#1. A tensile specimen shows σ = 320 MPa at ε = 0.08 (true values) and σ = 450 MPa at ε = 0.20. Using σ = Kε^n, the strain hardening exponent n is:
#2. A material follows σ = Kε^n with n = 0.25. According to the Considère criterion, necking begins at true strain:
#3. A biaxial stress state has σ₁ = 250 MPa, σ₂ = 150 MPa, σ₃ = 0. Calculate the von Mises equivalent stress.
#4. For the same biaxial state (σ₁ = 250 MPa, σ₂ = 150 MPa, σ₃ = 0), the maximum shear stress (Tresca) is:
#5. A steel block (E = 200 GPa, ν = 0.30) under hydrostatic pressure of 120 MPa. Calculate the volumetric strain.
#6. A material has K = 800 MPa and n = 0.20 in σ = Kε^n. At ε = 0.15, the true stress is:
#7. In plane stress (σz = 0), the through-thickness strain εz for σx = 200 MPa, σy = 100 MPa with ν = 0.3 and E = 200 GPa is:
#8. A material with σy = 350 MPa and E = 200 GPa. The strain at yield under uniaxial tension, and the elastic strain energy at yield per unit volume are:
#9. A component experiences stress ratio R = σmin/σmax = -0.5. If σmax = 300 MPa, what are σmin and the stress amplitude?
#10. The resolved shear stress on a slip system is τ = σcos(φ)cos(λ). If σ = 200 MPa, φ = 40°, λ = 50°, then τ =
#11. A wire is drawn from 12 mm to 9 mm diameter. If σy = 350 MPa, the ideal drawing force F = σy·Af·ln(A₀/Af) is:
#12. A sheet metal formability test shows width strain = -0.18 and thickness strain = -0.12. The R-value (plastic strain ratio) is:
#13. Springback increases with which combination?
#14. A material has G = 80 GPa and K = 160 GPa. Calculate E and ν.
#15. Under torsion, a shaft experiences τmax = 150 MPa with G = 80 GPa. The shear strain and strain energy density are:
#16. Using Basquin’s equation σa = σf'(2Nf)^b with σf’ = 1200 MPa and b = -0.12, find Nf for σa = 350 MPa.
#17. A component operates at three stress levels: 6000 cycles at N₁ = 15,000; 12,000 cycles at N₂ = 40,000; 25,000 cycles at N₃ = 150,000. Remaining life fraction is:
#18. Using modified Goodman with Se = 280 MPa, Su = 700 MPa, if σa = 200 MPa, maximum allowable mean stress is:
#19. A rotating beam specimen has Se’ = 350 MPa. With ka = 0.75 (machined), kb = 0.82 (size), kc = 0.85 (loading), kd = 0.95 (temp), ke = 0.868 (reliability 95%), corrected Se =
#20. The Paris law parameters are C = 3 × 10⁻⁸ mm/cycle and m = 2.5. If ΔK increases from 15 to 30 MPa√m, by what factor does da/dN increase?
#21. A component with Kt = 3.0 is made of a steel with notch sensitivity q = 0.85. The fatigue notch factor Kf is:
#22. For a material with zero mean stress, Gerber parabola gives σa/Se + (σm/Su)² = 1. If Se = 300 MPa, Su = 600 MPa, σm = 200 MPa, find σa:
#23. A shaft has 10⁷ cycle endurance limit of 400 MPa and 10³ cycle strength of 900 MPa. Using log-log interpolation, estimate fatigue strength at 10⁵ cycles.
#24. In corrosion fatigue compared to air fatigue, the S-N curve shows:
#25. A component undergoes completely reversed loading (R = -1) with σa = 250 MPa. If the material has Se = 300 MPa and Su = 600 MPa, what is the safety factor using Goodman (σm = 0)?
#26. A plate has KIC = 55 MPa√m. For an edge crack (Y = 1.12) at σ = 180 MPa, the critical crack length is:
#27. Using Paris law da/dN = C(ΔK)^m, if C = 1.5 × 10⁻⁸ mm/cycle, m = 3, and ΔK = 22 MPa√m, the crack growth rate is:
#28. For valid KIC testing, B ≥ 2.5(KIC/σy)². A material with KIC = 70 MPa√m and σy = 500 MPa requires minimum thickness:
#29. A ceramic with E = 350 GPa, γs = 0.8 J/m², and internal crack half-length a = 100 μm. Using Griffith, fracture stress is:
#30. The strain energy release rate G = K²/E. If K = 45 MPa√m and E = 210 GPa, G under plane stress is:
#31. A center-cracked panel (Y = 1.0) with 2a = 15 mm under σ = 100 MPa. If KIC = 35 MPa√m, is fracture expected?
#32. For a material with KIC = 40 MPa√m, E = 200 GPa, ν = 0.3, calculate G under plane strain using G = K²(1-ν²)/E.
#33. In leak-before-break design, the critical crack size for a pressure vessel with KIC = 50 MPa√m at operating stress 150 MPa (Y = 1.12) must be:
#34. Paris law integration: If da/dN = C(ΔK)^m where ΔK = Yσ√(πa), m = 2, the cycles to grow from a₁ to a₂ is proportional to:
#35. A material with threshold stress intensity ΔKth = 5 MPa√m is loaded with σ = 50 MPa. Below what crack size (Y = 1.0) will fatigue cracks not propagate?
#36. A hypereutectoid steel (1.0% C) slowly cooled from austenite. At 750°C (α at 0.02% C, γ at 0.77% C), the proeutectoid phase and its weight fraction are:
#37. For Al-Cu alloys, the aging sequence is: SSSS → GP zones → θ” → θ’ → θ. Which provides maximum strengthening?
#38. The critical cooling rate to avoid pearlite formation in a steel is determined from:
#39. A 0.60% C steel is normalized. The resulting microstructure compared to full annealing will have:
#40. In a eutectoid steel, the Lever rule at 726°C (just below eutectoid) gives the phase fractions as:
#41. The driving force for recrystallization in cold-worked metals is:
#42. Ausforming is a thermomechanical treatment where deformation occurs:
#43. For a binary isomorphous system, the partition coefficient k = CS/CL. If k < 1, during solidification the first solid to form:
#44. In martempering (marquenching), the steel is:
#45. Tempering of martensite involves carbon diffusion. At ~200°C, the reaction is:
#46. A carbon fiber/epoxy (Ef = 230 GPa, Em = 3.5 GPa, Vf = 0.60) under longitudinal stress σ = 500 MPa. The stress in the fibers is:
#47. For minimum weight design of a stiff tie rod (tension), the performance index to maximize is:
#48. For minimum weight design of a stiff beam in bending, the performance index is:
#49. For minimum weight design of a stiff plate in bending, the performance index is:
#50. Compare steel (E = 210 GPa, ρ = 7.8 g/cm³) and aluminum (E = 70 GPa, ρ = 2.7 g/cm³) for a stiff beam. Which has higher E^(1/2)/ρ?
#51. A laminate composite [0/90/0] has 0° plies with E₁ = 140 GPa and 90° plies with E₁ = 140 GPa, E₂ = 10 GPa. The extensional stiffness is dominated by:
#52. The critical fiber length for load transfer Lc = σf·d/(2τi), where σf = 3500 MPa, d = 7 μm, τi = 50 MPa. Lc =
#53. For fibers shorter than the critical length, the composite strength:
#54. Intermetallic compounds like Ni₃Al are attractive for high-temperature applications but limited by:
#55. The Weibull modulus m for ceramics affects reliability. If m increases from 5 to 15:
#56. The Larson-Miller parameter LMP = T(C + log₁₀tr). If LMP = 22,500 at a stress, and C = 20, find tr at 900 K.
#57. In power-law creep ε̇ = Aσⁿexp(-Q/RT), if temperature increases from 800 K to 900 K with Q = 250 kJ/mol, the creep rate ratio is approximately:
#58. The Monkman-Grant relationship states that ε̇_s × tr = C (constant). If steady-state creep rate doubles, rupture time:
#59. For a material with n = 5 in power-law creep, reducing stress by 20% reduces creep rate by what factor?
#60. Superplasticity requires which conditions?
#61. Stress corrosion cracking requires:
#62. Hydrogen embrittlement is most severe in:
#63. The EMF series predicts galvanic corrosion. If ΔE between two metals increases:
#64. Intergranular corrosion in austenitic stainless steels is caused by:
#65. Environmental stress cracking in polymers occurs due to:
#66. For carbon diffusion in γ-Fe, D₀ = 2.3 × 10⁻⁵ m²/s, Q = 148 kJ/mol. At 1100°C (R = 8.314 J/mol·K), D is:
#67. If diffusion distance x ∝ √(Dt) and D doubles while time is halved, x:
#68. Case hardening by carburizing typically produces a surface carbon content of:
#69. The error function solution for diffusion gives (Cx – C₀)/(Cs – C₀) = 1 – erf(x/2√Dt). For x/2√Dt = 0.5, erf(0.5) ≈ 0.52, so Cx/Cs ≈:
#70. Fick’s second law ∂C/∂t = D(∂²C/∂x²) applies to:
#71. A casting with V/A = 3 cm solidifies in 5 minutes. Using Chvorinov, a similar casting with V/A = 6 cm solidifies in:
#72. In wire drawing, the maximum area reduction per pass without fracture is limited to approximately:
#73. The extrusion constant k in ram force F = kA₀ln(A₀/Af) accounts for:
#74. In hot rolling, the coefficient of friction between roll and workpiece typically:
#75. Residual stresses in a welded joint are primarily caused by:
#76. A single crystal under tension (σ = 180 MPa) with slip plane normal at 50° and slip direction at 45° to tensile axis. Critical resolved shear stress τcrss to initiate slip must be less than:
#77. The Peach-Koehler force on a dislocation under applied stress is F = τb per unit length. If τ = 50 MPa and b = 0.25 nm, F =
#78. For edge dislocations, the stress field creates a compressive region:
#79. In BCC metals, the slip direction is and slip can occur on {110}, {112}, and {123} planes. The number of slip systems is:
#80. The equilibrium vacancy concentration nv/N = exp(-Qv/kT). If Qv = 0.9 eV, T = 800 K (kT ≈ 0.069 eV), nv/N ≈:
#81. Point defect clusters affect mechanical properties primarily by:
#83. Mechanical twinning as a deformation mechanism is more prevalent in:
#84. The critical thickness for epitaxial thin films before misfit dislocation formation depends on:
#85. Grain boundary sliding contributes significantly to deformation at:
#86. For ultrasonic testing, the relationship between transducer frequency, wavelength λ, and velocity v is v = fλ. In steel (v ≈ 5900 m/s) with f = 5 MHz, wavelength is:
#87. In radiographic testing, the sensitivity to detect flaws depends on:
#88. Eddy current testing is particularly effective for detecting:
#89. The acoustic emission technique detects:
#90. For dye penetrant testing, the most critical step for reliable detection is:
#91. A metal’s thermal conductivity K and electrical conductivity σ are related by Wiedemann-Franz: K/σT = L (Lorenz number). At 300 K with σ = 6 × 10⁷ S/m and L = 2.44 × 10⁻⁸ WΩ/K², K ≈:
#92. A bimetallic strip of steel (α = 12 × 10⁻⁶/°C) and brass (α = 19 × 10⁻⁶/°C) is heated. It bends toward:
#93. In n-type semiconductors doped with phosphorus (Group V), the majority carriers are:
#94. Ferromagnetic materials lose their magnetic ordering above the:
#95. Piezoelectric materials (like PZT) generate:
#96. The Maxwell model (spring and dashpot in series) exhibits:
#97. The Kelvin-Voigt model (spring and dashpot in parallel) shows:
#98. For polymers, the time-temperature superposition principle states:
#99. The storage modulus E’ and loss modulus E” of a viscoelastic material are related to energy dissipation by:
