Introduction and synopsis

Though they lacked the means to prove it, the ancient Greeks suspected that solids were made of discrete atoms that were packed in a regular, orderly way to give crystals. Today, with the 21st century techniques of X-ray and electron diffraction and lattice-resolution microscopy available to us, we know that all solids are indeed made of atoms, and that most (but not all) are crystalline. The common engineering metals and ceramics are made of many small crystals, or grains, stuck together at grain boundaries to make polycrystalline aggregates. The properties of the material—its strength, stiffness, toughness, conductivity and so forth—are strongly influenced by the underlying crystallinity. So it is important to be able to describe it.

Crystallography is a language for describing the 3-dimensional arrangement of atoms or molecules in crystals. Simple crystalline and non-crystalline structures were introduced earlier in the book. Here we go a little further, exploring a wider range of structures, interstitial space (important in understanding the hardness of steel), and ways of describing planes and directions in crystals (crucial in the production of turbine blades and in the manufacture of microchips).

Exercises are provided in Parts 1–4, so do these as you go along; solutions are provided at the end of the unit.

PART 1: Crystal structures

Three-dimensional crystal lattices may be characterised by the repeating geometry of the atomic arrangement they contain, particularly their degree of symmetry. In fact it may be shown that there are 14 distinguishable three-dimensional lattices. If you are a crystallographer you need to know about all of them¹. But engineering materials, for the most part, have simple structures. We shall concentrate on these.

The great majority of the 92 stable elements are metallic, and of these, the majority (68 in all) have one of the three simple structures introduced in Chapter 4: face-centred cubic (FCC), body-centred cubic (BCC) and hexagonal close packed (HCP or CPH). We start by revisiting them briefly.

The face-centred cubic (FCC) structure

The FCC structure is described by a cubic unit cell with one atom at each corner and one at the centre of each face. Atoms touch along the diagonals of the cube faces. From a packing viewpoint, it consists of close-packed planes stacked in an ABCABC … sequence (Figure GL1.1).

Figure GL1.1 An FCC structure of packed spheres showing the ABC stacking.

Among the metallic elements, 17 have the FCC structure. Engineering materials with this structure include the following.

Material	Typical uses
Aluminum and its alloys	Airframes, space frames and bodies of trains, trucks, cars, drink cans
Nickel and its alloys	Turbine blades and disks
Copper and α-brass	Conductors, bearings
Lead	Batteries, roofing, cladding of buildings
Austenitic stainless steels	Stainless cookware, chemical and nuclear engineering, cryogenic engineering
Silver, gold, platinum	Jewelry, coinage, electrical contacts

FCC metals have the following characteristics:

• They are very ductile when pure, work hardening rapidly but softening again when annealed, allowing them to be rolled, forged, drawn or otherwise shaped by deformation processing.
• They are generally tough, resistant to crack propagation (as measured by their fracture toughness, K_1c).
• They retain their ductility and toughness to absolute zero, something very few other structures allow.

The body-centred cubic (BCC) structure

The BCC structure is described by a cubic unit cell with one atom at each corner and one in the middle of the cube (Figure GL1.2). The atoms touch along the internal diagonal of the cube.

Figure GL1.2 A BCC structure of non-close-packed spheres.

Of the metallic elements, 21 have this structure (most are rare earths). They include the following.

Material	Typical uses
Iron, mild steel	The most important metal of engineering: construction, cars, cans and more
Carbon steels, low alloy steels	Engine parts, tools, pipelines, power generation
Tungsten	Lamp filaments
Chromium	Electroplated coatings

BCC metals have the following characteristics:

• They are ductile, particularly when hot, allowing them to be rolled, forged, drawn or otherwise shaped by deformation processing.
• They are generally tough, resistant to crack propagation (as measured by their fracture toughness, K_1c) at and above room temperature.
• They become brittle at low temperatures. The change happens at the ‘ductile-brittle transition temperature’, limiting their use below this.
• Their strength depends on temperature, even at low temperatures.
• They can generally be hardened with interstitial solutes.

The close-packed hexagonal (HCP) structure

The HCP structure is usually described by a hexagonal unit cell with an atom at each corner, one at the centre of the hexagonal faces and three in the middle. From a packing viewpoint, it consists of close-packed planes stacked in an ABAB … sequence (Figure GL1.3).

Figure GL1.3 An HCP structure of packed spheres showing the ABA stacking.

Of the metallic elements, 30 have this structure. They include the following:

Material	Typical uses
Zinc	Die-castings, plating
Magnesium	Lightweight structures
Titanium, its alloys	Light, strong components for airframes and engines, biomedical and chemical engineering
Cobalt	High temperature superalloys, bone-replacement implants
Beryllium	The lightest of the light metals; its use is limited by expense and potential toxicity

HCP metals have the following characteristics.

• They are ductile, allowing them to be forged, rolled and drawn, but in a more limited way than FCC metals.
• Their structure makes them more anisotropic than FCC or BCC metals.

Exercises

• E.1 Copper has an FCC structure. Draw the third layer on Figure GL1.4, below left, to make it.
• E.2 Magnesium has an HCP structure. Draw the third layer on Figure GL1.4, above right, to create this structure.
• E.3
  • (a) How many atoms are contained within the FCC unit cell? (Hint: consider how many adjoining cells ‘share’ each corner atom and face atom in the cell.)
  • (b) Determine the atomic packing factor (the fractional volume occupied) for an FCC structure, assuming that the atoms behave as hard spheres. (Note that in FCC, atoms touch along the diagonals of the face of the unit cell.)
  • (c) The atomic diameter of an atom of nickel is 0.2492 nm. Calculate the lattice constant (the edge-length of the unit cell) of FCC nickel.
  • (d) The atomic weight of nickel is 58.71 kg/kmol. Calculate the density of nickel. (The number of atoms in 1 kmol is Avogadro’s number = 6.022 × 10²⁶.)
• E.4 Follow the same procedures as in Exercise E.3 to answer the following:
  • (a) Determine the atomic packing factor for BCC, and comment on the value compared with that for FCC. (Note that in BCC, atoms touch along the internal diagonal of the unit cell.)
  • (b) The atomic diameter of BCC iron is 0.2482 nm. Calculate the lattice constant of iron.
  • (c) The atomic weight of iron is 55.85 kg/kmol. Calculate the density of iron.

Figure GL1.4 Two layers of close-packed planes.

PART 2: Interstitial space

Interstitial space is the unit of space between the atoms of molecules. The FCC and HCP structures both contain interstitial space of two sorts: tetrahedral and octahedral. They are shown for the FCC structure in Figure GL1.5(a) and (b).

Figure GL1.5 (a) and (b) The two types of interstitial hole in the FCC structure, (c) the tetrahedral hole of the BCC structure and (d) the octahedral hole in the CPH structure.

The holes are important because foreign atoms, if small enough, can fit into them. For both structures, the tetrahedral hole can accommodate, without strain, a sphere with a radius of 0.22 of that of the host. The octahedral holes are larger: they can hold a sphere that is … but wait, that is Exercise E.5. The atoms of which crystals are made, in reality, are slightly squashy, so that foreign atoms that are larger than the holes can be squeezed into the interstitial space.

This is particularly important for carbon steel. Carbon steel is iron with carbon in some of the interstitial holes. Iron is BCC, and, like the FCC structure, it contains holes (Figure GL1.5(c)). They are tetrahedral. They can hold a sphere with a radius 0.29 times that of the host without distortion. Carbon goes into these holes, but, because it is too big, it distorts the structure. It is this distortion that gives carbon steels much of their strength.

The CPH structure, like the FCC, has both octahedral and tetrahedral interstitial holes. The larger of the two, the octahedral hole, is shown in Figure GL1.5(d). The holes have the same sizes as those in the FCC structure.

We shall encounter these interstitial holes in another context later: they provide a way of understanding the structures of many oxides, carbides and nitrides.

Now more Exercises, in the interstitial space before the next subsection.

Exercises

• E.5 Calculate the diameter of the largest sphere that will fit into the octahedral hole in the FCC structure. Take the diameter of the host-spheres to be unity.
• E.6 The HCP structure contains tetrahedral interstitial holes as well as octahedral ones. Identify a tetrahedral hole on the HCP lattice of Figure GL1.6.

Figure GL1.6 The HCP structure.

PART 3: Describing planes

The properties of crystals depend on direction. The elastic modulus of hexagonal titanium (HCP), for example, is greater along the hexagonal axis than normal to it. This difference can be used in engineering design. In making titanium turbine blades, for example, it is helpful to align the hexagonal axis along a turbine blade to use the extra stiffness. Silicon (which also has a cubic structure) oxidises in a more uniform way on the cube faces than on other planes, so silicon ‘wafers’ are cut with their faces parallel to a cube face. For these and many other reasons, a way of describing planes and directions in crystals is needed. Miller indices provide it.

First, let’s describe planes. The Miller indices of a plane are the reciprocals of the intercepts the plane makes with the three axes that define the edges of the unit cell, reduced to the smallest integers.

Figure GL1.7 illustrates how the Miller indices of a plane are found. Draw the plane in the unit cell so that it does not contain the origin—if it does, displace it along one axis until it doesn’t. Extend the plane until it intersects the axes. Measure the intercepts, in units of the cell edge-length, and take the reciprocals. (If the plane is parallel to an axis, its intercept is at infinity and the reciprocal is zero.) Reduce the result to the smallest set of integers by multiplying through to get rid of fractions or dividing to remove common factors. The six little sketches show, in order, the (100), the (110), the (111), the (211), the and the (112) planes; the means that the plane intercepts the y-axis at the point –1. Check that you get the same indices.

Figure GL1.7 The procedure for finding the Miller indices of a plane. The indices of four common planes are shown on the right.

The Miller indices of a plane are always written in round brackets: (111). But there are many planes of the 111-type; two are shown in Figure GL1.7. The complete family is described by putting the indices in curly brackets, thus:

Exercises

• E.7 What are the Miller indices of the planes shown in the six sketches of Figure GL1.8(a), (b) and (c)? (Remember that, to get the indices, the plane must not pass through the origin, and remember too to get rid of fractions or common factors.)
• E.8 When iron is cold it cleaves (fractures in a brittle way) on the (100) planes. What will be the angle between cleavage facets?
• E.9 Mark, on the cells of Figure GL1.8(d), (e) and (f), the following planes:, (120), .

Figure GL1.8 Identify the planes shown in the upper three figures. Draw those required in the question in the lower three.

PART 4: Describing directions

The Miller indices of a direction are the components of a vector (not reciprocals) that starts from the origin, along the direction, reduced to the smallest integer set.

Figure GL1.9 shows how to find the indices of a direction. Draw a line from the origin, parallel to the direction, extending it until it hits a cell edge or face. Read off the coordinates of the point of intersection. Get rid of any fraction or common factor by multiplying all the components by the same constant. The six sketches show the [010], the [011], the [111], the [021], the and the directions. The means that the y-coordinate of the intersection point is –1.

Figure GL1.9 Miller indices of directions. Always translate the direction so that it starts at the origin.

The Miller indices of a direction are always written in square brackets: [100]. As with planes, there are several directions of the 100-type. The complete family is described by putting the indices in angle-brackets, thus:

Exercises

• E.10 Identify the Miller indices of the three directions shown in the sketches of Figure GL1.10(a), (b) and (c).
• E.11 Mark, on the cells of Figure GL1.10(d), (e) and (f), the following directions: , [210] and .

Figure GL1.10 Identify the directions shown in the upper three figures. Draw those required in the question in the lower three.

PART 5: Ceramic crystals

Technical ceramics give us the hardest, most refractory materials of engineering. Those of greatest economic importance include:

• Alumina, Al₂O₃ (spark plug insulators, substrates for microelectronic devices).
• Magnesia, MgO (refractories).
• Zirconia, ZrO₂ (thermal barrier coatings, ceramic cutting tools).
• Uranium dioxide, UO₂ (nuclear fuels).
• Silicon carbide, SiC (abrasives, cutting tools).
• Diamond, C (abrasives, cutting tools, dies, bearings).

The ceramic family also gives us many of the functional materials—those that are semiconductors, are ferromagnetic, show piezo-electric behaviour and so on. Their structures often look complicated, but when deconstructed, so to speak, a large number turn out to be comprehensible (and for good reasons) as atoms of one type arranged on a simple FCC, HCP or BCC lattice with the atoms of the second type (and sometimes a third) inserted into the interstices of the structure of the first.

The diamond cubic (DC) structure

We start with the hardest ceramic of the lot—diamond—of major engineering importance for cutting tools, abrasives, polishes and scratch-resistant coatings, and at the same time as a valued gemstone for jewelry. Silicon and germanium, the foundation of semiconductor technology, have the same structure.

Figure GL1.11 shows the unit cell. Think of it as an FCC lattice with an additional atom in each of its tetrahedral interstices, labeled 1, 2, 3 and 4 in the figure. The tetrahedral hole is far too small to accommodate a full-sized atom, so the others are pushed farther apart, lowering the density. The cause is the 4-valent nature of the carbon, silicon and germanium atoms—they are happy only when each has 4 nearest neighbors, symmetrically placed around them. That is what this structure does.

Figure GL1.11 The diamond-cubic (DC) structure.

Silicon carbide, like diamond, is very hard, and it too is widely used for abrasives and cutting tools. The structures of the two materials are closely related. Carbon lies directly above silicon in the periodic table. Both have the same crystal structure and are chemically similar. So it comes as no surprise that a compound of the two with the formula SiC has a structure like that of diamond, with half the carbon atoms replaced by silicon, as in Figure GL1.12 (it is the atoms marked 1, 2, 3 and 4 of Figure GL1.11 that are replaced).

Figure GL1.12 The structure of silicon carbide.

Materials with the diamond-cubic structure	Comment
Carbon, as diamond	Cutting and grinding tools, jewelry
Silicon, germanium	Semiconductors
Silicon carbide	Abrasives, cutting tools

Oxides with the Rocksalt (Halite) structure

These oxides all have the formula MO, where M is a metal ion. Oxygen ions are large, usually bigger than those of the metal. When this is so the oxygen packs in an FCC structure, the metal atoms occupy the octahedral holes in this lattice. The resulting structure is sketched in Figure GL1.13. This is known as the Rocksalt (or Halite) structure because it is that of sodium chloride, NaCl, with chlorine where the oxygens are and sodium where the magnesiums are in the figure.

Figure GL1.13 The structure of MgO, typical of many simple oxides.

Sodium chloride (common salt) itself is a material of engineering importance. It has been used for road beds and buildings, there are plans to bury nuclear waste in rocksalt deposits; and large single crystals of rocksalt are used for the windows of high-powered lasers.

Materials with the Rocksalt structure	Comment
Magnesia, MgO	A refractory, and ceramic with useful strength
Ferrous oxide, FeO	One of several oxides of iron
Nickel oxide, NiO	Ceramic superconductors
All alkali halides, including salt, NaCl	Feedstock for chemical industry, nuclear waste storage

Oxides with the Corundum structure

A number of oxides have the formula M₂O₃, among them alumina, Al₂O₃. The oxygen ions, the larger of the two, are close packed in an HCP stacking. The M atoms occupy two-thirds of the octahedral holes in this lattice, one of which is shown for alumina, filled, in Figure GL1.14.

Figure GL1.14 The M atoms (here aluminum) of the Corundum structure lie in the octahedral holes of an HCP oxygen lattice.

Materials with the Corundum structure	Comment
Alumina, Al2O3	The most widely used technical ceramic
Iron oxide, Fe2O3	The oxide from which iron is extracted
Chromium oxide, Cr2O3	The oxide that gives chromium its protective coating
Titanium oxide, Ti2O3	The oxide that gives titanium its protective coating

Oxides with the Fluorite structure

The metal atoms in the important technical ceramic zirconium dioxide, ZrO₂, and the nuclear fuel uranium dioxide, UO₂, being far down in the periodic table, are large in size. Unlike the oxides described earlier, the M of the MO₂ in these is larger than the O. In these it is the M atoms that form a close-packed FCC structure and it is the oxygens that fit into the tetrahedral interstices in it, as shown in Figure GL1.15.

Figure GL1.15 The M atoms of fluorite-structured oxides pack in an FCC array, with oxygen in the tetrahedral interstices.

Materials with the Fluorite structure	Comment
Zirconia, ZrO2 (slightly distorted fluorite)	The toughest high-temperature ceramic
Urania, UO2	Nuclear fuel
Thoria, ThO2	Nuclear fuel
Plutonia, Pu2O3	A product of nuclear fuel reprocessing, a fuel in itself

PART 6: Polymer crystals

Many polymers crystallise. The long chains line up and pack to give an ordered, repeating structure, just like any other crystal. The low symmetry of the individual molecules means that the choice of lattice is limited. Figure GL1.16 is a typical example; it is polyethylene.

Figure GL1.16 The structure of a crystalline polyethylene.

Few engineering polymers are completely crystalline, but many have as much as 90% crystallinity. Among those of engineering importance are: