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Practice SetStructural Analysis

Structural Analysis MCQ Practice Set — 46 Questions with Answers

46 exam-oriented Structural Analysis multiple-choice questions with the correct answer and a clear explanation for each. Frequently asked in AE Level Civil Engineering, JE Level Civil Engineering. Solve the full set below for free — no login required.

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  1. 1
    Original practiceMEDIUM

    For a circular shaft, torque T=3.0 kN-m, polar moment J=2.0×10^6 mm⁴ and radius 30 mm. Maximum shear stress is

    A67.5 N/mm²
    B90 N/mm²
    C22.5 N/mm²
    D45 N/mm²

    Answer: D. 45 N/mm²

    Explanation: Torsion formula: shear stress tau = T*R/J, where T = applied torque, J = polar second moment of area (pi*D^4/32 for solid shaft), R = outer radius. Full torsion equation: tau/R = T/J = G*theta/L (G = modulus of rigidity, theta = angle of twist, L = shaft length). Valid for circular sections in elastic range. Using τ/R = T/J, τ = TR/J = 45 N/mm².

  2. 2
    Structural AnalysisMEDIUM

    The three governing equation-types in structural analysis are (i) equilibrium, (ii) compatibility, (iii) constitutive. To solve an indeterminate structure, the equations used are

    A(i) and (ii)
    B(i) and (iii)
    C(ii) and (iii)
    D(i), (ii) and (iii)

    Answer: D. (i), (ii) and (iii)

    Explanation: Solving an indeterminate structure requires all three types of equations: equilibrium equations (static), compatibility equations (kinematic), and constitutive relations (material properties).

  3. 3
    Structural AnalysisMEDIUM

    The static indeterminacy of the frame shown (with 3 internal hinges, plus hinged and fixed supports) is not calculable without the provided figure. Assuming the question refers to a standard frame problem with 15 as the intended answer.

    A12
    B13
    C14
    D15

    Answer: D. 15

    Explanation: The static indeterminacy calculation depends on the specific geometry and connectivity of the frame. Without the figure, the problem is incomplete.

  4. 4
    Moment Distribution MethodEASY

    Fixed end moments (FEM) for a beam of span L with central concentrated load W:

    AFEM = WL/8 at each end (opposite signs)
    BFEM = WL/12
    CFEM = WL/6
    DFEM = 5WL/48

    Answer: A. FEM = WL/8 at each end (opposite signs)

    Explanation: FEM for central load W: at near end = +WL/8, far end = −WL/8. For UDL w: FEM = ±wL²/12. Sign convention: sagging positive.

  5. 5
    Plastic AnalysisEASY

    Shape factor S for a rectangular cross-section in plastic analysis:

    AS = 1.0
    BS = 1.5
    CS = 1.12
    DS = 2.0

    Answer: B. S = 1.5

    Explanation: S = Mp/Me = Z_p/Z_e. For rectangle: Z_e = bd²/6; Z_p = bd²/4. S = (bd²/4)/(bd²/6) = 6/4 = 1.5. For circle: 1.7. For I-section: ~1.12–1.15.

  6. 6
    Influence LinesMEDIUM

    Muller-Breslau principle states: the influence line for any response function (reaction, SF, BM) has a shape equal to:

    AThe bending moment diagram under unit load
    BThe deflected shape of the structure when the constraint corresponding to the response is removed and a unit displacement/rotation is given
    CThe elastic curve of the beam
    DThe shear force diagram

    Answer: B. The deflected shape of the structure when the constraint corresponding to the response is removed and a unit displacement/rotation is given

    Explanation: Muller-Breslau: IL shape = deflected shape of released structure. Use for qualitative ILs. For quantitative: place unit load and compute response.

  7. 7
    Strength of Materials and Structural AnalysisMEDIUM

    Fixed end moment for full-span UDL w on fixed beam is:

    AwL/2
    BwL squared/12 at each end
    CwL squared/24 only at one end
    DwL squared/8 at each end

    Answer: B. wL squared/12 at each end

    Explanation: UDL on fixed beam gives hogging end moments wL^2/12.

  8. 8
    Strength of Materials and Structural AnalysisMEDIUM

    A pin-jointed plane truss is perfect if:

    Am = 3j - 6
    Bm = 2j + 3
    Cm = 2j - 3
    Dm = j - 3

    Answer: C. m = 2j - 3

    Explanation: For a perfect plane truss, members m and joints j satisfy m = 2j - 3.

  9. 9
    Shape Factor I-sectionMEDIUM

    The shape factor for a standard I-section (W/H section) is approximately:

    A1.12 to 1.15
    B1.5
    C2.0
    D1.0

    Answer: A. 1.12 to 1.15

    Explanation: For a standard I-section, most material is in the flanges. The shape factor typically ranges from 1.12 to 1.15 (less than 1.5 for solid rectangular because flanges are far from NA and carry most of the moment in both elastic and plastic cases).

  10. 10
    Collapse Load Propped CantileverMEDIUM

    For a propped cantilever of span L (fixed at A, prop at B) with plastic moment M_p, the collapse load for a concentrated load W at midspan is:

    A4M_p/L
    B6M_p/L
    C8M_p/L
    D2M_p/L

    Answer: B. 6M_p/L

    Explanation: Two hinges form: at fixed end A and at midspan C. Virtual work: W × (L/2)θ = M_p×2θ + M_p×θ = 3M_pθ. Hence W = 6M_p/L.

  11. 11
    Hollow Circular Shape FactorMEDIUM

    The shape factor for a thin-walled hollow circular cross-section is:

    A1.0
    B16/(3π) ≈ 1.70
    C4/π ≈ 1.27
    D1.5

    Answer: C. 4/π ≈ 1.27

    Explanation: For a thin-walled circular tube: Z_p = 4r²t and Z_e = πr²t (where r = radius, t = wall thickness). Shape factor = 4r²t/πr²t = 4/π ≈ 1.27.

  12. 12
    Kani Method of AnalysisMEDIUM

    In the Kani method of analyzing multi-storey frames, rotation contributions are calculated at each joint using the rotation factor (mu). For a joint with n members meeting, the rotation factor for member ij is:

    Amu_ij = -K_ij / (sum of K at joint j) = -1/2 x relative stiffness factor
    Bmu_ij = -0.5 x K_ij / sum(K) = -0.5 x distribution factor
    Cmu_ij = K_ij / (2 x sum K)
    Dmu_ij = EI/(2L) for each member

    Answer: B. mu_ij = -0.5 x K_ij / sum(K) = -0.5 x distribution factor

    Explanation: Kani method: rotation factor mu_ij = -0.5 x (K_ij / sum K_j), where K_ij = relative stiffness of member ij. Rotation contribution M_ij = 2 mu_ij x sum(rotation contributions) + fixed end moments contribution. Iteration converges faster than moment distribution for frames with sway. Sway is handled by storey sway factors.

  13. 13
    Portal Frame SwayMEDIUM

    A symmetric portal frame with fixed bases is analysed for a horizontal sway force H at the beam level. The horizontal shear in each column is:

    AH
    BH/2
    CH/4
    D2H

    Answer: B. H/2

    Explanation: By symmetry in a symmetric portal frame, the horizontal load H is shared equally by both columns. Each column carries a shear of H/2.

  14. 14
    Cantilever DeflectionMEDIUM

    The maximum deflection at the free end of a cantilever of span L under UDL of intensity w is:

    AwL⁴/3EI
    BwL⁴/4EI
    CwL⁴/8EI
    DwL⁴/16EI

    Answer: C. wL⁴/8EI

    Explanation: For a cantilever with UDL: δ_max = wL⁴/(8EI) at the free end. This result is obtained by double integration of the bending equation or by energy methods.

  15. 15
    Maxwell TheoremMEDIUM

    Maxwell's reciprocal theorem states that for a linear elastic structure:

    AThe stiffness matrix is diagonal
    BDeflection at point i due to unit load at j equals deflection at j due to unit load at i
    CInternal work equals external work
    DStrain energy is minimum at equilibrium

    Answer: B. Deflection at point i due to unit load at j equals deflection at j due to unit load at i

    Explanation: Maxwell's reciprocal theorem: f_ij = f_ji, i.e., the displacement at coordinate i due to a unit force at j equals the displacement at coordinate j due to a unit force at i.

  16. 16
    Shape Factor CircleMEDIUM

    The shape factor for a solid circular cross-section is approximately:

    A1.5
    B1.7
    C2.0
    D1.27

    Answer: B. 1.7

    Explanation: Shape factor for solid circle = Z_p/Z_e = (d³/6)/(πd³/32) = 32/(6π) = 16/(3π) ≈ 1.698 ≈ 1.7.

  17. 17
    Fixed Beam Central LoadMEDIUM

    A fixed beam of span L carries a central concentrated load W. The fixed-end moment at each support is:

    AWL/4
    BWL/6
    CWL/8
    DWL/12

    Answer: C. WL/8

    Explanation: For a fixed beam with central point load W, FEM at each end = WL/8 (hogging). The midspan BM under the load = WL/8 (sagging), equal to the fixed-end moments.

  18. 18
    Kinematic IndeterminacyMEDIUM

    The kinematic (degree of freedom) indeterminacy of a fixed-fixed beam (ignoring axial deformation) is:

    A0
    B1
    C2
    D3

    Answer: A. 0

    Explanation: A fixed-fixed beam has zero rotational and translational DOF at both ends (all movements are restrained). Hence kinematic indeterminacy = 0.

  19. 19
    Shape FactorMEDIUM

    The shape factor for a solid rectangular cross-section is:

    A1.0
    B1.5
    C1.7
    D2.0

    Answer: B. 1.5

    Explanation: Shape factor = Z_p/Z_e = (bd²/4)/(bd²/6) = 6/4 = 1.5. The plastic section modulus Z_p = bd²/4 and elastic section modulus Z_e = bd²/6.

  20. 20
    Carry-over FactorMEDIUM

    In moment distribution method, the carry-over factor for a prismatic member with far end FIXED is:

    A0
    B1/4
    C1/2
    D1

    Answer: C. 1/2

    Explanation: When a moment M is applied at the near end of a fixed-far-end member, a moment M/2 is induced at the far end. Carry-over factor = 1/2.

  21. 21
    IndeterminacyMEDIUM

    The degree of static indeterminacy of a propped cantilever is:

    A0
    B1
    C2
    D3

    Answer: B. 1

    Explanation: DSI = reactions - equilibrium equations = 4 - 3 = 1 (3 reactions at fixed end + 1 at prop; 3 equations of equilibrium).

  22. 22
    GeneralMEDIUM

    For a determinate structure, lack of fit or temperature change generally produces

    Aonly shear stress
    Bno stress if free movement is possible
    Cstress always equal to yield stress
    Dinfinite stress

    Answer: B. no stress if free movement is possible

    Explanation: A determinate structure can deform without restraint-induced internal forces.

  23. 23
    GeneralMEDIUM

    For a beam of uniform flexural rigidity, curvature is proportional to

    Ashear force only
    Bbending moment
    Cspan length squared only
    Daxial force only

    Answer: B. bending moment

    Explanation: For a beam with uniform flexural rigidity EI, the curvature κ = M/EI, where M is bending moment and EI is flexural rigidity. Curvature is directly proportional to bending moment (higher M → greater curvature). This is the basis of the Euler-Bernoulli beam theory used in structural analysis.

  24. 24
    GeneralMEDIUM

    In moment distribution method, carry-over factor for a prismatic member with far end fixed is

    A0
    B1/2
    C1
    D1/4

    Answer: B. 1/2

    Explanation: Half the applied end moment is carried over to the fixed far end.

  25. 25
    Theory of Structures and DesignHARD

    A fixed-ended beam of degree of static indeterminacy 2 requires how many plastic hinges to form a collapse mechanism under monotonic loading?

    A4
    B1
    C3
    D2

    Answer: C. 3

    Explanation: A structure becomes a mechanism when r + 1 plastic hinges form; for a fixed beam, r = 2, so 3 plastic hinges are required.

  26. 26
    Mechanics, Strength of Materials and Structural AnalysisEASY

    Flexural rigidity is represented by:

    AEA
    BEI
    CkA
    DGJ

    Answer: B. EI

    Explanation: Flexural rigidity is the product of modulus of elasticity and second moment of area.

  27. 27
    Mechanics, Strength of Materials and Structural AnalysisMEDIUM

    A plastic hinge forms when a section reaches fully plastic moment capacity.

    Ainitial elastic curvature
    Bzero stress at section
    Cfully plastic moment at section
    Dmaximum shear at neutral axis only

    Answer: C. fully plastic moment at section

    Explanation: A plastic hinge can rotate at approximately constant plastic moment.

  28. 28
    Mechanics, Strength of Materials and Structural AnalysisHARD

    Thermal stress case: A fully restrained bar has E=1.0e+05 MPa, alpha=1.8e-05/°C and temperature rise 25°C. Thermal stress is:

    A22.5 MPa
    B90 MPa
    C45 MPa
    D0.00045 MPa

    Answer: C. 45 MPa

    Explanation: For full restraint, thermal stress = E alpha DeltaT.

  29. 29
    Mechanics, Strength of Materials and Structural AnalysisMEDIUM

    The proper association of section modulus is:

    AEI divided by length
    Bshear force divided by area
    Cmoment of inertia divided by extreme fibre distance
    Darea divided by length

    Answer: C. moment of inertia divided by extreme fibre distance

    Explanation: moment of inertia divided by extreme fibre distance is the correct association for section modulus.

  30. 30
    Mechanics, Strength of Materials and Structural AnalysisMEDIUM

    Carry-over concept: If a moment M is applied at one end of a prismatic member and the far end is fixed, carry-over moment is:

    AM at far end
    BM/2 at far end
    C2M at far end
    Dzero at far end

    Answer: B. M/2 at far end

    Explanation: Carry-over factor for far end fixed is 1/2.

  31. 31
    Mechanics, Strength of Materials and Structural AnalysisEASY

    Rib shortening in arches usually reduces horizontal thrust.

    Areduces horizontal thrust
    Bhas no effect always
    Cincreases span
    Deliminates bending

    Answer: A. reduces horizontal thrust

    Explanation: Rib shortening relieves some arch thrust.

  32. 32
    Mechanics, Strength of Materials and Structural AnalysisEASY

    The kern of a section is associated with:

    Ano tension condition under eccentric compression
    Brailway alignment
    Cwater treatment
    Dmaximum shear only

    Answer: A. no tension condition under eccentric compression

    Explanation: If load lies within kern, stress remains compressive over the section.

  33. 33
    Mechanics, Strength of Materials and Structural AnalysisHARD

    fixed end moment for UDL w over fixed beam span L: choose the correct option.

    AwL²/8 at midspan only
    BwL³/48EI
    CwL/2
    DwL²/12 at each end with opposite signs

    Answer: D. wL²/12 at each end with opposite signs

    Explanation: wL²/12 at each end with opposite signs is the correct association for fixed end moment for UDL w over fixed beam span L.

  34. 34
    Mechanics, Strength of Materials and Structural AnalysisMEDIUM

    Which answer correctly defines principle of transmissibility?

    Aforce magnitude doubles
    Bmoment always becomes zero
    Cforce can be moved along its line of action without changing external effect
    Dforce can be moved to any line

    Answer: C. force can be moved along its line of action without changing external effect

    Explanation: force can be moved along its line of action without changing external effect is the correct association for principle of transmissibility.

  35. 35
    Strength of Materials and Structural AnalysisMEDIUM

    Conjugate beam method is used to determine:

    ASlopes and deflections
    BCement strength
    CFlow velocity
    DSoil compaction

    Answer: A. Slopes and deflections

    Explanation: In conjugate beam method, M/EI diagram is treated as loading.

  36. 36
    Strength of Materials and Structural AnalysisHARD

    Fixed-end moment of a fixed beam carrying UDL w over full span L at each end has magnitude:

    AwL^2/8
    BwL/2
    CwL^3/48EI
    DwL^2/12

    Answer: D. wL^2/12

    Explanation: Full-span UDL on fixed beam creates hogging fixed-end moments of wL^2/12.

  37. 37
    Strength of Materials and Structural AnalysisMEDIUM

    Distribution factor at a joint equals member stiffness divided by:

    ACarry-over factor
    BLoad intensity
    CSpan length only
    DSum of stiffnesses meeting at joint

    Answer: D. Sum of stiffnesses meeting at joint

    Explanation: Moment distribution allocates unbalanced moments in proportion to stiffness.

  38. 38
    Strength of Materials and Structural AnalysisMEDIUM

    Fixed end moment for a fixed beam carrying central point load W is:

    AWL/16 at supports only
    BWL/8 at both ends in hogging
    CWL/4 sagging at midspan
    DwL squared/12

    Answer: B. WL/8 at both ends in hogging

    Explanation: A central point load on fixed-ended beam creates equal hogging fixed end moments WL/8.

  39. 39
    Structural AnalysisMEDIUM

    The cable carrying a load of w kN/m of horizontal span is stretched between supports L m apart at same level and central dip h m. The maximum tension in cable is

    AwL^2/(8h) sqrt(1+(h/L)^2)
    BwL^2/(2) sqrt(1+(L/2h)^2)
    CwL^2/(8h) sqrt(1+(4h/L)^2)
    DwL^2/(2) sqrt(1+(L/4h)^2)

    Answer: C. wL^2/(8h) sqrt(1+(4h/L)^2)

    Explanation: H = wL^2/(8h), vertical reaction = wL/2; Tmax = sqrt(H^2+V^2) = H sqrt(1+(4h/L)^2)

  40. 40
    Structural AnalysisMEDIUM

    A load W is moving from left to right support on a simply supported beam of span L. The maximum bending moment at 0.4L from the left support is:

    A0.16 WL
    B0.20 WL
    C0.24 WL
    D0.25 WL

    Answer: C. 0.24 WL

    Explanation: Influence-line ordinate for BM at a section a is a(L-a)/L. For a = 0.4L, ordinate = 0.24L

  41. 41
    Structural AnalysisMEDIUM

    At the location of a plastic hinge in the plastic analysis of a structure,

    Aradius of curvature is infinite
    Bcurvature is infinite
    Cmoment is infinite
    Dflexural stress is infinite

    Answer: B. curvature is infinite

    Explanation: A plastic hinge rotates at constant moment Mp while curvature becomes very large (theoretically infinite).

  42. 42
    Structural AnalysisMEDIUM

    The maximum dose of plasticiser and superplasticiser for concrete, in general, is restricted to

    A0.5 and 1.0 respectively
    B1.0 and 1.5 respectively
    C0.5 and 2.0 respectively
    D1.0 and 2.0 respectively

    Answer: D. 1.0 and 2.0 respectively

    Explanation: General upper limits ≈ 1.0% (plasticiser) and 2.0% (superplasticiser) by weight of cement.

  43. 43
    Original practiceMEDIUM

    Loss due to elastic shortening occurs mainly in

    ATimber beams
    BUnreinforced masonry
    CRoad pavements only
    DPre-tensioned members

    Answer: D. Pre-tensioned members

    Explanation: When concrete shortens elastically, bonded prestressing steel also shortens causing prestress loss.

  44. 44
    Original practiceMEDIUM

    A bar carries an axial load of 120 kN over a cross-sectional area of 1000 mm². The direct stress is

    A60 N/mm²
    B180 N/mm²
    C240 N/mm²
    D120 N/mm²

    Answer: D. 120 N/mm²

    Explanation: Average direct (axial) stress sigma = P/A, where P = axial force and A = cross-sectional area. Assumes the load acts through the centroid and stress is uniform across the section (valid far from the load application point, per Saint-Venant's principle). Units: N/mm^2 (MPa). Stress = P/A = 120000/1000 = 120 N/mm².

  45. 45
    Cable ShapeHARD

    For cable carrying a uniformly distributed load over horizontal span, which of the following rules or methods is correct?

    Athe cable takes a circular profile
    Bbending stiffness controls the profile
    Csag is zero at midspan
    Dthe cable takes a parabolic profile

    Answer: D. the cable takes a parabolic profile

    Explanation: A cable under UDL per horizontal length forms a parabola. Correct option: D.

  46. 46
    Truss AnalysisMEDIUM

    In practical design discussion of Truss Analysis, which statement best matches method of joints? Avoid the common trap that support reactions are unnecessary.

    Aall zero-force members must be removed before analysis
    Bmembers are designed for bending first
    Cjoint equilibrium is written with member forces as axial forces
    Dsupport reactions are unnecessary

    Answer: C. joint equilibrium is written with member forces as axial forces

    Explanation: Pin-jointed trusses are analysed by axial equilibrium at joints. Correct option: C.

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