Engineering Mechanics and Strength of Materials MCQs 2026

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

📚 Civil Engineering📄 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 1Questions 1–10 of 90
  1. Q1Past Paper · PPSC/FPSC/NTSeasy

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

    1. Athe free end
    2. Bmid-span
    3. Cthe fixed support
    4. Dpoint of load only if interior
    💡 Explanation:

    Cantilever: max BM at fixed end.

  2. Q2medium

    Deflection of a beam is related to curvature by

    1. Ady/dx = M/EI
    2. B1/R ≈ d²y/dx² = M/(EI)
    3. Cy = M/EI
    4. Dcurvature equals shear
    💡 Explanation:

    Elastic curve equation uses second derivative.

  3. Q3medium

    Euler-Bernoulli beam theory assumes

    1. Alarge deflections always
    2. Bplane sections remain plane and small deformations
    3. Cshear deformation dominates always
    4. Dmaterial is rigid-plastic
    💡 Explanation:

    Classical beam theory neglects shear deformation.

  4. Q4Past Paper · PPSC/FPSC/NTSeasy

    Moment of inertia of area is used to compute

    1. Abending stress and deflection
    2. Bonly axial stress
    3. Conly torsional shear in all sections
    4. Donly thermal expansion
    💡 Explanation:

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

  5. Q5Past Paper · PPSC/FPSC/NTSeasy

    Parallel axis theorem states I about any axis equals

    1. AI_c − Ad²
    2. BI_c only
    3. CAd² only
    4. DI_c + Ad²
    💡 Explanation:

    Shift from centroidal axis by distance d.

  6. Q6Past Paper · PPSC/FPSC/NTSeasy

    Section modulus Z is defined as

    1. Ay_max/I
    2. BA/I
    3. CI/y_max
    4. DI/A
    💡 Explanation:

    Z = I/c used in σ = M/Z.

  7. Q7Past Paper · PPSC/FPSC/NTSeasy

    Bending stress in a beam varies

    1. Auniformly across depth
    2. Blinearly with distance from neutral axis
    3. Cparabolically always
    4. Donly at supports
    💡 Explanation:

    σ = My/I gives linear variation.

  8. Q8Past Paper · PPSC/FPSC/NTSeasy

    Neutral axis in pure bending passes through

    1. Atop fibre always
    2. Bcentroid of cross-section for homogeneous beams
    3. Cbottom fibre always
    4. Dshear centre always
    💡 Explanation:

    Zero normal strain at NA; passes through centroid.

  9. Q9easy

    Maximum bending stress occurs at

    1. Aextreme fibres farthest from neutral axis
    2. Bneutral axis
    3. Cshear centre only
    4. Dcentroid only
    💡 Explanation:

    y is maximum at extreme fibres.

  10. Q10Past Paper · PPSC/FPSC/NTSmedium

    Shear stress distribution in rectangular beam is

    1. Auniform
    2. Btriangular with max at top
    3. Czero at NA
    4. Dparabolic with maximum at neutral axis
    💡 Explanation:

    τ max at NA for rectangular section.

  11. Q11Past Paper · PPSC/FPSC/NTSmedium

    Principal stresses at a point are

    1. Aalways equal to σ_x and σ_y
    2. Balways zero
    3. Cnormal stresses on planes with zero shear stress
    4. Donly shear stresses
    💡 Explanation:

    Principal planes have τ = 0.

  12. Q12Past Paper · PPSC/FPSC/NTSmedium

    Mohr circle for plane stress is used to find

    1. Aprincipal stresses and maximum shear stress
    2. Bdeflection of beams
    3. Cbuckling load only
    4. Dreinforcement area
    💡 Explanation:

    Mohr circle graphically solves stress transformation.

  13. Q13medium

    Maximum in-plane shear stress equals

    1. Asum of principal stresses
    2. Bdifference of normal stresses without halving
    3. Cradius of Mohr circle
    4. Dzero at principal planes
    💡 Explanation:

    τ_max = (σ1 − σ2)/2 = radius.

  14. Q14hard

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

    1. Aσ and −σ
    2. Bzero and σ
    3. C2σ and 0
    4. Dσ and σ (both tensile)
    💡 Explanation:

    Equal normal stresses: both principals equal σ.

  15. Q15hard

    Von Mises yield criterion is important for

    1. Aductile materials under combined stress
    2. Bonly brittle fracture
    3. Conly concrete in tension
    4. Donly soil consolidation
    💡 Explanation:

    Distortion energy theory for yielding.

  16. Q16medium

    Torsional shear stress in circular shaft varies

    1. Auniformly across section
    2. Binversely with radius
    3. Clinearly with radius
    4. Donly at surface zero
    💡 Explanation:

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

  17. Q17medium

    Polar moment of inertia J for solid circle is

    1. Aπd⁴/64
    2. Bπd⁴/32
    3. Cbh³/12
    4. Dπr²
    💡 Explanation:

    J = πd⁴/32 for torsion of solid shaft.

  18. Q18medium

    Power transmitted by shaft is

    1. ATω where T is torque and ω angular velocity
    2. BT/ω
    3. Cω/T
    4. DTω² only
    💡 Explanation:

    P = 2πNT/60 in rpm form.

  19. Q19hard

    Strain energy due to axial load is

    1. APL/AE
    2. BP²L/(2AE)
    3. CAE/L
    4. Dσ/E only
    💡 Explanation:

    U = ½ Pδ = P²L/(2AE).

  20. Q20hard

    Castigliano theorem states that

    1. Astress equals strain energy
    2. Bmoment equals shear
    3. Cdeflection equals partial derivative of strain energy with respect to load
    4. Dload equals stiffness squared
    💡 Explanation:

    Useful for indeterminate structures.

  21. Q21easy

    A statically determinate structure has

    1. Ainfinite solutions
    2. Bno equilibrium equations
    3. Conly compatibility needed
    4. Dunique reactions solvable from equilibrium alone
    💡 Explanation:

    Determinate: equations = unknowns.

  22. Q22easy

    Three equations of static equilibrium in 2D are

    1. Afour moment equations
    2. BΣFx=0, ΣFy=0, ΣM=0
    3. Conly ΣF=0
    4. DΣF and compatibility
    💡 Explanation:

    Two force and one moment equation in plane.

  23. Q23easy

    Method of joints in truss analysis uses

    1. Acompatibility of deflections only
    2. BMohr circle
    3. Cmoment distribution only
    4. Dequilibrium of concurrent forces at each joint
    💡 Explanation:

    Joint equilibrium solves member forces.

  24. Q24medium

    Method of sections in trusses uses

    1. Aonly joint summation
    2. Bequilibrium of a cut free-body
    3. Conly energy method
    4. Donly plastic analysis
    💡 Explanation:

    Cut through members and use ΣF, ΣM.

  25. Q25medium

    Zero-force members in trusses occur when

    1. Aall members always carry load
    2. Bonly at supports
    3. Ctwo non-collinear members at unloaded joint carry no force in third member
    4. Donly in roof trusses
    💡 Explanation:

    Classic zero-force rules at simple joints.

  26. Q26medium

    Deflection of cantilever with end load P is

    1. APL³/(48EI)
    2. BPL³/(3EI)
    3. C5wL⁴/(384EI)
    4. DPL/4AE
    💡 Explanation:

    Standard cantilever end-load formula.

  27. Q27medium

    Deflection of simply supported beam centre load P is

    1. APL³/(3EI)
    2. BwL⁴/(8EI)
    3. CPL/2AE
    4. DPL³/(48EI)
    💡 Explanation:

    Mid-span deflection for central point load.

  28. Q28hard

    Macaulay bracket method helps in

    1. Afinding centroid only
    2. Bsoil bearing capacity
    3. Cmix design only
    4. Dwriting deflection equation with discontinuous loading
    💡 Explanation:

    Singularity functions for beam loading.

  29. Q29medium

    Stress concentration occurs at

    1. Auniform prismatic bars only
    2. Bsudden changes in cross-section
    3. Ccentroid only
    4. Dneutral axis only
    💡 Explanation:

    Notches and holes raise local stress.

  30. Q30medium

    Fatigue failure is caused by

    1. Asingle overload only
    2. Bonly corrosion
    3. Conly high temperature creep
    4. Drepeated fluctuating stresses below static strength
    💡 Explanation:

    Cyclic loading leads to crack growth.

  31. Q31Past Paper · PPSC/FPSC/NTSeasy

    The SI unit of force is

    1. Ajoule
    2. Bpascal
    3. Cwatt
    4. Dnewton
    💡 Explanation:

    Force is measured in newtons (N) in SI.

  32. Q32Past Paper · PPSC/FPSC/NTSeasy

    A body is in static equilibrium when

    1. Aonly forces sum to zero
    2. Bvelocity is maximum
    3. Csum of forces and sum of moments are both zero
    4. Dacceleration is constant
    💡 Explanation:

    Equilibrium requires both force and moment balance.

  33. Q33Past Paper · PPSC/FPSC/NTSeasy

    The principle of transmissibility of force states that

    1. Aforce direction may be changed freely
    2. Bmoment about any point changes
    3. Cforce must act only at centroid
    4. Da force may be moved along its line of action without changing external effects
    💡 Explanation:

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

  34. Q34Past Paper · PPSC/FPSC/NTSmedium

    Varignon theorem relates

    1. Astress to strain only
    2. Bwork to power only
    3. Cmoment of a resultant force to sum of moments of components
    4. Dmass to weight only
    💡 Explanation:

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

  35. Q35hard

    The centroid of a uniform semicircular area lies at

    1. Ar/2 from the diameter
    2. Br from the diameter
    3. Cat the centre of full circle
    4. D4r/(3π) from the diameter
    💡 Explanation:

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

  36. Q36Past Paper · PPSC/FPSC/NTSeasy

    Coefficient of friction is defined as the ratio of

    1. Anormal force to weight
    2. Bshear stress to normal stress
    3. Celastic modulus to shear modulus
    4. Dlimiting friction to normal reaction
    💡 Explanation:

    μ = F/N at impending motion.

  37. Q37Past Paper · PPSC/FPSC/NTSeasy

    Angle of friction is the angle whose tangent equals

    1. Acoefficient of friction
    2. BPoisson ratio
    3. Cmodulus of elasticity
    4. Dfactor of safety
    💡 Explanation:

    tan φ = μ.

  38. Q38Past Paper · PPSC/FPSC/NTSmedium

    Lami theorem applies to

    1. Aparallel forces only
    2. Bfour forces in space
    3. Cthree coplanar concurrent forces in equilibrium
    4. Ddistributed loads only
    💡 Explanation:

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

  39. Q39Past Paper · PPSC/FPSC/NTSeasy

    A couple produces

    1. Apure rotation without translation
    2. Bpure translation only
    3. Czero moment everywhere
    4. Donly normal stress
    💡 Explanation:

    Couple has zero resultant force but nonzero moment.

  40. Q40medium

    The moment of a couple is independent of

    1. Amagnitude of forces
    2. Bperpendicular distance
    3. Cplane of action
    4. Dthe point about which moment is taken
    💡 Explanation:

    Couple moment is constant about any point in the plane.

  41. Q41Past Paper · PPSC/FPSC/NTSeasy

    Hooke law in elastic range states that

    1. Astress equals ultimate strength
    2. Bstrain is always zero
    3. Cload is independent of deformation
    4. Dstress is proportional to strain
    💡 Explanation:

    σ = Eε within proportional limit.

  42. Q42Past Paper · PPSC/FPSC/NTSeasy

    Young modulus E is defined as

    1. Aratio of longitudinal stress to longitudinal strain
    2. Bshear stress to shear strain
    3. Clateral strain to longitudinal strain
    4. Dload to deflection
    💡 Explanation:

    E = σ/ε for uniaxial tension/compression.

  43. Q43Past Paper · PPSC/FPSC/NTSeasy

    Poisson ratio is defined as

    1. Alateral strain divided by longitudinal strain with proper sign
    2. Blongitudinal to lateral strain without sign
    3. Cshear to normal strain
    4. Dstress to strain at failure
    💡 Explanation:

    ν = −ε_lateral/ε_longitudinal.

  44. Q44Past Paper · PPSC/FPSC/NTSeasy

    Modulus of rigidity G relates shear stress and shear strain as

    1. Aσ = Gε
    2. Bτ = Eγ
    3. Cσ = Gγ
    4. Dτ = Gγ
    💡 Explanation:

    G is the shear modulus.

  45. Q45medium

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

    1. Aisotropic homogeneous elastic materials
    2. Bonly plastic materials
    3. Conly orthotropic timber always
    4. Donly fluids
    💡 Explanation:

    Standard elastic constant relation for isotropic solids.

  46. Q46Past Paper · PPSC/FPSC/NTSeasy

    Normal stress on a plane is defined as

    1. Aforce perpendicular to the plane divided by area
    2. Btangential force divided by area
    3. Cmoment divided by area
    4. Ddeflection divided by length
    💡 Explanation:

    σ = P/A for axial/normal loading.

  47. Q47Past Paper · PPSC/FPSC/NTSeasy

    Shear stress is defined as

    1. Anormal force divided by volume
    2. Btangential force divided by area
    3. Cbending moment divided by area
    4. Dtorque divided by length only
    💡 Explanation:

    τ = V/A in simple shear.

  48. Q48Past Paper · PPSC/FPSC/NTSeasy

    Factor of safety is generally

    1. Aallowable stress divided by ultimate strength
    2. Bload divided by deflection
    3. Cultimate strength divided by allowable stress
    4. Dstrain divided by stress
    💡 Explanation:

    FOS = σ_ult/σ_allow or similar strength ratio.

  49. Q49easy

    A prismatic bar under axial tension has constant

    1. Anormal stress if load is concentric
    2. Bshear stress throughout always
    3. Cbending moment
    4. Dtorsion
    💡 Explanation:

    Uniform cross-section and axial load give uniform σ.

  50. Q50Past Paper · PPSC/FPSC/NTSmedium

    Thermal stress in a fully restrained bar is

    1. Azero always
    2. BαΔTE (compressive if temperature rises in restraint)
    3. Cindependent of E
    4. Dequal to yield stress always
    💡 Explanation:

    σ = αΔTE when expansion is prevented.

  51. Q51Past Paper · PPSC/FPSC/NTSeasy

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

    1. APL/8
    2. BPL/2
    3. Czero at centre
    4. DPL/4
    💡 Explanation:

    BM max at mid-span for central point load.

  52. Q52Past Paper · PPSC/FPSC/NTSeasy

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

    1. AwL²/8
    2. BwL²/2
    3. CwL²/4
    4. DwL/8
    💡 Explanation:

    Standard SSB with full-span UDL.

  53. Q53Past Paper · PPSC/FPSC/NTSeasy

    Shear force at any section equals

    1. Aalgebraic sum of vertical forces on one side of the section
    2. Bsum of bending moments
    3. Cproduct of stress and area only
    4. Dderivative of moment with respect to time
    💡 Explanation:

    SF is vertical force resultant to one side.

  54. Q54Past Paper · PPSC/FPSC/NTSeasy

    Bending moment at a section equals

    1. Asum of vertical forces only
    2. Bshear force times span always
    3. Calgebraic sum of moments of forces on one side about the section
    4. Darea of SFD
    💡 Explanation:

    BM is moment of forces about the section.

  55. Q55Past Paper · PPSC/FPSC/NTSmedium

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

    1. AdM/dx = w
    2. BdV/dx = M
    3. CdM/dx = V and dV/dx = −w
    4. DM = V/w always
    💡 Explanation:

    Differentiation links load, shear, and moment diagrams.

  56. Q56Past Paper · PPSC/FPSC/NTSmedium

    Point of contraflexure is where

    1. Abending moment is zero and changes sign
    2. Bshear force is maximum
    3. Cdeflection is maximum
    4. Dslope is maximum
    💡 Explanation:

    BM crosses zero between positive and negative regions.

  57. Q57easy

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

    1. Aat the supports
    2. Bat mid-span
    3. Cat quarter span only
    4. Dnowhere
    💡 Explanation:

    SF max at supports; zero at centre.

  58. Q58medium

    Creep is

    1. Ainstant elastic strain
    2. Btime-dependent deformation under constant stress at high temperature
    3. Conly impact loading
    4. Donly buckling
    💡 Explanation:

    Creep matters in metals and concrete at sustained load.

  59. Q59medium

    Buckling load for pin-ended slender column by Euler is

    1. AπEI/L
    2. B4π²EI/L²
    3. Cπ²EI/L²
    4. DAE/L
    💡 Explanation:

    P_cr = π²EI/L² for effective length L.

  60. Q60medium

    Effective length of column fixed at both ends is

    1. A2L
    2. BL
    3. C0.7L
    4. D0.5L
    💡 Explanation:

    K = 0.5 for fixed-fixed ideal case.

  61. Q61easy

    Slenderness ratio is defined as

    1. Aradius over length
    2. Beffective length divided by radius of gyration
    3. Carea over I
    4. Dload over area
    💡 Explanation:

    λ = Le/r.

  62. Q62easy

    Radius of gyration r equals

    1. AI/A
    2. BA/I
    3. C√(I/A)
    4. DI×A
    💡 Explanation:

    r = √(I/A).

  63. Q63hard

    Pure shear stress state has principal stresses

    1. Aboth tensile
    2. Bequal in magnitude and opposite in sign
    3. Cboth zero
    4. Done zero one tensile
    💡 Explanation:

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

  64. Q64medium

    Complementary shear stresses on perpendicular planes are

    1. Aequal in magnitude
    2. Bzero always
    3. Cdouble each other
    4. Dindependent
    💡 Explanation:

    τ_xy = τ_yx equilibrium requirement.

  65. Q65easy

    Resultant of parallel forces equals

    1. Aalgebraic sum of forces
    2. Bvector sum perpendicular only
    3. Cproduct of forces
    4. Dmoment about any point
    💡 Explanation:

    Parallel forces: R = ΣF.

  66. Q66medium

    Wedge friction problem uses

    1. Alimiting equilibrium on inclined plane
    2. Bonly energy method
    3. Conly Mohr-Coulomb for soil always
    4. Donly truss analysis
    💡 Explanation:

    Block on incline classic statics.

  67. Q67medium

    Centre of gravity of composite body uses

    1. Aonly largest part centroid
    2. Bonly volume without density
    3. Conly perimeter
    4. Dmoment summation of parts divided by total weight
    💡 Explanation:

    x̄ = ΣW_ix_i/ΣW_i.

  68. Q68easy

    A frame differs from truss in that

    1. Aframes have only pin joints always
    2. Btrusses carry bending primarily
    3. Cframes have no moments
    4. Dframes resist loads by bending in members and joints may be rigid
    💡 Explanation:

    Frames: rigid/fixed joints allow BM.

  69. Q69easy

    Strain is defined as

    1. Astress divided by area
    2. Bforce times displacement
    3. Cmoment divided by EI
    4. Dchange in dimension divided by original dimension
    💡 Explanation:

    ε = ΔL/L.

  70. Q70easy

    Proportional limit is

    1. Astress at fracture
    2. Bultimate stress
    3. Cstress below which stress-strain is linear
    4. Dfatigue limit
    💡 Explanation:

    Hooke law valid up to proportional limit.

  71. Q71easy

    Yield point is where

    1. Aelastic modulus changes to zero instantly
    2. Bspecimen fractures
    3. Cmaterial begins large inelastic deformation
    4. Dcreep starts always
    💡 Explanation:

    Yield marks onset of plastic flow in ductile metals.

  72. Q72easy

    Ultimate tensile strength is

    1. Astress at elastic limit only
    2. Bmaximum stress on engineering stress-strain curve
    3. Cminimum stress
    4. Dshear at NA
    💡 Explanation:

    UTS is peak load/original area.

  73. Q73medium

    Ductility is measured by

    1. Aonly elastic modulus
    2. Bpercentage elongation or reduction in area
    3. Conly hardness number
    4. Donly density
    💡 Explanation:

    Elongation indicates ductility.

  74. Q74easy

    Brittle materials fail

    1. Aonly after large necking
    2. Bsuddenly with little plastic deformation
    3. Conly in compression always
    4. Donly under torsion
    💡 Explanation:

    Brittle fracture without much yield.

  75. Q75hard

    Modulus of resilience is

    1. Atotal area to fracture
    2. Byield stress only
    3. Chardness index
    4. Darea under elastic portion of stress-strain curve
    💡 Explanation:

    Energy stored elastically per unit volume.

  76. Q76hard

    Modulus of toughness is

    1. Aonly elastic area
    2. Btotal area under stress-strain curve to fracture
    3. Cσ_y/E
    4. Dshear modulus
    💡 Explanation:

    Energy absorbed until failure.

  77. Q77hard

    For thin cylinder under internal pressure, hoop stress is

    1. Atwice longitudinal stress in thin-wall approximation
    2. Bhalf longitudinal stress
    3. Cequal to shear only
    4. Dzero
    💡 Explanation:

    σ_h = pr/t; σ_l = pr/(2t).

  78. Q78medium

    Bending moment diagram for cantilever with UDL is

    1. Alinear
    2. Bconstant
    3. Cparabolic
    4. Dtriangular only at supports
    💡 Explanation:

    M varies as x² for UDL on cantilever.

  79. Q79easy

    Shear force diagram for beam with UDL is

    1. Aparabolic
    2. Blinear
    3. Cconstant non-zero throughout
    4. Dcircular
    💡 Explanation:

    V varies linearly when w is constant.

  80. Q80hard

    Combined bending and axial tension shifts neutral axis

    1. Ato centroid always
    2. Bto compression side always
    3. Ctoward tension side
    4. Doutside section always
    💡 Explanation:

    Superposition shifts NA from centroid.

  81. Q81medium

    Eccentric axial load on short column produces

    1. Acombined axial stress and bending stress
    2. Bonly uniform axial stress
    3. Conly torsion
    4. Donly shear
    💡 Explanation:

    e = M/P type eccentricity effects.

  82. Q82hard

    Shear centre is point through which

    1. Amoment is zero always
    2. BBM is maximum
    3. Cshear load must act to avoid twisting
    4. Daxial stress is zero
    💡 Explanation:

    Loads through shear centre prevent torsion.

  83. Q83hard

    Saint-Venant principle states that

    1. Alocal effects decay away from application region
    2. Bstress is uniform everywhere
    3. Cequilibrium is unnecessary
    4. Dcompatibility is zero
    💡 Explanation:

    Disturbances die out at distance ~ characteristic dimension.

  84. Q84easy

    Work of a force is

    1. Aforce times time
    2. Bforce times displacement in direction of force
    3. Cmoment times angle only
    4. Dmass times acceleration
    💡 Explanation:

    W = F·d cosθ.

  85. Q85easy

    Kinetic energy of particle is

    1. A½mv²
    2. Bmv
    3. C½mv
    4. Dmgh always
    💡 Explanation:

    KE = ½mv².

  86. Q86easy

    Potential energy due to gravity is

    1. A½mgh
    2. Bmv²
    3. Cmgh
    4. DFh always
    💡 Explanation:

    PE = weight × height.

  87. Q87hard

    Virtual work principle requires

    1. Aplastic collapse only
    2. Bequilibrium and virtual displacements compatible
    3. Conly dynamic loads
    4. Donly truss zero force
    💡 Explanation:

    External virtual work = internal.

  88. Q88medium

    A propped cantilever is

    1. Astatically determinate
    2. Bunstable
    3. Cstatically indeterminate to first degree
    4. Dthree degrees indeterminate always
    💡 Explanation:

    Extra prop adds one redundant.

  89. Q89medium

    Compatibility condition in indeterminate structures ensures

    1. Aonly ΣF=0
    2. Bonly concrete cracks
    3. Conly steel yields
    4. Ddeformations fit together
    💡 Explanation:

    Compatibility supplements equilibrium.

  90. Q90easy

    Flexural rigidity of beam is

    1. AEA
    2. BEI
    3. CGJ
    4. DEI/L only
    💡 Explanation:

    EI resists bending curvature.