Mechanical Properties: Elasticity, Slip, Strengthening, Fracture, Fatigue, Creep and Composites

Practice questions (18)

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.

1. Crystals A and B have the same bonding energy per atom (the same depth of the potential well U(r)), but the well of A is narrower, with a larger curvature (d²U/dr²) at the equilibrium separation r₀ than the well of B. Their equilibrium spacings are equal. Which statement about their Young’s moduli is correct?

   1. E_B > E_A, because a wider well lets the atoms vibrate over a larger range and so holds them more firmly
   2. E_A = E_B, because the modulus depends only on the depth of the well, which is the same for both
   3. E_A > E_B, because E ≈ S₀/r₀ and the bond stiffness S₀ is the curvature of U(r) at r₀
   4. No comparison is possible, because the modulus is fixed by the dislocation density of each crystal
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   Answer: C — E_A > E_B, because E ≈ S₀/r₀ and the bond stiffness S₀ is the curvature of U(r) at r₀

   For small displacements the force is F ≈ −S₀(r − r₀) with S₀ = (d²U/dr²) at r₀, and E ≈ S₀/r₀ for equal spacings. The narrower well has the larger S₀ and therefore the larger modulus. The tempting answer that the depth fixes E confuses bond energy with bond stiffness; the two correlate in real materials only because deep wells are usually steep as well. The modulus is structure-insensitive, so dislocation density does not fix it.

2. Frenkel’s model for a perfect crystal predicts a shear strength of about G/2π, but annealed pure metal crystals yield at a stress thousands of times smaller. What is the accepted resolution of this discrepancy?

   1. Slip occurs by dislocation glide, which strains only the bonds along the line at any instant, not a whole plane at once
   2. The shear modulus G of a real crystal is several thousand times smaller than the value that was used in the Frenkel estimate
   3. Frenkel’s calculation describes tensile cleavage of bonds, which has no connection with shear of the lattice planes at all
   4. Vacancies weaken every bond in the crystal, so all the bonds across a slip plane break together at a much lower stress
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   Answer: A — Slip occurs by dislocation glide, which strains only the bonds along the line at any instant, not a whole plane at once

   The Frenkel estimate assumes the whole plane slides over its neighbour at once. Dislocations let the slipped region advance one atom row at a time, so the stress needed is controlled by the lattice friction of the moving line and is orders of magnitude lower. G of real crystals is what it is (tens of GPa for metals) and vacancies are far too dilute to weaken all bonds.

3. At room temperature a polycrystal of an HCP metal such as magnesium is much less ductile than a polycrystal of an FCC metal such as copper. Which reason is correct?

   1. Slip in HCP metals can occur only along directions that are not close-packed, which raises the critical resolved shear stress
   2. HCP crystals contain no close-packed planes, so dislocations cannot glide on any plane at all, at any temperature
   3. The Burgers vector of every dislocation in HCP is larger than the atomic spacing, so the Peierls stress is infinite
   4. Easy slip is confined to three basal systems, fewer than the five independent systems an arbitrary shape change needs
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   Answer: D — Easy slip is confined to three basal systems, fewer than the five independent systems an arbitrary shape change needs

   The basal plane (0001) with three ⟨11-20⟩ directions is close-packed and easy, but gives only two independent systems; a polycrystal needs five to follow arbitrary shape changes of grains without cracking, and the non-basal systems have a much higher critical stress at room temperature. HCP does have close-packed planes and glide along close-packed directions, so the other reasons misstate the structure.

4. A single crystal is loaded in uniaxial tension along an axis perpendicular to the slip direction of a given slip system (so λ = 90°). Which statement about slip on that system is correct?

   1. Slip begins at σ = 2τ_CRSS, the lowest yield stress any orientation can give, since the Schmid factor has its maximum
   2. No slip occurs on that system at any tensile stress, because cos λ = 0 makes the resolved shear stress zero
   3. Slip begins at σ = τ_CRSS, because the whole load then acts along the slip plane as a pure shear
   4. Slip is only delayed, because τ_CRSS itself rises towards infinity as λ approaches 90°
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   Answer: B — No slip occurs on that system at any tensile stress, because cos λ = 0 makes the resolved shear stress zero

   τ_R = σ cos φ cos λ, and cos 90° = 0, so τ_R = 0 for every σ; the system is inactive and other systems must operate. The maximum Schmid factor of 0.5 needs φ = λ = 45°. τ_CRSS is a material constant for the given temperature and purity; it does not depend on orientation, and a load perpendicular to the slip direction produces no shear force along it.

5. Among the four classical routes of strengthening a metal (strain hardening, solid-solution hardening, precipitation hardening and grain-size refinement), which is generally credited with raising the yield strength and the toughness together?

   1. Strain hardening, because the extra dislocations also blunt the tips of cracks and so resist their extension
   2. Precipitation hardening, because coherent particles make the crack paths much longer at peak ageing
   3. Grain-size refinement, because more boundaries block both dislocation motion and crack extension
   4. Solid-solution hardening, because the solute strain fields smooth out the stress concentrations at flaws
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   Answer: C — Grain-size refinement, because more boundaries block both dislocation motion and crack extension

   Finer grains raise σ_y through σ₀ + k_y d^(−1/2) and also make cracks change direction at boundaries, so toughness generally rises along with strength. Strain hardening and precipitation hardening, and usually solid-solution hardening, reduce ductility and toughness to some degree; none of them removes stress concentrations.

6. An age-hardenable alloy is held at a fixed ageing temperature and its hardness is measured against time. It rises to a peak and then falls. Which explanation of the fall (over-ageing) is correct?

   1. The precipitates coarsen and their spacing L grows, so dislocations bow between them more easily, Δτ ≈ Gb/L
   2. The precipitates dissolve completely back into the matrix at the ageing temperature, restoring a single-phase solid solution
   3. Prolonged ageing multiplies the dislocation density, and the Taylor term then reduces the hardness
   4. The elastic modulus of the matrix falls with ageing time, and hardness is proportional to the modulus
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   Answer: A — The precipitates coarsen and their spacing L grows, so dislocations bow between them more easily, Δτ ≈ Gb/L

   At constant temperature below the solvus the particles grow (coarsen) while their number density falls, so L increases; once particles are large enough that dislocations bypass them by Orowan bowing, the strengthening Gb/L falls with increasing L. Dissolution would need a temperature above the solvus, ageing does not multiply dislocations, and modulus is nearly unaffected by ageing.

7. A glass plate with an internal crack of total length 2a fractures in tension at stress σ_f. An identical plate has an internal crack of total length 8a. Using the Griffith criterion for the same plane-stress conditions, its fracture stress is

   1. σ_f/4, because the stress concentration at the tip grows in proportion to the crack length
   2. σ_f/16, because the elastic energy released varies as the square of the crack length
   3. σ_f/√2, because only the half-length of the crack enters the energy balance
   4. σ_f/2, because σ_f ∝ a^(−1/2) and the crack length has increased fourfold
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   Answer: D — σ_f/2, because σ_f ∝ a^(−1/2) and the crack length has increased fourfold

   σ_f = √(2Eγ_s/(πa)) ∝ a^(−1/2). Going from 2a to 8a multiplies a by 4, so σ_f becomes σ_f/√4 = σ_f/2. The energy released grows as a² but the surface energy grows as a, and balancing them gives the inverse square root; the stress concentration argument alone would give a different, wrong, dependence.

8. Continuous aligned carbon fibres (E_f = 230 GPa) occupy a volume fraction of 0.60 in an epoxy matrix (E_m = 3.0 GPa). The composite is loaded perpendicular to the fibres and the two phases carry the same stress (iso-stress). Its modulus in that direction, in GPa, is closest to

   1. 139, the weighted sum E_f V_f + E_m V_m
   2. 7.4, from 1/E = V_f/E_f + V_m/E_m
   3. 5.0, from the same expression with the two volume fractions swapped
   4. 117, the simple average of the two moduli
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   Answer: B — 7.4, from 1/E = V_f/E_f + V_m/E_m

   Iso-stress gives 1/E⊥ = 0.60/230 + 0.40/3.0 = 0.002609 + 0.13333 = 0.13594 GPa⁻¹, so E⊥ = 7.36 GPa, about 7.4. The 139 GPa value is the iso-strain (axial) bound, the most tempting wrong choice; transverse loading is dominated by the soft matrix, so the modulus stays close to that of the epoxy.

9. A component is cycled between a maximum stress of 240 MPa (tension) and a minimum stress of −60 MPa (compression). Its stress amplitude σ_a and stress ratio R = σ_min/σ_max are

   1. σ_a = 300 MPa and R = −0.25, because the amplitude is the full range
   2. σ_a = 150 MPa and R = −4, because R is the ratio of the larger stress to the smaller
   3. σ_a = 150 MPa and R = −0.25, because the amplitude is half the range
   4. σ_a = 90 MPa and R = −0.25, because the amplitude is the mean stress
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   Answer: C — σ_a = 150 MPa and R = −0.25, because the amplitude is half the range

   The range is 240 − (−60) = 300 MPa, the amplitude is half of it, 150 MPa, the mean stress is (240 − 60)/2 = 90 MPa, and R = σ_min/σ_max = −60/240 = −0.25. The 300 MPa choice is the range, the 90 MPa choice is the mean, and R = −4 inverts the definition.

10. A fine-grained ceramic creeps at a moderate temperature and low stress by Coble creep, in which atoms diffuse along grain boundaries and the strain rate is linear in stress. If the grain size is doubled with temperature and stress unchanged, the creep rate becomes

    1. 1/8 of its value, since Coble creep rate varies as d⁻³
    2. 1/4 of its value, since the rate varies as d⁻² for all diffusional creep
    3. 1/2 of its value, since the number of boundaries per unit length halves
    4. Unchanged, since creep depends only on stress and temperature, not on the microstructure
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    Answer: A — 1/8 of its value, since Coble creep rate varies as d⁻³

    Coble creep (grain-boundary diffusion) has ε̇ ∝ σ D_gb/d³, so doubling d divides the rate by 2³ = 8. The d⁻² law belongs to Nabarro-Herring creep (lattice diffusion), the tempting alternative; the stem names grain-boundary diffusion. Only dislocation (power-law) creep is independent of grain size.

11. Which of the following statements about strengthening mechanisms are correct?

    1. Plastic deformation raises the dislocation density, and the resulting strain hardening makes σ_y rise roughly as √ρ
    2. In solid-solution hardening the strain field around a misfitting solute atom interacts elastically with dislocations and hinders their glide
    3. Grain boundaries strengthen because dislocations pile up against them and cannot pass easily into a neighbouring grain of different orientation
    4. A precipitation-hardened alloy is strongest when the precipitates are as large as possible, so ageing for longer always increases the strength
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    Answer: A — Plastic deformation raises the dislocation density, and the resulting strain hardening makes σ_y rise roughly as √ρ; B — In solid-solution hardening the strain field around a misfitting solute atom interacts elastically with dislocations and hinders their glide; C — Grain boundaries strengthen because dislocations pile up against them and cannot pass easily into a neighbouring grain of different orientation

    Taylor hardening gives Δτ = αGb√ρ; solute fields and grain boundaries obstruct glide as described. The precipitation statement is false: strength peaks at a critical particle size and falls on over-ageing as particles coarsen and the spacing L grows, so longer ageing does not always strengthen.

12. Which of the following statements about creep of a crystalline metal at constant stress and high temperature are correct?

    1. In primary creep the strain rate falls with time because strain hardening outpaces recovery
    2. In secondary (steady-state) creep the nearly constant rate reflects a balance between hardening and recovery
    3. Dislocation creep has a stress exponent n of exactly 1 and a strong dependence on grain size
    4. Tertiary creep is a stage of falling strain rate in which the metal becomes stronger until the test is stopped
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    Answer: A — In primary creep the strain rate falls with time because strain hardening outpaces recovery; B — In secondary (steady-state) creep the nearly constant rate reflects a balance between hardening and recovery

    The statements on the primary and secondary stages are the standard description. Dislocation (power-law) creep has n of about 3 to 8 and no grain-size dependence; n = 1 and grain-size dependence belong to diffusional creep. Tertiary creep is an accelerating stage caused by necking, voids and grain-boundary cracking, ending in rupture.

13. Which of the following are expected to lengthen the fatigue life of a steel shaft cycled at a fixed stress amplitude?

    1. Cutting a keyway with sharp internal corners instead of generous fillet radii
    2. Shot peening the surface to introduce compressive residual stress
    3. Leaving a tensile residual stress in the surface after welding or grinding
    4. Polishing the surface to remove scratches and machining marks
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    Answer: B — Shot peening the surface to introduce compressive residual stress; D — Polishing the surface to remove scratches and machining marks

    Fatigue cracks start at the surface at stress raisers. Compressive residual stress lowers the effective tensile part of the cycle, and polishing removes the micro-notches where cracks start. Sharp corners raise the local stress, and tensile residual stress adds to the mean stress, so both shorten life.

14. A single crystal is pulled in uniaxial tension. The angle between the tensile axis and the slip-plane normal is 60° and the angle between the tensile axis and the slip direction is 45°. The critical resolved shear stress is 12.0 MPa. The tensile stress at which slip begins is ____ MPa (answer to one decimal place).

    Numerical answer — type the value.

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    Answer: 33.9

    Schmid factor m = cos 60° × cos 45° = 0.5 × 0.7071 = 0.3536. Slip begins when σ m = τ_CRSS, so σ = 12.0/0.3536 = 33.94 MPa, which is 33.9 to one decimal place.

15. A polycrystalline metal obeys the Hall-Petch relation σ_y = σ₀ + k_y d^(−1/2) with d the mean grain diameter in mm. Its yield stress is 160 MPa for d = 0.0625 mm and 250 MPa for d = 0.0100 mm. Its yield stress for d = 0.0016 mm (1.6 µm) is ____ MPa (answer as a whole number).

    Numerical answer — type the value.

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    Answer: 475

    d^(−1/2) = 4 mm^(−1/2) for 0.0625 mm and 10 for 0.0100 mm. Then k_y = (250 − 160)/(10 − 4) = 15 MPa·mm^(1/2) and σ₀ = 160 − 15 × 4 = 100 MPa. For d = 0.0016 mm, d^(−1/2) = 1/0.04 = 25, so σ_y = 100 + 15 × 25 = 475 MPa.

16. A thin glass plate (plane stress) contains an internal through-thickness crack of total length 2a = 2.0 µm. Take E = 70 GPa and surface energy γ_s = 1.0 J/m², and use the Griffith criterion σ_f = √(2Eγ_s/(πa)) with a the half-length of the crack. The fracture stress is ____ MPa (answer as a whole number).

    Numerical answer — type the value.

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    Answer: 211

    The half-length is a = 1.0 µm = 1.0 × 10⁻⁶ m. σ_f = √(2 × 70 × 10⁹ × 1.0/(π × 1.0 × 10⁻⁶)) = √(4.456 × 10¹⁶) = 2.111 × 10⁸ Pa = 211 MPa. Using the full length 2.0 µm for a would give 149 MPa, the common slip.

17. In the finite-life region of a steel’s S-N curve, log σ_a varies linearly with log N. The stress amplitude is 400 MPa at N = 10⁴ cycles and 300 MPa at N = 10⁶ cycles. The base-10 logarithm of the fatigue life N at a stress amplitude of 340 MPa is ____ (answer to two decimal places).

    Numerical answer — type the value.

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    Answer: 5.13

    Slope: log σ_a falls by log(400/300) = 0.12494 over 2 decades of N, so the exponent is b = 0.06247 per decade. From 400 MPa to 340 MPa, log(400/340) = 0.07058. Additional decades of life = 0.07058/0.06247 = 1.1298. So log N = 4 + 1.1298 = 5.13 (N ≈ 1.35 × 10⁵ cycles).

18. Continuous aligned glass fibres (E_f = 70 GPa) make up 40 percent by volume of a polymer-matrix composite (E_m = 3.0 GPa). The composite is loaded along the fibre direction and the fibres and matrix strain equally. Its elastic modulus is ____ GPa (answer to one decimal place).

    Numerical answer — type the value.

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    Answer: 29.8

    Iso-strain loading gives E_∥ = E_f V_f + E_m V_m = 70 × 0.40 + 3.0 × 0.60 = 28.0 + 1.8 = 29.8 GPa.