Materials Science Practice Exam — Free

Theoretical density is the mass of the atoms in one unit cell divided by the volume of the cell:

$$\rho = \frac{nA}{V_C N_A}$$

where $n$ is the number of atoms per unit cell, $A$ the atomic weight, $V_C$ the unit-cell volume and $N_A$ Avogadro's number.

For FCC, $n = 4$ (8 corner atoms at 1/8 each plus 6 face atoms at 1/2 each). The atoms touch along the face diagonal, which has length $4R$ and also equals $a\sqrt{2}$, so

$$a = 2\sqrt{2}\,R = 2\sqrt{2}(0.143\,\text{nm}) = 0.4045\,\text{nm} = 4.045 \times 10^{-8}\,\text{cm}$$ $$V_C = a^3 = (4.045 \times 10^{-8})^3 = 6.618 \times 10^{-23}\,\text{cm}^3$$ $$\rho = \frac{4(26.98)}{(6.618 \times 10^{-23})(6.022 \times 10^{23})} = \frac{107.9}{39.85} = 2.71\,\text{g/cm}^3$$

This matches the measured density of aluminum (2.71 g/cm³), as expected for a close-packed metal with few defects.

The tempting answer 1.35 g/cm³ uses $n = 2$, the atom count of a body-centred cubic cell, with the FCC lattice parameter; 0.677 g/cm³ uses $n = 1$ (simple cubic). 7.66 g/cm³ comes from setting $a = 2R$, the simple-cubic relation between radius and lattice parameter, while keeping $n = 4$.

$\rho = 2.71\,\text{g/cm}^3$

At equilibrium in a two-phase region the compositions of the phases are fixed by the ends of the tie line at that temperature, not by the overall alloy composition. So the $\alpha$ phase contains 10 wt% B and the $\beta$ phase contains 90 wt% B, regardless of how much of each is present.

The amounts of the phases follow from a mass balance on component B, which is the lever rule. Let $W_\alpha$ and $W_\beta$ be the mass fractions of the phases, with $W_\alpha + W_\beta = 1$. The B in the alloy must equal the B in the two phases combined:

$$C_0 = W_\alpha C_\alpha + W_\beta C_\beta \;\Rightarrow\; 40 = 10\,W_\alpha + 90(1 - W_\alpha)$$ $$W_\alpha = \frac{C_\beta - C_0}{C_\beta - C_\alpha} = \frac{90 - 40}{90 - 10} = \frac{50}{80} = 0.625$$ $$W_\beta = \frac{C_0 - C_\alpha}{C_\beta - C_\alpha} = \frac{40 - 10}{80} = 0.375$$

Each phase fraction uses the opposite arm of the lever: the alloy composition (40) is closer to the $\alpha$ end (10) than to the $\beta$ end (90), so $\alpha$ is the majority phase. The fractions are consistent with the overall composition, since $0.625(10) + 0.375(90) = 6.25 + 33.75 = 40$ wt% B.

$W_\alpha = 0.625$ and $W_\beta = 0.375$; $\alpha$ is 10 wt% B and $\beta$ is 90 wt% B

The stress is needed for the elastic check, so it comes first. Engineering stress uses the original cross-sectional area:

$$A_0 = \frac{\pi d_0^2}{4} = \frac{\pi(10.0 \times 10^{-3})^2}{4} = 7.854 \times 10^{-5}\,\text{m}^2$$ $$\sigma = \frac{F}{A_0} = \frac{20.0 \times 10^3}{7.854 \times 10^{-5}} = 254.6 \times 10^6\,\text{Pa} = 255\,\text{MPa}$$

(a) Since 255 MPa is below the yield strength of 350 MPa, the specimen is still on the linear portion of the stress–strain curve, Hooke's law applies, and the deformation recovers fully on unloading.

(b) Hooke's law gives the axial strain:

$$\varepsilon = \frac{\sigma}{E} = \frac{254.6 \times 10^6}{200 \times 10^9} = 1.273 \times 10^{-3}$$

(c) Strain is elongation per unit original length, so

$$\Delta L = \varepsilon L_0 = (1.273 \times 10^{-3})(50.0\,\text{mm}) = 0.0637\,\text{mm}$$

(d) Poisson's ratio relates the lateral strain to the axial strain, $\varepsilon_x = -\nu\varepsilon_z$; the minus sign records that the specimen gets thinner as it stretches:

$$\varepsilon_x = -0.30(1.273 \times 10^{-3}) = -3.82 \times 10^{-4}$$ $$\Delta d = \varepsilon_x d_0 = (-3.82 \times 10^{-4})(10.0\,\text{mm}) = -0.00382\,\text{mm}$$

The diameter shrinks by about 4 micrometres, roughly 6% of the elongation, which is the expected ratio since $\nu\,d_0/L_0 = 0.30(10/50) = 0.06$.

(a) Elastic, since $\sigma = 255\,\text{MPa} < \sigma_y = 350\,\text{MPa}$; (b) $\sigma = 255$ MPa, $\varepsilon = 1.27 \times 10^{-3}$; (c) $\Delta L = 0.0637$ mm; (d) $\Delta d = -0.00382$ mm (a decrease)