Structural Analysis MCQs 2026

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

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Page 1 of 1Questions 1–10 of 90
  1. Q1Past Paper · PPSC/FPSC/NTShard

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

    1. AD_s = m + r − 2j
    2. BD_s = 2m − j
    3. CD_s = m − r + j
    4. DD_s = 3m + r − 3j
    💡 Explanation:

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

  2. Q2hard

    Degree of kinematic indeterminacy of a frame equals

    1. Anumber of members only
    2. Bnumber of supports only
    3. Cnumber of independent joint displacements
    4. Dalways zero
    💡 Explanation:

    Kinematic indeterminacy counts unknown displacements/rotations.

  3. Q3Past Paper · PPSC/FPSC/NTSmedium

    A two-hinged arch is statically

    1. Adeterminate
    2. Bindeterminate to three degrees
    3. Cunstable
    4. Dindeterminate to one degree externally
    💡 Explanation:

    One redundant hinge reaction beyond three equilibrium equations.

  4. Q4Past Paper · PPSC/FPSC/NTSmedium

    Three-hinged arch is statically

    1. Aindeterminate to one degree
    2. Bindeterminate to two degrees
    3. Cdeterminate
    4. Dunstable always
    💡 Explanation:

    Extra hinge provides release making structure determinate.

  5. Q5Past Paper · PPSC/FPSC/NTSmedium

    Influence line for reaction at support shows

    1. Aonly BM diagram under UDL
    2. Bonly shear at mid-span fixed
    3. CMohr circle
    4. Dvariation of reaction with unit load position
    💡 Explanation:

    ILD plots response to moving unit load.

  6. Q6Past Paper · PPSC/FPSC/NTShard

    Muller-Breslau principle states that

    1. AILD equals SFD always
    2. BILD equals axial force only
    3. Cdeflection is zero
    4. Dinfluence line equals deflected shape from removing corresponding constraint
    💡 Explanation:

    Qualitative ILD from released structure deflection.

  7. Q7Past Paper · PPSC/FPSC/NTSmedium

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

    1. Atriangular with max ordinate at mid-span
    2. Brectangular
    3. Cparabolic fixed
    4. Dzero at mid-span
    💡 Explanation:

    Unit load at centre gives max BM.

  8. Q8hard

    Slope-deflection method accounts for

    1. Aonly axial forces
    2. Bonly truss pin joints
    3. Cjoint rotations and displacements in member end moments
    4. Donly soil settlement without stiffness
    💡 Explanation:

    Stiffness method variant using slope-deflection equations.

  9. Q9Past Paper · PPSC/FPSC/NTSmedium

    Moment distribution method was developed by

    1. AHardy Cross
    2. BEuler
    3. CMohr
    4. DCastigliano
    💡 Explanation:

    Hardy Cross iterative carry-over method.

  10. Q10Past Paper · PPSC/FPSC/NTSmedium

    Carry-over factor in moment distribution for prismatic member is

    1. A1.0
    2. B0.25
    3. Czero
    4. D0.5
    💡 Explanation:

    Half the applied moment carries to far fixed end.

  11. Q11Past Paper · PPSC/FPSC/NTSmedium

    Stiffness factor for prismatic member with far end fixed is

    1. A3EI/L
    2. BEI/L
    3. C6EI/L
    4. D4EI/L
    💡 Explanation:

    K = 4EI/L fixed far end.

  12. Q12Past Paper · PPSC/FPSC/NTSmedium

    Stiffness factor with far end pinned is

    1. A3EI/L
    2. B4EI/L
    3. C2EI/L
    4. DEI/L only
    💡 Explanation:

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

  13. Q13Past Paper · PPSC/FPSC/NTSmedium

    Distribution factor at a joint equals

    1. Amember stiffness divided by sum of stiffnesses at joint
    2. Balways 0.5
    3. Calways 1
    4. Dzero for all members
    💡 Explanation:

    DF_i = K_i/ΣK.

  14. Q14Past Paper · PPSC/FPSC/NTShard

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

    1. AwL²/8
    2. BwL/2
    3. Czero at both ends
    4. DwL²/12 at each end (hogging sign convention as per method)
    💡 Explanation:

    Standard FEM for UDL on fixed-fixed beam.

  15. Q15hard

    Clapeyron theorem of three moments applies to

    1. Acontinuous beams on supports
    2. Bonly trusses
    3. Conly cables
    4. Donly columns
    💡 Explanation:

    Three-moment equation links BM at three supports.

  16. Q16Past Paper · PPSC/FPSC/NTSmedium

    Unit load method for deflection uses

    1. Avirtual work with unit force at desired displacement
    2. Bonly Mohr circle
    3. Conly moment distribution
    4. Donly soil bearing
    💡 Explanation:

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

  17. Q17hard

    Graphical integration of M/(EI) diagram gives

    1. Aaxial force
    2. Bslope or deflection depending on first or second area-moment
    3. Cshear only
    4. Dstress at fibre
    💡 Explanation:

    Area-moment method for beam deflection.

  18. Q18hard

    Conjugate beam method maps

    1. Atruss to frame
    2. Bsoil to concrete
    3. Creal beam supports to conjugate beam supports
    4. Dsteel to timber
    💡 Explanation:

    M/EI diagram as load on conjugate beam.

  19. Q19hard

    Williot-Mohr diagram is used for

    1. Atruss deflections graphically
    2. BRCC design
    3. Csoil classification
    4. Dmix design
    💡 Explanation:

    Graphical truss displacement analysis.

  20. Q20Past Paper · PPSC/FPSC/NTSmedium

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

    1. Am + r = 3j
    2. Bm + r = 2j for plane truss
    3. Cm = j
    4. Dm + r = j
    💡 Explanation:

    Simple counting criterion for plane truss.

  21. Q21medium

    Internal hinge in structure

    1. Aincreases indeterminacy always
    2. Bhas no effect
    3. Cmakes structure unstable always
    4. Dreduces indeterminacy
    💡 Explanation:

    Hinge releases moment, adds equation.

  22. Q22Past Paper · PPSC/FPSC/NTSeasy

    Statically indeterminate beams have

    1. Amore unknown reactions/internal forces than equilibrium equations
    2. Bfewer unknowns
    3. Cno compatibility
    4. Donly axial loads
    💡 Explanation:

    Redundants need compatibility conditions.

  23. Q23medium

    Compatibility equation in indeterminate analysis expresses

    1. Aequality of deformations
    2. Bsum of forces zero only
    3. Cconcrete grade
    4. Dsteel yield only
    💡 Explanation:

    Deformations must be consistent.

  24. Q24hard

    Force method selects

    1. Adisplacements only
    2. Bconcrete cover
    3. Credundant forces as unknowns
    4. Dbolt diameter
    💡 Explanation:

    Flexibility/force method solves redundants.

  25. Q25hard

    Displacement method selects

    1. Aredundant forces only
    2. Bonly reactions
    3. Cjoint displacements as unknowns
    4. Donly stresses
    💡 Explanation:

    Stiffness method solves displacements.

  26. Q26hard

    Stiffness matrix method assembles

    1. Aglobal stiffness from element matrices
    2. Bonly soil moduli
    3. Conly truss zero force
    4. Donly Mohr circle
    💡 Explanation:

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

  27. Q27Past Paper · PPSC/FPSC/NTSeasy

    For determinate truss, member forces found by

    1. Amethod of joints and sections
    2. Bonly moment distribution
    3. Conly limit state design
    4. Donly ILD only
    💡 Explanation:

    Equilibrium suffices for determinate truss.

  28. Q28medium

    Deflection of truss joint found by

    1. Aonly Euler buckling
    2. Bonly IS 456
    3. Cunit load method or graphical methods
    4. Donly bolt shear only
    💡 Explanation:

    Virtual work on truss members.

  29. Q29Past Paper · PPSC/FPSC/NTSmedium

    Temperature rise in determinate truss member causes

    1. Aalways compression force
    2. Bfree expansion without induced force
    3. Calways tension force
    4. Dmoment at joints
    💡 Explanation:

    Determinate: no thermal stress if free to expand.

  30. Q30Past Paper · PPSC/FPSC/NTSmedium

    Temperature change in indeterminate structure causes

    1. Ano stress ever
    2. Bstresses due to restraint
    3. Conly vertical deflection zero always
    4. Donly plastic hinge
    💡 Explanation:

    Redundancy restrains expansion.

  31. Q31hard

    Support settlement in determinate beam causes

    1. Arigid body rotation/deflection without internal stress
    2. Binternal stress always
    3. Czero deflection
    4. Dbuckling
    💡 Explanation:

    Determinate: settlement causes geometry change only.

  32. Q32hard

    Support settlement in indeterminate beam causes

    1. Ano internal forces
    2. Bonly axial truss forces
    3. Conly shear in bolts
    4. Dbending stresses
    💡 Explanation:

    Restraint induces member forces.

  33. Q33Past Paper · PPSC/FPSC/NTSmedium

    ILD for shear at section x in SSB is

    1. Asingle parabola always
    2. Bhorizontal line only
    3. Ctwo straight lines with sudden change at load point
    4. Dcircle
    💡 Explanation:

    Shear ILD piecewise linear.

  34. Q34Past Paper · PPSC/FPSC/NTShard

    Maximum BM in moving load series occurs when

    1. Aloads at supports only
    2. Bload group is placed per ILD ordinates
    3. Cloads at contraflexure only
    4. Dloads zero
    💡 Explanation:

    Position loads where ILD product maximized.

  35. Q35medium

    Concentrated moving load on SSB for max BM places load at

    1. Aalways mid-span for any section
    2. Bsection where ILD ordinate is maximum
    3. Calways support
    4. Dalways quarter span
    💡 Explanation:

    Load at peak ILD ordinate for that effect.

  36. Q36hard

    Uniformly distributed moving load shorter than span for max BM

    1. Acovers maximum positive ILD area
    2. Bcovers minimum area
    3. Cplaced at one support only
    4. Dplaced off span
    💡 Explanation:

    UDL length covers beneficial ILD region.

  37. Q37Past Paper · PPSC/FPSC/NTSmedium

    Arches carry load primarily by

    1. Apure bending only
    2. Bpure tension only
    3. Cshear only
    4. Dcompression along thrust line
    💡 Explanation:

    Arch action reduces bending moments.

  38. Q38Past Paper · PPSC/FPSC/NTShard

    Horizontal thrust in two-hinged parabolic arch under UDL equals

    1. AwL/2
    2. Bzero always
    3. CwL²/(8h) for complete UDL over span L rise h
    4. DwLh
    💡 Explanation:

    Thrust formula for parabolic arch UDL.

  39. Q39medium

    Cable structure under point loads forms

    1. Apolygonal funicular shape
    2. Bparabola always
    3. Ccircle
    4. Dstraight line only
    💡 Explanation:

    Cable edges between loads.

  40. Q40medium

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

    1. Ashear force only
    2. Baxial force only
    3. Ccurvature
    4. Dtemperature only
    💡 Explanation:

    Curvature κ = M/(EI).

  41. Q41easy

    Symmetry in structure and loading allows

    1. Adoubling all forces incorrectly
    2. Bignoring equilibrium
    3. Cno boundary conditions
    4. Danalysing half structure
    💡 Explanation:

    Symmetry reduces unknowns.

  42. Q42medium

    Antisymmetric loading on symmetric structure can use

    1. Ahalf structure with antisymmetric boundary conditions
    2. Bfull structure always required
    3. Conly truss method
    4. Donly RCC limit state
    💡 Explanation:

    Antisymmetry simplification.

  43. Q43Past Paper · PPSC/FPSC/NTSmedium

    Portal frame under lateral load develops

    1. Aonly axial forces
    2. Bonly torsion in all members
    3. Cbase moments and storey shears
    4. Dzero reactions
    💡 Explanation:

    Lateral load resisted by frame action.

  44. Q44medium

    Sway in frame occurs when

    1. Aonly vertical loads act
    2. Blateral displacement develops under unsymmetric stiffness/loading
    3. Conly determinate truss
    4. Donly cables
    💡 Explanation:

    Sidesway is lateral joint displacement.

  45. Q45hard

    Sidesway in moment distribution is prevented by

    1. Aextra equilibrium equation or symmetry
    2. Bignoring joint balance
    3. Conly increasing cover
    4. Donly bolt pretension
    💡 Explanation:

    Sway frames need additional condition.

  46. Q46Past Paper · PPSC/FPSC/NTSeasy

    Release of moment at hinge means

    1. Aslope is zero
    2. Bbending moment is zero at that point
    3. Cshear is zero
    4. Daxial force zero
    💡 Explanation:

    Internal hinge: M=0.

  47. Q47medium

    Relative stiffness of member in distribution depends on

    1. Aonly cross-section area
    2. Bonly concrete grade
    3. CI/L and end conditions
    4. Donly bolt grade
    💡 Explanation:

    EI/L with fixity affects K.

  48. Q48hard

    Carry-over in non-prismatic member

    1. Aalways 0.5
    2. Balways 1
    3. Calways zero
    4. Dis not necessarily 0.5
    💡 Explanation:

    Prismatic assumption gives 0.5 carry-over.

  49. Q49Past Paper · PPSC/FPSC/NTSeasy

    Fixed support provides

    1. Aonly vertical reaction
    2. Brestraint against translation and rotation
    3. Conly moment release
    4. Dno reactions
    💡 Explanation:

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

  50. Q50Past Paper · PPSC/FPSC/NTSeasy

    Pin support in 2D provides

    1. Amoment and two forces
    2. Bno reactions
    3. Conly moment
    4. Dreaction force but no moment restraint
    💡 Explanation:

    Pin: 2 force components.

  51. Q51Past Paper · PPSC/FPSC/NTSeasy

    Roller support provides

    1. Atwo force components and moment
    2. Bsingle reaction perpendicular to rolling surface
    3. Cmoment only
    4. Dno reaction
    💡 Explanation:

    Roller: one reaction typically vertical.

  52. Q52Past Paper · PPSC/FPSC/NTSeasy

    Stable structure requires

    1. Aonly more members than joints
    2. Bzero reactions
    3. Cadequate supports and non-mechanism geometry
    4. Donly internal hinges everywhere
    💡 Explanation:

    Stability prevents mechanisms.

  53. Q53medium

    Mechanism in structure means

    1. Astable indeterminate frame
    2. Bunstable geometry allowing rigid-body motion
    3. Conly elastic deformation
    4. Donly creep
    💡 Explanation:

    Mechanism: insufficient restraint.

  54. Q54Past Paper · PPSC/FPSC/NTSmedium

    Redundant reaction in beam is

    1. Aalways zero
    2. Balways at mid-span load
    3. Conly in truss diagonals
    4. Dextra support beyond determinate requirements
    💡 Explanation:

    Redundant cannot be found by ΣF, ΣM alone.

  55. Q55hard

    Consistent deformation method is another name for

    1. Aonly method of joints
    2. Bonly Mohr circle
    3. Conly Winkler soil
    4. Dslope-deflection or stiffness approach family
    💡 Explanation:

    Displacement-based analysis.

  56. Q56hard

    Flexibility matrix relates

    1. Aforces to plastic moment
    2. Bdisplacements to forces
    3. Cstress to strain in concrete only
    4. Dbolt area to cover
    💡 Explanation:

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

  57. Q57hard

    Stiffness matrix relates

    1. Adisplacements to time only
    2. Bonly thermal strain
    3. Conly cover depth
    4. Dforces to displacements
    💡 Explanation:

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

  58. Q58medium

    Linear elastic analysis assumes

    1. Aplastic hinges throughout
    2. Bsuperposition valid and Hookean material
    3. Clarge deflection mandatory
    4. Dcracking ignored in concrete always required
    💡 Explanation:

    Elastic analysis uses proportionality.

  59. Q59Past Paper · PPSC/FPSC/NTShard

    Plastic analysis uses

    1. Aonly elastic ILD
    2. Bonly Mohr circle
    3. Cformation of plastic hinges for collapse load
    4. Donly bolt bearing
    💡 Explanation:

    Limit analysis for steel/plastic design.

  60. Q60hard

    Collapse mechanism in plastic frame has

    1. Ano hinges
    2. Bonly elastic behaviour
    3. Conly one hinge always enough
    4. Dsufficient plastic hinges for mechanism
    💡 Explanation:

    Mechanism needs enough hinges.

  61. Q61hard

    Shape factor for plastic section modulus is

    1. AI/Z
    2. Bratio of plastic to elastic section modulus
    3. CZ/I only without plastic
    4. Dyield stress over E
    💡 Explanation:

    S = Z_p/Z_e.

  62. Q62Past Paper · PPSC/FPSC/NTSeasy

    Influence line ordinate at point equals

    1. Aalways unity everywhere
    2. Bzero always
    3. Cvalue of function when unit load at that point
    4. DUDL magnitude
    💡 Explanation:

    Definition of ILD ordinate.

  63. Q63medium

    Absolute maximum BM in SSB with moving concentrated load is

    1. APL/8 always at support
    2. BPL/4 when load at mid-span
    3. Czero
    4. DPL/2 at support
    💡 Explanation:

    Moving single load max at centre.

  64. Q64medium

    Envelope of BM from moving loads gives

    1. Amaximum and minimum BM at each section
    2. Bonly shear
    3. Conly axial
    4. Donly deflection without sign
    💡 Explanation:

    Design for worst-case moving load.

  65. Q65easy

    Continuous beam is typically

    1. Astatically indeterminate
    2. Balways determinate
    3. Cunstable
    4. Dtruss type
    💡 Explanation:

    Multiple spans create redundants.

  66. Q66medium

    Moment redistribution in RCC allows

    1. Alimited adjustment of elastic moments per code
    2. Bunlimited increase anywhere
    3. Czero tension in steel
    4. Dignoring equilibrium
    💡 Explanation:

    IS 456 permits limited redistribution.

  67. Q67easy

    Shear force sign convention often takes

    1. Adownward positive always on left
    2. Bmoment positive always hogging only
    3. Caxial zero
    4. Dupward positive on left face of section
    💡 Explanation:

    Sign conventions vary; consistency matters.

  68. Q68medium

    Hogging bending moment makes beam

    1. Aalways sagging
    2. Bzero curvature
    3. Conly shear
    4. Dconcave downward (sagging negative/top in compression in sagging convention)
    💡 Explanation:

    Convention: hogging negative BM at supports in continuous beams.

  69. Q69easy

    Sagging bending moment in span makes beam

    1. Aconcave downward always
    2. Bstraight only
    3. Conly axial
    4. Dconcave upward in beam convention
    💡 Explanation:

    Mid-span positive sagging typical in continuous beam.

  70. Q70medium

    Point of inflection in deflected beam corresponds to

    1. Amax SF point
    2. Bmax deflection always
    3. Cpoint of contraflexure in BM diagram
    4. Dsupport only
    💡 Explanation:

    Zero curvature where M=0.

  71. Q71hard

    Elastic critical load for frame buckling depends on

    1. Aonly concrete cover
    2. Bonly bolt threads
    3. Cmember stiffness and geometry
    4. Donly UDL on slab
    💡 Explanation:

    Frame stability: GJ, EI, lengths, bracing.

  72. Q72medium

    Bracing in steel frame reduces

    1. Aonly concrete shrinkage
    2. Beffective length and lateral sway
    3. Conly development length
    4. Donly w/c ratio
    💡 Explanation:

    Bracing increases lateral stability.

  73. Q73hard

    K-value for column effective length in frames depends on

    1. Aonly section depth
    2. Bend restraint stiffness ratios
    3. Conly paint coating
    4. Donly aggregate size
    💡 Explanation:

    Alignment charts give K factors.

  74. Q74hard

    Rayleigh-Ritz method is approximate for

    1. Aexact truss forces only
    2. Bvibration or buckling eigenvalues
    3. Cconcrete mix proportion
    4. Dbolt pretension only
    💡 Explanation:

    Energy-based approximate analysis.

  75. Q75medium

    Damping in structures dissipates

    1. Astatic dead load
    2. Bonly axial force
    3. Conly yield stress
    4. Dvibration energy
    💡 Explanation:

    Damping reduces dynamic response.

  76. Q76hard

    Dynamic amplification in moving load relates to

    1. Aonly static ILD
    2. Bspeed and natural frequency
    3. Conly concrete grade
    4. Donly lap length
    💡 Explanation:

    Resonance/impact increases response.

  77. Q77hard

    Space truss has member force determined with

    1. A3D equilibrium equations
    2. Bonly 2D joints
    3. Conly Mohr circle
    4. Donly moment distribution
    💡 Explanation:

    3D truss: ΣF_x,y,z.

  78. Q78medium

    Compound truss combines

    1. Aonly RCC and steel
    2. Bonly soil layers
    3. Csimple trusses linked by connecting members
    4. Donly one bar
    💡 Explanation:

    Compound structures need full analysis.

  79. Q79hard

    Method of consistent deformations solves redundants by

    1. Aonly joints in 2D
    2. Bcompatibility integrals with flexibility coefficients
    3. Conly IS 800 bolts
    4. Donly w/c
    💡 Explanation:

    Force method canonical equations.

  80. Q80hard

    Beam on elastic foundation uses

    1. Aonly truss zero force
    2. Bonly Mohr for plane stress
    3. CWinkler model with modulus of subgrade reaction k
    4. Donly parabolic arch thrust only
    💡 Explanation:

    Soil as springs: p = kx.

  81. Q81easy

    ILD for support reaction in determinate beam is

    1. Alinear between supports for unit load
    2. Bparabolic between supports always
    3. Czero everywhere
    4. Dconstant 1 everywhere
    💡 Explanation:

    Reaction ILD piecewise linear.

  82. Q82medium

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

    1. Amid-span of adjacent spans loaded only for min
    2. Bno load on bridge
    3. Cspans loaded to create hogging at support
    4. Dload off structure
    💡 Explanation:

    Loading pattern for hogging envelope.

  83. Q83easy

    Prime structural analysis objective is to find

    1. Ainternal forces, reactions and deformations
    2. Bonly concrete mix
    3. Conly paint colour
    4. Donly bolt length only
    💡 Explanation:

    Analysis yields forces and displacements for design.

  84. Q84medium

    Linear superposition applies when

    1. Alarge plastic collapse
    2. Bmaterial linear and small displacements
    3. Ccracking nonlinear always
    4. Dgeometric nonlinearity dominates
    💡 Explanation:

    Superpose load cases in elastic range.

  85. Q85easy

    Shear diagram jump at concentrated load equals

    1. Amagnitude of load
    2. Bmoment at load
    3. Czero always
    4. DEI value
    💡 Explanation:

    SF discontinuity = applied point load.

  86. Q86medium

    BM diagram has sudden change in slope at

    1. Auniform load region
    2. Bzero shear point only
    3. Cconcentrated load (SF jump)
    4. Dcentroid
    💡 Explanation:

    dM/dx = V; V jump changes slope.

  87. Q87medium

    Deflection limit serviceability check uses

    1. Aonly ultimate load factor
    2. Bspan/250 or code-specified limits often
    3. Conly bolt bearing
    4. Donly yield only
    💡 Explanation:

    Serviceability limits on deflection/crack.

  88. Q88hard

    Influence surface is extension of ILD to

    1. Aonly axial bars
    2. Bonly bolts
    3. Ctwo-way slab or plate loading
    4. Donly soil classification
    💡 Explanation:

    ILD generalization for plates.

  89. Q89hard

    Graphical statics uses force polygons for

    1. Aonly RCC shear design
    2. Bonly creep coefficient
    3. Conly surge tank
    4. Dequilibrium of coplanar forces
    💡 Explanation:

    Force polygon closes for equilibrium.

  90. Q90hard

    Reciprocal theorem (Betti-Maxwell) relates

    1. Aonly thermal expansion
    2. Bwork done by one load system on displacements from another
    3. Conly mix design
    4. Donly zero force members
    💡 Explanation:

    Linear elastic reciprocity between load cases.