Introduction to Inorganic Chemistry/Metals and Alloys: Mechanical Properties

Chapter 7: Metals and Alloys: Mechanical Properties

How much do the mechanical properties of metals and alloys vary with processing? The answer is, a great deal. Consider the following hypothetical situation: Upon graduation, you go to work as an engineer for Boeing. Your job is to work with aluminum companies to help them produce high strength alloys. Why? A large jet airplane weighs a total of 500 tons. Of that total, 50 tons is cargo, 150 tons is the plane structure, and the remainder is fuel. If you can triple the strength of the materials in the structure (aluminum), you can reduce the mass of the structure to 50 tons and increase the cargo to 150 tons. Look at what has been done already:

Material	Tensile strength yield (psi)
pure (99.45%) annealed Al	4 x 103
pure (99.45%) cold drawn Al	24 x 103
Al alloy - precipitated, hardened	50 x 103

By chemical and physical manipulation we have already increased the yield strength 12 times over annealed Al. Yet the yield strength of a "perfect" single crystal of pure Al is ca. 10⁶ psi. We still have 3 orders of magnitude to go. This just shows that there will still be plenty to do on this project between now and graduation!

Learning goals for Chapter 7:

• Understand the structure of dislocations and grain boundaries and their role in controlling the mechanical properties of solids.
• Explain why the mechanical properties of bcc metals and alloys differ from those with close packed structures.
• Explain the effects of work hardening and annealing on structure and mechanical properties.
• Explain the mechanical properties of steel in terms of its phase behavior.
• Understand the structure and mechanical properties of amorphous metals.

  7.1 Defects in metallic crystals

Grains and grain boundaries in a polycrystalline material

“Crystals are like people, it is the defects in them which tend to make them interesting!” - Colin Humphreys.
Metals, by virtue of their non-directional bonding, are more energetically tolerant of defects than are covalent network or ionic solids. Because there is no strong preference for one atomic position over another, the energy of a metallic crystal is not greatly impaired by the vacancy of a single atom or by the dislocation of a group of atoms. These kinds of "mistakes" in the packing of metal atoms within crystals are collectively called defects. The deformability of metals is the direct result of defects in the crystal structure. Defects in metals such as Al and Fe are responsible for the three orders of magnitude difference between the yield stress of annealed polycrystalline samples (i.e., normal articles of commerce) and perfect single crystals.

There are several different kinds of defects that can be found in metallic crystals. One kind is called a vacancy, i.e., a place where an atom is missing in the structure. A dislocation, on the other hand, is a line defect; it runs somewhat like a string through the crystal. A dislocation is the result of one atom or group of atoms being pulled slightly out of position with respect to perfect crystal packing. A third kind of defect is called a grain boundary; it is a two-dimensional interface between two different crystal grains in a solid sample. Since the two crystallites have in general different orientation, the structures do not match up exactly at the interface. While point vacancy defects do not significantly affect the mechanical properties of metallic crystals, both dislocations and grain boundaries have large effects, as explained below.

  7.2 Work hardening, alloying, and annealing

One of the questions we would like to ask is, why are the yield stresses of normal (polycrystalline) metal samples so much lower (by a factor of 1000) than they are in perfect single crystals? The answer has to do with the motion of dislocations. Consider the picture below, which shows planes of metal atoms near a dislocation (the individual atoms are numbered to help you see which bonds are broken and which are formed). The arrows indicate force applied under shear stress. Notice how the dislocation moves by breaking/making metal-metal bonds.

The key point here is that we can induce plastic deformation (shear) by breaking only one line of metal-metal bonds at a time along the dislocation line. This involves far less force than breaking an entire plane of bonds, as we would need to do to shear a perfect crystal. In a given polycrystalline sample, there are many dislocation lines that run perpendicular to all possible shear directions, so their motion can be used to "tear" the metal apart. Turbine rotors on large jets are made of very expensive single crystal nickel-titanium alloys, so that these shearing deformations can be avoided.

We can see that motion of dislocations is basically bad news if we want a metal to be strong and hard (e.g., if we want a structural material, or a knife that can hold a decent edge). There are several ways we can overcome (to some extent) this problem:

1. Use single crystals (expensive - especially with large items like turbine blades, and impossible with very large items like airplane wings or bridges).

2. Work hardening of the metal - this moves all dislocations to grain boundaries (the dislocation essentially becomes part of the grain boundary). Since a grain boundary is a planar defect, it is much less responsive to stress than a line defect.

3. Introduce impurity atoms (that is alloying elements) or impurity phases that "pin" the motion of defects. An impurity atom stops the motion because it is a different size, or makes stronger bonds, than the other metal atoms; the line defect has a hard time moving away from rows of such atoms. An impurity phase (like Fe₃C in iron) makes extra grain boundaries that can stop the motion of defects. This effect is analogous to the graphite fibers in fiber-reinforced cross-linked polymers (used, e.g., in tennis rackets) that stop the propagation of cracks.

Blacksmith, 1606

A simple illustration of work hardening can be done with a piece of copper wire. When struck many times with a hammer, the copper wire becomes stiffer, and it is possible to hang a weight from it. Dislocations move to the crystal grain boundaries during work hardening, effectively halting their motion and at the same time making the individual crystal grains smaller. Because the crystal grains are now smaller, the amount of grain boundary area has increased, and with it the free energy of the material. Annealing reverses the process by lowering the free energy. When the wire is annealed in a flame (heated so that atoms can move and rearrange), the crystal grains grow, and the dislocations reappear. The copper again becomes ductile, and bends easily. Cold-working (work hardening) of metals is important for strengthening structural materials (e.g., iron beams) and for making brittle, hard edges (this is why blacksmiths hammer on knives and swords when they are making them. If you have ever watched them, they do the same thing to horseshoes, when they cool down, to make them stiff).
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  7.3 Malleability of metals and alloys

Bronze age weapons from Romania.

β-brass is an ordered BCC alloy of Cu and Zn

Metals with close-packed structures (HCP and FCC) such as copper, gold, silver, zinc, magnesium, etc. are in general more malleable than those with the BCC structure (tungsten, vanadium, chromium, etc.). Why? In the close-packed structure, there is relatively little corrugation between sheets of metal atoms. This means that these planes can slip past each other relatively easily. In the BCC structure, there are no close-packed planes, and much greater corrugation between atoms at different levels. This makes it much harder for one row to slide past another.

This effect explains the hardness of alloys like brass (CuZn, which has the BCC structure), which are made by combining two soft metals (Cu and Zn, which are respectively FCC and HCP as pure metals, are both soft and ductile). Bronzes - originally made as alloys of copper and arsenic, but later as alloys of copper and tin - are harder than either of the constituent metals for the same reason.

The history of bronzes and brass dates to pre-historic times, with the earliest brasses made by smelting copper-zinc ores. In the Bronze Age, possession of these hard alloys provided a tactical advantage in warfare (see image at right), that was later supplanted when the technology for smelting iron was developed.

  7.4 Iron and steel

The iron–iron carbide (Fe–Fe₃C) phase diagram. Below 912 °C, pure iron exists as the alpha phase, ferrite, which has the BCC structure. Between 912 and 1,394 °C, pure iron exists as the gamma phase, austenite, which has the FCC structure. Carbon is more soluble in the FCC phase, which occupies area "γ" on the phase diagram, than it is in the BCC phase. The percent carbon determines the type of iron alloy that is formed upon cooling from the FCC phase, or from liquid iron: alpha iron, carbon steel (pearlite), or cast iron.

Upon cooling, high carbon steels phase segregate into a mixture of bcc iron (light gray) and Fe3C (dark gray) microscopic grains.

One other very important place where the difference between the hardness of a BCC and a close-packed metal is important is in steelmaking. Between room temperature and 912^oC, iron has the BCC structure, and is a tough, hard metal ("tough as nails"). Above 912^oC, pure iron switches over to the FCC (austenite) structure, which is much more ductile. So hot iron can be bent and worked into a variety of shapes when it is very hot but still solid (it melts at 1535^oC). Rapid quenching of hot iron - e.g., when the blacksmith plunges a red hot piece directly into cold water - cools it to room temperature, but doesn't allow time for the FCC --> BCC phase transition to occur; therefore, such pieces are still relatively malleable and can be shaped.

Carbon is added (about 1% by weight) to iron to make "carbon steel", which is a very hard material. Carbon is rather soluble in the FCC phase of iron, but not in the BCC phase. Therefore, when the ductile FCC phase cools and turns into BCC ("tempering" the steel, which means cooling it slowly enough so the FCC to BCC transformation can occur), the iron can no longer dissolve the excess carbon. The carbon forms layers or grains of an extra phase, Fe₃C ("cementite" - a very hard material) which are layered or dotted throughout the matrix of BCC iron grains. The effect of all these little grains of Fe₃C is to stop the motion of dislocations, making for a harder but (with higher carbon content) increasingly brittle material. This is why knives and swords are quenched from the FCC phase, cold worked into the appropriate shapes, and then heated up again and tempered (before they are sharpened) when they are made. Cast iron objects (frying pans, radiators, etc) have a higher carbon content and are therefore very strong, but tend to fracture rather than bend because of the larger fraction of the brittle Fe₃C phase in the alloy.

  7.5 Amorphous alloys

Amorphous metals, also called bulk metallic glasses, are materials of growing technological importance. Because their glassy structures do not support the movement of dislocations, they are stronger and more wear-resistant than crystalline metals of similar composition.

All pure metals and most simple alloys crystallize easily. Bulk samples of metals prepared by ordinary chemical or electrochemical methods are polycrystalline, with grain sizes ranging from tens of nanometers to tens of microns in size. Larger crystals of metals can be made by very slow crystal growth, e.g., by the Czochralski process. In contrast, glassy or amorphous metals can be prepared by very rapid cooling from the melt. With pure metals and simple alloys the cooling rate needed is so fast - on the order of 10⁶ K/s - that amorphous samples can only be made as very thin films.

Alloys of metals with more complex stoichiometries can be made in amorphous form by slower cooling from the melt. These alloys have been prepared and studied since the 1960s, and since the 1990s amorphous alloys have been discovered that can be prepared in bulk form at cooling rates on the order of 1 deg/s, similar to the cooling rates of other kinds of glasses.

Currently amorphous metals (marketed under the tradenames Vitreloy and Liquidmetal) are used commercially in golf clubs, watches, USB flash drives, and other applications where very high elasticity, yield strength, and/or wear resistance are needed.

Year	Alloy	Cooling Rate (K/s)
1960	Au75Si24	106 - thin films & ribbons
1969	Pd-Cu-Si	100-1000
1980s	La-Al-Cu & others	1-100
1990s	Zr-Ti-Cu-Ni-Be	~1 (similar to oxide glasses)

  7.6 Discussion questions

• Discuss the thermodynamics of work hardening and annealing in terms of the microscopic picture of defects in metallic crystals.
• In your pocket or purse, you may have a brass key, which is an alloy of Cu and Zn. How do the mechanical properties of this alloy depend on its structure, and why don't we make keys out of pure Cu or Zn?
• Cooling carbon steels along the eutectic lines (A and B) in the iron-carbon phase diagram above results in the formation of pearlite and ledeburite. How does the microstructure of these two iron alloys differ, and how does the microstructure affect their mechanical properties?

  7.7 Problems

1. Use a drawing to show how a dislocation moves through a metallic crystal.

2. What happens to the dislocations in a metal (a) when it is work hardened, and (b) when it is annealed? Explain which process results in smaller crystal grains and why.

3. Why are alloys such as bronze mechanically stronger than their constituent metals (copper and tin)? How did the discovery of these alloys change civilization?

4. The drawing below shows three sections along the z-axis of the unit cell of a fcc metal (e.g., Cu or Au). The opposite corners of the cell are numbered 1 and 12. Perpendicular to the line joining those two atoms is close-packed layer that contains atoms 3, 7, 8, 11, 13, and 14.

(a) In the unit cell there are other close-packed atomic planes parallel to the one described above. Sort the remaining eight numbered atoms into groups by the parallel planes that contains the atoms.

(b) How many body diagonals does this cell have? Identify the other body diagonals by the numbers of atoms at the corners of the cell.

5. The images below show four cubic unit cells. How many atoms are in each of these cells, and which of the 14 Bravais lattices does each one correspond to? (Hint: The origin of the unit cell is not necessarily at the corner of the cells in the drawing).