Mechanical Behaviour of Materials: Elasticity, Plasticity, Dislocations, Strengthening, Fracture, Fatigue and Creep
2. Tensile testing, yield criteria, and slip and twinning
A uniaxial tensile test records load against extension; dividing by the original area and gauge length gives the engineering stress-strain curve, and true values follow while deformation is uniform: σ_t = σ(1 + ε) and ε_t = ln(1 + ε), with volume constancy. Metals show a linear elastic region, yield (a sharp yield point with Lüders bands in low-carbon steel, a 0.2% offset proof stress otherwise), uniform strain hardening often fitted by Hollomon’s law σ_t = Kε_tⁿ, a maximum load at the ultimate tensile strength, necking and fracture. The Considère criterion dσ_t/dε_t = σ_t places the onset of necking at ε_t = n, so the strain-hardening exponent is the uniform true strain. Ductility is reported as percent elongation and percent reduction in area; the area under the curve is the toughness and the elastic area σ_y²/(2E) the modulus of resilience. Ceramics are linear to fracture at small strain and much stronger in compression than in tension, so they are tested in bending (modulus of rupture) and their strengths scatter, described by Weibull statistics. Polymers are viscoelastic and strongly temperature- and rate-dependent: a glassy polymer is brittle, a semicrystalline one above T_g yields, necks and cold-draws as the neck propagates along the specimen, and an elastomer extends hundreds of percent reversibly.
Yield under a multiaxial stress is predicted by a criterion calibrated against the uniaxial yield stress σ_y. Tresca (maximum shear stress): yielding when σ₁ − σ₃ = σ_y, so the shear yield stress is k = σ_y/2. Von Mises (distortion energy, or octahedral shear stress): yielding when √(½[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]) = σ_y, so k = σ_y/√3 = 0.577σ_y. Both ignore the hydrostatic stress, as plasticity by dislocation glide does; von Mises fits most ductile metals slightly better, and Tresca is the more conservative (its hexagon lies inside the von Mises ellipse, touching at six points, with the largest difference, 15%, in pure shear).
Plastic deformation in crystals occurs mainly by slip: dislocations glide on slip planes (usually the most densely packed) in slip directions (the most densely packed), the pair being a slip system. FCC slips on {111}⟨110⟩, twelve systems, and is ductile down to the lowest temperatures; BCC slips in ⟨111⟩ on {110}, {112} and {123} planes and has a strongly temperature-dependent flow stress; HCP metals with c/a near ideal slip mainly on the basal plane, three systems, which is why magnesium and zinc are hard to form cold. Slip starts when the resolved shear stress on a system reaches the critical resolved shear stress: Schmid’s law, τ_R = σ cos φ cos λ, where φ and λ are the angles of the tensile axis to the slip-plane normal and to the slip direction; cos φ cos λ, the Schmid factor, has a maximum of 0.5. Twinning is a homogeneous shear of a region into the mirror orientation of the parent across a twin plane; it produces only a small strain but reorients the lattice so that slip can continue, and it matters in HCP metals and in BCC metals at low temperatures or high strain rates. Deformation twins are distinct from the annealing twins of FCC metals.
3. Dislocation theory
A dislocation is the boundary, inside a crystal, between a region that has slipped and one that has not; its Burgers vector b is the closure failure of a circuit around it and is constant along its length. An edge dislocation has b perpendicular to its line and can be pictured as an extra half-plane; it glides only on the one plane containing b and the line, and moves out of that plane by climb, which needs vacancies to diffuse and so happens only at high temperature. A screw dislocation has b parallel to its line; since b and the line do not define a unique plane it can cross-slip from one slip plane to another. A curved dislocation loop is edge in some places, screw in others and mixed in between. The stress field of a screw dislocation is pure shear, τ = Gb/(2πr); that of an edge dislocation has both shear and normal components, compressive above the slip plane (on the side of the extra half-plane) and tensile below, which is why solute atoms larger than the host segregate below an edge dislocation and smaller ones above it (Cottrell atmospheres). The elastic energy per unit length is about Gb²/2, so dislocations adopt the shortest Burgers vectors available and behave as if they had a line tension of about Gb²/2.
Annealed metals contain about 10¹⁰-10¹² m of dislocation per m³ and heavily cold-worked ones about 10¹⁵-10¹⁶, so dislocations must multiply during deformation. The Frank-Read source is a segment of length L pinned at both ends; under a shear stress of about Gb/L it bows out, passes a semicircle (the critical configuration), wraps around the pins and releases a complete loop while regenerating the segment, repeatedly. Cross-slip of screw segments creates new sources (multiple cross-slip). The plastic strain rate follows from Orowan’s equation, γ̇ = ρ_m b v, with ρ_m the mobile dislocation density and v their velocity. In FCC metals a perfect dislocation a/2[1̄10] can lower its energy by dissociating into two Shockley partials, a/2[1̄10] → a/6[2̄11] + a/6[1̄21̄] — the reaction is favourable by Frank’s b² criterion, since a²/2 > a²/6 + a²/6 = a²/3 — separated by a ribbon of stacking fault whose width is inversely proportional to the stacking-fault energy. Wide partials (low SFE, as in austenitic stainless steel and brass) cross-slip with difficulty, giving planar slip and high work hardening; narrow partials (high SFE, as in aluminium) cross-slip easily and recover readily. Dislocations interact through their stress fields — like signs on the same plane repel, opposite signs attract and annihilate — and by intersecting, which creates jogs (steps out of the slip plane, which impede screw dislocations) and kinks (steps within it). Reactions between dislocations on intersecting {111} planes form sessile Lomer-Cottrell locks, a source of work hardening in FCC.
4. Strengthening mechanisms
Every strengthening mechanism is a way of making dislocation motion harder. Work (strain) hardening raises the dislocation density, and dislocations obstruct one another: the Taylor relation τ = τ₀ + αGb√ρ, with α about 0.2-0.5, ties flow stress to the square root of density; it is removed by annealing. Grain boundaries stop slip because the slip systems do not continue across them, so dislocations pile up against a boundary until the stress concentration at the head of the pile-up activates a source in the next grain; the result is the Hall-Petch relation, σ_y = σ₀ + k_y d^(−½), where σ₀ is the friction stress and k_y the locking parameter. Grain refinement is the only mechanism that raises strength and lowers the ductile-to-brittle transition temperature at the same time. Solid-solution strengthening comes from the interaction of dislocations with solute atoms through their size misfit and modulus misfit; the increment rises roughly as the square root of concentration for dilute solutions, and interstitials that produce tetragonal distortions (C and N in BCC iron) are far more potent than substitutionals.
Precipitation and dispersion strengthening use second-phase particles. A dislocation meeting a small, coherent, shearable precipitate cuts through it, and the resistance — from the new interface created, from coherency strains, from the antiphase boundary in an ordered particle, from modulus mismatch — rises with particle size. A large or incoherent particle cannot be cut, and the dislocation bows between particles and bypasses them by the Orowan mechanism, leaving a loop around each; the stress needed is about τ = Gb/λ, with λ the interparticle spacing, and it FALLS as particles coarsen at constant volume fraction because λ grows. The crossover from cutting to bowing at an optimum particle size is peak ageing, and the fall beyond it is overageing. Dispersion strengthening uses hard, insoluble, incoherent particles — oxides such as Y₂O₃ in ODS alloys, ThO₂ in TD-nickel, Al₂O₃ in sintered aluminium — introduced by powder metallurgy or internal oxidation; because they neither dissolve nor coarsen, the strengthening survives to temperatures at which precipitates would have dissolved.
5. Fracture: Griffith, LEFM, fracture toughness, fractography and the ductile-brittle transition
Griffith explained why brittle solids break at a small fraction of their theoretical strength: they contain cracks, and a crack grows when the elastic energy it releases exceeds the surface energy it must create. For a through crack of length 2a in a large plate under remote stress σ, this gives σ_f = √(2Eγ_s/(πa)) in plane stress. For metals the energy absorbed in plastic work at the crack tip, γ_p, dwarfs γ_s, and Orowan’s modification replaces 2γ_s with 2(γ_s + γ_p), which is the fracture energy G_c. Linear elastic fracture mechanics characterises the crack-tip field by the stress intensity factor K = Yσ√(πa), where Y is a geometry factor (1 for a central crack in an infinite plate, about 1.12 for a small edge crack). Three loading modes exist — I (opening), II (sliding), III (tearing) — and mode I governs most failures. Fracture occurs when K reaches the material’s fracture toughness K_c; its lower, thickness-independent value under plane-strain constraint is K_IC, a material property, related to G_c by K_IC² = EG_c/(1 − ν²). A valid K_IC test needs plane strain: thickness and crack length at least 2.5(K_IC/σ_y)². The size of the plastic zone at the crack tip, r_p ≈ (1/2π)(K/σ_y)² in plane stress (about a third of that in plane strain), must be small against the crack and the specimen for LEFM to apply.
Fractography reads the mechanism from the surface. Ductile fracture proceeds by microvoid nucleation at inclusions and second-phase particles, void growth and coalescence, leaving equiaxed dimples under tension (elongated ones under shear) and, in a tensile bar, the cup-and-cone profile. Cleavage is transgranular separation along low-index planes ({100} in BCC iron), leaving flat facets with river patterns that converge in the direction of local crack propagation, and chevron marks pointing back to the origin on a large surface. Intergranular fracture follows grain boundaries weakened by segregation (temper embrittlement by P, Sb, Sn), precipitates or hydrogen. Fatigue leaves beach marks visible to the eye and striations, one per cycle, in the SEM. BCC and HCP metals, whose flow stress rises steeply as temperature falls, show a ductile-to-brittle transition: below the transition temperature cleavage becomes easier than yielding. It is measured by the Charpy V-notch impact energy against temperature. FCC metals show no such transition. The transition temperature is lowered by grain refinement, nickel and manganese, and a low carbon content, and raised by phosphorus, coarse grains, a sharper notch, higher loading rate and neutron irradiation.
Ceramics have low fracture toughness (typically below about 5 MPa√m) because they cannot deform plastically at the crack tip, so they are toughened by mechanisms that shield the tip instead. Transformation toughening in partially stabilised zirconia uses metastable tetragonal particles that transform to monoclinic in the crack-tip stress field with a volume expansion of about 4%, putting the crack faces into compression. Microcrack toughening, crack deflection and crack bowing around second-phase particles, and crack bridging by grains, whiskers or fibres in the wake of the crack also raise the toughness, and give a rising crack-resistance (R-curve) behaviour.
6. Fatigue: low- and high-cycle fatigue and crack growth
Fatigue is failure under a fluctuating stress well below the static strength, in three stages: crack initiation (often at persistent slip bands, which extrude and intrude at a free surface, or at inclusions and notches), stable crack growth, and final fast fracture when K_max reaches K_c. The S-N curve plots stress amplitude against cycles to failure. Steels and titanium alloys show an endurance limit below which life is effectively infinite (for steels roughly half the tensile strength); aluminium and copper alloys do not, and a fatigue strength at, say, 10⁷ cycles is quoted instead. A tensile mean stress shortens life, as captured by the Goodman line σ_a/σ_e + σ_m/σ_u = 1. High-cycle fatigue (more than about 10⁴-10⁵ cycles, nominally elastic) is stress-controlled and follows Basquin’s law σ_a = σ′_f(2N_f)^b; low-cycle fatigue (fewer cycles, with plastic strain every cycle) is strain-controlled and follows the Coffin-Manson law Δε_p/2 = ε′_f(2N_f)^c, with c about −0.5 to −0.7.
Once a crack exists, its growth per cycle in the intermediate regime follows the Paris law, da/dN = C(ΔK)^m, where ΔK = YΔσ√(πa) and m is typically 2-4 for metals. Below a threshold ΔK_th cracks do not grow; at the upper end growth accelerates as K_max approaches K_c. Integrating the Paris law from an initial crack a₀ (often the largest undetectable flaw) to the critical crack a_c (from K_IC) gives the remaining life — the basis of damage-tolerant design, with inspection intervals set so that a crack is found before it becomes critical. Fatigue life is improved by smooth surfaces and generous radii, and by compressive residual surface stresses from shot peening, surface rolling, carburising or nitriding.
7. High-temperature deformation: creep, its mechanisms and stress rupture
Above roughly 0.4 T_m metals deform continuously under a constant stress: creep. The creep curve has an instantaneous strain, a primary stage of decreasing rate (hardening outpaces recovery), a secondary or steady-state stage in which hardening and recovery balance and the rate ε̇_s is minimum and constant, and a tertiary stage of accelerating rate as necking, cavitation and microstructural degradation lead to rupture. The steady-state rate follows a power law with Arrhenius temperature dependence: ε̇_s = Aσⁿ exp(−Q_c/RT). From two tests at the same temperature the stress exponent is n = ln(ε̇₂/ε̇₁)/ln(σ₂/σ₁); from two tests at the same stress the activation energy is Q_c = R ln(ε̇₂/ε̇₁)/(1/T₁ − 1/T₂). For pure metals Q_c is usually close to the activation energy for self-diffusion, showing that diffusion controls creep.
| Mechanism | Stress exponent n | Grain-size dependence | Regime |
|---|---|---|---|
| Dislocation (power-law) creep: glide plus climb | About 3-8 (often 4-5 for pure metals) | Little | High stress |
| Nabarro-Herring: vacancy diffusion through the grains | 1 | ε̇ ∝ 1/d² | Low stress, high temperature |
| Coble: vacancy diffusion along grain boundaries | 1 | ε̇ ∝ 1/d³ | Low stress, lower temperature, fine grains |
| Grain-boundary sliding | About 2 (superplasticity) | Strong | Fine, stable grains, intermediate stress |
Ashby’s deformation-mechanism maps plot normalised stress σ/G against homologous temperature T/T_m and show which mechanism gives the fastest strain rate in each region. Since diffusional creep speeds up with finer grains, creep-resistant alloys are coarse-grained or single crystals (turbine blades), strengthened by stable coherent precipitates (γ′ in nickel superalloys) and dispersoids, with solutes that lower diffusivity and carbides that pin boundaries. Stress rupture tests measure the time to fracture at a constant load and temperature; for design they are extrapolated with the Larson-Miller parameter LMP = T(C + log₁₀ t_r), with T in kelvin, t_r in hours and C about 20, so that a short test at a higher temperature stands in for a long service life at a lower one.
Key takeaways
- Tresca: σ₁ − σ₃ = σ_y (shear yield σ_y/2); von Mises: √(½Σ(σᵢ − σⱼ)²) = σ_y (shear yield σ_y/√3). Schmid: τ_R = σ cos φ cos λ, maximum factor 0.5. Necking at ε_t = n.
- Edge: b ⊥ line, glides on one plane, climbs; screw: b ∥ line, cross-slips. Frank-Read source operates at τ ≈ Gb/L; a/2⟨110⟩ → two a/6⟨112⟩ Shockley partials (b² falls from a²/2 to a²/3).
- Taylor τ ∝ √ρ; Hall-Petch σ_y = σ₀ + k_y d^(−½); Orowan τ ≈ Gb/λ for bypassed particles, cutting for small coherent ones — the crossover is peak ageing.
- Griffith σ_f = √(2Eγ_s/(πa)); K = Yσ√(πa); fracture at K = K_IC; plane strain needs B ≥ 2.5(K_IC/σ_y)². FCC metals have no ductile-brittle transition; fine grains lower it.
- Paris da/dN = C(ΔK)^m; Coffin-Manson for LCF, Basquin for HCF. Creep ε̇ = Aσⁿexp(−Q/RT): n from two stresses, Q from two temperatures; Nabarro-Herring ∝ d⁻², Coble ∝ d⁻³; Larson-Miller T(20 + log t_r).
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
At a point the principal stresses are 200 MPa, 50 MPa and −100 MPa. What is the von Mises equivalent stress, in MPa (to one decimal place)?
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Answer: 259.8
σ_vm = √(½[(200 − 50)² + (50 + 100)² + (−100 − 200)²]) = √(½[22 500 + 22 500 + 90 000]) = √67 500 = 259.8 MPa. Check with the invariant form √(σ₁² + σ₂² + σ₃² − σ₁σ₂ − σ₂σ₃ − σ₃σ₁) = √(40 000 + 2500 + 10 000 − 10 000 + 5000 + 20 000) = √67 500. The Tresca equivalent σ₁ − σ₃ = 300 MPa is larger, as it always is.In a tensile test the engineering stress is 400 MPa at an engineering strain of 0.20, before necking. What is the true stress, in MPa?
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Answer: 480
With volume constancy A = A₀/(1 + ε), so σ_t = σ(1 + ε) = 400 × 1.20 = 480 MPa. The matching true strain is ln 1.20 = 0.182. Dividing by 1.2 instead (333 MPa) puts the area change the wrong way — the bar gets thinner, so the true stress is higher.An FCC single crystal with a critical resolved shear stress of 10 MPa is pulled along [001]. Considering the (111)[1̄01] slip system, what tensile stress, in MPa, is needed to start slip (to two decimal places)?
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Answer: 24.49
cos φ = [001]·[111]/(1 × √3) = 1/√3 and cos λ = [001]·[1̄01]/(1 × √2) = 1/√2, so the Schmid factor is 1/√6 = 0.408. σ = τ_CRSS/(cos φ cos λ) = 10 × √6 = 24.49 MPa. Check: 24.49 × 0.577 × 0.707 = 10.0 MPa resolved. Using the maximum Schmid factor 0.5 gives 20 MPa, which this orientation cannot reach.In an FCC metal, which dissociation of a perfect dislocation lying on a (111) plane is correct and energetically favourable?
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Answer: A — a/2[1̄10] → a/6[1̄21̄] + a/6[2̄11]
Vector sum: a/6[1̄21̄] + a/6[2̄11] = a/6[3̄30] = a/2[1̄10], and both partials lie in (111) because −1 + 2 − 1 = 0 and −2 + 1 + 1 = 0. Energy: b² falls from a²/2 to a²/6 + a²/6 = a²/3, so it is favourable. Option B does not sum to a/2[1̄10] (it gives a/6[3̄10]); option C sums correctly but b² RISES from a²/2 to a²/2 + a²/2 = a², so it is unfavourable; option D sums correctly but b² is unchanged (a²/4 + a²/4) and the components are not FCC slip vectors.Which type of dislocation can change its slip plane by cross-slip?
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Answer: A — Screw
A screw dislocation has b parallel to its line, so every plane containing the line also contains b and is a possible glide plane — it can cross-slip. An edge dislocation’s b and line define one plane; it leaves that plane only by climb, which needs diffusion and high temperature. A Frank partial is sessile and cannot glide at all.For a steel the Hall-Petch constants are σ₀ = 70 MPa and k_y = 0.74 MPa m^½. What is the yield strength for a grain size of 25 µm, in MPa?
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Answer: 218
σ_y = σ₀ + k_y d^(−½) = 70 + 0.74/√(25 × 10⁻⁶) = 70 + 0.74/0.005 = 70 + 148 = 218 MPa. Check the units: MPa m^½ divided by m^½ gives MPa. Putting d in µm (√25 = 5) gives 70.15 MPa — the grain-size term is then almost lost.A metal has a yield strength of 200 MPa at a grain size of 100 µm and 300 MPa at 25 µm. Assuming Hall-Petch behaviour, what is its yield strength at a grain size of 4 µm, in MPa?
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Answer: 600
d^(−½) is 100 m^(−½) at 100 µm and 200 m^(−½) at 25 µm. So k_y = (300 − 200)/(200 − 100) = 1.0 MPa m^½ and σ₀ = 200 − 100 = 100 MPa. At 4 µm, d^(−½) = 500 m^(−½), so σ_y = 100 + 500 = 600 MPa. Check the 25 µm point: 100 + 200 = 300 MPa. Interpolating linearly in d, rather than in d^(−½), gives a strength that barely changes.In an aluminium alloy (G = 26 GPa, b = 0.286 nm) non-shearable particles are spaced 0.2 µm apart. Taking the Orowan stress as Gb/λ, what shear stress, in MPa, is needed for dislocations to bypass them (to one decimal place)?
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Answer: 37.2
τ = Gb/λ = 26 × 10⁹ × 0.286 × 10⁻⁹/(0.2 × 10⁻⁶) = 7.436/(2 × 10⁻⁷) = 3.718 × 10⁷ Pa = 37.2 MPa. Scaling check: halving the spacing doubles the stress, and coarsening at fixed volume fraction widens λ and lowers it — overageing. Leaving b in nm and λ in µm without converting gives a result off by a factor of 1000.A glass plate (E = 70 GPa, surface energy 1 J/m²) contains an internal through crack of total length 2 µm. What is its Griffith fracture stress under plane stress, in MPa (to the nearest whole number)?
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Answer: 211
The half-length is a = 1 µm. σ_f = √(2Eγ_s/(πa)) = √(2 × 70 × 10⁹ × 1/(π × 10⁻⁶)) = √(4.456 × 10¹⁶) = 2.111 × 10⁸ Pa = 211 MPa. As K: this corresponds to K_c = σ√(πa) = 2.111 × 10⁸ × 1.772 × 10⁻³ = 0.374 MPa√m = √(2Eγ_s), consistent. Using the full length 2 µm as a gives 149 MPa.A steel plate with K_IC = 50 MPa√m carries a remote tensile stress of 400 MPa. Taking Y = 1 for a central through crack, what is the critical crack half-length, in mm (to two decimal places)?
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Answer: 4.97
K_IC = σ√(πa_c) gives a_c = (1/π)(K_IC/σ)² = (1/π)(50/400)² = 0.015625/π = 4.974 × 10⁻³ m = 4.97 mm. Check: 400 × √(π × 0.004974) = 400 × 0.125 = 50 MPa√m. Forgetting the 1/π gives 15.6 mm; the total critical crack length is 2a = 9.95 mm.A steel has K_IC = 50 MPa√m and a yield strength of 1000 MPa. What is the minimum specimen thickness, in mm, for a valid plane-strain K_IC test (to two decimal places)?
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Answer: 6.25
B ≥ 2.5(K_IC/σ_y)² = 2.5 × (50/1000)² = 2.5 × 0.0025 = 6.25 × 10⁻³ m = 6.25 mm. The ratio K_IC/σ_y is in √m, so its square is in metres. A tougher or weaker steel needs a much thicker specimen: halving σ_y quadruples the thickness to 25 mm.Which of the following statements about the ductile-to-brittle transition are correct?
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Answer: A — FCC metals such as copper and austenitic stainless steel do not show a sharp ductile-to-brittle transition; B — Refining the grain size lowers the transition temperature of a ferritic steel; D — Below the transition, fracture surfaces of BCC steels typically show cleavage facets with river patterns
FCC metals have a flow stress that depends only weakly on temperature, so they stay ductile (A). Grain refinement raises the cleavage stress more than the yield stress and lowers the transition (B). Below the transition, BCC steels cleave, leaving facets and river patterns (D). C is reversed: phosphorus embrittles ferrite and segregates to boundaries, RAISING the transition temperature.A wide plate with a central crack (Y = 1) is cycled with a stress range of 100 MPa. The Paris law is da/dN = 1 × 10⁻¹¹ (ΔK)³, with da/dN in m/cycle and ΔK in MPa√m. How many cycles, in units of 10⁵, does the crack take to grow from a half-length of 1 mm to 10 mm (to two decimal places)?
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Answer: 7.77
N = ∫da/(C(Δσ√(πa))³) = [1/(C Δσ³ π^1.5)] × 2(a₀^(−½) − a_f^(−½)). Here C Δσ³ π^1.5 = 10⁻¹¹ × 10⁶ × 5.568 = 5.568 × 10⁻⁵, and 2(1/√0.001 − 1/√0.01) = 2(31.62 − 10.00) = 43.25. N = 43.25/5.568 × 10⁻⁵ = 7.77 × 10⁵ cycles. Check the dominance of the early crack: the first millimetre-to-two costs 2(31.62 − 22.36)/5.568 × 10⁻⁵ = 3.3 × 10⁵ cycles, over 40% of the life. Evaluating ΔK at the final crack only gives about 1.6 × 10⁵ cycles.At a fixed temperature a metal creeps at a steady-state rate of 1 × 10⁻⁸ s⁻¹ under 50 MPa and 1.6 × 10⁻⁷ s⁻¹ under 100 MPa. What is the stress exponent n?
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Answer: 4
n = ln(ε̇₂/ε̇₁)/ln(σ₂/σ₁) = ln 16/ln 2 = 2.773/0.693 = 4. Check: 2⁴ = 16. An exponent of 4 lies in the 3-8 band of dislocation (power-law) creep; diffusional creep would give n = 1, i.e. a doubling of the rate.At a constant stress the steady-state creep rate of an alloy is 1 × 10⁻⁸ s⁻¹ at 800 K and 5 × 10⁻⁸ s⁻¹ at 850 K. What is the apparent activation energy for creep, in kJ/mol (to the nearest whole number)? Take R = 8.314 J/(mol K).
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Answer: 182
Q = R ln(ε̇₂/ε̇₁)/(1/T₁ − 1/T₂) = 8.314 × ln 5/(1/800 − 1/850) = 8.314 × 1.6094/(7.353 × 10⁻⁵) = 13.381/7.353 × 10⁻⁵ = 181 980 J/mol ≈ 182 kJ/mol. Check: exp(181 980 × 7.353 × 10⁻⁵/8.314) = exp(1.609) = 5.0. Using log₁₀ 5 in place of ln 5 gives 79 kJ/mol.An alloy ruptures in 1000 h at 800 K under a given stress. Using the Larson-Miller parameter with C = 20, what rupture life, in hours, is predicted at 900 K under the same stress (to two decimal places)?
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Answer: 2.78
LMP = T(C + log t_r) = 800 × (20 + 3) = 18 400. At 900 K: 20 + log t_r = 18 400/900 = 20.444, so log t_r = 0.444 and t_r = 10^0.444 = 2.78 h. Check: 900 × (20 + log 2.78) = 900 × 20.444 = 18 400. Using natural logarithms, or dropping C, gives meaningless lives; a 100 K rise costs a factor of about 360 in life here.Which of the following statements about creep mechanisms are correct?
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Answer: A — Nabarro-Herring creep has a stress exponent of about 1 and a rate inversely proportional to the square of the grain size; B — Coble creep occurs by diffusion along grain boundaries and its rate varies as the inverse cube of grain size; D — Dislocation creep involves climb of dislocations over obstacles
Nabarro-Herring creep is lattice diffusion with n = 1 and ε̇ ∝ d⁻² (A); Coble creep is boundary diffusion with ε̇ ∝ d⁻³ (B); dislocation creep is glide controlled by climb, which is why its activation energy is close to that of self-diffusion (D). C is reversed: finer grains shorten the diffusion paths and multiply the boundaries, so diffusional creep gets FASTER — creep-resistant parts use coarse grains or single crystals.