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Precipitates and their Characteristics
IIθ precipitates are under non-equilibrium condition, because the way they are produced is under a non-equilibrium condition. When they are small, they're named IIθ= After overaging they become large and incoherent. They are named Iθ= after excessive overaging we get the phase. Precipitates will have similar characteristics to the ones we observe in equilibrium phase diagrams. (Lowest strength at this phase)
A mechanical engineer should design the material according to the maximum yield strength. If we perform the aging at a larger temperature, the whole process happens faster, but the strength becomes lower. Natural aging will give you better mechanical properties, but it will be more expensive because you will spend more time in a furnace, hence more energy.
What about the cutting stress and bowing stress lines? The answer is given by theory.
SLIDE 33
We summarize the 2 mechanisms which are in force during precipitation. What are the 2
mechanisms which are acting during aging? According to coherency, semi-coherency, and incoherency we have understood that dislocations may be hindered. The two opposite mechanisms are named shearing (cutting) and bowing (bending). Which is sheared and which is bowed? Sheared are particles, bowed are the dislocations. Bowing of the dislocation happens because a shear stress pushes it (look at the annotations). Why will it bow? Because the particles are quite large. The dislocation cannot cut the particles; therefore, it will bow. The bowing mechanism holds only for precipitates which are large in size and incoherent but also semi-incoherent. When they're coherent, the other mechanism will come into play. (See the other figures, particles under shearing) During shearing, the dislocation cuts the precipitate during the application of the shear stress. In the previous diagram, we have a maximal point. Let's measure the shear stress due to the cutting of dislocations. Let's see this.phenomenon by increasing the radius of the particles. We can make this comparison by assuming 3 different kinds of particle volume fractions. (Given by the 3 lines)
What happens to the shear mechanism? If we have 100 particles in the matrix and we increase the shear stress. Assuming they are 10 nm, they will be sheared. If they will be sheared, dislocations will suffer some hindering. (Which identifies with a certain value of the shear stress).
If the number of particles will increase, do you think the shear stress will be lower or higher? It will be higher because more particles will oppose the dislocation. With increasing the radius, the shear stress will be larger and larger.
Let's see with bowing. It is trickier. We have seen in the F-R source that the separation of particles is important for bowing. If the particles are far from one another, I need smaller shear stress, but the inflection point will be small. Small shear stress, small bowing. If I want the same bowing but the particles
are less separated, the shear stress will be larger. Since our comparison also considers the radius of the curvature, this distance x depends also on the radius. If the radius is larger now, the separation distance x will decrease. We are considering two factors in the same observation. Making the distance between particles smaller means we are increasing the volume fractions of the particles. So, it is now a problem because we have 2 factors working in this phenomenon which we must separate. Before we made the comparison with constant volume fraction, by changing the radius. And we saw when the radius increases, shearing is more difficult, and when the volume fraction increases, shearing is more difficult (increasing strength by changing these 2 factors). Now, if I want to see what happens when x increases, I should keep volume fraction constant. If the radius of particles is increasing, inevitably the distance between the particles is decreasing, but this means I am increasing the volume fraction,
which I want to be constant. Therefore, if we are increasing the radius, inevitably x decreases, the stress increasing too. But I don't want to increase volume fraction. Therefore, the behavior of the distance at constant volume fraction will bring the material to decrease the stress. Increasing volume fraction due to increasing radius, increased stress. Constant volume fraction means decreased stress. In order to get best precipitation condition in any alloy, we must ensure the largest population of particles to have semi-coherent interfaces, which means that the particle size shouldn't be too small or too large.
SLIDE 34
Here we have experimental evidence of the 2 phenomena. TEM shows that the black spots are precipitates. You can see that these particles have been sheared by the dislocations. Shearing also depends on the hardness of the precipitate, not just the size. On the other hand, the other figure shows a bowing process. Here the particles are harder and larger than the previous.
The dislocation bows and dislocation rings are created around the particles.
The strengthening mechanism by bowing can be summarized by this equation.
L is the same as x before.
The square root of the particle volume fraction comes into discussion again. (Square root of the concentration in solid solution strengthening)
Question: Dislocations is a mechanism which shows plastic flow.
Because of dislocations we have b, characteristic measure of a dislocation.
If the material is flowing plastically, why do we have an elastic property (like G) in the equation?
The point is dislocations, when they move in the crystal, they move by jumping plane to plane while breaking the bonds.
But when the dislocations step forward, the material recovers.
That is a characteristic of metals. Metals do not break completely the bonds like in ceramics.
In ceramics, when a dislocation cuts the bond, the bonding between atoms remains broken.
Here metals have metallic bonding which is established between ions and electrons.
Electrons allow the dislocations to break the bond, but once the dislocation continues over the crystal, the bonding is re-established, therefore the motion of dislocation in the crystals requires an elastic property. The stronger the bonding of the metallic alloy, the larger the stress we need to apply from outside.be in the grain boundaries. Where are they locatedin this figure? Grain boundaries. Therefore, this material was subjectedat hot rolling but mild cooling.
If precipitates are at the boundary, do they have a strengtheningeffect? No direct effect. Indirect effect yes. The indirect effect is relatedto the fact that during hot processes, the grain boundaries tend toexpand. Consider the boundaries of the grains always made of half freeatoms, because half bonding with the atoms below*. Half bonding isbroken, therefore they are more free to vibrate. And if they can vibratefreely in high temperature means that the boundaries can expand.
Grain growth is not good, grain refinement yes. Having precipitates inthe grain boundaries is good, because they will hinder the grainboundaries to grow. Precipitates at grain boundaries will promotestrength by grain size refinement.
If you see more carefully, some of those precipitates are also located attriple points. These points in regular materials are not essential.
But when you face very small grain material like nanostructured materials, then triple points make up for a higher fraction compared to polycrystalline grains. Triple points then become very important.
SLIDE 36 (STRENGTHENING BY PHASE TRANSFORMATION)
Steels have a large variability in properties. When steel will be cooled down very rapidly, a strong phase will be formed. They are not in fact phases, but micro constituents.
A grain in a polycrystalline material of ferrite is a phase or not a phase? It is a phase because alpha is a solid solution of iron and carbon.
Macroscopically speaking, we may also have a polycrystalline phase with no change in properties within a range into the material, because by changing direction, the properties will not change. That will be a phase.
When we have pearlite, is pearlite a phase or not a phase? Not a phase because pearlite is a mixture of 2 phases: ferrite and cementite. By moving in different directions, we may jump in ferrite, or we may jump in cementite.
(different properties). Therefore, pearlite is not a phase, but a micro-constituent. In steels, when steels are cooled very rapidly, we may get bainite or martensite. They are similar in morphology with pearlite. But also, martensite is made of 2 phases, which is distorted ferrite and cementite, therefore cannot be a phase. If we have an alloy made of ferrite, and we promote phase transformation in order to produce martensite, what do we expect will happen with the mechanical properties? Ferrite is a soft material, martensite is hard. The final material will be harder. If we want to compare the mechanical properties of austenite and the mixture of ferrite and pearlite, the latter is stronger because pearlite is a mixture of 2 phases, alpha soft but cementite is hard. Pearlite will give the final alloy a stronger property. How do we compare in terms of strength alpha and gamma? Which is stronger? We can compare their crystal structures, supposing everything else is the same (carbon content, precipitate).Gamma has FCC, while alpha has BCC. Which can slip easier? FCC. So the resistance will be lower in austenite, therefore ferrite is stronger. Gamma will be softer (higher ductility).
Now I change a little bit the question: what phase among both will have a higher ability to be stronger? FCC will have the larger ability for strain hardening. Why? FCC will get stronger than BCC especially when stacking fault energy will be small (it means the material will have ability to rotate, therefore if dislocations will move on a slip plane of FCC, and while they become crowded and slow down, the crystal can rotate, and more dislocations can flow. Therefore, the ability to pack the maximum number of dislocations will be higher in FCC).
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