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21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2016

Question 3 of 7: Question III — Crystal Structure I (20 marks)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2016 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Seven questions of 20 marks each (Roman numerals I–VII); the rubric asks for any five, with only the first five in the answer book marked. All seven are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and an error-function table are provided in the exam's own appendix (reproduced where used below).

Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only Question VI (dislocation theory) is genuinely deformation/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — electron structure, bonding, crystal structure, crystallographic directions/planes, solid solubility, XRD, microscopy and phase diagrams — and is answered as such below.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:

Question III — Crystal Structure I (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

III.1 — Density of BCC iron

Given.

QuantityValue
Lattice parameter, $a$0.2866 nm
Molar mass, $A_w$55.847 g/mol
StructureBCC ($n=2$ atoms/cell)
Avogadro's number, $N_A$$6.023\times10^{23}$ /mol

Find. Theoretical density $\rho$.

Approach. $\rho=nA_w/(a^3N_A)$, with $a$ converted to cm.

  1. Unit-cell volume. $$a=0.2866\times10^{-7}\ \text{cm} \quad\Rightarrow\quad a^3=2.354\times10^{-23}\ \text{cm}^3$$
  2. Density. $$\rho=\frac{nA_w}{a^3N_A}=\frac{2(55.847)}{(2.354\times10^{-23})(6.023\times10^{23})}=\boxed{7.88\ \text{g/cm}^3}$$

This matches the accepted density of α-iron (7.87 g/cm³) essentially exactly, confirming the given lattice parameter and BCC assumption are self-consistent.

III.2 — Ideal HCP packing factor

Given. HCP lattice, ideal $c/a=1.633$; atomic radius $r$ (cancels out).

Find. Atomic packing factor, APF.

Approach. $\text{APF}=\dfrac{n\cdot\frac{4}{3}\pi r^3}{V_{cell}}$, with $n=6$ atoms/cell, $a=2r$, and $V_{cell}=\dfrac{3\sqrt3}{2}a^2c$ for the full hexagonal prism.

  1. Cell volume in terms of $r$. $$V_{cell}=\frac{3\sqrt3}{2}a^2c=\frac{3\sqrt3}{2}(2r)^2(1.633)(2r)=12\sqrt3(1.633)\,r^3$$
  2. Atomic packing factor. $$\text{APF}=\frac{6\left(\frac{4}{3}\pi r^3\right)}{12\sqrt3(1.633)\,r^3}=\frac{8\pi}{12\sqrt3(1.633)}=\frac{\pi}{3\sqrt2}=\boxed{0.740}$$

0.740 is the maximum possible packing fraction for equal spheres (identical to the FCC value) — expected, since ideal HCP ($c/a=1.633$) and FCC are the two close-packed stackings of identical spheres (ABAB… vs. ABCABC…), differing only in stacking sequence, not packing density.

III.3 — Drawing planes and directions

(a) Cubic planes. $(1\bar10)$: intercepts at $x=1$, $y=-1$, parallel to $z$ — drawn (equivalently, shifted by one cell) as the diagonal rectangle through $(0,0,0)$-$(0,0,1)$-$(1,1,1)$-$(1,1,0)$. $(221)$: intercepts $a/h=\tfrac12$, $a/k=\tfrac12$, $a/l=1$, giving the triangle joining $(\tfrac12,0,0)$-$(0,\tfrac12,0)$-$(0,0,1)$.

x y z Plane: (1̄10) diagonal plane through (0,0,0)-(0,0,1)-(1,1,1)-(1,1,0)
(a-i) $(1\bar10)$ plane in the cubic unit cell — the diagonal rectangle containing the cube's $z$-edges through $(0,0,0)$ and $(1,1,0)$.
x y z Plane: (221) intercepts a/2, a/2, a: triangle (½,0,0)-(0,½,0)-(0,0,1)
(a-ii) $(221)$ plane in the cubic unit cell — the triangle through the half-intercepts on $x,y$ and the full intercept on $z$.

(b) Hexagonal directions. Using the Miller–Bravais axes $a_1,a_2,a_3=-(a_1+a_2),c$, a direction $[uvtw]$ is the vector $u\vec a_1+v\vec a_2+t\vec a_3+w\vec c$ (with $t=-(u+v)$ automatically). $[1\bar100]$ ($u{=}1,v{=}{-}1,t{=}0,w{=}0$) lies entirely in the basal plane along $\vec a_1-\vec a_2$. $[112\bar0]$ ($u{=}1,v{=}1,t{=}{-}2,w{=}0$) also lies in the basal plane, along $\vec a_1+\vec a_2$ — this is the close-packed $\langle11\bar20\rangle$-family direction, of length exactly $a$.

a₁ a₂ c Direction [1̄100] u=1, v=−1, t=0, w=0
(b-i) Direction $[1\bar100]$ in the HCP (rhombic sub-cell) unit cell, along $\vec a_1-\vec a_2$.
a₁ a₂ c Direction [112̄0] u=1, v=1, t=−2, w=0
(b-ii) Direction $[112\bar0]$ — the close-packed basal direction, along $\vec a_1+\vec a_2$.