Technology of metallic materials - Strengthening methods of alloys

Strengthening methods of alloys

Slide 1 & 2

There are a few strengthening methods:

• Strengthening by s.s.s. and i.s.s (substitutional solid solutions, interstitial solid solutions)
• Strengthening by work hardening
• Strengthening by grain refinement
• Strengthening by precipitation (aging)
• Strengthening by dispersion of hard second phase
• Strengthening by hard phase (e.g., martensite)
• Strengthening by fiber orientation (composites, texture)

Slide 3

The most important factors in the strengthening of metals are dislocations. Metals can deform plastically, only a portion of them will perform elastically. Larger plasticity, larger strengthening. Scientists, in the past tried to combine the theory with practice. They introduced a fictitious defect, dislocations, in order to decrease the theoretical strength, that they calculated by common theories available at that time. They had never seen such dislocations.

After some time (1960), the transmission electron microscope was invented. Now they could test the material under a tensile test for the first time. It was very surprising to see that dislocations were real and not fictitious. They could move under the action of an external force, which acts under the plane in which the dislocations move. We will say that planes where dislocations move are named sliding planes (slip planes). Dislocations must move over a plane and along a specific direction, therefore we must know the slip system. The slip system is just the combination of a plane and a direction.

So, metals are made of lattices, which can be packed or closely packed. If the dislocations must move in metals (consider pure metals only) into single crystals (one grain of polycrystalline materials). Dislocations move easier in a continuous crystal structure, so the continuity of the crystal structure favors the movement of interstitials. That is why dislocations can move better in packed and close-packed crystals, BCC and FCC respectively. So, the continuity and the density of the crystal lattice are responsible for easy movement of dislocations.

BCC: One blue plane, 2 directions. It cuts the cube along face diagonal. The directions we identify are along the diagonals of the rectangle. Why is this the favorite direction for dislocations to move? We know that in BCC, since the atom at the center of the cube characterizes the denser plane and the denser direction, we can say that the plane must cut the cell in the center of the cube. The direction must be also dense, so it must cross the sphere in the center. How many planes and directions do we have in BCC? The most common family of planes are {110} with <111> directions. We call them family because we have a few planes identical to this, and they behave in the same way. The direction <111> must cross the diagonal of the cube. (2 directions shown in the picture). Just like the planes, we have a set of directions too. If you consider the symmetry of the cube, totally we have 6 planes and 12 directions (12 systems). But in practice, through the transmission electron microscope, it has been observed that dislocations in BCC can also move in other planes, {112} and {123} while directions are the same, so <111>. Totally we have 48 slip systems possible in BCC.

FCC: In the case of FCC, we have very close packing. The atoms are arranged very tightly. This means that if the packing factor is higher, dislocations can move easily in FCC compared to BCC, and this is true. What are planes in which the atoms are better packed? In FCC, we have atoms in the center of each face. So, the densest plane is {111} (diagonal plane of the cube). The slip directions are the diagonals of the faces of the cube. By symmetry, we have 4 planes like this and 3 directions. So, 12 slip systems in FCC.

HCP: In the hexagonal close packing, we have high anisotropy, especially because the vertical planes of the hexagonal structure are poorly dense and if a shear force is applied on them, we can observe material failure. The other planes are not very good in plastic deformation. It is much more limited than in BCC or FCC crystal structures. Slip happens only in the basal {0001} planes along the <1120> directions. That’s the reason why alloys are not produced in hcp.

Which of these structures will allow easy plastic deformation? Or which of these crystals will quickly start to slip? The answer is BCC. It is the easiest to slip because it has many slip systems. The probability for a slip system to be activated is higher in BCC.

How do we consider FCC then? Since FCC will not slip the easiest, is FCC more effective in anything compared to BCC? What is the peculiarity of FCC? The reason why FCC is peculiar is because it can accumulate dislocations more than BCC. We know that when dislocations move, the material is plastically flowing, so it will elongate plastically. If I see macroscopically that a metal is elongating, that means dislocations are moving. In some metals, you have experienced bending of a steel wire. When you bend several times a material, you see that the material becomes stronger and stronger. This is what we want to do with the dislocations. We want to make the metals stronger by plastically deforming them, then we stop at a certain level before applications. It may happen that in the application the material may experience during serving condition a further plastic deformation. The problem is that if the material saturates under service, it will break down. In that case, the material will reach the tensile strength, neglecting the second part of the curve where the material continues to deform but in an unstable manner. We are not interested in that part for design purposes; therefore, we may consider the material once reaching the tensile strength, it reaches its maximum bearing load. After that, even if the material does not break, it becomes unstable, and we cannot consider it reliable anymore. In other terms, if we have BCC and we bend that material, and then we have FCC and we do the same, in your opinion, which of both metals can be bent in a larger number of bending operations? FCC. It will experience a larger number of bending.

And which of both will experience the largest strength after many bending operations before fracture? FCC, because it has only one system and because it is densest in atoms, the dislocations can more effectively flow even if the dislocations will encounter difficulties. FCC will allow the largest flowing ability and since large flowing means strengthening, also the largest strengthening. FCC will strengthen by plastic deformation more than BCC.

How will BCC behave during bending? BCC has many systems, but no one is very densely packed, whenever the dislocations flow, they will encounter a difficulty to flow because the dislocations will interact with each other and then another system will be activated. When that other system also encounters difficulty, another system will be activated. So, BCC, because of its many slip systems, it can activate many of them as soon as some dislocation encounters difficulty in flowing. But this easy way to change slip systems means that the dislocations cannot be effectively packed in every system like FCC does. So, at the end the strength that BCC can reach is always lower than FCC, for the same element. Therefore, FCC will more effectively let the dislocations move, and more effectively work-harden (strain-harden) it than BCC. FCC will saturate better the plasticity available in the metal. The plasticity is the area by a tensile test curve, in range between the yield strain (corresponding to the yield strength) and the tensile strain (maximal strain). (Plastic region of material) FCC will always have the largest area compared to BCC.

Slide 4 (Stacking fault)

This figure shows you something that may happen in plastic deformation. It may happen when the metal transforms into the solid state from liquid. The crystal lattice may fail to be produced, or it may not properly form in some regions. FCC can be arranged (piled up) in planes which is a sequence of ABC, ABC, etc. You can see that the B plane is located over the interstitials. The same happens with the planes B and C (interstitials located between them). In case of FCC stacking or piling of planes, this sequence will allow the regular crystal lattice to be formed, either during plastic deformation processes or during casting. Sometimes, if you do plastic deformation, something may go wrong. We can compare it with the stacking in HCP. You can see that HCP in which we have a high density of atoms, the stacking is ABABAB…. What happens when plastic deformation occurs? The sequence of these planes may fail. For example, the C plane can disappear in a single region. Therefore, rather than having ABCABC, C isn’t there and therefore ABAB, which resembles HCP crystal structure. So, when this happens (it is very likely to happen) in FCC, then we say that a stacking fault is produced (a mistake in the arrangement of the crystal). Dislocations will temporarily cut the crystal, and usually the crystal is recomposed after that. But sometimes, when the dislocation passes through, the crystal is not perfectly recomposed, and 1 plane will miss, and this will create a defect, which is named stacking fault. In the stacking fault, the sequence will be like that of the HCP. This fault will extend into the lattice depending on the condition of the lattice when the dislocation passes through. To make you understand better, a stacking fault may be performed if there is some impurity, especially when it is very large or very small (interstitial). The impurity distorts the FCC crystal, and the dislocation cuts the crystal where the interstitial is located, it may be possible that the crystal will not recompose after the dislocation passes.

Stacking fault energy is the energy required to produce this defect. (Copper has low energy, aluminum has very high energy, so it is less prone to stacking faults and twins)

Slide 5

Now, we start all the methods of strengthening. Assumption: Regardless of the way the material will strain-harden by using 1 or more of those methods, we assume that all these methods can be summed linearly. (Physically not very rigorous to sum them linearly, but it seems that in practical engineering it can be done.) The first method I want to tell you is the simplest one. If we do not have any defect, a single crystal. (It may be extended to polycrystals too) (applies to all crystal structures) Let’s assume that a dislocation will move in 1 single crystal lattice. Is there any resistance by the crystal to the movement of the dislocation? Yes. Because atoms are linked by interatomic forces, and if a dislocation wants to pass through the crystal it has to break these bonds. This means that the dislocation requires some energy from outside to pass. If I have no defect on the lattice, just the lattice, and apply an external force, to be efficient and to properly move the dislocation, this force must be a shear stress and it must be parallel to the slip plane. So even if there are not any other defects, I need to apply a non-zero external force to make the dislocation cut through the crystal. This kind of stress is named friction stress (it is named this way because there are no other defects, and the stress can be viewed as a friction of the lattice) B is the burger vector. What can be seen is that all the unknowns have an elastic character. This is the minimal condition of the external stress we must apply to make the dislocation "beat" the resistance of the crystal.

Slide 6 (Solid Solutions)

Now we consider the first strengthening mechanism, the presence of a solid solution. Now we are putting some alloying elements in the lattice, and we want to see how the dislocations will face the obstacle in their motion, created by impurities. The alloys are usually stronger than pure metals. Why? Substitutional and interstitial alloys cause lattice distortions that interfere with the strain field created by the dislocations, by hindering their motion. Impurities tend to diffuse and be trapped by the dislocation core (dislocation line) to find more suitable sites which better fit with their dimensions. The impurities will act as an obstacle to the dislocation movement due to their own stress field. The impurities may also change location, because we know that atoms can jump from one site or another. Impurities may diffuse and may jump where the dislocation is moving, by slowing down the dislocation. When the atoms will interact with the dislocations, we call this kind of interaction "anchoring", and when the dislocation passes over the atom, it is called "disanchoring". Anchoring means that the material has difficulty to flow. Disanchoring means that the material restarts to flow.

The dislocations attract impurities and pull them into higher-energy lattice regions of higher strain, where other dislocations are located. This makes the motion of the dislocation easier. The strains induced by the latter are not compensated anymore by impurities (metastable configuration). It will not be in equilibrium anymore. Partial relaxation of the lattice strains is observed.

Slide 9

Here I want to introduce one phenomenon which occurs when there is an interaction between dislocations and interstitials. Why? Because interstitials, contrary to other kinds of defects, like twins, staking faults, etc., may move. Interstitials, due to their small size, since they only need interstitial sites to jump to, they can diffuse at room temp in these sites. Therefore, we have both dislocations and interstitials that may move. Notice that dislocations can move under the application of shear stress, and they can move along their slip plane. Dislocations can move from left to right by cutting these atomic planes. You can see that in these slip planes, you may find an array of interstitials. According to what we have said before, interactions between dislocations and interstitials may bring a strengthening. But here there is another phenomenon too, Cottrell atmosphere. Interstitials can diffuse at room temperature too and can interact with dislocations. They try to collect in a special region of the dislocation, which is favored by a lower stress. So, interstitials will go to this region and will act as an obstacle to the motion of the dislocation. This kind of obstacle is named pinning of the dislocation, so the dislocation is pinned by interstitials in their motion, which means hindered (slowed down). As a result of this pinning, stress will increase from outside to let the dislocation move. So, this request of increasing the external stress from outside is a signal of the increase of the resistance of the material. So, this kind of strengthening is just temporary, just while the dislocation is interacting with the interstitial. If we apply a sufficient stress from outside, the dislocation may overcome the obstacle created by the interstitials, and therefore the dislocation becomes unpinned and then continues to move. Because the interstitial will collect like a cloud, we call it an atmosphere. So, there is an atmosphere of interstitials around the dislocation that will create a temporary strengthening of the material.

Slide 10

When we make a tensile test of iron with a minimal number of interstitials, (this kind of alloy is named mild steels), then you will discover that at the beginning of the plastic region, the material will experience an upper and lower yield strength. Now I can understand why this happens (schematically shown). In the initial phase, nothing happens. We have some initial dislocations and C concentration.

As soon as we subject the material in a tensile test (external stress is applied), the dislocations start to move. When they start to move in the slip plane, the dislocation acts like a brusher, and it brushes all the interstitials around. So, as you can see all the interstitials around are collected around the dislocation, and this will create a kind of atmosphere, until the dislocation becomes pinned. The dislocation will struggle to move, and an increase in strength is required. (Upper yield strength) After increasing the strength enough, the dislocation will be unpinned (disanchored) and will continue its motion. (When it moves, we have macroscopic plastic flow)

Slide 11 (Work Hardening)

Second mechanism of strengthening, work hardening (strain hardening). Paradoxically, if a metal is highly ductile, then it has also high ability to become stronger (by exploiting strain ability). When we have large plastic strain, the material has the largest ability to strain-harden. Microscopically, in the material, larger dislocations larger interactions between them. From a lattice point of view, strain-hardening is just the increase of interactions between dislocations. Larger interactions between the dislocation lines and their strain fields will cause strengthening.

Macroscopically, when a material begins to be plastically deformed like in rolling, initial materials are polycrystalline with polygonal grains. Whenever material is plastically deformed, the grain will be squeezed and aligned along the rolling direction (if we are rolling the material). This kind of plastic deformation which decreases the thickness of the material, will cause the creation of a large amount of dislocation. By increasing the dislocation density, we increase the probability of dislocations interacting with each other into every grain. This will translate into an increased strength.

Slide 12

In this slide, you can see the strain field created by dislocation. When it is moving along its slip plane, it will bring its own strain field. And it is of different types. Below the dislocation it is of tensile type, while above it is of compression type. (Static situation of strain field around dislocation). But now we know the dislocation can move, and it will bring with itself its own strain field. What happens if 2 dislocations of the same sign move in the same slip plane? Tensile...