y = 0) is only contributed to by the component of the force in the third member. For this sum to equal zero, the force in the third member must be zero (assuming it is not collinear with the other two). Thus, the third member is a zero-force member.

10A symmetric roof truss of span 12 m and height 3 m is subjected to a vertical point load of 40 kN at the apex (ridge joint). What is the force in the horizontal bottom tie member at the center of the span?

A.20 kN (Tension)

B.40 kN (Tension)

C.20 kN (Compression)

D.40 kN (Compression)

Explanation: Let the truss be a simple triangular truss ABC, with A and B at the supports (12m apart) and C at the apex (3m high). The support reactions are R_A = R_B = 20 kN. Consider the joint A: the inclined member AC has a horizontal run of 6m and vertical rise of 3m. Angle theta with horizontal is tan(theta) = 3/6 = 0.5. The vertical reaction is balanced by the vertical component of force in AC: F_AC * sin(theta) = 20 kN. F_AC = 20 / sin(theta). Since sin(theta) = 3/sqrt(3^2 + 6^2) = 3/sqrt(45) = 1/sqrt(5). F_AC = 20 * sqrt(5) (Compression). The horizontal force in the bottom tie member AB is balanced by the horizontal component of F_AC: F_AB = F_AC * cos(theta) = [20 * sqrt(5)] * [2/sqrt(5)] = 40 kN (Tension). Therefore, the force in the horizontal bottom tie is 40 kN in tension.

About the IESL Part II Civil Exam

The IESL Qualifying Examination Part II Civil Engineering is a comprehensive assessment covering structural analysis, hydraulics, soil mechanics, surveying, and construction materials. It forms a key milestone for non-accredited degree holders aiming to qualify for Associate Membership (AMIESL) and the Chartered Engineer (C.Eng) pathway in Sri Lanka.

Questions

100 scored questions

Time Limit

3 hours per paper (historical)

Passing Score

40%

Exam Fee

Historic registration (contact IESL for modern equivalence paths) (Institution of Engineers, Sri Lanka (IESL))

IESL Part II Civil Exam Content Outline

20%

Structural Analysis

Analysis of determinate and indeterminate structures, bending moment and shear force diagrams, deflections, truss analysis, and force/displacement methods.

20%

Hydraulics & Water Resources

Fluid mechanics principles, hydrostatic pressure, pipe flow networks, open-channel hydraulics, Bernoulli's equation, and basic hydrology.

20%

Soil Mechanics & Geotechnical Engineering

Phase relationships, soil classification, flow of water through soils (seepage), consolidation, shear strength, bearing capacity, and lateral earth pressures.

20%

Surveying

Levelling operations, traverse adjustments (Bowditch method), curve setting (circular and transition curves), areas and volumes, and modern instruments (Total Stations/GPS).

20%

Construction Materials & Concrete Technology

Cement chemistry, concrete properties and mix design, workability, aggregates testing, steel reinforcement, timber technology, and construction technology.

How to Pass the IESL Part II Civil Exam

What You Need to Know

Passing score: 40%
Exam length: 100 questions
Time limit: 3 hours per paper (historical)
Exam fee: Historic registration (contact IESL for modern equivalence paths)

Keys to Passing

Complete 500+ practice questions
Score 80%+ consistently before scheduling
Focus on highest-weighted sections
Use our AI tutor for tough concepts

IESL Part II Civil Study Tips from Top Performers

1For Structural Analysis, practice identifying determinacy and using the method of sections for quick truss analysis under point loads.

2In Hydraulics, master Bernoulli's equation and pipe head loss calculations using Darcy-Weisbach and Manning's formula for open-channel flows.

3Soil Mechanics requires a strong grasp of the phase diagram (void ratio, porosity, saturation) and calculation of effective vertical stress at depth.

4Review the rise-and-fall and height-of-collimation booking methods in Surveying, and practice computing curve deflection angles.

5Understand the chemical composition of cement (Bogue's compounds) and factors influencing the water-cement ratio and concrete workability.

Frequently Asked Questions

What is the IESL Qualifying Examination Part II Civil Engineering?

Historically, the IESL Part II was a written examination administered by the Institution of Engineers, Sri Lanka. It served as a critical qualifying step for engineering diplomates and non-accredited degree holders to satisfy the academic requirements for Associate Membership (AMIESL).

Is the IESL Part II exam still active?

No. The IESL officially phased out the Part I, Part II, and Part III written examinations, concluding the final sessions in December 2022. Candidates seeking recognition now undergo the IESL Academic Qualification Assessment route or study through the accredited IESL College of Engineering.

What topics were covered in the Civil Engineering syllabus?

The syllabus required mastery of five core subjects: Structural Analysis (determinacy, truss analysis, beams, and indeterminate frames), Hydraulics/Water Resources (fluid mechanics, pipe networks, and open channel flow), Soil Mechanics (classification, permeability, consolidation, and shear strength), Surveying (traverse, levelling, and circular curves), and Construction Materials (cement chemistry, concrete design, steel, and timber).

What was the passing mark for the IESL Part II exam?

The passing mark was 40% per subject paper. If a candidate passed a majority of the papers, they could be referred in the remaining papers and retake them in subsequent sessions.

Does practicing these questions help with the current IESL assessment?

Yes. The current Academic Qualification Assessment and the examinations conducted by the IESL College of Engineering cover the exact same engineering science fundamentals. Practicing these core civil engineering problems is highly beneficial.