Mechanical Properties of Materials Questions and Answers

Answer: B

The lowest specific heat capacity belongs to Mercury.

Approximate specific heats at room temperature, per unit mass:

• water: about 4.18 J/(g K)
• copper: about 0.385 J/(g K)
• silver: about 0.235 J/(g K)
• mercury: about 0.14 J/(g K)

The molar heat capacity of metals is roughly the same (the Dulong-Petit value, about 25 J/(mol K)), so on a per-gram basis the heaviest atoms have the lowest specific heat. Mercury, with an atomic mass of about 201, comes out lowest; silver (108) and copper (64) are higher. Water, with its hydrogen bonding, has one of the highest specific heats of any common substance.

Answer: A

The Pilling-Bedworth ratio is the ratio of volume of oxide to volume of metal.

It compares the volume of the oxide formed with the volume of the metal consumed to form it:

\[ \text{PBR} = \frac{M_{ox}\,\rho_{m}}{n\,M_{m}\,\rho_{ox}} \]

where n is the number of metal atoms per oxide formula unit. A ratio below 1 (as for magnesium) gives a porous, non-protective oxide; between about 1 and 2 (as for aluminium) a dense protective film; and well above 2 an oxide that cracks and spalls.

The formula contains densities and masses, which may tempt one towards the weight or density options, but the quantity defined is a volume ratio.

Answer: B

The match is P-2, Q-4, R-5, S-1.

• Thermal diffusivity \(\alpha = k/(\rho c_{p})\) has units m²s^-1.
• Fracture toughness \(K_{IC} = Y\sigma\sqrt{\pi a}\) is stress times root length, Pa m^1/2 = N m^-2 m^1/2 = N m^-3/2.
• Surface energy is energy per unit area, J m^-2.
• Magnetic permeability has units H m^-1, as in \(\mu_{0} = 4\pi \times 10^{-7}\) H/m.

F m^-1 is the unit of electrical permittivity, a distractor for permeability. Options that give fracture toughness J m^-2 confuse it with the critical strain energy release rate.

Answer: C

The match is P-4, Q-5, R-2, S-1.

• Toughness is the energy absorbed up to fracture, the whole area under the stress-strain curve.
• Resilience is the energy stored elastically, the area under the elastic part of the curve.
• Creep is time-dependent permanent deformation under constant load, usually at high temperature.
• Hardness is resistance to plastic deformation, typically local indentation.

'Total elongation at failure' is ductility, which is not in Column I, so options that use item 3 are wrong. Swapping toughness and resilience, as two of the options do, confuses the whole curve with its elastic part.

Answer: C

The UTS is the maximum stress on the engineering stress-strain curve.

Engineering stress is load divided by the original area. It rises through yielding and strain hardening to a peak, the ultimate tensile strength, where necking begins; after that the load, and so the engineering stress, falls until fracture.

The proportional limit is the end of the straight elastic line, and the yield stress is where plastic deformation starts; both are well below the UTS for mild steel. The fracture stress on the engineering curve is lower than the UTS, because the load drops during necking.

Answer: A

The true stress increases linearly with the true strain.

A linear hardening (bilinear) material model has two straight segments: elastic loading up to yield, then a plastic segment of smaller but positive slope, the tangent or hardening modulus. Beyond yield the stress keeps rising in proportion to the plastic strain.

A constant stress after yield describes an elastic-perfectly plastic material, with no hardening. A decreasing stress would be softening, and "first increases then decreases" describes necking in an engineering stress-strain curve, not a linear hardening law.

Answer: B

UTS is the maximum load attained divided by the original cross-sectional area.

The ultimate tensile strength is an engineering stress: the highest point of the engineering stress-strain curve. Engineering stress always divides the load by the original area \(A_{0}\), so \(\text{UTS} = P_{max}/A_{0}\). After this point the specimen necks and the load falls.

The other definitions describe different quantities. The transition from elastic to plastic behaviour is the yield point (yield strength). The maximum load divided by the instantaneous area is the true stress at maximum load, which is higher than the UTS because the area has already shrunk. The stress at which the specimen breaks is the fracture (breaking) stress, which for a ductile metal lies below the UTS on the engineering curve.

Answer: C

The stress ratio is -1.

The stress ratio is R = σ_min/σ_max. In completely reversed loading the stress swings between equal tensile and compressive values, σ_min = -σ_max, so R = -1 and the mean stress is zero.

R = 0 describes zero-to-tension (pulsating) loading, R = 1 a static load with no cycling, and -1/2 a cycle whose compressive peak is half the tensile peak.

Answer: D

The sample loaded along [123], (iv), yields first.

Aluminium is FCC and slips on {111}\(\langle 110 \rangle\) systems. Yield begins when the resolved shear stress on the most favourable system reaches the critical value, \(\tau_{c} = \sigma\cos\phi\cos\lambda\), so the load needed is smallest for the orientation with the largest Schmid factor \(m = \cos\phi\cos\lambda\).

• [100] and [101]: maximum m is about 0.41.
• [111]: maximum m is about 0.27.
• [123]: a single-slip orientation near the middle of the standard triangle, with m close to the ideal 0.5 (about 0.47).

So [123] needs the lowest load. [111] is the hardest orientation, and [100] and [101] lie in between.

Answer: C

The inference is BHN of 1 \(\lt\) BHN of 2.

Point E lies on the linear elastic part of the curve, so sample 1 recovers completely on unloading and its microstructure is unchanged. Point P lies beyond the 0.2% offset yield point, so sample 2 has deformed plastically: the dislocation density has risen and the alloy has work hardened. After unloading, sample 2 keeps that strain hardening and so shows a higher Brinell hardness.

Polishing removes only the surface, not the bulk work hardening, so equal hardness is wrong; and enough information is given, since the two points sit on opposite sides of yield.