Structural Analysis MCQs 2026
90 questions with detailed answers · 31 from past papers · 9 quiz batches available
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- Q1 Past Paper · PPSC/FPSC/NTS easy
For determinate truss, member forces found by
💡 Explanation:Equilibrium suffices for determinate truss.
- Q2 medium
Deflection of truss joint found by
💡 Explanation:Virtual work on truss members.
- Q3 Past Paper · PPSC/FPSC/NTS medium
Temperature rise in determinate truss member causes
💡 Explanation:Determinate: no thermal stress if free to expand.
- Q4 Past Paper · PPSC/FPSC/NTS medium
Temperature change in indeterminate structure causes
💡 Explanation:Redundancy restrains expansion.
- Q5 hard
Support settlement in determinate beam causes
💡 Explanation:Determinate: settlement causes geometry change only.
- Q6 hard
Support settlement in indeterminate beam causes
💡 Explanation:Restraint induces member forces.
- Q7 Past Paper · PPSC/FPSC/NTS medium
ILD for shear at section x in SSB is
💡 Explanation:Shear ILD piecewise linear.
- Q8 Past Paper · PPSC/FPSC/NTS hard
Maximum BM in moving load series occurs when
💡 Explanation:Position loads where ILD product maximized.
- Q9 medium
Concentrated moving load on SSB for max BM places load at
💡 Explanation:Load at peak ILD ordinate for that effect.
- Q10 hard
Uniformly distributed moving load shorter than span for max BM
💡 Explanation:UDL length covers beneficial ILD region.
- Q11 Past Paper · PPSC/FPSC/NTS medium
Arches carry load primarily by
💡 Explanation:Arch action reduces bending moments.
- Q12 Past Paper · PPSC/FPSC/NTS hard
Horizontal thrust in two-hinged parabolic arch under UDL equals
💡 Explanation:Thrust formula for parabolic arch UDL.
- Q13 medium
Cable structure under point loads forms
💡 Explanation:Cable edges between loads.
- Q14 medium
Deflected shape of elastic curve has M/(EI) proportional to
💡 Explanation:Curvature κ = M/(EI).
- Q15 easy
Symmetry in structure and loading allows
💡 Explanation:Symmetry reduces unknowns.
- Q16 medium
Antisymmetric loading on symmetric structure can use
💡 Explanation:Antisymmetry simplification.
- Q17 Past Paper · PPSC/FPSC/NTS medium
Portal frame under lateral load develops
💡 Explanation:Lateral load resisted by frame action.
- Q18 medium
Sway in frame occurs when
💡 Explanation:Sidesway is lateral joint displacement.
- Q19 hard
Sidesway in moment distribution is prevented by
💡 Explanation:Sway frames need additional condition.
- Q20 Past Paper · PPSC/FPSC/NTS easy
Release of moment at hinge means
💡 Explanation:Internal hinge: M=0.
- Q21 medium
Relative stiffness of member in distribution depends on
💡 Explanation:EI/L with fixity affects K.
- Q22 hard
Carry-over in non-prismatic member
💡 Explanation:Prismatic assumption gives 0.5 carry-over.
- Q23 Past Paper · PPSC/FPSC/NTS easy
Fixed support provides
💡 Explanation:Fixed: 3 reactions in 2D (2 force + moment).
- Q24 Past Paper · PPSC/FPSC/NTS easy
Pin support in 2D provides
💡 Explanation:Pin: 2 force components.
- Q25 Past Paper · PPSC/FPSC/NTS easy
Roller support provides
💡 Explanation:Roller: one reaction typically vertical.
- Q26 Past Paper · PPSC/FPSC/NTS easy
Stable structure requires
💡 Explanation:Stability prevents mechanisms.
- Q27 medium
Mechanism in structure means
💡 Explanation:Mechanism: insufficient restraint.
- Q28 Past Paper · PPSC/FPSC/NTS medium
Redundant reaction in beam is
💡 Explanation:Redundant cannot be found by ΣF, ΣM alone.
- Q29 easy
Prime structural analysis objective is to find
💡 Explanation:Analysis yields forces and displacements for design.
- Q30 medium
Linear superposition applies when
💡 Explanation:Superpose load cases in elastic range.
- Q31 easy
Shear diagram jump at concentrated load equals
💡 Explanation:SF discontinuity = applied point load.
- Q32 medium
BM diagram has sudden change in slope at
💡 Explanation:dM/dx = V; V jump changes slope.
- Q33 medium
Deflection limit serviceability check uses
💡 Explanation:Serviceability limits on deflection/crack.
- Q34 hard
Influence surface is extension of ILD to
💡 Explanation:ILD generalization for plates.
- Q35 hard
Graphical statics uses force polygons for
💡 Explanation:Force polygon closes for equilibrium.
- Q36 hard
Reciprocal theorem (Betti-Maxwell) relates
💡 Explanation:Linear elastic reciprocity between load cases.
- Q37 medium
Maximum negative BM at interior support of continuous beam often occurs with
💡 Explanation:Loading pattern for hogging envelope.
- Q38 easy
ILD for support reaction in determinate beam is
💡 Explanation:Reaction ILD piecewise linear.
- Q39 hard
Beam on elastic foundation uses
💡 Explanation:Soil as springs: p = kx.
- Q40 hard
Method of consistent deformations solves redundants by
💡 Explanation:Force method canonical equations.
- Q41 medium
Compound truss combines
💡 Explanation:Compound structures need full analysis.
- Q42 hard
Space truss has member force determined with
💡 Explanation:3D truss: ΣF_x,y,z.
- Q43 hard
Dynamic amplification in moving load relates to
💡 Explanation:Resonance/impact increases response.
- Q44 medium
Damping in structures dissipates
💡 Explanation:Damping reduces dynamic response.
- Q45 hard
Rayleigh-Ritz method is approximate for
💡 Explanation:Energy-based approximate analysis.
- Q46 hard
K-value for column effective length in frames depends on
💡 Explanation:Alignment charts give K factors.
- Q47 medium
Bracing in steel frame reduces
💡 Explanation:Bracing increases lateral stability.
- Q48 hard
Elastic critical load for frame buckling depends on
💡 Explanation:Frame stability: GJ, EI, lengths, bracing.
- Q49 medium
Point of inflection in deflected beam corresponds to
💡 Explanation:Zero curvature where M=0.
- Q50 easy
Sagging bending moment in span makes beam
💡 Explanation:Mid-span positive sagging typical in continuous beam.
- Q51 medium
Hogging bending moment makes beam
💡 Explanation:Convention: hogging negative BM at supports in continuous beams.
- Q52 easy
Shear force sign convention often takes
💡 Explanation:Sign conventions vary; consistency matters.
- Q53 medium
Moment redistribution in RCC allows
💡 Explanation:IS 456 permits limited redistribution.
- Q54 easy
Continuous beam is typically
💡 Explanation:Multiple spans create redundants.
- Q55 medium
Envelope of BM from moving loads gives
💡 Explanation:Design for worst-case moving load.
- Q56 medium
Absolute maximum BM in SSB with moving concentrated load is
💡 Explanation:Moving single load max at centre.
- Q57 Past Paper · PPSC/FPSC/NTS easy
Influence line ordinate at point equals
💡 Explanation:Definition of ILD ordinate.
- Q58 hard
Shape factor for plastic section modulus is
💡 Explanation:S = Z_p/Z_e.
- Q59 hard
Collapse mechanism in plastic frame has
💡 Explanation:Mechanism needs enough hinges.
- Q60 Past Paper · PPSC/FPSC/NTS hard
Plastic analysis uses
💡 Explanation:Limit analysis for steel/plastic design.
- Q61 medium
Linear elastic analysis assumes
💡 Explanation:Elastic analysis uses proportionality.
- Q62 hard
Stiffness matrix relates
💡 Explanation:Stiffness {F}=[k]{u}.
- Q63 hard
Flexibility matrix relates
💡 Explanation:Flexibility {u}=[f]{F}.
- Q64 hard
Consistent deformation method is another name for
💡 Explanation:Displacement-based analysis.
- Q65 hard
Stiffness matrix method assembles
💡 Explanation:FE/stiffness assembly K{u}={F}.
- Q66 hard
Displacement method selects
💡 Explanation:Stiffness method solves displacements.
- Q67 hard
Force method selects
💡 Explanation:Flexibility/force method solves redundants.
- Q68 medium
Compatibility equation in indeterminate analysis expresses
💡 Explanation:Deformations must be consistent.
- Q69 Past Paper · PPSC/FPSC/NTS easy
Statically indeterminate beams have
💡 Explanation:Redundants need compatibility conditions.
- Q70 medium
Internal hinge in structure
💡 Explanation:Hinge releases moment, adds equation.
- Q71 Past Paper · PPSC/FPSC/NTS medium
Truss with m members, j joints, r reactions is determinate if
💡 Explanation:Simple counting criterion for plane truss.
- Q72 hard
Williot-Mohr diagram is used for
💡 Explanation:Graphical truss displacement analysis.
- Q73 hard
Conjugate beam method maps
💡 Explanation:M/EI diagram as load on conjugate beam.
- Q74 hard
Graphical integration of M/(EI) diagram gives
💡 Explanation:Area-moment method for beam deflection.
- Q75 Past Paper · PPSC/FPSC/NTS medium
Unit load method for deflection uses
💡 Explanation:Castigliano/virtual work: δ = ∫(M m)/(EI) dx.
- Q76 hard
Clapeyron theorem of three moments applies to
💡 Explanation:Three-moment equation links BM at three supports.
- Q77 Past Paper · PPSC/FPSC/NTS hard
Fixed-end moment in beam due to uniform load w on span L is
💡 Explanation:Standard FEM for UDL on fixed-fixed beam.
- Q78 Past Paper · PPSC/FPSC/NTS medium
Distribution factor at a joint equals
💡 Explanation:DF_i = K_i/ΣK.
- Q79 Past Paper · PPSC/FPSC/NTS medium
Stiffness factor with far end pinned is
💡 Explanation:K = 3EI/L when far end is pinned.
- Q80 Past Paper · PPSC/FPSC/NTS medium
Stiffness factor for prismatic member with far end fixed is
💡 Explanation:K = 4EI/L fixed far end.
- Q81 Past Paper · PPSC/FPSC/NTS medium
Carry-over factor in moment distribution for prismatic member is
💡 Explanation:Half the applied moment carries to far fixed end.
- Q82 Past Paper · PPSC/FPSC/NTS medium
Moment distribution method was developed by
💡 Explanation:Hardy Cross iterative carry-over method.
- Q83 hard
Slope-deflection method accounts for
💡 Explanation:Stiffness method variant using slope-deflection equations.
- Q84 Past Paper · PPSC/FPSC/NTS medium
For simply supported beam, ILD for mid-span moment is
💡 Explanation:Unit load at centre gives max BM.
- Q85 Past Paper · PPSC/FPSC/NTS hard
Muller-Breslau principle states that
💡 Explanation:Qualitative ILD from released structure deflection.
- Q86 Past Paper · PPSC/FPSC/NTS medium
Influence line for reaction at support shows
💡 Explanation:ILD plots response to moving unit load.
- Q87 Past Paper · PPSC/FPSC/NTS medium
Three-hinged arch is statically
💡 Explanation:Extra hinge provides release making structure determinate.
- Q88 Past Paper · PPSC/FPSC/NTS medium
A two-hinged arch is statically
💡 Explanation:One redundant hinge reaction beyond three equilibrium equations.
- Q89 hard
Degree of kinematic indeterminacy of a frame equals
💡 Explanation:Kinematic indeterminacy counts unknown displacements/rotations.
- Q90 Past Paper · PPSC/FPSC/NTS hard
Degree of static indeterminacy of a rigid jointed plane frame is given by
💡 Explanation:Frame formula: 3m + r − 3j for internal + support redundancies.