ENG1051
Materials for Energy and Sustainability
Week 10: Atomic-Scale Structure
Complete questions marked with * for homework
Workshop Questions Section 1: Short easy ones to get you started
1. Put the following bond types in order of strength from strongest to weakest (on average – there is
overlap). Covalent, Ionic, Metallic, van der Waals.
Based on Callister (Table 2.3): Ionic, Covalent, Metallic, van der Waals. But as long as van der Waals is
last, there’s overlap between the other three types.
2. What structure arises from stacking close-packed spheres in an ABABAB... sequence? HCP
Describe what the difference is between this and the ABCABC... sequence. This is FCC
3.
a) How many other carbon atoms is each carbon atom bonded to in polyethylene? 2
b) How many other carbon atoms is each carbon atom bonded to in diamond? 4
c) Describe the bonding of carbon atoms in graphite. Each carbon is bound to 3 other carbons in a
2D hexagonal pattern
Section 2: Packing Efficiency
4. Calculate the 2-D packing efficiency of square-packed circles.
Lattice is a square, area = a2
Filled area = πa2/4
Efficiency = filled area/total area = π/4 = 78.5%
ENG1051
Materials for Energy and Sustainability
5. Calculate the 2-D packing efficiency of close-packed circles.
Lattice = rhombus, area = a2.sin(60)
Filled area = πa2/4
Efficiency = filled area/total area = π/(4 × 0.866) = 90.7%
ENG1051
Materials for Energy and Sustainability
6. Calculate the 3-D packing efficiency of a simple cubic array of spheres
Efficiency = (4/3)πr3 / (2r)3 = 52%
Calculate the 3-D packing efficiency of a cubic close-packed array of spheres.
a x sqrt(2)
4r = 𝑎√2 → a = 2.82 r
Vcube = (2.82r)3
Vfilled = No. of spheres x 4/3 πr3 = (8 x 1/8 + 6 x ½) x 4/3 πr3 = 16/3 πr3
Eff. = (16π/3) / (16√2) = 74%
7.
a) Calculate the 3-D packing efficiency of a body-centred cubic array of spheres.
a x sqrt(3)
ENG1051
Materials for Energy and Sustainability
4r = 𝑎√3
No. of atoms in unit cell = 8 x 1/8 + 1 = 2
Eff. = Vfilled / Vcube = (2 x (4/3) x πr3) / (4r/√3)3 = (8π/3) / (64/(3√3)) = 68%
b) Use your answer to the previous question to predict the expected density for pure copper metal,
which has an FCC structure. Copper is generally considered to have an atomic mass of 63.5 g/mol
(1 mol = 6.022 × 1023 atoms) , and an atomic radius of 0.128 nm.
6.022 × 1023 atoms / 4 = 1.5055 × 1023 FCC unit cells and weighs 63.5 g
4r = 𝑎√2 → a = 0.362 nm
Volume of 1.5055 × 1023 FCC unit cells = 1.5055 × 1023 × a3 = 7.144 x 10-6 m3
Density = mass/volume = 63.5 g / 7.144 x 10-6 m3 = 8889 kg/m3
c) How would this density change if copper instead had a BCC structure? 68% packing efficiency vs
74%, so the new density would be 8168 kg/m3 – quite a large difference
Section 3: Metallic structures
8. *In the graphs below can be found the work hardening charts for 1040 steel
You may assume the cold work is given by the following equation:
%CW =
A0 - Ad
´100%
A0
Where A0 is the original area and Ad is the deformed area.
A sample of 1040 steel with 0% cold work is plastically stretched from being 1 m in length to 1.3 m
in length.
(a) Predict the change in area (as a %) Ad = A0/1.3 = 77% of A0
(b) Predict the change in yield stress and ductility This means %CW = 23%, so yield stress will
now be ~750MPa and ductility ~ 12%
(c) Predict the change in the force that would need to be applied to make the steel start yielding
again (compared to the original force for yielding, as a %) F = sigma*A, so ratio of the forces is
(750 * 0.77)/(450*1) = 128%
ENG1051
Materials for Energy and Sustainability
9. Describe some differences you might expect to see in the properties between a FCC metal and a HCP
metal. Explain your predictions by relating them back to the structures and theories of deformation that
we’ve discussed. The main one we discussed was FCC = more ductile, due to higher number of slip
systems. They will also work harden more slowly (which is closely related) and tend to have lower yield
strengths.
Section 4: Structure-Property relationships
10. *Explain what differences you would expect to see in the properties of a semi-crystalline polymer
compared to an amorphous polymer with the same chemistry. Relate your predictions back to the
structures that are present within the material. Semi-crystalline = chains will be closely packed
together → secondary bonds holding them will be stronger throughout the crystalline regions, and the
strength overall will likely be increased vs amorphous. This will also mean that any plastic
deformation is likely to be concentrated in the remaining amorphous regions, so overall the semicrystalline polymer will also be more brittle than the amorphous one. You may have also learned in
earlier weeks that amorphous polymers tend to be transparent while crystalline ones are opaque (due
to light interactions with crystal planes) – this is also acceptable.