Engineering Mechanics and Strength of Materials MCQs 2026
90 questions with detailed answers · 31 from past papers · 9 quiz batches available
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- Q1Past Paper · PPSC/FPSC/NTSeasy
A cantilever with point load at free end has maximum BM at
💡 Explanation:Cantilever: max BM at fixed end.
- Q2medium
Deflection of a beam is related to curvature by
💡 Explanation:Elastic curve equation uses second derivative.
- Q3medium
Euler-Bernoulli beam theory assumes
💡 Explanation:Classical beam theory neglects shear deformation.
- Q4Past Paper · PPSC/FPSC/NTSeasy
Moment of inertia of area is used to compute
💡 Explanation:I appears in σ = My/I and EI in deflection.
- Q5Past Paper · PPSC/FPSC/NTSeasy
Parallel axis theorem states I about any axis equals
💡 Explanation:Shift from centroidal axis by distance d.
- Q6Past Paper · PPSC/FPSC/NTSeasy
Section modulus Z is defined as
💡 Explanation:Z = I/c used in σ = M/Z.
- Q7Past Paper · PPSC/FPSC/NTSeasy
Bending stress in a beam varies
💡 Explanation:σ = My/I gives linear variation.
- Q8Past Paper · PPSC/FPSC/NTSeasy
Neutral axis in pure bending passes through
💡 Explanation:Zero normal strain at NA; passes through centroid.
- Q9easy
Maximum bending stress occurs at
💡 Explanation:y is maximum at extreme fibres.
- Q10Past Paper · PPSC/FPSC/NTSmedium
Shear stress distribution in rectangular beam is
💡 Explanation:τ max at NA for rectangular section.
- Q11Past Paper · PPSC/FPSC/NTSmedium
Principal stresses at a point are
💡 Explanation:Principal planes have τ = 0.
- Q12Past Paper · PPSC/FPSC/NTSmedium
Mohr circle for plane stress is used to find
💡 Explanation:Mohr circle graphically solves stress transformation.
- Q13medium
Maximum in-plane shear stress equals
💡 Explanation:τ_max = (σ1 − σ2)/2 = radius.
- Q14hard
For equal biaxial tension σ_x = σ_y = σ, principal stresses are
💡 Explanation:Equal normal stresses: both principals equal σ.
- Q15hard
Von Mises yield criterion is important for
💡 Explanation:Distortion energy theory for yielding.
- Q16medium
Torsional shear stress in circular shaft varies
💡 Explanation:τ = Tr/J; linear from centre to surface.
- Q17medium
Polar moment of inertia J for solid circle is
💡 Explanation:J = πd⁴/32 for torsion of solid shaft.
- Q18medium
Power transmitted by shaft is
💡 Explanation:P = 2πNT/60 in rpm form.
- Q19hard
Strain energy due to axial load is
💡 Explanation:U = ½ Pδ = P²L/(2AE).
- Q20hard
Castigliano theorem states that
💡 Explanation:Useful for indeterminate structures.
- Q21easy
A statically determinate structure has
💡 Explanation:Determinate: equations = unknowns.
- Q22easy
Three equations of static equilibrium in 2D are
💡 Explanation:Two force and one moment equation in plane.
- Q23easy
Method of joints in truss analysis uses
💡 Explanation:Joint equilibrium solves member forces.
- Q24medium
Method of sections in trusses uses
💡 Explanation:Cut through members and use ΣF, ΣM.
- Q25medium
Zero-force members in trusses occur when
💡 Explanation:Classic zero-force rules at simple joints.
- Q26medium
Deflection of cantilever with end load P is
💡 Explanation:Standard cantilever end-load formula.
- Q27medium
Deflection of simply supported beam centre load P is
💡 Explanation:Mid-span deflection for central point load.
- Q28hard
Macaulay bracket method helps in
💡 Explanation:Singularity functions for beam loading.
- Q29medium
Stress concentration occurs at
💡 Explanation:Notches and holes raise local stress.
- Q30medium
Fatigue failure is caused by
💡 Explanation:Cyclic loading leads to crack growth.
- Q31Past Paper · PPSC/FPSC/NTSeasy
The SI unit of force is
💡 Explanation:Force is measured in newtons (N) in SI.
- Q32Past Paper · PPSC/FPSC/NTSeasy
A body is in static equilibrium when
💡 Explanation:Equilibrium requires both force and moment balance.
- Q33Past Paper · PPSC/FPSC/NTSeasy
The principle of transmissibility of force states that
💡 Explanation:External effects depend on magnitude, direction, and line of action.
- Q34Past Paper · PPSC/FPSC/NTSmedium
Varignon theorem relates
💡 Explanation:The moment of R equals the sum of moments of its components.
- Q35hard
The centroid of a uniform semicircular area lies at
💡 Explanation:Standard result: ȳ = 4r/(3π) for semicircle.
- Q36Past Paper · PPSC/FPSC/NTSeasy
Coefficient of friction is defined as the ratio of
💡 Explanation:μ = F/N at impending motion.
- Q37Past Paper · PPSC/FPSC/NTSeasy
Angle of friction is the angle whose tangent equals
💡 Explanation:tan φ = μ.
- Q38Past Paper · PPSC/FPSC/NTSmedium
Lami theorem applies to
💡 Explanation:Each force is proportional to sine of angle between other two.
- Q39Past Paper · PPSC/FPSC/NTSeasy
A couple produces
💡 Explanation:Couple has zero resultant force but nonzero moment.
- Q40medium
The moment of a couple is independent of
💡 Explanation:Couple moment is constant about any point in the plane.
- Q41Past Paper · PPSC/FPSC/NTSeasy
Hooke law in elastic range states that
💡 Explanation:σ = Eε within proportional limit.
- Q42Past Paper · PPSC/FPSC/NTSeasy
Young modulus E is defined as
💡 Explanation:E = σ/ε for uniaxial tension/compression.
- Q43Past Paper · PPSC/FPSC/NTSeasy
Poisson ratio is defined as
💡 Explanation:ν = −ε_lateral/ε_longitudinal.
- Q44Past Paper · PPSC/FPSC/NTSeasy
Modulus of rigidity G relates shear stress and shear strain as
💡 Explanation:G is the shear modulus.
- Q45medium
The relationship E = 2G(1+ν) is valid for
💡 Explanation:Standard elastic constant relation for isotropic solids.
- Q46Past Paper · PPSC/FPSC/NTSeasy
Normal stress on a plane is defined as
💡 Explanation:σ = P/A for axial/normal loading.
- Q47Past Paper · PPSC/FPSC/NTSeasy
Shear stress is defined as
💡 Explanation:τ = V/A in simple shear.
- Q48Past Paper · PPSC/FPSC/NTSeasy
Factor of safety is generally
💡 Explanation:FOS = σ_ult/σ_allow or similar strength ratio.
- Q49easy
A prismatic bar under axial tension has constant
💡 Explanation:Uniform cross-section and axial load give uniform σ.
- Q50Past Paper · PPSC/FPSC/NTSmedium
Thermal stress in a fully restrained bar is
💡 Explanation:σ = αΔTE when expansion is prevented.
- Q51Past Paper · PPSC/FPSC/NTSeasy
In a simply supported beam with central point load P, maximum bending moment is
💡 Explanation:BM max at mid-span for central point load.
- Q52Past Paper · PPSC/FPSC/NTSeasy
In a simply supported beam with UDL w over span L, maximum bending moment is
💡 Explanation:Standard SSB with full-span UDL.
- Q53Past Paper · PPSC/FPSC/NTSeasy
Shear force at any section equals
💡 Explanation:SF is vertical force resultant to one side.
- Q54Past Paper · PPSC/FPSC/NTSeasy
Bending moment at a section equals
💡 Explanation:BM is moment of forces about the section.
- Q55Past Paper · PPSC/FPSC/NTSmedium
The relationship between load w, shear force V and bending moment M is
💡 Explanation:Differentiation links load, shear, and moment diagrams.
- Q56Past Paper · PPSC/FPSC/NTSmedium
Point of contraflexure is where
💡 Explanation:BM crosses zero between positive and negative regions.
- Q57easy
Maximum shear in a simply supported beam with central point load occurs
💡 Explanation:SF max at supports; zero at centre.
- Q58medium
Creep is
💡 Explanation:Creep matters in metals and concrete at sustained load.
- Q59medium
Buckling load for pin-ended slender column by Euler is
💡 Explanation:P_cr = π²EI/L² for effective length L.
- Q60medium
Effective length of column fixed at both ends is
💡 Explanation:K = 0.5 for fixed-fixed ideal case.
- Q61easy
Slenderness ratio is defined as
💡 Explanation:λ = Le/r.
- Q62easy
Radius of gyration r equals
💡 Explanation:r = √(I/A).
- Q63hard
Pure shear stress state has principal stresses
💡 Explanation:τ_max planes carry σ1 = τ, σ2 = −τ.
- Q64medium
Complementary shear stresses on perpendicular planes are
💡 Explanation:τ_xy = τ_yx equilibrium requirement.
- Q65easy
Resultant of parallel forces equals
💡 Explanation:Parallel forces: R = ΣF.
- Q66medium
Wedge friction problem uses
💡 Explanation:Block on incline classic statics.
- Q67medium
Centre of gravity of composite body uses
💡 Explanation:x̄ = ΣW_ix_i/ΣW_i.
- Q68easy
A frame differs from truss in that
💡 Explanation:Frames: rigid/fixed joints allow BM.
- Q69easy
Strain is defined as
💡 Explanation:ε = ΔL/L.
- Q70easy
Proportional limit is
💡 Explanation:Hooke law valid up to proportional limit.
- Q71easy
Yield point is where
💡 Explanation:Yield marks onset of plastic flow in ductile metals.
- Q72easy
Ultimate tensile strength is
💡 Explanation:UTS is peak load/original area.
- Q73medium
Ductility is measured by
💡 Explanation:Elongation indicates ductility.
- Q74easy
Brittle materials fail
💡 Explanation:Brittle fracture without much yield.
- Q75hard
Modulus of resilience is
💡 Explanation:Energy stored elastically per unit volume.
- Q76hard
Modulus of toughness is
💡 Explanation:Energy absorbed until failure.
- Q77hard
For thin cylinder under internal pressure, hoop stress is
💡 Explanation:σ_h = pr/t; σ_l = pr/(2t).
- Q78medium
Bending moment diagram for cantilever with UDL is
💡 Explanation:M varies as x² for UDL on cantilever.
- Q79easy
Shear force diagram for beam with UDL is
💡 Explanation:V varies linearly when w is constant.
- Q80hard
Combined bending and axial tension shifts neutral axis
💡 Explanation:Superposition shifts NA from centroid.
- Q81medium
Eccentric axial load on short column produces
💡 Explanation:e = M/P type eccentricity effects.
- Q82hard
Shear centre is point through which
💡 Explanation:Loads through shear centre prevent torsion.
- Q83hard
Saint-Venant principle states that
💡 Explanation:Disturbances die out at distance ~ characteristic dimension.
- Q84easy
Work of a force is
💡 Explanation:W = F·d cosθ.
- Q85easy
Kinetic energy of particle is
💡 Explanation:KE = ½mv².
- Q86easy
Potential energy due to gravity is
💡 Explanation:PE = weight × height.
- Q87hard
Virtual work principle requires
💡 Explanation:External virtual work = internal.
- Q88medium
A propped cantilever is
💡 Explanation:Extra prop adds one redundant.
- Q89medium
Compatibility condition in indeterminate structures ensures
💡 Explanation:Compatibility supplements equilibrium.
- Q90easy
Flexural rigidity of beam is
💡 Explanation:EI resists bending curvature.