Engineering Mechanics and Strength of Materials MCQs 2026

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

📚 Civil Engineering Mcqs 📄 31 Past-Paper Qs ✓ Free · No Login Needed
🎯 Mock Test

Read each question, think about the answer, then click Show Answer to reveal the correct option and explanation. Load 10 at a time so it stays manageable — perfect for one-topic study sessions on the bus or during a break.

Page 1 of 1 Questions 110 of 90
  1. Q1 easy

    Yield point is where

    1. A elastic modulus changes to zero instantly
    2. B specimen fractures
    3. C material begins large inelastic deformation
    4. D creep starts always
    💡 Explanation:

    Yield marks onset of plastic flow in ductile metals.

  2. Q2 easy

    Proportional limit is

    1. A stress at fracture
    2. B ultimate stress
    3. C stress below which stress-strain is linear
    4. D fatigue limit
    💡 Explanation:

    Hooke law valid up to proportional limit.

  3. Q3 easy

    Strain is defined as

    1. A stress divided by area
    2. B force times displacement
    3. C moment divided by EI
    4. D change in dimension divided by original dimension
    💡 Explanation:

    ε = ΔL/L.

  4. Q4 easy

    A frame differs from truss in that

    1. A frames have only pin joints always
    2. B trusses carry bending primarily
    3. C frames have no moments
    4. D frames resist loads by bending in members and joints may be rigid
    💡 Explanation:

    Frames: rigid/fixed joints allow BM.

  5. Q5 medium

    Centre of gravity of composite body uses

    1. A only largest part centroid
    2. B only volume without density
    3. C only perimeter
    4. D moment summation of parts divided by total weight
    💡 Explanation:

    x̄ = ΣW_ix_i/ΣW_i.

  6. Q6 Past Paper · PPSC/FPSC/NTS easy

    A cantilever with point load at free end has maximum BM at

    1. A the free end
    2. B mid-span
    3. C the fixed support
    4. D point of load only if interior
    💡 Explanation:

    Cantilever: max BM at fixed end.

  7. Q7 medium

    Deflection of a beam is related to curvature by

    1. A dy/dx = M/EI
    2. B 1/R ≈ d²y/dx² = M/(EI)
    3. C y = M/EI
    4. D curvature equals shear
    💡 Explanation:

    Elastic curve equation uses second derivative.

  8. Q8 medium

    Euler-Bernoulli beam theory assumes

    1. A large deflections always
    2. B plane sections remain plane and small deformations
    3. C shear deformation dominates always
    4. D material is rigid-plastic
    💡 Explanation:

    Classical beam theory neglects shear deformation.

  9. Q9 Past Paper · PPSC/FPSC/NTS easy

    Moment of inertia of area is used to compute

    1. A bending stress and deflection
    2. B only axial stress
    3. C only torsional shear in all sections
    4. D only thermal expansion
    💡 Explanation:

    I appears in σ = My/I and EI in deflection.

  10. Q10 Past Paper · PPSC/FPSC/NTS easy

    Parallel axis theorem states I about any axis equals

    1. A I_c − Ad²
    2. B I_c only
    3. C Ad² only
    4. D I_c + Ad²
    💡 Explanation:

    Shift from centroidal axis by distance d.

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

    Section modulus Z is defined as

    1. A y_max/I
    2. B A/I
    3. C I/y_max
    4. D I/A
    💡 Explanation:

    Z = I/c used in σ = M/Z.

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

    Bending stress in a beam varies

    1. A uniformly across depth
    2. B linearly with distance from neutral axis
    3. C parabolically always
    4. D only at supports
    💡 Explanation:

    σ = My/I gives linear variation.

  13. Q13 Past Paper · PPSC/FPSC/NTS easy

    Neutral axis in pure bending passes through

    1. A top fibre always
    2. B centroid of cross-section for homogeneous beams
    3. C bottom fibre always
    4. D shear centre always
    💡 Explanation:

    Zero normal strain at NA; passes through centroid.

  14. Q14 easy

    Maximum bending stress occurs at

    1. A extreme fibres farthest from neutral axis
    2. B neutral axis
    3. C shear centre only
    4. D centroid only
    💡 Explanation:

    y is maximum at extreme fibres.

  15. Q15 Past Paper · PPSC/FPSC/NTS medium

    Shear stress distribution in rectangular beam is

    1. A uniform
    2. B triangular with max at top
    3. C zero at NA
    4. D parabolic with maximum at neutral axis
    💡 Explanation:

    τ max at NA for rectangular section.

  16. Q16 Past Paper · PPSC/FPSC/NTS medium

    Principal stresses at a point are

    1. A always equal to σ_x and σ_y
    2. B always zero
    3. C normal stresses on planes with zero shear stress
    4. D only shear stresses
    💡 Explanation:

    Principal planes have τ = 0.

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

    Mohr circle for plane stress is used to find

    1. A principal stresses and maximum shear stress
    2. B deflection of beams
    3. C buckling load only
    4. D reinforcement area
    💡 Explanation:

    Mohr circle graphically solves stress transformation.

  18. Q18 medium

    Maximum in-plane shear stress equals

    1. A sum of principal stresses
    2. B difference of normal stresses without halving
    3. C radius of Mohr circle
    4. D zero at principal planes
    💡 Explanation:

    τ_max = (σ1 − σ2)/2 = radius.

  19. Q19 hard

    For equal biaxial tension σ_x = σ_y = σ, principal stresses are

    1. A σ and −σ
    2. B zero and σ
    3. C 2σ and 0
    4. D σ and σ (both tensile)
    💡 Explanation:

    Equal normal stresses: both principals equal σ.

  20. Q20 hard

    Von Mises yield criterion is important for

    1. A ductile materials under combined stress
    2. B only brittle fracture
    3. C only concrete in tension
    4. D only soil consolidation
    💡 Explanation:

    Distortion energy theory for yielding.

  21. Q21 medium

    Torsional shear stress in circular shaft varies

    1. A uniformly across section
    2. B inversely with radius
    3. C linearly with radius
    4. D only at surface zero
    💡 Explanation:

    τ = Tr/J; linear from centre to surface.

  22. Q22 medium

    Polar moment of inertia J for solid circle is

    1. A πd⁴/64
    2. B πd⁴/32
    3. C bh³/12
    4. D πr²
    💡 Explanation:

    J = πd⁴/32 for torsion of solid shaft.

  23. Q23 medium

    Power transmitted by shaft is

    1. A Tω where T is torque and ω angular velocity
    2. B T/ω
    3. C ω/T
    4. D Tω² only
    💡 Explanation:

    P = 2πNT/60 in rpm form.

  24. Q24 hard

    Strain energy due to axial load is

    1. A PL/AE
    2. B P²L/(2AE)
    3. C AE/L
    4. D σ/E only
    💡 Explanation:

    U = ½ Pδ = P²L/(2AE).

  25. Q25 hard

    Castigliano theorem states that

    1. A stress equals strain energy
    2. B moment equals shear
    3. C deflection equals partial derivative of strain energy with respect to load
    4. D load equals stiffness squared
    💡 Explanation:

    Useful for indeterminate structures.

  26. Q26 easy

    A statically determinate structure has

    1. A infinite solutions
    2. B no equilibrium equations
    3. C only compatibility needed
    4. D unique reactions solvable from equilibrium alone
    💡 Explanation:

    Determinate: equations = unknowns.

  27. Q27 easy

    Three equations of static equilibrium in 2D are

    1. A four moment equations
    2. B ΣFx=0, ΣFy=0, ΣM=0
    3. C only ΣF=0
    4. D ΣF and compatibility
    💡 Explanation:

    Two force and one moment equation in plane.

  28. Q28 easy

    Method of joints in truss analysis uses

    1. A compatibility of deflections only
    2. B Mohr circle
    3. C moment distribution only
    4. D equilibrium of concurrent forces at each joint
    💡 Explanation:

    Joint equilibrium solves member forces.

  29. Q29 medium

    Method of sections in trusses uses

    1. A only joint summation
    2. B equilibrium of a cut free-body
    3. C only energy method
    4. D only plastic analysis
    💡 Explanation:

    Cut through members and use ΣF, ΣM.

  30. Q30 medium

    Zero-force members in trusses occur when

    1. A all members always carry load
    2. B only at supports
    3. C two non-collinear members at unloaded joint carry no force in third member
    4. D only in roof trusses
    💡 Explanation:

    Classic zero-force rules at simple joints.

  31. Q31 medium

    Deflection of cantilever with end load P is

    1. A PL³/(48EI)
    2. B PL³/(3EI)
    3. C 5wL⁴/(384EI)
    4. D PL/4AE
    💡 Explanation:

    Standard cantilever end-load formula.

  32. Q32 medium

    Deflection of simply supported beam centre load P is

    1. A PL³/(3EI)
    2. B wL⁴/(8EI)
    3. C PL/2AE
    4. D PL³/(48EI)
    💡 Explanation:

    Mid-span deflection for central point load.

  33. Q33 hard

    Macaulay bracket method helps in

    1. A finding centroid only
    2. B soil bearing capacity
    3. C mix design only
    4. D writing deflection equation with discontinuous loading
    💡 Explanation:

    Singularity functions for beam loading.

  34. Q34 medium

    Wedge friction problem uses

    1. A limiting equilibrium on inclined plane
    2. B only energy method
    3. C only Mohr-Coulomb for soil always
    4. D only truss analysis
    💡 Explanation:

    Block on incline classic statics.

  35. Q35 easy

    Resultant of parallel forces equals

    1. A algebraic sum of forces
    2. B vector sum perpendicular only
    3. C product of forces
    4. D moment about any point
    💡 Explanation:

    Parallel forces: R = ΣF.

  36. Q36 medium

    Complementary shear stresses on perpendicular planes are

    1. A equal in magnitude
    2. B zero always
    3. C double each other
    4. D independent
    💡 Explanation:

    τ_xy = τ_yx equilibrium requirement.

  37. Q37 hard

    Pure shear stress state has principal stresses

    1. A both tensile
    2. B equal in magnitude and opposite in sign
    3. C both zero
    4. D one zero one tensile
    💡 Explanation:

    τ_max planes carry σ1 = τ, σ2 = −τ.

  38. Q38 easy

    Radius of gyration r equals

    1. A I/A
    2. B A/I
    3. C √(I/A)
    4. D I×A
    💡 Explanation:

    r = √(I/A).

  39. Q39 easy

    Slenderness ratio is defined as

    1. A radius over length
    2. B effective length divided by radius of gyration
    3. C area over I
    4. D load over area
    💡 Explanation:

    λ = Le/r.

  40. Q40 medium

    Effective length of column fixed at both ends is

    1. A 2L
    2. B L
    3. C 0.7L
    4. D 0.5L
    💡 Explanation:

    K = 0.5 for fixed-fixed ideal case.

  41. Q41 medium

    Buckling load for pin-ended slender column by Euler is

    1. A πEI/L
    2. B 4π²EI/L²
    3. C π²EI/L²
    4. D AE/L
    💡 Explanation:

    P_cr = π²EI/L² for effective length L.

  42. Q42 medium

    Creep is

    1. A instant elastic strain
    2. B time-dependent deformation under constant stress at high temperature
    3. C only impact loading
    4. D only buckling
    💡 Explanation:

    Creep matters in metals and concrete at sustained load.

  43. Q43 Past Paper · PPSC/FPSC/NTS easy

    The SI unit of force is

    1. A joule
    2. B pascal
    3. C watt
    4. D newton
    💡 Explanation:

    Force is measured in newtons (N) in SI.

  44. Q44 easy

    Ultimate tensile strength is

    1. A stress at elastic limit only
    2. B maximum stress on engineering stress-strain curve
    3. C minimum stress
    4. D shear at NA
    💡 Explanation:

    UTS is peak load/original area.

  45. Q45 medium

    Ductility is measured by

    1. A only elastic modulus
    2. B percentage elongation or reduction in area
    3. C only hardness number
    4. D only density
    💡 Explanation:

    Elongation indicates ductility.

  46. Q46 easy

    Brittle materials fail

    1. A only after large necking
    2. B suddenly with little plastic deformation
    3. C only in compression always
    4. D only under torsion
    💡 Explanation:

    Brittle fracture without much yield.

  47. Q47 hard

    Modulus of resilience is

    1. A total area to fracture
    2. B yield stress only
    3. C hardness index
    4. D area under elastic portion of stress-strain curve
    💡 Explanation:

    Energy stored elastically per unit volume.

  48. Q48 hard

    For thin cylinder under internal pressure, hoop stress is

    1. A twice longitudinal stress in thin-wall approximation
    2. B half longitudinal stress
    3. C equal to shear only
    4. D zero
    💡 Explanation:

    σ_h = pr/t; σ_l = pr/(2t).

  49. Q49 hard

    Modulus of toughness is

    1. A only elastic area
    2. B total area under stress-strain curve to fracture
    3. C σ_y/E
    4. D shear modulus
    💡 Explanation:

    Energy absorbed until failure.

  50. Q50 medium

    Bending moment diagram for cantilever with UDL is

    1. A linear
    2. B constant
    3. C parabolic
    4. D triangular only at supports
    💡 Explanation:

    M varies as x² for UDL on cantilever.

  51. Q51 easy

    Shear force diagram for beam with UDL is

    1. A parabolic
    2. B linear
    3. C constant non-zero throughout
    4. D circular
    💡 Explanation:

    V varies linearly when w is constant.

  52. Q52 hard

    Combined bending and axial tension shifts neutral axis

    1. A to centroid always
    2. B to compression side always
    3. C toward tension side
    4. D outside section always
    💡 Explanation:

    Superposition shifts NA from centroid.

  53. Q53 medium

    Eccentric axial load on short column produces

    1. A combined axial stress and bending stress
    2. B only uniform axial stress
    3. C only torsion
    4. D only shear
    💡 Explanation:

    e = M/P type eccentricity effects.

  54. Q54 hard

    Shear centre is point through which

    1. A moment is zero always
    2. B BM is maximum
    3. C shear load must act to avoid twisting
    4. D axial stress is zero
    💡 Explanation:

    Loads through shear centre prevent torsion.

  55. Q55 hard

    Saint-Venant principle states that

    1. A local effects decay away from application region
    2. B stress is uniform everywhere
    3. C equilibrium is unnecessary
    4. D compatibility is zero
    💡 Explanation:

    Disturbances die out at distance ~ characteristic dimension.

  56. Q56 easy

    Work of a force is

    1. A force times time
    2. B force times displacement in direction of force
    3. C moment times angle only
    4. D mass times acceleration
    💡 Explanation:

    W = F·d cosθ.

  57. Q57 easy

    Kinetic energy of particle is

    1. A ½mv²
    2. B mv
    3. C ½mv
    4. D mgh always
    💡 Explanation:

    KE = ½mv².

  58. Q58 easy

    Potential energy due to gravity is

    1. A ½mgh
    2. B mv²
    3. C mgh
    4. D Fh always
    💡 Explanation:

    PE = weight × height.

  59. Q59 hard

    Virtual work principle requires

    1. A plastic collapse only
    2. B equilibrium and virtual displacements compatible
    3. C only dynamic loads
    4. D only truss zero force
    💡 Explanation:

    External virtual work = internal.

  60. Q60 medium

    Stress concentration occurs at

    1. A uniform prismatic bars only
    2. B sudden changes in cross-section
    3. C centroid only
    4. D neutral axis only
    💡 Explanation:

    Notches and holes raise local stress.

  61. Q61 Past Paper · PPSC/FPSC/NTS easy

    A body is in static equilibrium when

    1. A only forces sum to zero
    2. B velocity is maximum
    3. C sum of forces and sum of moments are both zero
    4. D acceleration is constant
    💡 Explanation:

    Equilibrium requires both force and moment balance.

  62. Q62 easy

    Flexural rigidity of beam is

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

    EI resists bending curvature.

  63. Q63 medium

    Compatibility condition in indeterminate structures ensures

    1. A only ΣF=0
    2. B only concrete cracks
    3. C only steel yields
    4. D deformations fit together
    💡 Explanation:

    Compatibility supplements equilibrium.

  64. Q64 medium

    A propped cantilever is

    1. A statically determinate
    2. B unstable
    3. C statically indeterminate to first degree
    4. D three degrees indeterminate always
    💡 Explanation:

    Extra prop adds one redundant.

  65. Q65 easy

    Maximum shear in a simply supported beam with central point load occurs

    1. A at the supports
    2. B at mid-span
    3. C at quarter span only
    4. D nowhere
    💡 Explanation:

    SF max at supports; zero at centre.

  66. Q66 Past Paper · PPSC/FPSC/NTS medium

    Point of contraflexure is where

    1. A bending moment is zero and changes sign
    2. B shear force is maximum
    3. C deflection is maximum
    4. D slope is maximum
    💡 Explanation:

    BM crosses zero between positive and negative regions.

  67. Q67 Past Paper · PPSC/FPSC/NTS medium

    The relationship between load w, shear force V and bending moment M is

    1. A dM/dx = w
    2. B dV/dx = M
    3. C dM/dx = V and dV/dx = −w
    4. D M = V/w always
    💡 Explanation:

    Differentiation links load, shear, and moment diagrams.

  68. Q68 Past Paper · PPSC/FPSC/NTS easy

    Bending moment at a section equals

    1. A sum of vertical forces only
    2. B shear force times span always
    3. C algebraic sum of moments of forces on one side about the section
    4. D area of SFD
    💡 Explanation:

    BM is moment of forces about the section.

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

    Shear force at any section equals

    1. A algebraic sum of vertical forces on one side of the section
    2. B sum of bending moments
    3. C product of stress and area only
    4. D derivative of moment with respect to time
    💡 Explanation:

    SF is vertical force resultant to one side.

  70. Q70 Past Paper · PPSC/FPSC/NTS easy

    In a simply supported beam with UDL w over span L, maximum bending moment is

    1. A wL²/8
    2. B wL²/2
    3. C wL²/4
    4. D wL/8
    💡 Explanation:

    Standard SSB with full-span UDL.

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

    In a simply supported beam with central point load P, maximum bending moment is

    1. A PL/8
    2. B PL/2
    3. C zero at centre
    4. D PL/4
    💡 Explanation:

    BM max at mid-span for central point load.

  72. Q72 Past Paper · PPSC/FPSC/NTS medium

    Thermal stress in a fully restrained bar is

    1. A zero always
    2. B αΔTE (compressive if temperature rises in restraint)
    3. C independent of E
    4. D equal to yield stress always
    💡 Explanation:

    σ = αΔTE when expansion is prevented.

  73. Q73 easy

    A prismatic bar under axial tension has constant

    1. A normal stress if load is concentric
    2. B shear stress throughout always
    3. C bending moment
    4. D torsion
    💡 Explanation:

    Uniform cross-section and axial load give uniform σ.

  74. Q74 Past Paper · PPSC/FPSC/NTS easy

    Factor of safety is generally

    1. A allowable stress divided by ultimate strength
    2. B load divided by deflection
    3. C ultimate strength divided by allowable stress
    4. D strain divided by stress
    💡 Explanation:

    FOS = σ_ult/σ_allow or similar strength ratio.

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

    Shear stress is defined as

    1. A normal force divided by volume
    2. B tangential force divided by area
    3. C bending moment divided by area
    4. D torque divided by length only
    💡 Explanation:

    τ = V/A in simple shear.

  76. Q76 Past Paper · PPSC/FPSC/NTS easy

    Normal stress on a plane is defined as

    1. A force perpendicular to the plane divided by area
    2. B tangential force divided by area
    3. C moment divided by area
    4. D deflection divided by length
    💡 Explanation:

    σ = P/A for axial/normal loading.

  77. Q77 medium

    The relationship E = 2G(1+ν) is valid for

    1. A isotropic homogeneous elastic materials
    2. B only plastic materials
    3. C only orthotropic timber always
    4. D only fluids
    💡 Explanation:

    Standard elastic constant relation for isotropic solids.

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

    Modulus of rigidity G relates shear stress and shear strain as

    1. A σ = Gε
    2. B τ = Eγ
    3. C σ = Gγ
    4. D τ = Gγ
    💡 Explanation:

    G is the shear modulus.

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

    Poisson ratio is defined as

    1. A lateral strain divided by longitudinal strain with proper sign
    2. B longitudinal to lateral strain without sign
    3. C shear to normal strain
    4. D stress to strain at failure
    💡 Explanation:

    ν = −ε_lateral/ε_longitudinal.

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

    Young modulus E is defined as

    1. A ratio of longitudinal stress to longitudinal strain
    2. B shear stress to shear strain
    3. C lateral strain to longitudinal strain
    4. D load to deflection
    💡 Explanation:

    E = σ/ε for uniaxial tension/compression.

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

    Hooke law in elastic range states that

    1. A stress equals ultimate strength
    2. B strain is always zero
    3. C load is independent of deformation
    4. D stress is proportional to strain
    💡 Explanation:

    σ = Eε within proportional limit.

  82. Q82 medium

    The moment of a couple is independent of

    1. A magnitude of forces
    2. B perpendicular distance
    3. C plane of action
    4. D the point about which moment is taken
    💡 Explanation:

    Couple moment is constant about any point in the plane.

  83. Q83 Past Paper · PPSC/FPSC/NTS easy

    A couple produces

    1. A pure rotation without translation
    2. B pure translation only
    3. C zero moment everywhere
    4. D only normal stress
    💡 Explanation:

    Couple has zero resultant force but nonzero moment.

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

    Lami theorem applies to

    1. A parallel forces only
    2. B four forces in space
    3. C three coplanar concurrent forces in equilibrium
    4. D distributed loads only
    💡 Explanation:

    Each force is proportional to sine of angle between other two.

  85. Q85 medium

    Fatigue failure is caused by

    1. A single overload only
    2. B only corrosion
    3. C only high temperature creep
    4. D repeated fluctuating stresses below static strength
    💡 Explanation:

    Cyclic loading leads to crack growth.

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

    Angle of friction is the angle whose tangent equals

    1. A coefficient of friction
    2. B Poisson ratio
    3. C modulus of elasticity
    4. D factor of safety
    💡 Explanation:

    tan φ = μ.

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

    Coefficient of friction is defined as the ratio of

    1. A normal force to weight
    2. B shear stress to normal stress
    3. C elastic modulus to shear modulus
    4. D limiting friction to normal reaction
    💡 Explanation:

    μ = F/N at impending motion.

  88. Q88 hard

    The centroid of a uniform semicircular area lies at

    1. A r/2 from the diameter
    2. B r from the diameter
    3. C at the centre of full circle
    4. D 4r/(3π) from the diameter
    💡 Explanation:

    Standard result: ȳ = 4r/(3π) for semicircle.

  89. Q89 Past Paper · PPSC/FPSC/NTS medium

    Varignon theorem relates

    1. A stress to strain only
    2. B work to power only
    3. C moment of a resultant force to sum of moments of components
    4. D mass to weight only
    💡 Explanation:

    The moment of R equals the sum of moments of its components.

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

    The principle of transmissibility of force states that

    1. A force direction may be changed freely
    2. B moment about any point changes
    3. C force must act only at centroid
    4. D a force may be moved along its line of action without changing external effects
    💡 Explanation:

    External effects depend on magnitude, direction, and line of action.