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Page 1 of 1Questions 1–10 of 90
Q1Past Paper · PPSC/FPSC/NTSeasy
Normal stress on a plane is defined as
Aforce parallel to area divided by volume✓
Bmoment divided by area✓
Cstrain times modulus only✓
Dforce perpendicular to area divided by area✓
💡 Explanation:
σ = P/A for axial loading.
Q2Past Paper · PPSC/FPSC/NTSeasy
Shear stress τ is given by
Anormal force divided by area✓
Bforce times moment arm✓
Ctangential force divided by area✓
Dstrain divided by time✓
💡 Explanation:
τ = V/A for direct shear.
Q3easy
Hooke's law in elastic range states
Astress equals strain always✓
Bstrain is independent of stress✓
Cstress equals modulus only without strain✓
Dstress is proportional to strain✓
💡 Explanation:
σ = E ε within proportional limit.
Q4easy
Modulus of elasticity E has units of
Adimensionless only✓
Bm/s✓
CPa or N/m²✓
DN·m✓
💡 Explanation:
E relates stress and strain; same units as stress.
Q5Past Paper · PPSC/FPSC/NTSmedium
Poisson's ratio ν is defined as
Aaxial strain divided by lateral strain✓
Bshear strain divided by normal strain always✓
Cstress divided by strain squared✓
Dlateral strain divided by axial strain with opposite sign✓
💡 Explanation:
ν = −ε_lateral/ε_axial for uniaxial loading.
Q6easy
For a bar in simple tension, maximum normal stress occurs on
Aplane at 45° to load always for max✓
Bplane parallel to load✓
Cany plane equally✓
Dplane perpendicular to load✓
💡 Explanation:
Axial stress is maximum on cross section normal to force.
Q7medium
Shear strain γ is approximately
Achange in right angle between originally perpendicular lines✓
Bratio of normal stresses✓
Cchange in volume only✓
Dstress divided by E only✓
💡 Explanation:
γ measures angular distortion.
Q8Past Paper · PPSC/FPSC/NTSmedium
Modulus of rigidity G relates
Anormal stress and lateral strain only✓
Bbulk modulus and pressure only✓
Cshear stress and shear strain✓
Dthermal expansion and temperature✓
💡 Explanation:
τ = G γ in elastic shear.
Q9medium
Bulk modulus K measures resistance to
Auniform volumetric compression✓
Bbending only✓
Ctorsion only✓
Dfatigue cracking only✓
💡 Explanation:
K = −p / (ΔV/V).
Q10easy
Factor of safety is generally defined as
Aworking stress divided by failure strength✓
Balways equal to 1✓
Cload divided by deflection✓
Dfailure strength divided by allowable or working stress✓
💡 Explanation:
FOS > 1 provides design margin.
Q11Past Paper · PPSC/FPSC/NTSeasy
In bending, neutral axis is the locus of points where
Ashear stress is maximum always✓
Bnormal stress is maximum always✓
Cdeflection is maximum✓
Dlongitudinal strain is zero✓
💡 Explanation:
NA separates tension and compression zones.
Q12medium
Flexure formula σ = My/I applies to
Aelastic homogeneous beam in pure bending✓
Bplastic collapse always✓
Cany shape without NA✓
Dtorsion of circular shaft✓
💡 Explanation:
Bending stress varies linearly with y from NA.
Q13medium
Section modulus Z equals
AI divided by distance to extreme fiber c✓
BI times c✓
CI plus c✓
Dc divided by I✓
💡 Explanation:
Z = I/c used in σ = M/Z.
Q14Past Paper · PPSC/FPSC/NTSmedium
Maximum shear stress in rectangular beam occurs at
Atop fiber only✓
Bbottom fiber only✓
Cquarter depth from NA always for all sections✓
Dneutral axis✓
💡 Explanation:
Parabolic shear distribution peaks at NA for rectangle.
Q15hard
Deflection of simply supported beam with central point load is proportional to
AL³/(E I)✓
BL only✓
C1/I only without E✓
DE I only without L✓
💡 Explanation:
δ ∝ PL³/(E I) for given loading.
Q16Past Paper · PPSC/FPSC/NTSmedium
Torsion formula τ = T r / J applies to
Acircular shafts in elastic range✓
Bany non-circular section without correction✓
Cbeams in pure bending✓
Dthick cylinders under internal pressure only✓
💡 Explanation:
Circular cross-section with J polar moment.
Q17medium
Angle of twist for shaft is φ =
AT G / (L J)✓
BT L / (G J)✓
CG J L / T✓
DJ / (T L G)✓
💡 Explanation:
Elastic torsion relation.
Q18easy
Solid circular shaft polar moment J equals
Aπ d²/4✓
Bπ d³/16✓
Cπ d⁴/32✓
Dbh³/12✓
💡 Explanation:
Standard torsion constant.
Q19Past Paper · PPSC/FPSC/NTSmedium
Hollow shaft is preferred when
Aweight reduction with similar torsional strength is needed✓
Bonly bending dominates always✓
Cno shear stress exists✓
Dmaterial is brittle only✓
💡 Explanation:
Material farther from axis carries more shear.
Q20medium
Mohr's circle for plane stress plots
Aonly principal strains without stress✓
Bnormal and shear stress on various planes✓
Conly thermal gradients✓
Donly fatigue cycles✓
💡 Explanation:
Graphical tool for stress transformation.
Q21medium
Principal stresses are stresses on planes where
Anormal stress is zero✓
Bboth stresses are equal always✓
Cstrain is maximum always✓
Dshear stress is zero✓
💡 Explanation:
Principal planes have τ = 0.
Q22Past Paper · PPSC/FPSC/NTShard
Maximum shear stress in plane stress equals
Aradius of Mohr's circle✓
Bsum of principal stresses✓
Cdifference of principal strains✓
Dzero always✓
💡 Explanation:
τ_max = (σ1 − σ2)/2.
Q23hard
Euler column buckling load for pinned-pinned column is
Aπ² E I / L²✓
BE I / L✓
Cπ E I / L✓
D4 π² E I / L² always✓
💡 Explanation:
Critical load P_cr = π²EI/L² for end condition factor 1.
Q24medium
Effective length of column fixed at one end and free at other is
AL/2✓
BL✓
C2L✓
D0.7 L✓
💡 Explanation:
Equivalent pinned length 2L for standard end condition.
Q25Past Paper · PPSC/FPSC/NTSmedium
Slenderness ratio for column is
Adiameter divided by length✓
Bstress divided by strain✓
Ceffective length divided by least radius of gyration✓
Dload divided by area only✓
💡 Explanation:
λ = Le/k governs buckling mode.
Q26easy
Fatigue failure occurs at stress
Abelow static yield strength under cyclic loading✓
Bonly above ultimate strength always✓
Conly in single static overload✓
Donly at zero mean stress always✓
💡 Explanation:
Progressive damage under repeated loads.
Q27medium
Endurance limit on S-N curve for ferrous materials in reversed bending often occurs near
A10⁶ cycles✓
B10³ cycles✓
C10⁹ cycles always for all materials✓
Done cycle only✓
💡 Explanation:
Horizontal asymptote of S-N curve for steel.
Q28Past Paper · PPSC/FPSC/NTSmedium
Stress concentration factor Kt is ratio of
Anominal to maximum stress✓
Bfatigue limit to yield✓
Cmaximum local stress to nominal stress✓
Dshear to normal stress always✓
💡 Explanation:
Kt > 1 at geometric discontinuities.
Q29hard
Goodman line in fatigue relates
Athermal stress and strain only✓
Bmean and alternating stress for failure✓
Cbuckling load and length only✓
Dtorque and power only✓
💡 Explanation:
Modified Goodman criterion for fatigue.
Q30hard
Bolt subjected to axial external load shares load with
Aonly nut always✓
Bconnected members due to joint stiffness✓
Conly washer friction without members✓
Dair gap only✓
💡 Explanation:
Load sharing depends on bolt and member stiffness.
Q31Past Paper · PPSC/FPSC/NTSmedium
Initial tightening torque on bolt creates
Apreload in shank✓
Bonly shear in plate without axial force✓
Czero stress always✓
Donly bending always✓
💡 Explanation:
Torque induces axial tension via thread friction.
Q32medium
Square key transmits torque between shaft and hub by
Atension in key only✓
Btorsion in key as primary mode always✓
Cfriction only without bearing✓
Dshear and bearing on key sides✓
💡 Explanation:
Key fails in shear/bearing if overloaded.
Q33hard
Woodruff key is
Asquare parallel key only✓
Bsplined shaft only✓
Csemicircular disk key for tapered hubs✓
Dset screw only✓
💡 Explanation:
Used with tapered hubs on shafts.
Q34Past Paper · PPSC/FPSC/NTShard
Shaft design for combined bending and torsion often uses
Aonly axial stress formula✓
Bequivalent bending or torsion theories✓
Conly thermal expansion✓
Donly Bernoulli equation✓
💡 Explanation:
Equivalent moment/te torque per ASME or other codes.
Q35medium
Hollow shaft compared to solid of same weight has
Alower torsional stiffness always✓
Bhigher polar moment of inertia✓
Csame J always✓
Dzero shear stress✓
💡 Explanation:
Material placed farther from axis increases J.
Q36hard
Helical compression spring rate k equals
AG d⁴ / (8 D³ N) for active coils N✓
BE I / L✓
CP/A only✓
DT/J only✓
💡 Explanation:
Spring stiffness from wire and coil geometry.
Q37Past Paper · PPSC/FPSC/NTSmedium
Spring index C is ratio of
Awire diameter to coil diameter✓
Bfree length to solid length always✓
Cload to deflection only✓
Dmean coil diameter to wire diameter✓
💡 Explanation:
C = D/d affects stress and manufacturability.
Q38hard
Wahl factor accounts for
Aonly buckling in columns✓
Bonly radiation heat transfer✓
Ccurvature and direct shear in spring wire stress✓
Donly entropy generation✓
💡 Explanation:
Corrects τ = 16T/(πd³) for springs.
Q39medium
Thin cylinder with internal pressure p and radius r has hoop stress
Ap r / t✓
Bp t / r✓
Cp r t✓
Dp / (r t) only without relation✓
💡 Explanation:
σ_h = pr/t for thin wall t << r.
Q40Past Paper · PPSC/FPSC/NTSmedium
Longitudinal stress in thin cylinder is
Ap r / t✓
B2 p r / t✓
Cp r / (2t)✓
Dzero always✓
💡 Explanation:
σ_l = pr/(2t) half of hoop for thin cylinder.
Q41hard
Thick cylinder analysis uses
Aonly thin wall formula always✓
Bonly Euler buckling formula✓
Conly Fourier law✓
DLamé equations with radial and hoop stress variation✓
💡 Explanation:
Lamé solutions account for radial stress gradient.
Q42hard
Maximum shear stress in thin cylinder under internal pressure occurs at
Ainner surface for thick wall; approximately uniform for thin✓
Bouter surface only always✓
Cmid-wall only for thin always incorrectly always✓
Dzero everywhere✓
💡 Explanation:
Inner radius sees highest hoop stress in thick cylinders.
Q43Past Paper · PPSC/FPSC/NTSmedium
Strain energy per unit volume in elastic uniaxial stress is
Aσ E✓
BE/σ✓
Cσ²/(2E)✓
Dσ E²✓
💡 Explanation:
U = ½ σ ε = σ²/(2E).
Q44hard
Castigliano's theorem relates deflection to
Athermal expansion only✓
Bpartial derivative of strain energy with respect to load✓
Centropy only✓
D Reynolds number only✓
💡 Explanation:
δ = ∂U/∂P for linear elastic systems.
Q45medium
Maximum normal stress theory (Rankine) is suitable for
Abrittle materials in tension✓
Bductile combined loading always better with von Mises✓
Cany rubber behavior✓
Dfluid flow only✓
💡 Explanation:
Failure when max principal stress reaches ultimate.
Q46Past Paper · PPSC/FPSC/NTShard
von Mises yield criterion is based on
Amaximum principal stress only for all materials always✓
Bdistortion energy✓
Cvolume change only✓
Dthermal stress only✓
💡 Explanation:
Equivalent stress from distortion energy theory.
Q47medium
Eccentric loading on column introduces
Aonly torsion✓
Bonly shear without axial✓
Ccombined axial stress and bending stress✓
Dzero stress✓
💡 Explanation:
M = P e adds bending to axial.
Q48Past Paper · PPSC/FPSC/NTShard
Shear center of open thin-walled channel section lies
Aat geometric centroid always✓
Bat farthest fiber always✓
Cat infinity always✓
Doutside the cross-section on web side✓
💡 Explanation:
Loads through shear center avoid twisting.
Q49medium
Deflection curve slope equals
Asecond derivative only always✓
Bfirst derivative of deflection with respect to x✓