Structural Analysis MCQs 2026

90 questions with detailed answers · 31 from past papers · 9 quiz batches available

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Page 1 of 1 Questions 110 of 90
  1. Q1 Past Paper · PPSC/FPSC/NTS easy

    For determinate truss, member forces found by

    1. A method of joints and sections
    2. B only moment distribution
    3. C only limit state design
    4. D only ILD only
    💡 Explanation:

    Equilibrium suffices for determinate truss.

  2. Q2 medium

    Deflection of truss joint found by

    1. A only Euler buckling
    2. B only IS 456
    3. C unit load method or graphical methods
    4. D only bolt shear only
    💡 Explanation:

    Virtual work on truss members.

  3. Q3 Past Paper · PPSC/FPSC/NTS medium

    Temperature rise in determinate truss member causes

    1. A always compression force
    2. B free expansion without induced force
    3. C always tension force
    4. D moment at joints
    💡 Explanation:

    Determinate: no thermal stress if free to expand.

  4. Q4 Past Paper · PPSC/FPSC/NTS medium

    Temperature change in indeterminate structure causes

    1. A no stress ever
    2. B stresses due to restraint
    3. C only vertical deflection zero always
    4. D only plastic hinge
    💡 Explanation:

    Redundancy restrains expansion.

  5. Q5 hard

    Support settlement in determinate beam causes

    1. A rigid body rotation/deflection without internal stress
    2. B internal stress always
    3. C zero deflection
    4. D buckling
    💡 Explanation:

    Determinate: settlement causes geometry change only.

  6. Q6 hard

    Support settlement in indeterminate beam causes

    1. A no internal forces
    2. B only axial truss forces
    3. C only shear in bolts
    4. D bending stresses
    💡 Explanation:

    Restraint induces member forces.

  7. Q7 Past Paper · PPSC/FPSC/NTS medium

    ILD for shear at section x in SSB is

    1. A single parabola always
    2. B horizontal line only
    3. C two straight lines with sudden change at load point
    4. D circle
    💡 Explanation:

    Shear ILD piecewise linear.

  8. Q8 Past Paper · PPSC/FPSC/NTS hard

    Maximum BM in moving load series occurs when

    1. A loads at supports only
    2. B load group is placed per ILD ordinates
    3. C loads at contraflexure only
    4. D loads zero
    💡 Explanation:

    Position loads where ILD product maximized.

  9. Q9 medium

    Concentrated moving load on SSB for max BM places load at

    1. A always mid-span for any section
    2. B section where ILD ordinate is maximum
    3. C always support
    4. D always quarter span
    💡 Explanation:

    Load at peak ILD ordinate for that effect.

  10. Q10 hard

    Uniformly distributed moving load shorter than span for max BM

    1. A covers maximum positive ILD area
    2. B covers minimum area
    3. C placed at one support only
    4. D placed off span
    💡 Explanation:

    UDL length covers beneficial ILD region.

  11. Q11 Past Paper · PPSC/FPSC/NTS medium

    Arches carry load primarily by

    1. A pure bending only
    2. B pure tension only
    3. C shear only
    4. D compression along thrust line
    💡 Explanation:

    Arch action reduces bending moments.

  12. Q12 Past Paper · PPSC/FPSC/NTS hard

    Horizontal thrust in two-hinged parabolic arch under UDL equals

    1. A wL/2
    2. B zero always
    3. C wL²/(8h) for complete UDL over span L rise h
    4. D wLh
    💡 Explanation:

    Thrust formula for parabolic arch UDL.

  13. Q13 medium

    Cable structure under point loads forms

    1. A polygonal funicular shape
    2. B parabola always
    3. C circle
    4. D straight line only
    💡 Explanation:

    Cable edges between loads.

  14. Q14 medium

    Deflected shape of elastic curve has M/(EI) proportional to

    1. A shear force only
    2. B axial force only
    3. C curvature
    4. D temperature only
    💡 Explanation:

    Curvature κ = M/(EI).

  15. Q15 easy

    Symmetry in structure and loading allows

    1. A doubling all forces incorrectly
    2. B ignoring equilibrium
    3. C no boundary conditions
    4. D analysing half structure
    💡 Explanation:

    Symmetry reduces unknowns.

  16. Q16 medium

    Antisymmetric loading on symmetric structure can use

    1. A half structure with antisymmetric boundary conditions
    2. B full structure always required
    3. C only truss method
    4. D only RCC limit state
    💡 Explanation:

    Antisymmetry simplification.

  17. Q17 Past Paper · PPSC/FPSC/NTS medium

    Portal frame under lateral load develops

    1. A only axial forces
    2. B only torsion in all members
    3. C base moments and storey shears
    4. D zero reactions
    💡 Explanation:

    Lateral load resisted by frame action.

  18. Q18 medium

    Sway in frame occurs when

    1. A only vertical loads act
    2. B lateral displacement develops under unsymmetric stiffness/loading
    3. C only determinate truss
    4. D only cables
    💡 Explanation:

    Sidesway is lateral joint displacement.

  19. Q19 hard

    Sidesway in moment distribution is prevented by

    1. A extra equilibrium equation or symmetry
    2. B ignoring joint balance
    3. C only increasing cover
    4. D only bolt pretension
    💡 Explanation:

    Sway frames need additional condition.

  20. Q20 Past Paper · PPSC/FPSC/NTS easy

    Release of moment at hinge means

    1. A slope is zero
    2. B bending moment is zero at that point
    3. C shear is zero
    4. D axial force zero
    💡 Explanation:

    Internal hinge: M=0.

  21. Q21 medium

    Relative stiffness of member in distribution depends on

    1. A only cross-section area
    2. B only concrete grade
    3. C I/L and end conditions
    4. D only bolt grade
    💡 Explanation:

    EI/L with fixity affects K.

  22. Q22 hard

    Carry-over in non-prismatic member

    1. A always 0.5
    2. B always 1
    3. C always zero
    4. D is not necessarily 0.5
    💡 Explanation:

    Prismatic assumption gives 0.5 carry-over.

  23. Q23 Past Paper · PPSC/FPSC/NTS easy

    Fixed support provides

    1. A only vertical reaction
    2. B restraint against translation and rotation
    3. C only moment release
    4. D no reactions
    💡 Explanation:

    Fixed: 3 reactions in 2D (2 force + moment).

  24. Q24 Past Paper · PPSC/FPSC/NTS easy

    Pin support in 2D provides

    1. A moment and two forces
    2. B no reactions
    3. C only moment
    4. D reaction force but no moment restraint
    💡 Explanation:

    Pin: 2 force components.

  25. Q25 Past Paper · PPSC/FPSC/NTS easy

    Roller support provides

    1. A two force components and moment
    2. B single reaction perpendicular to rolling surface
    3. C moment only
    4. D no reaction
    💡 Explanation:

    Roller: one reaction typically vertical.

  26. Q26 Past Paper · PPSC/FPSC/NTS easy

    Stable structure requires

    1. A only more members than joints
    2. B zero reactions
    3. C adequate supports and non-mechanism geometry
    4. D only internal hinges everywhere
    💡 Explanation:

    Stability prevents mechanisms.

  27. Q27 medium

    Mechanism in structure means

    1. A stable indeterminate frame
    2. B unstable geometry allowing rigid-body motion
    3. C only elastic deformation
    4. D only creep
    💡 Explanation:

    Mechanism: insufficient restraint.

  28. Q28 Past Paper · PPSC/FPSC/NTS medium

    Redundant reaction in beam is

    1. A always zero
    2. B always at mid-span load
    3. C only in truss diagonals
    4. D extra support beyond determinate requirements
    💡 Explanation:

    Redundant cannot be found by ΣF, ΣM alone.

  29. Q29 easy

    Prime structural analysis objective is to find

    1. A internal forces, reactions and deformations
    2. B only concrete mix
    3. C only paint colour
    4. D only bolt length only
    💡 Explanation:

    Analysis yields forces and displacements for design.

  30. Q30 medium

    Linear superposition applies when

    1. A large plastic collapse
    2. B material linear and small displacements
    3. C cracking nonlinear always
    4. D geometric nonlinearity dominates
    💡 Explanation:

    Superpose load cases in elastic range.

  31. Q31 easy

    Shear diagram jump at concentrated load equals

    1. A magnitude of load
    2. B moment at load
    3. C zero always
    4. D EI value
    💡 Explanation:

    SF discontinuity = applied point load.

  32. Q32 medium

    BM diagram has sudden change in slope at

    1. A uniform load region
    2. B zero shear point only
    3. C concentrated load (SF jump)
    4. D centroid
    💡 Explanation:

    dM/dx = V; V jump changes slope.

  33. Q33 medium

    Deflection limit serviceability check uses

    1. A only ultimate load factor
    2. B span/250 or code-specified limits often
    3. C only bolt bearing
    4. D only yield only
    💡 Explanation:

    Serviceability limits on deflection/crack.

  34. Q34 hard

    Influence surface is extension of ILD to

    1. A only axial bars
    2. B only bolts
    3. C two-way slab or plate loading
    4. D only soil classification
    💡 Explanation:

    ILD generalization for plates.

  35. Q35 hard

    Graphical statics uses force polygons for

    1. A only RCC shear design
    2. B only creep coefficient
    3. C only surge tank
    4. D equilibrium of coplanar forces
    💡 Explanation:

    Force polygon closes for equilibrium.

  36. Q36 hard

    Reciprocal theorem (Betti-Maxwell) relates

    1. A only thermal expansion
    2. B work done by one load system on displacements from another
    3. C only mix design
    4. D only zero force members
    💡 Explanation:

    Linear elastic reciprocity between load cases.

  37. Q37 medium

    Maximum negative BM at interior support of continuous beam often occurs with

    1. A mid-span of adjacent spans loaded only for min
    2. B no load on bridge
    3. C spans loaded to create hogging at support
    4. D load off structure
    💡 Explanation:

    Loading pattern for hogging envelope.

  38. Q38 easy

    ILD for support reaction in determinate beam is

    1. A linear between supports for unit load
    2. B parabolic between supports always
    3. C zero everywhere
    4. D constant 1 everywhere
    💡 Explanation:

    Reaction ILD piecewise linear.

  39. Q39 hard

    Beam on elastic foundation uses

    1. A only truss zero force
    2. B only Mohr for plane stress
    3. C Winkler model with modulus of subgrade reaction k
    4. D only parabolic arch thrust only
    💡 Explanation:

    Soil as springs: p = kx.

  40. Q40 hard

    Method of consistent deformations solves redundants by

    1. A only joints in 2D
    2. B compatibility integrals with flexibility coefficients
    3. C only IS 800 bolts
    4. D only w/c
    💡 Explanation:

    Force method canonical equations.

  41. Q41 medium

    Compound truss combines

    1. A only RCC and steel
    2. B only soil layers
    3. C simple trusses linked by connecting members
    4. D only one bar
    💡 Explanation:

    Compound structures need full analysis.

  42. Q42 hard

    Space truss has member force determined with

    1. A 3D equilibrium equations
    2. B only 2D joints
    3. C only Mohr circle
    4. D only moment distribution
    💡 Explanation:

    3D truss: ΣF_x,y,z.

  43. Q43 hard

    Dynamic amplification in moving load relates to

    1. A only static ILD
    2. B speed and natural frequency
    3. C only concrete grade
    4. D only lap length
    💡 Explanation:

    Resonance/impact increases response.

  44. Q44 medium

    Damping in structures dissipates

    1. A static dead load
    2. B only axial force
    3. C only yield stress
    4. D vibration energy
    💡 Explanation:

    Damping reduces dynamic response.

  45. Q45 hard

    Rayleigh-Ritz method is approximate for

    1. A exact truss forces only
    2. B vibration or buckling eigenvalues
    3. C concrete mix proportion
    4. D bolt pretension only
    💡 Explanation:

    Energy-based approximate analysis.

  46. Q46 hard

    K-value for column effective length in frames depends on

    1. A only section depth
    2. B end restraint stiffness ratios
    3. C only paint coating
    4. D only aggregate size
    💡 Explanation:

    Alignment charts give K factors.

  47. Q47 medium

    Bracing in steel frame reduces

    1. A only concrete shrinkage
    2. B effective length and lateral sway
    3. C only development length
    4. D only w/c ratio
    💡 Explanation:

    Bracing increases lateral stability.

  48. Q48 hard

    Elastic critical load for frame buckling depends on

    1. A only concrete cover
    2. B only bolt threads
    3. C member stiffness and geometry
    4. D only UDL on slab
    💡 Explanation:

    Frame stability: GJ, EI, lengths, bracing.

  49. Q49 medium

    Point of inflection in deflected beam corresponds to

    1. A max SF point
    2. B max deflection always
    3. C point of contraflexure in BM diagram
    4. D support only
    💡 Explanation:

    Zero curvature where M=0.

  50. Q50 easy

    Sagging bending moment in span makes beam

    1. A concave downward always
    2. B straight only
    3. C only axial
    4. D concave upward in beam convention
    💡 Explanation:

    Mid-span positive sagging typical in continuous beam.

  51. Q51 medium

    Hogging bending moment makes beam

    1. A always sagging
    2. B zero curvature
    3. C only shear
    4. D concave downward (sagging negative/top in compression in sagging convention)
    💡 Explanation:

    Convention: hogging negative BM at supports in continuous beams.

  52. Q52 easy

    Shear force sign convention often takes

    1. A downward positive always on left
    2. B moment positive always hogging only
    3. C axial zero
    4. D upward positive on left face of section
    💡 Explanation:

    Sign conventions vary; consistency matters.

  53. Q53 medium

    Moment redistribution in RCC allows

    1. A limited adjustment of elastic moments per code
    2. B unlimited increase anywhere
    3. C zero tension in steel
    4. D ignoring equilibrium
    💡 Explanation:

    IS 456 permits limited redistribution.

  54. Q54 easy

    Continuous beam is typically

    1. A statically indeterminate
    2. B always determinate
    3. C unstable
    4. D truss type
    💡 Explanation:

    Multiple spans create redundants.

  55. Q55 medium

    Envelope of BM from moving loads gives

    1. A maximum and minimum BM at each section
    2. B only shear
    3. C only axial
    4. D only deflection without sign
    💡 Explanation:

    Design for worst-case moving load.

  56. Q56 medium

    Absolute maximum BM in SSB with moving concentrated load is

    1. A PL/8 always at support
    2. B PL/4 when load at mid-span
    3. C zero
    4. D PL/2 at support
    💡 Explanation:

    Moving single load max at centre.

  57. Q57 Past Paper · PPSC/FPSC/NTS easy

    Influence line ordinate at point equals

    1. A always unity everywhere
    2. B zero always
    3. C value of function when unit load at that point
    4. D UDL magnitude
    💡 Explanation:

    Definition of ILD ordinate.

  58. Q58 hard

    Shape factor for plastic section modulus is

    1. A I/Z
    2. B ratio of plastic to elastic section modulus
    3. C Z/I only without plastic
    4. D yield stress over E
    💡 Explanation:

    S = Z_p/Z_e.

  59. Q59 hard

    Collapse mechanism in plastic frame has

    1. A no hinges
    2. B only elastic behaviour
    3. C only one hinge always enough
    4. D sufficient plastic hinges for mechanism
    💡 Explanation:

    Mechanism needs enough hinges.

  60. Q60 Past Paper · PPSC/FPSC/NTS hard

    Plastic analysis uses

    1. A only elastic ILD
    2. B only Mohr circle
    3. C formation of plastic hinges for collapse load
    4. D only bolt bearing
    💡 Explanation:

    Limit analysis for steel/plastic design.

  61. Q61 medium

    Linear elastic analysis assumes

    1. A plastic hinges throughout
    2. B superposition valid and Hookean material
    3. C large deflection mandatory
    4. D cracking ignored in concrete always required
    💡 Explanation:

    Elastic analysis uses proportionality.

  62. Q62 hard

    Stiffness matrix relates

    1. A displacements to time only
    2. B only thermal strain
    3. C only cover depth
    4. D forces to displacements
    💡 Explanation:

    Stiffness {F}=[k]{u}.

  63. Q63 hard

    Flexibility matrix relates

    1. A forces to plastic moment
    2. B displacements to forces
    3. C stress to strain in concrete only
    4. D bolt area to cover
    💡 Explanation:

    Flexibility {u}=[f]{F}.

  64. Q64 hard

    Consistent deformation method is another name for

    1. A only method of joints
    2. B only Mohr circle
    3. C only Winkler soil
    4. D slope-deflection or stiffness approach family
    💡 Explanation:

    Displacement-based analysis.

  65. Q65 hard

    Stiffness matrix method assembles

    1. A global stiffness from element matrices
    2. B only soil moduli
    3. C only truss zero force
    4. D only Mohr circle
    💡 Explanation:

    FE/stiffness assembly K{u}={F}.

  66. Q66 hard

    Displacement method selects

    1. A redundant forces only
    2. B only reactions
    3. C joint displacements as unknowns
    4. D only stresses
    💡 Explanation:

    Stiffness method solves displacements.

  67. Q67 hard

    Force method selects

    1. A displacements only
    2. B concrete cover
    3. C redundant forces as unknowns
    4. D bolt diameter
    💡 Explanation:

    Flexibility/force method solves redundants.

  68. Q68 medium

    Compatibility equation in indeterminate analysis expresses

    1. A equality of deformations
    2. B sum of forces zero only
    3. C concrete grade
    4. D steel yield only
    💡 Explanation:

    Deformations must be consistent.

  69. Q69 Past Paper · PPSC/FPSC/NTS easy

    Statically indeterminate beams have

    1. A more unknown reactions/internal forces than equilibrium equations
    2. B fewer unknowns
    3. C no compatibility
    4. D only axial loads
    💡 Explanation:

    Redundants need compatibility conditions.

  70. Q70 medium

    Internal hinge in structure

    1. A increases indeterminacy always
    2. B has no effect
    3. C makes structure unstable always
    4. D reduces indeterminacy
    💡 Explanation:

    Hinge releases moment, adds equation.

  71. Q71 Past Paper · PPSC/FPSC/NTS medium

    Truss with m members, j joints, r reactions is determinate if

    1. A m + r = 3j
    2. B m + r = 2j for plane truss
    3. C m = j
    4. D m + r = j
    💡 Explanation:

    Simple counting criterion for plane truss.

  72. Q72 hard

    Williot-Mohr diagram is used for

    1. A truss deflections graphically
    2. B RCC design
    3. C soil classification
    4. D mix design
    💡 Explanation:

    Graphical truss displacement analysis.

  73. Q73 hard

    Conjugate beam method maps

    1. A truss to frame
    2. B soil to concrete
    3. C real beam supports to conjugate beam supports
    4. D steel to timber
    💡 Explanation:

    M/EI diagram as load on conjugate beam.

  74. Q74 hard

    Graphical integration of M/(EI) diagram gives

    1. A axial force
    2. B slope or deflection depending on first or second area-moment
    3. C shear only
    4. D stress at fibre
    💡 Explanation:

    Area-moment method for beam deflection.

  75. Q75 Past Paper · PPSC/FPSC/NTS medium

    Unit load method for deflection uses

    1. A virtual work with unit force at desired displacement
    2. B only Mohr circle
    3. C only moment distribution
    4. D only soil bearing
    💡 Explanation:

    Castigliano/virtual work: δ = ∫(M m)/(EI) dx.

  76. Q76 hard

    Clapeyron theorem of three moments applies to

    1. A continuous beams on supports
    2. B only trusses
    3. C only cables
    4. D only columns
    💡 Explanation:

    Three-moment equation links BM at three supports.

  77. Q77 Past Paper · PPSC/FPSC/NTS hard

    Fixed-end moment in beam due to uniform load w on span L is

    1. A wL²/8
    2. B wL/2
    3. C zero at both ends
    4. D wL²/12 at each end (hogging sign convention as per method)
    💡 Explanation:

    Standard FEM for UDL on fixed-fixed beam.

  78. Q78 Past Paper · PPSC/FPSC/NTS medium

    Distribution factor at a joint equals

    1. A member stiffness divided by sum of stiffnesses at joint
    2. B always 0.5
    3. C always 1
    4. D zero for all members
    💡 Explanation:

    DF_i = K_i/ΣK.

  79. Q79 Past Paper · PPSC/FPSC/NTS medium

    Stiffness factor with far end pinned is

    1. A 3EI/L
    2. B 4EI/L
    3. C 2EI/L
    4. D EI/L only
    💡 Explanation:

    K = 3EI/L when far end is pinned.

  80. Q80 Past Paper · PPSC/FPSC/NTS medium

    Stiffness factor for prismatic member with far end fixed is

    1. A 3EI/L
    2. B EI/L
    3. C 6EI/L
    4. D 4EI/L
    💡 Explanation:

    K = 4EI/L fixed far end.

  81. Q81 Past Paper · PPSC/FPSC/NTS medium

    Carry-over factor in moment distribution for prismatic member is

    1. A 1.0
    2. B 0.25
    3. C zero
    4. D 0.5
    💡 Explanation:

    Half the applied moment carries to far fixed end.

  82. Q82 Past Paper · PPSC/FPSC/NTS medium

    Moment distribution method was developed by

    1. A Hardy Cross
    2. B Euler
    3. C Mohr
    4. D Castigliano
    💡 Explanation:

    Hardy Cross iterative carry-over method.

  83. Q83 hard

    Slope-deflection method accounts for

    1. A only axial forces
    2. B only truss pin joints
    3. C joint rotations and displacements in member end moments
    4. D only soil settlement without stiffness
    💡 Explanation:

    Stiffness method variant using slope-deflection equations.

  84. Q84 Past Paper · PPSC/FPSC/NTS medium

    For simply supported beam, ILD for mid-span moment is

    1. A triangular with max ordinate at mid-span
    2. B rectangular
    3. C parabolic fixed
    4. D zero at mid-span
    💡 Explanation:

    Unit load at centre gives max BM.

  85. Q85 Past Paper · PPSC/FPSC/NTS hard

    Muller-Breslau principle states that

    1. A ILD equals SFD always
    2. B ILD equals axial force only
    3. C deflection is zero
    4. D influence line equals deflected shape from removing corresponding constraint
    💡 Explanation:

    Qualitative ILD from released structure deflection.

  86. Q86 Past Paper · PPSC/FPSC/NTS medium

    Influence line for reaction at support shows

    1. A only BM diagram under UDL
    2. B only shear at mid-span fixed
    3. C Mohr circle
    4. D variation of reaction with unit load position
    💡 Explanation:

    ILD plots response to moving unit load.

  87. Q87 Past Paper · PPSC/FPSC/NTS medium

    Three-hinged arch is statically

    1. A indeterminate to one degree
    2. B indeterminate to two degrees
    3. C determinate
    4. D unstable always
    💡 Explanation:

    Extra hinge provides release making structure determinate.

  88. Q88 Past Paper · PPSC/FPSC/NTS medium

    A two-hinged arch is statically

    1. A determinate
    2. B indeterminate to three degrees
    3. C unstable
    4. D indeterminate to one degree externally
    💡 Explanation:

    One redundant hinge reaction beyond three equilibrium equations.

  89. Q89 hard

    Degree of kinematic indeterminacy of a frame equals

    1. A number of members only
    2. B number of supports only
    3. C number of independent joint displacements
    4. D always zero
    💡 Explanation:

    Kinematic indeterminacy counts unknown displacements/rotations.

  90. Q90 Past Paper · PPSC/FPSC/NTS hard

    Degree of static indeterminacy of a rigid jointed plane frame is given by

    1. A D_s = m + r − 2j
    2. B D_s = 2m − j
    3. C D_s = m − r + j
    4. D D_s = 3m + r − 3j
    💡 Explanation:

    Frame formula: 3m + r − 3j for internal + support redundancies.